Answer:
[tex]FOS=2.39[/tex]
Explanation:
Given:
maximum gauge pressure in the steel , [tex]p=10\ MPa[/tex]
outer diameter of the vessel, [tex]d_o=200\ mm[/tex]
wall thickness of the vessel, [tex]t=6\ mm[/tex]
ultimate stress of the steel, [tex]\sigma_u=400\ MPa[/tex]
We first check the type of shell:
[tex]\frac{t}{d_o} =\frac{6}{200}<\frac{1}{20}[/tex]
so the vessel is a thin shell.
Now the maximum tensile stress in the vessel due to maximum pressure is given as:
[tex]\sigma=\frac{p.d_o}{2t}[/tex]
[tex]\sigma=\frac{10\times 200}{2\times 6}[/tex]
[tex]\sigma=166.67\ MPa[/tex]
Now the factor of safety:
[tex]FOS=\frac{\sigma_u}{\sigma}[/tex]
[tex]FOS=\frac{400}{166.67}[/tex]
[tex]FOS=2.39[/tex]