Question 5 of 7: Beam–column check for AB, and the welded corner at B

Question 5: Beam–column check for AB, and the welded corner at 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 Question 4 column, a welded I with flanges \(200 \times 20\) and web \(400 \times 10\) (\(A = 12\,000\ \text{mm}^2\), \(d = 440\ \text{mm}\), \(Z = 2.08 \times 10^{6}\ \text{mm}^3\), \(r_y = 47.2\ \text{mm}\)), carrying at collapse an axial force \(C_f = 900 + 607.5 = 1507.5\ \text{kN}\) and a moment \(M_f = M_p = 607.5\) kN·m at B. Lateral support exists at A, at the load point (5 m) and at B.

Find. (a) whether that section satisfies CSA S16 Cl. 13.8.2 as a beam–column, and if not what section does; (b) a welded corner detail at B.

Approach. Apply all three parts of Cl. 13.8.2 — cross-sectional strength, overall member strength and lateral–torsional buckling — using \(KL_y = 5\ \text{m}\) out of plane and \(KL_x = 0.8(10) = 8\ \text{m}\) in plane; then design the corner for the flange couple that the plastic hinge delivers.

1. (a) Compressive resistance of the Question 4 column. Out of plane the unbraced length is 5 m and \(KL/r_y = 5000/47.2 = 106\), so \(\lambda = 1.412\) and $$C_r = \phi AF_y\left(1 + \lambda^{2n}\right)^{-1/n} = 3780(0.391) = 1478\ \text{kN}$$ with \(n = 1.34\). Already \(C_f = 1507.5\ \text{kN} > C_r\): the section cannot even carry its axial load, let alone the plastic hinge moment.
2. (a) The three interaction checks, for the record. With \(C_{ex} = \pi^2EI_x/(KL_x)^2 = 12\,534\ \text{kN}\), \(U_{1x} = 1.137\), \(M_r = \phi ZF_y = 655.2\) and \(M_r^{LTB} = 507.8\ \text{kN}\cdot\text{m}\): $$\text{(a) cross-section: } \frac{1507.5}{3780} + 0.85(1.137)\frac{607.5}{655.2} = \boxed{1.295}$$ $$\text{(b) overall member: } 1.916, \qquad \text{(c) lateral--torsional: } 2.176$$ All three exceed 1.0, so the answer to the question as posed is no — the Question 4 section is not adequate for the beam–column AB. This is the intended finding: the plastic method sizes members on \(M_p\) alone, and a column carrying 1500 kN over a 5 m unbraced length is not governed by \(M_p\).
3. (a) Revised section. Deepen and widen the column to flanges \(280 \times 22\) with a \(480 \times 12\) web (\(A = 18\,080\ \text{mm}^2\), \(d = 524\ \text{mm}\), \(Z = 3.784 \times 10^{6}\ \text{mm}^3\), \(r_y = 66.8\ \text{mm}\)). Now \(KL/r_y = 74.9\), \(C_r = 3404\ \text{kN}\), \(M_r = 1192\ \text{kN}\cdot\text{m}\) and, from \(M_u = 1913\ \text{kN}\cdot\text{m}\) over the 5 m segment, \(M_r^{LTB} = 1105\ \text{kN}\cdot\text{m}\). With \(U_{1x} = 1.058\) the three checks become $$0.723, \qquad 0.901, \qquad \boxed{0.937 \le 1.0}$$ and the section remains Class 1 under combined loading (\(h/w = 40 \le 1100/\sqrt{F_y}\,(1 - 0.39C_f/\phi C_y) = 52.7\)), so the collapse mechanism assumed in Question 4 can still form. Lateral–torsional buckling governs, as it usually does for a beam–column braced only at 5 m centres.
4. (b) Forces on the corner at B. The joint must transfer the hinge moment \(M_p = 607.5\ \text{kN}\cdot\text{m}\) from the beam into the column. Taking the lever arm between the beam flange centroids, \(d_b - t_f = 604 - 22 = 582\ \text{mm}\): $$T_f = \frac{M_p}{d_b - t_f} = \frac{607.5 \times 10^{6}}{582} = \boxed{1044\ \text{kN}}$$ in each flange, together with the beam end shear of 742.5 kN in the web and a column shear at B of \(300 - 115.1 = 184.9\ \text{kN}\).
5. (b) Panel-zone shear. The web panel enclosed by the two beam flanges and the two column flanges carries the flange couple less the column shear: $$V_{\text{panel}} = T_f - V_{\text{col}} = 1044 - 184.9 = 859\ \text{kN}$$ $$V_r = 0.55\,\phi F_y d_c w = 0.55(0.9)(350)(524)(12) \times 10^{-3} = \boxed{1089\ \text{kN}}$$ The ratio is 0.79, so the 12 mm column web suffices and no diagonal stiffener is needed — one of the direct benefits of the heavier column adopted in step 3.
6. (b) Welds and stiffeners. Join the beam flanges to the column flange with complete-joint-penetration groove welds using matching E49XX electrode; a CJP weld develops \(\phi b t F_y = 1733\ \text{kN} > 1044\ \text{kN}\) and needs no further calculation. The beam web is attached with 8 mm fillets on both faces, which supply \(2(0.67\phi_wA_wX_u)(560) = 1393\ \text{kN}\) against the 742.5 kN web shear and satisfy the Cl. 13.13 minimum leg size for a 22 mm thicker part. Finally, transverse continuity stiffeners opposite each beam flange need \(A_{st} \ge T_f/(\phi F_y) = 3314\ \text{mm}^2\); a pair of \(130 \times 14\) plates gives 3640 mm² with \(b/t = 9.3 \le 200/\sqrt{F_y} = 10.7\).