Beam support reactions

Engineering · First-year engineering · about 10 min

A beam sits on supports, and those supports push back. Finding how hard each one pushes, the support reactions, is the first step in almost every beam problem: you need them before you can draw a shear force or bending moment diagram, size the beam, or check a bearing.

The key idea

A beam that isn't moving is in equilibrium, so the forces balance and the moments balance:

\sum F_y = 0 \qquad \sum M = 0

Take moments about one support to find the other reaction in one line, then use vertical equilibrium for the first.

Try it yourself

Move the load along the beam and watch the two reactions trade off.

Pin and roller supports

A simply supported beam has a pin at one end and a roller at the other.

The pin stops the beam moving up, down and sideways, so it can push in two directions: $A_x$ and $A_y$ .
The roller only stops vertical movement, so it gives one vertical reaction, $B_y$ .

With only vertical loads, $A_x = 0$ , which leaves two unknowns and two equations. That's what makes the problem solvable.

A horizontal beam. Pin support under the left end A, roller support under the right end B. Short dark blue reaction arrows act on the beam ends: at A, one arrow pointing right and one pointing up; at B, one arrow pointing up. — A pin gives two reactions and a roller gives one.

Handling a distributed load

A uniformly distributed load (UDL) of $w$ kN/m over a length $\ell$ acts like a single force

W = w\,\ell

placed at the middle of the length it covers. Replace it with that one force when you take moments; never multiply $w$ by the beam length if the load only covers part of it.

Two identical horizontal beams, one above the other. Top beam: evenly spaced downward arrows, joined along their tops by a thin line, covering the middle part of the beam only, over a length marked with a dimension line. Bottom beam: the distributed load is replaced by one large downward arrow at the middle of that same length, with two equal dimensions either side of it. — A uniform load acts like one force at the middle of the length it covers.

The method, step by step

Draw a free-body diagram: the beam, every load, and the unknown reactions as arrows.
Turn each distributed load into its resultant force at its centre.
Take moments about support $A$ (anticlockwise positive, or your own convention, just stay consistent). The unknown $A_y$ passes through the point, so it drops out and you solve for $B_y$ directly.
Use $\sum F_y = 0$ to find $A_y$ .
Check by taking moments about $B$ : you should get the same $A_y$ .

Worked example: a point load and a UDL

A simply supported beam $AB$ is $6$ m long. It carries a point load of $12$ kN at $2$ m from $A$ , and a uniformly distributed load of $3$ kN/m over its whole length. Find the reactions at $A$ and $B$ .

A horizontal beam 6 m long. Pin support at the left end A, roller support at the right end B. Evenly spaced small downward arrows along the whole beam, joined along their tops by a thin line. One large downward force arrow at 2 m from A. Dimensions under the beam on two rows: 2 m and 4 m on the first row, 6 m on the second. — The beam in this example.

Show the solution

The UDL's resultant is $W = 3 \times 6 = 18$ kN, acting at the middle, $3$ m from $A$ .

Moments about $A$ (clockwise loads, anticlockwise reaction):

B_y \times 6 = 12 \times 2 + 18 \times 3 = 24 + 54 = 78

B_y = 13 \text{ kN}

Vertical equilibrium:

A_y = 12 + 18 - 13 = 17 \text{ kN}

Check, moments about $B$ : $A_y \times 6 = 12 \times 4 + 18 \times 3 = 48 + 54 = 102$ , so $A_y = 17$ kN. It matches.

Common mistakes

Putting a distributed load's force at the end of the beam

The resultant acts at the centre of the loaded length. For a full-span UDL on a $6$ m beam that's $3$ m from either end.

Mixing up which support gets the bigger reaction

The support closer to a load carries more of it. A load at $2$ m from $A$ on a $6$ m beam puts $\tfrac{4}{6}$ of it on $A$ .

Forgetting to check the answer

Taking moments about the other support takes ten seconds and catches almost every sign or arithmetic slip.

Check your understanding

1. A $10$ kN point load sits at the middle of a simply supported beam. What is each reaction?

2. A UDL of $4$ kN/m covers the whole of an $8$ m beam. What single force replaces it, and where?

3. Why take moments about a support rather than some other point?

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