Question 8.13: A bracket is welded to the vertical plate by means of two fi...

\text { Given } P=50 kN \quad \tau=70 N / mm ^{2} .

Step I Primary shear stress
The total area of two vertical welds is given by,

A=2(400 t)=(800 t) mm ^{2} .

The primary shear stress in the weld is given by,

\tau_{1}=\frac{P}{A}=\frac{50000}{800 t}=\frac{62.5}{t} N / mm ^{2}             (i).

Step II Bending stress
The moment of inertia of two welds about the X-axis is given by,

I=2\left[\frac{t(400)^{3}}{12}\right]=\left(10.67 \times 10^{6} t\right) mm ^{4} .

From Eq. (8.28),

\sigma_{b}=\frac{M_{b} y}{I}               (8.28).

\sigma_{b}=\frac{M_{b} y}{I}=\frac{\left(50 \times 10^{3} \times 300\right)(200)}{\left(10.67 \times 10^{6} t\right)} .

=\frac{281.16}{t} N / mm ^{2}              (ii).

Step III Maximum shear stress
The maximum principal shear stress in the weld is given by,

\tau=\sqrt{\left(\frac{\sigma_{b}}{2}\right)^{2}+\left(\tau_{1}\right)^{2}}=\sqrt{\left(\frac{281.16}{2 t}\right)^{2}+\left(\frac{62.5}{t}\right)^{2}} .

=\frac{153.85}{t} N / mm ^{2} .

Step IV Size of weld
The permissible shear stress in the weld is 70 N/mm². Therefore,

\frac{153.85}{t}=70 \quad \text { or } \quad t=2.2 mm .

h=\frac{t}{0.707}=\frac{2.2}{0.707}=3.11 \quad \text { or } \quad 4 mm .