Question 2 of 7: Welded corner at joint B and the beam-column check

Question 2: Welded corner at joint B and the beam-column check (10 + 5 + 2 + 3 = 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 sections selected in Question 1 — beam W410×67 (d = 410, flange 179 × 14.4, web 8.8) framing into column W530×92 (d = 533, flange 209 × 15.6, web 10.2) — and the collapse actions at that joint: a beam hinge moment of 375 kN·m, a beam end shear of 150 kN and a column shear of 56.2 kN. Both columns carry C_f = 750 kN. Electrodes are E49xx (X_u = 490 MPa), φ_w = 0.67.

Find. A welded knee detail that can deliver the beam’s plastic moment into the column, and the CSA S16 Cl 13.8 verification of AB and DF as beam-columns.

Approach. Resolve the beam moment into a flange couple, check the column panel zone in shear and the column web opposite each flange, add stiffeners for the shortfall, size the welds; then run the three Cl 13.8.2 interaction checks on each column using the collapse moments and the bracing implied by the paper’s note.

1. Part (a) — convert the beam moment into a flange couple. At the collapse moment the beam delivers $$T = C = \frac{M_f}{d_b - t_{fb}} = \frac{375\times10^6}{410 - 14.4} = 948\ \text{kN}$$ Because this is a plastic-design joint, the connection is proportioned for the beam’s full plastic moment rather than the demand, so that the hinge forms in the member and not in the connection: $$T_{cap} = \frac{F_y Z_x}{d_b - t_{fb}} = \frac{350(1.339\times10^6)}{395.6} = 1185\ \text{kN}$$
2. Check the panel zone in shear. The horizontal shear carried by the column web within the knee is the flange force less the column shear: $$V_f = 1185 - 56.2 = 1128\ \text{kN},\qquad V_r = 0.55\phi F_y w_c d_c = 0.55(0.9)(350)(10.2)(533)/10^3 = 942\ \text{kN}$$ The web alone is 187 kN short, so a diagonal stiffener is required across the panel.
3. Size the diagonal stiffener. The diagonal runs from the re-entrant corner to the outer corner, at $\theta = \tan^{-1}(410/533) = 37.6^{\circ}$ to the horizontal, so the axial force needed to supply 187 kN of horizontal shear is $$F_d = \frac{187}{\cos 37.6^{\circ}} = 235\ \text{kN},\qquad A_{req} = \frac{235\times10^3}{0.9(350)} = 747\ \text{mm}^2$$ Provide a pair of 90 × 12 mm plates ($A = 2160\ \text{mm}^2$), one each side of the column web.
4. Transverse (continuity) stiffeners. Opposite each beam flange the column web must carry the concentrated force. Taking a bearing length $N = t_{fb} + 2(8) = 30$ mm, CSA S16 Cl 14.3.2 gives $$B_r = \phi_{bi} w (N + 10t)F_y = 0.80(10.2)(30 + 156)(350)/10^3 = 531\ \text{kN} < 1185\ \text{kN}$$ so stiffeners must take the balance, $1185 - 531 = 654$ kN, requiring $A = 654\times10^3/(0.9\times350) = 2076\ \text{mm}^2$. The same pair of 90 × 12 mm plates ($A = 2160\ \text{mm}^2$) is used opposite both flanges, giving one consistent plate size for the whole knee.
5. Weld the joint. Each beam flange is joined to the column with a complete-joint-penetration groove weld, which develops the flange in tension and needs no separate calculation. The beam web is fillet welded both sides; a 6 mm fillet in E49xx over the 381 mm clear web depth gives $$V_r = 2(0.67)\phi_w (0.707 \times 6)(X_u)L = 2(0.67)(0.67)(4.242)(490)(381)/10^3 = 711\ \text{kN} \gg 150\ \text{kN}$$ so 6 mm fillets both sides are ample. The stiffeners are fillet welded to the web and to the flanges they bear against, and the re-entrant corner is given a generous radius to avoid a notch at the point of highest strain.
6. Part (b) — actions on the two beam-columns. From the collapse state of Question 1, both columns carry $C_f = 750$ kN. Column AB has end moments 74.3 and 375 kN·m; column DF has 375 kN·m at D, 150.4 kN·m at E and 434.3 kN·m at F. The paper’s note “lateral support is provided where necessary” is taken as bracing at mid-height of each column, so $K_yL_y = 4000$ mm and the unbraced lengths for lateral-torsional buckling are also 4000 mm.
7. Compression resistances (W530×92). With $A = 11\,639\ \text{mm}^2$, $r_x = 216.2$ mm and $r_y = 45.2$ mm, $$\lambda = \frac{KL}{r}\sqrt{\frac{F_y}{\pi^2E}},\qquad C_r = \phi A F_y\left(1+\lambda^{2n}\right)^{-1/n},\ n = 1.34$$ giving $\lambda_x = 0.493 \Rightarrow C_{rx} = 3303$ kN over the 8 m in-plane length and $\lambda_y = 1.178 \Rightarrow C_{ry} = 1822$ kN over the 4 m braced length. The cross-sectional axial resistance is $\phi AF_y = 3666$ kN and $M_{rx} = \phi Z F_y = 733.6\ \text{kN}\cdot\text{m}$. Because the frame is unbraced against sway, $U_{1x} = 1.0$ is used in all three checks (Cl 13.8.4).
8. Column AB — the three Cl 13.8.2 checks. $$\text{(a) cross-section: } \frac{750}{3666} + \frac{0.85(375)}{733.6} = 0.205 + 0.434 = 0.639$$ $$\text{(b) overall member: } \frac{750}{1822} + 0.434 = 0.412 + 0.434 = 0.846$$ For (c) the worst segment is the upper 4 m, whose moments 150.4 and 375 kN·m give $\kappa = -0.401$, $\omega_2 = 1.377$, $M_u = 1182$ kN·m and hence $M_r = 680.8$ kN·m, so $$\text{(c) lateral-torsional: } \frac{750}{3303} + \frac{0.85(375)}{680.8} = 0.227 + 0.468 = 0.695$$ Governing ratio 0.846.
9. Column DF. The same resistances apply, with the larger moment 434.3 kN·m at the base: $$\text{(a) } \frac{750}{3666} + \frac{0.85(434.3)}{733.6} = 0.708,\qquad \text{(b) } \frac{750}{1822} + 0.503 = 0.915$$ For (c) the lower segment E–F is bent in double curvature, $\kappa = +0.346$, $\omega_2 = 2.150$, $M_u = 1845$ kN·m and $M_r = 733.6$ kN·m (the full plastic value), so $$\text{(c) } \frac{750}{3303} + \frac{0.85(434.3)}{733.6} = 0.227 + 0.503 = 0.730$$ Governing ratio 0.915.
10. Verdict. Both members satisfy every Cl 13.8.2 check, DF with the smaller margin because the 60 kN horizontal load is applied to that column. The sections chosen in Question 1 are therefore adequate as beam-columns, and no change is needed. Had the bracing assumption been relaxed to a single 8 m unbraced length, $C_{ry}$ would fall to about 760 kN and check (b) would rise past 1.4 — which is precisely why the paper permits lateral support to be assumed where necessary.