How do I calculate an amount of thermal expansion for a given shape if you
have Coefficient of Linear Thermal Expansion (7.2x105 in my case)
Thanks,
Alex

you

Depends on the units... is the coeff. in cm/°C or inches/°F or..?

In any case it's a differential thing, if it's 5 inches at 20° and 5.1" at 50°...

Tim

-- In the immortal words of Ned Flanders: "No foot longs!" Website @ http://webpages.charter.net/dawill/tmoranwms

Think of your lump of plastic as expanding by 80 parts per million of a dimension per degree C

Then a dimension of 50 length units if subject to 20 degrees C temp rise will expand by 50 X 80/1000000 X 20 = 0.08 so its dimension grows from 50 to 50.08 length units

You're welcome

Brian W

length x coefficent x temp change

Alex wrote:

Alex wrote:

Alex wrote:

First you must know the units for the coefficient. ie... in/ft/deg. C in/in/deg F. depending upon the material. The material expands exactly as stated in the coefficient. Take the distance from point one to point two and apply the coefficient to it.

John

First you must know the units for the coefficient. ie... in/ft/deg. C in/in/deg F. depending upon the material. The material expands exactly as stated in the coefficient. Take the distance from point one to point two and apply the coefficient to it.

John

Forget the math formulas. Use 1/16" per foot per 100 degrees F. An eight foot
length can expand 1/2" if temperature moves from 0 to 100. If one end is
fastened, it expands to the unfinished end. Don't forget--the width expands
proportionately also. Humidity or moisture also plays a part. As an example,
close tolerance nylon parts will seize if immersed.

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