Shell thickness calculation by One Foot Method for Joint Efficiency of 0.85

[TheEverest]

How do we calculate thickness of shell according to One Foot Method (5.6.3), if Joint Efficiency of 0.85 is given?
Can anyone please help me?

 

[IFRs]

Use 0.85 in the denominator of the 1-foot equation

 

[HTURKAK]

API 650

(5.6.2.3 Appendix A permits an alternative shell design with a fixed allowable stress of 145 MPa (21,000 lbf/in.2) and a joint efficiency factor of 0.85 or 0.70. This design may only be used for tanks with shell thicknesses less than or equal to 13 mm (1/2 in.).)

You can find the required wall thicknesses based on maximum permissible shell-plate thickness 13 mm, allowable design stress of 145 MPa, a joint efficiency of 0.85 at Table A-2a.


The use of 1-foot method with joint efficiency for SI units td =(4.9D(H – 0.3)G)/(Ej*Sd) + CA

 

[JStephen]

If it's an Annex A tank, use the equation in Annex A, which includes joint efficiency.
If it's not an Annex A tank, include the 0.85 in the denominator per IFRs' statement.
Note that API-650, in the main body of the code, assumes a joint efficiency of 1.0 (which is why joint efficiency doesn't appear in the equation) based on spot radiography, which is different from the vessel codes.

 

[TheEverest]

Thank you for your replies guys. These replies mean a lot to me.