Principal stress at surface of thin walled pipe

AI Thread Summary
The discussion revolves around calculating principal stress in a thin-walled pipe under axial load, specifically focusing on determining σx and σy. The original poster seeks guidance on how to find these stresses, given that internal pressure should be ignored. After several calculations, they initially miscalculate the cross-sectional area and axial stress, leading to confusion. Corrections are suggested by other participants, emphasizing the importance of accurately calculating the area of the pipe's cross-section. Ultimately, the original poster resolves the issue and submits the assignment.
musicmar
Messages
95
Reaction score
0

Homework Statement


See attached jpg for problem statement and diagram.

I know we didn't discuss this type of problem in class. The rest of this homework set has been solving stress transformations with Mohr's circles for a given state of stress. I know how to find τxy (that was supposed to be "tau"). So, once I find σx and σy, I know what to do with Mohr's circle to find the principal stress.

If someone could point me in the right direction for how to find σx and σy, that would be great.

Thank you!
Any hints would be greatly appreciated.

P.S.
This is due in about 12 hours.***Just remembered that my professor told us to ignore the internal pressure. So, to find sigma x and sigma y, I only use the 200 lb axial force?
 

Attachments

  • IMG00166-20111101-2220.jpg
    IMG00166-20111101-2220.jpg
    34.1 KB · Views: 785
Last edited:
Physics news on Phys.org
Have you set up an axis system and drawn a free-body diagram of the problem yet?
Do you know how to determine the axial stress?
Do you know how to determine the stress in the pipe wall due to internal pressure?
 
1. Yes, I have.
2. I believe so:
A = ∏(0.5)2-∏(0.5-0.025)^2 = 0.0766 in2
σax=P/A = 200 lb/ (0.0766 in2)
= 2611.77 psi
3. just remembered that we were told to ignore the internal pressure

Is my axial stress correct? And how do I get σ1,2 from here?
 
Your calculation of A is incorrect. Re-read the description of the pipe carefully.
 
Oops. I read it as the outer diameter. Now I get A = 0.0805 in^2 and σ = 2484.37 psi.
 
Sorry, A is still incorrect. You should review how to calculate the area of a circle.
 
Wow. Maybe I should try doing homework when I am actually awake. diameter...radius
So, A = 0.0131 in^2, and σ= 15238.1 psi.
 
Sorry, you are just not calculating the correct A. you have forgotten to multiply by pi.
 
Well, I think I've got it now, and in any event, I've now turned in the assignment.
 
Back
Top