A shear force and bending moment diagram calculator for simply supported, overhanging and cantilever beams. Enter the beam length, support type and loads (point loads, uniform, triangular and polynomial distributed loads, applied moments) and the tool draws the shear force diagram (SFD) and bending moment diagram (BMD), solves the support reactions, and lists the piecewise expressions for V(x) and M(x) on each load interval.
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When a beam carries transverse loads, internal forces develop inside the material to maintain equilibrium at every cross-section. The two key internal quantities are shear force and bending moment.
Shear force V(x) at a cross-section is the net vertical force to one side of that cut. At a point load, V jumps by the magnitude of that load. Under a UDL, V changes linearly.
Bending moment M(x) is the net moment about the cross-section from all forces to one side. Large bending moments cause large bending stresses, which is usually the limiting factor in beam design. Where V is constant, M is linear; where V is linear (under a UDL), M is parabolic.
The relationship: dV/dx = -w(x) and dM/dx = V(x). M is maximized where V crosses zero, which the tool marks with a dashed line.
Positive shear when the left face of a cut section has an upward force; positive moment when the beam is concave upward (sagging). Matches Hibbeler's Mechanics of Materials and most US undergraduate texts.
Problem: A 10 m simply supported beam carries 5 kN/m over its full length.
Reactions. By symmetry: R_A = R_B = (5 x 10) / 2 = 25 kN upward.
SFD. At x = 0, V = +25 kN. V decreases linearly at 5 kN/m, crossing zero at x = 5 m.
BMD. M = 0 at both supports, parabolic, peaking at midspan: M_max = R_A(5) - 5(5)(2.5) = 62.5 kN·m.
| Case | Max Shear | Max Moment | Location of M_max |
|---|---|---|---|
| Simply supported, point load P at midspan | P/2 | PL/4 | Midspan |
| Simply supported, full-span UDL w | wL/2 | wL^2/8 | Midspan |
| Simply supported, point load P at position a | Pb/L | Pab/L | x = a |
| Cantilever, point load P at free end | P (constant) | PL | Fixed wall |
| Cantilever, full-span UDL w | wL (at wall) | wL^2/2 | Fixed wall |
Reactions, V(x) and M(x) are compared against a separate solver written in Python with sympy. It finds the reactions from the two equilibrium equations, then integrates the load exactly over the free body on one side of each cut. The test set is 18 beams covering every support type and every load type, five of them overhanging at one or both ends, including the worked example above, sampled at three points in every load interval: 284 values, all agreeing to within 1 part in 109.
Limits: one span, statically determinate supports only. Axial load and deflection are not computed; the beam formula calculator covers deflection for standard cases.
A triangular BMD results from constant shear, which happens under a point load only. A UDL gives linear shear and parabolic moment. Both are correct behavior.
Not currently. The calculator handles statically determinate beams: simply supported, with or without overhangs (two unknowns) and cantilever (three unknowns, three equations). Continuous and propped beams also need a compatibility condition, which this tool does not solve. The beam formula calculator has closed-form fixed-fixed and propped cantilever cases.
Hibbeler's Mechanics of Materials and Engineering Mechanics: Statics. Positive shear is upward on the left face, positive moment is sagging (concave up).