Question 5 of 7: Beam-Column AB and the Welded Corner at B

Question 5: Beam-Column AB and the Welded Corner 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. From the collapse state of Question 4: $C_f = 1507.5$ kN with $M_B = 607.5$, $M_{5\text{m}} = 316$ and $M_A = 260$ kN·m. Lateral support at all joints and load locations means the column is braced out of plane at 0, 5 and 10 m; in plane it is a non-sway member of length 10 m ($K = 1.0$). Provisional section: W310 × 143.

Find. (a) whether W310 × 143 satisfies all three Clause 13.8.2 checks, and if not what does; (b) the welds and stiffening of the knee joint at B.

Approach. Run the three Clause 13.8.2 interaction checks — cross-sectional strength, overall member strength and lateral–torsional buckling — on the provisional section, revise if any fails, then design the knee panel for the shear that the corner moment imposes on it.

1. Part (a) — set up the amplification factor. The column carries a concentrated transverse load between its supports, so Clause 13.8.5(c) gives $\omega_1 = 0.85$. With $C_e = \pi^2EI_x/L^2 = 6792$ kN for the W310 × 143, $$U_{1x} = \frac{\omega_1}{1 - C_f/C_e} = \frac{0.85}{1 - 1507.5/6792} = 1.093$$
2. Check (a) — cross-sectional strength. With $C_{r0} = \phi AF_y = 5680$ kN and $M_{rx} = \phi Z_xF_y = 754$ kN·m, $$\frac{C_f}{C_{r0}} + \frac{0.85U_{1x}M_{fx}}{M_{rx}} = 0.265 + 0.749 = 1.014 \;>\; 1.0 \quad \textbf{(fails, marginally)}$$
3. Check (b) — overall member strength. In the plane of bending the unbraced length is the full 10 m, giving $KL/r_x = 72.4$ and $C_r = 3510$ kN: $$\frac{1507.5}{3510} + 0.749 = 0.430 + 0.749 = \boxed{1.178 \;>\; 1.0 \quad \textbf{fails}}$$ Check (c), lateral–torsional buckling, returns the same 1.178 because with $\kappa = +0.52$ the segment moment gradient gives $\omega_2 = 2.38$ and $M_u = 4544$ kN·m, so the LTB resistance is capped at the full $\phi M_p$. The answer to the question as asked is therefore no — the section chosen on plastic-moment grounds is 18 % overstressed once in-plane column buckling is accounted for.
4. Revise the section. The deficiency is stiffness about the strong axis over a 10 m unbraced length, not area, so go wider rather than heavier. W360 × 147 has almost the same mass but $I_x = 458 \times 10^6$ mm⁴ against $344 \times 10^6$:

   Clause 13.8.2 check	W310 × 143	W360 × 147
   (a) cross-sectional strength	1.014	0.853
   (b) overall member strength	1.178	0.968
   (c) lateral–torsional buckling	1.178	0.968

   so AB: W360 × 147 is adequate, with 3 % in hand. Note that increasing the column capacity cannot lower the collapse load — the beam mechanism of Question 4 still governs, and $M_p = 607.5$ kN·m stands.
5. Part (b) — what the knee joint must do. In a plastically designed frame the connection must develop the plastic moment of the weaker member, here the column: $M_p = Z_xF_y = 982.8$ kN·m. That moment arrives as a couple in the beam flanges, $$F_{\text{flange}} = \frac{M_p}{d_b - t_f} = \frac{982.8\times10^6}{617 - 22.2} = 1652\ \text{kN}$$ which is transferred by complete-joint-penetration groove welds from each beam flange to the column flange, with matching E49XX electrodes; a CJP weld develops the base metal, so no weld sizing is needed beyond specifying it.
6. Check the knee panel for shear. The flange couple pushes the corner panel into pure shear. Requiring the panel to yield in shear no earlier than the member yields in bending gives the classical plastic-design requirement $$w_{\text{req}} = \frac{\sqrt{2}\,M_p}{\phi F_y d_b d_c} = \frac{1.414(982.8\times10^6)}{0.9(350)(617)(360)} = 19.9\ \text{mm}$$ against a column web of only 12.3 mm. The panel is therefore under-strength in shear, and either a doubler plate or a diagonal stiffener is required.
7. Design the diagonal stiffener. The panel shear is $V = M_p/d_b = 1593$ kN; the web itself resists $\phi(0.55F_y)d_c w = 767$ kN, leaving 826 kN. A stiffener on the diagonal of the panel, of length $\sqrt{d_b^2 + d_c^2} = 714$ mm, must carry that balance resolved along its own axis: $$F_{st} = 826\left(\frac{714}{617}\right) = 956\ \text{kN} \quad\Longrightarrow\quad A_{st} = \frac{F_{st}}{\phi F_y} = 3035\ \text{mm}^2$$ Use 2 plates 130 × 14 (3640 mm²), one each side of the web, with $b/t = 9.3 \le 200/\sqrt{F_y} = 10.7$. Fit horizontal continuity stiffeners opposite both beam flanges to the same thickness.
8. Size the web welds. The beam delivers a shear of 607.5 kN at B over a clear web depth of 573 mm. A double fillet needs $$\frac{607.5\times10^3}{2(573)} = 530\ \text{N/mm}$$ An 8 mm fillet with E49XX electrodes provides $0.67\phi_w(0.707D)X_u = 1244$ N/mm, so 8 mm fillets both sides are ample and are the practical minimum for a 12.3 mm web.

Final Results