Answer:
a. [tex]S_{e}[/tex]= 220 MPa
b. [tex](S_{f} )_{1000} = 220.6 Mpa[/tex]
c. [tex](S_{f} )_{200000} = 91.06 MPa[/tex]
Explanation:
AISI 1018 Cold drawn steel mechanical properties
[tex]S_{ut}=440 MPa[/tex]
For Steel
[tex]S_{e}[/tex]= rotating beam specimen endurance limit = 0 .15[tex]S_{e}[/tex]
a.
for Endurance limit,
[tex]S_{e}[/tex]= 0.5 x 440
[tex]S_{e}[/tex]= 220 MPa
b. For Fatigue strength at 1000 stress cycles,
[tex]S_{e}[/tex]= 220 MPa
[tex]S_{ut}=440 MPa[/tex]
N = 1000
Ф.f=[tex](S_{ut} +345)Mpa=785Mpa[/tex]
[tex]b=\frac{ -log(785/220)}{log(2 .1000)}[/tex]
b=-0.167
Now equation for S-N line
[tex](S_{f} )_{N} = \phi_{f} (2N)^{b} =785(2N)^{-0.167}[/tex]
[tex](S_{f} )_{1000} =785(2\times1000)^{-0.167}[/tex]
[tex](S_{f} )_{1000} = 220.6 Mpa[/tex]
c.
[tex](S_{f} )_{200000} =785(2\times200000)^{-0.167}[/tex]
[tex](S_{f} )_{200000} = 91.06 MPa[/tex]