Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thin-walled pressure vessel, hoop stress Question

  1. Aug 14, 2016 #1
  2. jcsd
  3. Aug 16, 2016 #2
    I did not see a p*2pi*r^2 anywhere. Is this a case of a cylinder or a sphere?
     
  4. Aug 17, 2016 #3
    They split the cylinder in half (including the air inside), and determine the forces on half the cylinder, either axially or in the hoop direction. Axially, the forces on the half-cylinder are ##p\pi R^2## and ##2\pi R\sigma_{axial} t##, where ##\sigma_{axial}## is the axial stress in the shell. So, for equilibrium, $$2\pi R t\sigma_{axial}=p\pi R^2$$
    Similarly, for the hoop direction, the forces on half the half-cylinder are ##p(2R)L## and ##2Lt\sigma_{hoop}##. So, for equilibrium,
    $$2Lt\sigma_{hoop}=p(2R)L$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Thin-walled pressure vessel, hoop stress Question
  1. Stress-strain question (Replies: 0)

Loading...