# Anyone know about stress tests?

sonhouse
Science 19 May '15 12:19
1. sonhouse
Fast and Curious
19 May '15 12:19
We deposit SiO2 of X depth followed by an SIC of Y depth and one way of measuring the stress is to put in a test strip of thin metal, 0.001 inch (0.03mm) thick and 0.6000 inch wide (15.24 mm) and measure the subsequent bowing with micrometer. In this case, the bowing came out at 0.011 inch and 0.5895 inch wide, giving a radius of curvature of just under 4 inches. Anyone know how to calculate stress from those numbers?
2. 19 May '15 12:47
I'm pretty sure that you need the force applied as well.

Otherwise you have too many unknowns.
3. sonhouse
Fast and Curious
19 May '15 13:151 edit
I'm pretty sure that you need the force applied as well.

Otherwise you have too many unknowns.
This measurement was before Vs after a deposition run on a plasma deposition machine. Not much way to measure the force, the actual vacuum level was about 8 microns and the temperature of the gas about 200 degrees C. That would be only a whisper of a force on the metal. The strip starts out flat and ends up with that radius of curvature. The curvature starts out at zero and ends up with that number, also, a 9 degree arc in the bend after the deposition is finished, about 3/4 micron thick layer total.
4. DeepThought
19 May '15 17:43
Originally posted by sonhouse
We deposit SiO2 of X depth followed by an SIC of Y depth and one way of measuring the stress is to put in a test strip of thin metal, 0.001 inch (0.03mm) thick and 0.6000 inch wide (15.24 mm) and measure the subsequent bowing with micrometer. In this case, the bowing came out at 0.011 inch and 0.5895 inch wide, giving a radius of curvature of just under 4 inches. Anyone know how to calculate stress from those numbers?
What that measures is the strain (deformation over original length). To get the stress you need the Young's modulus, and the relevant formula is:

stress = Y * strain.

http://en.wikipedia.org/wiki/Young's_modulus
5. 19 May '15 21:38
Originally posted by DeepThought
What that measures is the strain (deformation over original length). To get the stress you need the Young's modulus, and the relevant formula is:

stress = Y * strain.

http://en.wikipedia.org/wiki/Young's_modulus
Calculating stress requires knowing the force, or you have to know the Young's Modulus of the materiel in question.

Neither of which are given.

I'm not even sure about strain, as I don't know if you can tell from the information given whether the materiel
stretched as well as deformed laterally.
6. DeepThought
20 May '15 00:10
Calculating stress requires knowing the force, or you have to know the Young's Modulus of the materiel in question.

Neither of which are given.

I'm not even sure about strain, as I don't know if you can tell from the information given whether the materiel
stretched as well as deformed laterally.
I made the natural assumption that they know what the material is and hence the Young's modulus. They should be able to get the strain correctly from their measurements. On the other hand the stress is just the applied force over the relevant area, so it should be calculable from the information they have.
7. 20 May '15 00:49
Originally posted by DeepThought
I made the natural assumption that they know what the material is and hence the Young's modulus. They should be able to get the strain correctly from their measurements. On the other hand the stress is just the applied force over the relevant area, so it should be calculable from the information they have.
Ah, I was mistakenly thinking about the plasma deposited laminate as the materiel being measured.

The thin strip of metal should have a known Young's Modulus [YM].

I know stress is force over area. Which is why I asked if they knew the force being applied.

If they know the strain and YM then they can calculate stress.

However, I still have reservations about calculating the strain with the given information.
As if the strip has stretched or compressed in addition to being bent, then it will have been
subjected to different stress than if it has simply been bent.
I can't tell from the description what it is they are actually doing, so I can't tell how relevant that is.
8. sonhouse
Fast and Curious
20 May '15 14:222 edits
Ah, I was mistakenly thinking about the plasma deposited laminate as the materiel being measured.

The thin strip of metal should have a known Young's Modulus [YM].

I know stress is force over area. Which is why I asked if they knew the force being applied.

If they know the strain and YM then they can calculate stress.

However, I still have re ...[text shortened]... l from the description what it is they are actually doing, so I can't tell how relevant that is.
As far as I can determine, the dimensions of the thin metal strip is unchanged before and after deposition of the SiO2/SIC layers. If the dimensions changed it is below the ability of my calipers to measure (1/10,000 of an inch)
I wonder if you can't simulate what the force is by just squeezing the unmodified metal strip and see how much it deforms with X amount of force?
9. 20 May '15 14:34
Originally posted by sonhouse
As far as I can determine, the dimensions of the thin metal strip is unchanged before and after deposition of the SiO2/SIC layers. If the dimensions changed it is below the ability of my calipers to measure (1/10,000 of an inch)
I wonder if you can't simulate what the force is by just squeezing the unmodified metal strip and see how much it deforms with X amount of force?
At what temperature did this deformation occur?

Because the YM of a materiel is going to vary with temperature.
10. sonhouse
Fast and Curious
20 May '15 18:29
At what temperature did this deformation occur?

Because the YM of a materiel is going to vary with temperature.
The temperature inside the vacuum plasma chamber is under 200 C.
11. joe shmo
Strange Egg
21 May '15 01:461 edit
Originally posted by sonhouse
As far as I can determine, the dimensions of the thin metal strip is unchanged before and after deposition of the SiO2/SIC layers. If the dimensions changed it is below the ability of my calipers to measure (1/10,000 of an inch)
I wonder if you can't simulate what the force is by just squeezing the unmodified metal strip and see how much it deforms with X amount of force?
Generally if the deflections are small relative to the member, and its cross section is constant (the thin metal strip in your case) The deflection is typically modeled by:

D = F*L/(A*E)

D = Deflection
F = Axial force applied to the member
L = Original length of the member
A = Cross-sectional Area
E = Youngs Modulous ( A.K.A Modulus of Elasticity) @ the appropriate temperature

So the normal stress would be given by:

Stress = F/A = D*E/L = ε*E

Also, their were induced lateral strains on the member ( albeit relatively small in comparison) with the axial strain.

If you want to know what those are you must use Poisson's Ratio( another material property). The strain in each of the directions orthogonal to the load are:

ε_y = -v*F/(A*E)

ε_z = -v*F/(A*E)

As you can see the signs in front are negative, meaning a positive strain in the axial direction creates a negative strain in the y & z directions of the member.
12. joe shmo
Strange Egg
21 May '15 02:352 edits
Originally posted by sonhouse
As far as I can determine, the dimensions of the thin metal strip is unchanged before and after deposition of the SiO2/SIC layers. If the dimensions changed it is below the ability of my calipers to measure (1/10,000 of an inch)
I wonder if you can't simulate what the force is by just squeezing the unmodified metal strip and see how much it deforms with X amount of force?
I should read first some more of the post first... It seems that your loading isn't uni-axial tension . From what I can surmise, it is similar to the loading conditions of column buckling that would model your problem.

However, I don't think there is an analytic function ( at least one that is practical for something like this) that can determine what you ask, from what you've given. Everything I've ever seen with column buckling is aimed at the determination of the critical stress at which a column will buckle, not what stress is in the column at any given radius of curvature. What you are after "may" be derivable, but it wouldn't be easy (and that is an understatement).

Theoretically ( with something this small, and potentially curved) you would have to derive the "equation of the elastic curve" without the simplifying assumptions that make it manageable in an analytic sense. I believe you would need to solve the nonlinear DE following equation as is:

d²y/dx²/( 1 + (dy/dx)² ) ^(3/2) = M(x)/(E*I)

Have your resident Physicist get started as soon as possible.
13. sonhouse
Fast and Curious
21 May '15 13:10
Originally posted by joe shmo
I should read first some more of the post first... It seems that your loading isn't uni-axial tension . From what I can surmise, it is similar to the loading conditions of column buckling that would model your problem.

However, I don't think there is an analytic function ( at least one that is practical for something like this) that can determine what y ...[text shortened]... (dy/dx)² ) ^(3/2) = M(x)/(E*I)

Have your resident Physicist get started as soon as possible.
Sounds difficult for sure. It seems to me a more practical approach would be to just take the unaltered metal strip and make a chart of bending, say, 100 milligrams produces radius A, 200 milligrams produces curve B and so forth, where the 100 mg is an inward pinch of the edges of the metal.

Wouldn't that result in the same thing as a theoretical approach?
14. joe shmo
Strange Egg
21 May '15 20:401 edit
Originally posted by sonhouse
Sounds difficult for sure. It seems to me a more practical approach would be to just take the unaltered metal strip and make a chart of bending, say, 100 milligrams produces radius A, 200 milligrams produces curve B and so forth, where the 100 mg is an inward pinch of the edges of the metal.

Wouldn't that result in the same thing as a theoretical approach?
I suppose...if you can precisely mimic/control the loading caused by the deposition. Btw, are there any links that could give a hint to show how exactly this deposition process applies a force to the strip for clarification or perhaps you can try and explain the process in more detail?
15. sonhouse
Fast and Curious
21 May '15 22:521 edit
Originally posted by joe shmo
I suppose...if you can precisely mimic/control the loading caused by the deposition. Btw, are there any links that could give a hint to show how exactly this deposition process applies a force to the strip for clarification or perhaps you can try and explain the process in more detail?
The deposition is done in a vacuum chamber with argon as background gas. There is a target of the material to be deposited, and the volume between the target and the platen is charged with RF at about 500 to 1000 watts, which ionizes the argon. Now argon, at 40 AMU is a fairly heavy atom and when ionized like that it has a degree of kinetic energy that will allow it to blast off the surface layer of the target, whatever that is.

Aluminum, SiO2, SIC, titanium, chrome, whatever, the argon slams into the target and it is like a molecular bead blaster or sand blaster and the molecules of the target gets transported to the platen where the substrate to be deposited is sliding by back and forth, slowly getting coated by the target material. So the layers are built up bit by bit, from about 25 angstroms per pass of the platen to maybe 50 or more depending on the RF energy given to the argon gas which is at something like 10 microns of pressure, pretty low medium pressure.

So the layers go on the substrate and the metal strip layer after layer and in the process, there is a curling force, small but present, in that process. So the metal strip, which is one thousands of an inch thick and about 0.6 inch wide gets somewhat curled. The more the curl. the higher the stress level of the layers building up and the problem there is if the layer is subjected to forces such as heat or friction, the layers can delaminate, not a good thing. So the layers have to be built up with as little stress as possible.
Don't know if I can put it much better than that.