Energy Methods — Unit Load, Castigliano & Reciprocal Theorems — Structural Analysis Notes

Last reviewed 16 Sept 2026 · 6 min read

Why energy methods

Geometric methods (double integration, moment-area) become tedious for frames, curved members and trusses. Energy methods find displacements from the work done by forces and the strain energy stored, using simple integrals or sums. They also give the extra equations needed for statically indeterminate structures.

All results here assume linear elastic behaviour and small deformations.

Strain energy expressions

Action	Strain energy
Axial force $N$	$\int \frac{N ^{2}}{2 A E} d x$
Bending moment $M$	$\int \frac{M ^{2}}{2 E I} d x$
Shear force $V$	$\int \frac{k V ^{2}}{2 G A} d x$ ( $k$ = shape factor for shear, 1.2 for rectangles)
Torsion $T$	$\int \frac{T ^{2}}{2 G J} d x$

For beams and frames, bending energy dominates; axial and shear terms are usually ignored. For trusses, only axial energy exists.

Principle of virtual work

For a deformable body in equilibrium, if a system of forces in equilibrium undergoes a compatible set of small virtual displacements, the external virtual work equals the internal virtual work.

Two applications:

Virtual displacements applied to real forces → equilibrium equations (e.g. finding reactions, plastic collapse loads).
Virtual forces applied through real displacements → deflections (the unit load method).

Unit load (dummy load) method

To find the displacement $Δ$ at a point in a given direction:

Analyse the structure for the real loads: find $M$ (or $N$ for trusses).
Remove the real loads and apply a unit load at the point in the direction of the required displacement: find $m$ (or $k$ ).
Equate the external work of the unit load with the internal work:

FormulaUnit load method

1 \cdot Δ = \int \frac{M m}{E I} d x (beams and frames)

1 \cdot Δ = \sum \frac{N k L}{A E} (trusses)

For a rotation, apply a unit couple and use the moments it produces. A positive answer means the displacement is in the direction of the unit load.

Product integrals (Vereshchagin's rule)

When one of the two diagrams is linear, $\int M m d x = A_{M} \times m_{c}$ , where $A_{M}$ is the area of the $M$ diagram and $m_{c}$ is the ordinate of the linear $m$ diagram under the centroid of the $M$ diagram. This speeds up frame problems greatly.

Temperature and lack of fit in trusses

Member heated by $T$ : $Δ = \sum k α T L$ .
Member fabricated too long by $λ$ : $Δ = \sum k λ$ .
Beam with temperature gradient (top $T_{1}$ , bottom $T_{2}$ , depth $d$ ): curvature $= α (T_{2} - T_{1}) / d$ , and $Δ = \int m \frac{α ( T _{2} - T _{1} )}{d} d x$ .

Castigliano's theorems

First theorem: $\frac{\partial U}{\partial Δ _{i}} = P_{i}$ (valid for non-linear elastic too).

Second theorem (linear structures):

Δ_{i} = \frac{\partial U}{\partial P _{i}} = \int \frac{M}{E I} \frac{\partial M}{\partial P _{i}} d x, θ_{i} = \frac{\partial U}{\partial M _{i}}

Note that $\partial M / \partial P_{i}$ is exactly the moment $m$ due to a unit load at $i$ — so Castigliano's second theorem and the unit load method give identical working. Where there is no load at the point, apply a fictitious load $Q$ , differentiate, and set $Q = 0$ .

Theorem of least work

For an indeterminate structure with redundant $R$ (and supports that do not yield), the redundant takes the value that makes the strain energy a minimum:

\frac{\partial U}{\partial R} = 0

With several redundants, one equation per redundant. If a support yields by $δ$ in the direction of $R$ , use $\partial U / \partial R = - δ$ (sign as per direction).

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Energy Methods — Unit Load, Castigliano & Reciprocal Theorems