×
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

Problem : Tank lifting as per appendix J

Problem : Tank lifting as per appendix J

Problem : Tank lifting as per appendix J

(OP)
I refer to Par. J.3.8.3 of API650
The required safety factor of 4 is referred to Tensile strength or Yield Strength of material.
Why I must double the load acting on a lug ?
If a customer require 1.5 or 1.25 I may assume this value or I must assume 2 ?
Many thanks in advance
 
 



 

RE: Problem : Tank lifting as per appendix J

Your primary concern is to provide a nominal safety factor against failure due to excessive plastic deformation of the lifting lug.  Thus, to prevent plastic deformation of the lifting lug you do not want to exceed the yield stress.  API 8A and ASME B30.20 both limit the allowable stress based on a safety factor applied to the yield strength of the material.  This is a common approach and widely used.  For "normal" sized loads, these standard require a safety factor of approximately 3:1 on yield stress.  On the other hand, ANSI N14.6 requires a SF of 3 based on yield and 5 based on tensile strength.  For A-36 material, the end result is the same value for allowable stress (12 ksi).

So, a safety factor of 4 doesn't really agree with any of the three codes listed above.  I would consult someone who is very familiar with the requirements before making a decision.  Lifting attachments are a critical part of the design and should not be taken lightly.  Lifting lugs are typically designed very conservatively since safety is paramount and steel is cheap.

I am not sure why the code requires that the load be doubled for design purposes.  My only idea is that this is meant to be an impact factor.  However, I have never seen a code impose an impact factor.  This is usually a client requirement.

If your client is requiring a 1.25 or a 1.5 impact factor, you must also ensure that you meet code requirements.  If the code says 2.0, then you must use 2.0 for an impact factor.  The client requirements are in addition to the code requirements.  However, if they are both impact factors, then only meet the most stringent requirement.

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