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Shear and moment diagrams: the ultimate guide.

If you are having trouble calculating the shear and moment diagrams of a beam, look no further. This article will give you an in-depth understanding on how the internal forces of beams work, and how to calculate them.

In order to undestand what shear and bending moment are, you need to have a clear grasp of internal stresses first.

What are internal stresses?

When a material is subject to some kind of deformation, it develops internal stresses. The bigger the deformation, the higher the internal stresses. Think of them as a force applied to an area:\[\sigma=\cfrac{F}{A}\]

In a beam, there are three types of deformation that cause internal stresses:

  • Axial deformation
  • Shear deformation
  • Bending deformation

The three deformations create three distinct types of internal stresses. By calculating the resultant of the internal stresses, we get the three internal forces of a beam:

  • normal force
  • Shear force
  • Bending moment

Let's discuss each one in detail.

What is a normal force?

The normal force is the resultant of the stresses caused by axial deformation. This kind of deformation happens when a beam is subject to an axial load:

The force F must be in equilibrium with the internal stresses that develop inside the beam. If the beam has a cross-section with area A, the normal stresses are perpendicular to the plane of the section: \[\sigma=\cfrac{F}{A}\]

We can calculate the resultant of the internal stresses to get the normal force in the beam: \[N=\sigma\cdot A=\cfrac{F}{A}\cdot{A}=F\] Unsurprisingly, the normal force is equal to F, because the beam must be in equilibrium with the external forces.

What is a shear force?

The shear force is the resultant of the stresses caused by shear deformation:

Shear deformation causes stresses that are parallel to the plane of the section:

The average value of the shear stress is given by \[\tau=\cfrac{F}{A}\] So the resultant shear force V is given by \[V= \tau\cdot A=\cfrac{F}{A}\cdot{A}=F\]

What is a bending moment?

When a beam is subject to a perpendicular load, it bends causing the top part of the section to compress and the bottom part of the section to extend.

This creates internal stresses that are perpendicular to the plane of the section, distributed with the typical "butterfly" shape:

The resultants of the two wings are equal in magnitude, and are separated by a lever arm z. If the cross-section has a height equal to d, the lever arm is equal to \[z=\cfrac{2}{3}\cdot \cfrac{d}{2}+ \cfrac{2}{3}\cdot \cfrac{d}{2}=\cfrac{2}{3}\cdot d\]

The bending moment is given by: \[M=F_c\cdot z=F_t\cdot z\] where \[F_c=F_t=\cfrac{\sigma}{2}\cdot\cfrac{A}{2}=\cfrac{\sigma\cdot A}{4}\]

Shear and bending moment diagrams: what you need to calculate them

Now that we know where internal forces come from and how they are related to internal stresses, we can start talking about how to calculate them.

But what are internal force diagrams? They are just a representation of the values of the internal forces at every point along the beam.

In order to be able to calculate them, you need to know the values of all the external forces acting on the beam, including the reactions at the supports.

If you don't know how to calculate support reactions, check out this article for an in-depth guide on that subject.

The method of cuts

The method of cuts is a general method that engineers use when they want to plot the internal force diagrams of a beam. It might look complicated at first, but it's actually very straight-forward once you learn how to use it.

Let's consider a simply supported beam of length \(L\) subject to a distributed load \(q\) and an axial load \(P\):

By applying the equilibrium equations to the free body diagram we get the following values:

\[ \begin{cases} M_{1,1}=M_{1,2}=R_{1,h}=R_{2,h}=0\\ N_{1,1}=N_{1,2}=P\\ V_{1,1}=+\cfrac{L\cdot q}{2}\\ V_{1,0}=-\cfrac{L\cdot q}{2}\\ R_{1,v}=\cfrac{q\cdot L}{2}\\ R_{1,h}=-P\\ R_{2,v}=\cfrac{q\cdot L}{2}\\ \end{cases} \]

If you don't know how to calculate the support reactions of a beam, check out this article.

By slicing the beam at an arbitrary point along its length, you end up with two separate "chunks". In the section where you made the cut, you will find the three resultants of the internal stresses discussed before: normal force, shear force and bending moment. These are the unknowns that you are trying to calculate with the method of cuts.

Notice how the sign of the internal forces is chosen following the usual convention:

The resultants N, V and M must be in equilibrium with all the other forces acting on the two chunks. This means that you can write three equilibrium equations (two translational, one rotational) that allow you to calculate the three unknowns at the cut.

How to calculate the equations of the internal forces of a beam

Imagine you are walking backward along the beam, going left to right. Consider the chunk of beam that goes from the left support to where you are.

All the forces acting on this chunk must be in equilibrium with the internal forces acting on the section of the beam under your feet.

As you keep walking, the length of the chunk grows, so the internal forces change to preserve equilibrium in the new configuration.

This means that the normal force, shear and bending moment along the beam are a function of the distance from the left support \(s\).

How to calculate the normal force

Going from left to right, we cut the beam at a distance \(s\).

The normal force \(N\) acting on the cut must be in equilibrium with all the other normal forces acting on the chunk: \[N(s)-P=0\] therefore \[N(s)=P\]

How to calculate the shear force

The shear force \(V(s)\) acting on the cut must be in equilibrium with all the forces perpendicular to the beam.

In this case we have \(V_{1,1}\) and the resultant of the distributed load: \[V(s)+\cfrac{q\cdot L}{2}-\cfrac{q\cdot s}{2}=0\] therefore \[V(s)=\cfrac{q\cdot s}{2}-\cfrac{q\cdot L}{2}\]

How to calculate the bending moment

The value of the bending moment \(M(s)\) must be in equilibrium with the value of the bending moment given by all the forces acting on the chunk.

In order to write the rotational equilibrium equation, we need to choose a reference point. You can choose any point you want on the plane, but in order to simplify the calculation people usually choose the location of the section cut as the reference point. In case you don't recall, the moment is calculated by multiplying a force with a distance, often called moment arm. In our case, the moment arm is the distance \(s\) of the section cut from the left support.

So what are the forces that produce a moment about the section cut? The lever arm of \(N(s)\) and \(V(s)\) is zero, so they don't generate moments. \(N_{1,1}\) also doesn't generate moment because it doesn't have a perpendicular component to the lever arm.

So we are left with the load \(q\) and \(V_{1,1}\). We can write the rotational equilibrium equation as follows: \[M(s)+q\cdot s\cdot \cfrac{s}{2}-\cfrac{q\cdot L}{2}\cdot s\] therefore \[M(s)=\cfrac{q\cdot L}{2}\cdot s-\cfrac{q\cdot s^2}{2}\] Notice that for \(s=0\) and \(s=L\) the value of the bending moment is 0 as it should be.


And there you have it. You should now have a better understanding on how internal force diagrams work and how to calculate them. The difficult part is calculating the support reactions, not the diagrams. If you get the support reactions right with this method, calculating the equations of the normal force, shear force and bending moment is a breeze.

Remember to always keep in mind the sign convention to remain consistent.

For more guides on similar subjects, check out our resources page.