# How to calculate support reactions: the ultimate guide

Welcome to our ultimate guide on support reactions. Reading through this article you will learn:

- What support reactions are;
- How to draw the free body diagram of a structure;
- How to calculate the equilibrium equations of a beam;
- How to calculate the equilibrium equations of a joint;
- How to calculate support reactions of a simply supported beam;
- How to calculate support reactions of a cantilever beam;
- How to calculate support reactions in a frame;

The goal is to teach you a general method that you will be able to apply to any determinate structure you will ever encounter.

getting the support reactions is the first step if you want to ultimately calculate the bending moment, shear force and deflection of the members in the structure. If you get this step wrong, then the rest of you analysis will be wrong as well!

Here at beamsolver.com we had to develop a general method to solve this problem for our online beam calculator. So of course it had to work for every possible beam configuration imaginable.

Note

In this article we will only consider
**determinate structures**. Indeterminate structures are a bit more complicated because you have to take boundary
conditions into account in order to solve them. We will upload more information on indeterminate structures in the future. If you
don't know how to calculate the degree of determinacy of a structure, check out
this article.

Let's dive in now, and understand what support reactions actually are.

## What are support reactions?

Structures are made up by **members** connected together by **joints**. When subject to loading,
internal stresses are created inside the members. The internal stresses "carry" the forces acting on the structure to the
supports, i.e. the places where the members are anchored to the ground.

At the supports, the ground applies an equal and opposite force to the structure. This ensures that the
**free body diagram** of the structure is in equilibrium, and everything stays fixed in place.

The forces that the ground applies to the structure are called
**support reactions**, and are what this article will focus on.

In order to calculate support reactions, we need to go through the following steps:

- Draw the free body diagram of every beam in the structure;
- Draw the free body diagram of every joint in the structure;
- Write the equilibrium equations for the beams and the joints;
- Solve the system of equilibrium equations and get the support reactions;

So let's start by learning how to draw the free body diagram of a structure and its members.

## How to draw the free body diagram of a structure

The purpose of free body diagrams is to understand what the components of a structure are and what forces are acting on them. This will allow us to easily write a system of equilibrium equations that we can solve.

So, think of any given structure as a collection of
**members** connected together by **joints**. A member can have three distinct internal forces:

- Axial force;
- Shear force;
- Bending moment.

In order to calculate support reactions, we have to know the values of the internal forces at the supports. The ground will apply an identical and opposite force to the structure so that the resultant of the internal forces and the reactions is zero.

When drawing the free body of beams and trusses, the following sign convention is used:

As we said, beams are connected together by joints. Therefore, in order to preserve equilibrium, there must be equal and opposite
forces acting on the joints as well. Consider The next example. We can draw an
**exploded** diagram of the structure and write down all the forces acting on every entity:

It's important to be consistent with the names you give to the variables. For joint \(j\) connected to beam \(i\), we can write the internal forces as \[N_{i,j},V_{i,j},M_{i,j}\].

For every joint where some kind of support is present, we have to include the support reactions in the free body diagram. The names of the reactions for a joint \(j\) are these:\[R_{j,h},R_{j,v},R_{j,r}\]

All you have to do is enumerate the beams and joints from 1 to n and be consistent with the sign convention. By doing it this way, we have named every force present in the structure. This will make our life a lot easier when writing the equilibrium equations.

## How to calculate the equilibrium equations of a beam

If you draw the free body diagram correctly, writing the equilibrium equations is easy. For a beam laying on a 2d plane, we can write two translational equations and one rotational equation.

So all we have to do is go through all the beams in the structure one by one and write the three equilibrium equations. The free body diagram is going to look something like this:

Note that you will never find support reactions inside the beam equilibrium equations, because support reactions always act on the
**joints** of a structure.

So, let's see how to write the equilibrium equations of a beam subject to various types of loads.

### Equilibrium equations of a beam subject to a point load

The horizontal equilibrium equation can be written as follow:

\[-N_{1,1}+N_{1,2}=0\]

We can write the vertical equilibrium equation in the same manner, taking into consideration the point load \(P\) as well. Remember to be consistent with the sign!

\[V_{1,1}-V_{1,2}-P=0\]

The rotational equation is a bit more tricky. We have to choose a reference location, let's say joint 1, and calculate the moment given by all the forces relative to that point. Clockwise moments are considered positive. If the beam has length \(L\), we can write:

\[M_{1,1}-M_{1,2}+V_{1,2}\cdot L+P\cdot \frac{L}{2}=0\]

Notice that the moment given by \(N_{1,1}\) and \(V_{1,1}\) is zero because moment arm is zero, and the moment given by \(N_{1,2}\) is also zero because \(N_{1,2}\) does not have a perpendicular component.

### Equilibrium equations of a beam subject to a moment load

In this case, the load will only appear in the rotational equation:

\[ \begin{cases} -N_{1,1}+N_{1,2}=0\\ V_{1,1}-V_{1,2}=0\\ M_{1,1}-M_{1,2}+V_{1,2}\cdot L+Q=0 \end{cases} \]

### Equilibrium equations of a beam subject to a distributed load

In order to write the equilibrium equations of members subject to a distributed load, we need to calculate the
**resultant force** first. We also need to know where that resultant force is, so that we can calculate the moment
arm.

For uniform distributed loads and triangular loads, the configuration is as follows:

So, for a beam of length \(L\) subject to an uniform load \(q\), we can write the equilibrium equations like this:

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

## How to calculate the equilibrium equations of a joint

Joints also have two translational equations and one rotational equation. Joints are where the members of a structure come together, so the same internal forces that we used to write the equilibrium equations of the beams appear here as well, except they are flipped.

Joints are also where the structure is attached to the ground, so the support reactions are also going to be part of the equations.

Since a joint is a one-dimensional entity, only point and moment loads are permitted.

Consider this free body diagram, with one point load P, one moment load Q, a support and the internal forces of one beam:

The equilibrium equations can be easily written as follows:

\[ \begin{cases} R_{1,h}+N_{1,1}+P=0\\ R_{1,v}-V_{1,1}=0\\ R_{1,r}-M_{1,1}-Q=0 \end{cases} \]

This is where we need to start talking about
**support types**. When drawing the free body diagram of a joint, you don't always need to include all the support
reactions. For a roller, you already know that \(R_h\) and \(R_r\) must be zero, so it's pointless to include them in the
equations.

This table summarizes the various support types, and their contribution in terms of reaction forces:

icon | name | reactions |
---|---|---|

Fixed | \(R_h,R_v,R_r\) | |

Pin | \(R_h,R_v\) | |

Horizontal skater | \(R_v, R_m\) | |

Vertical skater | \(R_h, R_m\) | |

Horizontal roller | \(R_v\) | |

Vertical roller | \(R_h\) | |

Fixed rotation | \(R_r\) |

Let's tie all this together by looking at some examples.

## How to calculate support reactions of a simply supported beam

Let's consider a simply supported beam of length \(L\) subject to an uniform distributed load \(q\). This configuration has three degrees of freedom and three restraints, so it is statically determinate. First, we draw the free body diagram:

More experienced engineers are able to solve simple configurations like this without doing any kind of calculations, just relying on common sense. However, if this is one of your first times solving static structures, it's useful to go through all the steps nonetheless.

So, let's write the equilibrium equations of the beam:

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

Next, we write the equilibrium equations of joint 1:

\[ \begin{cases} R_{1,h}+N_{1,1}=0\\ R_{1,v}-V_{1,1}=0\\ -M_{1,1}=0 \end {cases} \]

Now let's write the equilibrium equations of joint 2:

\[ \begin{cases} N_{1,2}=0\\ R_{1,v}+V_{1,1}=0\\ M_{1,2}=0 \end {cases} \]

We have 9 equations and 9 unknowns. We can solve the system by substitution and get the following result:

## How to calculate support reactions of a cantilever beam

Let's consider a cantilever beam of length \(L\) subject to an uniform distributed load \(q\). This configuration has three degrees of freedom and three restraints, so it is statically determinate. First, we draw the free body diagram:

The equilibrium equations of the beam are going to be the same as the previous example:

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

The first joint gives us the following equations:

\[ \begin{cases} R_{1,h}+N_{1,1}=0\\ R_{1,v}-V_{1,1}=0\\ R_{1,r}-M_{1,1}=0 \end {cases} \]

In the second joint, since there is no support, everything must be zero:

\[ \begin{cases} N_{1,2}=0\\ V_{1,2}=0\\ M_{1,2}=0 \end {cases} \]

Solving the system of equations by substitution we get:

## Conclusion

You should now have a basic understanding of how to calculate support reactions in beams. The most important aspect is to
**draw the free body diagram correctly**.

If you get that right, everything else comes automatically. Just remember to be consistent with the sign convention, and of course don't make mistakes when you solve the system of equations!

Once you have calculated the support reactions, the next step is to draw the shear and moment diagrams. Check out our ultimate guide on shear and moment diagrams to learn how to do it.

Read more articles similar to this one by browsing our resources section.