Question 2 of 7: Welded corner at joint B, and the beam–columns

Question 2: Welded corner at joint B, and the beam–columns (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 and forces established in Question 1: beam 2–180 × 14 flanges + 350 × 10 web (\(d = 378\ \text{mm}\), \(Z = 1.224\times10^6\ \text{mm}^3\)); columns 2–240 × 16 flanges + 460 × 10 web (\(d = 492\ \text{mm}\), \(A = 12\,280\ \text{mm}^2\), \(r_x = 205\), \(r_y = 54.8\ \text{mm}\), \(M_r = 742.4\ \text{kN}\cdot\text{m}\)); factored column axial load \(C_f = 754.1\ \text{kN}\) with a coincident moment of 380.1 kN·m; E49XX electrodes (\(X_u = 490\ \text{MPa}\)).

Find. (a) the weld sizes, panel-zone reinforcement and stiffeners that let the knee at \(B\) develop the beam's plastic moment; (b) whether the columns satisfy CSA S16 Cl. 13.8.2 as beam–columns.

Approach. A plastic-design connection must transmit the nominal plastic moment of the weaker member, \(M_p = ZF_y = 428.2\ \text{kN}\cdot\text{m}\), not merely the factored demand, otherwise the hinge cannot rotate and the assumed mechanism never forms. Size the flange and web welds first, then check the two things that actually fail at a knee: the panel zone in shear and the column flange in local bending. Part (b) is the three interaction equations of Cl. 13.8.2.

1. Force to be transmitted. With the flanges fully yielded,
   $$T_f = A_f F_y = 180(14)(350) = 882\ \text{kN} \qquad V_{\text{panel}} = \frac{M_p}{d_b - t_{fb}} = \frac{428.2\times10^6}{364} = 1176\ \text{kN}$$
2. Flange welds. Complete-joint-penetration groove welds made with matching E49XX electrodes develop the full base metal, so the beam flanges are joined to the column with CJP welds and no weld calculation is needed — the governing check is the base metal itself, already satisfied in Question 1. Run-off tabs and back gouging are specified because the joint sits at a plastic hinge.
3. Web weld. CSA S16 Cl. 13.13.2.2 carries its own 0.67 coefficient in addition to \(\phi_w = 0.67\):
   $$V_r = 0.67\,\phi_w (0.707D) X_u\left(1.00 + 0.50\sin^{1.5}\theta\right)M_w$$

   For a 6 mm longitudinal fillet, \(V_r = 0.4489(0.707 \times 6)(490) = 933\ \text{N/mm}\). Two lines over an effective 330 mm give \(2(933)(330)/10^3 = 615.8\ \text{kN}\), comfortably above the 154.1 kN beam shear, and 6 mm satisfies the minimum size for a 16 mm thicker part.

4. Panel zone. The bare column web offers
   $$V_r = 0.55\,\phi\, d_c w F_y = 0.55(0.90)(492)(10)(350)/10^3 = 852.4\ \text{kN} \ <\ 1176\ \text{kN}$$

   a shortfall of 324 kN, so the panel must be reinforced. A diagonal stiffener across the knee is the neater solution for a corner. The diagonal is \(\sqrt{492^2 + 378^2} = 620\ \text{mm}\) long, inclined so that \(\cos\theta = 492/620 = 0.793\):

   $$F_{\text{diag}} = \frac{324.1}{0.793} = 408.7\ \text{kN} \qquad A_{st} \ge \frac{408.7\times10^3}{0.90(350)} = 1297\ \text{mm}^2$$

   Provide a pair of 110 × 12 plates, one each side of the web (\(A_{st} = 2640\ \text{mm}^2\), \(b/t = 9.2 \le 200/\sqrt{350} = 10.7\)). The alternative is a doubler plate thickening the web to \(10(1176/852) = 13.8\ \text{mm}\), i.e. one 6 mm doubler, which is heavier to fabricate.

5. Continuity stiffeners. The 882 kN flange force is delivered to the column flange, whose local bending resistance is
   $$T_r = 7\phi t_c^2 F_y = 7(0.90)(16)^2(350)/10^3 = 564.5\ \text{kN} \ <\ 882\ \text{kN}$$

   so horizontal stiffeners are mandatory opposite both beam flanges, carrying \((882 - 564.5) \times 10^3/(0.90 \times 350) = 1008\ \text{mm}^2\). The same 2–110 × 12 plates serve, welded to the column flanges with CJP welds and to the web with 6 mm fillets.

6. Part (b): effective lengths for the columns. The frame is unbraced in its plane. With \(G_A = 1.0\) at the fixed base and
   $$G_B = \frac{\sum I_c/L_c}{\sum I_g/L_g} = \frac{516.3\times10^6/8000}{202.8\times10^6/10\,000} = 3.18$$

   the sway alignment chart gives \(K = 1.57\). Out of plane the columns are braced only where we choose to brace them; take a brace at \(E\), the level of the 60 kN load, so \(K_yL_y = 4000\ \text{mm}\).

7. Amplification factor. With \(C_e = \pi^2EI/L^2 = 15\,924\ \text{kN}\) and \(\omega_1 = 0.6 - 0.4\kappa = 1.0\) for the uniform moment in the segment,
   $$U_{1x} = \frac{\omega_1}{1 - C_f/C_e} = \frac{1.0}{1 - 754/15\,924} = 1.050$$
8. The three Cl. 13.8.2 checks. Each has the form \(C_f/C_r + 0.85\,U_{1x}M_{fx}/M_{rx} \le 1.0\) for a Class 1 section:

   Check	\(KL/r\)	\(C_r\) (kN)	\(M_r\) (kN·m)	Utilisation
   (a) cross-sectional strength (\(K = 1\), in plane)	39.0	3434	742.4	0.676
   (b) overall member strength (\(K = 1.57\), in plane)	61.2	2753	742.4	0.731
   (c) lateral–torsional buckling, braced at \(E\)	73.0	2372	690.4	0.809
   (c) lateral–torsional buckling, no brace at \(E\)	145.9	912	341.8	1.819

   For the braced case Cl. 13.6 gives \(M_u = 1207\ \text{kN}\cdot\text{m} > 0.67M_p\), hence \(M_r = 1.15\phi M_p(1 - 0.28M_p/M_u) = 690.4\ \text{kN}\cdot\text{m}\). Removing the brace drops \(M_u\) to 380 kN·m, below \(0.67M_p\), so \(M_r = \phi M_u = 341.8\) kN·m and the axial resistance collapses at the same time.

9. Verdict. The sections chosen in Question 1 are adequate as beam–columns — but only if the out-of-plane brace at \(E\) is actually provided. Left unbraced over the full 8 m storey height the governing check reaches 1.82, an 82 % overstress, and the frame could not reach the collapse load assumed in Question 1. The paper's "assume lateral support is provided where necessary" is doing real work here, and the calculation is what tells you where "necessary" is.