Deflection by the principle of virtual work: 07-Str-A1, May 2014, Q8

Question 8: Deflection by the principle of virtual work (22 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 6 m beam with a 1 m overhang at the left end. Measuring $x$ from the free end ①:

Question 8 — data
Quantity	Value
Nodes	① at $x = 0$ (free), ② at $x = 1$ m (pin), ③ at $x = 4$ m, ④ at $x = 6$ m (roller)
Point loads	5 kN downward at ①; 7 kN downward at ③
Distributed load	2 kN/m over the whole 6 m
Flexural rigidity	$EI = 1.7 \times 10^{3}$ kN·m², constant

Find. The vertical deflection of point ③, by virtual work.

Question 8 — the beam with its reactions, the real bending moment diagram M and the virtual moment diagram m produced by a unit downward load at point 3. The overhang contributes nothing to the integral because m is identically zero there.

Approach. Build the real bending moment function, then the virtual moment function for a unit downward load at point ③, and evaluate $\delta = \int M m \, dx / EI$ piecewise between the discontinuities at $x = 1$ m and $x = 4$ m.

Real reactions. Moments about the pin at $x = 1$ m, with the 12 kN of distributed load acting at $x = 3$ m: $$-5(1) + 12(2) + 7(3) - 5\,R_{4} = 0 \;\Rightarrow\; R_{4} = \frac{40}{5} = \boxed{8.00\ \text{kN} \uparrow}$$ and vertical equilibrium gives $R_{2} = 24 - 8 = 16.0$ kN upward.
Real moment function. Working from the left, $$M = -5x - x^{2} \quad (0 \le x \le 1)$$ $$M = -5x - x^{2} + 16(x - 1) \quad (1 \le x \le 4)$$ $$M = -5x - x^{2} + 16(x - 1) - 7(x - 4) \quad (4 \le x \le 6)$$ Its key values are $M(1) = -6.00$ kN·m over the pin, $M(4) = +12.0$ kN·m under the 7 kN load and $M(6) = 0$, which closes at the roller. The shear changes sign at the 7 kN load, from $+3.00$ kN to $-4.00$ kN, so that $+12.0$ kN·m at point ③ is also the largest sagging moment anywhere on the beam.
Virtual system. Remove every real load and apply a single downward unit force at point ③. Its reactions are $0.4$ at the pin and $0.6$ at the roller, so $$m = 0 \quad (0 \le x \le 1), \qquad m = 0.4(x - 1) \quad (1 \le x \le 4), \qquad m = 0.6(6 - x) \quad (4 \le x \le 6)$$ with $m(4) = 1.20$ m. That $m$ vanishes on the overhang is worth noticing: the overhang contributes nothing at all to the integral even though it carries both a point load and its share of the distributed load.
Integrate the first interval. On $1 \le x \le 4$ the product is $$M m = \left[-x^{2} + 11x - 16\right](0.4)(x - 1)$$ $$\int_{1}^{4} M m \, dx = 0.4\int_{1}^{4}\left(-x^{3} + 12x^{2} - 27x + 16\right)dx = 13.5\ \text{kN}\cdot\text{m}^{3}$$
Integrate the second interval. On $4 \le x \le 6$ the real moment is $-x^{2} + 4x + 12$, so $$\int_{4}^{6} M m \, dx = 0.6\int_{4}^{6}\left(-x^{2} + 4x + 12\right)(6 - x)\,dx = 10.4\ \text{kN}\cdot\text{m}^{3}$$
Assemble the deflection. Adding the two contributions and dividing by the constant rigidity, $$\delta_{3} = \frac{13.5 + 10.4}{1.7 \times 10^{3}} = \frac{23.9}{1.7 \times 10^{3}} = \boxed{0.01406\ \text{m} = 14.06\ \text{mm} \downarrow}$$ The result is positive, so the deflection is in the direction of the unit load, downward, as expected under gravity loading between the supports.
Proportion check. Point ③ is 3 m from the pin and 2 m from the roller in a 5 m span, and 14.06 mm is roughly $L/356$ — large for a serviceability limit but entirely reasonable for a beam whose rigidity is only $1.7 \times 10^{3}$ kN·m², which is a small member.

Question 8 — results
Quantity	Value
Reaction at ② (pin)	16.0 kN up
Reaction at ④ (roller)	8.00 kN up
Real moment at ②	$-6.00$ kN·m
Real moment at ③	$+12.0$ kN·m
Virtual moment at ③	1.20 m
$\int M m \, dx$	23.9 kN·m³
Deflection at ③	14.06 mm downward

Topic. The principle of virtual work (unit-load method) applied to a beam with an overhang.

Key relations. $1 \cdot \delta = \displaystyle\int \frac{M m}{EI}\,dx$, where $M$ is the real bending moment and $m$ the bending moment produced by a unit load applied at the point and in the direction of the required displacement. Break the integral wherever either function changes its algebraic form.

Why this works. The principle of virtual forces states that for any virtual force system in equilibrium, the external virtual work equals the internal virtual work done on the real deformations. Choosing the virtual system to be a single unit load at the point of interest makes the external work numerically equal to the required displacement, and for slender flexural members the internal work reduces to the bending term alone; shear and axial contributions are typically well under one per cent for a span-to-depth ratio of this order. Because the virtual system need only satisfy equilibrium, it can be as simple as the geometry allows — here it is just two reactions and two straight lines.

Common pitfall. Integrating over the overhang. The virtual moment is identically zero from the free end to the pin, so that part of the beam contributes nothing, however heavily it is loaded; carrying it into the integral by mistake typically adds three or four millimetres. The second pitfall is dropping the $-7(x-4)$ term in the last interval, which leaves the moment function open at the roller instead of closing at zero — always check $M(L) = 0$ before integrating.

Source. Hibbeler, Structural Analysis 10th ed., §9.1–9.4; Kassimali, Structural Analysis 6th ed., §7.3–7.5; Leet, Uang and Gilbert, Fundamentals of Structural Analysis 5th ed., Ch. 9.

Question 8 of 8: Deflection by the principle of virtual work

Question 8: Deflection by the principle of virtual work (22 marks)