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## Q. 4.8

An offset link subjected to a force of 25 kN is shown in Fig. 4.27. It is made of grey cast iron FG300 and the factor of safety is 3. Determine the dimensions of the cross-section of the link.

## Verified Solution

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

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,

$t \cong 25.5 mm$.