Question 4.8: An offset link subjected to a force of 25 kN is shown in Fig...

\text { Given } P=25 kN \quad S_{u t}=300 N / mm ^{2} \quad(f s)=3 .

Step I Calculation of permissible tensile stress for the link

\sigma_{t}=\frac{S_{u t}}{(f s)}=\frac{300}{3}=100 N / mm ^{2}         (a).

Step II Calculation of direct tensile and bending stresses
The cross-section is subjected to direct tensile stress and bending stresses. The stresses are maximum at the top fibre. At the top fibre,

\sigma_{t}=\frac{P}{A}+\frac{M_{b} y}{I} .

=\frac{25 \times 10^{3}}{t(2 t)}+\frac{25 \times 10^{3}(10+t)(t)}{\left[\frac{1}{12} t(2 t)^{3}\right]}             (b).

Step III Calculation of dimensions of cross-section
Equating (a) and (b),

\frac{12500}{t^{2}}+\frac{37500(10+t)}{t^{3}}=100 .

\text { or, } t^{3}-500 t-3750=0 .

Solving the above equation by trial and error method,

t \cong 25.5 mm .