Page 114 - Trenchless Technology Piping Installation and Inspection

80 Cha pte r T w o

⎛ maximum diameter – mean diameter ⎞
q = 100 × ⎜ ⎟
⎝ mean diameteer ⎠
v = Poisson’s ratio (0.30 typical for liner pipe)
N = safety factor (typically 1.5 to 2)
a. Determine load

.
h × 62 4 12 × 62 4 .
=
=
=
P w 5.20 psi
144 144
b. Calculate thickness
K = enhancement factor = 7.00
v = Poisson’s ratio = 0.30
t = 48 1
⎡ 2 × 7 × 125 000 × 0 84 ⎤ 3
,
.

⎢ ⎥ + 1
2
.
⎣ 520 × 2 × 1 ( – . ) ⎦
03
t = 0.88 in. (external pressure buckling)
IV. Pressure limited due to stress
15. × q × ⎜ ⎛ 1 + q ⎞ ⎟ × DR 2 – 05 × ⎜ ⎛ 1 + q ⎞ ⎟ × DR = S
.
100 ⎝ 100 ⎠ ⎝ 100 ⎠ P PN
where S = flexural strength of liner pipe, psi
Solve for t,

⎛ 20 ⎞ ⎛ 20 . ⎞ ⎛ 48 ⎞ 2 ⎛ ⎞ ⎛ 48 ⎞
.
15. × ⎜ . ⎟ × ⎜ 1 + ⎟ × ⎜ ⎟ –. × + 20 ⎟ × ⎜ ⎟ = 4500
05 × ⎜ 1
×
⎝ 100 ⎠ ⎝ 100 ⎠ ⎝ t ⎠ ⎝ 100 ⎠ ⎝ t ⎠ 25 20
.
.
70 5024 24 48
.
– = 432 7
.
t 2 t
432.69t + 24.48t – 70.5 = 0
2
t = 0.376 in. (maximum compressive hoop stress)
V. Select minimum CIPP thickness
a. Thickness limitations
1.01 in. : External pressure, deteriorated conduit (AWWA)
0.790 in. : Minimum stiffness limitation, deteriorated conduit
0.878 in. : External pressure buckling
0.376 in. : Maximum compressive hoop stress
b. Minimum design thickness = 1.01 in. (25.6 mm) DR = 47.6
Design case: external pressure, deteriorated conduit (AWWA)