Answer:
55.908 KN
Explanation:
The ratio D/d=60/40=1.5
The ratio r/d=10/40=0.25
From the curve attached as missing part of the question and using the above two ratios we get concentration factor, k=1.6
Therefore, [tex]\sigma_{ave}=\frac {\sigma_{max}}{k}=\frac {\sigma_{max}}{1.6}[/tex]
Since factor of safety, FS is given by
[tex]FS=\frac {Ultimate strength}{allowable strength}[/tex]
Allowable strength=[tex]\frac {Ultimate}{FS}=\frac {246}{1.1}=223.63 Mpa[/tex]
Substituting 223.63 Mpa for[tex] \sigma_{max}[/tex] then for d=40 the stress is given by [tex]\frac {223.63}{1.6}=139.77[/tex]
Also, [tex]stress=\frac {P}{A}[/tex] hence
[tex]P=stress\times A=40*10*139.77= 55908 N[/tex]
P=55.908 KN
