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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
Influence Line
Unit moving load
A
B
2
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
SOLUTION
• Reaction at A
1
x
C
A
Ay
x
Ay
0
1
4
0.5
8
0
12
-0.5
By
8m
4m
Ay (8) 1(8 x) 0,
+ SMB = 0:
1
Ay 1 x
8
Ay
1
0.5
8m
12 m
x
4m
-0.5
4
• Reaction at B
1
x
C
A
Ay
x
By
0
0
4
0.5
8
1
12
1.5
8m
+ SMA = 0:
By
4m
By (8) 1x 0,
By
1
x
8
1.5
By
1
0.5
x
4m
8m
12 m
5
• Shear at C
0 x4
4 x 12
1
x
C
A
Ay
4m
4m
By
4m
1
x
0 x4
A
1
Ay 1 x
8
4m
x
4 x 12
C
MC
1
1 x 1 VC 0
8
1
VC x
8
+ SFy = 0:
1
1 x VC 0
8
1
VC 1 x
8
VC
C
A
1
Ay 1 x
8
4m
+ SFy = 0:
MC
VC
6
0 x4
4 x 12
1
x
1
VC x
8
C
A
Ay
4m
4m
VC
0.5
4m
1
VC x
8
-0.5
By
1
VC 1 x
8
4m
x
VC
0
0
4-
-0.5
4+
0.5
8
0
12
-0.5
1
VC 1 x
8
8m
12 m
x
-0.5
7
• Bending moment at C
0 x4
4 x 12
1
x
C
A
Ay
4m
4m
By
4m
1
x
0 x4
A
1
Ay 1 x
8
4m
x
4 x 12
C
+ SMC = 0:
1
M C 1(4 x) (1 x)( 4) 0
8
1
MC x
2
MC
+ SMC = 0:
1
M C (1 x)( 4) 0
8
1
MC 4 x
2
VC
C
A
1
Ay 1 x
8
4m
MC
VC
8
0 x4
4 x 12
1
x
1
x
2
0
0
4
2
MC 4
1
x
2
8
0
12
-2
C
MC
4m
MC 4
4m
1
x
2
2
By
MC 4
MC
MC
A
Ay
x
4m
1
x
2
8m
12 m
x
4m
-2
9
Or using equilibrium conditions:
• Reaction at A
1
x
C
A
Ay
x
Ay
0
1
4
0.5
8
0
12
-0.5
By
8m
4m
Ay (8) 1(8 x) 0,
+ SMB = 0:
1
Ay 1 x
8
Ay
1
0.5
8m
12 m
x
4m
-0.5
10
• Shear at C
0 x4
4 x 12
1
x
C
A
Ay
4m
4m
By
4m
1
x
0 x4
A
1
Ay 1 x
8
4m
x
4 x 12
C
MC
Ay 1 VC 0
VC Ay 1
VC
C
A
1
Ay 1 x
8
4m
+ SFy = 0:
MC
VC
+ SFy = 0:
Ay VC 0
VC Ay
11
C
A
B
4m
4m
4m
Ay
1
0.5
8m
12 m
x
4m
VC Ay 1
-0.5
VC Ay
VC
0.5
4m
-0.5
8m
12 m
x
-0.5
12
• Bending moment at C
0 x4
4 x 12
1
x
C
A
Ay
4m
4m
By
4m
1
x
0 x4
A
1
Ay 1 x
8
4m
x
4 x 12
C
MC
Ay (4) 1(4 x) MC 0
M C 4 Ay (4 x)
VC
C
A
1
Ay 1 x
8
4m
+ SMC = 0:
MC
VC
+ SMC = 0:
Ay (4) M C 0
M C 4 Ay
13
C
A
B
4m
4m
4m
Ay
1
0.5
8m
12 m
x
4m
-0.5
M C 4 Ay (4 x) M C 4 Ay
MC
2
8m
12 m
x
4m
-2
14
• Shear before support B
x
C
A
Ay
1
4m
4m
By
4m
1
x
MB
8m
MB
8m
VB-
Ay
VB-
Ay
VB- = Ay
VB- = Ay-1
Ay 1
0.5
8m
12 m
x
VB-
4m
-0.5
x
-0.5
-1.0
-0.5
15
• Shear after support B
x
C
A
Ay
1
4m
4m
By
4m
1
MB
MB
VB+
4m
VB+
VB+ = 0
Ay 1
4m
VB+ = 1
0.5
8m
12 m
x
4m
VB+
1
-0.5
x
16
• Moment at support B
x
C
A
Ay
1
4m
4m
By
4m
1
x
MB
8m
MB
8m
VB-
Ay
MB = 8Ay
MB = 8Ay-(8-x)
Ay 1
VB-
Ay
0.5
8m
12 m
x
MB
4m
-0.5
x
1
-4
17
Influence Line for Beam
• Reaction
P=1
C
A
B
x'
L
P=1
dy = 1
A
C
d y'
sB
dy
L
1
L
B
Ay
By
Ay (1) 1(d y ' ) By (0) 0
Ay d y '
18
P=1
C
A
B
x'
L
P=1
d y'
C
A
Ay
sA
dy
L
1
L
dy = 1
B
By
Ay (0) 1(d y ' ) By (1) 0
Ay d y '
19
- Pinned Support
C
A
B
b
a
L
B
A
RA
RA
1
b
L
x
20
- Fixed Support
A
B
a
b
L
A
B
RA
RA
1
1
x
21
• Shear
P=1
C
A
B
b
a
L
P=1
VC
dy=1
A
dyR
dyL
Ay
sA
1
L
d y'
sB
1
L
B
By
VC
Ay (0) VC (d yL ) VC (d yR ) 1(d y ' ) By (0) 0
VC (d yL d yR ) d y '
dy=1
VC d y '
slopes: sA sB
22
- Pinned Support
C
A
B
b
a
L
VC
B
A
VC
VC
1
1
Slope sB =
1
L
b
L
x
-a
1
Slope sA =
L
L
-1
Slope at A = Slope at B
23
- Fixed Support
A
B
a
b
L
VB
A
B
VB
VB
1
1
x
24
• Bending Moment
P=1
C
A
B
b
a
L
A B 1
P=1
h
MC
MC
d y'
B
A
Ay
A
h
a
Ay (0) MC A MC B 1(d y ' ) By (0) 0
1
MC ( A B ) d y '
M C d y'
B
h
b
By
h h
( ) 1
a b
h( a b)
1,
ab
h
ab
ab
( a b) L
25
- Pinned Support
C
A
B
b
a
L
Hinge
B
A
MC
MC
MC
a
C = A + B = 1
b
ab
L
A =
x
b
L
B=
a
L
26
- Fixed Support
A
B
a
b
L
A
B
MC
MC
MB
1
x
-b
27
• General Shear
C
A
L/4
D
L/4
E
L/4
B
F
L/4
L/4
G
L/4
H
L/4
L
VC
VD
3/4
1
x
-1/4
2/4
1
VE
x
-2/4
1/4
x
1
VBL
-3/4
x
-1
28
C
A
L/4
VBL
D
L/4
E
L/4
B
F
L/4
L/4
G
L/4
H
L/4
L
x
-1
VBR
1
x
VF
1
x
VG
1
x
29
• General Bending Moment
C
A
L/4
D
L/4
E
L/4
B
L/4
F
L/4
G
L/4
H
L/4
L
MC
3/16
= sA + sB = 1
sA = 3/4
sB = 1/4
4/1
6
MD
= sA + sB = 1
sA = 1/2
ME
x
sB = 1/2
3/1
6
x
= sA + sB = 1
x
sA = 1/4
sB = 3/4
30
C
A
L/4
D
L/4
E
L/4
B
L/4
F
G
L/4
H
L/4
L/4
L
MB
x
1
3L/4
MF
x
1
2L/4
MG
1
x
L/4
31
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
2m
Hinge
2m
D
C
2m
E
4m
32
SOLUTION
B
D
C
E
A
2m
RA
RA
2m
2m
4m
1
x
33
B
D
C
A
2m
2m
2m
E
4m
RC
8/6
RC
1
4/6
x
34
A
B
D
C
E
2m
2m
2m
4m
RE
1
2/6
RE
x
-2/6
35
VD
A
B
D
C
2m
2m
VD
2m
1
E
4m
4/6
VD
=
2/6
1
x
=
sE = 1/6
sC = 1/6
-1
-2/6
• sE = sC
36
1
Or using equilibrium conditions:
A
Hinge
B
2m
2m
VD
MD
D
C
2m
4m
E
4m
1
VD
MD
RE
VD = 1 -RE
VD = -RE
x
4m
RE
1
2/6
RE
x
-2/6
VD
4/6
2/6
x
-2/6
37
A
B
MD
C
MD
E
D
2m
2m
2m
4m
(2)(4)/6 = 1.33
D = C+E = 1
2
MD
C = 4/6
-1.33
2/6 = E
4
x
38
1
Or using equilibrium conditions:
A
Hinge
B
2m
2m
VD
MD
D
C
E
2m
4m
4m
1
VD
MD
RE
MD = 4RE
x
4m
RE
MD = -(4-x)+4RE
1
2/6
RE
x
-2/6
MD
8/6
x
-8/6
39
VCL
A
C
E
B
D
VCL
2m
2m
2m
4m
VCL
x
-1
-1
40
Or using equilibrium conditions:
1
A
B
2m
2m
1
RA
RA
D
C
VCL = RA - 1
2m
E
4m
MB
MB
VCL
RA
VCL
VCL = RA
1
x
VCL
x
-1
-1
41
VCR
A
C
E
B
2m
D
VCR
2m
2m
4m
1
VCR
0.333
0.667
x
42
Or using equilibrium conditions:
A
1
B
2m
D
C
2m
E
2m
4m
1
MC
MC
VCR
VCR
RE
VCR = -RE
VCR = 1 -RE
RE
1
2/6=0.33
RE
x
-2/6 = -0.333
1
VCR
0.333
0.667
x
43
A
MC C MC
B
2m
2m
D
2m
E
4m
MC
x
1
-2
44
Or using equilibrium conditions:
A
1
B
2m
D
C
2m
E
2m
4m
1
x'
MC
MC
VCR
6m
VCR
RE
MC = 6RE
6m
M C 6RA x'
RE
1
2/6=0.33
RE
x
-2/6 = -0.333
MC
x
1
-2
45
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
A
B
D
2m
2m
Hinge
C
E
2m
2m
F
2m
2m
46
SOLUTION
B
D
C
E
F
A
2m
RA
1
RA
2m
2m
2m
2m
2m
1
0.5
x
-0.5
-1
47
A
B
D
E
C
F
RC
2m
2m
2m
2m
2m
2m
1.5
2
1
0.5
RC
x
48
VD
A
D
B
C
E
F
VD
2m
2m
=
1
VD
2m
2m
2m
2m
1
0.5
=
x
-0.5
-1
49
VE
A
B
D
E
C
F
VE
2m
2m
2m
2m
2m
2m
0.5
1
=
VE
x
=
-0.5
-0.5
-1
50
VF
A
B
D
C
E
F
VF
2m
2m
2m
2m
2m
2m
=
1
VF
=
x
51
A
B
D
MD
2m
C
E
F
MD
2m
2m
2m
2m
2m
2
MD
1
x
D = 1
-1
-2
52
A
2m
ME
B ME
D
2m
ME C
E
2m
2m
(2)(2)/4 = 1
F
2m
2m
E = 1
x
B = 0.5
C = 0.5
-1
-2
53
A
B
D
2m
2m
C ME
E
2m
2m
2m
F
ME
2m
MF
F = 1
x
-2
54
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
4m
B
4m
4m
55
SOLUTION
8000 N
3000 N/m
1000 N/m
A
B
C
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
56
3000 N/m
3000 N/m
8000 N
1000 N/m
A
B
C
4m
4m
4m
VC
0.5
0.5(4)(0.5) = 1
0.5(4)(-0.5) = -1
0.5(4)(-0.5) = -1
-0.5
(VC)max
x
-0.5
= (1000)(-2+1) + (3000)(-2) + (8000)(-0.5)
= -11000 N = 11 kN
57
8000 N
3000 N/m
1000 N/m
A
B
C
4m
4m
MC
4m
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
58
Influence Line for Girder
• Forces Apply to the Girder
P2
P1
B
C
A
D
E
F
H
G
a
b
L
P1
A
P2
B
(b/L)P1
A
(a/L)P1
B
H
D
P2
C
F
D
E
G
59
• Reaction
1
A
B
0.5
0.5
H
C
D
E
F
G
RH
RG
1
RH
60
1
B
A
0.5
0.5
C
D
E
F
H
G
RH
RG
1
RG
61
• Shear
1
1
B
A
0.5
C
0.5
D
1
E
F
H
G
a
b
b
b
L
VC
VC
2b/L
VC
1
x
-(a+b)/L
62
1
A
0.5
1
B
C
0.5
0.5
D
E
0.5
F
H
G
a
b
L
VCD
VCD
b/L
VCD
x
-a/L
63
• Bending Moment
1
1
B
A
0.5
C
0.5
D
1
E
F
H
G
a
b
L
MC
MC
ab/L
MC
x
64
1
1
B
A
0.5
C
0.5
D
0.5
E
0.5
F
H
G
a
b
L
MF
MF
ab/L
MF
x
65
Example 6-5
Draw the influence for
- the Reaction RG and RF
- Shear VCD
- the moment MC and MH
B
C
D
A
E
H
F
1.5 m
1.5 m
G
3m
3m
3m
3m
66
B
C
D
A
E
H
F
1.5 m
G
3m
1.333
RG
RG
3m
1.5 m
3m
3m
1
0.667
0.333
x
67
B
C
D
A
E
H
F
1.5 m
G
3m
RF
1.5 m
RF
3m
3m
0.333
3m
0.667
1
x
-0.333
68
1
B
C
0.5
A
D
0.5
H
E
F
1.5 m
G
3m
3m
1.5 m
3m
3m
VCD
VCD
VCD
4.5/9
0.333
0.333
x
-0.333
-4.5/9
69
1
B
C
D
1
A
E
H
F
1.5 m
G
3m
3m
1.5 m
3m
MC
3m
MC
(3)(6)/9 = 2
1
MC
x
-2
70
1
B
C
0.5
A
D
0.5
H
E
F
1.5 m
G
3m
3m
1.5 m
3m
MH
3m
MH
(4.5)(4.5)/9 = 2.25
1.5
MH
1.5
x
-1.5
71
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.
2m
A
B
C
D
E
H
F
G
6 @ 4 m = 24 m
72
SOLUTION
1
2m
1
A
B
C
D
E
RC6 @ 4 m = 24 m
1.25
1
0.75
H
0.50
F
G
0.25
RC
73
1
2m
1
A
B
C
D
E
H
F
RG
6 @ 4 m = 24 m
0.25
G
0.50
0.75
1
RG
-0.25
74
1
1
1
A
2m
1
B
C
D
H
E
F
G
6 @ 4 m = 24 m
8m
8m
VE
VE
8/16 = 0.5
0.25
0.25
VE
-0.25
-8/16 = 0.5
75
1
1
1
A
2m
0.5
0.5
B
C
D
E
H
F
G
6 @ 4 m = 24 m
6m
10 m
VH
VH
6/16 = 0.375
0.25
0.25
VH
-0.25
-0.50
-10/16 = 0.625
76
1
1
1
A
2m
0.5
0.5
B
C
D
E
H
F
G
6 @ 4 m = 24 m
6m
10 m
MH
MH
(10)(6)/16 = 3.75
1.50
3
2.50
MH
-1.50
77
Example 6-7
B
C
D
A
E
F
H
2m
2m
2m
4m
G
2m
4m
4m
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
78
1
B
A
0.5
C
D
0.5
F
H
2m
2m
RH
2m
E
4m
G
2m
4m
4m
10/14
RG
1
6/14
1/14
2/14
RG
79
1
B
A
0.5
C
D
0.5
F
H
2m
R2H m
2m
1
6/14
E
4m
G
2m
4m
4m
RG
12/14
8/14
4/14
RH
80
1
1
B
A
0.5
C
0.5
1
H
2m
D
F
2m
2m
4m
E
G
2m
4m
4m
VC
VC
8/14
1
VC
-1/14 -2/14
4/14
x
-6/14
81
1
A
1
0.5
B
C
0.5
0.5
H
2m
D
0.5
F
2m
2m
4m
E
G
2m
4m
4m
VF
VF
6/14
VCD
4/14
x
-1/14
-2/14
-6/14
-8/14
82
1
1
B
A
0.5
C
0.5
1
H
2m
2m
4m
G
2m
4m
(2/6)(3.43)
E
F
2m
MC
MC
D
4m
MC
(6)(8)/14 = 3.43
(4/8)(3.43)
x
83
1
1
B
A
0.5
C
D
0.5
2m
1
F
H
2m
2m
4m
MD
G
2m
4m
MD
(2/10)(2.86)
E
(6/10)(2.86)
4m
MD
(10)(4)/14 = 2.86
x
84
1
1
B
A
0.5
C
0.5
0.5
H
2m
D
2m
4m
0.429
0.857
G
2m
4m
MF
MF
0.5
F
2m
E
4m
MF
(8)(6)/14=3.428
2.571
2.285
x
85
Maximum Reaction
50 kN
5 kN/m
B
C
D
A
E 2 kN/m
F
H
2m
2m
4m
2m
G
2m
4m
4m
10/14
1
6/14
1/14
2/14
[0.5×4 (2/14)] + [(0.5)(2/14 + 1)(12) = 7.143
RG
(RG)max =
(2)(7.143) + (5)(7.143) + (50)(1)
= 100 kN
86
Maximum Reaction
50 kN
5 kN/m
B
C
D
A
E 2 kN/m
F
H
2m
2m
4m
2m
G
2m
4m
4m
12/14
8/14
6/14
(0.5)(16)(12/14) = 6.857
4/14
RH
(RH)max =
(2)(6.857) + (5)(6.857) + (50)(12/14)
= 90.86 kN
87
Maximum Shear
50 kN
5 kN/m
B
C
E2 kN/m
D
A
F
H
2m
2m
2m
4m
G
2m
4m
4m
4/14
2.4 m
VCD
-1/14
-2/14
-6/14
+ (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
(VCD)max = (2)(-1.943 + 0.8) (5)(-1.943) + (50)(-6/14)
= -33.43 kN
88
Influence Line for Trusses
H
G
F
h
E
A
B
1
C
4@xm
(3/4)
RA
(2/4)
D
(1/4)
x
RE
(1/4)
(2/4)
(3/4)
1
x
89
H
Consider the right hand
FGH
G
F
FGB
h
FCB
A
RE
B 1 kN
(1/4)
E
C
4@xm
(2/4)
D
RE
(3/4)
1
+ SMG = 0:
RE (2x) FBC (h) 0
x
FBC
2x
RE
h
(2x/h)(1/4)
FBC
x
90
Consider the left hand
H FHG
G
F
FBG
h
FBC
A
C 1 kN
4@xm
B
RA
1
E
(3/4)
RA
D
+ SMG = 0:
(2/4)
FBC (h) RA (2x) 0
(1/4)
x
(2x/h)(2/4)
(2x/h)(1/4)
FBC
FBC
2x
RA
h
(2x/h)(1/4)
x
91
Example 6-8
H
G
F
4m
E
A
3m
4 kN
2m
B
1.5 kN
3m
C
3m
D
3m
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
92
SOLUTION
H
G
F
4m
E
A
3m
1
B
3m
0.75
RA
C
3m
0.5
D
3m
0.25
x
RE
0.25
0.5
0.75
1
x
93
Influence Line for FBC
H
FGH
Consider the right hand
G
F
FGB
4m
FCB
A
RE
B 1 kN
3m
3m
0.25
E
C
3m
0.5
D
3m
0.75
RE
1 + SM = 0:
G
FBC = (6/4) RE
= 1.5 RE
x
1.5(0.25) = 0.375
FBC
x
94
Consider the left hand
H FHG
G
F
FBG
4m
FBC
A
3m
RA
B
1
3m
0.75
RA
E
C 1 kN
3m
0.5
D
3m
+ SMG = 0:
0.25
x
FBC
FBC = 1.5 RA
1.5(0.5) = 0.75
0.375
0.375
x
95
Influence Line for FBG
H
FGH
Consider the right hand
G
F
FGB
4m
FCB
A
RE
B 1 kN
3m
3m
E
C
0.25
3m
0.5
D
3m
0.75
RE
1
x
SFy = 0 ;
FBG cos = RE
FBG (4/5) = RE
1.25(0.25) = 0.3125
FBG = 1.25 RE
FBG
x
96
Consider the left hand
H FHG
G
F
FBG
4m
FBC
A
3m
RA
B
1
3m
0.75
RA
E
C 1 kN
3m
0.5
D
3m
SFy = 0 ;
FBG cos + RA = 0
0.25
x
FBG (4/5) = -RA
FBG = -1.25 RA
0.3125
FBG
x
-0.3125
-1.25(0.5) = -0.625
97
Influence Line for FCG
H
G
F
4m
0
E
A
FCG
B 1 kN
3m
3m
C
3m
D 1 kN
3m
x
98
H
G
F
4m
1
E
A
3m
B
3m
C 1 kN
3m
D
3m
1
FCG
x
99
(FBC)max by Loads
H
G
F
2m
4 kN
1.5 kN
4m
E
A
B
3m
FBC
3m
C
3m
0.75
0.375
D
0.5
3m
0.375
x
(FBC)max =
(4)(0.75) + (1.5)(0.5)
= 3.75 kN #
100
(FBG)max by Loads
H
G
4 kN
F
2m
1.5 kN
4m
E
A
3m
B
3m
C
3m
D
3m
0.3125
FBG
x
-0.625
(FBG)max = (4)(-0.625)
-0.417
+ (1.5)(-0.417)
-0.3125
= -3.126 kN #
101
(FCG)max by Loads
H
G
4 kN
F
2m
1.5 kN
4m
E
A
3m
B
3m
C
3m
D
3m
1
0.333
FCG
x
(FCG)max = (4)(1) + (1.5)(0.333)
= 4.50 kN #
102
Maximum Influence at a Point Due to a Series of Concentrated Loads
F3
F2
F1
x1
A
x2
B
C
a
VC
b
(b/L)
x
(-a/L)
ab/L
MC
x
103
F3
F2
F1
x1
x2
A
B
C
a
VC
b
(b/L)
V2'
V3'
x
(-a/L)
ab/L
MC
M 2'
M 3'
x
(VCR )' F1 (b / L) F2 (V2' ) F3 (V3' )
(VCL )' F1 (a / L) F2 (V2' ) F3 (V3' )
M ' F1 (ab / L) F2 (M 2' ) F3 (M 3' )
104
F3
F2
F1
x1
x2
A
B
C
a
b
(b/L)
VC
V3''
x
V1''
MC
M1''
(-a/L)
ab/L
M 3''
x
(VCR )'' F1V1'' F2 (b / L) F3 (V3'' )
(VCL )'' F1V1'' F2 (a / L) F3 (V3'' )
M ' ' F1M1'' F2 (ab / L) F3 (M 3'' )
105
F3
F2
F1
x1
x2
A
B
C
a
b
(b/L)
VC
x
V2 '''
(-a/L)
ab/L
MC
M 2'''
x
(VCR )'' ' F2V2''' F3 (b / L)
(VCL )'' ' F2V2''' F3 (a / L)
M ' ' ' F2 M 2''' F3 (ab / L)
106
• Shear
4 kN
1 kN
1m
A
6 kN
3m
B
2m
C
6m
0.75
VC
x
-0.25
107
Case 1
4 kN
1 kN
1m
6 kN
3m
A
B
2m
C
0.75
VC
6m
0.625
0.25
x
-0.25
(VC)1 = 1(0.75) + 4(0.625) + 6(0.25) = 4.75 kN
108
Case 2
4 kN
1 kN
1m
6 kN
3m
A
B
2m
C
6m
0.75
VC
0.375
x
-0.125
-0.25
(VC)2 = 1(-0.125) + 4(0.75) + 6(0.375) = 5.125 kN
109
Case 3
4 kN
1 kN
1m
6 kN
3m
A
B
2m
C
6m
0.75
VC
x
-0.25
(VC)3 = 6(0.75) = 4.5 kN
(VC)max = (VC)2 = 5.125 kN
110
The critical position of the loads can be determined in a more direct manner by finding
the change in shear, DV, which occurs when the load are moved from Case 1 to Case 2,
then from Case 2 to Case 3, and so on.If the slope of the influence line is s, then
(y2 - y1) = s(x2 - x1), and therefore
DV = Ps(x2 - x1)
----------(6-1)
Sloping Line
If the load moves past a point where there is a discontinuity or “jump” in the influence
line, as point C, then the change in shear is simply
DV = P(y2 - y1)
----------(6-2)
Jump
111
4 kN
1 kN
1m
6 kN
3m
A
B
2m
C
0.75
VC
6m
s = 1/8
x
-0.25
DV1-2 = 1(-0.25 - 0.75) + 1(1/8)(1) + 4(1/8)(1) + 6(1/8)(1) = 0.375 kN
(VC)2 = (VC)1 + DV1-2 = 4.75 + 0.375 = 5.125 kN
112
4 kN
1 kN
1m
6 kN
3m
A
B
2m
C
0.75
VC
6m
s = 1/8
x
-0.25
DV2-3 = 1(1/8)(1) + 4(-0.25-0.75) + 4(1/8)(2) + 6(1/8)(3) = -0.625 kN
(VC)3 = (VC)2 + DV2-3 = 5.125 -0.625 = 4.5 kN
113
4 kN
1 kN
1m
• Moment
A
3m
B
2m
MC
6 kN
C
6m
(2)(6)/8 = 1.5
x
114
4 kN
1 kN
1m
Case 1
6 kN
3m
A
B
2m
MC
C
1.5
6m
1.25
0.5
x
(MC)1 = 1(1.5) + 4(1.25) + 6(0.5) = 9.5 kN•m
115
4 kN
1 kN
1m
Case 2
6 kN
3m
A
B
2m
MC
0.75
C
1.5
6m
0.75
x
(MC)2 = 1(0.75) + 4(1.5) + 6(0.75) = 11.25 kN•m
116
6 kN
Case 3 4 kN
1 kN
1m
3m
A
B
2m
MC
C
6m
1.5
x
(MC)3 = 6(1.5) = 9 kN•m
(MC)max = (MC)2 = 11.25 kN•m
117
We can also use the foregoing methods to determine the critical position of a series of
concentrated forces so that they create the largest internal moment at a specific point in
a structure. Of course, it is first necessary to draw the influence line for the moment at
the point and determine the slopes s of its line segments. For a horizontal movement
(x2 - x1) of a concentrated force P, the change in moment, DM, is equivalent to the
magnitude of the force times the change in the influence-line ordinate under the load,
that is
DM = Ps(x2 - x1)
----------(6-3)
Sloping Line
118
4 kN
1 kN
1m
6 kN
3m
A
B
2m
MC
C
6m
1.5
x
1.5
1.5
DM 1 2 1( )(1) (4 6)( )(1) = 1.75 kN•m
2
6
(M C)2 (M C)1 DM12 9.5 1.75 = 11.25 kN•m
119
4 kN
1 kN
1m
6 kN
3m
A
B
2m
MC
C
6m
1.5
x
1.5
1.5
1.5
DM 23 1( )(1) 4( )( 2) 6( )( 3) = -2.25 kN•m
2
2
6
(M C)3 (M C)2 DM 23 11.25 2.25= 9 kN•m
120
Example 6-9
Determine the maximum shear created at point B in the beam shown in the figure
below due to the wheel loads of the moving truck.
15 kN
9 kN
4 kN
B
A
4m
10 kN
C
4m
1m
2m
2m
121
SOLUTION
4 kN
1m
B
A
2m
10 kN
2m
C
4m
4m
VB
15 kN
9 kN
Case 1
0.5
0.375
0.125
x
-0.5
VBR = 4(0.5) + 9(0.375) + 15(0.125) = 7.25 kN
VBL = 4(-0.5) + 9(0.375) + 15(0.125) = 3.25 kN
122
Case 2
15 kN
9 kN
4 kN
1m
B
A
2m
2m
C
4m
4m
0.5
VB
10 kN
0.25
x
-0.375
-0.5
VBR = 4(-0.375) + 9(0.5) + 15(0.25) + 10(0) = 6.75 kN
VBL = 4(-0.375) + 9(-0.5) + 15(0.25) + 10(0) = -2.25 kN
123
Case 3
15 kN
9 kN
4 kN
1m
2m
10 kN
2m
B
A
C
4m
4m
0.5
VB
0.25
x
-0.125
-0.25
-0.5
VBR = 4(-0.125) + 9(-0.25) + 15(0.5) + 10(0.25) = 7.25 kN
VBL = 4(-0.125) + 9(-0.25) + 15(-0.5) + 10(0.25) = -7.75 kN
124
Case 4
15 kN
9 kN
4 kN
1m
2m
10 kN
2m
B
A
C
4m
4m
0.5
VB
x
-0.25
-0.5
VBR = 9(0) + 15(-0.25) + 10(0.5) = 1.25 kN
VBL = 9(0) + 15(-0.25) + 10(-0.5) = -8.75 kN
The maximum shear created at point B is -8.75 kN
125
Summary
A
4 kN
1m
B
2m
10 kN
2m
C
4m
4m
VB
15 kN
9 kN
0.5
x
-0.5
Case 1
Case 2
VBR = 7.25 kN
1-m Movement of 4 kN, DVB = 6.75-7.25 = - 0.5 kN
VBR = 6.75 kN
Case 3
VBR = 7.25 kN
Case 4
VBR = 1.25 kN
2-m Movement of 9 kN, DVB = 7.25-6.75 = + 0.5 kN
2-m Movement of 15 kN, DVB = 1.25-7.25 = - 6 kN
126
Or using the sloping line
15 kN
9 kN
4 kN
1m
B
A
2m
10 kN
2m
C
4m
4m
0.5
VB
x
-0.5
s
0.5
4
127
15 kN
9 kN
2m
2m
4 kN 1 m
B
C
A
4m
9 kN
2m
4 kN 1 m
4m
15 kN
10 kN
10 kN DV Pjump ( y2 y1 ) Pn s( x2 x1 )
2m
1-m Movement of 4 kN
DV1 2 4(1) (4 9 15)(
15 kN
9 kN
2m
2m
4 kN 1 m
0.5
)(1)
4
= -0.5 kN
10 kN
2-m Movement of 9 kN
DV23 9(1) (4 9 15 10)(
4 kN 9 kN
1m
2m
A
VB
= +0.5 kN
15 kN
2m
4m
0.5
)( 2)
4
10 kN
B
0.5
-0.5
4m
2-m Movement of 15 kN
0.5
DV3 4 15(1) 4( )(1)
C
4
0.5
(9 15 10)( )( 2)
4
= -6 kN
128
x The correct position of the loads.
15 kN
4 kN 9 kN
1m
2m
2m
10 kN
B
A
C
4m
4m
0.5
VB
x
(-0.25)
-0.5
VBR = 9(0) + 15(-0.25) + 10(0.5) = 1.25 kN
VBL = 9(0) + 15(-0.25) + 10(-0.5) = -8.75 kN
The maximum shear created at point B is -8.75 kN
129
Example 6-10
Determine the maximum positive moment and negative moment created at point
B in the beam shown in the figure below due to the wheel loads of the crane.
8 kN
4 kN
3 kN
3m
2m
B
A
C
2m
3
m
2m
130
SOLUTION
• Positive moment
4 kN
8 kN
3 kN
3m
2m
A
CASE I
B
2m
MB
C
3
m
2m
2(3)/5 = 1.2
x
-0.8
MB = 3(1.2) + 8(0) = 3.6 kN•m
131
8 kN
4 kN
3 kN
3m
2m
A
CASE II
B
C
2m
MB
2m
3
m
1.2
0.4
x
-0.8
MB = 8(1.2) + 3(0.4) = 10.8 kN•m
132
8 kN
4 kN
3 kN
3m
A
CASE III
B
C
2m
MB
2m
3
m
2m
1.2
x
-0.8
MB = 4(1.2) + 8(0) + 3(-0.8) = 2.4 kN•m
The maximum positive moment created at point B is 10.8 kN•m
133
• Negative moment
8 kN
4 kN
3 kN
3m
A
B
C
2m
MB
2m
CASE I
2m
3
m
1.2
0.4
x
-0.8
MB = 8(-0.8) + 4(0.4) = -4.8 kN•m
134
8 kN
4 kN
3 kN
3m
A
B
C
2m
MB
2m
3
m
CASE II
2m
1.2
x
-0.8
MB = 4(-0.8) = -3.2 kN•m
The maximum negative moment created at point B is 4.8 kN•m
135
Absolute Maximum Shear and Moment
FR
F2 CL
x x' x
F1
F3
A
B
2
x'
d1
d2
Ay
By
L/2
L/2
L
MB 0:
Ay
1
L
( FR )[ ( x' x)]
L
2
136
F1
d1
M2
A
(L/2 - x)
2
V2
1
L
Ay ( FR )[ ( x' x)]
L
2
M2 0 :
L
M 2 Ay ( x) F1d1
2
1
L
L
( FR )[ ( x' x)]( x) F1d1
L
2
2
FR L FR x' FR x 2 FR x x'
M2
F1d1
4
2
L
L
For maximum M2 we require
dM 2
2 F x F x'
R R 0
dx
L
L
x
x'
2
137
FR
CL
F1
F2
x
A
B
L/2
L/2
x1
x2 x
F3
x2
x1 x
L
138
FR
F2 b
F1
x1
A
FR
x2
F3
x
2
B
bL
a
F2 b
F1
x1
x2
x
2
F3
b
b
ab / L M 2''
M1''
M 3''
M
x
M S1 F1M1'' F2 M 2'' F3M 3''
139
FR
F1
A
FR
a F
3
F2
x1
F1
x2 x
2
x2
x1
B
La
a
M
M1'''
M 2'''
a
F2
F3
x2 x
2
x2
a
b
ab / L M 3'''
x
M S 2 F1M1''' F2 M 2''' F3M 3'''
The absolute maximum moment is comparison by MS1 and MS2.
140
Example 6-11
Determine the absolute maximum moment on the simply supported beam cased
by the wheel loads.
1200 kg
A
400 kg
B
30 m
8m
141
1600 kg
CL
1200 kg
A
400 kg
x
B
8m
15 m
15 m
SM@1200 kg = 0 :
1600x 1200(0) 400(8)
x2
142
CL
a 1600 kg
1200 kg
1m
400 kg
1m
A
B
1
2m
6m
Or using equilibrium conditions:
a
14 m
Global: + SMB = 0:
1600(14) Ay (30) 0
16 m
Ay = 746.67 kg
(14)(16)/30 = 7.47
14 m
3.74
M1200
A
1
x
MS = (1200)(7.47) + (400)(3.74)
= 10460 kg•m
M1
V1
Ay = 746.67 kg
+ SM1 = 0:
M1= 10453.38 kg•m
143
CL
1600
kg
1600
kg
b
1200 kg
3m
b
1200 kg
400 kg
3m
3m
A
400 kg
3m
B
b
6m
2m
18 m
2m
b
6m
12 m
(18)(12)/30 = 7.2
4
M400
x
MS = (1200)(4) + (400)(7.2)
By comparison,
= 7680 kg•m
Mmax = 10460 kg•m
144
Example 6-12
Determine the absolute maximum moment on the simply supported beam cased
by the wheel loads.
4.6 T
A
8.2T 8.2T
B
20 m
4.2 m
1.2 m
145
FR=21 T
4.6 T
A
8.2T 8.2T
B
4.2 m
20 m
1.2 m
x
4.2 x 0.45
SM@ 4.6 T = 0 :
21x 4.6(0) 8.2(4.2) 8.2(5.4)
x 3.75
146
CL FR=21 T
4.6 T
8.2T 8.2T
A
B
4.2 m
10 m
a = 8.125 m
1.875 m
10 m
b = 11.875 m
4.82
M4.6T
1.2 m
3.12
2.63
x
MS = (4.6)(4.82)+ (8.2)(3.12)+ (8.2)(2.63) = 69.32 T•m
147
4.6 T
FR=21 T
CL
8.2T 8.2T
A
B
2
4.2 m
Or using equilibrium conditions:
1.2 m
Global: + SMB = 0:
21(10.225) Ay (20) 0
0.225 m
10 m
10 m
a = 10.225 m
Ay = 10.74 T
4.6 T
4.2 m
b = 9.775 m
M2
A
5
2.95
10.225 m
4.39
M8.2T
x
Mmax = 90.57 T•m
V2
Ay = 10.74 T
MS = (4.6)(2.95)+ (8.2)(5) + (8.2)(4.39) = 90.57 T•m
By comparison,
2
+ SM2 = 0:
M2= 90.50 T•m
148
Example 6-13
Determine the absolute maximum moment on the simply supported beam cased
by the wheel loads.
15 kN
9 kN
4 kN
A
10 kN
B
4m
1m
2m
2m
149
FR=38 kN
9 kN
15 kN 10 kN
4 kN 1.74 m
0.26 m
SOLUTION
A
B
1m
4m
2m
2m
x
SM@ 4 kN = 0 :
38x 4(0) 9(1) 15(3) 10(5)
x 2.74
150
FR=38 kN
15 kN
9 kN
4 kN
0.87 0.87
A
B
1m
M9 kN
10 kN
2m
2m
4m
4m
a = 3.13 m
b = 4.87 m
1.30
(3.13)(4.87)/8 = 1.91
1.13
0.34
x
MS = 4(1.30) + 9(1.91) + 15(1.13) + 10(0.34) = 42.74 kN•m
151
9 kN FR 15 kN 10 kN
4 kN 0.13 m
0.13 m
A
B
3
1m
2m
Or using equilibrium conditions:
2m
Global: + SMA = 0:
4m
4m
a = 4.13 m
b = 3.87 m
M15 kN
0.55
1.03
38(3.87) By (8) 0
By = 18.38 kN
15 kN
(4.13)(3.87)/8 = 2.0
M3
0.97
2m
x
V3
MS = 4(0.55) + 9(1.03) + 15(2.0) + 10(0.97) = 51.17 kN•m
By comparison,
Mmax = 51.17 kN•m
10 kN
3
B
3.87 m
+ SM3 = 0:
18.38 kN
-M3 -10(2) + 18.38(3.87) = 0
M3= 51.13 kN•m
152