# Q about Thermal expansion of steel tube

• posted

Dear All,

I'm a student of physics and I have two Q.

First, When the temperature of steel tube is changed, How can I calculate the change of outer diameter and inner diameter of tube? the oute diameter is 40mm, the inner diameter is 30 mm, the length of tube is 100mm

Next, When the temperature of steel tube is changed, How can I calculate the change of length of tube? the oute diameter is 40mm, the inner diameter is 30 mm, the length of tube is 300mm

Best Regards, Aref, snipped-for-privacy@yahoo.com

• posted

You need to have the coefficient of linear expansion for the material. and the appropriate "length"

delta_L = alpha * delta_T * L

alpha = coeff. linear expansion delta_T = change in temp L = length

all in a consistent set of units, of course. Determining alpha is a typical high school or college introductory physics experiment.

-- Ed Ruf Lifetime AMA# 344007 ( snipped-for-privacy@EdwardG.Ruf.com)

• posted

Change in circumference due to heating = pi*d*a*delta_T a=as above (alpha) d=original diameter delta_T=as above

You can see that it is tha same formula just for axial expansion (contraction).

• posted

Both the previous replies assume a large diameter and very thin wall and are an approximation. In your case the wall thickness is over 16% of the hole diameter, so these approaches would have a considerable error. Fortuantely the correct answer is also fairly simple.

Actually, the OD of the tube will expand volumetrically the same as if it were a solid cylinder with no hole. The hole will expand in volume the same as if it were a solid cylinder of the same material.

There are two types of expansion to be consdered: volumetric expansion and linear expansion. The change in volume is simply

deltaV = B * V0 * deltaT

where B is the coefficient of volumetric expansion (.000036/C for steel), V0 is the inital volume and deltaT is the change in temperature in C. This will tell you how much the volumes of your cylinders change.

There will also be a change in length of the cylinders

deltaL = A * L0 * deltaT

Where A is the coefficient of "linear" expansion (.000012/C for steel)* and L0 is the initial length. From this you can find out how much the cylinder will elongate.

If you know the new length and you know the new volumes, it is simple to figure out what the new diameters will be.

Now, go do you homework and stop screwing around on the Internet!

Don Kansas City

*note that A is always exactly one-third of B for a material.
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Don,

Won't the inside diameter shrink? The material expands with heat, thus causing the inside diameter to get smaller.

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Nope. This is freshman physics.

Don Kansas City

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No, just as any hole gets larger. This is why when you have a fitted joint you can't get apart heating it can help.

-- Ed Ruf Lifetime AMA# 344007 ( snipped-for-privacy@EdwardG.Ruf.com)

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"Don A. Gilmore" wrote in news:_Glkf.3429\$Dk.2016 @tornado.rdc-kc.rr.com:

True, but the fundamental reason is that the 3d matrix of atoms are all getting pushed further apart in all directions. If you think about it that way, this mysterious behaviour is much simplified.

Cheers

Greg

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A good explanation for 1-d, 2-d and 3-d thermal expansion is found here:

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Let us suppose as a concrete example, that the temperature change creates a change in linear dimensions of X 1.01 - the linear thermal expansion as it's sometimes called.

so the length becomes 100mm X 1.01 = 101 mm the circumference of pi X 35mm X 1.01 = pi X 35.35 The wall thickness of 5 mm becomes 5mm X 1.01 = 5.05 mm This leads to a new OD of 35.35 + 5.05 mm = 40.40 mm a new ID of 35.35 - 5.05 mm = 30.30 mm

[You'll notice that is just the same as expanding the inner and outer diameter by the given factor - a method that sometimes seems strange to people who may think that we are expanding the inner diameter of AIR by the steel's expansion factor.]

As a cross check, we could find the new volume of the tube: and discover that is is very nearly the original volume times a volumetric thermal expansion factor of 1.03

Brian Whatcott Altus OK

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