Simple Calculation for Pressure Vessel Strain | 69.6 kPa Solution

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Homework Statement

Homework Equations

The Attempt at a Solution

Not sure if I'm on the right track here at all.

p=40*10^-6*6*10^-3*290*10⁹/1=69600 Pa=69.6 kPa

 

[Chestermiller]

You used the equation ε=σ/E to calculate the hoop stress, but this approach is not correct because the state of stress is biaxial. You need to calculate the axial stress in terms of the pressure, and then use Hooke's law in its tensorial form to determine the pressure.

Chet

 

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Chet, I might not understand it correct. The reason why I've used the hoop stress is that in the question the strain gauges are installed on the vessel to measure the hoop stress as hoop stress = 2x axial stress. Should I use the axial stress instead, why?

 

[Chestermiller]

No. The problem is with the equation you used to get the hoop strain. In terms of the (unknown) pressure, your equation for the hoop stress is correct. You indicated that the axial stress is half the hoop stress. So, in terms of the (unknown) pressure p, what is the axial stress? What are the Hooke's law equations for the hoop stain and the axial strain in terms of the hoop stress and the axial stress (both equations involve both stresses)?

Chet

 

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Chet, looks like I may need a bit more help.

 

[Chestermiller]

Chet, looks like I may need a bit more help.

Have you learned about the general form(s) of the Hooke's law relationship for multiaxial loading?

Chet

 

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Will the hoop strain for a closed-end cylindrical pressure vessel be ε=(p*r_i/t*E)*(1-v/2)? How do I get v?

 

[Chestermiller]

Will the hoop strain for a closed-end cylindrical pressure vessel be ε=(p*r_i/t*E)*(1-v/2)? How do I get v?

Yes. Nice job. You need to look up the Poisson ratio for the pipe material. They give you the Young's modulus, but not the Poisson ratio?

Chet

 

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Chet, thanks for you help with this. I've contacted the tutor and he accidentally forgot to include the Poisons ratio, it's v=0.3. It should be straight forward now, just need to make p the subject of the formula. I got p=(εEt)/[r(1-0.5v)].