Expansion of Holes Question 3

[mecheng19777]

I have a 39 ft piece of steel, which will expand/contract 5/16" over an operating temperature of -20C (-4F) to 40C (104F),
using the steel coefficient of expansion.

In that piece of steel, I have 1 3/16" (1.1875") holes, and I'd like to know how much the holes will expand/contract.
Is it only a portion of the total expansion. For example 1.1875//468 = 0.25% of 5/16" = 0.08"

Any help would be appreciated.

 

[RVAmeche]

I would draw it in cad (2D, simple), and scale it by the appropriate factor. That would give you your answer assuming one edge/point of the plate is fixed

 

[JJPellin]

0.25 percent, not 25 percent. 0.0008" not 0.08".

Johnny Pellin

 

[dvd]

The coefficient of linear expansion indicates just that - the linear dimensions change according to the length x coefficient x delta-T. So your hole diameter will change in accordance.

 

[handleman]

Less than you can measure.

 

[dhengr]

Mecheng19777:
What are you really trying to accomplish or compare? The change in the center to center spacing of the holes or what? Don’t forget, you are probably not measuring these dimensions with a micrometer/caliper which measure to .001”. Also, the 39’ piece of steel was probably fabricated at about 60-70°F, so the length may grow by the expansion caused by a 44 or 34°F change, or shrink by the contraction caused by a 64 to 74°F change. Then, you have some tolerances in your length and hole dia. tolerances, and the mating materials will also shrink or grow due to the same temperature changes, whatever their coef. of expansion may be. I noticed the same error that Johnny pointed out, but what significance does “1.1875//468” have, in any case? You have to define your problem, and what you are trying to accomplish better. Usually, our tolerances and things like oversized bolt holes, etc. take care of these things. I have seen (heard of) cases where they went to great lengths to manufacture and assemble things at about the same temps. because of special tolerance conditions. But that’s not very common.

 

[mecheng19777]

Yes my mistake, 0.0008".

I am trying to make sure I have enough clearance in my bolt holes. The bolt is a track bolt which has a oval locking head. The bolt is 1" diameter; using a 1 3/16" hole,
but the oval slot has only 1/16" clearance.

So I would like to know how the bolt hole expands/contracts. Thank you

 

[rb1957]

I think it's more complicated.

It's easy to calc the linear change in a linear dimension (like 5/16" in you 39' plate).
But the change in a hole diameter won't be 5/16/468, at least I don't think it will be.
Eg, does heating a plate enlarge the holes (as you'd think) or does the surrounding material expand (causing the hole to contract) ?

I think you need text books on thermal analysis, or thermal FEA.

another day in paradise, or is paradise one day closer ?

 

[GregLocock]

Heating a plate with a hole in it moves all the atoms further apart, so the hole grows in diameter in proportion to all the other dimensions.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[dgallup]

Depends a bit on what the bolt is make from and if it's at the same temperature as the steel part. If the materials and temps are similar you will have clearance at all temps.

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The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.

 

[Compositepro]

***groan*** I am amazed at how many engineers get this wrong. Just as Greg said, the hole will expand with temperature increase, exactly the same as the steel that was removed from the hole would expand.

 

[mecheng19777]

I took RVAmeche (Mechanical) advice and drew it in cad, scaled it by an appropriate factor to get 5/16" length and the hole expanded by 0.0008". Basically, by a blond one.

My theory gave the same result:

1.1875 / (39*12) = 0.0025%
0.0025% x 5/16 = 0.0008"

Thanks all.

 

[rb1957]

if I assume CTE = 7E-6, then for 39' = 468in and +36degF gives 0.118in (the change from RT).
for 108degF (the total range given) = 0.354in … near enough to your 5/16"

the point is 5/16" isn't the right dimension to use. From assembly (presumably at RT) the plate heats up 36deg F and cools 72degF (assuming RT = 20degC).

but the point is also, as noted above, if the bolt fits at RT then there should be no problem with fitment … assuming the bolt material and the plate are similar.
Actually, the problem is different to what you're thinking about … the change in thickness (from thermal effects) impacts the bolt load.

another day in paradise, or is paradise one day closer ?