What is strain energy?

Strain energy is the energy stored in a body due to elastic deformation under load - the work done by the load in deforming the body. For a gradually applied axial load it equals P^2 L/(2AE); for bending it is the integral of M^2/(2EI) along the member.

What does Castigliano's second theorem state?

Castigliano's second theorem states that the partial derivative of the total strain energy with respect to a load equals the displacement at the point of and in the direction of that load: delta = dU/dP. The derivative with respect to a moment gives the rotation.

How do you find deflection where there is no load using Castigliano's method?

Apply a fictitious dummy load Q at the point and direction of the required deflection, compute strain energy in terms of Q, take dU/dQ, then set Q = 0. The result is the deflection there.

Strain Energy and Castigliano's Theorem — Deflection by Energy Methods with Examples

Energy methods give an elegant alternative to integrating the elastic curve: instead of solving differential equations, you compute the energy stored in a structure and differentiate. Castigliano's theorem is the workhorse, and it handles beams, trusses and frames with the same recipe. This article shows how.

Strain Energy Stored in a Member

When a load is applied gradually, the work it does is stored as recoverable strain energy U. For the three basic actions:

Action	Strain Energy U
Axial force P	P²L / (2AE)
Bending moment M(x)	∫ M² / (2EI) dx
Torque T	T²L / (2GJ)

Strain energy per unit volume (resilience) for direct stress is σ²/2E — the area under the elastic stress-strain line.

Castigliano's Second Theorem

δ = ∂U/∂P     (deflection at, and in the direction of, load P)
φ = ∂U/∂M     (rotation at, and in the direction of, moment M)

The procedure: write U in terms of the load(s), differentiate with respect to the load at the point of interest, and evaluate. If no real load acts there, apply a dummy load Q, differentiate, then set Q = 0.

Worked Example 1 — Cantilever End Deflection

Find the free-end deflection of a cantilever length L carrying an end load W, using Castigliano's theorem.

Bending moment at distance x from the free end: M = −Wx.
U = ∫₀ᴸ M²/(2EI) dx = ∫₀ᴸ (W²x²)/(2EI) dx = W²L³/(6EI)
δ = ∂U/∂W = WL³/(3EI)

This matches the standard formula exactly — derived in three lines with no boundary conditions.

Worked Example 2 — Axial Strain Energy

A steel bar carries an axial pull of 50 kN. Length 2 m, area 500 mm², E = 200 GPa. Find the strain energy stored and the elongation.

U = P²L/(2AE) = (50 000² × 2000) / (2 × 500 × 200 000) = 5×10¹² / 2×10⁸ = 25 000 N·mm = 25 J
δ = ∂U/∂P = PL/AE = (50 000 × 2000)/(500 × 200 000) = 1.0 mm

Worked Example 3 — Deflection Using a Dummy Load

Find the mid-span deflection of a simply supported beam under a UDL using a dummy point load Q at mid-span.

Apply dummy Q at centre; write M(x) including both the UDL and Q.
Compute U = ∫M²/(2EI)dx, then δ = ∂U/∂Q.
Set Q = 0 in the result; the integral evaluates to the known 5wL⁴/384EI.

The dummy-load trick is what makes the method universal — you can find a displacement anywhere, even with no real load there.

Where Energy Methods Win

Deflection of trusses (sum of P²L/2AE over members, then ∂/∂P).
Curved members, frames and redundant structures, where integrating the elastic curve is awkward.
Analysis of statically indeterminate structures (Castigliano's first theorem / theorem of least work).

Common Mistakes

Differentiating before integrating when it is easier to differentiate the integrand first (Leibniz) — both are valid, but keep it consistent.
Forgetting to set the dummy load to zero at the end.
Omitting axial or shear strain energy when it is significant (usually small in slender beams, but not always).