N.Y. Comp. Codes R. & Regs. tit. 16, Appendices, app 14-G
EXAMPLES ILLUSTRATING THE REINFORCEMENT OF WELDED
BRANCH CONNECTIONS
Example 1
An 8 inch outlet is welded into a 24 inch header. The header material is API 5LX 46 with 5/16 inch wall. The outlet is API 5L Grade B (Seamless) Sched. 40 with 0.322 inch wall. The working pressure is 650 PSI. The Class Location is 1. The joint efficiency is 1.00. The temperature is 100° F. Design Factors F=0.60, E=1.00, T=1.00. For dimensions see Figure for Example 1, Appendix 14-G.
Header:
Nominal wall thickness:
t =(PD/2 SFET)(650 x 24/2 x 46000 x 0.60 x 1.00 x 1.00) = 0.283 inch
Excess thickness in header wall (H-t) =0.312 - 0.283 = 0.029 inch
Outlet:
Nominal wall thickess:
t b = ( 650 x 8.625/2 x 35000 x 0.60 x 1.00 x 1.00) = 0.133 inch
Excess thickness i outlet wall (B-t b) =0.322-0.133 = 0.189 inch
d = diameter of opening = 8.625-(2 x 0.322) =7.981 inch
Reinforcement required.
A R = d x t = 7.981 x 0.283 = 2.26 square inch
Reinforcement provided:
A 1 = (H-t)d = 0.029 x 7.981 = 0.23 square inch
Effective area in outlet:
Height (L) 21/2 B + M (assume 1/4 inch pad) = 21/2 x 0.322 + 0.25 = 1.05 inches or 21/2 H = 2.5 x 0.312 = 0.78 inch. Use 0.78 inch.
A 2 = 2(B-t b)L = 2 x 0.189 x 0.78 = 0.295 sq. in.
This must be multiplied by 35000/46000
Effective A 2 = 0.295 x (35000/46000) = 0.22 sq. in.
Required area A 3 = A R - A 1 - A 2 = 2.26 - 0.23 - 0.22 = 1.81 sq. in.
Use reinf. pl. 1/4 inch thick (minimum practicable) x 15.5 inch diameter.
Area (15.50 - 8.62) x 0.25 = 1.72 sq. in.
Fillet welds (assuming two 1/4 inch welds each side)
0.25 x 0.25 x 0.50 x 2 x 2 = 0.12 sq. in.
Total A 3 provided 1.84 sq. in.
Example 2
A 16 inch outlet is welded into a 24 inch header. The header material is API 5LX46 with 5/16 inch wall. The outlet is API 5L Grade B (Seamless) Sched. 20 with 0.312 wall. The working pressure is 650 PSI. The Class Location is 1. The reinforcement must be of the complete encirclement type. The joint efficiency is 1.00. The temperature is 100° F. Design Factors F=0.60, E=1.00, T=1.00. For dimensions see Figure for Example 2 in Appendix 14-G.
Header:
Nominal wall thickness:
t =(PD/2 SFET)(650 x 24/2 x 46000 x 0.60 x 1.00 x 1.00) = 0.283 inch
Excess thickness is header wall (H-t) =0.020 inch
N.Y. Comp. Codes R. & Regs. tit. 16, Appendices, app 14-G