Theory Of Structure 130 Questions Exercises Book

130Questions
ExerciseBook

TSTHRUECOTURREYof

ChiaSoiLee
SetiSuhaidanibintiMohammed
ZawanibintiAbuRais

Theory of Structure
(130 Questions Exercises Book)

Chia Soi Lee
Seti Suhadaini Binti Mohammed

Zawani Binti Abu Rais

Publisher
Politeknik Sultan Idris Shah

Sungai Lang
Selangor Darul Ehsan

2020

First Edition : 2020
All Rights Reserved
This book or part thereof cannot be translated or reproduced in any form (except
for review or criticism) without the written permission of the Authors and the
Publishers)

Theory of Structure (130 Questions Exercise Book)
eISBN ????

Published by :
Politeknik Sultan Idris Shah
Sungai Lang,
45100 Sungai Air Tawar,
Selangor Darul Ehsan,
MALAYSIA

ii

PREFACE

We would like to thank the Almighty God that this book was successfully
completed in 2020. This Theory of Structure (130 Questions Exercise Book)
provides many exercises according to topics focused on structural analysis on
Beams, Frames and Trusses. The purpose of this book is to strengthen the
understanding among students on related topics after going through the process
of teaching and learning in the classroom. This book contains eight chapters
which include Slope Deflection Method, Moment Distribution Method for beams
and portal frame, Joint Method, Section Method as well as Virtual Load Method
for trusses and Influence Lines Method for beam. Exercise questions
constructed according to these topics are arranged according to the level of
difficulty and suitable to be practiced by each student in improving their
knowledge and strengthening their understanding of related topics. All units
used in this book are SI System units. Each question posed has an answer at
the end of this book as a review of answer. Since this course is very important
to the science of engineering especially Civil Engineering, the author has tried
to complete this book so that it can be referred by all students and lecturers of
Higher Education Institution. However, the author welcomes any feedback
related to the content this book for future improvement.

Chia Soi Lee
Seti Suhadaini Binti Mohammed
Zawani Binti Abu Rais
2020

iii

ACKNOWLEDGEMENT

The author would like to take this opportunity to thank the Sultan Idris Shah
Polytechnic for providing a platform for us to publish this written book. Many
thanks are also extended to the Library Unit of Sultan Idris Shah Polytechnic
for the effort to obtain the International Standard Book Number (ISBN) to
produce scholarly books for the use of students and lecturers in the field of Civil
Engineering. We also would like to hereby acknowledge Ms. Zanariah Zaini for
her valuable suggestions, comments and feedback during the development and
writing of this book. Not to be forgotten is the high appreciation for Ms. Angela
Kwon Mei Jun for the language review and Ms. Amisha Adnan for helping with
the interesting book cover design. Thanks also to our fellow lecturers who have
taught the Theory of Structure and Structural Analysis courses and contributed
in any form to this book. Finally, thanks are also extended to all parties directly
or indirectly involved, especially our families who are willing to be patient with
the authors’ busyness while completing this book. Hopefully this book can
benefit all of us.

iv

CONTENT

Preface iii
Acknowledgement iv
Content v

Chapter Page
1 Slope Defelction Method for Statically Indeterminate Beams 1
12
2 Slope Deflection Method for Statically Indeterminate Portal
Frame 24

3 Moment Distribution Method for Statically Indeterminate 35
Beams
47
4 Moment Distribution Method for Statically Indeterminate 59
Portal Frame
70
5 Analysis Of Statically Determinate 2D Pin-Jointed Trusses 83

6 The Joint Displacement of Statically Determinate 2D Pin- 91
Jointed Trusses 93

7 Analysis of Statically Indeterminate 2D Pin-Jointed Trusses

8 Influence Line of Beam

List of Formular
Reference

v

CHAPTER 1

SLOPE DEFELCTION METHOD FOR STATICALLY INDETERMINATE BEAMS

EXERCISES

Q1. The statically indeterminate beam carries a point load and uniformly distribution load as
shown in Figure 1.1. Calculate the value of Fixed End Moment for each support.

45 kN

EI 1m B 50 kN/m C
A 2m EI
7m

Figure 1.1

Q2. The continuous beam carries a uniformly distribution load and a point load as shown in
Figure 1.2 below. Determine Fixed End Moment for each support.

20 kN 15 kN

30 kN/m

EI B EI
A 10 m 3m
C
3m 3m

Figure 1.2

Q3. Determine the value of Fixed End Moment at each support for the indeterminate beam as
shown in Figure 1.3 below.

30 kN 30 kN 30 kN

EI EI
1m
A 2m 1m B 2m 2m C

Figure 1.3

1

Q4. Determine the value of Fixed End Moment at each support for the indeterminate beam as
shown in Figure 1.4 below.

30 kN

15 kN/m

1.5EI 3EI
6m
A 2m 2m B C

Figure 1.4

Q5. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 1.5 below. Determine the value of Fixed End Moment at each support.

50 kN/m EI 50 kN
EI B 5m C 3m D

A 5m Figure 1.5

Q6. Determine the Moment Equation for the indeterminate beam as shown in Figure 1.6 by
using the Slope Deflection Method.

40 kN 40 kN 40 kN

10 kN/m

EI EI
2m 2m C
A 2m 2 m B 2m

Figure 1.6

2

Q7. Determine the Moment Equation for the indeterminate beam as shown in Figure 1.7 by
using the Slope Deflection Method.

50 kN

30 kN/m

1.5EI EI 2EI
4m
A 3m B C D
2m 4m

Figure 1.7

Q8. Determine the Moment Equation for the indeterminate beam as shown in Figure 1.8 by
using the Slope Deflection Method.

10 kN 10 kN

20 kN/m B EI
2m
EI C
A Figure 1.8 2m 2m

6m

Q9. Determine the Moment Equation for the indeterminate beam as shown in Figure 1.9 by
using the Slope Deflection Method.

30 kN/m

2EI 3EI EI D
A 4m B 4m C 4m

Figure 1.9

3

Q10. Based on Moment Equation given below, determine the value of Slope ƟA and ƟB.

∑MA =0
MAB
MAB =0

= 4 + 2 − 26.67
6 6

∑MB =0
MBA + MBC
=0
MBA
= 2 + 4 + 33.33
MBC 6 6
4
= 6 − 48.75

Q11. Based on Moment Equation given below, determine the value of Slope ƟB and ƟC.

∑MB =0
MBA + MBC
=0
MBA
= 8 2
MBC 4 5
4
= 5 + − 145.83

∑MC =0

MCB + MCD + MCE = 0

MCB = 2 + 4 + 145.83
5 5
= 8
MCD 4

MCE = 4
2

4

Q12. Based on Moment Equation given below, determine the value of Slope ƟB and ƟC.

∑MB =0
MBA + MBC
=0
MBA
= 4 + 10
MBC 4
4 2
= 6 + 6 − 30

∑MC =0
MCB
MCB =0

= 2 + 4 + 30
6 6

Q13. Based on Moment Equation given below, determine the value of Slope ƟB and ƟC.

∑MB =0
MBA + MBC
=0
MBA
= 8 + 13.33
MBC 4
6 3
= 6 + 6 − 15

∑MC =0
MCB + MCD
=0
MCB
= 3 + 6 + 15
MCD 6 6
= 4
2

Q14. An indeterminate beam is subjected to a load as shown in Figure 1.10. By using Slope
Deflection Method, determine the value of internal moment for each support.

25 kN

10 kN/m

EI EI
6m
A 4m 4m B C

Figure 1.10

5

Q15. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 1.11. By using Slope Deflection Method, determine the value of internal
moment for each support.

5 kN

20 kN/m

EI

A 5m B 2m C

Figure 1.11

Q16. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 1.12. By using Slope Deflection Method, determine the value of internal
moment for each support.

20 kN 5 kN 5 kN
10 kN/m

2EI 3EI
1m 2m
A 2m 1m B 2m C

Figure 1.12

Q17. An indeterminate beam is subjected to a uniformly distribution load of 50 kN/m and a point
load of 100 kN as shown in Figure 1.13. By using Slope Deflection Method, determine the
value of internal moment for each support.

100 kN

50 kN/m

EI EI EI

A 2m B C 5m D
2m 5m

Figure 1.13

6

Q18. Draw Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for the
indeterminate beam as shown in Figure 1.14.

20 kN

A EI B
2m 2m

Figure 1.14

Q19. Sketch Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for the
indeterminate beam as shown in Figure 1.15.

– 250 kNm 20 kN/m 75 kN
EI B
125 kN
A 10 m
Figure 1.15

Q20. Sketch Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for the
indeterminate beam as shown in Figure 1.16.

– 78.75 kNm 30 kN
EI
10 kN/m

58.125 kN 3m 3m 31.875 kN
A B

Figure 1.16

7

ANSWER

A1.
FEMAB = – 10 kNm

FEMBA = 20 kNm
FEMBC = – 204.167 kNm

FEMCB = 204.167 kNm

A2.
FEMAB = – 250 kNm

FEMBA = 250 kNm
FEMBC = – 36.667 kNm

FEMCB = 33.333 kNm

A3.
FEMAB = – 6.667 kNm

FEMBA = 13.333 kNm
FEMBC = – 36 kNm

FEMCB = 36 kNm

A4.
FEMAB = – 27.5 kNm

FEMBA = 27.5 kNm
FEMBC = – 45 kNm

FEMCB = 45 kNm

A5.
FEMAB = – 104.167 kNm

FEMBA = 104.167 kNm

FEMBC = 0

FEMCB = 0
MCD = – 150 kNm

A6. = − 33.333
MAB 2
MBA = + 33.333
MBC
MCB = 2 − 83.333
3
A7. =
MAB 3 + 83.333
MBA
= 3 − 24
MBC 5
MCB = 6
MCD 5 + 36
MDC
= +
2
=
2 +

= 2 − 40

= + 40

8

A8.

MAB = − 60
3
MBA = 2
3 + 60

MBC = 2 + − 13.333
3 3
MCB = 2
3 + 3 + 13.333

A9. = − 40
MAB
MBA = 2 + 40
MBC
= 3 + 3 − 40
MCB 2
= 3
MCD 2 + 3 + 40
MDC
=

=
2

A10.
= 39.111


= 1.787



A11.

= 57.102


= 35.139
−

A12.

= 23.333


= 56.667
−

A13.

= 1.429


= 5.238
−

A14. = – 88.69 kNm
MAB = 57.62 kNm
MBA
= – 57.62 kNm
MBC = 16.19 kNm
MCB

9

A15. = – 57.5 kNm
MAB = 10 kNm
MBA = – 10 kNm
MBC
=0
A16. = 17.398 kNm
MAB = – 19.287 kNm
MBA = 30.606 kNm
MBC
MCB = – 25.106 kNm
= 99.788 kNm
A17. = – 99.788 kNm
MAB = 53.178 kNm
MBA = – 53.178 kNm
MBC =0
MCB
MCD
MDC

A18. 13.75 kN
+ve
SFD 0
-ve 0
–6.25 kN

–15 kNm

-ve

BMD 0 0

+ve
12.5 kNm

A19.

125 kN

+ve

SFD 0 0
-ve
– 75 kN

–250 kNm

-ve

BMD 0 0

+ve 10

×

140.625 kNm

A20.

58.125 kN

+ve 28.125 kN
– 1.875 kN
SFD 0 0
-ve

– 31.875 kN

– 78.75 kNm 0

-ve

BMD 0

+ve

×

50.625 kNm

11

CHAPTER 2

SLOPE DEFLECTION METHOD FOR STATICALLY INDETERMINATE PORTAL
FRAME

EXERCISES

Q1. The statically indeterminate portal frame carries a uniformly distribution load and a point
load as shown in Figure 2.1. Calculate the value of Fixed End Moment for each support.

20 kN

10 kN/m B
A

8m

C
3m 3m

Figure 2.1

Q2. A statically indeterminate portal frame carries a uniformly distribution load and point loads
as shown in Figure 2.2. Determine Fixed End Moment for each support.

5 kN

8 kN/m

A 2EI B EI C

EI 10 kN 3m
D 1.5 m

6m 2m

Figure 2.2

12

Q3. Determine the Moment Equation for the indeterminate portal frame as shown in Figure 2.3
by using the Slope Deflection Method.

40 kN

15 kN/m C

B

6m

A 4m
2m Figure 2.3

Q4. Determine the Moment Equation for the indeterminate portal frame as shown in Figure 2.4
by using the Slope Deflection Method.

100 kN/m

A 2EI B 2EI C

3EI 4 m

D
5m 2m

Figure 2.4

13

Q5. Determine the Moment Equation for each support for the indeterminate portal frame as
shown in Figure 2.5 by using the Slope Deflection Method.

70 kN/m

B EI C EI E

4 m 2EI 2EI

AD
5m 3m

Figure 2.5

Q6. Determine the value of Slope ƟB for the indeterminate portal frame as shown in Figure 2.6
by using the Slope Deflection Method.

25 kN/m B
A

4m

C
5m

Figure 2.6

14

Q7. Determine the value of Slope ƟB and ƟC for the indeterminate portal frame as shown in
Figure 2.7 by using the Slope Deflection Method.

40 kN 40 kN

BC

30 kN 2.5 m
2.5 m

A

2m 2m 2m
Figure 2.7

Q8. Determine the value of Slope ƟB, ƟC and ƟE for the indeterminate portal frame as shown
in Figure 2.8 by using the Slope Deflection Method.

20 kN 20 kN

5 kN/m

B EI C EI E

2EI 2EI 4 m

A D 3m
2m 2m 2m

Figure 2.8

15

Q9. An indeterminate portal frame is subjected to a load as shown in Figure 2.9. By using Slope
Deflection Method determine the value of internal moment for each support.

10 kN/m

AB
1m

5 kN

3m

C
5m

Figure 2.9

Q10. An indeterminate portal frame is subjected to a load as shown in Figure 2.10. By using
Slope Deflection Method determine the value of internal moment for each support.

20 kN 8 kN/m C
A 2EI B 2EI

EI 6 m

D

2m 5m 4m

Figure 2.10

16

Q11. An indeterminate portal frame is subjected to a load as shown in Figure 2.11. By using
Slope Deflection Method determine the value of internal moment for each support.

10 kN/m
BC

2EI

EI EI 4 m

AD
6m

Figure 2.11

Q12. An indeterminate portal frame is subjected to a load as shown in Figure 3.18. By using
Slope Deflection Method determine the value of reaction force for each support.

B 8 kN/m C
2EI 2EI

2m
15 kN

3m

A
8m

Figure 2.12

17

Q13. An indeterminate portal frame is subjected to a load as shown in Figure 3.13. By using
Slope Deflection Method determine the value of reaction force for each support.

B 10 kN/m C
2m 2EI

15 kN EI EI 15 kN

2m D
A

6m

Figure 2.15

Q14. Draw Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for the
indeterminate portal frame as shown in Figure 2.14.

20 kN/m B
A 2EI

3EI 4 m

C
3m

Figure 2.14

18

Q15. Draw Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for the
indeterminate portal frame as shown in Figure 2.15.

30 kN

50 kN/m

A 2EI B 2EI C

3EI 4 m

D
3m 3m 2m

Figure 2.15

19

ANSWER

A1.
FEMAB = – 45 kNm

FEMBA = 45 kNm
FEMBC = 0
FEMCB = 0

A2.
FEMAB = – 24 kNm

FEMBA = 24 kNm
MBC = – 10 kNm
FEMCB = – 3.333 kNm

FEMCB = 6.667 kNm

A3.

MAB =

3
MBA = 2

3
MBC = 2 3
3 + − 80.556

MCB = + 2 + 62.778
3 3

A4.

MAB = 4 − 208.333
5
MBA = 8
5 + 208.333

MBC = 4 + 2 − 33.333

MCB = 2 + 4 + 33.333
MBD = 2

MDB = 3
2

A5.

MAB =

MBA = 2 2
4 5
MBC = 5 + − 145.833

MCB = 2 + 4 + 145.333
5 5

MCE = 4 + 2
3 3

MEC = 2 + 4
3 3

MAB = 2
MBA =

20

A6.

= 28.935
−

A7.

= 47.115


= 103.557
−

A8.

= 16.466


= 6.724
−
= 7.888
−

A9.
MAB = 0

MBA = 20.586 kNm
MBC = – 20.586 kNm
MCB = – 7.949 kNm

A10. = – 57.806 kNm
MAB
MBA = 31.367 kNm
MBC = – 27.227 kNm
MCB
MBD = 2.387 kNm
MDB = – 4.14 kNm

=0

A11. = 9 kNm
MAB
MBA = 18 kNm
MBC = – 18 kNm
MCB
MCD = 18 kNm
MDC = – 18 kNm

= 9 kNm

A12. = 0.04 kN
FAX = 14.96 kN
FBX = 29.59 kN
FBY = 34.41 kN
FCY

21

A13. = 2.653 kN
FAX = 12.347 kN
FBX = 30 kN
FBY = 30 kN
FCY = 12.347 kN
FCX = 2.653 kN
FDX
33.529
A14.

SFD 33.429(3)
= (33.529 + 26.471)

= 1.675

– 26.471

2.978

– 18.529

BMD – 7.941
– 7.941

9.552

3.971

22

A15. 171.70 kN

92.66

SFD 21.70 – 7.34
– 8.30
92.66(2)
= (92.66 + 7.34)
– 158.30
= 1.853
22.63

– 185.91 – 145.71
– 85.35
BMD
0.499
– 60.33

104.59

30.19

23

CHAPTER 3

MOMENT DISTRIBUTION METHOD FOR STATICALLY INDETERMINATE
BEAMS

EXERCISE

Q1. The statically indeterminate beam carries a point load and uniformly distribution load as
shown in Figure 3.1. Calculate the value of Fixed End Moment for each support.

15 kN

30 kN/m

1.5EI 1m B 2EI C
A 2m Figure 3.1 5m

Q2. The continuous beam carries a uniformly distribution load and point loads as shown in

Figure 3.2. Determine the value of Fixed End Moment for each support.

15 kN 15 kN

40 kN/m

EI B EI
1m
A C
3m 1m 1m

Figure 3.2

Q3. Figure 3.3 shows an indeterminate beam subjected with point load and uniformly
distribution load. Determine the value of stiffness factor for each span of the beam.

10 kN

25 kN/m

2EI 2EI

A 3m 3m B 7m C

Figure 3.3

24

Q4. Figure 3.4 shows an indeterminate beam subjected with point loads and uniformly
distribution load. Determine the value of stiffness factor for each span of the beam.

30 kN 30 kN 30 kN

40 kN/m

2EI 3EI
1m
A 2m 1m B 2m 2m C

Figure 3.4

Q5. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 3.5. Determine the value of stiffness factor for each span of the beam.

25 kN/m

1.5EI 1.5EI C EI
A 4m B 4m D

4m
Figure 3.5

Q6. Determine the distribution factor for the indeterminate beam as shown in Figure 3.6.

50 kN 50 kN 50 kN

2.5EI 3EI
2m
A 2m 2m B 2m 2m C

Figure 3.6

25

Q7. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 3.7. Determine the value of distribution factor for each span of the beam.

20 kN

20 kN/m

1.5EI 2EI 2EI
3m 3m
A 2m B C D
1m

Figure 3.7

Q8. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 3.8. Determine the value of distribution factor for each span of the beam
using Moment Distribution Method with modify distribution.

10 kN 10 kN

15 kN/m

A EI B EI C
2m 2m 2m 6m

Figure 3.8

Q9. Fill in the box with the value of carry over factor at each span of the indeterminate beam as
shown in Figure 3.9.

COFBA = COFBC =

COFAB = B COFCB =
A Figure 3.9
C

26

Q10. Fill in the box with the value of carry over factor at each span of the indeterminate beam as
shown in Figure 3.10 using Moment Distribution Method with modify distribution.

COFBA = COFBC = COFDC = D

COFAB = COFCB = COFCD =
A B C

Figure 3.10

Q11. Fill in the box with the value of carry over factor at each span of the indeterminate beam as
shown in Figure 3.11 using Moment Distribution Method with modify distribution.

COFBA = COFBC =

COFAB = COFCB = COFCD =
A B CD

Figure 3.11

Q12. An indeterminate beam is subjected to a load as shown in Figure 3.12. By using Moment
Distribution Method, determine the value of internal moment at each support.

35 kN/m

EI
A 5m B

Figure 3.12

27

Q13. An indeterminate beam is subjected to a load as shown in Figure 3.13. By using Moment
Distribution Method, determine the value of internal moment at each support.

15 kN

5 kN/m

EI EI
6m C
A 2m 2m B

Figure 3.13

Q14. An indeterminate beam is subjected to a point loads and distribution load as shown in
Figure 3.14. By using Moment Distribution Method, determine the value of final moment at
each support.

10 kN 50 kN

40 kN/m

2EI 3EI 1.5EI
5m
A 3m 2m B C D
2m 2m

Figure 3.14

Q15. An indeterminate beam is subjected to point loads as shown in Figure 3.15. By using
Moment Distribution Method with modify distribution, determine the value of final moment
at each support.

30 kN 10 kN 10 kN

2EI 2EI
1m
A 2m 1m B 2m 2m C

Figure 3.15

28

Q16. An indeterminate beam is subjected to loads as shown in Figure 3.16. By using Moment
Distribution Method with modify distribution, determine the value of final moment at each
support.

10 kN

5 kN/m

EI EI EI

A 2m B C 5m D
2m 5m

Figure 3.16

Q17. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 3.17. By using Moment Distribution Method determine the value of
internal moment for each support.

2 kN

10 kN/m

A 5m B 2m C
Figure 3.17

Q18. An indeterminate beam is subjected to loads as shown in Figure 3.18. By using Moment
Distribution Method determine the value of reaction force for each support.

20 kN

10 kN/m

EI EI

A 2m 2m B 7m C

Figure 3.18

29

Q19. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 3.19. By using Moment Distribution Method determine the value of
internal moment and reaction force at each support.

100 kN

50 kN/m

EI EI EI

A BC D

2m 2m 5m 5m

Figure 3.19

Q20. An indeterminate beam is subjected to a uniformly distribution load and a point load as
shown in Figure 3.20. By using Moment Distribution Method determine the value of
internal moment at each support and Draw Shear Force Diagram and Bending Moment
Diagram.

20 kN 10 kN/m
8 kN/m

EI B EI C
2.5 m 7m
A
2.5 m

Figure 3.20

30

ANSWER

A1.
FEMAB = – 25.833 kNm
FEMBA = 29.167 kNm
FEMBC = – 62.5 kNm
FEMCB = 62.5 kNm

A2.
FEMAB = – 30 kNm
FEMBA = 30 kNm
FEMBC = – 40 kNm
FEMCB = 40 kNm

A3.
KAB = 1.333EI
KBA = 1.333EI
KBC = 1.143EI
KCB = 1.143EI

A4.
KAB = 2EI
KBA = 2EI
KBC = 2.4EI
KCB = 2.4EI

A5.
KAB = 1.5EI
KBA = 1.5EI
KBC = 1.5EI
KCB = 1.5EI
KCD = EI
KDC = EI

A6. =0
DFAB = 0.56
DFBA = 0.44
DFBC =0
DFCB

A7. =0
DFAB = 0.429
DFBA = 0.571
DFBC = 0.5
DFCB = 0.5
DFCD =0
DFDC

31

A8. =0
DFAB = 0.571
DFBA = 0.429
DFBC =1
DFCB

A9.
COFAB = 0
COFBA = 0.5
COFBC = 0.5
COFCB = 0

A10.
COFAB = 0
COFBA = 0.5
COFBC = 0.5
COFCB = 0.5
COFCD = 0
COFDC = 1

A11.
COFAB = 0
COFBA = 0.5
COFBC = 0
COFCB = 0.5
COFCD = 0

A12. = – 109.38 kNm
MAB =0

MBA

A13. = – 5.25 kNm
MAB = 12 kNm
MBA = – 12 kNm
MBC = 16.5 kNm
MCB

A14. = – 88.8 kNm
MAB = 89.22 kNm
MBA = – 89.22 kNm
= 60.23 kNm
MBC = – 60.23 kNm
MCB = 49.88 kNm
MCD
MDC

32

A15. =0
MAB = 14.06 kNm
MBA = – 14.06 kNm
MBC = 10.96 kNm
MCB
= – 25.106 kNm
A16. = 99.788 kNm
MAB = – 99.788 kNm
MBA = 53.178 kNm
MBC = – 53.178 kNm
MCB =0
MCD
MDC = – 29.25 kNm
= 4 kNm
A17. = – 4 kNm
MAB
MBA = 25.824 kNm
MBC = 34.176 kNm
= 33.636 kNm
A18. = 36.364 kNm
Ay
By1 = 31.33 kN
By2 = 68.67 kN
Cy = 134.322 kN
= 155.678 kN
A19. = 10.636 kN
Ay = – 10.636 kN
By1
By2
Cy1
Cy2
Dy

33

A20.

23.73 42.15
-36.27
3.72

SFD

-16.27

-50.09 -27.85
38.74
BMD - 18.73

15.58

34

CHAPTER 4

MOMENT DISTRIBUTION METHOD FOR STATICALLY INDETERMINATE
PORTAL FRAME

EXERCISE

Q1. The statically indeterminate portal frame carries a point load and uniformly distribution
load as shown in Figure 4.1. Calculate the value of Fixed End Moment for each support.

20 kN/m
AB

1m
10 kN

2m

C
4m

Figure 4.1

Q2. The statically indeterminate portal frame carries a uniformly distribution load and point loads
as shown in Figure 4.2. Determine the value of Fixed End Moment for each support.

40 kN 40 kN 40 kN

10 kN/m

AB C

4m

D
2m 2m 2m 2m 2m

Figure 4.2

35

Q3. Determine the value of stiffness factor for an indeterminate portal frame subjected with
point loads and uniformly distribution load as shown in Figure 4.3.

20 kN 20 kN

A 2EI B 15 kN/m C
2EI

3EI 4 m

1m 1m D 3m
1m

Figure 4.3

Q4. Figure 4.4 shows an indeterminate portal frame subjected with 60 kN/m uniformly
distribution load. Determine the value of stiffness factor for each span of the beam and
column.

AB 60 kN/m C EI D
EI EI 2EI

2EI

5m

E 2m F 3m
3m 2m 2m

Figure 4.4

36

Q5. Determine the distribution factor for the indeterminate portal frame as shown in
Figure 4.5.

25 kN/m

B 2EI C
3m

30 kN 3EI

3m

A

5m

Figure 4.5

Q6. Determine the distribution factor for the indeterminate portal frame as shown in
Figure 4.6.

35 kN 30 kN

B EI C EI E
2EI 2EI
4m

A 1.5 m D 2m
1.5 m 1.5 m

Figure 4.6

37

Q7. Fill in the box with the value of carry over factor (COF) of the indeterminate portal frame
as shown in Figure 4.7.

COFBA = COFBC = COFCB =
B C COFCD =

COFAB = COFDC =
D
A

Figure 4.7

Q8. Fill in the box with the value of carry over factor (COF) of the indeterminate portal frame
as shown in Figure 4.8.

COFAB = COFBA = COFBC = COFCB = COFCD =
COFAE = COFCG = C COFDC =
A B COFBF = D

COFEA = E F COFFB = COFGC = G

Figure 4.8

38

Q9. An indeterminate portal frame is subjected to point loads as shown in Figure 4.9. By using
Moment Distribution Method, determine the value of internal moment at each support.

20 kN 30 kN

BC
3m

10 kN

3m

A 2m 2m
2m

Figure 4.9

Q10. An indeterminate portal frame is subjected to a point loads and an uniformly distribution
load as shown in Figure 4.10. By using Moment Distribution Method, determine the value
of final moment at each support.

30 kN 30 kN

A 2EI 25 kN/m
B 2EI C

3EI 4 m

D
2m 2m 2m 2m

Figure 4.10

39

Q11. Determine the value of internal moment at each support by using Moment Distribution
Method for the indeterminate portal frame as shown in Figure 4.11.

50 kN
20 kN/m

B 3EI C 2EI E

2EI 2EI
4m

A D

2.5 2.5 m 2.5 m
m

Figure 4.11

Q12. Determine the value of reaction force for each support by using Moment Distribution
Method for the indeterminate portal frame as shown in Figure 4.12.

55 kN/m

A 3EI B 3EI C
2.5 m

2EI 15 kN
2.5 m

D
5m 3m

Figure 4.12

40

Q13. An indeterminate portal frame is subjected to a load as shown in Figure 4.13. By using
Moment Distribution Method, determine the value of reaction force for each support.

10 kN 10 kN

A 2EI B 20 kN/m C
2EI

3EI 4 m

D 3m
1m 1m 1m

Figure 4.13

Q14. Draw Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for the
indeterminate portal frame as shown in Figure 4.14.

B 8 kN/m C
EI 2EI 2m

15 kN 3m

A
8m

Figure 4.14

41

Q15. Draw Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for the
indeterminate portal frame as shown in Figure 4.15.

2m B 30 kN/m C
10 kN 2EI 3EI 2EI 10 kN

2m A 6m D

Figure 4.15

42

ANSWER

A1.
FEMAB = – 26.667 kNm

FEMBA = 26.667 kNm
FEMBC = – 4.444 kNm

FEMCB = 2.222 kNm

A2.
FEMAB = – 33.333 kNm
FEMBA = 33.333 kNm
FEMBC = – 66.667 kNm
FEMCB = 66.667 kNm

FEMBD = 0
FEMDB = 0

A3.
KAB = 2EI
KBA = 2EI
KBC = 2EI
KCB = 2EI
KBD = 3EI
KDB = 3EI

A4.
KAB = EI
KBA = EI
KBC = 0.667EI
KCB = 0.667EI
KCD = EI
KDC = EI
KBE = 1.6EI
KEB = 1.6EI
KDB = 1.6EI
KDB = 1.6EI

A5. =0
DFAB = 0.625
DFBA = 0.325
DFBC =1
DFCB

A6. =0
DFAB = 0.692
DFBA = 0.308
DFBC = 0.202
DFCB = 0.456
DFCD =0
DFDC = 0.342
DFCE =1
DFEC

43

A7.
COFAB = 0
COFBA = 0.5
COFBC = 0.5
COFCB = 0.5
COFCD = 0.5
COFDC = 0

A8.
COFAB = 0.5
COFBA = 0.5
COFBC = 0.5
COFCB = 0.5
COFCD = 0
COFDC = 0.5
COFAE = 0.5
COFEA = 0
COFBF = 0
COFFB = 0.5
COFCG = 0.5
COFGC = 0

A9.

MAB = 4.326 kNm

MBA = 31.151 kNm
MBC = – 31.151 kNm

MCB =0

A10. = – 127.737 kNm
MAB = 89.527 kNm
MBA = – 46.542 kNm
MBC = – 10.772 kNm
MCB = – 42.985 kNm
MBD =0
MDB
= 19.657 kNm
A11. = 39.321 kNm
MAB = – 39.321 kNm
MBA = 69.319 kNm
MBC = – 22.655 kNm
MCB = – 46.665 kNm
MCD = – 7.701 kNm
MDC = – 11.324 kNm
MCE
MEC

44