# ES3C7 Structural Design and Analysis SELF-LEARNING EXERCISE

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

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. hows

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

If you drew (a) go to Frame 7.
Assuming elastic (small displacement) behaviour, draw the deflected shape for the following beam If 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.

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

FRAME 3 Go to Frame 8.

Yes, the beam will deflect as shown:
FRAME 4 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.

Yes, the top side of the beam is in tension near the fixed
FRAME 5 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.

For the simply-supported beam the side of the beam in tension
FRAME 6 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, (b) go to Frame 10, (c) go to Frame 13, (d) go to Frame 9.

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 8 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 9 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.

It cannot hinge like this:
FRAME 10 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.

The supports for this beam
FRAME 11 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 12 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.

Yes, the beam will deflect as shown:
FRAME 13 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. If you drew: (a) go to Frame 8, (b) go to Frame 3, (c) go to Frame 5, (d) go to Frame 15.

To resist this, the moment at B must be clockwise:
FRAME 14 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 15 Go to Frame 8.

If a cantilever is loaded by an end point force:
FRAME 16 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 14 Go to Frame 20.

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

The bending moment is the maximum
FRAME 18 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, (b) maximum in the centre, zero at the ends; linear variation, go to Frame 24, (c) some other variation, go to Frame 22.

The bending moment diagrams are:
FRAME 19 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.

Yes, a clockwise moment must be applied at B.
FRAME 20 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 21 Go to Frame 27.

FRAME 22 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.

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

FRAME 24 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 25 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.

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

No, this is incorrect. The lengths of the side in tension are:
FRAME 27 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: Don’t understand why, go to Frame 16 and work through again. Otherwise go to Frame 25.

FRAME 28 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): 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.

(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.

FRAME 30 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.

This is the correct deflected shape: The b.m. diagram for
FRAME 31 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.

Yes, the correct shape is: Now draw the b.m. diagram. If you drew:
FRAME 32 Yes, 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.

As this joint compels one member end to rotate
FRAME 33 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. (b) go to Frame 38. (c) go to Frame 31.

FRAME 34 Go to Frame 52.

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 36 Go to Frame 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

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.

the deflected shape of this structure
FRAME 39 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. Provides support Free to move

Always remember to check the stability of structures!
FRAME 40 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 41 Go to Frame 46.

FRAME 42 Go to Frame 53.

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

Yes. You have chosen the correct b.m. diagram.
FRAME 44 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, (b) (c) go to Frame 42, (d) go to Frame 47. (e) Something else, go to Frame 54.

FRAME 45 Go to Frame 40.

The lengths of side in tension are:
FRAME 46 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 47 Go to Frame 53.

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.

FRAME 49 Go to Frame 46.

FRAME 50 Go to Frame 40.

Yes, you have chosen the correct deflected shape:
FRAME 51 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, (b) go to Frame 49, (c) go to Frame 41, (d) Something else, go to Frame 46.

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.

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

FRAME 54 Go to Frame 53.

Yes, this is the correct b.m. diagram:
FRAME 55 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 56 If 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.

FRAME 57 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. (See Frames 33 and 31.)

FRAME 58 Go to Frame 64.

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

FRAME 60 Go to Frame 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.

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

Now draw the b.m. diagram for this beam problem:
FRAME 63 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, (b) Something else, look again at Frame 55 again.

It is known that the deflected shape is:
FRAME 64 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.

Now draw the deflected shape and b.m. diagram for this structure:
FRAME 65 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.

You should get either: or
FRAME 66 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.

The structure will deflect as: and the b.m. diagram will be
FRAME 67 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 ‘VD‘ will lead to b.m. in the column, which will therefore bend Now go to Frame 61. H = 0

FRAME 68 Go to Frame 64.

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.

The END Acknowledgments:
FRAME 70 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 