16-Civ-A2 Elementary Structural Design · May 2017

Question 6 of 7: B3 — Moment and shear resistance of a reinforced concrete culvert

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 6: B3 — Moment and shear resistance of a reinforced concrete culvert (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.

Quantity	Value
Overall width	2500 mm (0.5 + 1.5 + 0.5 m; the top slab cantilevers 500 mm each side of the box)
Overall depth	2000 mm
Wall and slab thickness	200 mm (assumed — not dimensioned; see callout)
Longitudinal steel	5–30M in the bottom slab (3500 mm²); 4–25M in the top slab (2000 mm²)
Transverse steel	20M closed ties at 200 mm in the walls; 20M at 150 mm in the top slab
Materials and cover	f_c' = 35 MPa, f_y = 400 MPa, 65 mm cover (earth-exposed)

Find. The factored moment resistance M_r and shear resistance V_r of the section spanning longitudinally as a beam.

Figure B3 — reinforced concrete culvert cross-section with its longitudinal reinforcement.

Approach. Treat the whole 2500 × 2000 mm box as one flexural member. Find the neutral axis by strain compatibility, checking whether the compression block stays inside the solid top slab — if it does, the voids are invisible in flexure. For shear, take b_w as the sum of the webs only and count the tie legs that actually cross the shear plane.

Effective depths. With 65 mm cover, a 20M tie and the bar itself, $$d = 2000 - 65 - 19.5 - \tfrac{29.9}{2} = 1900.6\ \text{mm}, \qquad d' = 65 + 19.5 + \tfrac{25.2}{2} = 97.1\ \text{mm}$$
Locate the neutral axis. The section carries no axial load, so the bar forces and the concrete block must sum to zero. Solving by strain compatibility (extreme fibre at 0.0035) gives $$c = 46.7\ \text{mm}, \qquad a = \beta_1 c = 41.2\ \text{mm}$$ Two consequences follow at once. First, a = 41 mm is far less than the 200 mm slab thickness, so the compression block lies entirely within the solid top slab and the section computes as a 2500 mm-wide rectangle — the voids play no part in flexure. Second, c = 46.7 mm is above the top bars at d' = 97.1 mm, so the 4–25M are below the neutral axis and act in tension, not compression.
Moment resistance. Taking moments of the concrete block and both bar groups about mid-depth, $$M_r = \boxed{2289\ \text{kN}\cdot\text{m}}$$ It is worth quantifying what the top bars are worth. Discarding them entirely and solving again gives c = 29.7 mm and M_r = 2246 kN·m — only 1.9 % less. They sit so close to the compression face that their large tensile force is offset by a short lever arm and by the deeper block their presence demands. Including them is correct, but the section is overwhelmingly governed by the 5–30M at the bottom.
Ductility check. With c/d = 46.7/1900.6 = 0.025, the section is extraordinarily tension-controlled: the bottom steel is many times past yield before the concrete crushes. That is characteristic of a lightly reinforced deep box, and it means M_r is essentially φ_sf_yA_s times the lever arm and barely depends on f_c'.
Shear — the effective web width. Here the voids matter. Only the two walls cross a horizontal shear plane, so $$b_w = 2 \times 200 = 400\ \text{mm}$$ not the 2500 mm used in flexure. With d_v = max(0.9d, 0.72h) = 1710.5 mm and β = 0.18, $$V_c = 0.65(1.0)(0.18)\sqrt{35}(400)(1710.5) = 474\ \text{kN}$$
Shear carried by the ties. Each wall carries one closed 20M tie, and a closed tie presents two legs, so four legs cross the shear plane: A_v = 4(300) = 1200 mm² at s = 200 mm. With θ = 35°, $$V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = \frac{0.85(1200)(400)(1710.5)(1.428)}{200} = 4983\ \text{kN}$$
Apply the crushing cap. That number cannot be delivered. The diagonal-compression limit is $$V_{r,\max} = 0.25\phi_c f_c' b_w d_v = 0.25(0.65)(35)(400)(1710.5) = \boxed{3891\ \text{kN}}$$ and V_c + V_s = 5457 kN exceeds it, so web crushing governs and V_r = 3891 kN. This is the decisive lesson of the section: with only 400 mm of web across a 1.71 m shear depth, the concrete struts run out of capacity long before the ties do. Adding more tie steel would buy nothing at all; only thicker walls would raise V_r.

Quantity	Value
Effective depths d / d'	1900.6 / 97.1 mm
Neutral axis c, block depth a	46.7 / 41.2 mm (inside the 200 mm top slab)
Top bars	Below the neutral axis — in tension, worth +1.9 % on M_r
*M_r*	2289 kN·m
Effective web width b_w	400 mm (two 200 mm walls — not 2500 mm)
d_v	1710.5 mm
V_c / V_s (uncapped)	474 / 4983 kN
Crushing limit 0.25φ_cf_c'b_wd_v	3891 kN — governs
*V_r*	3891 kN

Check — wall and slab thickness assumed. Figure B3 dimensions only the 2.5 m width and the 2 m depth; no member thickness is labelled, and the hand sketch is not to scale. The 200 mm adopted here is derived from the bar layout the figure does give: 65 mm cover to each face, two 20M tie legs and one 25M longitudinal bar require 65 + 65 + 2(19.5) + 25.2 = 194 mm, so 200 mm is the thinnest buildable wall consistent with the detailing shown. M_r is insensitive to this choice because the 41 mm stress block fits inside any plausible slab, but V_r scales directly with it — 150 mm walls would give V_r = 2918 kN and 250 mm walls 4864 kN. This assumption is stated under page-1 NOTE 1.

Topic. Flexural and shear resistance of a voided (box) reinforced concrete section.

Key relations. Strain compatibility with ε_cu = 0.0035 and a = β₁c; V_c = φ_cλβ√f_c'b_wd_v; V_s = φ_sA_vf_yd_vcotθ/s; V_r ≤ 0.25φ_cf_c'b_wd_v.

Why this works. A void changes nothing that lies outside the compression block, because concrete in the tension zone is already assumed cracked and carries no load. So as long as the block depth is less than the top slab thickness, a box behaves in flexure exactly like a solid rectangle of the same overall width — and in a deep, lightly reinforced section the block is always shallow, so this is the normal case rather than the exception. Shear is the opposite story: the diagonal compression field must physically pass through the webs, so the voids remove capacity in direct proportion to the concrete they take away. Splitting the section this way — full width in flexure, webs only in shear — is the single idea the question is testing.

Common pitfall. Using b = 2500 mm in the shear equations, which overstates V_r by a factor of six. Almost as common is treating the top bars as compression steel because they are near the top face: with c = 47 mm they are 50 mm below the neutral axis and are yielding in tension. Finally, reporting V_c + V_s = 5457 kN without applying the 0.25φ_cf_c'b_wd_v cap overstates the real resistance by 40 % and would let a designer "solve" a shear deficiency by adding ties that cannot help.

Source. CSA A23.3:19 Clauses 10.1, 11.3.3 (d_v and b_w), 11.3.4, 11.3.5.1 and 11.3.3 (maximum shear).

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