Question

A welded joint has '$ l $' mm length and '$ t $' mm thickness fillet carries a steady load of 'F' N along the weld. Then the maximum shear stress induced in the weld in terms of N/mm$ ^2 $ will be

• A. \( \frac{F}{0.707 tl} \)
• B. \( \frac{F}{tl} \)
• C. \( \frac{0.707 F}{tl} \)
• D. \( \frac{F}{1.414 tl} \)

Hint:

For a fillet weld subjected to a load, the critical area for shear stress is the throat area. The throat thickness of a standard fillet weld is approximately 0.707 times the leg thickness. Maximum shear stress is then calculated as the load divided by the throat area.

Answer 1

The Correct Option is A

Step 1: Understand the concept of fillet welds and throat thickness.
For a fillet weld, the critical section is the throat, which is the shortest distance from the root to the face of the weld. When a load is applied along the weld, the weld experiences shear stress. The maximum shear stress occurs on this throat area.
Step 2: Determine the throat thickness.
Let \( t \) be the leg thickness (or size) of the fillet weld. The throat thickness, \( t_t \), for a standard fillet weld (where the angle between the fusion faces is 90 degrees) is given by:
\( t_t = t \sin(45^\circ) \) Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707 \), \( t_t = 0.707t \)
Step 3: Calculate the throat area.
The area of the weld that resists the shear load is the throat area, \( A_t \).
Given the length of the weld as \( l \), the throat area is:
\( A_t = t_t \times l = 0.707tl \)
Step 4: Calculate the maximum shear stress.
Shear stress (\( \tau \)) is defined as force per unit area.
Given the steady load \( F \), the maximum shear stress induced in the weld is: \[ \tau_{max} = \frac{\text{Load}}{\text{Throat Area}} = \frac{F}{A_t} = \frac{F}{0.707tl} \] The final answer is $\boxed{\text{1}}$.