Beam reactions: one load at the centre
A simply supported beam rests on a pin and a roller. With one downward load exactly in the middle, the two supports share the vertical load equally.
What the formula is saying
Vertical equilibrium requires RA + RB = P. Taking moments about the left support gives RB × L = P × L/2. Therefore RB = P/2, and RA is the same.
Read the symbols in plain language
- P
- Centre point loadN
Sort out the units first
Enter P in N. 10 kN = 10,000 N. The answer is the upward reaction at each support, not their sum.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
- P · Centre point load
- 10000 N
Write the total downward load
There is only one applied vertical load in this model.
(10000) = 10000 NShare it equally between two supports
Equal distances to the load give equal reactions.
(10000) ÷ 2 = 5000 N
Does this worked answer make sense?
Add the two reactions: 5,000 + 5,000 = 10,000 N. The upward forces balance the downward load.
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Use these new values. Work it out first, then check your answer.
- P · Centre point load
- 18000 N
Find: Reaction at each support (RA = RB)
A hint, not the answer
Vertical equilibrium requires RA + RB = P. Taking moments about the left support gives RB × L = P × L/2. Therefore RB = P/2, and RA is the same.
Enter P in N. 10 kN = 10,000 N. The answer is the upward reaction at each support, not their sum.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Write the total downward load
There is only one applied vertical load in this model.
(18000) = 18000 NShare it equally between two supports
Equal distances to the load give equal reactions.
(18000) ÷ 2 = 9000 N
Avoid the common trap
P/2 is not correct for an off-centre point load. Do not report P/2 as the sum of both reactions.
When this method applies — and when it does not
An ideal straight beam with the supports and load stated here. Loads are magnitudes; self-weight is omitted unless already included. This is an equilibrium result, not a check of strength, deflection or stability. Enter the nonnegative magnitude of the downward load; sign conventions for internal moments are explained separately.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
One idea understood. Keep going.
This optional checkmark is saved only in this browser. It is your own progress note, not a certificate.
Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Lesson updated: · Worked examples checked against the implemented formula; not an independent engineering certification.