Presentation on theme: "ES3C7 Structural Design and Analysis SELF-LEARNING EXERCISE"— Presentation transcript:
1 ES3C7 Structural Design and Analysis SELF-LEARNING EXERCISE Deflected Shape and Bending MomentsEngineering Year 3: Civil Engineering
2 IntroductionThe 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.hows
3 Summary (for later reference) To study the deflected shape of simply-supported beam begin at FRAME 1To study the deflected shape of a propped cantilever beam begin at FRAME 6To study bending moments in a cantilever beam begin at FRAME 16To study bending moment diagrams of a simplysupported beam begin at FRAME 18To study effect of rigid joints begin at FRAME 30To study effect of pinned joints begin at FRAME 37To study effect of roller supports begin at FRAME 39To study bending moment diagrams of a portal frame begin at FRAME 44To study uniformly distributed loading begin at FRAME 55To study several point loads begin at FRAME 29To study more complex structures and loading begin at FRAME 67
4 If you drew (a) go to Frame 7. Assuming elastic (small displacement) behaviour, draw the deflected shape for the following beamIf you drew (a) go to Frame 7.If you drew (b) go to Frame 9.If you drew (c) go to Frame 4.If you drew (d) some other shape, go to Frame 12.
5 FRAME 2This is incorrect. If the member is elastic it will NOT ‘kink’ at any point along its length.Return to Frame 6.
7 Yes, the beam will deflect as shown: FRAME 4Yes, the beam will deflectas shown:In this position the beam is compressed along the top edge and stretched along the bottom edge.Now go to Frame 6.
8 Yes, the top side of the beam is in tension near the fixed FRAME 5Yes, the top side of the beamis in tension near the fixedsupport, and on the undersideelsewhere:Now returning to the previousexample;it can be seen that when the beam deflects, neither support resists the bending (unlike the propped cantilever beam).Now go to Frame 11.
9 For the simply-supported beam the side of the beam in tension FRAME 6For the simply-supported beamthe side of the beam in tensioncan be labelled as shown, i.e. with a ‘T’ on the edge that would be convex when bent.Now draw the deflected shapefor the propped cantilever beam:If you drew (a) go to Frame 2,(b) go to Frame 10,(c) go to Frame 13,(d) go to Frame 9.
10 Now return to Frame 1, and try again! 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!
11 FRAME 8You 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.
12 FRAME 9No, 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.
13 It cannot hinge like this: FRAME 10No, 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.
14 The supports for this beam FRAME 11The supports for this beamare knife-edges which do not restrain rotation at the ends of the beam.So as it bends, the endswill deflect thus:Now got to Frame 16.
15 FRAME 12No, 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.
16 Yes, the beam will deflect as shown: FRAME 13Yes, the beam will deflect asshown:Now label those lengths of side that will be in tension, as shown for the simple supported beam in Frame 6.If you drew: (a) go to Frame 8,(b) go to Frame 3,(c) go to Frame 5,(d) go to Frame 15.
17 To resist this, the moment at B must be clockwise: FRAME 14No, 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.
19 If a cantilever is loaded by an end point force: FRAME 16If a cantilever is loaded by an endpoint force:then to keep end B ‘fixed’, a momentM has to be applied by the support tothe 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 14 Go to Frame 20.
20 FRAME 17Now draw the bending moment diagrams for the structure shown in Frame 56, and then go to Frame 19.
21 The bending moment is the maximum FRAME 18The bending moment can be measured at any point along a member. In the case of the cantilever:The bending moment is the maximumat 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,(b) maximum in the centre, zero at the ends; linear variation,go to Frame 24,(c) some other variation, go to Frame 22.
22 The bending moment diagrams are: FRAME 19The bending moment diagrams are:If you did not draw these correctlyand are still unsure of why you werewrong, you may benefit fromreading the whole programme again; otherwise go to Frame 29.
23 Yes, a clockwise moment must be applied at B. FRAME 20Yes, a clockwise moment must beapplied 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.
25 FRAME 22You 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.
26 Now continue with Frame 25. Yes, this is correct:Now continue with Frame 25.
27 FRAME 24Yes, 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.
28 FRAME 25In 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.
29 The deflected shape, and lengths of side in tension are: If you drew: FRAME 26The deflected shape, and lengthsof side in tension are:If you drew:(a) go to Frame 23,go to Frame 21,(c) go to Frame 27.
30 No, this is incorrect. The lengths of the side in tension are: FRAME 27No, this is incorrect. The lengthsof 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:Don’t understand why,go to Frame 16 and workthrough again.Otherwise go to Frame 25.
31 FRAME 28The 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):Here the beam is always in tension on the underside, and the b.m. diagram is drawn below 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.
32 (c) Something else, go to Frame 52. 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,(b) go to Frame 34,(c) Something else, go to Frame 52.
33 FRAME 30Rigid joints. A perfectly rigid joint keeps the angles, between the members meeting at that joint, constant. It is denotedthus: However, the joint may of course rotate when the deforms, forcing the member ends to rotate by the same amount:Angle at jointbetween membersremains constant.Now go to Frame 33.
34 This is the correct deflected shape: The b.m. diagram for FRAME 31This is the correct deflected shape:The b.m. diagram forthis 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.
35 Yes, the correct shape is: Now draw the b.m. diagram. If you drew: FRAME 32Yes, the correct shape is:Now draw the b.m. diagram.If you drew:(a) go to Frame 55,(b) go to Frame 60,(c) go to Frame 64,(d) Something else, go to Frame 68.
36 As this joint compels one member end to rotate FRAME 33As this joint compels one member end to rotateby the same amount as the other, it transmitsmoment from one member to the other, just as ifthey 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.(b) go to Frame 38.(c) go to Frame 31.
40 If you drew: (a) go to Frame 40, (b) go to Frame 36, 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,(b) go to Frame 36,(c) go to Frame 50,(d) Something else, go to Frame 45.denotes pinned joint
41 Now return to Frame 33 and try again. 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.
42 the deflected shape of this structure FRAME 39Roller 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 drawthe deflected shape of this structureand label the side lengths in tension.Then turn to Frame 43.Provides supportFree to move
43 Always remember to check the stability of structures! FRAME 40Due 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.
46 If you drew: (a) go to Frame 48, (b) go to Frame 51,(c) go to Frame 57.
47 Yes. You have chosen the correct b.m. diagram. FRAME 44Yes. 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,(b)(c)go to Frame 42,(d)go to Frame 47.(e) Something else, go to Frame 54.
49 The lengths of side in tension are: FRAME 46The lengths of side in tension are:Since the b.m. diagram must be drawn at the tension sideit must cross-over the horizontal member, but stayon the right of the vertical Thus: prop. The moment at the roller is zero.Now go to Frame 44.
51 Otherwise return to Frame 43. 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.Otherwise return to Frame 43.
54 Yes, you have chosen the correct deflected shape: FRAME 51Yes, you have chosen the correct deflected shape:Now use the deflected shape to sketch the b.m.diagramIf you drew: (a) go to Frame 44,(b) go to Frame 49,(c) go to Frame 41,(d) Something else, go to Frame 46.
55 Now return to Frame 29 and try again. 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.
56 The correct solution is: FRAME 53The correct solution is:The deflected shape and length of sides are:Now go to Frame 56.
58 Yes, this is the correct b.m. diagram: FRAME 55Yes, this is the correct b.m. diagram:Now if the number of point loads is increased, the diagram becomes:Now go to Frame 63.
59 FRAME 56If arrived here from Frame 17 go to Frame 19 after drawing b.m.Now draw the deflected shape for these two structures, labelling those lengths of the side in tension.Then turn to Frame 59.
60 FRAME 57No, 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.(See Frames 33and 31.)
64 If arrived here from Frame 62 go to Frame 69 after drawing b.m. If arrived here from Frame 67 draw the deflected shape and label those lengths of side in tension in this final example:Now turn to Frame 66.
65 Now draw the b.m. diagram for the structure shown in Frame 61, If b.m diagram is drawn go to Frame 69.
66 Now draw the b.m. diagram for this beam problem: FRAME 63Now 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,(b) Something else, look again at Frame 55 again.
67 It is known that the deflected shape is: FRAME 64It 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.
68 Now draw the deflected shape and b.m. diagram for this structure: FRAME 65Correct. 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.
69 You should get either: or FRAME 66You should get either:orThe 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.
70 The structure will deflect as: and the b.m. diagram will be FRAME 67The structure will deflect as:and the b.m. diagram will beThe 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 ‘VD‘ will lead to b.m. in the column, which will therefore bend Now go to Frame 61.H = 0
72 To correspond to the deflected shapes in Frame 66. The b.m. diagrams are:To correspond to the deflected shapes in Frame 66.If you have already seen Frame 66 the exercise has ended, and so you go to Frame 70.
73 The END Acknowledgments: FRAME 70The ENDAcknowledgments: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