Beam support reactions: practice problems with step-by-step solutions
Free Beam support reactions practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: 1 point load, 2 point loads, point load + UDL.
Free practice problemEasy
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.
Draw a free-body diagram and replace each support by its reactions.
Replace every distributed load by a single force at its centroid.
Take moments about one support to find the reaction at the other.
Use ΣFy = 0 to find the remaining reaction, and check with a second moment equation.
Common mistakes
Taking the moment arm from the wrong point.
Forgetting to convert a distributed load into its resultant.
Mixing up which support carries which reaction.
Key formulas
Equilibrium
ΣFx = 0, ΣFy = 0, ΣM = 0
Uniform load
resultant w·L at the middle of the load
Moment
M = F·d
Learn Beam support reactions from scratch
1What support reactions are
A beam is a long straight member that carries loads across a gap, like a shelf, a floor joist or a small bridge.
The beam rests on supports. When loads push the beam down, the supports push back up. These upward forces are the support reactions.
Here a 10 kN load sits exactly in the middle. By symmetry each support carries half of it: RA = RB = 5 kN.
When the load is not in the middle, the nearer support carries more. Finding exactly how much is the topic of this lesson.
You need the reactions first for almost every other beam question: shear force, bending moment, stress and deflection.
Self-check A beam on two supports carries 10 kN exactly at midspan. What is the reaction at each support?
10 kN
2.5 kN
20 kN
5 kN
Answer: 5 kN. The two reactions together must carry the whole 10 kN, and by symmetry they are equal: 10/2 = 5 kN. 10 kN at each support would carry the load twice.
2Supports, moments and equilibrium
Each type of support can push back in its own way.
A roller (B in the figure) can only push at right angles to the surface it rolls on. It gives one vertical reaction.
A pin (A) stops the beam moving in any direction. It gives a vertical and a horizontal reaction. With only vertical loads, the horizontal one is zero.
A fixed support, such as a wall, also stops the beam from rotating. It gives a vertical force and a reaction moment.
A moment is the turning effect of a force about a point. It is the force times its perpendicular distance from that point:
M = F · d
A beam at rest does not move and does not rotate. So three equilibrium equations hold:
ΣFx = 0 ΣFy = 0 ΣM = 0
The moment equation holds about any point. Pick a support as that point: its reaction has zero distance, so it drops out of the equation.
Self-check A 6 kN force acts 2 m from point A. What is its moment about A?
12 kN·m
3 kN·m
6 kN·m
8 kN·m
Answer: 12 kN·m. M = F·d = 6·2 = 12 kN·m. Dividing (3) or adding (8) the force and distance are the usual slips.
3Distributed loads
Some loads are spread along the beam, such as the weight of a floor or of snow. A uniformly distributed load (UDL) has the same value w on every metre, in kN/m.
For the reactions you can replace a UDL by one single force. Its size is the load per metre times the loaded length, and it acts at the middle of the loaded part:
W = w · ℓ
In the figure, ℓ = 3 m, so W = 4 · 3 = 12 kN. It acts at the middle of the load, at x = 1 + 1.5 = 2.5 m from A.
A cantilever is fixed at one end only, like a balcony. The wall gives an upward force and a moment. For an end load P at distance L: RA = P and MA = P·L.
In this cantilever, RA = 5 kN and MA = 5 · 3 = 15 kN·m.
Self-check A UDL of 4 kN/m acts from x = 1 m to x = 4 m. What single force replaces it, and where?
12 kN at x = 1.5 m
12 kN at x = 2 m
12 kN at x = 2.5 m
4 kN at x = 2.5 m
Answer: 12 kN at x = 2.5 m. W = 4·3 = 12 kN at the middle of the load, 1 + 1.5 = 2.5 m from A. 1.5 m is measured from the start of the load, not from A. 2 m is the middle of the whole span, not of the loaded part.
4The method step by step
Draw the beam and replace each support by its reactions: RA at the pin and RB at the roller.
Replace each UDL by a single force w·ℓ at the middle of the loaded part.
Take moments about A. RA drops out, so you get RB directly: RB·L = Σ(each load × its distance from A).
Use ΣFy = 0: RA = (total load) − RB.
Check: take moments about B. This should give the same RA.
Sign rule for step 3: the reaction RB turns the beam one way about A, the downward loads turn it the other way. Their moments must balance.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A simply supported beam of span 6 m carries P1 = 12 kN at 2 m from A and P2 = 6 kN at 4 m from A. Find RA and RB.
1
Draw the free-body diagram
The pin at A and the roller at B each give one upward reaction. There are two unknowns, so two equations are enough.
2
Moments about A
RA passes through A, so it has no moment about A. The loads turn the beam clockwise, RB turns it anticlockwise:
RB · 6 = 12 · 2 + 6 · 4 = 48
RB = 48 / 6 = 8 kN
3
Vertical force balance
The two reactions together carry the total load of 12 + 6 = 18 kN.
RA = 18 − 8 = 10 kN
4
Check the answer
Take moments about B instead. The distances to B are 4 m for P1 and 2 m for P2.
RA · 6 = 12 · 4 + 6 · 2 = 60 ⇒ RA = 10 kN
Both routes agree. RA is the larger one, which makes sense: the heavier load is closer to A.
Answer
RA = 10 kN and RB = 8 kN.
6Common mistakes
Measuring a distance from the wrong point. In a moment equation about A, every distance is measured from A.
Swapping the reactions. Taking moments about A gives RB, not RA.
Forgetting to turn a UDL into one force w·ℓ, or placing that force at the middle of the span instead of the middle of the load.
Mixing up the turning directions, so that a load and a reaction get the same sign in the moment equation.
Not checking. A second moment equation about B catches most slips in a few seconds.
Self-check A 4 m beam on two supports carries 8 kN at 1 m from A. What is RA?
2 kN
4 kN
6 kN
8 kN
Answer: 6 kN. Moments about A: RB·4 = 8·1, so RB = 2 kN and RA = 8 − 2 = 6 kN. 2 kN is RB: the support nearer the load carries more.
7How exam questions look
Easy: a simply supported beam with one point load. You find RA or RB.
Medium: two point loads, as in the worked example.
Hard: a point load together with a UDL over part or all of the span. Replace the UDL by w·ℓ at its middle first.
Other versions use the same three equations: a cantilever (find the wall moment MA), an inclined load (use only its vertical component, P·sin θ), an applied couple, a beam that overhangs a support, or a triangular load (resultant ½·w·ℓ at one third of its length from the high end).
Read the dimensions under the beam carefully and write each distance from A before you start.
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.
L = 4 mP = 55 kN at 2 m
Find RB(kN)
Show hints
Use static equilibrium: ΣFy = 0 and ΣM = 0. Two unknowns, two equations.
Take moments about A. RA passes through A, so it drops out and you get RB directly.
Total downward load is 55 kN, so RA + RB = 55 kN.
Show the answer and the step-by-step solution
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.
Free-body diagram
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·4 = 55·2
RB = 110/4 = 27.5 kN
ΣFy = 0: RA = 55 − 27.5 = 27.5 kN
Answer
RB = 27.5 kN
Example 2Medium
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.
L = 6 mP1 = 15 kN at 2 mP2 = 30 kN at 3 m
Find RB(kN)
Show hints
Use static equilibrium: ΣFy = 0 and ΣM = 0. Two unknowns, two equations.
Take moments about A. RA passes through A, so it drops out and you get RB directly.
Total downward load is 45 kN, so RA + RB = 45 kN.
Show the answer and the step-by-step solution
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.
Free-body diagram
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·6 = 15·2 + 30·3
RB = 120/6 = 20 kN
ΣFy = 0: RA = 45 − 20 = 25 kN
Answer
RB = 20 kN
Example 3Hard
The simply supported beam has a pin at A and a roller at B. A load P acts at 3 m from A, inclined at 30° to the horizontal, and a vertical load Q acts at 2 m from A. Find the magnitude of the total reaction force RA at the pin A.
L = 4 mP = 25 kN at 3 m, θ = 30°Q = 25 kN at 2 m
Find RA(kN)
Show hints
Split P into P·cos θ (horizontal) and P·sin θ (vertical). The roller takes no horizontal force.
Find RB from moments about A, then Ay from ΣFy = 0 and Ax from ΣFx = 0.
Ax = 21.65 kN, Ay = 15.63 kN.
Show the answer and the step-by-step solution
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation.
Components of P
Pv = 25·sin 30° = 12.5 kN
Ph = 25·cos 30° = 21.65 kN
Reactions
ΣMA = 0: RB·4 = 12.5·3 + 25·2 → RB = 21.88 kN
Ay = 12.5 + 25 − 21.88 = 15.63 kN
Ax = Ph = 21.65 kN
Resultant at A
RA = √(21.65² + 15.63²) = 26.7 kN
Answer
RA = 26.7 kN
Questions and answers
What is Beam support reactions?
A beam is a long straight member that carries loads across a gap, like a shelf, a floor joist or a small bridge.
The beam rests on supports. When loads push the beam down, the supports push back up. These upward forces are the support reactions.
How do I solve Beam support reactions problems step by step?
Draw a free-body diagram and replace each support by its reactions.
Replace every distributed load by a single force at its centroid.
Take moments about one support to find the reaction at the other.
Use ΣFy = 0 to find the remaining reaction, and check with a second moment equation.
What are common mistakes in Beam support reactions?
Taking the moment arm from the wrong point.
Forgetting to convert a distributed load into its resultant.
Mixing up which support carries which reaction.
What do the Beam support reactions exercises cover?
Easy: 1 point load
Medium: 2 point loads
Hard: point load + UDL
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.