Question 4 of 7: Design of a Reinforced Concrete T-Beam

Question B1: Design of a Reinforced Concrete T-Beam (6 + 8 + 6 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. From Figure B1 the beam is 6.0 m long overall: a pin support and a roller support 4.0 m apart, with a 1.0 m overhang beyond each. Specified point loads are 80 kN at each overhang tip and 300 kN at midspan. Materials $f_c' = 35$ MPa, $f_y = 400$ MPa, normal-density concrete at 24 kN/m³, 40 mm cover, 10M stirrups.

Find. T-section dimensions, the flexural reinforcement for both the sagging and the hogging regions, and the shear reinforcement.

Approach. Choose a trial T-section, compute the self-weight, build the factored bending moment and shear envelopes, then size the tension steel at midspan (flange in compression) and over the supports (web in compression, because the overhangs hog), and finish with stirrups.

1. Select trial dimensions. Take an overall depth $h = 600$ mm (about span/7), a web $b_w = 300$ mm, a flange thickness $h_f = 125$ mm and an effective flange width $b = 1000$ mm. A23.3 permits an overhang of one-twelfth of the span each side, $6000/12 = 500$ mm, so 1000 mm total is available. Gross area $A_g = 1000(125) + 300(475) = 0.2675$ m² and $$w_{DL} = 0.2675(24) = 6.42 \text{ kN/m}, \qquad w_f = 1.25(6.42) = 8.03 \text{ kN/m}$$
2. Factored load effects. With $P_{tip} = 1.5(80) = 120$ kN and $P_{mid} = 1.5(300) = 450$ kN, symmetry gives each reaction $$R = \frac{2(120) + 450 + 8.03(6.0)}{2} = 369.1 \text{ kN}$$ Hogging over each support comes from the overhang alone, while the sagging peak is at midspan: $$M_{hog} = 120(1.0) + \tfrac{8.03(1.0)^2}{2} = \boxed{124.0 \text{ kN}\cdot\text{m}}$$ $$M_{sag} = 369.1(2.0) - 120(3.0) - \tfrac{8.03(3.0)^2}{2} = \boxed{342.0 \text{ kN}\cdot\text{m}}$$ and the shear just inside the support is $V_f = 369.1 - 120 - 8.03(1.0) = 241.1$ kN.
3. Stress-block parameters. For $f_c' = 35$ MPa, $$\alpha_1 = 0.85 - 0.0015 f_c' = 0.7975, \qquad \beta_1 = 0.97 - 0.0025 f_c' = 0.8825$$
4. Sagging reinforcement. Try 3 — 30M in one layer ($A_s = 2100$ mm²), giving $d = 600 - 40 - 10 - 15 = 535$ mm. Equating the steel and concrete forces over the flange width, $$a = \frac{\phi_s A_s f_y}{\alpha_1 \phi_c f_c' b} = \frac{0.85(2100)(400)}{0.7975(0.65)(35)(1000)} = 39.4 \text{ mm}$$ Since $a = 39.4 < h_f = 125$ mm the compression block lies wholly within the flange and the section behaves as a 1000 mm wide rectangle. The moment of resistance is $$M_r = \phi_s A_s f_y \left(d - \tfrac{a}{2}\right) = 0.85(2100)(400)(535 - 19.7) = \boxed{368 \text{ kN}\cdot\text{m}} \; > \; 342.0$$ The neutral axis depth $c = a/\beta_1 = 44.6$ mm gives $c/d = 0.083$, far below the 0.6 ductility limit, so the section is comfortably tension-controlled and will warn before failing. Clear bar spacing is $(300 - 100 - 3 \times 29.9)/2 = 55$ mm, exceeding the 1.4$d_b$ = 42 mm minimum.
5. Hogging reinforcement. Over each support the flange is in tension and the 300 mm web carries the compression, so the section is a plain rectangle. Providing 4 — 20M top steel ($A_s = 1200$ mm², $d = 540$ mm), $$a = \frac{0.85(1200)(400)}{0.7975(0.65)(35)(300)} = 75.0 \text{ mm}, \qquad M_r = 0.85(1200)(400)(540 - 37.5) = 205.0 \text{ kN}\cdot\text{m}$$ which exceeds the 124.0 kN·m demand. The minimum reinforcement $A_{s,min} = 0.2\sqrt{f_c'}\,b_w h / f_y = 532$ mm² is satisfied by both layers. The top bars must run the full length of each overhang and be developed past the point of contraflexure into the span.
6. Shear design. The effective shear depth is $d_v = \max(0.9d,\, 0.72h) = 481.5$ mm. With at least minimum stirrups the simplified method takes $\beta = 0.18$ and $\theta = 35^\circ$: $$V_c = \phi_c \lambda \beta \sqrt{f_c'}\, b_w d_v = 0.65(1.0)(0.18)(5.916)(300)(481.5)/10^3 = 100.0 \text{ kN}$$ $$V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = \frac{0.85(200)(400)(481.5)(1.428)}{150} = 311.7 \text{ kN}$$ so $V_r = 100.0 + 311.7 = \boxed{412 \text{ kN}} \; > \; V_f = 241$ kN using 10M double-leg stirrups at 150 mm. The spacing satisfies $s \le \min(0.7 d_v, 600) = 337$ mm, and the upper bound $V_{r,max} = 0.25\phi_c f_c' b_w d_v = 822$ kN confirms the web is not over-reinforced in shear. Stirrups may be relaxed to 300 mm beyond midspan where $V_f$ drops below $V_c$.
7. Final layout. The design is a 1000 × 125 mm flange on a 300 mm web, 600 mm deep overall: 3 — 30M bottom bars continuous through midspan, 4 — 20M top bars over both supports and through the overhangs, and 10M closed stirrups at 150 mm from each support to midspan.