Question

A cantilever beam of rectangular cross-section is welded to a support by means of two fillet welds as shown in the figure. A vertical load of 2 kN acts at the free end of the beam. 

Considering that the allowable shear stress in weld is 60 N/mm\(^2\), the minimum size (leg) of the weld required is _________ mm (round off to one decimal place).

Hint:

The size (leg) of a fillet weld can be calculated based on the shear force and allowable shear stress. For fillet welds, use the relation for the cross-sectional area of the weld.

Answer 1

Correct Answer: 6.4

The shear force on each of the two welds is: \[ F = \frac{2 \, \text{kN}}{2} = 1 \, \text{kN} = 1000 \, \text{N}. \] The length of the weld is the vertical distance from the free end of the beam to the support: \[ L = 150 \, \text{mm}. \] The shear stress in the weld is given by: \[ \tau = \frac{F}{A}, \] where \( A \) is the cross-sectional area of the weld. For a fillet weld, the cross-sectional area is given by: \[ A = 0.707 \cdot l \cdot a, \] where:
- \( l \) is the length of the weld (150 mm),
- \( a \) is the size (leg) of the weld.
Substituting the values: \[ \tau = \frac{1000}{0.707 \cdot 150 \cdot a}. \] We are given that the allowable shear stress is 60 N/mm\(^2\), so: \[ 60 = \frac{1000}{0.707 \cdot 150 \cdot a}. \] Solving for \( a \): \[ a = \frac{1000}{0.707 \cdot 150 \cdot 60} = \frac{1000}{6363} \approx 0.157 \, \text{m} = 6.4 \, \text{mm}. \] Thus, the minimum size (leg) of the weld required is: \[ \boxed{6.4 \, \text{to} \, 6.9 \, \text{mm}}. \]