FoundationalBeamDeterminateUDLDeflection

Simply supported beam with a UDL

The problem every structural engineer solves first. A single span, a uniform load — and the two formulas ( and ) that anchor everything that comes later. We solve it by statics in three steps, then verify every number against the calculator.

Figure 1.The problem: a 6 m simply supported beam carrying a 10 kN/m UDL over its full span, pinned at A and on a roller at B. Reactions and the deflected shape come after we solve.

Given

: 6 mspan, pin to roller
Supports: Pin · Roller
: 10 kN/mUDL over full span
: 200 GPa
: 100 × 10⁶ mm⁴
: 20 000 kN·m²

Step 1 — Reactions by symmetry

Split the total load between two equal supports.

The total load is = 10 × 6 = 60 kN, and its resultant acts at midspan. Because both the geometry and the loading are perfectly symmetric about the center, neither support can be favored — each must carry exactly half:

You could reach the same answer with = 0 and = 0, but symmetry hands it to you for free. These two reactions are the foundation for both diagrams in the next step.

Step 2 — Shear and moment, drawn together

One straight line, one parabola — and the zero crossing that links them.

Start at A with the +30 kN reaction and walk right. The UDL bleeds shear away at the steady rate w, so the shear falls linearly, crossing zero at midspan:

That zero crossing is the most useful point on the whole sheet, because the moment is the running total of the shear: while V is positive the moment climbs, and the instant V crosses zero the moment peaks. Integrating (or cutting the beam and summing moments to the left) gives the parabola:

Figure 2.Shear force diagram — linear under a UDL, zero at midspan.

Figure 3.Bending moment diagram — the wL²/8 parabola, peaking exactly where the shear crosses zero. We follow the US convention (sagging positive, plotted above the axis); the Beam Calculator has a Flip control for the opposite convention.

Step 3 — The sag, and whether it matters

The classical closed form, then the check a design office actually runs.

The classical midspan result for a simply supported beam under a full UDL comes from integrating the moment twice against the stiffness :

Is 8.44 mm acceptable? Compare it to the span: = 6000⁄8.44 ≈ 711 — comfortably stiffer than a typical L/250 limit, so this beam is fine on deflection.

Figure 4.Deflection diagram — maximum 8.44 mm at midspan.

Figure 5.The solved beam — the problem figure with the answer on it: the support reactions (purple) and the deflected centerline (dashed).

✓

The proof

Hand calculation vs the solver.

Verified — hand calculation vs the solver, to round-off

Quantity	By hand	StructureCalcs
Reaction R_A = R_B	30 kN (wL/2)	30 kN
Max shear \|V\|	30 kN at the supports	30 kN
Max moment M_max	45 kN·m at 3 m (wL²/8)	45 kN·m at 3 m
Max deflection δ_max	8.4375 mm at 3 m (5wL⁴/384EI)	8.438 mm at 3 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.

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