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Beam Load Calculator

Created by Kenneth Alambra
Reviewed by Dominik Czernia, PhD and Jack Bowater
Last updated: Jan 18, 2024

This beam load calculator will help you determine the reactions at the supports of a simply-supported beam due to vertical point loads or forces. In this calculator, we will explore the following:

  • What support reaction is;
  • How to calculate support reactions in a beam;
  • Sample calculation of how to find support reactions; and
  • How to calculate beam load capacity using this beam load calculator.

Knowing how to find support reactions is a great place to start when analyzing beams, such as when determining beam deflection. Keep on reading to learn more.

🙋 If you're curious about how to calculate beam deflections, we have our beam deflection calculator you can check out, or our feature-packed wood beam span calculator for in-depth calculation on wood beam design.

What is a support reaction?

According to Newton's third law of motion, every force on an object has an equal and opposite reaction. If you try to push onto something, let's say a wall, it feels like the wall is also pushing back at you. That is exactly the phenomena Newton's third law of motion describes.

In engineering, structural members such as beams and columns interact with each other at the points where they meet. Imagine a beam that is being supported in place by two columns. The beam's weight pushes down on the columns, and because of Newton's third law of motion, we can then also say that the columns exert an equivalent opposite reacting force back on the beam. We call these reacting forces support reactions.

Animation of a beam load pushing forces on two columns and vice versa.

You can imagine a door header as the beam shown above and the two jack studs (the vertical studs where the door header rests) as the two supporting structures of the door header. Learn more about door header sizes by checking out our door header size calculator.

On a simply-supported beam, the support reactions at each end of the beam can either be equal to each other or of different values. Their values depend on the beam's applied loads. If more loads are found at a closer distance to one support, that support-structure experiences more force and therefore experiences a greater reaction.

How to calculate support reactions in a beam?

Since support reactions act in the opposite direction as the force, we can say that the whole system is in equilibrium. That means the beam is not moving, and the summation of forces and moments result in zero. By equating the moments due to load to the moments due to support reactions, we can then determine the reactions at the supports.

Diagram of a simply-supported beam and point loads at distance x from support A (support at the left).

Just like when calculating torque, we can also perform a summation of moments at each supports to solve for the reactions. Below, we express a summation, \small \sum, of the moments at support A to find the reaction at support B, denoted as RB\small R_\text{B}, as shown below:

i=1n(Fi×xi)(RB×span)=0\small \sum_{i=1}^n(F_i\times x_i) - (R_\text{B}\times \text{span}) = 0


(F1×x1) ⁣+ ⁣(F2×x2) ⁣+ ⁣(F3×x3) ⁣+... ⁣+ ⁣(Fn×xn) ⁣ ⁣(RB×span)=0,\small (F_1\times x_1)\! +\! (F_2\times x_2)\! +\! (F_3\times x_3)\! +\\ ...\! +\!(F_n\times x_n)\! -\! (R_\text{B}\times \text{span}) = 0,


  • F1F_1, F2F_2, F3F_3, and FnF_n – The point loads on the beam at distances x1x_1, x2x_2, x3x_3, and xnx_n, from support A, respectively;
  • RBR_\text{B} – Reaction at support B; and
  • span\text{span} – Length of the beam between support A and support B.

By rearranging the equation, we can isolate RBR_\text{B} as follows:

RB ⁣× ⁣span= (F1×x1) ⁣+ ⁣(F2×x2)+ (F3×x3)+... ⁣+ ⁣(Fn×xn)\small \begin{align*} R_\text{B}\! \times\! \text{span} =\ & (F_1\times x_1)\! +\! (F_2\times x_2)\\ &+\ (F_3\times x_3)\\ & +...\! +\!(F_n\times x_n)\\ \end{align*}
RB= [(F1×x1) ⁣+ ⁣(F2×x2) ⁣+(F3×x3)+...+(Fn×xn)]/span ✔\small \begin{align*} R_\text{B} =\ &[(F_1\times x_1)\! +\! (F_2\times x_2)\!\\ &+ (F_3\times x_3)\\ & +... +(F_n\times x_n)] / \text{span}\ \text{✔} \end{align*}

Now that we have an expression to find RB\small R_\text{B}, and since we know that the total applied forces are equal to the reactions' total, we can now also find the reaction at support A, RA\small R_\text{A}, using the following equations:

i=1nFi=RA+RB\small \sum_{i=1}^n F_i = R_\text{A} + R_\text{B}
RA=i=1nFiRB ✔\small R_\text{A} = \sum_{i=1}^n F_i - R_\text{B}\ \text{✔}

Now that we know how to calculate support reactions in beams, let us now consider an example calculation to strengthen our knowledge about it. 🙂

Sample calculation of how to find support reactions

Suppose we have a 4.0-meter long simply-supported beam with an applied 10.0 kilonewtons (kN) point load 2.0 meters from support A and another applied 3.5 kN point load 1.5 meters from support B, as shown below:

Diagram of a 4-meter long simply-supported beam with 2 applied loads for example.

To calculate for RB\small R_\text{B}, we formulate the equation of moment equilibrium as follows:

RB=(F1×x1)+(F2×x2)span=(10 kN×2.0 m)+(3.5 kN×(4.0 m1.5 m))4.0 m=20 kN-m+8.75 kN-m4.0 m=7.1875 kN\scriptsize \begin{align*} &R_\text{B} = \frac{(F_1\times x_1)\! +\! (F_2\times x_2)}{\text{span}}\\[1.2em] &= \frac{(10\ \text{kN}\!\times\! 2.0\ \text{m})\! +\! (3.5\ \text{kN}\!\times\! (4.0\ \text{m}\! -\! 1.5\ \text{m}))}{4.0\ \text{m}}\\[1.2em] &= \frac{20\ \text{kN-m} + 8.75\ \text{kN-m}}{4.0\ \text{m}}\\[1.2em] &= 7.1875\ \text{kN} \end{align*}

By performing a summation of forces, we obtain:

 ⁣F=0F1+F2+(RA)+(RB)=010+3.5+(RA)+(7.1875)=0RA=10+3.57.1875RA=6.3125 kN\!\scriptsize \begin{align*} \sum F\! &= 0\\ F_1\! +\! F_2\! +\! (-R_\text{A})\! +\! (-R_\text{B})\! &= 0\\ 10\! +\! 3.5\! +\! (-R_\text{A})\! +\! (-7.1875)\! &= 0\\ R_\text{A}\! &= 10\! +\! 3.5\! -7.1875\\ R_\text{A}\! &= 6.3125\ \text{kN} \end{align*}

Please note that for this summation, we have considered all downward forces as positive and all upward forces as negative. Based on our calculations above, we have now obtained the reactions at supports A and B to be 6.3125 kN and 7.1875 kN, respectively.

Also, please take note that in this example of how to find support reactions and in the beam load calculator, we assumed that the beam is weightless. However, if the beam's weight is indicated, you can consider the beam's weight as another downward point load at the center, or centroid, of the beam.

Using our beam load calculator

Our calculator is easy and simple to use. All you have to do is input the span of the beam, the magnitude of the point loads, and their distances from support A. At first, you will only see fields for two loads (Load 1 and Load 2), but once you enter a value for x2\small x_2, the fields for Load 3 will show up, and so on.

If you want to enter an upward load, simply enter a negative value for the load magnitude. In total, you can input up to 11 point loads in our beam load calculator.


What's a simply supported beam?

A simply supported beam is a beam that has two supports located at each end. One support is a pinned support, which allows only one degree of freedom, the rotation around the z-axis (perpendicular to the paper). At the other end, there's a roller support, which enables two degrees of freedom, the horizontal movement along the x-axis and rotation around the perpendicular z-axis.

How do I determine the support reactions on a simply supported beam?

To determine the reactions on a simply supported beam:

  1. Draw the beam's free body diagram.

  2. Sum the moments at one support, such as pin support A:

    Σ(Fi × xi) - RyB × beam_span = 0


    • Fi – Vertical forces;
    • xi – Distance between A and the point load; and
    • RyB – Vertical reaction at roller support B.
  3. To find the vertical reaction at A, RyA, sum the vertical forces:

    ΣFy = ΣFi - RyB - RyA = 0.

  4. Sum the horizontal forces to find the horizontal reaction at A RxA:

    ΣFy = ΣFi - RxA = 0.

What's the importance of calculating the support reactions?

Calculating the support reactions is essential for understanding the internal forces in a beam. In beam design, we need to understand how the internal shear forces and bending moments behave. By knowing how these loads vary along the beam, we can accurately analyze and predict the resulting stresses and deformations, helping us to design beams capable of withstanding these conditions, ensuring structural integrity and safety.

What are the reactions of a 6 m simply supported beam?

For a beam with a uniformly distributed weight load of 5 kN/m, the reactions at A and B are both 15 kN. To calculate this:

  1. Draw the beam's free body diagram, including the equivalent point load for the weight.

  2. Sum the moments at pin support A to find the vertical reaction at roller support B:

    RyB = 30 kN × 3 m/6 m = 15 kN.

  3. Sum the vertical forces to find the vertical reaction at A:

    RyA = 30 kN - 15 kN = 15 kN.

  4. The horizontal reaction at support A, RxA, is zero.

Kenneth Alambra
Related calculators
Beam detail
Span, L
Illustration of a simply-supported beam of length, L, and showing the loads applied on the beam with their corresponding distances from support A.
Load 1
Distance from support A, x₁
Load 2
Distance from support A, x₂
Output values
Show values in...
pounds-force (lbf)
With the beam fully loaded:

▪ Support reaction at A is Rᴀ = 3.3333 lbf
▪ Support reaction at B is Rʙ = 1.6667 lbf
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