Problem in Pipe Design

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.
1. DEAD LOAD, We
(Check for the maximum fill height.)
Fill Height:
H = 10 ft
Soil Unit Weight: w = 120 pcf
Pipe O.D.:
BC = 49.5 in.
Earth Load
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