State and prove Maxwell's reciprocal theorem

Statement

In a linearly elastic structure, the deflection at point A due to a unit load at point B equals the deflection at B due to a unit load at A:

$$\delta_{AB} = \delta_{BA}$$

Here $\delta_{ij}$ is the displacement at $i$ caused by a unit load at $j$.

Proof

Let loads $P_A$ and $P_B$ act at A and B. Let $\delta_{AA}, \delta_{BA}$ be the deflections at A and B caused by a unit load at A, and $\delta_{AB}, \delta_{BB}$ those caused by a unit load at B.

Case 1: apply $P_A$ first, then $P_B$.

• $P_A$ gradually applied: work $= \tfrac12 P_A(P_A\delta_{AA})$.
• $P_B$ then applied: its own work $= \tfrac12 P_B(P_B\delta_{BB})$. While $P_B$ is applied, $P_A$ (already at full value) moves through the extra displacement $P_B\delta_{AB}$ at A and does work $P_A(P_B\delta_{AB})$.

$$U_1 = \tfrac12 P_A^2\delta_{AA} + \tfrac12 P_B^2\delta_{BB} + P_AP_B\,\delta_{AB}$$

Case 2: apply $P_B$ first, then $P_A$. In the same way,

$$U_2 = \tfrac12 P_B^2\delta_{BB} + \tfrac12 P_A^2\delta_{AA} + P_BP_A\,\delta_{BA}$$

Strain energy depends only on the final loads, not on the order of loading, so $U_1 = U_2$. Hence

$$P_AP_B\,\delta_{AB} = P_AP_B\,\delta_{BA} \ \Rightarrow\ \delta_{AB} = \delta_{BA}$$

Example

A cantilever of length $L$: the tip deflection due to a unit load at mid-span is $\dfrac{5L^3}{48EI}$, which equals the mid-span deflection due to a unit load at the tip. The theorem also holds for rotations and for a force-couple pair, so the flexibility matrix is symmetric: $f_{ij} = f_{ji}$.