A settled pier in a continuous truss
Zero load — and over 100 kN locked into the chords. When the center pier of a continuous truss drops just 5 mm, the redundant support fights the geometry and every member feels it. The same settlement under a determinate truss would do nothing at all. We solve it properly by the flexibility method — releasing the redundant, solving the unit-load case for the n-forces, tabulating δ₁₁ member by member, and superposing — with every number coming from the solver, not asserted.
- Configuration
- Warren, 2 × 6 m bays
- Panels
- four 3 m × 2.5 m high
- Supports
- Pin (1) · Roller (3) · Roller (5)
- Chords
- 2000 mm²
- Diagonals
- 1500 mm²
- 200 GPa
- Settlement
- 5 mm entered −5, at pier node 3
- Indeterminacy
- degree 1m + r = 19 > 2j = 18
- Applied load
- none
Step 1 — Pick the redundant, build the primary structure
One support too many — release it and make the truss determinate.
Count the unknowns: m + r = 15 members + 4 reaction components = 19, against 2j = 18 equations. One redundant — the truss is indeterminate to the first degree, so statics alone can’t solve it. The flexibility method’s move is to choose a redundant and remove it, leaving a determinate primary structure we can solve.
The natural choice here is the extra support — the center roller at node 3 (the pier that settles). Remove it and the truss becomes simply supported over 12 m on the pin at node 1 and the roller at node 5: fully determinate. Call the force that roller carried the redundant X. We’ll find X by demanding the released truss deflect at node 3 by exactly the pier’s settlement.
Step 2 — Solve the primary structure under a unit load
The n-forces: each member’s share of a 1 kN push at node 3.
Apply X = 1 kN downward at node 3 and solve the released truss by the method of joints. By symmetry each support carries 0.5 kN. Walking the joints from the pin:
Joint 1 (pin, reaction 0.5 kN up). The end diagonal 1–6 climbs at sin θ = 0.8575:
Continue joint by joint and the pattern emerges: the bottom chord runs +0.30, +0.90, +0.90, +0.30 toward the center; the top chord −0.60, −1.20, −0.60; the diagonals alternate ±0.583. Rather than write out all nine joints, the figure below draws the released truss solved — these colored forces ARE the n-values.
Step 3 — The flexibility coefficient, member by member
How far node 3 moves per kN — a virtual-work sum over all 15 members.
The deflection at node 3 per unit force there is the virtual-work integral, which for a pin-jointed truss is a simple sum of over every member (the unit-load system and the real system are the SAME n-forces here, so N·n becomes n²):
Tabulating all fifteen members with their solved n-values:
| Member | Type | (m) | (kN) | ||
|---|---|---|---|---|---|
| 1–2 | bottom | +0.3000 | 3.000 | 400,000 | 0.675×10⁻⁶ |
| 2–3 | bottom | +0.9000 | 3.000 | 400,000 | 6.075×10⁻⁶ |
| 3–4 | bottom | +0.9000 | 3.000 | 400,000 | 6.075×10⁻⁶ |
| 4–5 | bottom | +0.3000 | 3.000 | 400,000 | 0.675×10⁻⁶ |
| 6–7 | top | -0.6000 | 3.000 | 400,000 | 2.700×10⁻⁶ |
| 7–8 | top | -1.2000 | 3.000 | 400,000 | 10.800×10⁻⁶ |
| 8–9 | top | -0.6000 | 3.000 | 400,000 | 2.700×10⁻⁶ |
| 1–6 | diagonal | -0.5831 | 2.915 | 300,000 | 3.304×10⁻⁶ |
| 6–2 | diagonal | +0.5831 | 2.915 | 300,000 | 3.304×10⁻⁶ |
| 2–7 | diagonal | -0.5831 | 2.915 | 300,000 | 3.304×10⁻⁶ |
| 7–3 | diagonal | +0.5831 | 2.915 | 300,000 | 3.304×10⁻⁶ |
| 3–8 | diagonal | +0.5831 | 2.915 | 300,000 | 3.304×10⁻⁶ |
| 8–4 | diagonal | -0.5831 | 2.915 | 300,000 | 3.304×10⁻⁶ |
| 4–9 | diagonal | +0.5831 | 2.915 | 300,000 | 3.304×10⁻⁶ |
| 9–5 | diagonal | -0.5831 | 2.915 | 300,000 | 3.304×10⁻⁶ |
| δ₁₁ = Σ n²L/EA | 56.134×10⁻⁶ m/kN | ||||
Every above is the force that member carries in the released truss under the 1 kN unit load — read straight from the solver, not assumed. Chords = 200 × 2000 = 400 000 kN; diagonals 200 × 1500 = 300 000 kN.
The three member families contribute in the ratio the classic hand solution gives — bottom chords 1.35×10⁻⁵, top chords 1.62×10⁻⁵, diagonals 2.64×10⁻⁵ — summing to δ₁₁ = 5.613×10⁻⁵ m/kN.
Step 4 — Compatibility: make the deflection equal the settlement
The one equation that fixes X.
Here is the physical condition. In the real structure the roller is present, so the truss can only move at node 3 as far as the pier lets it: it must drop exactly Δ = 5 mm. In the released structure, the redundant force X alone drives node 3 down by X·δ₁₁. Setting that equal to the settlement:
So the pier pushes up on the truss with X ≈ 89 kN (and by equilibrium the two outer supports react 44.5 kN ↑ each). All from a support that merely dropped five millimeters.
Step 5 — Superpose: the real member forces
Every member force is n × X.
With no external load, the real force in each member is simply the redundant scaled by that member’s n-value — the primary structure carried nothing else:
The center bottom chord (n = 0.9): 0.9 × 89.1 = +80.2 kN (T). The center top chord (n = −1.2): −1.2 × 89.1 = −106.9 kN (C). The diagonals (n = ±0.583): ±51.9 kN. The results figure below draws them all.
The proof
Hand calculation vs the solver.
| Quantity | By hand | StructureCalcs | |
|---|---|---|---|
| Pier reaction (node 3) | 89.07 kN (X = Δ/δ₁₁) | −89.073 kN | |
| Outer reactions (1, 5) | 44.54 kN each | 44.537 kN each | |
| Bottom chord center (2-3, 3-4) | +80.16 kN (T) (n = 0.9) | +80.166 kN | |
| Top chord center (7-8) | −106.9 kN (C) (n = −1.2) | −106.888 kN | |
| Diagonals | ±51.94 kN (n = ±0.5831) | ±51.938 kN | |
| 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.
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.
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