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Castigliano's theorem and unit load method

Short answer

Castigliano's second theorem states that the partial derivative of strain energy with respect to a load gives the corresponding displacement. The unit load method applies a unit load at the point of interest and computes deflection as ∫(M m / EI) dx. Both methods are used for deflections in beams, frames, trusses, and for solving statically indeterminate structures.

State Castigliano's second theorem and prove it for a linearly elastic body. Explain the unit (dummy) load method for finding the deflection of a beam or frame and state where each method is used in engineering.

Strain energy expressions

For a linear elastic member (energy stored equals work done by gradually applied loads):

ActionStrain energy
AxialU=∫N22AE dxU = \displaystyle\int \frac{N^2}{2AE}\,dx
BendingU=∫M22EI dxU = \displaystyle\int \frac{M^2}{2EI}\,dx
TorsionU=∫T22GJ dxU = \displaystyle\int \frac{T^2}{2GJ}\,dx
ShearU=∫αV22GA dxU = \displaystyle\int \frac{\alpha V^2}{2GA}\,dx (α\alpha = shear form factor)

For slender beams the bending term dominates; the others are usually neglected.

Castigliano's second theorem

Statement: For a linearly elastic structure in equilibrium, the partial derivative of the total strain energy (written in terms of the loads) with respect to a load PiP_i equals the displacement δi\delta_i of the point of application of that load, in the direction of the load.

δi=∂U∂Pi,θi=∂U∂Mi\delta_i = \frac{\partial U}{\partial P_i},\qquad \theta_i = \frac{\partial U}{\partial M_i}

Proof: Let loads P1,…,PnP_1,\dots,P_n act and produce energy UU. Increase PiP_i by dPidP_i; the energy becomes U+∂U∂PidPiU + \dfrac{\partial U}{\partial P_i}dP_i.

Now apply the loads in the reverse order: first dPidP_i alone, then P1…PnP_1\ldots P_n. The energy is

12 dPi dδi+U+dPi δi\tfrac12\,dP_i\,d\delta_i + U + dP_i\,\delta_i

The first term is the work of dPidP_i while it is applied alone. The last term is the work of dPidP_i (already acting at its full value) as the other loads are applied and move its point of application through δi\delta_i. The final state, and so the stored energy, does not depend on the order of loading. Equating the two expressions and neglecting the second-order term dPi dδidP_i\,d\delta_i:

∂U∂Pi dPi=δi dPi⇒δi=∂U∂Pi\frac{\partial U}{\partial P_i}\,dP_i = \delta_i\,dP_i \quad\Rightarrow\quad \delta_i = \frac{\partial U}{\partial P_i}

If no load exists at the point of interest, apply a fictitious load QQ (or moment M0M_0), differentiate, then set Q=0Q = 0.

For beams, differentiating under the integral sign:

δ=∫MEI∂M∂P dx\delta = \int \frac{M}{EI}\frac{\partial M}{\partial P}\,dx

Unit (dummy) load method

Apply a unit load (or unit moment for slope) at the point and in the direction of the required deflection, and let mm be the bending moment due to it. By the principle of virtual work (external work 1⋅δ1\cdot\delta = internal work):

1⋅δ=∫M mEI dx1\cdot\delta = \int \frac{M\,m}{EI}\,dx

For trusses: δ=∑F f LAE\delta = \sum \dfrac{F\,f\,L}{AE}, where FF is the member force from the real load and ff from the unit load. Because the derivation uses only virtual work, the method can also be used for non-linear material by replacing M/EIM/EI by the actual curvature.

Use

  • Deflections of beams, bent frames, hooks, curved bars and trusses, where double integration is lengthy.
  • Deflection at a point where no load acts, or when the loading is discontinuous.
  • Solution of statically indeterminate structures: choose the redundant RR and impose a compatibility condition, e.g. ∂U/∂R=0\partial U/\partial R = 0 when the support does not move.

✓ Answer checked by Claude Sonnet 5.5

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