07-Str-A1 · December 2018

Question 3 of 8: Vertical Deflections at B and D of a Stepped Beam

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

Notes on this paper

Paper format. National Exams, December 2018 — 07-Str-A1 Elementary Structural Analysis; 3 hours, CLOSED BOOK (an approved Casio or Sharp calculator is permitted). Six questions constitute a complete paper: answer all of Questions 1–5, then one only of Questions 6, 7 or 8. Marks are shown in the left margin (6, 18, 18, 18, 18, then 22 for the optional question). Every question is worked below, including all three alternatives, because the full set is the more useful study resource.

Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2 determinacy and stability; Ch. 4 influence lines; Ch. 6–9 deflections and virtual work; Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 6, 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 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 — two figure readings that the printed figures leave open, both stated where they are used. (i) In Question 2(c) the hatched plane under the upper roller is drawn parallel to the inclined member, so the reaction is taken normal to the member. That reading makes every result an integer (R = 28 kN, shears 32 / −4 / −28 kN, axial forces 25 / 10 / 0 kN), whereas a vertical-plane reading gives nothing round; and because the two readings differ only by a force acting along the member axis, the shear and moment diagrams are identical either way. (ii) In Question 7 both wall supports are drawn as triangles on hatching. The truss is analysable only if one of them releases a component; taking the pin at U₂ and the horizontal-only roller at L₃ gives δ = 29.0 mm, and the same value follows if the thin vertical line between the two wall joints is a dimension witness line rather than a member. The alternative (pin at L₃) would give 31.8 mm.

Question 3: Vertical Deflections at B and D of a Stepped Beam (18 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
Span A to C (pin to roller)	12.0 m, in two 6 m halves with B at mid-span
Overhang C to D	1.5 m
Tip load at D	24 kN downward
Flexural rigidity, A to B	$EI_0 = 18\,000$ kN·m$^2$
Flexural rigidity, B to D	$2EI_0 = 36\,000$ kN·m$^2$

Find. The vertical deflection at mid-span B and at the loaded tip D, each with its direction.

Question 3: the thicker line from B onwards marks the 2EI₀ length; A is a pin, C a roller and D the loaded free end.

Approach. The beam is determinate ($r = 3$), so find the real moment diagram from statics, then apply the unit-load (virtual work) method $\delta = \int M m / (EI)\,dx$ twice, with the stepped rigidity handled by splitting the integral at B.

Reactions from statics, and note the sign of the one at the pin. Moments about A give $$12C_y = (24)(13.5) = 324 \qquad C_y = \boxed{27.0\ \text{kN}\ \uparrow}$$ $$A_y = 24 - 27.0 = -3.0\ \text{kN}, \ \text{i.e.}\ \boxed{3.0\ \text{kN}\ \downarrow}$$ The pin holds the beam down: the tip load is trying to lever the left end upwards about C.
Write the real bending moment. Taking sagging positive, the span carries only the downward pin reaction, so $$M(x) = -3.0x \quad (0 \le x \le 12) \qquad M(x) = -24(13.5 - x) \quad (12 \le x \le 13.5)$$ with $M(12) = -36.0$ kN·m at the roller. The whole diagram is hogging — a single straight line from zero at A down to $-36.0$ kN·m at C and back to zero at D — which is worth stating because it already tells us the span must bow upwards.
Set up the virtual system for B. A unit downward load at mid-span of the simply supported 12 m span gives half reactions and leaves the overhang unstressed: $$m_B(x) = 0.5x \ \ (0 \le x \le 6) \qquad m_B(x) = 0.5(12 - x) \ \ (6 \le x \le 12) \qquad m_B = 0 \ \ (x > 12)$$
Integrate for the deflection at B, splitting at the change of section. Over the first 6 m, with $EI_0$, $$\int_0^6 \frac{(-3x)(0.5x)}{EI_0}\,dx = \frac{-1.5}{EI_0}\!\left[\frac{x^3}{3}\right]_0^6 = \frac{-108}{EI_0}$$ and over the second 6 m, with $2EI_0$, $$\int_6^{12} \frac{(-3x)\bigl[0.5(12-x)\bigr]}{2EI_0}\,dx = \frac{-1.5}{2EI_0}\int_6^{12}(12x - x^2)\,dx = \frac{-1.5(144)}{2EI_0} = \frac{-108}{EI_0}$$ The two halves contribute equally, which is a neat consequence of the doubled stiffness exactly offsetting the larger $Mm$ product. Adding them, $$\delta_B = \frac{-216}{EI_0} = \frac{-216}{18\,000} = -0.0120\ \text{m} = \boxed{12.0\ \text{mm}\ \text{upward}}$$ The negative sign means the deflection opposes the assumed downward unit load, so mid-span lifts — exactly what the all-hogging moment diagram predicted.
Repeat with the unit load at the tip D. Now $R_C = 13.5/12 = 1.125$ and $R_A = -0.125$, so $$m_D(x) = -0.125x \ \ (0 \le x \le 12) \qquad m_D(x) = -(13.5 - x) \ \ (12 \le x \le 13.5)$$ Because $M$ and $m_D$ now have the same sign everywhere, all three contributions are positive: $$\int_0^6 \frac{(-3x)(-0.125x)}{EI_0}dx = \frac{27}{EI_0} \qquad \int_6^{12} \frac{(-3x)(-0.125x)}{2EI_0}dx = \frac{94.5}{EI_0}$$ $$\int_{12}^{13.5} \frac{24(13.5-x)^2}{2EI_0}dx = \frac{12}{EI_0}\!\left[\frac{u^3}{3}\right]_0^{1.5} = \frac{13.5}{EI_0}$$ Summing, $$\delta_D = \frac{27 + 94.5 + 13.5}{EI_0} = \frac{135}{18\,000} = \boxed{7.5\ \text{mm}\ \text{downward}}$$ Only 10 % of that comes from the cantilever's own flexure; the rest is the rigid-body rotation the hogged span imposes on the overhang, which is why the answer is dominated by what happens between A and C.

Question 3: the real moment diagram is entirely hogging, so the span between A and C deflects upwards while the loaded tip goes down.

Quantity	Value
Reaction at the roller C	27.0 kN ↑
Reaction at the pin A	3.0 kN ↓ (holds the beam down)
Bending moment at C	−36.0 kN·m (hogging)
Deflection at B	$\delta_B = 216/EI_0 = $ 12.0 mm upward
Deflection at D	$\delta_D = 135/EI_0 = $ 7.5 mm downward

Topic. Deflection of a determinate stepped beam by the unit-load method

Key relations. $1 \cdot \delta = \int_0^L \dfrac{M m}{EI}\,dx$, with $M$ from the real loading and $m$ from a unit load applied at, and in the direction of, the required displacement. A negative result means the displacement opposes the unit load. Overhang moment $M = P a$; simple-span unit-load moment $m = ab/L$ at the load point.

Why this works. The unit-load method is Betti's reciprocal theorem in working dress: the internal virtual work done by the virtual moments $m$ riding on the real curvatures $M/EI$ equals the external virtual work $1 \times \delta$. Because the integrand carries $EI$ in the denominator, a step change in section is handled simply by splitting the integral at the step — no new theory is needed, and the beam does not have to be prismatic. Choosing the virtual system wisely also saves work: for point B the unit load stresses only the 12 m span, so the overhang drops out of that integral entirely.

Common pitfall. Two slips are common. First, forgetting that the overhang contributes to $\delta_D$ through its own curvature as well as through the rotation of the span, or conversely computing only the cantilever term $PL^3/3EI$ and missing the much larger rotation effect. Second, quoting a magnitude without a direction: B moves 12 mm up, and reporting it as a downward sag reverses the physical behaviour that the hogging diagram makes obvious.

Source. Hibbeler, Structural Analysis, §8.4 and §9.4 (method of virtual work for beams); Kassimali, §7.4.

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