Question 6 of 7: Design of a square reinforced concrete column

Question B3: Design of a square reinforced concrete column (20 marks: 4 + 4 + 12)

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 6 m column built in at both ends, carrying an axial load at its head and a horizontal wind load at mid-height.

Find. A square cross-section with its longitudinal and tie reinforcement.

Approach. Form the two governing NBCC load combinations, obtain the first-order moments from the fixed-fixed frame action, magnify them for slenderness, then check a trial section against its own axial-moment interaction diagram computed by strain compatibility.

(a) Load combinations and first-order moments (4 marks)

1. Frame action. For a member built in at both ends with a point load W at mid-length, $$M_{\text{end}} = M_{\text{mid}} = \frac{WL}{8}$$ in double curvature, with the mid-height moment sagging and both end moments hogging.
2. The two combinations that matter. Treating the 300 kN as a specified live load and the 80 kN as wind:

   NBCC factored load cases

   Case	Axial Pf	Lateral Wf	First-order M0
   2: 1.25D + 1.5L + 0.4W	1.5(300) + 1.25(17.6) = 472.1 kN	0.4(80) = 32 kN	24.0 kN·m
   4: 1.25D + 1.4W + 0.5L	0.5(300) = 150 kN	1.4(80) = 112 kN	84.0 kN·m

   The 17.6 kN in case 2 is the self-weight of the trial 350 mm column; it is omitted from case 4, where extra axial load would raise the moment capacity of so lightly loaded a section and the omission is therefore conservative. Case 4 has the smaller axial force and three and a half times the moment, so it is the one that will govern.

(b) Slenderness (4 marks)

1. Effective length and slenderness ratio. Fixed at both ends and braced against sidesway, $k = 0.5$, and for a square section $r = 0.3h$. For the trial 350 mm section $$\frac{kl_u}{r} = \frac{0.5(6\,000)}{0.3(350)} = 28.6$$ which is small but not negligible, so rather than argue about the neglect criterion the moments are simply magnified.
2. Non-sway moment magnifier (Clause 10.15.3), with $E_c = 4\,500\sqrt{f_c'} = 26\,622$ MPa, $I_g = 350^4/12 = 1\,251\times10^{6}\ \text{mm}^4$, $EI = 0.4E_cI_g/(1+\beta_d)$ with $\beta_d = 1.0$ for a fully sustained axial load, and $C_m = 1.0$ because the member carries a transverse load between its supports: $$P_c = \frac{\pi^2EI}{(kl_u)^2} = 7\,302\ \text{kN}, \qquad \delta_b = \frac{C_m}{1 - P_f/(0.75P_c)} \ge 1.0$$ This gives $\delta_b = 1.094$ for case 2 and 1.028 for case 4, hence design moments $$\boxed{M_f = 26.3\ \text{kN}\cdot\text{m at } P_f = 472.1\ \text{kN}; \quad M_f = 86.4\ \text{kN}\cdot\text{m at } P_f = 150\ \text{kN}}$$

(c) Section design (12 marks)

1. Trial section. Take 350 × 350 mm with 8 – 20M vertical bars, giving $A_{st} = 2\,400\ \text{mm}^2$ and a reinforcement ratio $\rho = 2\,400/122\,500 = 1.96\%$, within the 1 to 8 per cent range of Clause 10.9.1 and 10.9.2. With 40 mm cover and 10M ties the outer bars sit 59.8 mm from each face; three bars go on each of the two faces perpendicular to the wind and two at mid-depth.
2. Pure axial capacity, as a sanity check (Clause 10.10.4): $$P_{r,\max} = 0.80\left[\alpha_1\phi_cf_c'(A_g - A_{st}) + \phi_sf_yA_{st}\right] = 2\,396\ \text{kN}$$ Both cases sit far below this, confirming that the design is bending-driven rather than compression-driven.
3. Interaction check. The resistance is found by strain compatibility: for a trial neutral-axis depth $c$, the concrete block contributes $\alpha_1\phi_cf_c'ba$ with $a = \beta_1c$, each bar layer contributes $\phi_sA_{si}f_{si}$ with $f_{si} = 0.0035E_s(c-y_i)/c$ capped at $\pm f_y$, and $c$ is adjusted until the axial resistance equals the applied $P_f$; the moment resistance is then the sum of first moments about the plastic centroid. This gives

   Interaction check of the 350 × 350 section with 8 – 20M

   Case	Pf (kN)	c (mm)	Mr (kN·m)	Mf (kN·m)	Utilisation
   2	472.1	121.4	147.3	26.3	18 %
   4	150.0	89.2	121.5	86.4	71 %

   so the wind case governs and $$\boxed{M_r = 121.5\ \text{kN}\cdot\text{m} > M_f = 86.4\ \text{kN}\cdot\text{m} \quad (71\%\ \text{utilised})}$$ Both cases lie below the balanced point, so the section fails in a ductile, tension-controlled manner.
4. Ties. Clause 7.6.5 limits tie spacing to the least of 16 longitudinal bar diameters ($16 \times 19.5 = 312$ mm), 48 tie diameters ($48 \times 11.3 = 542$ mm) and the least column dimension (350 mm). Adopt 10M ties at 300 mm, with the first tie 50 mm from the face of each end restraint and every corner bar held by a tie corner.
5. Why not a smaller column. A 300 mm section with the same eight bars reaches 94.5 kN·m against a magnified demand of 88.5 kN·m, so it is arithmetically adequate — but at 94 per cent utilisation it has no margin at all, its reinforcement ratio rises to 2.7 per cent, and eight 20M bars plus ties in a 300 mm square leave very little room where the column steel meets the beam steel at the built-in ends. The 350 mm section is the better buildable answer and is the one carried into Question C1.