Lessons · Engineering · reactions of a simple beam
What each end of a simple beam carries
For a beam on two supports, a moment sum about one support gives the other reaction, and the vertical force sum gives the first.
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In beta. This lesson was written for Hone and has not yet been checked by a licensed engineer. Practice material, not professional advice. What that means.
What it is for
Every floor joist, every crane runway, every pipe bridge is a beam on two supports, and the first thing anyone sizes is the connection at each end. The reactions are the loads on those connections.
How to think about it
Replace a distributed load with one force at the centre of the length it covers. Take moments about A to find R_B. Then up equals down for R_A. Then the sanity check: the support nearer the load carries more.
Worked example
Simply supported, span L = 5 m, point load P = 20 kN at 2 m from AThe picture in one line.
ΣM_A = 0: R_B × 5 − 20 × 2 = 0 → R_B = 40 / 5 = 8 kNAbout A. The far reaction comes out alone.
ΣF_y = 0: R_A + R_B − 20 = 0 → R_A = 12 kNUp equals down.
Check: the load is nearer A, so A carries more. 12 > 8 OKIf the bigger reaction is at the far support, an arm is wrong.
A uniform load w = 6 kN/m over the whole span acts like 6 × 5 = 30 kN at midspan: R_A = R_B = 15 kNA distributed load becomes one force at the centre of the length it covers.
Your turn
Span 8 m, a 24 kN load 7 m from A. Write the moment equation about A.
R_B × 8 − 24 × = 0
Solve one, graded on the server
The trap
Putting a distributed load at the middle of the beam. It sits at the middle of the length it covers, and when it covers only part of the span the difference is the whole answer.