Beam Displacement by an Energy Method: 22-Mec-A7 Advanced Strength of Materials, May 2013, Q1

Question 1: Beam Displacement by an Energy Method (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 CA (pin at C, roller at A), span 7 m, carrying the data below.

Given data
Flexural rigidity	E = 200 GPa, I = 185×10⁶ mm⁴ ⇒ EI = 37 000 kN·m²
Uniform load (over C–B)	w = 5 kN/m, on 0 ≤ x ≤ 4 m
Applied couple at A	M = 11 kN·m (clockwise, at x = 7 m)
Point of interest	B at x = 4 m (distributed-load cut-off)

Find. The vertical displacement of point B, using an energy (unit-load / Castigliano) method.

Simply supported beam: UDL over span C–B (4 m) and an applied couple at roller A; deflection sought at B.

Approach. Compute the reactions, write the real bending moment M(x); apply a virtual unit downward load at B to obtain m(x); the deflection follows from the unit-load form of the strain-energy theorem, δ_B = ∫ M m /(EI) dx.

Support reactions. With the clockwise couple taken as negative, vertical equilibrium and moments about C give $$R_A=\frac{w\!\cdot\!4\,(2)-M}{7}=\frac{20(2)-11}{7}=7.286\ \text{kN},\qquad R_C=20-R_A=12.714\ \text{kN}.$$
Real bending moment. Measuring x from C, $$M(x)=\begin{cases}12.714\,x-2.5\,x^{2}, & 0\le x\le 4\\[2pt]12.714\,x-20\,(x-2), & 4\le x\le 7\end{cases}$$ which correctly returns $M(7)=-11$ kN·m at the applied couple.
Virtual system. Remove the real loads and apply a unit downward force at B (x = 4). Its reactions are $\bar r_C=\tfrac{3}{7}$, $\bar r_A=\tfrac{4}{7}$, so $$m(x)=\begin{cases}\tfrac{3}{7}\,x, & 0\le x\le 4\\[2pt]\tfrac{4}{7}\,(7-x), & 4\le x\le 7.\end{cases}$$
Evaluate the integral. Combining the two spans, $$\int_0^{7}\! M\,m\;dx=\int_0^{4}\!\Big(12.714x-2.5x^{2}\Big)\tfrac{3}{7}x\,dx+\int_4^{7}\!\Big(-7.286x+40\Big)\tfrac{4}{7}(7-x)\,dx=56.86\ \text{kN}\cdot\text{m}^{3}.$$
Deflection. Dividing by the flexural rigidity, $$\delta_B=\frac{1}{EI}\int_0^{7}M\,m\,dx=\frac{56.86}{37\,000}=1.537\times10^{-3}\ \text{m}.$$ $$\boxed{\delta_B \approx 1.54\ \text{mm (downward)}}$$

Question 1 — results
Quantity	Value
Reactions R_C, R_A	12.71 kN, 7.29 kN
∫ M m dx	56.86 kN·m³
Displacement of B	1.54 mm downward

Check: the applied couple is read from the figure as clockwise; a counter-clockwise couple would give R_A = 4.14 kN and δ_B ≈ 2.6 mm. The clockwise reading is used throughout.

Topic: Deflection of a beam by the unit-load (virtual-work) energy method.

Key relations: Unit-load theorem $1\!\cdot\!\delta=\int M m/(EI)\,dx$; equivalently Castigliano’s second theorem $\delta=\partial U/\partial Q$ with a dummy load Q at the point of interest.

Why this works: The virtual unit load creates a self-equilibrated moment field m(x). The internal virtual work of that field acting through the real curvatures M/EI equals the external virtual work of the unit load moving through the true deflection. Integrating the product of the real and virtual moment diagrams therefore extracts exactly the displacement conjugate to the applied unit load. The method needs only static (equilibrium) moment fields, so it applies to any statically determinate beam without solving the differential equation of the elastic curve.

Common pitfall: forgetting that a couple applied at a support still enters the reaction equations and leaves a non-zero moment ordinate at that end (here M(7) = −11 kN·m, not zero); and mismatching the segment limits of M(x) and m(x).

Source: Hibbeler, Mechanics of Materials, Ch. 14 (Energy Methods); Boresi & Schmidt, Advanced Mechanics of Materials, Ch. 5.

Question 1 of 8: Beam Displacement by an Energy Method

Question 1: Beam Displacement by an Energy Method (20 marks)