Question 2 of 7: Welded rigid splice in an overhanging W610x241 beam

Question A2: Welded rigid splice in an overhanging W610x241 beam (4 + 16 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. W610x241 of G40.21 350W ($F_y = 350$ MPa, $F_u = 450$ MPa) with $d = 635$ mm, $b = 329$ mm, flange $t = 31.0$ mm, web $w = 17.9$ mm, $Z_x = 7\,650\times10^{3}$ mm³. Specified loads 60, 60 and 50 kN as shown; electrodes E49XX ($X_u = 490$ MPa). Resistance factors $\phi = 0.90$ for the member and $\phi_w = 0.67$ for welds.

Find. Part (i): the factored moment and shear that the joint at B must carry. Part (ii): a welded connection detail that transfers both, with plate sizes and weld sizes and lengths.

Approach. Find the reactions, read $M$ and $V$ at B off the diagrams and factor them; resolve the moment into a couple of flange forces; size a pair of flange splice plates and their fillet welds for that force; size a pair of web splice plates and their weld group for the shear, allowing for the eccentricity of the weld group from the splice line.

1. Part (i) — reactions and internal actions at B. Taking moments about A for the whole beam, $$R_B(5.0)=60(1.5)+60(3.5)+50(7.0)=90+210+350=650\ \Rightarrow\ R_B=130\ \text{kN}$$ and $R_A = 60+60+50-130 = 40$ kN. The cantilever BC then fixes the moment at B directly: $M_B = 50(2.0) = 100$ kN·m hogging. The shear changes across the support, being 50 kN just to the right of B and $|40-120| = 80$ kN just to the left, so the splice must be detailed for the larger of the two.
2. Factor the actions. With the loads treated as specified live load, $$M_f = 1.5(100)=\boxed{150\ \text{kN}\cdot\text{m}}\qquad V_f = 1.5(80)=120\ \text{kN}$$ For reference the member itself is far stronger than this: $M_r = \phi Z_x F_y = 0.90(7\,650\times 10^{3})(350) = 2\,410$ kN·m and $V_r = \phi A_w (0.66 F_y) = 0.90(635)(17.9)(231) = 2\,363$ kN, so the W610x241 is working at about 6 per cent of its flexural capacity. The section has evidently been chosen for stiffness or for a different load case, and the splice is designed here for the forces the question specifies.
3. Part (ii) — resolve the moment into flange forces. A rigid splice is made by carrying the moment on the flanges and the shear on the web. Taking the couple to act between flange centroids, $$T_f = C_f = \frac{M_f}{d-t}=\frac{150\times10^{6}}{635-31.0}=\frac{150\times10^{6}}{604} =248\ \text{kN}$$
4. Size the flange splice plates. Try one plate 250 mm x 12 mm on the outer face of each flange (250 mm is narrower than the 329 mm flange, so both long edges stay accessible for welding). Gross-section yielding governs a welded plate because there are no holes: $$T_r=\phi A_g F_y = 0.90(250\times12)(350)=945\ \text{kN}\ \gg\ 248\ \text{kN}\quad\checkmark$$
5. Size the flange welds. For a fillet weld loaded parallel to its axis, CSA S16 Cl 13.13.2.2 gives $V_r = 0.67\,\phi_w A_w X_u$ with $A_w = 0.707 D$ per millimetre of run. For a 6 mm fillet and E49XX electrode, $$v_r = 0.67(0.67)(0.707)(6)(490)=933\ \text{N/mm}$$ The fusion face on 350W base metal gives $0.67(0.67)(6)(450)=1\,212$ N/mm, so the weld metal governs. The run required on each side of the splice is $L = 248\,000/933 = 266$ mm; carried on the two long edges of the plate that is 133 mm per edge. Provide 150 mm of 6 mm fillet on each edge, each side, so each flange plate is $2(150)+10 = 310$ mm long, say 320 mm.
6. Size the web splice plates for shear. Use one plate 350 mm x 10 mm on each face of the web, lapped 150 mm onto each beam length with a 10 mm gap at the joint. Their shear resistance is $$V_r=\phi A_g(0.66F_y)=0.90\,[2(350)(10)](231)=1\,455\ \text{kN}\ \gg\ 120\ \text{kN}\quad\checkmark$$
7. Check the web weld group for shear plus eccentricity. Each plate is welded to the web with a C-shaped run: two horizontal legs 150 mm long and one vertical leg 350 mm long at the far edge. The group is 650 mm long and its centroid lies 120.4 mm from the splice line, so the 60 kN carried by one plate acts at that eccentricity and twists the group. With $J = I_x + I_y = 12.76\times10^{6}+1.47\times10^{6}=14.23\times10^{6}$ mm³ per unit throat, the direct component is $60\,000/650 = 92.3$ N/mm and the torsional components at the extreme corner are $88.8$ N/mm horizontally and $17.6$ N/mm vertically, giving a resultant $$v_f=\sqrt{88.8^{2}+(92.3+17.6)^{2}}=141\ \text{N/mm}$$ This needs a fillet of only 0.9 mm, so the minimum size for a 17.9 mm thick part — 6 mm, CSA S16 Table 13.1 — governs and is ample.
8. Assemble the detail. The connection is therefore: flange plates 250 x 12 x 320 long, top and bottom, with 6 mm fillet welds 150 mm long on each edge each side; web plates 350 x 10 each side of the web, C-welded with 6 mm fillets over a 150 mm lap; a 10 mm root gap at the joint. All welds E49XX, made with the beam supported so that no rotation occurs before the flange plates are complete.