16-Civ-A2 Elementary Structural Design · December 2015

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, December 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour closed-book paper (textbooks and design handbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and the one question in Part C — five solutions, all of equal value. All seven are solved here, because the set is a study resource rather than an exam script.

Reference texts.

CSA S16, Design of Steel Structures — Clauses 11 (classification), 13.3 (compression), 13.5–13.6 (flexure), 13.8 (axial + bending), 13.11–13.12 (bolts, block shear).
CISC, Handbook of Steel Construction — section property tables for W-shapes and hollow structural sections.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial), 11 (shear), 10.15 (slender columns).
CSA O86, Engineering Design in Wood — Clause 6 (modification factors), Clause 7 (sawn lumber, bending and shear).
Canadian Wood Council, Wood Design Manual — specified strengths for Beam and Stringer grades.

Check — load factoring. Note 6 on page 1 states “all loads shown are unfactored”, so every load taken from a figure is a specified load and is factored here as 1.25D + 1.5L (NBCC Table 4.1.3.2, Case 2), treating the drawn point loads as live and member self-weight as dead. Where a question states a load directly in its text without a dead/live split (B3), that load is taken as already factored; this is stated again at that question.

Question 5: B2 — Reinforced concrete beam with an overhang (12 + 8 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. Span $L = 6$ m between supports with a 2 m overhang beyond the right-hand support; specified live load 50 kN/m over the full 8 m; $f_c' = 35$ MPa, $f_y = 400$ MPa; concrete density 24 kN/m³.

Find. A beam size, longitudinal reinforcement for both sagging and hogging, stirrup arrangement, and a reinforcing sketch.

Loading, bending moment diagram and the adopted 350 × 700 section. The 2 m overhang produces a hogging peak at B that must be reinforced on the top face.

Approach. Assume a trial section, compute the factored load including self-weight, find both the sagging peak and the hogging peak at the interior support, size the flexural steel for each, then design stirrups from the shear at a distance $d_v$ from the support face.

Trial section and factored load. Take $b = 350$ mm, $h = 700$ mm with 40 mm cover, 10M stirrups and 25M bars, giving $$d=700-40-11.3-\tfrac{25.2}{2}=636.2\ \text{mm}$$ The self-weight is $0.35(0.70)(24) = 5.88$ kN/m, so $$w_f=1.25(5.88)+1.5(50)=\boxed{82.35\ \text{kN/m}}$$
Reactions. Taking moments about A for the whole 8 m of loaded length, $$R_B=\frac{w_f(8)(4)}{6}=439.2\ \text{kN},\qquad R_A=82.35(8)-439.2=219.6\ \text{kN}$$
Design moments. Shear vanishes at $x = R_A/w_f = 2.667$ m, where the sagging moment peaks: $$M^{+}=219.6(2.667)-\frac{82.35(2.667)^{2}}{2}=\boxed{292.8\ \text{kN}\cdot\text{m}}$$ The overhang cantilevers back over B and produces $$M^{-}=\frac{82.35(2)^{2}}{2}=164.7\ \text{kN}\cdot\text{m}$$ Both must be reinforced, on opposite faces.
Bottom steel for the sagging region. Trying 3–25M ($A_s = 1500\ \text{mm}^{2}$), $$a=\frac{\phi_sA_sf_y}{\alpha_1\phi_cf_c'b}=\frac{0.85(1500)(400)}{0.7975(0.65)(35)(350)}=80.3\ \text{mm}$$ $$M_r=\phi_sA_sf_y\left(d-\frac{a}{2}\right)=0.85(1500)(400)(636.2-40.2)=304.0\ \text{kN}\cdot\text{m}\;>\;292.8\ \checkmark$$ with $c/d = 0.143$, so the steel yields well before the concrete crushes.
Top steel over the support. Trying 2–25M ($A_s = 1000\ \text{mm}^{2}$) gives $M_r = 207.2\ \text{kN}\cdot\text{m} > 164.7$. Both choices clear the minimum $$A_{s,\min}=\frac{0.2\sqrt{f_c'}\,b_th}{f_y}=\frac{0.2\sqrt{35}(350)(700)}{400}=725\ \text{mm}^{2}$$ Three 25M bars fit comfortably in a 350 mm web: the clear spacing works out at 86 mm against a required minimum of 35 mm.
Design shear. The largest shear is just left of B, $$V_f=\left|R_A-w_fL\right|=\left|219.6-494.1\right|=274.5\ \text{kN}$$ Sections within $d_v$ of the support face may be designed for the shear at $d_v$, where $d_v=\max(0.9d,0.72h)=572.6$ mm: $$V_{f,d}=274.5-82.35(0.573)=227.4\ \text{kN}$$
Stirrups. The concrete contribution is $$V_c=\phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v=0.65(1.0)(0.18)\sqrt{35}(350)(572.6)=138.7\ \text{kN}$$ Adopting 10M double-leg stirrups ($A_v = 200\ \text{mm}^{2}$) at $s = 250$ mm, $$V_s=\frac{0.85(200)(400)(572.6)\cot 35^\circ}{250}=222.4\ \text{kN},\qquad V_r=\boxed{361\ \text{kN}\;>\;227.4\ \checkmark}$$ The spacing satisfies $s\le\min(0.7d_v,600) = 401$ mm and is well inside the minimum-area limit of 644 mm. Crushing is not close: $V_{r,\max} = 1140$ kN.
Reinforcing details. Run 3–25M along the bottom, with at least one third continued into the supports and anchored; run 2–25M along the top through the support and to the end of the overhang, developed a full anchorage length into the span past the point of contraflexure. Use 10M stirrups at 250 mm through the shear-critical zones each side of B and out to the overhang tip, relaxing to 400 mm near midspan where $V_f < V_c$. Provide two 15M hanger bars in the top of the sagging region to support the stirrup cage.

Item	Design
Section	350 × 700 mm, $d = 636$ mm, 40 mm cover
Factored load $w_f$	82.35 kN/m
Reactions $R_A$ / $R_B$	219.6 kN / 439.2 kN
Sagging $M^{+}$ at 2.67 m / capacity	292.8 / 304.0 kN·m (3–25M bottom)
Hogging $M^{-}$ at B / capacity	164.7 / 207.2 kN·m (2–25M top)
Design shear at $d_v$ / capacity	227.4 / 361 kN
Stirrups	10M double-leg @ 250 mm (400 mm at midspan)

Topic. Flexural and shear design of a determinate beam with a cantilever overhang.

Key relations. $a=\phi_sA_sf_y/(\alpha_1\phi_cf_c'b)$; $M_r=\phi_sA_sf_y(d-a/2)$; $A_{s,\min}=0.2\sqrt{f_c'}b_th/f_y$; $V_r=V_c+V_s$ with $V_c=\phi_c\lambda\beta\sqrt{f_c'}b_wd_v$.

Why this works. A beam with an overhang has two distinct critical sections and they load opposite faces of the member, so a design that only reinforces the bottom will fail in tension at the top over the support. Locating the sagging peak by setting the shear to zero is exact for a uniformly loaded determinate span, and the cantilever moment follows directly from the overhang statics without any analysis of the span. The rectangular stress block converts the real parabolic compression distribution into an equivalent rectangle of depth $a = \beta_1c$, which is what lets the lever arm be written simply as $d - a/2$. In shear, the permission to design for the value at $d_v$ from the face reflects that a diagonal crack forming at the support is restrained by direct strut action into the bearing.

Common pitfall. Forgetting the self-weight, which the question explicitly requires; here it adds 7.35 kN/m factored, about 9 % of the total, and moves the required steel by a full bar. A second pitfall is stopping the top steel at the support instead of developing it into the span past the point of contraflexure.

Source. CSA A23.3, Design of Concrete Structures, Clauses 10.1–10.5 (flexure and minimum steel), 11.3 (shear) and 12 (development and anchorage).

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