Thermal expansion on a brass ring
Thermal expansion on a brass ring
(OP)
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,
thanks in advance,





RE: Thermal expansion on a brass ring
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
RE: Thermal expansion on a brass ring
thanks
Mark
RE: Thermal expansion on a brass ring
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.
RE: Thermal expansion on a brass ring
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,
RE: Thermal expansion on a brass ring
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 = 5x10-6in/in/°F.
Say now we heat up the ring by 300°F, this increases the perimeter by;
DelPerim = 5x10-6in/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".
RE: Thermal expansion on a brass ring
RE: Thermal expansion on a brass ring
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.
RE: Thermal expansion on a brass ring
Thought I was going nuts or something ! (This site will do that to you)!
Cheers
RE: Thermal expansion on a brass ring
Regards,
RE: Thermal expansion on a brass ring
But depends on the school - and on who "you" happens to be !
RE: Thermal expansion on a brass ring
RE: Thermal expansion on a brass ring
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'