QB_AE_STRUC_NEW_013 Section Question Bank MCQ · 1 Mark
Aerospace Structures • Energy Methods - Castigliano's Theorem Deflection

A cantilever beam of length $L$ and uniform flexural rigidity $EI$ is subjected to a uniformly distributed load of intensity $w$ over its entire length, together with a concentrated tip load $P$ acting downward. Using Castigliano's second theorem, the vertical downward deflection $\delta$ at the free tip of the beam is given by:

(A) $\delta = \frac{P L^3}{3 EI} + \frac{w L^4}{8 EI}$
(B) $\delta = \frac{P L^3}{3 EI} + \frac{w L^4}{6 EI}$
(C) $\delta = \frac{P L^3}{8 EI} + \frac{w L^4}{24 EI}$
(D) $\delta = \frac{P L^3}{2 EI} + \frac{w L^4}{8 EI}$
Answer Key:
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Step-by-Step Verified Solution

### 1. Strategy & Theoretical Principles By Castigliano's second theorem, the deflection $\delta$ at the point of application and in the direction of a concentrated force $P$ equals the partial derivative of total complementary strain energy $U$ with respect to $P$: $$\delta = \frac{\partial U}{\partial P} = \int_0^L \frac{M(x)}{EI} \frac{\partial M(x)}{\partial P} \, dx$$ ### 2. Analytical Formulation Let $x$ be measured from the free end ($x = 0$ at tip, $x = L$ at fixed root). The bending moment at section $x$ is: $$M(x) = P x + \frac{w x^2}{2}$$ The partial derivative with respect to tip load $P$ is: $$\frac{\partial M(x)}{\partial P} = x$$ ### 3. Step-by-Step Calculation Substitute into Castigliano's integral: $$\delta = \frac{1}{EI} \int_0^L \left(P x + \frac{w x^2}{2}\right) (x) \, dx = \frac{1}{EI} \int_0^L \left(P x^2 + \frac{w x^3}{2}\right) \, dx$$ Integrating term by term: $$\delta = \frac{1}{EI} \left[ P \frac{x^3}{3} + \frac{w}{2} \frac{x^4}{4} \right]_0^L = \frac{P L^3}{3 EI} + \frac{w L^4}{8 EI}$$ ### 4. Distractor Forensics - **Option A:** Correct result matching linear superposition. - **Option B:** Uses factor $1/6$ from slope expression rather than deflection. - **Option C:** Inverted coefficients between tip and distributed loads. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Note that integrating $\frac{M}{EI} \frac{\partial M}{\partial P}$ directly is far faster than computing $U = \int \frac{M^2}{2EI} dx$ first and differentiating afterwards. - **Standard Reference:** Megson, T.H.G., *Aircraft Structures for Engineering Students*, Chapter 5 (Energy Methods).
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