1. GOVERNMENT ENGINEERING COLLEGE RAJKOT CIVIL ENGINEERING DEPARTMENT

2. INFLUNCE LINES FOR DETERMINATE STRUCTURE Prepared By : 130200106010 Chavda Sagar 130200106016 Gohel Maulik 130200106020 Godalaka Jaimin 130200106038 Mer Yogesh 130200106043 Pawar Divyam 130200106053 Sarpadadiya Darshan STRUCTURAL ANALYSIS - II

3. INFLUNCE LINE An influence for any given point or section of structure is a curve or graph whose ordinate represent to scale the variation of a function, such as shear force, bending moment, reaction at a support deflection etc. at the point or section as unit load moves across the structure. An influence line for any given point C on a structure is such a curve that its ordinate at any point D gives the shear force, bending moment, reaction or deflection at C. when a unit load is placed at D. For statically determinate structure, the influence lines for S.F., B.M. are composed of straight line while curvilinear for statically indeterminate structure.

4. DIFFERENCE BETWEEN S.F. OR B.M. DIAGRAME AND INFLUNCE LINE DIAGRAME Difference between S.F. or B.M. diagrams and influence line diagrams is that the ordinate of S.F. or B.M. diagram gives the values of the S.F. or B.M. at the section, where the ordinate has been drawn. While influence line, the ordinate at any point gives the value of the S.F. or B.M. at the given section only and not at the point at which the ordinate has been drawn. Also, there is only one single S.F. or B.M. curve under the action of a given set or train of loads, which there are infinite number of influence lines.

5. IMPORTANCE OF INFLUENCE LINE DIAGRAM In practice, structures are subjected to live loads (moving), dead load (fixed), and other loads. As the loads are moving, they will produce different shear, moments, etc. due to vibration in structure for their different positions. Any structure or structural element shall have to be designed for the most severe stress condition that can occur on it.

6. IMPORTANCE OF INFLUENCE LINE DIAGRAM Structural designer find out the load positions to produce the maximum stress in a structure. That can be done by influence line diagram. The influence lines are very useful in the speedy determination of the value of a function at the given section under any complex system of loading.

7. INFLUENCE LINE FOR REACTIONS

8. Let us consider a simply supported beam AB of span L. Consider the unit load, at a distance y from A. and I.L. for R a : At y = 0. R a = 1 At Y = L. R b = 0 I.L. for R b : At y = 0. R b = 0 At y = L. R a = 1 For intermediate values, the graph is straight line.

9. INFLUENCE LINE FOR SHEAR FORCE

10. Let us consider a simply supported beam AB of span L, and construct the influence line for S.F. at distance C distance x from A. The position of the section is fixed while the unit load moves from left to right. When unit load is in AC : when unit load is in portion AC, S.F. at c is given by, At C, y = x

11. INFLUENCE LINE FOR SHEAR FORCE

12. When unit load is in CB: When unit load is in portion C, S.F. at C is given by, At C, y = x, Thus, influence line for V c (S.F. at C) follows R b for potion AC and follows Ra for portion CB. When there is point load (W1) and y1 is the influence line ordinate under the load, we have When there is u.d.l. of length ‘a’, V c = intensity of u.d.l. X area of I.L. dia. Under u.d.l.

13. INFLUENCE LINE FOR BENDING MOMENT

14. Let us construct the I.L. for B.M. at C. Consider a section C of a beam AB of span L, at a distance of x from A. Let the unit load move across the beam from A to B. When unit load is in AC : Let unit load is at distance y from A : When y = x

15. INFLUENCE LINE FOR BENDING MOMENT When unit load in CB : When y = x, Thus, the I.L. diagram for Mc is a triangle having maximum ordinate under the section. When there is a single point W1 and y 1 is the influence line ordinate under the load. We have M c = W 1. y 1

16. INFLUENCE LINE FOR BENDING MOMENT If there are several point load W 1, W 2, W 3 …W n and corresponding I.L. ordinates are y 1, y 2, y 3 …..y n, we have M c = W 1. y 1 + W 2. y 2 + W 3. y 3 +…….. + W n.y n When there is an u.d.l. of length ‘a’. M c = intencity of u.d.l. * area of I.L. dia. Under u.d.l.

17. LOAD POSITION FOR MAXIMUM SF AT A SECTION

19. LOAD POSITION FOR MAXIMUM SF AT A SECTION 2) U.D.L. Longer than span: Max. Negative SF will occur when span AC is loaded and span CB is empty. Vc ( max – Ve) = -w. Area of I.L. Diameter under u.d.l. = -w. ½.x.x/L = -wx2/2L Max. positive SF will occur when span CB is loaded and span AC is empty. Vc ( max + Ve) = w.1/2(L-x).(L-x)/L = w (L-x)2/2L

20. LOAD POSITION FOR MAXIMUM SF AT A SECTION 3) U.D.L. Shorter than span: let u.d.l. Of length ‘a’ travel from left to right a<L. Max. Negative SF at C will occur when the head of the load reaches C. Max. positive SF at C will occur when the tail of the load is at C.

21. LOAD POSITION FOR MAXIMUM SF AT A SECTION

22. 4) Several point loads : a) Max. Negative SF at C: To get max. Negative SF at C,let the first load W1 be at the section C, and let another load W2 be at d behind it. If the loads are moved by a distance d to the right, so that W2 comes over C. the change δVc = W.d/L – W1 Where, δVc = change in SF W = sum of all the point load d = distance between load at C and the next load W1 = load at C L = span of the beam

23. LOAD POSITION FOR MAXIMUM SF AT A SECTION If δFc is positive rolling to the left will increase +ve SF in such a case rolling must be continue till δVc becomes –ve. Thus the load position which gives –ve δVc will be the correct position of loads to get the max. +ve SF at C.

24. LOAD POSITION FOR MAXIMUM SF AT A SECTION

25. b) Maximum +ve SF at C: To get the max. +ve SF at C, let the last loads W4 be at the section C and let another load W3 be at distance d a head of W4. If loads are moved by a distance d to the left so that W3 comes over C. the change δFc is given by δVc = W. d/L –W4

26. LOAD POSITION FOR MAXIMUM B.M. AT A SECTION :

28. LOAD POSITION FOR MAXIMUM B.M. AT A SECTION : 3) u.d.l shorter than span :

29. Let the u.d.l. of length 'a' rolls from left to right a < L. For maximum B.M. at C, the u.d.l. must be so placed over the beam that ordinates aa 1 and bb 1 are equal. Maximum area under u.d.l. will be obtained only if aa 1 = bb 1. For aa 1 = bb 1, the following condition must be satisfied.

31. POSITION AND AMOUNT OF MAXIMUM BM FOR SEVERAL POINT LOADS

32. We studied how to find out the absolute max. BM at a section when a series of point load crosses the girder. For the design of the beam, the absolute max. BM occurring anywhere in the beam should be found out. If the span of the beam is very large, we may assume that the absolute max. BM occur at the centre of the beam.

33. Preposition When a series of wheel loads crosses a girder, simply supported at the ends, the max. BM under any given will load occurs when the centre of the span is mid way between the CG of the load system and the wheel load under consideration. In figure the centre of the span (C ) is mid way between the resultant R and wheel load W3. this arrangement will give max BM below load W3.

34. Preposition The following points must be kept in mined to reduce the no. Of trials 1. The max. BM always occurs under a wheel load, and a not anywhere between two wheel loads. 2. The wheel load should be so selected that the centre of the span is midway between the CG of the load system and wheel load under consideration. 3. Absolute max. BM always occur at a section near the centre of the span. 4. The absolute max. BM generally occurs under the heavier wheel load which is near to the CG of the load system.

35. POSITION AND AMOUNT OF MAXIMUM S.F. FOR SEVERAL POINT LOADS Place the load (W 1 ) at B and the value of

36. POSITION AND AMOUNT OF MAXIMUM S.F. FOR SEVERAL POINT LOADS If value of δV c is +ve. Place the second load(w 2 ) at B and find the value of Where, P is the load leaving the girder here, P =W 1. The load place at B which gives the value of δV c = -ve is the critical load. This critical load placed at B will give maximum S.F. at B i.e.R b

37. POSITION AND AMOUNT OF MAXIMUM S.F. FOR SEVERAL POINT LOADS Similarly the place the last load (W 4 ) at A and find the value of δV c. If value of δV c is negative W 4 is the critical load and it should be placed at A to find the maximum S.F. at A. i.e. R a If W 4 at A gives +ve value of δV c try load W 3 at A and find δV c. Thus, maximum shear will occur at either end A or B ( i.e. R a or R b ).

40. (b) I.L. for B.M. at 3m from left :