Discontinuous Deformation Analysis Enriched by the Bonding Block Model

20
Research Article Discontinuous Deformation Analysis Enriched by the Bonding Block Model Yue Sun, 1 Qian Chen, 2 Xiangchu Feng, 1 and Ying Wang 3 1 Department of Applied Mathematics, Xidian University, Xi’an 710126, China 2 School of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, China 3 University of Chinese Academy of Science, Beijing 100049, China Correspondence should be addressed to Yue Sun; yue [email protected] Received 21 January 2015; Accepted 5 March 2015 Academic Editor: Fabio Tramontana Copyright © 2015 Yue Sun et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. e discontinuous deformation analysis (DDA) has been extensively applied in geotechnical engineering owing to its salient merits in the modeling of discontinuities. However, this method assumes a constant stress field within every block and hence cannot provide reliable estimation for block deformations and stresses. is paper proposes a novel scheme to improve the accuracy of the DDA. In our method, advanced subdivision is introduced to represent a block as an assembly of triangular or quadrilateral elements, in which overlapped element edges are separated from each other and are glued together by bonding springs. e accuracy and the effectiveness of the proposed method are illustrated by three numerical experiments for both continuous and discontinuous problems. 1. Introduction e discontinuous deformation analysis (DDA) [1, 2] is developed for the modeling of the statics and dynamics problems in geological engineering. is method represents a jointed rock mass as an assembly of blocks with constantly changing deformations and contact status. At every contact point, a normal spring and a shear spring are applied. Such contact springs must satisfy no tension in an open contact and no penetration in a close contact, which are fulfilled through repeated open-close iterations. In the literature [38], the augmented Lagrangian method is also introduced into the DDA to increase the accuracy for contact computation. e DDA uses an incremental procedure to solve block movements and deformations on a given time mesh. In every time step, the simultaneous equilibrium equations are formed by the submatrices derived from minimizing all sources of potential energies. Due to the above unique merits, this method has been extensively applied in the analysis of seismic landslides [911], crack propagations [1214], and hydraulic fractures [15, 16]. However, the DDA employs the first-order polynomial to approximate block displacements, and hence the stress within every block keeps invariable. Such assumption leads to the difficulty to implement this method to simulate block cracks, which require accurate stress estimation. In order to improve the accuracy of the DDA, a multitude of methods have been developed. A direct way is to use second- or higher-order polynomial functions to approximate block displacements [1722]. But the accuracy of this method is only adequate for regularly shaped blocks. Another way is the so-called subblock DDA [3, 13, 16, 23], which subdivides blocks into smaller ones by preassumed artificial joints and applies contact springs along the interfaces to prevent the relative movements. Since contact detections and open- close iterations are also required for the contacts between subblocks, the computational burden of this method is very heavy, especially when adopting finer subdivisions. e accuracy of the DDA can also be improved by coupling with the FEM. In the literature [24, 25], the DDA and the FEM codes are integrated into a computer program to alternately figure out block deformations and contact forces. Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2015, Article ID 723263, 19 pages http://dx.doi.org/10.1155/2015/723263

Transcript of Discontinuous Deformation Analysis Enriched by the Bonding Block Model

Research ArticleDiscontinuous Deformation Analysis Enrichedby the Bonding Block Model

Yue Sun1 Qian Chen2 Xiangchu Feng1 and Ying Wang3

1Department of Applied Mathematics Xidian University Xirsquoan 710126 China2School of Mathematics and Information Science Shaanxi Normal University Xirsquoan 710062 China3University of Chinese Academy of Science Beijing 100049 China

Correspondence should be addressed to Yue Sun yue sun163com

Received 21 January 2015 Accepted 5 March 2015

Academic Editor Fabio Tramontana

Copyright copy 2015 Yue Sun et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The discontinuous deformation analysis (DDA) has been extensively applied in geotechnical engineering owing to its salient meritsin the modeling of discontinuities However this method assumes a constant stress field within every block and hence cannotprovide reliable estimation for block deformations and stressesThis paper proposes a novel scheme to improve the accuracy of theDDA In ourmethod advanced subdivision is introduced to represent a block as an assembly of triangular or quadrilateral elementsin which overlapped element edges are separated from each other and are glued together by bonding springs The accuracy andthe effectiveness of the proposed method are illustrated by three numerical experiments for both continuous and discontinuousproblems

1 Introduction

The discontinuous deformation analysis (DDA) [1 2] isdeveloped for the modeling of the statics and dynamicsproblems in geological engineering This method representsa jointed rock mass as an assembly of blocks with constantlychanging deformations and contact status At every contactpoint a normal spring and a shear spring are applied Suchcontact springs must satisfy no tension in an open contactand no penetration in a close contact which are fulfilledthrough repeated open-close iterations In the literature [3ndash8] the augmented Lagrangianmethod is also introduced intothe DDA to increase the accuracy for contact computationThe DDA uses an incremental procedure to solve blockmovements and deformations on a given timemesh In everytime step the simultaneous equilibrium equations are formedby the submatrices derived from minimizing all sources ofpotential energies Due to the above unique merits thismethod has been extensively applied in the analysis of seismiclandslides [9ndash11] crack propagations [12ndash14] and hydraulicfractures [15 16]

However the DDA employs the first-order polynomialto approximate block displacements and hence the stresswithin every block keeps invariable Such assumption leadsto the difficulty to implement this method to simulate blockcracks which require accurate stress estimation In order toimprove the accuracy of the DDA a multitude of methodshave been developed A direct way is to use second- orhigher-order polynomial functions to approximate blockdisplacements [17ndash22] But the accuracy of this method isonly adequate for regularly shaped blocks Another way isthe so-called subblock DDA [3 13 16 23] which subdividesblocks into smaller ones by preassumed artificial joints andapplies contact springs along the interfaces to prevent therelative movements Since contact detections and open-close iterations are also required for the contacts betweensubblocks the computational burden of this method is veryheavy especially when adopting finer subdivisions

The accuracy of the DDA can also be improved bycoupling with the FEM In the literature [24 25] the DDAand the FEMcodes are integrated into a computer program toalternately figure out block deformations and contact forces

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 723263 19 pageshttpdxdoiorg1011552015723263

2 Mathematical Problems in Engineering

Besides a FEM postprocessing technique is introduced torecover block stresses from DDA solution in [14 26] Adisadvantage of these two methods lies in that in everytime step both DDA equations and FEMpostprocessingequations have to be solved and their results also need to bepassed to each other The additional calculation tasks reducethe computing efficiency Different from the above couplingschemes the method presented in [27] divides a problemzone into two domains on which the DDA and the FEM areimplemented respectively Along the domain interface theso-called line blocks are introduced to transfer the contactforces computed by the DDA and the deformation of FEMdomain However this method can only analyze statics prob-lems and need complicated displacement control algorithmto ensure the convergence of the computation Alternativelythe nodal-based DDA [28ndash33] incorporates finite elementtype mesh on every block and uses nodal displacements asthe unknowns to be determined But there still exists thedifficulty due to the constraint on nodal displacements whichmust continue across element interfaces especially in largedeformation problems

In this paper we propose a novel scheme referred to here-after as the bonding block model (BBM) in the frameworkof the DDA to improve the accuracy of block deformationsand stresses Compared with the standard DDA our methodrepresents every nonrigid block as an assembly of triangu-lar or quadrilateral elements as shown in Figure 1 whereoverlapped element edges are separated from each otherand are glued together by the introduced bonding springsContacts are allowed to occur along the element edges thatcompose block boundaries At every contact point a normalcontact spring and a shear contact spring are applied and ourmethod retains the open-close iteration procedure developedin the standard DDA to handle their installation Bench-mark numerical experiments involving both continuousand discontinuous problems are conducted The results arecompared with analytical solutions and ANSYS simulationsto illustrate the accuracy and effectiveness of the proposedmethod

Although both of our method and the subblock DDAsubdivide blocks into smaller partitions the main differencelies in that we connect the elements within a block bybonding springs rather than the contact springs applied onsubblock interfaces In this way it is avoided to implementcontact detections and time consuming open-close iterationsalong artificial joints Compared with the coupling DDA-FEM schemes our method is a dynamics one and in everytime step block deformations and contact effects can besimultaneously figured out without alternately executing theDDA subroutine and the FEMpostprocessing subroutineThe proposed method is also distinguished from the nodal-basedDDAby the unique discretization in which overlappedelement edges are separated from each other but are notsharing a common grid line and adjacent elements areconnected by bonding springs instead of continuous elementshape functions

The rest of the paper is organized as follows In thenext section basic concepts of the proposed method areintroduced In Section 3 detailed formulations are derived

Real joints

Block 1

Block 2

Figure 1 Block subdivision adopted in the BBM

In Section 4 three numerical experiments are conducted toillustrate the accuracy and the efficiency of our method forthe analyses of both continuous and discontinuous problemsSection 5 concludes this paper

2 Concepts of the BBM

21 Subdivision For a jointed rock mass assume it containsΛ blocks occupying the volumesΩ

1 Ω

2 Ω

Λ As shown in

Figure 1 the BBM implements an advanced subdivision overblocks to enhance their flexibility The subdivision processcan be schematized from Figure 2(a) to Figure 2(b) wherethe block Ω

120572is partitioned into the triangular elements I

119894

I119894+1

I119895 As illustrated in Figure 2(c) the overlapped

element edges are separated from each other Proceed suchsubdivision over all blocks and assume 119872 elements aregenerated including I

1I

2 I

119872 On each element say

I119894 119894 = 1 2 119872 the displacement at any point (119909 119910) can

be approximated by the following function

U (119909 119910) = [119879119894(119909 119910)] 119863

119894 (1)

where [119879119894(119909 119910)] is the displacement mode matrix

[119879119894(119909 119910)] = (

1 0 119910 minus 119910 119909 minus 119909 0119910 minus 119910

2

0 1 119909 minus 119909 0 119910 minus 119910119909 minus 119909

2

) (2)

in which (119909 119910) denotes the barycenter coordinate of theelement I

119894 119863

119894 is composed of the unknowns to be

determined of the element that is

119863119894 = (1199060

V0

1199030

120576119909

120576119910

120574119909119910)

T (3)

where 1199060 V

0are the translational displacements of the

barycenter 1199030is the rotation of the centroid and 120576

119909 120576

119910 and

120574119909119910

denote the strain components ofI119894

Since the approximation introduced above is indepen-dent of mesh nodes arbitrarily shaped elements can beadopted such as triangles quadrilaterals and even originalblocks without subdivision Although the formulations in

Mathematical Problems in Engineering 3

Ω120572

(a)

119973j

119973i

119973i+1

119973i+2

(b)

119973j

119973i

119973i+1

119973i+2

(c)

Figure 2 The subdivision process implemented on the block Ω120572

P4P3 P2

P1P6P5

119973j

119973i

(a)

P4

P4

P3 P2

P1P6

119973j

119973i

(b)

Figure 3 Overlapped element edges within a block are glued together by a pair of bonding springs applied between the coincided vertexes(a) The initial status of bonding springs (b) Bonding spring elongations

Section 3 are derived with the illustration figures for trian-gular elements they can be directly employed in the analysisby using other element shape options

22 Element Connections For an element system obtainedthrough the process introduced above the element edgescan be classified into two types the ldquoboundary edgesrdquo thatare parts of block boundaries where contacts may occurand the ldquointernal edgesrdquo including the element edges insideoriginal blocks As shown in Figure 2 I

119894and I

119895are two

adjacent elements of the block Ω120572 From Figure 3(a) it can

be recognized that 1198752119875

3and 119875

4119875

5are boundary edges while

1198751119875

2 119875

3119875

1 119875

5119875

6 and 119875

6119875

4are internal edges Just like the

DDA the BBM applies a normal contact spring and a shearcontact spring at every contact point and controls theirinstallation through open-close iterations

In order to preserve the displacement consistencybetween each overlapped internal edge a couple of bondingsprings are applied between the coincided vertexes Asillustrated in Figure 3(b) the overlapped edges 119875

3119875

1and 119875

6119875

4

are glued together by two bonding springs applied between119875

1and 119875

6and between 119875

3and 119875

4 respectively The bonding

springs between every two elements have the same stiffnessAnd we use the following formulation to determine thebonding spring stiffness between the elementsI

119894andI

119895

120581119894119895

=120581119866

(12) (119904119894+ 119904

119895)

(4)

where 120581 is a constant 119866 is the average shear modulus ofthe two elements and 119904

119894and 119904

119895denote the areas of I

119894

and I119895 respectively By using 120599 to denote the ratio of the

maximum element area with the minimum element area fora subdivision we can take the value of 120581 ranging from 10 (for120599 asymp 1) to 1000 (for 120599 ≫ 1)

The stiffness given by (4) is the minimum requirementfor the proper implementation of bonding springs Althoughlarger stiffness is also suitable in theory very large one isdetrimental to the computing convergence and even leads toill-conditioned global stiffness matrix For the simplicity weuse 120581 instead of 120581

119894119895to denote bonding spring stiffness in the

following sections

3 Computing Formulations

The BBM implements an incremental iteration process tosolve both dynamics and statics problems of the elementsystem I

1I

2 I

119872 on a given time mesh 0 = 119905

0lt

1199051

lt sdot sdot sdot lt 119905119899

lt sdot sdot sdot lt 119905119873 In particular time steps used

in statics problems represent the loading steps Below the119899th time step indicates the interval [119905

119899minus1 119905

119899] Obviously it

is equivalent to say the value of a variable at the end of the(119899 minus 1)th step and the one at the beginning of the 119899th step

For the element system I1I

2 I

119872 generated in

the previous section the total potential energy Π is thesummation of all sources of potential energies includingthe strain energies inertia effects external loads volumeforces bonding loadings and contact forces Minimizingthe potential energy Π in the 119899th step leads to the globalsimultaneous equilibrium equations

(

11987011

11987012

sdot sdot sdot 1198701119872

11987021

11987022

sdot sdot sdot 1198702119872

d

1198701198721

1198701198722

sdot sdot sdot 119870119872119872

)(

119863119899

1

119863119899

2

119863119899

119872

) = (

1198651

1198652

119865119872

) (5)

where the submatrices [119870119894119895] and 119865

119894 are of the sizes 6 times 6 and

6times 1 respectively while 119863119899

119894 denotes the unknown vector to

4 Mathematical Problems in Engineering

be determined 119894 119895 = 1 2 119872 The detailed formulationsof these submatrices [119870

119894119895] and 119865

119894 are derived as follows

31 Element Elastic Submatrices The strain energy of the 119894thelementI

119894can be expressed as

Π119894

119890=

1

2∬

I119894

120576119899

119894T(120590

119899

119894 + 2 120590

119899

119894) 119889119909 119889119910 (6)

where the vectors 120576119899

119894 120590119899

119894 and 120590

119899

119894 denote the strain stress

and the initial stress in the 119899th time step respectively Forisotropic linearly elastic materials 120590

119899

119894 and 120576

119899

119894 have the

following relationship

120590119899

119894 =

119864

1 minus ]2(

1 ] 0

] 1 0

0 0(1 minus ])

2

)120576119899

119894 = [119864

119894] 120576

119899

119894 (7)

in which 119864 and ] indicate Youngrsquos modulus and Poissonrsquosratio respectively The strain 120576

119899

119894 can be expressed by the

displacement as

120576119899

119894 = (

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

) 119863119899

119894 = [Θ] 119863

119899

119894 (8)

Substituting (7) and (8) into (6) we have

Π119894

119890=

1

2119863

119899

119894T(∬

I119894

[Θ]T[119864

119894] [Θ]) 119863

119899

119894

+ 119863119899

119894T∬

I119894

120590119899

119894 119889119909 119889119910

(9)

Minimizing Π119894

119890with respect to 119863

119899

119894 leads to

1205972Π

119894

119890

120597 119863119899

119894T120597 119863

119899

119894

= ∬I119894

[Θ]T[119864

119894] [Θ] 119889119909 119889119910 997888rarr [119870

119894119894]

minus120597Π

119894

119890(0)

120597 119863119899

119894T

= minus∬I119894

[Θ]T120590

119899

119894 119889119909 119889119910 997888rarr 119865

119894

(10)

which are added to the stiffness matrix and the right item of(5) respectively

32 Element Inertia Submatrices The potential energy con-tributed by the inertia force acting on the element I

119894at the

moment 119905 can be expressed as follows

Π119894

120588= ∬

I119894

119880119894(119905 119909 119910)

T(120588

1205972

1205971199052119880

119894(119905 119909 119910)) 119889119909 119889119910 (11)

where 119880119894(119905 119909 119910) and 120588 indicate the displacement and the

mass per area respectively and the multiplication betweenthe parentheses in the integrand of the equation above

thereby computes the density of the inertia force on I119894

Substituting (1) into (11) leads to

Π119894

120588= 119863

119894(119905)

T∬

I119894

120588 [119879119894]T[119879

119894]120597

2119863

119894(119905)

1205971199052119889119909 119889119910 (12)

Using the Newmark time integrating method we have thefollowing finite difference approximation

1205972119863

119894(119905)

1205971199052= (

2 119863119899

119894

Δ2

119899

minus

2 119881119899minus1

119894

Δ119899

) (13)

whereΔ119899

= 119905119899minus119905

119899minus1is the 119899th stepsize and119881

119899minus1

119894indicates the

initial velocity of the 119899th time step and can be updated by theiteration formulation below

119881119899

119894 =

2 119863119899minus1

119894

Δ119899

minus 119881119899minus1

119894 119899 ⩾ 2 (14)

Substituting (13) into (12) we then have

Π119894

120588=

2 119863119899

119894T

Δ2

119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119863

119899

119894

minus2 119863

119899

119894T

Δ119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119881

119899minus1

119894

(15)

Minimizing the right items of the equation abovewith respectto 119863

119899

119894 leads to

2

Δ2

119899

∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910 997888rarr [119870

119894119894] (16)

2

Δ119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119881

119899minus1

119894 997888rarr 119865

119894 (17)

where the integrals can be solved by the simplex integralmethod [34] The submatrices equation (16) and equation(17) are added to the stiffness matrix and the right itemof the global simultaneous equilibrium equations in (5)respectively

33 Bonding Submatrices In order to preserve the dis-placement consistency along the element interfaces withina block each pair of initially overlapped element edges isglued together by two bonding springs applied between thecoincided vertexes As shown in Figure 4(a) the points119875

1and

1198752are a pair of initially overlapped vertexes of the elementsI

119894

and I119895which are within the same block A bonding spring

with the stiffness 120581 is applied between 1198751and 119875

2to glue I

119894

and I119895together Denote the coordinate of 119875

1and 119875

2at the

time node 119905119899by (119909

119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) respectively

Firstly we compute the elongation 119889119899of the bonding

spring between 1198751and 119875

2at the time node 119905

119899 As illustrated

in Figure 4(b) 119889119899is actually the length of 997888997888997888rarr

1198751119875

2and can be

computed from the following equality

1198892

119899= 119897

119899T119897

119899 (18)

Mathematical Problems in Engineering 5

119973i

119973j

120581120581

P2

P1

(a)

119973i

119973j

dn

120581

120581

P2

P1

(b)

Figure 4 Bonding springs applied along overlapped element edges (a) Bonding spring length is zero at 1199050 (b) The bonding spring between

1198751and 119875

2has the elongation 119889

119899at 119905

119899

where 119897119899 denotes the vector 997888997888997888rarr

1198751119875

2at 119905

119899 that is

119897119899 = (

119909119899

2

119910119899

2

) minus (

119909119899

1

119910119899

1

) (19)

Assume the displacements of 1198751and 119875

2in the 119899th time step

are (119906119899

1 V119899

1) and (119906

119899

2 V119899

2) respectivelyThen the coordinates of

1198751and 119875

2at 119905

119899can be decomposed into the incremental form

below

(119909119899

1 119910

119899

1) = (119909

119899minus1

1+ 119906

119899

1 119910

119899minus1

1+ V119899

1) (20a)

(119909119899

2 119910

119899

2) = (119909

119899minus1

2+ 119906

119899

2 119910

119899minus1

2+ V119899

2) (20b)

Substituting (20a) and (20b) into (19) leads to

119897119899 = 119897

119899minus1 + (

119906119899

1

V119899

1

) minus (

119906119899

2

V119899

2

) (21)

Substitute (3) and (1) into the equation above and then wehave

119897119899 = 119897

119899minus1 + [119879

1

119894] 119863

119899

119894 minus [119879

2

119895] 119863

119899

119895 (22)

where [1198791

119894] = [119879

119894(119909

1 119910

1)] and [119879

2

119895] = [119879

119895(119909

2 119910

2)] Substitut-

ing (22) into (18) then 1198892

119899can be expanded as the following

summation

1198892

119899= 119897

119899minus1T119897

119899minus1 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119863119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

T[119879

1

119894] 119863

119899

119894

+ [1198791

119894]T119863

119899

119894T119897

119899minus1 minus 119897

119899minus1T[119879

2

119895] 119863

119899

119895

minus [1198792

119895]T119863

119899

119895T119897

119899minus1

(23)

Next we compute the strain energy Π120581

119892of the bonding

spring between 1198751and 119875

2in the 119899th time step [119905

119899minus1 119905

119899] Π120581

119892

can be expressed as follows

Π120581

119892=

1

2120581119889

2

119899 (24)

Substitute (23) into the expression above andΠ120581

119892becomes the

following sum

Π120581

119892= 120587

0+ 120587

119894+ 120587

119895+ 120587

119894119894+ 120587

119894119895+ 120587

119895119894+ 120587

119895119895 (25)

where

1205870=

120581

2119897

119899minus1T119897

119899minus1 (26a)

120587119894119894

=120581

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (26b)

120587119895119895

=120581

2119863

119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895 (26c)

120587119894119895

= minus120581

2119863

119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895 (26d)

120587119895119894

= minus120581

2119863

119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894 (26e)

120587119894=

120581

2119897

119899minus1T[119879

1

119894] 119863

119899

119894 +

120581

2[119879

1

119894]T119863

119899

119894T119897

119899minus1 (26f)

120587119895= minus

120581

2119897

119899minus1T[119879

2

119895] 119863

119899

119895 minus

120581

2[119879

2

119895]T119863

119899

119895T119897

119899minus1 (26g)

Then minimize Π120581

119892with respect to the unknowns As

119897119899minus1

is known in the 119899th time step the derivative of 1205870is 0

6 Mathematical Problems in Engineering

Minimize 120587119894119894 120587

119894119895 120587

119895119894 and 120587

119895119895with respect to 119863

119894 and 119863

119895

and we have

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 120581 [1198791

119894]T[119879

1

119894] 997888rarr [119870

119894119894] (27a)

1205972120587

119894119895

120597 119863119899

119894T120597 119863

119899

119895

= 120581 [1198791

119894]T[119879

2

119895] 997888rarr [119870

119894119895] (27b)

1205972120587

119895119894

120597 119863119899

119895T120597 119863

119899

119894

= 120581 [1198792

119895]T[119879

1

119894] 997888rarr [119870

119895119894] (27c)

1205972120587

119895119895

120597 119863119899

119895T120597 119863

119899

119895

= 120581 [1198792

119895]T[119879

2

119895] 997888rarr [119870

119895119895] (27d)

These 6 times 6 submatrices in the equation above are added intothe global stiffness matrix in (5) Minimizing 120587

119894and 120587

119895with

respect to 119863119894 and 119863

119895 leads to

minus120597120587

119894(0)

120597 119863119894

= minus120581 [1198791

119894]T119897

119899minus1 997888rarr 119865

119894 (28a)

minus

120597120587119895(0)

120597 119863119895

= 120581 [1198792

119895]T119897

119899minus1 997888rarr 119865

119895 (28b)

which are added to the right side of (5)

34 Bonding Submatrices at Fixed Points As the boundaryconditions some elements are fixed at specific points whichare called fixed points in the original DDA As shown inFigure 5(a) the point 119875 of the element I

119894is a fixed point

119894 = 1 2 119872 Denote the coordinates of 119875 at the timenode 119905

119899by (119909

119899

119875 119910

119899

119875) 119899 = 1 2 119873 Assume there exists

a fictitious element I0which is immersed in the support

wall and always keeps a statics state during a computationSuch fictitious element does not contribute potential energyto the element system Assume the vertex 119875

0of I

0overlaps

with the fixed point 119875 at the initial time 1199050 In order to resist

the displacement of the point 119875 a bonding spring with thestiffness 120581

0is applied between 119875 and 119875

0 The stiffness 120581

0can

be determined by the following formulation

1205810= 120581119864 (29)

where 120581 is the same factor used in (4)As shown in Figure 5(b) the bonding spring between 119875

and 1198750has the elongation 119889

119899at the time node 119905

119899 Denote

the coordinate of 1198750by (119909 119910) which keeps constant during

a computation Since 119889119899is actually the length of 997888997888rarr119875

0119875 we have

1198892

119899= 119897

119899

119875T119897

119899

119875 (30)

where 119897119899

119875 denotes the vector 997888997888rarr

1198750119875 at 119905

119899 that is

119897119899

119875 = (

119909119899

119875

119910119899

119875

) minus (

119909

119910) (31)

Assume the displacement of 119875 in the 119899th time step is (119906119899

119875 V119899

119875)

Then the coordinates of 119875 at 119905119899can be written as (119909

119899

119875 119910

119899

119875) =

(119909119899minus1

119875+ 119906

119899

119875 119910

119899minus1

119875+ V119899

119875) which is substituted into (31) that is

119897119899

119875 = (

119909119899minus1

119875+ 119906

119899

119875

119910119899minus1

119875+ V119899

119875

) minus (

119909

119910) = 119897

119899minus1 + (

119906119899

119875

V119899

119875

) (32)

Substitute (3) and (1) into the equation above and then wehave

119897119899

119875 = 119897

119899minus1

119875 + [119879

119875

119894] 119863

119899

119894 (33)

where [119879119875

119894] indicates [119879

119894(119909

119899

119875 119910

119899

119875)] Substituting (33) into (30)

leads to

1198892

119899= 119897

119899minus1

119875T119897

119899minus1

119875 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

119875T[119879

1

119894] 119863

119899

119894 + [119879

1

119894]T119863

119899

119894T119897

119899minus1

119875

(34)

Then we can express the strain energy of the bonding springbetween 119875 and 119875

0as

Π1205810

119875=

1

2120581

0119889

2

119899= 120587

0+ 120587

119894119894+ 120587

119894 (35)

where

1205870=

1205810

2119897

119899minus1

119875T119897

119899minus1

119875 (36a)

120587119894119894

=120581

0

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (36b)

120587119894= 120581

0[119879

1

119894]T119863

119899

119894T119897

119899minus1

119875 (36c)

Obviously theminimization of1205870with respect to 119863

119899

119894 is zero

Then we minimize 120587119894119894and 120587

119894

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 1205810[119879

119875]T[119879

119875] 997888rarr [119870

119894119894] (37a)

120597120587119894

120597 119863119899

119894T

= 1205810119897

119899minus1

119875T[119879

119875] 997888rarr 119865

119894 (37b)

where [119870119894119894] and 119865

119894 are added to the stiffness matrix and the

right item of (5)

35 Line Loading Submatrices Assume a loading 119865 =

(119865119909(119909 119910) 119865

119910(119909 119910))

T is applied along a line upon the bound-ary of the elementI

119894 119894 = 1 2 119872 The coordinate of any

point on the line at the timemoment 119905119899can be determined by

the following parametric equations

119909119899

= 119909119899(120585) = (119909

119899

2minus 119909

119899

1) 120585 + 119909

119899

1

119910119899

= 119910119899(120585) = (119910

119899

2minus 119910

119899

1) 120585 + 119910

119899

1

0 ⩽ 120585 ⩽ 1

(38)

Mathematical Problems in Engineering 7

119973i

1205810 1205810

P0

P

1199730

(a)

119973i

dn

12058101205810

P0

P

1199730

(b)

Figure 5 Bonding springs are applied at fixed points (a) The length of the bonding spring applied between the fixed point 119875 of the elementI

119894and the vertex 119875

0of the fictitious support elementI

0is zero at 119905

0 (b) At 119905

119899 the bonding spring between 119875

0and 119875 has the elongation 119889

119899

where (119909119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) are the coordinates of the ending

points of the line whose length at the end of the last time stepcan be computed by

119904119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

1)

2

+ (119910119899minus1

2minus 119910

119899minus1

1)

2

(39)

where time step 119899 = 1 2 119873 Then we can obtain thepotential energy contributed by the loading 119865

Π119904= minusint

1

0

(

119906119899(119909 (120585) 119910 (120585))

V119899(119909 (119904) 119910 (119904))

)

T

(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (40)

Substituting (3) and (1) into (40) leads to

Π119904= minus 119863

119899

119894Tint

1

0

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (41)

Minimize Π119904with respect to 119863

119899

120585 and we have

120597Π119904(0)

120597 119863119899

119894T

= int

0

1

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 997888rarr 119865119894 (42)

The 6 times 1 submatrix is added to the right item of the globalequation (5) If 119865 = (119865

119909 119865

119910) which is constant (42) has the

analytical form

120597Π119904(0)

120597 119863119899

119894T

= 119904119899minus1

((((((((((

(

119865119909

119865119910

21199100minus 119910

2119865

119909+

119909 minus 21199090

2119865

119910

119909 minus 21199090

2119865

119909

119910 minus 21199100

2119865

119910

21199100minus 119910

4119865

119909+

119909 minus 21199090

4119865

119910

))))))))))

)

997888rarr 119865119894

(43)

where 119909 = 1199092+ 119909

1 119910 = 119910

2+ 119910

1and (119909

0 119910

0) is the coordinate

of the barycenter of the elementI119894

36 Volume Force Submatrix When the elementI119894bears the

body loading 119891(119909 119910) = (119891119909(119909 119910) 119891

119910(119909 119910))

T 119894 = 1 2 119872

the potential energy Π119908due to the force 119891 is calculated as

Π119908

= minus∬I119894

(

119906119899(119909 119910)

V119899(119909 119910)

)

T

119891 (119909 119910) 119889119909 119889119910 (44)

Substituting (1) into the equations above leads to

Π119908

= minus 119863119899

119894T∬

I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 (45)

Minimize Π119908with respect to 119863

119899

120585 and we have

120597Π119908

(0)

120597 119863119899

119894T

= minus∬I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 997888rarr [119865

119894]

(46)

The 6 times 1 submatrix above is added to the right side of theglobal simultaneous equation (5) When the body force isconstant say 119891(119909 119910) = (119891

119909 119891

119910)T (46) becomes

120597Π119908

(0)

120597 [119863119899

119894]T

= minus [119863119899

119894]T(119891

119909119860

119894 119891

119910119860

119894 0 0 0 0)

T997888rarr [119865

119894] (47)

where 119860119894denotes the area of the elementI

119894

37 Contact Submatrices The element edges along theboundaries of different blocks may contact with each otherThis kind of interactions is allowed in three modes vertex-to-vertex edge-to-edge and vertex-to-edge contacts In two-dimensional problems both vertex-to-vertex and edge-to-edge contacts can be decomposed into couples of vertex-to-edge ones The BBM adopts the ldquorigidrdquo contact model Inorder to prevent penetration and relative sliding along the

8 Mathematical Problems in Engineering

119973j

P3

120581perp

120581

P1119973i

P2

(a)

P3

P0P1

119973i

hperp

h

P2

(b)

Figure 6 Contact springs between the elementsI119894andI

119895 (a) The normal contact spring with the stiffness 120581

perpand the shear contact spring

with the stiffness 120581 (b) The normal contact distance ℎ

perpand the shear contact distance ℎ

interfaces of a close contact a normal contact spring and ashear contact spring have to be applied at the contact point

During each time step the simultaneous equilibriumequations in (5) have to be assembled and solved repeat-edly whenever the implementation of any contact springchanges All of the contact springs must satisfy the followingconstraints no tension appears in an open contact and nopenetration occurs in a close one These two conditions canbe fulfilled by an open-close iteration process to controlthe installation and uninstallation of every contact springin each time step Readers are recommended to refer tothe dissertation [1] for the implementation of open-closeiterations

As illustrated in Figure 6(a) the vertex 1198751of the element

I119894and the edge 119875

2119875

3of the element I

119895are in a contact

where a normal contact spring with the stiffness 120581perpand a

shear contact spring with the stiffness 120581are applied As

shown in Figure 6(b) assume the point 1198750on the edge 119875

2119875

3

is the contact point 1198751is a vertex of the block which is

different from the block where 1198750 119875

2 and 119875

3locate Denote

the coordinate of 119875120572at 119905

119899by (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 and 119899 =

1 2 119873 The displacement of 119875120572in the 119899th time step can

be expressed as

U (119909119899

120572 119910

119899

120572) = (

119906119899

120572

V119899

120572

) = (

119909119899

120572minus 119909

119899minus1

120572

119910119899

120572minus 119910

119899minus1

120572

) (48)

Substitute (1) into the above equation and we have

(

119906119899

1

V119899

1

) = [1198791

119894] 119863

119899

119894 (49)

(

119906119899

120572

V119899

120572

) = [119879120573

119895] 119863

119899

119895 120573 = 0 2 3 (50)

where [119879120572

120585] is the abbreviation of [119879

120585(119909

120572 119910

120572)] 120572 = 0 1 2 3

In every iteration step the normal contact spring and theshear contact spring contribute the strain energiesΠ

perpandΠ

respectively and they can be expressed as

Πperp

=1

2120581

perpℎ

2

perp (51a)

Π=

1

2120581

2

(51b)

where ℎperpand ℎ

denote the normal and the shear contact

displacement respectively As shown in Figure 6(b) ℎperpis the

distance from 1198751to the edge 119875

2119875

3and can be computed as

follows

ℎperp

=119860

119899

119871119899 (52)

where119860119899 is the area of the triangleΔ

119875111987521198753and 119871

119899 denotes thelength of the edge 119875

2119875

3 Since ℎ

is the projection of the line

1198750119875

1on the edge 119875

2119875

3or its extension line we have

ℎ=

997888997888997888rarr119875

2119875

3sdot997888997888997888rarr119875

0119875

1

10038161003816100381610038161003816119875

2119875

3

10038161003816100381610038161003816

=1

119871119899(

119909119899

3minus 119909

119899

2

119910119899

3minus 119910

119899

2

)

T

(

119909119899

1minus 119909

119899

0

119910119899

1minus 119910

119899

0

)

(53)

As follows the minimizations of the potential energy itemsΠ

perpand Π

are derived respectively

371 Normal Contact Submatrices To obtain the derivativeof Π

perpin (51a) we compute the normal contact displacement

ℎperpfirstly 119860119899 in (52) can be deduced by using (48) as

119860119899

=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899

1119910

119899

1

1 119909119899

2119910

119899

2

1 119909119899

3119910

119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1+ 119906

119899

1119910

119899minus1

1+ V119899

1

1 119909119899minus1

2+ 119906

119899

2119910

119899minus1

2+ V119899

2

1 119909119899minus1

3+ 119906

119899

3119910

119899minus1

3+ V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1119910

119899minus1

1

1 119909119899minus1

2119910

119899minus1

2

1 119909119899minus1

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1V119899

1

1 119906119899

2V119899

2

1 119906119899

3V119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(54)

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

Besides a FEM postprocessing technique is introduced torecover block stresses from DDA solution in [14 26] Adisadvantage of these two methods lies in that in everytime step both DDA equations and FEMpostprocessingequations have to be solved and their results also need to bepassed to each other The additional calculation tasks reducethe computing efficiency Different from the above couplingschemes the method presented in [27] divides a problemzone into two domains on which the DDA and the FEM areimplemented respectively Along the domain interface theso-called line blocks are introduced to transfer the contactforces computed by the DDA and the deformation of FEMdomain However this method can only analyze statics prob-lems and need complicated displacement control algorithmto ensure the convergence of the computation Alternativelythe nodal-based DDA [28ndash33] incorporates finite elementtype mesh on every block and uses nodal displacements asthe unknowns to be determined But there still exists thedifficulty due to the constraint on nodal displacements whichmust continue across element interfaces especially in largedeformation problems

In this paper we propose a novel scheme referred to here-after as the bonding block model (BBM) in the frameworkof the DDA to improve the accuracy of block deformationsand stresses Compared with the standard DDA our methodrepresents every nonrigid block as an assembly of triangu-lar or quadrilateral elements as shown in Figure 1 whereoverlapped element edges are separated from each otherand are glued together by the introduced bonding springsContacts are allowed to occur along the element edges thatcompose block boundaries At every contact point a normalcontact spring and a shear contact spring are applied and ourmethod retains the open-close iteration procedure developedin the standard DDA to handle their installation Bench-mark numerical experiments involving both continuousand discontinuous problems are conducted The results arecompared with analytical solutions and ANSYS simulationsto illustrate the accuracy and effectiveness of the proposedmethod

Although both of our method and the subblock DDAsubdivide blocks into smaller partitions the main differencelies in that we connect the elements within a block bybonding springs rather than the contact springs applied onsubblock interfaces In this way it is avoided to implementcontact detections and time consuming open-close iterationsalong artificial joints Compared with the coupling DDA-FEM schemes our method is a dynamics one and in everytime step block deformations and contact effects can besimultaneously figured out without alternately executing theDDA subroutine and the FEMpostprocessing subroutineThe proposed method is also distinguished from the nodal-basedDDAby the unique discretization in which overlappedelement edges are separated from each other but are notsharing a common grid line and adjacent elements areconnected by bonding springs instead of continuous elementshape functions

The rest of the paper is organized as follows In thenext section basic concepts of the proposed method areintroduced In Section 3 detailed formulations are derived

Real joints

Block 1

Block 2

Figure 1 Block subdivision adopted in the BBM

In Section 4 three numerical experiments are conducted toillustrate the accuracy and the efficiency of our method forthe analyses of both continuous and discontinuous problemsSection 5 concludes this paper

2 Concepts of the BBM

21 Subdivision For a jointed rock mass assume it containsΛ blocks occupying the volumesΩ

1 Ω

2 Ω

Λ As shown in

Figure 1 the BBM implements an advanced subdivision overblocks to enhance their flexibility The subdivision processcan be schematized from Figure 2(a) to Figure 2(b) wherethe block Ω

120572is partitioned into the triangular elements I

119894

I119894+1

I119895 As illustrated in Figure 2(c) the overlapped

element edges are separated from each other Proceed suchsubdivision over all blocks and assume 119872 elements aregenerated including I

1I

2 I

119872 On each element say

I119894 119894 = 1 2 119872 the displacement at any point (119909 119910) can

be approximated by the following function

U (119909 119910) = [119879119894(119909 119910)] 119863

119894 (1)

where [119879119894(119909 119910)] is the displacement mode matrix

[119879119894(119909 119910)] = (

1 0 119910 minus 119910 119909 minus 119909 0119910 minus 119910

2

0 1 119909 minus 119909 0 119910 minus 119910119909 minus 119909

2

) (2)

in which (119909 119910) denotes the barycenter coordinate of theelement I

119894 119863

119894 is composed of the unknowns to be

determined of the element that is

119863119894 = (1199060

V0

1199030

120576119909

120576119910

120574119909119910)

T (3)

where 1199060 V

0are the translational displacements of the

barycenter 1199030is the rotation of the centroid and 120576

119909 120576

119910 and

120574119909119910

denote the strain components ofI119894

Since the approximation introduced above is indepen-dent of mesh nodes arbitrarily shaped elements can beadopted such as triangles quadrilaterals and even originalblocks without subdivision Although the formulations in

Mathematical Problems in Engineering 3

Ω120572

(a)

119973j

119973i

119973i+1

119973i+2

(b)

119973j

119973i

119973i+1

119973i+2

(c)

Figure 2 The subdivision process implemented on the block Ω120572

P4P3 P2

P1P6P5

119973j

119973i

(a)

P4

P4

P3 P2

P1P6

119973j

119973i

(b)

Figure 3 Overlapped element edges within a block are glued together by a pair of bonding springs applied between the coincided vertexes(a) The initial status of bonding springs (b) Bonding spring elongations

Section 3 are derived with the illustration figures for trian-gular elements they can be directly employed in the analysisby using other element shape options

22 Element Connections For an element system obtainedthrough the process introduced above the element edgescan be classified into two types the ldquoboundary edgesrdquo thatare parts of block boundaries where contacts may occurand the ldquointernal edgesrdquo including the element edges insideoriginal blocks As shown in Figure 2 I

119894and I

119895are two

adjacent elements of the block Ω120572 From Figure 3(a) it can

be recognized that 1198752119875

3and 119875

4119875

5are boundary edges while

1198751119875

2 119875

3119875

1 119875

5119875

6 and 119875

6119875

4are internal edges Just like the

DDA the BBM applies a normal contact spring and a shearcontact spring at every contact point and controls theirinstallation through open-close iterations

In order to preserve the displacement consistencybetween each overlapped internal edge a couple of bondingsprings are applied between the coincided vertexes Asillustrated in Figure 3(b) the overlapped edges 119875

3119875

1and 119875

6119875

4

are glued together by two bonding springs applied between119875

1and 119875

6and between 119875

3and 119875

4 respectively The bonding

springs between every two elements have the same stiffnessAnd we use the following formulation to determine thebonding spring stiffness between the elementsI

119894andI

119895

120581119894119895

=120581119866

(12) (119904119894+ 119904

119895)

(4)

where 120581 is a constant 119866 is the average shear modulus ofthe two elements and 119904

119894and 119904

119895denote the areas of I

119894

and I119895 respectively By using 120599 to denote the ratio of the

maximum element area with the minimum element area fora subdivision we can take the value of 120581 ranging from 10 (for120599 asymp 1) to 1000 (for 120599 ≫ 1)

The stiffness given by (4) is the minimum requirementfor the proper implementation of bonding springs Althoughlarger stiffness is also suitable in theory very large one isdetrimental to the computing convergence and even leads toill-conditioned global stiffness matrix For the simplicity weuse 120581 instead of 120581

119894119895to denote bonding spring stiffness in the

following sections

3 Computing Formulations

The BBM implements an incremental iteration process tosolve both dynamics and statics problems of the elementsystem I

1I

2 I

119872 on a given time mesh 0 = 119905

0lt

1199051

lt sdot sdot sdot lt 119905119899

lt sdot sdot sdot lt 119905119873 In particular time steps used

in statics problems represent the loading steps Below the119899th time step indicates the interval [119905

119899minus1 119905

119899] Obviously it

is equivalent to say the value of a variable at the end of the(119899 minus 1)th step and the one at the beginning of the 119899th step

For the element system I1I

2 I

119872 generated in

the previous section the total potential energy Π is thesummation of all sources of potential energies includingthe strain energies inertia effects external loads volumeforces bonding loadings and contact forces Minimizingthe potential energy Π in the 119899th step leads to the globalsimultaneous equilibrium equations

(

11987011

11987012

sdot sdot sdot 1198701119872

11987021

11987022

sdot sdot sdot 1198702119872

d

1198701198721

1198701198722

sdot sdot sdot 119870119872119872

)(

119863119899

1

119863119899

2

119863119899

119872

) = (

1198651

1198652

119865119872

) (5)

where the submatrices [119870119894119895] and 119865

119894 are of the sizes 6 times 6 and

6times 1 respectively while 119863119899

119894 denotes the unknown vector to

4 Mathematical Problems in Engineering

be determined 119894 119895 = 1 2 119872 The detailed formulationsof these submatrices [119870

119894119895] and 119865

119894 are derived as follows

31 Element Elastic Submatrices The strain energy of the 119894thelementI

119894can be expressed as

Π119894

119890=

1

2∬

I119894

120576119899

119894T(120590

119899

119894 + 2 120590

119899

119894) 119889119909 119889119910 (6)

where the vectors 120576119899

119894 120590119899

119894 and 120590

119899

119894 denote the strain stress

and the initial stress in the 119899th time step respectively Forisotropic linearly elastic materials 120590

119899

119894 and 120576

119899

119894 have the

following relationship

120590119899

119894 =

119864

1 minus ]2(

1 ] 0

] 1 0

0 0(1 minus ])

2

)120576119899

119894 = [119864

119894] 120576

119899

119894 (7)

in which 119864 and ] indicate Youngrsquos modulus and Poissonrsquosratio respectively The strain 120576

119899

119894 can be expressed by the

displacement as

120576119899

119894 = (

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

) 119863119899

119894 = [Θ] 119863

119899

119894 (8)

Substituting (7) and (8) into (6) we have

Π119894

119890=

1

2119863

119899

119894T(∬

I119894

[Θ]T[119864

119894] [Θ]) 119863

119899

119894

+ 119863119899

119894T∬

I119894

120590119899

119894 119889119909 119889119910

(9)

Minimizing Π119894

119890with respect to 119863

119899

119894 leads to

1205972Π

119894

119890

120597 119863119899

119894T120597 119863

119899

119894

= ∬I119894

[Θ]T[119864

119894] [Θ] 119889119909 119889119910 997888rarr [119870

119894119894]

minus120597Π

119894

119890(0)

120597 119863119899

119894T

= minus∬I119894

[Θ]T120590

119899

119894 119889119909 119889119910 997888rarr 119865

119894

(10)

which are added to the stiffness matrix and the right item of(5) respectively

32 Element Inertia Submatrices The potential energy con-tributed by the inertia force acting on the element I

119894at the

moment 119905 can be expressed as follows

Π119894

120588= ∬

I119894

119880119894(119905 119909 119910)

T(120588

1205972

1205971199052119880

119894(119905 119909 119910)) 119889119909 119889119910 (11)

where 119880119894(119905 119909 119910) and 120588 indicate the displacement and the

mass per area respectively and the multiplication betweenthe parentheses in the integrand of the equation above

thereby computes the density of the inertia force on I119894

Substituting (1) into (11) leads to

Π119894

120588= 119863

119894(119905)

T∬

I119894

120588 [119879119894]T[119879

119894]120597

2119863

119894(119905)

1205971199052119889119909 119889119910 (12)

Using the Newmark time integrating method we have thefollowing finite difference approximation

1205972119863

119894(119905)

1205971199052= (

2 119863119899

119894

Δ2

119899

minus

2 119881119899minus1

119894

Δ119899

) (13)

whereΔ119899

= 119905119899minus119905

119899minus1is the 119899th stepsize and119881

119899minus1

119894indicates the

initial velocity of the 119899th time step and can be updated by theiteration formulation below

119881119899

119894 =

2 119863119899minus1

119894

Δ119899

minus 119881119899minus1

119894 119899 ⩾ 2 (14)

Substituting (13) into (12) we then have

Π119894

120588=

2 119863119899

119894T

Δ2

119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119863

119899

119894

minus2 119863

119899

119894T

Δ119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119881

119899minus1

119894

(15)

Minimizing the right items of the equation abovewith respectto 119863

119899

119894 leads to

2

Δ2

119899

∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910 997888rarr [119870

119894119894] (16)

2

Δ119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119881

119899minus1

119894 997888rarr 119865

119894 (17)

where the integrals can be solved by the simplex integralmethod [34] The submatrices equation (16) and equation(17) are added to the stiffness matrix and the right itemof the global simultaneous equilibrium equations in (5)respectively

33 Bonding Submatrices In order to preserve the dis-placement consistency along the element interfaces withina block each pair of initially overlapped element edges isglued together by two bonding springs applied between thecoincided vertexes As shown in Figure 4(a) the points119875

1and

1198752are a pair of initially overlapped vertexes of the elementsI

119894

and I119895which are within the same block A bonding spring

with the stiffness 120581 is applied between 1198751and 119875

2to glue I

119894

and I119895together Denote the coordinate of 119875

1and 119875

2at the

time node 119905119899by (119909

119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) respectively

Firstly we compute the elongation 119889119899of the bonding

spring between 1198751and 119875

2at the time node 119905

119899 As illustrated

in Figure 4(b) 119889119899is actually the length of 997888997888997888rarr

1198751119875

2and can be

computed from the following equality

1198892

119899= 119897

119899T119897

119899 (18)

Mathematical Problems in Engineering 5

119973i

119973j

120581120581

P2

P1

(a)

119973i

119973j

dn

120581

120581

P2

P1

(b)

Figure 4 Bonding springs applied along overlapped element edges (a) Bonding spring length is zero at 1199050 (b) The bonding spring between

1198751and 119875

2has the elongation 119889

119899at 119905

119899

where 119897119899 denotes the vector 997888997888997888rarr

1198751119875

2at 119905

119899 that is

119897119899 = (

119909119899

2

119910119899

2

) minus (

119909119899

1

119910119899

1

) (19)

Assume the displacements of 1198751and 119875

2in the 119899th time step

are (119906119899

1 V119899

1) and (119906

119899

2 V119899

2) respectivelyThen the coordinates of

1198751and 119875

2at 119905

119899can be decomposed into the incremental form

below

(119909119899

1 119910

119899

1) = (119909

119899minus1

1+ 119906

119899

1 119910

119899minus1

1+ V119899

1) (20a)

(119909119899

2 119910

119899

2) = (119909

119899minus1

2+ 119906

119899

2 119910

119899minus1

2+ V119899

2) (20b)

Substituting (20a) and (20b) into (19) leads to

119897119899 = 119897

119899minus1 + (

119906119899

1

V119899

1

) minus (

119906119899

2

V119899

2

) (21)

Substitute (3) and (1) into the equation above and then wehave

119897119899 = 119897

119899minus1 + [119879

1

119894] 119863

119899

119894 minus [119879

2

119895] 119863

119899

119895 (22)

where [1198791

119894] = [119879

119894(119909

1 119910

1)] and [119879

2

119895] = [119879

119895(119909

2 119910

2)] Substitut-

ing (22) into (18) then 1198892

119899can be expanded as the following

summation

1198892

119899= 119897

119899minus1T119897

119899minus1 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119863119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

T[119879

1

119894] 119863

119899

119894

+ [1198791

119894]T119863

119899

119894T119897

119899minus1 minus 119897

119899minus1T[119879

2

119895] 119863

119899

119895

minus [1198792

119895]T119863

119899

119895T119897

119899minus1

(23)

Next we compute the strain energy Π120581

119892of the bonding

spring between 1198751and 119875

2in the 119899th time step [119905

119899minus1 119905

119899] Π120581

119892

can be expressed as follows

Π120581

119892=

1

2120581119889

2

119899 (24)

Substitute (23) into the expression above andΠ120581

119892becomes the

following sum

Π120581

119892= 120587

0+ 120587

119894+ 120587

119895+ 120587

119894119894+ 120587

119894119895+ 120587

119895119894+ 120587

119895119895 (25)

where

1205870=

120581

2119897

119899minus1T119897

119899minus1 (26a)

120587119894119894

=120581

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (26b)

120587119895119895

=120581

2119863

119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895 (26c)

120587119894119895

= minus120581

2119863

119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895 (26d)

120587119895119894

= minus120581

2119863

119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894 (26e)

120587119894=

120581

2119897

119899minus1T[119879

1

119894] 119863

119899

119894 +

120581

2[119879

1

119894]T119863

119899

119894T119897

119899minus1 (26f)

120587119895= minus

120581

2119897

119899minus1T[119879

2

119895] 119863

119899

119895 minus

120581

2[119879

2

119895]T119863

119899

119895T119897

119899minus1 (26g)

Then minimize Π120581

119892with respect to the unknowns As

119897119899minus1

is known in the 119899th time step the derivative of 1205870is 0

6 Mathematical Problems in Engineering

Minimize 120587119894119894 120587

119894119895 120587

119895119894 and 120587

119895119895with respect to 119863

119894 and 119863

119895

and we have

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 120581 [1198791

119894]T[119879

1

119894] 997888rarr [119870

119894119894] (27a)

1205972120587

119894119895

120597 119863119899

119894T120597 119863

119899

119895

= 120581 [1198791

119894]T[119879

2

119895] 997888rarr [119870

119894119895] (27b)

1205972120587

119895119894

120597 119863119899

119895T120597 119863

119899

119894

= 120581 [1198792

119895]T[119879

1

119894] 997888rarr [119870

119895119894] (27c)

1205972120587

119895119895

120597 119863119899

119895T120597 119863

119899

119895

= 120581 [1198792

119895]T[119879

2

119895] 997888rarr [119870

119895119895] (27d)

These 6 times 6 submatrices in the equation above are added intothe global stiffness matrix in (5) Minimizing 120587

119894and 120587

119895with

respect to 119863119894 and 119863

119895 leads to

minus120597120587

119894(0)

120597 119863119894

= minus120581 [1198791

119894]T119897

119899minus1 997888rarr 119865

119894 (28a)

minus

120597120587119895(0)

120597 119863119895

= 120581 [1198792

119895]T119897

119899minus1 997888rarr 119865

119895 (28b)

which are added to the right side of (5)

34 Bonding Submatrices at Fixed Points As the boundaryconditions some elements are fixed at specific points whichare called fixed points in the original DDA As shown inFigure 5(a) the point 119875 of the element I

119894is a fixed point

119894 = 1 2 119872 Denote the coordinates of 119875 at the timenode 119905

119899by (119909

119899

119875 119910

119899

119875) 119899 = 1 2 119873 Assume there exists

a fictitious element I0which is immersed in the support

wall and always keeps a statics state during a computationSuch fictitious element does not contribute potential energyto the element system Assume the vertex 119875

0of I

0overlaps

with the fixed point 119875 at the initial time 1199050 In order to resist

the displacement of the point 119875 a bonding spring with thestiffness 120581

0is applied between 119875 and 119875

0 The stiffness 120581

0can

be determined by the following formulation

1205810= 120581119864 (29)

where 120581 is the same factor used in (4)As shown in Figure 5(b) the bonding spring between 119875

and 1198750has the elongation 119889

119899at the time node 119905

119899 Denote

the coordinate of 1198750by (119909 119910) which keeps constant during

a computation Since 119889119899is actually the length of 997888997888rarr119875

0119875 we have

1198892

119899= 119897

119899

119875T119897

119899

119875 (30)

where 119897119899

119875 denotes the vector 997888997888rarr

1198750119875 at 119905

119899 that is

119897119899

119875 = (

119909119899

119875

119910119899

119875

) minus (

119909

119910) (31)

Assume the displacement of 119875 in the 119899th time step is (119906119899

119875 V119899

119875)

Then the coordinates of 119875 at 119905119899can be written as (119909

119899

119875 119910

119899

119875) =

(119909119899minus1

119875+ 119906

119899

119875 119910

119899minus1

119875+ V119899

119875) which is substituted into (31) that is

119897119899

119875 = (

119909119899minus1

119875+ 119906

119899

119875

119910119899minus1

119875+ V119899

119875

) minus (

119909

119910) = 119897

119899minus1 + (

119906119899

119875

V119899

119875

) (32)

Substitute (3) and (1) into the equation above and then wehave

119897119899

119875 = 119897

119899minus1

119875 + [119879

119875

119894] 119863

119899

119894 (33)

where [119879119875

119894] indicates [119879

119894(119909

119899

119875 119910

119899

119875)] Substituting (33) into (30)

leads to

1198892

119899= 119897

119899minus1

119875T119897

119899minus1

119875 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

119875T[119879

1

119894] 119863

119899

119894 + [119879

1

119894]T119863

119899

119894T119897

119899minus1

119875

(34)

Then we can express the strain energy of the bonding springbetween 119875 and 119875

0as

Π1205810

119875=

1

2120581

0119889

2

119899= 120587

0+ 120587

119894119894+ 120587

119894 (35)

where

1205870=

1205810

2119897

119899minus1

119875T119897

119899minus1

119875 (36a)

120587119894119894

=120581

0

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (36b)

120587119894= 120581

0[119879

1

119894]T119863

119899

119894T119897

119899minus1

119875 (36c)

Obviously theminimization of1205870with respect to 119863

119899

119894 is zero

Then we minimize 120587119894119894and 120587

119894

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 1205810[119879

119875]T[119879

119875] 997888rarr [119870

119894119894] (37a)

120597120587119894

120597 119863119899

119894T

= 1205810119897

119899minus1

119875T[119879

119875] 997888rarr 119865

119894 (37b)

where [119870119894119894] and 119865

119894 are added to the stiffness matrix and the

right item of (5)

35 Line Loading Submatrices Assume a loading 119865 =

(119865119909(119909 119910) 119865

119910(119909 119910))

T is applied along a line upon the bound-ary of the elementI

119894 119894 = 1 2 119872 The coordinate of any

point on the line at the timemoment 119905119899can be determined by

the following parametric equations

119909119899

= 119909119899(120585) = (119909

119899

2minus 119909

119899

1) 120585 + 119909

119899

1

119910119899

= 119910119899(120585) = (119910

119899

2minus 119910

119899

1) 120585 + 119910

119899

1

0 ⩽ 120585 ⩽ 1

(38)

Mathematical Problems in Engineering 7

119973i

1205810 1205810

P0

P

1199730

(a)

119973i

dn

12058101205810

P0

P

1199730

(b)

Figure 5 Bonding springs are applied at fixed points (a) The length of the bonding spring applied between the fixed point 119875 of the elementI

119894and the vertex 119875

0of the fictitious support elementI

0is zero at 119905

0 (b) At 119905

119899 the bonding spring between 119875

0and 119875 has the elongation 119889

119899

where (119909119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) are the coordinates of the ending

points of the line whose length at the end of the last time stepcan be computed by

119904119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

1)

2

+ (119910119899minus1

2minus 119910

119899minus1

1)

2

(39)

where time step 119899 = 1 2 119873 Then we can obtain thepotential energy contributed by the loading 119865

Π119904= minusint

1

0

(

119906119899(119909 (120585) 119910 (120585))

V119899(119909 (119904) 119910 (119904))

)

T

(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (40)

Substituting (3) and (1) into (40) leads to

Π119904= minus 119863

119899

119894Tint

1

0

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (41)

Minimize Π119904with respect to 119863

119899

120585 and we have

120597Π119904(0)

120597 119863119899

119894T

= int

0

1

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 997888rarr 119865119894 (42)

The 6 times 1 submatrix is added to the right item of the globalequation (5) If 119865 = (119865

119909 119865

119910) which is constant (42) has the

analytical form

120597Π119904(0)

120597 119863119899

119894T

= 119904119899minus1

((((((((((

(

119865119909

119865119910

21199100minus 119910

2119865

119909+

119909 minus 21199090

2119865

119910

119909 minus 21199090

2119865

119909

119910 minus 21199100

2119865

119910

21199100minus 119910

4119865

119909+

119909 minus 21199090

4119865

119910

))))))))))

)

997888rarr 119865119894

(43)

where 119909 = 1199092+ 119909

1 119910 = 119910

2+ 119910

1and (119909

0 119910

0) is the coordinate

of the barycenter of the elementI119894

36 Volume Force Submatrix When the elementI119894bears the

body loading 119891(119909 119910) = (119891119909(119909 119910) 119891

119910(119909 119910))

T 119894 = 1 2 119872

the potential energy Π119908due to the force 119891 is calculated as

Π119908

= minus∬I119894

(

119906119899(119909 119910)

V119899(119909 119910)

)

T

119891 (119909 119910) 119889119909 119889119910 (44)

Substituting (1) into the equations above leads to

Π119908

= minus 119863119899

119894T∬

I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 (45)

Minimize Π119908with respect to 119863

119899

120585 and we have

120597Π119908

(0)

120597 119863119899

119894T

= minus∬I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 997888rarr [119865

119894]

(46)

The 6 times 1 submatrix above is added to the right side of theglobal simultaneous equation (5) When the body force isconstant say 119891(119909 119910) = (119891

119909 119891

119910)T (46) becomes

120597Π119908

(0)

120597 [119863119899

119894]T

= minus [119863119899

119894]T(119891

119909119860

119894 119891

119910119860

119894 0 0 0 0)

T997888rarr [119865

119894] (47)

where 119860119894denotes the area of the elementI

119894

37 Contact Submatrices The element edges along theboundaries of different blocks may contact with each otherThis kind of interactions is allowed in three modes vertex-to-vertex edge-to-edge and vertex-to-edge contacts In two-dimensional problems both vertex-to-vertex and edge-to-edge contacts can be decomposed into couples of vertex-to-edge ones The BBM adopts the ldquorigidrdquo contact model Inorder to prevent penetration and relative sliding along the

8 Mathematical Problems in Engineering

119973j

P3

120581perp

120581

P1119973i

P2

(a)

P3

P0P1

119973i

hperp

h

P2

(b)

Figure 6 Contact springs between the elementsI119894andI

119895 (a) The normal contact spring with the stiffness 120581

perpand the shear contact spring

with the stiffness 120581 (b) The normal contact distance ℎ

perpand the shear contact distance ℎ

interfaces of a close contact a normal contact spring and ashear contact spring have to be applied at the contact point

During each time step the simultaneous equilibriumequations in (5) have to be assembled and solved repeat-edly whenever the implementation of any contact springchanges All of the contact springs must satisfy the followingconstraints no tension appears in an open contact and nopenetration occurs in a close one These two conditions canbe fulfilled by an open-close iteration process to controlthe installation and uninstallation of every contact springin each time step Readers are recommended to refer tothe dissertation [1] for the implementation of open-closeiterations

As illustrated in Figure 6(a) the vertex 1198751of the element

I119894and the edge 119875

2119875

3of the element I

119895are in a contact

where a normal contact spring with the stiffness 120581perpand a

shear contact spring with the stiffness 120581are applied As

shown in Figure 6(b) assume the point 1198750on the edge 119875

2119875

3

is the contact point 1198751is a vertex of the block which is

different from the block where 1198750 119875

2 and 119875

3locate Denote

the coordinate of 119875120572at 119905

119899by (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 and 119899 =

1 2 119873 The displacement of 119875120572in the 119899th time step can

be expressed as

U (119909119899

120572 119910

119899

120572) = (

119906119899

120572

V119899

120572

) = (

119909119899

120572minus 119909

119899minus1

120572

119910119899

120572minus 119910

119899minus1

120572

) (48)

Substitute (1) into the above equation and we have

(

119906119899

1

V119899

1

) = [1198791

119894] 119863

119899

119894 (49)

(

119906119899

120572

V119899

120572

) = [119879120573

119895] 119863

119899

119895 120573 = 0 2 3 (50)

where [119879120572

120585] is the abbreviation of [119879

120585(119909

120572 119910

120572)] 120572 = 0 1 2 3

In every iteration step the normal contact spring and theshear contact spring contribute the strain energiesΠ

perpandΠ

respectively and they can be expressed as

Πperp

=1

2120581

perpℎ

2

perp (51a)

Π=

1

2120581

2

(51b)

where ℎperpand ℎ

denote the normal and the shear contact

displacement respectively As shown in Figure 6(b) ℎperpis the

distance from 1198751to the edge 119875

2119875

3and can be computed as

follows

ℎperp

=119860

119899

119871119899 (52)

where119860119899 is the area of the triangleΔ

119875111987521198753and 119871

119899 denotes thelength of the edge 119875

2119875

3 Since ℎ

is the projection of the line

1198750119875

1on the edge 119875

2119875

3or its extension line we have

ℎ=

997888997888997888rarr119875

2119875

3sdot997888997888997888rarr119875

0119875

1

10038161003816100381610038161003816119875

2119875

3

10038161003816100381610038161003816

=1

119871119899(

119909119899

3minus 119909

119899

2

119910119899

3minus 119910

119899

2

)

T

(

119909119899

1minus 119909

119899

0

119910119899

1minus 119910

119899

0

)

(53)

As follows the minimizations of the potential energy itemsΠ

perpand Π

are derived respectively

371 Normal Contact Submatrices To obtain the derivativeof Π

perpin (51a) we compute the normal contact displacement

ℎperpfirstly 119860119899 in (52) can be deduced by using (48) as

119860119899

=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899

1119910

119899

1

1 119909119899

2119910

119899

2

1 119909119899

3119910

119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1+ 119906

119899

1119910

119899minus1

1+ V119899

1

1 119909119899minus1

2+ 119906

119899

2119910

119899minus1

2+ V119899

2

1 119909119899minus1

3+ 119906

119899

3119910

119899minus1

3+ V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1119910

119899minus1

1

1 119909119899minus1

2119910

119899minus1

2

1 119909119899minus1

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1V119899

1

1 119906119899

2V119899

2

1 119906119899

3V119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(54)

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

Ω120572

(a)

119973j

119973i

119973i+1

119973i+2

(b)

119973j

119973i

119973i+1

119973i+2

(c)

Figure 2 The subdivision process implemented on the block Ω120572

P4P3 P2

P1P6P5

119973j

119973i

(a)

P4

P4

P3 P2

P1P6

119973j

119973i

(b)

Figure 3 Overlapped element edges within a block are glued together by a pair of bonding springs applied between the coincided vertexes(a) The initial status of bonding springs (b) Bonding spring elongations

Section 3 are derived with the illustration figures for trian-gular elements they can be directly employed in the analysisby using other element shape options

22 Element Connections For an element system obtainedthrough the process introduced above the element edgescan be classified into two types the ldquoboundary edgesrdquo thatare parts of block boundaries where contacts may occurand the ldquointernal edgesrdquo including the element edges insideoriginal blocks As shown in Figure 2 I

119894and I

119895are two

adjacent elements of the block Ω120572 From Figure 3(a) it can

be recognized that 1198752119875

3and 119875

4119875

5are boundary edges while

1198751119875

2 119875

3119875

1 119875

5119875

6 and 119875

6119875

4are internal edges Just like the

DDA the BBM applies a normal contact spring and a shearcontact spring at every contact point and controls theirinstallation through open-close iterations

In order to preserve the displacement consistencybetween each overlapped internal edge a couple of bondingsprings are applied between the coincided vertexes Asillustrated in Figure 3(b) the overlapped edges 119875

3119875

1and 119875

6119875

4

are glued together by two bonding springs applied between119875

1and 119875

6and between 119875

3and 119875

4 respectively The bonding

springs between every two elements have the same stiffnessAnd we use the following formulation to determine thebonding spring stiffness between the elementsI

119894andI

119895

120581119894119895

=120581119866

(12) (119904119894+ 119904

119895)

(4)

where 120581 is a constant 119866 is the average shear modulus ofthe two elements and 119904

119894and 119904

119895denote the areas of I

119894

and I119895 respectively By using 120599 to denote the ratio of the

maximum element area with the minimum element area fora subdivision we can take the value of 120581 ranging from 10 (for120599 asymp 1) to 1000 (for 120599 ≫ 1)

The stiffness given by (4) is the minimum requirementfor the proper implementation of bonding springs Althoughlarger stiffness is also suitable in theory very large one isdetrimental to the computing convergence and even leads toill-conditioned global stiffness matrix For the simplicity weuse 120581 instead of 120581

119894119895to denote bonding spring stiffness in the

following sections

3 Computing Formulations

The BBM implements an incremental iteration process tosolve both dynamics and statics problems of the elementsystem I

1I

2 I

119872 on a given time mesh 0 = 119905

0lt

1199051

lt sdot sdot sdot lt 119905119899

lt sdot sdot sdot lt 119905119873 In particular time steps used

in statics problems represent the loading steps Below the119899th time step indicates the interval [119905

119899minus1 119905

119899] Obviously it

is equivalent to say the value of a variable at the end of the(119899 minus 1)th step and the one at the beginning of the 119899th step

For the element system I1I

2 I

119872 generated in

the previous section the total potential energy Π is thesummation of all sources of potential energies includingthe strain energies inertia effects external loads volumeforces bonding loadings and contact forces Minimizingthe potential energy Π in the 119899th step leads to the globalsimultaneous equilibrium equations

(

11987011

11987012

sdot sdot sdot 1198701119872

11987021

11987022

sdot sdot sdot 1198702119872

d

1198701198721

1198701198722

sdot sdot sdot 119870119872119872

)(

119863119899

1

119863119899

2

119863119899

119872

) = (

1198651

1198652

119865119872

) (5)

where the submatrices [119870119894119895] and 119865

119894 are of the sizes 6 times 6 and

6times 1 respectively while 119863119899

119894 denotes the unknown vector to

4 Mathematical Problems in Engineering

be determined 119894 119895 = 1 2 119872 The detailed formulationsof these submatrices [119870

119894119895] and 119865

119894 are derived as follows

31 Element Elastic Submatrices The strain energy of the 119894thelementI

119894can be expressed as

Π119894

119890=

1

2∬

I119894

120576119899

119894T(120590

119899

119894 + 2 120590

119899

119894) 119889119909 119889119910 (6)

where the vectors 120576119899

119894 120590119899

119894 and 120590

119899

119894 denote the strain stress

and the initial stress in the 119899th time step respectively Forisotropic linearly elastic materials 120590

119899

119894 and 120576

119899

119894 have the

following relationship

120590119899

119894 =

119864

1 minus ]2(

1 ] 0

] 1 0

0 0(1 minus ])

2

)120576119899

119894 = [119864

119894] 120576

119899

119894 (7)

in which 119864 and ] indicate Youngrsquos modulus and Poissonrsquosratio respectively The strain 120576

119899

119894 can be expressed by the

displacement as

120576119899

119894 = (

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

) 119863119899

119894 = [Θ] 119863

119899

119894 (8)

Substituting (7) and (8) into (6) we have

Π119894

119890=

1

2119863

119899

119894T(∬

I119894

[Θ]T[119864

119894] [Θ]) 119863

119899

119894

+ 119863119899

119894T∬

I119894

120590119899

119894 119889119909 119889119910

(9)

Minimizing Π119894

119890with respect to 119863

119899

119894 leads to

1205972Π

119894

119890

120597 119863119899

119894T120597 119863

119899

119894

= ∬I119894

[Θ]T[119864

119894] [Θ] 119889119909 119889119910 997888rarr [119870

119894119894]

minus120597Π

119894

119890(0)

120597 119863119899

119894T

= minus∬I119894

[Θ]T120590

119899

119894 119889119909 119889119910 997888rarr 119865

119894

(10)

which are added to the stiffness matrix and the right item of(5) respectively

32 Element Inertia Submatrices The potential energy con-tributed by the inertia force acting on the element I

119894at the

moment 119905 can be expressed as follows

Π119894

120588= ∬

I119894

119880119894(119905 119909 119910)

T(120588

1205972

1205971199052119880

119894(119905 119909 119910)) 119889119909 119889119910 (11)

where 119880119894(119905 119909 119910) and 120588 indicate the displacement and the

mass per area respectively and the multiplication betweenthe parentheses in the integrand of the equation above

thereby computes the density of the inertia force on I119894

Substituting (1) into (11) leads to

Π119894

120588= 119863

119894(119905)

T∬

I119894

120588 [119879119894]T[119879

119894]120597

2119863

119894(119905)

1205971199052119889119909 119889119910 (12)

Using the Newmark time integrating method we have thefollowing finite difference approximation

1205972119863

119894(119905)

1205971199052= (

2 119863119899

119894

Δ2

119899

minus

2 119881119899minus1

119894

Δ119899

) (13)

whereΔ119899

= 119905119899minus119905

119899minus1is the 119899th stepsize and119881

119899minus1

119894indicates the

initial velocity of the 119899th time step and can be updated by theiteration formulation below

119881119899

119894 =

2 119863119899minus1

119894

Δ119899

minus 119881119899minus1

119894 119899 ⩾ 2 (14)

Substituting (13) into (12) we then have

Π119894

120588=

2 119863119899

119894T

Δ2

119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119863

119899

119894

minus2 119863

119899

119894T

Δ119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119881

119899minus1

119894

(15)

Minimizing the right items of the equation abovewith respectto 119863

119899

119894 leads to

2

Δ2

119899

∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910 997888rarr [119870

119894119894] (16)

2

Δ119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119881

119899minus1

119894 997888rarr 119865

119894 (17)

where the integrals can be solved by the simplex integralmethod [34] The submatrices equation (16) and equation(17) are added to the stiffness matrix and the right itemof the global simultaneous equilibrium equations in (5)respectively

33 Bonding Submatrices In order to preserve the dis-placement consistency along the element interfaces withina block each pair of initially overlapped element edges isglued together by two bonding springs applied between thecoincided vertexes As shown in Figure 4(a) the points119875

1and

1198752are a pair of initially overlapped vertexes of the elementsI

119894

and I119895which are within the same block A bonding spring

with the stiffness 120581 is applied between 1198751and 119875

2to glue I

119894

and I119895together Denote the coordinate of 119875

1and 119875

2at the

time node 119905119899by (119909

119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) respectively

Firstly we compute the elongation 119889119899of the bonding

spring between 1198751and 119875

2at the time node 119905

119899 As illustrated

in Figure 4(b) 119889119899is actually the length of 997888997888997888rarr

1198751119875

2and can be

computed from the following equality

1198892

119899= 119897

119899T119897

119899 (18)

Mathematical Problems in Engineering 5

119973i

119973j

120581120581

P2

P1

(a)

119973i

119973j

dn

120581

120581

P2

P1

(b)

Figure 4 Bonding springs applied along overlapped element edges (a) Bonding spring length is zero at 1199050 (b) The bonding spring between

1198751and 119875

2has the elongation 119889

119899at 119905

119899

where 119897119899 denotes the vector 997888997888997888rarr

1198751119875

2at 119905

119899 that is

119897119899 = (

119909119899

2

119910119899

2

) minus (

119909119899

1

119910119899

1

) (19)

Assume the displacements of 1198751and 119875

2in the 119899th time step

are (119906119899

1 V119899

1) and (119906

119899

2 V119899

2) respectivelyThen the coordinates of

1198751and 119875

2at 119905

119899can be decomposed into the incremental form

below

(119909119899

1 119910

119899

1) = (119909

119899minus1

1+ 119906

119899

1 119910

119899minus1

1+ V119899

1) (20a)

(119909119899

2 119910

119899

2) = (119909

119899minus1

2+ 119906

119899

2 119910

119899minus1

2+ V119899

2) (20b)

Substituting (20a) and (20b) into (19) leads to

119897119899 = 119897

119899minus1 + (

119906119899

1

V119899

1

) minus (

119906119899

2

V119899

2

) (21)

Substitute (3) and (1) into the equation above and then wehave

119897119899 = 119897

119899minus1 + [119879

1

119894] 119863

119899

119894 minus [119879

2

119895] 119863

119899

119895 (22)

where [1198791

119894] = [119879

119894(119909

1 119910

1)] and [119879

2

119895] = [119879

119895(119909

2 119910

2)] Substitut-

ing (22) into (18) then 1198892

119899can be expanded as the following

summation

1198892

119899= 119897

119899minus1T119897

119899minus1 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119863119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

T[119879

1

119894] 119863

119899

119894

+ [1198791

119894]T119863

119899

119894T119897

119899minus1 minus 119897

119899minus1T[119879

2

119895] 119863

119899

119895

minus [1198792

119895]T119863

119899

119895T119897

119899minus1

(23)

Next we compute the strain energy Π120581

119892of the bonding

spring between 1198751and 119875

2in the 119899th time step [119905

119899minus1 119905

119899] Π120581

119892

can be expressed as follows

Π120581

119892=

1

2120581119889

2

119899 (24)

Substitute (23) into the expression above andΠ120581

119892becomes the

following sum

Π120581

119892= 120587

0+ 120587

119894+ 120587

119895+ 120587

119894119894+ 120587

119894119895+ 120587

119895119894+ 120587

119895119895 (25)

where

1205870=

120581

2119897

119899minus1T119897

119899minus1 (26a)

120587119894119894

=120581

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (26b)

120587119895119895

=120581

2119863

119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895 (26c)

120587119894119895

= minus120581

2119863

119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895 (26d)

120587119895119894

= minus120581

2119863

119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894 (26e)

120587119894=

120581

2119897

119899minus1T[119879

1

119894] 119863

119899

119894 +

120581

2[119879

1

119894]T119863

119899

119894T119897

119899minus1 (26f)

120587119895= minus

120581

2119897

119899minus1T[119879

2

119895] 119863

119899

119895 minus

120581

2[119879

2

119895]T119863

119899

119895T119897

119899minus1 (26g)

Then minimize Π120581

119892with respect to the unknowns As

119897119899minus1

is known in the 119899th time step the derivative of 1205870is 0

6 Mathematical Problems in Engineering

Minimize 120587119894119894 120587

119894119895 120587

119895119894 and 120587

119895119895with respect to 119863

119894 and 119863

119895

and we have

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 120581 [1198791

119894]T[119879

1

119894] 997888rarr [119870

119894119894] (27a)

1205972120587

119894119895

120597 119863119899

119894T120597 119863

119899

119895

= 120581 [1198791

119894]T[119879

2

119895] 997888rarr [119870

119894119895] (27b)

1205972120587

119895119894

120597 119863119899

119895T120597 119863

119899

119894

= 120581 [1198792

119895]T[119879

1

119894] 997888rarr [119870

119895119894] (27c)

1205972120587

119895119895

120597 119863119899

119895T120597 119863

119899

119895

= 120581 [1198792

119895]T[119879

2

119895] 997888rarr [119870

119895119895] (27d)

These 6 times 6 submatrices in the equation above are added intothe global stiffness matrix in (5) Minimizing 120587

119894and 120587

119895with

respect to 119863119894 and 119863

119895 leads to

minus120597120587

119894(0)

120597 119863119894

= minus120581 [1198791

119894]T119897

119899minus1 997888rarr 119865

119894 (28a)

minus

120597120587119895(0)

120597 119863119895

= 120581 [1198792

119895]T119897

119899minus1 997888rarr 119865

119895 (28b)

which are added to the right side of (5)

34 Bonding Submatrices at Fixed Points As the boundaryconditions some elements are fixed at specific points whichare called fixed points in the original DDA As shown inFigure 5(a) the point 119875 of the element I

119894is a fixed point

119894 = 1 2 119872 Denote the coordinates of 119875 at the timenode 119905

119899by (119909

119899

119875 119910

119899

119875) 119899 = 1 2 119873 Assume there exists

a fictitious element I0which is immersed in the support

wall and always keeps a statics state during a computationSuch fictitious element does not contribute potential energyto the element system Assume the vertex 119875

0of I

0overlaps

with the fixed point 119875 at the initial time 1199050 In order to resist

the displacement of the point 119875 a bonding spring with thestiffness 120581

0is applied between 119875 and 119875

0 The stiffness 120581

0can

be determined by the following formulation

1205810= 120581119864 (29)

where 120581 is the same factor used in (4)As shown in Figure 5(b) the bonding spring between 119875

and 1198750has the elongation 119889

119899at the time node 119905

119899 Denote

the coordinate of 1198750by (119909 119910) which keeps constant during

a computation Since 119889119899is actually the length of 997888997888rarr119875

0119875 we have

1198892

119899= 119897

119899

119875T119897

119899

119875 (30)

where 119897119899

119875 denotes the vector 997888997888rarr

1198750119875 at 119905

119899 that is

119897119899

119875 = (

119909119899

119875

119910119899

119875

) minus (

119909

119910) (31)

Assume the displacement of 119875 in the 119899th time step is (119906119899

119875 V119899

119875)

Then the coordinates of 119875 at 119905119899can be written as (119909

119899

119875 119910

119899

119875) =

(119909119899minus1

119875+ 119906

119899

119875 119910

119899minus1

119875+ V119899

119875) which is substituted into (31) that is

119897119899

119875 = (

119909119899minus1

119875+ 119906

119899

119875

119910119899minus1

119875+ V119899

119875

) minus (

119909

119910) = 119897

119899minus1 + (

119906119899

119875

V119899

119875

) (32)

Substitute (3) and (1) into the equation above and then wehave

119897119899

119875 = 119897

119899minus1

119875 + [119879

119875

119894] 119863

119899

119894 (33)

where [119879119875

119894] indicates [119879

119894(119909

119899

119875 119910

119899

119875)] Substituting (33) into (30)

leads to

1198892

119899= 119897

119899minus1

119875T119897

119899minus1

119875 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

119875T[119879

1

119894] 119863

119899

119894 + [119879

1

119894]T119863

119899

119894T119897

119899minus1

119875

(34)

Then we can express the strain energy of the bonding springbetween 119875 and 119875

0as

Π1205810

119875=

1

2120581

0119889

2

119899= 120587

0+ 120587

119894119894+ 120587

119894 (35)

where

1205870=

1205810

2119897

119899minus1

119875T119897

119899minus1

119875 (36a)

120587119894119894

=120581

0

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (36b)

120587119894= 120581

0[119879

1

119894]T119863

119899

119894T119897

119899minus1

119875 (36c)

Obviously theminimization of1205870with respect to 119863

119899

119894 is zero

Then we minimize 120587119894119894and 120587

119894

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 1205810[119879

119875]T[119879

119875] 997888rarr [119870

119894119894] (37a)

120597120587119894

120597 119863119899

119894T

= 1205810119897

119899minus1

119875T[119879

119875] 997888rarr 119865

119894 (37b)

where [119870119894119894] and 119865

119894 are added to the stiffness matrix and the

right item of (5)

35 Line Loading Submatrices Assume a loading 119865 =

(119865119909(119909 119910) 119865

119910(119909 119910))

T is applied along a line upon the bound-ary of the elementI

119894 119894 = 1 2 119872 The coordinate of any

point on the line at the timemoment 119905119899can be determined by

the following parametric equations

119909119899

= 119909119899(120585) = (119909

119899

2minus 119909

119899

1) 120585 + 119909

119899

1

119910119899

= 119910119899(120585) = (119910

119899

2minus 119910

119899

1) 120585 + 119910

119899

1

0 ⩽ 120585 ⩽ 1

(38)

Mathematical Problems in Engineering 7

119973i

1205810 1205810

P0

P

1199730

(a)

119973i

dn

12058101205810

P0

P

1199730

(b)

Figure 5 Bonding springs are applied at fixed points (a) The length of the bonding spring applied between the fixed point 119875 of the elementI

119894and the vertex 119875

0of the fictitious support elementI

0is zero at 119905

0 (b) At 119905

119899 the bonding spring between 119875

0and 119875 has the elongation 119889

119899

where (119909119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) are the coordinates of the ending

points of the line whose length at the end of the last time stepcan be computed by

119904119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

1)

2

+ (119910119899minus1

2minus 119910

119899minus1

1)

2

(39)

where time step 119899 = 1 2 119873 Then we can obtain thepotential energy contributed by the loading 119865

Π119904= minusint

1

0

(

119906119899(119909 (120585) 119910 (120585))

V119899(119909 (119904) 119910 (119904))

)

T

(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (40)

Substituting (3) and (1) into (40) leads to

Π119904= minus 119863

119899

119894Tint

1

0

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (41)

Minimize Π119904with respect to 119863

119899

120585 and we have

120597Π119904(0)

120597 119863119899

119894T

= int

0

1

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 997888rarr 119865119894 (42)

The 6 times 1 submatrix is added to the right item of the globalequation (5) If 119865 = (119865

119909 119865

119910) which is constant (42) has the

analytical form

120597Π119904(0)

120597 119863119899

119894T

= 119904119899minus1

((((((((((

(

119865119909

119865119910

21199100minus 119910

2119865

119909+

119909 minus 21199090

2119865

119910

119909 minus 21199090

2119865

119909

119910 minus 21199100

2119865

119910

21199100minus 119910

4119865

119909+

119909 minus 21199090

4119865

119910

))))))))))

)

997888rarr 119865119894

(43)

where 119909 = 1199092+ 119909

1 119910 = 119910

2+ 119910

1and (119909

0 119910

0) is the coordinate

of the barycenter of the elementI119894

36 Volume Force Submatrix When the elementI119894bears the

body loading 119891(119909 119910) = (119891119909(119909 119910) 119891

119910(119909 119910))

T 119894 = 1 2 119872

the potential energy Π119908due to the force 119891 is calculated as

Π119908

= minus∬I119894

(

119906119899(119909 119910)

V119899(119909 119910)

)

T

119891 (119909 119910) 119889119909 119889119910 (44)

Substituting (1) into the equations above leads to

Π119908

= minus 119863119899

119894T∬

I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 (45)

Minimize Π119908with respect to 119863

119899

120585 and we have

120597Π119908

(0)

120597 119863119899

119894T

= minus∬I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 997888rarr [119865

119894]

(46)

The 6 times 1 submatrix above is added to the right side of theglobal simultaneous equation (5) When the body force isconstant say 119891(119909 119910) = (119891

119909 119891

119910)T (46) becomes

120597Π119908

(0)

120597 [119863119899

119894]T

= minus [119863119899

119894]T(119891

119909119860

119894 119891

119910119860

119894 0 0 0 0)

T997888rarr [119865

119894] (47)

where 119860119894denotes the area of the elementI

119894

37 Contact Submatrices The element edges along theboundaries of different blocks may contact with each otherThis kind of interactions is allowed in three modes vertex-to-vertex edge-to-edge and vertex-to-edge contacts In two-dimensional problems both vertex-to-vertex and edge-to-edge contacts can be decomposed into couples of vertex-to-edge ones The BBM adopts the ldquorigidrdquo contact model Inorder to prevent penetration and relative sliding along the

8 Mathematical Problems in Engineering

119973j

P3

120581perp

120581

P1119973i

P2

(a)

P3

P0P1

119973i

hperp

h

P2

(b)

Figure 6 Contact springs between the elementsI119894andI

119895 (a) The normal contact spring with the stiffness 120581

perpand the shear contact spring

with the stiffness 120581 (b) The normal contact distance ℎ

perpand the shear contact distance ℎ

interfaces of a close contact a normal contact spring and ashear contact spring have to be applied at the contact point

During each time step the simultaneous equilibriumequations in (5) have to be assembled and solved repeat-edly whenever the implementation of any contact springchanges All of the contact springs must satisfy the followingconstraints no tension appears in an open contact and nopenetration occurs in a close one These two conditions canbe fulfilled by an open-close iteration process to controlthe installation and uninstallation of every contact springin each time step Readers are recommended to refer tothe dissertation [1] for the implementation of open-closeiterations

As illustrated in Figure 6(a) the vertex 1198751of the element

I119894and the edge 119875

2119875

3of the element I

119895are in a contact

where a normal contact spring with the stiffness 120581perpand a

shear contact spring with the stiffness 120581are applied As

shown in Figure 6(b) assume the point 1198750on the edge 119875

2119875

3

is the contact point 1198751is a vertex of the block which is

different from the block where 1198750 119875

2 and 119875

3locate Denote

the coordinate of 119875120572at 119905

119899by (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 and 119899 =

1 2 119873 The displacement of 119875120572in the 119899th time step can

be expressed as

U (119909119899

120572 119910

119899

120572) = (

119906119899

120572

V119899

120572

) = (

119909119899

120572minus 119909

119899minus1

120572

119910119899

120572minus 119910

119899minus1

120572

) (48)

Substitute (1) into the above equation and we have

(

119906119899

1

V119899

1

) = [1198791

119894] 119863

119899

119894 (49)

(

119906119899

120572

V119899

120572

) = [119879120573

119895] 119863

119899

119895 120573 = 0 2 3 (50)

where [119879120572

120585] is the abbreviation of [119879

120585(119909

120572 119910

120572)] 120572 = 0 1 2 3

In every iteration step the normal contact spring and theshear contact spring contribute the strain energiesΠ

perpandΠ

respectively and they can be expressed as

Πperp

=1

2120581

perpℎ

2

perp (51a)

Π=

1

2120581

2

(51b)

where ℎperpand ℎ

denote the normal and the shear contact

displacement respectively As shown in Figure 6(b) ℎperpis the

distance from 1198751to the edge 119875

2119875

3and can be computed as

follows

ℎperp

=119860

119899

119871119899 (52)

where119860119899 is the area of the triangleΔ

119875111987521198753and 119871

119899 denotes thelength of the edge 119875

2119875

3 Since ℎ

is the projection of the line

1198750119875

1on the edge 119875

2119875

3or its extension line we have

ℎ=

997888997888997888rarr119875

2119875

3sdot997888997888997888rarr119875

0119875

1

10038161003816100381610038161003816119875

2119875

3

10038161003816100381610038161003816

=1

119871119899(

119909119899

3minus 119909

119899

2

119910119899

3minus 119910

119899

2

)

T

(

119909119899

1minus 119909

119899

0

119910119899

1minus 119910

119899

0

)

(53)

As follows the minimizations of the potential energy itemsΠ

perpand Π

are derived respectively

371 Normal Contact Submatrices To obtain the derivativeof Π

perpin (51a) we compute the normal contact displacement

ℎperpfirstly 119860119899 in (52) can be deduced by using (48) as

119860119899

=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899

1119910

119899

1

1 119909119899

2119910

119899

2

1 119909119899

3119910

119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1+ 119906

119899

1119910

119899minus1

1+ V119899

1

1 119909119899minus1

2+ 119906

119899

2119910

119899minus1

2+ V119899

2

1 119909119899minus1

3+ 119906

119899

3119910

119899minus1

3+ V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1119910

119899minus1

1

1 119909119899minus1

2119910

119899minus1

2

1 119909119899minus1

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1V119899

1

1 119906119899

2V119899

2

1 119906119899

3V119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(54)

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

be determined 119894 119895 = 1 2 119872 The detailed formulationsof these submatrices [119870

119894119895] and 119865

119894 are derived as follows

31 Element Elastic Submatrices The strain energy of the 119894thelementI

119894can be expressed as

Π119894

119890=

1

2∬

I119894

120576119899

119894T(120590

119899

119894 + 2 120590

119899

119894) 119889119909 119889119910 (6)

where the vectors 120576119899

119894 120590119899

119894 and 120590

119899

119894 denote the strain stress

and the initial stress in the 119899th time step respectively Forisotropic linearly elastic materials 120590

119899

119894 and 120576

119899

119894 have the

following relationship

120590119899

119894 =

119864

1 minus ]2(

1 ] 0

] 1 0

0 0(1 minus ])

2

)120576119899

119894 = [119864

119894] 120576

119899

119894 (7)

in which 119864 and ] indicate Youngrsquos modulus and Poissonrsquosratio respectively The strain 120576

119899

119894 can be expressed by the

displacement as

120576119899

119894 = (

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

) 119863119899

119894 = [Θ] 119863

119899

119894 (8)

Substituting (7) and (8) into (6) we have

Π119894

119890=

1

2119863

119899

119894T(∬

I119894

[Θ]T[119864

119894] [Θ]) 119863

119899

119894

+ 119863119899

119894T∬

I119894

120590119899

119894 119889119909 119889119910

(9)

Minimizing Π119894

119890with respect to 119863

119899

119894 leads to

1205972Π

119894

119890

120597 119863119899

119894T120597 119863

119899

119894

= ∬I119894

[Θ]T[119864

119894] [Θ] 119889119909 119889119910 997888rarr [119870

119894119894]

minus120597Π

119894

119890(0)

120597 119863119899

119894T

= minus∬I119894

[Θ]T120590

119899

119894 119889119909 119889119910 997888rarr 119865

119894

(10)

which are added to the stiffness matrix and the right item of(5) respectively

32 Element Inertia Submatrices The potential energy con-tributed by the inertia force acting on the element I

119894at the

moment 119905 can be expressed as follows

Π119894

120588= ∬

I119894

119880119894(119905 119909 119910)

T(120588

1205972

1205971199052119880

119894(119905 119909 119910)) 119889119909 119889119910 (11)

where 119880119894(119905 119909 119910) and 120588 indicate the displacement and the

mass per area respectively and the multiplication betweenthe parentheses in the integrand of the equation above

thereby computes the density of the inertia force on I119894

Substituting (1) into (11) leads to

Π119894

120588= 119863

119894(119905)

T∬

I119894

120588 [119879119894]T[119879

119894]120597

2119863

119894(119905)

1205971199052119889119909 119889119910 (12)

Using the Newmark time integrating method we have thefollowing finite difference approximation

1205972119863

119894(119905)

1205971199052= (

2 119863119899

119894

Δ2

119899

minus

2 119881119899minus1

119894

Δ119899

) (13)

whereΔ119899

= 119905119899minus119905

119899minus1is the 119899th stepsize and119881

119899minus1

119894indicates the

initial velocity of the 119899th time step and can be updated by theiteration formulation below

119881119899

119894 =

2 119863119899minus1

119894

Δ119899

minus 119881119899minus1

119894 119899 ⩾ 2 (14)

Substituting (13) into (12) we then have

Π119894

120588=

2 119863119899

119894T

Δ2

119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119863

119899

119894

minus2 119863

119899

119894T

Δ119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119881

119899minus1

119894

(15)

Minimizing the right items of the equation abovewith respectto 119863

119899

119894 leads to

2

Δ2

119899

∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910 997888rarr [119870

119894119894] (16)

2

Δ119899

(∬I119894

120588 [119879119894]T[119879

119894] 119889119909 119889119910) 119881

119899minus1

119894 997888rarr 119865

119894 (17)

where the integrals can be solved by the simplex integralmethod [34] The submatrices equation (16) and equation(17) are added to the stiffness matrix and the right itemof the global simultaneous equilibrium equations in (5)respectively

33 Bonding Submatrices In order to preserve the dis-placement consistency along the element interfaces withina block each pair of initially overlapped element edges isglued together by two bonding springs applied between thecoincided vertexes As shown in Figure 4(a) the points119875

1and

1198752are a pair of initially overlapped vertexes of the elementsI

119894

and I119895which are within the same block A bonding spring

with the stiffness 120581 is applied between 1198751and 119875

2to glue I

119894

and I119895together Denote the coordinate of 119875

1and 119875

2at the

time node 119905119899by (119909

119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) respectively

Firstly we compute the elongation 119889119899of the bonding

spring between 1198751and 119875

2at the time node 119905

119899 As illustrated

in Figure 4(b) 119889119899is actually the length of 997888997888997888rarr

1198751119875

2and can be

computed from the following equality

1198892

119899= 119897

119899T119897

119899 (18)

Mathematical Problems in Engineering 5

119973i

119973j

120581120581

P2

P1

(a)

119973i

119973j

dn

120581

120581

P2

P1

(b)

Figure 4 Bonding springs applied along overlapped element edges (a) Bonding spring length is zero at 1199050 (b) The bonding spring between

1198751and 119875

2has the elongation 119889

119899at 119905

119899

where 119897119899 denotes the vector 997888997888997888rarr

1198751119875

2at 119905

119899 that is

119897119899 = (

119909119899

2

119910119899

2

) minus (

119909119899

1

119910119899

1

) (19)

Assume the displacements of 1198751and 119875

2in the 119899th time step

are (119906119899

1 V119899

1) and (119906

119899

2 V119899

2) respectivelyThen the coordinates of

1198751and 119875

2at 119905

119899can be decomposed into the incremental form

below

(119909119899

1 119910

119899

1) = (119909

119899minus1

1+ 119906

119899

1 119910

119899minus1

1+ V119899

1) (20a)

(119909119899

2 119910

119899

2) = (119909

119899minus1

2+ 119906

119899

2 119910

119899minus1

2+ V119899

2) (20b)

Substituting (20a) and (20b) into (19) leads to

119897119899 = 119897

119899minus1 + (

119906119899

1

V119899

1

) minus (

119906119899

2

V119899

2

) (21)

Substitute (3) and (1) into the equation above and then wehave

119897119899 = 119897

119899minus1 + [119879

1

119894] 119863

119899

119894 minus [119879

2

119895] 119863

119899

119895 (22)

where [1198791

119894] = [119879

119894(119909

1 119910

1)] and [119879

2

119895] = [119879

119895(119909

2 119910

2)] Substitut-

ing (22) into (18) then 1198892

119899can be expanded as the following

summation

1198892

119899= 119897

119899minus1T119897

119899minus1 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119863119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

T[119879

1

119894] 119863

119899

119894

+ [1198791

119894]T119863

119899

119894T119897

119899minus1 minus 119897

119899minus1T[119879

2

119895] 119863

119899

119895

minus [1198792

119895]T119863

119899

119895T119897

119899minus1

(23)

Next we compute the strain energy Π120581

119892of the bonding

spring between 1198751and 119875

2in the 119899th time step [119905

119899minus1 119905

119899] Π120581

119892

can be expressed as follows

Π120581

119892=

1

2120581119889

2

119899 (24)

Substitute (23) into the expression above andΠ120581

119892becomes the

following sum

Π120581

119892= 120587

0+ 120587

119894+ 120587

119895+ 120587

119894119894+ 120587

119894119895+ 120587

119895119894+ 120587

119895119895 (25)

where

1205870=

120581

2119897

119899minus1T119897

119899minus1 (26a)

120587119894119894

=120581

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (26b)

120587119895119895

=120581

2119863

119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895 (26c)

120587119894119895

= minus120581

2119863

119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895 (26d)

120587119895119894

= minus120581

2119863

119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894 (26e)

120587119894=

120581

2119897

119899minus1T[119879

1

119894] 119863

119899

119894 +

120581

2[119879

1

119894]T119863

119899

119894T119897

119899minus1 (26f)

120587119895= minus

120581

2119897

119899minus1T[119879

2

119895] 119863

119899

119895 minus

120581

2[119879

2

119895]T119863

119899

119895T119897

119899minus1 (26g)

Then minimize Π120581

119892with respect to the unknowns As

119897119899minus1

is known in the 119899th time step the derivative of 1205870is 0

6 Mathematical Problems in Engineering

Minimize 120587119894119894 120587

119894119895 120587

119895119894 and 120587

119895119895with respect to 119863

119894 and 119863

119895

and we have

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 120581 [1198791

119894]T[119879

1

119894] 997888rarr [119870

119894119894] (27a)

1205972120587

119894119895

120597 119863119899

119894T120597 119863

119899

119895

= 120581 [1198791

119894]T[119879

2

119895] 997888rarr [119870

119894119895] (27b)

1205972120587

119895119894

120597 119863119899

119895T120597 119863

119899

119894

= 120581 [1198792

119895]T[119879

1

119894] 997888rarr [119870

119895119894] (27c)

1205972120587

119895119895

120597 119863119899

119895T120597 119863

119899

119895

= 120581 [1198792

119895]T[119879

2

119895] 997888rarr [119870

119895119895] (27d)

These 6 times 6 submatrices in the equation above are added intothe global stiffness matrix in (5) Minimizing 120587

119894and 120587

119895with

respect to 119863119894 and 119863

119895 leads to

minus120597120587

119894(0)

120597 119863119894

= minus120581 [1198791

119894]T119897

119899minus1 997888rarr 119865

119894 (28a)

minus

120597120587119895(0)

120597 119863119895

= 120581 [1198792

119895]T119897

119899minus1 997888rarr 119865

119895 (28b)

which are added to the right side of (5)

34 Bonding Submatrices at Fixed Points As the boundaryconditions some elements are fixed at specific points whichare called fixed points in the original DDA As shown inFigure 5(a) the point 119875 of the element I

119894is a fixed point

119894 = 1 2 119872 Denote the coordinates of 119875 at the timenode 119905

119899by (119909

119899

119875 119910

119899

119875) 119899 = 1 2 119873 Assume there exists

a fictitious element I0which is immersed in the support

wall and always keeps a statics state during a computationSuch fictitious element does not contribute potential energyto the element system Assume the vertex 119875

0of I

0overlaps

with the fixed point 119875 at the initial time 1199050 In order to resist

the displacement of the point 119875 a bonding spring with thestiffness 120581

0is applied between 119875 and 119875

0 The stiffness 120581

0can

be determined by the following formulation

1205810= 120581119864 (29)

where 120581 is the same factor used in (4)As shown in Figure 5(b) the bonding spring between 119875

and 1198750has the elongation 119889

119899at the time node 119905

119899 Denote

the coordinate of 1198750by (119909 119910) which keeps constant during

a computation Since 119889119899is actually the length of 997888997888rarr119875

0119875 we have

1198892

119899= 119897

119899

119875T119897

119899

119875 (30)

where 119897119899

119875 denotes the vector 997888997888rarr

1198750119875 at 119905

119899 that is

119897119899

119875 = (

119909119899

119875

119910119899

119875

) minus (

119909

119910) (31)

Assume the displacement of 119875 in the 119899th time step is (119906119899

119875 V119899

119875)

Then the coordinates of 119875 at 119905119899can be written as (119909

119899

119875 119910

119899

119875) =

(119909119899minus1

119875+ 119906

119899

119875 119910

119899minus1

119875+ V119899

119875) which is substituted into (31) that is

119897119899

119875 = (

119909119899minus1

119875+ 119906

119899

119875

119910119899minus1

119875+ V119899

119875

) minus (

119909

119910) = 119897

119899minus1 + (

119906119899

119875

V119899

119875

) (32)

Substitute (3) and (1) into the equation above and then wehave

119897119899

119875 = 119897

119899minus1

119875 + [119879

119875

119894] 119863

119899

119894 (33)

where [119879119875

119894] indicates [119879

119894(119909

119899

119875 119910

119899

119875)] Substituting (33) into (30)

leads to

1198892

119899= 119897

119899minus1

119875T119897

119899minus1

119875 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

119875T[119879

1

119894] 119863

119899

119894 + [119879

1

119894]T119863

119899

119894T119897

119899minus1

119875

(34)

Then we can express the strain energy of the bonding springbetween 119875 and 119875

0as

Π1205810

119875=

1

2120581

0119889

2

119899= 120587

0+ 120587

119894119894+ 120587

119894 (35)

where

1205870=

1205810

2119897

119899minus1

119875T119897

119899minus1

119875 (36a)

120587119894119894

=120581

0

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (36b)

120587119894= 120581

0[119879

1

119894]T119863

119899

119894T119897

119899minus1

119875 (36c)

Obviously theminimization of1205870with respect to 119863

119899

119894 is zero

Then we minimize 120587119894119894and 120587

119894

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 1205810[119879

119875]T[119879

119875] 997888rarr [119870

119894119894] (37a)

120597120587119894

120597 119863119899

119894T

= 1205810119897

119899minus1

119875T[119879

119875] 997888rarr 119865

119894 (37b)

where [119870119894119894] and 119865

119894 are added to the stiffness matrix and the

right item of (5)

35 Line Loading Submatrices Assume a loading 119865 =

(119865119909(119909 119910) 119865

119910(119909 119910))

T is applied along a line upon the bound-ary of the elementI

119894 119894 = 1 2 119872 The coordinate of any

point on the line at the timemoment 119905119899can be determined by

the following parametric equations

119909119899

= 119909119899(120585) = (119909

119899

2minus 119909

119899

1) 120585 + 119909

119899

1

119910119899

= 119910119899(120585) = (119910

119899

2minus 119910

119899

1) 120585 + 119910

119899

1

0 ⩽ 120585 ⩽ 1

(38)

Mathematical Problems in Engineering 7

119973i

1205810 1205810

P0

P

1199730

(a)

119973i

dn

12058101205810

P0

P

1199730

(b)

Figure 5 Bonding springs are applied at fixed points (a) The length of the bonding spring applied between the fixed point 119875 of the elementI

119894and the vertex 119875

0of the fictitious support elementI

0is zero at 119905

0 (b) At 119905

119899 the bonding spring between 119875

0and 119875 has the elongation 119889

119899

where (119909119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) are the coordinates of the ending

points of the line whose length at the end of the last time stepcan be computed by

119904119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

1)

2

+ (119910119899minus1

2minus 119910

119899minus1

1)

2

(39)

where time step 119899 = 1 2 119873 Then we can obtain thepotential energy contributed by the loading 119865

Π119904= minusint

1

0

(

119906119899(119909 (120585) 119910 (120585))

V119899(119909 (119904) 119910 (119904))

)

T

(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (40)

Substituting (3) and (1) into (40) leads to

Π119904= minus 119863

119899

119894Tint

1

0

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (41)

Minimize Π119904with respect to 119863

119899

120585 and we have

120597Π119904(0)

120597 119863119899

119894T

= int

0

1

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 997888rarr 119865119894 (42)

The 6 times 1 submatrix is added to the right item of the globalequation (5) If 119865 = (119865

119909 119865

119910) which is constant (42) has the

analytical form

120597Π119904(0)

120597 119863119899

119894T

= 119904119899minus1

((((((((((

(

119865119909

119865119910

21199100minus 119910

2119865

119909+

119909 minus 21199090

2119865

119910

119909 minus 21199090

2119865

119909

119910 minus 21199100

2119865

119910

21199100minus 119910

4119865

119909+

119909 minus 21199090

4119865

119910

))))))))))

)

997888rarr 119865119894

(43)

where 119909 = 1199092+ 119909

1 119910 = 119910

2+ 119910

1and (119909

0 119910

0) is the coordinate

of the barycenter of the elementI119894

36 Volume Force Submatrix When the elementI119894bears the

body loading 119891(119909 119910) = (119891119909(119909 119910) 119891

119910(119909 119910))

T 119894 = 1 2 119872

the potential energy Π119908due to the force 119891 is calculated as

Π119908

= minus∬I119894

(

119906119899(119909 119910)

V119899(119909 119910)

)

T

119891 (119909 119910) 119889119909 119889119910 (44)

Substituting (1) into the equations above leads to

Π119908

= minus 119863119899

119894T∬

I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 (45)

Minimize Π119908with respect to 119863

119899

120585 and we have

120597Π119908

(0)

120597 119863119899

119894T

= minus∬I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 997888rarr [119865

119894]

(46)

The 6 times 1 submatrix above is added to the right side of theglobal simultaneous equation (5) When the body force isconstant say 119891(119909 119910) = (119891

119909 119891

119910)T (46) becomes

120597Π119908

(0)

120597 [119863119899

119894]T

= minus [119863119899

119894]T(119891

119909119860

119894 119891

119910119860

119894 0 0 0 0)

T997888rarr [119865

119894] (47)

where 119860119894denotes the area of the elementI

119894

37 Contact Submatrices The element edges along theboundaries of different blocks may contact with each otherThis kind of interactions is allowed in three modes vertex-to-vertex edge-to-edge and vertex-to-edge contacts In two-dimensional problems both vertex-to-vertex and edge-to-edge contacts can be decomposed into couples of vertex-to-edge ones The BBM adopts the ldquorigidrdquo contact model Inorder to prevent penetration and relative sliding along the

8 Mathematical Problems in Engineering

119973j

P3

120581perp

120581

P1119973i

P2

(a)

P3

P0P1

119973i

hperp

h

P2

(b)

Figure 6 Contact springs between the elementsI119894andI

119895 (a) The normal contact spring with the stiffness 120581

perpand the shear contact spring

with the stiffness 120581 (b) The normal contact distance ℎ

perpand the shear contact distance ℎ

interfaces of a close contact a normal contact spring and ashear contact spring have to be applied at the contact point

During each time step the simultaneous equilibriumequations in (5) have to be assembled and solved repeat-edly whenever the implementation of any contact springchanges All of the contact springs must satisfy the followingconstraints no tension appears in an open contact and nopenetration occurs in a close one These two conditions canbe fulfilled by an open-close iteration process to controlthe installation and uninstallation of every contact springin each time step Readers are recommended to refer tothe dissertation [1] for the implementation of open-closeiterations

As illustrated in Figure 6(a) the vertex 1198751of the element

I119894and the edge 119875

2119875

3of the element I

119895are in a contact

where a normal contact spring with the stiffness 120581perpand a

shear contact spring with the stiffness 120581are applied As

shown in Figure 6(b) assume the point 1198750on the edge 119875

2119875

3

is the contact point 1198751is a vertex of the block which is

different from the block where 1198750 119875

2 and 119875

3locate Denote

the coordinate of 119875120572at 119905

119899by (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 and 119899 =

1 2 119873 The displacement of 119875120572in the 119899th time step can

be expressed as

U (119909119899

120572 119910

119899

120572) = (

119906119899

120572

V119899

120572

) = (

119909119899

120572minus 119909

119899minus1

120572

119910119899

120572minus 119910

119899minus1

120572

) (48)

Substitute (1) into the above equation and we have

(

119906119899

1

V119899

1

) = [1198791

119894] 119863

119899

119894 (49)

(

119906119899

120572

V119899

120572

) = [119879120573

119895] 119863

119899

119895 120573 = 0 2 3 (50)

where [119879120572

120585] is the abbreviation of [119879

120585(119909

120572 119910

120572)] 120572 = 0 1 2 3

In every iteration step the normal contact spring and theshear contact spring contribute the strain energiesΠ

perpandΠ

respectively and they can be expressed as

Πperp

=1

2120581

perpℎ

2

perp (51a)

Π=

1

2120581

2

(51b)

where ℎperpand ℎ

denote the normal and the shear contact

displacement respectively As shown in Figure 6(b) ℎperpis the

distance from 1198751to the edge 119875

2119875

3and can be computed as

follows

ℎperp

=119860

119899

119871119899 (52)

where119860119899 is the area of the triangleΔ

119875111987521198753and 119871

119899 denotes thelength of the edge 119875

2119875

3 Since ℎ

is the projection of the line

1198750119875

1on the edge 119875

2119875

3or its extension line we have

ℎ=

997888997888997888rarr119875

2119875

3sdot997888997888997888rarr119875

0119875

1

10038161003816100381610038161003816119875

2119875

3

10038161003816100381610038161003816

=1

119871119899(

119909119899

3minus 119909

119899

2

119910119899

3minus 119910

119899

2

)

T

(

119909119899

1minus 119909

119899

0

119910119899

1minus 119910

119899

0

)

(53)

As follows the minimizations of the potential energy itemsΠ

perpand Π

are derived respectively

371 Normal Contact Submatrices To obtain the derivativeof Π

perpin (51a) we compute the normal contact displacement

ℎperpfirstly 119860119899 in (52) can be deduced by using (48) as

119860119899

=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899

1119910

119899

1

1 119909119899

2119910

119899

2

1 119909119899

3119910

119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1+ 119906

119899

1119910

119899minus1

1+ V119899

1

1 119909119899minus1

2+ 119906

119899

2119910

119899minus1

2+ V119899

2

1 119909119899minus1

3+ 119906

119899

3119910

119899minus1

3+ V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1119910

119899minus1

1

1 119909119899minus1

2119910

119899minus1

2

1 119909119899minus1

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1V119899

1

1 119906119899

2V119899

2

1 119906119899

3V119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(54)

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

119973i

119973j

120581120581

P2

P1

(a)

119973i

119973j

dn

120581

120581

P2

P1

(b)

Figure 4 Bonding springs applied along overlapped element edges (a) Bonding spring length is zero at 1199050 (b) The bonding spring between

1198751and 119875

2has the elongation 119889

119899at 119905

119899

where 119897119899 denotes the vector 997888997888997888rarr

1198751119875

2at 119905

119899 that is

119897119899 = (

119909119899

2

119910119899

2

) minus (

119909119899

1

119910119899

1

) (19)

Assume the displacements of 1198751and 119875

2in the 119899th time step

are (119906119899

1 V119899

1) and (119906

119899

2 V119899

2) respectivelyThen the coordinates of

1198751and 119875

2at 119905

119899can be decomposed into the incremental form

below

(119909119899

1 119910

119899

1) = (119909

119899minus1

1+ 119906

119899

1 119910

119899minus1

1+ V119899

1) (20a)

(119909119899

2 119910

119899

2) = (119909

119899minus1

2+ 119906

119899

2 119910

119899minus1

2+ V119899

2) (20b)

Substituting (20a) and (20b) into (19) leads to

119897119899 = 119897

119899minus1 + (

119906119899

1

V119899

1

) minus (

119906119899

2

V119899

2

) (21)

Substitute (3) and (1) into the equation above and then wehave

119897119899 = 119897

119899minus1 + [119879

1

119894] 119863

119899

119894 minus [119879

2

119895] 119863

119899

119895 (22)

where [1198791

119894] = [119879

119894(119909

1 119910

1)] and [119879

2

119895] = [119879

119895(119909

2 119910

2)] Substitut-

ing (22) into (18) then 1198892

119899can be expanded as the following

summation

1198892

119899= 119897

119899minus1T119897

119899minus1 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119863119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895

minus 119863119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

T[119879

1

119894] 119863

119899

119894

+ [1198791

119894]T119863

119899

119894T119897

119899minus1 minus 119897

119899minus1T[119879

2

119895] 119863

119899

119895

minus [1198792

119895]T119863

119899

119895T119897

119899minus1

(23)

Next we compute the strain energy Π120581

119892of the bonding

spring between 1198751and 119875

2in the 119899th time step [119905

119899minus1 119905

119899] Π120581

119892

can be expressed as follows

Π120581

119892=

1

2120581119889

2

119899 (24)

Substitute (23) into the expression above andΠ120581

119892becomes the

following sum

Π120581

119892= 120587

0+ 120587

119894+ 120587

119895+ 120587

119894119894+ 120587

119894119895+ 120587

119895119894+ 120587

119895119895 (25)

where

1205870=

120581

2119897

119899minus1T119897

119899minus1 (26a)

120587119894119894

=120581

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (26b)

120587119895119895

=120581

2119863

119899

119895T[119879

2

119895]T[119879

2

119895] 119863

119899

119895 (26c)

120587119894119895

= minus120581

2119863

119899

119894T[119879

1

119894]T[119879

2

119895] 119863

119899

119895 (26d)

120587119895119894

= minus120581

2119863

119899

119895T[119879

2

119895]T[119879

1

119894] 119863

119899

119894 (26e)

120587119894=

120581

2119897

119899minus1T[119879

1

119894] 119863

119899

119894 +

120581

2[119879

1

119894]T119863

119899

119894T119897

119899minus1 (26f)

120587119895= minus

120581

2119897

119899minus1T[119879

2

119895] 119863

119899

119895 minus

120581

2[119879

2

119895]T119863

119899

119895T119897

119899minus1 (26g)

Then minimize Π120581

119892with respect to the unknowns As

119897119899minus1

is known in the 119899th time step the derivative of 1205870is 0

6 Mathematical Problems in Engineering

Minimize 120587119894119894 120587

119894119895 120587

119895119894 and 120587

119895119895with respect to 119863

119894 and 119863

119895

and we have

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 120581 [1198791

119894]T[119879

1

119894] 997888rarr [119870

119894119894] (27a)

1205972120587

119894119895

120597 119863119899

119894T120597 119863

119899

119895

= 120581 [1198791

119894]T[119879

2

119895] 997888rarr [119870

119894119895] (27b)

1205972120587

119895119894

120597 119863119899

119895T120597 119863

119899

119894

= 120581 [1198792

119895]T[119879

1

119894] 997888rarr [119870

119895119894] (27c)

1205972120587

119895119895

120597 119863119899

119895T120597 119863

119899

119895

= 120581 [1198792

119895]T[119879

2

119895] 997888rarr [119870

119895119895] (27d)

These 6 times 6 submatrices in the equation above are added intothe global stiffness matrix in (5) Minimizing 120587

119894and 120587

119895with

respect to 119863119894 and 119863

119895 leads to

minus120597120587

119894(0)

120597 119863119894

= minus120581 [1198791

119894]T119897

119899minus1 997888rarr 119865

119894 (28a)

minus

120597120587119895(0)

120597 119863119895

= 120581 [1198792

119895]T119897

119899minus1 997888rarr 119865

119895 (28b)

which are added to the right side of (5)

34 Bonding Submatrices at Fixed Points As the boundaryconditions some elements are fixed at specific points whichare called fixed points in the original DDA As shown inFigure 5(a) the point 119875 of the element I

119894is a fixed point

119894 = 1 2 119872 Denote the coordinates of 119875 at the timenode 119905

119899by (119909

119899

119875 119910

119899

119875) 119899 = 1 2 119873 Assume there exists

a fictitious element I0which is immersed in the support

wall and always keeps a statics state during a computationSuch fictitious element does not contribute potential energyto the element system Assume the vertex 119875

0of I

0overlaps

with the fixed point 119875 at the initial time 1199050 In order to resist

the displacement of the point 119875 a bonding spring with thestiffness 120581

0is applied between 119875 and 119875

0 The stiffness 120581

0can

be determined by the following formulation

1205810= 120581119864 (29)

where 120581 is the same factor used in (4)As shown in Figure 5(b) the bonding spring between 119875

and 1198750has the elongation 119889

119899at the time node 119905

119899 Denote

the coordinate of 1198750by (119909 119910) which keeps constant during

a computation Since 119889119899is actually the length of 997888997888rarr119875

0119875 we have

1198892

119899= 119897

119899

119875T119897

119899

119875 (30)

where 119897119899

119875 denotes the vector 997888997888rarr

1198750119875 at 119905

119899 that is

119897119899

119875 = (

119909119899

119875

119910119899

119875

) minus (

119909

119910) (31)

Assume the displacement of 119875 in the 119899th time step is (119906119899

119875 V119899

119875)

Then the coordinates of 119875 at 119905119899can be written as (119909

119899

119875 119910

119899

119875) =

(119909119899minus1

119875+ 119906

119899

119875 119910

119899minus1

119875+ V119899

119875) which is substituted into (31) that is

119897119899

119875 = (

119909119899minus1

119875+ 119906

119899

119875

119910119899minus1

119875+ V119899

119875

) minus (

119909

119910) = 119897

119899minus1 + (

119906119899

119875

V119899

119875

) (32)

Substitute (3) and (1) into the equation above and then wehave

119897119899

119875 = 119897

119899minus1

119875 + [119879

119875

119894] 119863

119899

119894 (33)

where [119879119875

119894] indicates [119879

119894(119909

119899

119875 119910

119899

119875)] Substituting (33) into (30)

leads to

1198892

119899= 119897

119899minus1

119875T119897

119899minus1

119875 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

119875T[119879

1

119894] 119863

119899

119894 + [119879

1

119894]T119863

119899

119894T119897

119899minus1

119875

(34)

Then we can express the strain energy of the bonding springbetween 119875 and 119875

0as

Π1205810

119875=

1

2120581

0119889

2

119899= 120587

0+ 120587

119894119894+ 120587

119894 (35)

where

1205870=

1205810

2119897

119899minus1

119875T119897

119899minus1

119875 (36a)

120587119894119894

=120581

0

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (36b)

120587119894= 120581

0[119879

1

119894]T119863

119899

119894T119897

119899minus1

119875 (36c)

Obviously theminimization of1205870with respect to 119863

119899

119894 is zero

Then we minimize 120587119894119894and 120587

119894

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 1205810[119879

119875]T[119879

119875] 997888rarr [119870

119894119894] (37a)

120597120587119894

120597 119863119899

119894T

= 1205810119897

119899minus1

119875T[119879

119875] 997888rarr 119865

119894 (37b)

where [119870119894119894] and 119865

119894 are added to the stiffness matrix and the

right item of (5)

35 Line Loading Submatrices Assume a loading 119865 =

(119865119909(119909 119910) 119865

119910(119909 119910))

T is applied along a line upon the bound-ary of the elementI

119894 119894 = 1 2 119872 The coordinate of any

point on the line at the timemoment 119905119899can be determined by

the following parametric equations

119909119899

= 119909119899(120585) = (119909

119899

2minus 119909

119899

1) 120585 + 119909

119899

1

119910119899

= 119910119899(120585) = (119910

119899

2minus 119910

119899

1) 120585 + 119910

119899

1

0 ⩽ 120585 ⩽ 1

(38)

Mathematical Problems in Engineering 7

119973i

1205810 1205810

P0

P

1199730

(a)

119973i

dn

12058101205810

P0

P

1199730

(b)

Figure 5 Bonding springs are applied at fixed points (a) The length of the bonding spring applied between the fixed point 119875 of the elementI

119894and the vertex 119875

0of the fictitious support elementI

0is zero at 119905

0 (b) At 119905

119899 the bonding spring between 119875

0and 119875 has the elongation 119889

119899

where (119909119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) are the coordinates of the ending

points of the line whose length at the end of the last time stepcan be computed by

119904119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

1)

2

+ (119910119899minus1

2minus 119910

119899minus1

1)

2

(39)

where time step 119899 = 1 2 119873 Then we can obtain thepotential energy contributed by the loading 119865

Π119904= minusint

1

0

(

119906119899(119909 (120585) 119910 (120585))

V119899(119909 (119904) 119910 (119904))

)

T

(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (40)

Substituting (3) and (1) into (40) leads to

Π119904= minus 119863

119899

119894Tint

1

0

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (41)

Minimize Π119904with respect to 119863

119899

120585 and we have

120597Π119904(0)

120597 119863119899

119894T

= int

0

1

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 997888rarr 119865119894 (42)

The 6 times 1 submatrix is added to the right item of the globalequation (5) If 119865 = (119865

119909 119865

119910) which is constant (42) has the

analytical form

120597Π119904(0)

120597 119863119899

119894T

= 119904119899minus1

((((((((((

(

119865119909

119865119910

21199100minus 119910

2119865

119909+

119909 minus 21199090

2119865

119910

119909 minus 21199090

2119865

119909

119910 minus 21199100

2119865

119910

21199100minus 119910

4119865

119909+

119909 minus 21199090

4119865

119910

))))))))))

)

997888rarr 119865119894

(43)

where 119909 = 1199092+ 119909

1 119910 = 119910

2+ 119910

1and (119909

0 119910

0) is the coordinate

of the barycenter of the elementI119894

36 Volume Force Submatrix When the elementI119894bears the

body loading 119891(119909 119910) = (119891119909(119909 119910) 119891

119910(119909 119910))

T 119894 = 1 2 119872

the potential energy Π119908due to the force 119891 is calculated as

Π119908

= minus∬I119894

(

119906119899(119909 119910)

V119899(119909 119910)

)

T

119891 (119909 119910) 119889119909 119889119910 (44)

Substituting (1) into the equations above leads to

Π119908

= minus 119863119899

119894T∬

I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 (45)

Minimize Π119908with respect to 119863

119899

120585 and we have

120597Π119908

(0)

120597 119863119899

119894T

= minus∬I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 997888rarr [119865

119894]

(46)

The 6 times 1 submatrix above is added to the right side of theglobal simultaneous equation (5) When the body force isconstant say 119891(119909 119910) = (119891

119909 119891

119910)T (46) becomes

120597Π119908

(0)

120597 [119863119899

119894]T

= minus [119863119899

119894]T(119891

119909119860

119894 119891

119910119860

119894 0 0 0 0)

T997888rarr [119865

119894] (47)

where 119860119894denotes the area of the elementI

119894

37 Contact Submatrices The element edges along theboundaries of different blocks may contact with each otherThis kind of interactions is allowed in three modes vertex-to-vertex edge-to-edge and vertex-to-edge contacts In two-dimensional problems both vertex-to-vertex and edge-to-edge contacts can be decomposed into couples of vertex-to-edge ones The BBM adopts the ldquorigidrdquo contact model Inorder to prevent penetration and relative sliding along the

8 Mathematical Problems in Engineering

119973j

P3

120581perp

120581

P1119973i

P2

(a)

P3

P0P1

119973i

hperp

h

P2

(b)

Figure 6 Contact springs between the elementsI119894andI

119895 (a) The normal contact spring with the stiffness 120581

perpand the shear contact spring

with the stiffness 120581 (b) The normal contact distance ℎ

perpand the shear contact distance ℎ

interfaces of a close contact a normal contact spring and ashear contact spring have to be applied at the contact point

During each time step the simultaneous equilibriumequations in (5) have to be assembled and solved repeat-edly whenever the implementation of any contact springchanges All of the contact springs must satisfy the followingconstraints no tension appears in an open contact and nopenetration occurs in a close one These two conditions canbe fulfilled by an open-close iteration process to controlthe installation and uninstallation of every contact springin each time step Readers are recommended to refer tothe dissertation [1] for the implementation of open-closeiterations

As illustrated in Figure 6(a) the vertex 1198751of the element

I119894and the edge 119875

2119875

3of the element I

119895are in a contact

where a normal contact spring with the stiffness 120581perpand a

shear contact spring with the stiffness 120581are applied As

shown in Figure 6(b) assume the point 1198750on the edge 119875

2119875

3

is the contact point 1198751is a vertex of the block which is

different from the block where 1198750 119875

2 and 119875

3locate Denote

the coordinate of 119875120572at 119905

119899by (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 and 119899 =

1 2 119873 The displacement of 119875120572in the 119899th time step can

be expressed as

U (119909119899

120572 119910

119899

120572) = (

119906119899

120572

V119899

120572

) = (

119909119899

120572minus 119909

119899minus1

120572

119910119899

120572minus 119910

119899minus1

120572

) (48)

Substitute (1) into the above equation and we have

(

119906119899

1

V119899

1

) = [1198791

119894] 119863

119899

119894 (49)

(

119906119899

120572

V119899

120572

) = [119879120573

119895] 119863

119899

119895 120573 = 0 2 3 (50)

where [119879120572

120585] is the abbreviation of [119879

120585(119909

120572 119910

120572)] 120572 = 0 1 2 3

In every iteration step the normal contact spring and theshear contact spring contribute the strain energiesΠ

perpandΠ

respectively and they can be expressed as

Πperp

=1

2120581

perpℎ

2

perp (51a)

Π=

1

2120581

2

(51b)

where ℎperpand ℎ

denote the normal and the shear contact

displacement respectively As shown in Figure 6(b) ℎperpis the

distance from 1198751to the edge 119875

2119875

3and can be computed as

follows

ℎperp

=119860

119899

119871119899 (52)

where119860119899 is the area of the triangleΔ

119875111987521198753and 119871

119899 denotes thelength of the edge 119875

2119875

3 Since ℎ

is the projection of the line

1198750119875

1on the edge 119875

2119875

3or its extension line we have

ℎ=

997888997888997888rarr119875

2119875

3sdot997888997888997888rarr119875

0119875

1

10038161003816100381610038161003816119875

2119875

3

10038161003816100381610038161003816

=1

119871119899(

119909119899

3minus 119909

119899

2

119910119899

3minus 119910

119899

2

)

T

(

119909119899

1minus 119909

119899

0

119910119899

1minus 119910

119899

0

)

(53)

As follows the minimizations of the potential energy itemsΠ

perpand Π

are derived respectively

371 Normal Contact Submatrices To obtain the derivativeof Π

perpin (51a) we compute the normal contact displacement

ℎperpfirstly 119860119899 in (52) can be deduced by using (48) as

119860119899

=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899

1119910

119899

1

1 119909119899

2119910

119899

2

1 119909119899

3119910

119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1+ 119906

119899

1119910

119899minus1

1+ V119899

1

1 119909119899minus1

2+ 119906

119899

2119910

119899minus1

2+ V119899

2

1 119909119899minus1

3+ 119906

119899

3119910

119899minus1

3+ V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1119910

119899minus1

1

1 119909119899minus1

2119910

119899minus1

2

1 119909119899minus1

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1V119899

1

1 119906119899

2V119899

2

1 119906119899

3V119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(54)

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

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6 Mathematical Problems in Engineering

Minimize 120587119894119894 120587

119894119895 120587

119895119894 and 120587

119895119895with respect to 119863

119894 and 119863

119895

and we have

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 120581 [1198791

119894]T[119879

1

119894] 997888rarr [119870

119894119894] (27a)

1205972120587

119894119895

120597 119863119899

119894T120597 119863

119899

119895

= 120581 [1198791

119894]T[119879

2

119895] 997888rarr [119870

119894119895] (27b)

1205972120587

119895119894

120597 119863119899

119895T120597 119863

119899

119894

= 120581 [1198792

119895]T[119879

1

119894] 997888rarr [119870

119895119894] (27c)

1205972120587

119895119895

120597 119863119899

119895T120597 119863

119899

119895

= 120581 [1198792

119895]T[119879

2

119895] 997888rarr [119870

119895119895] (27d)

These 6 times 6 submatrices in the equation above are added intothe global stiffness matrix in (5) Minimizing 120587

119894and 120587

119895with

respect to 119863119894 and 119863

119895 leads to

minus120597120587

119894(0)

120597 119863119894

= minus120581 [1198791

119894]T119897

119899minus1 997888rarr 119865

119894 (28a)

minus

120597120587119895(0)

120597 119863119895

= 120581 [1198792

119895]T119897

119899minus1 997888rarr 119865

119895 (28b)

which are added to the right side of (5)

34 Bonding Submatrices at Fixed Points As the boundaryconditions some elements are fixed at specific points whichare called fixed points in the original DDA As shown inFigure 5(a) the point 119875 of the element I

119894is a fixed point

119894 = 1 2 119872 Denote the coordinates of 119875 at the timenode 119905

119899by (119909

119899

119875 119910

119899

119875) 119899 = 1 2 119873 Assume there exists

a fictitious element I0which is immersed in the support

wall and always keeps a statics state during a computationSuch fictitious element does not contribute potential energyto the element system Assume the vertex 119875

0of I

0overlaps

with the fixed point 119875 at the initial time 1199050 In order to resist

the displacement of the point 119875 a bonding spring with thestiffness 120581

0is applied between 119875 and 119875

0 The stiffness 120581

0can

be determined by the following formulation

1205810= 120581119864 (29)

where 120581 is the same factor used in (4)As shown in Figure 5(b) the bonding spring between 119875

and 1198750has the elongation 119889

119899at the time node 119905

119899 Denote

the coordinate of 1198750by (119909 119910) which keeps constant during

a computation Since 119889119899is actually the length of 997888997888rarr119875

0119875 we have

1198892

119899= 119897

119899

119875T119897

119899

119875 (30)

where 119897119899

119875 denotes the vector 997888997888rarr

1198750119875 at 119905

119899 that is

119897119899

119875 = (

119909119899

119875

119910119899

119875

) minus (

119909

119910) (31)

Assume the displacement of 119875 in the 119899th time step is (119906119899

119875 V119899

119875)

Then the coordinates of 119875 at 119905119899can be written as (119909

119899

119875 119910

119899

119875) =

(119909119899minus1

119875+ 119906

119899

119875 119910

119899minus1

119875+ V119899

119875) which is substituted into (31) that is

119897119899

119875 = (

119909119899minus1

119875+ 119906

119899

119875

119910119899minus1

119875+ V119899

119875

) minus (

119909

119910) = 119897

119899minus1 + (

119906119899

119875

V119899

119875

) (32)

Substitute (3) and (1) into the equation above and then wehave

119897119899

119875 = 119897

119899minus1

119875 + [119879

119875

119894] 119863

119899

119894 (33)

where [119879119875

119894] indicates [119879

119894(119909

119899

119875 119910

119899

119875)] Substituting (33) into (30)

leads to

1198892

119899= 119897

119899minus1

119875T119897

119899minus1

119875 + 119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894

+ 119897119899minus1

119875T[119879

1

119894] 119863

119899

119894 + [119879

1

119894]T119863

119899

119894T119897

119899minus1

119875

(34)

Then we can express the strain energy of the bonding springbetween 119875 and 119875

0as

Π1205810

119875=

1

2120581

0119889

2

119899= 120587

0+ 120587

119894119894+ 120587

119894 (35)

where

1205870=

1205810

2119897

119899minus1

119875T119897

119899minus1

119875 (36a)

120587119894119894

=120581

0

2119863

119899

119894T[119879

1

119894]T[119879

1

119894] 119863

119899

119894 (36b)

120587119894= 120581

0[119879

1

119894]T119863

119899

119894T119897

119899minus1

119875 (36c)

Obviously theminimization of1205870with respect to 119863

119899

119894 is zero

Then we minimize 120587119894119894and 120587

119894

1205972120587

119894119894

120597 119863119899

119894T120597 119863

119899

119894

= 1205810[119879

119875]T[119879

119875] 997888rarr [119870

119894119894] (37a)

120597120587119894

120597 119863119899

119894T

= 1205810119897

119899minus1

119875T[119879

119875] 997888rarr 119865

119894 (37b)

where [119870119894119894] and 119865

119894 are added to the stiffness matrix and the

right item of (5)

35 Line Loading Submatrices Assume a loading 119865 =

(119865119909(119909 119910) 119865

119910(119909 119910))

T is applied along a line upon the bound-ary of the elementI

119894 119894 = 1 2 119872 The coordinate of any

point on the line at the timemoment 119905119899can be determined by

the following parametric equations

119909119899

= 119909119899(120585) = (119909

119899

2minus 119909

119899

1) 120585 + 119909

119899

1

119910119899

= 119910119899(120585) = (119910

119899

2minus 119910

119899

1) 120585 + 119910

119899

1

0 ⩽ 120585 ⩽ 1

(38)

Mathematical Problems in Engineering 7

119973i

1205810 1205810

P0

P

1199730

(a)

119973i

dn

12058101205810

P0

P

1199730

(b)

Figure 5 Bonding springs are applied at fixed points (a) The length of the bonding spring applied between the fixed point 119875 of the elementI

119894and the vertex 119875

0of the fictitious support elementI

0is zero at 119905

0 (b) At 119905

119899 the bonding spring between 119875

0and 119875 has the elongation 119889

119899

where (119909119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) are the coordinates of the ending

points of the line whose length at the end of the last time stepcan be computed by

119904119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

1)

2

+ (119910119899minus1

2minus 119910

119899minus1

1)

2

(39)

where time step 119899 = 1 2 119873 Then we can obtain thepotential energy contributed by the loading 119865

Π119904= minusint

1

0

(

119906119899(119909 (120585) 119910 (120585))

V119899(119909 (119904) 119910 (119904))

)

T

(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (40)

Substituting (3) and (1) into (40) leads to

Π119904= minus 119863

119899

119894Tint

1

0

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (41)

Minimize Π119904with respect to 119863

119899

120585 and we have

120597Π119904(0)

120597 119863119899

119894T

= int

0

1

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 997888rarr 119865119894 (42)

The 6 times 1 submatrix is added to the right item of the globalequation (5) If 119865 = (119865

119909 119865

119910) which is constant (42) has the

analytical form

120597Π119904(0)

120597 119863119899

119894T

= 119904119899minus1

((((((((((

(

119865119909

119865119910

21199100minus 119910

2119865

119909+

119909 minus 21199090

2119865

119910

119909 minus 21199090

2119865

119909

119910 minus 21199100

2119865

119910

21199100minus 119910

4119865

119909+

119909 minus 21199090

4119865

119910

))))))))))

)

997888rarr 119865119894

(43)

where 119909 = 1199092+ 119909

1 119910 = 119910

2+ 119910

1and (119909

0 119910

0) is the coordinate

of the barycenter of the elementI119894

36 Volume Force Submatrix When the elementI119894bears the

body loading 119891(119909 119910) = (119891119909(119909 119910) 119891

119910(119909 119910))

T 119894 = 1 2 119872

the potential energy Π119908due to the force 119891 is calculated as

Π119908

= minus∬I119894

(

119906119899(119909 119910)

V119899(119909 119910)

)

T

119891 (119909 119910) 119889119909 119889119910 (44)

Substituting (1) into the equations above leads to

Π119908

= minus 119863119899

119894T∬

I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 (45)

Minimize Π119908with respect to 119863

119899

120585 and we have

120597Π119908

(0)

120597 119863119899

119894T

= minus∬I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 997888rarr [119865

119894]

(46)

The 6 times 1 submatrix above is added to the right side of theglobal simultaneous equation (5) When the body force isconstant say 119891(119909 119910) = (119891

119909 119891

119910)T (46) becomes

120597Π119908

(0)

120597 [119863119899

119894]T

= minus [119863119899

119894]T(119891

119909119860

119894 119891

119910119860

119894 0 0 0 0)

T997888rarr [119865

119894] (47)

where 119860119894denotes the area of the elementI

119894

37 Contact Submatrices The element edges along theboundaries of different blocks may contact with each otherThis kind of interactions is allowed in three modes vertex-to-vertex edge-to-edge and vertex-to-edge contacts In two-dimensional problems both vertex-to-vertex and edge-to-edge contacts can be decomposed into couples of vertex-to-edge ones The BBM adopts the ldquorigidrdquo contact model Inorder to prevent penetration and relative sliding along the

8 Mathematical Problems in Engineering

119973j

P3

120581perp

120581

P1119973i

P2

(a)

P3

P0P1

119973i

hperp

h

P2

(b)

Figure 6 Contact springs between the elementsI119894andI

119895 (a) The normal contact spring with the stiffness 120581

perpand the shear contact spring

with the stiffness 120581 (b) The normal contact distance ℎ

perpand the shear contact distance ℎ

interfaces of a close contact a normal contact spring and ashear contact spring have to be applied at the contact point

During each time step the simultaneous equilibriumequations in (5) have to be assembled and solved repeat-edly whenever the implementation of any contact springchanges All of the contact springs must satisfy the followingconstraints no tension appears in an open contact and nopenetration occurs in a close one These two conditions canbe fulfilled by an open-close iteration process to controlthe installation and uninstallation of every contact springin each time step Readers are recommended to refer tothe dissertation [1] for the implementation of open-closeiterations

As illustrated in Figure 6(a) the vertex 1198751of the element

I119894and the edge 119875

2119875

3of the element I

119895are in a contact

where a normal contact spring with the stiffness 120581perpand a

shear contact spring with the stiffness 120581are applied As

shown in Figure 6(b) assume the point 1198750on the edge 119875

2119875

3

is the contact point 1198751is a vertex of the block which is

different from the block where 1198750 119875

2 and 119875

3locate Denote

the coordinate of 119875120572at 119905

119899by (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 and 119899 =

1 2 119873 The displacement of 119875120572in the 119899th time step can

be expressed as

U (119909119899

120572 119910

119899

120572) = (

119906119899

120572

V119899

120572

) = (

119909119899

120572minus 119909

119899minus1

120572

119910119899

120572minus 119910

119899minus1

120572

) (48)

Substitute (1) into the above equation and we have

(

119906119899

1

V119899

1

) = [1198791

119894] 119863

119899

119894 (49)

(

119906119899

120572

V119899

120572

) = [119879120573

119895] 119863

119899

119895 120573 = 0 2 3 (50)

where [119879120572

120585] is the abbreviation of [119879

120585(119909

120572 119910

120572)] 120572 = 0 1 2 3

In every iteration step the normal contact spring and theshear contact spring contribute the strain energiesΠ

perpandΠ

respectively and they can be expressed as

Πperp

=1

2120581

perpℎ

2

perp (51a)

Π=

1

2120581

2

(51b)

where ℎperpand ℎ

denote the normal and the shear contact

displacement respectively As shown in Figure 6(b) ℎperpis the

distance from 1198751to the edge 119875

2119875

3and can be computed as

follows

ℎperp

=119860

119899

119871119899 (52)

where119860119899 is the area of the triangleΔ

119875111987521198753and 119871

119899 denotes thelength of the edge 119875

2119875

3 Since ℎ

is the projection of the line

1198750119875

1on the edge 119875

2119875

3or its extension line we have

ℎ=

997888997888997888rarr119875

2119875

3sdot997888997888997888rarr119875

0119875

1

10038161003816100381610038161003816119875

2119875

3

10038161003816100381610038161003816

=1

119871119899(

119909119899

3minus 119909

119899

2

119910119899

3minus 119910

119899

2

)

T

(

119909119899

1minus 119909

119899

0

119910119899

1minus 119910

119899

0

)

(53)

As follows the minimizations of the potential energy itemsΠ

perpand Π

are derived respectively

371 Normal Contact Submatrices To obtain the derivativeof Π

perpin (51a) we compute the normal contact displacement

ℎperpfirstly 119860119899 in (52) can be deduced by using (48) as

119860119899

=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899

1119910

119899

1

1 119909119899

2119910

119899

2

1 119909119899

3119910

119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1+ 119906

119899

1119910

119899minus1

1+ V119899

1

1 119909119899minus1

2+ 119906

119899

2119910

119899minus1

2+ V119899

2

1 119909119899minus1

3+ 119906

119899

3119910

119899minus1

3+ V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1119910

119899minus1

1

1 119909119899minus1

2119910

119899minus1

2

1 119909119899minus1

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1V119899

1

1 119906119899

2V119899

2

1 119906119899

3V119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(54)

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

119973i

1205810 1205810

P0

P

1199730

(a)

119973i

dn

12058101205810

P0

P

1199730

(b)

Figure 5 Bonding springs are applied at fixed points (a) The length of the bonding spring applied between the fixed point 119875 of the elementI

119894and the vertex 119875

0of the fictitious support elementI

0is zero at 119905

0 (b) At 119905

119899 the bonding spring between 119875

0and 119875 has the elongation 119889

119899

where (119909119899

1 119910

119899

1) and (119909

119899

2 119910

119899

2) are the coordinates of the ending

points of the line whose length at the end of the last time stepcan be computed by

119904119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

1)

2

+ (119910119899minus1

2minus 119910

119899minus1

1)

2

(39)

where time step 119899 = 1 2 119873 Then we can obtain thepotential energy contributed by the loading 119865

Π119904= minusint

1

0

(

119906119899(119909 (120585) 119910 (120585))

V119899(119909 (119904) 119910 (119904))

)

T

(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (40)

Substituting (3) and (1) into (40) leads to

Π119904= minus 119863

119899

119894Tint

1

0

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 (41)

Minimize Π119904with respect to 119863

119899

120585 and we have

120597Π119904(0)

120597 119863119899

119894T

= int

0

1

[119879119894(120585)]

T(

119865119909(120585)

119865119910(120585)

) 119904119899minus1

119889120585 997888rarr 119865119894 (42)

The 6 times 1 submatrix is added to the right item of the globalequation (5) If 119865 = (119865

119909 119865

119910) which is constant (42) has the

analytical form

120597Π119904(0)

120597 119863119899

119894T

= 119904119899minus1

((((((((((

(

119865119909

119865119910

21199100minus 119910

2119865

119909+

119909 minus 21199090

2119865

119910

119909 minus 21199090

2119865

119909

119910 minus 21199100

2119865

119910

21199100minus 119910

4119865

119909+

119909 minus 21199090

4119865

119910

))))))))))

)

997888rarr 119865119894

(43)

where 119909 = 1199092+ 119909

1 119910 = 119910

2+ 119910

1and (119909

0 119910

0) is the coordinate

of the barycenter of the elementI119894

36 Volume Force Submatrix When the elementI119894bears the

body loading 119891(119909 119910) = (119891119909(119909 119910) 119891

119910(119909 119910))

T 119894 = 1 2 119872

the potential energy Π119908due to the force 119891 is calculated as

Π119908

= minus∬I119894

(

119906119899(119909 119910)

V119899(119909 119910)

)

T

119891 (119909 119910) 119889119909 119889119910 (44)

Substituting (1) into the equations above leads to

Π119908

= minus 119863119899

119894T∬

I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 (45)

Minimize Π119908with respect to 119863

119899

120585 and we have

120597Π119908

(0)

120597 119863119899

119894T

= minus∬I119894

[119879119894(119909 119910)]

T119891 (119909 119910) 119889119909 119889119910 997888rarr [119865

119894]

(46)

The 6 times 1 submatrix above is added to the right side of theglobal simultaneous equation (5) When the body force isconstant say 119891(119909 119910) = (119891

119909 119891

119910)T (46) becomes

120597Π119908

(0)

120597 [119863119899

119894]T

= minus [119863119899

119894]T(119891

119909119860

119894 119891

119910119860

119894 0 0 0 0)

T997888rarr [119865

119894] (47)

where 119860119894denotes the area of the elementI

119894

37 Contact Submatrices The element edges along theboundaries of different blocks may contact with each otherThis kind of interactions is allowed in three modes vertex-to-vertex edge-to-edge and vertex-to-edge contacts In two-dimensional problems both vertex-to-vertex and edge-to-edge contacts can be decomposed into couples of vertex-to-edge ones The BBM adopts the ldquorigidrdquo contact model Inorder to prevent penetration and relative sliding along the

8 Mathematical Problems in Engineering

119973j

P3

120581perp

120581

P1119973i

P2

(a)

P3

P0P1

119973i

hperp

h

P2

(b)

Figure 6 Contact springs between the elementsI119894andI

119895 (a) The normal contact spring with the stiffness 120581

perpand the shear contact spring

with the stiffness 120581 (b) The normal contact distance ℎ

perpand the shear contact distance ℎ

interfaces of a close contact a normal contact spring and ashear contact spring have to be applied at the contact point

During each time step the simultaneous equilibriumequations in (5) have to be assembled and solved repeat-edly whenever the implementation of any contact springchanges All of the contact springs must satisfy the followingconstraints no tension appears in an open contact and nopenetration occurs in a close one These two conditions canbe fulfilled by an open-close iteration process to controlthe installation and uninstallation of every contact springin each time step Readers are recommended to refer tothe dissertation [1] for the implementation of open-closeiterations

As illustrated in Figure 6(a) the vertex 1198751of the element

I119894and the edge 119875

2119875

3of the element I

119895are in a contact

where a normal contact spring with the stiffness 120581perpand a

shear contact spring with the stiffness 120581are applied As

shown in Figure 6(b) assume the point 1198750on the edge 119875

2119875

3

is the contact point 1198751is a vertex of the block which is

different from the block where 1198750 119875

2 and 119875

3locate Denote

the coordinate of 119875120572at 119905

119899by (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 and 119899 =

1 2 119873 The displacement of 119875120572in the 119899th time step can

be expressed as

U (119909119899

120572 119910

119899

120572) = (

119906119899

120572

V119899

120572

) = (

119909119899

120572minus 119909

119899minus1

120572

119910119899

120572minus 119910

119899minus1

120572

) (48)

Substitute (1) into the above equation and we have

(

119906119899

1

V119899

1

) = [1198791

119894] 119863

119899

119894 (49)

(

119906119899

120572

V119899

120572

) = [119879120573

119895] 119863

119899

119895 120573 = 0 2 3 (50)

where [119879120572

120585] is the abbreviation of [119879

120585(119909

120572 119910

120572)] 120572 = 0 1 2 3

In every iteration step the normal contact spring and theshear contact spring contribute the strain energiesΠ

perpandΠ

respectively and they can be expressed as

Πperp

=1

2120581

perpℎ

2

perp (51a)

Π=

1

2120581

2

(51b)

where ℎperpand ℎ

denote the normal and the shear contact

displacement respectively As shown in Figure 6(b) ℎperpis the

distance from 1198751to the edge 119875

2119875

3and can be computed as

follows

ℎperp

=119860

119899

119871119899 (52)

where119860119899 is the area of the triangleΔ

119875111987521198753and 119871

119899 denotes thelength of the edge 119875

2119875

3 Since ℎ

is the projection of the line

1198750119875

1on the edge 119875

2119875

3or its extension line we have

ℎ=

997888997888997888rarr119875

2119875

3sdot997888997888997888rarr119875

0119875

1

10038161003816100381610038161003816119875

2119875

3

10038161003816100381610038161003816

=1

119871119899(

119909119899

3minus 119909

119899

2

119910119899

3minus 119910

119899

2

)

T

(

119909119899

1minus 119909

119899

0

119910119899

1minus 119910

119899

0

)

(53)

As follows the minimizations of the potential energy itemsΠ

perpand Π

are derived respectively

371 Normal Contact Submatrices To obtain the derivativeof Π

perpin (51a) we compute the normal contact displacement

ℎperpfirstly 119860119899 in (52) can be deduced by using (48) as

119860119899

=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899

1119910

119899

1

1 119909119899

2119910

119899

2

1 119909119899

3119910

119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1+ 119906

119899

1119910

119899minus1

1+ V119899

1

1 119909119899minus1

2+ 119906

119899

2119910

119899minus1

2+ V119899

2

1 119909119899minus1

3+ 119906

119899

3119910

119899minus1

3+ V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1119910

119899minus1

1

1 119909119899minus1

2119910

119899minus1

2

1 119909119899minus1

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1V119899

1

1 119906119899

2V119899

2

1 119906119899

3V119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(54)

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

119973j

P3

120581perp

120581

P1119973i

P2

(a)

P3

P0P1

119973i

hperp

h

P2

(b)

Figure 6 Contact springs between the elementsI119894andI

119895 (a) The normal contact spring with the stiffness 120581

perpand the shear contact spring

with the stiffness 120581 (b) The normal contact distance ℎ

perpand the shear contact distance ℎ

interfaces of a close contact a normal contact spring and ashear contact spring have to be applied at the contact point

During each time step the simultaneous equilibriumequations in (5) have to be assembled and solved repeat-edly whenever the implementation of any contact springchanges All of the contact springs must satisfy the followingconstraints no tension appears in an open contact and nopenetration occurs in a close one These two conditions canbe fulfilled by an open-close iteration process to controlthe installation and uninstallation of every contact springin each time step Readers are recommended to refer tothe dissertation [1] for the implementation of open-closeiterations

As illustrated in Figure 6(a) the vertex 1198751of the element

I119894and the edge 119875

2119875

3of the element I

119895are in a contact

where a normal contact spring with the stiffness 120581perpand a

shear contact spring with the stiffness 120581are applied As

shown in Figure 6(b) assume the point 1198750on the edge 119875

2119875

3

is the contact point 1198751is a vertex of the block which is

different from the block where 1198750 119875

2 and 119875

3locate Denote

the coordinate of 119875120572at 119905

119899by (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 and 119899 =

1 2 119873 The displacement of 119875120572in the 119899th time step can

be expressed as

U (119909119899

120572 119910

119899

120572) = (

119906119899

120572

V119899

120572

) = (

119909119899

120572minus 119909

119899minus1

120572

119910119899

120572minus 119910

119899minus1

120572

) (48)

Substitute (1) into the above equation and we have

(

119906119899

1

V119899

1

) = [1198791

119894] 119863

119899

119894 (49)

(

119906119899

120572

V119899

120572

) = [119879120573

119895] 119863

119899

119895 120573 = 0 2 3 (50)

where [119879120572

120585] is the abbreviation of [119879

120585(119909

120572 119910

120572)] 120572 = 0 1 2 3

In every iteration step the normal contact spring and theshear contact spring contribute the strain energiesΠ

perpandΠ

respectively and they can be expressed as

Πperp

=1

2120581

perpℎ

2

perp (51a)

Π=

1

2120581

2

(51b)

where ℎperpand ℎ

denote the normal and the shear contact

displacement respectively As shown in Figure 6(b) ℎperpis the

distance from 1198751to the edge 119875

2119875

3and can be computed as

follows

ℎperp

=119860

119899

119871119899 (52)

where119860119899 is the area of the triangleΔ

119875111987521198753and 119871

119899 denotes thelength of the edge 119875

2119875

3 Since ℎ

is the projection of the line

1198750119875

1on the edge 119875

2119875

3or its extension line we have

ℎ=

997888997888997888rarr119875

2119875

3sdot997888997888997888rarr119875

0119875

1

10038161003816100381610038161003816119875

2119875

3

10038161003816100381610038161003816

=1

119871119899(

119909119899

3minus 119909

119899

2

119910119899

3minus 119910

119899

2

)

T

(

119909119899

1minus 119909

119899

0

119910119899

1minus 119910

119899

0

)

(53)

As follows the minimizations of the potential energy itemsΠ

perpand Π

are derived respectively

371 Normal Contact Submatrices To obtain the derivativeof Π

perpin (51a) we compute the normal contact displacement

ℎperpfirstly 119860119899 in (52) can be deduced by using (48) as

119860119899

=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899

1119910

119899

1

1 119909119899

2119910

119899

2

1 119909119899

3119910

119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1+ 119906

119899

1119910

119899minus1

1+ V119899

1

1 119909119899minus1

2+ 119906

119899

2119910

119899minus1

2+ V119899

2

1 119909119899minus1

3+ 119906

119899

3119910

119899minus1

3+ V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1119910

119899minus1

1

1 119909119899minus1

2119910

119899minus1

2

1 119909119899minus1

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1V119899

1

1 119906119899

2V119899

2

1 119906119899

3V119899

3

100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(54)

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

For a small enough time step the last item in (54) will be asecond-order infinitesimal and hence can be ignored Thenwe have

119860119899

asymp 119860119899minus1

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119906119899

1119910

119899minus1

1

1 119906119899

2119910

119899minus1

2

1 119906119899

3119910

119899minus1

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1 119909119899minus1

1V119899

1

1 119909119899minus1

2V119899

2

1 119909119899minus1

3V119899

3

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(55)

In the same case 119871119899 can be approximated as 119871119899minus1

119871119899

asymp 119871119899minus1

= radic(119909119899minus1

2minus 119909

119899minus1

3)

2

+ (119910119899minus1

2minus 119910

119899minus1

3)

2

(56)

Substituting (50) (55) and (56) into (52) leads to

ℎperp

=119860

119899minus1

119871119899minus1+

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895] 119863

119899

119895

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198793

119895] 119863

119899

119895

(57)

Substitute (57) into (51a) and thenminimizeΠperpwith respect

to 119863119899

119894 and 119863

119899

119895

1205972Π

perp

(120597 119863119899

119894T120597 119863

119899

119894)

= 120581perp

[Φperp]T[Φ

perp] 997888rarr [119870

119894119894] (58a)

1205972Π

perp

120597 119863119899

119894T120597 119863

119899

119895

= 120581perp

[Φperp]T[Ψ

perp] 997888rarr [119870

119894119895] (58b)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119894

= 120581perp

[Ψperp]T[Φ

perp] 997888rarr [119870

119895119894] (58c)

1205972Π

perp

120597 119863119899

119895T120597 119863

119899

119895

= 120581perp

[Ψperp]T[Ψ

perp] 997888rarr [119870

119895119895] (58d)

minus120597Π

perp(0)

120597 119863119899

119894

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119894 (58e)

minus120597Π

perp(0)

120597 119863119899

119895

= 120581perp119860

119899minus1[Φ

perp] 997888rarr 119865

119895 (58f)

where

[Φperp] =

1

119871119899minus1(

119910119899minus1

2minus 119910

119899minus1

3

119909119899minus1

3minus 119909

119899minus1

2

)

T

[1198791

119894] (59a)

[Ψperp] =

1

119871119899minus1(

119910119899minus1

3minus 119910

119899minus1

1

119909119899minus1

1minus 119909

119899minus1

3

)

T

[1198792

119895]

+1

119871119899minus1(

119910119899minus1

1minus 119910

119899minus1

2

119909119899minus1

2minus 119909

119899minus1

1

)

T

[1198792

119895]

(59b)

The submatrices in (58a) (58b) (58c) and (58d) are addedto the global stiffness matrix in (5) while the submatrices in(58e) and (58f) are added to the right item of the equilibriumequations in (5)

372 Shear Contact Submatrices In order to obtain theminimization of the strain energy Π

in (51b) we compute

the detailed formulation of the shear contact displacement ℎ

first To this end substitute (48) into (53) and we have

ℎ=

1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

+1

119871119899(

119906119899

3minus 119906

119899

2

V119899

3minus V119899

2

)

T

(

119906119899

1minus 119906

119899

0

V119899

1minus V119899

0

)

(60)

With a small enough stepsize the last two items in theequation above are infinitesimals and hence can be ignoredwhile 119871

119899 can be approximately replaced by 119871119899minus1 Therefore

substituting (50) into (60) leads to

ℎ=

119861119899minus1

119871119899minus1+

1

119871119899minus1(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

)

T

[1198791

119894] 119863

119899

119894

+1

119871119899minus1(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

)

T

[1198790

119895] 119863

119899

119895

(61)

where

119861119899minus1

= (

119909119899minus1

1minus 119909

119899minus1

0

119910119899minus1

1minus 119910

119899minus1

0

)

T

(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (62)

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

Now substitute (61) into (51b) and then minimize Πwith

respect to 119863119899

119894 and 119863

119899

119895 and we obtain

1205972Π

120597 119863119899

119894T120597 119863

119899

119894

= 120581[Φ

]⊤

[Φ] 997888rarr [119870

119894119894] (63a)

1205972Π

120597 119863119899

119894T120597 119863

119899

119895

= 120581[Φ

]⊤

[Ψ] 997888rarr [119870

119894119895] (63b)

1205972Π

120597 119863119899

119895T120597 119863

119899

119894

= 120581[Ψ

]⊤

[Φ] 997888rarr [119870

119895119894] (63c)

1205972Π

120597 119863119899

119895T120597 119863

119899

119895

= 120581[Ψ

]⊤

[Ψ] 997888rarr [119870

119895119895] (63d)

minus120597Π

0

120597 119863119899

119894

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119894 (63e)

minus120597Π

0

120597 119863119899

119895

= 120581119861

119899minus1[Φ

] 997888rarr 119865

119895 (63f)

which are added to the global stiffness matrix and the rightitem in (5) respectively and here

[Φ] =

1

119871119899minus1(119910

119899minus1

2minus 119910

119899minus1

3 119909

119899minus1

3minus 119909

119899minus1

2) [119879

1

119894]

[Ψ] =

1

119871119899minus1(119909

119899minus1

2minus 119909

119899minus1

3 119910

119899minus1

2minus 119910

119899minus1

3) [119879

0

119895]

(64)

38 Friction Submatrices When the Mohr-Coulomb failurecriterion is satisfied along the interface of two contactelements relative sliding will occur In this case a normalcontact spring needs to be applied at the contact point to resistthe penetration along the sliding surface and the frictioneffect has to be taken into account if the friction angle of thejoint is not zero In the BBM frictions are considered as a kindof external forces acting on the elements on the two sides ofthe sliding surfaces

As shown in Figure 7(a) the vertex 1198751of the element

I119894is sliding along the edge 119875

2119875

3of the element I

119895 119894 119895 =

1 2 119872 A normal contact spring with the stiffness 120581perpis

implemented to attempt to push the element I119894out of the

element I119895 As shown in Figure 7(b) the point 119875

0on 119875

2119875

3

is the contact point 1199040and 119904

1are the sliding displacements

of 1198750and 119875

1along the directed edge 997888997888997888rarr

1198752119875

3 respectively and

ℎperpis the normal contact distance which has the computing

formulation given in (57) If the friction angle 120593 is not zerothe normal contact force N and the friction force F can becomputed as follows

N = 120581perpℎ

perp (65a)

F = sgn (997888997888997888rarr119875

0119875

1sdot997888997888997888rarr119875

2119875

3)N tan (120593) (65b)

where the sign function sgn(119909) takes the values of 1 0 andminus1 in the cases of 119909 gt 0 119909 = 0 and 119909 lt 0 respectively As

the friction forceF acts on both sides of the sliding surfacesits potential energy ΠF can be calculated as the followingsummation

ΠF = F1199040+ F119904

1 (66)

At the time node 119905119899 119899 = 1 2 119873 assume the coordinate

of the point 119875120572is (119909

119899

120572 119910

119899

120572) 120572 = 0 1 2 3 In this way the

displacements 1199040and 119904

1during the 119899th time step can be

computed by

1199040=

1

119871119899(119906

119899

0 V119899

0)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

0 V119899

0)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67a)

1199041=

1

119871119899(119906

119899

1 V119899

1)997888997888997888rarr119875

2119875

3=

1

119871119899(119906

119899

1 V119899

1)(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (67b)

where 119871119899 denotes the length of 119875

2119875

3at 119905

119899 For a small enough

time interval [119905119899minus1

119905119899]119871119899 can be approximated by119871

119899minus1 whichcan be computed by (56) In this case substituting (1) and (3)into (67a) and (67b) leads to

1199040=

1

119871119899minus1119863

119899

119895T[119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68a)

1199041=

1

119871119899minus1119863

119899

119894T[119879

1

119894]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) (68b)

Substitute (68a) and (68b) into (66) and then minimize ΠF

with respect to 119863119899

119894 and 119863

119899

119895 respectively and we have

minus120597ΠF (0)

120597 119863119899

119894

=1

119871119899minus1F [119879

1

119894]T(

119909119899minus1

2minus 119909

119899minus1

3

119910119899minus1

2minus 119910

119899minus1

3

) 997888rarr 119865119894

(69a)

minus120597ΠF (0)

120597 119863119899

119895

=1

119871119899minus1F [119879

0

119895]T(

119909119899minus1

3minus 119909

119899minus1

2

119910119899minus1

3minus 119910

119899minus1

2

) 997888rarr 119865119895

(69b)

The above 6times1 submatrices are added to the right item of (5)

4 Numerical Examples

The computer program for the BBM has been coded in theC language whose flowchart is illustrated in Figure 8 Thesimultaneous equilibrium equations formed in each time stepare solved by the symmetric successive overrelaxation pre-conditioned conjugate gradient (SSOR-PCG)method [35] Inthis section three elaborate benchmark numerical examplesare conducted to verify the formulations derived in theprevious sections The following stress contour figures areplotted based on nodal stresses and the stress on every meshnode is the mean stress of the elements which overlap witheach other at the grid point

41 Parameter Settings As shown in Table 1 the parametersused in the experiments are given including time step

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 11

Table 1 The values of the parameters used in the experiments

Number Δ119905 119873 120598119905

120596 120598 120581 120581perp

120581

I 10 100 10 10 1119890 minus 8 10 mdash mdashII 10 1 10 10 1119890 minus 8 10

2 mdash mdashIII 1119890 minus 6 40 10 10 1119890 minus 8 10 10

2119864

dagger10

2119864

dagger119864 denotes Youngrsquos modulus of the material in Section 44

119973j

119973i

P1

P3

P2

120581perp

(a)

119973i

P1

P3

P2

s1

s0P0

hperp

(b)

Figure 7 The vertex 1198751of the elementI

119894is sliding along the edge 119875

2119875

3of the elementI

119895 (a) A normal contact spring with the stiffness 120581

perp

is applied (b)

Start

Contact detection on boundary edges

Add all submatrices to global equations except contact submatrices

Open-close iteration converges

Input data

Installuninstall contact springs and add contact submatrices to global equations

SSOR-PCG solver

Update element displacements stresses and velocities

End

Reduce time stepsize

No

Yes

Yes

No

Ope

n-clo

se it

erat

ion

proc

ess

Tim

e ite

ratio

n pr

oces

s

No Yes

Time steps ge N

Loops gt 6

Figure 8 The flowchart of the computer program for the BBM

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Mathematical Problems in Engineering

Mass density 001 kgm2

Youngrsquos modulus 300MPaPoissonrsquos ratio 03

y

xo

A B

P = 1kN

h = 1m

L = 10m

(a)

(b)

Figure 9 A cantilever subjected to a concentrated load at the free end (a) Problem illustration on the geometry configuration and the givenmechanical parameters (b) Triangular mesh used in both the BBM and the FEM

interval Δ119905 total time steps 119873 the error tolerance 120598119905for time

iterations the SSOR relaxation factor 120596 the error tolerance120598 for SSOR-PCG iterations the coefficient 120581 used in (4) and(29) the normal contact spring stiffness 120581

perp and the shear

contact spring stiffness 120581

42 Experiment I The first experiment is designed to exam-ine the capability of the BBM to analyze elasticity problemsAs shown in Figure 9(a) a cantilever beam subjected to aconcentrated load 119875 = 1KN at the free end is consideredThe length and the height of the beam are 119871 = 10m and ℎ =

1m respectively The material parameters of the beam are asfollows Youngrsquos modulus 119864 = 300MPa Poissonrsquos ratio ] =

03 and the mass density 120588 = 001Kgm2 The displacementof this problem has the following analytical solution [36]

119880119909

=119875119910

6119864119868[(6119871 minus 3119909) 119909 + (2 + ]) (119910

2minus

ℎ2

4)]

119880119910

=119875119910

6119864119868[3]1199102

(119871 minus 119909) + (4 + 5])ℎ

2119910

4+ (3119871 minus 119909) 119909

2]

(70)

where 119868 = ℎ312 is the moment of inertia

The DDA the BBM and the PLANE2 element (PLANE2is defined by six nodes having two degrees of freedom at eachnode and possesses a quadratic displacement behavior) of thecommercial software ANSYS are implemented to solve thisproblem The computation model of the DDA contains onlyone block that is the beam so there are 6 degrees of freedom(DOFs) involved Both of the computations of the BBM andthe FEM adopt the unstructured triangular mesh as shown inFigure 9(b) where there exist totally 2202 triangle elementsThe DOFs consumed by the three methods are compared inTable 2 in which the DOFs used by the FEM are two timesthose by the BBM

This statics problem is solved through the Newton-Raphson incremental method Here time step indicates theiteration counter The stopping criterion is to exceed the

Table 2 Comparing the DOFs involved in the DDA the FEM andthe BBM for Section 42

Models DDA FEM BBMEntity Blocks Element BBM elementDOFsentity 6 12 6Total entities 1 2202 2202Total DOFs 6 26424 13202

maximum iteration number 119873 or to satisfy the followinginequality constraint

( sum

(119909119910)isincup119872

119894=1I119894

10038171003817100381710038171003817119880

119899 minus 119880

119899minus110038171003817100381710038171003817

2

2)

12

⩽ 120598119873119877

(71)

where time step 119899 = 1 2 100 element number 119872 =

2202 119880119899 = 119880

119899(119909 119910) denotes the computed displacement

at (119909 119910) in the 119899th time step sdot 2is the square norm 120598

119873119877=

10minus8 is the error tolerance for Newton-Raphson iterations In

this example the computation of the BBM converges at 119899 = 3

Comparative Task 1 Figure 10 plots the computed displace-ments on the nodes distributed along the line 119860119861 = (119909 119910)

0 ⩽ 119909 ⩽ 119871 119910 = minusℎ2 According to the figure both of thedisplacement curves of the BBM and the FEM coincide withthat of the analytical solution while the results obtained bythe DDA are low in accuracy So the accuracy of the originalDDA indeed can be significantly improved by our method

Comparative Task 2 In Table 3 the displacement components119880

119909and119880

119910at the point 119861(119871 minusℎ2) computed by the BBM and

the FEM are compared with the theoretical solutions Fromthe table it can be found that both the BBM and the FEM arevery close to the analytical displacements

Comparative Task 3 Figures 11 and 12 plot the computed con-tours of the displacement components119880

119909and119880

119910throughout

the considered cantilever respectively According to thesepictures we can find that BBM results agree with ANSYSsolutions very well in spite of the introduction of the bondingsprings to glue elements

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 13

Table 3 The displacements at 119861(119871 minusℎ2) obtained by the analytical solution the BBM and the FEM

Displm Theory BBM Accuracy FEM Accuracy119880

119909(119861) minus0001 minus0000993 9930 minus0000998 9980

119880119910(119861) minus0013425 minus0013417 9994 minus0013410 9989

00000

minus00015

minus00030

minus00045

Ux

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalytx

UFEMx

UDDAx

UBBMx

(a)

0000

minus0005

minus0010

minus0015

Uy

(m)

0 1 2 3 4 5 6 7 8 9 10

x (m)

UAnalyty

UFEMy

UDDAy

UBBMy

(b)

Figure 10 The computed displacement components 119880119909and 119880

119910of the nodes along the line 119860119861

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(a)

minus00009

minus00007

minus00005

minus00003

minus00001 0

00001

00003

00005

00007

00009

(b)

Figure 11 The contour plots of the displacement 119880119909(m) throughout the beam (a) BBM result (b) FEM result

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(a)

minus0013

minus0012

minus0011

minus001

minus0009

minus0008

minus0007

minus0006

minus0005

minus0004

minus0003

minus0002

minus0001

(b)

Figure 12 The contour plots of the displacement 119880119910(m) throughout the beam (a) BBM result (b) FEM result

Comparative Task 4 As shown in Figure 13 the von-Misesstress distributions figured out by the BBM and the FEMwithPLANE2 element are illustrated One can observe that theresults obtained by the two methods are very close Based onthis comparative we can see that the BBM is capable of givingaccurate enough stress analysis for this elasticity problem

43 Experiment II The objective of the second exampleis to demonstrate the capability of the BBM to analyzethe elastostatics problem defined on an irregularly shapeddomain As shown in Figure 14(a) a thin plate with a circularhole at the center is pressured by a uniform horizontal stress120590

infin= 1MPa Assume that Youngrsquos modulus Poissonrsquos ratio

the thickness and the mass density of the plate are 2900GPa03 02m and 001 Kgm2 respectively The radium of thehole is 2119886 the length and thewidth of the plate are 2119871 and 2119882

respectively where 119886 = 05m119871 = 10m and119882 = 5m Set theorigin of the Cartesian coordinate system at the point 119900 thecenter of the hole and the 119900119909-axis toward the right As shownin Figure 14(a) we also use the polar coordinate system (119903 120579)where the pole locates at the origin 119900 120579 is the polar angle inradian and 119903 is the distance in meter away from 119900 In the caseof infinite 119871 and 119882 this problem has the following analyticalsolution derived in [37 38]

120590119903=

120590infin

2(1 minus

1198862

1199032)[1 + (1 minus 3

1198862

1199032) cos 2120579] (72a)

120590120579

=120590

infin

2(1 +

1198862

1199032) minus

120590infin

2(1 minus 3

1198862

1199032) cos 2120579 (72b)

We implement both the BBM and PLANE2 element ofANSYS to numerically solve this issue One-quarter of the

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

14 Mathematical Problems in Engineering

Table 4 The stresses 120590119909computed by the BBM and the FEM are compared with the analytical solutions at the points 119860(0 05) 119861(05 0)

119862(10 0) 119863(10 5) and 119864(0 5)

StressMPa Theory BBM Accuracy FEM Accuracy120590

119909(119860) 30 2949130 98304 303450 98850

120590119909(119861) 00 minus261091119890 minus 2 mdash minus113856119890 minus 3 mdash

120590119909(119862) 0993759 0999961 99376 1000023 99370

120590119909(119863) 0998758 0999944 99881 0999960 99880

120590119909(119864) 1005150 0989241 98417 0985517 98047

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(a)

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

55000

60000

65000

(b)

Figure 13 The contour plots of the von-Mises stress (Pa) throughout the beam (a) BBM result (b) FEM result

120590infin

=1

MPa

120590infin

=1

MPa

a = 1m

E

y

x

rA

oB

120579

L = 10m

W = 5m

D

CMass density 001 kgm

Thickness 02m

Youngrsquos modulus 2900GPaPoissonrsquos ratio 03

(a) (b)

Figure 14 A finite-width plate with a central circular hole subjected to horizontal uniform tensions (a)The geometry configuration and themechanical constants of the plate (b) The mesh used in the computations of the BBM and the FEM

domain that is the area 119860119861119862119863119864 is analyzed due to thesymmetry and the symmetrical boundary constraints areapplied along the119860119864 and 119861119862 respectivelyThemesh adoptedin both the BBM and the FEM is plotted in Figure 14(b)where 3253 triangles are generated During the computationthe BBMand the FEMconsume totally 19518DOFs and 39036DOFs respectively Since this example involves only smallelastic deformations it is not necessary to use incrementaliterations for the solution So we use a large time interval andonly one time step in the BBM code

Comparative Task 1 In Table 4 the horizontal stresses com-puted by the BBM and the FEM are compared with theanalytical solutions at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) According to this table both the BBMand the FEM can give accurate enough horizontal stresses 120590

119909

at 119860 119862 119863 and 119864 while 120590119909(119861) computed by the two methods

have some difference from the theoretical solution Actuallysuch errors are caused due to the fact that the plate sizes 119871 and119882 are assumed to be infinite in the analytical formulations in(72a) and (72b) but they have the specific values of 119871 = 10

and 119882 = 5 in the computational models of the BBM and theFEM So it still can be noticed that the stress analysis by theBBM agrees with that by the FEM very well

Table 5 The displacements 119880119909and 119880

119910computed by the BBM and

the FEM are compared at the points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5)

Displm FEM BBM Relative errordagger

119880119909(119860) 00 169968119890 minus 11 mdash

119880119910(119860) minus178046119890 minus 07 minus175791119890 minus 07 1266527

119880119909(119861) 523535119890 minus 07 522699119890 minus 07 0159684

119880119910(119861) 00 minus32779119890 minus 12 mdash

119880119909(119862) 349394119890 minus 06 349639119890 minus 06 0070121

119880119910(119862) 00 9402119890 minus 13 mdash

119880119909(119863) 348394119890 minus 06 34873119890 minus 06 0096443

119880119910(119863) minus509281119890 minus 07 minus509625119890 minus 07 0067546

119880119909(119864) 00 693519119890 minus 11 mdash

119880119910(119864) minus563371119890 minus 07 minus56286624119890 minus 07 0089596

daggerThe relative error between the BBM result 119880FEM120572

and the FEM solution119880FEM120572

is determined by 10038161003816100381610038161003816119880FEM120572

minus 119880BBM120572

10038161003816100381610038161003816119880

FEM 120572 = 119909 119910

Comparative Task 2 In Table 5 the displacement components119880

119909and119880

119910computed by the BBM and the FEM are compared

at the five measurement points 119860(0 05) 119861(05 0) 119862(10 0)119863(10 5) and 119864(0 5) From the table we can find that the

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 15

4

3

2

1

0

Ux

(m)

times10minus6

0510 20 30 40 50 60 70 80 90 100

2

1

0

minus1

times10minus10

Uy

(m)

x (m)

0510 20 30 40 50 60 70 80 90 100

x (m)

The FEM resultThe BBM result

Figure 15 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM and the FEM on the mesh nodes along the symmetric line119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

Ux

(m)

times10minus7

times10minus8

Uy

(m)

05 10 20 2515 30 35 40 45 50

x (m)

05 10 20 2515 30 35 40 45 50

x (m)

The FEM resultThe BBM result

10

05

00

minus05

minus1

minus2

minus3

minus4

minus5

minus6

Figure 16 The displacements 119880119909(m) and 119880

119910(m) computed by the

BBM the FEM and the analytical formulations on the mesh nodesalong the symmetric line 119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

displacements solved out by the BBM are very close tothe FEM results The subtle differences between the resultsof the two methods on 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864)

are due to the fact that the BBM imports the symmetricboundary constraints by using penalty-like bonding springs

12100806040200

minus020510 20 30 40 50 60 70 80 90 100

x (m)

120590x

(Pa)

times106

The FEM resultAnalytical solution

The BBM result

Figure 17 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line 119861119862 = (119909 119910) 05 ⩽ 119909 ⩽ 50 119910 = 0

120590x

(Pa)

times106

30

25

20

15

10

05 10 15 20 25 30 35 40 45 50

y (m)

The FEM resultAnalytical solution

The BBM result

Figure 18 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the symmetric line119860119864 = (119909 119910) 119909 = 0 05 ⩽ 119910 ⩽ 50

120590x

(Pa)

times106

30252015100500

000 020 040 060 080 100 120 140 157

120579 (radian)

The FEM resultAnalytical solution

The BBM result

Figure 19 The horizontal stress 120590119909(Pa) computed by the BBM the

FEM and the analytical formulations in (72a) and (72b) on the meshnodes along the arc 119861119860 = (119903 120579) 119903 = 05 0 ⩽ 120579 ⩽ 1205872

Since such small displacements (less than 10minus11 m) can be

interpreted physically as the local elastoplastic deformationof the plate near the partially fixed boundaries the relativeerrors between ourmethod and the FEM on the computationof 119880

119909(119860) 119880

119910(119861) 119880

119910(119862) and 119880

119909(119864) are acceptable

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

16 Mathematical Problems in Engineering

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(a)

28E + 06252E + 06224E + 06196E + 06168E + 0614E + 06112E + 068400005600002800000

(b)

Figure 20 The contour plots of the computed distributions of the horizontal stress 120590119909(Pa) throughout the plate (a) The result of the BBM

(b) The result of the FEM

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(a)

40000027000014000010000minus120000minus250000minus380000minus510000minus640000minus770000minus900000

(b)

Figure 21 The contour plots of the computed distributions of the horizontal normal stress 120590119910(Pa) throughout the plate (a) The result of the

BBM (b) The result of the FEM

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(a)

3E minus 06275E minus 0625E minus 06225E minus 062E minus 06175E minus 0615E minus 06125E minus 061E minus 0675E minus 075E minus 07

(b)

Figure 22 The contour plots of the computed distributions of the displacement component 119880119909(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

Comparative Task 3 Then we compare the computed defor-mations along the 119900119909- and 119900119910-axis with the theoretical resultsFigure 15 plots the displacements119906

119909and119906

119910obtained from the

BBM and the FEM results on the nodes along the boundary119861119862 And in Figure 16 119906

119909and 119906

119910solved by the two methods

on the nodes along 119860119864 are drawn From the figures it canbe observed that the BBM results are in accord with the FEMsolutions

Comparative Task 4 As shown in Figures 17 18 and 19 thecomputed horizontal stresses 120590

119909on the nodes along 119861119862 119860119864

and the arc 119861119860 are illustrated respectively According to thethree plots one can find that both of the results of the BBMand the FEM are consistent with the analytical solutions

Comparative Task 5 Next we compare the BBM and the FEMon analyzing the deformation distributions of the quarter

of the plate As shown in Figures 20 and 21 the computedcontour figures of the stresses 120590

119909throughout the domain are

illustrated and it can be observed that the BBM results agreewith the FEM very well

Comparative Task 6 In Figures 22 and 23 the distributionsof the displacements obtained by the BBM and the FEM aredemonstrated According to these contour plots we can seethat the result of the BBM conforms with the solution of theFEM

44 Experiment III The third numerical example is todemonstrate the accuracy and the effectiveness of the BBMin the analysis of a frictionless sliding problem As shownin Figure 24(a) a small rectangular block subjected to ahorizontal traction force 119875 slides along the top surface of afixed flat plate The origin of the coordinate system locates at

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 17

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(a)

minus5E minus 08minus1E minus 07minus15E minus 07minus2E minus 07minus25E minus 07minus3E minus 07minus35E minus 07minus4E minus 07minus45E minus 07minus5E minus 07minus55E minus 07

(b)

Figure 23 The contour plots of the computed distributions of the displacement component 119880119910(m) throughout the plate (a) The result of

the BBM (b) The result of the FEM

P = 2kNA y

xo

Mass density 1kgm2

Youngrsquos modulus 21GPaPoissonrsquos ratio 03

01

m

02m 02m

2m

(a) (b)

Figure 24 A rectangular block subjected to the traction 119875 slides along the surface of a fixed flat plate (a) The geometry configuration andthe mechanical parameters (b) The subdivision

the plate centroid and the119909direction points to the right Boththe upper block and the lower plate are all linear isotropicelastic and are with the material constants in Figure 24(a)

We implement the BBM to solve this problem by usingthe quadrilateral mesh as plotted in Figure 24(b) wherethe upper rectangular block and the lower flat plate aresubdivided into 8 and 160 elements respectively In thecomputation set the uniform time step length as 10

minus6 sBelow the computed displacement at the point 119860(minus1 03) asmarked in Figure 24(a) is concerned The analytical solution(119880

Analyt119909

119880Analyt119910

) can be deduced based on kinematics theo-ries that is 119880Analyt

119909= 50000 sdot 119905

2 and 119880Analyt119910

= 0 As shown inFigure 25 the computed displacement components 119880

119909and

119880119910versus the time are plotted From the figure it can be

observed that the results of the BBM agree with the analyticalsolutions very well

5 Conclusion

In this paper we extended the concept of DDA blockswhich are separated by sets of geological joints Our methodreconstructs a block system as an assembly of block-likeelements through an advanced subdivision process Theseelements can be triangles quadrilaterals and even the orig-inal DDA blocks without partitioning Overlapped elementedges within a block are glued together by bonding springswhile contact springs are implemented on the element edgesthat compose the boundaries of original blocks In this wayblock displacement and stress fields can be refined and theefficient contact algorithms developed in the standard DDAare also retained Such scheme is easy to use and can be

UAnalytx

UAnalyty UBBM

y

UBBMx

00005

00004

00003

00002

00001

00000

Disp

lace

men

t (m

)

000 001 002 003 004 005 006 007 008 009 010

Time (ms)

Figure 25The computed displacements measured at the point119860 inthe 119909 and 119910 directions respectively

adopted in both the 2-dimensional and the 3-dimensionalDDA although it was performed in the 2-dimensional caseonly here Numerical experiments indicate that the proposedmethod is accurate and effective for both continuous and dis-continuous problems Therefore this procedure is potentialto be applied in crack propagation analysis However suchproblem exceeds the main concern of this work We willcontinue the task in the foreseeable future

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

18 Mathematical Problems in Engineering

Acknowledgments

The authors sincerely thank the referees and the editor fortheir comments and suggestions This research is supportedby the National Natural Science Foundation of China (Grantsnos 60902098 and 1110222)

References

[1] G-H ShiDiscontinuous deformation analysis a new numericalmodel for the statics and dynamics of block systems [PhD thesis]University of California Berkeley Calif USA 1988

[2] G-H Shi and R E Goodman ldquoGeneralization of two-dimensional discontinuous deformation analysis for forwardmodellingrdquo International Journal for Numerical and AnalyticalMethods in Geomechanics vol 13 no 4 pp 359ndash380 1989

[3] C T Lin B Amadei J Jung and J Dwyer ldquoExtensions ofdiscontinuous deformation analysis for jointed rock massesrdquoInternational Journal of Rock Mechanics and Mining Sciences ampGeomechanics Abstracts vol 33 no 7 pp 671ndash694 1996

[4] S A R Beyabanaki R G Mikola S O R Biabanaki andS Mohammadi ldquoNew point-to-face contact algorithm for 3-Dcontact problems using the augmented Lagrangian method in3-D DDArdquo Geomechanics and Geoengineering vol 4 no 3 pp221ndash236 2009

[5] S A R Beyabanaki B Ferdosi and SMohammadi ldquoValidationof dynamic block displacement analysis and modification ofedge-to-edge contact constraints in 3-D DDArdquo InternationalJournal of Rock Mechanics and Mining Sciences vol 46 no 7pp 1223ndash1234 2009

[6] YNing J YangGMa andPChen ldquoContact algorithmmodifi-cation of DDA and its verificationrdquo inAnalysis of DiscontinuousDeformation NewDevelopments and Applications GMa and YZhou Eds pp 73ndash81 2010

[7] S Amir Reza Beyabanaki and A C Bagtzoglou ldquoNon-rigiddisk-based DDA with a new contact modelrdquo Computers andGeotechnics vol 49 pp 25ndash35 2013

[8] H Bao Z Zhao and Q Tian ldquoOn the implementation ofaugmented lagrangian method in the two-dimensional dis-continuous deformation analysisrdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no6 pp 551ndash571 2014

[9] J-H Wu ldquoSeismic landslide simulations in discontinuousdeformation analysisrdquo Computers and Geotechnics vol 37 no5 pp 594ndash601 2010

[10] J-H Wu and C-H Chen ldquoApplication of DDA to simu-late characteristics of the tsaoling landsliderdquo Computers andGeotechnics vol 38 no 5 pp 741ndash750 2011

[11] Y Zhang X Fu and Q Sheng ldquoModification of the discontinu-ous deformation analysis method and its application to seismicresponse analysis of large underground cavernsrdquoTunnelling andUnderground Space Technology vol 40 pp 241ndash250 2014

[12] C J Pearce A Thavalingam Z Liao and N Bicanic ldquoCom-putational aspects of the discontinuous deformation analysisframework for modelling concrete fracturerdquo Engineering Frac-ture Mechanics vol 65 no 2-3 pp 283ndash298 2000

[13] Y-J Ning J Yang X M An and G W Ma ldquoModelling rockfracturing and blast-induced rock mass failure via advanceddiscretisation within the discontinuous deformation analysisframeworkrdquo Computers and Geotechnics vol 38 no 1 pp 40ndash49 2011

[14] Q Tian Z Zhao and H Bao ldquoBlock fracturing analysisusing nodal-based discontinuous deformation analysis withthe double minimization procedurerdquo International Journal forNumerical and Analytical Methods in Geomechanics vol 38 no9 pp 881ndash902 2013

[15] Y X Ben Y Wang and G-H Shi ldquoDevelopment of a modelfor simulating hydraulic fracturing with DDArdquo in Proceedingsof the 11th International Conference on Analysis of DiscontinuousDeformation (ICADD rsquo13) pp 169ndash175 Fukuoka Japan August2013

[16] Y-Y Jiao H-Q Zhang X-L Zhang H-B Li and Q-H JiangldquoA two-dimensional coupled hydromechanical discontinuummodel for simulating rock hydraulic fracturingrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 39 no 5 pp 457ndash481 2015

[17] C Y Koo and J C Chern ldquoThe development of DDA withthird order displacement functionrdquo in Proceedings of the 1stInternational Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R SalamiandD Banks Eds pp 342ndash349 TSI Press Berkeley Calif USA1996

[18] M Y Ma M Zaman and J H Zhu ldquoDiscontinuous defor-mation analysis using the third order displacement functionrdquoin Proceedings of the 1st International Forum on DiscontinuousDeformation Analysis (DDA) and Simulations of DiscontinuousMedia M R Salami and D Banks Eds pp 383ndash394 TSI PressBerkeley Calif USA June 1996

[19] SMHsiung ldquoDiscontinuous deformation analysis (DDA)withnth order polynomial displacement functionsrdquo in Proceedings ofthe 38th US Rock Mechanics Symposium on Rock Mechanics inthe National Interest pp 1413ndash1420 American Rock MechanicsAssociation Balkema Rotterdam The Netherlands Washing-ton DC USA 2001

[20] S A R Beyabanaki A Jafari and M R Yeung ldquoHigh-order three-dimensional discontinuous deformation analysis(3-D DDA)rdquo International Journal for Numerical Methods inBiomedical Engineering vol 26 no 12 pp 1522ndash1547 2010

[21] A Q Wu Y Zhang and S-Z Lin ldquoComplete and high orderpolynomial displacement approximation and its applicationto elastic mechanics analysis based on DDArdquo in Advances inDiscontinuous Numerical Methods and Applications in Geome-chanics and Geoengineering T Sasaki Ed pp 43ndash54 2012

[22] B Lu A Wu and X Ding ldquoMixed higher-order discontinuousdeformation analysisrdquo in Frontiers of Discontinuous NumericalMethods and Practical Simulations in Engineering and DisasterPrevention G Chen Y Ohnishi L Zheng and T Sasaki Edspp 243ndash248 CRC Press 2013

[23] T C Ke ldquoArtificial joint based DDArdquo in Proceedings of the1st International Forum on Discontinuous Deformation Analysis(DDA) and Simulations of Discontinuous Media M R Salamiand D Banks Eds pp 326ndash333 TSI Press Albuquerque NMUSA 1996

[24] Y H Zhang and Y M Cheng ldquoCoupling of FEM and DDAmethodsrdquoThe International Journal of Geomechanics vol 2 no4 pp 503ndash517 2002

[25] M S Khan Investigation of discontinuous deformation analysisfor application in jointed rock masses [PhD thesis] Departmentof Civil Engineering of University of Toronto 2010

[26] Z-Y Zhao and J Gu ldquoStress recovery procedure for discontin-uous deformation analysisrdquo Advances in Engineering Softwarevol 40 no 1 pp 52ndash57 2009

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 19

[27] M Zhang H Yang and Z K Li ldquoA coupling model ofthe discontinuous deformation analysis method and the finiteelement methodrdquo Tsinghua Science and Technology vol 10 no2 pp 221ndash226 2005

[28] K K Shyu Nodal-based discontinuous deformation analysis[PhD thesis] Department of Civil Engineering University ofCalifornia Berkeley Calif USA 1993

[29] Q Chang Non-linear dynamic discontinuous deformation anal-ysis with finite element meshed block systems [PhD thesis]University of California Berkeley Calif USA 1994

[30] R Grayeli and A Mortazavi ldquoDiscontinuous deformationanalysis with second-order finite element meshed blockrdquo Inter-national Journal for Numerical and Analytical Methods inGeomechanics vol 30 no 15 pp 1545ndash1561 2006

[31] R Grayeli andKHatami ldquoImplementation of the finite elementmethod in the three-dimensional discontinuous deformationanalysis (3D-DDA)rdquo International Journal for Numerical andAnalytical Methods in Geomechanics vol 32 no 15 pp 1883ndash1902 2008

[32] S A R Beyabanaki A Jafari S O R Biabanaki and M RYeung ldquoNodal-based three-dimensional discontinuous defor-mation analysis (3-D DDA)rdquo Computers and Geotechnics vol36 no 3 pp 359ndash372 2009

[33] H R Bao and Z Y Zhao ldquoModeling brittle fracture with thenodal-based discontinuous deformation analysisrdquo InternationalJournal of Computational Methods vol 10 no 6 Article ID1350040 26 pages 2013

[34] G-H Shi ldquoModeling dynamic rock failure by discontinuousdeformation analysis with simplex integrationsrdquo in Proceedingsof the 1st North American Rock Mechanics Symposium pp 591ndash598 Austin Tex USA 1994

[35] X Chen and K K Phoon ldquoSome numerical experienceson convergence criteria for iterative finite element solversrdquoComputers and Geotechnics vol 36 no 8 pp 1272ndash1284 2009

[36] C E Augarde and A J Deeks ldquoThe use of Timoshenkorsquosexact solution for a cantilever beam in adaptive analysisrdquo FiniteElements in Analysis and Design vol 44 no 9-10 pp 595ndash6012008

[37] S P Timoshenko and J N Goodier Theory of ElasticityMcGraw-Hill New York NY USA 3rd edition 1970

[38] J C Jaeger N G W Cook and R W Zimmerman Fundamen-tals of Rock Mechanics John Wiley amp Sons 4th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of