Question 6 of 7: Reinforced concrete design of member AC

Question 6: Reinforced concrete design of member AC (14 + 6 = 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. The flexural reinforcement (top and bottom), the shear reinforcement, and a drawn layout for member \(AC\).

Approach. Unlike Question 4, a concrete frame is designed elastically, so re-analyse the same geometry with the actual relative stiffnesses of the concrete members (the beam is 2.6 times stiffer than the column) and with the members' own weight included. Design the peak sagging section by the rectangular stress block, provide continuous top steel through the joint for the hogging moment, and size stirrups by the A23.3 simplified method.

1. Section trial and stiffnesses. A 16 m span suggests a depth near \(L/15\); take the beam as 500 × 1100 mm and the column as 500 × 800 mm:
   $$I_{beam} = \frac{500(1100)^3}{12} = 5.546\times10^{10}, \qquad I_{col} = \frac{500(800)^3}{12} = 2.133\times10^{10}\ \text{mm}^4$$
   Their self-weights are 13.20 and 9.60 kN/m. Gross (uncracked) sections are used throughout, as permitted for a first-order design frame analysis.
2. Elastic frame analysis. Solving the same one-redundant force problem with these stiffnesses, and factoring the applied loads at 1.5 and self-weight at 1.25,
   $$M_{f,max} = \boxed{1634.9\ \text{kN}\cdot\text{m}\ \text{(sagging, at } x = 7.26\ \text{m)}}, \qquad M_{f,C} = 195.2\ \text{kN}\cdot\text{m}\ \text{(hogging)}$$
   $$V_{f,A} = 419.7\ \text{kN}, \qquad V_{f,C} = 444.1\ \text{kN}$$
   Because the beam is 2.6 times stiffer than the column, joint \(C\) behaves almost as a pin: the hogging moment there is only 12 % of the peak sagging moment, and the span acts nearly as a simply supported beam.
3. Sagging reinforcement. With two layers of 30M bars, cover 40 mm and 10M stirrups, take \(d = 1000\ \text{mm}\). Equating the internal couple to \(M_f\),
   $$M_r = \phi_s A_s f_y\left(d - \frac{a}{2}\right), \qquad a = \frac{\phi_s A_s f_y}{\alpha_1\phi_c f'_c b}$$
   Solving the resulting quadratic gives \(A_{s,req} = 5452\ \text{mm}^2\). Provide 8 – 30M (\(A_s = 5600\ \text{mm}^2\)) in two layers of four:
   $$a = 242.6\ \text{mm}, \quad c = \frac{a}{\beta_1} = 271.1\ \text{mm}, \quad \frac{c}{d} = 0.271 \le 0.6 \;\text{(tension-controlled)}$$
   $$M_r = 0.85(5600)(400)(1000 - 121.3)\times10^{-6} = \boxed{1673.1\ \text{kN}\cdot\text{m}} \;\ge\; 1634.9\ \text{kN}\cdot\text{m}$$
   The steel ratio is \(\rho = 0.0112\), well above \(\rho_{min} = 0.2\sqrt{f'_c}/f_y = 0.00274\) and comfortably below the tension-controlled ceiling.
4. Hogging reinforcement at C. The factored hogging moment needs only 582 mm², so the minimum-steel rule governs (\(A_{s,min} = 1370\ \text{mm}^2\)). Provide 4 – 25M continuous top steel (2000 mm²), anchored through the joint and lapped into the column:
   $$M_r = 0.85(2000)(400)(1000 - 21.7)\times10^{-6} = 650.5\ \text{kN}\cdot\text{m} \;\ge\; 195.2\ \text{kN}\cdot\text{m}$$
   Continuous top steel is required in any case to carry the hanger and confinement demands at the joint and to control cracking over the support.
5. Shear design. With \(d_v = \max(0.9d,\,0.72h) = 900\ \text{mm}\) and the simplified method for a member containing at least minimum stirrups (\(\beta = 0.18\), \(\theta = 35^{\circ}\)),
   $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_w d_v = 0.65(0.18)\sqrt{30}(500)(900)\times10^{-3} = 288.4\ \text{kN}$$
   $$V_s = V_f - V_c = 444.1 - 288.4 = 155.7\ \text{kN}$$
   $$\frac{A_v}{s} = \frac{V_s}{\phi_s f_y d_v \cot\theta} = \frac{155.7\times10^{3}}{0.85(400)(900)(1.428)} = 0.356\ \text{mm}^2/\text{mm}$$
   Two-legged 10M stirrups give \(A_v = 200\ \text{mm}^2\), so strength alone would allow \(s = 561\ \text{mm}\).
6. Stirrup spacing limits. Three limits apply and the minimum-steel rule governs:
   $$s \le \frac{A_v f_y}{0.06\sqrt{f'_c}\,b_w} = 487\ \text{mm}, \qquad s \le 0.7d_v = 630\ \text{mm}, \qquad s \le 600\ \text{mm}$$
   (the last two apply because \(V_f = 444\ \text{kN} \le 0.125\lambda\phi_c f'_c b_w d_v = 1097\ \text{kN}\)). Adopt 10M stirrups at 400 mm throughout, closed to 200 mm over the first 2 m from each end and either side of the 200 kN load points, where diagonal cracking initiates.
7. Layout and detailing. Bottom steel: 8 – 30M in two layers of four, with the outer layer extended full length into the supports and the inner layer curtailed where \(M_f\) falls below the four-bar capacity plus a development length. Top steel: 4 – 25M continuous, hooked into the column at \(C\) for a full development length \(l_d\). Side-face crack-control bars 15M at 300 mm each face over the lower half of the 1100 mm web, as required for members deeper than 750 mm. Clear cover 40 mm.