Question 8.14: A beam of rectangular cross-section is welded to a support b...

\text { Given } P=25 kN \quad \tau=75 N / mm ^{2} .

Step I Primary shear stress
The total area of the horizontal and vertical welds is given by

A=2[100 t+150 t]=(500 t) mm ^{2} .

The primary shear stress in the welds is given by

\tau_{1}=\frac{P}{A}=\frac{25000}{(500 t)}=\left(\frac{50}{t}\right) N / mm ^{2}         (i).

Step II Bending stress
Referring to Fig. 8.35, the moment of inertia of four welds about the X-axis is given by

I_{x x}=2\left[\frac{b t^{3}}{12}+(b t) \times\left(\frac{d}{2}\right)^{2}\right]+2\left[\frac{t d^{3}}{12}\right] .

Assuming b and d to be large as compared to the throat dimension t and neglecting the terms containing t³, we have

I_{x x}=t\left[\frac{b d^{2}}{2}+\frac{d^{3}}{6}\right] .

Substituting the values,

I_{x x}=t\left[\frac{(100)(150)^{2}}{2}+\frac{(150)^{3}}{6}\right]=\left[(75)(150)^{2}\right] t mm ^{4} .

From Eq. (8.28),

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

\sigma_{b}=\frac{M_{b} y}{I}=\frac{(25000 \times 500)(75)}{(75)(150)^{2} t} .

=\left(\frac{555.55}{t}\right) N / mm ^{2}              (ii).

Step III Maximum shear stress
From Eq. (8.29), the maximum 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{555.55}{2 t}\right)^{2}+\left(\frac{50}{t}\right)^{2}} .

=\frac{282.24}{t} N / mm ^{2} .

Step IV Size of weld
Since the permissible shear stress in the weld is 75 N/mm².

\left(\frac{282.24}{t}\right)=75 \quad \text { or } \quad t=3.76 mm .

\text { and } \quad h=\frac{t}{0.707}=\frac{3.76}{0.707}=5.32 \cong 6 mm .