Page 284 - Buried Pipe Design

Rigid Pipe Products 255

H 5.0
1.97 (from Fig. 2.2, C d 1.4)
B d 2.54

K 0.165

2
Load W d 1.4 (120) (2.54) 1084 lb/ft

(not 1404, as determined previously). Alternatively the load can be
obtained by using Fig. 2.5 for projecting conduits.

H
4.37 (from Fig. 2.5, C c 7.0)
B c
13.74 2
2
W c C c B c 7.0 (120) 1111 lb/ft
12
The 1111 lb/ft is essentially the same as 1084 lb/ft as previously calcu-
lated. The error is due to graphical interpolations for C d and C c .
3. Determine the live load.
W L 340 lb/ft (from Fig. 2.19)

4. Determine the total load.

W T W c W L 1111 340 1451

5. Determine the internal pressure requirement.
Case I (live load is zero):

p (100) (4.0) 400 lb/in 2

1111 (2.5)
w 2136 lb/ft
1.3
Case II (surge is zero):

p (100) (2.5) 250 lb/in 2

1451 (2.5)
w 2790 lb/ft
1.3

By use of the Schlick formula, we can now determine the P and W or wall
thickness of the pipe which is required. For AC pipe, the performance crite-
ria P and W may be solved by trial and error as follows (class 100, P 490
2
lb/in , W 5200 lb/ft). See Fig. 5.3.