Answer:
Max. normal stress, [tex]\sigma = \farc{3}{0.589} = 1.67F ksi[/tex]
Given:
Outer diameter, d = 3 in
thickness, t = [tex]\frac{3}{4} in[/tex]
Solution:
Now, we know that in case of axial force, the normal stress is maximum, therefore:
Max. normal stress, [tex]\sigma = \farc{force, F }{Cross -sectional area, A}[/tex] Â Â Â Â Â Â (1)
The axial load is F kips (say)
Now, cross sectional area, A = [tex]\frac{\pi }{4}(d'^{2} - d^{2})[[/tex] Â Â (2)
where
d = inner diameter
Now,
d = d' - 2t
d = 3 - [tex]2\fra{3}{4} = \frac{3}{2} in[/tex]
Now, using the value of 'd' in eqn (2):
A = [tex]\frac{\pi }{4}(3^{2} - (\frac{3}{2})^{2}) = 0.589 in^{2}[/tex]
Now, using eqn (1):
Max. normal stress, [tex]\sigma = \farc{F}{0.589} = 1.67F[/tex]
