A forged steel link with uniform diameter of 30 mm at the centre is subjected to an axial force that varies from 40 kN in compression to 160 kN in tension.
$\text{The tensile}\left(S_u\right)\text{, yield}\left(S_y\right)\text{and corrected endurance}\left(S_e\right)$ strengths of the steel material are 600 MPa, 420 MPa and 240 MPa respectively. The factor of safety against fatigue endurance as Soderberg's criterion is

The correct option is B 1.26
Diameter; d = 30 mm

$F_{max}=+160kN\left(Tension\right)$

$F_{min}=-40kN\left(Compression\right)$

$Tensilestrength:S_u=600MPa$

$Yieldstrength:S_y=430MPa$

Corrected endurance,

$S_e=240MPa$

Maximum stress,

${\sigma }_{max}=\frac{F_{max}}{A}=\frac{160\times {10}^{3}N}{\frac{\pi }{4}(30{)}^{2}m{m}^2}$

= 226.47 MPa (Tensile)

Minimum stress,

${\sigma }_{min}=\frac{F_{min}}{A}=\frac{-40\times {10}^{3}N}{\frac{\pi }{4}(30{)}^{2}m{m}^2}$

= -56.62 MPa (Compression)

Stress amplitude,

${\sigma }_a=\frac{1}{2}({\sigma }_{max}-{\sigma }_{min})$

$=\frac{1}{2}[226.47-(-56.62\left)\right]$

= 141.54 MPa

Mean stress,

${\sigma }_m=\frac{1}{2}({\sigma }_{max}+{\sigma }_{min})$

$=\frac{1}{2}[226.47+(-56.62\left)\right]$

= 84.925 MPa

Assume factor of safety is n then

$S_a=n{\sigma }_a=141.54n$

$S_m=n{\sigma }_m=84.925n$

The equation of the Soderberg line is as follows

$\frac{S_a}{S_e}+\frac{S_m}{S_{y}t}=1$

$\Rightarrow \frac{141.54n}{240}+\frac{84.925n}{420}=1$

$\Rightarrow n=1.26$