Question 6 of 7: Beam-column CD and the rigid connection at B (14 + 6 marks)

Question 6: Beam-column CD and the rigid connection at B (14 + 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. The reinforced concrete frame of Question 5, now braced at joints A, B, C and D. From the factored elastic analysis the column CD carries 1341 kN of axial compression with 1907 kN·m at its head, and the joint at B transfers 710 kN·m between the 6 m column and the beam.

Find. (a) whether the BC section also serves as the beam-column CD, and (b) the reinforcement at the rigid connection B.

Approach. Build the axial-force / moment interaction diagram for the section with symmetric steel and locate the design point on it, check that slenderness may be neglected, then treat joint B as a corner subjected to an opening moment and detail it for joint shear and bar anchorage.

1. Part (a) — make the section symmetric. A beam needs steel where the tension is; a column reverses under load pattern and sway, so the same concrete section is reused with 9-25M on both faces. Strain compatibility with $\varepsilon_{cu}=0.0035$ then gives the interaction diagram: for a chosen neutral axis $c$, $$P_{r}=\alpha_{1}\phi_{c}f^{\prime}_{c}b a+\phi_{s}A_{s}^{\prime}f_{s}^{\prime}-\phi_{s}A_{s}f_{s}$$ with the companion moment taken about mid-depth.
2. Part (a) — where the design point sits. The balanced point of this section is near $P_{b}=6993\ \text{kN}$, so a factored load of 1341 kN is deep in the tension-controlled region, where axial compression increases the moment capacity by closing the flexural crack. Interpolating the diagram at $C_{f}=1341\ \text{kN}$ gives $$\boxed{M_{r}=2882\ \text{kN}\cdot\text{m}\ \ge\ M_{f}=1907\ \text{kN}\cdot\text{m}}$$ a utilisation of 0.66, against 0.94 for the same section in pure flexure. The section is adequate as the beam-column CD.
3. Part (a) — slenderness. In the plane of the frame $r=0.3h=450\ \text{mm}$, so $kl_{u}/r=8000/450=17.8$, below the braced-frame threshold $25-10(M_{1}/M_{2})=25$ with $M_{1}=0$ at the pin. Slenderness may be neglected in plane. About the weak axis $kl_{u}/r=44.4$, so either the floor system must brace the column at mid-height or the small minimum-eccentricity moment $C_{f}(15+0.03h)=80\ \text{kN}\cdot\text{m}$ must be magnified and combined; a mid-height tie beam is the cheaper answer and is assumed here.
4. Part (b) — what kind of corner joint B is. At B the beam hogs and the column carries the same moment, so the tension face runs continuously around the outside of the corner: this is an opening-moment corner, the detail that fails at a fraction of the members’ capacity if the bars are simply lapped around the re-entrant corner, because the resultant of the two bar forces pulls straight out of the concrete.
5. Part (b) — joint shear. The beam steel delivers $T=A_{s}f_{y}=2500(400)=1000\ \text{kN}$ at yield and the column shear relieves 118 kN, so the joint carries $V_{j}=882\ \text{kN}$. For a corner joint confined on two faces only, $$V_{r}=1.0\lambda\phi_{c}\sqrt{f^{\prime}_{c}}\,A_{j}=1.0(0.65)\sqrt{30}(600)(1500)=3204\ \text{kN}$$ so the joint is at 28 percent of its shear capacity; the 1500 mm column depth is what makes it so comfortable.
6. Part (b) — anchorage and the corner detail. The 25M beam bars are bent down into the column with standard hooks; $l_{dh}=100d_{b}/\sqrt{f^{\prime}_{c}}=460\ \text{mm}$ against 1 420 mm available inside the column, so the hooks develop easily. Loop the outer bars around the corner with a generous bend radius, add $\boxed{3\text{-}25\text{M}\ \text{diagonal bars}}$ across the re-entrant corner (at least $0.5A_{s}=1250\ \text{mm}^{2}$) to carry the outward resultant back into the section, and continue the column ties, 10M at 300 mm, through the joint.