Question 6 of 7: The beam-column CD, and the reinforcement at the rigid connection B

Question 6: The beam-column CD, and the reinforcement at the rigid connection B (14 + 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. The 600 × 1500 mm section of Question 5 with 9 – 25M near each of the two 600 mm faces; member CD is 8.0 m long, pinned at D; from the elastic analysis $C_f=1341$ kN and $M_f=1907$ kN·m at C, zero moment at D; the frame is braced at all four joints. $f'_c=30$ MPa, $f_y=400$ MPa.

Find. (a) Whether the section satisfies A23.3 as a slender/short column under this axial-moment pair; (b) the reinforcement needed to make joint B behave as the rigid connection the analysis assumed.

(a) Member CD as a beam-column

1. Slenderness. The clear height is $l_u=8000-750=7250$ mm and, for a rectangular section, $r=0.3h=450$ mm. The frame is braced, so $k=1.0$ and $$\frac{kl_u}{r}=\frac{7250}{450}=16.1\ <\ 34-12\frac{M_1}{M_2}=34-0=34$$ Slenderness effects may be neglected: CD is a short column and no moment magnification is required.
2. Strain-compatibility check at $C_f=1341$ kN. Take the extreme-fibre concrete strain as 0.0035 and locate the neutral axis $c$ so that the axial resistance equals the applied 1341 kN. With $d=1386$ mm, $d'=114$ mm and $A_s=A'_s=4500$ mm$^2$, the balance $$P_r=\alpha_1\phi_cf'_cb\beta_1c+\phi_sA'_sf'_s-\alpha_1\phi_cf'_cA'_s-\phi_sA_sf_s$$ is satisfied at $c=207$ mm. There $\varepsilon'_s=0.0035(207-114)/207=0.00157$, so the compression steel is just below yield at $f'_s=314$ MPa, while the tension steel is far past yield.
3. Moment resistance at that axial load. Taking moments of the three forces about the mid-depth, $$M_r=1742\left(750-\frac{185}{2}\right)+1530\left(1386-750\right)+1130\left(750-114\right)\times10^{-3}$$ $$M_r=\boxed{2836\ \text{kN}\cdot\text{m}}\ \ge\ M_f=1907\ \text{kN}\cdot\text{m}$$ The point $\left(C_f,M_f\right)=(1341,\ 1907)$ lies well inside the interaction diagram. The axial load helps: at 1341 kN the section is below its balance point, so compression closes the flexural crack and raises $M_r$ from the 1996 kN·m of the singly reinforced beam to 2836 kN·m.
4. Column detailing rules. The section must also meet the rules that apply to a member classified as a column. Reinforcement ratio: $$\rho=\frac{2(4500)}{600(1500)}=0.0100$$ which is exactly the A23.3 minimum of 0.01 and far below the 0.08 maximum; add 2 – 25M on each 1500 mm face to give $\rho=0.0122$ and to satisfy the requirement that every alternate bar be laterally supported. Provide 10M ties at the lesser of $16d_b=403$ mm, $48d_{tie}=538$ mm and the least dimension 600 mm — use 10M ties at 400 mm, closed to 150 mm over a distance $h=1500$ mm below joint C. Finally, the section is nowhere near its squash limit: $P_{r,max}=0.80P_{ro}=13\,637$ kN against $C_f=1341$ kN.
5. Conclusion. The 600 × 1500 mm section of Question 5 is adequate for CD without change of dimensions; only the bar layout is adjusted (side bars and ties) to satisfy the column detailing clauses.

(b) Reinforcement at the rigid connection B

1. What the joint must do. Joint B is a knee: column AB arrives from below and beam BO leaves horizontally. The elastic analysis assumed the joint transfers $M_B=710$ kN·m without rotating relative to the members, so the joint must (i) anchor the beam and column bars, (ii) resist the horizontal shear that the two opposing bar forces impose on the panel, and (iii) be confined so the concrete inside stays intact.
2. Joint shear. Anchored beam bars are taken at yield. With 9 – 25M and the column shear $V_{col}=118$ kN from the analysis, $$V_j=A_sf_y-V_{col}=4500(400)\times10^{-3}-118=\boxed{1682\ \text{kN}}$$
3. Joint shear resistance. A knee joint is confined on only two adjacent faces, so it takes the lowest of the A23.3 Cl. 21.7.4 coefficients: $$V_r=1.3\lambda\phi_c\sqrt{f'_c}\,A_j=1.3(1)(0.65)\sqrt{30}\left(600\times1500\right)\times10^{-3}=4165\ \text{kN}\ \gg\ V_j$$ The joint is not shear-critical — the 1500 mm column depth is doing the work.
4. Anchorage of the beam bars. Take the nine top bars of the beam around the corner and down into the column, or terminate them in standard 90° hooks bent into the joint. The development length of a hooked 25M bar is $$l_{dh}=\frac{f_yd_b}{5.4\sqrt{f'_c}}=\frac{400(25.2)}{5.4\sqrt{30}}=341\ \text{mm}\ \ (\ge 8d_b=202\ \text{mm},\ \ge150\ \text{mm})$$ The available depth inside the column is $1500-40=1460$ mm, so the hooks fit four times over; bend them into the joint so the hook tail bears against the confined core, never outwards where it would spall the cover.
5. Joint hoops. Continue the column ties through the full depth of the joint — this is the detail most often left out on drawings. Provide 10M hoops with cross-ties (four legs) at 150 mm through the joint, satisfying the requirement that transverse reinforcement in a joint be at least that required at the ends of the column.
6. Diagonal corner bar. Because a knee joint under a closing moment tends to split along its outer diagonal, add 2 – 20M diagonal bars across the outside corner, lapped with both members' outer-face steel. This is cheap insurance and standard practice for a frame corner carrying 710 kN·m.