Question

A circular tube of thickness 10 mm and diameter 250 mm is welded to a flat plate using 5 mm fillet weld along the circumference. Assume Fe410 steel and shop welding.
As per IS 800:2007, the torque that can be resisted by the weld (in kN.m) is __________ (round off to one decimal place).
 

• A. 65.1
• B. 78.1
• C. 156.2
• D. 130.2

Hint:

For welded connections under torque, always use the correct formula for section modulus and torque capacity as per IS standards. Ensure unit consistency throughout the calculation.

Answer 1

The Correct Option is A

Given:
\( d = 250 \) mm,   \( S = 5 \) mm,   \( t = 10 \) mm,   Thickness of flat plate = \( 5 \) mm,
\( f_u = 410 \) N/mm²,   \( \gamma_{{mw}} = 1.25 \)

Step 1: Throat thickness of the weld (\( t_t \))
\[ t_t = 0.75 \times 5 = 3.5 \text{ mm} \]
Step 2: Calculation of Section Modulus (\( Z_p \))
\[ Z_p = \frac{J}{r} = \frac{A r^2}{r} = A r \]
Using geometry for a circular section:
\[ Z_p = \left(\pi \times t \times \frac{d}{2}\right) \times \frac{d^2 t}{2} \]
\[ Z_p = \left(\pi \times 10 \times \frac{250}{2}\right) \times \frac{250^2 \times 3.5}{2} \]
Step 3: Torque that can be applied on the plate
The maximum torque is given by:
\[ T = (f_s \times Z_p) = \left(\frac{f_u}{\sqrt{3} \gamma_{{mw}}}\right) \times \left(\pi \times \left(\frac{d^2 t}{2}\right)\right) \]
Substituting the known values:
\[ T = \left(\frac{410}{\sqrt{3} \times 1.25}\right) \times \pi \times \frac{250^2 \times 3.5}{2} \]
\[ T = 65.07 \text{ kN.m} \approx 65.1 \text{ kN.m} \]
Thus, the torque that can be resisted by the weld is \( \boxed{65.1} \) kN.m.