Pitched-roof portal frame — the real industrial shed
A real shed is not flat-roofed. The pitch changes how much load there is, how the rafters carry it, and where the frame’s peak moment sits — which is exactly where the haunches go.
- Span
- 12 m
- Eaves height
- 4 m
- Apex height
- 6 mabout 18° pitch
- 8 kN/malong each rafter
- Bases
- pinned
Part 1 — A roof with a pitch
Five nodes: the apex is the one that matters.
A real industrial shed is not flat-roofed. This one spans 12 m with eaves at 4 m and an apex at 6 m — a pitch of about 18°:
N 0,0 N 0,4 N 6,6 N 12,4 N 12,0
Two columns and two inclined rafters meeting at the apex. Draw both rafters up toward the apex so their local axes mirror one another:
M 1 2 M 2 3 M 4 3 M 5 4
S 1 P S 5 P L 2 -8 L 3 -8
The 8 kN/m acts along each rafter, which is longer than its horizontal projection — so the total load is more than 8 × 12.
Part 2, step 1 — The pitch changes the total
A sloping member is longer than the span it covers.
Each rafter runs from the eaves to the apex — 6 m across and 2 m up:
The solver reports the member length as 6.325 m. With 8 kN/m along each of the two rafters:
And the solver returns 101.2 kN. Had the load been specified on plan it would have been 8 × 12 = 96 kN — a 5% difference from the pitch alone, on a roof that is comparatively shallow.
Part 2, step 2 — The apex is a moment connection
And the peak moment in the whole frame is right beside it.
The frame’s peak moment is 73.96 kN·m, and it lives in the rafters rather than the columns. The apex has to transfer that moment from one rafter to the other — it is not a hinge, it is one of the most heavily worked connections in the building.
The pitch is doing real structural work: it lets the rafters carry load partly in compression along their length rather than purely in bending, which is why a pitched shed spans further than a flat one for the same steel.
Hand calculation vs solver
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Rafter length | √(6² + 2²) = 6.325 m | 6.325 m | |
| Total vertical reaction | 2 × 8 × 6.325 = 101.2 kN | 101.2 kN | |
| Same load taken on plan | 8 × 12 = 96 kN | — (not what was modeled) | |
| Total horizontal reaction | 0 (symmetric gravity) | 0 kN | |
| Peak member moment | indeterminate | 73.96 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.