CIPP Pressure Pipe Thickness

These detailed calculations shall provide the input data as well as the actual calculations for Eqs X1.1, X1.3, X1.4 and X1.7 of Appendix X1. of ASTM F1216. The design submittal shall also clearly identify the physical properties used for design.

NOTE - Full Det. Calc is ok, others need work

Refer to LD's calc's for OC-FM-05-01, these appear complete

Neptune Specs for Fully Deteriorated Pipe

The CIPP shall be designed as per ASTM F1216, Appendix X1.3.1 for the Fully Deteriorated Pressure Pipe condition. These detailed calculations shall provide the input data as well as the actual calculations for Eqs X1.1, X1.3, X1.4 and X1.7 of Appendix X1. of ASTM F1216.

ASTM1216:

Partially Deteriorated Pressure Condition—A CIPP installed in an existing underground pipe is designed to support external hydrostatic loads due to groundwater as well aswithstand the internal pressure in spanning across any holes in the original pipe wall. The results of Eq X1.1 are compared to those from Eq X1.6 or Eq X1.7, as directed by Eq X1.5, and the largest of the thicknesses is selected.

Fully Deteriorated Pressure Pipe Condition—A CIPP to be installed in an underground condition is designed to withstand all external loads and the full internal pressure. The design thicknesses are calculated from Eq X1.1, Eq X1.3, EqX1.4, and Eq X1.7, and the largest thickness is selected.

For the internal pressure design in Appendix X.1 of ASTM F1216, the design shall bebased on factor of safety of 2.0 and a long-term tensile strength equal to 1/3 of the designinitial tensile strength.

properties of the finished CIPP shall meet or exceed the following structural standards:MINIMUM PHYSICAL PROPERTIES

ASTM Polyester Filled Polyes Vinyl EsterProperty Test Method System System SystemFlexural Strength D790 4,500psi 4,500psi 5,000psiFlexural Modulus (Initial) D790 250,000psi 400,000psi 300,000psiFlexural Modulus (50'Yr) D790 125,000psi 200,000psi 150,000psiTensile Strength D638 3,000psi 3,000psi 4,000psi

Minimum ValuesProperty

Need to confirm what value should be used for the long term (time corrected) tensile strength.

Systems

300,000 psi

(2070 MPa)

- 150,000 psi

Test Method

Thermoplastic

Systems

Polyester Resin

Epoxy and Vinylester

Resins

Corrosion Resistance

ASTM F1216 Section

X2

Green Book Sec. 210-

2.3.3

Flexural Modulus (Initial)

ASTM D790

136,000 psi

250,000 psi

(940 MPa)

(1720 MPa)

Flexural Modulus (Long Term)

ASTM 2990

125,000 psi

-

(1030 MPa)

- 4500 psi 5000 psi(31 MPa) (34 MPa)

3200 psi 3000 psi 4000 psi(22 MPa) (21 MPa) (28 MPa)

- 250,000 psi

(1720 MPa)

- 125,000 psi

(860 MPa)

210 ft-lb - -

- -

15 psi - -

- -

Flexural Modulus (Long Term)

ASTM 2990

(860 MPa)

Flexural Strength

ASTM D790

Tensile Strength (Yield)

ASTM D638

Tensile Modulus (Initial)

ASTM D638

300,000 psi

(2070 MPa)

Tensile Modulus (Long Term)

ASTM D638

150,000 psi

(1030 MPa)

Impact Resistance

ASTM D2444(1) (29 m-

kg)

Pipe Flattening

ASTM D3034(2)

60% deflectio

n

Pipe Stiffness

ASTM D2412 (103

kPa)

Environmental Stress-Crack Resistance

ASTM D1693

Condition C

2000 hours



Partially Deteriorated Pressure Condition—A CIPP installed in an existing underground pipe is designed to support external hydrostatic loads due to groundwater as well aswithstand the internal pressure in spanning across any holes in the original pipe wall. The results of Eq X1.1 are compared to those from Eq X1.6 or Eq X1.7, as directed by Eq X1.5, and the largest of the thicknesses is selected.

Fully Deteriorated Pressure Pipe Condition—A CIPP to be installed in an underground condition is designed to withstand all external loads and the full internal pressure. The design thicknesses are calculated from Eq X1.1, Eq X1.3, Eq

http://www.vendor.buyboard.com/bid_specs/2009_bid_specifications/cured_in_place_pipe_326-09/cipp_specifications.pdf

http://www.nassco.org/publications/specs/spec_guidelines/cipp-nassco.pdf

Need to confirm what value should be used for the long term (time corrected) tensile strength.



withstand the internal pressure in spanning across any holes in the original pipe wall. The results of Eq X1.1 are compared to those from Eq X1.6 or Eq X1.7, as directed by Eq X1.5, and the largest of the thicknesses is selected.

Fully Deteriorated Pressure Pipe Condition—A CIPP to be installed in an underground condition is designed to withstand all external loads and the full internal pressure. The design thicknesses are calculated from Eq X1.1, Eq X1.3, Eq

http://www.vendor.buyboard.com/bid_specs/2009_bid_specifications/cured_in_place_pipe_326-09/cipp_specifications.pdf

Per ASTM F1216, using eqn X1. (re-arranging to sovle for t):

thickness required:t= OD/[2*K*EL*C/N*Pw*(1-μ^2)^1/3 +1]

where:Pw = groundwater load, psi (MPa), 4.33 psiK = enhancement factor of the soil and existing pipe 7adjacent to the new pipe (a minimum value of 7.0 isrecommended where there is full support of theexisting pipe),EL = long-term (time corrected) modulus of elasticity for 1/2*300,000psi=CIPP, psi (MPa) 150,000.00 psiμ = Poisson’s ratio (0.3 average), 0.3SDR = standard dimension ratio of CIPP,C = ovality reduction factor 1q = percentage ovality of original pipe 0N = factor of safety 2OD =host pipe ID = 2.08 ft orq = percentage ovality of original pipe =

= 100 * (Mean Inside Diameter - Minimum Inside Diameter)/Mean Inside DiameterOR= 100 * (Maximum Inside Diameter - Mean Inside Diameter)/Mean Inside Diameter

For our purposes (and per spec) q = 0

C = ovality reduction factor = ([1-q/100]/[1 + q/100]^2)^3

Which for our purposes = 1

t= OD/[2*K*EL*C/N*Pw*(1-μ^2)^1/3 +1]0.032 ft or0.383 inch

Can withstand buckling due to hydrostatic?The physical properties used in the design submittal shall be clearly identified. These physical properties shall be the basis for the acceptance of submittals of field samples and the acceptance of the final product. At a minimum, the pipe lining shall have the following physical properties:

Initial Fle 300,000 PSIInitial Flexural Strength ASTM D7 4,500 PSIInitial Tensile Strength ASTM D63 3,000 PSI*Value are for design conditions @ 75ºF (25ºC)

The external hydrostatic load design (as per Eq. X1.1 of ASTM F1216) shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 0.3, and a factor of safety of 2.0. The long-term (time-corrected) flexural modulus of elasticity shall be determined by multiplying the design initial flexural modulus of elasticity by a creep retention factor (CL). At a minimum, a creep retention factor of 50% shall be applied.

The pipe lining shall also be capable of withstanding instantaneous transient vacuum occurrences. For the instantaneous transient vacuum load condition, the design shall also be based on Eq. X1.1 of ASTM F1216. It is assumed that the internal vacuum effect is similar to the external loading of groundwater. The design shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 3.0, the initial flexural modulus of elasticity, and a total design factor of safety of 3.0, which consists of a cyclic vacuum loading design factor of 2.0 and a additional factor of safety of 1.5.

Pw = groundwater load, psi (MPa),Pw= Hw(ft) *62.4 pcf (density of water)/144 in^2/ft^2Hw = height of water table 10 ft

Pw= 4.33 psi 0.43333325 in

The physical properties used in the design submittal shall be clearly identified. These physical properties shall be the basis for the acceptance of submittals of field samples and the acceptance of the final product. At a minimum, the pipe lining shall have the following physical properties:



Per ASTM F1216, Section X1.3:x1.3.1 Partially Deteriorated Pressure Pipe Condition

Assumed hole dia dLTD found req'd cipp wall thickness using Table X1.1 first, then checked if 1" (assumed) hole would satisfy Eqn X1.5

Eqn X1.5 d/D <= 1.83 (t/D)^.5d = dia hole (in)D = mean ID of original pipe 25 int = thickness of CIPP inch 0.591 in

For part det pipe using table x1.1, t= 0.6 in 0.591d/D <= 0.281368d <= 7.034202 in

So for a 7.03 inch dia hole, part deteriorated is good

0.281368 <= 0.281368

Therefore Eqn X1.6 can be used to check thickness required for maximum hole sizeOtherwise must use Eqn 1.7, which is equation for fully deteriorated

Eqn 1.6 P = 5.33/(SDR-1)^2 * (D/d)^2 *'σL/NP= internal pressure, or external hydrostatic loadd = dia hole (in) 7.03 inD = mean ID of original pipe 25 inσL psi long term (time corrected) flexural strength (typ 50 yr)SDR = OD/wall thk 41.67N= factor of safety 2P= 30.53245 psi

LTD found req'd cipp wall thickness using Table X1.1 first, then checked if 1" (assumed) hole would satisfy Eqn X1.5

3.9 in hole max for 100 psi

Use 4500 divided by 3 1500Should be one-third

LTD used 4500 (.5)= 2250 psi This does NOT comport with the specifications used by CMCMUA)The external hydrostatic load design (as per Eq. X1.1 of ASTM F1216) shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 0.3, and a factor of safety of 2.0. The long-term (time-corrected) flexural modulus of elasticity shall be determined by multiplying the design initial flexural modulus of elasticity by a creep retention factor (CL). At a minimum, a creep retention factor of 50% shall be applied.The pipe lining shall be designed to span over any small holes that exist in the pipeline (as per Eq. X1.6 of ASTM F1216), under the normal internal pressure design conditions. For the hole spanning condition, the design shall be based on a factor of safety of 2.0 and a flexural strength, reduced to account for long-term effects, equal to 1/3 of the initial design flexural strength

45.79868

This does NOT comport with the specifications used by CMCMUA)The external hydrostatic load design (as per Eq. X1.1 of ASTM F1216) shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 0.3, and a factor of safety of 2.0. The long-term (time-corrected) flexural modulus of elasticity shall be determined by multiplying the design initial flexural modulus of elasticity by a creep retention factor (CL). At a minimum, a creep retention factor of 50% shall be applied.The pipe lining shall be designed to span over any small holes that exist in the pipeline (as per Eq. X1.6 of ASTM F1216), under the normal internal pressure design conditions. For the hole spanning condition, the design shall be based on a factor of safety of 2.0 and a flexural strength, reduced to account for long-term effects, equal to 1/3 of the initial design flexural strength

The external hydrostatic load design (as per Eq. X1.1 of ASTM F1216) shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 0.3, and a factor of safety of 2.0. The long-term (time-corrected) flexural modulus of elasticity shall be determined by multiplying the design initial flexural modulus of elasticity by a creep retention factor (CL). At a minimum, a creep retention factor of 50% shall be applied.The pipe lining shall be designed to span over any small holes that exist in the pipeline (as per Eq. X1.6 of ASTM F1216), under the normal internal pressure design conditions. For the hole spanning condition, the design shall be based on a factor of safety of 2.0 and a flexural strength, reduced to account for long-term effects, equal to 1/3 of the initial design flexural strength

qt = C/N [32R wB' E's (EL * I /D^3)]^1/2 (X1.3)Rearranging and solving for t:t = 0.721125 D [N^2 qt^2 / (C^2 *EL * Rw *B' *Es')]^1/3

where:16.94

= 0.433Hw+ wHRw/144 + Ws, (English Units),0.00981Hw+ wHRw/1000 + Ws, (Metric Units)

0.67

1401010

40.24

0.0325216743

12

7001/2*300,000psi=

and 150,000.00

25

t = 0.721125 D [N^2 qt^2 / (C^2 *EL * Rw *B' *Es')]^1/3

t= 0.730777 in

X1.2.2.1 The CIPP design from Eq X1.3 should have a minimum thickness as calculated by the following formula:EI/D^3 = E/ (12*(sdr)^3 >= 0.093 (in-lb)EI/D^3 = 0.624416 >= 0.093 OK??E/ (12*(sdr)^3 = 0.624416 >= 0.093 OK??

where:E = initial modulus of elasticity, psi (MPa) 300000sdr 34.210174894

X1.2.2 Fully Deteriorated Gravity Pipe Condition—The CIPP is designed to support hydraulic, soil, and live loads. The groundwater level, soil type and depth, and live load should be determined by the purchaser, and the following equation should be used to calculate the CIPP thickness required to withstand these loads without collapsing:

qt = total external pressure on pipe, psi (MPa),

Rw = water buoyancy factor (0.67 min) = 1 − 0.33 (Hw/H)

w = soil density, lb/ft3(KN/m3),Ws = live load, psi (Mpa),Hw = height of water above top of pipe, ft (m)H = height of soil above top of pipe, ft (m),B’ = coefficient of elastic support = 1/(1 + 4e−0.065H)inch-pound units, (1/(1 + 4e−0.213H) SI unitsI = moment of inertia of CIPP, in.4/in. (mm4/mm) = t^3/12,t = thickness of CIPP, in. (mm),C = ovality reduction factor (see X1.2.1),N = factor of safety,E' s = modulus of soil reaction, psi (MPa) (see Note X1.4),EL = long-term modulus of elasticity for CIPP, psi (MPa),

D = mean inside diameter of original pipe, in. (mm)

t = 0.721 D0 [(N qt / C)^2 / (EL * Rw *B' *E')] Parralell pipe design - live loadsBuried Pipe Design

psi qt= qw + qs + qlEL = long-term (time corrected) modulus of elasticity for

qw= 4.33 CIPP, psi (MPa) (0.67 min) qs= 2.605556

ql= 10pcfpsi Rwcalc 0.175ftftin-lb

psi1/2*300,000psi=

psi

in

X1.2.2.1 The CIPP design from Eq X1.3 should have a minimum thickness as calculated by the following formula:

psi

The CIPP is designed to support hydraulic, soil, and live loads. The groundwater level, soil type and depth, and live load should be determined by the purchaser, and the following equation should be used to calculate the CIPP thickness required to withstand these loads

By A. P. Moser, Steven L. Folkman

EL = long-term (time corrected) modulus of elasticity for 1/2*300,000psi=### psi

Per ASTM F1216, using eqn X1.7:Fully Deteriorated Pressure Pipe ConditionP = 2σTL/[(SDR-2)N]P= internal pressureσTL psi long term (time corrected) tensile strength (typ 50 yr)SDR = OD/wall thkn= factor of safety

So for 100 psi working pressure and safety/surge factor of 1.5, the required SDR can be found:P 150 psiN 2σTL 3000 psi long term time corrected tensile strength (typ 50 yr)

SDR = 2σTL/[PxN] - 2SDR= 18

18

SDR = OD/wall thkFor 25 in OD (=ID of Host)

Wall Thk = 1.39 in or 35.28 mm44.44 /32 in

5.56 /4 in

Resultant ID 22.22222 In

cipp psi long term time corrected tensile strength x1.7p2

So for 100 psi working pressure and safety/surge factor of 1.5, the required SDR can be found: Work P 100=des p for oc fm rehabRAA Spec requires SF = 2.5σTL 3000 psi Neptune Spec INITIAL Minimum

RAA Minimum 'Cured Liner Standard Results 'Tensile Stress ASTM-D-638 3,000 psi

Per ASTM F1216, Section X1.3:x1.3.1 Partially Deteriorated Pressure Pipe Condition

Assumed hole dia dLTD found req'd cipp wall thickness using Table X1.1 first, then checked if 1" (assumed) hole would satisfy Eqn X1.5

Eqn X1.5 d/D <= 1.83 (t/D)^.5d = dia hole (in)D = mean ID of original pipe 25 int = thickness of CIPP inch 0.6 in

d/D <= 0.283502d <= 7.08756 in

So for a 7.09 inch dia hole, part deteriorated is good

0.283502 <= 0.283502

Therefore Eqn X1.6 can be used to check thickness required for maximum hole sizeOtherwise must use Eqn 1.7, which is equation for fully deteriorated

Eqn 1.6 P = 5.33/(SDR-1)^2 * (D/d)^2 *'σL/NP= internal pressure, or external hydrostatic loadd = dia hole (in) 3.89 inD = mean ID of original pipe 25 inσL psi long term (time corrected) flexural strength (typ 50 yr)SDR = OD/wall thk 41.67N= factor of safety 2P= 99.83714 psi

LTD found req'd cipp wall thickness using Table X1.1 first, then checked if 1" (assumed) hole would satisfy Eqn X1.5

1.388889

3.89 in hole max for 100 psiwith t= 0.6

9.30 in hole max for 100 psiwith t= 1.389

Use 4500 divided by 3 1500Should be one-third

LTD used 4500 (.5)= 2250 psi This does NOT comport with the specifications used by CMCMUA)The external hydrostatic load design (as per Eq. X1.1 of ASTM F1216) shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 0.3, and a factor of safety of 2.0. The long-term (time-corrected) flexural modulus of elasThe pipe lining shall be designed to span over any small holes that exist in the pipeline (as per Eq. X1.6 of ASTM F1216), under the normal internal pressure design conditions. For the hole spanning condition, the design shall be based on a factor of safe

149.7557

This does NOT comport with the specifications used by CMCMUA)The external hydrostatic load design (as per Eq. X1.1 of ASTM F1216) shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 0.3, and a factor of safety of 2.0. The long-term (time-corrected) flexural modulus of elasThe pipe lining shall be designed to span over any small holes that exist in the pipeline (as per Eq. X1.6 of ASTM F1216), under the normal internal pressure design conditions. For the hole spanning condition, the design shall be based on a factor of safe

The external hydrostatic load design (as per Eq. X1.1 of ASTM F1216) shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 0.3, and a factor of safety of 2.0. The long-term (time-corrected) flexural modulus of elasThe pipe lining shall be designed to span over any small holes that exist in the pipeline (as per Eq. X1.6 of ASTM F1216), under the normal internal pressure design conditions. For the hole spanning condition, the design shall be based on a factor of safe

Per ASTM F1216, using eqn X1.7:Fully Deteriorated Pressure Pipe ConditionP = 2σTL/[(SDR-2)N]P= internal pressureσTL psi long term (time corrected) tensile strength (typ 50 yr)SDR = OD/wall thkn= factor of safety

So for 46 psi working pressure and safety/surge factor of 1.5, the required SDR can be found:P 69 psiN 2σTL 3000 psi long term time corrected tensile strength (typ 50 yr)

SDR = 2σTL/[PxN] - 2SDR= 41.47826

41.47826

SDR = OD/wall thkFor 25 in OD (=ID of Host)

Wall Thk = 0.602725 in19.29 /32 in

2.41 /4 in

Resultant ID 23.79455 In

cipp psi long term time corrected tensile strength x1.7p2

So for 46 psi working pressure and safety/surge factor of 1.5, the required SDR can be found: Work P 46=des p for oc fm rehabRAA Spec requires SF = 2.5σTL 3000 psi Neptune Spec INITIAL Minimum

RAA Minimum 'Cured Liner Standard Results 'Tensile Stress ASTM-D-638 3,000 psi



where:Pw = groundwater load, psi (MPa), 3.47K = enhancement factor of the soil and existing pipe 7adjacent to the new pipe (a minimum value of 7.0 isrecommended where there is full support of theexisting pipe),EL = long-term (time corrected) modulus of elasticity for 1/2*350,000psi=CIPP, psi (MPa) 175,000.00μ = Poisson’s ratio (0.3 average), 0.3SDR = standard dimension ratio of CIPP,C = ovality reduction factor 0.715q = percentage ovality of original pipe 3.75N = factor of safety 2OD =host pipe ID = 2.00q = percentage ovality of original pipe =






6.30933333331751750277645.2307765.237414288=G20/(((2*G9*G14*G17)/(G19*G8*(1-G15^2)))^1/3 +1)

Can withstand buckling due to hydrostatic?The physical properties used in the design submittal shall be clearly identified. These physical properties shall be the basis for the acceptance of submittals of field samples and the acceptance of the final product. At a minimum, the pipe lining shall h

Initial Fle 300,000 PSIInitial Flexural Strength ASTM D7 4,500 PSIInitial Tensile Strength ASTM D63 3,000 PSI

psi *Value are for design conditions @ 75ºF (25ºC)

The external hydrostatic load design (as per Eq. X1.1 of ASTM F1216) shall be based on an enhancement factor (K) of 7.0, an ovality (q) of 0%, a Poisson’s ratio of 0.3, and a factor of safety of 2.0. The long-term (time-corrected) flexural modulus of elas

The pipe lining shall also be capable of withstanding instantaneous transient vacuum occurrences. For the instantaneous transient vacuum load condition, the design shall also be based on Eq. X1.1 of ASTM F1216. It is assumed that the internal vacuum effec1/2*350,000psi=

psiPw = groundwater load, psi (MPa),Pw= Hw(ft) *62.4 pcf (density of water)/144 in^2/ft^2Hw = height of water table 8 ft

%Pw= 3.47 psi

ft or 24 in

= 100 * (Mean Inside Diameter - Minimum Inside Diameter)/Mean Inside Diameter

= 100 * (Maximum Inside Diameter - Mean Inside Diameter)/Mean Inside Diameter

The physical properties used in the design submittal shall be clearly identified. These physical properties shall be the basis for the acceptance of submittals of field samples and the acceptance of the final product. At a minimum, the pipe lining shall h


The pipe lining shall also be capable of withstanding instantaneous transient vacuum occurrences. For the instantaneous transient vacuum load condition, the design shall also be based on Eq. X1.1 of ASTM F1216. It is assumed that the internal vacuum effec



where:Pw = groundwater load, psi (MPa), 9.88 psiK = enhancement factor of the soil and existing pipe 7adjacent to the new pipe (a minimum value of 7.0 isrecommended where there is full support of theexisting pipe),EL = long-term (time corrected) modulus of elasticity for 1/2*250,000psi=CIPP, psi (MPa) 125,000.00 psiμ = Poisson’s ratio (0.3 average), 0.3SDR = standard dimension ratio of CIPP,C = ovality reduction factor 0.64q = percentage ovality of original pipe 0N = factor of safety 2OD =host pipe ID = 2.00 ft orq = percentage ovality of original pipe =






Can withstand buckling due to hydrostatic?The physical properties used in the design submittal shall be clearly identified. These physical properties shall be the basis for the acceptance of submittals of field samples and the acceptance of the final product. At a minimum, the pipe lining shall h

Initial Fle 300,000 PSIInitial Flexural Strength ASTM D7 4,500 PSIInitial Tensile Strength ASTM D63 3,000 PSI*Value are for design conditions @ 75ºF (25ºC)



Pw = groundwater load, psi (MPa),Pw= Hw(ft) *62.4 pcf (density of water)/144 in^2/ft^2Hw = height of water table 22.8 ft

Pw= 9.88 psi24 in

The physical properties used in the design submittal shall be clearly identified. These physical properties shall be the basis for the acceptance of submittals of field samples and the acceptance of the final product. At a minimum, the pipe lining shall h



qt = C/N [32R wB' E's (EL * I /D^3)]^1/2 (X1.3)Rearranging and solving for t:t = 0.721125 D [N^2 qt^2 / (C^2 *EL * Rw *B' *Es')]^1/3

where: 14.6214.62

= 0.433Hw+ wHRw/144 + Ws, (English Units),0.00981Hw+ wHRw/1000 + Ws, (Metric Units)

0.824

13008

150.40

0.10

0.842

7001/2*350,000psi=

and 175,000.00

48

t = 0.721125 D [N^2 qt^2 / (C^2 *EL * Rw *B' *Es')]^1/3

t= 1.077033 in

X1.2.2.1 The CIPP design from Eq X1.3 should have a minimum thickness as calculated by the following formula:EI/D^3 = E/ (12*(sdr)^3 >= 0.093 (in-lb)EI/D^3 = 0.282425 >= 0.093 OK??E/ (12*(sdr)^3 = 0.282425 >= 0.093 OK??

where:E = initial modulus of elasticity, psi (MPa) 300000sdr 44.566860325

X1.2.2 Fully Deteriorated Gravity Pipe Condition—The CIPP is designed to support hydraulic, soil, and live loads. The groundwater level, soil type and depth, and live load should be determined by the purchaser, and the following equation should be used to

qt = total external pressure on pipe, psi (MPa),

Rw = water buoyancy factor (0.67 min) = 1 − 0.33 (Hw/H)

w = soil density, lb/ft3(KN/m3),Ws = live load, psi (Mpa),Hw = height of water above top of pipe, ft (m)H = height of soil above top of pipe, ft (m),B’ = coefficient of elastic support = 1/(1 + 4e−0.065H)inch-pound units, (1/(1 + 4e−0.213H) SI unitsI = moment of inertia of CIPP, in.^4/in. (mm4/mm) = t^3/12,t = thickness of CIPP, in. (mm),C = ovality reduction factor (see X1.2.1),N = factor of safety,E' s = modulus of soil reaction, psi (MPa) (see Note X1.4),EL = long-term modulus of elasticity for CIPP, psi (MPa),

D = mean inside diameter of original pipe, in. (mm)

t = 0.721 D0 [(N qt / C)^2 / (EL * Rw *B' *E')] Parralell pipe design - live loadsBuried Pipe Design

psi qt= qw + qs + qlEL = long-term (time corrected) modulus of elasticity for

qw= 3.464 CIPP, psi (MPa) (0.67 min) qs= 11.15833333333

ql= 0pcfpsi Rwcalc 0.8240ftftin-lb 0.4

28389721.11003855.2505284444

psi 3.01253586E-051/2*350,000psi= 0.031115544722

psi 1.077033465011

in =0.721125*G32*(((G28^2*G14^2)/(G27^2*G31*G17*G23*G29))^(1/3))

34.81507936508

1212.089751197

40234865.51875

X1.2.2.1 The CIPP design from Eq X1.3 should have a minimum thickness as calculated by the following formula: 3.01253586E-05

0.0311155447221.077033465011

8.952380952480.145124717

403760001.984969E-06

psi 0.01256756880.4349384221

The CIPP is designed to support hydraulic, soil, and live loads. The groundwater level, soil type and depth, and live load should be determined by the purchaser, and the following equation should be used to

By A. P. Moser, Steven L. Folkman

EL = long-term (time corrected) modulus of elasticity for 1/2*300,000psi=### psi

CIPP Pressure Pipe Thickness

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