A cantilever beam of rectangular cross-section is welded to a support by means of two fillet welds as shown in figure. A vertical load of 2 kN acts at free end of the beam.

Considering that the allowable shear stress in weld is 60 N/mm² , the minimum size (leg) of the weld required is __________ mm (round off to one decimal place)

The correct answer is 6.655.

1. Given Data from the Diagram:
From the given images, we can identify the following parameters of the cantilever beam and the fillet welds:
- Vertical load acting at the free end, P = 2 kN = 2000 N
- Distance of the load from the welded support (eccentricity), L = 150 mm
- Dimensions of the rectangular cross-section of the beam: width = 50 mm, depth = 40 mm
- There are two vertical fillet welds, each of length d = 40 mm, separated by a distance of 50 mm.
- Allowable shear stress in the weld, ${\tau }_p=60{\text{N/mm}}^2$
Let t be the throat thickness of the weld, and h be the size (leg) of the weld, where $t=hcos\left({45}^{\circ }\right)=\frac{h}{\sqrt{2}}\approx 0.707h$.

2. Properties of the Weld Cross-Section:
The total throat area of the two vertical welds is:
$A_t=2\times (t\times 40)=80t{\text{mm}}^2$
The moment of inertia of the throat section about the neutral axis (horizontal centroidal axis) is:
$I_{\text{NA}}=2\times \left(\frac{t\times {40}^3}{12}\right)=\frac{t\times {40}^3}{6}{\text{mm}}^4$

3. Primary Shear Stress (${\tau }^{\prime }$):
The primary shear stress due to the vertical load P is distributed uniformly over the throat area:
${\tau }^{\prime }=\frac{P}{A_t}=\frac{2000}{80t}=\frac{25}{t}{\text{N/mm}}^2$

4. Bending Stress (${\sigma }_b$):
The bending moment acting on the weld section is:
$M=P\times L=2000\times 150=300000\text{N·mm}$
The maximum bending stress occurs at the outermost fiber, where $y_{\text{max}}=\frac{40}{2}$:
${\left({\sigma }_b\right)}_{\text{max}}=\frac{M\times y_{\text{max}}}{I_{\text{NA}}}=\frac{300000\times 20}{\frac{t\times {40}^3}{6}}=\frac{300000\times 20\times 6}{64000t}=\frac{562.5}{t}{\text{N/mm}}^2$

5. Maximum Shear Stress (${\tau }_{\text{max}}$):
Using the shear stress formula for combined bending and torsion/shear:
${\tau }_{\text{max}}=\sqrt{{\left(\frac{{\sigma }_b}{2}\right)}^2+{\left({\tau }^{\prime }\right)}^2}$
Substituting the calculated stress values:
${\tau }_{\text{max}}=\sqrt{{\left(\frac{562.5}{2t}\right)}^2+{\left(\frac{25}{t}\right)}^2}=\frac{1}{t}\sqrt{{281.25}^2+{25}^2}=\frac{282.359}{t}{\text{N/mm}}^2$

6. Calculation of Weld Size (h):
For a safe design, the maximum shear stress must not exceed the allowable shear stress:
${\tau }_{\text{max}}\le {\tau }_p$
$\frac{282.359}{t}\le 60\Rightarrow t\ge \frac{282.359}{60}\approx 4.706\text{mm}$
Now, find the minimum size of the weld leg (h):
$h=\sqrt{2}\times t\approx 1.4142\times 4.706\approx 6.655\text{mm}$