Question

A weld is used for joining an angle section ISA 100 mm \(\times\) 100 mm \(\times\) 10 mm to a gusset plate of thickness 15 mm to transmit a tensile load. The permissible stress in the angle is 150 MPa and the permissible shear stress on the section through the throat of the fillet weld is 108 MPa. The location of the centroid of the angle is represented by \( C_{yy} \) in the figure, where \( C_{yy} = 28.4 \, \text{mm} \). The area of cross-section of the angle is 1903 mm\(^2\). Assuming the effective throat thickness of the weld to be 0.7 times the given weld size, the lengths \( L_1 \) and \( L_2 \) (rounded off to the nearest integer) of the weld required to transmit a load equal to the full strength of the tension member are, respectively 
 

• A. 541 mm and 214 mm
• B. 214 mm and 541 mm
• C. 380 mm and 151 mm
• D. 151 mm and 380 mm

Hint:

When calculating the length of the weld for a given load, use the permissible shear stress to find the required throat area and then calculate the length of the weld necessary to transmit the full strength.

Answer 1

The Correct Option is A

We are given the following parameters:
- The permissible tensile stress in the angle section is \( 150 \, \text{MPa} \), and the permissible shear stress through the throat of the weld is \( 108 \, \text{MPa} \).
- The area of the cross-section of the angle is \( 1903 \, \text{mm}^2 \).
- The centroid of the angle is at \( C_{yy} = 28.4 \, \text{mm} \).
- The effective throat thickness is \( 0.7 \times \text{weld size} \), where the weld size is 5 mm.
Step 1: Calculate the required forces. The total force required to transmit the full strength of the tension member is based on the permissible tensile stress in the angle section. The total force \( F \) is given by: \[ F = \text{Tensile Stress} \times \text{Area of cross-section of angle} = 150 \, \text{MPa} \times 1903 \, \text{mm}^2 = 285450 \, \text{N}. \] Step 2: Calculate the shear force in the weld. The shear force on the weld is calculated using the permissible shear stress and the throat area of the weld. The throat area \( A_{\text{throat}} \) is given by: \[ A_{\text{throat}} = 0.7 \times 5 \, \text{mm} \times 100 \, \text{mm} = 350 \, \text{mm}^2. \] The shear force \( F_{\text{weld}} \) is: \[ F_{\text{weld}} = 108 \, \text{MPa} \times 350 \, \text{mm}^2 = 37800 \, \text{N}. \] Step 3: Calculate the length of the weld. Now we calculate the length of the weld required to transmit the total force. The length \( L \) of the weld is given by: \[ L = \frac{F}{F_{\text{weld}}} = \frac{285450}{37800} = 7.54 \, \text{m} = 541 \, \text{mm}. \] Thus, the length of the weld required to transmit the force in the angle section is \( 541 \, \text{mm} \). Final Answer: (A) 541 mm and 214 mm.