Question 6 of 7: Beam-Column Checks on the Concrete Column BC

Question 6: Beam-Column Checks on the Concrete Column BC (6 + 5 + 5 + 4 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 450 mm by 900 mm section designed in Question 5, now to be used for the 10 m column BC of the same frame.

Find. Whether the 450 by 900 section, suitably reinforced as a column, satisfies every check A23.3 requires of a compression member: minimum reinforcement, slenderness, moment magnification, axial-flexural interaction, shear and ties.

Approach. Work through the checks in the order the code imposes them, because each one can change the next: first the reinforcement the section must carry as a column, then whether it is slender, then the magnified moment, then the interaction diagram, and finally shear and confinement.

1. Provide the minimum column reinforcement. A23.3 Cl 10.9.1 requires $\rho\ge0.01$ for a compression member, so $A_{st}\ge0.01\times450\times900 = 4050$ mm$^{2}$. The beam steel of Question 5 (3000 mm$^{2}$ on one face) is not enough on its own, so the column is detailed with 8-30M, $A_{st} = 5600$ mm$^{2}$ and $\rho = 0.0138$ — three bars in each 450 mm face plus one at each mid-depth. The section survives, but only when re-reinforced as a column.
2. Check slenderness (Cl 10.15.2). The radius of gyration of a rectangle is $r = 0.3h = 270$ mm, so with $k = 1.0$ for a braced frame $$\frac{kl_{u}}{r} = \frac{9550}{270} = 35.4$$ The end moments at B and C have the same sign, so the column is in single curvature with $M_{1}/M_{2} = 72.4/420.1 = 0.172$ and the limit is $34 - 12(0.172) = 31.9$. Since $35.4 \gt 31.9$ the column is slender and its moments must be magnified — had it been in double curvature, the same geometry would have been classed as short.
3. Magnify the moment (Cl 10.15.3). The flexural stiffness for stability is $EI = 0.4E_{c}I_{g}/(1+\beta_{d}) = 0.4(24\,975)(2.734\times10^{10})/1.6 = 1.707\times10^{14}$ N.mm$^{2}$, so $$P_{c} = \frac{\pi^{2}EI}{(kl_{u})^{2}} = \frac{\pi^{2}\times1.707\times10^{14}}{9550^{2}} = 18\,471\ \text{kN}$$ Because a concentrated transverse load acts between the ends, $C_{m} = 1.0$, giving $$\delta_{b} = \frac{C_{m}}{1 - P_{f}/(\phi_{m}P_{c})} = \frac{1.0}{1 - 1028/(0.75\times18\,471)} = 1.080$$ and a design moment $M_{c} = 1.080\times420.1 = \boxed{454\ \text{kN.m}}$. The magnification is only 8 per cent because the column is barely past the slenderness threshold.
4. Build the interaction diagram. Sweeping the neutral-axis depth and summing the concrete stress block with the three bar rows gives the factored envelope plotted below. Two landmarks: the pure-compression cap is $$P_{r,\max} = 0.80\left[\alpha_{1}\phi_{c}f'_{c}(A_{g} - A_{st}) + \phi_{s}f_{y}A_{st}\right] = 6539\ \text{kN}$$ and the balanced point sits at 3233 kN with 1259 kN.m.
5. Check the design point against the envelope. At $P_{f} = 1028$ kN the envelope offers $M_{r} = 1068$ kN.m, so $$\frac{M_{c}}{M_{r}} = \frac{454}{1068} = \boxed{0.425}$$ The point lies far below the balanced load, in the tension-controlled branch where extra axial force would increase the moment resistance. The section is therefore satisfactory as a beam-column with a wide margin, and its size is set by the stiffness the Question 5 frame analysis assumed rather than by strength — trimming it would invalidate that analysis.
6. Check shear. The column carries only $V_{f} = 109.8$ kN in the upper half, against a concrete contribution alone of $$V_{c} = \phi_{c}\lambda\beta\sqrt{f'_{c}}b_{w}d_{v} = 0.65(0.18)\sqrt{30}(450)(756)\times10^{-3} = 218\ \text{kN}$$ Axial compression only improves this, so no shear reinforcement is needed beyond the ties, and the check is not close.
7. Detail the ties. Cl 7.6.5 caps tie spacing at the least of 16 longitudinal bar diameters ($16\times29.9 = 478$ mm), 48 tie diameters ($48\times11.3 = 542$ mm) and the least column dimension (450 mm), so 450 mm governs; adopt 10M ties at 400 mm centres to match the beam and to keep every corner bar and every alternate bar restrained by a tie corner.
8. Collect the verdict. Every check passes: reinforcement 0.0138 against a 0.01 minimum, slenderness handled by an 8 per cent magnifier, interaction at 0.425, shear at 0.50 on the concrete alone, and ties inside the spacing limits. The 450 by 900 section chosen for flexure in Question 5 is satisfactory for the beam-column BC, provided it is re-reinforced with 8-30M to meet the column minimum.