Resistances of the reinforced concrete culvert: 07-Str-A2, May 2014, Q6

Question B3: Resistances of the reinforced concrete culvert (20 marks: 10 + 10)

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 box	2.0 m wide × 1.5 m deep
Wall thickness	250 mm, constant all round (void 1500 × 1000)
Bottom steel	4–30M ($A_s = 2800$ mm$^2$)
Top steel	4–25M ($A_s' = 2000$ mm$^2$)
Transverse steel	20M closed ties at 200 mm ($A_v = 2 \times 300 = 600$ mm$^2$)
Cover to bar centres	70 mm typical, so $d = 1430$ mm and $d' = 70$ mm
Materials	$f_c' = 35$ MPa, $f_y = 400$ MPa

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

Figure B3: a closed box culvert 2.0 m by 1.5 m with 250 mm walls, reinforced as a longitudinally spanning flexural member. The strain diagram shows how shallow the neutral axis is — only 47 mm into a 250 mm top slab.

Check — how the culvert spans. The reinforcement, not the outline, settles this. Four bars concentrated in the top face, four in the bottom face, and a closed tie at 200 mm running right round the perimeter is a beam cage: it is the detail for a member spanning along the culvert, a hollow box girder whose top and bottom slabs are the flanges and whose two side walls are the webs. A culvert designed to span transversely between its walls would instead show mats of distributed steel in each slab, near both faces. The resistances below are therefore about the horizontal axis of the box, per culvert, not per metre of length. Should the transverse frame action be wanted instead, the walls and slabs would have to be analysed as a closed frame under earth and water pressure — a different question with a different answer.

Approach. For flexure, note that the compression zone will lie inside the 250 mm top slab so the box behaves as a rectangle 2000 mm wide, then solve strain compatibility for the neutral axis with both bar layers included. For shear, take the two side walls as the webs and add the concrete and tie contributions by the simplified method.

Set up the flexural equilibrium. Assume sagging (heavier steel in the bottom face, so that is the tension side) and a compression block within the top slab, so $b = 2000$ mm. Then $$\alpha_1\phi_c f_c' b = 0.7975(0.65)(35)(2000) = 36\,286\ \text{N per mm of } a$$ and the tension force at yield is $$T = \phi_s f_y A_s = 0.85(400)(2800) = 952\ \text{kN}$$
Find the neutral axis, including the top bars. A first pass ignoring the top steel gives $a = 952\,000/36\,286 = 26.2$ mm and $c = 29.7$ mm — which is above the top bars at $d' = 70$ mm. Those bars therefore lie below the neutral axis and are in tension, not compression, and must be carried through strain compatibility: $$f_s' = \frac{0.0035(c - 70)}{c}E_s \quad (\text{negative, i.e. tensile})$$ $$\alpha_1\phi_c f_c' b \beta_1 c + \phi_s A_s' f_s' = \phi_s f_y A_s$$ Solving gives $$\boxed{c = 47.4\ \text{mm}, \qquad a = \beta_1 c = 41.9\ \text{mm}, \qquad f_s' = -333\ \text{MPa (tension, not yielded)}}$$ Since $a = 41.9 \ll 250$ mm, the block really is inside the top slab and the rectangular assumption stands. This is an extremely lightly reinforced section: 2800 mm$^2$ of steel against a 2000 mm wide flange.
Moment resistance. Taking moments about the centroid of the bottom steel, $$M_r = C_c\left(d - \frac{a}{2}\right) - T'\left(d - d'\right) = 1518.5(1.409) - 566.6(1.360)$$ $$\boxed{M_r = 1369\ \text{kN}\cdot\text{m}}$$ Had the top bars simply been ignored, the answer would have been 1349 kN·m — a 1.5 per cent difference, because the tensile force they contribute acts very close to the concrete compression resultant and its net moment effect nearly cancels.
Shear: geometry of the webs. The two side walls resist the vertical shear, each 250 mm thick measured perpendicular to the shear, so $$b_v = 2(250) = 500\ \text{mm}, \qquad d_v = \max(0.9d,\ 0.72h) = \max(1287,\ 1080) = 1287\ \text{mm}$$
Concrete contribution. The 20M ties at 200 mm greatly exceed the minimum, so the simplified method with $\beta = 0.18$ and $\theta = 35^{\circ}$ applies: $$V_c = \phi_c \beta \sqrt{f_c'}\, b_v d_v = 0.65(0.18)\sqrt{35}(500)(1287) \times 10^{-3} = 445.4\ \text{kN}$$
Tie contribution. The closed tie crosses each wall once, so $A_v = 2(300) = 600$ mm$^2$ per 200 mm of length: $$V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = \frac{0.85(600)(400)(1287)(1.428)}{200} \times 10^{-3} = 1874.5\ \text{kN}$$
Total shear resistance, and the crushing ceiling. $$V_r = V_c + V_s = 445.4 + 1874.5$$ $$\boxed{V_r = 2320\ \text{kN}}$$ against the diagonal-crushing limit $$V_{r,\max} = 0.25\phi_c f_c' b_v d_v = 0.25(0.65)(35)(500)(1287) \times 10^{-3} = 3660\ \text{kN}$$ so the section is not web-crushing limited and the full 2320 kN may be quoted. The tie steel dominates by a factor of four — 20M at 200 mm in a 1.5 m deep member is very heavy shear reinforcement.
Report the minimum-reinforcement finding. The section as detailed does not satisfy A23.3 Cl.10.5.1. The gross second moment is $$I_g = \frac{2000(1500)^3 - 1500(1000)^3}{12} = 4.375 \times 10^{11}\ \text{mm}^4$$ and with $f_r = 0.6\lambda\sqrt{f_c'} = 3.55$ MPa the cracking moment is $$M_{cr} = \frac{f_r I_g}{y_t} = \frac{3.55(4.375 \times 10^{11})}{750} \times 10^{-6} = 2071\ \text{kN}\cdot\text{m}$$ $$\boxed{M_r = 1369\ \text{kN}\cdot\text{m} \;<\; M_{cr} = 2071\ \text{kN}\cdot\text{m}}$$ so the box would fail suddenly at first cracking rather than yielding first. (The prescriptive form of the same rule, $A_{s,\min} = 0.2\sqrt{f_c'}b_t h/f_y = 8874$ mm$^2$, is likewise far above the 2800 mm$^2$ provided, though it is conservative for a hollow section.) The resistances asked for are as computed, but the answer is incomplete without saying that the section is under-reinforced and needs roughly 4800 mm$^2$ in the bottom face to be code compliant.

Result	Value
Effective depth $d$ / $d'$	1430 / 70 mm
Neutral-axis depth $c$	47.4 mm (block $a = 41.9$ mm, inside the 250 mm top slab)
Stress in the 4–25M top bars	333 MPa tensile (below the neutral axis)
Moment resistance $M_r$	1369 kN·m
$M_r$ ignoring the top bars, for comparison	1349 kN·m
Shear geometry	$b_v = 500$ mm, $d_v = 1287$ mm
$V_c$ / $V_s$	445.4 / 1874.5 kN
Shear resistance $V_r$	2320 kN (ceiling $V_{r,\max} = 3660$ kN)
Minimum-reinforcement check	$M_r < M_{cr} = 2071$ kN·m — Cl.10.5.1 not satisfied

Topic. Flexural and shear resistance of a hollow box section, and the minimum-reinforcement (ductility) requirement.

Key relations. Rectangular stress block with strain compatibility for both bar layers; $M_r$ by moments about the tension steel; $b_v$ measured perpendicular to the shear, $d_v = \max(0.9d, 0.72h)$; $V_c = \phi_c\beta\sqrt{f_c'}b_v d_v$; $V_s = \phi_s A_v f_y d_v\cot\theta/s$; $V_{r,\max} = 0.25\phi_c f_c' b_v d_v$; $M_{cr} = f_r I_g/y_t$ with $f_r = 0.6\lambda\sqrt{f_c'}$.

Why this works. A hollow box in sagging is just a very deep flanged beam: the top slab supplies the compression flange and the void is removed from the tension side, where the concrete was doing nothing anyway. Because the flange is 2000 mm wide and the steel only 2800 mm$^2$, the compression block is astonishingly shallow, and that shallowness is what puts the top bars into tension — a result that looks wrong until the strain diagram is drawn. The two width choices in the shear calculation are the mirror of each other and both follow one rule: use the width perpendicular to the force being resisted. Vertical shear crosses the two vertical walls, so $b_v = 2t$. Finally, the $M_r$ against $M_{cr}$ comparison is the physical statement behind Cl.10.5.1: if the steel cannot carry more than the concrete carried before it cracked, the first crack is also the failure, with no warning and no ductility.

Common pitfall. Assuming the top bars are in compression because they are near the compression face. With $c = 47$ mm and $d' = 70$ mm they are not, and treating them as a compression force adds it to the wrong side of the equilibrium equation. The second is computing $M_r$ and $V_r$, finding them large, and stopping — the section as drawn is below its own cracking moment, which is the most important thing to be said about it.

Source. CSA A23.3:19 Cl.10.1 (flexural assumptions), Cl.10.5.1 (minimum reinforcement), Cl.11.3.4 to 11.3.6 (shear, simplified method); MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Ch.4 and 6.

Question 6 of 7: Resistances of the reinforced concrete culvert

Question B3: Resistances of the reinforced concrete culvert (20 marks: 10 + 10)