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

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:

\[ \text{Weld Strength} = \sigma_{\text{weld}} \times A_{\text{weld}} \] where:
- \( \sigma_{\text{weld}} = 150 \, \text{MPa} \)
- \( A_{\text{weld}} = \frac{1}{2} \times \text{leg size}^2 \)

For a standard fillet weld, the leg size equals the plate thickness (10 mm):
\[ A_{\text{weld}} = \frac{1}{2} \times (10)^2 = 50 \, \text{mm}^2 \]
Step 3: Using the factor of safety.
The allowable stress is given by:
\[ \sigma_{\text{allowable}} = \frac{\sigma_{\text{weld}}}{\text{Factor of Safety}} = \frac{150}{3} = 50 \, \text{MPa} \]
The required weld strength is:
\[ F = 5 \, \text{kN} = 5000 \, \text{N} \]
Step 4: Calculation of the minimum weld length.
Using the formula:
\[ L = \frac{F}{\sigma_{\text{allowable}} \times A_{\text{weld}}} \] Substitute the values:
\[ L = \frac{5000}{50 \times 50} = \frac{5000}{2500} = 2.0 \, \text{mm} \] Correction:
There was an inconsistency in the units. Since area is in mm² and stress is in MPa (N/mm²), the final length is:
\[ L = 2.0 \, \text{mm} \]

Step 5: Conclusion.
The minimum length of the weld required to resist the applied force is 2.0 mm, not 20.0 mm. The original rounding appears to have a decimal misplacement.