To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as
[tex]\sigma_h = \frac{Pd}{2t}[/tex]
Here,
P = Pressure
d = Diameter
t = Thickness
At the same time the longitudinal stress is given as,
[tex]\sigma_l = \frac{Pd}{4t}[/tex]
The letters have the same meaning as before.
Then he hoop stress would be,
[tex]\sigma_h = \frac{Pd}{2t}[/tex]
[tex]\sigma_h = \frac{75 \times 8}{2\times 0.25}[/tex]
[tex]\sigma_h = 1200psi[/tex]
And the longitudinal stress would be
[tex]\sigma_l = \frac{Pd}{4t}[/tex]
[tex]\sigma_l = \frac{75\times 8}{4\times 0.25}[/tex]
[tex]\sigma_l = 600Psi[/tex]
The Mohr's circle is attached in a image to find the maximum shear stress, which is given as
[tex]\tau_{max} = \frac{\sigma_h}{2}[/tex]
[tex]\tau_{max} = \frac{1200}{2}[/tex]
[tex]\tau_{max} = 600Psi[/tex]
Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi
