Why is strain also normalized to the parameters of engineering strain?

Question

I get that you normalize stress to engineering stress, so that it is independent of the cross sectional area. This means you can calculate (roughly) the stress needed to elongate the specimen to the degree you want. But why normalize the strain aswell?

I would have thought that the orginal length of the specimen does not matter. Lets say you want to make the specimen 2 mm bigger, then I would have thought that you would need the same engineering stress no matter how long the specimen is. Thus that elongation is only dependant on cross sectional area and load.

For example you have a specimen, with orginal length, of 60 mm and 100 mm and then you want (l-lo) to be 2 mm. I would have thought that it would take same amount of engineering stress. And that a 2% strain would lead to different stresses needed, since in one this means an elongation of 1.2 mm and with the other an elongation of 2 mm.

Does this mean that the length of a specimen also determines how 'easily' a specimen elongates? That it more difficult to elongate a 100 mm to 102 mm than a 60 to 62 mm. Because the tensile stress-strain curve implies that to elongate a 60 mm specimen to 61.2 takes same amount of stress as elongating a specimen to 62 mm.

If the length of the specimen relates to how easily you can elongate it, does this then also explain why the gauge length should be specified when giving the percent elongation (%EL) when talking about ductility? Because most of the plastic deformation is at the neck, so not uniform and thus independant of the length of the specimen orginally.

So before necking the elongation is dependant of orginal length and engineering stress and after necking is independant of orginal length?

Answer 1

From what I understand of the question, there are three interlinked concepts that are causing confusion, each of which could be a separate question.

1. Is axial stress dependent on length of a specimen?

Axial stress is independent of length. @Wasabi's answer goes into more detail, but the core concept is that a stress of $\sigma$ on a specimen of length $L$ has the same effect on a specimen per unit length as a stress of $\sigma$ on a specimen of length $2L$, or indeed of any length.

2. Is axial strain dependent on length of a specimen?

Axial strain is independent of length. This follows from the answer to (1.) above, and from Hooke's law, specifically that

$$ \sigma = E \varepsilon $$

so that if $\sigma$ is independent of length, so must be $\varepsilon$ since they are linearly related.

3. Is axial deformation dependent on length of a specimen?

Axial deformation is dependent on length. It also depends on cross-sectional area (assuming constant cross-section) and tensile modulus. This follows from a quick and dirty derivation of the axial deformation formula. Consider a constant-cross-section tensile member of area $A$, length $L$, experiencing a uniform load $P$, with modulus $E$. The member will experience a deformation of $\delta$. Note that by definition $\varepsilon = \delta / L$ and $\sigma = P / A$.

From Hooke's law,

$$ \sigma = E \varepsilon $$

and substituting our notes gives

$$ \frac{P}{A} = E \frac{\delta}{L} \\ $$

which after some rearranging yields

$$ \delta = \frac{PL}{AE} $$

which is the equation for axial deformation of constant-cross-section tensile members. As should be obvious from the equation, axial deformation depends on length, as well as area.

How Can I Think About This Intuitively?

Elasticity is just like springs, and indeed Hooke's law is also used for linear springs. Now consider two springs such that one has twice the number of coils as the other, but the rest of their parameters are the same. Assume they are also weightless. If you held just the end of each spring, hooked identical weights to the other ends and let them uncoil toward the floor, the spring with twice the number of coils would extend twice as far. As a result, the change in length per per unit original length, equivalent to the strain, would be the same.

The same is true of two tension rods, one twice as long, experiencing the same stress with the same cross-sectional area. The one that is twice as long would have double the deformation, but they would have the same strain.

How Can I Use This Information?

As an example, consider a tensile rod. Suppose you must minimize axial deformation. Then you would choose a material with high tensile modulus so that stress has less effect on strain, and thus less effect on deformation.

Now suppose instead you have a family of tensile rods all made of the same material. If each must not deform more than 1 mm, then the longer rods must have a larger cross-sectional area to make up for the additional "slack" from their added length.

What About Plastic Deformation?

The above three questions only apply in the elastic regime. In the plastic regime, volume is practically conserved in most materials, which means that any instantaneous extension of a tensile specimen must cause an instantaneous reduction of area such that the volume gained by increasing length is equal to the volume lost by reducing the cross-section area.

However, there is an instability inherent to tensile plastic deformation called necking, which you noted. Namely, if one length-wise segment of a specimen is slightly narrower than the rest of the specimen, stresses will be concentrated at the narrow region and it will reach the elastic limit slightly before the rest of the specimen. Plastic deformation will thus begin at the slightly narrower section, causing a greater decrease in area than the rest of the specimen, further concentrating stresses, which causes further and greater deformation. The specimen will tend to fail wherever it is narrowest, and will do so unstably in a smooth, constant-cross-section specimen. That is, it is impossible to predict where failure will occur without intentionally notching the specimen.

Beware! Necking does not necessarily proceed the same way in polymers. In polymers, necking is associated with polymer chain alignment, which locally strengthens the material so much it prevents further deformation there, unlike metals. Instead, the neighboring regions begin to neck causing them to strengthen, and so on, until the entire specimen becomes aligned.

Answer 2

Indeed, deformation is a function of an element's length. The longer the element, the lower the stress required to obtain an equal deformation. This is because stress is proportional to strain (adimensional, such as 2%), not to deformation (such as 2 mm).

Imagine you have two objects: a crystaline structure composed of nine iron atoms ($\alpha$ allotrope), and a twelve-$\alpha$-iron-crystal-long strand. Both can be seen below. You apply the same tensile force $F$ to each of the objects in the horizontal direction.

Now, what does it mean when an element is stretched? It means that all of their constituent particles and their chemical bonds are stretched. The single crystal deforms by some very tiny value $\delta$. The "chain" however, transmits this same force $F$ along each of its constituent blocks, such that each of them also deform by the same value $\delta$ as the single crystal.

So, you applied a force $F$ to each of the objects. The single crystal deformed such that its length is now equal to $\ell + \delta$. The chain, however, deformed such that it is now $L + 12\delta$.

And so you see that for the same force (or stress), two objects of equal cross-section but different lengths present different elongations (in linear units). However, they do present the same deformations $\epsilon$ (adimensional), since $\epsilon = \dfrac{\delta}{\ell} = \dfrac{12\delta}{L} = \dfrac{12\delta}{12\ell}$.