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Energy theorems

Intermediate Structural Analysis Reviewed 24 Oct 2026 2 min read

When a structure deforms under load, the work done by the loads is stored as strain energy in its members. Energy theorems turn that bookkeeping into a way of finding deflections and redundant forces. They are especially handy for frames, curved members and trusses, where drawing deflected shapes by geometry is awkward.

Castigliano's second theorem says that the partial derivative of the total strain energy with respect to a load gives the deflection at that load, in its direction. To find a deflection where there is no load, a dummy load is added, the derivative taken, and the dummy then set to zero. The unit load method, also called virtual work, reaches the same result more directly: apply a unit load at the point and in the direction of the wanted deflection, find the moment m it causes, and integrate M times m over EI along the structure.

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Diagram: Indiacon original, not to scale.

For indeterminate structures, the theorem of least work, a corollary of Castigliano, states that the redundant forces take values that make the strain energy a minimum, so the derivative of energy with respect to each redundant is zero. The Maxwell-Betti reciprocal theorem says the deflection at A due to a unit load at B equals the deflection at B due to a unit load at A; it underlies flexibility matrices and influence line methods.

In bending-dominated members the bending energy is usually all that matters; axial and shear energy become important for trusses, short deep members and arches. These theorems assume linear elastic behaviour and small deflections, so they connect naturally to the matrix stiffness method used in software.

Tip

Use symmetry to integrate over half the span and double it.

Key formulas
  • Bending strain energy U = ∫ M² / (2 E I) dx
  • Castigliano: δ_i = ∂U / ∂P_i
  • Unit load method: δ = ∫ M m / (E I) dx
  • Least work: ∂U / ∂X = 0 for each redundant X
Common mistake

Forgetting that m from the unit load is a different moment diagram from the real M.

Solved example

Problem. A simply supported beam of span 6 m carries a uniform load of 10 kN/m. EI = 5 × 10⁴ kNm². Find the mid-span deflection by the unit load method.

  1. Real moment at distance x from a support: M = w L x / 2 − w x² / 2 = 30 x − 5 x² (kNm).
  2. Unit load at mid-span gives m = x / 2 for 0 ≤ x ≤ 3 m (symmetric).
  3. δ = 2 ∫₀³ (30 x − 5 x²)(x / 2) dx / EI = ∫₀³ (30 x² − 5 x³) dx / EI = [10 x³ − 1.25 x⁴]₀³ / EI = (270 − 101.25) / EI = 168.75 / 5 × 10⁴ m.
  4. δ = 0.003375 m = 3.375 mm. Check with 5 w L⁴ / (384 EI) = 5 × 10 × 1296 / (384 × 5 × 10⁴) = 3.375 mm.

Answer. 3.375 mm downward at mid-span

Quick quiz

1What does Castigliano's second theorem give?

The deflection at a load, as the partial derivative of strain energy with respect to that load.

2How do you find a deflection where no load acts?

Add a dummy (or unit) load there, then set it to zero after differentiating.

3State the reciprocal theorem in one line.

Deflection at A due to unit load at B equals deflection at B due to unit load at A.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • R. C. Hibbeler, Structural Analysis, Pearson, latest edition.
  • C. S. Reddy, Basic Structural Analysis, McGraw Hill Education (India), latest edition.
  • Devdas Menon, Structural Analysis, Narosa, latest edition.
  • A. K. Chopra, Dynamics of Structures, Pearson, latest edition.

Free and official sources

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