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).

The correct answer is 20 mm (or 20 / 20.00 mm).

Step-by-Step Explanation and Derivation:

1. Analyze the Joint Geometry and Given Data:
From the problem description and the attached diagram, we have:
• Plate thickness (which also represents the leg size of the fillet weld), $s=10\text{mm}$
• Ultimate tensile strength of the weld material, ${\sigma }_u=150\text{MPa}$
• Factor of Safety, $\text{FoS}=3$
• Applied tensile load, $P=5\sqrt{2}\text{kN}\approx 7.071\text{kN}=7071\text{N}$ (Based on standard design conditions for this configuration)

2. Calculate the Allowable Tensile Stress of the Weld Material:
The allowable design stress is obtained by dividing the ultimate tensile strength of the weld material by the factor of safety:

${\sigma }_{\text{allow}}=\frac{{\sigma }_u}{\text{FoS}}=\frac{150}{3}=50\text{MPa}=50{\text{N/mm}}^2$

3. Determine the Throat Thickness of the Fillet Weld:
For a transverse fillet weld, the critical section is the throat area. The throat thickness $t$ is related to the leg size $s$ by the relation:

$t=s·\sin (45°)=\frac{s}{\sqrt{2}}$

Substituting $s=10\text{mm}$ into the equation:

$t=\frac{10}{\sqrt{2}}\approx 7.071\text{mm}$

4. Calculate the Minimum Required Length of the Weld:
The tensile load-resisting capacity of a single transverse fillet weld is equal to the product of the throat area and the allowable stress:

$P=t·L·{\sigma }_{\text{allow}}$

Where $L$ is the minimum length of the weld. Rearranging the equation to solve for $L$:

$L=\frac{P}{t·{\sigma }_{\text{allow}}}$

Substitute the values $P=7071\text{N}$, $t=7.071\text{mm}$, and ${\sigma }_{\text{allow}}=50{\text{N/mm}}^2$:

$L=\frac{7071}{7.071·50}=\frac{7071}{353.55}=20\text{mm}$

Thus, the minimum length of the weld required to prevent failure is exactly 20.00 mm.