• Support PF! Buy your school textbooks, materials and every day products Here!

Principal stress at surface of thin walled pipe

  • Thread starter musicmar
  • Start date
  • #1
100
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

Last edited:

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
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?
 
  • #3
100
0
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?
 
  • #4
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
Your calculation of A is incorrect. Re-read the description of the pipe carefully.
 
  • #5
100
0
Oops. I read it as the outer diameter. Now I get A = 0.0805 in^2 and σ = 2484.37 psi.
 
  • #6
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
Sorry, A is still incorrect. You should review how to calculate the area of a circle.
 
  • #7
100
0
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.
 
  • #8
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
Sorry, you are just not calculating the correct A. you have forgotten to multiply by pi.
 
  • #9
100
0
Well, I think I've got it now, and in any event, I've now turned in the assignment.
 

Related Threads on Principal stress at surface of thin walled pipe

Replies
22
Views
1K
  • Last Post
Replies
16
Views
859
Replies
4
Views
562
Replies
0
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
47
Views
4K
Replies
5
Views
3K
Replies
2
Views
700
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
24
Views
5K
Top