Calculator · Structural Mechanics
Shear Force Calculator
The internal shear force V at a chosen station of a determinate beam. V is taken from the left free body, so dV/dx = q and a concentrated force produces a true jump — this page reports both sides of it rather than picking one.
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
| Shear just left of x, V(x⁻) | 5 kN |
|---|---|
| Shear just right of x, V(x⁺) | -5 kN |
| Jump ΔV = V(x⁺) − V(x⁻) = ΣFy | -10 kN |
| Maximum shear | 5 kN at 0 m |
| Minimum shear | -5 kN at 3 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 |
A concentrated force or a support reaction acts exactly at this station, so the shear is discontinuous there: both one-sided limits are reported and neither is “the” value.
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 whole shear diagram with every jump preserved.
Technical reference
V(x) = Σ Fy over the left free body
dV/dx = q_y(x)
V(x₀⁺) − V(x₀⁻) = Fy(x₀)
- V
- internal shear force on the cut, from the left segment[N]
- q_y
- distributed load intensity, negative downward[N/m]
- Fy
- concentrated force or support reaction at the station[N]
Because dV/dx = q, the shear is constant where there is no distributed load, linear under a uniform load, and quadratic under a linearly varying one.
A concentrated force — applied or reactive — makes V discontinuous. The jump equals the force itself, so asking for “the” shear at that exact coordinate is not a well-posed question; the two limits are.