Thermal expansion on a brass ring 1

[drax]

I have a brass ring, about 5.0" OD ring about .25" thick. I am right to say that during thermal expansion this ring would increase in OD but not in thickness?

thanks in advance,

 

[desertfox]

Hi drax

No if you heat the ring up it will increase in thickness or length too.What is the i.d. of the ring or are you saying the 0.25 is the wall thickness of the cylinder? and therefore the i.d. is 4.5"

regards desertfox

 

[drax]

the ring is actually a cast brass window bezel. It has a complicated shape but the OD is 5.31 and the ID 3.25, the thickness of the narrow part of the casting is .25". Under heat will the entire unit expand in both directions? I'm trying to figure out how to apply the thermal co. formula to get the change in length. Do I use the volume of the casting? and original volume, change in temp.?

thanks
Mark

 

[Scipio]

The ring will increase in thickness, but will usually be negligible depending on your tolerances. Significant change will be in the length (perimeter) of the ring. As the brass grows in that direction, it increases the length of the perimeter, increasing diameter.

Just because I love the origin of the phrase, the reverse is where the old expression 'cold enough to freeze the balls off a brass monkey' comes from. A brass monkey was a brass ring used to hold the bottom row of cannon balls in place in stacks besides the cannons on board old man-of-war ships. When the weather got cold enough, the brass ring contracted enough that the cannon balls were squeezed out of the ring and went rolling across the deck.

 

[EnglishMuffin]

I've always wanted to get the brass ring. When a ring expands, assuming it remains at a uniform temperature, its increase in radius at any radius will be

Delr = e*r*delT

Wher e = coefficient of linear expansion
r = radius
delT = temperature rise

Assuming the material is isotropic and the measurements are made under uniform isothermal conditions, then after a change in temperature, the radial displacement from an arbitrary origin at any point within an object of arbitrary shape is independent of that shape and is proportional to the radial distance from the origin,

 

[Scipio]

EnglishMuffin,

Actually, the radius of a disk would expand with your formula, however the material in a ring will behave differently.

Disregarding thickness for the time being, the thermal effect on the linear length of the perimeter of the ring changes per your formula, if you substitute your radius term for perimeter, so

DelPerim = e * Perimeter * DelT

The resulting change in the length of the perimeter will translate back into a change in radius. In this instance, with an OD of 5 inches and a thickness of 0.25 inches, we've got a nominal diameter of 4.75", and a perimeter of 14.9226". I don't know the coefficient for your particular material, but just for the sake of an example arbitrarily say

e = 5x10[sup]-6[/sup]in/in/°F.

Say now we heat up the ring by 300°F, this increases the perimeter by;

DelPerim = 5x10[sup]-6[/sup]in/in/°F * 14.92in * 300°F
= 0.02238"

Rounding off, the new perimeter is 14.945". The fractionally larger perimeter produces an even smaller increase in nominal diameter to 4.757".

At the same time, the thickness of your ring will also increase, that one's a little more straight forward as it's just multiplying the temperature change and coefficienty by the thickness. In the case of 0.25", the change in thickness would be 0.000375", for a total thickness of 0.250375". Given our new nominal diameter, that means a new inner and outer diameter of 4.5066" and 5.0074".

 

[EnglishMuffin]

Scipio : The material in a ring will NOT behave differently! It maked no difference whether you do it with perimeters or radii (although I don't know why you want to bring perimeters into it). If you use perimeters, the perimeter is always 2*pi times the radius, so you will get exactly the same result. The increase in "thickness" will also be e*thk*delT where thk is the thickness.

 

[Scipio]

EnglishMuffin,
You're absolutely right, I was thinking of the mechanism behind the increase in diameter, but the fact of the matter when I thought of it some more is the difference between the two is simply pi, the ratio is still linear regardless of thickness, diameter, circumference, sides of a triangle or surface area of a sphere.

 

[EnglishMuffin]

Thanks - Even the increase in Radial thickness is e*Radthk*delT. Its the only formula you need !
Thought I was going nuts or something ! (This site will do that to you)!
Cheers

 

[drax]

Thanks for all of your help! I believe you can learn more practically here than school!

Regards,

 

[EnglishMuffin]

You got that right.
But depends on the school - and on who "you" happens to be !

 

[Scipio]

Not to mention whether you're reading/posting before or after your morning coffee ;-)

 

[GregSmith]

All linear dimensions of the ring will increase by the same percentage.

delta L = L * alpha * delta T
where
delta L = change in linear dimension,
L = original value of linear dimension (at the reference temperature),
alpha = coefficient of thermal expansion,
dalta T = change of temperature (relative to reference temperature)

examples of linear dimensions - radius, thickness, diameter, circumference - anything measured with units of 'length'