Theory Of Structure 130 Questions Exercises Book

A12. = 143.115 kN
FAY = 131.885 kN
FBY1 = 98.097 kN
FBY2 = 66.903 kN
FCY = 12.184 kN
FBX = 2.816 kN
FDX
= 39.855 kN
A13. = 40.145 kN
FAY = 37.428 kN
FBY1 = 22.572 kN
FBY2 = – 0.122 kN
FCY = 0.122 kN
FBX
FDX 28.69

A14

SFD 28.69(8)
= (28.69 + 35.31)
– 14
= 3.586

– 35.31

1

– 51.37

BMD – 25
– 25

3

26.50

45

A15. 90 90(6)
= (90 + 90)
SFD
= 3
– 26.25
26.25
– 16.25
– 90

16.25

61.67 61.67
– 61.67 – 61.67

BMD 73.33

23.33 23.33

46

CHAPTER 5

ANALYSIS OF STATICALLY DETERMINATE 2D PIN-JOINTED TRUSSES

EXERCISE

Q1. Refer to Figure 5.1, determine force in members and its magnitude by using the method of
joint.

50 kN C D

4m

A B

4m
Figure 5.1

Q2. Refer to the truss as shown in Figure 5.2,
a. Calculate the reaction at joint E and D.
b. Determine the forces in member and its magnitude at joints A, E, D and C by using the
method of joint.

C 22.5kN

15m

B

15m

A ED

20kN 12m
Figure 5.2
12m

47

Q3. Determine the forces in truss members AB, BC and CD as shown in Figure 5.3 by using
method of section.

30 kN

10 kN C D

2m

A B

2m 2m
Figure 5.3

Q4. Refer to Figure 5.4, determine force in members and its magnitude by using the method
of joint.

B CD

6m

AG E
F

20 kN 60 kN
8m 8m 8m

Figure 5.4

48

Q5. Refer to the truss as shown in Figure 5.5,
a. Calculate the reaction at joints A and F.
b. Determine the forces in members and its magnitude by using the method of joint.

BD

6m

AC F
E

6 kN 3 kN

8m 8m 8m

Figure 5.5

Q6. Refer to Figure 5.6, determine force in members and its magnitude by using the method
of joint.

F 15 kN

D E 4m
4m
A 20 kN

5m B C

5m
Figure 5.6

49

Q7. Refer to the truss as shown in Figure 5.7,
a. Calculate the reaction at joints D and E.
b. Determine the forces in members and its magnitude by using the method of joint.

AB C

100 kN 3m

D E

4m 4m

Figure 5.7

Q8. Refer to Figure 5.8, determine force in members and its magnitude by using the method
of joint.

30 kN F G H

Determine force in Determine force in 3m
members and its members and its
magnitude by using magnitude by using E
the method of joints the method of joints

A D
BC

40 kN 40 kN 40 kN
4m 4m
4m 4m

Figure 5.8

50

Q9. Refer to Figure 5.9, determine force in members and state whether members are in
compression or tension by using the method of joint.

AB C

200 kN D 1m
2m
Figure 5.9 E

2m

Q10. Determine the forces in truss members AB, BE and DE as shown in Figure 5.10 by using
method of section.

20 kN

AB

4m

E DC

10 kN 10 kN

2m 2m

Figure 5.10

51

Q11. Determine the forces in truss members CD, CG and GH as shown in Figure 5.11 by using
method of section.

400N 400N

CD E

Determine force in Determine force in 3m
members and its members and its
magnitude by using magnitude by using B
the method of joints the method of joints
3m
A G F
H
3m
3m 3m

Figure 5.11

Q12. Determine the forces in truss members HC, HG and BC as shown in Figure 5.12 using
method of section

30 kN H G F

Determine force in Determine force in 20m
members and its members and its
magnitude by using magnitude by using E
the method of joints the method of joints

A D
BC

40 kN 40 kN 40 kN
15m 15m
15m 15m

Figure 5.12

52

Q13. Truss as shown in Figure 5.13 is supported at joints A and E. The load of 100kN is
suspended at point D. By using section methods, calculate the forces in members BC,
CF and FE and state whether the members experience compression or tension.

BC D

100 kN 4m

A F E

4m 4m 4m

Figure 5.13

Q14. Identify the forces in truss members BC and BE as shown in Figure 5.14 by using method
of section. State whether the internal forces is tension or compression.

20 kN

BC

1m

A E D

2m 35 kN 15 kN
2m

Figure 5.14

53

Q15. By using method of joints, determine the internal forces of all truss members as shown in
Figure 5.15 due to the external loads. Member FE and ED have an equal length. State
whether the internal forces is tension or compression.

15 kN

10 kN F 3m
E
G 3m

A D
B
C

20 kN 30 kN

5m 5m 5m

Figure 5.15

54

ANSWER

A1.
FAB = -50 kN
FAC = 0
FCD = -50 kN
FAD = 70.71 kN
FDB = -50 kN

A2(a). = 96.25 kN
RE = -76.25 kN
RD = 22.5 kN
HD

A2(b). = 25.61 kN
FAB = -16 kN
FAE = -16 kN
FED = -96.25 kN
FEB = 61.63 kN
FBD = -36.02 kN
FBC = 28.125 kN
FCD

A3.
FAB = -40 kN
FCD = 30 kN
FCB = -40 kN

A4.
FAB = 20 kN
FBC = 22.67 kN
FBG = -33.33 kN
FAG = 0
FCG = 0
FCD = 22.67 kN
FGF = -80 kN
FDF = -100 kN
FGD = 33.33 kN
FDE = -100 kN
FFE = -80 kN

A5(a). = 5 kN
RA = 4 kN
RF = 0 kN
HA

55

A5(b). = -8.33 kN
FAB = 6.67 kN
FAC = 5 kN
FBC = -6.67 kN
FBD = 1.67 kN
FCD = 5.34 kN
FCE = 3 kN
FDE = -6.67 kN
FDF = 5.34 kN
FEF
= -10 kN
A6. = -12 kN
FAB = 12.81 kN
FAD = -25 kN
FAE = -16 kN
FBC = 32.02 kN
FBE = 15 kN
FCE = -19.21 kN
FDE = 12 kN
FDF
FEF = 200 kN
= -100 kN
A7(a). = 0 kN
RD
RE = 133.33 kN
HE = -166.66 kN
=0
A7(b). = 133.33 kN
FAB = -166.66 kN
FAD =0
FBD = 100 kN
FBC
FDC
FDE
FCE

56

A8. = 67.5 kN
FAB = -62.5 kN
FAF = 67.5 kN
FBC = 40 kN
FBF = 52.5 kN
FCD = 12.5 kN
FCF =0
FCG = 37.5 kN
FCH = 52.5 kN
FDE = 40 kN
FDH = -87.5 kN
FEH = -75 kN
FFG = -75 kN
FGH
= 400 kN (Tension)
A9. = 447.2 kN (Compression)
FAB =0
FAD = 400 kN (Tension)
FBD = 447.2 kN (Compression)
FBC = 200 kN (Tension)
FCD
FCE 0
FED
= 25 kN
A10. = -44.721 kN
FAB = -5 kN
FBE
FDE = -600 N
= 424.26 N
A11. = 600 N
FCD
FCG = 12.5 kN
FED = -75 kN
= 67.5 kN
A12.
FHC = 50 kN (Tension)
FHG = 70.7 kN (Tension)
FBC = 100 kN (Compression)

A13.
FBC
FCF
FFE

57

A14. = 30 kN (Tension)
FBC
FBE = 156.52 kN (Tension)

A15. = 58.33 kN (Tension)
FAB
FAG = 68.02 kN (Compression)
FGB
FBC = 20 kN (Tension)
FGC
FGF = 58.33 kN (Tension)
FFC
FFE = 29.15 kN (Compression)
FCE
FCD = 38.87 kN (Compression)
FED
= 45 kN (Tension)

= 52.07 kN (Compression)

=0

= 33.33 kN (Tension)

= 52.07 kN (Compression)

58

CHAPTER 6

THE JOINT DISPLACEMENT OF STATICALLY DETERMINATE 2D PIN-
JOINTED TRUSSES

EXERCISE

Q1. Determine the vertical displacement (Δv) of joint B in the frame as shown in Figure 6.1. The
cross-sectional area of all members, A = 1000 mm2 and E = 210 GPa.

30 kN C

4m

A C
B

80 kN
3m 3m

Figure 6.1

Q2. Determine the vertical displacement (Δv) of joint D in the frame as shown in Figure 6.2. The
cross-sectional area (mm2) of all members are as shown and E = 200 kN/mm2.

B (200) D

80 kN

(200) (200)

(200) (200) 1.5m

A (150) (150) E

C

150 kN 2m

2m

Figure 6.2

59

Q3. A statically determinate truss is subjected to the external loads as shown in Figure 6.3.
Given the cross sectional area, A = 400 mm2 and modulus of elasticity, E = 200 kN/mm2.
a. Prove that the truss is statically determinate.
b. Determine the internal forces in each member of the truss due to the external load and
a vertical load at joint B.
c. Calculate the vertical displacement (Δv) of joint B.

B

40 kN

AD 3m

C

80 kN

4m 4m

Figure 6.3

Q4. The cross sectional area of each member of a truss as shown in Figure 6.4, is A = 400mm2
and E = 200 GPa. Determine the vertical displacement (Δv) of joint C if a 4 kN force is
applied to the truss at C.

C 4 kN

3m

AB

4m 4m

Figure 6.4

60

Q5. Determine the vertical displacement (Δv) of joint C of the steel truss as shown in Figure 6.5.
The cross sectional area of each member is A = 300mm2 and E = 200 GPa.

FE

A 3m
BC
D

20 kN 20 kN

3m 3m 3m

Figure 6.5

Q6. Figure 6.6 shows a truss that is supported at A and B. The truss is subjected to a horizontal
load of 80 kN at joint F. Given cross-sectional area, A for all members are 1200mm2 and

Young Modulus, E is 210 GPa.

a. Identify the internal force in each member of the truss due to external forces.
b. Determine the value of internal forces due to the vertical 1-unit load at joint E.
c. Calculate total vertical displacement (Δv) at joint E.

DE F 80 kN

Determine force in Determine force in

members and its members and its

8m magnitude by using magnitude by using

the method of joints the method of joints

AC B

6m 6m

Figure 6.6

61

Q7. A plane truss is loaded as shown in Figure 6.7. All the members have the cross sectional
area of 350 mm2 and Young Modulus, E of 210 kN/mm2.
a. Calculate the internal forces in the entire member of the truss due to the external loads
and virtual unit loads.
b. Determine the horizontal displacement (Δh) of joint C.

45 kN

15 kN B C

5m

A D
E
5m
20 kN

3m

Figure 6.7

Q8. Determine the horizontal displacement (Δh) of joint D in the frame as shown in Figure 6.8.
The cross-sectional area (mm2) of all members are as shown and E = 200 kN/mm2.

B (200) D
(200)
80 kN
A
(200) (200)

(200) 1.5m

(150) (150) E

C

150 kN

2m 2m

Figure 6.8

62

Q9. Determine the vertical displacement (Δv) at joint E of the truss as shown in Figure 6.9. The
modulus of elasticity for each member is E = 210 kN/mm2 and the data given is as follows:

Member L ( x 102 mm) A ( x 104 mm2 ) Member L ( x 102 mm) A ( x 104 mm2 )
AH 180 2 DE 202 3
HG 180 2 BH 90 1
GF 180 1 BG 202 1.5
FE 180 1 CG 180 3
AB 202 3 GD 202 1.5
BC 202 4 DF 90 1
CD 202 4

C

BD

18 m

A F E
HG

20 kN 20 kN 20 kN

18m 18m 18m 18m

Figure 6.9

63

Q10. Determine the vertical displacement (Δv) at joint E of the truss as shown in Figure 6.10. The
modulus of elasticity of each member is E = 200 kN/mm2 and the cross sectional area of
each member is A = 1000 mm2.

40 kN

20 kN E C A

40 kN

3m

D

2m 2m B

Figure 6.10

Q11. Referring to Figure 6.11, the modulus of elasticity of each member is E = 20 kN/mm2 and
the cross sectional area of each member is 1000 mm2, except for members EB and EC,
which are 40 mm2. Determine;
a. The vertical displacement (Δv) at joint C
b. The horizontal displacement (Δh) at joint C

20 kN

B 20 kN

A 1m
E
C

1m

D

1m 1m

Figure 6.11

64

Q12. Determine the vertical displacement (Δv) at joint C of the truss as shown in Figure 6.12.
The modulus of elasticity of each member is 200 x 106 kN/m2 and the cross sectional area
for all members is 100 mm2.

D

20 kN C 6m

B E

45 kN 6m

A F
6m
8m
Figure 6.12

Q13. Determine the horizontal displacement (Δh) of joint B in the frame as shown in Figure 6.13.
The cross-sectional area of all members, A = 1000 mm2 and E = 210 GPa.

30 kN D

4m

A C
B

80 kN

3m 3m

Figure 6.13

65

Q14. Determine the vertical displacement (Δv) and horizontal displacement (Δh) of joint C in the
frame as shown in Figure 6.14. Assume that, all members have the same axial rigidity.

10 kN

B C

5 kN

4m

AD

4m
Figure 6.14

Q15. Determine the horizontal displacement (Δh) and vertical displacement (Δv) of joint A in the
frame as shown in Figure 6.15. Assume that, all members have the same axial rigidity.

AB C

200 kN 1m

D E
1m
1m
Figure 6.15

66

ANSWER

A1.
ΔvB = 0.00317 m ( ↓ )

A2.
ΔvD = 20.12 mm ( ↓ )

A3(a).
m + r = 2j
8=8

A3(b).
Due to external loads:
FAB = -41.66kN
FAD = 73.33kN
FBD = 80kN
FBC = -41.66kN
FDC = 73.33kN

Due to vertical load at joint B
FAB = -0.836kN
FAD = 0.67kN
FBD = 0
FBC = -0.836kN
FDC = 0.67kN

A3(c).
ΔvB = 0.00927 m ( ↓ )

A4.
ΔvC = 0.133 mm ( ↓ )

A5.
Δcv = 6.16 mm ( ↓ )

A6(a).
Due to external loads:
FAD = -53.33kN
FAC = -80kN
FDE = -40kN
FDC = 66.66kN
FEC = 0
FEF = -40kN
FCF = -66.66kN
FCB = 0
FFB = 53.33kN

67

A6(b). = -0.5kN
=0
FAD = -0.375kN
FAC = 0.625kN
FDE = -1kN
FDC = -0.375kN
FEC = 0.625kN
FEF =0
FCF = -0.5kN
FCB
FFB

A6(c).
ΔvE = -1.504 mm ( ↑ )

A7(a).
Due to external loads:

FAB = -3.644kN
FAE = 1.875kN
FBE = 3.125kN
FBC = -16.875kN
FEC = 23.865kN
FED = -15kN
FCD = -61.875kN

Due to horizontal load at joint C:

FAB = -0.729kN
FAE = 0.375kN
FBE = 0.625kN
FBC = -0.375kN
FEC = -0.884kN
FED = 1kN
FCD = 0.625kN

A7(b).
ΔhD = -4.88 mm ( → )

A8.
ΔhD = 10.30 mm ( → )

A9.
ΔvE = 3.76 mm ( ↓ )

A10.
ΔvE = 6.28 mm ( ↓ )

68

A11(a).
ΔvC = 1 mm ( ↓ )

A11(b).
ΔhD = 26 mm ( → )

A12.
ΔvC = 0.185 m ( ↓ )

A13.
ΔhB = 0.000214 m ( → )

A14.

 vC = 116.569 (→)

AE

( )hC= 60 
AE

A15.

 hA = 1165.509 ()
AE

= 1731.018
AE
( )vC 

69

CHAPTER 7

ANALYSIS OF STATICALLY INDETERMINATE 2D PIN-JOINTED TRUSSES

EXERCISE

Q1. A simple frame structure is subjected to load as shown in Figure 7.1. The cross-sectional
area and young modulus are constant for all members. Assume AD as redundant member.
a. Identify the type of structure with respect to indeterminacy
b. Determine the member forces

30 kN 30 kN

50 kN C D

3m

AB

4m
Figure 7.1

Q2. A simple frame structure is subjected to load as shown in Figure 7.2. The cross-sectional
area and young modulus are constant for all members. Assume DF as a redundant
member.
a. Identify the type of structure with respect to indeterminacy
b. Determine the member forces

30 kN 30 kN

CD

3.5m

A FE B
4m 4m 4m

Figure 7.2

70

Q3. The cross sectional area (cm2) for members are given as shown in Figure 7.3. E is constant.
Assume DF as redundant member. Determine the internal forces for the frame structure.

40 kN (15) 30 kN

C D

(20) (20) (20) (20)

(15) (15) 3.5m

A B
(15) F (15)
E (15)
4m 4m 4m

Figure 7.3

Q4. A simply supported truss is subjected to a horizontal axial load as shown in Figure 7.4. The
cross sectional area, A is 15 cm2 and the modulus elasticity, E is 200 kN/mm2 in each
member of the truss.
a. Identify the value of reactions at supports A and B
b. Compute the internal force for all members of the truss due to the external load if AC
member is a redundant.
c. Calculate the internal forces for all members of the truss due to the virtual unit load at
AC member.
d. Calculate the internal force for all members of the truss by using magnitude of
redundant, R.

50 kN D C

12m

A B

16m

Figure 7.4

71

Q5. The cross sectional area (cm2) for members are given as shown in Figure 7.5. The modulus
of elasticity (E) of each member is constant. Assume CF as redundant member.
a. Classify the trusses
b. Determine the internal forces.

B (2) C

(2) (1) (1) (2) 28.8m

(1) (1)

A D

(2) F (2) E (2)

10 kN 25 kN

28.8m 28.8m 28.8m

Figure 7.5

Q6. A truss carries a load as shown in Figure 7.6. The modulus of elasticity (E) of each member
is 200 kN/mm2.
a. Prove that the truss is internally statically indeterminate.
b. Determine the internal forces of each member assuming CE member as redundant.

Member AB BC CD DE EF FA BF CF CE DF

Area (cm2) 10 12 12 10 12 8 8 10 8 8

B CD

4m 3m

A E

F

20 kN

2m 2m

Figure 7.6

72

Q7. The cross sectional area (cm2) for members are given as shown in Figure 7.7. The modulus
of elasticity (E) of each member is constant. Assume support G as redundant.
a. Prove the truss is externally statically indeterminate
b. Determine the reactions and internal forces of the trusses.

40 kN

B (4) C (4) D

Determine force in Determine force in

(4) members and its (3) members and its (4) 24m
(3) magni(t3u)de by using magnitude b(3y) using (3)
the method of joints the method of joints

A E

(4) H (4) G (4) F (4)

30 kN 30 kN

30m 30m 30m 30m

Figure 7.7

Q8. The Figure 7.8 shows the indeterminate truss is pinned supported at A and D. Modulus
elasticity, E and cross sectional area, A are given.
a. Determine the classification of the truss.
b. Determine the reaction and member forces for all members using the method of virtual
work if the horizontal reaction at A is selected as the redundant.

Member E (kN/mm2) A (mm2)
AB 200 625
BC 200 500
CD 200 625
AC 30 400
BD 30 400

BC

50 kN

4m

AD

3m
Figure 7.8

73

Q9. The cross sectional areas of the members (cm2) are given as shown in Figure 7.9. Assume
E = 2.0 x 105 N/mm2. Calculate reactions and member forces of the truss by using the
method of virtual work if the horizontal reaction at E is selected as the redundant.

60 kN

F (15) G 30 kN

(25) (10) (25) (25) (25) 4m
(10)
A B (15) C (15) E
3m 3m D (15)
(15) 3m
Figure 7.9
3m

Q10. Determine the internal forces and reactions for the frame structure as shown in Figure 7.10.
The cross sectional area (A) and E is constant. Assume the vertical reaction at D as
redundant.

35 kN 25 kN

AB C

4.5m

9m

E

D 6m
6m

Figure 7.10

74

Q11. Determine the internal forces and reactions for the frame structure as shown in Figure 7.11.
The cross sectional area (A) and E is constant. Assume support B as redundant.

30 kN D

9m

AB C

12m 12m

Figure 7.11

Q12. Determine the internal forces and reactions for the frame structure as shown in Figure 7.12.
The cross sectional area (A) of the members given are in 10-3 m2 and E is constant. Assume
the reaction at F as redundant.

B (2.7) C (2.7) D (2.7) E
(3.6)
(0.9) Determine force in Determine force in (3.6)
(2.25) membe(1r.s8)and its member(1s.8a) nd its
A
6m (0.9) magnitude by usin(0g.9) magnitude by usin(0g.9) 8m
the method of joints the method of joints
F

G (2.25) H (2.25) I (2.25)

150 kN 200 kN 150 kN

6m 6m 6m

Figure 7.12

75

Q13. A statically indeterminate truss is subjected to an external load as shown in Figure 7.13.
Given the cross sectional area, A and the modulus of elasticity, E are constant for each
member of the truss.
a. Prove that the truss is statically indeterminate structure.
b. Determine the internal forces in all members of the truss which was caused by the
external loads and virtual unit load if BD member is a redundant.
c. Calculate the internal forces in all members of the truss by using magnitude of
redundant, R.

60 kN

75 kN B C

4m

115 kN A 40 kN

60 kN D

5m

Figure 7.13

Q14. A statically indeterminate truss is subjected to an external load as shown in Figure 7.14.
The cross sectional area, A and Modulus of Young, E are constant for each member.
Assume member BF as a redundant. Given the value reaction force at support A, RA = -
40kN, HA = -20kN and at support F, RF = 60kN. Calculate the internal forces in all members
of the truss by using magnitude of redundant, R.

B C

20 kN

D

4m

AE
F

4m 4m 20 kN

Figure 7.14

76

Q15. The truss as shown in Figure 7.15 is pinned supported at A and rolled at E and F. The cross
sectional areas for all members are given (in unit mm2) and E= 3 X 102 kN/mm2. Determine
all the members force using method of virtual work if vertical reaction at E is taken as the
redundant.

B (600) C 9 kN
(500)
(500) (200) (500)

(200) 20 m

A F

(600) D (600) E (600)

18 kN

15 m 15 m 15 m

Figure 7.15

77

ANSWER

A1(a).
Internally statically indeterminate structure, Di = 1

A1(b). = 15kN
FAB = -18.75kN
FAC = 6.25kN
FAD = -18.75kN
FBC = -33.75kN
FBD = -5kN
FCD

A2(a).
Internally statically indeterminate structure, Di = 1

A2(b). = -45.54kN
FAC = 34.27kN
FAF = -45.54kN
FBD = 34.27kN
FBE = -34.27kN
FCD =0
FCE =0
FCF =0
FDE =0
FDF = 34.27kN
FEF

A3. = -55.67kN
FAC = 41.89kN
FAF = -50.6kN
FBD = 38.07kN
FBE = -39.97kN
FCD = -2.54kN
FCE = -1.67kN
FCF = 1.68kN
FDE = 2.54kN
FDF = 39.98kN
FEF

A4(a).
RA = -37.5 KN
RB = 37.5 KN
HB = -50 KN

A4(b). =0
FAB = 37.5kN
FAD =0
FDC = -62.5kN
FDB =0
FCB

78

A4(c). = -0.8kN
FAB = -0.6kN
FAD = -0.8kN
FDC = 1kN
FDB = -0.6kN
FCB = 1kN
FAC

A4(d). = 16.3kN
FAB = 17.1kN
FAD = 16.3kN
FDC = -50.3kN
FDB = -20.4kN
FCB = 12.2kN
FAC

A5(a).
Internally statically indeterminate structure, Di = 1

A5(b). = 15kN
FAF = 13.47kN
FFE = 20kN
FED = -21.2kN
FAB = -21.53kN
FBC = -28.3kN
FCD = 8.47kN
FBF = 9.26kN
FBE = 2.16kN
FFC = 18.47kN
FCE

A6(a).
m = 10
r =3
j =6
m + r > 2j
Internally statically indeterminate structure, Di = 1

A6(b). = -10kN
FAB = -6.67kN
FBC = -3.36kN
FCD = -5kN
FDE = 3.31kN
FEF = 0kN
FFA = 12kN
FBF = 5kN
FCF = -6.02kN
FCE = 5.98kN
FDF

79

A7(a).
m = 13
r =4
j =8
m + r > 2j
Externally statically indeterminate structure, Do = 1

A7(b). = 15kN
FAH = 15kN
FHG = 40kN
FGF = 40kN
FFE = -19.2kN
FAB = 7.5kN
FBC = 7.5kN
FCD = -51.2kN
FDE = 30kN
FBH = -28.8kN
FBG = 0kN
FCG = -60.8kN
FGD = 30kN
FDF

A8(a).
Externally statically indeterminate structure Do = 1

A8(b).
RA = 64 kN
RD = 64 kN
HA = -24 kN
HD = -24 kN

FAB = 32kN
FBC = 24kN
FCD = -32kN
FAC = 40kN
FBD = -40kN

A9.
Reaction
RA = 5 kN
HA = 7.5 kN
RE = 55 kN
HE = -37.5 kN

FAB = -3.75kN
FBC = -3.75kN
FCD = 3.75kN
FDE = 3.75kN
FFG = -7.5kN
FFB = 0
FGD = 0
FAF = -6.25kN
FFC = 6.25kN
FCG = -6.25kN
FGE = -68.75kN

80

A10.
Reaction
RA = 17.5 kN
HA = -56.67 kN
RD = 42.5 kN
HD = -56.67 kN

FAB = 33.33kN
FAD = 0kN
FAE = 29.18kN
FBC = 33.33kN
FBE = -35kN
FCE = -41.66kN
FDE = -70.84kN

A11.
Reaction
RA = -14.20 kN
HA = 30 kN
RC = 8.31 kN
RB = 5.89 kN

FAB = 11.05kN
FAD = 23.70kN
FBC = 11.05kN
FBD = -5.89kN
FCD = -13.80kN

A12.
Reaction
HA = -228.39 kN
HB = 228.39 kN
RB = 326.13 kN
RF = 173.87 kN

FAB = 326.13kN
FAG = 16.21kN
FCD = -148.31kN
FCH = 220.16kN
FDH = 0
FEI = 150kN
FGH = 16.21kN
FIF = 130.40kN
FAC = -407.66kN
FBC = 228.39kN
FCG = 150kN
FDE = -148.31kN
FEF = -217.34kN
FEH = 29.84kN
FHI = 130.40kN

81

A13(a).
m=6
r =3
j =4
m + r > 2j
Statically indeterminate structure

A13(b).
Due to external load :
FAB = 0
FAD = -40kN
FAC = -96.05kN
FBD = 0
FBC = 75kN
FCD = 0

Due to virtual unit load :
FAB = -0.625kN
FAD = -0.781kN
FAC = 1kN
FBD = 1kN
FBC = -0.781kN
FCD = -0.625kN

A13(c). = -21.32kN
FAB = -66.64kN
FAD = -61.93kN
FAC = 34.12kN
FBD = 48.36kN
FBC = -21.32kN
FCD

A14. = 20kN
FAB =0
FAF = 28.28kN
FAC = -28.29kN
FBF = 20kN
FBC = -40kN
FCF = 28.28kN
FCD =0
FFD = 28.28kN
FDE = -20kN
FFE

A15. = -4.89kN
FAB = 11.93kN
FAD = 18kN
FBD = 7.73kN
FBC = -17.76kN
FBE = 11.93kN
FDE = 1.83kN
FCE = -2.13kN
FCF = 1.27kN
FEF

82

CHAPTER 8

INFLUENCE LINE OF BEAM

EXERCISES

Q1. Draw the Influence Line of the followings for an overhanging beam as shown in Figure 8.1
with corresponding value when the unit load is at 3 m distance from point A.
a. Reaction RA and RB
b. Shear force of C
c. Bending moment of C

A C BD
6m 8m 3m

Figure 8.1

Q2. Referring to Figure 8.2, calculate the shear force at point C.

40 kN

15 kN/m

A 2m C 1m 1m 2m B

Shear Force Diagram 0.67

0.50

0.33

- 0.33

Figure 8.2

83

Q3. Referring to Figure 8.3, calculate the bending moment at point C.

25 kN

10 kN/m

A 5m C 4m B
3m

Bending Moment Diagram

1.67 2.67

Figure 8.3

Q4. Figure 8.4 shows a beam subjected to a point load and uniformly distribution load. By using
the Influence Line Method, calculate:
a. Reaction RB
b. Negative shear force of C

40 kN

15 kN/m

A 7m C 3m B
Figure 8.4

84

Q5. A combination of point loads and uniformly distribution load crosses on a simply supported
beam as shown in Figure 8.5. By using Influence Line Method, calculate the followings for
the load position,
a. Reaction RA
b. Shear force of C (right)
c. Bending moment of C

20 kN 50 kN 10 kN
5 kN/m
2m
1m 3m

C B
A

9m

Figure 8.5

Q6. A simply supported beam with 25 m long is subjected to 1 kN/m uniformly distribution load
of 8 m long. Calculate the maximum shear force and bending moment at a section of 5 m
from one end of that beam.

Q7. A beam has span of 20 m. Determine the maximum shear force and bending moment for a
section of 8 m from the left support which is due to two point loads of 8 kN and 4 kN at a
fixed distance of 2 m apart rolling from left to right with either of the loads leading.

Q8. A uniform load of 4000 kN/m with 6 m long crosses a girder of 20 m span. Calculate the
maximum shear force and bending moment for a section of 10 m from left support.

Q9. A train of loads as shown in Figure 8.6 crosses a girder of 20 m span from left to right.
Determine the maximum shear force and bending moment at quarter of span.

4 kN 6 kN 6 kN 2 kN 4 kN

2m 2m 2m 1m

Figure 8.6

85

Q10. A beam which is supported at points A and B as shown in Figure 8.11 is subjected to a
series of moving load of 50 kN, 70 kN and 60 kN respectively. By using Influence Line
Method,
a. Draw influence line for shear force and bending moment at point C.
b. Calculate the maximum shear force at point C due to the series of loads moving from
right to left.
c. Determine the maximum bending moment due to the series of loads moving from right
to left.

50 kN 70 kN 60 kN

3m 7m

A C B
5m 5m

Figure 8.11

Q11. A series of concentrated loads moving on 20 m long of simply supported beams as shown
in Figure 8.7. Identify the location of the resultant force.

60 kN 50 kN 80 kN

3m 5m

A B
20 m

Figure 8.7

Q12. Referring to Figure 8.8, calculate the Absolute Maximum Moment of a beam which is
subjected to a series of moving loads.

15 kN 60 kN

2.5 m

A B
10 m

Figure 8.8

86

Q13. A 25 m long of overhanging beam carries a series of moving concentrated loads as shown
in Figure 8.9
a. Calculate the resultant force of the series of concentrated loads
b. Identify the location of resultant force
c. Determine the Absolute Maximum Moment due to the concentrated loads

12 kN 30 kN 30 kN 8 kN

2.5 m 3m 1m

AB 20 m C
5m Figure 8.9

Q14. A simply supported beam with 40 m span is subjected to a series of moving concentrated
loads as shown in Figure 8.10.
a. Calculate the resultant force of the series of concentrated loads
b. Identify the location of resultant force
c. Analyze the Absolute Maximum Moment due to the concentrated loads

44 kN 38 kN 52 kN 16 kN

7m 8m 4m

A B
40 m

Figure 8.10

Q15. A series of point loads pass through a 30 m long of simply supported beam of bridge as
shown in Figure 8.12. Calculate:
a. Maximum bending moment at 8 m distance from left support
b. Absolute Maximum Moment

5 kN 20 kN 20 kN 15 kN 10 kN 15 kN

2m 1m 2m 2m 1m

Figure 8.12

87

ANSWER 1 0.79
A1(a). 0 3m 1 − 14

RA
14
14m 17m
-0.21

1.21
1

RB 0.21 14m 17m
0 3m

A1(b).

0.57
1 − 14
6m
SFD 3m -0.43 17m
-0.21 -0.21
0 14m


− 14

A1(c). BMD 8 1.71 3.43 6 (1 − )
14 3m 6m
A2. 14 17m
VC 0 -1.29
14m
A3.
MC = 17.03kN

A4(a). = 160.25 kNm
RB
= 103 kN
A4(b).
VC,-ve = 58 kN

A5(a). = 64.44 kN
RA
= -25.56 kN
A5(b).
VC = 157.64 kNm

A5(c).
MC

88

A6. = 5.12 kN
Vmax = 26.70 kNm

Mmax

A7. = 6.80 kN
Vmax = 54.40 kNm

Mmax

A8. = 8.40 kN
Vmax = 102 kNm

Mmax

A9. = 12.70 kN
Vmax = 66.50 kNm

Mmax

A10(a).

SFD 0.5
5m 1 − 10
0
-0.5 10m

2.5
− 10 5m

BMD 5 (1 − )
5 ( 10 )
10

A10(b). 0 10m

Vmax = -45 kN

A10(c).

Mmax = 225 kNm

A11. = 3.84 m
x

A12. = 169.05 kNm
MC’

A13(a).
P = 80 kN

A13(b).
x = 2.51 m

A13(c).
MC’ = 337.48 kNm

89

A14(a).
P = 150 kN

A14(b).
x = 10 m

A14(c).
MC’ = 1045.75 kNm

A15(a).

Mmax = 436.80 kNm

A15(b).
MC’ = 542.85 kNm

90

LIST OF FORMULAR

1. Fixed End Moment

P

A
= − 8 B = 8

² L/2 L/2 ²
= − ² 2 b B = ²

P
a

A

L

² A w ²
= − 12 L B = 12

2 A PP B 2
= − 9 L/3 L/3 L/3 = 9

2. Slope Deflection Method

= 2 3∆
= [2 + − ] ±

2 3∆
[2 + − ] ±

91

3. Moment Distribution Method
i. Stiffness Factor
K = 4EI / L (for Fixed or Continuous)
K = 3EI / L (for Pinned or Roller)
ii. Distribution Factor
DF = K / ∑ K

4. Statically Indeterminate Truss
i. Redundant Force
R = – ∑ PµL / AE
∑ µ2L / AE
ii. Internal Force
F = P + µR

5. Displacement
i. External Load
∆ = ∑ PµL / AE
ii. Temperature Changers
∆ = ∑ µcLt
iii. Fabrication Error
∆ = ∑ µλ

6. Influence Lines
i. RA = 1 – x/L, RB = x/L
ii. x < a, VC = – x/L
x > a, VC = 1 – x/L
iii. x < a, MC = bx/L
x > a, MC = a (1 – x/L)
x = a, MC = ab/L

92

REFERENCE

1. Hibbeler, R.C. (2009). Structural Analysis Seventh Edition in SI Units. Singapore: Prentice
Hall.

2. Punmia, B.C., Ashok, K.J. and Arun, K.J. (2004). Theory of Structure. Delhi: Laxmi
Publications (P) Ltd.

3. Yusof Ahmad (2004). Teori Struktur (Cetakan Pertama). Skudai Johor: Universiti Teknologi
Malaysia.

4. Vazirani, V. N. and Ratwani, M.M. (2005). Analysis of Structure Vol. II (Theory, Design &
Details of Structures). Delhi: Khanna Publishers.

93