Beam Deflection at B by Castigliano’s Theorem: 22-Mec-A7 Advanced Strength of Materials, December 2013, Q4

Question 4: Beam Deflection at B 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.

Given. A simply supported beam C–A (span 9 m) with a full-span UDL and an end couple, deflection sought at the interior point B.

Given data
Flexural rigidity	E = 210 GPa, I = 285×10⁶ mm⁴ ⇒ EI = 59 850 kN·m²
Uniform load	w = 10 kN/m over the entire span 0 ≤ x ≤ 9 m
Applied couple at A	M = 20 kN·m (clockwise, at x = 9 m)
Point of interest	B at x = 5 m from C

Find. The vertical displacement of point B.

Full-span UDL plus an end couple at roller A; deflection sought at interior point B (x = 5 m).

Approach. Castigliano’s deflection theorem δ_B = ∫ (M/EI)(∂M/∂Q) dx is applied with a dummy load Q at B. Since no real point load acts at B, the operation is equivalent to the unit-load method: form M(x), the sensitivity ∂M/∂Q, integrate, then set Q = 0.

Support reactions. Taking the clockwise couple as negative and moments about C, $$R_A=\frac{w L^{2}/2+M}{L}=\frac{10(9)^2/2+20}{9}=47.22\ \text{kN},\qquad R_C=wL-R_A=42.78\ \text{kN}.$$
Real bending moment. Measuring x from C (the end couple enters only through the reactions), $$M(x)=R_Cx-\tfrac{w}{2}x^{2}=42.78\,x-5\,x^{2},$$ which correctly returns $M(9)=-20$ kN·m at the applied couple.
Sensitivity to a dummy load at B. A downward Q at x = 5 raises $\partial R_C/\partial Q=\tfrac{4}{9}$, giving $$\frac{\partial M}{\partial Q}=\begin{cases}\tfrac{4}{9}\,x, & 0\le x\le 5\\[2pt]\tfrac{4}{9}\,x-(x-5)=\tfrac{5}{9}(9-x), & 5\le x\le 9.\end{cases}$$
Evaluate the integral. Combining the two spans, $$\int_0^{9} M\,\frac{\partial M}{\partial Q}\,dx=\int_0^{5}(42.78x-5x^{2})\tfrac{4}{9}x\,dx+\int_5^{9}(42.78x-5x^{2})\tfrac{5}{9}(9-x)\,dx=738.0\ \text{kN}\cdot\text{m}^{3}.$$
Deflection. Dividing by EI, $$\boxed{\delta_B=\frac{738.0}{59\,850}=0.01233\ \text{m}=12.3\ \text{mm (downward).}}$$

Results — Question 4
Quantity	Value
Reaction R_C (pin)	42.78 kN ↑
Reaction R_A (roller)	47.22 kN ↑
∫ M(∂M/∂Q) dx	738.0 kN·m³
Displacement of B, δ_B	12.3 mm downward

Topic: Castigliano’s second theorem (unit-load method) for the deflection of a statically determinate beam under distributed load and an applied couple.

Key relations: δ = ∂U/∂Q = ∫ (M/EI)(∂M/∂Q) dx. When no real load sits at the target point, a dummy load Q is introduced, differentiated, then set to zero — identical to multiplying the real M(x) by the virtual-unit-load moment m(x).

Why this works: the complementary strain energy of a beam is ∫ M²/(2EI) dx; its derivative with respect to a load applied at a point equals the displacement of that point in the load’s direction. Only bending energy is retained, as is standard for slender beams.

Common pitfall: trying to insert the end couple into M(x) as a step at x = 9 — it is a boundary action and enters solely through the reactions; and mismatching units (keep kN, m so that EI = 59 850 kN·m² and δ comes out in metres).

Source: Hibbeler, Mechanics of Materials 10th ed., §14.7–14.9 (Castigliano’s theorem and the method of virtual work for beams).

Question 4 of 7: Beam Deflection at B by Castigliano’s Theorem

Question 4: Beam Deflection at B by Castigliano’s Theorem (20 marks)