A member fabricated short: lack of fit
The workshop cut a diagonal 5 mm short, and the crew pulled it into place anyway. In a determinate truss that would cost nothing — the geometry would simply shift. In a redundant one the member fights back: over 100 kN of self-stress locks in, with no load on the structure and every support reaction exactly zero. Here we’ll pin down exactly how much.
- Configuration
- 3 m × 3 m X-braced panel
- Nodes / members
- 4 nodes · 6 membersindeterminate, degree 1
- Area (all)
- 1500 mm²
- 200 GPa
- 300 000 kN
- Diagonal 1–3 length
- 3√2 = 4.243 m
- Misfit
- 5 mm shortentered −5
- Applied load
- none
Step 1 — Count the members: one too many
One more member than statics can resolve — and that is the whole story.
Before we touch the misfit, count. A plane truss offers two equilibrium equations at every joint — 2j = 8 here — against nine unknowns: six member forces and three reaction components. So m + r = 6 + 3 = 9 > 2j = 8, and the panel is indeterminate to the first degree — one member more than statics can resolve.
That one surplus member decides everything that follows. A determinate truss (m + r = 2j) has exactly one geometry compatible with whatever member lengths the workshop delivers: arrive with a diagonal 5 mm short and the truss simply assembles into a very slightly different shape, force-free. This panel has no such freedom. Its members must all agree on one shared geometry, so a diagonal that shows up short can only be connected by stretching itself and squeezing its neighbors. Force is coming — the next steps compute exactly how much.
Step 2 — Cut the misfit member, apply a unit pair
Release the redundant, load it with a unit tension.
The flexibility method turns that surplus member into the unknown. We cut the redundant — member M5 (the diagonal 1–3) — leaving a determinate truss statics can handle, and pull its two cut ends apart with a unit tension pair. The member forces this pair produces are the mode shape n. The classic X-braced-square answer, solvable at any joint:
Read the shape: a unit pull across one diagonal squeezes the square’s four sides (M1–M4) and stretches the opposite diagonal M6 (2–4) — entirely internally. The support reactions under this self-equilibrated pair are already zero, which is the whole reason the finished self-stress state will read zero at every support too.
Step 3 — The flexibility of the cut, by virtual work
One sum condenses the whole panel into a single number.
How hard is it to close a gap at that cut? Virtual work answers with one sum. The flexibility of the cut — how far the cut ends separate per unit of the pair — is summed over every member: the four 3 m sides at , the two 4.243 m diagonals at :
Notice that n enters squared: every member adds flexibility whether its mode force is tension or compression — the squeezed sides help the cut open just as the stretched diagonal does. From here on, nothing else about the truss matters; the whole panel has been condensed into this one number.
Step 4 — Five millimeters becomes 103.55 kN
Five millimeters → over 100 kN, by virtual work.
Member M5 (the diagonal 1–3) must close a ΔL = 5 mm gap, so the redundant force is the misfit divided by that flexibility:
Every other member force follows from :
That is roughly 69 MPa of locked-in stress from five millimeters of mis-cut — before the structure carries a single newton of real load. Fabrication tolerance on a redundant frame is not pedantry.
Step 5 — Why every support reaction is zero
Statics sees nothing; strain gauges see 69 MPa.
A self-stress state is internally balanced: the member set pushes and pulls against itself. At each joint the member forces close into equilibrium without help from a support, so ΣF at every support is exactly zero. Statics, looking only at the boundary, sees an unloaded structure — yet a strain gauge on any member would read a real, permanent force.
The proof
Hand calculation vs the solver.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Short diagonal 1-3 | +103.55 kN (T) — X = ΔL/δ₁₁ | +103.553 kN | |
| Other diagonal 2-4 | +103.55 kN (T) — n = +1 | +103.553 kN | |
| All four sides | −73.22 kN (C) — n = −1/√2 | −73.223 kN | |
| Support reactions | zero (self-stress is internal) | 0 / 0 | |
| 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.
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