GRE Pipe Strain and Thrust Load
GRE Pipe Strain and Thrust Load
(OP)
I am working on an 8" GRE pipeline operating at 1150 psi (7.9 MPa) and trying to determine the growth due to internal pressure and thermal expansion and the forces required to restrain the pipe. The data I have found from different sources conflicts and I would like to find which is correct.
The GRE material is 245 mm OD and 22.7 mm WT so should be treated as thick wall pipe and not thin wall.
The material is anisotropic so has a different elastic modulus and Poisson ratio in the axial and hoop directions [Ea, Ua, Eh, Uh] and I have all these from the manufacturer.
Roark provides me with calcs for isotropic materials for internal pressure, thick wall (Tbl 32, 1b) and axial load, thin wall (Tbl 28, 1a) with formulas for both length and diameter change. The trouble is which poisson ratio to use?
1) dL = p.L/E.r^2.(1-2.U)/(R^2-r^2)
The Ameron Bondstrand Calculation Manual provides a formula which when rationalised gives
2) dL = p.L/Ea.r^2.(1-2.Ua.Eh/Ea)/(2.t.(R+r)/2)
(While the denominator in brackets looks different it is in fact numerically the same as the Roark formula.)
From the Coade Engineering News of June 1998 an article provides two things.
First Coade provides a statement that
3) Uh = Ua.Ea/Eh.
This suggest to me that perhaps either Ameron or Coade have a mistake e.g. Uh = Ua.Eh/Ea which would mean that the Ameron formula would become
4) dL = p.L/Ea.r^2.(1-2.Uh)/(2.t.(R+r)/2)
(Note the combination of an axial modulus with a hoop ratio)
Second Coade provides a formula for thin walled pipes
5) dL = L.(Sa/Ea-Ua.Sh/Eh) which when rationalised becomes
6) dL = p.L.r/(t.Ea).(0.5-Uh) which matches Roark Tbl 28, 1c
(Again the combination of an axial modulus with a hoop ratio)
Can someone please advise me if my formula 4 is correct and whether it is Coade or Ameron who have it wrong?
Note that I am also solving for diameter change and wall thickness change so have applied the same combination of axial and hoop properties to these formulas. I have assumed that the pipe material volume will be unchanged by pressure or axial load in order to determine the wall thickness.
I have made a spreadsheet of my work if it could be useful to others. Any help would be appreciated.
Rgds
Dennis Kirk-Burnnand
The GRE material is 245 mm OD and 22.7 mm WT so should be treated as thick wall pipe and not thin wall.
The material is anisotropic so has a different elastic modulus and Poisson ratio in the axial and hoop directions [Ea, Ua, Eh, Uh] and I have all these from the manufacturer.
Roark provides me with calcs for isotropic materials for internal pressure, thick wall (Tbl 32, 1b) and axial load, thin wall (Tbl 28, 1a) with formulas for both length and diameter change. The trouble is which poisson ratio to use?
1) dL = p.L/E.r^2.(1-2.U)/(R^2-r^2)
The Ameron Bondstrand Calculation Manual provides a formula which when rationalised gives
2) dL = p.L/Ea.r^2.(1-2.Ua.Eh/Ea)/(2.t.(R+r)/2)
(While the denominator in brackets looks different it is in fact numerically the same as the Roark formula.)
From the Coade Engineering News of June 1998 an article provides two things.
First Coade provides a statement that
3) Uh = Ua.Ea/Eh.
This suggest to me that perhaps either Ameron or Coade have a mistake e.g. Uh = Ua.Eh/Ea which would mean that the Ameron formula would become
4) dL = p.L/Ea.r^2.(1-2.Uh)/(2.t.(R+r)/2)
(Note the combination of an axial modulus with a hoop ratio)
Second Coade provides a formula for thin walled pipes
5) dL = L.(Sa/Ea-Ua.Sh/Eh) which when rationalised becomes
6) dL = p.L.r/(t.Ea).(0.5-Uh) which matches Roark Tbl 28, 1c
(Again the combination of an axial modulus with a hoop ratio)
Can someone please advise me if my formula 4 is correct and whether it is Coade or Ameron who have it wrong?
Note that I am also solving for diameter change and wall thickness change so have applied the same combination of axial and hoop properties to these formulas. I have assumed that the pipe material volume will be unchanged by pressure or axial load in order to determine the wall thickness.
I have made a spreadsheet of my work if it could be useful to others. Any help would be appreciated.
Rgds
Dennis Kirk-Burnnand
Dennis Kirk Engineering
www.ozemail.com.au/~denniskb





RE: GRE Pipe Strain and Thrust Load
Since I don't have the Ameron equation to analyze, the best I can do is discuss COADE's thinking. Using the terminology of the newsletter article, we defined Va/h as "the Poisson's ratio relating axial strain to hoop strain". So it should make sense when we say that the longitudinal growth due to the pressure is:
dL = L*(Sa/Ea-Va/h * Sh/Eh)
This equation says that the total longitudinal growth is equal to the total length times [(the longitudinal strain rate) minus (the longitudinal strain induced by the hoop strain)]. If we rearrange it we get:
dL = L*(Sa - (Va/h * Ea/Eh) * Sh) / Ea
This provides a very convenient parameter (Va/h * Ea/Eh) which can be used for relating axial strain to hoop stress, where again Va/h is defined as above.
In the article, we just happen to point out the fact (coincidence?) that the two different Poisson's values are related by:
Vh/a = Va/h * Ea/Eh and therefore, Va/h = Vh/a * Eh/Ea
(if you check your Ameron data, you should find that this is true). When presented with two Poisson's values, this fact can help you find which one you should use.
Anyway, if that is true, than the equation can (by coincidence?) be converted to:
dL = L*(Sa - Vh/a * Sh) / Ea
which, as you point out, seems strange (since Vh/a, under the 1998 notation, indicates strain in the hoop direction given axial stress-induced strain). But if that relationship is discomforting, just stick with the previous equation:
dL = L*(Sa - (Va/h * Ea/Eh) * Sh) / Ea
which should not seem to be a conflict.
Richard Ay
COADE, Inc.
RE: GRE Pipe Strain and Thrust Load
Thank you, your discussion confirms exactly what I had assumed to be the correct solution, which is good because I already processed the work.
The error in the Ameron calc Eqn 2 is that the term in the brackets (1-2.Ua.Eh/Ea) should have been (1-2.Ua.Ea/Eh) which can then be rationalised to (1-2.Uh).
The revised definition of Uh/a per BS 7159 does help.
The mathematical relationship between the four parameters must ensure that the material volume remains constant as I'm sure the bulk modulus will be an order of magnitude higher than the tesile modulus.
For information the formulas I have now are as follows:
Pressure:
dr=p.r/Eh*(R^2.(1+Ua)+r^2.(1-2.Ua))/(R^2-r^2) - internal radius
dR=p.R/Eh.r^2*(2-Ua)/(R^2-r^2) - external radius
dL=p.L/Ea.r^2*(1-2.Uh)/(R^2-r^2) - length
Temperature:
dr=C.r.dT
dR=C.R.dT
L=C.L.dT
Axial Load:
dr=-F.Ua/(2.Pi.Eh.t)
rf=r+dr
dL=F.L/(2.Pi.R.Ea.t)
Lf=L=dL
Rf=sqrt(L/Lf.(R^2-r^2)+rf^2)
Dennis Kirk Engineering
www.ozemail.com.au/~denniskb