--- title: "State and prove Castigliano's theorem" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-second-theorem-prove" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-second-theorem-prove site: BCT Engineering --- # State and prove Castigliano's theorem Theory of Structures II · Chapter 2: Force method - Asked in: 2078 Bhadra, 6 marks; 2067 Asar, 5 marks (asked 2 times) - Video: [CASTIGLIANO'S THEOREM in Just Over 10 Minutes!](https://www.youtube.com/watch?v=JH6iOhOe8h0) — Less Boring Lectures - 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/castigliano-second-theorem-prove - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question State and prove Castigliano's second theorem (theorem for determination of displacement in a structural system). ## Short answer Castigliano’s second theorem states that for a linearly elastic structure the partial derivative of strain energy U with respect to an applied force P_i gives the corresponding displacement Δ_i: Δ_i = ∂U/∂P_i. It is proved by equating the total strain energy for two loading sequences (applying all loads then a small increment dP_i, versus applying dP_i first) and neglecting second‑order terms. ## Answer #### 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 $\partial U/\partial R = -\delta$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Castigliano's second theorem says the displacement in the direction of P_i equals: A. dU/dP_i B. U/P_i C. P_i dU D. integral of U dP_i Answer: A. The partial derivative of strain energy with respect to the load gives the displacement. (Key point: Statement) 2. The proof equates two loading orders because: A. Loads are equal B. Strain energy depends only on the final state, not the order of loading C. Displacements are zero D. Work is not conserved Answer: B. U is a state function, so both sequences give the same energy. (Key point: Proof idea) 3. In Case 2 the extra term dP_i x Delta_i has no 1/2 because: A. dP_i is already acting at full value while the loads displace point i B. Delta_i is small C. Work is doubled D. Loads are removed Answer: A. dP_i acts fully while the displacement Delta_i occurs, so the work is dP_i Delta_i. (Key point: Case 2 term) 4. To find displacement at an unloaded point, we: A. Cannot use the theorem B. Apply a dummy load P, differentiate, then set P = 0 C. Use 2U D. Differentiate with respect to EI Answer: B. A dummy load is applied at the point and set to zero afterwards. 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