Question 3 of 7: Welded corner connection and the beam-column AB (14 + 6 marks)

Question 3: Welded corner connection and the beam-column AB (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 same frame and the same W690x152 section, with the knee at B joining the 6 m column AB to the 16 m beam BOC at a right angle. At the factored collapse state the column carries 982 kN of compression with a moment of 384 kN·m at its top, and the section develops a plastic moment of 1745 kN·m. Lateral support is provided at all joints and load points.

Find. (a) a welded corner connection at B able to transmit the members’ plastic moment, and (b) confirmation that the Question 2 section works as the beam-column AB.

Approach. Treat the knee as a capacity-design problem — the corner must be at least as strong as the members it joins, or the assumed redistribution cannot happen — which reduces to a web-panel shear check that the panel fails and a diagonal stiffener then rescues. The beam-column follows the three CSA S16 Clause 13.8.2 interaction checks.

1. Part (a) — the flange force to be carried. Plastic design assumes moments redistribute freely, so the joint must develop the members rather than merely the moment they happen to carry at collapse. The flange couple at $M_{p}$ is $$T=\frac{M_{p}}{d-t}=\frac{1745\times10^{6}}{688-21.4}=2618\ \text{kN}$$ delivered horizontally by the beam flange and vertically by the column flange.
2. Part (a) — web panel shear. Those two perpendicular flange forces shear the panel enclosed by them. Deducting the column shear of $1.5(42.7)=64\ \text{kN}$, the panel demand is $2554\ \text{kN}$ against a resistance $V_{r}=\phi(0.66F_{y})wd_{c}=0.9(231)(13.1)(688)=1874\ \text{kN}$. The panel is 680 kN short, which is entirely normal: the 13.1 mm web of a rolled beam is chosen for beam shear, not for the far larger shear a rigid corner imposes.
3. Part (a) — diagonal stiffener. Because the panel is square ($d_{b}=d_{c}=688\ \text{mm}$) the stiffener runs at 45 degrees and carries $$F_{d}=\frac{680}{\cos45^{\circ}}=962\ \text{kN},\qquad A_{st}\ \ge\ \frac{962\times10^{3}}{0.9(350)}=3055\ \text{mm}^{2}$$ Two plates $120\times14$ give $3360\ \text{mm}^{2}$ with $b/t=8.57\le200/\sqrt{F_{y}}=10.69$, so the stiffener cannot buckle before it yields. A 6 mm doubler plate is the alternative, since the panel would otherwise need to be 17.9 mm thick; the diagonal is lighter and easier to weld.
4. Part (a) — welds. The outer flanges are made continuous around the corner with complete-joint-penetration groove welds, which develop the flange without calculation. The diagonal stiffener delivers 481 kN at each end over 240 mm of fillet weld (both sides of a 120 mm plate), that is 2005 N/mm. A 10 mm transverse fillet gives $0.67\phi_{w}(0.707D)X_{u}(1.5)=2333\ \text{N/mm}$, so 10 mm fillets are adequate. The webs are joined by 8 mm fillets sized on the beam shear.
5. Part (b) — section classification under axial load. With $C_{f}/(\phi C_{y})=0.161$ the Class 1 web limit falls to $(1100/\sqrt{F_{y}})(1-0.39C_{f}/\phi C_{y})=55.1$, and $h/w=49.3$ clears it; the flange at $b/2t=5.93$ clears $145/\sqrt{F_{y}}=7.75$. The section stays Class 1 with the axial load on it, which plastic design requires.
6. Part (b) — compressive and flexural resistances. Out of plane the column is held only at its ends, so $KL/r_{y}=6000/55.05=109$, $\lambda=1.451$ and $C_{r}=2287\ \text{kN}$; in plane $KL/r_{x}=21.5$ gives $C_{r}=5932\ \text{kN}$ and the squash value is $\phi AF_{y}=6087\ \text{kN}$. With the moment falling to zero at the pinned base, $\kappa=0$ and $\omega_{2}=1.75$, so $M_{u}=2265\ \text{kN}\cdot\text{m}$ and $M_{r}=1.15\phi M_{p}(1-0.28M_{p}/M_{u})=1417\ \text{kN}\cdot\text{m}$.
7. Part (b) — interaction. Applying the three checks with $U_{1x}=1.0$: cross-sectional strength $982/6087+0.85(384)/1571=0.369$; overall member strength $982/5932+0.207=0.373$; lateral–torsional buckling strength $$\frac{982}{2287}+0.85\frac{384}{1417}=0.430+0.230=\boxed{0.660\ \le\ 1.0}$$ The weak-axis case governs, as it always does for a pin-based column braced only at its ends, and the W690x152 chosen for the beam is comfortably adequate as the beam-column AB.