Pairing negative thermal expansion ALLVAR Alloys with material with known positive thermal expansion means you can calculate how to achieve an overall Zero thermal expansion or chosen CTE across the entire assembly, rod, strut or other component. Choose your own CTE to create your athermal assembly.

Transcript Below

### Tailor thermal expansion to a chosen value, in this case zero CTE.

Hello again, in this video we’re going to be talking about the practical application of the combination of a positive thermal expansion and negative thermal expansion material developed in the previous video. So in this case we have aluminum and ALLVAR. If we’re interested in creating for example zero CTE this doesn’t necessarily need to be zero [CTE], it can be any constant, but zero [CTE] seems to be quite popular, for stability’s sake, which is important. You start to combine aluminum, this is the strain versus temperature between -40°C to 80°C, and if you look at aluminum 6061 with a T6 heat treatment you see that it increases with increasing temperature. The ALLVAR Alloy 30 does the opposite. You’ll also notice that the ALLVAR alloy 30 has a curvature to it, whereas the aluminum 6061 is very straight. We’ll actually get into the implications of that curvature versus that straight line.

### Tuning to zero CTE using the average coefficient of thermal expansion

When we combine a positive thermal expansion aluminum and a negative thermal expansion ALLVAR, we have the equation we developed to understand how long the ALLVAR piece of material needs to be compared to the total length of the beam, that is at equal to α_{total} minus α_{aluminum} divided by α_{ALLVAR} minus α_{aluminum. }If we want zero coefficient of thermal expansion this term [α_{total}] goes to zero, and we’re left with negative α_{aluminum} divided by α_{ALLVAR} minus α_{aluminum}. If we want zero coefficient of thermal expansion, it’s this simple ratio of the aluminum thermal expansion [α_{aluminum}] divided by the difference between the ALLVAR and the aluminum coefficients of thermal expansion [α_{ALLVAR} – αaluminum].

In the previous video, we developed this for thermal expansion coefficient or the average thermal expansion coefficient. It kind of brings up a question well what is Temperature 1 [T_{1}] for the average coefficient of thermal expansion? Well, the average coefficient of thermal expansion is given across a temperature range, so if you had a temperature range for example from -40 to 80[°C], your T_{1} or your reference temperature is going to be equal to your low temperature plus your high temperature [T_{low} + T_{high}] divided by 2. It’s effectively the average of your temperature extremes. In this case if you have -40°C plus 80°C divided by 2, you’re looking at 20°C, which is halfway between -40 and 80°C.

If I were to take the average thermal expansion for these two alloys between -40 and 80°C, we can look at what this equation would come out to. This ratio of ALLVAR to the total length at T_{1} equal to 20°C is equal to negative 22.x 10^-6 divided by -30.2×10^-6 minus 22.3×106. The average CTE – if you don’t remember – you take the strain at one temperature at your lower temperature extreme and the strain at the upper temperature extreme you draw a straight line, and that slope of that straight line is equal to your average CTE. So this is the average CTE from -40 to 80°C of ALLVAR Alloy 30 which is -30.2[ppm°C] This is real data from a real piece of material, and so it’s not exactly -30, but it gives you that ballpark range which is really good. So this is true actual data collected on these materials. What you see is for this piece of ALLVAR [Alloy 30] and this piece of aluminum that was tested, you get an L_{1}^{ALLVAR} of our divided by L^{1}_{total} of 0.42. If you have, let’s say, a 1 meter-long piece of material, 0.42 meters will be ALLVAR and the balance will be aluminum to achieve zero coefficient of the average coefficient of thermal expansion from -40 to positive 80.

But this curvature that the ALLVAR alloy 30 has can make it to where the extreme temperatures will experience differences in strain. So if you actually take the average CTE, or the you take this ratio, and you actually put it into the strain values you see that the strain versus temperature will look something like this.

This is for L_{1}^{ALLVAR} divided by L^{1}_{total} at T_{1} equals to 20°C is equal to 0.42. This would be for alpha average [CTE Average], so this is the average thermal expansion you see that at the extremes there’s going to be there’s going to be a shrinking over here and a shrinking down here. At 20°C it’ll be 0 CTE, which is the slope of this curve. That discrepancy at the extreme temperatures is due to the curvature compared to the straight line of the ALLVAR Alloy 30 compared to the aluminum 6061.

### Setting a zero-point crossing, or zero CTE at a particular temperature, using instantaneous CTE

This curvature can give us almost a hidden or a secret weapon. So what happens is if we take a look at the instantaneous CTE, and use this equation – I know it was developed for average CTE – but what happens when we start to use the instantaneous CTE? What we see here that T_{1} is equal to a constant value – a single temperature. We can start to use this CTE, this unique curvature, to dial in where the CTE is. If I wanted to set α_{total} instantaneous to equal to zero at T_{1} equal to 20°C then I would use L_{1}^{ALLVAR} divided by L^{1}_{total} and use the instantaneous CTE instead of the average CTE here. It comes out to something quite similar.

In this case, the instantaneous CTE at 20°C for aluminum is 23.4×10^-6 divided by ALLVAR being negative 29. x10^-6 minus 23. x10^-6. If you calculate that, out it comes to 0.43. So in this instance the average CTE and the instantaneous CTE for zero CTE at room temperature is quite similar 0.42 versus 0.43. If we take a look at the instantaneous CTE versus temperature, this being aluminum 6061 and this being ALLVAR Alloy 30, we see that these two produce a curve, an instantaneous CTE curve, that looks something like this. Where this is α_{total}, so that’s an effective CTE of this total bar.

Now here’s where the interesting thing comes in. Because the ALLVAR CTE is changing with temperature, we can actually create a zero CTE crossing – zero instantaneous CTE crossing – at any temperature that we want. So for example, if I want the total coefficient of thermal expansion to be 0 at T_{1 }equal to -40°C, what I need to do is take L_{1}^{ALLVAR} at -40°C divided by L^{1}_{total} at -40°C. Notice you actually have to measure these at the low temperature, but what that gives you is negative 21.4×10^-6, so that’s the instantaneous CTE, this is -40. It’s the instantaneous CTE of the aluminum 6061 at -40°C. You divide that by negative 23.4×10^-6, and then you subtract the Aluminum’s coefficient of thermal expansion as well times 10 to the negative six, and that is equal to 0.48. So by adding a little bit more aluminum, what you get is an instantaneous CTE curve that looks something like this, where your zero point CTE crossing is at -40°C instead of 20°C, so this would be α_{total }set to T equals -40, is equal to zero.

So now you can, actually if you added a little bit less or you add a little bit more aluminum, took away some of the ALLVAR, then you can actually get the zero CTE crossing to increase in temperature. You can actually dial it in across a very wide temperature range, and this can be important if you’re not concerned about very large temperature swings, but if you’re concerned about smaller temperature swings maybe plus or minus 10 or 20°C, where this discrepancy that’s caused by the curvature at the extreme temperatures isn’t that big of a deal. You can actually dial it into whatever the operating temperature of the system you’re working with is going to be working at. This is a really exciting and interesting thing that you can do with ALLVAR that you can’t do with any other material, because it doesn’t have the curvature and it doesn’t have the negative thermal expansion coefficient.