Question 4 of 7: Design of an overhanging reinforced concrete beam

Question B1: Design of an overhanging reinforced concrete beam (4 + 12 + 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. Live loads of 40 kN at the tip of a 2.0 m overhang and 200 kN at the centre of the adjoining 7.0 m span; overall length 9.0 m; $f'_c = 35$ MPa, $f_y = 400$ MPa, normal-density concrete at 24 kN/m³. Resistance factors $\phi_c = 0.65$, $\phi_s = 0.85$; stress-block factors $\alpha_1 = 0.85 - 0.0015 f'_c = 0.7975$ and $\beta_1 = 0.97 - 0.0025 f'_c = 0.8825$ (CSA A23.3 Cl 10.1.7).

Find. Part (i): the factored design actions. Part (ii): rectangular section dimensions and the flexural reinforcement. Part (iii): the shear reinforcement.

Approach. Assume a trial section so that self-weight can be included, analyse the determinate beam for factored actions, size the section from the sagging moment, choose bars for the sagging and hogging regions, check minimum steel and ductility, then design the stirrups by the simplified method of CSA A23.3 Cl 11.3.6.3.

1. Choose a trial section and its self-weight. Take $b = 400$ mm and $h = 800$ mm, which gives a span-to-depth ratio of $7\,000/800 = 8.8$, comfortably stiffer than the $L/16$ of Table 9.2. Its self-weight is $0.4(0.8)(24) = 7.68$ kN/m, factored $1.25(7.68) = 9.6$ kN/m. With 40 mm cover, 10M stirrups and one layer of 25M bars, the effective depth is $d = 800 - 70 = 730$ mm.
2. Factor the loads and find the reactions. The live point loads become $1.5(40) = 60$ kN and $1.5(200) = 300$ kN. Measuring $x$ from the free end, the supports are at $x = 2.0$ and $x = 9.0$ m. Taking moments about the right-hand support, $$R_1(7.0)=60(9.0)+9.6(9.0)(4.5)+300(3.5)=540+388.8+1\,050=1\,978.8$$ $$R_1 = 282.7\ \text{kN},\qquad R_2 = 60+300+9.6(9.0)-282.7=163.7\ \text{kN}$$
3. Part (i) — the design actions. The overhang gives the hogging moment at the first support directly, and the sagging peak sits under the 300 kN load because the shear changes sign there: $$M_f^{-}=60(2.0)+\frac{9.6(2.0)^2}{2}=139.2\ \text{kN}\cdot\text{m}$$ $$M_f^{+}=282.7(3.5)-60(5.5)-\frac{9.6(5.5)^2}{2}=989.4-330-145.2 =\boxed{514.2\ \text{kN}\cdot\text{m}}$$ with the largest shear just inside the first support, $V_f = 282.7 - 60 - 9.6(2.0) = 203.5$ kN.
4. Part (ii) — flexural steel for the sagging moment. With $M_r = \phi_s f_y A_s\,(d - a/2)$ and $a = \phi_s f_y A_s/(\alpha_1\phi_c f'_c b)$, substituting the trial dimensions gives the quadratic $7.964\,A_s^{2}-248\,200\,A_s+514.2\times10^{6}=0$, whose smaller root is $$A_s = 2\,232\ \text{mm}^2$$ Provide 5-25M ($A_s = 2\,500$ mm²). Five 25M bars need $5(25.2)+4(35.3)=267$ mm of clear width against the $400-2(40)-2(11.3)=297$ mm available, so they fit in one layer.
5. Check the sagging section. With $A_s = 2\,500$ mm², $$a=\frac{0.85(400)(2\,500)}{0.7975(0.65)(35)(400)}=117.1\ \text{mm},\qquad c=\frac{a}{\beta_1}=132.7\ \text{mm},\qquad \frac{c}{d}=0.18$$ which is far below the balanced value $700/(700+f_y)=0.636$, so the section is tension-controlled and ductile. The resistance is $$M_r=0.85(400)(2\,500)\left(730-\frac{117.1}{2}\right)=\boxed{571\ \text{kN}\cdot\text{m}} \;>\;514.2\ \text{kN}\cdot\text{m}\quad\checkmark$$
6. Hogging steel over the first support. The same quadratic with $M_f = 139.2$ kN·m returns $A_s = 571$ mm², but minimum reinforcement governs: $$A_{s,min}=\frac{0.2\sqrt{f'_c}}{f_y}b_t h=\frac{0.2\sqrt{35}}{400}(400)(800)=947\ \text{mm}^2$$ (the $4/3$ rule of Cl 10.5.1.3 would allow 762 mm², which is still more than the computed area). Provide 2-25M top ($1\,000$ mm²), giving $M_r = 240$ kN·m against 139.2 kN·m. The top bars are carried through the support and lapped into the span to anchor the hogging moment, and two of the bottom bars are run continuous into the overhang for buildability.
7. Part (iii) — shear design. Take $d_v = \max(0.9d,\,0.72h) = \max(657,\,576) = 657$ mm. The critical section lies $d_v$ from the support face, where $V_f = 203.5 - 9.6(0.657) = 197.2$ kN. With at least minimum stirrups the simplified method allows $\beta = 0.18$ and $\theta = 35^{\circ}$, so $$V_c=\phi_c\lambda\beta\sqrt{f'_c}\,b_w d_v = 0.65(1.0)(0.18)\sqrt{35}(400)(657)=181.9\ \text{kN}$$ leaving only 15.3 kN for the stirrups — minimum steel therefore governs.
8. Choose the stirrups. Because $V_f = 197\ \text{kN} < 0.125\phi_c f'_c b_w d_v = 747$ kN, the spacing limit is $s \le \min(0.7 d_v,\,600) = 460$ mm; the minimum-area rule $A_v \ge 0.06\sqrt{f'_c}\,b_w s/f_y$ permits $s \le 563$ mm for 10M double-leg stirrups. Provide 10M closed stirrups at 400 mm throughout, which delivers $$V_s=\frac{\phi_s A_v f_y d_v \cot\theta}{s} =\frac{0.85(200)(400)(657)(1.428)}{400}=159.5\ \text{kN}$$ $$V_r = V_c+V_s = 181.9+159.5=\boxed{341\ \text{kN}}\;>\;197.2\ \text{kN}\quad\checkmark$$ and is well below the crushing limit $0.25\phi_c f'_c b_w d_v = 1\,495$ kN.