16-Civ-A1 Elementary Structural Analysis · Undated paper

Question 8 of 8: Horizontal deflection of a frame by virtual work

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Engineers Canada / EGBC national examination 16-Civ-A1 Elementary Structural Analysis, 3 hours, CLOSED BOOK. Six questions constitute a complete paper: answer Q1–Q5 and one of Q6/Q7/Q8. For completeness all eight questions are worked here.

Reference texts: R. C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy and stability (Ch. 2), method of joints and sections (Ch. 3), shear and bending-moment diagrams (Ch. 4), influence lines and moving loads (Ch. 6), virtual-work deflections (Ch. 8–9), slope-deflection and moment distribution (Ch. 10–11); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged frames, Maxwell’s law of reciprocal deflections.

Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside of a beam), hogging negative; member axial force tension positive (T), compression negative (C). Shear is positive when the resultant of the forces on the left of a section acts upward.

Check: figure reading. All geometry, loads and support symbols used below were read directly from the printed figures. Where a support symbol governs the answer it is called out explicitly in the Given.

Question 8: Horizontal deflection of a frame by 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 vertical member runs from a pin support at joint 1 (0, 0) through joint 2 (0, 4) to a free top at joint 3 (0, 8). A horizontal beam 8 m long runs from joint 2 to a roller support at joint 4 (8, 4). A uniformly distributed horizontal load of 6 kN/m pushes to the right over the whole 8 m height of the vertical member (resultant 48 kN at $y = 4\text{ m}$). All members are inextensible with $EI = 1.8 \times 10^{5}\text{ kN}\cdot\text{m}^{2}$.

Find. The horizontal deflection of joint 2.

Q8 — the frame. The wind load runs the full height of the vertical member, including the free cantilever above joint 2.

Approach. Confirm the frame is determinate, write the real bending-moment field $M$, then apply a unit horizontal load at joint 2 and write the virtual field $m$. Because the members are inextensible, only the flexural term of the virtual-work integral survives.

Determinacy. With $m = 3$ members, $j = 4$ joints and $r = 3$ (pin plus roller), $$\text{DSI} = 3(3) + 3 - 3(4) = 0$$ so the frame is determinate and $M$ can be written from statics alone.
Real reactions. Horizontal equilibrium fixes the pin thrust, and moments about joint 1 fix the roller: $$A_x = 48\ \text{kN} \leftarrow, \qquad R_4(8) = 48(4) \;\Rightarrow\; R_4 = \boxed{24\ \text{kN} \uparrow}$$ and vertical equilibrium then gives $A_y = 24\text{ kN} \downarrow$.
Real moment field. On the lower column, measuring $y$ up from the pin and taking the free body below the cut, $$M(y) = -48y + 3y^{2} \qquad (0 \le y \le 4)$$ which is $-144\text{ kN}\cdot\text{m}$ at joint 2. On the cantilever above joint 2, working from the free top, $$M(y) = -3(8 - y)^{2} \qquad (4 \le y \le 8)$$ which is $-48\text{ kN}\cdot\text{m}$ at joint 2. On the beam, working from the roller, $$M(x) = 24(8 - x)$$ which is $+192\text{ kN}\cdot\text{m}$ at joint 2. Joint equilibrium confirms the set: $-144 - 48 + 192 = 0$.
Virtual system. Remove the real load and apply a single 1 kN horizontal force at joint 2, in the direction of the deflection sought (to the right). Then $a_x = 1\text{ kN} \leftarrow$ and, by moments about joint 1, $r_4 = 1(4)/8 = 0.5\text{ kN} \uparrow$.
Virtual moment field. On the lower column $m(y) = -y$, reaching $-4\text{ kN}\cdot\text{m}$ at joint 2. On the cantilever above joint 2 there is no virtual load and no virtual reaction, so $$m = 0 \qquad (4 \le y \le 8)$$ On the beam $m(x) = 0.5(8 - x)$, reaching $+4\text{ kN}\cdot\text{m}$ at joint 2.
Assemble the virtual-work integral. For inextensible members with shear deformation neglected, $$1 \cdot \Delta = \int \frac{M m}{EI}\,ds$$ Member 1–2: $$\int_0^{4}\left(-48y + 3y^{2}\right)(-y)\,dy = \int_0^{4}\left(48y^{2} - 3y^{3}\right)dy = 1024 - 192 = 832$$ Member 2–3 contributes nothing because $m = 0$ there, even though $M$ is large. Member 2–4: $$\int_0^{8} 24(8 - x)\cdot 0.5(8 - x)\,dx = 12\int_0^{8}(8 - x)^{2}dx = 12 \times \frac{512}{3} = 2048$$
Deflection. Summing and dividing by the flexural rigidity, $$\Delta_2 = \frac{832 + 2048}{1.8 \times 10^{5}} = \frac{2880}{180\,000} = 0.0160\ \text{m}$$ $$\Delta_2 = \boxed{16.0\ \text{mm to the right}}$$ The result is positive, so joint 2 moves in the direction of the unit load, i.e. downwind. An independent direct-stiffness model of the same frame returns 16.000 mm.

Q8 — the real moment field M (from the 6 kN/m wind) and the virtual field m (from a 1 kN horizontal load at joint 2), plotted on the developed axis 3–2–1 and then 2–4.

Quantity	Value
Resultant wind load	48 kN at y = 4 m
Roller reaction at joint 4	24 kN ↑
Pin reaction at joint 1	48 kN ←, 24 kN ↓
M at joint 2 (column / cantilever / beam)	−144 / −48 / +192 kN·m
∫Mm ds over member 1–2	832 kN²·m³
∫Mm ds over member 2–3	0 (m = 0)
∫Mm ds over member 2–4	2048 kN²·m³
Horizontal deflection of joint 2	16.0 mm to the right

Topic. Deflection of a determinate frame by the principle of virtual work (unit-load method).

Key relations. $1 \cdot \Delta = \displaystyle\int \frac{Mm}{EI}ds + \int \frac{Nn}{AE}ds + \int \frac{f_s Vv}{GA}ds$; for inextensible members with slender proportions the last two terms vanish and only the flexural term is retained. Both $M$ and $m$ must be measured from the same side of each cut and with the same sign convention.

Why this works. Virtual work equates the external work done by a fictitious unit load moving through the real displacement to the internal work its stress field does on the real strains. Because the unit load is applied at the point and in the direction of interest, the external work is numerically the deflection itself. The great economy is that $m$ need only satisfy equilibrium — it is not required to be compatible — and for a determinate frame both fields come straight from statics. The vanishing of $m$ on member 2–3 illustrates the point vividly: the cantilever above joint 2 is heavily bent by the real load, but the unit load at joint 2 produces no moment there at all, so it contributes nothing to this particular deflection.

Common pitfall. Loading only the lower half of the column: the 6 kN/m acts over the entire 8 m height, and the top 4 m of load is what gives the cantilever its $-48\text{ kN}\cdot\text{m}$ and, through equilibrium, sets the beam moment at 192 kN·m. The other classic error is flipping the sign of $M$ or $m$ on one member by computing one from the left free body and the other from the right; the product $Mm$ is invariant only if the same side is used for both.

Source. Hibbeler, Structural Analysis, 10th ed., §9.4–9.5 (virtual work for beams and frames); Kassimali, Structural Analysis, 6th ed., §7.4–7.6.

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