barlow's equation and lame's equation
barlow's equation and lame's equation
(OP)
Hello,
I am comparing the results obtained from barlow's formula and lame#s formula to decide which is more accurate.
Barlow's equation is for thin walled pipes and lame's equation is used for thick walled.
My question is, how to decide whether which equation can hold good to calculate the stress or pressure to with given dimensions.
one of the checking point is if the thickness is less than the one tenth of inside radius then it is said to use barlow's eq and
when the thickness is more than the one tenth of inside radius then it is said to use lame's eq.
For ex.
1) ID = 2,0 ; t = 0,125 ; OD = 2,25 ; S = 10000 (stress)
In this case when we apply barlows eq we get Pressure = 1250.
and when we use same lame's eq we get pressure as = 1172,41
In this case the thickness is less than the one tenth of the dia so we use barlow,s ey we say that this pipe is thin walled
2) ID = 2,0 ; t = 0,375 ; OD = 2,75 ; S = 10000 (stress)
In this case when we apply barlows eq we get Pressure = 3750.
and when we use same lame's eq we get pressure as = 3081
In this case the thickness is more than the one tenth of the dia so we use lame's eq i.e we say that this pipe is thick walled
is there any other hypothesis wherein we can decide which formula to be used with given diameters.
and also can anyone tell me whether "thickness is less than the one tenth of inside radius" is correct or "thickness is less than the one tenth of inside diameter" is correct, because different book says this differently.
Please help me
I am comparing the results obtained from barlow's formula and lame#s formula to decide which is more accurate.
Barlow's equation is for thin walled pipes and lame's equation is used for thick walled.
My question is, how to decide whether which equation can hold good to calculate the stress or pressure to with given dimensions.
one of the checking point is if the thickness is less than the one tenth of inside radius then it is said to use barlow's eq and
when the thickness is more than the one tenth of inside radius then it is said to use lame's eq.
For ex.
1) ID = 2,0 ; t = 0,125 ; OD = 2,25 ; S = 10000 (stress)
In this case when we apply barlows eq we get Pressure = 1250.
and when we use same lame's eq we get pressure as = 1172,41
In this case the thickness is less than the one tenth of the dia so we use barlow,s ey we say that this pipe is thin walled
2) ID = 2,0 ; t = 0,375 ; OD = 2,75 ; S = 10000 (stress)
In this case when we apply barlows eq we get Pressure = 3750.
and when we use same lame's eq we get pressure as = 3081
In this case the thickness is more than the one tenth of the dia so we use lame's eq i.e we say that this pipe is thick walled
is there any other hypothesis wherein we can decide which formula to be used with given diameters.
and also can anyone tell me whether "thickness is less than the one tenth of inside radius" is correct or "thickness is less than the one tenth of inside diameter" is correct, because different book says this differently.
Please help me





RE: barlow's equation and lame's equation
The only design code I know which actually has two different equations for calculating thickness depending on D/t ratio is BS PD 8010 pt 1, which quotes a d/t limit of 20, not 10.
It says that for a given pipe, the lame equation gives a lower stress, but errors at D/t <20 is less than 5%.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
RE: barlow's equation and lame's equation
it is unusual in my experience to calculate a pressure based on a stress (normal for me is to calculate the stress based on the pressure load) but i can see how/why things might get turned around).
with 10% thickness i get very small difference between lame and barlow (10 vs 10.5); based on r1 = 1 and r2 = 1.1. any lame gives a (peak) inner surface stress of 10.52, almost identical to barlow.
It comes down to lame is always going to give you a varying stress across the thickness and barlow a constant. when is the difference big enough to matter ? why not always use lame as a more accurate model (lame applied to a thin walled pipe will only give three stresses very close to barlow (one slightly higher, one slightly lower, and the mid-plane almost identical)) ?
another day in paradise, or is paradise one day closer ?
RE: barlow's equation and lame's equation
Ted
RE: barlow's equation and lame's equation
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
RE: barlow's equation and lame's equation
lame calc's a bending stress field, a non-linear varying stress across the thickness.
the two equations are very different.
another day in paradise, or is paradise one day closer ?
RE: barlow's equation and lame's equation
In highly ductile material Barlow is closer (even when heavy wall), in less ductile materials Lame is a much better option. The other difference is with materials that work harden a lot (austenitic stainless steel) Barlows will underestimate the burst pressure quite a bit.
= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube
RE: barlow's equation and lame's equation
so i used the ratio d/t < or > 10. And also i need to ask u a question, u said barlow equation uses diameter - OD, is it for thick cylinders we have to use the OD in the equation? and for thin cylinders we have to use ID or mean diameter as the thickness is very small compared to inside diameter?
RE: barlow's equation and lame's equation
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: barlow's equation and lame's equation
another day in paradise, or is paradise one day closer ?
RE: barlow's equation and lame's equation
That is, you either have a force P*ID per unit length, or (P*ID/D)*D per unit length.
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: barlow's equation and lame's equation
@rb1957 (Aerospace): thanks for the reply sir, even i am using these formula to calculate stresses with given pressure load say 500 MPA for example. According to equation its clar that the stress obtained from the lame is varying and from barlow it is constant. But, for example consider thick cylinder, and when i use both equations to calculate the stresses for pressure 500MPA and OD=47,625 and ID=38,1 and thickness= 4,7625. I get
Stress from barlow equation: 2500 MPA (when i use OD in the equation as it's thick walled cylinder)
Stress form lame equation: 2277,77 MPA.
So in this case which is the better answer for stress for thick cylinders (or thin cylinders)?
y is that barlow's equation preferred for Thin cylinders and not the lame' equation. Is it because the thickness in the thin cylinders is very much less than the inside diameter ?
RE: barlow's equation and lame's equation
For what purpose is this calculation? ASME has equations for this type of problem if you are interested. I think they gutted the lame eqn years ago from section viii.
RE: barlow's equation and lame's equation
RE: barlow's equation and lame's equation
no, it has more to do with how the structures react to the applied pressure. under pressure the internal face wants to expand. in a thick shell it is prevented by the bulk of the shell. in a thin shell it can easily expand. because there is little restraint the outer face is assumed to expand the same, that the pressure is reacted all by "membrane" in-plane load, the well known hoop tension. In truth (and as i keep saying at work, there is no truth in engineering, just good approximations) the outer face almost certainly deflects different to the inner face, that under the initial (very small) pressure even a thin walled shell will behave like it's thick (generating a bending stress field across the thickness) but as pressure increases the outer face displacement will get closer to the inner face (and it'll approach the ideal thin-walled shell).
another day in paradise, or is paradise one day closer ?