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6-Influence Lines II

This document discusses determining the maximum shear and moment effects on structures from moving concentrated loads through influence lines. It presents a trial and error method as well as a more direct approach using the change in influence line ordinates. The direct approach calculates the change in shear or moment as the load moves from one position to the next. An example application to a beam with moving wheel loads is provided.

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Steven Kua
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188 views17 pages

6-Influence Lines II

This document discusses determining the maximum shear and moment effects on structures from moving concentrated loads through influence lines. It presents a trial and error method as well as a more direct approach using the change in influence line ordinates. The direct approach calculates the change in shear or moment as the load moves from one position to the next. An example application to a beam with moving wheel loads is provided.

Uploaded by

Steven Kua
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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CE 3155 Structural Analysis

Influence lines for statically determinate structures II

Dr Poh Leong Hien


Department of Civil and Environmental Engineering
E1A-07-13
Email: ceeplh@nus.edu.sg
2

Maximum Influence at a point due to a series of concentrated loads

• The maximum effect caused by a live concentrated force is determined by multiplying the
peak ordinate of the influence line by the magnitude of the force.
• In some cases, a series of concentrated loadings is placed on the structure. How do we
determine the maximum effect then?
3

Trial and Error

Consider the simply supported beam with the associated influence line for shear at point C. A
series of concentrated loads (e.g. wheels) moves from right to left. Determine the maximum
positive shear at C.

Critical loading occurs when one of the


loads is placed just to the right of C.
To consider all 3 possibilities – trial and
error method.
4

(VC )1  4.5(0.75)  18(0.625)  18(0.5)


 23.63kN

(VC ) 2  4.5(0.125)  18(0.75)  18(0.625)


 24.19kN
5

(VC ) 3  4.5(0)  18(0.125)  18(0.75)


 11.25kN

Critical loading is thus Case 2.

However, the trial and error method can be tedious at times. A more direct approach is to find V
when the loads are moved from Case 1 to 2, and from Case 2 to 3.

Idea: If V > 0, then the new position is more critical case then the previous position.
6

Shear

Recall: The change in shear V when load P moves from position x1 to x2 is determined by
multiplying P by the change in ordinate of influence line (y2 – y1).

Let s be the slope of the influence line. Then,


Note: Whether s is positive or negative
V  Ps ( x2  x1 ) depends on the direction of approach of the
concentrated loads.

If the load moves past a point when there is a discontinuity in the influence line, the change
in shear is

V  P( y2  y1 )
7

Case 1 to Case 2
• 4.5kN load jumps down (-1)
• All loads move up the slope of
influence line

V12 
8

Case 2 to Case 3
• 18kN load jumps down (-1)
• All loads move up the slope of
influence line

V23 
9

Moment

The change in moment M is equivalent to the magnitude of force multiply by the change in
the influence-line ordinate,

M  Ps( x2  x1 )
Example
Consider the bending moment at point C for the beam below, when subjected to the loading
shown.
10

 2.25   2.25 
M 1 2  9 (1.2)  (18  13.5) (1.2)
 3   12  3 
 1.35 kNm
11

 2.25   2.25 
M 23  (9  18) (1.8)  13.5 (1.8)
 3   12  3 
 30.38 kNm  0
12

Hence, critical loading from Case 2.

( M C ) max  9(1.35)  18(2.25)  13.5(18)  77.0 kNm


13

Example

Determine the maximum positive shear at point B due to the wheel loads of the moving
truck.

Influence line for VB


14

Case1:
Imagine that the 18 kN load acts just to the right of
point B so that we obtain its maximum positive
influence.
18kN 40.5kN 67.5kN 45kN
0.9m 1.8m 1.8m When the truck moves 0.9 m to the left, the 18 kN
load jumps downward on the influence line by 1 unit.

Case 1 From Case 1 to 2,


A B C

 0.5 
18kN 40.5kN 67.5kN 45kN VB  18(1)  (18  40.5  67.5) 0.9
0.9m 1.8m 1.8m
 3 
 0.9 kN

A B C Case 2
15

Case 2 - 3:
The 40.5kN load, initially just to the right of B, moves
1.8m to the left.

18kN 40.5kN 67.5kN 45kN Note: the 45kN load has moved only 1.2m on the
0.9m 1.8m 1.8m beam.

Case 2 From Case 2 to 3,


A B C
 0 .5 
VB  40.5(1)  (18  40.5  67.5) (1.8)
18kN 40.5kN 67.5kN 45kN  3 
0.9m 1.8m 1.8m
 0 .5 
 4.5 (1.2)
 3 
A B C Case 3  6.3 kN
16

Case 3 - 4:
The 67.5kN load, initially just to the right of B, moves
1.8m to the left.

18kN 40.5kN 67.5kN 45kN Note: the 18kN load has moved only 0.3m on the
0.9m 1.8m 1.8m beam; the 40.5kN load has moved only 1.2m on the
beam.

Case 3 From Case 3 to 4,


A B C
 0 .5   0.5 
VB  67.5(1)  18  ( 0 . 3)  40 . 5 (1.2)
40.5kN 67.5kN 45kN  3   3 
1.8m 1.8m
 0 .5 
 (67.5  45) (1.8)
 3 
A B C Case 4  24.8 kN  0
17

Hence, the most critical loading is Case 3.

(VB ) max  18(0.05)  40.5(0.2)  67.5(0.5)  45(0.2)


 33.8 kN
18kN 40.5kN 67.5kN 45kN
0.9m 1.8m 1.8m

A B C Case 3

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