Moment and Shear Resistance of a Concrete Culvert Section: 16-Civ-A2 Elementary Structural Design, December 2013, Q5

Question B2: Moment and Shear Resistance of a Concrete 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. From Figure B2 the culvert is an open trapezoidal trough: 4 m across the top, 3 m across the base, 2 m deep, with walls sloping at 70° to the horizontal and a wall and base thickness of 300 mm constant. Reinforcement is 6 — 30M in the base slab, 15M at 200 mm centres in the sloping walls, and 2 — 20M at the top of each wall. Materials $f_c' = 35$ MPa, $f_y = 400$ MPa, 40 mm cover.

Find. The factored moment of resistance and shear resistance of the section, which for a trough of this form means the 3 m wide base slab that carries the 6 — 30M main steel, with the sloping wall checked as a companion strip.

Figure B2 — reinforced concrete culvert: 300 mm walls at 70°, 3 m base slab with 6 — 30M, 15M @ 200 in the walls and 2 — 20M at the wall tops.

Approach. Treat the base slab as a singly reinforced rectangular section 3000 mm wide and 300 mm thick, find the stress-block depth from force equilibrium, then take moments about the compression resultant; apply the simplified shear method for a member without stirrups. Repeat on a one-metre strip of wall.

Base slab geometry and steel. Six 30M bars give $A_s = 6(700) = 4200$ mm² distributed across $b = 3000$ mm. With 40 mm cover to a 29.9 mm bar, $$d = 300 - 40 - \tfrac{29.9}{2} = 245 \text{ mm}$$
Depth of the compression block. Force equilibrium between the yielding steel and the rectangular stress block, with $\alpha_1 = 0.7975$ and $\beta_1 = 0.8825$, gives $$a = \frac{\phi_s A_s f_y}{\alpha_1\phi_c f_c' b} = \frac{0.85(4200)(400)}{0.7975(0.65)(35)(3000)} = 26.2 \text{ mm}$$ The neutral axis is at $c = a/\beta_1 = 29.7$ mm, so $c/d = 0.121$. This is far below the 0.6 limit, confirming that all six bars reach $f_y$ and the section is under-reinforced.
Moment of resistance. Taking moments about the centroid of the compression block, $$M_r = \phi_s A_s f_y\left(d - \tfrac{a}{2}\right) = 0.85(4200)(400)(245 - 13.1)/10^6 = \boxed{331 \text{ kN}\cdot\text{m}}$$ across the full 3 m width, equivalently 110 kN·m per metre of culvert length. The steel comfortably exceeds $A_{s,min} = 0.2\sqrt{35}(3000)(300)/400 = 2662$ mm².
Shear resistance of the base slab. The slab has no stirrups, and its effective depth is under 350 mm, so the simplified method takes $\beta = 0.21$. The effective shear depth is $d_v = \max(0.9 \times 245,\, 0.72 \times 300) = 220.5$ mm: $$V_r = \phi_c \lambda \beta \sqrt{f_c'}\, b\, d_v = 0.65(1.0)(0.21)\sqrt{35}(3000)(220.5)/10^3 = \boxed{534 \text{ kN}}$$ or 178 kN per metre of length. Since $\sqrt{f_c'} = 5.92 < 8$ MPa, no cap on the concrete contribution applies.
Sloping wall strip. The wall carries 15M at 200 mm centres, which is 1000 mm² per metre of wall, at $d = 300 - 40 - 8 = 252$ mm. On a one-metre strip, $$a = \frac{0.85(1000)(400)}{0.7975(0.65)(35)(1000)} = 18.7 \text{ mm}, \qquad M_r = 0.85(1000)(400)(252 - 9.4)/10^6 = 82.5 \text{ kN}\cdot\text{m/m}$$ and the same simplified shear expression gives $V_r = 183$ kN per metre. The 2 — 20M bars at the wall tops are detailing steel that closes the reinforcement cage and controls the free edge; they do not contribute to the flexural resistances above.
Interpretation. The base slab spans the 3 m clear width between the wall feet and is the critical flexural element; the walls act as propped cantilevers retaining the fill and are governed by the earth-pressure moment at their base. Because $V_r$ for the base is 534 kN while the entire self-weight and fill reaction on a metre of slab would be an order of magnitude less, the shear check is never critical for a culvert of these proportions — flexure and crack control govern.

Element	Quantity	Value
Base slab (3000 × 300)	$A_s$ / $d$ / $a$	4200 mm² / 245 mm / 26.2 mm
Base slab	$c/d$	0.121 (tension-controlled)
Base slab	$M_r$	331 kN·m (110 kN·m/m)
Base slab	$V_r$	534 kN (178 kN/m)
Sloping wall (per metre)	$M_r$ / $V_r$	82.5 kN·m/m / 183 kN/m

Topic. Flexural and shear capacity of singly reinforced slab sections in a buried structure.

Key relations. $a = \phi_s A_s f_y/(\alpha_1\phi_c f_c' b)$; $M_r = \phi_s A_s f_y (d - a/2)$; $c = a/\beta_1$ with $c/d \le 0.6$ for tension control; $V_c = \phi_c\lambda\beta\sqrt{f_c'}b\,d_v$ with $\beta = 0.21$ when no stirrups are present and $d \le 350$ mm.

Why this works. Culvert base slabs are wide and shallow, so the concrete needs only a sliver of its depth to balance the steel force — here 26 mm out of 300. That makes the lever arm almost the full effective depth and means $M_r$ is very nearly proportional to $A_s d$, which is why practical designs are governed by bar size and cover rather than by concrete strength. On the shear side, a slab without stirrups relies entirely on aggregate interlock and dowel action, captured empirically by $\beta$; the code raises $\beta$ from 0.18 to 0.21 for shallow members precisely because thin slabs suffer less of the size effect that reduces interlock in deep beams.

Common pitfall. Including the 2 — 20M top bars in the tension steel of the base slab, when they belong to a different element and a different action; and using $\beta = 0.18$ (the value for members with minimum stirrups) on a slab that has none. Watch also that $d$ is measured to the bar centroid, not to the face of the bar.

Source. Cement Association of Canada, Concrete Design Handbook, CSA A23.3 Cl. 10.1 (flexure) and Cl. 11.3.4–11.3.6 (simplified shear); MacGregor & Bartlett, Ch. 4 and Ch. 6.

Question 5 of 7: Moment and Shear Resistance of a Concrete Culvert Section

Question B2: Moment and Shear Resistance of a Concrete Culvert Section (10 + 10 marks)