Question

Two plates, subjected to direct tension, each of 10 mm thickness and having widths of 100 mm and 175 mm, respectively are to be fillet welded with an overlap of 200 mm. Given that the permissible weld stress is 110 MPa and the permissible stress in steel is 150 MPa, the length of the weld required using the maximum permissible weld size as per IS : 800-1984 is _____.

Answer 1

Permissible weld stress
= 110 MPa
Permissible tensile stress in steel
= 150 MPa
Maximum force the smaller plate can resist

$P_{max}=crosssectionalareaofplate\times permissibletensilestress$
= 100 x 10 x 150 N
= 150 kN
Thickness of plate, t = 10 mm
For square edge, maximum size of weld,
s = t - 1.5
= 8.5 mm

Strength of weld,
= 0.7s x permissible weld stress
= 0.7 x 8.5 x 110
= 654.5 N/mm

Length of weld required for resisting $P_{max}$ force
$L=\frac{P_{max}}{Strengthofweld}$

$=\frac{150\times {10}^3}{654.5}mm$

= 229.18 mm
$\simeq$ 229.2 mm