SeempleSTEM sites
Engimations

Engineering Mechanics · Forces and Equilibrium

Equilibrium and support reactions

Use the three equations of 2-D equilibrium to find the support reactions of a beam.

45 minYear 1Prerequisites: Resultant of a force system

By the end of this topic you will be able to

  • Identify the reactions that pin, roller and fixed supports can provide.
  • Replace a distributed load by an equivalent point load.
  • Apply ΣFx = 0, ΣFy = 0 and ΣM = 0 to find support reactions of a determinate beam.

Equilibrium in two dimensions

A body that is not accelerating is in equilibrium. For a body loaded in a plane, that gives three independent equations:

∑Fx=0,∑Fy=0,∑MA=0\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum M_A = 0

The third equation can take moments about any point; choose a point where one or more unknown forces act, so that they drop out of the equation.

What can a support provide?

Support Symbol Reaction components
Roller triangle on wheels one force, perpendicular to the surface
Pin (hinge) triangle two forces, horizontal and vertical
Fixed (built-in) wall two forces and a moment

A beam on one pin and one roller has 2+1=32 + 1 = 3 unknown reactions, so the three equations are enough. Such a beam is statically determinate.

Distributed loads

A uniformly distributed load (UDL) of ww kN/m over a length LL is replaced, for the equilibrium equations only, by a single force

W=wLW = wL

acting at the middle of the loaded length.

Method

  1. Draw the free-body diagram with every load and an unknown arrow for each reaction.
  2. Take moments about one support to find the other reaction directly.
  3. Use ∑Fy=0\sum F_y = 0 to find the remaining reaction.
  4. Check with a moment equation about the other support.
Interactive modelChange the loads and span; the reactions and the equilibrium checks update.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

Full screen

Practice problems

  1. A simply supported beam of span 8 m carries a single 40 kN load 3 m from the left support A. Find the reactions.
    Show answer
    Moments about A: RB×8=40×3R_B \times 8 = 40 \times 3, so RB=15R_B = 15 kN. Then RA=40−15=25R_A = 40 - 15 = 25 kN.
  2. A cantilever of length 3 m carries a UDL of 10 kN/m over its whole length. Find the reaction force and moment at the wall.
    Show answer
    Total load =30= 30 kN at 1.5 m from the wall. Wall force =30= 30 kN upward; wall moment =30×1.5=45= 30 \times 1.5 = 45 kN·m (resisting the load).
  3. A 10 m beam on supports A and B carries 24 kN/m over its left half (5 m). Find the reactions.
    Show answer
    Equivalent load =120= 120 kN at 2.5 m from A. RB=120×2.5/10=30R_B = 120 \times 2.5 / 10 = 30 kN; RA=90R_A = 90 kN.
  4. Why is the reaction at a roller always perpendicular to the surface it rolls on?
    Show answer
    A roller can only push or pull along the line normal to the surface; any sideways (tangential) force would make it roll, so it can carry none.

Key formulae

Force equilibrium
∑Fx=0,∑Fy=0\sum F_x = 0, \quad \sum F_y = 0
Moment equilibrium
∑MA=0\sum M_A = 0
Equivalent load of a UDL
W=wL,acting at mid-lengthW = wL, \quad \text{acting at mid-length}
Reaction of a simply supported beam
RB=∑(load×distance from A)LR_B = \frac{\sum (\text{load} \times \text{distance from A})}{L}
Lecturer notes: suggested plan, activity and assessment

Suggested plan (45 min). 10 min: three support types with real photos (bridge bearings, pin-jointed gate, cantilevered balcony); 10 min: method on the board; 15 min: worked example; 10 min: pairs use the model to check problem 1 and 3.

Activity. "Move the load": ask students to predict which reaction grows when the point load moves toward B, then confirm in the model.

Discussion. What changes if both supports are pinned? (Four unknowns, three equations: statically indeterminate.)

Assessment idea. Provide a different beam and ask for the reactions and a moment check about the second support.