Question 4 of 7: Reinforced concrete section for member ABC

Question B1: Reinforced concrete section for member ABC (20 marks: 6 + 8 + 6)

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. Overall dimensions $b \times h$ for member ABC, plus the amount, position and layout of longitudinal steel and stirrups that satisfy flexure and shear along its whole length.

Approach. Use the hinge at B as the fourth equation to get the reaction at A by inspection, draw the shear and moment diagrams for ABC, factor them, then size a rectangular section for the peak hogging moment at C and detail the rest of the member around it.

1. Reaction at A from the hinge condition. The segment AB carries only the 250 kN load and the reaction at A, and the hinge transmits no moment, so taking moments about B for AB alone: $$\sum M_B^{AB} = 0: \quad A_y (3.0) - 250(1.5) = 0 \;\Rightarrow\; \boxed{A_y = 125\ \text{kN (up)}}$$ The roller at A takes no horizontal force, so member ABC carries no axial load — it is a pure beam, and the 80 kN horizontal load lives entirely in the column.
2. Shear and moment along ABC. With $s$ measured from A: $$V = +125\ \text{kN}\ (0 \leq s < 1.5), \qquad V = -125\ \text{kN}\ (1.5 < s \leq 6.0)$$ $$M(1.5) = 125(1.5) = +187.5\ \text{kN}\cdot\text{m}\ \text{(sagging)}$$ $$M(3.0) = 125(3.0) - 250(1.5) = 0 \quad \text{(the hinge — a useful check)}$$ $$M(6.0) = 125(6.0) - 250(4.5) = -375\ \text{kN}\cdot\text{m}\ \text{(hogging at C)}$$ The moment falls to zero at the hinge and then grows linearly to its peak at the column, because beyond B the beam is simply a cantilever from C loaded by the 125 kN passing through the hinge.
3. Factored actions. $$M_f^{-} = 1.5(375) = 562.5\ \text{kN}\cdot\text{m}\ \text{at C},\qquad M_f^{+} = 1.5(187.5) = 281.3\ \text{kN}\cdot\text{m},\qquad V_f = 1.5(125) = 187.5\ \text{kN}$$
4. Choose trial dimensions. Aim for a lightly enough reinforced section that it is ductile and buildable. Take $b = 400$ mm and $h = 800$ mm; with 40 mm cover, 10M stirrups and one layer of 30M bars, $d = 800 - 40 - 11.3 - 15 \approx 730$ mm. The depth is about $L/8$ of the 6 m member, generous but appropriate to a 250 kN point load on a determinate frame.
5. Steel for the hogging moment at C. With $C = \alpha_1\phi_c f_c' b a$ and $T = \phi_s f_y A_s$, equilibrium gives $a = \phi_s f_y A_s / (\alpha_1\phi_c f_c' b)$, and $$M_r = \phi_s f_y A_s\left(d - \frac{a}{2}\right)$$ Setting $M_r = M_f^{-} = 562.5$ kN·m and solving the resulting quadratic gives $A_s = 2461$ mm$^2$. Provide 4–30M = 2800 mm$^2$ in the top face (a single layer needs $2(40) + 2(11.3) + 4(29.9) + 3(40) = 342$ mm against the 400 mm width, so it fits).
6. Check the section as detailed. $$a = \frac{0.85(400)(2800)}{0.7975(0.65)(35)(400)} = \frac{952\,000}{7257} = 131.2\ \text{mm}, \qquad c = \frac{a}{\beta_1} = 148.6\ \text{mm}$$ $$\frac{c}{d} = \frac{148.6}{730} = 0.204 \;<\; \frac{700}{700 + f_y} = 0.636$$ so the steel yields comfortably and the section is ductile. Then $$M_r = 0.85(400)(2800)\left(730 - \frac{131.2}{2}\right) \times 10^{-6}$$ $$\boxed{M_r = 632.5\ \text{kN}\cdot\text{m} \;>\; 562.5\ \text{kN}\cdot\text{m} \quad (\text{utilisation } 0.89)}$$
7. Steel for the sagging region. Repeating the same solution with $M_f^{+} = 281.3$ kN·m gives $A_s = 1178$ mm$^2$; provide 3–25M = 1500 mm$^2$ in the bottom face ($M_r = 354$ kN·m). Both faces exceed the minimum $$A_{s,\min} = \frac{0.2\sqrt{f_c'}}{f_y}b_t h = \frac{0.2\sqrt{35}}{400}(400)(800) = 947\ \text{mm}^2$$
8. Shear. With $d_v = \max(0.9d,\ 0.72h) = \max(657,\ 576) = 657$ mm and the simplified method ($\beta = 0.18$, $\theta = 35^{\circ}$, valid because minimum stirrups are provided): $$V_c = \phi_c \beta \sqrt{f_c'}\, b_w d_v = 0.65(0.18)\sqrt{35}(400)(657) \times 10^{-3} = 181.9\ \text{kN}$$ $V_c = 181.9 < V_f = 187.5$ kN, so stirrups are required — only just, but required. Trying 10M double-leg stirrups ($A_v = 200$ mm$^2$) at $s = 300$ mm: $$V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = \frac{0.85(200)(400)(657)(1.428)}{300} \times 10^{-3} = 212.7\ \text{kN}$$ $$\boxed{V_r = V_c + V_s = 181.9 + 212.7 = 394.6\ \text{kN} \;>\; 187.5\ \text{kN}}$$ Spacing limits: $V_f = 187.5 < 0.125\phi_c f_c' b_w d_v = 747$ kN, so $s \leq \min(0.7 d_v,\ 600) = 460$ mm — 300 mm is comfortable; and $A_v = 200 > A_{v,\min} = 0.06\sqrt{f_c'}b_w s/f_y = 107$ mm$^2$. The crushing ceiling $0.25\phi_c f_c' b_w d_v = 1495$ kN is nowhere near.
9. Detailing. The 4–30M top steel must run the full length from C back past the hinge at B (the moment is hogging over the whole 3 m of BC) and be fully anchored into the column at C — the joint is monolithic, so the bars hook into the column and lap with its ties. The 3–25M bottom steel covers AB and can be curtailed a development length beyond the hinge. Two 15M bars run the length of the top face over AB and the bottom face over BC as stirrup hangers.