Answer :
Answer:
a) Factor of safety ≤ 1.689
b) Factor of safety ≤ 1.944
Explanation:
Given data:
Yield strength,[tex]S_{yr}=S_{ye}=170\ \textup{kpsi}[/tex]
True strain factor = 0.55
[tex]\sigma_x=30\ \textup{kpsi}[/tex]
[tex]\sigma_y=-15\ \textup{kpsi}[/tex]
[tex]\tau_{xy}=-45\ \textup{kpsi}[/tex]
Now, principle stress is given as:
[tex]\sigma_1,\sigma_2 = \frac{\sigma_x+\sigma_y}{2}\pm\sqrt{(\frac{\sigma_x-\sigma_y}{2})^2+\tau_{xy}^2[/tex]
on substituting the values we get
[tex]\sigma_1,\sigma_2 = \frac{30-15}{2}\pm\sqrt{(\frac{30-(-15)}{2})^2+(-45)^2[/tex]
[tex]\sigma_1,\sigma_2 =7.5\pm50.312[/tex]
or
[tex]\sigma_1 =7.5+50.312=57.812\ \textup{kpsi}[/tex]
and
[tex]\sigma_2 =7.5-50.312=-42.81\ \textup{kpsi}[/tex]
a) By maximum shear stress theory
we have
[tex]\sigma_1-\sigma_2\leqslant \frac{\textup{Yield\ strength}}{\textup{Factor\ of\ safety}}[/tex]
on substituting the values, we get
[tex]57.812-(-42.81)\leqslant\frac{170}{\textup{Factor\ of\ safety}}[/tex]
or
Factor of safety ≤ 1.689
b) By the distortion energy theory
[tex]\frac{1}{2}[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2]\leqslant\frac{S_{yr}}{\textup{Factor\ of\ safety}}[/tex]
since no force is acting in the z- direction, thus [tex]\sigma_3 = 0[/tex]
on substituting the values, we get
[tex]\frac{1}{2}[(30-(-15))^2+(-15-0)^2+(0-30)^2]\leqslant\frac{170}{\textup{Factor\ of\ safety}}[/tex]
or
Factor of safety ≤ 1.944