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’ V c 0.75

30’ -0.25

10’

ILD for Shear at Point C

40’ x 4

Case 1

A 10’ C 30’ B 1K 4K 4K 5’ 5’ 5

Case 1

A 1K 4K 4K C B 10’ 30’ V c

0.75

0.625

0.5

10’ 15’ 20’ 40’ (Vc) 1

-0.25

= 1(0.75) + 4(0.625) + 4(0.5) = 5.25 k x 6

Case 2

A 10’ C 30’ B 1K 4K 4K 5’ 5’ 7

Case 2

1K 4K 4K A C B 10’ 30’ V c

0.75

0.625

(Vc) 2 5’ x 10’

-0.125

-0.25

15’ 40’ = 1(-0.125) + 4(0. 75) + 4(0.625) = 5.375 k 8

Case 3

A C 10’ 30’ B 1K 4K 4K 5’ 5’ 9

Case 3

1K 4K 4K A C B 10’ 30’

0.75

V c 5’ 10’

-0.125

-0.25

15’ 40’ (Vc) 3 = 1(0) + 4(-0.125) + 4(0.75) = 2.5 k x 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.

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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.

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Method Based on Change in Function …

  The change in shear  V for a load P that moves from position x be determined by 1 to x 2 over a beam can  multiplying P by the change in the ordinate of the influence line, that is, (y 2 -y 1 ). If the slope of the influence line is s, then (y 2 y 1 )=s(x 2 -x 1 )  V = Ps(x 2 -x 1 ) 14

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 (y 2 -y 1 ) 15

Case 1-2

A 10’ C 30’ B 1K 4K 4K 5’ 5’ 16

Case 1-2

1K 4K 4K A C B 10’ 30’ V c

0.75

0.625

S = 0.75/(40-10) = 0.25/10 = 0.025

5’ 10’

-0.125

-0.25

 V 1-2 15’ 40’ = 1(-1) + [1+4+4] (0.025)(5)= +0.125 k x 17

Case 2-3

A C 10’ 30’ B 1K 4K 4K 5’ 5’ 18

Case 2-3

1K 4K 4K A C B 10’ 30’

0.75

V c 5’ 10’

-0.125

-0.25

 V 2-3 15’ 40’ = 4(-1) + [1+4+4] (0.025)(5)= -2.875 k x 19

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

C A 10’ 7.5

M c 30’ 10’

ILD for Moment at Point C

40’ B x 21

Critical Position of Loads

 M = Ps(x 2 -x 1 )  M 1-2 = -2(7.5/10)(4) + (4+3)(7.5/(40-10))(4) = 1.0 k. ft  M 2-3 = -(2+4)(7.5/10)(6) + (3)(7.5/(40-10))(6) = -22.5 k. ft 22

Change in M

C A

Case 1

10’ A C

Case 2

10’ A C

Case 3

10’ 30’ 30’ 30’ B 2K 4K 3K B 2K 4’ 4K 6’ 3K B 4’ 2K 4K 6’ 3K 4’ 6’ 23

Change in M

2K 4K 3K A C B

Case 1

A 10’ 2K 4K C

Case 2

2K A 10’ 4K 3K C 30’  M1-2 = -2(7.5/10)(4) + (4+3)(7.5/(40-10))(4) = 1.0 k. ft B

Case 3

10’ 3K 30’ 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.

M c 4.5

7.5

6 6’ 10’ 16’ 40’ x 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

F 1 F 2 x x’-x F R F 3 A Ay x’ d 1 d 2 L/2 L/2 

M b A y

  0 1

L

 

R

 

L

2  

x

' 

x

   B By 32

Example…

A F 1 L/2 - x d 1 V 2 M 2 

M b M

2   0

A y L

2

x



F

1

d

1  1

L

 

R

 

L

2  

x

' 

x

  

L

2

x



F

1

d

1 

F R L

 4

F R x

'  2

F R x

2

L



F R xx

' 

L F

1

d

1 33

Example…

A F 1 L/2 - x d 1 M 2 V 2 For Maximum M 2 we require

dM

2   2

dx F R L x



x



x

' 2

F R x

'

L

 0 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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