# Why do we even use engineering stress?

Surprisingly this hasn't been asked before, so I must be missing something simple.

We use engineering stress and engineering strain in this eq. Stress = (Young's modulus) × (strain). This eq. is used in analysis of bending beams, twisting shafts and in buckling. So the final equation of bending $(\frac{M}{I} = \frac{\sigma}{y})$ and torsion $(\frac{T}{I} = \frac{\tau}{r})$ will give us value of engineering stress but not the value of stress.

Why are we considering engineering stress instead of true stress while we know it will not give correct value of stress?

Some things I read are:

1. Difficult to measure.
2. Not that much of a difference and we can just apply a Factor of Safety.
3. "We don't consider materials to change their cross-sectional area after loading, since we design to have no plastic deformation the elastic region is most important, therefore what happens after the proportional limit is not important"

Firstly, 1 and 2 are not real reasons for me. Number 3 seems plausible since we always design in the elastic region, but is this it? Does engineering strain even give valid information after the proportional limit?

• Approximations abound in engineering. The prudent engineer knows the applicability and limitations to the approximations.
– Paul
Nov 3, 2015 at 19:28

## 2 Answers

We use engineering strain even though it is not the "correct" value because in most cases, specifically in the elastic regime, engineering strain differs negligibly from true strain.

For linear elastic, Hookean materials, it is generally the case strain at the elastic limit is very small. Even the strongest steels, for example, have an upper limit when cold worked of about $\sigma_{\textrm{el}}=1\times 10^{9}\ \textrm{Pa}$. The modulus of steel is approximately $E=200\times 10^{9}\ \textrm{Pa}$. Thus $\varepsilon_{\textrm{el}}=0.005=0.5\%$ for the strongest steels. So at the onset of plastic deformation, engineering strain is $0.5\%$. Many useful elastic materials have much lower engineering strain at their elastic limits.

For an isotropic, Hookean elastic solid, the following is true

$$\varepsilon_{x_{1}} = \frac{1}{E}\left[\sigma_{x_{1}}-\nu\left(\sigma_{x_{2}}+\sigma_{x_{3}}\right)\right]$$

without loss of generality in choice of $x_{i}$. So in uniaxial tension at the elastic limit, $\sigma_{x_{2}}=\sigma_{x_{3}}=0$ assuming the material is free to contract. Thus $\varepsilon_{x_{2}}=\varepsilon_{x_{3}}=-\frac{\sigma_{\textrm{el}}\nu}{E}=-\nu\varepsilon_{\textrm{el}}$. Since the Poisson ratio $\nu$ is approximately 0.3 for steels in the elastic regime, the cross-sectional linear compressive strain is $0.0015$. The cross-sectional area at the elastic limit is thus $(1-0.0015)^{2}A_{0}$, or very close to $0.997$ times the original area.

Thus the true strain is $\frac{1}{0.997}$ times larger than engineering strain at the elastic limit, or about $1.003$ times, or about $0.3\%$ larger. Keep in mind this is at the elastic limit of an exceptionally strong linearly elastic material, and so is a reasonably conservative estimate of the difference between true strain and engineering strain in the elastic regime.

While the above analysis is reasonably useful for linearly elastic, Hookean solids, it doesn't hold quite as well for polymers and biological materials. Such materials are usually viscoelastic (or another class of material entirely), and thus obey different rules in their behavior. True strain also diverges quite wildly from engineering strain in the plastic regime, as evidenced in the following plot (found here)

As for your points:

1. Measuring changes in cross-sectional area during deformation is difficult. It requires careful placement of calibrated instrumentation on precisely machined test samples. One could use strain gages placed on the sides of a tensile bar to measure lateral strain in uniaxial tension and compression in tensile testing equipment. Obtaining statistically meaningful results takes many samples, as well as significant time, effort and cost.

2. There is little difference. I hope I have adequately explained above how small the difference is: I calculated approximately $0.3\%$ difference in a conservative case.

3. The idea that we can ignore anything beyond the end of the elastic regime, or that we always design for the elastic regime, is not true. Plastic deformation can often be worth studying. Modeling continuous shape-forming processes such as rolling, drawing, extruding, etc. requires a deep understanding of the mechanics of plastic deformation to perform successfully, and to that end true stress and true strain are invaluable. Specifically for wire drawing, see (this pdf) and find equation 7. Plastic deformation is also useful for modeling materials that must permanently deform in some expected use cases, such as car body panels and frame components during a collision. The plastic deformation is useful because it absorbs kinetic energy.

Edit: I apologize, I didn't actually answer the question for stress. However, it should be fairly clear that the same points apply to stress as apply to strain given their linear relationship in the elastic regime. Again, in the plastic regime, there can be large variations.

Adding to @starrise's answer:

With regards to your dismissal of reasons 1 and 2, you are forgetting to consider the cost-benefit analysis regarding them. As @starrise showed in their answer, the difference is usually not substantial (though other materials will usually have larger differences).

On the other hand, the materials always show intrinsic variation of their values. Steel's modulus of elasticity has a range defined in different articles from $\pm6\%$ [a] to $\pm15\%$ [b] (for the 95% confidence interval).

So, what's the point of considering the true strain in everyday engineering practice if all the other properties (including yield strength and cross-section dimensions) will have random fluctuations that are all but certain to drown out the "error" due to the use of engineering strain instead of true strain?