Constraint preserving discontinuous Galerkin method for ideal ...
Discontinuous Deformation Analysis Enriched by the Bonding Block Model
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
ℎ
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
ℎ
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 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
ℎ
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
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
ℎ
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 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
ℎ
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
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
ℎ
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
ℎ
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
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
ℎ
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 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
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 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