Portal frame under lateral wind — sway and the windward/leeward split
Push the same portal sideways and it stops being a mirror image of itself. One base lifts, the other presses down, and the frame leans — the mode that governs most real portal designs.
- Span
- 6 m
- Height
- 4 m
- 15 kNlateral, at the windward eaves
- 200 GPa
- 100 × 10⁶ mm⁴
- Bases
- pinned
Part 1 — Build the frame and push it sideways
A nodal load, not a member load — wind arrives at the frame, not along it.
Same portal geometry as the gravity examples:
N 0,0 N 0,4 N 6,4 N 6,0 M 1 2 M 2 3 M 4 3 S 1 P S 4 P
Now the load. Wind on a shed is resisted by the cladding and delivered into the frame at the eaves, so it is applied as a nodal load pushing in the +x direction at node 2 — not as a UDL along a member:
P 2 15 0
Read as “at node 2, a horizontal force of 15 kN and a vertical force of zero”. In a real design this comes from the wind pressure times the tributary area of one frame — the spacing between frames, times the height they catch.
Part 2, step 1 — Horizontal equilibrium, and what it does not tell you
The total is free; the split is not.
The 15 kN has to be resisted by the two bases:
The solver returns 15 kN total, exactly as required. But statics stops there. How the 15 kN divides between the two columns depends on their relative stiffness, and with identical columns the split is even — change one column’s section and it stops being.
There is also no net vertical load, so — and the solver agrees, returning 0 kN. But the two bases are individually not zero, which is the interesting part.
Part 2, step 2 — One base lifts, the other presses down
The overturning couple, and the reason wind rips buildings off their footings.
The lateral load tries to tip the frame over, and this part is pure statics. Take moments about one base: the 15 kN acts 4 m up, and the only thing available to resist it is a vertical couple acting across the 6 m span.
The solver returns -10 kN and 10 kN — the hand value exactly. They sum to zero, because no net vertical load was applied, but individually they are equal and opposite, which means one of the feet is in uplift.
Under gravity alone this frame put 30 kN of compression into each base. Under wind alone, one base wants to lift off the ground entirely.
Part 2, step 3 — Sway
The deflection that usually governs, and it is not the one people check first.
The frame leans. The peak horizontal movement is 14.08 mm, and the deflected shape shows the whole frame moving with the load rather than any one member sagging.
Sway is usually checked as a fraction of height — is a common serviceability limit for a portal, which for these 4 m columns would be about 27 mm. That is the number a real design is measured against, and it very often decides the section long before strength does.
Hand check vs solver
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Total horizontal reaction | 15 kN (applied load) | 15 kN | |
| Total vertical reaction | 0 (no vertical load) | 0 kN | |
| Vertical at node 1 (uplift) | −Hh/L = −10 kN | -10 kN | |
| Vertical at node 4 | +Hh/L = +10 kN | 10 kN | |
| Peak sway | indeterminate | 14.08 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.