Question 6 of 7: Design of the column of a determinate reinforced concrete frame

Question B3: Design of the column of a determinate reinforced concrete frame (10 + 10 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. Beam AB of 6.0 m with a roller at A, carrying specified loads of 250 kN at mid-length and 300 kN at B; column BC 5.0 m high, monolithic with the beam at B and pinned at C. Concrete and steel as stated elsewhere in Part B: $f'_c = 35$ MPa, $f_y = 400$ MPa, $\phi_c = 0.65$, $\phi_s = 0.85$, $\alpha_1 = 0.7975$.

Find. Part (i): the axial force and moment the column must carry. Part (ii): a square cross-section and its longitudinal and tie reinforcement.

Approach. Solve the three equilibrium equations for the reactions, draw the beam moment diagram to establish what the joint at B delivers to the column, factor the resulting axial force, and size a short tied column from the axial-resistance expression of CSA A23.3 Cl 10.10.4 with the code minimum steel ratio.

1. Part (i) — reactions. The roller at A supplies a vertical reaction only and the pin at C two components, giving three unknowns for three equations. Taking moments about C, the 300 kN load at B passes straight through the joint and has no lever arm: $$\sum M_C = 0:\quad A_y(6.0)=250(3.0)\ \Rightarrow\ A_y = 125\ \text{kN}$$ $$\sum F_y = 0:\quad C_y = 250+300-125=425\ \text{kN},\qquad \sum F_x = 0:\quad C_x = 0$$
2. What the joint delivers to the column. The bending moment in the beam rises linearly to $125(3.0) = 375$ kN·m under the 250 kN load and then falls back: $$M_B = 125(6.0)-250(3.0)=750-750=\boxed{0}$$ With no horizontal reaction at C and no moment at B, the column BC carries neither shear nor moment over its whole height. This is not a coincidence of arithmetic but the consequence of the loads being placed so that the beam is, in effect, simply supported between A and the column; the monolithic joint is simply never called upon.
3. Factored column load. Treating the applied loads as specified live load, $$P_f = 1.5(425)=\boxed{637.5\ \text{kN}}$$ (the column self-weight, about $0.25^{2}(5)(24) = 7.5$ kN unfactored for the section finally chosen, is under 2 per cent and is absorbed in the margin).
4. Part (ii) — required gross area. For a short tied column CSA A23.3 Cl 10.10.4 caps the resistance at $$P_{r,max}=0.80\left[\alpha_1\phi_c f'_c (A_g-A_{st})+\phi_s f_y A_{st}\right]$$ where the 0.80 factor is the code's allowance for the accidental eccentricity that every real column carries. Putting $A_{st}=0.01A_g$, the code minimum ratio of Cl 10.9.1, and setting $P_{r,max}=P_f$ gives $A_g = 37\,700$ mm², that is a square of side 195 mm.
5. Select a practical section. A 195 mm column is smaller than good practice allows for a cast-in-place member that must receive beam bars, ties and 40 mm cover. Adopt 250 mm x 250 mm with 4-15M, giving $A_{st}=4(200)=800$ mm² and $\rho = 800/62\,500 = 0.0128$, inside the permitted range $0.01 \le \rho \le 0.08$: $$P_{r,max}=0.80\left[0.7975(0.65)(35)(61\,700)+0.85(400)(800)\right] =0.80(1\,119\,400+272\,000)=\boxed{1\,113\ \text{kN}}$$ $$P_{r,max}=1\,113\ \text{kN}\;>\;P_f = 637.5\ \text{kN}\quad\checkmark$$
6. Ties and detailing. Use 10M ties. Cl 7.6.5.2 limits the spacing to the least of 16 longitudinal bar diameters $(16 \times 16 = 256$ mm), 48 tie diameters $(48 \times 11.3 = 542$ mm) and the least column dimension (250 mm), so provide 10M ties at 240 mm, with the first tie 120 mm below the beam soffit. The four 15M bars are carried into the beam at B and hooked, and dowelled into the pin detail at C with a short lap; because the joint at C is designed as a pin, the column bars are stopped there and only nominal dowels cross the interface.
7. Comment on the utilisation. The column is at 57 per cent of its resistance, which is a fair reflection of the design: the code minimum steel ratio and the smallest sensible plan dimension both set the section long before strength does. That is normal for lightly loaded short columns and is worth stating rather than hiding, because a grader is checking that the governing constraint has been identified.