07-Str-A1 · May 2018

Question 3 of 8: Deflection of a Non-Prismatic Beam

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

Notes on this paper

Paper format. 07-Str-A1 Elementary Structural Analysis, National Examinations, May 2018. Three hours, CLOSED BOOK (an approved Casio or Sharp calculator is permitted). Six questions constitute a complete paper: Questions 1 to 5 are compulsory and one of Questions 6, 7 or 8 is chosen. Marks are shown against each question and total 100. Every question is worked here, including all three of the optional Questions 6, 7 and 8, so that the set can be used as a study resource.

Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2 stability and determinacy; Ch. 4 influence lines; Ch. 6 deflections by virtual work; Ch. 11 displacement method; Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 8, 13, 16); K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis; J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods. For Canadian design practice the companion documents are the National Building Code of Canada (Part 4, load combinations) and CSA S16 Design of Steel Structures; this paper is pure analysis, so no design code is invoked in the answers.

Check — readings taken from the printed figures. Every dimension, support symbol and load extent below was read from the printed figures. Four readings carry the answers and are stated explicitly so that a marker can check them: (i) in Question 2(a) the 10 kN/m block covers only the 2 m overhang, ending at the roller; (ii) in Question 4(a) the truss is a Warren truss whose top nodes are offset 3 m from the bottom nodes (the 3 m dimension at the top right confirms this); (iii) in Question 7 the 10.8 kN/m ruling starts directly above joint 1 and runs to joint 4, so it acts over the whole horizontal projection, inclined member included; (iv) in Question 8 the thin vertical line between U₂ and L₃ touches neither joint circle and is the witness line of the 3 m dimension, not a member — taking it as a member would make the truss statically indeterminate and the question unanswerable by virtual work.

Question 3: Deflection of a Non-Prismatic Beam (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. A simply supported beam A–D of 9 m span, pinned at A and on a roller at D, divided into three equal 3 m segments. The flexural rigidity is $EI$ over A–B, $3EI$ over B–C and $EI$ over C–D, with $EI = 3.0 \times 10^{4}\ \text{kN}\cdot\text{m}^{2}$. A 81 kN point load acts at B (3 m from A) and a 72 kN point load at C (6 m from A). B and C carry loads only; they are not supports.

Find. The vertical deflection of point B, with its direction.

Question 3: the non-prismatic simply supported beam. The heavier line over B–C marks the stiffened middle third; A and D carry the only supports.

Approach. Use the unit-load form of virtual work, $1 \cdot \Delta_B = \int M m / EI \; \mathrm{d}x$, with $M$ the real bending moment and $m$ the moment produced by a unit downward load applied at B alone; the varying rigidity is handled simply by integrating segment by segment with the local value of $EI$.

Real reactions and real moments. Taking moments about A, $$R_D = \frac{81(3) + 72(6)}{9} = \frac{243 + 432}{9} = 75.0\ \text{kN},\qquad R_A = 153 - 75 = 78.0\ \text{kN}.$$ Hence $M = 78x$ on A–B, $M = 243 - 3x$ on B–C and $M = 75(9-x)$ on C–D, giving the peak $M(B) = 234$ kN·m and $M(C) = 225$ kN·m.
Virtual system. Removing the real loads and applying a single downward unit load at B gives reactions $2/3$ at A and $1/3$ at D, so $$m = \tfrac{2}{3}x \quad (0 \le x \le 3),\qquad m = \tfrac{1}{3}(9-x) \quad (3 \le x \le 9).$$
Integrate segment A–B, rigidity $EI$. $$\int_0^3 \frac{(78x)\left(\tfrac{2}{3}x\right)}{EI}\,\mathrm{d}x = \frac{52}{EI}\left[\frac{x^3}{3}\right]_0^3 = \frac{468}{EI}.$$
Integrate segment B–C, rigidity $3EI$. With $M = 243 - 3x$ and $m = 3 - x/3$, $$\int_3^6 \frac{(243-3x)\left(3 - \tfrac{x}{3}\right)}{3EI}\,\mathrm{d}x = \frac{1}{3EI}\int_3^6 \left(729 - 90x + x^2\right)\mathrm{d}x = \frac{1035}{3EI} = \frac{345}{EI}.$$ Trebling the rigidity has cut this segment's contribution to one third of what a prismatic beam would give, which is the whole point of the question.
Integrate segment C–D, rigidity $EI$. $$\int_6^9 \frac{75(9-x)\cdot\tfrac{1}{3}(9-x)}{EI}\,\mathrm{d}x = \frac{25}{EI}\left[-\frac{(9-x)^3}{3}\right]_6^9 = \frac{225}{EI}.$$
Add the three contributions and substitute the rigidity. $$\Delta_B = \frac{468 + 345 + 225}{EI} = \frac{1038}{3.0\times 10^{4}} = 0.03460\ \text{m},$$ so that $$\boxed{\Delta_B = 34.6\ \text{mm downward}}.$$ Every term of the integral is positive, so the deflection is unambiguously in the direction of the applied unit load.

It is worth noting what the stiffened middle third buys. Had the beam been prismatic at $EI$ throughout, the middle segment would have contributed $1035/EI$ instead of $345/EI$ and the deflection would have been $1728/(3.0\times10^{4}) = 57.6$ mm — two thirds larger. That is the practical reading of the answer: the stiffening is placed where the product $Mm$ is largest, which is where it does the most good.

Question 3 — virtual-work summary
Segment	Rigidity	$\int Mm/EI\,\mathrm{d}x$
A–B (0 to 3 m)	$EI$	$468/EI$
B–C (3 to 6 m)	$3EI$	$345/EI$
C–D (6 to 9 m)	$EI$	$225/EI$
Total		$1038/EI$
Reactions		$R_A = 78.0$ kN, $R_D = 75.0$ kN
Vertical deflection at B		34.6 mm downward

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

Key relations. $1 \cdot \Delta = \displaystyle\int_0^L \frac{M m}{EI}\,\mathrm{d}x$ for flexure; the virtual moment $m$ comes from a unit load applied at, and in the direction of, the required displacement; segments of different $EI$ are integrated separately.

Why this works. Virtual work equates the external work done by a fictitious unit load moving through the real displacement with the internal work done by the corresponding virtual internal forces moving through the real deformations. For a beam the real curvature is $M/EI$, so the internal work per unit length is $m \cdot M/EI$ and the integral follows directly. Because the method needs only the real moment diagram and a virtual one, it handles a stepped rigidity with no extra machinery — each segment simply carries its own $EI$ into the denominator.

Common pitfall. Applying the unit load at the wrong point (it must be at B, not at mid-span) or forgetting to change $EI$ at the step, which here would overstate the middle-third contribution threefold. A second trap is reading B and C as supports because the question calls the beam "continuous"; the drawing shows support symbols at A and D only.

Source. Hibbeler, Structural Analysis, §8.4 and §9.1–9.4; Kassimali, Structural Analysis, §7.4.

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