Question

Two plates are connected by fillet welds of size 10 mm and subjected to tension, as shown in the sketch. The thickness of each plate is 12 mm. The yield stress and the ultimate tensile stress of steel are 250 MPa and 410 MPa respectively.

The welding is done in the workshop $({\gamma }_{mw}=1.25)$. As per the Limit State Method of IS 800 : 2007, the minimum length (rounded off to the nearest higher multiple of 5 mm) of each weld to transmit a force P equal to 270 kN is ______.

Answer 1

Let length of weld on either side be 'l'
Size of weld (S) = 10 mm
$Forceresistedbyweld=\frac{0.7S}{1.25}\times \left(\frac{f_u}{\sqrt{3}}\right)\times l$

$=\frac{0.7\times 10}{1.25}\times \frac{410}{\sqrt{3}}\times l=270\times {10}^3$
$\Rightarrow l=203.68mm$
$\therefore$ Length required on each side
$=\frac{l}{2}=\frac{203.68}{2}$
= 101.84 mm
= 105 mm