Strain Energy and Castigliano
The area under a load-extension line is a triangle, which is where the one-half comes from — and why an impact deflects twice as far.
Skip to the animationThe work done by a gradually applied load is stored as strain energy equal to the area under the load-extension line — a triangle, hence the one-half — and differentiating that energy with respect to a load gives the deflection at it.
Where the one-half comes from
As a bar stretches, the load doing the stretching rises with it. The work done is the area under the load-extension curve — a triangle — so U = ½·P·δ, not P·δ.
Apply the full load suddenly and the work is the rectangle P·δ, twice as much, so the structure deflects twice as far. That is the origin of the impact factor of 2: it is geometry, not a safety margin.
Expressions for each action
| Action | Strain energy | Note |
|---|---|---|
| Axial | U = P²L / 2AE | The only term for a truss |
| Bending | U = ∫ M² dx / 2EI | Dominant in beams and frames |
| Torsion | U = T²L / 2GJ | Shafts and grid structures |
| Shear | U = ∫ V² dx / 2AG | Usually negligible; matters for deep beams |
Every one has the form (action)² × length over twice the stiffness. They are energies, so they simply add.
Castigliano's theorems
- Second theorem
δᵢ = ∂U/∂Pᵢ— the deflection at a load, in that load's direction. This is the one used constantly.- First theorem
Pᵢ = ∂U/∂δᵢ— the force required to produce a given displacement. The basis of the stiffness method.
The second theorem works because increasing Pᵢ slightly stores extra energy equal to the deflection times that increase — so the derivative *is* the deflection. It replaces integrating a differential equation with differentiating an expression.
To find a deflection where no load acts, apply a dummy load Q there, carry it symbolically through, differentiate, then set Q = 0. The load never existed; it exists solely to create something to differentiate with respect to.
Maxwell's reciprocal theorem
The deflection at B due to a load at A equals the deflection at A due to the same load at B, in any linear elastic structure however complicated. It follows because strain energy is a function of the loads and mixed partial derivatives commute.
Betti's theorem generalises it to whole load systems. Maxwell is also what makes the flexibility matrix symmetric in the force method, and what underlies Müller-Breslau's principle for influence lines.
Where energy methods sit
They give one deflection, at one point, in one direction — exactly what a design check asks for, and useless for a full deflected shape. They scale to trusses and frames where double integration is impractical.
The assumptions are linear elastic behaviour and small deflections throughout: no yielding, no buckling, no geometry changes large enough to matter.
The numbers you will be asked for
- Strain energy
U = ½·P·δ
The area of a triangle — the load rose from zero.
- Axial
U = P²L / 2AE
Sum over members for a truss.
- Bending
U = ∫ M² dx / 2EI
Integrate over the length.
- Castigliano II
δᵢ = ∂U / ∂Pᵢ
Deflection at a load, in its direction.
- Maxwell
δ_AB = δ_BA
Any linear elastic structure.
- Impact factor
δ_sudden = 2 · δ_gradual
Rectangle against triangle.
Advantages and disadvantages
Advantages
- Handles trusses and frames where integration methods are impractical.
- Gives a deflection directly, with no differential equation to solve.
- The dummy load trick reaches points with no load on them.
- Maxwell's theorem falls out for free, and is used everywhere afterwards.
Disadvantages
- One deflection at a time — no deflected shape.
- The dummy load must be carried symbolically, which gets messy on large structures.
- Assumes linear elastic behaviour and small deflections.
- Differentiating an integral is error-prone by hand, which the unit load method tidies up.
Watch it work
Check yourself
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One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.