Question 4 of 7: Design of a reinforced concrete T-section

Question B1: Design of a reinforced concrete T-section (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. A 6.0 m beam supported 1.0 m in from each end, so that a 4.0 m span carries a central load while each 1.0 m overhang carries a load at its tip.

Find. T-section dimensions and the amount and layout of longitudinal and shear reinforcement, including the beam's own weight.

Approach. Choose a trial T-section, compute its self-weight, obtain the factored moment and shear envelopes, then proportion the sagging steel with the flange in compression, the hogging steel with the web acting alone, and finally the stirrups by the CSA A23.3 simplified method.

(a) Trial section and factored actions (6 marks)

1. Trial dimensions. Take a flange 1 000 mm wide and 150 mm thick on a web 400 mm wide, overall depth 700 mm — a span-to-depth ratio of about 6, appropriate for the heavy concentrated load. The area is $$A_c = 1\,000(150) + 400(550) = 0.370\ \text{m}^2 \quad\Rightarrow\quad w = 0.370(24) = 8.88\ \text{kN/m}$$
2. Factored loads. With 1.5 on the specified point loads and 1.25 on self-weight, $$P_{f,\text{tip}} = 120\ \text{kN}, \qquad P_{f,\text{mid}} = 450\ \text{kN}, \qquad w_f = 11.10\ \text{kN/m}$$ By symmetry each reaction is $R = \left[2(120) + 450 + 11.10(6.0)\right]/2 = 378.3\ \text{kN}$.
3. Moment and shear envelope. At a support the overhang delivers a hogging moment $$M_{f,\text{sup}} = -\left[120(1.0) + \tfrac{11.10(1.0)^2}{2}\right] = -125.6\ \text{kN}\cdot\text{m}$$ and at mid-span $$M_{f,\text{mid}} = 378.3(2.0) - 120(3.0) - \tfrac{11.10(3.0)^2}{2} = +346.7\ \text{kN}\cdot\text{m}$$ The largest shear is immediately inside a support, $$V_{f} = 378.3 - 120 - 11.10(1.0) = 247.2\ \text{kN}$$ $$\boxed{M_f^{+} = 346.7,\quad M_f^{-} = -125.6\ \text{kN}\cdot\text{m},\quad V_f = 247.2\ \text{kN}}$$

(b) Flexural reinforcement (8 marks)

1. Sagging steel at mid-span. With 40 mm cover, 10M stirrups and one layer of 25M bars, $d = 700 - 40 - 10 - 12.6 = 637\ \text{mm}$. Assuming the flange is in compression and the stress block stays within it, equilibrium and moment give $$a = \frac{\phi_s A_s f_y}{\alpha_1\phi_c f_c' b}, \qquad M_r = \phi_s A_s f_y\left(d - \frac{a}{2}\right)$$ Iterating for $M_r = 346.7\ \text{kN}\cdot\text{m}$ yields $A_{s,\text{req}} = 1\,639\ \text{mm}^2$ at $a = 30.7$ mm, which is well inside the 150 mm flange, so the rectangular assumption holds.
2. Choose the bars and check. Provide 4 – 25M ($A_s = 2\,000\ \text{mm}^2$) in one layer; the clear web width of $400 - 2(40) - 2(11.3) = 297\ \text{mm}$ leaves 65 mm between bars, comfortably above the minimum. Then $$a = \frac{0.85(2\,000)(400)}{0.7975(0.65)(35)(1\,000)} = 37.5\ \text{mm}, \qquad c = a/\beta_1 = 42.5\ \text{mm}$$ $$M_r = 0.85(2\,000)(400)(637 - 18.7) = 420.7\ \text{kN}\cdot\text{m} > 346.7\ \text{kN}\cdot\text{m}$$ with $c/d = 0.067$, so the section is strongly under-reinforced and the steel yields long before the concrete crushes. Minimum steel, $0.2\sqrt{f_c'}\,b_wh/f_y = 828\ \text{mm}^2$, is satisfied.
3. Hogging steel over the supports. Here the flange is in tension and only the 400 mm web resists compression, so the section is designed as a rectangle of that width with $d = 640\ \text{mm}$ to the 20M top bars. For $M_f = 125.6\ \text{kN}\cdot\text{m}$ the same iteration gives $A_{s,\text{req}} = 590\ \text{mm}^2$. Provide 3 – 20M ($900\ \text{mm}^2$) spread across the flange width, which is 1.5 times the amount analysis requires and therefore satisfies Clause 10.5.1.3 in place of the minimum-steel rule; the resulting $$\boxed{M_r^{-} = 189.5\ \text{kN}\cdot\text{m} > 125.6\ \text{kN}\cdot\text{m}}$$ These bars must run the full length of each overhang and be developed past the point of contraflexure into the span.

(c) Shear reinforcement (6 marks)

1. Design shear and effective shear depth. Taking a 300 mm wide support, the critical section is $d$ from its face, at 1.79 m from the beam end, where $V_f = 238.5\ \text{kN}$. The effective shear depth is $$d_v = \max(0.9d,\ 0.72h) = \max(573,\ 504) = 573.7\ \text{mm}$$
2. Concrete contribution (simplified method, Clause 11.3.6.3, valid because the beam carries at least minimum transverse reinforcement, so $\beta = 0.18$ and $\theta = 35^\circ$): $$V_c = \phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v = 0.65(1.0)(0.18)\sqrt{35}(400)(573.7) = 158.8\ \text{kN}$$ so the stirrups must carry at least $238.5 - 158.8 = 79.7\ \text{kN}$.
3. Stirrups. With 10M double-leg stirrups, $A_v = 200\ \text{mm}^2$, and $$V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = \frac{0.85(200)(400)(573.7)(1.4281)}{s}$$ Spacing is limited to $0.7d_v = 402$ mm by Clause 11.3.8 and to 563 mm by the minimum-steel rule, so adopt 10M stirrups at 300 mm throughout. Then $V_s = 185.7$ kN and $$\boxed{V_r = 158.8 + 185.7 = 344.5\ \text{kN} > V_f = 247.2\ \text{kN}}$$ comfortably above even the un-reduced support shear. Web crushing is not an issue: $0.25\phi_cf_c'b_wd_v = 1\,305\ \text{kN}$.
4. Detailing that completes the design. Close the stirrups (they must enclose both the top and the bottom steel, since the beam is hogging over the supports); carry them through the overhangs; anchor the bottom bars with a standard hook or a 90-degree bend past the support centre-line; and lap the top bars only in the mid-span region where they are least stressed.