ANALYSIS OF STEEL BUILDING USING MOMENT DISTRIBUTION AND SAP2000 COMPUTER SOFTWARE

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COLLEGE of ENGINEERING & COMPUTER SCIENCE DEPERTMENT of CIVIL ENGINEERNG Class Project Finite Element Analysis: ENGR 5600 ANALYSIS OF STEEL BUILDING USING MOMENT DISTRIBUTION AND SAP2000 COMPUTER SOFTWARE By Khalafalla, Mohamed (MSc in Engineering: Civil) Class Instructor: Dr. Wu, Weidong.., PhD A project submitted in fulfillment of the requirements for the class, Indeterminate Structural Analysis, Spring 2014. The University of Tennessee at Chattanooga April, 2014

Transcript of ANALYSIS OF STEEL BUILDING USING MOMENT DISTRIBUTION AND SAP2000 COMPUTER SOFTWARE

COLLEGE of ENGINEERING & COMPUTER SCIENCE

DEPERTMENT of CIVIL ENGINEERNG

Class Project –Finite Element Analysis: ENGR 5600

ANALYSIS OF STEEL BUILDING USING MOMENT DISTRIBUTION AND

SAP2000 COMPUTER SOFTWARE

By

Khalafalla, Mohamed

(MSc in Engineering: Civil)

Class Instructor: Dr. Wu, Weidong.., PhD

A project submitted in fulfillment of the requirements for the class, Indeterminate

Structural Analysis, Spring 2014.

The University of Tennessee at Chattanooga

April, 2014

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DECLARATION, CERTIFICATION

AND

COPYRIGHT

I, Khalafalla Mohamed declare that this poject is my original work and that it has not been

presented and will not be pesented in any other University or institution for a similar way

unless I’m concerted.

Khalafalla, Mohamed: Signature: ..................................

Date: ..........................................

Dr. Wu, Weidong.., PhD

(Class Instructor)

Date: ……………………

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ABSTRACT

The main purpose of this project is to analyze a beam part of a steel building indeterminate

structural analysis. The geometry and loading conditions of the building are sized to mimic

a real world environment. The analysis is performed using SAP2000, and Moment

distribution hand calculation.

Analysis was performed using the above mentioned methods. The results was

approximately the same.

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ACKNOWLEDGEMENTS

I’m Mohamed would like to thank God Almighty who make us alive up to date, and

enabled us to join the University of Tennessee at Chattanooga.

My special thanks go out to our class instructor, Dr. Weidong Wu, for his guidance

throughout the spring semester while taking this class. He has been very kind, and nice

to us at every step of this class.

I would like to thank everyone who has contributed in this project for one way or another

throughout the duration of this project.

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DEDICATION

Finally, the authors would like to dedicate this project work to all people who strive for

intellectual achievements especially at Engineering.

CONTENTS

Acknowledgements……………………………………………..

Abstract…………………………………………………………

Contents…………………………………………………………

1.0 Introduction………………………………………………………………………………..6

2.0 Moment Distribution Method……………………………………………………………..6

3.0 CSI…………………………………………………………………………………………7

4.0 About the Project …………………………………………………………………….……8

5.0 Objective and Scope……………………………………………………………………….8

6.0 Project Problem: Geometry………………………………………..………………………8

7.0 Loading…………………………………………………………………………………..11

8.0 Hand calculation using moment distribution method……………………………………16

9.0 SAP2000 Results…………………………………………………………………………21

10.0 General Results…………………………………………………………………………23

10.0 Conclusion and recommendations……..………………………….……………………23

References ……………………..……………………………………………………………24

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1.0 Introduction

It is customary to divide structures into statically determinate and statically indeterminate.

By a statically determinate system we mean a system for which all the reactions of supports

can be determined by means of equations of equilibrium, and the internal forces also can

be found by method of sections.

A statically indeterminate system means that the reactions and internal forces cannot be

analyzed by the application of the equations of static alone.

The indeterminacy of the structure may be either external, internal, or both. The space

structure is externally indeterminate if the number of the reaction components is more than

six. The corresponding number in a plane structure is three.

2.0 Moment distribution method

The moment distribution method is a structural analysis method for statically indeterminate

beams and frames developed by Hardy Cross. It was published in 1930 in an ASCE journal.

The method only accounts for flexural effects and ignores axial and shear effects. From the

1930s until computers began to be widely used in the design and analysis of structures, the

moment distribution method was the most widely practiced method.

In the moment distribution method, every joint of the structure to be analyzed is fixed so

as to develop the fixed-end moments. Then each fixed joint is sequentially released and

the fixed-end moments are distributed to adjacent members until equilibrium is achieved.

The moment distribution method in mathematical terms can be demonstrated as the

process of solving a set of simultaneous equations by means of iteration.

The moment distribution method falls into the category of displacement method of

structural analysis.

In order to apply the moment distribution method to analyze a structure, the following

things must be considered.

Fixed end moments

Fixed end moments are the moments produced at member ends by external loads

when the joints are fixed.

Flexural stiffness

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The flexural stiffness (EI/L) of a member is represented as the product of the

modulus of elasticity (E) and the second moment of area (I) divided by the length (L) of

the member. What is needed in the moment distribution method is not the exact value but

the ratio of flexural stiffness of all members.

Distribution factors

When a joint is released and begins to rotate under the unbalanced moment,

resisting forces develop at each member framed together at the joint. Although the total

resistance is equal to the unbalanced moment, the magnitudes of resisting forces

developed at each member differ by the members' flexural stiffness. Distribution factors

can be defined as the proportions of the unbalanced moments carried by each of the

members. In mathematical terms, distribution factor of member k framed at joint j is

given as:

3.0 CSI

Computers and Structures, Inc. (CSI) is a structural and earthquake engineering Software

Company founded in 1975 and based in Berkeley, California. Additional office locations

include New York, Houston, Madrid, New Delhi, and the Caribbean. The structural

analysis and design software CSI produce include SAP2000, CSiBridge, ETABS, SAFE,

PERFORM-3D, and CSiCOL.

One of Computer and Structure, Inc.'s software, ETABS, was used to create the

mathematical model of the Burj Khalifa, currently the world's tallest building, designed by

Chicago, Illinois-based Skidmore, Owings & Merrill LLP (SOM). In the Structural

analysis section of their December 2009 Structural Engineer magazine article entitled

Design and construction of the world's tallest building: The Burj Dubai, since renamed to

Burj Khalifa, William F. Baker, S.E. and James J. Pawlikowski, S.E. mention that gravity,

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wind, and seismic response were all characterized using ETABS. Further, ETABS'

geometric nonlinear capability provided for P-Delta Effect consideration.

4.0 About the project

The project is to re analyze existing building of steel composed of 20 stories placed in

Khartoum, Sudan which is the Head Quarter of Arabic company for production of meat

and chicken products (ARAPPCO).

5.0 Objective and Scope

The objective of this project is to analyze the beam element using the indeterminate

structural Approach so as to obtain all the parameters needed for the design.

The project will be limited to analyzing the problem using hand calculation and SAP2000

computer software based on the indeterminate structural approach. Finally, the project

concluded with the overview discussion and recommendation on the analysis of the beam

element by the two approaches in performing the structural analysis.

6.0 Project Problem: Geometry

The main purpose of the project is to design so we will take the steel strength of

275 N/mm2

The second moment of area and cross-sectional area will be unknown but we

assume that the section is the same in all members.

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A B C D E F G H

1

2

3

4

5

6

7

5.0 5.0 5.0 5.0 5.0 5.0 5.0

5.0

6.0

4.0

4.0

6.0

5.0

2.0

2.0

30.0

35.0

LIFT

PIT

LIFT

PIT

1.5

1.5

Ground Floor

All Dimension in (m)

Mean Beam

Secondary Beam

Column

Stair

Braced Wall

N

10

Lift 1

Lift 2

Lift 3

Lift 4

5.05.05.05.05.05.05.01.5

1.5

1.5

5.0

6.0

4.0

4.0

6.0

5.0

1.51.5

1.51.5

1.51.5

Typical Floor

38.0

33.0

Mean Beam

Secondary Beam

Column

Stair

Braced Wall

All Dimension in (m)

Void

A B C D E F G H

1

2

3

4

5

6

7

N

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7.0 Loading:

Dead load:

30mm

100mm

100mm

120° Corrugated deck

Concrete

2 Dia10mm

steel bar

Secondary Beam

100mm130mm

Deck Slab

Corrugated deck (thickness =0.5mm) with concrete (thickness= 130mm).

Concrete density = 24 KN/m3

Steel density = 78 KN/m3

(100+130)/2 * 24 + 78*0.5 = 2.8 KN/m2

Screed = 25*10-3*(3.8KN/m3) = 0.095 KN/m2

Marble = 10*10-3*(26.7KN/m3) =0.267 KN/m2

Partition =1.0 KN/m2

Wires & air conditioning = 0.5 KN/m2

Total dead load for all floors & roof = 4.7 KN/m2

Live load:

for roof = 1.5 KN/m2

for floors = 2.5 KN/m2

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1.0

5.0

Secondary Beam

Mean Beam

Plan

5.0 5.0

1.0

1.0

1.0

1.0

1.0

Secondary beam:

dead load =4.7×1.0 = 4.7 KN/m

live load =2.5 ×1.0 = 2.5 KN/m

factored load = 1.4 D.L + 1.6 L.L

= 1.4×4.7 +1.6×2.5 = 10.68 KN/m

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10.68 KN/m

26.7 KN26.7 KN 5.0

B.M.D

S.F.D

Max Moment

=33.38 KN.m

Max Shear

=26.7 KN

Max. shear = 10.68×5/2 = 26.7 KN

Max. bending moment = 10.68×52/8 = 33.38 KN.m

Try section 254×146 UB 43

Properties:

D= 259.6 mm B= 147.3mm t= 7.3mm T=12.7mm r= 7.6mm d=218.9mm Sx=568

cm3 rmin=ry= 3.51 cm A= 55.1 cm2

b/T = 5.8 d/t = 30.0 u= 0.889 X= 21.1

147.3

12.7

259.6

7.3

218.9

R7.6

12.7

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Loading on Main beam:

5.05.0

1.5

5.0

6.0

4.0

4.0

2.0

5.0

Secondary Beam

Mean Beam

Plan

Weight of secondary beams = (43*9.81*10-3)*5

= 2.11 KN

Dead load= 4.7 KN/m2 = (4.7*5) = 23.5 KN/m

Total dead load = 25.61 KN

Live load = 2.5*5 = 12.5 KN/m

Factored load= 1.4*25.61 + 1.6*12.5 = 55.854 KN/m ؞

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1.5 5.0 6.0 4.0 4.0 2.0

1.5

1.5

c1 c3

A B C D E

c5 c7 c9

c10c8c6c4c2

55.854 KN/m

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8.0 Hand calculation using moment distribution method:

Fixed end moments:

MF = WL2/12 (for distributed load)

MFAB= - 55.854*52/12 = - 116.36 KN.m

MFBA = 55.854*52/12 = 116.36 KN.m

MFBC =- 55.854*52/12 = -167.56 KN.m

MFCB = 55.854*52/12 = 167.56 KN.m

MFCD = - 55.854*52/12 = -74.47 KN.m

MFDC = 55.854*52/12 = 74.47 KN.m

MFDE = - 55.854*52/12 = -74.47 KN.m

MFED = 55.854*52/12 = 74.47 KN.m

MFcant.L.= 55.854*1.52/2 = 62.84 KN.m

MFcant.R.= 55.854*22/2 = 111.71 KN.m

Distribution factors:

...

.

2

2

1

1

1

1

1

LI

LI

LI

FD

Joint A:

13.0

5.11

5.11

51

51

.

ABFD

43.0

5.11

5.11

51

5.11

. 1

ACFD

43.0

5.11

5.11

51

5.11

. 2

ACFD

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Joint B:

12.0

61

5.11

5.11

51

51

.

BAFD

1.0

61

5.11

5.11

51

61

.

BCFD

39.0

61

5.11

5.11

51

5.11

.. 43

BCBC FDFD

Joint C:

1.0

61

5.11

5.11

41

61

.

CBFD

14.0

61

5.11

5.11

41

41

.

CDFD

38.0

61

5.11

5.11

41

5.11

.. 65

CCCC FDFD

Joint D:

14.0

41

5.11

5.11

41

41

..

DEDC FDFD

36.0

41

5.11

5.11

41

5.11

.. 87

DCDC FDFD

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Joint E:

16.0

5.11

5.11

41

41

.

EDFD

42.0

5.11

5.11

41

5.11

.. 109

ECEC FDFD

D.F for:

C1A = C2A = C3B =C4B = C5C =C6C= C7D =C8D =C9E =C10E = 1.0 (pin end)

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20

62.84 104.3 128.3 166.7 154.7 90.0 76.4 70.2 83.5 111.17

58.5 90.833.4 35.0

1.5 5.0 6.0 4.0 4.0 2.0

1.5

1.5

c1 c3

A B C D E

c5 c7 c9

c10c8c6c4c2

55.854 KN/m

20.4 19.2 32.4 1.5 14.2

218.61 314.01 283.0 214.44 226.74

83.8

134.83 169.57

144.4

117.45

165.56

108.47

105.97

111.71

114.95

S.F.D

in (KN)

(B.M.D) in

Columns in

(KN.m)

(B.M.D) in

Beams in

(KN.m)

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9.0 SAP2000 Results

Moment diagram

Shear force diagram

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11.0 Summary of the results:

Hand

Calculations

SAP2000

Max Moments -166.7 kN.m -165.93 kN.m

Max Shear -169.57 kN -168.197 kN

12.0 Conclusion & Recommendations

The results found from hand calculation and the computer software are very

close to other. So I recommend the use of the software but with the full knowledge

of the theory because sometimes you may get wrong results from the software due

to wrong inputs but unless you know the theory behind that you will not discover

the problem.

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References

1. Aristizabal-Ochoa, J.D., Elastic Stability and Second-Order Analysis of Three-

Dimensional Frames: Effects of Column Orientation, J. Engrg. Mech., Volume 129,

Issue 11, pp. 1254-1267 (November 2003)

2. Salonikios, T., Karakostas, C., Lekidis, V., Anthoine, A., Comparative Inelastic Pushover

Analysis of Masonry Frames, Engineering Structures, Volume 25, Issue 12, pp. 1515-

1523, October 2003.

3. Romero, M.L., and Museros, P., Structural Analysis Education through Model

Experiments and Computer Simulation, J. Profl. Issues in Engrg. Educ. and Pract.,

Volume 128, Issue 4, pp. 170-175 (October 2002)

4. Mabsout, M.E., Baddaf, I.Y., Tarhini, K.M., and Frederick, G.R., Load Reduction in Steel

Girder Bridges, Pract. Periodical on Struct. Des. and Constr., Volume 7, Issue 1, pp. 37-

43 (February 2002)

5. Mylonakis, G., Papastamatiou, D., Psycharis, J, and Mahmoud, K., Simplified Modeling

of Bridge Response on Soft Soil to Nonuniform Seismic Excitation, J. Bridge Engrg.,

Volume 6, Issue 6, pp. 587-597 (November/December 2001