Question 5 of 7: B2 — Reinforced-concrete column ABC of the determinate frame

Question 5: B2 — Reinforced-concrete column ABC of the determinate frame (8 + 6 + 6 = 20 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.

Find. Cross-section dimensions and longitudinal and transverse reinforcement for column ABC.

Approach. The frame has three reaction components (pin plus roller) on one rigid body, so it is determinate: solve the reactions, read the column moment diagram, check slenderness, then verify a trial section by strain compatibility at the actual axial load.

1. Factored loads and reactions. All loads become $1.5 \times$ their drawn values: 300 kN and 150 kN on the beam, 150 kN horizontally at B. Moments about A (with D at $x$ = 6 m, $y$ = 8 m):$$6R_D = 300(2.0)+150(4.0)+150(4.0)\;\Longrightarrow\;R_D = 300\ \text{kN}$$ and then $A_y = 300+150-300 = 150$ kN upward, $A_x = 150$ kN acting to the left.
2. Column actions, checked two ways. Reading up the column, the moment at B is $A_x(4.0) = 600$ kN·m. Above B the applied 150 kN cancels $A_x$ exactly, so the shear in segment BC is zero and the moment stays constant at 600 kN·m all the way to C. Reading instead along the beam from D confirms the joint:$$M_C = 300(6.0)-300(2.0)-150(4.0)=600\ \text{kN}\cdot\text{m}\;\checkmark$$ The axial force in the column is the beam reaction, $P_f = A_y = 150$ kN. So$$P_f = 150\ \text{kN},\qquad M_f = 600\ \text{kN}\cdot\text{m},\qquad e = \frac{M_f}{P_f} = \boxed{4000\ \text{mm}}$$ An eccentricity of four metres means this “column” is really a flexural member that happens to carry a little compression — it must be designed by strain compatibility, never with the pure-axial formula $P_{r,\max}=0.80[\alpha_1\phi_cf_c'(A_g-A_{st})+\phi_sf_yA_{st}]$, which for the section below would return 6307 kN, forty-two times the applied load.
3. Slenderness and the sway magnifier. Because the far end of the beam sits on a roller, nothing restrains the frame horizontally at C: the column is a sway member with a pinned base, so $k = 2.0$ and, for a trial 400 × 900 section, $k\ell_u/r = 2(8000)/(0.3\times900) = 59.3$. That is well past the A23.3 10.15.2 sway threshold of 22, so second-order effects must be included. With $E_c = 4500\sqrt{35} = 26\,622$ MPa and $EI = 0.4E_cI_g$,$$P_c = \frac{\pi^2EI}{(k\ell_u)^2} = 9977\ \text{kN},\qquad \delta = \frac{1}{1-P_f/(0.75P_c)} = 1.020$$ so $M_f = 600(1.020) = 612$ kN·m. The magnifier is small only because the axial load is tiny relative to $P_c$; the slenderness itself is severe.
4. Trial section and reinforcement. Take 400 × 900 mm with the 900 mm dimension in the plane of bending, reinforced symmetrically with 3-30M on each 400 mm face ($A_{st} = 4200$ mm²). The steel ratio is $\rho = 4200/(400\times900) = 1.17\%$, inside the A23.3 10.9.1 limits of 1 % and 8 % for a tied column. Cover 40 mm plus 10M ties plus half a 30M bar puts the steel at 66 mm from each face.
5. Resistance by strain compatibility. Assume the extreme concrete fibre reaches $\varepsilon_{cu} = 0.0035$, take a linear strain profile, and adjust the neutral-axis depth $c$ until the net axial force equals $P_f = 150$ kN, with the concrete force from the rectangular stress block ($\alpha_1 = 0.7975$, $\beta_1 = 0.8825$) and each bar force from $\phi_sA_sf_s$, $f_s = E_s\varepsilon_s$ capped at $f_y$. Convergence gives $c = 89.8$ mm, so the far bars are far past yield in tension and the near bars are only lightly stressed. Taking moments of all forces about mid-depth:$$M_r = \boxed{621\ \text{kN}\cdot\text{m}}\;>\;M_f = 612\ \text{kN}\cdot\text{m}\quad(\text{utilisation }0.99)$$ The section is very nearly fully utilised, which is what a well-proportioned answer to this question should look like.
6. Shear and ties. The column shear is $A_x = 150$ kN below B and zero above it. With $d_v = \max(0.9d, 0.72h) = 750$ mm, the concrete alone gives $V_c = 208$ kN, so no shear reinforcement is required for strength. Confinement still governs the detailing: A23.3 7.6.5 caps the tie spacing at $\min(16d_b, 48d_{tie}, \text{least dimension}) = \min(478, 542, 400) = 400$ mm. Provide 10M ties at 400 mm, reduced to 200 mm within one section depth of the joints at B and C where the bars are lapped.