CivilSolve

Statics & Mechanics of Materials

Beam Shear & Moment Diagrams

Support reactions plus complete shear V(x) and bending moment M(x) diagrams 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 maximum shear and maximum/minimum bending moments with their locations. Use for: 'draw the shear and moment diagrams', 'find the maximum bending moment', V and M at a section. Use beam-reactions instead when ONLY reactions are asked.

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 prismatic and rigid for equilibrium; x is measured from the left end.
  • Sign conventions: shear positive = net upward force on the left segment; moment positive = sagging.
  • All loads act in the vertical plane; self-weight is not included unless entered as a load.

Solution steps

  1. Reaction at B from moment equilibrium about A

    Summing moments about support A eliminates the reaction at A.

    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 load.

    RA = ΣW − RB

    RA = 5 kN

    = 5 kN

  3. Construct the shear diagram

    V starts at zero, jumps up at upward reactions, jumps down at downward point loads, and slopes at −w under distributed loads.

    dV/dx = −w(x)

    V(0 m⁺) = 5 kN; V(3 m⁺) = -5 kN; V(6 m⁺) = 0 kN

    = 5 kN

  4. Locate the moment extrema

    M slopes at V (dM/dx = V), so extreme moments occur where the shear crosses zero or at concentrated loads/supports.

    dM/dx = V(x)

    M_max = 15 kN·m at x = 3 m; M_min = 0 kN·m at x = 0 m

    = 15 kN·m

Beam loading schematic
Shear diagramShear diagram5-5006x (m)V (kN)V max (0, 5)
Moment diagramMoment diagram150006x (m)M (kN·m)M max (3, 15)

Results

Reaction at A (left support)

5kN

Reaction at B (right support)

5kN

Maximum shear (by magnitude, signed)

5kN

Location of maximum shear

0m

Maximum bending moment

15kN·m

Location of maximum moment

3m

Minimum (most negative) bending moment

0kN·m

Location of minimum moment

0m

Where this answer was checked
source
NCEES FE Reference Handbook, simply supported beam formulas — R = P/2, Mmax = PL/4 at midspan
verified by
hand-recomputed
derivation
R = 10/2 = 5 kN; Mmax = 10·6/4 = 15 kN·m at x = 3 m; Vmax = ±5 kN.

Example 2

simply supported, full uniform load (w=3 kN/m, L=4 m)

Given

span
4 m
supports
  1. 1.type pinposition 0 m
  2. 2.type rollerposition 4 m
loads
  1. 1.kind udlmagnitude 3 kN/mstart 0 mend 4 m

Assumptions

  • Beam is prismatic and rigid for equilibrium; x is measured from the left end.
  • Sign conventions: shear positive = net upward force on the left segment; moment positive = sagging.
  • All loads act in the vertical plane; self-weight is not included unless entered as a load.

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: 12 kN at x = 2 m

  2. Reaction at B from moment equilibrium about A

    Summing moments about support A eliminates the reaction at A.

    ΣM about A = 0 ⇒ RB · (xB − xA) = Σ(load moments about A)

    RB = 6 kN (supports at x = 0 m and x = 4 m)

    = 6 kN

  3. Reaction at A from vertical equilibrium

    The remaining reaction balances the total load.

    RA = ΣW − RB

    RA = 6 kN

    = 6 kN

  4. Construct the shear diagram

    V starts at zero, jumps up at upward reactions, jumps down at downward point loads, and slopes at −w under distributed loads.

    dV/dx = −w(x)

    V(0 m⁺) = 6 kN; V(4 m⁺) = 0 kN

    = 6 kN

  5. Locate the moment extrema

    M slopes at V (dM/dx = V), so extreme moments occur where the shear crosses zero or at concentrated loads/supports.

    dM/dx = V(x)

    Zero shear at x = 2 m; M_max = 6 kN·m at x = 2 m; M_min = 0 kN·m at x = 0 m

    = 6 kN·m

Beam loading schematic
Shear diagramShear diagram6-6004x (m)V (kN)V max (0, 6)
Moment diagramMoment diagram60004x (m)M (kN·m)M max (2, 6)

Results

Reaction at A (left support)

6kN

Reaction at B (right support)

6kN

Maximum shear (by magnitude, signed)

6kN

Location of maximum shear

0m

Maximum bending moment

6kN·m

Location of maximum moment

2m

Minimum (most negative) bending moment

0kN·m

Location of minimum moment

0m

Where this answer was checked
source
NCEES FE Reference Handbook, simply supported beam formulas — R = wL/2, Mmax = wL²/8 at midspan
verified by
hand-recomputed
derivation
R = 3·4/2 = 6 kN; Mmax = 3·16/8 = 6 kN·m at x = 2 m; Vmax = 6 kN at supports.

Example 3

cantilever, tip point load (P=8 kN, L=3 m, fixed at left)

Given

span
3 m
supports
  1. 1.type fixedposition 0 m
loads
  1. 1.kind pointmagnitude 8 kNposition 3 mdirection down

Assumptions

  • Beam is prismatic and rigid for equilibrium; x is measured from the left end.
  • Sign conventions: shear positive = net upward force on the left segment; moment positive = sagging.
  • All loads act in the vertical plane; self-weight is not included unless entered as a load.

Solution steps

  1. Wall reactions from equilibrium

    The fixed support supplies a vertical force and a reaction moment.

    ΣFy = 0 and ΣM about the wall = 0

    R = 8 kN; internal bending moment at the wall M = -24 kN·m

    = 8 kN

  2. Construct the shear diagram

    V starts at zero, jumps up at upward reactions, jumps down at downward point loads, and slopes at −w under distributed loads.

    dV/dx = −w(x)

    V(0 m⁺) = 8 kN; V(3 m⁺) = 0 kN

    = 8 kN

  3. Locate the moment extrema

    M slopes at V (dM/dx = V), so extreme moments occur where the shear crosses zero or at concentrated loads/supports.

    dM/dx = V(x)

    M_max = 0 kN·m at x = 0 m; M_min = -24 kN·m at x = 0 m

    = 0 kN·m

Beam loading schematic
Shear diagramShear diagram80003x (m)V (kN)V max (0, 8)
Moment diagramMoment diagram0-24003x (m)M (kN·m)M max (0, 0)M min (0, -24)

Results

Vertical reaction at the wall

8kN

Bending moment at the wall (sagging positive)

-24kN·m

Maximum shear (by magnitude, signed)

8kN

Location of maximum shear

0m

Maximum bending moment

0kN·m

Location of maximum moment

0m

Minimum (most negative) bending moment

-24kN·m

Location of minimum moment

0m

Where this answer was checked
source
NCEES FE Reference Handbook, cantilever beam formulas — R = P, Mwall = −PL
verified by
hand-recomputed
derivation
R = 8 kN up; wall moment = −8·3 = −24 kN·m (hogging, sagging-positive convention).