Stiffness (F=Kx) is the extent to which an object resists deformation in response to an applied force. Elastic Modulus (E=Stress/Strain) is a quantity that measures an object or substance's resistance to being deformed elastically when a stress is applied to it. In Solid Mechanics, We can relate these K=AE/L. I am confused in these. Both resist deformations when load is applied on it. Is K constant like E is constant. Another thing which is confusing is hardness which is the same (resists deformation on application of load).


3 Answers 3


"$E$", "$K$", and "hardness", all indicate stiffness but are measured and used in different manners.

$E$ - Elastic modulus is defined as the slope of the tangent line to the stress-strain (elastic) curve. It is a material-specific quantity that measures the resistance to being deformed elastically when stress is applied to it, in turn, it indicates the stiffness of the material.

$K$ - Spring constant, or member stiffness factor, is a factor/constant characterized by the geometric terms of a single member/element - $A$, "$L$", "$I$", the material property "$E$", and the load "$P$" and "$M$". It measures the force required to produce a unit displacement/rotation and indicates its strength relative to other members/elements. Also, in a structural joint, it indicates its capability of sharing the force applied to the joint by proportionality.

Worth noting is that in FEM programming, there is a "global stiffness matrix", sometimes nicknamed big-K, which is an assembly of the members/elements stiffness (small-K) in an integral structure. Big-K is usually unique for a given structure.

"Hardness" - Hardness (antonym: softness) is a measure of the resistance to localized plastic deformation induced by either mechanical indentation or abrasion. In general, different materials differ in their hardness. Its usage is limited to the cases stated above.

  • $\begingroup$ Thanks. But I am still confused. In plastic region in stress strain curves, does stiffness lie? (K=AE/L since E is directly proportional to K) and E is no longer present in plastic region. $\endgroup$ Jan 3 at 9:26
  • $\begingroup$ Theoretically, beyond yield, stiffness no longer exists, as the element will deform under constant stress. $\endgroup$
    – r13
    Jan 3 at 11:57
  • 1
    $\begingroup$ +1. An additional way of describing the difference between $E$ and $K$ is that they operate on different units and orders of magnitude: $E$ describes the relationship between stress and strain (so basically how individual atoms move under load), while $K$ describes the relationship between force and deflection (how a beam or structure behaves under load). $\endgroup$
    – Wasabi
    Jan 26 at 19:19
  • $\begingroup$ @Wasabi Agree that's a better explanation to distinguish E & K. $\endgroup$
    – r13
    Jan 26 at 20:22

(Spring) Stiffness $K$ is a property of a structure which includes geometric and material effects.

On the other hand, Young's Modulus $E$ is a property of the material.

Bottom line is that given the same material (i.e. same Young's modulus), changing the cross-section A or the length L could result in different deformation.


  • for a given structure K is constant.
  • for all structures made from the same material E is constant (or approximately).
  • $\begingroup$ In plastic zone, what will be the stiffness (K)? since K=AE/L. What is meant by material effects? How can we define hardness then? $\endgroup$ Dec 30, 2021 at 18:02
  • 1
    $\begingroup$ in the plastic zone, neither the elastic modulus nor stiffness are applicable. Also hardness is also a different quantity (which is indirectly affected by elastic modulus). $\endgroup$
    – NMech
    Dec 30, 2021 at 18:20
  • $\begingroup$ So we should call it zero stiffer material in plastic zone? $\endgroup$ Jan 3 at 9:27
  • $\begingroup$ I'm not sure I understand what you mean by your comment. However, the stiffness is not zero in the plastic zone in the general case. $\endgroup$
    – NMech
    Jan 3 at 9:30
  • $\begingroup$ In elastic region, we have both E & K present, in plastic region E is absent since plastic zone is started. Does stiffness is also absent? If it is present then how? $\endgroup$ Jan 3 at 10:46

Check out this link, it might help.

Should we use Hooke's Law (that linearly relates stresses to strains) if the stiffness of body is changing during deformation?

Now, the stiffness equation i.e. K = EA/L is only used for axial loading conditions. It is derived by dividing the load applied by max deflection. However, for bending cases, the bending stiffness equation becomes: K = EI/L. This is only bending stiffness, and not deformation stiffness due to bending. Check out the difference here:

Why is the beam bending stiffness taken as EI/L?

For plasticity cases, you cannot use these equations to determine the stiffness of the beam or any other object. The reason being: since the stiffness equation (bending or deformation, doesn't matter) is obtained basically by dividing the transverse load by the max deflection in the direction of transverse load, and the max deflection itself is calculated by using either Euler-Bernoulli Beam theory or Timoshenko Beam theory (both of which assumes that the deformations should be small), so using it for regions beyond yield strength means that the deformations are no longer small (since strains are too large). The prediction of deformation from these two theories are no longer valid in plasticity region, and hence cannot be used to find the stiffness of the beam in plasticity. You have to go for Finite Element Analysis (FEA) to numerically approximate the deformations and hence the stiffness of the structure undergoing plasticity.


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