True Stress-Strain vs Engineering Stress-Strain – Materials Science & Engineering

So, now you know all about engineering stress-strain curves. Normally I write these articles to stand alone, but in this case, I’ll assume you’re here because you googled a homework question 🙂 If you don’t understand the basics of the stress-strain curve, I recommend reading that one first.

So, what is the difference between engineering and true stress-strain curves?

When deforming a sample, engineering stress simplifies by neglecting cross-sectional change. True stress correctly accounts for the changing cross-sectional area.

The most obvious thing you may notice is that the true stress-strain curve never decreases. That is because the material never gets weaker! The decrease in the engineering stress is an illusion created because the engineering stress doesn’t consider the decreasing cross-sectional area of the sample.

Let’s start by mathematically defining the true and engineering stress-strain curves, talk about why you might want to use one versus the other, and then dive into the math and show how to convert from one to the other.

Outline

Definitions of Engineering and True Stress-Strain Curves
Why Should You Use an Engineering vs. True Stress Strain Curve?
Converting between the Engineering and True Stress-Strain Curves
Summary
References and For Further Reading

Definitions of Engineering and True Stress-Strain Curves

If you want the origins of these definitions, I explained the math in my previous article. Otherwise, be a good engineer and accept this as our starting point!

$\sigma_{eng} =\frac{F}{A_0}$

$\varepsilon_{eng} = \frac{\Delta L}{L_0}=\frac{L_f-L_0}{L_0}$

$\sigma _{True} =\frac{F}{A_i}$

$\varepsilon_{True} =ln\left(\frac{L_i}{L_0}\right)$

Remember that $\sigma$ is stress, $\varepsilon$ is strain, $F$ is load, $L$ is the length of the specimen in a tensile test, and the subscripts $i$ , $0$ , and $f$ mean instantaneous, original, and final.

The data for these equations would come from a tensile test. The engineering stress-strain curve plots engineering strain on the x-axis and engineering stress on the y-axis. The true stress-strain curve plots true strain on the x-axis and true stress on the y-axis.

Why Should You Use an Engineering vs. True Stress Strain Curve?

The engineering stress-strain curve is ideal for performance applications. The true stress-strain curve is ideal for material property analysis.

For everyone except (some) materials scientists, the engineering stress-strain curve is simply more useful than the true stress-strain curve.

When an engineer designs a part, he or she knows the original size of the part and the forces the part will experience. That’s exactly how engineering stress is calculated.

$\sigma_{eng} =\frac{F}{A_0}$

where $\sigma$ is the stress, $F$ is the applied force, and $A_0$ is the original cross-sectional area.

Most values (such as toughness) are also easier to calculate from an engineering stress-strain curve. Because engineering stress and strain are calculated relative to an unchanging reference, I prefer to say that engineering stress is “normalized force” and engineering strain is “normalized displacement.”

Thus, any calculations involving force or displacement–such as toughness or ultimate tensile strength–can be done directly from an engineering stress-strain curve.

The ultimate strength is completely obscured in a true stress-strain curve. However, the engineering stress-strain curve hides the true effect of strain hardening.

The true stress-strain curve is ideal for showing the actual strain (and strength) of the material.

Some materials scientists may be interested in fundamental properties of the material. In this case, the true stress-strain curve is better. This curve tells the actual state of stress in the material at any point. It also shows strain hardening without being affected by the changing area of the sample.

For example, many metals show strain-hardening behavior that can be modeled as:

$\sigma _{True}=K\varepsilon ^n_{True}$

Where $K$ is a constant and $n$ is the strain-hardening exponent. $n$ is always less than 1.

Click here to see examples of $K$ and $n$ values.

Material	n	K (MPa)
Aluminum 1100-O	0.20	180
Aluminum 2024-T4	0.16	690
Aluminum 6061-O	0.20	205
Aluminum 6061-T6	0.05	410
Aluminum 7075-O	0.17	400
Brass 70-30, annealed	0.49	895
Brass 85-15, cold rolled	0.34	580
Cobalt-base alloy, heat treated	0.50	2,070
Copper, annealed	0.54	315
Steel Low-carbon, annealed	0.26	530
Steel 4135, annealed	0.17	1015
Steel 4135, cold rolled	0.14	1100
Steel 4340, annealed	0.15	640
Steel 304 stainless, annealed	0.45	1275
Steel 410 stainless, annealed	0.10	960

Adapted from Manufacturing Engineering and Technology [1]

If you were doing research on a new alloy and needed to determine the strain-hardening constants yourself, you would need to plot true stress-strain curves and fit them to the above equation.

If you want to play with some parameters yourself, try

But remember, this strain hardening expression is only valid between the yield strength and ultimate tensile strength. Before the yield strength, the curve will be a straight line with slope = Young’s modulus. After the ultimate tensile strength, the true stress-strain curve can only be determined experimentally.

This empirical equation only works in the region of plastic deformation, before necking occurs (i.e. between the yield point and maximum point on an engineering stress-strain curve).

Converting between the Engineering and True Stress-Strain Curves

I usually hide the math in sections like this, but I’m guessing that most people who find this page are specifically looking for this section. (Yes, I sometimes scoured the internet for help on my homework, too).

To convert from true stress and strain to engineering stress and strain, we need to make two assumptions.

First, we assume that the total volume is constant. In other words,

Second, we need to assume that the strain is evenly distributed across the sample gauge length. This means that we can not convert between true and engineering stresses after necking begins.

$\sigma _{True} =\frac{F}{A_i}=\frac{F}{A_0}\cdot \frac{A_0}{A_i}$

Because there is no volume change, $\frac{A_0}{A_i}=\frac{L_i}{L_0}$$ , so

$\sigma _{True} =\frac{F}{A_0}\cdot \frac{A_0}{A_i}= \frac{F}{\:A_0}\cdot \frac{L_i}{{L_0}}$

And remembering

$\varepsilon_{eng} = \frac{\Delta L}{L_0}=\frac{L_i-L_0}{L_0}=\frac{L_i}{{L_0}}-1$

$\sigma _{eng} =\frac{F}{A_0}$

$\sigma _{True} =\sigma _{eng}(\varepsilon _{eng}+1)$

Continuing to use $(\varepsilon _{eng}+1)=\frac{L_i}{L_0}$ , we see that

$\varepsilon_{True} =ln\left(\frac{L_i}{L_0}\right)=ln\left(\varepsilon _{eng}+1\right)$

\sigma_{True} =\sigma_{Eng}(\varepsilon _{Eng}+1)

The difference between these values increases with plastic deformation. In principle, you could plot two entirely separate curves for true and engineering stress and strain, but in practice, they will be essentially the same until the proportional limit.

Also remember, these equations are only valid before necking begins. Beyond the ultimate strength, you would need actual experimental data (gauge cross section, gauge length, load) to manually compute the true stress-strain curve.

There’s also another problem with graphing the true stress-strain curve: the uniaxial stress correction. This is why the equation doesn’t work after necking.

See, when a tensile specimen is pulled, all of the stress is in one direction: tension.

All the force is along a single axis, so the stress also acts in that axis.

However, once a neck develops, the gauge is no longer homogenous. The material that is necked experiences a more complex stress state, which involves other stress components–not just the tension along the axis!

This requires a correction factor because the component of stress in the axial direction (what you’re trying to measure, because you are only measuring strain in the axial direction) is smaller than the total stress on the specimen.

The graph above shows the engineering stress-strain curve in blue, the calculated true stress-strain curve in red, and the corrected stress-strain curve in red dashes.

You can see why the engineering stress-strain curve is so much more convenient!

Summary

If you understood all of this, congratulations! You know more about the true stress-strain curve than most PhD students!

In summary,

The engineering stress-strain curve is better:

Because it’s easy to calculate and is always more the convenient option if both work
For designing parts in products
For determining toughness or ultimate tensile strength (UTS)
For determining fracture strain or percent elongation

Both curves are equal:

For determining the yield point

The true stress-strain curve is better:

For strain-hardening effects

Additionally, you can convert an engineering stress-strain curve into a true stress-strain curve in the region between the yield point and UTS with the equations:

References and Further Reading

[1] Kalpakjian, Serope and Steven R. Schmid (2014), Manufacturing Engineering and Technology (6th ed.), New York: Pearson Education, p. 62

If you somehow got to the end of this article and didn’t read my general article on stress-strain curves, you probably already know everything in that article. But just in case: here it is.

For more on mechanical properties, check out this presentation from UPenn’s Materials Science Program.

Check out this presentation from National Chung Hsing University to learn more about strain hardening of metals and necking.

Read this publication if you want to know more about strain hardening.

This article was part of a series about mechanical properties.

Engineering Stress To True Stress	$\sigma_{True} =\sigma_{Eng}(\varepsilon _{Eng}+1)$
Engineering Strain To True Strain	$\varepsilon_{True} =ln\left(\frac{L_i}{L_0}\right)=ln\left(\varepsilon _{Eng}+1\right)$

Engineering Stress To True Stress	$\sigma_{True} =\sigma_{Eng}(\varepsilon _{Eng}+1)$
Engineering Strain To True Strain	$\varepsilon_{True} =ln\left(\frac{L_i}{L_0}\right)=ln\left(\varepsilon _{Eng}+1\right)$