16-Civ-A2 Elementary Structural Design · May 2013

Question 4 of 7: B1 — Design of an Overhanging Reinforced Concrete Beam

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: 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. Overhanging beam: free end at the left carrying a 40 kN live point load, first support 2.0 m to its right, then a 7.0 m span to the second support, with a 200 kN live point load at mid-span (3.5 m + 3.5 m). $f_c' = 35$ MPa, $f_y = 400$ MPa, $\gamma_{concrete} = 24$ kN/m³, $\phi_c = 0.65$, $\phi_s = 0.85$. Stress-block parameters $\alpha_1 = 0.85-0.0015f_c' = 0.7975$ and $\beta_1 = 0.97-0.0025f_c' = 0.8825$.

Find. Rectangular section dimensions $b$ and $h$, the flexural reinforcement for both the sagging and hogging regions, and the shear reinforcement.

Approach. Assume a trial section, add its factored self-weight to the factored live loads, solve the statics for the hogging moment at the first support and the sagging moment under the 200 kN load, size the tension steel from the rectangular stress block, then check bar fit and design the stirrups by the A23.3 simplified shear method.

1. Trial section and factored loads. A span-to-depth ratio of about ten suits a heavily loaded 7 m beam, so try $b = 400$ mm, $h = 700$ mm. Its self-weight is $0.4(0.7)(24) = 6.72$ kN/m, dead load, so $$w_f=1.25(6.72)=8.4\ \text{kN/m},\qquad P_{f,tip}=1.5(40)=60\ \text{kN},\qquad P_{f,mid}=1.5(200)=300\ \text{kN}$$
2. Reactions. Measuring $x$ from the free tip, the supports are at $x = 2.0$ m (A) and $x = 9.0$ m (B); the uniform load totals $8.4(9.0) = 75.6$ kN acting at $x = 4.5$ m. Taking moments about A, $$R_B=\frac{-60(2.0)+75.6(2.5)+300(3.5)}{7.0}=\frac{-120+189+1\,050}{7.0}=160\ \text{kN}$$ $$R_A=60+75.6+300-160=276\ \text{kN}$$
3. Design moments. The hogging moment at support A comes from the overhang alone: $$M_f^-=-\left[60(2.0)+\frac{8.4(2.0)^2}{2}\right]=-137\ \text{kN}\cdot\text{m}$$ The shear changes sign at the 200 kN load, so the peak sagging moment is there, at $x = 5.5$ m: $$M_f^+=276(3.5)-60(5.5)-\frac{8.4(5.5)^2}{2}=965-330-127$$ $$\boxed{M_f^+=508\ \text{kN}\cdot\text{m}\ \ \text{(sagging)},\qquad M_f^-=137\ \text{kN}\cdot\text{m}\ \ \text{(hogging)}}$$
4. Effective depth and required bottom steel. With 40 mm cover, 10M stirrups and one layer of 25M bars, $d = 700-40-11.3-25.2/2 = 636$ mm; take $d = 635$ mm. Writing the internal couple with $T = \phi_sA_sf_y$ and $a = T/(\alpha_1\phi_cf_c'b)$, where $\alpha_1\phi_cf_c'b = 0.7975(0.65)(35)(400) = 7\,258$ N/mm, $$M_r=T\left(d-\frac{T}{2(7\,258)}\right)=508\times10^6$$ Solving the quadratic gives $T = 885$ kN, hence $$A_s=\frac{T}{\phi_sf_y}=\frac{885\,000}{0.85(400)}=2\,603\ \text{mm}^2$$
5. Select the bottom bars and check the section. Use 6–25M ($A_s = 3\,000$ mm²). Then $T = 0.85(3\,000)(400) = 1\,020$ kN, $a = 1\,020\,000/7\,258 = 141$ mm and $$M_r=1\,020\left(0.635-\frac{0.141}{2}\right)=576\ \text{kN}\cdot\text{m}\;>\;508\ \text{kN}\cdot\text{m}\quad\checkmark$$ The neutral axis lies at $c = a/\beta_1 = 159$ mm, so the steel strain is $\varepsilon_s = 0.0035(635-159)/159 = 0.0105$, five times the yield strain — a properly under-reinforced, ductile section. Six 25M bars in one layer need $2(40)+2(11.3)+6(25.2)+5(25.2) = 380$ mm of width, which fits inside $b = 400$ mm.
6. Top steel over the support. Repeating the same solution for $M_f^- = 137$ kN·m gives $T = 221$ kN and $A_s = 649$ mm², but the minimum reinforcement rule controls: $$A_{s,min}=\frac{0.2\sqrt{f_c'}\,b_th}{f_y}=\frac{0.2\sqrt{35}(400)(700)}{400}=828\ \text{mm}^2$$ Use 3–20M ($A_s = 900$ mm²) in the top, giving $M_r = 188$ kN·m against the demand of 137 kN·m. These bars run through the overhang and are anchored a development length into the span.
7. Shear demand. The largest shear is on the span side of support A: $$V_f=R_A-60-8.4(2.0)=276-60-16.8=199\ \text{kN}$$ Because the load is applied to the top of the beam and the reaction is compressive, the section may be taken at $d$ from the support face: $$V_f=199-8.4(0.635)=194\ \text{kN}$$
8. Concrete contribution (A23.3 simplified method). The effective shear depth is $d_v = \max(0.9d,\,0.72h) = \max(571.5,\,504) = 571.5$ mm. With at least minimum stirrups and 20 mm aggregate, $\beta = 0.18$ and $\theta = 35^\circ$: $$V_c=\phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v=0.65(1.0)(0.18)\sqrt{35}(400)(571.5)=158\ \text{kN}$$ so the stirrups must supply at least $194-158 = 36$ kN.
9. Stirrups. Try 10M double-leg stirrups, $A_v = 200$ mm². The code spacing ceilings are $s \le \min(0.7d_v,\,600) = 400$ mm and, from the minimum-area rule $A_{v,min} = 0.06\sqrt{f_c'}b_ws/f_y$, $s \le 564$ mm. Both are far above the strength requirement, so the maximum permitted spacing governs. At $s = 400$ mm, $$V_s=\frac{\phi_sA_vf_yd_v\cot\theta}{s}=\frac{0.85(200)(400)(571.5)(1.428)}{400}=139\ \text{kN}$$ $$\boxed{V_r=V_c+V_s=158+139=297\ \text{kN}\;>\;194\ \text{kN}}$$ The upper bound $V_{r,max} = 0.25\phi_cf_c'b_wd_v = 1\,300$ kN is not approached, and $V_f$ is well under $0.125\lambda\phi_cf_c'b_wd_v = 650$ kN, which is the condition that permits the 400 mm spacing.

A 400 × 700 mm section therefore works comfortably in both flexure and shear, with the bottom steel at 1.18 % of the gross area — a normal, buildable ratio. Practically, the stirrups would be closed and detailed at 400 mm through the span but tightened near the support and through the overhang where the top steel is anchored.