80/3848–12FE Reference Handbook 10.6 · Statics · p. 95 Equilibrium (ΣFy = 0, ΣM = 0). Pin and roller are not given a separate support table.
1. Moment & Equilibrium2. Centroid of a Composite Area3. Coulomb Friction4. Force Resultant5. Rectangle Centroidal Moment of Inertia6. Parallel-Axis Theorem7. Distributed Load Resultant8. Method of Joints9. Two-Force Member10. Ideal Pulley11. Moment of a Force12. Simply Supported Beam Reactions13. Load–Shear–Moment Relations14. Method of Sections (Truss)15. Sagging Cable (Small Sag)16. Overhanging-Beam Reactions17. Radius of Gyration18. Triangle Area Moment of Inertia19. Triangular Distributed Load Resultant20. Three-Force Member21. Couple Moment22. Frames: Multi-Force Members23. Zero-Force Truss Members

Handbook formula

Simply Supported Beam Reactions

A simply supported beam has a pin and a roller (3 unknown scalars in 2D with no axial load often 2). Take moments about one support to find the other reaction, then ΣFy. Check with the opposite moment.

(N)
Vertical reactions
(m)
Span
(N)
Point load

Step-by-step solved example

Beam length 8 m, pin at A, roller at B. Point load 20 kN at 2 m from A. Find reactions.

pin Aroller B20 kN ↓L = 8 m2 m6 mRARB20 kN × 2 mRB × 8 mReplay
Simply supported beam: pin + roller. ΣM about A eliminates RA.

  1. 1. ΣM_A

    8 RB = 20×2 → RB = 5 kN.

  2. 2. ΣFy

    RA + 5 = 20 → RA = 15 kN.

Answer: RA = 15 kN, RB = 5 kN

10 practice questions

0/10 correct

1.Midspan point load P on span L: each reaction is

2.Uniform w on simple span: each reaction is

3.A roller on a horizontal beam (2D) cannot resist

4.Cantilever fixed at A, load P at the free end. RA =

5.Taking ΣM about A eliminates

6.P = 12 kN at 3 m from A on a 9 m simple beam. RB =

7.Negative reaction means the support

8.Internal hinge in a beam adds

9.Check: RA + RB must equal

10.A couple M applied at midspan of a simple beam produces reactions