Question 6 of 7: Square Concrete Column under Axial Load and Wind

Question B3: Square Concrete Column under Axial Load and Wind (4 + 4 + 12 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. Column length $L = 6.0$ m, fixed at both ends, carrying a specified axial load of 300 kN at the top and a specified horizontal wind load of 80 kN at mid-height. Materials $f_c' = 35$ MPa, $f_y = 400$ MPa; 40 mm cover, 10M ties.

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

Approach. Combine the loads under 1.25D + 1.4W, obtain the first-order moment from the fixed-fixed beam result $WL/8$, test slenderness, magnify the moment if required, then check a trial section on its $P$–$M$ interaction curve.

1. Factored load combination. The axial load is gravity (dead) and the transverse load is wind, so NBCC combination 1.25D + 1.4W applies: $$P_f = 1.25(300) = 375 \text{ kN}, \qquad W_f = 1.4(80) = 112 \text{ kN}$$
2. First-order moment. A member fixed at both ends with a point load at mid-height develops equal end moments and an equal mid-height moment of $$M_f = \frac{W_f L}{8} = \frac{112(6.0)}{8} = \boxed{84.0 \text{ kN}\cdot\text{m}}$$ The eccentricity $e = M_f/P_f = 224$ mm is large relative to any sensible column size, so this is a genuine beam-column and not a nominally axial member.
3. Trial section and slenderness. Try a 350 mm square. With fixed ends in a braced frame $k = 0.5$, so $k l_u = 3000$ mm and, using $r = 0.3h$, $$\frac{k l_u}{r} = \frac{3000}{0.3(350)} = 28.6$$ The end moments act in double curvature, giving $M_1/M_2 = +1$ and a slenderness threshold of $34 - 12(1) = 22$. Since 28.6 exceeds 22, slenderness effects must be included — they cannot be waved away.
4. Moment magnification. With $E_c = 4500\sqrt{35} = 26\,622$ MPa, $I_g = 350^4/12 = 1.251 \times 10^9$ mm⁴ and $\beta_d = 1.0$ (the axial load is entirely sustained), $$EI = \frac{0.4 E_c I_g}{1+\beta_d} = 6.66 \times 10^{12} \text{ N}\cdot\text{mm}^2, \qquad P_c = \frac{\pi^2 EI}{(k l_u)^2} = 7302 \text{ kN}$$ $$\delta_{ns} = \frac{C_m}{1 - P_f/(0.75 P_c)} = \frac{1.0}{1 - 375/5477} = 1.074 \;\Rightarrow\; M_c = 1.074(84.0) = \boxed{90.2 \text{ kN}\cdot\text{m}}$$ The magnification is modest because the column, though slender enough to require the check, is loaded to only 5 % of its buckling load.
5. Reinforcement and interaction check. Provide 4 — 25M ($A_{st} = 2000$ mm²), two bars in each face at $d' = 40 + 10 + 12.6 = 62.6$ mm from the faces, giving $\rho = 2000/350^2 = 1.63$ %, within the 1 % to 4 % range of A23.3 Cl. 10.9.1. Solving strain compatibility for the neutral axis that equilibrates $P_f = 375$ kN gives $c = 94.7$ mm, with the compression bars at a strain of 0.00119 (stress 238 MPa) and the tension bars yielded. Taking moments about the section centroid, $$M_r = 129.7 \text{ kN}\cdot\text{m} \; > \; M_c = 90.2 \text{ kN}\cdot\text{m} \quad \text{OK}$$ a utilisation of 0.70. The point (375, 90.2) lies below the balance point, in the tension-controlled region of the interaction diagram, which is the desirable place for a wind-governed column to sit.
6. Pure axial and tie checks. The squash limit is $$P_{r,max} = 0.80\left[\alpha_1\phi_c f_c'(A_g - A_{st}) + \phi_s f_y A_{st}\right] = 2293 \text{ kN} \; \gg \; 375 \text{ kN}$$ Ties are 10M at the smaller of 16 longitudinal-bar diameters (403 mm), 48 tie diameters (542 mm) and the least column dimension (350 mm), so 10M ties at 350 mm. Corner bars are held by tie corners as required.
7. Final design. 350 mm square column, 4 — 25M longitudinal bars, 10M ties at 350 mm, 40 mm cover. The section is governed by the wind moment rather than the axial load, and by the 1 % minimum-steel rule rather than by strength — a smaller 300 mm section would satisfy the interaction check only marginally while pushing the slenderness ratio to 33.