thermal expansion steel and similar

[Micheletecnospazio]

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?

I was at work. I just came back... Sorry I'm late. .
So the measures are of the order and about... because these are actually orders of greatness... I couldn't give you precise 'cause I didn't have precise... and especially because I didn't want the precise calculation... I didn't want you to solve multiplications or equations but ask for advice and a manual formula if ever. I don't waste time doing math tasks. steel is a material... of course I did not specify what kind of mild alloy steel not tied.. inconel or other etc. always for the reason why I didn't want to make you count.. There are no special processing such as welding laminatures etc. that I can inflate on our speeches
By convention
obvious that the formula of linch is worth a t0 to 0 degrees centigrade if I don't remember badly not to 20... anyway vabo... the thing that is not really clear to me linch of this formula + if I can use it just for a ring.... not being full as you say I calculate the thermal expansion so
1500x0.000012x20= 0.36 mm just as if all 1500 were full,... i.e. as if I had a meter and a half of steel with molecules that by thermal upswing increase their reciprocal distances and therefore lead to a total variation of 0.36....

That's not what's coming back. Anyway, if you say that I try to do as you say... even if I think here molecules that increase their reciprocal distances are not for a meter and a half effective but only for external diament less internal diameter... .

 

[Micheletecnospazio]

http://www.cad3d.it/forum1/showthread.php?t=2582&page=3reading this thread maybe it would be better to treat my ring as a tubular shape duct with a very reduced height... practically in which the inner diameter dicomposes and the outer diameter increases. . .
Whatever. . linch try to use your formula in a raw way on the average diameter and see if I am for n measurements


goods to all

 

[Micheletecnospazio]

I'll take this Internet trip


Note that in relationships expressing linear, superficial and cubic dilations it would be more correct to write δ t instead of t, since the cause of expansion is a temperature variation; in this case it appears t because it was chosen as reference temperature that of 0 oc, for which δ t = t – 0 = t.
if the reference temperature is not 0 oc, the rigorous procedure would imply the calculation of the various lengths referred to this temperature; in practice, since the error you make is negligible, you prefer to use the formulas before deduced in the form:

l = l (1 + λ δ t)

S = so (1 + λ δ t)

v = vo (1 + λ δ t)

 

[sampom]

... practically where the inner diameter dimunisce and the outer diameter increases.. .

It seems to me that you do not respect reality.
that is a "big" ring "very" thin.. also the inner diameter will increase.
moving molecules move towards "more free" directions; the ring can be a long folded bar (so forced) on itself. the greatest "freedom of movement" is all outward.
the expansions towards the inside will be resolved with a "overgulating" of the circle (see that the possibility of expansion is reduced by the "close" of the circular form), but all the diametral dimensions will be increased.
and experience shows it; What do you do to stock/extract the inner ring of large roller bearings for example?

greetings
Marco:smile:

 

[Micheletecnospazio]

It seems to me that you do not respect reality.
that is a "big" ring "very" thin.. also the inner diameter will increase.
moving molecules move towards "more free" directions; the ring can be a long folded bar (so forced) on itself. the greatest "freedom of movement" is all outward.
the expansions towards the inside will be resolved with a "overgulating" of the circle (see that the possibility of expansion is reduced by the "close" of the circular form), but all the diametral dimensions will be increased.
and experience shows it; What do you do to stock/extract the inner ring of large roller bearings for example?

greetings
Marco:smile:

I've said a cabbage. .

 

[linch]

obvious that the formula of linch is worth a t0 to 0 degrees centigrade if I don't remember badly not to 20... anyway vabo... the thing that is not really clear to me linch of this formula + if I can use it just for a ring.... not being full as you say I calculate the thermal expansion so
1500x0.000012x20= 0.36 mm just as if all 1500 were full,... i.e. as if I had a meter and a half of steel with molecules that by thermal upswing increase their reciprocal distances and therefore lead to a total variation of 0.36....

That's not what's coming back. Anyway, if you say that I try to do as you say... even if I think here molecules that increase their reciprocal distances are not for a meter and a half effective but only for external diament less internal diameter... .

I tried to help you, but I see you persevere with your illogical doubts, (the inner diameter that you contract, etc.). I could cite countless examples, but I won't do it because I'm convinced that you have no intention of cooperating, or maybe it's not your strong learning.
Good job
I'm sorry.

 

[sail]

Excuse me, but the correct procedure is not to drop the piece and scale a value = (1+(coefficient thermal expansion* delta t) ?

then at that point one goes to measure internal, external, median diameters and what else at least with numbers to the hand understands.

 

[Micheletecnospazio]

I tried to help you, but I see you persevere with your illogical doubts, (the inner diameter that you contract, etc.). I could cite countless examples, but I won't do it because I'm convinced that you have no intention of cooperating, or maybe it's not your strong learning.
Good job
I'm sorry.

I'm not going to cooperate. ... illogical doubts... is not your strong learning. . .
I don't know
I thanked you 100 times and if you read well I said despite my doubts I will do as you said..
good work for you too