Answer:
a) [tex]\mathbf{\sigma _ 1 = 4800 psi}[/tex]
[tex]\mathbf{ \sigma _2 = 0}[/tex]
b)[tex]\mathbf{\sigma _ 1 = 6000 psi}[/tex]
[tex]\mathbf{ \sigma _2 = 3000 psi}[/tex]
Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;
[tex]\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}[/tex]
[tex]\sigma_{long} = \sigma _2 = 0[/tex]
[tex]\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}[/tex]
[tex]\mathbf{\sigma _ 1 = 4800 psi}[/tex]
b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi
[tex]\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}[/tex]
[tex]\sigma_{long} = \sigma _2 = \frac{Pd}{4t}[/tex]
[tex]\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}[/tex]
[tex]\mathbf{\sigma _ 1 = 6000 psi}[/tex]
[tex]\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}[/tex]
[tex]\mathbf{ \sigma _2 = 3000 psi}[/tex]
The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below
