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hoop stress calculation

hoop stress calculation

hoop stress calculation

(OP)
Hi,

I have a question:
When calculating the hoop (cylinder) stress in a pipe you use static equilibrium and equate the stress in the wall to the gauge pressure:
2*T*S*1 where S is the hoop stress, T is the wall thickness, 1 is the unit length and then equate it to the gauge pressure times the area so  2*T*S*1=gauge pressure*D*1 where D is the diameter and 1 is the unit length.
they do the derivation here:
http://www.google.ca/url?sa=t&rct=j&q=mit%20pressure%20vessels&source=web&cd=1&ved=0CFYQFjAA&url=http%3A%2F%2Focw.mit.edu%2Fcourses%2Fmaterials-science-and-engineering%2F3-11-mechanics-of-materials-fall-1999%2Fmodules%2Fpv.pdf&ei=3dgeT7OSDaTfiALo7sXVCw&usg=AFQjCNF0TaEc7tX1ALYWE5mj-ODIo80u0A&cad=rja
My question is : why use the projection of the area (D*1) rather than the actual area: D*pi/2. Is it because you're projecting the area in the same direction as the wall stress, to get the component of the pressure in the same direction as the wall stress for the static equilibrium calculation?

Really simple, dumb question, i know, but I'm not a mechanical engineer and it's been a while since I've done this.

thanks
 

RE: hoop stress calculation

Its the horizontal resultant force of all the pressure acting (actually radially) on the inside surface of a free body of 1/2 a piece of pipe.  If you set up a polar coordinate system and integrate the pressure, P, over Θ=0 to Π/2 of * cos(Θ to get the horizontal summation, dR,  Lo and behold it = P*D

From "BigInch's Extremely simple theory of everything."

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