Given: For offset link, d = 30 mm , $\sigma_{Max} = 80 N/mm^{2}$

Eccentricity = e = 40 + $\frac{d}{2}$ = 40 + $\frac{30}{2}$ = 55mm.

Solution:

$c/s Area = A = \frac{\pi}{4} \times d^{2} = \frac{\pi}{4} \times 30^{2} = 706.86 mm^{2}$

$M.I = I = \frac{\pi}{64} \times d^{4} = \frac{\pi}{64} \times 30^{4} = 39.76 \times 10^{3} mm^{4}$

Section Modulus = Z = I/Y max = $\frac{39.76 \times 10^{3}}{15} = 2650.7 mm^{3}$

Using the relation,

$\sigma_{max} = \sigma_{o} + \sigma_{b} = \frac{P}{A} + \frac{P.e}{4}$

$80 = \frac{P}{706.86} + \frac{P \times 55}{2650.7}$

$\therefore P = 3610.1N$