Question

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. The $tensile\left(S_u\right),yield\left(S_y\right)andcorrectedendurance\left(S_e\right)$ 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

• A. 1.45
• B. 1.26
• C. 1.37
• D. 2.00

Answer 1

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 Soderberg line is as follows
$\frac{S_a}{S_e}+\frac{S_m}{S_{y}t}=1$

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

$\Rightarrow n=1.26$