Question 6 of 7: B3 — Design of a reinforced concrete beam with an overhang
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A2 Elementary Structural Design, May 2019 sitting, 3 hours, closed book with handbooks and textbooks permitted. Seven questions of 20 marks each: Part A (A1–A3, structural steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and one from Part C — all seven are solved here.
Reference texts. CSA S16 Design of Steel Structures and the CISC Handbook of Steel Construction (section properties, Tables 4-4 and 7); CSA A23.3 Design of Concrete Structures with Brzev & Pao, Reinforced Concrete Design: A Practical Approach; CSA O86 Engineering Design in Wood with the Canadian Wood Council Wood Design Manual; NBCC Part 4 for load combinations.
Question 6: B3 — Design of a reinforced concrete beam with an overhang (20 marks)
Given. A beam pinned at A and on a roller at B 4 m away, continuing as a 1 m overhang past B, carrying a uniform load stated on the drawing as already factored.
Given data — B3
Quantity
Symbol
Value
Back span A to B
$L_a$
4.0 m
Overhang beyond B
$L_o$
1.0 m
Uniform load (stated factored)
$w_f$
230 kN/m
Concrete
$f_c'$
35 MPa
Reinforcement
$f_y$
400 MPa
Find. Beam dimensions, longitudinal reinforcement for both the sagging and hogging regions, and transverse reinforcement.
Figure 6 — B3 beam: 4 m back span plus a 1 m overhang, uniform factored load of 230 kN/m over the whole 5 m.
Check: self-weight. The drawing labels the load “230 kN/m factored”, so it is taken to already include the beam's own weight (about 7 kN/m unfactored for the section adopted, i.e. under 4 % of the total). If the 230 kN/m were superimposed load only, $M^{+}$ would rise to about 420 kN·m and the sagging steel would go to 5–25M.
Approach. Find the reactions and the complete moment and shear diagrams, size the section for the larger (sagging) moment, provide top steel for the overhang hogging moment subject to the minimum-steel rule, and design stirrups at the critical section using the general shear method.
Find the reactions. The total load is $230(5)=1150$ kN acting at the centroid of the full length, 2.5 m from A. Taking moments about A,
$$B_y=\frac{1150(2.5)}{4}=718.75\ \text{kN},\qquad A_y=1150-718.75=431.25\ \text{kN}$$
Locate and evaluate the maximum sagging moment. Shear vanishes where $A_y=w_fx$:
$$x=\frac{431.25}{230}=1.875\ \text{m}$$
$$M^{+}=A_yx-\frac{w_fx^{2}}{2}=431.25(1.875)-\frac{230(1.875)^{2}}{2}=\boxed{404.3\ \text{kN}\cdot\text{m}}$$
Evaluate the hogging moment and peak shear. The overhang is a simple cantilever about B:
$$M^{-}=\frac{w_fL_o^{2}}{2}=\frac{230(1)^{2}}{2}=115\ \text{kN}\cdot\text{m}$$
$$V_{\max}=\left|A_y-w_fL_a\right|=|431.25-920|=488.75\ \text{kN}\ \text{(just left of B)}$$
The sagging moment is 3.5 times the hogging moment, so the bottom steel sizes the section.
Adopt a trial section. Take 400 mm wide by 750 mm deep, giving with 40 mm cover, 10M stirrups and 25M bars
$$d=750-40-11.3-\frac{25.2}{2}=686\ \text{mm}$$
The stress-block parameters at 35 MPa are $\alpha_1=0.7975$ and $\beta_1=0.8825$.
Size the sagging reinforcement. Estimating the lever arm at $0.93d$ gives $A_s\approx1960$ mm²; adopt 4–25M ($A_s=2000$ mm²) in one layer. Checking by equilibrium,
$$T=0.85(2000)(400)=680\ \text{kN},\qquad a=\frac{680\times10^{3}}{0.7975(0.65)(35)(400)}=93.7\ \text{mm}$$
$$M_r=T\left(d-\frac{a}{2}\right)=680\times10^{3}(686-46.9)=\boxed{434.7\ \text{kN}\cdot\text{m}\ >\ 404.3}$$
Utilisation 0.93, and $c/d=106/686=0.155$ confirms a strongly tension-controlled, ductile section.
Size the hogging reinforcement, checking minimum steel. The 115 kN·m at B needs only about 548 mm², but Cl. 10.5.1.2 demands
$$A_{s,\min}=\frac{0.2\sqrt{35}}{400}(400)(750)=887\ \text{mm}^2$$
Cl. 10.5.1.3 relieves this to four thirds of the amount required, $\tfrac{4}{3}(548)=731$ mm², which still exceeds the analysis requirement. Adopt 2–25M top bars ($A_s=1000$ mm²) run continuously over B and anchored into the back span:
$$M_r^{-}=225.3\ \text{kN}\cdot\text{m}\ >\ 115\ \text{kN}\cdot\text{m}\quad\checkmark$$
Design the shear reinforcement. The effective shear depth is
$$d_v=\max(0.9d,\ 0.72h)=\max(617.5,\ 540)=617.5\ \text{mm}$$
The critical section is $d_v$ from the face of B, at $x=4-0.617=3.383$ m:
$$V_f=|431.25-230(3.383)|=346.7\ \text{kN}$$
With minimum stirrups present ($\beta=0.18$, $\theta=35^{\circ}$),
$$V_c=0.65(0.18)\sqrt{35}(400)(617.5)=171.0\ \text{kN}$$
so the stirrups must supply $346.7-171.0=175.7$ kN. Using 10M double-leg stirrups,
$$s\ \le\ \frac{\phi_sA_vf_yd_v\cot\theta}{V_s}=\frac{0.85(200)(400)(617.5)(1.428)}{175.7\times10^{3}}=341\ \text{mm}$$
Adopt and confirm the stirrup detail. Provide 10M double-leg stirrups at 300 mm throughout, which satisfies the Cl. 11.3.8.1 spacing limit $0.7d_v=432$ mm and delivers
$$V_s=199.9\ \text{kN},\qquad V_r=171.0+199.9=\boxed{370.9\ \text{kN}\ >\ 346.7\ \text{kN}}$$
The crushing cap $0.25\phi_cf_c'b_wd_v=1405$ kN is nowhere near critical, so the ties are fully effective.
Design summary: 400 × 750 mm, 4–25M bottom, 2–25M top continuous over the support and through the overhang, 10M double-leg stirrups at 300 mm. The detail worth stressing is the top steel: an inexperienced designer looks at a 115 kN·m hogging moment against a 404 kN·m sagging moment and provides token top reinforcement, but the minimum-steel rule — which exists to prevent a brittle failure at first cracking — nearly doubles what the analysis alone would call for.
