Two-bay portal frame — what the interior column really carries
Add a second bay and something exact appears: the interior column carries about two and a half times the axial load of its neighbours and precisely no moment, because the two rafters cancel each other. It is also the most fragile result in these examples — load one bay only and it vanishes.
- Bays
- 2 × 6 m12 m overall
- Height
- 4 m
- 10 kN/mon each rafter
- 200 GPa
- 100 × 10⁶ mm⁴
- Bases
- all pinned
Part 1 — Two bays, three columns
The interior column is shared, so it only gets modeled once.
Six nodes — three column lines, each with a base and an eaves node:
N 0,0 N 0,4 N 6,4 N 6,0 N 12,4 N 12,0
Then three columns and two rafters. Note that node 3 — the interior eaves — is shared by both rafters and the interior column:
M 1 2 M 2 3 M 4 3 M 3 5 M 6 5
Pin all three bases and load both bays equally:
S 1 P S 4 P S 6 P L 2 -10 L 4 -10
Part 2, step 1 — The interior column takes MORE than twice the load
And tributary area is not the reason why.
Total gravity load across both bays:
The solver returns 120 kN, shared between three columns — but not equally, and not the way a tributary-area sketch would suggest. The exterior columns each carry 26.58 kN; the interior one carries 66.85 kN. That is a ratio of about 2.515 : 1, not 2 : 1.
Tributary area says the interior column collects half a bay from each side and should take 60 kN against 30 kN — but the rafters are continuous over the interior column, not two separate simply-supported spans. Continuity drags load toward the interior support, the same effect that gives a two-span continuous beam its familiar reactions:
On a bare continuous beam that would be 75 kN and 22.5 kN. The frame lands between the two estimates — 66.85 and 26.58 kN — because the columns supply some rotational restraint at the outer ends that a beam on knife-edge supports does not have.
Part 2, step 2 — And carries no moment at all
Exactly zero, and not by accident.
Now the surprise. The exterior columns carry 16.12 kN·m of moment — but the interior column carries 0 kN·m. Exactly zero.
The reason is symmetry, and it is exact rather than approximate. The left rafter arrives at node 3 with an end moment; the right rafter arrives with an equal and opposite one. They cancel, leaving the column head with nothing to resist. The rafters peak at 36.67 kN·m each, and the columns beneath them feel none of it.
An exterior column has no such partner — there is nothing on its outer side to balance the rafter pushing in — so it takes the full unbalanced moment.
Hand calculation vs solver
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Total vertical reaction | w × 12 = 120 kN | 120 kN | |
| Interior column moment | 0 by symmetry | 0 kN·m | |
| Exterior column moment | indeterminate | 16.12 kN·m | |
| Rafter peak moment | indeterminate | 36.67 kN·m | |
| Interior : exterior axial | 2 : 1 tributary — WRONG; continuity raises it | 66.85 : 26.58 kN |
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.