Question 7 of 7: Design of member AB as a reinforced-concrete beam-column

Question 7: Design of member AB as a reinforced-concrete beam-column (12 + 4 + 4 = 20 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. The column AB of the Figure 4 frame, 8 m from the fixed base A to joint B, carrying the actions delivered by the analysis of Question 6.

Find. A column section and reinforcement that satisfy the axial-flexural interaction, with the slenderness question settled and the detailing shown.

Approach. Keep the 600 × 1600 mm section used for the beam so the equal-stiffness assumption holds, test whether slenderness may be neglected, build the interaction diagram for a trial reinforcement, read the resistance at the factored axial load, then check shear and set out the ties.

1. Eccentricity and section choice. The load eccentricity is $$e = \frac{M_f}{C_f} = \frac{2421.6}{1585.8} = 1.53\ \text{m}$$ almost the full member depth, so AB is far out on the flexural side of its interaction diagram and behaves much more like a beam than a column. Retain 600 × 1600 mm with the 1600 mm dimension in the plane of the frame.
2. Slenderness. The frame is braced against sway by its own symmetry, and the base is fixed while the top is restrained by the beam, so $k \approx 0.7$. With $r = 0.3h = 480$ mm and a clear height $l_u = 8000 - 800 = 7200$ mm, $$\frac{kl_u}{r} = \frac{0.7(7200)}{480} = 10.5$$ against the Cl 10.15.2 threshold $34 - 12(M_1/M_2) = 34 - 12(0.5) = 28$. Since $10.5 < 28$, slenderness effects may be neglected and the member is designed for the first-order moments.
3. Minimum longitudinal steel. CSA A23.3 Cl 10.9.1 requires $\rho \ge 0.01$, so $$A_{st,min} = 0.01(600)(1600) = 9600\ \text{mm}^2 \rightarrow 14\ \text{–}\ 30M = 9800\ \text{mm}^2\ (\rho = 1.02\%)$$ arranged as five bars in each 600 mm face and two on each side face.
4. Interaction resistance at the design axial load. Building the $P_r$–$M_r$ curve from strain compatibility (concrete strain 0.0035 at the compression face, rectangular stress block $\alpha_1 = 0.805$, $\beta_1 = 0.895$, steel elastic-perfectly plastic at 400 MPa), the point corresponding to $C_f = 1586$ kN is reached at a neutral-axis depth of about 308 mm and gives $$\boxed{M_r = 3403\ \text{kN}\cdot\text{m} \ \ge\ M_f = 2421.6\ \text{kN}\cdot\text{m}}$$ a utilisation of 0.71. The balance point of the same section is at $P \approx 7650$ kN with $M_r = 4700$ kN·m, and the maximum axial resistance is $P_{r,max} = 0.80\left[\alpha_1\phi_cf'_c(A_g-A_{st}) + \phi_sf_yA_{st}\right] = 14\,598$ kN, so the design point sits well below the balance load in the tension-controlled region.
5. Why the column is not fully worked. The section size is fixed by the stiffness the beam analysis assumed, not by strength: shrinking it would change the frame stiffness ratio and invalidate the analysis, while the 1% minimum steel already delivers more resistance than the load demands. This is normal for the columns of a hinged portal, and the reserve is welcome because it keeps the column elastic while any redistribution happens in the beam.
6. Shear. The column shear is constant at 454 kN. Ignoring the beneficial effect of the axial compression, $$V_c = \phi_c\lambda\beta\sqrt{f'_c}b_wd_v = 0.65(0.18)\sqrt{30}(600)(1350)/10^3 = 519\ \text{kN} > 454\ \text{kN}$$ so no shear reinforcement is required by calculation and the ties are set by the detailing rules.
7. Ties and detailing. Cl 7.6.5 limits the tie spacing to the least of 16 longitudinal bar diameters ($16 \times 29.9 = 478$ mm), 48 tie diameters ($48 \times 11.3 = 542$ mm) and the least member dimension (600 mm), so 10M ties at 300 mm are adopted for the full height, tightened to 150 mm within the top and bottom 1600 mm where the moments peak. Every corner bar and every alternate bar is restrained by the corner of a tie, and cross-ties engage the intermediate bars in the 1600 mm faces. The base at A is detailed with all 14 bars continuing into the footing and lapped with dowels of the same size, since the base moment of 1211 kN·m requires the fixity the analysis assumed. At the top, the column bars pass through the joint core and the beam bars hook around them.
8. Consistency check on the analysis. Because the column and the beam are both 600 × 1600 mm, the gross moment of inertia ratio $I_b/I_c = 1.0$, exactly the “same stiffness for all members” the question stipulates. No re-analysis with cracked-section stiffnesses is therefore required, though in practice a designer would confirm that a cracked beam and a lightly cracked column do not shift the thrust at C materially — here they would not, because the hinge makes the beam moments statically determinate.