Find support reactions for simply supported beams and cantilevers under point loads, distributed loads, and moments, with equilibrium steps.
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Worked examples
These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
Example 1
simply supported, midspan point load (P=10 kN, L=6 m)
Given
span
6 m
supports
1.type pinposition 0 m
2.type rollerposition 6 m
loads
1.kind pointmagnitude 10 kNposition 3 mdirection down
Assumptions
Beam is rigid for equilibrium purposes; x is measured from the left end.
All loads act in the vertical plane; self-weight is not included unless entered as a load. Reactions are positive upward; the cantilever wall moment is reported sagging-positive.
Solution steps
Reaction at B from moment equilibrium about A
Summing moments about support A eliminates the reaction at A, leaving one equation for RB.
ΣM about A = 0 ⇒ RB · (xB − xA) = Σ(load moments about A)
RB = 5 kN (supports at x = 0 m and x = 6 m)
= 5 kN
Reaction at A from vertical equilibrium
The remaining reaction balances the total applied load.
RA = ΣW − RB
RA = 10 kN − 5 kN = 5 kN
= 5 kN
Results
Reaction at A (left support)
5kN
Reaction at B (right support)
5kN
Total applied downward load
10kN
Where this answer was checked
source
NCEES FE Reference Handbook, simply supported beam formulas — R = P/2 each for a midspan point load
verified by
hand-recomputed
derivation
Symmetry: RA = RB = 10/2 = 5 kN; total load 10 kN.
Example 2
cantilever with full uniform load (w=2 kN/m, L=4 m, fixed at left)
Given
span
4 m
supports
1.type fixedposition 0 m
loads
1.kind udlmagnitude 2 kN/mstart 0 mend 4 m
Assumptions
Beam is rigid for equilibrium purposes; x is measured from the left end.
All loads act in the vertical plane; self-weight is not included unless entered as a load. Reactions are positive upward; the cantilever wall moment is reported sagging-positive.
Solution steps
Replace distributed loads by resultants
For reactions, each distributed load acts as its total force at its centroid.
W = area under the load diagram, at the centroid
Resultants: 8 kN at x = 2 m
Wall reactions from equilibrium
The fixed support supplies a vertical force and a reaction moment; no other support shares the load.
ΣFy = 0 and ΣM about the wall = 0
R = 8 kN; internal bending moment at the wall M = -16 kN·m
= 8 kN
Results
Vertical reaction at the wall
8kN
Bending moment at the wall (sagging positive)
-16kN·m
Total applied downward load
8kN
Where this answer was checked
source
NCEES FE Reference Handbook, cantilever beam formulas — R = wL, Mwall = −wL²/2
verified by
hand-recomputed
derivation
R = 2·4 = 8 kN; wall moment = −2·16/2 = −16 kN·m (hogging, sagging-positive convention).
Example 3
US units: simply supported, off-center point load (P=6 kip at a=5 ft, L=20 ft)
Given
span
20 ft
supports
1.type pinposition 0 ft
2.type rollerposition 20 ft
loads
1.kind pointmagnitude 6 kipposition 5 ftdirection down
Assumptions
Beam is rigid for equilibrium purposes; x is measured from the left end.
All loads act in the vertical plane; self-weight is not included unless entered as a load. Reactions are positive upward; the cantilever wall moment is reported sagging-positive.
Solution steps
Reaction at B from moment equilibrium about A
Summing moments about support A eliminates the reaction at A, leaving one equation for RB.
ΣM about A = 0 ⇒ RB · (xB − xA) = Σ(load moments about A)
RB = 1.5 kip (supports at x = 0 ft and x = 20 ft)
= 1.5 kip
Reaction at A from vertical equilibrium
The remaining reaction balances the total applied load.
RA = ΣW − RB
RA = 6 kip − 1.5 kip = 4.5 kip
= 4.5 kip
Results
Reaction at A (left support)
4.5kip
Reaction at B (right support)
1.5kip
Total applied downward load
6kip
Where this answer was checked
source
NCEES FE Reference Handbook, simply supported beam formulas — RA = Pb/L, RB = Pa/L