barlow's equation and lame's equation 6

[ppppraxxx]

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

 

[LittleInch]

Are you sure that's the right way around?

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.

 

[rb1957]

are you comparing like with like ? ie lame at mid-thickness with barlow ?

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 ?

 

[hydtools]

Since D = 2r, either one is the same as the other.

Ted

 

[LittleInch]

Barlow only use one D ( the outer one), lame uses two different r's. That's why you get the difference.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[rb1957]

barlow calc's one stress, assumes constant stress across the thickness.

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 ?

 

[EdStainless]

Not to make things less clear but the material matters also.
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

 

[ppppraxxx]

@LittleInch (Petroleum): thanks for the reply sir, the criteria which i have mentioned is given in two text books and also from some video lectures for thin and thick cylinders.
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?

 

[GregLocock]

Barlow is just the hoop stress, that is it assumes the walls are in pure tangential stress. So from first principles pD=2ts, and strictly speaking D is the ID of the tube.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[rb1957]

nah, consult the FBD ... strictly speaking D is the mid thickness diameter

another day in paradise, or is paradise one day closer ?

 

[GregLocock]

Should you not then discount the pressure by ID/D? Otherwise the force is wrong.

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

http://eng-tips.com/market.cfm?

 

[ppppraxxx]

@GregLocock (Automotive) thanks for your reply sir.

@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 ?

 

[csk62]

The easiest way to conceptualize your question is to ask whether the structure should be modelled as thin shell with constant stress throughout the "thickness", or as a solid cylinder with a radially dependent stress.

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.

 

[ppppraxxx]

@csk62 thanks you sir for your reply, I thank all for their support.

 

[rb1957]

"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 ?" ...

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 ?