Simply supported beam with a triangular load
The first load that refuses to be uniform. A ramp of load — zero at one end, 15 kN/m at the other — and one classical trick: collapse the triangle to a single force at its centroid. The reactions fall out by the thirds rule, the shear becomes a parabola, and the peak moment slides away from midspan to . Four steps by statics, then every number verified against the calculator.
- 6 mspan, pin to roller
- Supports
- Pin · Roller
- 0 → 15 kN/mtriangular, peak at B
- 200 GPa
- 100 × 10⁶ mm⁴
Step 1 — Replace the triangle by its resultant
One force at the centroid stands in for the whole ramp of load — that is the trick.
The load ramps from nothing at A to = 15 kN/m at B. For the reactions we do not need the ramp — only its total and where it acts. The total is the area of the load triangle, and it acts through the triangle’s centroid, which sits two-thirds of the way toward the heavy end:
That single 45 kN force, placed at 4 m, is statically equivalent to the whole distributed triangle — for the purpose of finding reactions. Hold that caveat; it returns in Step 4.
Step 2 — Reactions by the thirds rule
The support under the heavy end carries exactly twice the other.
With the resultant placed, statics is two lines. Moments about A — the 45 kN acts at 4 m, so:
The split is worth memorising: a triangular load sends one third of the total to the support at the zero end and two thirds to the support under the peak — the heavy end takes exactly twice the light end, because the resultant sits twice as close to it. Check: 15 + 30 = 45 kN ✓.
Step 3 — A parabolic shear and a cubic moment
The growing load bends every curve one degree up from the UDL case.
Start at A with the +15 kN reaction and walk right. The load intensity at x is — it keeps growing — so the load accumulated by x is a little triangle of area . The shear therefore falls along a parabola, not a straight line: gently at first where there is almost no load, then ever steeper toward B:
The zero crossing is where the moment peaks — and note it is not midspan. Integrating the shear once more gives a cubic moment curve:
The whole hierarchy moves up one degree from the UDL example: constant load gave linear shear and a parabolic moment; linearly growing load gives parabolic shear and a cubic moment. And the peak slides toward the heavy end — to — because that is where the accumulated load finally cancels the reaction.
Step 4 — The shape tells the same story
The sag leans toward the heavy end — and the resultant trick shows its limit.
The deflected shape says qualitatively what the diagrams said numerically: the beam sags everywhere, but the low point sits just past midspan, nudged toward B — the heavy end pulls the whole curve its way. No closed form needed here; the picture carries the idea.
The proof
Hand calculation vs the solver.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Reaction R_A | 15 kN (W/3) | 15 kN | |
| Reaction R_B | 30 kN (2W/3) | 30 kN | |
| Shear zero V = 0 | at L/√3 = 3.464 m | 3.464 m | |
| Max moment M_max | 34.64 kN·m at 3.464 m (w₀L²/9√3) | 34.64 kN·m |
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.