How to Calculate Beam Reactions

Support reactions are the first step of every beam analysis. Get them wrong and every diagram downstream is wrong too. For a statically determinate beam — one pin and one roller — two equations of equilibrium are all you need.

The two equations

For a planar beam loaded vertically:

• ΣM = 0 — the sum of moments about any point is zero
• ΣF = 0 — the sum of vertical forces is zero

Take moments about one support and the reaction at that support drops out of the equation, leaving the other reaction directly.

Worked example: point load

A 6 m beam, pin at A (x = 0), roller at B (x = 6), carrying a 10 kN load at x = 2 m.

Moments about A (counterclockwise positive):

ΣM_A = R_B · 6 − 10 · 2 = 0
R_B = 20 / 6 = 3.333 kN

Vertical equilibrium:

ΣF_y = R_A + R_B − 10 = 0
R_A = 10 − 3.333 = 6.667 kN

Check: moments about B: R_A · 6 − 10 · 4 = 40 − 40 = 0 ✓

Worked example: uniform load

A uniform load w over the full span L acts as a single force wL at midspan. Both reactions are therefore wL/2 by symmetry.

For a partial uniform load from x = a to x = b, replace it with the resultant w(b − a) placed at (a + b)/2 and repeat.

Applied moments

An applied (concentrated) moment enters the moment equation directly and does not appear in the force equation. A clockwise applied moment M₀ adds −M₀ to ΣM about any point — it shifts load between the supports without changing their sum.

Common mistakes

1. Forgetting the moment arm. The arm is the perpendicular distance from the point you are taking moments about to the load’s line of action.
2. Sign convention drift. Pick counterclockwise-positive and stick to it through the whole problem.
3. Uniform loads. Use the resultant at the centroid of the loaded length, not at the start of it.
4. Not checking. Always verify with moments about the other support. It costs ten seconds and catches most arithmetic slips.

Next step

With the reactions in hand you can build the shear and moment diagrams by inspection. For beams with more than two supports, equilibrium alone is not enough — see the three-moment equation.