16-Civ-A2 Elementary Structural Design · May 2015

Question 4 of 7: B1 — Moment and shear resistances of the r.c. culvert section

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

Notes on this paper

Paper format. National Examinations, May 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, closed book (textbooks and design handbooks permitted; one approved Casio or Sharp calculator). Seven questions in three parts: Part A (A1–A3, structural 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, 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”, but the paper never splits the figure loads into dead and live components. Every applied figure load in Parts A and B is therefore treated as live load and factored by 1.5; concrete self-weight, where it is not expressly excluded, would be dead load at 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). Part C names its components, so 1.25D + 1.5L is applied directly there. If an examiner intended a different split, the method below is unchanged — only the numerical factor moves.

Check — figure dimensions. All four figures are hand sketches on page 3; every dimension below was read from the printed figures. Figure A2 reads 80 – 300 – 80 across the bottom plate (overall 460 mm wide) with the 300 mm dimension spanning the two web centrelines, and 250 mm overall depth. Figure B1 reads 3.0 m top, 2.5 m bottom, 2.0 m deep, walls “300 constant” and cover “70 typical”. Figure B2 reads 5 m + 5 m span, 4 m + 4 m column height, roller at A and hinge at D.

Question 4: B1 — Moment and shear resistances of the r.c. culvert 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.

Quantity	Value
Overall geometry	trapezoidal box: 3.0 m wide at the top, 2.5 m at the base, 2.0 m deep
Wall thickness	300 mm constant (top slab, base slab and both side walls)
Cover	70 mm typical
Longitudinal steel	5–25M top, 5–30M bottom, 4–20M in each side wall
Transverse steel	15M closed perimeter tie at 200 mm centres
Materials	f'_c = 35 MPa, f_y = 400 MPa

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

Figure B1 — trapezoidal r.c. culvert cross-section. In flexure the void is irrelevant because the 21.9 mm stress block stays inside the 300 mm top slab; in shear only the two 300 mm side walls count.

Approach. Treat the box as a single flexural member 2.0 m deep spanning along the culvert axis, sagging so that the heavier bottom steel is in tension. Find the depth of the compression block, check it lies inside the 300 mm top slab so the section computes as a rectangle 3000 mm wide, then apply the A23.3 simplified shear method with the side walls as the webs.

Stress-block parameters (A23.3 10.1.7). $$ \alpha_1 = 0.85 - 0.0015 f'_c = 0.85 - 0.0525 = 0.7975 $$ $$ \beta_1 = 0.97 - 0.0025 f'_c = 0.97 - 0.0875 = 0.8825 $$ with $\phi_c = 0.65$ and $\phi_s = 0.85$.
Effective depth. The 5–30M bottom bars sit inside the 70 mm cover and the 15M tie: $$ d = 2000 - 70 - 16 - \frac{29.9}{2} = 1899~\text{mm} $$ with $A_s = 5(700) = 3500~\text{mm}^2$.
Depth of the compression block. The tensile force the bottom steel can develop is $$ T = \phi_s A_s f_y = 0.85(3500)(400) = 1190~\text{kN} $$ Equating it to the compressive force in a block of width equal to the 3000 mm top slab: $$ a = \frac{T}{\alpha_1 \phi_c f'_c b} = \frac{1\,190\,000}{0.7975(0.65)(35)(3000)} = 21.9~\text{mm} $$ Since 21.9 mm is far less than the 300 mm slab thickness, the entire compression zone lies within the solid top slab and the void has no effect on flexure: the box computes exactly as a 3000 mm wide rectangular section.
Confirm ductile behaviour. The neutral-axis depth is $c = a/\beta_1 = 24.8$ mm, so $c/d = 0.013$ and the steel strain at crushing is $$ \varepsilon_s = 0.0035\,\frac{d - c}{c} = 0.0035\,\frac{1874}{24.8} = 0.264 \;\gg\; \frac{f_y}{E_s} = 0.002 $$ The section is very heavily under-reinforced and will warn extensively before failure. The 5–25M top bars sit in the compression zone but are so close to the neutral axis that they are barely stressed; ignoring them is both conservative and negligible.
Moment resistance. $$ M_r = T\left(d - \frac{a}{2}\right) = 1190\left(1899 - 10.9\right)\times10^{-3} = \boxed{2248~\text{kN}\cdot\text{m}} $$
Shear: effective web width and effective shear depth. In flexure the void was invisible; in shear it is decisive. Only the two side walls cross a horizontal shear plane, so $$ b_w = 2(300) = 600~\text{mm} $$ and, from A23.3 Clause 2.3, $$ d_v = \max(0.9d,\; 0.72h) = \max(1709,\; 1440) = 1709~\text{mm} $$
Check that minimum transverse steel is present (A23.3 11.2.8.2). The closed perimeter tie crosses the shear plane once in each side wall, so $A_v = 2(200) = 400~\text{mm}^2$ at $s = 200$ mm: $$ A_{v,\min} = 0.06\sqrt{f'_c}\,\frac{b_w s}{f_y} = 0.06(5.92)\frac{600(200)}{400} = 107~\text{mm}^2 \;<\; 400~\text{mm}^2 $$ Minimum steel is comfortably exceeded, so the simplified method may use $\beta = 0.18$ and $\theta = 35^\circ$.
Concrete and steel contributions. $$ V_c = \phi_c \lambda \beta \sqrt{f'_c}\, b_w d_v = 0.65(1.0)(0.18)(5.92)(600)(1709)\times10^{-3} = 710~\text{kN} $$ $$ V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = \frac{0.85(400)(400)(1709)(1.428)}{200}\times10^{-3} = 1660~\text{kN} $$ $$ V_r = V_c + V_s = \boxed{2370~\text{kN}} $$
Check the crushing limit (A23.3 11.3.3). $$ V_{r,\max} = 0.25\,\phi_c f'_c b_w d_v = 0.25(0.65)(35)(600)(1709)\times10^{-3} = 5834~\text{kN} $$ The computed 2370 kN is only 41 percent of the diagonal-crushing limit, so the stirrup contribution is fully available and the section is not over-reinforced in shear.

Check — minimum flexural reinforcement. A23.3 Clause 10.5.1.2 requires $A_{s,\min} = 0.2\sqrt{f'_c}\,b_t h/f_y = 0.2(5.92)(2500)(2000)/400 = 14\,790~\text{mm}^2$ for a 2.5 m wide tension face on a 2.0 m deep member. The 3500 mm² provided is far below that, so the section relies on Clause 10.5.1.3, which waives 10.5.1.2 where the steel provided is at least one third greater than that required by analysis — satisfied whenever the applied moment does not exceed about 0.75 M_r. The question asks only for resistances, so this is recorded rather than resolved; a designer using this section must confirm the applied moment against that limit.

Check — sloping walls. The side walls lean 250 mm over the 2.0 m height, so their horizontal width is 300/cos(7.1°) = 302 mm. Using the perpendicular thickness of 300 mm for b_w is conservative by less than one percent.

Quantity	Value
Effective depth, d	1899 mm
Tension steel	5–30M = 3500 mm²
Depth of stress block, a	21.9 mm (inside the 300 mm top slab)
c/d	0.013 — strongly under-reinforced
Moment resistance, M_r	2248 kN·m
Effective web width, b_w	600 mm (two side walls only)
Effective shear depth, d_v	1709 mm
Concrete contribution, V_c	710 kN
Tie contribution, V_s	1660 kN
Shear resistance, V_r	2370 kN (limit 5834 kN)

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

Key relations. $a = \phi_s A_s f_y/(\alpha_1\phi_c f'_c b)$; $M_r = \phi_s A_s f_y(d - a/2)$; $d_v = \max(0.9d, 0.72h)$; $V_c = \phi_c\lambda\beta\sqrt{f'_c}b_w d_v$ with $\beta = 0.18$; $V_s = \phi_s A_v f_y d_v\cot\theta/s$ with $\theta = 35^\circ$.

Why this works. A box section behaves completely differently in the two actions, and the reason is where each action needs material. Flexural compression is carried in a thin zone near the extreme fibre, so as long as the stress block fits inside the solid top slab the void might as well not exist and the section is analysed as a solid rectangle of the full flange width. Shear, by contrast, is carried by diagonal compression fields that must run through the depth of the member, and those fields can only exist where there is concrete on a vertical plane — that is, in the webs. Using the full 3000 mm width for shear would overstate the resistance five-fold. The same closed tie serves both walls, which is why one tie at 200 mm centres delivers two effective legs.

Common pitfall. Carrying the flange width forward into the shear calculation, or counting four tie legs because the drawing shows a closed loop. A single perimeter tie crosses any given horizontal plane exactly twice, once in each wall. A third trap is forgetting to confirm that a is less than the slab thickness before treating the section as rectangular — if the block dipped into the void, a T-section analysis with the web widths would be required.

Source. CSA A23.3, Design of Concrete Structures, Clauses 10.1, 10.5 and 11.3; CAC Concrete Design Handbook, Chapters 2 and 4; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Chapters 4 and 6.

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