thermal expansion steel and similar

[Micheletecnospazio]

Hello everyone.. .
I would like your opinion on a matter in general very simple, in my case a little more complex
calculate the thermal expansion of a steel.... equipped with tables of linear expansion coefficients superf and volumetric and formulas
.δs = σs1δt with δs expansion, σ coeffic superfic, s1 initial sup and δt temperature difference
δl = σl1δt the respective for the linear
δv = kv1δt for cubica

I wondered... for my case what is the best to use?
I have to measure the expansion of a steel cable ring in which the height is much less preponderant than the diameter and diameter of the quarry

I can measure the volumetrics. then the linear as I use it? as if the ring was stretched out and therefore I mix the difference of circumferences but so I do not maintain myself with excessive safety?


Maybe someone who came out of some examination of railway mechanics or bridge joints knows more....

Thank you in advance for the solutions that will arrive:smile:

 

[PierArg]

Hi.

I would like to ask that I am the least suitable for answering you for this question, I hope that someone can deepen.

I, personally, would take into account the variation of volume and the whole (or at least I would transcur the variation of circumference) on the increase of thickness.
I believe, and I would not say a boiata, that the material finds much more ease in expanding in the direction of the thickness in both directions (increasing it) and therefore the circumference will result greater cause increase thickness.

 

[Micheletecnospazio]

Hi.

I, personally, would take into account the variation of volume and the whole (or at least I would transcur the variation of circumference) on the increase of thickness.
I believe, and I would not say a boiata, that the material finds much more ease in expanding in the direction of the thickness in both directions (increasing it) and therefore the circumference will result greater cause increase thickness.

Hi pier arg... Thanks for the answer.. .
So you tell me to calculate the volumetric... and charge it all perhaps with a coefficient slightly lower than one to the variation of thickness?
perhaps considering that the outer diameter is increased more than the inner one?
actually your reasoning does not bend... quite qualitative. . .
found that it changes we say of a dm cube... to what extent should I transform this deformation into variations of internal and external diameter?

 

[PierArg]

from the change of volume, assuming unchanged the circumference, I calculate the new thickness.

According to me, always to be taken with extreme caution, it increases out as it decreases the diameter of the whole ring. However, if the latter is very large (and I believe your case), you could idealize it as a straight. Therefore the thickness increases in both directions in the same way.

 

[Fulvio Romano]

Obviously the three formulas must give you the same result*, so the choice depends on the convenience.
If the ring is approximate to a closed wire on itself, I would use the first, calculate the new length of the circumference and therefore the new diameter.

(*) ok, less than the square term that is unclear

 

[Micheletecnospazio]

Obviously the three formulas must give you the same result*, so the choice depends on the convenience.
If the ring is approximate to a closed wire on itself, I would use the first, calculate the new length of the circumference and therefore the new diameter.

(*) ok, less than the square term that is unclear

Hello fulvio
So at the bottom you and pier arg recommend me the lagic of the railway track
in which I take the main measure as the one that takes all the deformation
pier arg let me get the diameter from the volume variation ... you instead proponi directly linearly. . verifying so much that the results are similar

I hope when I come to work to check these conditions with zeiss lighthouse... I do a couple of experimental tests and compare theoretical approx.


thanks to both

I still hope that some good civil ing has some empirical formula for no more doubt


hi pier arg ciao fulvio

 

[linch]

empirical formula:
diam. x 0.000012 x delta t°c.
applied in the three axes.
Where's the problem?

 

[Fulvio Romano]

Yes, but I didn't understand what the problem is.
Those "three" formulas are actually one.
only that if a dimension is preponderant compared to the others, then it can "simplify". in the development of the formula, from a linear measure to a square and cubic, you get out of squares and cubes of a binomium. Spend the top grade term and find the simplified formulas.

in fact l, s and k are tied between them (obviously with factors 2 and 3), they are not and can not be independent.

What do you care? Are you interested in how much the radius increases, or are you interested in moving any point of volume? If you use that volume, it's okay anyway. It is enough to take into account that thermal expansion is a kind of transformation (here, we are in a cad forum, don't ask me what it is!) and from the volume you can get the moves.

 

[Micheletecnospazio]

Yes, but I didn't understand what the problem is.
Those "three" formulas are actually one.
only that if a dimension is preponderant compared to the others, then it can "simplify". in the development of the formula, from a linear measure to a square and cubic, you get out of squares and cubes of a binomium. Spend the top grade term and find the simplified formulas.

in fact l, s and k are tied between them (obviously with factors 2 and 3), they are not and can not be independent.

What do you care? Are you interested in how much the radius increases, or are you interested in moving any point of volume? If you use that volume, it's okay anyway. It is enough to take into account that thermal expansion is a kind of transformation (here, we are in a cad forum, don't ask me what it is!) and from the volume you can get the moves.

It is clear that the three formulas are equal... and that the volumetric is the most exhausting. . It is also true that the linear takes into account the variation of only one value (one size ) that volumetric takes into account the variation of all three dimensions... Now if I consider that volumetric (to be more precise) can I obtain the variations of the measurement of the "new" diameter directly?

the core of the problem is this... if for a bar I know that the preponderant measure is the length.. for a ring shall I consider the measure of external circumference and internal as main? (as if it were a bar bent on itself? )
or for greater precision there is an empirical formula that minimizes the approximations for rings and tells me:

look at the linear deformation of the greater diameter is tot
that of the smaller diameter is tot
or of the volumetric deformation I know that the diameter is deformed with a coefficient tot according to the other dimensions (that is, the volumetric deformation with an excellent approximation both for the tot per cent to be detected in the diameters)
In short, I wanted to know if there was any empirical formula from manual
because it is I'm sorry. (of course I can be wrong for this I ask you) as linch says

empirical formula:
diam. x 0.000012 x delta t°c.
applied in the three axes.
Where's the problem?
It is not exactly correct at least because I do not speak of a cylinder but of a ring... so not of a completely full volume but empty in the center... so I can't use the formula in the three axes the same way... the height let it lose... the thickness I take the thickness as an external radius less internal and as a measure along the third axis I take the average circumference as if it were stretched out and from this then I echo the new circumference then the new diameters? ? ?I thank you for the answers that continue to clarify the problem

 

[linch]

It is clear that the three formulas are equal... and that the volumetric is the most exhausting. . It is also true that the linear takes into account the variation of only one value (one size ) that volumetric takes into account the variation of all three dimensions... Now if I consider that volumetric (to be more precise) can I obtain the variations of the measurement of the "new" diameter directly?

the core of the problem is this... if for a bar I know that the preponderant measure is the length.. for a ring shall I consider the measure of external circumference and internal as main? (as if it were a bar bent on itself? )
or for greater precision there is an empirical formula that minimizes the approximations for rings and tells me:

look at the linear deformation of the greater diameter is tot
that of the smaller diameter is tot
or of the volumetric deformation I know that the diameter is deformed with a coefficient tot according to the other dimensions (that is, the volumetric deformation with an excellent approximation both for the tot per cent to be detected in the diameters)
In short, I wanted to know if there was any empirical formula from manual
because it is I'm sorry. (of course I can be wrong for this I ask you) as linch says

empirical formula:
diam. x 0.000012 x delta t°c.
applied in the three axes.
Where's the problem?
It is not exactly correct at least because I do not speak of a cylinder but of a ring... so not of a completely full volume but empty in the center... so I can't use the formula in the three axes the same way... the height let it lose... the thickness I take the thickness as an external radius less internal and as a measure along the third axis I take the average circumference as if it were stretched out and from this then I echo the new circumference then the new diameters? ? ?I thank you for the answers that continue to clarify the problem

feel, be patient, place a design of the element to calculate and the delta t°c applied, tell us what thermal/dimensional variations you want to know, will be indicated with extreme precision.

 

[Fulvio Romano]

It is clear that the three formulas are equal... and that the volumetric is the most exhausting. . It is also true that the linear takes into account the variation of only one value (one size ) that volumetric takes into account the variation of all three dimensions... Now if I consider that volumetric (to be more precise) can I obtain the variations of the measurement of the "new" diameter directly?

depends on whether you can get the size of a bull knowing its volume.

I'm sorry, but I feel like you're completely off the road. in my previous post you find the complete solution to the problem:

- precise solution: use the volume formula. calculate the new volume. know the original shape, you know that the transformation is affine (read: "scale" function of the cad) and from the volume you recalculate the new dimensions

- Approximate solution: I approach the ring with a circumference. of course not internal and external, but average. If you have to consider internal and external means that the approximate formula is not applicable. with the calculation formula the new length of the circumference.

and then, the "empiric" formula of linch does not seem so "empiric". It is simply explicit the expansion coefficient, and I would like to understand how to do and express a dilation coefficient without knowing:
- material
- allotropic state (you know that metals have different coefficients for each crystalline state? )
- temperature (you know that coefficients are temperature function? )

 

[linch]

depends on whether you can get the size of a bull knowing its volume.

I'm sorry, but I feel like you're completely off the road. in my previous post you find the complete solution to the problem:

- precise solution: use the volume formula. calculate the new volume. know the original shape, you know that the transformation is affine (read: "scale" function of the cad) and from the volume you recalculate the new dimensions

- Approximate solution: I approach the ring with a circumference. of course not internal and external, but average. If you have to consider internal and external means that the approximate formula is not applicable. with the calculation formula the new length of the circumference.

and then, the "empiric" formula of linch does not seem so "empiric". It is simply explicit the expansion coefficient, and I would like to understand how to do and express a dilation coefficient without knowing:
- material
- allotropic state (you know that metals have different coefficients for each crystalline state? )
- temperature (you know that coefficients are temperature function? )

I remind you that the initial question was:
"steel and similar thermal expansion"
So my formula is compatible or not?

 

[Fulvio Romano]

I remind you that the initial question was:
"steel and similar thermal expansion"
So my formula is compatible or not?

"steel" is not a material.
We have not even understood whether it is isotropic or not, and this, if we know what steel we are talking about, depends on the type of processing.
we have not even understood what precision we are looking for, so much to know if it is the case of going for the thin (specifying the type of steel) or if it is not worth it.

However, I did not express myself on the compatibility of the formula, but on whether or not it is empirical. "empirical" means "invented to marry experimental evidence, but without anything analytical behind".

 

[linch]

"steel" is not a material.
We have not even understood whether it is isotropic or not, and this, if we know what steel we are talking about, depends on the type of processing.
we have not even understood what precision we are looking for, so much to know if it is the case of going for the thin (specifying the type of steel) or if it is not worth it.

However, I did not express myself on the compatibility of the formula, but on whether or not it is empirical. "empirical" means "invented to marry experimental evidence, but without anything analytical behind".

dear fulvio, that formula has been used for hot calettare thousands of components (cuscinets, gears, flanges, etc.) and even the wheels on the axles of the railway carts of some metropolitan.
I'll let you imagine how many times his truth has been verified to the hundredth of a millimeter.
isotropic? No material can be, even if the steel is very close.
type of processing? I don't understand the connection between these and the calculated expansions.
Hi.

 

[Fulvio Romano]

dear fulvio, that formula has been used for hot calettare thousands of components (cuscinets, gears, flanges, etc.) and even the wheels on the axles of the railway carts of some metropolitan.
I'll let you imagine how many times his truth has been verified to the hundredth of a millimeter.

So what? is it empirical?
right, it'll be right. Who says otherwise?

isotropic? No material can be, even if the steel is very close.
type of processing? I don't understand the connection between these and the calculated expansions.
Hi.

a rolled bar can be strongly anisotropa. a welded piece can " distort" for thermal expansion.
but we're going ot.

 

[Micheletecnospazio]

feel, be patient, place a design of the element to calculate and the delta t°c applied, tell us what thermal/dimensional variations you want to know, will be indicated with extreme precision.

diameters are of the order of 1500mm
order thickness of 150mm (so outer diameter less internal=30)
We say height of order of 150mm also it...
Thermal excursion about 10-20 degrees and material we say a generic steel
we say therefore 0.000011 as expansion coefficient for example. .

 

[Fulvio Romano]

But you got it in 3d?
Excuse me, but if the steps (1 + l * dt) don't get the dilated piece?
(use linear expansion in this case, because the scaling refers to distances)

 

[linch]

diameters are of the order of 1500mm
order thickness of 150mm (so outer diameter less internal=30)
We say height of order of 150mm also it...
Thermal excursion about 10-20 degrees and material we say a generic steel
we say therefore 0.000011 as expansion coefficient for example. .

Well, "order odds", excursion t "about", you are not very precise in the description.
still apply the formula, and you will have the linear expansion you look for.
Keep in mind that the thermal excursion to be considered is equal to the actual temperature of the element minus the ambient temperature established in the standard in 20°c.
example:temperature detected on element 40°c, delta t°c =20
dtl= 1500x0.000012x20= 0.36 mm.

 

[Fulvio Romano]

Keep in mind that the thermal excursion to be considered is equal to the actual temperature of the element minus the ambient temperature established in the standard in 20°c.

linch, I don't want inisister, you probably used these formulas more than me.
But if I want thermal expansion between two temperatures, I have to use those two temperatures. If I have no idea what the starting temperature is, then I can try to refer to the 20°c standard.

If I have to calculate the expansion of that piece, employed in Greenland, with this approximation I would look great. You think?

 

[linch]

linch, I don't want inisister, you probably used these formulas more than me.
But if I want thermal expansion between two temperatures, I have to use those two temperatures. If I have no idea what the starting temperature is, then I can try to refer to the 20°c standard.

If I have to calculate the expansion of that piece, employed in Greenland, with this approximation I would look great. You think?

fulvio excuse me, you are quite prepared to know that there are no formulas valid exclusively in some parts of the globe, it takes as standard that temperature because generally certain works are performed in the workshop.I remind you that the metrological laboratories are kept constantly at 20°c.
If you work on the pack in Greenland, you should obviously take into account the ambient temperature that will already give you different parameters on the diameters, but the formula will remain still valid, the linear expansion will be proportional to the delta t obtained.
But the direct interested party is gone?