Support settlement in a continuous beam
No load anywhere on the beam — and yet it carries 36 kN·m of moment. When a support of an indeterminate beam settles, the other supports refuse to let it follow freely, and the fight between them bends the beam. A determinate beam would simply tilt and feel nothing. This is why a few millimeters of settlement matter so much more for continuous structures — and here we’ll see exactly how much.
- Configuration
- 2 spans × 5 m
- Supports
- Pin · Roller · Roller
- 200 GPa
- 150 × 10⁶ mm⁴
- 30 000 kN·m²
- Settlement
- 10 mm entered −10
- Applied load
- none
Step 1 — Release the redundant support
One support too many for statics — so take it away and ask a geometric question instead.
The beam is indeterminate to the first degree — one support too many for statics alone. The compatibility method turns that extra support into the star of the show: release the middle support B, and what remains is a simply supported beam over 2L = 10 m, which statics can handle.
Now watch what the released beam does. Unloaded, it stays perfectly straight. But the real support is no longer on that straight line — it has settled to sit 10 mm below the chord, and the real beam must pass through it. So the question that replaces our missing equation is purely geometric: what downward force F at midspan produces exactly that 10 mm deflection?
Step 2 — The force the settlement demands
Find the force the settled support must carry.
A central point load deflects a simply supported beam by , so setting that equal to Δ:
That force is the reaction the settled support pulls with. Equilibrium of the whole beam then splits it between the two ends:
Look at the directions. The middle support pulls down: forced through the dropped support, the beam would spring back up if it could, and B must hold it there. In the calculator’s down-is-negative convention that reaction reports as −14.4 kN — the same convention the settlement itself is entered with (10 mm down → −10).
Step 3 — The moment it bends into the beam
No load, yet 36 kN·m at the settled support.
Take moments from the left end to the settled support — only acts over that span:
Step 4 — The picture the compatibility condition draws
The beam is dragged down to meet the support.
The deflected shape makes the whole argument visible: the beam bends down to exactly −10 mm at B, meeting the settled support, and rises back to zero at the two ends it’s still pinned to. That single geometric condition — “the beam must pass through the dropped support” — is what generated every force above.
The proof
Hand calculation vs the solver.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Induced moment at B | 36 kN·m (3EIΔ/L²) | 36 kN·m @ 5 m | |
| End reactions R_A, R_C | 7.2 kN | 7.2 kN | |
| Support reaction R_B | 14.4 kN | −14.4 kN | |
| Deflection at B | −10 mm | −10 mm @ 5 m | |
| Applied load | zero | zero |
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.