1. 1 Influence Line Diagram -III Theory of Structures-I

2. 2 Contents 1. Maximum Influence at a point due to a series of concentrated loads 2. Absolute Maximum Shear and Moment

3. 3 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

4. 4 Shear 10’30’ CAB x VcVc -0.25 0.75 10’ 40’ ILD for Shear at Point C

5. 5 Case 1 10’30’ CAB 1K4K 5’

6. 6 Case 1 10’30’ CAB 1K4K x VcVc -0.25 0.75 10’ 40’ 15’20’ 0.625 0.5 (Vc) 1 = 1(0.75) + 4(0.625) + 4(0.5) = 5.25 k

7. 7 Case 2 10’30’ CAB 1K4K 5’

8. 8 Case 2 10’30’ CAB 1K4K x VcVc -0.25 0.75 10’ 40’ 15’ 0.625 (Vc) 2 = 1(-0.125) + 4(0. 75) + 4(0.625) = 5.375 k 5’ -0.125

9. 9 Case 3 10’30’ CAB 1K4K 5’

10. 10 Case 3 10’30’ CAB 1K4K x VcVc -0.25 0.75 10’ 40’ 15’ (Vc) 3 = 1(0) + 4(-0.125) + 4(0.75) = 2.5 k 5’ -0.125

11. 11 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

12. 12 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.

13. 13 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.

14. 14 Method Based on Change in Function … The change in shear  V for a load P that moves from position x 1 to x 2 over a beam can be determined by 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 )

15. 15 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 )

16. 16 Case 1-2 10’30’ CAB 1K4K 5’

17. 17 Case 1-2 10’30’ CAB 1K4K x VcVc -0.25 0.75 10’ 40’ 15’ 0.625  V 1-2 = 1(-1) + [1+4+4] (0.025)(5)= +0.125 k 5’ -0.125 S = 0.75/(40-10) = 0.25/10 = 0.025

18. 18 Case 2-3 10’30’ CAB 1K4K 5’

19. 19 Case 2-3 10’30’ CAB 1K4K x VcVc -0.25 0.75 10’ 40’ 15’ 5’ -0.125  V 2-3 = 4(-1) + [1+4+4] (0.025)(5)= -2.875 k

20. 20 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.

21. 21 Moment 10’30’ CAB x McMc 10’ 40’ ILD for Moment at Point C 7.5

22. 22 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

23. 23 Change in M CAB 2K4K3K 4’6’ Case 1 CAB 2K4K3K 4’6’ Case 2 CAB 2K4K3K 4’6’ Case 3 10’30’ 10’30’ 10’30’

24. 24 Change in M CAB 2K4K3K Case 1 CAB 2K4K3K Case 2 CAB 2K4K3K Case 3 10’30’ 10’30’ 10’30’  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

25. 25 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. x McMc 10’ 40’ 7.5 4.5 6 6’ 16’

26. 26 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.

27. 27 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

28. 28 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

29. 29 Moment in Cantilever Beam M abs max

30. 30 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. AB V abs max

31. 31 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.

32. 32 Example AB C LFRFR F3F3 F2F2 F1F1 L/2 x x’-x x’ d1d1 d2d2 AyBy

33. 33 Example… A F1F1 L/2 - x d1d1 V2V2 M2M2

34. 34 Example… A F1F1 L/2 - x d1d1 V2V2 M2M2 For Maximum M 2 we require

35. 35 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.

36. 36