When I read about stress analysis and specifically in the context of strain gauge, I often see the phrase resistance per unit strain or voltage per unit strain. I understand the electronics and how some of the equations are derived or come about. I have a basic understanding of what stress and strain are, but I don't get what per unit strain is and why we're using it.

Take the simplest example - a strain gauge connected in series with a constant current source, where $ E $ is the voltage drop across a strain gauge, $ R_g $ is the resistance of the strain gauge, $ G_F $ the gage factor, and $ \epsilon $ the strain:

$$ E + \Delta E = I(R_g + \Delta R_g) $$ $$ \Delta E = I \Delta R_g = I R_g {\Delta R_g \over R_g} = I R_g G_F \epsilon$$

Everything makes sense until the equation is rearranged and call it the potential drop per unit of strain:

$$ {\Delta E \over \epsilon} = I R_g G_F $$

One unit is a measure of unity. One unit of strain therefore is $ \epsilon = \Delta L / L_0 = unity = 1 $. That implies $ \Delta L = L_0 $, or the change in length equates its initial length. If the object is under tensile stress, $ \Delta L = L $ is a 100% elongation. So the term potential drop per unit of strain would logically mean the voltage drop per each 100% elongation! Most strain being measured in the elastic region is well below a few percent. 100% elongation (or my take on per unit strain) feels .... out of context here.

  • $\begingroup$ Much more common is defining things "per unit length" - this doesn't imply that the length is 1, but that the units of length don't matter (e.g. mm, km, inches). Strain, being made up of two length values, it doesn't matter what units of length you choose - those units will cancel out when you divide them. $\endgroup$ Feb 5 '21 at 7:53
  • $\begingroup$ Isn't per unit length....really per unit length of 1? Say, if a cable has resistance 10ohm per unit meter, that does mean one meter length of cable has 10ohm resistance... Same logic implies potential drop per unit strain is the voltage drop per 100% elongation... $\endgroup$
    – KMC
    Feb 5 '21 at 8:16
  • 1
    $\begingroup$ No. In your example, the resistance would be quoted as 10Ω/m - that's Ohms per meter, not per unit length, because the units don't cancel out. In the case of Strain, however, it doesn't matter what the units are, as the % change is identical $\endgroup$ Feb 5 '21 at 12:51
  • $\begingroup$ The key point is "per unit" does not mean "per 1". $\endgroup$ Feb 5 '21 at 12:52
  • $\begingroup$ "Same logic implies potential drop per unit strain is the voltage drop per 100% elongation" -- yes. You could phrase it this way, but then you'd have to apply the x100 factor and say "per percent strain" $\endgroup$
    – Dave X
    Feb 5 '21 at 16:02

While I completely agree with @NMech's answer, let me take a different approach.

As I see it, the problem is that you're forgetting something: look at the right-hand side of that equation: $IR_gG_F$. Now, I don't know anything about the subject you're looking at about voltage drops or what have you. But I'm willing to bet that if you calculate that, you'll get an absurdly high value for the drop $\Delta E$.

But then again, that makes perfect sense. After all, you're right: a "unit strain" such that $\epsilon = 1$ implies 100% elongation, which is also an absurdly high strain!

But since the left-hand side of the function is a simple proportion $\dfrac{\Delta E}{\epsilon}$, we can conclude that for reasonable values of $\epsilon$ (such as 0.002), you'll get a reasonable value for $\Delta E$ as well. After all, the right-hand side merely gives the slope of the relationship between $\Delta E$ and $\epsilon$:

$$\frac{\text{absurd }\Delta E}{\text{absurd }\epsilon} = \frac{\text{reasonable }\Delta E}{\text{reasonable }\epsilon}$$

If it helps clear things out, replace the word "unit" with "step". Then define what your "strain step" will be. For every step of $\epsilon = 1$, you'll get a really big step up in the value of the drop. But if you take steps of a more reasonable $\epsilon = 0.0001$, then you'll get reasonable steps up in the drop.

  • $\begingroup$ replacing unit with "step" magically recovers my intuition in associating ratio relation to be a slope where an increment can be any fraction of a step. Thanks! $\endgroup$
    – KMC
    Feb 9 '21 at 4:08

If I understand correctly, your question is "Why do we use a unit strain instead of something more meaningful like a percent strain?".

The reason is that we are normalising with respect to a well-defined (and easily comprehensible) quantity. This is a very common theme in material science.

If you look at the definition of the modulus of elasticity, the quantity is defined as $\frac{\Delta \sigma}{\Delta \epsilon}$, so again you have stress per unit strain. The physical interpretation is that the modulus of elasticity is the stress developed in a material when the length doubles in tension (or if it vanishes out of existence in compression—tongue in cheek statement). However, even in tension, the modulus of elasticity (i.e., the rate the stress changes with respect to strain) does not remain constant for more than a few percent.

enter image description here

The graph above exhibits the relationship of the modulus of elasticity (dotted line) with the actual stress–strain behaviour of low carbon steel.

My point is: although the physical interpretation behind the modulus of elasticity value could be described as nonsense, it is (in all probability) the most sought-after property in structural analysis problems (even more so than stress because there are so many variation of stress: yield stress, failure stress, shear failure stress, etc.).

NOTE: I am not certain that I 100% got your question (excuse the pun), so please comment if I misunderstood, so that I delete or update.

NOTE 2: when dealing with such high strains, the concept of true strain (vs. the commonly used engineering strain) can alleviate/correct some of the problems I highlight (see erasure from existence and negative matter in compression).

UPDATE Upon seeing Jonathan R Swift's comment, the following thing came to me, which might be relevant to the question, i.e., the use of per unit strain.

Strain is a nondimensional number. As a result, when you are dividing the voltage drop $\Delta V$ by strain, the units are the same as resistance. Therefore, *the use of "per unit strain" emphasizes that the resulting value $\frac{\Delta V}{\epsilon}$ is not in [volts] but in [volts per unit strain]**.

  • $\begingroup$ And engineers often analyse or check things by comparing dimensionless terms... $\endgroup$
    – Solar Mike
    Feb 5 '21 at 6:56
  • $\begingroup$ Exactly. The resulting value is not just "Volts divided by 1"! $\endgroup$ Feb 5 '21 at 15:28
  • $\begingroup$ Funny, can one seamingly straight and honest question, can have so many different interpretations. @KMC it would be helpful, if you commented on whether any of this answer resonates with your original question. $\endgroup$
    – NMech
    Feb 5 '21 at 15:32
  • $\begingroup$ I was confused with this answer until I read@Wasabi ♦ 's and realized I forgot what per unit of something fundamentally means. Then the rest make sense. $\endgroup$
    – KMC
    Feb 9 '21 at 4:13

I don't get what per unit strain is and why we're using it.

When you have a quantity expressed as "per unit (of) X", you get two things.

First, you can recover the value of the quantity for any X by just multiplying (assuming that the relationship is linear). That's what multiplication does - it just scales some base value up or down. So you can express some relationship that's relative to X in a more universal way by fixing the quantity to the value it would have if X were 1, and then using that as a multiplication factor (the slope). So you're designating a particular value to be the reference value.

enter image description here

This is what the term "scalar" should be evocative of.

So the term potential drop per unit of strain would logically mean the voltage drop per each 100% elongation! Most strain being measured in the elastic region is well below a few percent.

It doesn't matter if X actually obtains the value of 1 in practice. What matters is that the mathematical relationship is correct.

Take the spring force equation (Hooke's law): $F = -kx$. The spring constant describes force magnitude per unit displacement - letting you recover the value for any displacement. It doesn't matter whether or not displacements of +/- 1m make sense for that particular spring; there's a range of values where the model is applicable.

enter image description here

You can also invert it (take the reciprocal) and use it as "displacement per unit Force", to infer the actual displacement if the force is what you can measure. So "potential drop per unit of strain" and it's inverse let you recover one or the other depending on what you can measure, by just plugging in the measured number - as long as you can assume the linear relationship.

The second thing you get is the ability to do comparisons in a normalized way. E.g., you're not interested in absolute values, but instead you want to compare two things relative to each other, or to some "standard" - "per unit X" can fulfill that role. E.g., you can compare the strength of the gravitational force (of two different stars) per unit mass (a 1kg "test mass"). Or the stiffness of two springs. Or the sensitivity of two strain gauges.

  • $\begingroup$ although technically not an answer by itself, it serves as a great primer before grasping the other two answers $\endgroup$
    – KMC
    Feb 9 '21 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.