Sciencemadness Discussion Board

Nitinol calculation check?

elementcollector1 - 8-2-2018 at 23:42

Due to lack of access to my usual testing equipment, I've lately resorted to trying to calculate how much nitinol shape-memory wire I need to put into a robotic joint to produce a given force. The objective of this robot would eventually be to lift a human, as well as itself (total weight assumed to be 90 kg). It would have four legs, each with one nitinol-based actuator at the knee joint to exert force against the weight. Peltier modules will be used to cool the nitinol wires when they are no longer actuating, and the size of these determines the dimensions of the overall actuator (hence why they're referred to in the calculations).

I recently redid my calculations with some improved numbers, but came up with some unusual results - specifically, that nitinol is apparently very strong. Could someone check my math, just to make sure I didn't goof somewhere? I suspect the problem is that I'm misunderstanding the concept of nitinol's 'restoration pressure.' To quote from Wikipedia:
Quote:
A great deal of pressure can be produced by preventing the reversion of deformed martensite to austenite — from 35,000 psi to, in many cases, more than 100,000 psi (689 MPa).


In my calculations, to get a conservative estimate, I assumed this intense pressure was only acting on the cross-section of a wire. This still produces insanely high numbers - a wire with a diameter of just 1 millimeter can apparently output over 500 N of force! That can't be right... can it?

Attachment: nitinolforcecalculator3.xlsx (12kB)
This file has been downloaded 285 times

Fulmen - 9-2-2018 at 02:12

Seems a little high. Here's some relevant data:
http://www.imagesco.com/articles/nitinol/04.html

elementcollector1 - 9-2-2018 at 12:41

Quote: Originally posted by Fulmen  
Seems a little high. Here's some relevant data:
http://www.imagesco.com/articles/nitinol/04.html



The problem with reports like these is that they only focus on Flexinol, which has a different actuation mechanism and microstructure than the SMA nitinol I'm working with. Flexinol actuates by changing its length, whereas the wire I'm using actuates by physically bending and straightening. Still, the 0.1524-mm diameter wire mentioned in the paper can produce a lifting force of about 3.06 N (11 ounces), so maybe my numbers are within the range of possibility?

Fulmen - 9-2-2018 at 14:45

Still, it's real data. I haven't looked too close at your spreadsheet, but if you tell me what you want to accomplish I could always run through the calculations myself. Haven't used statics in a couple of years now so it'll be a good exercise.

elementcollector1 - 9-2-2018 at 17:48

Quote: Originally posted by Fulmen  
Still, it's real data. I haven't looked too close at your spreadsheet, but if you tell me what you want to accomplish I could always run through the calculations myself. Haven't used statics in a couple of years now so it'll be a good exercise.


The spreadsheet tries to calculate two things: The force required to bend a single nitinol wire, using the bending moment and distance from one end (the bend is centered, so it's just half the length of the wire); and the force the nitinol wire would produce during the martensite-to-austenite phase during electrical heating (estimated from the 'restoration pressure' given by various sources as 30 to 100 ksi and the cross-sectional area of the wire in square inches). Finally, it compares the force of one wire restoring to its normal shape to the force required for a single leg of a quadruped to lift 90 kg of weight against the force of gravity, thereby estimating how many wires are needed to accomplish this.

The module itself would consist of several layers of nitinol wires stacked atop a few Peltier modules, grouped on each side of the bend. When bent, it would retain the angle of the bend, and when electrically stimulated, it would straighten again, pushing against any resistance with a given force (which would presumably increase with current?)

I suppose the only concrete answer would be to get a wire, attach it to a weight, and test it. I'm still skeptical at how strong these are proclaimed to be, but if they do work out that way, it'll save a lot of work for me in the long run.

Fulmen - 10-2-2018 at 01:30

I've taken a quick peek at your math and one thing instantly pops out: The use of multiple measuring units. Start by converting the restoring pressures to MPa rather than converting the units mid-calculation. The correct answer should then be obvious, the bending force and restoring force will be the exact same calculation except for the material data.

elementcollector1 - 10-2-2018 at 13:01

Quote: Originally posted by Fulmen  
I've taken a quick peek at your math and one thing instantly pops out: The use of multiple measuring units. Start by converting the restoring pressures to MPa rather than converting the units mid-calculation. The correct answer should then be obvious, the bending force and restoring force will be the exact same calculation except for the material data.


Good idea! I put the restoring pressure in MPa, and removed the fudge factor in the restoring force calculation. This gives me the exact same answers as before, except for a few numbers' ending digits changing slightly.

Can you clarify what you mean about the bending and restoring force being 'the same calculation'? Restoring pressure appears to be independent of the yield strength of the martensitic phase.

EDIT: Nevermind. Just noticed the yield strength of nitinol in the austenitic phase matches up exactly with the restoring pressure. This is why you take time to read the data, kids. : P

Using this calculation, the numbers work out much more reasonably - I'd need at least 31 wires per joint in the best-case scenario. Thanks for pointing that out, Fulmen!

Attachment: nitinolforcecalculator3a.xlsx (13kB)
This file has been downloaded 270 times

[Edited on 2/10/2018 by elementcollector1]

Fulmen - 10-2-2018 at 13:25

Maybe I wasn't as clear as I could be.
You have calculated the bending force correctly using moment of inertia and bending moment. The restoring force should be calculated the exact same way, the only change will be the restoring pressure in MPa. So for the 1mm wire minimum restoring force should be 3,7N*207MPa/70MPa=10,8N while the max should be 7.3N*689MPa/140MPa=35,9N.

elementcollector1 - 10-2-2018 at 13:36

Quote: Originally posted by Fulmen  
Maybe I wasn't as clear as I could be.
You have calculated the bending force correctly using moment of inertia and bending moment. The restoring force should be calculated the exact same way, the only change will be the restoring pressure in MPa. So for the 1mm wire minimum restoring force should be 3,7N*207MPa/70MPa=10,8N while the max should be 7.3N*689MPa/140MPa=35,9N.



I realized the moment after I made the post... the updated spreadsheet has the calculation you suggested, and similar answers. I got 36.1 N for the 1 mm wire, but I suppose that's a matter of significant digits.

The only thing I'm still confused by is how electrical current factors into things. Increased current reduces the time to contraction, but it's unclear if it has any effect on the force produced - maybe it produces a lower impulse instead, because F(t2-t1) works out to be lower?

It may be that electrical current only affects time, and therefore I could simply vary it to be fast or slow depending on how fast I want the muscle to move, but I want to be absolutely sure I'm not increasing or decreasing the force as I'm doing so.

Fulmen - 10-2-2018 at 14:20

The current shouldn't have any effect by itself, it's the heating effect you must consider. As far as I understand it the transition from martensite to austenite occurs within a temperature range (As to Af), so you should be able to control the force produced by controlling the temperature within this range. The maximum force at Af should be the same regardless of heating rate.