Overhanging beam with a partial UDL and an applied moment
Three exam classics in one beam: a partial UDL (the resultant trick), an applied moment (which steps the moment diagram without touching the shear), and a point of contraflexure — the spot where sagging turns to hogging and, in concrete, where you switch the reinforcement face. The beam is determinate, so every force falls straight out of statics.
- Span
- A (pin, x = 0) → B (roller, x = 6 m)
- Overhang
- B → free end at x = 8 m
- UDL
- 8 kN/m over 1 m ≤ x ≤ 4 m
- Applied moment
- 15 kN·m (clockwise) at x = 7 m
- 120 × 10⁶ mm⁴span segment
- 60 × 10⁶ mm⁴overhang segment
Step 1 — Read the joint at B before you touch the numbers
A full-height end plate bolted through both sections: moment crosses, so the beam is continuous.
Look closely at the drawing at B (6 m), where the beam changes section: the stiffer span segment (I₁ = 120×10⁶ mm⁴) meets the shallower overhang segment (I₂ = 60×10⁶ mm⁴) directly over the roller. The two pieces are joined end-to-end through full-height plates bolted together — an end-plate moment connection. Because the plates run the full depth and the bolts clamp them at the flange levels, the flanges — the parts of an I-section that carry bending — have a continuous path across the joint. Moment crosses: structurally this is one beam with a stiffness step, not two.
Contrast the alternative: a small plate on the webs alone (a fin plate) passes shear but gives the flanges no path, so it behaves as a hinge. Here that would be fatal — the hinge and the roller would sit at the same point, and the overhang would be free to spin about B under the 15 kN·m applied at x = 7 m: a mechanism, not a structure. The drawn connection is not decoration; the hogging we are about to compute over B has to travel through those bolts.
Step 2 — Reactions by statics
Collapse the UDL to a resultant, then take moments.
The beam is determinate — a pin and a roller, three unknowns, three equations — so the reactions come straight from statics with no compatibility needed. First replace the partial UDL by its resultant: w = 8 kN/m acting over the 3 m from x = 1 to x = 4 gives a single downward force
Now take moments about A. A couple contributes its full magnitude wherever it sits, so the applied moment M₀ = 15 kN·m enters as itself:
and vertical equilibrium closes the other reaction:
Step 3 — The shear diagram and the sagging peak
Where V = 0 the moment peaks — and the whole overhang carries no shear at all.
The shear starts at +R_A = +11.5 kN, slides down through the UDL to 11.5 − 24 = −12.5 kN at x = 4, then holds that value to B, where the reaction closes it exactly to zero. The entire overhang therefore carries no shear at all. The sagging moment peaks where the shear passes through zero, inside the UDL at
Step 4 — Hogging over B, and the point of contraflexure
Where M = 0 the curvature flips — and the overhang is the reason the crossing exists.
Over the support at B the moment has swung the other way:
Somewhere between the peak and B the moment must therefore cross zero. On that stretch , and setting it to zero gives
Past B the moment sits constant at −15 kN·m along the overhang until the applied moment at x = 7 cancels it, leaving the last meter to the free tip completely unstressed.
Step 5 — Deflection with two stiffness segments
Halving I on the overhang moves the shape, not the forces.
Because the beam is determinate, the forces above were found without ever mentioning EI — halving the second moment of area on the overhang (I₂ = 60×10⁶ mm⁴ versus I₁ = 120×10⁶ mm⁴ on the span) changes no reaction and no moment. It only changes the deflected shape. The span dips about 2.5 mm inside the loaded region, and the free tip rides up as the span rotates back over the support at B.
The proof
Hand calculation vs the solver.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| RA | 11.5 kN | 11.5 kN | |
| RB | 12.5 kN | 12.5 kN | |
| Max sagging moment | 19.766 kN·m at 2.4375 m | 19.766 kN·m at 2.437 m | |
| Hogging at B | −15 kN·m | −15 kN·m at 6 m | |
| Max shear |V| | 12.5 kN (x = 4 m B) | 12.5 kN | |
| Contraflexure | M = 0 at x = 4.8 m | BMD crosses zero at 4.8 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.
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