16-Civ-A2 Elementary Structural Design · December 2014

Question 4 of 7: B1 — Moment and shear resistance of a double-cell hollow section

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

Notes on this paper

Paper format. National Examinations, December 2014 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel, CSA S16), Part B (B1–B3, reinforced concrete, CSA A23.3) and Part C (C1, timber, CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here so the set works as a complete study resource.

Reference texts. CSA S16, Design of Steel Structures, with the CISC Handbook of Steel Construction; CSA A23.3, Design of Concrete Structures, with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, with the CWC Wood Design Manual; National Building Code of Canada (load combinations); Kulak & Grondin, Limit States Design in Structural Steel; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design.

Check — load factors. Note 6 on page 1 states “All loads shown are unfactored, unless otherwise specified”, but the paper never splits the figure loads into dead and live components. Throughout Parts A and B the applied figure loads are therefore treated as live load, factored by 1.5, and self-weight of concrete members as dead load, factored by 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). In Part C the paper does name the components, so 1.25D + 1.5S is used directly. If an examiner intended a different split the method is unchanged — only the numerical factor moves.

Check — figure dimensions. Every figure on this paper is a hand sketch on page 3; all geometry below was read from the printed figures. Figure A1 is not drawn to scale horizontally (its printed 600 and 300 dimensions are authoritative, not the drawn proportions). In Figure B1 the dimension “1 m × 1 m” runs from the section centreline to the outer face, so each cell is 1 m square and the box is 2000 mm wide × 1000 mm deep; the “600 × 600” dimension line brackets the void, giving 200 mm walls and a 400 mm central web. That reading is confirmed independently by the “65 typical” cover note: 65 + 65 cover plus two 15M tie legs plus one 30M bar needs exactly 200 mm of wall.

Question 4: B1 — Moment and shear resistance of a double-cell hollow section (10 + 10 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. A two-cell hollow box, 2000 mm wide × 1000 mm deep overall, each cell 1 m square with a 600 × 600 void, giving 200 mm outer walls and a 400 mm central web. Cover 65 mm typical.

Quantity	Value
Overall width × depth	2000 × 1000 mm
Voids (two)	600 × 600 mm each
Wall / central web thickness	200 mm / 400 mm
Longitudinal steel	24-30M (A_b = 700 mm²)
Ties	15M @ 250 mm (A_b = 200 mm²)
Clear cover, typical	65 mm
f'_c / f_y	35 / 400 MPa

Find. The factored moment resistance M_r and the factored shear resistance V_r of the section.

Figure B1 — the double-cell box. One closed 15M tie encircles each cell, putting one vertical leg in each outer wall and two in the central web. Dimensions in millimetres.

Approach. Locate the bars from the 65 mm cover, then find the neutral-axis depth that balances the equivalent rectangular compression block against the forces in all four bar layers; take moments for M_r. For shear, use the A23.3 simplified method with the summed web widths.

Locate the reinforcement. With 65 mm cover to a 15M tie (16 mm) and a 30M bar (29.9 mm), $$c_{\text{bar}} = 65 + 16 + \tfrac{29.9}{2} = 95.95\ \text{mm} \;\Rightarrow\; d = 1000-95.95 = 904\ \text{mm}$$ Counting the dots on the figure gives 8 bars in the top face, 8 in the bottom face and 8 distributed up the two outer walls and the central web — 24 in all, on four levels equally spaced at $(904-96)/3 = 269.4$ mm.
Stress-block parameters (A23.3 10.1.7). $$\alpha_1 = 0.85-0.0015f'_c = 0.7975 \qquad \beta_1 = 0.97-0.0025f'_c = 0.8825$$
Solve for the neutral axis. The section is a box: the top 200 mm is solid over the full 2000 mm width, below which only the 800 mm of web remains. Requiring zero net axial force, with each bar layer taking $f_s = E_s\varepsilon_s$ limited to $\pm f_y$ and $\varepsilon_{cu} = 0.0035$, gives $$c = 107.6\ \text{mm} \qquad a = \beta_1 c = 95.0\ \text{mm}$$ The block therefore lies entirely within the solid top flange, so the compression resultant is $$C_c = \alpha_1\phi_c f'_c\,b\,a = 0.7975(0.65)(35)(2000)(95.0) = 3447\ \text{kN}$$ Since $c/d = 0.119 \ll 0.5$, all three tension layers yield and the section is comfortably ductile.
Sum the bar forces. The eight top bars sit 96 mm from the compressed face, just below the neutral axis, so they are in mild compression ($f_s = 75.5$ MPa); the two intermediate layers and the bottom layer all yield in tension: $$F_s = 0.85\left[5600(75.5) - 2800(400) - 2800(400) - 5600(400)\right] = -3449\ \text{kN}$$ which balances $C_c$ to within rounding, confirming the neutral-axis depth.
Moment resistance. Taking moments of every force about the compression resultant: $$M_r = \sum F_i\,z_i = 1631 + 302 + 559 - 17 \ \text{kN}\cdot\text{m}$$ $$\boxed{M_r = 2475\ \text{kN}\cdot\text{m}}$$
Effective shear depth. $$d_v = \max(0.9d,\ 0.72h) = \max(813.6,\ 720) = 813.6\ \text{mm}$$
Concrete contribution (A23.3 11.3.4, simplified method). Only the webs resist shear, so $b_w$ is the sum of the two 200 mm walls and the 400 mm central web, i.e. 800 mm. The section carries at least minimum transverse reinforcement, so $\beta = 0.18$ and $\theta = 35^\circ$: $$V_c=\phi_c\lambda\beta\sqrt{f'_c}\,b_w d_v = 0.65(1.0)(0.18)\sqrt{35}\,(800)(813.6)$$ $$\boxed{V_c = 451\ \text{kN}}$$
Steel contribution. One closed 15M tie encircles each cell, so a vertical shear plane is crossed by four legs — one in each outer wall and two in the central web — giving $A_v = 4(200) = 800$ mm²: $$V_s=\frac{\phi_s A_v f_y d_v \cot\theta}{s}=\frac{0.85(800)(400)(813.6)\cot 35^\circ}{250}$$ $$\boxed{V_s = 1264\ \text{kN}}$$
Total shear resistance, with the crushing check. $$V_r = V_c + V_s = 451 + 1264 = \boxed{1715\ \text{kN}}$$ $$V_{r,\max}=0.25\phi_c f'_c b_w d_v = 0.25(0.65)(35)(800)(813.6) = 3702\ \text{kN} \; > \; 1715 \quad\checkmark$$ Diagonal crushing of the web concrete does not control, so the computed value stands.

Two features of this section are worth noticing. First, the neutral axis falls inside the 200 mm top wall, so the voids never enter the flexural calculation at all — the box behaves exactly like a 2000 mm wide solid slab for bending, and the hollow cores simply remove weight that was doing nothing. Second, in shear the picture reverses completely: the voids cut the effective width from 2000 mm to 800 mm, and the concrete term drops in the same proportion. It is the closely spaced ties, not the concrete, that supply nearly three-quarters of the shear capacity.

Quantity	Value
Effective depth, d	904 mm
Neutral-axis depth, c	107.6 mm (c/d = 0.119, ductile)
Moment resistance, M_r	2475 kN·m
Effective shear depth, d_v	813.6 mm
Web width for shear, b_w	800 mm
Concrete contribution, V_c	451 kN
Steel contribution, V_s	1264 kN
Shear resistance, V_r	1715 kN (< V_r,max = 3702 kN)

Topic: Flexural and shear capacity of a multi-cell reinforced concrete box section.

Key relations: $\alpha_1 = 0.85-0.0015f'_c$, $\beta_1 = 0.97-0.0025f'_c$; strain compatibility with $\varepsilon_{cu} = 0.0035$; $C_c = \alpha_1\phi_cf'_c b a$; $d_v=\max(0.9d,0.72h)$; $V_c=\phi_c\lambda\beta\sqrt{f'_c}b_wd_v$; $V_s=\phi_sA_vf_yd_v\cot\theta/s$; $V_{r,\max}=0.25\phi_cf'_cb_wd_v$.

Why this works: Flexure and shear ask different questions of the same geometry. Flexure only cares about the material between the extreme compression fibre and the neutral axis, plus the steel; because the axis here lands 107.6 mm down, inside a solid 200 mm wall, the voids are irrelevant and the calculation reduces to a wide singly reinforced rectangle with three tension layers. Shear, by contrast, is a diagonal-field problem that must be carried across the full depth, so only the vertical webs count — and any wall a shear crack must cross adds its thickness to $b_w$. The variable-angle truss model of A23.3 then splits the resistance into a concrete term (aggregate interlock and the compression chord) and a stirrup term proportional to $\cot\theta$, which at $35^\circ$ multiplies the stirrup contribution by 1.43.

Common pitfall: Using the full 2000 mm width as $b_w$ in the shear equations — that alone would overstate V_c by 150 %. The mirror-image mistake is deducting the voids from the flexural compression block when the neutral axis has not reached them. A third slip is counting only two tie legs; a closed tie around each cell puts two legs in the shared central web.

Source: CSA A23.3, Design of Concrete Structures, Clauses 10.1 and 11.3; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Ch. 4 and 6.

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