From This page it can be seen that young's modulus is given by, E = (Δstress) / (Δstrain)

Which means E is a point function, that changes as the strain in member is increased.

I am feeling skeptical about this definition, as I haven't read E as derivative anywhere before.

enter image description here

  • $\begingroup$ Perhaps this may help : engineering.stackexchange.com/a/10220/10902 $\endgroup$
    – Solar Mike
    Jan 2, 2019 at 6:18
  • $\begingroup$ Yes and no, in the elastic region for some materials yes, the relationship is pure linear, so the slope is equal to the derivative. But this is not true for polymers for example, the relationship usually described by an exponent later in plastic region, it's not shown in your diagram. $\endgroup$ Jan 2, 2019 at 10:15
  • 1
    $\begingroup$ The bottom line is, treating E as a derivative isn't very useful in practice except for linear materials where it is just a constant, for two reasons. (1) Most stress and strain fields are not uniaxial, so what you are calling "E" is really a parameter in a (nonlinear) fourth-order tensor, not a scalar. (2) It is usually easier to work from a numerical definition of the stress-strain graph itself, rather than by fitting a mathematical function to it. $\endgroup$
    – alephzero
    Jan 2, 2019 at 11:01

4 Answers 4


In the theory of elasticity, the constitutive relation in three dimensions is

$\sigma_{ij} = C_{ijkl}\epsilon_{kl}$

This gives the ${\bf linear}$ relationship between stress and strain.

The fourth order tensor - $C_{ijkl}$ takes 81 components in its own form and reduces to 21 if it is a symmetric stress, and 9 for planar stress. Finally it takes only one component for one dimensional problem such as uni-axial tensile problem.

i.e., $\sigma = C\epsilon$ or, $\sigma = E\epsilon$.

Since the relation between stress & strain is linear (with in elastic limit) it is sufficient having just $\sigma $ and $ \epsilon$ values to get $E$.

Then when why we need $\delta\sigma $ and $ \delta\epsilon$ to calculate $E$?

When you are testing the specimen in UTM, will it show zero deformation for zero load applied? at least for me during my undergraduate studies the load was non-zero when the displacement/deformation was zero. This is due to the friction between the model and fixtures/Jigs.

Also within the elastic region, the material tends to show $hysteresis$ effects if you are using the same sample again and again. This leads to non-zero deflection for zero load.

Lastly, the calculation of $E$ at any specific $\sigma$ or $\epsilon$ is prone to error if the $\sigma$ - $\epsilon$ curve have local error due to the instrumentation/observation.

To avoid these errors it is better to take the infinitesimal changes in stress and strain or slope of the $\sigma$ - $\epsilon$ curve.

Hope this helps!


The Young's modulus (sometimes refered to as elastic modulus) is only relevant for the elastic region of the material and doesn't apply after plastic deformation begins.

Rather than a derivative, the Young's modulus is better defined as the gradient of the linear (elastic) section at the begining of the graph. Or in real life terms it is the gradient of a (linear) line of best fit applied to the data in the elastic (linear) region.

This aproach works fine for most solid materials. But things can start to get complicated for some materials like rubbers.


I dont see any problem with that definition. You can check ie. ISO 527. Where E is calculated from a tensile test.

To further understand the meaning or defining E I'd recommend you to read something about anisotropic materials. You can check for instance long fiber composites - which is orthotropic material, look at the E matrix/tensor.


As has been mentioned in other answers, the Young's modulus, E, is the ratio of stress to strain in the elastic range of stress/strain curve and is a constant, not a derivative.

It is a constant as opposed to what a derivative implies which is a variable depending on the local curvature.

Analysis and design of structures and members is based heavily on the assumption that E is a constant and will remain constant for the periods of loading and unloading a beam, column, or a whole structure.

The fact that E is a slope of straight line and a constant as such, at least assumed, allows all our methods of analysis of the behavior of a structural member work. Otherwise we had to go to unlimited number of trial and errors, assuming a different E for a different stress configuration.

** Edit** If we don't handle Youngs modulus as a constant, we can not assume E constant the entire system of analysing beam stresses and bending collapses.

We calculate I, second area moment which is fundamental in structural analysis based on the assumption that E is constantan and cross section of beams and columns remain planar after bending. If the E is not constant the strain is not linear and I will be meaning less.


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.