What is the stress in the steel ring?

[JustinLiang]

Homework Statement

I need help specifically with b.

The inner diameter of a steel ring is 2.0000 cm, and the diameter of an aluminum
disk is 2.0100 cm. Both are at 430 C. At what common temperature will the disk fit
precisely into the hole in the steel ring? b) If after the aluminum disk is fitted precisely
into the hole the two metals are the temperature is changed to 200 C, what is the stress
in the steel ring?

Homework Equations

deltaL=a*Lo*deltaT
deltaL=F*Lo/Y*A

The Attempt at a Solution

I have attached my attempt. I would like to know if it makes sense to have the change in length of the steel=change in length of the aluminum. Thus we can solve for the stress in between.

What my friend did was have the change in length of the steel=negative change in length of the aluminum. Meaning that he would have two negative signs, one for each of the change in lengths in my first equation. I told him this is wrong because by doing so, he is assuming the the overall change in length of the system is zero. This is not the case because there are no rigid supports preventing the rings from expanding. Therefore, the ring and disk will be bigger.

 

[JustinLiang]

bump, need some help ASAP!

 

[JustinLiang]

Wow no one knows how to do this?

By the way, I made a mistake, the answer should be negative since I mixed up the a for steel with the a for aluminum.

 

[JustinLiang]

helllo

 

[JustinLiang]

what theeeeeeeeeeeeeeeeeeeeeeeeeeeeeee hhhhhhhhhhhhhhhhhhhhhhhhhek :O

 

[JustinLiang]

last bump... disappointing.

 

[omoplata]

Bumping the thread might have worked against you. When looking for interesting problems to solve, I usually look for posts with 0 replies. I guess the others do the same. If there is at least one reply to a post that means the OP is being helped.

change in length of the steel=change in length of the aluminum

I agree with this, as long as you are talking about the TOTAL change in length of steel. The logic, as you've written down in the attachment, is that the ring and the disk are in contact when the temperature change starts, and they are in contact when the temperature change finishes.

I also agree with these.

Total change of length of the steel ring = change of length due to expansion of the ring + change of length due to stress on the ring

Total change of length of the aluminium disk = change of length due to expansion of the disk + change of length due to stress of the disk

Or using your notation,
\Delta L_{\mathrm{s-total}} = \Delta L_{\mathrm{s-thermal}} + \Delta L_{\mathrm{s-tensile}}
\Delta L_{\mathrm{A-total}} = \Delta L_{\mathrm{A-thermal}} + \Delta L_{\mathrm{A-tensile}}

But of the four terms \Delta L_{\mathrm{s-thermal}}, \Delta L_{\mathrm{s-tensile}}, \Delta L_{\mathrm{A-thermal}} and \Delta L_{\mathrm{A-tensile}}, some might be positive and some might be negative.

If you consider \Delta L_{\mathrm{s-thermal}} and \Delta L_{\mathrm{A-thermal}}, it's easy to see whether they are positive or negative since the temperature change is an increase.

To find whether \Delta L_{\mathrm{s-tensile}} and \Delta L_{\mathrm{A-tensile}} are positive or negative, consider an already expanded ring and disk, which had the same outer and inner diameters respectively at a lower temperature. So the final outer diameter of the disk will be larger than the final inner diameter of the ring. To insert the ring to the disk, the ring will have to be stretched forcibly and the disk will have to be compressed forcibly. So which one of the terms \Delta L_{\mathrm{s-tensile}} and \Delta L_{\mathrm{A-tensile}} is positive and which one is negative? And how would that change the answer that you've attached?