×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

ASME B31.3 Chapter IX High Pressure Piping

ASME B31.3 Chapter IX High Pressure Piping

ASME B31.3 Chapter IX High Pressure Piping

(OP)
tongueHi all,

I am wondering if someone could kindly help me with this problem I am facing here. It's with regards ASME B31.3 Chapter IX High Pressure Piping. The formula for the minimum pressure design thickness is given as t = (D - 2co)/2 * [1 - exp (-p/s)]. From my understanding, the reason behind this formula is the fact that we are no longer treating the pipe using thin wall theory, rather we are looking at it as a thick walled pipe. So, I am trying to derive the formula by integrating the hoop stress from T = 0 -> T = t. Such that: s dA = P[D - 2(T + co)]. However, from my derivation it appears that the final formula will be t = (D - 2co)/2 * [1 - exp (-2p/s)] (With the hope that my knowledge on integration is still correct. :P ). When I refer to the Charles Becht book (Complete guide to ASME B31.3), the comment is the formula is based on limit load pressure rather than the hoop stress - or am I misunderstanding it? Could anyone please kindly help me with it?


Secondly, the footnote 5 says that the intent of the equation is to provide a factor of not less than 1.732 (SQRT 3). Is that based on the fact that according to Von-Mises failure criterion, the failure will be 1.732 times lower in a pure shear case than a pure tensile case. In a thick walled cylinder, the inclusion of radial stress in the consideration will give rise to shear stress and hence making the case of pure shear a valid consideration. So is that the reason behind the 1.732? And if so, how is that built-in to the equation?


Many thanks!


Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources