Question 1 of 7: A1. Welded moment-and-shear connection (5 + 10 + 5 marks)

Question 1 — A1. Welded moment-and-shear connection (5 + 10 + 5 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.

Find. The factored moment and shear delivered to the connection at B, and a welded detail (weld type, size and length) that develops both, together with the checks on the supporting column that the detail implies.

Approach. Factor the load, read the fixed-end moment and end shear from the standard beam diagram in the Handbook, convert the moment into a flange force couple that the flange welds carry, size the web welds for the shear alone, then confirm the column flange and web can accept the concentrated flange force.

1. Factor the applied load. The load is specified (unfactored) live load, so $$P_f = 1.5 P = 1.5(300) = 450\ \text{kN}$$
2. Read the end actions from the Handbook beam diagram. For a prismatic beam built in at both ends with a central concentrated load, the tabulated end moment and end shear are $M = PL/8$ and $V = P/2$. Substituting the factored load, $$M_f = \frac{P_f L}{8} = \frac{450(6.0)}{8} = 337.5\ \text{kN}\cdot\text{m}, \qquad V_f = \frac{P_f}{2} = \frac{450}{2} = 225\ \text{kN}$$ These are the actions the connection at B must transfer. As a sanity check the beam itself is comfortably adequate: $M_r = \phi Z_x F_y = 0.90(2360\times10^{3})(350) = 743\ \text{kN}\cdot\text{m}$, so $M_f/M_r = 0.45$ and the connection, not the member, is the design problem.
3. Convert the moment to a flange force couple. In a directly-welded moment connection the moment is carried almost entirely by the flanges, acting at the flange centroids a distance $(d - t)$ apart: $$T_f = C_f = \frac{M_f}{d - t} = \frac{337.5\times10^{6}}{533 - 15.6} = 652\,300\ \text{N}$$ $$\boxed{T_f = 652\ \text{kN}\ \text{tension in the top flange, equal compression in the bottom}}$$
4. Size the flange welds. Welding each flange on both its upper and lower surface gives an effective length per flange of $L_w = 2b - w = 2(209) - 10.2 = 408$ mm. The weld runs transverse to the force, so $\theta = 90^\circ$ and S16 Clause 13.13.2.2 gives, per millimetre of leg size $D$, $$v_r = 0.67\,\phi_w (0.707 D) X_u (1 + 0.50\sin^{1.5}\theta) = 0.67(0.67)(0.707D)(490)(1.5) = 233.3\,D\ \text{N/mm}$$ Equating supply to demand, $$D_{req} = \frac{T_f / L_w}{233.3} = \frac{652\,300/408}{233.3} = 6.86\ \text{mm}$$ Rounding up to the next standard size, $D = 8$ mm, which supplies $V_r = 233.3(8)(408) = 761\ \text{kN} > 652\ \text{kN}$. $$\boxed{\text{Flanges: 8 mm fillet welds, both faces, full flange width}}$$ A complete-joint-penetration groove weld is the equally valid alternative: it develops the full flange, $T_r = \phi b t F_y = 0.90(209)(15.6)(350) = 1027\ \text{kN}$, and is usually preferred where the flange force approaches the flange capacity.
5. Size the web welds for the shear. Fillet welds down both sides of the web run parallel to the shear, so $\theta = 0^\circ$ and $v_r = 0.67(0.67)(0.707D)(490) = 155.5\,D$ N/mm. Taking a practical weld length of 460 mm clear of the flange fillets, the demand is $225\,000/(2\times460) = 245$ N/mm, requiring only $D = 1.6$ mm. The minimum fillet size for the 19.6 mm column flange (S16 Table 7) governs instead: $$\boxed{\text{Web: 6 mm fillet welds, both sides, 460 mm long}\ \ (V_r = 858\ \text{kN} > 225\ \text{kN})}$$ The beam web itself is not critical: $V_r = \phi d w (0.66F_y) = 0.90(533)(10.2)(231) = 1130$ kN.
6. Check the supporting column. The 652 kN flange force arrives as a concentrated load on the column flange. Local flange bending gives $B_r = 0.90(6.25)t_{fc}^2 F_y = 0.90(6.25)(19.6)^2(350) = 756\ \text{kN} > 652\ \text{kN}$, and column web yielding over the dispersed length gives $B_r = \phi(t + 10k)F_y w_c = 0.90(15.6 + 300)(350)(11.9) = 1183\ \text{kN}$. Both exceed the applied flange force, so the W610 × 125 accepts the connection without transverse stiffeners — though the flange-bending margin is only 16 %, and horizontal stiffeners opposite each beam flange remain the prudent detail if the beam is ever re-loaded.

The finished detail is therefore a directly-welded rigid connection: flanges welded across their full width to the column face, web welded down both sides for the shear, with the two welds sized independently because the flange couple and the web shear act on essentially independent parts of the section.