Question 4 of 7: B1 — T-section design for the beam with an overhang

Question 4: B1 — T-section design for the beam with an overhang (8 + 8 + 4 = 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. T-section proportions plus the amount and layout of flexural and shear reinforcement.

Approach. Choose trial T-section proportions that satisfy A23.3 10.3.4 for an isolated T-beam, add the resulting self weight to the factored load case, resolve the determinate statics, then design the sagging region as a flanged section, the hogging region over B as a rectangular web section, and finally the stirrups.

1. Trial section and self weight. Take a flange $b = 1200$ mm by $h_f = 200$ mm on a web $b_w = 400$ mm, overall depth $h = 900$ mm. A23.3 10.3.4 requires an isolated T-beam to have $h_f \ge b_w/2 = 200$ mm and $b \le 4b_w = 1600$ mm — both satisfied. The gross area is $1200(200)+400(700)=0.520$ m², so$$w = 0.520(24) = 12.48\ \text{kN/m},\qquad w_f = 1.25(12.48)= 15.60\ \text{kN/m}$$ Dead load takes 1.25 and the two point loads take 1.5.
2. Factored reactions. With $P_A = 150$ kN at the tip, $P = 450$ kN at midspan and the UDL over the full 7 m, moments about B give$$R_C = \frac{450(3.0)+15.60(7.0)(2.5)-150(1.0)}{6.0}=245.5\ \text{kN},\qquad R_B = 463.7\ \text{kN}$$
3. Design actions. Over the overhang the moment is hogging; at midspan it is sagging, and the shear changes sign under the 450 kN load, confirming that is where the maximum sagging moment sits:$$M_B = -\left[150(1.0)+\tfrac{15.60(1.0)^2}{2}\right]=-158\ \text{kN}\cdot\text{m}$$$$M_{\text{sag}} = 463.7(3.0)-150(4.0)-\tfrac{15.60(4.0)^2}{2}=\boxed{666\ \text{kN}\cdot\text{m}}$$ The largest shear is immediately right of the roller, $V_f = 463.7-150-15.6 = 298$ kN.
4. Sagging reinforcement (flange in compression). With 40 mm cover, 10M stirrups and one layer of 30M bars, $d = 900-40-11.3-15 = 834$ mm. The stress-block parameters are $\alpha_1 = 0.85-0.0015f_c' = 0.7975$ and $\beta_1 = 0.97-0.0025f_c' = 0.8825$, giving $\alpha_1\phi_cf_c' = 18.14$ MPa. Try 4-30M ($A_s = 2800$ mm²):$$a=\frac{\phi_sA_sf_y}{\alpha_1\phi_cf_c'b}=\frac{0.85(2800)(400)}{18.14(1200)}=43.7\ \text{mm}$$ Since $a = 43.7 \ll h_f = 200$ mm, the compression block lies wholly inside the flange and the section behaves as a 1200 mm wide rectangle:$$M_r = \phi_sA_sf_y\left(d-\tfrac{a}{2}\right)= 0.85(2800)(400)(834-21.9)/10^6 = \boxed{773\ \text{kN}\cdot\text{m}}$$ against $M_f = 666$ kN·m, a utilisation of 0.86. With $c/d = 49.5/834 = 0.059$ the section is strongly tension-controlled, so the steel is yielding and failure would be ductile.
5. Bar layout in the web. Four 30M bars in a 400 mm web leave a clear spacing of $(400-2\times40-2\times11.3-4\times29.9)/3 = 59$ mm, which exceeds the A23.3 7.4.1.2 minimum of $\max(1.4d_b, 1.4\times$ aggregate, 30 mm) = 42 mm. The bars fit in a single layer, so the assumed $d$ stands.
6. Hogging reinforcement over B (flange in tension). Now only the 400 mm web is in compression, and $M_f = 158$ kN·m requires just 568 mm². Minimum steel governs instead:$$A_{s,\min}=\frac{0.2\sqrt{f_c'}}{f_y}b_th=\frac{0.2\sqrt{35}}{400}(400)(900)=1065\ \text{mm}^2$$ Provide 4-20M top ($A_s = 1200$ mm²), continuous over the support and anchored into the span, which gives $M_r = 329$ kN·m — more than double the demand, but that is what the minimum-steel rule costs on a deep member.
7. Shear design. The effective shear depth is $d_v = \max(0.9d, 0.72h) = 750$ mm. Using the A23.3 11.3 simplified method with $\beta = 0.18$ and $\theta = 35^\circ$ (minimum stirrups present):$$V_c = \phi_c\beta\sqrt{f_c'}b_wd_v = 0.65(0.18)\sqrt{35}(400)(750)/10^3= 208\ \text{kN}$$ The stirrups must carry the remaining 90 kN, which needs only a very wide spacing, so the A23.3 11.3.8 maximum spacing $\min(0.7d_v, 600) = 525$ mm governs. Provide 10M double-leg stirrups at 500 mm ($A_v = 200$ mm², above $A_{v,\min} = 178$ mm²):$$V_r = V_c+V_s = 208+146 = \boxed{354\ \text{kN}}\;>\;298\ \text{kN}$$ Tighten the spacing to 250 mm over the first metre either side of B where the shear peaks, for crack control and to suit the bar-cut-off detailing.