07-Str-A2 · May 2018

Question 6 of 7: B3 — Design of an overhanging reinforced concrete beam

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

Notes on this paper

Paper format. National Examinations, May 2018 — 07-Str-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). A candidate submits five solutions — two from Part A, two from Part B and the one question in Part C — all of equal value. All seven questions are solved below, because this set is a study resource rather than an exam script. Page 1 states that all loads shown are unfactored, so every load case is factored here before any resistance is compared against it.

Reference texts.

CSA S16:19, Design of Steel Structures — and CISC, Handbook of Steel Construction, 11th ed. (section tables, Table 2 class limits, Table 7 minimum fillet sizes).
CSA A23.3:19, Design of Concrete Structures — and Cement Association of Canada, Concrete Design Handbook, 4th ed.
CSA O86:19, Engineering Design in Wood — and Canadian Wood Council, Wood Design Manual.
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 2015, Table 4.1.3.2 (load combinations).

Check — one dimension is missing from the source. Question A2 describes a stub cantilever welded to a column but the paper contains no Figure A2 and never states the cantilever projection. The projection is therefore taken as L = 1.5 m from the column face throughout question A2; every result below is also given in the general form so any other projection can be substituted directly. All other data are read from the printed text and from Figure A3, B1, B2 and B3 on page 3.

Question 6: B3 — Design of an overhanging reinforced concrete beam (10 + 5 + 5 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. From Figure B3: overhangs AB = CD = 1 m either side of a 4 m span BC, pin support at B and roller at C, and a uniformly distributed load of 50 kN/m acting over the whole 6 m length. Specified load, so it is factored at 1.5; self-weight is dead load at 1.25 with γ_concrete = 24 kN/m³. Materials f_c′ = 35 MPa, f_y = 400 MPa.

Find. Cross-section dimensions b and h, the longitudinal reinforcement for the sagging and hogging moments, and the stirrup arrangement.

Figure 6.1 — Loading and the resulting factored bending moment diagram: hogging peaks at each support, sagging peaks at mid-span.

Approach. Assume a trial section, compute the self-weight and hence the factored load, obtain the two critical moments and the support shear, then size the steel; iterate once if the trial section proves unsuitable.

Trial section and factored load. Take b = 300 mm and h = 400 mm as a first trial (span/h = 10, comfortably above the A23.3 Table 9.2 minimum of L/16 = 250 mm). Then $$w_{sw}=0.30(0.40)(24)=2.88\ \text{kN/m},\qquad w_{f}=1.5(50)+1.25(2.88)=\boxed{78.6\ \text{kN/m}}$$
Analysis. With the load over the full length and supports 1 m in from each end, symmetry gives $$R_{B}=R_{C}=\frac{w_{f}(4+2)}{2}=3w_{f}=235.8\ \text{kN}$$ The hogging moment at each support comes from the cantilevered overhang alone, and the sagging moment at mid-span follows from the left free body: $$M_{B}=-\frac{w_{f}(1)^{2}}{2}=-39.3\ \text{kN}\cdot\text{m},\qquad M_{\text{mid}}=3w_{f}(2)-\frac{w_{f}(3)^{2}}{2}=1.5w_{f}=+117.9\ \text{kN}\cdot\text{m}$$ The overhangs relieve the span substantially: without them the mid-span moment would be 2w_f = 157 kN·m.
Effective depth. With 40 mm cover, 10M stirrups and 20M main bars, $$d=400-40-11.3-\tfrac{19.5}{2}=339\ \text{mm}$$
Bottom steel for the sagging moment. Solving φ_sf_yA_s(d − a/2) = M_mid with a = φ_sf_yA_s/(α₁φ_cf_c′b) gives the quadratic $$10.62A_{s}^{2}-115\,260A_{s}+117.9\times10^{6}=0\ \Longrightarrow\ A_{s}=1144\ \text{mm}^{2}$$ Provide 4–20M (1200 mm²) in one layer; the bars need 4(19.5) + 3(25) + 2(50) = 253 mm against 300 mm available, so they fit. The block depth is a = 71.4 mm, c = 81.0 mm and c/d = 0.239, well below the 0.5 tension-control threshold.
Top steel for the hogging moment. The same quadratic with 39.3 kN·m returns A_s = 352 mm², but the minimum of Cl. 10.5.1.2 governs: $$A_{s,\min}=\frac{0.2\sqrt{f_{c}'}}{f_{y}}bh=\frac{0.2\sqrt{35}}{400}(300)(400)=355\ \text{mm}^{2}$$ Provide 2–20M (600 mm²) continuous over each support, extended past the point of contraflexure into the span by at least d or 12 bar diameters.
Shear design. The critical section is d_v from the support face; with a 300 mm bearing and d_v = max(0.9d, 0.72h) = max(305, 288) = 305 mm, $$V_{f}=R_{B}-w_{f}\left(1.0+0.15+0.305\right)=121.4\ \text{kN}$$ With at least minimum stirrups the simplified method gives β = 0.18, θ = 35°: $$V_{c}=0.65(0.18)\sqrt{35}(300)(305)=63.4\ \text{kN}$$ so V_s = 121.4 − 63.4 = 58 kN is required. Stirrup spacing is in fact capped by detailing at s ≤ 0.7d_v = 214 mm, so adopt 10M double-leg stirrups at 200 mm throughout: $$V_{s}=\frac{0.85(200)(400)(305)(1.428)}{200}=148.1\ \text{kN},\qquad V_{r}=63.4+148.1=\boxed{211.5\ \text{kN}}$$ against V_f = 121.4 kN. The crushing cap 0.25φ_cf_c′b d_v = 520 kN is nowhere near critical.
Confirm the trial section. Flexural utilisation is 1144/1200 = 0.95 on the provided steel, shear utilisation 121.4/211.5 = 0.57, and the reinforcement ratio ρ = 1200/(300 × 339) = 1.18 % lies in the economical band. The trial section is retained; no second iteration is needed.

The overhangs do real work in this beam. They halve the negative-moment demand relative to a fixed-ended member while cutting the mid-span moment by a quarter, which is why a 300 × 400 section carries a 50 kN/m service load over 4 m with only four 20M bars.

Item	Design
Cross-section	300 mm × 400 mm (d = 339 mm)
Factored load, w_f	78.6 kN/m (self-weight 2.88 kN/m included)
M_mid / M_B / V_f	+117.9 / −39.3 kN·m / 121.4 kN
Bottom steel (span)	4–20M (1200 mm²; 1144 required)
Top steel (over B and C)	2–20M (600 mm²; minimum 355 governs)
Stirrups	10M double-leg @ 200 mm (V_r = 211.5 kN)
c/d	0.239 — tension-controlled

Topic. Proportioning and reinforcing a determinate overhanging beam for flexure and shear, including self-weight.

Key relations. For a UDL over a beam with equal overhangs a and span L: R = w(L + 2a)/2, M_support = −wa²/2, M_mid = R(L/2) − w(L/2 + a)²/2. Flexure: φ_sf_yA_s(d − a/2) = M_f. Minimum steel A_s,min = 0.2√f_c′bh/f_y. Shear: V_c = φ_cβ√f_c′b_wd_v, V_s = φ_sA_vf_yd_vcotθ/s, s ≤ 0.7d_v.

Why this works. Design is a fixed-point problem because the self-weight depends on the very dimensions being sought. Choosing a slenderness-based trial section makes the loop converge in a single pass: the self-weight is a small fraction of the total, so a reasonable first guess barely shifts the answer. The overhangs then act as partial fixity — the cantilever moment at each support is a genuine hogging demand that must be reinforced on the top face, and it simultaneously reduces the sagging moment in the span.

Common pitfall. Forgetting top steel over the supports, or terminating it at the support centreline. The hogging moment does not vanish at the support; the bars must run past the point of contraflexure. A second pitfall is checking shear at the support centreline rather than d_v from the face, which overstates V_f by about 25 % here.

Source. CSA A23.3:19 Cl. 9.8, 10.1, 10.5.1.2, 11.3; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Chs. 4–6.

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