Weld size for bracket on ISMB 400

Assumptions: the bracket plate is welded to the column flange by fillet welds on three sides: two horizontal welds of 250 mm (top and bottom) and one vertical weld of 300 mm. The factored load $P=200$ kN acts parallel to the flange at $e=100$ mm from the weld plane, so the weld carries direct shear and bending $M=Pe$. ISMB 400: $t_f=16$ mm. Shop weld, E250 ($f_u=410$).

Weld group (per mm of throat)

• $L=2\times250+300=800$ mm
• Neutral axis at mid-depth. $I_{xx}=\dfrac{300^3}{12}+2\times250\times150^2=13500000$ mm³
• $M=200\times10^3\times100=20\times10^6$ N·mm

Force per mm of weld at the extreme weld (y = 150 mm)

• Shear: $q_s=\dfrac{P}{L}=\dfrac{200\times10^3}{800}=250.0$ N/mm
• Bending: $q_b=\dfrac{My}{I_{xx}}=\dfrac{20\times10^6\times150}{13500000}=222.2$ N/mm
• Resultant: $q=\sqrt{q_s^2+q_b^2}=334.5$ N/mm

Size of weld

$$f_{wd}=\frac{410}{\sqrt3\times1.25}=189.4\ \text{N/mm}^2,\quad t_t=\frac{q}{f_{wd}}=\frac{334.5}{189.4}=1.77\ \text{mm},\quad s=\frac{t_t}{0.7}=2.52\ \text{mm}$$

The calculated size is small, so the minimum size of IS 800:2007 Table 21 governs: thicker part 16 mm (10 < t ≤ 20) gives 5 mm.

Answer: provide 5 mm fillet weld all round the three sides. Check: capacity of 5 mm weld $=662.8$ N/mm against demand $334.5$ N/mm (utilisation $0.50$).