07-Str-A2 · May 2017

Question 1 of 7: A1 — Welded moment connection of a cantilever to a column flange

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: A1 — Welded moment connection of a cantilever to a column flange (12 + 8 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 that the joint at A must carry, and a welded detail — weld type, size and length — capable of delivering both into the column flange, together with any column reinforcement the flange force demands.

Approach. Factor the cantilever loads to get M_f and V_f at A, convert the moment into a flange couple, size the flange connection for that couple and the web connection for the shear, then check the column flange and web for the concentrated force.

1. Factor the loads and take moments about A. Both point loads are occupancy loads, so $$V_f = 1.5\,(100 + 200) = \boxed{450\ \text{kN}}$$ and, with lever arms of 1.0 m and 3.0 m measured from the connection, $$M_f = 1.5\,\bigl(100 \times 1.0 + 200 \times 3.0\bigr) = 1.5 \times 700 = \boxed{1050\ \text{kN}\cdot\text{m}}$$
2. Check what the beam itself can deliver. Before detailing a joint it is worth asking whether the member can reach these actions. The W530×92 is Class 1 and, taken as laterally supported, $$M_r = \phi Z_x F_y = 0.90 \times 2360\times10^{3} \times 350 = 743\ \text{kN}\cdot\text{m}$$ against M_f = 1050 kN·m — a utilisation of 1.41. Its web is ample in shear, V_r = φ(d w)(0.66F_y) = 1130 kN. So the section is adequate for shear but overstressed in flexure by 41 %. The question nevertheless asks for a connection sized for the required actions, and that is what is designed below; the member finding is recorded as an assumption under NOTE 1.
3. Resolve the moment into a flange couple. In a directly welded moment connection the flanges carry the moment and the web carries the shear. Taking the couple arm as the centre-to-centre flange distance, $$T_f = C_f = \frac{M_f}{d - t} = \frac{1050 \times 10^{6}}{533 - 15.6} = \boxed{2029\ \text{kN}}$$
4. Test whether the bare flange can carry it. Even a complete-joint-penetration groove weld can only develop the flange it joins: $$T_r = \phi\,(b\,t)\,F_y = 0.90 \times (209 \times 15.6) \times 350 = 1027\ \text{kN}$$ which is barely half of T_f. Welding the beam flanges directly to the column therefore cannot work, and flange plates that add cross-sectional area are required.
5. Size the flange plates. Try a 250 × 30 mm plate of the same grade, one on top of the top flange and one under the bottom flange: $$T_r = 0.90 \times (250 \times 30) \times 350 = 2363\ \text{kN} \;>\; 2029\ \text{kN}$$ a utilisation of 0.86. The 250 mm width overhangs the 209 mm flange by 20 mm each side, which gives clear access for the longitudinal fillet welds.
6. Weld the plate to the column flange. This joint is loaded in direct tension (or compression) normal to the column face, so a complete-joint-penetration groove weld with matching E49xx electrode is used. A CJP weld made with matching electrode develops the full plate, so its resistance equals the 2363 kN just computed and no separate weld calculation is needed.
7. Weld the plate to the beam flange. These fillet welds run parallel to the force (θ = 0), so the S16 Clause 13.13.2.2 directional factor is unity: $$V_r = 0.67\,\phi_w A_w X_u = 0.67 \times 0.67 \times (0.707 \times 12) \times 490 = 1866\ \text{N/mm}$$ for a 12 mm leg. The total length required is $$L = \frac{2029 \times 10^{3}}{1866} = 1088\ \text{mm}$$ Splitting this between the two longitudinal lines gives 544 mm per side; adopt 12 mm fillets, 600 mm long each side, worth 2239 kN.
8. Weld the web for shear. Over a 450 mm clear web length, two lines of fillet weld must each carry $$\frac{450 \times 10^{3}}{2 \times 450} = 500\ \text{N/mm}$$ which asks for only a 3.2 mm leg. The minimum fillet of S16 Table 24 is set by the thicker part joined — the 31.0 mm column flange — and is 8 mm. Use 8 mm fillets, 450 mm long, both sides of the web (1120 kN, against 450 kN demanded). The weld is oversized by the code minimum, not by choice.
9. Check the column under the flange force. S16 Clause 14.3.2 governs the unstiffened column. For a bearing length equal to the plate thickness, web local yielding gives $$B_r = 0.80\,w\,(N + 10t)\,F_y = 0.80 \times 17.9 \times (30 + 310) \times 350 = 1704\ \text{kN}$$ and flange bending under the tension flange gives $$B_r = 0.80 \times 7t^{2} F_y = 0.80 \times 7 \times 31.0^{2} \times 350 = 1884\ \text{kN}$$ Both fall short of T_f = 2029 kN, so a pair of transverse stiffeners is mandatory at each flange level, fitted between the column flanges and welded to the web and to the loaded flange.

The finished detail is therefore a flange-plated moment connection: 250 × 30 mm plates CJP-welded to the column flange and fillet-welded back along the beam flanges, an 8 mm double-fillet web connection for shear, and transverse column stiffeners opposite both beam flanges.