Influence Line
Diagram -III
Theory of Structures-I
1
Contents
1.
Maximum Influence at a point due to a
series of concentrated loads
2.
Absolute Maximum Shear and Moment
2
Maximum Influence at a Point
Due to a Series of
Concentrated Loads

General Case: Develop ILD for a function
and then maximum effect is calculated by
Maximum Effect (Point Load)= Magnitude of force x
Peak ordinate of ILD

In some cases, several concentrated forces
must be placed on structure e.g. truck
loading or train loading on a bridge
3
Shear
A
C
B
10’
Vc
30’
0.75
x
10’
-0.25
40’
ILD for Shear at Point C
4
Case 1
1K
A
C
4K
B
5’
10’
4K
5’
30’
5
Case 1
1K
A
4K
4K
C
B
10’
Vc
30’
0.75
0.625 0.5
x
10’
15’
20’
40’
-0.25
(Vc)1 = 1(0.75) + 4(0.625) + 4(0.5) = 5.25 k
6
Case 2
1K
A
C
4K
B
5’
10’
4K
5’
30’
7
Case 2
1K
A
4K
4K
C
B
10’
30’
0.75
Vc
0.625
5’
x
10’
-0.125
-0.25
15’
40’
(Vc)2 = 1(-0.125) + 4(0. 75) + 4(0.625) = 5.375 k
8
Case 3
1K
A
C
4K
B
5’
10’
4K
5’
30’
9
Case 3
1K
4K
A
4K
C
B
10’
30’
0.75
Vc
5’
x
10’
-0.125
-0.25
15’
40’
(Vc)3 = 1(0) + 4(-0.125) + 4(0.75) = 2.5 k
10
Comparison
Case 1:
(Vc)1 = 1(0.75) + 4(0.625) + 4(0.5) = 5.25 k
Case 2:
(Vc)1 = 1(-0.125) + 4(0. 75) + 4(0.625) = 5.375 k
Case 3:
(Vc)1 = 1(0) + 4(-0.125) + 4(0.75) = 2.5 k
11
Method Based on Change in
Function

When many concentrated loads act on the span, the
trial-and-error computations used above can be
tedious.

Critical position of loads can be determined in a more
direct manner by finding the change in shear v when
loads move from case 1 to case 2 and case 3.

As long as each computed v is positive, the new
position will yield a larger shear in the beam at C than
the previous position.
12
Method Based on Change in
Function …

Each movement is investigated until a
negative change in shear is computed.

When this occurs, the previous position of the
loads will give the critical value.
13
Method Based on Change in
Function …

The change in shear V for a load P that
moves from position x1 to x2 over a beam can
be determined by


multiplying P by the change in the ordinate of the
influence line, that is, (y2-y1).
If the slope of the influence line is s, then (y2y1)=s(x2-x1)
V = Ps(x2-x1)
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Method Based on Change in
Function …

If the load moves past a point where there is
a discontinuity or jump in the influence line,
as point c in previous examples, then change
in shear is simply
V = P (y2-y1)
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Case 1-2
1K
A
C
4K
B
5’
10’
4K
5’
30’
16
Case 1-2
1K
A
4K
4K
C
B
10’
30’
0.75
Vc
0.625
S = 0.75/(40-10) = 0.25/10 = 0.025
5’
x
10’
-0.125
-0.25
15’
40’
V1-2 = 1(-1) + [1+4+4] (0.025)(5)= +0.125 k
17
Case 2-3
1K
A
C
4K
B
5’
10’
4K
5’
30’
18
Case 2-3
1K
4K
A
4K
C
B
10’
30’
0.75
Vc
5’
x
10’
-0.125
-0.25
15’
40’
V2-3 = 4(-1) + [1+4+4] (0.025)(5)= -2.875 k
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Moment


Use the same method to calculate the critical
position of series of concentrated forces so
that they create largest internal moment at a
specified position in the structure.
First Draw ILD of moment for the given point
and then proceed with the calculations.
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Moment
A
C
10’
Mc
B
30’
7.5
x
10’
40’
ILD for Moment at Point C
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Critical Position of Loads
M = Ps(x2-x1)
M1-2 = -2(7.5/10)(4) + (4+3)(7.5/(40-10))(4) = 1.0 k. ft
M2-3 = -(2+4)(7.5/10)(6) + (3)(7.5/(40-10))(6) = -22.5 k. ft
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Change in M
A
Case 1
2K
C
10’
4K
B
4’
30’
2K
A
C
4K
4’
A
C
10’
3K
6’
30’
2K
Case 3
6’
B
Case 2
10’
3K
4K
3K
B
30’
4’
6’
23
Change in M
2K
A
Case 1
4K
3K
C
10’
2K
A
B
30’
4K
3K
C
B
Case 2
2K
10’
4K
A
Case 3
30’
3K
M1-2 = -2(7.5/10)(4) + (4+3)(7.5/(40-10))(4) = 1.0 k. ft
C
10’
B
30’
24
M2-3 = -(2+4)(7.5/10)(6) + (3)(7.5/(40-10))(6) = -22.5 k. ft
Maximum Moment

From the results we can conclude that case 2
will produce the maximum moment.
(Mc) max = 2(4.5) + 4(7.5) + 3(6.0) = 57 k. ft.
Mc
7.5
6
4.5
x
6’
10’
16’
40’
25
Absolute Maximum Shear and
Moment

We developed the methods for computing
maximum shear and moment at a specified
point due to series of concentrated moving
loads.

Now to determine both the location of the
point in the beam and the position of loading
on the beam so that one can obtain the
absolute maximum shear and moment
caused by the loads.
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Shear in Cantilever Beam

For a cantilevered beam the absolute
maximum shear will occur at a point located
just next to the fixed support. Loads will be
positioned closed to the support.
V abs max
27
Moment in Cantilever Beam

For a cantilevered beam the absolute
maximum moment will occur at a same point
where absolute maximum shear occur but the
loads will be located at far end of the beam.
M abs max
28
Moment in Cantilever Beam
M abs max
29
Shear in Simply Supported Beam

For simply supported beams the absolute
maximum shear will occur just next to one of
the supports. Loads are positioned such that
first load is near the support.
A
B
V abs max
30
Moment at Simply Supported
Beam

For simply supported beam the critical
position of the loads and the associated
absolute maximum moment cannot, in
general, be determined by inspection.

We can determine the position analytically.
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Example
F1
C
L
F2
FR
F3
x’-x
x
A
B
d1
x’
d2
L/2
Ay
M
L/2
By
b
0
1
L

Ay  FR    x' x 
L
2

32
Example…
M
F1
M2
A
d1
L/2 - x
V2
b
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 xx'




 F1d1
4
2
L
L
33
Example…
F1
M2
A
d1
L/2 - x
V2
For Maximum M2 we require
dM 2
 2 FR x
F x'

 R 0
dx
L
L
x'
x
2
34
Conclusion: Simply supported
beams

Absolute maximum moment in a simply
supported beam occurs under one of the
concentrated forces, such that this force is
positioned on the beam so that it and the
resultant force of the system are equidistant
from the beam’s centerline.
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