# Shear Force and Bending Moment Diagrams

This article is part of the solid mechanics course, aimed at engineering students. Please leave feedback in the discussion section above.

## Contents

## What is shear force?[edit]

The shear force at a section of a beam is the force that shears off the section and is obtained as the algebraic sum of all forces including the reactions acting normal to the axis of the beam either to the left or right of the beam. 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 transferred internally all along the beam. This internal force is known as a shear force. The shear force between point A and B is usually plotted on a shear force diagram. As the shear force is 10 N all along the beam, the plot is just a straight line in this example.

The idea of shear force might seem odd, so this example is included to help clarify the concept. Imagine pushing an object along a kitchen table, with a 10 N force. Even though you're applying the force only at one point on the object, it's not just that point of the object that moves forward. The whole object moves forward, which tells you that the force must be acting along the entire object, such that every atom of the object is experiencing this 10 N force.

Please note that this is not the full explanation of what shear force actually is.

## Basic shear diagram[edit]

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

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 force 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 = 30Newtons

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:

Shear force at B = 10N

Now place your paper JUST after point B:

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

(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[edit]

Bending moment refers to the internal moment that causes something to bend. When you bend a ruler, even though you've applied 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:

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 = 10Nx3m - 20Nx2m = -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:

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... (So what if you take consider of moment in equilibrium?) - answer: there should be reactions (forces that counters the red ones) at the suports. These forces can be found by assuming the moment to end in zero.

## Point moments[edit]

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: R_{1} + R_{2} = 40

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

R_{1} = 30N , R_{2} = 10N

If instead you were to take moments about D you would get: - 20 - 40·4 + 6·R_{1} = 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 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 R_{1} on the beam. However it is also capable of exerting a point moment M_{1} on the beam.

Force equilibrium: R_{1} = 10N

Taking moments about A: -M_{1} + 10·2 = 0 → M_{1} = 20Nm

## Uniformly Distributed Load (UDL)[edit]

Below is a brick lying on a beam. The weight of the brick is uniformly distributed on the beam (shown in diagram 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

**[Oh no, this is wrong, the placement of the force along the x-axis will influence the deflection of the beam. With your logic (whoever wrote this), we could spread out the load over the entire beam and it would be equivalent, but everyone knows a force applied directly above or near the supports will not bend the beam anywhere near as much as if you apply the same force at the center. 30N distributed over an area of the beam will not deflect it as much as 30N concentrated at the center...]**

**[To whoever wrote the message above: Yes, You're right. However this all gets accounted for in the analysis shown on this page. So consider 2 cases - In the first, a 30N is concentrated in the centre, and in the second the 30N force is spread over the whole beam. If you draw the bending moment diagrams for both cases, you will see that the bending moment (at any point on the beam), would be lower in the second case. And a lower bending moment would result in less bending/deflection. Still, I probably should not have written that these 2 diagrams 'show the same thing', but rather that being able to convert from a type A diagram to a type B is a useful hack for doing these types of questions.]**

**[Can you please unhack the convention and explain the actual scenario if the load is spread across?]**

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 R_{1} and R_{2}. 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.

Force equilibrium: R_{1} + R_{2} = 50

Take moments about A: 4·30 + 5·20 + 40 - 10·R_{2} = 0

R_{1} = 24N , R_{2}= 26N

Update original diagram:

**Shear force diagram**[edit]

**point A:**

**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

**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:

**Bending moment diagram**[edit]

**Point A**

Bending moment at A: 0Nm

**Point B**

Bending moment at B: 24·1 = 24Nm

**point C:**

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

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

**point D:**

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

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

**point E:**

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

**point F:**

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

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.

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**[edit]

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.

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.

Back to the solid mechanics course

Taltastic 14:11, 15 September 2010 (UTC)