Loads That Are Not Applied at the Joints
The stiffness equation has one place for load: F, the forces at the joints. But almost nothing in a real structure is loaded at a joint. Slabs put uniform load along beams, wind puts pressure along columns, and a point load lands wherever the machine sits. The method handles this with a trick that costs nothing and is exact: solve a different problem, then add back the difference.
Clamp every joint, find the forces the clamps carry (the fixed end forces), apply the reverse of those forces to the real structure, and add the two together. The joints move as they should, and each member's own load is added back at the end.
The two-problem superposition
Because the analysis is linear, any load can be split into two problems whose sum is the original.
Problem 1 — the restrained structure
Imagine every joint clamped: no translation, no rotation, anywhere. Each loaded member is now a fully fixed-ended beam under its own load, and the forces its ends exert on the clamps are the standard fixed end forces, f̄0. They are read from a table — the same table used in moment distribution — and no joint has moved, so this problem produces no displacements at all.
Problem 2 — the equivalent joint loads
The clamps do not exist, so their forces have to be removed. Sum each joint's clamping forces, reverse the sign, and apply that as a joint load on the real structure:
where f0 is each member's fixed end force vector rotated into global axes, and the sum runs over the members meeting at that joint. This is the problem the stiffness equation actually solves, and its displacements are the true ones — because in Problem 1 nothing moved.
The table every program carries
For a prismatic member of length L, ends i and j, with sagging moment at the near end taken as positive:
| Uniform load w | Shears wL/2 at each end; moments wL²/12 at each end, opposite in sense. |
|---|---|
| Point load P at a from i, b = L − a | Moment Pab²/L² at i and Pa²b/L² at j; shears Pb²(L+2a)/L³ and Pa²(L+2b)/L³. |
| Point load at mid-span | Shears P/2 at each end; moments PL/8 at each end. (The case above with a = b = L/2.) |
| Triangular load, zero at i to w at j | Moments wL²/30 at i and wL²/20 at j; shears 3wL/20 and 7wL/20. |
| Applied moment M at a | End moments Mb(2a−b)/L² and Ma(2b−a)/L²; shears 6Mab/L³, opposite in sense. |
| Uniform temperature rise ΔT | Axial force EAαΔT, compressive, with no shear or moment. |
A member with a released end uses its own row of the table — a propped cantilever's fixed end moment is wL²/8 at the fixed end and zero at the pin, not wL²/12 at both — which is the same condensation described in the guide on member stiffness, applied to the load instead of to the stiffness.
Recovering the truth afterwards
Here is where analyses go wrong by hand. The equivalent joint loads were a device; they are not the load the member carries. Once the displacements u are known, each member's end forces are
— the forces caused by its ends moving, plus the fixed end forces put back. Forget the second term and a uniformly loaded beam comes out with no mid-span moment at all, which is the classic symptom.
The same applies to the diagrams. Between the two ends, a member under distributed load has a shear that varies linearly and a moment that varies parabolically, and neither is visible in the end forces alone. The diagram is drawn by starting from the end force and integrating the span load along the member — which is why a program can show the true parabola between two joints that have nothing between them.
A check that never fails
A simply supported beam, span L, uniform w, modelled as one member. The fixed end moments wL²/12 are reversed onto the joints; the joints rotate under them; the end moments from those rotations come back as −wL²/12, and the sum at each support is zero, as a pin demands. Mid-span, the parabola built from the end shear reaches wL²/8. If a program gives you that, its superposition is right.
Two things worth watching
- Displacements are exact, in between is interpolated. The joint displacements from this procedure are the exact solution of the beam equations for these loads. The deflected shape drawn along a member is built from the end displacements and the span load — accurate for the standard load types, and the reason a program subdivides a member when the loading is unusual.
- Projected loads are not member loads. Snow quoted per metre of plan and wind quoted per metre of elevation act on the horizontal or vertical projection of a sloping member, so the intensity along the member is smaller: wmember = wplan cos θ. Get this wrong on a roof and every rafter is out by that factor.
Next: the members are ready — each with its stiffness in global axes and its share of the load. Assembly is what makes them a structure.