Should the external pressure calculation use a welding joint coefficient of 1 or 0.85?
Thread Content
It is subject to internal and external pressures; the welding joint coefficient is 0.85 for internal pressures, but is it still 0.85 or does it change to 1 for external pressures?
It makes no difference; it has no impact on the calculations.
The welding joint coefficient is independent of external and internal pressures...
For externally pressurized equipment, the weld joint factor has no relation to the calculations; it is only related to the proportion of non-destructive testing. The flaw detection ratio shall comply with the requirements of GB150.4.
It has nothing to do with external or internal pressure! It is only related to the proportion of non-destructive testing!
I understand it should be 1; the principles of force application are different – the main factors affecting external pressure include diameter, thickness, length, and ellipticity – while the coefficient related to the welded joint isn’t that significant.
Butt welds on opposite sides and fully penetrative butt joints equivalent to double-sided welding: 100% non-destructive testing, 1.00:； Local non-destructive testing: 0.85; Single-sided welded butt joints: 100% non-destructive testing: 0.9 ； Local non-destructive testing for 0.8 diameter cylindrical vessels: The cylinder thickness is calculated using the coefficient for longitudinal weld joints; for cylinders without longitudinal weld joints, this value is 1.00. For flange joints: The coefficient for longitudinal weld joints of the shell is used as a reference.