Question 1 of 7: Reinforced concrete design of member BCD

Question 1: Reinforced concrete design of member BCD

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. The frame of Figure 1: columns AB and FD each 8 m between the base and the beam axis, beam BCD spanning 12 m with C at mid-span, both bases built in. Unfactored loads are 300 kN at B, 200 kN at C, 300 kN at D (all vertical) and a 50 kN horizontal load acting to the left at E, the mid-height of the right-hand column. Materials $f'_c = 30$ MPa, $f_y = 400$ MPa. The question directs that all members be given the same stiffness.

Find. Flexural and shear reinforcement for member BCD to CSA A23.3:19, with a reinforcing sketch.

[Figure not reproduced: Figure 1 as printed. The 300 kN loads sit directly over the columns; only the 200 kN at C and the beam self weight bend the beam. See the official exam paper.]

Approach. Size the beam from span/depth rules, run an elastic frame analysis at factored load with equal $EI$ throughout as instructed, then design the three critical sections for flexure and the ends for shear.

1. Choose trial member sizes. A23.3 Table 9.2 allows deflections to be waived when a beam continuous at both ends is at least $\ell_n/21$ deep; with $\ell_n \approx 11.1$ m that is 529 mm. A 12 m span carrying a 200 kN point load needs considerably more, so take the beam as $$b_w \times h = 400 \times 900 \text{ mm}$$ and the columns as $500 \times 900$ mm with the 900 mm dimension in the plane of the frame. Their gross second moments are $I_b = 0.02430$ and $I_c = 0.03037$ m$^4$, a ratio of 1.25 — close enough that the paper's "same stiffness" instruction is nearly satisfied by the real sections, which is checked in step 6.
2. Assemble the factored loads. The beam self weight is $w = 0.400 \times 0.900 \times 24 = 8.64$ kN/m, factored to $1.25 \times 8.64 = 10.80$ kN/m. The printed loads are factored by 1.5, giving 450 kN at B, 300 kN at C, 450 kN at D and 75 kN horizontally at E.
3. Run the elastic frame analysis. With both bases built in the frame is three times redundant; a stiffness solution with one $EI$ throughout returns the factored moments $$\begin{gathered}M_B = -473.1 \text{ kN}\cdot\text{m}, \\ M_C = +651.2 \text{ kN}\cdot\text{m}, \\ M_D = -413.2 \text{ kN}\cdot\text{m}\end{gathered}$$ (hogging negative) and the end shears $V_B = 219.8$ kN and $V_D = 209.8$ kN. The beam also carries an axial thrust of 94.6 kN, which is 0.7 % of $f'_c A_g$ and is neglected in the flexural design.
4. Check the moment field against statics. The two 300 kN loads act at the joints and therefore go straight into the columns; the beam is bent by the 300 kN factored load at C plus its own 10.80 kN/m. The simple-span moment is $300 \times 12/4 + 10.80 \times 12^2/8 = 1094.4$ kN·m, and subtracting the mean end moment $(473.1 + 413.2)/2 = 443.2$ kN·m leaves 651.2 kN·m at C, which is what the solver returns. The 50 kN lateral load is what makes B and D unequal.
Factored bending moments. The beam hogs 473.1 kN·m at B and 413.2 kN·m at D, and sags 651.2 kN·m at C.
5. Design the mid-span section for flexure. With 40 mm cover, 10M stirrups and 30M bars, $d = 830$ mm. Using $\alpha_1 = 0.805$, $\beta_1 = 0.895$, $\phi_c = 0.65$ and $\phi_s = 0.85$, the required tension steel follows from $$M_r = \phi_s A_s f_y \left(d - \frac{a}{2}\right), \qquad a = \frac{\phi_s A_s f_y}{\alpha_1 \phi_c f'_c b_w}$$ Solving for $M_f = 651.2$ kN·m gives $A_s = 2514$ mm$^2$. Provide 4-30M $= 2800$ mm$^2$, for which $a = 151.6$ mm and $$\boxed{M_r = 718.0 \text{ kN}\cdot\text{m} \gt M_f = 651.2 \text{ kN}\cdot\text{m}}$$ The neutral-axis depth is $c/d = 0.204$, comfortably below the 0.5 ductility limit, and $A_s$ far exceeds the A23.3 Cl 10.5.1.2 minimum of 986 mm$^2$.
6. Design the two support sections. The same expression at $M_f = 473.1$ kN·m needs $A_s = 1780$ mm$^2$ and at $M_f = 413.2$ kN·m needs 1542 mm$^2$. Provide 4-25M $= 2000$ mm$^2$ top steel over both B and D, giving $M_r = 527.6$ kN·m. Repeating the analysis with the true gross stiffnesses ($I_c/I_b = 1.25$ rather than unity) raises the support moments to 493.0 and 436.0 kN·m and lowers mid-span to 629.9 kN·m; the chosen bars still cover both analyses, so the paper's simplification is safe here.
7. Design for shear. Take $d_v = \max(0.9d,\, 0.72h) = 747$ mm and use the A23.3 Cl 11.3.6.3 simplified method ($\beta = 0.18$, $\theta = 35°$), which is permitted because minimum stirrups are supplied throughout: $$V_c = \phi_c \lambda \beta \sqrt{f'_c}\, b_w d_v = 191.5 \text{ kN}, \qquad V_{r,\max} = 0.25 \phi_c f'_c b_w d_v = 1457 \text{ kN}$$ The largest factored shear, 219.8 kN at B, is well below $V_{r,\max}$, so the web is not over-stressed. Two-leg 10M stirrups at 300 mm supply $$V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = 241.8 \text{ kN} \;\Rightarrow\; \boxed{V_r = 433.3 \text{ kN} \gt V_f = 219.8 \text{ kN}}$$ That spacing also clears the Cl 11.3.8.1 limit $\min(0.7 d_v,\, 600) = 523$ mm and the Cl 11.2.8.2 minimum area of 99 mm$^2$ per 300 mm.
8. Detail the reinforcement. Run the 4-30M bottom bars the full length of BCD and anchor them into the columns; cut off nothing in the sagging region because the point load leaves a sharp moment peak. Carry the 4-25M top bars a development length plus $d$ past each point of contraflexure, and lap two of them through the span as compression steel — that pair is what makes the long-term deflection of Question 3 acceptable. Use 10M closed stirrups at 300 mm throughout.