Calculator · Structural Mechanics
Bending Moment Calculator
The internal bending moment M at a chosen station of a determinate beam, sagging positive. A concentrated couple makes M discontinuous, and the jump is exactly −Mz — this page shows it rather than smoothing over it.
Beam
Scroll the schematic sideways to see the whole beam; every result is also listed in the tables below.
Beam and loads
The beam runs from x = 0 to x = L.
Supports
Point loads
Point moments
Uniform distributed loads
Downward loads are negative (q < 0, Fy < 0); an applied couple is positive counter-clockwise. Internally, M > 0 is sagging, v > 0 is upward and θ > 0 is counter-clockwise.
Evaluation station
Measured from the left end of the beam.
Results
| Bending moment M(x) | 15 kN·m |
|---|---|
| Sense | sagging |
| Maximum (most sagging) moment | 15 kN·m at 3 m |
| Minimum (most hogging) moment | 0 kN·m at 0 m |
| A — position | 0 m |
| A — vertical reaction Ry | 5 kN (upward) |
| B — position | 6 m |
| B — vertical reaction Ry | 5 kN (upward) |
| Configuration | simply supported |
| Equilibrium residuals ΣFx / ΣFy / ΣMz | 0 N / 9.09e-13 N / 0 N·m — satisfied |
Assumptions
- · Statically determinate planar beam; reactions come from the shared equilibrium solver.
- · Small displacements, linear elasticity, plane sections remain plane (Euler-Bernoulli); shear deformation is neglected.
- · Uniform E and I along the beam when slope and deflection are reported.
Open in the Lab
The Lab draws the full bending-moment diagram alongside the shear diagram.
Technical reference
M(x) = Σ Fy (x − xᵢ) − Σ Mz over the left free body
dM/dx = V(x)
M(x₀⁺) − M(x₀⁻) = −Mz(x₀)
- M
- internal bending moment; positive is sagging[N·m]
- V
- internal shear force[N]
- Mz
- applied couple, positive counter-clockwise[N·m]
M > 0 is sagging: the beam is concave upward, the top fibre shortens and the bottom fibre stretches.
Because dM/dx = V, the bending moment is stationary exactly where the shear crosses zero — that is where the engine looks for the peak, not in a chart sample.
A concentrated force changes the slope of M but never makes it jump. Only an applied couple does, and the jump is −Mz because the internal sagging moment must fall to balance a counter-clockwise couple.