Thickness of a cylinder in a compound thin cylinder

[DevonZA]

That will be the reduction in circumference, yes. But as I wrote, it will be more helpful to consider the fractional change.
You are given a certain coefficient E of the material. If there is a certain fractional change in the length of the circumference, what does that tell you about the circumferential compression force?

It tells me that the circumferential compression force is increased thus reducing that circumference. I don't quite understand what you mean by a fractional change?

 

[Nidum]

Fractional change = ( new value - original value ) / ( original value )

I think that @haruspex is trying to steer you towards thinking about the idea of strain . Look it up .

 

[haruspex]

circumferential compression force is increased thus reducing that circumference.

Right, but what is the algebraic relationship? You are given a value for "E". Do you understand that this is the Young's modulus for the material? Do you know how this relates stress to strain?

Fractional change = ( new value - original value ) / ( original value )

Thanks Nidum.

 

[DevonZA]

Right, but what is the algebraic relationship? You are given a value for "E". Do you understand that this is the Young's modulus for the material? Do you know how this relates stress to strain?

Thanks Nidum.

E=σ/ε

 

[haruspex]

E=σ/ε

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

 

[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:

View attachment 198507
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View attachment 198509

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!