1. 1
! 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
INFLUENCE LINES FOR STATICALLY
DETERMINATE STRUCTURES

2. 2
Influence Line
Unit moving load
A
B

3. 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.
B
A
C
4 m 4 m 4 m

4. 4
SOLUTION
• Reaction at A
+ ΣMB = 0: ,0)8(1)8( =−+− xAy xAy
8
1
1−=
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
A
C
4 m8 m
Ay By
1
x
0
4
8
12
x
1
0.5
0
-0.5
Ay

5. 5
A
C
4 m
Ay By
8 m
• Reaction at B
+ ΣMA = 0: ,01)8( =− xBy xBy
8
1
=
1
x
4 m 8 m 12 m
By
x
1.5
1
0.5
0
4
8
12
x
0
0.5
1
1.5
By

6. 6
• Shear at C
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
01
8
1
1 =−−− CVx
xVC
8
1
−=
ΣFy = 0:+
0
8
1
1 =−− CVx
xVC
8
1
1−=
ΣFy = 0:+
VC
MC
VC
MC
xAy
8
1
1−=
A
Cx
4 m
124 ≤< x
A
C
1
x
4 m
40 ≤≤ x
xAy
8
1
1−=

7. 7
0
4-
4+
8
12
x VC
0
-0.5
0.5
0
-0.5
xVC
8
1
−=
xVC
8
1
1−=
4 m 8 m 12 m
VC
x
-0.5
0.5
-0.5
xVC
8
1
−=
xVC
8
1
1−=
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x

8. 8
• Bending moment at C
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
VC
MC
0)4)(
8
1
1()4(1 =−−−+ xxMC
xMC
2
1
=
VC
MC
0)4)(
8
1
1( =−− xMC
+ ΣMC = 0:
+ ΣMC = 0:
xAy
8
1
1−=
A
Cx
4 m
124 ≤< x
A
C
1
x
4 m
40 ≤≤ x
xAy
8
1
1−=
xMC
2
1
4 −=

9. 9
0
4
8
12
x MC
0
2
0
-2
xMC
2
1
4 −=
xMC
2
1
=
4 m
8 m 12 m
MC
x
2
-2
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
xMC
2
1
=
xMC
2
1
4 −=

10. 10
Or using equilibrium conditions:
• Reaction at A
+ ΣMB = 0: ,0)8(1)8( =−+− xAy xAy
8
1
1−=
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
A
C
4 m8 m
Ay By
1
x

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

12. 12
• Shear at C
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
01 =−− Cy VA
1−= yC AV
ΣFy = 0:+
0=− Cy VA
yC AV =
ΣFy = 0:+
VC
MC
xAy
8
1
1−=
A
Cx
4 m
124 ≤< x
VC
MC
A
C
1
x
4 m
40 ≤≤ x
xAy
8
1
1−=

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

14. 14
• Bending moment at C
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
0)4(1)4( =+−+ Cy MxA
)4(4 xAM yC −−=
+ ΣMC = 0:
VC
MC
xAy
8
1
1−=
A
Cx
4 m
124 ≤< x
VC
MC
A
C
1
x
4 m
40 ≤≤ x
xAy
8
1
1−=
0)4( =+− Cy MA+ ΣMC = 0:
yC AM 4=

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

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

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

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

19. 19
P = 1
'x
• Reaction
Influence Line for Beam
C
A B
P = 1
Ay By
δy = 1 'yδ
LL
s
y
B
1
==
δ
C
A B
L
0)0()(1)1( ' =+− yyy BA δ
'yyA δ=

20. 20
δy = 1
'yδC
A B
P = 1
Ay ByLL
s
y
A
1
==
δ
C
A B
L
P = 1
'x
'yyB δ=
0)1()(1)0( ' =+− yyy BA δ

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

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

23. 23
• Shear
CA B
P = 1
a b
L
L
sB
1
=
δy=1
δyL
δyR
'yδ
A B
VC
VC
P = 1
Ay By
δy=1
L
sA
1
=
0)0()(1)()()0( ' =+−++ yyyRCyLCy BVVA δδδ
')( yyRyLCV δδδ =+
'yCV δ=
BA ssslopes =:

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

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

26. 26
1=+= BA θθφ
• Bending Moment
a b
L
a
h
A =θ
'yδ
b
h
B =θ
1
A B
MC
MC
P = 1
Ay By
h
CA B
P = 1
0)0()(1)()()0( ' =++++ yyBCACy BMMA δθθ
')( yBACM δθθ =+
'yCM δ=
1)( =+
b
h
a
h
)(
,1
)(
ba
ab
h
ab
bah
+
==
+

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

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

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

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

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

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

33. 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 C D E
2 m 2 m 2 m 4 m
Hinge

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

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

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

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

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

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

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

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

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

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

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

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

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

47. 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 B CD E F
2 m 2 m 2 m 2 m2 m 2 m

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

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

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

51. 51
A B CD E
2 m 2 m 2 m 2 m2 m 2 m
F
VE
VE
VE
x
1
-0.5
-1
0.5
-0.5
=
=

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

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

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

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

56. 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
4 m 4 m 4 m
A BC

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

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

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