Question

Two plates of thickness 10 mm each are to be joined by a transverse fillet weld on one side and the resulting structure is loaded as shown in the figure below. If the ultimate tensile strength of the weld material is 150 MPa and the factor of safety to be used is 3, the minimum length of the weld required to ensure that the weld does NOT fail is _____________ mm (rounded off to 2 decimal places).} 

 

Hint:

When calculating weld lengths, use the ultimate tensile strength, the applied load, the weld throat area, and the factor of safety.

Answer 1

Correct Answer: 18

Step 1: Understanding the problem setup. 
We are given two plates, each with a thickness of 10 mm, to be joined using a transverse fillet weld. The structure is subjected to a force of 5 kN. The ultimate tensile strength of the weld material is 150 MPa, and the factor of safety is 3. 
Step 2: Formula for the weld strength. 
The strength of the weld can be calculated using the formula: \[ {Weld Strength} = \sigma_{{weld}} \times A_{{weld}} \] where:
\( \sigma_{{weld}} \) is the ultimate tensile strength of the weld material (150 MPa),
\( A_{{weld}} \) is the throat area of the fillet weld. The throat area \( A_{{weld}} \) of a fillet weld is calculated using the formula: \[ A_{{weld}} = \frac{1}{2} \times {leg size}^2 \] For a standard fillet weld, the leg size is equal to the plate thickness (10 mm). 
Step 3: Using the factor of safety.
The allowable weld strength is calculated by dividing the ultimate strength by the factor of safety: \[ \sigma_{{allowable}} = \frac{\sigma_{{weld}}}{{Factor of safety}} = \frac{150}{3} = 50 \, {MPa} \] The required weld strength to resist the applied force \( F \) (5 kN) is: \[ {Required weld strength} = F = 5 \, {kN} = 5000 \, {N} \] Step 4: Calculation of the minimum weld length. 
The length of the weld required can be determined using the formula: \[ L = \frac{F}{\sigma_{{allowable}} \times A_{{weld}}} \] Substituting the known values: \[ L = \frac{5000}{50 \times \frac{1}{2} \times (10)^2} \] \[ L = \frac{5000}{50 \times 50} \] \[ L = \frac{5000}{2500} = 2 \, {m} = 20.0 \, {mm} \] Step 5: Conclusion. 
The minimum length of the weld required to ensure that the weld does not fail is 20.0 mm, rounded to 2 decimal places.