Question 4 of 7: Reinforced concrete cross-section for the Figure 4 beam

Question B1: Reinforced concrete cross-section for the Figure 4 beam (20 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.

Find. Cross-section dimensions in the ratio 1:3, the flexural steel for both the hogging region over B and the sagging region in the span, and the shear reinforcement.

[Figure not reproduced: Figure B1.1 — Figure 4 of the paper redrawn: a 3.0 m simple span with a 1.0 m overhang, W1 over the full length and P1 at the tip. The pin at A and the roller at B are drawn with the paper's own symbols. See the official exam paper.]

Approach. Choose a trial section in the required 1:3 ratio from the A23.3 span/depth rule, add its self weight, envelope the factored moments for the three live-load patterns a beam with an overhang requires, then check flexure, the minimum-steel rules and one-way shear.

1. Set the section from the deflection rule, not from strength. The loads here are very light, so the section will be governed by minimum dimensions. A23.3 Table 9.2 permits deflections to go unchecked if $h \ge L/16$ for a simple span and $h \ge L/8$ for a cantilever, i.e. $h \ge 3000/16 = 188\ \text{mm}$ and $h \ge 1000/8 = 125\ \text{mm}$. With the mandated 1:3 ratio the smallest practical section that also accommodates two bars and a stirrup across its width is $$\boxed{b = 150\ \text{mm},\quad h = 450\ \text{mm}}$$ which satisfies both depth rules by a wide margin.
2. Add the self weight. $$w_{sw} = 0.150 \times 0.450 \times 24 = 1.62\ \text{kN/m}$$ so the total dead load is $2.0 + 1.62 = 3.62\ \text{kN/m}$. Factoring, $$\begin{aligned} w_f &= 1.25(3.62) + 1.5(3.0) = 4.525 + 4.5 = 9.025\ \text{kN/m} \\ P_f &= 1.25(3.0) + 1.5(2.5) = 7.50\ \text{kN} \end{aligned}$$ and the dead-only factored UDL, needed for the patterned cases, is 4.525 kN/m.
3. Recognise that a beam with an overhang needs three load patterns. The hogging moment over B depends only on what stands on the cantilever, while the sagging moment in the span is reduced by anything on the cantilever. So the three cases are: live everywhere; live on the span only; live on the cantilever only. Loading everything is not the worst case for the span.
4. Compute the hogging moment at B. The cantilever is statically determinate, so with full load on it $$M_B = \frac{w_f a^2}{2} + P_f a = \frac{9.025(1.0)^2}{2} + 7.50(1.0) = 4.51 + 7.50 = \boxed{12.01\ \text{kN}\cdot\text{m}}$$ This value is independent of what stands on the span, which is why two of the three patterns give the same hogging moment.
5. Compute the sagging moment for each pattern. Taking moments about B, $R_A = (w_{span}L^2/2 - M_B)/L$, and the maximum sagging moment is $R_A^2/2w_{span}$ at $x = R_A/w_{span}$ from A. With live load on the span only the cantilever carries dead load alone, so $M_B$ drops to 6.01 and $R_A$ rises: $$\begin{aligned} R_A &= \frac{9.025(3.0)^2/2 - 6.01}{3.0} = 11.53\ \text{kN} \\ M_{sag} &= \frac{11.53^2}{2(9.025)} = \boxed{7.37\ \text{kN}\cdot\text{m}} \end{aligned}$$ at 1.278 m from A. Loading everything gives only 5.04 kN·m, confirming that patterning matters.
6. Collect the design shears. With full load the shear just left of B is $|R_A - w_f L| = |9.53 - 27.08| = 17.54\ \text{kN}$ and just right of B it is $w_f a + P_f = 9.03 + 7.50 = 16.53\ \text{kN}$. The governing value is $V_f = 17.54\ \text{kN}$.
7. Design the flexural steel. With 40 mm cover, 10M stirrups and 15M bars, $d = 450 - 40 - 11.3 - 8 = 390.7\ \text{mm}$. The minimum-steel rule of Clause 10.5.1.2 governs everything here: $$A_{s,min} = \frac{0.2\sqrt{f'_c}\,b_t h}{f_y} = \frac{0.2\sqrt{35}\,(150)(450)}{400} = 199.7\ \text{mm}^2$$ Provide 2–15M ($400\ \text{mm}^2$) top and bottom, continuous through the support so the same cage serves the hogging and sagging regions.
8. Check the moment resistance. With $\alpha_1 = 0.85 - 0.0015f'_c = 0.7975$, $$a = \frac{\phi_s A_s f_y}{\alpha_1 \phi_c f'_c b} = \frac{0.85(400)(400)}{0.7975(0.65)(35)(150)} = 50.0\ \text{mm}$$ $$M_r = \phi_s A_s f_y \left(d - \frac{a}{2}\right) = 0.85(400)(400)(390.7 - 25.0) = \boxed{49.74\ \text{kN}\cdot\text{m}}$$ so the utilisation is $12.01/49.74 = 0.242$. The section is four times stronger than it needs to be, purely because minimum steel and minimum dimensions dominate.
9. Check the cracking rule, which is the one that could have failed. Clause 10.5.1.3 requires $M_r \ge 1.2 M_{cr}$ so that the member does not fail the instant it cracks. With $f_r = 0.6\sqrt{f'_c} = 3.55\ \text{MPa}$ and $I_g = bh^3/12 = 1.139\times10^9\ \text{mm}^4$, $$\begin{aligned} M_{cr} &= \frac{f_r I_g}{y_t} = \frac{3.55(1.139\times10^9)}{225} = 17.97\ \text{kN}\cdot\text{m} \\ 1.2M_{cr} &= 21.56\ \text{kN}\cdot\text{m} \end{aligned}$$ and $49.74 \ge 21.56$  ✓. On a lightly loaded member this check, not the applied moment, is what justifies the steel provided.
10. Check one-way shear. With at least nominal transverse reinforcement the simplified method gives $\beta = 0.18$, and $d_v = \max(0.9d,\ 0.72h) = \max(351.6,\ 324) = 351.6\ \text{mm}$: $$V_c = \phi_c \lambda \beta \sqrt{f'_c}\, b_w d_v = 0.65(1.0)(0.18)\sqrt{35}\,(150)(351.6) = 36.51\ \text{kN}$$ so $V_f/V_c = 17.54/36.51 = 0.480$ and the concrete alone carries the shear. Since $h = 450 < 750\ \text{mm}$ and $V_f < V_c$, minimum stirrups are not strictly required; provide 10M closed ties at 200 mm throughout as a practical measure — they hold the cage, confine the compression zone at B and cost almost nothing.
11. Detail the reinforcement. Carry the two top bars from the tip of the cantilever right through B and a development length into the span; the hogging moment reaches zero about 0.6 m past B, so extending them 1.0 m beyond the support face is ample and simplifies the bar schedule. Carry the two bottom bars into the support and hook them, because the pin at A and the roller at B both deliver their reaction directly into the bottom of the section.

Final Results