How to calculate beam reactions, shear force and bending moment

Reactions, shear force diagrams (SFD) and bending moment diagrams (BMD) are the foundation of structural analysis. Get these right and the rest of the design follows. The process is the same for every beam — only the load case changes.

Step 1 — Draw the free-body diagram

Sketch the beam as a single line with supports as triangles (pin) or rollers, and loads as arrows. Label every load, every distance, and every reaction. A messy FBD is the single biggest cause of arithmetic mistakes.

Step 2 — Apply equilibrium for reactions

ΣFy = 0     ΣFx = 0     ΣM = 0

Take moments about one support to find the other reaction (one unknown in one equation). Then use ΣFy to find the remaining vertical reaction. Always check by taking moments about the other support — they should sum to zero.

Step 3 — Build the shear force diagram

Start at the left end with V = 0 (above the beam) and walk along the beam. At every point load, jump by that load's magnitude. Under a UDL, the SFD is a sloping line with slope = −w. Under a point moment, the SFD does not change.

• Point load downward → SFD jumps down
• Point load upward (reaction) → SFD jumps up
• UDL → SFD slopes linearly
• End of beam → SFD must return to zero (check)

Step 4 — Build the bending moment diagram

BMD = the integral (area) of the SFD. Walk from one end to the other adding up the area under the SFD. Where the SFD crosses zero, the BMD has a peak — this is usually where Mmax occurs.

dM ÷ dx = V     M(x) = ∫ V dx

Step 5 — Locate Mmax

Mmax is at the point where V = 0. For a UDL on a simply supported beam, that's at mid-span. For asymmetric loading, find x where V(x) = 0 by setting the shear equation to zero and solving.

Quick worked example

Simply supported 6 m beam, UDL w = 10 kN/m, point load P = 20 kN at 4 m from left. Reactions: ΣM about A: RB·6 = 10·6·3 + 20·4 → RB = 43.3 kN. ΣFy: RA = 60 + 20 − 43.3 = 36.7 kN. SFD starts at +36.7 kN, slopes down at 10 kN/m, crosses zero at x = 3.67 m, drops 20 kN at x = 4 m to −23.3 kN, slopes to −43.3 kN at right support. Mmax at x = 3.67 m: M = 36.7·3.67 − 10·3.67²/2 ≈ 67.3 kNm.