Question 6 of 7: Design of the Reinforced Concrete Column BC

Question 6: Design of the Reinforced Concrete Column BC (10 + 4 + 6 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 of Figure 3 runs 4.0 m from the monolithic joint at B down to a fixed base at C, in the frame analysed in Question 5. It carries the vertical reaction at B together with whatever moment the joint sheds into it.

Find. The column section, its longitudinal reinforcement and its ties, having allowed for the fact that the frame is unbraced against sway.

Approach. Take $N_{f}$ and the first-order moment from the Question 5 analysis, add the notional lateral load that A23.3 requires of an unbraced frame, magnify for sway using the stability index, and check the result against the section interaction diagram. Because the frame is unbraced the slenderness check is the part of this question that carries the marks.

1. Why the column moment is constant and small. No horizontal load acts on the frame and C is the only support that can supply a horizontal reaction, so $\sum H = 0$ forces the column shear to zero. A member with zero shear carries the same moment at both ends, and the rotational stiffness that joint B sees from the column is therefore $EI_{c}/h$, not the $4EI_{c}/h$ of a non-swaying member. That quarter-stiffness is why the joint sheds only 36.6 kN·m into a column of this size while the beam retains almost a thousand.
2. Notional lateral load. An unbraced frame must be checked against out-of-plumb, which Clause 10.16 handles through a notional horizontal load of 0.5 per cent of the factored gravity load: $$H_{n} = 0.005\sum P_{f} = 0.005(2016) = 10.08\ \text{kN}$$ applied at beam level. Acting on what is effectively a flagpole, it adds $H_{n}h = 10.08(4.0) = 40.3$ kN·m at the base.
3. Sway magnification. With $E_{c} = 4500\sqrt{30} = 24\,648$ MPa and $I_{c} = 0.70(0.6)^{4}/12 = 7.56 \times 10^{-3}\ \text{m}^{4}$, the notional sway is $\Delta = H_{n}h^{3}/(3EI_{c}) = 1.15$ mm, so the stability index is $$Q = \frac{\sum P_{f}\,\Delta}{H_{n}h} = \frac{2016(0.00115)}{10.08(4.0)} = 0.058$$ Just above the 0.05 that would let the storey be treated as braced, so the sway moments are magnified by $1/(1-Q) = 1.061$: $$M_{f} = (36.6 + 40.3)(1.061) = \boxed{81.7\ \text{kN}\cdot\text{m}}$$ Second-order effects therefore add six per cent, not sixty — worth computing precisely so that the answer can say so.
4. Slenderness limits. With $r = 0.3h = 180$ mm and $k = 2.0$ for a member that can sway, $$\frac{k l_{u}}{r} = \frac{2.0(3400)}{180} = 37.8$$ This exceeds the Clause 10.15.2 threshold of 22 for an unbraced member, which is why Steps 2 and 3 were necessary; but it is far below the Clause 10.13.2 ceiling of 100, above which a full second-order analysis would be mandatory. The magnifier approach is admissible.
5. Reinforcement and axial capacity. Clause 10.9.1 sets a floor of one per cent on the longitudinal steel. Provide 8-25M ($4000\ \text{mm}^{2}$, $\rho = 1.11$ per cent), three bars per face: $$P_{ro} = \alpha_{1}\phi_{c}f'_{c}(A_{g}-A_{st}) + \phi_{s}f_{y}A_{st} = 5588 + 1360 = 6948\ \text{kN}$$ $$P_{r,\max} = 0.80P_{ro} = 5559\ \text{kN} \;\Rightarrow\; \frac{N_{f}}{P_{r,\max}} = \frac{1374}{5559} = 0.25$$
6. Interaction check. Constructing the interaction diagram from strain compatibility (0.0035 at the compression face, the rectangular stress block, and the concrete displaced by the compression bars deducted), the point at $N_{r} = 1374$ kN corresponds to a neutral axis at $c = 201$ mm and $$M_{r} = 590\ \text{kN}\cdot\text{m} \;\Rightarrow\; \frac{M_{f}}{M_{r}} = \frac{81.7}{590} = 0.14$$ The point lies far inside the diagram, on the tension-controlled branch where extra axial load would still increase the moment capacity (the balance point is near 2850 kN).
7. Ties and detailing. Clause 7.6.5 sets the tie spacing at the least of 16 longitudinal bar diameters ($16 \times 25 = 400$ mm), 48 tie diameters ($48 \times 11.3 = 542$ mm) and the least column dimension (600 mm). Provide 10M ties at 400 mm, with every corner bar and every alternate bar restrained by a tie corner, closing to 100 mm within the joint as detailed in Question 5(b).

The column is plainly not strength-governed — a quarter of its axial capacity and a seventh of its moment capacity are used. Its size is set by three other things, and the answer should say so: the 600 mm dimension is the stiffness the Question 5 beam analysis assumed, so reducing it invalidates that analysis; one per cent is the minimum steel ratio A23.3 permits, so the bars cannot be trimmed either; and the section keeps the stability index near 0.05. A 450 mm square column would still carry the load, but its stiffness would fall by a factor of three, $Q$ would rise to about 0.18 and the sway magnifier to 1.22, and the frame would have to be re-analysed from the start.