For a given sustained load case report, how the program is arriving at the code stress from the bending stress and torsion stress given at the left side of SIFs(output report). Please check out the below typical sustained load case of a tee joint(stress units in kPa).
6020(Node no.) 3464.(bending) -853.(torsion) 1.450 / 1.450 (SIF) 5599.(code stress)
The axial sustained stress due to pressure is 2406 kPa.

Correct me if I am wrong, the bending stress mentioned here is resultant of in plane and out plane bending stresses. Using the formula as given in the code we do not arrive at the code stress given here.
For points having SIF=1.0, we get the correct code stress mentioned in the report by using the formula sqrt(bending**2+4xtorsion**2) and then adding the axial suatained stress. Please let me know why the program uses the old method of combining the stresses instead of the one available in the code. And also please let me know why the code stress is not correct when SIF is not equal to one. These calculations are applicable for code B31.1.