---
title: "State and prove Maxwell's reciprocal theorem"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:maxwell-reciprocal-theorem"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/maxwell-reciprocal-theorem
site: BCT Engineering
---

# State and prove Maxwell's reciprocal theorem

Theory of Structures II · Chapter 2: Force method

- Asked in: 2072 Chaitra, 4 marks
- Video: [Law of Reciprocal Deflection/ Maxwell Reciprocal theorem](https://www.youtube.com/watch?v=Y4cZU3Y4UFE) — KK  Sir ki classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/maxwell-reciprocal-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

State and prove Maxwell's Reciprocal theorem.

## Short answer

States deflection at A due to unit load at B equals deflection at B due to unit load at A (δ_AB = δ_BA). Proven by equating strain energy from two loading orders; work done by first load through displacement caused by second load is independent of loading sequence, proving symmetry of flexibility matrix.

## Answer

#### 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: $f_{ij} = f_{ji}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Maxwell's reciprocal theorem states that, in a linearly elastic structure:
   A. Deflection at A due to a unit load at B equals the deflection at B due to a unit load at A
   B. Deflection at A equals deflection at B under the same load
   C. Work done by two loads is always zero
   D. Deflection is inversely proportional to load
   Answer: A. The theorem gives delta_AB = delta_BA, the cross-deflections are equal. (Key point: Statement)

2. What fact allows the proof to equate U1 and U2?
   A. Loads are applied suddenly
   B. Strain energy depends only on the final loads, not on the order of loading
   C. The structure is rigid
   D. The loads are equal
   Answer: B. Strain energy is a state function for linear elastic systems, so applying P_A then P_B or P_B then P_A gives the same energy. (Key point: Key idea of the proof)

3. In Case 1, why does the term P_A P_B delta_AB appear with no factor 1/2?
   A. P_B acts gradually
   B. P_A is applied suddenly
   C. P_A is already at full value while it moves through the extra displacement caused by P_B
   D. Both loads are halved
   Answer: C. A constant full force P_A moves through P_B delta_AB, so its work is the full product. (Key point: Cross term in the work)

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