Answer:
factor of safety = 0.8
Explanation:
given data
maximum gauge pressure in steel p = 10 MPa
outer diameter do = 200 mm
wall thickness t = 2 mm
ultimate stress of the steel σU = 400 MPa
solution
We check the type of shell that is
[tex]\frac{t}{do} =\frac{6}{200}[/tex]
so we can see that that is [tex]<\frac{1}{20}[/tex]
so that it mean vessel is a thin shell now we can we get maximum pressure by the maximum tensile stress is
σ = [tex]\frac{p\times do}{2t}[/tex] .......................1
put here value and we get
σ = [tex]\frac{10\times 200}{2\times 2}[/tex]
σ = 500 MPa
so factor of safety will be express as
factor of safety = [tex]\frac{\sigma U}{\sigma}[/tex] ..........2
factor of safety = [tex]\frac{400}{500}[/tex]
factor of safety = 0.8
