Question 2 of 7: Member DEF as a reinforced concrete beam-column

Question 2: Member DEF as a reinforced concrete beam-column

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. Member DEF is the right-hand column of Figure 1: 8 m from the built-in base at F to the beam axis at D, with the 50 kN horizontal load applied at E, its mid-height. From the Question 1 analysis the factored actions are

Find. A cross-section and reinforcement for DEF, checked as a slender beam-column to CSA A23.3:19, with a reinforcing sketch.

Approach. Establish whether slenderness effects can be neglected by computing the storey stability index, then check the trial section on its factored axial-load / moment interaction diagram.

1. Note what governs this member. The axial load is only 659.8 kN against $0.1 f'_c A_g = 1350$ kN, i.e. 49 % of the value at which A23.3 would still let the member be designed as a pure flexural element. The moment at D, 413.2 kN·m, is the whole story: this is a beam that happens to carry some compression, and the eccentricity $e = M_f/C_f = 661$ mm is two-thirds of the section depth.
2. Test the frame for sway. The frame is unbraced. Re-running the analysis with the Cl 10.14.1.2 stiffnesses (0.70$I_g$ for columns, 0.35$I_g$ for the beam) gives a first-order storey drift $\Delta_o = 2.34$ mm under the 75 kN factored lateral load, with $\sum C_f = 1330$ kN over a storey height of 8 m. The Cl 10.14.4 stability index is $$Q = \frac{\sum C_f \, \Delta_o}{V_f h_s} = 0.0052 \ll 0.05$$ so the storey may be treated as non-sway and no moment magnification for sidesway is required. The Cl 10.16 notional lateral load, $0.005 \sum C_f = 6.6$ kN, is a tenth of the real 75 kN and does not govern either.
3. Test for member slenderness. The clear height below the beam is $\ell_u = 7550$ mm and $r = 0.3h = 270$ mm, so with $k = 1.0$ $$\frac{k \ell_u}{r} = 28.0$$ The moments at D and F are of opposite sign, so the member is in double curvature and the Cl 10.15.2 braced limit is $34 - 12(M_1/M_2) = 35.3$. Since 28.0 is below that limit, slenderness effects may be neglected and $\delta_b = 1.00$.
4. Select a trial section and reinforcement. Keep the $500 \times 900$ mm section assumed in Question 1 — trimming it would invalidate the stiffness on which the beam moments were computed. Cl 10.9.1 requires $\rho \geq 0.01$, so provide 10-25M $= 5000$ mm$^2$ ($\rho = 1.11$ %): four bars in each 500 mm face and one at mid-depth on each long face.
5. Check the section on its interaction diagram. Working from strain compatibility with $\varepsilon_{cu} = 0.0035$, a neutral axis at $c = 148$ mm gives $P_r = 662$ kN, which matches the applied 659.8 kN. At that axial load $$\boxed{M_r = 910 \text{ kN}\cdot\text{m} \; \gt \; M_f = 436 \text{ kN}\cdot\text{m}}$$ a utilisation of 0.479. The design moment used here is the larger of the equal-stiffness value (413.2) and the true-stiffness value (436.0 kN·m), so the section covers both analyses. For reference $P_{r,\max} = 6948$ kN and the balanced point sits far above the working point, confirming that the section is tension-controlled.
Factored interaction diagram. The working point lies deep in the tension-controlled region, where added moment capacity comes almost free.
6. Check shear and detail the ties. The largest column shear is 94.6 kN in segment DE, against $V_c = \phi_c \lambda \beta \sqrt{f'_c} b_w d_v = 241$ kN with the axial compression ignored, so shear reinforcement is a detailing requirement only. Cl 7.6.5 sets the tie spacing at $\min(16 d_b,\, 48 d_{tie},\, \text{least dimension}) = 403$ mm; use 10M ties at 400 mm, arranged so that every longitudinal bar is at a corner of a tie or held by a cross-tie.
7. Detail the reinforcement. Run all ten bars full height from the footing dowels at F to the beam-column joint at D and lap them above the base only, away from the peak moment. Close the tie spacing to 200 mm over a distance $h = 900$ mm below D and above F, the two regions where the moment gradient is steepest, and carry closed ties through the joint itself.