State and prove Castigliano's theorem

Statement

For a linearly elastic structure with supports that do not yield and no temperature effect, the partial derivative of the total strain energy $U$ with respect to any applied force $P_i$ (or couple) gives the displacement (or rotation) of the point of application in the direction of that force:

$$\Delta_i = \frac{\partial U}{\partial P_i}, \qquad \theta_i = \frac{\partial U}{\partial M_i}$$

Proof

Take an elastic body acted on by loads $P_1, P_2, \ldots, P_n$ producing displacements $\Delta_1, \Delta_2, \ldots, \Delta_n$ in the directions of the loads. The strain energy equals the work done by the loads:

$$U = \tfrac12 \left(P_1\Delta_1 + P_2\Delta_2 + \cdots + P_n\Delta_n\right)$$

Case 1. Apply all loads $P_1, \ldots, P_n$ gradually, then add a small extra load $dP_i$ at point $i$. The total strain energy is

$$U + \frac{\partial U}{\partial P_i}\,dP_i$$

Case 2. Apply $dP_i$ first, then the loads $P_1, \ldots, P_n$. The work done is the small energy of $dP_i$ alone, plus $U$ for the loads, plus the work of $dP_i$ moving through the displacement $\Delta_i$ that the loads produce at $i$. Since $dP_i$ is already acting at full value, the last term is $dP_i\,\Delta_i$. Neglecting the second-order term $\tfrac12 dP_i\,d\Delta_i$, the total is

$$U + dP_i\,\Delta_i$$

Comparison. Strain energy depends only on the final state of loading, not on the order. Equate Case 1 and Case 2:

$$U + \frac{\partial U}{\partial P_i}\,dP_i = U + dP_i\,\Delta_i$$ $$\Delta_i = \frac{\partial U}{\partial P_i}$$

Hence the displacement in the direction of a load equals the partial derivative of the strain energy with respect to that load.

Use

For bending, $U = \int \dfrac{M^2}{2EI}\,dx$, so

$$\Delta_i = \int \frac{M}{EI}\,\frac{\partial M}{\partial P_i}\,dx$$

At a point with no load, apply a dummy load $P$, differentiate, then set $P = 0$. For a redundant reaction $R$ at a rigid support, $\partial U/\partial R = 0$; if the support yields by $\delta$ in the direction opposite to $R$ (for example, settles while $R$ acts upward), then $\partial U/\partial R = -\delta$.