Question 4 of 7: T-beam design for an overhanging concrete beam

Question B1: T-beam design for an overhanging concrete beam (8 + 8 + 4 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. Overhang AB=1 m, main span BC=6 m (roller at B, pin at C); 100 kN at the free tip A; 300 kN at midspan of BC; f’_c=35 MPa, f_y=400 MPa. Trial T-section: overall depth h=600 mm, web b_w=300 mm, flange (slab) thickness h_f=150 mm, effective flange width b_E=1500 mm (governed by span/4=1.5 m, the smallest of the CSA A23.3 Cl. 10.4 limits with an assumed 2.4 m beam spacing); effective depth d=535 mm.

Find. (a) Factored design moments/shears including self-weight; (b) flexural reinforcement, positive and negative; (c) shear reinforcement.

Approach. Compute the self-weight UDL from the trial T-section, superpose it on the two point loads to get the reactions and the V/M envelope, then design the midspan (T-beam, flange in compression) and support (rectangular web, flange in tension) sections separately, and the stirrups at the critical shear section.

1. (a) Self-weight and reactions. Trial section area $$A=1500(150)+300(450)=360\,000\text{ mm}^2\Rightarrow w_{self}=0.36(24)=\boxed{8.64\text{ kN/m}}$$ over the full 7 m length. Solving equilibrium with the two point loads and the UDL: $$R_B=\boxed{302.0\text{ kN}},\qquad R_C=\boxed{158.5\text{ kN}}\ \text{(unfactored)}$$ Integrating V gives a hogging peak $$M(B)=-100(1)-8.64(1)^2/2=\boxed{-104.3\text{ kN}\cdot\text{m}}$$, a sagging peak at the 300 kN load point (x=4 m from A) of $$M=\boxed{+436.7\text{ kN}\cdot\text{m}}$$, and maximum shear just right of B, $$V_{max}=\boxed{193.3\text{ kN}}$$. Factoring by 1.5: $$M_f^+=655\text{ kN}\cdot\text{m},\quad M_f^-=156\text{ kN}\cdot\text{m},\quad V_f=290\text{ kN}$$
2. (b) Positive (midspan) steel – T-beam action. With $$\alpha_1=0.85-0.0015f_c'=0.7975$$ and compression in the b_E=1500 mm flange: $$M_r=\phi_sA_sf_y\left(d-\frac{A_sf_y}{2\alpha_1f_c'b_E}\right)=M_f^+\ \Rightarrow\ A_s=\boxed{3730\text{ mm}^2}$$ Select 6–30M (4200 mm², two per web-width row): stress-block depth $$a=A_sf_y/(\alpha_1f_c'b_E)=40.1\text{ mm} < h_f=150\text{ mm}\ \checkmark$$ (true T-behaviour confirmed) and $$M_r=\boxed{735\text{ kN}\cdot\text{m}} \ge 655\ \checkmark$$
3. Negative (support) steel – rectangular web. At B the flange is in tension, so only the b_w=300 mm web resists compression: $$M_r=\phi_sA_sf_y\left(d-\frac{A_sf_y}{2\alpha_1f_c'b_w}\right)=M_f^-\ \Rightarrow\ A_s=\boxed{896\text{ mm}^2}$$ Select 2–25M top bars (1000 mm²) extending from A across B into the span a full development length: $$a=47.8\text{ mm},\qquad M_r=\boxed{174\text{ kN}\cdot\text{m}} \ge 156\ \checkmark$$ Both exceed $$A_{s,min}=0.2\sqrt{f_c'}/f_y\, b_wh=532\text{ mm}^2$$.
4. (c) Shear design. $$d_v=\max(0.9d,0.72h)=481.5\text{ mm}$$ With the simplified method (β=0.18, λ=1.0): $$V_c=\beta\lambda\phi_c\sqrt{f_c'}\,b_wd_v=\boxed{100\text{ kN}}$$ Required steel contribution: $$V_s\ge(V_f-V_c)/\phi_s=224\text{ kN}$$ Using 2-legged 10M stirrups (A_v=200 mm²) at θ=35°: $$s=\frac{\phi_sA_vf_yd_v\cot\theta}{V_s}=\boxed{209\text{ mm}}\Rightarrow \text{use 10M @ 200 mm c/c}$$ near both supports, opening to a nominal 300 mm spacing beyond the point loads where V_f drops.