16-Civ-A1 Elementary Structural Analysis · Undated paper

Question 4 of 8: Deflection of a cable-propped beam and Maxwell’s theorem

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 4: Deflection of a cable-propped beam and Maxwell’s theorem (16 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.

Quantity	Value
Beam A–D, pin support at A	9 m, with B, C, D at 3, 6, 9 m
Vertical cable at D to a fixed anchor above	length 2 m
Flexural rigidity of the beam	EI = 21 000 kN·m²
Axial rigidity of the cable	AE = 2000 kN
Applied load	18 kN downward at B

Find. (a) the vertical deflection at C; (b) the vertical deflection at B when the same load is moved to C, with the supporting theorem named.

Q4 — beam pinned at A and hung at D from a 2 m cable. The cable is axially flexible, so it acts as a settling support.

Approach. The structure is determinate ($r = 3$: two components at the pin, one along the cable), so find the cable tension by statics, treat the cable stretch as a support settlement at D, and superpose the flexural deflection of a simply supported 9 m span on the rigid-body rotation caused by that settlement.

Cable tension, by moments about A. $$T(9) = 18(3) \;\Rightarrow\; T = \boxed{6\ \text{kN}}$$ and vertical equilibrium gives the pin reaction $A_y = 18 - 6 = 12\text{ kN} \uparrow$.
Cable elongation. The cable is a simple axial member of length 2 m: $$e = \frac{TL}{AE} = \frac{6 \times 2}{2000} = 0.006\ \text{m} = \boxed{6\ \text{mm}}$$ so D drops 6 mm. This is not a bending effect and must be added separately.
Flexural deflection, treating A and D as unyielding. For a simply supported span $L$ carrying a point load $P$ at distance $a$ from the left support ($b = L - a$), the deflection at a station $x$ beyond the load is $$\delta(x) = \frac{P a (L - x)}{6LEI}\left(2Lx - x^{2} - a^{2}\right)$$ With $P = 18\text{ kN}$, $a = 3\text{ m}$, $L = 9\text{ m}$ and $x = 6\text{ m}$: $$\delta_C^{\text{flex}} = \frac{18(3)(3)}{6(9)(21000)}\left(108 - 36 - 9\right) = \frac{162 \times 63}{1\,134\,000} = \boxed{9.0\ \text{mm} \downarrow}$$
Rigid-body contribution from the settling support. If D sinks 6 mm while A stays put, the whole beam rotates about A and any station at distance $x$ drops by $e\,x/L$: $$\delta_C^{\text{rigid}} = 6 \times \frac{6}{9} = \boxed{4.0\ \text{mm} \downarrow}$$
Superpose (part a). The system is linear-elastic, so the two effects simply add: $$\delta_C = 9.0 + 4.0 = \boxed{13.0\ \text{mm downward}}$$ An independent direct-stiffness model of the beam with the cable represented as a pin-ended axial bar returns 13.000 mm, confirming the hand solution.
Part (b) — Maxwell’s law of reciprocal deflections. For any linear-elastic structure, the deflection at point 1 caused by a load at point 2 equals the deflection at point 2 caused by the same load applied at point 1: $$\delta_{BC} = \delta_{CB}$$ The flexible cable does not spoil this — it is still a linear-elastic element of the same structure. Therefore, with the 18 kN load moved to C, $$\delta_B = \boxed{13.0\ \text{mm downward}}$$ by Maxwell’s law of reciprocal deflections (the special case of Betti’s theorem for a single pair of unit loads). No further analysis is required; the direct-stiffness re-run with the load at C also returns 13.000 mm.

Quantity	Value
Cable tension	6 kN
Reaction at A	12 kN ↑
Cable elongation	6.0 mm
Flexural part of δ_C	9.0 mm ↓
Support-settlement part of δ_C	4.0 mm ↓
(a) δ_C with the load at B	13.0 mm downward
(b) δ_B with the load at C	13.0 mm downward (Maxwell)

Topic. Deflection of a determinate beam with an axially flexible support, and reciprocal deflections.

Key relations. Axial extension $e = TL/AE$; simply supported point-load deflection $\delta(x) = \dfrac{Pa(L-x)}{6LEI}\left(2Lx - x^{2} - a^{2}\right)$ for $x \ge a$; rigid-body correction for a settling support $\delta = e\,x/L$; Maxwell’s law $\delta_{ij} = \delta_{ji}$; Betti’s theorem $\sum P_i \delta_i^{(2)} = \sum Q_j \delta_j^{(1)}$.

Why this works. Because the structure is determinate, the cable force is fixed by equilibrium and does not depend on the stiffnesses at all; the cable therefore behaves exactly like a support that settles a known amount. Superposition then lets you add a pure bending solution (rigid supports) to a pure rigid-body solution (no bending, one support lowered). Maxwell’s law follows from the symmetry of the flexibility matrix, which is itself a consequence of the strain energy being a quadratic form — so it holds for any linear-elastic assembly, bending plus axial plus shear, determinate or not.

Common pitfall. Ignoring the cable stretch and reporting 9 mm. The cable is deliberately soft (AE = 2000 kN is a very low axial rigidity) so that its 6 mm elongation contributes a third of the answer. The second pitfall in part (b) is to think that reciprocity fails because the load and the response points are not symmetric about midspan — symmetry is irrelevant, linearity is all that is required.

Source. Hibbeler, Structural Analysis, 10th ed., §8.2–8.4 (elastic curve, tabulated deflections) and §9.6 (Maxwell’s reciprocal theorem); Kassimali, Structural Analysis, 6th ed., §7.7.

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