Thickness of a cylinder in a compound thin cylinder

[DevonZA]

Ok, so express the strain, ε, in terms of r_i and R.
Then get an expression for the circumferential stress, σ.

Hi Haruspex

sorry for my delayed response.
Please see the answer provided by my lecturer below:

 

[haruspex]

Hi Haruspex

sorry for my delayed response.
Please see the answer provided by my lecturer below:

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Looks like the same approach I was trying to lead you through, just done in a different order.

 

[DevonZA]

Looks like the same approach I was trying to lead you through, just done in a different order.

As long as the answer is correct :)
Thank you for your help.

 

[jsp]

If I am too late, sorry. Just working on this question myself now.

σc=(Pd)/(2t)
the circumferential strain ε=σ/E = (Pd)/(2tE) but εc= (πΔd)/d
diametral strain εd=Δd/d
∴εc=εd the change in diameter = (Pd^2)/(2tE)
The sum of changes in the diameters of the two cylinders is equal to the interference between the diameters before shrinkage.
∴(Pd^2/2tE)inner + (Pd^2/2tE)outer = 4.305x10^-3
∴((200x10³)(0.095²))/((2(0.0025)(200x10⁹))+((200x10³)(0.1²))/((2(t)(200x10⁹))=4.305x10^-6
1.805x10^-6+(5x10^-9)/t=2.5x10^-6
t_o=0.002m = 2mm

b) Not sure about this part...
Now consider the inner cylinder:
inner pressure of 180 kPa and radial pressure of 200 kPa from outside
σ_c=pd/2t = ((180000-200000)(0.095))/(2x0.0025)
= -0.38 MPa

consider outer cylinder:
180 kPa+200kPa
σ_c=pd/2t = ((180000+200000)(0.1))/(2x0.002)
= 9.5 MPaThe first part I could confirm with answer in my textbook, but the second part I had to trust is correct..

 

[jsp]

Ah, I missed the second page of comments with all the answers on...
Well, confirmed my own answers, so thanks!