Weld ISA 100x100x10 to 12 mm gusset

The load is not given, so the weld is designed to develop the full design tensile strength of the angle (ISA 100×100×10, $A_g=1903$ mm², centroid $c_x=28.4$ mm from the back), which is the usual approach.

Load to be carried

$$T_{dg}=\frac{A_gf_y}{\gamma_{m0}}=\frac{1903\times250}{1.10}\times10^{-3}=432.5\ \text{kN}$$

Weld size (IS 800:2007 cl. 10.5)

• Thicker part $=12$ mm (10 < t ≤ 20): minimum size 5 mm.
• Maximum size along the rounded toe of the angle: $\tfrac34t=\tfrac34\times10=7.5$ mm. Use $s=6$ mm shop fillet weld.
• Throat $t_t=0.7\times6=4.2$ mm; $f_{wd}=\dfrac{f_u}{\sqrt3\gamma_{mw}}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²
• Strength per mm: $q=189.4\times4.2=795.4$ N/mm (= 0.795 kN/mm)

Weld length

Total effective length $L=\dfrac{432.5\times1000}{795.4}=544$ mm.

To avoid eccentricity, the welds are placed so that their resultant passes through the centroid of the angle. Let $L_1$ be the weld at the toe (far from the centroid) and $L_2$ the weld at the heel (back):

• $F_1$ (toe) $=T\dfrac{c_x}{100}=122.8$ kN → $L_1=154.4$ mm
• $F_2$ (heel) $=T\dfrac{100-c_x}{100}=309.7$ kN → $L_2=389.3$ mm

The effective length of a fillet weld is its actual length minus $2s$ (end craters), so add $2s=12$ mm to each weld.

Provide: 6 mm fillet weld, toe-side length $=170$ mm and heel-side length $=405$ mm (about 544 mm effective in total); the minimum length $4s=24$ mm is satisfied. The 12 mm gusset must be long enough to take the 405 mm weld; otherwise use an 8 mm weld at the heel or add an end (transverse) weld across the angle.