16-Civ-A2 Elementary Structural Design · December 2019

Question 6 of 7: B3 — Doubly overhanging concrete beam

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

Notes on this paper

Paper format. National Examinations, December 2019 — 16-Civ-A2 Elementary Structural Design. Three hours, CLOSED BOOK. Seven questions in three parts: Part A (A1–A3, steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). A candidate submits five solutions — two from Part A, two from Part B and the single Part C question — all of equal value. All seven are solved below, because this set is a study resource rather than an exam script. Note 6 on page 1 states that all loads shown are unfactored unless otherwise stated, so every load pattern is factored here before it meets a resistance; Note 7 fixes G40.21 300W steel and 400W reinforcement unless a question says otherwise.

Reference texts.

CSA S16:19, Design of Steel Structures, and CISC Handbook of Steel Construction, 11th ed. — Clauses 13.3 (compression), 13.13 (welds), 17 (composite beams), and the angle/W-shape section tables.
CSA A23.3:19, Design of Concrete Structures, and the Cement Association of Canada Concrete Design Handbook, 4th ed. — Clauses 10 (flexure and columns) and 11 (shear).
CSA O86:19, Engineering Design in Wood, and the Canadian Wood Council Wood Design Manual — Clause 6 (sawn lumber) and the service-condition and size-factor tables.
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition), Chs. 4, 6, 11.
National Building Code of Canada, Table 4.1.3.2 (load combinations).

Check — assumptions declared under Note 1 of the paper. (i) Figure A1 shows a solid vertical between F and C as well as the load arrow at C; that member is required, because without it the truss counts m + r = 12 + 3 = 15 against 2n = 16 and is a mechanism. With it, m + r = 16 = 2n and the truss is determinate, which is how it is solved here. (ii) In A3 the number of shear studs is reported as a bonus; the question asks only for the ultimate moment. (iii) In B1 the two stacked dimensions read as a sum, so the tee is 100 mm flange plus 400 mm of stem, i.e. 500 mm overall, and the beam spacing of 1250 mm governs the effective flange width because no span is given. (iv) In C1 the 1.5 kPa / 1.0 kPa area loads are converted to line loads on the 0.9 m tributary width, and the deflection limits adopted are L/240 on the back span and 2L/180 on the cantilever.

Question 6: B3 — Doubly overhanging concrete beam (20 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. The design starts from the beam of Figure B3 — 180 kN/m already factored over the whole 6 m length, with 1 m overhangs each side of a 4 m span — and an architecturally fixed 400 × 900 section in 35 MPa concrete.

Given data
Item	Value
Load	180 kN/m, already factored, over the whole 6 m length
Geometry	1 m overhang + 4 m span + 1 m overhang
Section (architect)	400 mm wide × 900 mm deep
Concrete / steel	f′_c = 35 MPa; 400W, f_y = 400 MPa
Cover	40 mm clear to a 10M stirrup

Find. The flexural reinforcement (top and bottom) and the shear reinforcement for the given 400 × 900 section.

Figure B3 rebuilt: 180 kN/m over the full 6 m, pin at A and roller at B with 1 m overhangs each side.

Approach. The structure is symmetric and determinate, so statics gives the reactions immediately; the overhangs hog over the supports while the interior span sags. Size the steel for each, check the A23.3 minimum, then design the stirrups for the shear just inside the supports.

Reactions. By symmetry, $$R_A = R_B = \frac{wL_{\text{tot}}}{2} = \frac{180(6)}{2} = \boxed{540\ \text{kN each}}$$
Hogging moment at the supports. Take the free body of one overhang: $$M_{\text{hog}} = \frac{w a^{2}}{2} = \frac{180(1)^{2}}{2} = 90\ \text{kN}\cdot\text{m}$$ acting to put the top fibre in tension.
Sagging moment at midspan. Measuring from the left free end and cutting at $x = 3$ m, $$M_{\text{sag}} = R_A(2) - \frac{w(3)^{2}}{2} = 1080 - 810 = \boxed{270\ \text{kN}\cdot\text{m}}$$ Notice how much the overhangs help: without them, a 4 m simple span under the same load would sag $180(4)^{2}/8 = 360$ kN·m, a third more.
Shear. The shear jumps by 540 kN at each support. Just outside A it is $-180(1) = -180$ kN and just inside it is $-180 + 540 = +360$ kN, so $$V_{f,\max} = 360\ \text{kN}$$
Effective depth. With 40 mm cover, a 10M stirrup and one layer of 30M bars, $$d = 900 - 40 - 11.3 - 14.95 = 833.8\ \text{mm}$$ The architect has given a very deep section for these moments, which is what makes the answer so lightly reinforced.
Steel required for strength. With $\alpha_1 = 0.85 - 0.0015(35) = 0.7975$, the compression per unit stress-block depth is $\alpha_1\phi_c f′_c b = 0.7975(0.65)(35)(400) = 7258$ N/mm. Solving $\phi_s A_s f_y\left(d - \phi_s A_s f_y/2k\right) = 270\times10^{6}$ gives $A_s = 979$ mm² for the sagging region and only 320 mm² for the hogging region.
Minimum steel governs both faces. A23.3 Clause 10.5.1.2 requires $$A_{s,\min} = \frac{0.2\sqrt{f′_c}}{f_y}\,b_t h = \frac{0.2\sqrt{35}}{400}(400)(900) = 1065\ \text{mm}^{2}$$ which exceeds both strength requirements. That is the real design driver here, and it is a direct consequence of the architect prescribing a 900 mm depth for a beam that structurally needs far less. Provide 2-30M top and 2-30M bottom, running continuously: $A_s = 1400$ mm².
Check the provided section. $$T = \phi_s A_s f_y = 0.85(1400)(400) = 476\ \text{kN},\qquad a = \frac{476\times10^{3}}{7258} = 65.6\ \text{mm}$$ $$M_r = T\left(d - \frac{a}{2}\right) = 476(833.8 - 32.8)/10^{3} = \boxed{381\ \text{kN}\cdot\text{m}}$$ against 270 kN·m sagging (utilisation 0.71) and 90 kN·m hogging (0.24). The section is very ductile, $c/d = 73/834 = 0.088$.
Shear design. $$d_v = \max(0.9d,\ 0.72h) = \max(750.4,\ 648) = 750.4\ \text{mm}$$ $$V_c = \phi_c\lambda\beta\sqrt{f′_c}\,b_w d_v = 0.65(1.0)(0.18)\sqrt{35}\,(400)(750.4)/10^{3} = 208\ \text{kN}$$ The concrete alone is short of the 360 kN demand, so stirrups are required over the whole length. Using 10M double-leg stirrups at 350 mm, $$V_s = \frac{\phi_s A_v f_y d_v \cot 35^\circ}{s} = \frac{0.85(200)(400)(750.4)(1.428)}{350}\cdot\frac{1}{10^{3}} = 208\ \text{kN}$$ $$\boxed{V_r = 208 + 208 = 416\ \text{kN} \;>\; 360\ \text{kN}}$$
Confirm the detailing limits. Minimum stirrup area $A_{v,\min} = 0.06\sqrt{35}(400)(350)/400 = 124$ mm² against the 200 mm² supplied; maximum spacing $\min(0.7d_v,\ 600) = 525$ mm against the 350 mm used; and the crushing cap $0.25\phi_c f′_c b_w d_v = 1707$ kN is nowhere near reached. A single stirrup spacing over the whole beam is the sensible detail at this size.

Final results
Quantity	Value
Reactions R_A = R_B	540 kN
Hogging moment at each support	90 kN·m
Sagging moment at midspan	270 kN·m
Maximum shear	360 kN (just inside each support)
Effective depth d	833.8 mm
A_s for strength / minimum	979 mm² / 1065 mm² — minimum governs
Flexural steel provided	2-30M top and 2-30M bottom (1400 mm² each face)
M_r provided	381 kN·m (utilisation 0.71 sagging)
Shear reinforcement	10M double-leg stirrups @ 350 mm
V_r	416 kN (V_c 208 + V_s 208)

Topic. Flexural and shear design of a determinate beam with symmetric overhangs.

Key relations. $R = wL_{\text{tot}}/2$ by symmetry; $M_{\text{hog}} = wa^{2}/2$ at each support; $M_r = \phi_s A_s f_y (d - a/2)$ with $a = \phi_s A_s f_y / (\alpha_1\phi_c f′_c b)$; $A_{s,\min} = 0.2\sqrt{f′_c}\,b_t h / f_y$; A23.3 simplified shear with $\beta = 0.18$, $\theta = 35^\circ$.

Why this works. Overhangs are structurally generous: the cantilever moment at each support is subtracted from the simple-span sagging moment, so a 1 m overhang on a 4 m span cuts the midspan moment by a quarter while introducing a hogging moment only a third as large. The price is that steel is needed in both faces and must be continuous through the supports, since the moment changes sign twice. The minimum-steel rule then does most of the work here, because the prescribed 900 mm depth is generous relative to the demand — a reminder that $A_{s,\min}$ is proportional to the gross section, not to the moment.

Common pitfall. Placing the top steel only over the supports and stopping it inside the span leaves the inflection region unreinforced against the real (if small) tension there; run both layers through. Another common error is taking the maximum shear as $wL/2$ on the 4 m span (360 kN happens to coincide here, but for asymmetric overhangs it does not), rather than reading it from the actual shear diagram.

Source. CSA A23.3:19 Clauses 10.5.1.2, 11.2.8.2 and 11.3; MacGregor & Bartlett, Chs. 4 and 6.

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