influence line of gader and truss

1. 60
A B C D EF
GH
Influence Line for Girder
• Forces Apply to the Girder
(b/L)P1 (a/L)P1 P2
P1
a b
P2
L
A B
P1
D
P2
A
B C D
F
GH

2. 61
A F
GH
• Reaction
1
0.5 0.5
RGRH
RH
1

3. 62
A F
GH
RG
1
0.5 0.5
RG
11
RH

4. 63
A F
GH
• Shear
L
a b b b
1
0.5 0.5
1
1
VC
VC
VC
x
2b/L
-(a+b)/L
1
C

5. 64
A F
GH
L
VCD
VCD
1
0.5 0.5
1
0.5 0.5
VCD x
-a/L
b/L
a b

6. 65
A
C
F
GH
• Bending Moment
L
a b
ab/L
1
0.5 0.5
MC
x
1
1
MC
MC
C

7. 66
A
B C D E
F
GH
L
MF
x
1
0.5 0.5
1
0.5 0.5
MF
MF
a b
ab/L

8. 67
Example 6-5
Draw the influence for
- the Reaction RG and RF
- Shear VCD
- the moment MC and MH
1.5 m 1.5 m
A
C D
H
F
3 m 3 m 3 m 3 m
G

9. 68
1
RG
x
RG
1.333
0.667
0.333
1.5 m 1.5 m
A H
F
3 m 3 m 3 m 3 m
G

10. 69
RF
x
RF
1
0.667
0.333
-0.333
1.5 m 1.5 m
A H
F
3 m 3 m 3 m 3 m
G

11. 70
1.5 m 1.5 m
A H
F
3 m 3 m 3 m 3 m
G
VCD
x
VCD
VCD
1
0.5 0.5
4.5/9
-4.5/9
0.333
-0.333
0.333
C D

12. 71
1.5 m 1.5 m
A H
F
3 m 3 m 3 m 3 m
G
MC
MC
1
1
MC
x
(3)(6)/9 = 2
-2
1
C

13. 72
MH
MH
1
0.5 0.5
MH
x
(4.5)(4.5)/9 = 2.25
1.51.5
-1.5
1.5 m 1.5 m
A H
F
3 m 3 m 3 m 3 m
G

14. 73
Example 6-6
Draw the influence line diagrams of girder for
- the reaction at C and G,
- shear at E and H,
- bending moment at H.
2 m
A
GB
6 @ 4 m = 24 m
C D E FH

15. 74
SOLUTION
RC
RC
1
0.75
0.50
0.25
1.25
A
GB
6 @ 4 m = 24 m
C D E FH
2 m

16. 75
A
GB
6 @ 4 m = 24 m
C D E FH
2 m
RG
RG
1
1
1
0.25
0.50
0.75
-0.25

17. 76
A
GB
6 @ 4 m = 24 m
C D E FH
2 m
VE
1
1
1
1
VE
VE
8/16 = 0.5
-8/16 = 0.5-0.25
0.250.25
8 m 8 m

18. 77
A
GB
6 @ 4 m = 24 m
C D E FH
2 m
6/16 = 0.375
-10/16 = 0.625
VH
1
1
VH
VH
1
0.5 0.5
-0.25
0.250.25
-0.50
10 m 6 m

19. 78
A
GB
6 @ 4 m = 24 m
C D E FH
2 m
MH
1
1
1
0.5 0.5
MH
MH
(10)(6)/16 = 3.75
1.50
2.50
-1.50
3
10 m 6 m

20. 79
Draw the influence for
- the Reaction RG and RH
- Shear VC and VCD
- the moment MC, MD, and MF
and determine the maximum for
- the Reaction (RG)max and (RH)max
- Shear (VCD)max
due to
- a uniform dead load 2 kN/m
- a uniform live load 5 kN/m
- a concentrated live load 50 kN
Example 6-7
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
F
GH

21. 80
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
F
GH
RG
1
0.5 0.5
10/14
6/14
2/14
1/14
RG
11
RH

22. 81
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
E
F
GH
1
0.5 0.5
RGRH
RH
1 12/14
8/14
4/146/14

23. 82
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
F
GH
VC
VC
1
0.5 0.5
VC
x
8/14
-6/14
1
1
1
-2/14
4/14
-1/14

24. 83
-8/14
6/14
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
F
GH
VF
VF
1
0.5 0.5
1
0.5 0.5
VCD x
-1/14 -2/14
-6/14
4/14

25. 84
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
F
GH
(6)(8)/14 = 3.43
1
0.5 0.5
MC
x
(4/8)(3.43)
(2/6)(3.43)
1
1
MC
MC

26. 85
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
E
F
GH
1
0.5 0.5
1
1
MD x
(10)(4)/14 = 2.86(6/10)(2.86)
(2/10)(2.86)
MD
MD

27. 86
(8)(6)/14=3.428
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
F
GH
MF
x
1
0.5 0.5
1
0.5 0.5
2.285
2.571
0.857
0.429
MF
MF

28. 87
[0.5×4 (2/14)] + [(0.5)(2/14 + 1)(12) = 7.143
RG
1
10/14
6/14
2/14
1/14
Maximum Reaction
2 kN/m
5 kN/m
50 kN
(RG)max = (2)(7.143) + (5)(7.143) + (50)(1)
= 100 kN
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D E
F
GH

29. 88
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D E
F
GH
(0.5)(16)(12/14) = 6.857
RH
12/14
8/14
4/146/14
Maximum Reaction
2 kN/m
5 kN/m
50 kN
(RH)max = (2)(6.857) + (5)(6.857) + (50)(12/14)
= 90.86 kN

30. 89
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D E
F
GH
+ (0.5)(5.6)(4/14) = + 0.8
-[0.5(4)(2/14) + 0.5(2/14) + (6/14)(4) + 0.5(2.4)(6/14)] = -1.943
2.4 m
VCD
-1/14 -2/14
-6/14
4/14
Maximum Shear
2 kN/m
5 kN/m
50 kN
(VCD)max = (2)(-1.943 + 0.8) (5)(-1.943) + (50)(-6/14)
= -33.43 kN
C D

31. 90
SP 6-1
Draw the influence line for
-the reaction at support A and D
-the shear and moment at E
- the moment at B and C .
2 m 2 m 2 m 2 m 2 m
A B E C D

32. 91
Influence Line for Trusses
RA
x
1
RE
x
1
(3/4)
(2/4)
(1/4)
(2/4)
(3/4)
(1/4)
A
B C
E
GH
D
F
4 @ x m
h

33. 92
A
B C
E
GH
D
F
4 @ x m
h
1 kN
Consider the section to the right
FCB
RE
FGB
FGH
FBC x
(2x/h)(1/4)
+ ΣMG = 0:
0)()2( =− hFxR BCE
EBC R
h
x
F
2
=
RE
x
1
(2/4)
(3/4)
(1/4)

34. 93
A
B C
E
GH
D
F
4 @ x m
h
1 kN
Consider section to the left
FHG
FBC
RA
FBG
+ ΣMG = 0:
0)2()( =− xRhF ABC
ABC R
h
x
F
2
=
FBC x
(2x/h)(1/4)
RA
x
1
(3/4)
(2/4)
(1/4)
(2x/h)(1/4)
(2x/h)(2/4)

35. 94
Determine the maximum force developed
in member BC , BG, and CG of the truss
due to the wheel loads of the car. Assume the
loads are applied directly to the truss.
the wheel loads of the car
4 kN 1.5 kN2 m
Example 6-8-1
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m

36. 95
RA
x
1
0.75
0.5
0.25
RE
x
0.5
0.75
0.25
1
SOLUTION
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m

37. 96
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
Influence Line for FBC
1 kN
Consider the section to the right
FBC = (6/4) RE
= 1.5 RE
FCB
RE
FGB
FGH
RE
x
0.5
0.75
0.25
1
FBC x
1.5(0.25) = 0.375
+ ΣMG = 0:

38. 97
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
0.375
1.5(0.5) = 0.75
1 kN
Consider the section to the left
FHG
FBC
RA
FBG
RA
x
1
0.75
0.5
0.25
FBC = 1.5 RA
FBC
x
0.375
+ ΣMG = 0:

39. 98
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
RE
x
0.5
0.75
0.25
1
Influence Line for FBG
1 kN
Consider the section to the right
ΣFy = 0 ;
FBG cos θ = RE
FGH
FGB
FCB
RE
θ
FBG = 1.25 RE
FBG (4/5) = RE1.25(0.25) = 0.3125
FBG x

40. 99
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
1 kN
Consider the left hand
ΣFy = 0 ;
FBG cos θ + RA = 0
RA
x
1
0.75
0.5
0.25
FHG
FBC
RA
FBG
θ
0.3125
FBG x
FBG = -1.25 RA
FBG (4/5) = -RA
-0.3125
-1.25(0.5) = -0.625

41. 100
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
FCG x
Influence Line for FCG
1 kN
0
1 kN
0

42. 101
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
FCG x
1 kN
1
1

43. 102
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
(FBC)max by Loads
(FBC)max = (4)(0.75) + (1.5)(0.5)
= 3.75 kN (T)
0.375
0.75
FBC
x
0.375
4 kN 1.5 kN2 m
0.5

44. 103
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
(FBG)max by Loads
(FBG)max = (4)(-0.625) + (1.5)(-0.417) = -3.126 kN (C)
FBG x
-0.3125
0.3125
-0.625
4 kN 1.5 kN2 m
-0.417

45. 104
A
B C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
(FCG)max by Loads
FCG x
1
0.333
(FCG)max = (4)(1) + (1.5)(0.333) = 4.50 kN (T)
4 kN 1.5 kN2 m

46. 105
Draw the influence line for FBC , FBH,and FCH.
Example 6-8-2
A
B C
E
HI
4 m
D
F
3 m 3 m 3 m 3 m
GJ

47. 106
RA
x
1
0.75
0.5
0.25
RE
x
0.5
0.75
0.25
1
SOLUTION
A
B C
E
HI
4 m
D
F
3 m 3 m 3 m 3 m
GJ

48. 107
A
B C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJInfluence Line for FBC
1 kN
Consider the section to the right
FBC = (6/4) RE
= 1.5 RE
FCB
RE
FHB
FHI
RE
x
0.5
0.75
0.25
1
FBC x
1.5(0.25) = 0.375
+ ΣMH = 0:

49. 108
A
B C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
0.375
1.5(0.5) = 0.75
1 kN
Consider the section to the left
FHI
FBC
RA
FBH
RA
x
1
0.75
0.5
0.25
FBC = 1.5 RA
FBC
x
0.375
+ ΣMH = 0:

50. 109
A
B C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
RE
x
0.5
0.75
0.25
1
Influence Line for FBH
1 kN
Consider the section to the right
ΣFy = 0 ;
FBH cos θ = RE
FHI
FBH
FCB
RE
θ
FBH = 1.25 RE
FBH (4/5) = RE1.25(0.25) = 0.3125
FBH x

51. 110
A
B C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
1 kN
Consider the section to the left
ΣFy = 0 ;
FBH cos θ + RA = 0
RA
x
1
0.75
0.5
0.25
FHI
FBC
RA
FBH
θ
0.3125
FBH x
FBH = -1.25 RA
FBH (4/5) = -RA
-0.3125
-1.25(0.5) = -0.625

52. 111
A
B C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
FCH x
Influence Line for FCH
1 kN
FGH
FCH
FCB
RE
x
0.5
0.75
0.25
1
RE
ΣFy = 0 ;
FCH = -RE
-0.25

53. 112
A
B C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
FCH x
1 kN
FGH
FCH
FCB
RA
ΣFy = 0 ;
FCH = RA
-0.25
RA
x
1
0.75
0.5
0.25
0.5
0.25

54. 113
Draw the influence line for FBC , FBG,and FHG.
Example 6-8-3
A
B C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m

55. 114
RA
x
1
0.75
0.5
0.25
RE
x
0.5
0.75
0.25
1
SOLUTION
A
B C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m

56. 115
A
B C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
Influence Line for FBC
1 kN
Consider the section to the right
FBC = (6/4.5) RE
= 1.33 RE
RE
x
0.5
0.75
0.25
1
FBC x
1.33(0.25) = 0.333
+ ΣMG = 0:
RE
FBC
FBG
FHG

57. 116
A
B C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
Consider the section to the left
FBC = (6/4.5) RA
= 1.33 RA
FBC x
1.33(0.25) = 0.333
+ ΣMG = 0:
RA
RA
x
1
0.75
0.5
0.25
FBC
FBG
FHG
1.33(0.5) = 0.667
1.33(0.25) = 0.333

58. 117
A
B C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
Influence Line for FHG
1 kN
Consider the section to the right
RE
x
0.5
0.75
0.25
1
FHG x
+ ΣMB = 0:
RE
FBC
FBG
FHG
26.57o
(FHG cos 26.57)(3) = -9RE
FHG = -3.35RE
-3.35(0.25) = -0.84

59. 118
A
B C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
Consider the section to the left
RA
RA
x
1
0.75
0.5
0.25
FBC
FBG
FHG
FHG x
-0.84
+ ΣMB = 0:
(FHG cos 26.57)(3) = -3RA
FHG = -1.12RA
-1.12(0.5) = -0.56
-1.12(0.25) = -0.28

60. 119
A
B C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
Influence Line for FBG
1 kN
Consider the section to the right
RE
x
0.5
0.75
0.25
1
FBG x
RE
FBC
FBG
FHG
26.57o
FHG x
-0.84
-0.56
-0.28
ΣFy = 0:+
-FBG cos 33.69
- FHGsin26.57 + RE = 0
FHG = +1.2RE - 0.54 FHG
33.69o
1.2(0.25)-0.54(-0.84) = 0.75

61. 120
FHG x
-0.84
-0.56
-0.28
ΣFy = 0:+
RA + FBG cos33.69
+ FHG sin 26.57 = 0
FBG = -1.2RA - 0.54FHG
A
B C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
RA
FBC
FBG
FHG
Consider the section to the left
RA
x
1
0.75
0.5
0.25
FBG x
0.75
-1.2(0. 5)-0.54(-0.56) = -0.30
-1.2(0. 25)-0.54(-0.28) = -0.15