Simply supported beam with a point load
The other first-principles beam. One concentrated load, off center — so symmetry is gone, moments about a support earn the reactions, the shear diagram jumps instead of sliding, and takes over from . We solve it by statics in three steps, then verify every number against the calculator.
- 6 mspan, pin to roller
- Supports
- Pin · Roller
- 24 kNpoint load at a = 2 m from A (b = 4 m)
- 200 GPa
- 100 × 10⁶ mm⁴
- 20 000 kN·m²
Step 1 — Reactions by moments about A
No symmetry to lean on this time — statics has to earn both numbers.
In the UDL example, symmetry handed us the reactions for free. Not here: the 24 kN load sits 2 m from A and 4 m from B, so the supports carry unequal shares and we go back to first principles. Take moments about A — the unknown passes through the pivot and vanishes, leaving alone:
Vertical equilibrium then delivers the other support — and notice it lands exactly on the closed form :
Read the lever arms in those formulas: each support’s share is proportional to the load’s distance from the other support. The load sits closer to A, so A carries twice as much. These two numbers drive both diagrams in the next step.
Step 2 — Shear and moment, drawn together
A jump in V and a kink in M — the point load’s signature.
Start at A with the +16 kN reaction and walk right. With nothing distributed along the span, nothing bleeds shear away — V holds perfectly flat at +16 kN all the way to the load. There it drops by the full 24 kN in a single step, to −8 kN, and stays flat again out to B, where the 8 kN reaction closes the diagram back to zero. Contrast this with the UDL example’s steady slide: a point load makes the shear jump.
The moment is still the running total of the shear. Constant positive shear means a straight climb from zero at A; constant negative shear means a straight descent to zero at B. The two lines meet directly under the load — where V changes sign — so the bending moment diagram is a triangle, and its peak is a sharp kink rather than the UDL’s smooth crest:
Step 3 — Deflection under the load
The sag where the load lands — and why the true maximum hides just beyond it.
For a simply supported beam with an off-center point load, the classical closed form for the deflection directly under the load is:
Is that the beam’s maximum sag? Almost, but not quite. With the load off center, the longer 4 m segment is the more flexible one, so the deepest point of the curve sits slightly toward the longer side — a little past the load, not under it. The difference is small; the deflection diagram below shows the shape.
The proof
Hand calculation vs the solver.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Reaction R_A | 16 kN (Pb/L) | 16 kN | |
| Reaction R_B | 8 kN (Pa/L) | 8 kN | |
| Max moment M_max | 32 kN·m at 2 m (Pab/L) | 32 kN·m at 2 m | |
| δ under the load | 4.267 mm (Pa²b²/3EIL) | 4.267 mm |
Every value was worked by hand with the classical method, then checked against this site’s solver — the same engine the Try it button opens. This agreement is re-run automatically on every build.
Now make it yours
Open this exact model in the calculator — then change a load, drag a support, and watch every diagram update in real time. The best way to build intuition is to break it and see what happens.
Take it with you
Export this worked example as a PDF, or download it as a .screport and open it in the Report Builder — the model travels inside the file, so you can reconstruct it, re-solve, and build your own report from it.