Question 4 of 7: B1 — Reinforced concrete beam with two overhangs

Question 4: B1 — Reinforced concrete beam with two overhangs (12 + 8 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. A 12 m beam: 2.5 m overhangs each side of a 7 m simple span, carrying 70 kN/m over the span between the supports and a 60 kN point load at each overhang tip, all unfactored. f_c' = 35 MPa, f_y = 400 MPa, 40 mm cover, 10M stirrups.

Find. Rectangular section dimensions plus top, bottom and shear reinforcement.

Approach. Trial a 400 × 800 section, add its own factored weight over the full 12 m, take the hogging moment at each support directly from the overhang and the sagging moment as the free simple-span parabola less that hogging value, then size the flexural steel and the stirrups.

1. Trial section and self-weight. Take b = 400 mm, h = 800 mm (span/depth ≈ 9, appropriate for the heavy loading): $$w_{sw}=0.400(0.800)(24)=7.68\ \text{kN/m}\ \Rightarrow\ 1.25(7.68)=9.60\ \text{kN/m}$$
2. Factored loads. The figure loads are live: $$\begin{aligned} &w_f=9.60+1.5(70)=114.6\ \text{kN/m}\ \text{(span)} \\ &P_f=1.5(60)=90.0\ \text{kN}\ \text{(each tip)} \end{aligned}$$ The overhangs carry only the factored self-weight, 9.60 kN/m.
3. Hogging moment at each support. Take the overhang as a cantilever: $$M^-=P_fL_{ov}+\frac{w_{sw,f}L_{ov}^2}{2}=90.0(2.5)+\frac{9.60(2.5)^2}{2}=\boxed{255.0\ \text{kN}\cdot\text{m}}$$
4. Sagging moment at midspan. Superimpose the free parabola on the end moments: $$M^+=\frac{w_fL^2}{8}-M^-=\frac{114.6(7)^2}{8}-255.0=701.9-255.0=\boxed{446.9\ \text{kN}\cdot\text{m}}$$
5. Reactions and shear. By symmetry each support carries $$R=90.0+9.60(2.5)+\frac{114.6(7)}{2}=515.1\ \text{kN}$$ so the shear jumps from −114.0 kN just outside the support to +401.1 kN just inside it.

With the moments settled the flexural design follows the rectangular stress block of A23.3 Clause 10.1.7, using \(\alpha_1=0.85-0.0015f_c'=0.7975\) and \(\beta_1=0.97-0.0025f_c'=0.8825\).

6. Effective depth and bottom steel. With 40 mm cover, 10M stirrups and 25M bars, \(d=800-40-11.3-12.6=736\) mm. Solving \(M_f=\phi_sA_sf_y(d-a/2)\) with \(a=\phi_sA_sf_y/(\alpha_1\phi_cf_c'b)\) gives \(A_s=1901\) mm² required. Provide 4–25M = 2000 mm²: $$\begin{aligned} &a=\frac{0.85(2000)(400)}{0.7975(0.65)(35)(400)}=93.7\ \text{mm} \\ &c/d=0.144 \end{aligned}$$ $$M_r=0.85(2000)(400)\left(736-\frac{93.7}{2}\right)=\boxed{468.7\ \text{kN}\cdot\text{m}}\ >\ 446.9$$
7. Top steel over the supports. For 255.0 kN·m the requirement is 1054 mm²; provide 3–25M = 1500 mm², giving \(M_r=357.5\) kN·m. Minimum flexural steel, Clause 10.5.1.2: $$A_{s,min}=\frac{0.2\sqrt{f_c'}}{f_y}b_th=\frac{0.2\sqrt{35}}{400}(400)(800)=947\ \text{mm}^2$$ which both arrangements satisfy. The value \(c/d=0.144\) confirms a strongly tension-controlled, ductile section.
8. Shear design. \(d_v=\max(0.9d,\,0.72h)=\max(662.5,\,576)=662.5\) mm. At \(d_v\) from the support the factored shear is $$V_f=401.1-114.6(0.6625)=325.2\ \text{kN}$$ With minimum stirrups present the simplified method gives \(\beta=0.18\), \(\theta=35^\circ\): $$V_c=\phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v=0.65(0.18)\sqrt{35}(400)(662.5)=183.4\ \text{kN}$$ $$V_s=\frac{\phi_sA_vf_yd_v\cot\theta}{s}=\frac{0.85(200)(400)(662.5)\cot35^\circ}{400}=160.8\ \text{kN}$$ $$V_r=183.4+160.8=\boxed{344.3\ \text{kN}}\ >\ 325.2\ \text{kN}$$ Maximum spacing is \(\min(0.7d_v,600)=464\) mm, so 10M closed stirrups at 400 mm are acceptable; the crushing limit \(0.25\phi_cf_c'b_wd_v=1507\) kN is nowhere approached.

The overhang shear of 114.0 kN is below \(V_c\), so nominal stirrups at the maximum spacing suffice out there; the same 10M at 400 mm is carried through for simplicity of detailing. The top bars must run continuously across each support and be developed a full anchorage length into the span, because the point of contraflexure lies about 1.4 m inside the support.