# Problem in Pipe Design

Topics: Pressure, Steel, Soil Pages: 15 (3393 words) Published: March 26, 2013
Therefore, t = 0.236 in. satisfies all internal pressure
conditions. Check this steel-wall thickness against
the external load due to 10 feet of fill.
STEP 2 EXTERNAL LOAD DESIGN
A . MODIFIED IOWA FORMULA
Prior to checking the anticipated horizontal
deflection of the pipe, the designer must evaluate
and determine the component parts to be used in the
modified Iowa formula.
(Check for the maximum fill height.)
Fill Height:
H = 10 ft
Soil Unit Weight:
w = 120 pcf
Pipe O.D.:
BC = 49.5 in.
We = 10 (120) 49.5/12
= 4950 lb/ft of pipe
= 413 lb/in. of pipe
For fill heights greater than 8 feet, the HS-20 live
load may be neglected; therefore, Wl = 0, meaning
W = We
2. PIPE STIFFNESS
Modulus of Elasticity of Steel:
Es = 30,000,000 psi
Modulus of Elasticity of Mortar:
El = 4,000,000 psi
Cement Mortar Lining Thickness:
Tl = 0.5 in.
Let Il = moment of inertia of the cement mortar lining
= ( 0 . 5 )3/ 1 2
= 0.0104 in.4/in. of pipe
Is = the moment of inertia of the steel cylinder
= ( 0 . 2 3 6 )3/ 1 2
= 0.0011 in.4/in. of pipe
Pipe stiffness:
E I = 30,000,000 (0.0011)+ 4,000,000 (0.0104)
= 74,600 lb-in.
DESIGN EXAMPLE
Design a 48-inch nominal I.D. pipeline for an operating
pressure of 200 psi and a transient pressure of 80 psi.
Additionally, this pipeline will be field pressure tested
to 250 psi. The pipe will have fill heights of 5-10 feet
and the pipe zone will consist of course-grained soils
with little or no fines, compacted to 90% of Standard
Proctor. Shop applied-cement mortar lining and tape
wrap coating will provide the corrosion protection.
STEP 1 INTERNAL PRESSURE
Nominal Pipe Size:
D = 48 in.
Steel Wall O.D.:
O . D . = 49.5 in.
Working Pressure:
Pw = 200 psi
Transient Pressure
Pt = 80 psi
Field Test Pressure:
Pf = 250 psi
Let t = Steel Wall Thickness, in.
Fs = Allowable Unit Steel Stress, psi
= 21,000 psi when P = Pw
= 31,500 psi when P = Pw + Pt or P = Pf
(based on ASTM A139 Grade C material
with minimum specified yield of the steel
= 42,000 psi)
A . WORKING PRESSURE CONDITION
4 9 . 5 ( 2 0 0 )
t = ÐÐÐÐÐÐÐÐÐÐÐÐÐÐ = 0.236 in.
2 ( 2 1 , 0 0 0 )
B. TRANSIENT PRESSURE CONDITION
49.5(200+ 80)
t = ÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐ = 0.220 in.
2 ( 3 1 , 5 0 0 )
C. FIELD TEST PRESSURE CONDITION
4 9 . 5 ( 2 5 0 )
t = ÐÐÐÐÐÐÐÐÐÐÐÐÐÐ = 0.196 in.
2 ( 3 1 , 5 0 0 )
The pipe designer must select the condition that
results in the maximum steel thickness. In this design
example, condition A for working pressure governs.
SAMPLE PROBLEM
1
C . EXTERNAL PRESSURE (VACUUM)
1 . Above ground pipelines subjected to vacuum
conditions require a wall thickness designed to
resist collapse. For such pipe, where collapse
pressure is less than 581 psi and t / D is greater
than 43, the collapse pressure is determined by the
Stewart formula:
w h e r e P c = collapse pressure
D = the pipe outside diameter in inches
t = the pipe wall thickness in inches
The Stewart Formula accounts for variation in wall
thickness (t), out-of-roundness, and other
manufacturing tolerances.
2 . Buried pipelines in granular soil with a friction
angle greater than 25¡, where ring deflection is
not less than 5%, will not collapse due to vacuum.
Shown below is a graph showing actual vacuum vs.
ring flexibility for different soils.1
PIPE: D = 10% RING DEFLECTION DUE TO
INSTALLATION
SOIL: UNSATURATED -- COHESIONLESS
H/D = 0.43
UNIT WEIGHT OF DRY SOIL = 15 kN/m3
(100 LB/FT3)
t
Pc=50,200,000 ÐÐ
D
solve for t=
t=D(Pc/50,200,000)
[ ]3
1/3
2
1 From ASCE paper by Watkins/Tupac, 1994, Ò V a c u u m
Design of Welded Steel Pipe in Buried Poor SoilÓ
If the water table is above the top of the pipe,
the soil will not liquefy if density of the
embedment is 90% Standard Proctor (ASTM D698 or
A A S H T O T-99). The height of water table, h, above
ground surface, must be added to the internal
vacuum. The worst case is an empty pipe with the
water table above ground...

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