Recent content by hideki

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    Thin walled pressure vessel with added forces

    Then the longitudinal stress would be: pr/2t - F/A where A is the area from the previous post
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    Thin walled pressure vessel with added forces

    Would the area be pi(.14^2-.008^2) = .06137 m^2?
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    Thin walled pressure vessel with added forces

    Sorry I don't understand. What would the area be then?
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    Thin walled pressure vessel with added forces

    Hi nvn, thanks for your reply. So the longitudinal stress due to the pressure is: longitudinal stress = pr/2t = 13.125 MPa [Tension] and the longitudinal stress due to the forces is: longutdinal stress = F/A = (30*10^3)/(pi*.14^2) = 0.487 MPa [Compression] Is that true?
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    Thin walled pressure vessel with added forces

    Hi Mapes, thankyou for your reply. I used a factor of two because there are two forces on each side of the cylinder. I guessed the cross sectional area was pi * (.14^2)...but this is probably incorrect because the end of the cylinder is curved and not a flat surface.
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    Thin walled pressure vessel with added forces

    Homework Statement Hi, this is my first time posting so I hope you can see the attachment. Homework Equations Hoop stress = pr/t Longitudinal stress = pr/2t The Attempt at a Solution I am stuck on part a), afterwards I think I should be fine. I am guessing you can "add" axial...