shear force and bending momentss(1)

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Engineering Mechanics – Statics For 1st Year Students – Mechanic Department29

th Week By : Eng . YOUNIS FAKHER 2010 ‐ 2011

Shear Force and Bending Moments Part 1

Example Problem

A beam is loaded and supported as shown in Fig. 1. For this beam

a. Draw complete shear force and bending moment diagrams.

b. Determine the equations for the shear force and the bending moment as functions of x.

Fig. 1

Solution

Overall Equilibrium

We start by drawing a free-body diagram (Fig. 2) of the beam and determining the support reactions. Summing moments about the left end of the beam

MA = 7RC - 2 [ 4 × 10 ] - 4(16) - 9(19) = 0 (1a)

gives RC = 45 kN (1b)

Then, summing forces in the vertical direction F = RA + RC - 4 × 10 - 16 - 19 = 0 (2a)

gives RA = 30 kN (2b)

Fig. 2

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Drawing the Shear Force Diagram

Sometimes we are not so much interested in the equations for the shear force and bending moment as we are in knowing the maximum and minimum values or the values at some particular point. In these cases, we want a quick and efficient method of generating the shear force and bending moment diagrams (graphs) so we can easily find the maximum and minimum values. That is the subject of this first part of the problem.

Concentrated Force

The 30-kN concentrated force (support reaction) at the left end of the beam causes the shear force graph to jump up (in the direction of the force) by 30 kN (the magnitude of the force) from 0 kN to 30 kN.

Fig. 3

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Distributed Load

The downward distributed load causes the shear force graph to slope downward (in the direction of the load). Since the distributed load is constant, the slope of the shear force graph is constant (dV/dx = w = constant).

The total change in the shear force graph between points A and B is 40 kN (equal to the area under the distributed load between points A and B) from +30 kN to -10 kN.

We also need to know where the shear force becomes zero. We know that the full 4 m of the distributed load causes a change in the shear force of 40 kN. So how much of the distributed load will it take to cause a change of 30 kN (from +30 kN to 0 kN)? Since the distributed load is uniform, the area (change in shear force) is just 10 × b = 30, which gives b = 3 m. That is, the shear force graph becomes zero at x = 3 m (3 m from the beginning of the uniform distributed load).

Concentrated Force

The 16-kN concentrated force at B causes the shear force graph to jump down (in the direction of the force) by 16 kN (the magnitude of the force) from -10 kN to -26 kN.

Fig. 4

Fig. 5

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No Loads

Since there are no loads between points B and C, the shear force graph is constant (the slope dV/dx = w = 0) at -26 kN.

Concentrated Force

The 45-kN concentrated force (support reaction) at C causes the shear force graph to jump up (in the direction of the force) by 45 kN (the magnitude of the force) from -26 kN to +19 kN.

Fig. 6

Fig. 7

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No Loads

Since there are no loads between points C and D, the shear force graph is constant (the slope dV/dx = w = 0) at +19 kN.

Concentrated Force

The 19-kN concentrated force at D causes the shear force graph to jump down (in the direction of the force) by 19 kN (the magnitude of the force) from +19 kN to 0 kN.

Fig. 8

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two points. The area of the rectangle is just M = (+19 × 2) = +38 kN·m. So the value of the bending moment at x = 7 m is M = -38 + 38 = 0 kN·m.

Determining the Shear Force and Bending Moment Equations

Sometimes we are not so much interested in the graphs of the shear force and bending moment as we are in knowing the equations. In particular, we need to integrate the equation for the bending moment to determine the shape of beam and how much the beam will bend as a result of the loads. That is the subject of the second part of this problem.

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Shear Force and Bending Moments Part 2

Solution (Cont.)

Determining the Shear Force and Bending Moment Equations

The easiest way to get the equations for the shear force and bending moment as functions of the position x is to use equilibrium.

0 m < x < 4 m

Figure 13 shows a free-body diagram of the left end of the beam to an arbitrary position, 0 m < x < 4 m. The right-hand portion of the beam that has been discarded exerts a shear force and a bending moment on the left-hand portion of the beam as shown. Summing forces in the vertical direction

F = 30 - (10x ) - V = 0 3agives

V = 30 - (10x ) kN (0 m < x < 4 m) 3b Summing moments about a point on the "cut end" of the beam

Mcut = M + (10x )(x/2) - 30x = 0 4agives

M = 30x - (5x 2) kN·m (0 m < x < 4 m) 4b 4 m < x < 7 m


Fig. 13

Fig. 14

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F = 30 - (10 × 4) - 16 - V = 0 5agives

V = -26 kN (4 m < x < 7 m) 5b Summing moments about a point on the "cut end" of the beam

Mcut = M + (10 × 4)(x - 2) + 16(x - 4) - 30x = 0 6agives

M = 144 - 26x kN·m (4 m < x < 7 m) 6b 7 m < x < 9 m


F = 30 - (10 × 4) - 16 - 45 - V = 0 7agives

V = 19 kN (7 m < x < 9 m) 7b Summing moments about a point on the "cut end" of the beam

Mcut = M + (10 × 4)(x - 2) + 16(x - 4) - 30x - 45(x - 7) = 0 8agives

M = 171 - 19x kN·m (7 m < x < 9 m) 8b

It is easily verified that these equations have the appropriate character to match the shear force and bending moment diagrams developed in the first part of this problem. It is also easily verified that these equations match the previous graphs at the points x = 0 m, x = 3 m, x = 4 m, x = 7 m, and x = 9 m.

Finally, note that these equations satisfy the load-shear force-bending moment relationships

dV/dx = w dM/dx = V

Fig. 15

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Shear Force and Bending Moment Diagrams

What is shear force? Below a force of 10N is exerted at point A on a beam. This is an external force. However because the beam is a rigid structure,the force will be internally transferred all along the beam. This internal force is known as shear force. The shear force between point A and B is usually plotted on a shear force diagram. As the shear force is 10N all along the beam, the plot is just a straight line:

The idea of shear force might seem odd, maybe this example will help clarify. Imagine pushing a 10cm cube along a kitchen table, with a 10N force. Even though you're applying the force only at one point on the cube, it's not just that point of the cube that moves forward. The whole cube moves forward, which tells you that the force must have transferred all along the cube, such that every atom of the cube is experiencing this 10N force.

Another way to think about shear force: If you slice the cube in half halfway up, you can push the top half off. The shear force is the force that must be applied by glue if you want to stick it back together.

Basic shear diagram What if there is more than one force, as shown in the diagram below, what would the shear force diagram look like then?

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The way you go about this is by figuring out the shear force at points A,B,C,E (as there is an external force acting at these points). The way you work out the shear stress at any point, is by covering (either with your hand or a piece of paper), everything to right of that point, and simply adding up the external forces. Then plot the point on the shear force diagram. For illustration purposes, this is done for point D:

Shear force at D = 10N - 20N + 40N = 30N

Now, let's do this for point B. BUT - slight complication - there's a force acting at point B, are you going to include it? The answer is both yes and no. You need to take 2 measurements. Firstly put your piece of paper, so it's JUST before point B:

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Shear force at B = 10N

Now place your paper JUST after point B:

Shear force at B = 10N - 20N = -10N

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(B' is vertically below B)

Now, do point A, D and E, and finally join the points. your diagram should look like the one below. If you don't understand why, leave a message on the discussion section of this page (its at the top), I will elaborate on the explanation:

Notice how nothing exciting happens at point D, which is why you wouldn't normally analyse the shear force at that point. For clarity, when doing these diagrams it is recommended you move you paper from left to right, and hence analyse points A,B, C, and E, in that order. You can also do this procedure covering the left side instead of the right, your diagram will be "upside down" though. Both diagrams are correct.

Basic bending moment diagram Bending moment refers to the internal moment that causes something to bend. When you bend a ruler, even though apply the forces/moments at the ends of the ruler, bending occurs all along the ruler, which indicates that there is a bending moment acting all along the ruler. Hence bending moment is shown on a bending moment diagram. The same case from before will be used here:

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To work out the bending moment at any point, cover (with a piece of paper) everything to the right of that point, and take moments about that point. (I will take clockwise moments to be positive). To illustrate, I shall work out the bending moment at point C:

Bending moment at C = 10x3 - 20x2 = -10Nm

Notice that there's no need to work out the bending moment "just before and just after" point C, (as in the case for the shear force diagram). This is because the 40N force at point C exerts no moment about point C, either way.

Repeating the procedure for points A,B and E, and joining all the points:

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Normally you would expect the diagram to start and end at zero, in this case it doesn't. This is my fault, and it happened because I accidentally chose my forces such that there is a moment disequilibrium. i.e. take moments about any point (without covering the right of the point), and you'll notice that the moments aren't balanced, as they should be. It also means that if you're covering the left side as opposed to the right, you will get a completely different diagram. Sorry about this...

Point moments Point moments are something that you may not have come across before. Below, a point moment of 20Nm is exerted at point C. Work out the reaction of A and D:

Force equilibrium: R1 + R2 = 40

Taking moments about A (clockwise is positive): 40·2 - 20 - 6·R2 = 0

R1 = 30N , R2 = 10N

If instead you were to take moments about D you would get: - 20 - 40·4 + 6·R1 = 0

I think it's important for you to see that wherever you take moments about, the point moment is always taken as a negative (because it's a counter clockwise moment).

So how does a point moment affect the shear force and bending moment diagrams? Well. It has absolutely no effect on the shear force diagram. You can just ignore point C when drawing

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the shear force diagram. When drawing the bending moment diagram you will need to work out the bending moment just before and just after point C:

Just before: bending moment at C = 3·30 - 1·40 = 50Nm

Just after: bending moment at C = 3·30 - 1·40 - 20 = 30Nm

Then work out the bending moment at points A, B and D (no need to do before and after for these points). And plot.

Cantilever beam

Until now, you may have only dealt with "simply supported beams" (like in the diagram above), where a beam is supported by 2 pivots at either end. Below is a cantilever beam, which means - a beam that rigidly attached to a wall. Just like a pivot, the wall is capable of exerting an upwards reaction force R1 on the beam. However it is also capable of exerting a point moment M1 on the beam.

Force equilibrium: R1 = 10N

Taking moments about A: -M1 + 10·2 = 0 → M1 = 20Nm

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Uniformly Distributed Load (UDL) Below is a brick lying on a beam. The weight of the brick is uniformly distributed on the beam (shown in digram A). The brick has a weight of 5N per meter of brick (5N/m). Since the brick is 6 meters long the total weight of the brick is 30N. This is shown in diagram B. So as you can see there are 2 different diagrams to show the same thing. You need to be able to convert from a type A diagram to a type B

To make your life more difficult I have added an external force at point C, and a point moment to the diagram below. This is the most difficult type of question I can think of, and I will do the shear force and bending moment diagram for it, step by step.

Firstly identify the key points at which you will work out the shear force and bending moment at. These will be points: A,B,C,D,E and F.

As you would have noticed when working out the bending moment and shear force at any given point, sometimes you just work it out at the point, and sometimes you work it out just before and after. Here is a summary: When drawing a shear force diagram, if you are dealing with a point force (points A,C and F in the above diagram), work out the shear force before and after the point. Otherwise (for points B and D), just work it out right at that point. When drawing a bending moment diagram, if you are dealing with a point moment (point E), work out the bending moment before and after the point. Otherwise (for points A,B,C,D, and F), work out the bending moment at the point.

After identifying the key points, you want to work out the values of R1 and R2. You now need to convert to a type B diagram, as shown below. Notice the 30N force acts right in the middle between points B and D.

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Force equilibrium: R1 + R2 = 50 Take moments about A: 4·30 + 5·20 + 40 - 10·R2 = 0 R1 = 24N , R2= 26N

Update original diagram:

Shear force diagram

point A:

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point B:

Notice that the uniformly distributed load has no effect on point B.

point C:

Just before C:

Now convert to a type B diagram. Total weight of brick from point B to C = 5x4 = 20N

Shear force before C: 24 - 20 = 4N

Shear force after C: 24 - 20 - 20 = -16N

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point D:

Shear force at D: 24 - 30 - 20 = -26N

point F:

(I have already converted to a type B diagram, below)

Finally plot all the points on the shear force diagram and join them up:

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Bending moment diagram

Point A

Bending moment at A: 0Nm

Point B

Bending moment at B: 24·1 = 24Nm

point C:


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Bending moment at C: 24·5 - 20·2 = 80Nm

point D:


Bending moment at D: 24·7 - 30·3 - 20·2 = 38Nm

point E:


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point F:


Bending moment at F: 24·10 - 30·6 - 20·5 + 40 = 0Nm

Finally, plot the points on the bending moment diagram. Join all the points up, EXCEPT those that are under the uniformly distributed load (UDL), which are points B,C and D. As seen below, you need to draw a curve between these points. Unless requested, I will not explain why this happens.

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Note: The diagram is not at all drawn to scale.

I have drawn 2 curves. One from B to C, one from C to D. Notice that each of these curves resembles some part of a negative parabola.

Rule: When drawing a bending moment diagram, under a UDL, you must connect the points with a curve. This curve must resemble some part of a negative parabola.

Note: The convention used throughout this page is "clockwise moments are taken as positive". If the convention was "counter-clockwise moments are taken as positive", you would need to draw a positive parabola.

Hypothetical scenario

For a hypothetical question, what if points B, C and D, were plotted as shown below. Notice how I have drawn the curves for this case.

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If you wanted to find the peak of the curve, how would you do it? Simple. On the original diagram (used at the start of the question) add an additional point (point G), centrally between point B and C. Then work out the bending moment at point G.

That's it! If you have found this article useful, please comment in the discussion section (at the top of the page), as this will help me decide whether to write more articles like this. Also please comment if there are other topics you want covered, or you would like something in this article to be written more clearly.

• Beams – structural members supporting loads at various points along the member • Transverse loadings of beams are classified as concentrated loads or distributed loads • Applied loads result in internal forces consisting of a shear force (from the shear stress

distribution) and a bending couple (from the normal stress distribution)

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Normal stress is often the critical design criteria

Requires determination of the location and magnitude of largest bending moment

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Classification of Beam Supports

Shear and Bending Moment Diagrams

• Determination of maximum normal and shearing stresses requires identification of maximum internal shear force and bending couple.

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• Shear force and bending couple at a point are determined by passing a section through the beam and applying an equilibrium analysis on the beam portions on either side of the section.

• Sign conventions for shear forces V and V’ and bending couples M and M’

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Sample Question

The structure shown is constructed of a W10x112 rolled-steel beam. (a) Draw the shear and bending-moment diagrams for the beam and the given loading. (b) determine normal stress in sections just to the right and left of point D.

SOLUTION:

• Treating the entire beam as a rigid body, determine the reaction forces • Section the beam at points near supports and load application points. Apply equilibrium analyses

on resulting free-bodies to determine internal shear forces and bending couples • Identify the maximum shear and bending-moment from plots of their distributions. • Apply the elastic flexure formulas to determine the corresponding maximum normal stress.

Treating the entire beam as a rigid body, determine the reaction forces

Section the beam and apply equilibrium analyses on resulting free-bodies

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Identify the maximum shear and bending-moment from plots of their distributions.

Apply the elastic flexure formulas to determine the corresponding maximum normal stress.

shear force and bending momentss(1)

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