Fillet welded channel tension member, 200 kN

Assumptions: factored tension 200 kN; Fe410 / E250 ($f_y=250$, $f_u=410$ N/mm²); shop welding; gusset plate 10 mm thick. The back of the channel web sits on the gusset and is welded along the two back edges (heels) of the channel and across the end of the web.

1. Section

$A_g\ge\dfrac{200\times10^3\times1.10}{250}=880$ mm². Try ISMC 125: $A=1619$ mm², $h=125$, $b_f=65$, $t_f=8.1$, $t_w=5.0$ mm.

• $T_{dg}=\dfrac{1619\times250}{1.10}\times10^{-3}=368.0$ kN $>200$ ✓

2. Weld size and length

• Thicker part (gusset) 10 mm → minimum size 3 mm (Table 21). The end weld runs along the square cut edge of the 5 mm web, so its size must not exceed $t-1.5=3.5$ mm. Use $s=3$ mm throughout.
• $f_{wd}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²; strength per mm $=189.4\times0.7\times3=397.7$ N/mm
• Required length $L_w=\dfrac{200\times10^3}{397.7}=503$ mm
• Provide: an end weld of 125 mm across the web plus two longitudinal welds of 190 mm along the heels: total $125+2\times190=505$ mm $>503$ mm ✓ (each side weld is longer than $4s=12$ mm and than the 125 mm distance between them ✓).

3. Net-section rupture with shear lag (cl. 6.3.3 and 6.3.4)

• $A_{nc}=125\times5=625$ mm² (no holes), $A_{go}=1619-625=994$ mm²
• Shear lag distance from the far edge of the flange to the weld line: $b_s=b_f=65$ mm; $L_c=190$ mm

$$\beta=1.4-0.076\times\frac{65}{8.1}\times\frac{250}{410}\times\frac{65}{190}=1.273$$ $$T_{dn}=\frac{0.9\times625\times410}{1.25}+\frac{1.273\times994\times250}{1.10}=472.0\ \text{kN}>200$$

The slenderness limit of 400 ($r_{yy}=19.2$ mm) allows $L\le7.7$ m.

Answer: ISMC 125 with 3 mm fillet welds: 2 × 190 mm along the heels and 125 mm across the end of the web.