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Pressure vessel's strain

  1. May 20, 2015 #1
    1. The problem statement, all variables and given/known data
    question.JPG

    question 2.JPG
    2. Relevant equations
    equations.JPG
    3. 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
     
  2. jcsd
  3. May 20, 2015 #2
    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
     
  4. May 21, 2015 #3
    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?
     
  5. May 21, 2015 #4
    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
     
  6. May 21, 2015 #5
    Chet, looks like I may need a bit more help.
     
  7. May 21, 2015 #6
    Have you learned about the general form(s) of the Hooke's law relationship for multiaxial loading?

    Chet
     
  8. May 21, 2015 #7
    Will the hoop strain for a closed-end cylindrical pressure vessel be ε=(p*ri/t*E)*(1-v/2)? How do I get v?
     
  9. May 21, 2015 #8
    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
     
  10. May 22, 2015 #9
    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)].
     
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