# Thin walled pressure vessel with added forces

by hideki
Tags: added, forces, pressure, vessel, walled
 P: 6 1. The problem statement, all variables and given/known data Hi, this is my first time posting so I hope you can see the attachment. 2. Relevant equations Hoop stress = pr/t Longitudinal stress = pr/2t 3. 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 stresses induced by the 30kN force...but I'm not sure? In this case I have: Circumferential stress = pr/t = 26.25MPa [Tension] Longitudinal stress = pr/2t - 2 * (F/A) = -285.29Mpa. [compression] The longitudinal stress is large which leads me to believe I have done something wrong. Any help or clarification would be greatly appreciated. Attached Thumbnails
 Sci Advisor HW Helper PF Gold P: 2,532 Hi hideki, welcome to PF! Your reasoning about superposing the stress from the 30 kN force is fine. But your calculations need checking. Where did the factor of 2 come from? What is the cross-sectional area of the material under stress in the cylindrical part of the vessel?
 P: 6 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.
 Sci Advisor HW Helper P: 2,124 Thin walled pressure vessel with added forces hideki: Both concepts in post 3 are incorrect. Try again. As Mapes mentioned, check your second calculation in post 1. Your circumferential stress in post 1 is correct. Always leave a space between a numeric value and its following unit symbol. E.g., 26.25 MPa, not 26.25MPa. See the international standard for writing units (ISO 31-0). Also, it is spelled MPa, not Mpa. Do not worry about the stress in the elliptical end caps, for now. Compute the stress in the cylindrical portion of the pressure vessel.
 P: 6 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?
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,525 In calculating the direct axial stress due to the 30 kN force, the stress is developed in the area of the 8 mm tank shell, not the entire cross section of the tank.
 P: 6 Sorry I don't understand. What would the area be then?
HW Helper
PF Gold
P: 2,532
 Quote by hideki Sorry I don't understand. What would the area be then?
If you were to look at a cross section of the cylindrical part, what is the area of the material region that you see? ($\pi r^2$ is the cross-sectional area taken up by air).
 P: 6 Would the area be pi(.14^2-.008^2) = .06137 m^2?
 P: 6 Then the longitudinal stress would be: pr/2t - F/A where A is the area from the previous post