CivilSolve

Statics & Mechanics of Materials

Beam Support Reactions

Support reactions for a determinate single-span beam (simply supported with optional overhangs, or a cantilever) under point loads, uniform/triangular/trapezoidal distributed loads, and applied moments. Reports the vertical reaction at each support and the fixed-end moment for a cantilever. Use only when reactions alone are asked; use beam-shear-moment for diagrams/max moment.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
supportsSupports. Simply supported = pin + roller (overhangs allowed). Cantilever = one fixed support at x = 0 or x = span.
supports 1
loadsApplied loads; magnitudes positive with explicit direction/sense
loads 1

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. 1.type pinposition 0 m
  2. 2.type rollerposition 6 m
loads
  1. 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

  1. 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

  2. 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

Beam loading schematic

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. 1.type fixedposition 0 m
loads
  1. 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

  1. 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

  2. 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

Beam loading schematic

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. 1.type pinposition 0 ft
  2. 2.type rollerposition 20 ft
loads
  1. 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

  1. 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

  2. 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

Beam loading schematic

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
verified by
hand-recomputed
derivation
RA = 6·15/20 = 4.5 kip; RB = 6·5/20 = 1.5 kip.