# How is non-uniaxial strain defined?

by scott_alexsk
Tags: defined, nonuniaxial, strain
 P: 353 Hello, How is non-uniaxial strain defined? How would Nitinol's 8% maximum strain before failure be defined as in terms of other than stretch? How does one define the 8% in terms of bending the wire? So in other words how much bending of the wire equates to an 8% strain limit? Thanks, -scott
 HW Helper P: 3,220 I know I'm not sounding very specific, but you may want to look at strain tensors.
 P: 353 Just, looking at that right now on Wikipedia, it does not seem to concern bending as a type of strain, just shear strain and uniaxial strain distortions. -scott
 HW Helper P: 3,220 How is non-uniaxial strain defined? Well, you'd need some laboratory tests to get the relation between stress and strain in the case of bending.
 P: 40 When a material bends, the outer curved suface is in tension and the inner curved suface is in compression. There is a neutral position between those two surfaces where there is no stress. For most materials the compressive strength is greater than the tensile strength. Therefore most bent materials fail in tension. Bending failure ususally occurs where tensile force is greatest. Stress-raisers, such as surface defects, can cause up to a three fold increase in stress. Failure in shear is less common, but that is what makes cutting operations possible. Failure in torsion is often a a shear failure. The measurement of stresses, the design of structures, and the analysis of their failures are the subjects included in courses on the mechanics of materials. I hope this points you in the right direction.
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P: 11,155
 Quote by scott_alexsk Hello, How is non-uniaxial strain defined? How would Nitinol's 8% maximum strain before failure be defined as in terms of other than stretch? How does one define the 8% in terms of bending the wire? So in other words how much bending of the wire equates to an 8% strain limit? Thanks, -scott
Flexural strain is defined (typically) as the strain in the outer (convex) surface, the surface which is under tension (for the reason stated by drachir, above).

Note: You can estimate the curvature that makes an 8% strain from the cross-sectional geometry and making the neutral axis lie along the cetroidal axis of the wire.

If you tell us the geometry of your wire, someone here can show you how to do the calculation.
 P: 353 The lenght of my wire is 25 cm and its diameter is 0.3175 cm. The 8 percent strain is actually not the point at which the Nitinol wire fails but the point that the wire is plastically deformed. I have also seen the point of plastic deformation written as 70 MPa. Does this basically mean that 70 MPa will create an 8% strain in most cases? I am going to be measuring the force differently treated samples of nitinol can pull with. The key is that I do not get above 70 MPa's pulling from the spring balance, so that the wire is not plastically deformed. The spring balance is measured in Newtons. How can I convert the 70 MPa's into newtons pulling uniaxially on the sample of Nitinol wire previously described? Thanks, -scott
 Emeritus Sci Advisor PF Gold P: 11,155 This is confusing! Are you looking at two different modes (i) flexure (bending), and (ii) tension (stretching) ? If you want to use the nitinol to "pull", then you are interested in tensile failure (yield point in tension). To convert from force in N to uniaxial stress in Pa, simply divide the force by the relevant area in sq. meters. A MPa is just 10^6 Pa. For tensile (stretching) loads, the area of interest is the cross-sectional area of the wire. PS: If 70MPa and 8% are the yield point and the strain at yield for the exact same alloy, then yes, the one produces the other (though only accurately when you have the same conditions as those for which the numbers are specified).
 P: 353 So the relevant area is simply the area is square meters of the single cross section? Sorry for being confusing, I am testing a lot of different things. Yes I am looking at two different modes. (I had a feeling I should not have posted that second set of questions) For one I am going to see what kind of force a straight annealed wire, strained less than 8%, will pull with. After that I intend to see what force a wire annealed in a curve will pull with. Using the information I find on calculating strains with bent materials, hopefully I can calculate what force from the spring will provide slightly less than an 8% strain for that sample also. Finally I hope to test what force another sample, trained with the two way memory effect, can pull with, also being strained less than the critical 8% (although testing this one will be messy.) The information I am asking about strain with curves most particularly applies to this third test, because when training the alloy for the two way memory effect it is critical when training to exceed the 8% strain. I have seen the two phrases of 8% strain and 70 MPa used interchangably to represent the point of plastic deformation and the point at which the memory properties are degraded. However looking on the fact sheet, 70 MPa is defined as the Yield Strength. Is it only when the wire is being uniaxially strained do the 8% and the 70 MPa equate? Otherwise when dealing with strain with curved surfaces do I simply go with the 8% strain for determining the point of plastic deformation? Thanks, -scott
 P: 353 So does anyone know how to find the strain of a curve or know where it might be discussed? Thanks, -scott
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 Quote by scott_alexsk So does anyone know how to find the strain of a curve or know where it might be discussed? Thanks, -scott
There's no such thing as "strain of a curve", so it's not clear what you're asking. Do you mean strain in flexure (bending)? Or are you simply talking about tensile strain in a curved (and constrained) wire?

Is the applied force trying to bend/unbend the wire or is it trying to stretch a wire bent around something?

http://www.diracdelta.co.uk/science/...e/image001.jpg <figure on left is flexure>

http://kosmoi.com/Science/Physics/Machines/pulley.gif <rope bent around pulley is in tension>
 P: 353 I am refering to flexural strain. -scott
 Emeritus Sci Advisor PF Gold P: 11,155 Let the angle (in radians) subtended at the center be $\theta$, when the strain is $\epsilon$. The length of the wire is L and its radius is r. Then, $\theta = \epsilon L /r$.
 P: 353 Thanks Gokul, but what unit is strain in? Is it percent lenght? Would 8% strain be 0.08, 1.08, 108, 8? I assume it is 0.08 but I am not sure. -scott
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 Quote by scott_alexsk I assume it is 0.08 but I am not sure.
Yup, that's the right one. It's a unitless ratio.
 P: 353 Thanks again. I get a pretty big number from that but from what I have seen, it seems true. -scott