Reinforced concrete column ABC in a determinate portal frame: 24-Bld-A2 Elementary Structural Design, December 2017, Q5

Question B2: Reinforced concrete column ABC in a determinate portal frame (8 + 6 + 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. Portal frame: pin at A, column A–B–C vertical (4 m + 4 m = 8 m), horizontal top beam C–D (6 m, roller at D); 100 kN lateral load at B; 200 kN and 100 kN vertical loads on the top beam at 2 m and 4 m from C; f’_c=35 MPa, f_y=400 MPa.

Find. (a) The factored axial force and moment governing column ABC; (b) longitudinal reinforcement; (c) tie reinforcement.

Figure B2: determinate portal frame, pin at A, roller at D; 100 kN lateral at B, 200 kN & 100 kN vertical on the top beam.

Approach. This is a determinate frame (pin + roller = 3 reactions, 3 equilibrium equations, no internal hinges) – solve reactions from statics alone, then cut sections along the column to get its N/V/M diagram, and design the governing section by strain-compatibility for combined axial load and bending.

(a) Reactions and column design forces. $\Sigma F_x=0:\ R_{Ax}=-100\text{ kN}$ Taking moments about A (200 kN and 100 kN act 2 m and 4 m from C, i.e. at 8 m height; the 100 kN lateral acts at 4 m height): $\Sigma M_A=0:\ 6R_{Dy}=100(4)+200(2)+100(4)=1200\Rightarrow R_{Dy}=\boxed{200\text{ kN}}$ $R_{Ay}=300-200=\boxed{100\text{ kN}}$ Cutting the column: axial force is constant at $N=R_{Ay}=100\text{ kN (compression)}$ throughout; shear is 100 kN in A–B but drops to zero above B once the 100 kN lateral load is applied there; moment grows linearly from 0 at A to $M(B)=100(4)=400\text{ kN}\cdot\text{m}$ and stays constant at 400 kN·m from B to C (zero shear there) – so the governing section is anywhere in B–C. Factoring by 1.5: $N_f=\boxed{150\text{ kN}},\qquad M_f=\boxed{600\text{ kN}\cdot\text{m}}$
(b) Trial section and longitudinal steel. At e=M_f/N_f=4.0 m (e/h≈5, tension-controlled), try a 400×800 mm column with symmetric reinforcement, d’=65 mm, d=735 mm. Solving strain compatibility (ε_cu=0.0035, α₁=0.7975, β₁=0.8825) for the neutral-axis depth c that gives P_r=N_f=150 kN with 5–25M each face (A_s=A_s’=2500 mm²): $c=93.9\text{ mm (tension steel yields, }f_s=f_y\text{)},\qquad P_r=150\text{ kN},\qquad M_r=\boxed{634\text{ kN}\cdot\text{m}} \ge 600\ \checkmark$
(c) Ties. Longitudinal bars are 25M (d_b=25.2 mm); minimum tie size 10M (d_b=11.3 mm) satisfies CSA A23.3 Cl. 7.6.5.2. Maximum spacing: $s_{max}=\min(16d_b,\,48d_{b,tie},\,\text{least dimension})=\min(403,\,542,\,400)=\boxed{400\text{ mm}}$ Use 10M ties @ 375 mm c/c.

Quantity	Value
R_Ax, R_Ay, R_Dy	−100 kN, 100 kN, 200 kN
N_f, M_f (governs B–C)	150 kN, 600 kN·m
Column section	400×800 mm, 10–25M (5 each face)
M_r at P_r=150 kN	634 kN·m
Ties	10M @ 375 mm c/c

Check: this is an unbraced (sway) frame with kL_u/r≈67 for the 8 m column – well above the CSA A23.3 Cl. 10.15.2 threshold (22) for neglecting slenderness in sway frames. The design above is first-order only; a complete design would apply the Cl. 10.16 moment-magnification (P–Δ) procedure, which would increase M_f further and likely require a larger section or additional steel.

Concept & Theory

Topic: Statics of a determinate sway frame and strain-compatibility design of an RC column under combined axial load and bending.

Key relations: $P_r=\phi_c\alpha_1f_c'ba+\phi_sA_s'(f_s'-\alpha_1f_c')-\phi_sA_sf_s,\quad M_r=\Sigma(\text{force})\times(\text{arm to centroid})$

Why this works: because the corner at C is rigid, the beam’s end shear becomes the column’s axial load and the beam’s end moment carries straight into the column at full magnitude – the column must be designed for that transferred moment even though no load is applied to the column itself above B. Because the eccentricity is very large relative to the section depth, tension steel yields well before the concrete crushes (tension-controlled behaviour), so the section behaves much like a doubly-reinforced beam with a small axial load superposed.

Common pitfall: treating the vertical member as if it only saw the lateral load directly applied to it (ignoring the moment transferred from the beam through the rigid corner at C), which would understate M_f and the required steel.

Source: CSA A23.3:19 Cl. 10.9–10.10 (columns), Cl. 7.6.5 (ties), Cl. 10.15–10.16 (slenderness); MacGregor & Bartlett, Reinforced Concrete, Ch. 11–12.

Question 5 of 7: Reinforced concrete column ABC in a determinate portal frame

Question B2: Reinforced concrete column ABC in a determinate portal frame (8 + 6 + 6 marks)