What is the significance of thermal expansion coefficients in material research?

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Thermal expansion coefficients are crucial in material research as they determine how different materials expand and contract with temperature changes, which can lead to stress and potential failure at interfaces. In the case of alumina on stainless steel, the significant difference in their coefficients (17.3 for alumina and 8.1 for stainless steel) suggests that they will experience different rates of expansion, potentially causing stress during heating. Understanding this interaction is essential for predicting material behavior, especially in thin film applications, where the film's stress levels must be compared to the material's failure stress. Resources for calculating thermal stress between dissimilar materials can provide valuable insights into managing these challenges. Overall, grasping the implications of thermal expansion is vital for successful material applications in engineering.
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Hi all,

I am very new to physics and have little to no background in engineering, however I am a biochemistry major whoe somehow wound up in material research lab.

One thing I am having trouble understanding the thermal expansion coefficients, how they relate to each other and which one I should be using. I am going to be putting a layer of alumina on stainless steel using atomic layer deposition. The heat this process is at is 180C. The thermal coefficients that I found for these two are 17.3 and 8.1. I don't really know what this means or how to decide if the difference between the two matters.

Can someone help explain it to me?

Thanks in advance
 
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Not a simple answer - the specifics take some work. Google something like 'calculating thermal stress between dissimilar materials' and take a look. The idea is that the joined materials expand at different rates when temperature changes, so they pull against each other, sometimes causing failure, especially when the temp changes many times in succession.

Generating a model to predict the specific outcome for a particular situation is often an involved procedure.
 
Thanks for your input I'l try starting there and see if I can't find an equation to compare the two.
 
This is a very thin film deposited on a relatively thick substrate ?
 
Nidum said:
This is a very thin film deposited on a relatively thick substrate ?
Sorry to mix units a bit here. It would be somewhere between 5-50nm thick film on a substrate that is about 0.015" thick.
 
Contraction for combination will be almost the same as that of the substrate alone .

Given the low temperatures involved I doubt whether there would be much out of plane distortion .

Can you see how to go from here to getting the actual stress level in the film and comparing it with the failure stress for the material ??
 
Nidum said:
Contraction for combination will be almost the same as that of the substrate alone .

Given the low temperatures involved I doubt whether there would be much out of plane distortion .

Can you see how to go from here to getting the actual stress level in the film and comparing it with the failure stress for the material ??

Not really no. Sorry, this isn't really my strong suit and I am trying to read and learn but I just feel way out of my depth on this project.
 
Nidum said:
Contraction for combination will be almost the same as that of the substrate alone .

Given the low temperatures involved I doubt whether there would be much out of plane distortion .

Can you see how to go from here to getting the actual stress level in the film and comparing it with the failure stress for the material ??

Is there any place in particular that I should go to look at how to get these numbers? I am trying to learn so I would prefer some place that gives steps and reasons. I am fine just plugging in numbers and getting answers but I would still like to learn why and how I am getting the results.
 
The below might help. They are not examples of joined dissimilar materials, but they might help you understand the basic material science at work.

Not sure how basic a discussion is helpful for you - below is a very basic discussion.

If you were to glue a sheet of aluminum to a sheet of stainless steel and you were to take the joined sheets and heat them over a stove burner, they would expand at different rates, and this would cause stresses at the glue-to-steel interface and the glue-to-aluminum interface. These stresses would result in strain, deformation of the steel, the glue, and the aluminum. If the stresses are great enough to cause any of the materials to go beyond their breaking points, then something in the steel / glue / aluminum stack will fail (break). Each of the three materials has a youngs modulus (basically a spring contant) to describe how much it deforms in response to stress, and a thermal expansion co-efficient to describe how much it expands when heated. You are calculating the expansion of the different materials and the resulting stresses that come from the materials not being able to move freely (they are becoming like compressed springs). The material stack can only expand as much as the material in the stack that is least sensitive to temperature. The other materials in the stack want to expand more, but they can't because they are joined to the least expansive material.

Hope that helps at least a little.

https://www.comsol.com/multiphysics/thermal-expansion-and-thermal-stresses

http://www.usc.edu/dept-00/dept/architecture/mbs/struct/Arch213A/213A-lectures/07-Thermal-stress.pdf
 
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  • #10
Grinkle said:
The below might help. They are not examples of joined dissimilar materials, but they might help you understand the basic material science at work.

Not sure how basic a discussion is helpful for you - below is a very basic discussion.

If you were to glue a sheet of aluminum to a sheet of stainless steel and you were to take the joined sheets and heat them over a stove burner, they would expand at different rates, and this would cause stresses at the glue-to-steel interface and the glue-to-aluminum interface. These stresses would result in strain, deformation of the steel, the glue, and the aluminum. If the stresses are great enough to cause any of the materials to go beyond their breaking points, then something in the steel / glue / aluminum stack will fail (break). Each of the three materials has a youngs modulus (basically a spring contant) to describe how much it deforms in response to stress, and a thermal expansion co-efficient to describe how much it expands when heated. You are calculating the expansion of the different materials and the resulting stresses that come from the materials not being able to move freely (they are becoming like compressed springs). The material stack can only expand as much as the material in the stack that is least sensitive to temperature. The other materials in the stack want to expand more, but they can't because they are joined to the least expansive material.

Hope that helps at least a little.

https://www.comsol.com/multiphysics/thermal-expansion-and-thermal-stresses

http://www.usc.edu/dept-00/dept/architecture/mbs/struct/Arch213A/213A-lectures/07-Thermal-stress.pdf

Hey Grinkle, thanks so much for the run down on what it is that I am looking at. Putting it that way helped me equate it to enzymes which I am much more comfortable with. I appreciate you helping me.
 
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