1. INFLUENCE LINES FOR STATICALLY
DETERMINATE STRUCTURES
! Influence Lines for Beams
! Influence Lines for Floor Girders
! Influence Lines for Trusses
! Maximum Influence at a Point Due to a Series
of Concentrated Loads
! Absolute Maximum Shear and Moment
1

2. Influence Line
Unit moving load
A
B
2

3. Example 6-1
Construct the influence line for
a) reaction at A and B
b) shear at point C
c) bending moment at point C
d) shear before and after support B
e) moment at point B
of the beam in the figure below.
C
A
B
4m 4m 4m
3

4. SOLUTION
• Reaction at A
1
x C
A
Ay By
8m 4m
x Ay
0 1 1
+ ΣMB = 0: − Ay (8) + 1(8 − x ) = 0, Ay = 1− x
4 0.5 8
8 0
Ay
12 -0.5
1
0.5
8m 12 m
x
4m
-0.5
4

5. • Reaction at B
1
x C
A
Ay By
8m 4m
x By
0 0
1
4 0.5 + ΣMA = 0: B y (8) − 1x = 0, By = x
8
8 1
12 1.5 By 1.5
1
0.5
x
4m 8m 12 m
5

6. • Shear at C
0≤ x<4 4 < x ≤ 12
1
x C
A
Ay By
4m 4m 4m
1
x C 1
+ ΣFy = 0: 1 − x − 1 − VC = 0
0≤ x≤4 A MC 8
1 1
Ay = 1− x VC = − x
8 VC 8
4m
x C
1
4 < x ≤ 12 A MC + ΣFy = 0: 1 − x − VC = 0
1 8
Ay = 1− x 1
8 VC VC = 1− x
4m 8
6

7. 0≤ x<4 4 < x ≤ 12 x VC
1 1 0 0
VC = − x
x C 8 4- -0.5
A 4+ 0.5
1
VC = 1− x 8 0
8
Ay By
4m 4m 4m 12 -0.5
1
VC VC = 1− x
8
0.5
4m 8m 12 m
x
-0.5 -0.5
1
VC = − x
8
7

8. • Bending moment at C
0≤ x<4 4 < x ≤ 12
1
x C
A
Ay By
4m 4m 4m
1
x C
1
0≤ x≤4 A MC + ΣMC = 0: M C + 1(4 − x) − (1 − x )(4) = 0
1 8
Ay = 1− x 1
8 VC MC = x
4m 2
x C
1
4 < x ≤ 12 A MC + ΣMC = 0: M C − (1 − x)( 4) = 0
1 8
Ay = 1− x 1
8 VC MC = 4 − x
4m 2
8

9. 0≤ x<4 4 < x ≤ 12
x MC
1
1 0 0
x MC = x
C 2 4 2
A
1 8 0
MC = 4 − x
Ay By 2 12 -2
4m 4m 4m
1
MC = x 1
2 MC = 4 − x
MC 2
2
8m 12 m
x
4m
-2
9

10. Or using equilibrium conditions:
• Reaction at A
1
x C
A
Ay By
8m 4m
1
+ ΣMB = 0: − Ay (8) + 1(8 − x ) = 0, Ay = 1− x
8
Ay
1
0.5
8m 12 m
x
4m
-0.5
10

11. • Reaction at B
1
x C
A
Ay By
8m 4m
Ay
+ ΣFy = 0: Ay + B y − 1 = 0
1
0.5 B y = 1 − Ay
8m 12 m
x
4m
-0.5
By B y = 1 − Ay 1.5
1
0.5
x
4m 8m 12 m
11

12. • Shear at C
0≤ x<4 4 < x ≤ 12
1
x C
A
Ay By
4m 4m 4m
1
x C
+ ΣFy = 0: Ay − 1 − VC = 0
0≤ x≤4 A MC
1 VC = Ay − 1
Ay = 1− x VC
8
4m
x C
4 < x ≤ 12 A MC + ΣFy = 0: Ay − VC = 0
1
Ay = 1− x
8 VC VC = Ay
4m
12

13. C
A
B
4m 4m 4m
Ay
1
0.5
8m 12 m
x
4m
VC = Ay − 1 VC = Ay -0.5
VC
0.5
4m 8m 12 m
x
-0.5 -0.5
13

14. • Bending moment at C
0≤ x<4 4 < x ≤ 12
1
x C
A
Ay By
4m 4m 4m
1
x C
0≤ x≤4 A MC + ΣMC = 0: Ay (4) + 1( 4 − x ) + M C = 0
1
Ay = 1− x VC M C = 4 Ay − (4 − x)
8
4m
x C
4 < x ≤ 12 A MC + ΣMC = 0: − Ay (4) + M C = 0
1
Ay = 1− x M C = 4 Ay
8 VC
4m
14

15. C
A
B
4m 4m 4m
Ay
1
0.5
8m 12 m
x
4m
M C = 4 Ay − (4 − x) M C = 4 Ay -0.5
MC
2
8m 12 m
x
4m
-2 15

16. • Shear before support B 1
x
C
A
Ay By
4m 4m 4m
1
x MB MB
8m 8m
Ay VB -
Ay VB-
VB- = Ay-1 VB- = Ay
Ay 1
0.5 8m 12 m
x
VB- 4m
-0.5
x
-0.5 -1.0 -0.5
16

17. • Shear after support B 1
x
C
A
Ay By
4m 4m 4m
1
MB MB
4m 4m
VB+ VB+
VB+ = 0 VB+ = 1
Ay 1
0.5 8m 12 m
x
4m
-0.5
VB+ 1
x
17

18. • Moment at support B 1
x
C
A
Ay By
4m 4m 4m
1
x MB MB
8m 8m
Ay VB -
Ay VB-
MB = 8Ay-(8-x) MB = 8Ay
Ay 1
0.5 8m 12 m
x
MB 4m
-0.5
x
1
-4
18

19. Influence Line for Beam
• Reaction P=1
C
A B
x'
L
P=1 δy 1
δ y' sB = =
δy = 1 L L
C
A B
Ay By
Ay (1) − 1(δ y ' ) + B y (0) = 0
Ay = δ y '
19

20. P=1
C
A B
x'
L
P=1
δ y' δy = 1
C
A B
δy 1
sA = =
Ay L L By
Ay (0) − 1(δ y ' ) + B y (1) = 0
By = δ y '
20

21. - Pinned Support
A C B
a b
L
B
A
RA
RA 1
b
L
x
21

22. - Fixed Support
A B
a b
L
A B
RA
RA 1 1
x
22

23. P=1
• Shear
A C B
a b
L
VC P=1
1
δ y' sB =
δy=1 δyR L
A B
δyL
Ay 1 By
sA =
L VC
Ay (0) + VC (δ yL ) + VC (δ yR ) − 1(δ y ' ) + B y (0) = 0
VC (δ yL + δ yR ) = δ y '
slopes : s A = sB
δy=1
VC = δ y '
23

24. - Pinned Support
A C B
a b
L
VC
A B
VC 1
VC 1 Slope s B =
L
b
1 L
x
-a
1 L
Slope s A =
L -1
Slope at A = Slope at B
24

25. - Fixed Support
A B
a b
L
VB
A B
VB
VB 1 1
x
25

26. • Bending Moment
P=1
A C B
a b
L
φ = θ A +θ B = 1
P=1
h MC MC δ y'
A B
h h
Ay
θA = θB =
a b By
Ay (0) + M C (θ A ) + M C (θ B ) + 1(δ y ' ) + B y (0) = 0 h h
( + ) =1
1 a b
M C (θ A + θ B ) = δ y '
h( a + b) ab
= 1, h=
ab ( a + b)
M C = δ y'
26

27. - Pinned Support
A C B
a b
L
Hinge
A B
MC MC
MC
a b
φC = θA + θB = 1
ab
a+b
x
b
θA = a
L θA =
L
27

28. - Fixed Support
A B
a b
L
A B
MC MC
MB
x
1
-b
28

29. • General Shear
C D E B F G H
A
L/4 L/4 L/4 L/4 L/4 L/4 L/4
L
VC
3/4
1
x
VD -1/4 2/4
1 x
VE -2/4
1/4
x
1
VBL
-3/4
x
-1 29

30. C D E B F G H
A
L/4 L/4 L/4 L/4 L/4 L/4 L/4
VBL L
x
-1
VBR 1
x
VF 1
x
VG 1
x
30

31. • General Bending Moment
C D E B F G H
A
L/4 L/4 L/4 L/4 L/4 L/4 L/4
L
MC 3L/16 φ = sA + sB = 1
x
θA = 3/4 θB = 1/4
MD 4L/16
φ = sA + sB = 1
x
θA = 1/2 θB = 1/2
ME 3L/16
φ = θA + θB = 1
x
θA = 1/4 θB = 3/4
31

32. C D E B F G H
A
L/4 L/4 L/4 L/4 L/4 L/4 L/4
L
MB
x
1
3L/4
MF
x
1
2L/4
MG
x
1
L/4
32

33. Example 6-2
Construct the influence line for
- the reaction at A, C and E
- the shear at D
- the moment at D
- shear before and after support C
- moment at point C
A B Hinge D
C E
2m 2m 2m 4m
33

34. SOLUTION
B C D E
A
2m 2m 2m 4m
RA
RA 1
x
34

35. B C D E
A
2m 2m 2m 4m
RC
8/6
1
RC 4/6
x
35

36. A B D
C
E
2m 2m 2m 4m
RE
1
RE 2/6
x
-2/6
36

37. VD
A B D
C E
VD
2m 2m 2m 4m
1 4/6
2/6
=
VD 1
x
sE = 1/6
=
sC = 1/6 -1
-2/6
• sE = sC
37

38. Or using equilibrium conditions: 1
A B Hinge D
C E
2m 2m 2m 4m
1
VD x
4m VD
MD 4m
MD
RE RE
VD = -RE VD = 1 -RE
1
RE 2/6
x
-2/6
4/6
VD 2/6
x
-2/6
38

39. A B MD MD
C E
D
2m 2m 2m 4m
(2)(4)/6 = 1.33
4
2 φD = θC+θE = 1
MD
θC = 4/6 2/6 = θE
x
-1.33
39

40. Or using equilibrium conditions: 1
A B Hinge D
C E
2m 2m 2m 4m
1
VD x
4m VD
MD 4m
MD
RE RE
MD = 4RE MD = -(4-x)+4RE
1
RE 2/6
x
-2/6
8/6
MD
x
-8/6
40

41. VCL
A
C E
B D
VCL
2m 2m 2m 4m
VCL
x
-1 -1
41

42. Or using equilibrium conditions:
A B 1 D
C E
2m 2m 2m 4m
1 MB MB
RA VCL RA VCL
VCL = RA - 1 VCL = RA
RA 1
x
VCL
x
-1 -1
42

43. VCR
A
C E
B D
VCR
2m 2m 2m 4m
1
0.667
VCR 0.333
x
43

44. Or using equilibrium conditions:
A B 1 D
C E
2m 2m 2m 4m
1
MC MC
VCR VCR
VCR = -RE RE VCR = 1 -RE RE
1
RE 2/6=0.33
x
-2/6 = -0.333
1
0.667
VCR 0.333
x
44

45. A B MC C MC D E
2m 2m 2m 4m
MC
x
1
-2
45

46. Or using equilibrium conditions:
A B 1 D
C E
2m 2m 2m 4m
1
x'
MC MC
6m 6m
VCR VCR
MC = 6RE RE M C = 6 RA − x' RE
1
RE 2/6=0.33
x
-2/6 = -0.333
MC
x
1
-2
46

47. Example 6-3
Construct the influence line for
- the reaction at A and C
- shear at D, E and F
- the moment at D, E and F
Hinge
A D B E C F
2m 2m 2m 2m 2m 2m
47

48. SOLUTION
D B E C F
A
RA 2m 2m 2m 2m 2m 2m
1 1
RA
0.5
x
-0.5
-1
48

49. A D B E F
C
RC
2m 2m 2m 2m 2m 2m
2
1.5
1
0.5
RC x
49

50. VD
A B E C F
D
VD
2m 2m 2m 2m 2m 2m
VD 1 1
=
0.5
x
=
-0.5 -1
50

51. VE
A D B E C F
VE
2m 2m 2m 2m 2m 2m
0.5
VE
=
1
x
=
-0.5 -0.5
-1
51

52. VF
A D B E C F
VF
2m 2m 2m 2m 2m 2m
1
=
VF
x
=
52

53. A D B E C F
MD MD
2m 2m 2m 2m 2m 2m
2
MD
1
x
θD = 1
-1
-2
53

54. A D B ME ME C
E F
2m 2m 2m 2m 2m 2m
(2)(2)/4 = 1
ME φE = 1
x
θB = 0.5 θC = 0.5
-1
-2
54

55. A D B C ME ME
E F
2m 2m 2m 2m 2m 2m
MF
x
θF = 1
-2
55

56. Example 6-4
Determine the maximum reaction at support B, the maximum shear at point C and
the maximum positive moment that can be developed
at point C on the beam shown due to
- a single concentrate live load of 8000 N
- a uniform live load of 3000 N/m
- a beam weight (dead load) of 1000 N/m
A C B
4m 4m 4m
56

57. SOLUTION 8000 N
3000 N/m
1000 N/m
A
C B
4m 4m 4m
RB
1.5
1
0.5
0.5(12)(1.5) = 9
x
(RB)max = (1000)(9) + (3000)(9) + (8000)(1.5)
= 48000 N = 48 kN
57

58. 8000 N
3000 N/m 3000 N/m
1000 N/m
A
C B
4m 4m 4m
VC
0.5
0.5(4)(0.5) = 1
x
0.5(4)(-0.5) = -1 0.5(4)(-0.5) = -1
-0.5 -0.5
(VC)max = (1000)(-2+1) + (3000)(-2) + (8000)(-0.5)
= -11000 N = 11 kN
58

59. 8000 N 3000 N/m
1000 N/m
A
C B
4m 4m 4m
MC
2
+(1/2)(8)(2) = 8
x
(1/2)(4)(2) = 4
-2
(MC)max positive = (8000)(2) + (3000)(8) + (8-4)(1000)
= 44000 N•m = 44 kN•m
59