B3. Moment and shear resistances of a box culvert (10 + 10 marks): 16-Civ-A2 Elementary Structural Design, May 2014, Q6

Question 6 — B3. Moment and shear resistances of a box culvert (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.

Figure B3 — box culvert cross-section: 2.0 m × 1.5 m overall, 250 mm walls throughout, 4–25M top, 4–30M bottom, 20M ties at 200 mm, 70 mm cover.

Given.

Quantity	Value
Overall section	2 000 mm wide × 1 500 mm deep, hollow
Wall thickness	250 mm constant, so the void is 1 500 × 1 000 mm
Bottom steel	4–30M ($A_s = 2\,800$ mm$^2$)
Top steel	4–25M ($A_s = 2\,000$ mm$^2$)
Transverse steel	20M ties at 200 mm ($A_v = 600$ mm$^2$, two legs)
Cover	70 mm typical
Materials	$f_c' = 35$ MPa, $f_y = 400$ MPa

Find. The factored moment resistance $M_r$ and shear resistance $V_r$ of the cross-section, treating the box as a flexural member spanning along the culvert axis.

Approach. Check whether the compression block stays within the 250 mm top slab; if it does the hollow section behaves exactly like a solid rectangle 2 000 mm wide for flexure. For shear, only the two 250 mm side walls are effective, so $b_w = 500$ mm.

Effective depths. With 70 mm cover and 20M ties, $$d_{bot} = 1500 - 70 - 19.5 - \tfrac{29.9}{2} = 1\,395.6\ \text{mm}, \qquad d_{top} = 1500 - 70 - 19.5 - \tfrac{25.2}{2} = 1\,397.9\ \text{mm}$$
Sagging moment resistance (bottom steel in tension). The tensile force is $$T = \phi_s A_s f_y = 0.85(2\,800)(400) = 952\,000\ \text{N}$$ and the depth of the equivalent rectangular stress block, taken across the full 2 000 mm width of the top slab, is $$a = \frac{T}{\alpha_1\phi_cf_c'b} = \frac{952\,000}{0.7975(0.65)(35)(2\,000)} = 26.2\ \text{mm}$$ Since $a = 26.2$ mm is far less than the 250 mm slab thickness, the compression zone lies entirely within the top slab and the section is analysed as a solid rectangle: $$\boxed{M_r = T\left(d - \frac{a}{2}\right) = 952\,000(1\,395.6 - 13.1) = 1\,316\ \text{kN}\cdot\text{m}}$$
Hogging moment resistance (top steel in tension). Repeating with the 4–25M top bars, $T = 680$ kN, $a = 18.7$ mm and $$M_r = 680\,000(1\,397.9 - 9.4) = 943.9\ \text{kN}\cdot\text{m}$$ A culvert is loaded in both senses as fill and traffic move across it, so both values are quoted; the sagging value governs where the barrel is supported at its ends.
Concrete contribution to shear. Only the two side walls carry vertical shear, so $b_w = 2(250) = 500$ mm, and $$d_v = \max(0.9d,\ 0.72h) = \max(1\,256,\ 1\,080) = 1\,256\ \text{mm}$$ With the 20M ties comfortably exceeding the minimum transverse reinforcement, the simplified method gives $\beta = 0.18$ and $$V_c = \phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v = 0.65(1.0)(0.18)(5.916)(500)(1\,256) = 434.7\ \text{kN}$$
Steel contribution and total. With $A_v = 600$ mm$^2$ at $s = 200$ mm and $\theta = 35^\circ$, $$V_s = \frac{\phi_sA_vf_yd_v\cot\theta}{s} = \frac{0.85(600)(400)(1\,256)(1.428)}{200} = 1\,829.6\ \text{kN}$$ $$\boxed{V_r = V_c + V_s = 434.7 + 1\,829.6 = 2\,264\ \text{kN}}$$ This is below the crushing limit $0.25\phi_cf_c'b_wd_v = 3\,572$ kN, so the value stands; the minimum transverse steel requirement, $A_{v,min} = 0.06\sqrt{f_c'}b_ws/f_y = 89$ mm$^2$, is met many times over.

Check: the reinforcement shown is light for a member of this size. A23.3 Clause 10.5.1.2 would require $A_{s,min} = 0.2\sqrt{f_c'}b_th/f_y = 8\,874$ mm$^2$ if the 2 000 × 1 500 outline were solid, against the 2 800 mm$^2$ provided. For a hollow culvert the minimum is normally satisfied through Clause 10.5.1.3 (one third more steel than the analysis requires) and through the distributed slab and wall reinforcement that the figure does not show. The resistances calculated above are those of the bars given; they should not be read as confirming that the reinforcement detail is code-complete.

Result	Value
$d$ to bottom / top steel	1 395.6 / 1 397.9 mm
Stress block depth (sagging)	26.2 mm — within the 250 mm slab
$M_r$, sagging (4–30M)	1 316 kN·m
$M_r$, hogging (4–25M)	943.9 kN·m
$b_w$, $d_v$	500 mm, 1 256 mm
$V_c$ / $V_s$	434.7 / 1 829.6 kN
$V_r$	2 264 kN (limit 3 572 kN)

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

Key relations. $a = \phi_sA_sf_y/(\alpha_1\phi_cf_c'b)$ with $b$ the full flange width provided $a \le$ flange thickness; $M_r = \phi_sA_sf_y(d - a/2)$; $V_r = V_c + V_s$ with $b_w$ the sum of the web widths only.

Why this works. Flexure and shear see completely different parts of a hollow section. Bending is resisted by a couple between the extreme fibres, so as long as the compression block fits inside the top slab the void is irrelevant and the full 2 000 mm width is available. Shear, by contrast, flows through the vertical elements, so the void removes 1 500 of the 2 000 mm and only the two 250 mm walls count. Checking $a$ against the slab thickness is the single test that decides whether the simple rectangular analysis is legitimate.

Common pitfall. Using $b_w = 2\,000$ mm in the shear calculation, which overstates $V_c$ four-fold. The mirror-image error is using the web width in flexure, which badly understates $M_r$. A third trap is counting four legs of tie steel: a single perimeter tie crosses the shear plane twice, once in each wall, so $A_v = 2 \times 300 = 600$ mm$^2$.

Source. CSA A23.3, Clauses 10.1, 10.5.1 and 11.3 (simplified method).

Question 6 of 7: B3. Moment and shear resistances of a box culvert (10 + 10 marks)

Question 6 — B3. Moment and shear resistances of a box culvert (10 + 10 marks)