16-Civ-A2 Elementary Structural Design · May 2017

Question 5 of 7: B2 — Reinforced concrete beam with an overhang

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

Notes on this paper

Paper format. National Examinations, May 2017 — 16-Civ-A2 Elementary Structural Design. Three hours, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, structural steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3) and Part C (C1, timber to CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions of equal value. All seven are solved here, because the set is a study resource rather than a sitting.

Reference texts.

CSA S16:19, Design of Steel Structures — Clauses 11 (section class), 13.3 (compression), 13.5 (bending), 13.8 (axial force and bending), 13.13 (welds), 14.3 (concentrated forces).
CISC, Handbook of Steel Construction — rolled-section dimensions and properties, weld tables.
CSA A23.3:19, Design of Concrete Structures — Clauses 10 (flexure and axial load), 11 (shear).
CSA O86:19, Engineering Design in Wood — Clause 6 (sawn lumber), Tables 6.3.1B and 6.4.5.
Canadian Wood Council, Wood Design Manual; National Building Code of Canada 2020 (load combinations).

Load factors. NOTE 6 on page 1 states that all loads shown are unfactored. The point loads drawn on Figures A1, B1 and B2 are occupancy loads, so they are factored by 1.5; the only dead load anywhere in this paper is the concrete self-weight of the Question B2 beam, which takes 1.25. This follows NBCC combination 2, 1.25D + 1.5L.

Check — figure page. Two figure readings are flagged where they arise: in Figure B1 the 350 kN load acts at midspan of beam BA (3.5 m from B), not at joint B; and the 80 kN horizontal load acts 4 m below B, i.e. 6 m above the pin at C.

Question 5: B2 — Reinforced concrete beam with an overhang (10 + 8 + 2 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 arrangement	Pin at A; 400 kN at 4 m; roller at B, 8 m from A; overhang B–C = 2 m
Tip load	70 kN downward at C (unfactored)
Materials	f_c' = 35 MPa, f_y = 400 MPa; concrete density 24 kN/m³
Self-weight	to be included — the only dead load, factored by 1.25

Find. Rectangular dimensions b × h, the flexural steel top and bottom, and the stirrup arrangement, all satisfying moment and shear.

Figure B2 — beam with overhang, and the factored bending-moment diagram.

Approach. Assume a trial section, add its self-weight as a UDL, solve the statically determinate beam for the factored envelope, then design the sagging steel at 4 m, the hogging steel over B and the stirrups at the critical shear section, confirming that the assumed self-weight is consistent.

Trial section and self-weight. The 400 kN point load on an 8 m span is heavy, so start at 500 × 1200 mm: $$w_{sw} = 0.500 \times 1.200 \times 24 = 14.4\ \text{kN/m} \;\Rightarrow\; w_f = 1.25(14.4) = 18.0\ \text{kN/m}$$ The point loads become 600 kN at 4 m and 105 kN at C.
Reactions. Taking moments about A over the full 10 m length, $$8R_B = 600(4) + 105(10) + 18.0(10)(5) = 2400 + 1050 + 900 = 4350$$ $$R_B = 543.8\ \text{kN}, \qquad R_A = 600 + 105 + 180 - 543.8 = 341.3\ \text{kN}$$
Bending-moment envelope. Between A and the point load, M(x) = 341.3x − 9x², whose stationary point lies at x = 19.0 m — well outside the span — so the maximum sagging moment is at the load itself: $$M_f^{+} = 341.3(4) - 9(16) = 1365 - 144 = \boxed{1221\ \text{kN}\cdot\text{m}}$$ Over the support, taking the free body to the right of B is quickest: $$M_f^{-} = -\bigl[105(2) + 18.0(2)(1)\bigr] = \boxed{-246\ \text{kN}\cdot\text{m}}$$ which the left-hand free body reproduces: 341.3(8) − 9(64) − 600(4) = −246 kN·m.
Shear envelope. Just to the left of B, $$V_f = 341.3 - 18.0(8) - 600 = -402.8\ \text{kN}$$ against only 141 kN just to the right, so the section immediately left of B governs. Design shear = 403 kN.
Bottom steel for the sagging moment. With 40 mm cover and 10M stirrups, d = 1200 − 40 − 11.3 − 14.95 = 1133.8 mm. Try 5–30M (A_s = 3500 mm²): $$T = \phi_s f_y A_s = 0.85(400)(3500) = 1190\ \text{kN}$$ $$a = \frac{T}{\alpha_1 \phi_c f_c' b} = \frac{1190 \times 10^{3}}{0.7975(0.65)(35)(500)} = 131.2\ \text{mm}$$ $$M_r = T\left(d - \frac{a}{2}\right) = 1190\left(1133.8 - 65.6\right) = \boxed{1271\ \text{kN}\cdot\text{m}} \;>\; 1221$$ a utilisation of 0.96. The section is tension-controlled (c/d = 0.13) and clears the minimum steel of Clause 10.5.1.2, A_s,min = 0.2√f_c'b_th/f_y = 1775 mm².
Confirm the bars fit in one layer. Five 30M bars across 500 mm, less 40 mm cover and 11.3 mm stirrup each side, leaves $$\frac{500 - 2(40) - 2(11.3) - 5(29.9)}{4} = 62.0\ \text{mm}$$ of clear spacing, against the required minimum of 1.4d_b = 41.9 mm. They fit, which is what preserves the 1133.8 mm effective depth assumed above — a second layer would cost roughly 40 mm of d and about 4 % of M_r, enough to fail the section.
Top steel over the support. The hogging demand is only 246 kN·m, so minimum steel governs. Use 4–25M (2000 mm² > 1775 mm²), giving a = 75.0 mm and $$M_r^{-} = 0.85(400)(2000)(1136.1 - 37.5) = 747\ \text{kN}\cdot\text{m} \;>\; 246\ \text{kN}\cdot\text{m}$$ These bars must run the full length of the overhang and lap far enough into the main span to develop past the point of contraflexure.
Stirrups. With d_v = max(0.9d, 0.72h) = 1020.4 mm and β = 0.18, $$V_c = 0.65(1.0)(0.18)\sqrt{35}(500)(1020.4) = 353\ \text{kN}$$ $$V_s = 403 - 353 = 50\ \text{kN}$$ Using 10M double-legged stirrups (A_v = 200 mm²) with θ = 35°, strength alone would allow a spacing of nearly 2 m. Three other limits are tighter: the minimum-area rule s ≤ A_vf_y/(0.06√f_c'b_w) = 451 mm; the spacing cap min(0.7d_v, 600) = 600 mm; and buildability. Use 10M stirrups at 400 mm, giving V_r = 601 kN. Since V_f = 403 kN is far below 0.125λφ_cf_c'b_wd_v = 1452 kN, the tighter 0.35d_v cap does not apply, and web crushing (2905 kN) is nowhere near.
Close the loop on self-weight. The adopted 500 × 1200 section is exactly the one assumed in step 1, so no second iteration is needed. It is worth seeing what it bought: dropping self-weight entirely gives R_B = 431.3 kN, M_f⁺ = 1095 kN·m and M_f⁻ = 210 kN·m, so ignoring it would understate the sagging demand by 10 % and the hogging demand by 15 %. At 1095 kN·m the 5–30M would look comfortable at 0.86; the beam only reaches 0.96 once its own weight is carried, which is exactly why the question insists on it.

Quantity	Value
Self-weight (factored)	14.4 kN/m → 18.0 kN/m
Reactions R_A / R_B	341.3 / 543.8 kN
Max sagging moment (at 4 m)	1221 kN·m
Hogging moment at B	246 kN·m
Max shear (left of B)	403 kN
Section	500 × 1200 mm, d = 1134 mm
Bottom steel	5–30M — M_r = 1271 kN·m (util. 0.96)
Top steel over B	4–25M — M_r = 747 kN·m (minimum steel governs)
V_c / V_r provided	353 / 601 kN
Stirrups	10M double-legged @ 400 mm

Topic. Flexural and shear design of a determinate reinforced concrete beam carrying an overhang, including self-weight.

Key relations. a = φ_sf_yA_s/(α₁φ_cf_c'b); M_r = φ_sf_yA_s(d − a/2); A_s,min = 0.2√f_c'b_th/f_y; V_r = V_c + V_s with V_s = φ_sA_vf_yd_vcotθ/s.

Why this works. A single-span beam with one overhang is determinate, so the whole envelope follows from two equilibrium equations regardless of how the section is finally sized — which is why the self-weight iteration converges in one pass provided the first trial is sensible. The two design moments are qualitatively different: the sagging moment comes almost entirely from the applied point load and needs a large tension area at the bottom, while the hogging moment over the support comes only from the short overhang and falls below the code's minimum-steel floor, so the top steel is set by crack control and robustness rather than by strength. That is a common and entirely legitimate outcome, and recognising it saves a great deal of arithmetic.

Common pitfall. Assuming the maximum sagging moment lies at midspan; with a point load at 4 m of an 8 m span and a UDL, the peak is at the load, and differentiating M(x) confirms the UDL cannot move it. A second trap is checking bar fit only after fixing d: five 30M bars just fit across 500 mm, and a 450 mm-wide beam would force two layers, cost 40 mm of effective depth and fail the section that had appeared to pass.

Source. CSA A23.3:19 Clauses 10.1, 10.5.1.2, 11.2.8.2 and 11.3.4–11.3.8.

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