Problem : Tank lifting as per appendix J

[edemarcos]

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

 

[ctmfab]

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.