18

Can we derive Hooke's law from the theory of elasticity? I know it is not a fundamental law and therefore can be derived from more basic considerations.

Qmechanic
  • 201,751
Solidification
  • 11,483

2 Answers2

16

Yes, we can derive Hooke's Law from more basic continuum conditions, provided that the material be stable and at equilibrium, so that the strain energy is smoothly minimized with respect to the distance between atoms. (This energy is sometimes called the pair potential and is modeled using functions such as the Lennard-Jones potential.)

Consider just a pair of atoms. For slight positional deviations $\delta$ (corresponding to the small-strain assumption of linear elasticity) around the smooth energy minimum $E(0)$, the energy $E$, regardless of its true functional form, can be expanded using a Taylor series:

$$E(\delta)=E(0)+E^{\prime}(0)\delta+\frac{E^{\prime\prime}(0)\delta^2}{2!}+\frac{E^{\prime\prime\prime}(0)\delta^3}{3!}+\cdots,$$

where the prime notation denotes derivatives with respect to position. Now let's set our energy reference to $E(0)=0$, note that $E^\prime(0)=0$ because we're at an energy minimum, and drop all but the very next term. This gives

$$E(\delta)\approx\frac{1}{2}E^{\prime\prime}(0)\delta^2,$$

which describes the energy of an idealized spring with spring constant $E^{\prime\prime}$ and displacement $\delta$; the derivative of this equation with respect to position provides the restoring force, which is

$$F(\delta)\approx E^{\prime\prime}(0)\delta.$$

Now define stress as $\sigma=F/A$ and engineering strain as $\epsilon=\delta/L$ and we have $$\sigma=\frac{LE^{\prime\prime}(0)}{A}\epsilon$$ or $$\sigma=Y\epsilon,$$

which is simple Hooke's Law for a corresponding elastic modulus $Y$.

Chemomechanics
  • 25,546
  • You actually derived Hooke's law from Lennard-Jones potential and then you derived elasticity from Hooke's law. – Diracology May 29 '18 at 18:53
  • 6
    I specifically noted that one does not need to use the Lennard-Jones potential or any specific potential; it's sufficient to assume that the substance is initially at a stable equilibrium. That's the beauty of a Taylor series expansion; it proves that every smooth minimum looks like a parabola from up close. A parabolic energy response is equivalent to a spring-like response. – Chemomechanics May 29 '18 at 18:59
  • That derivation works for materials that doesn't have any imperfections. Like perfect crystals. – nicoguaro May 29 '18 at 19:10
  • 1
    It works pretty well until vacancies/impurities start to make up a fair fraction of the atomic volume fraction. Small amounts of imperfections in crystals generally have much more of an effect on strength than on stiffness. – Chemomechanics May 29 '18 at 19:36
  • Great derivation! I have problems with the step "... provides the restoring force". You do not seem to differentiate w.r.t. the position, otherwise. $E''(0)$ would get an additional prime. Instead, you seem to divide the equation by $\frac{1}{2}\delta$ to get $$\frac{2E(\delta)}{\delta} \approx E''(0)\delta$$ at the LHS and then use $$\frac{2E(\delta)}{\delta} = F(\delta) \tag{$\star$}$$. But why does this last equation $(\star)$ hold? – user7427029 Nov 15 '22 at 13:00
  • I did differentiate with respect to position. $E^{\prime\prime}(0)$ is a constant. – Chemomechanics Nov 15 '22 at 15:26
  • I am not sure I follow the the "small-strain assumption" --- ie how do we know that the positional deviations $\delta$ are very small?. If a spring is stretched by several cm, couldn't the distance between particles increase quite a lot? Or do we also need to assume that the change in distance between particles is much smaller than the radius of an atom? Would a back-of-envelope calculation convince me? – Sanjay Manohar Nov 16 '23 at 23:19
  • If you're referring to a helical or coil spring, you might be surprised by how little shear strain it takes in a polycrystalline lattice to obtain enough torsional twist in a wire to produce large axial spring stretching. There are many spring models and calculators to explore online. – Chemomechanics Nov 16 '23 at 23:59
0

I like this particular derivation because it avoids the requirement to actually specify inter-atomic forces, such as the coulomb force and the Pauli force. Technically, however, you will have one relationship for compression and another for tension and therefore two elastic constants. On the macro-scale this method will yield the classic work and virtual work functions structural engineers are so familiar with. This is an excellent derivation.

Handsome stranger
  • 1