AdvancedTrussSelf-stressThermalVirtual work

A heated member in a redundant truss

Nothing touches this truss — no load, no settlement, no mis-cut member. The afternoon sun merely warms one diagonal by 40 °C. In a determinate truss that would cost nothing: the member grows 2 mm and the geometry quietly shifts. This frame — the fabrication example’s X-braced square, one member too many — refuses. The growth is contested, ≈28 MPa of compression locks into the heated diagonal, and every support reaction stays exactly zero. Here we’ll compute the whole self-stress state by hand.

Figure 1.The problem: the same 3 m × 3 m X-braced panel as the fabrication example — 4 nodes, 6 members, pin at node 1, roller at node 2. Member M5 (the diagonal 1–3) is heated +40 °C; nothing else touches the frame. No applied load — the self-stress comes after we solve.
Given
Frame
3 m X-braced squarethe fabrication example’s
Heated member
diagonal 1–3, +40 °CL = 3√2 = 4.243 m
1.2×10⁻⁵ /°C
Area (all)
1500 mm²
200 GPa
Load
none — that is the point
1

Step 1 — Temperature is a load

A 40 °C afternoon, expressed in millimeters.

Heat a bar and it wants to be longer. Free of any restraint, a member warmed by ΔT grows by — pure strain, zero stress. For the diagonal 1–3, at 4242.6 mm the longest member in the panel:

Two millimeters — and in a determinate truss that would be the end of the page. The geometry would simply shift to accommodate the longer member and no force would appear anywhere, exactly the force-free freedom the settlement example’s determinate spans enjoyed. But this panel has one member too many: m + r = 6 + 3 = 9 > 2j = 8, indeterminate to the first degree. Six members must agree on one shared geometry, and a diagonal that turns up 2.0365 mm too long — for any reason — can only stay connected by being squeezed. The growth is contested; the next steps compute the price.

2

Step 2 — The same cut as the fabrication page

δ₁₁ belongs to the frame and the cut — not to what caused the misfit.

This is deliberately the same frame and the same cut as the lack-of-fit worked example. We cut the heated diagonal — member M5 (1–3) — leaving a determinate truss, and pull the cut ends with a unit tension pair. The resulting mode shape n is the classic X-braced-square answer: in both diagonals, in all four sides. The flexibility of the cut is over every member:

Identical to the fabrication page’s — as it must be. Flexibility is a property of the frame and the cut, not of whatever caused the misfit: that page divided a 5 mm shortfall by this same number to get its 103.55 kN. Mis-cut steel or warm steel, the frame’s answer to “how hard is it to move this cut?” never changes.

3

Step 3 — Compatibility

The frame refuses the 2 mm — and charges 42 kN for it.

The member wants to grow 2.0365 mm; the frame won’t let it. Compatibility demands a redundant force that cancels the growth at the cut:

Read the sign. In the fabrication example the diagonal arrived short and had to be stretched into place — X came out positive, tension. Here the member tried to grow, so the frame pushes back on it: X is negative — the heated diagonal ends up in compression, squashed by the very members it is bolted to.

4

Step 4 — The locked-in state

28 MPa of compression from sunshine — with nothing applied.

Every member force follows from :

The frame squashes the member that tried to grow — and its untouched companion diagonal with it — while all four sides go into tension holding the square together. In the heated diagonal that is

(compression)

— from a 40 °C temperature rise and zero applied load. Sunshine, expressed in megapascals.

Figure 2.The locked-in thermal state — members colored red (tension) / blue (compression): all four sides at +29.82 kN tension, both diagonals at −42.18 kN compression, and every support reaction zero. A self-stress state with zero external load.
5

Step 5 — Zero reactions: a self-stress state

The supports feel nothing; the members feel everything.

The mode shape n was self-equilibrated — a unit pair across one diagonal balances entirely against the other five members — so the finished state is too. At every joint the member forces close into equilibrium without help from the ground: the pin at node 1 and the roller at node 2 both read exactly zero. Statics, watching the boundary, sees an unloaded structure; a strain gauge on any member reads a real, permanent force.

The proof

Hand calculation vs the solver.

Verified — hand calculation vs the solver, to round-off
QuantityBy handStructureCalcs
Both diagonals (heated 1–3 and companion 2–4)−42.18 kN (C) — X = −ΔL/δ₁₁, n = +1−42.18 kN
All four sides+29.82 kN (T) — n = −1/√2+29.82 kN
All reactions0 (self-stress)0

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.

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