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ES3C7 Structural Design and Analysis SELF-LEARNING EXERCISE Deflected Shape and Bending Moments Engineering Year 3: Civil Engineering.

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Presentation on theme: "ES3C7 Structural Design and Analysis SELF-LEARNING EXERCISE Deflected Shape and Bending Moments Engineering Year 3: Civil Engineering."— Presentation transcript:

1 ES3C7 Structural Design and Analysis SELF-LEARNING EXERCISE Deflected Shape and Bending Moments Engineering Year 3: Civil Engineering

2 Introduction The following self-learning exercise, of 69 frames, is branched so that you can be given further instructions on points you do not understand. It has been scrambled to help you resist the temptation to look at the next frame to find the answer to questions. You will therefore find that you jump from frame to frame. Your drawings are to be done free-hand and to reveal possible answers use the ENTER key for each fly-in from the left side. LEARNING OUTCOME: On successful completion of the exercise you should be able to draw the deflected shape and bending moment diagram for a variety of structures without calculation.

3 To study the deflected shape of simply-supported beam begin at FRAME 1FRAME 1 To study the deflected shape of a propped cantilever beam begin at FRAME 6FRAME 6 To study bending moments in a cantilever beam begin at FRAME 16FRAME 16 To study bending moment diagrams of a simply supported beam begin at FRAME 18FRAME 18 To study effect of rigid joints begin at FRAME 30FRAME 30 To study effect of pinned joints begin at FRAME 37FRAME 37 To study effect of roller supports begin at FRAME 39FRAME 39 To study bending moment diagrams of a portal frame begin at FRAME 44FRAME 44 To study uniformly distributed loading begin at FRAME 55FRAME 55 To study several point loads begin at FRAME 29FRAME 29 To study more complex structures and loading begin at FRAME 67FRAME 67 Summary ( for later reference )

4 Assuming elastic (small displacement) behaviour, draw the deflected shape for the following beam If you drew (a)go to Frame 7.Frame 7 If you drew (b)go to Frame 9.Frame 9 If you drew (c)go to Frame 4.Frame 4 If you drew (d) some other shape, go to Frame 12.Frame 12 FRAME 1

5 This is incorrect. If the member is elastic it will NOT kink at any point along its length. Return to Frame 6.Frame 6 FRAME 2

6 Go to Frame 8.Frame 8 FRAME 3

7 Yes, the beam will deflect as shown: In this position the beam is compressed along the top edge and stretched along the bottom edge. Now go to Frame 6.Frame 6 FRAME 4

8 Yes, the top side of the beam is in tension near the fixed support, and on the underside elsewhere: Now returning to the previous example; it can be seen that when the beam deflects, neither support resists the bending (unlike the propped cantilever beam). Now go to Frame 11.Frame 11 FRAME 5

9 For the simply-supported beam the side of the beam in tension can be labelled as shown, i.e. with a T on the edge that would be convex when bent. Now draw the deflected shape for the propped cantilever beam: If you drew (a)go to Frame 2,Frame 2 (b)go to Frame 10,Frame 10 (c)go to Frame 13,Frame 13 (d)go to Frame 9.Frame 9 FRAME 6

10 This is incorrect. You must assume linear elastic behaviour. For the beam to deflect like: it must be hinged in the middle, due to the presence there of a real hinge, or due to material yielding. If you are still uncertain, try flexing a ruler or strip of stiff card. Now return to Frame 1, and try again!Frame 1 FRAME 7

11 You have given the wrong distribution of tension. If we look at the deflected shape: Given the convex side of the beam is in tension, (i.e. the outside of the curve), tension can be labelled thus for this case: Now got to Frame 5.Frame 5 FRAME 8

12 No, this is not correct. The supports shown above are knife-edges which do not restrain rotation at the ends of the beam. So as it bends the ends will deflect so: Now return to Frame 1 (if you last attempted the simply- supported beam) or Frame 6 and try again.Frame 1 Frame 6 FRAME 9

13 No, this is not correct. At a fixed end the beam is held against rotation. It cannot hinge like this: if it is linear elastic. Instead, the beam will deflect thus: Now go to Frame 13.Frame 13 FRAME 10

14 The supports for this beam are knife-edges which do not restrain rotation at the ends of the beam. So as it bends, the ends will deflect thus: Now got to Frame 16.Frame 16 FRAME 11

15 No, the correct deflected shape is one of (a), (b), or (c). If you cannot see why, try each in turn and read through the relevant frames to find out why each is right or wrong. So now return to Frame 1.Frame 1 FRAME 12

16 Yes, the beam will deflect as shown: Now label those lengths of side that will be in tension, as shown for the simple supported beam in Frame 6.Frame 6 If you drew: (a) go to Frame 8,Frame 8 (b) go to Frame 3,Frame 3 (c) go to Frame 5,Frame 5 (d) go to Frame 15.Frame 15 FRAME 13

17 No, this is incorrect. The turning effect produced by the forces at A and B is anti-clockwise: To resist this, the moment at B must be clockwise: Now return to Frame 16 and try again.Frame 16 FRAME 14

18 Go to Frame 8.Frame 8 FRAME 15

19 If a cantilever is loaded by an end point force: then to keep end B fixed, a moment M has to be applied by the support to the cantilever, as shown. Which way should the end bending moment be applied to fix end B against rotation in this case: If you think (a)(b) Go to Frame 14Go to Frame 20.Frame 14Frame 20 FRAME 16

20 Now draw the bending moment diagrams for the structure shown in Frame 56, and then go to Frame 19.Frame 56 FRAME 17

21 The bending moment can be measured at any point along a member. In the case of the cantilever: The bending moment is the maximum at fixed end, and varies linearly to zero at the free end. How does the moment vary along this simply-supported beam? If you think it varies: (a) with maximum moment at the ends, and zero moment in the centre (linear variation between these points), go to Frame 22,Frame 22 (b) maximum in the centre, zero at the ends; linear variation, go to Frame 24,Frame 24 (c) some other variation, go to Frame 22.Frame 22 FRAME 18

22 The bending moment diagrams are: If you did not draw these correctly and are still unsure of why you were wrong, you may benefit from reading the whole programme again; otherwise go to Frame 29.Frame 29 FRAME 19

23 Yes, a clockwise moment must be applied at B. In the case of the simply-supported beam, the member can be visualised as TWO cantilevers joined together back-to-back in the middle, where the slope is zero. Now go to Frame 18.Frame 18 FRAME 20

24 FRAME 21 Go to Frame 27.Frame 27

25 You are wrong. If two cantilever moment diagrams are drawn back-to-back they show the diagram for the simply-supported beam. Now go to Frame 24.Frame 24 FRAME 22

26 Yes, this is correct: Now continue with Frame 25.Frame 25 FRAME 23

27 Yes, the bending moment will vary linearly from zero at the ends to a maximum at the centre. We can plot the moment against distance along the beam, using the beam line as the datum for moment, to obtain the bending moment diagram. For this case it is: Now go to Frame 28.Frame 28 FRAME 24

28 In this programme of directed learning we are considering two-dimensional structures which contain a variety of joints and support conditions. These have differing effects on the deflected shape and bending moment diagram. Now got to Frame 30.Frame 30 FRAME 25

29 The deflected shape, and lengths of side in tension are: If you drew: (a)go to Frame 23,Frame 23 (b)go to Frame 21,Frame 21 (c)go to Frame 27.Frame 27 FRAME 26

30 No, this is incorrect. The lengths of the side in tension are: Therefore bending moment diagram must cross from one side to the other, at some position to the left of the applied load. The correct shape is therefore: Dont understand why, go to Frame 16 and workFrame 16 through again. Otherwise go to Frame 25.Frame 25 FRAME 27

31 The sign convention is the moment diagram is always drawn on the side of the member that is in tension due to bending. This can be seen in the last slide (Frame 27):Frame 27 Here the beam is always in tension on the underside, and the b.m. diagram is drawnbelow the beam. However, if there are lengths of tension on the top side of the member, then the b.m. diagram would cross to that side. Now sketch the deflected shape, label the lengths of side in tension, and draw the b.m. diagram for this propped cantilever: and go to Frame 26.Frame 26 FRAME 28

32 So far, we have considered situations with only one point load. In the next few slides, more complex loading is considered. For example, sketch the deflected shape of the following two-span continuous beam, and label the lengths of side in tension. It should be assumed that the effect of loading on the left-hand span is similar to that on the right-hand span. If you drew: (a) go to Frame 32,Frame 32 (b) go to Frame 34,Frame 34 (c) Something else, go to Frame 52.Frame 52 FRAME 29

33 Rigid joints. A perfectly rigid joint keeps the angles, between the members meeting at that joint, constant. It is denoted thus:However, the joint may of course rotate when the deforms, forcing the member ends to rotate by the same amount: Angle at joint between members remains constant. Now go to Frame 33.Frame 33 FRAME 30

34 This is the correct deflected shape: The b.m. diagram for this condition is: Note that the b.m. diagram for each member is drawn with that member as datum. Obviously, for joint A to be in equilibrium the two member end moments must be equal in magnitude, but opposite direction. This is a useful check to see whether the diagram is correct or not. Note also that it is usual to determine b.m. diagrams on the underformed state of the structure. The moment in the horizontal member is therefore constant as the lever arm of the force is constant along this member. Now go to Frame 37.Frame 37 FRAME 31

35 Yes, the correct shape is: Now draw the b.m. diagram. If you drew: (a) go to Frame 55,Frame 55 (b) go to Frame 60,Frame 60 (c) go to Frame 64,Frame 64 (d) Something else, go to Frame 68.Frame 68 FRAME 32

36 As this joint compels one member end to rotate by the same amount as the other, it transmits moment from one member to the other, just as if they formed one continuous member. However, the b.m. diagram is still drawn perpendicular to each member. Draw the deflected shape for this: and label the lengths of side in tension. If you drew: (a)go to Frame 35.Frame 35 (b)go to Frame 38.Frame 38 (c)go to Frame 31.Frame 31 FRAME 33

37 FRAME 34 Go to Frame 52.Frame 52

38 No. For the frame to deflect in the way you have chosen, there would have to be a vertical upward force: Return to Frame 33 and try again.Frame 33 FRAME 35

39 FRAME 36 Go to Frame 40.Frame 40

40 FRAME 37 Pinned Joints. A pinned joint acts like a hinge, and the connected members can rotate about it. Pinned joints do not transmit moment, and so the moment at the pinned end(s) of a member must be zero. Draw the b.m. diagram for this structure: If you drew: (a)go to Frame 40,Frame 40 (b)go to Frame 36,Frame 36 (c)go to Frame 50,Frame 50 (d) Something else, go to Frame 45.Frame 45 denotes pinned joint

41 This is not correct. The rigid joint will transmit moment to the horizontal member which is also flexible and will therefore deflect in flexure. Now return to Frame 33 and try again.Frame 33 FRAME 38

42 Roller Supports. The roller support is a knife-edge free to move in one direction, but giving support in the perpendicular direction: In the exercises to follow, it should be assumed that such supports are also capable of holding-down the members to which they are attached. Now draw the deflected shape of this structure and label the side lengths in tension. Then turn to Frame 43.Frame 43 FRAME 39 Provides support Free to move

43 Due to the pinned joints, this structure is a mechanism, and is not in static equilibrium. Always remember to check the stability of structures! Now go to Frame 39.Frame 39 FRAME 40

44 FRAME 41 Go to Frame 46.Frame 46

45 FRAME 42 Go to Frame 53.Frame 53

46 If you drew: (a)go to Frame 48,Frame 48 (b)go to Frame 51,Frame 51 (c)go to Frame 57.Frame 57 FRAME 43

47 Yes. You have chosen the correct b.m. diagram. Following the same procedure, draw the b.m. diagram for: If you drew: (a) go to Frame 53,Frame 53 (b) go to Frame 53,Frame 53 (c) go to Frame 42,Frame 42 (d) go to Frame 47.Frame 47 (e) Something else, go to Frame 54.Frame 54 FRAME 44

48 FRAME 45 Go to Frame 40.Frame 40

49 The lengths of side in tension are: Since the b.m. diagram must be drawn at the tension side it must cross-over the horizontal member, but stay on the right of the vertical Thus:prop. The moment at the roller is zero. Now go to Frame 44.Frame 44 FRAME 46

50 FRAME 47 Go to Frame 53.Frame 53

51 No, the frame does not bend at the knee joint. This is a rigid joint as described in Frame 30. You may benefit from reading from Frame 30 again.Frame 30 Otherwise return to Frame 43.Frame 43 FRAME 48

52 FRAME 49 Go to Frame 46.Frame 46

53 FRAME 50 Go to Frame 40.Frame 40

54 Yes, you have chosen the correct deflected shape: Now use the deflected shape to sketch the b.m. diagram If you drew: (a)go to Frame 44,Frame 44 (b)go to Frame 49,Frame 49 (c)go to Frame 41,Frame 41 (d) Something else, go to Frame 46.Frame 46 FRAME 51

55 No, you are not correct. The beam is continuous over the central support so there should be no sharp changes of slope. As the effects of loading on the two spans are similar, both spans would have some lengths where the underside is in tension. [However, if, say, the left span was unloaded, then the shape would be:] Now return to Frame 29 and try again.Frame 29 FRAME 52

56 The correct solution is: The deflected shape and length of sides are: Now go to Frame 56.Frame 56 FRAME 53

57 FRAME 54 Go to Frame 53.Frame 53

58 Yes, this is the correct b.m. diagram: Now if the number of point loads is increased, the diagram becomes: Now go to Frame 63.Frame 63 FRAME 55

59 If arrived here from Frame 17 go to Frame 19 after drawing b.m.Frame 17 Frame 19 Now draw the deflected shape for these two structures, labelling those lengths of the side in tension. Then turn to Frame 59.Frame 59 FRAME 56

60 No, for the frame to deflect in the manner you have chosen, there would need to be no end support at the vertical member. However, the roller support on the inclined face forces the vertical member upwards when the horizontal force is applied. Now return to Frame 39 and try again.Frame 39 FRAME 57 (See Frames 33Frames 33 and 31.)31

61 FRAME 58 Go to Frame 64.Frame 64

62 The deflected shapes are: If you are puzzled, try reading through from Frame 30 again.Frame 30 Otherwise turn to Frame 17.Frame 17 FRAME 59

63 FRAME 60 Go to Frame 64.Frame 64

64 If arrived here from Frame 62 go to Frame 69 after drawing b.m.Frame 62 Frame 69 If arrived here from Frame 67 draw the deflected shape and label those lengths of side in tension in this final example:Frame 67 Now turn to Frame 66.Frame 66 FRAME 61

65 Now draw the b.m. diagram for the structure shown in Frame 61,Frame 61 If b.m diagram is drawn go to Frame 69.Frame 69 FRAME 62

66 Now draw the b.m. diagram for this beam problem: The loading on the right-hand span is uniformly distributed. If you drew: (a)go to Frame 65,Frame 65 (b) Something else, look again at Frame 55 again.Frame 55 FRAME 63

67 It is known that the deflected shape is: and also that the b.m. diagram is drawn on this side in tension. We also know that the b.m. is zero at the extreme ends of the beam, which are simply-supported, and that a linear variation occurs between supports and/or load points. So, by inspection, we can draw the b.m. diagram: Now go to Frame 55.Frame 55 FRAME 64

68 Correct. Uniformly distributed loading leads to a parabolic b.m. diagram. Now draw the deflected shape and b.m. diagram for this structure: Go to Frame 67.Frame 67 FRAME 65

69 You should get either: or The correct shape depends upon the relative effects of the horizontal and vertical loading. If vertical load dominates, (b) is correct. Now go to Frame 62.Frame 62 FRAME 66

70 The structure will deflect as: and the b.m. diagram will be The horizontal reaction at A is zero, and therefore if the moments are to be determined on the undeformed shape, then there is zero moment in the column. The beam then behaves as a propped cantilever. If the moments are determined on the deformed shape then moment V will lead to b.m. in the column, which will therefore bend. Now go to Frame 61.Frame 61 FRAME 67 H = 0

71 FRAME 68 Go to Frame 64.Frame 64

72 The b.m. diagrams are: To correspond to the deflected shapes in Frame 66.Frame 66 If you have already seen Frame 66 the exercise has ended, and so you go to Frame 70.Frame 70 FRAME 69

73 The END Acknowledgments: This programme of self-learning was developed pre-1987 by a third year undergraduate student as part of a third year individual project. Dr Mottram thanks this unknown student and supervisor (Ian May, now Professor at Heriot-Watt Univ.) for their contribution so that we have this excellent self-learning programme on deflected shape and bending moments of two- dimensional structures. J. T. Mottram Sept FRAME 70


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