Beam deflection by Castigliano’s theorem: 22-Mec-A7 Advanced Strength of Materials, December 2016, Q2

Question 2: Beam deflection by Castigliano’s theorem (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Q2 — simply supported span C–A (11 m). Triangular load 2 kN/m at C tapering to zero at B (6 m), point load P = 9 kN at B, and a clockwise couple M = 19 kN·m at the right pin A.

Given. A simply supported beam, pinned at C (x = 0) and A (x = 11 m), carries the loads tabulated below; B is at x = 6 m.

Given data
Span C–A	L = 11 m (C–B = 6 m, B–A = 5 m)
Triangular load	w(x) = 2(1 − x/6) kN/m on 0 ≤ x ≤ 6 (max at C)
Point load at B	P = 9 kN (down)
Couple at A	M = 19 kN·m (clockwise)
Section	E = 205 GPa, I = 165×10⁶ mm⁴ → EI = 3.383×10⁶ N·m²

Find. The vertical deflection of point B.

Approach. Because a real point load P acts exactly at B, Castigliano’s theorem gives the deflection there directly as $\delta_B=\partial U/\partial P=\int_0^L \frac{M}{EI}\,\frac{\partial M}{\partial P}\,dx$; equivalently, apply a unit load at B and integrate $\int M\,m\,dx/EI$.

Reactions as functions of P. With $\sum F_y=0$ and $\sum M_C=0$ (clockwise couple $-M$), $$R_A=\frac{12+6P+19}{11},\qquad R_C=6+P-R_A,$$ so at $P=9$ kN, $R_A=7.73$ kN and $R_C=7.27$ kN (the triangular resultant is 6 kN at $x=2$ m).
Bending moment and its sensitivity. Writing $M(x)$ from the left in each segment (distributed part for $x\le6$, adding $-P(x-6)$ beyond B), the factor $\partial M/\partial P$ is the moment $m(x)$ produced by a unit load at B: $m=\tfrac{5}{11}x$ for $x\le6$ and $m=\tfrac{5}{11}x-(x-6)$ for $x>6$.
Integrate over the span. Evaluating $\displaystyle \delta_B=\frac{1}{EI}\int_0^{11} M(x)\,m(x)\,dx$ with $EI=3.383\times10^{6}\ \text{N}\cdot\text{m}^2$ gives $$\boxed{\delta_B = 5.21\ \text{mm (downward).}}$$

Check: the couple at A is taken clockwise as drawn on the figure (arc sweeping over the top and down the right side). A counter-clockwise reading would change $R_A$ by $2\times19/11=3.45$ kN and reduce $\delta_B$ accordingly; solve with the drawn sense and state the assumption per the exam’s Note 1.

Final results — Q2
Quantity	Value
Reaction R_C	7.27 kN ↑
Reaction R_A	7.73 kN ↑
Deflection at B	5.21 mm (down)

Topic: Deflection of a beam by Castigliano’s second theorem.

Key relations: $\delta=\partial U/\partial P=\int M\,(\partial M/\partial P)\,dx/EI$; equivalently the unit-load method $\int Mm\,dx/EI$.

Why this works: Castigliano’s second theorem states the deflection at, and in the direction of, a load equals the partial derivative of the total strain energy with respect to that load. For bending-dominated members the strain energy is $\int M^2\,dx/2EI$, whose derivative is $\int M(\partial M/\partial P)\,dx/EI$. Since $\partial M/\partial P$ is exactly the moment from a unit load at B, the theorem and the unit-load method are the same integral. Superposing the triangular load, the point load and the end couple keeps $M(x)$ linear in P, so the derivative is clean.

Common pitfall: Forgetting that the applied couple M does not depend on P (so $\partial M/\partial P$ omits its contribution) or dropping the couple from the reactions — it shifts $R_A$ and therefore the whole moment diagram.

Source: Hibbeler, Mechanics of Materials §14.7–14.9 (Castigliano’s theorem); Boresi & Schmidt Ch. 5.

Question 2 of 8: Beam deflection by Castigliano’s theorem

Question 2: Beam deflection by Castigliano’s theorem (20 marks)