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The classical equation for the Young modulus in elasticity theory for a homogeneous isotropic material in one-dimension is commonly given in the formulation

$$ E = \frac{\sigma}{\epsilon} \quad,$$

with $\sigma$ as the uniaxial stress, and $\epsilon$ as the dimensionless strain parameter.

However, then I also discovered that $E$ can be rewritten in a very different form in terms of the Lamé constants $\mu$ and $\lambda$,

$$ E = \frac{\mu \left(3\lambda + 2\mu\right)}{\lambda + \mu} $$ (vide, e.g. wikipedia).

However, in an unpublished manuscript on stellar astrophysics, a variant of the bulk modulus $K$, which I know now decribes the volumetric elasticity, i.e. elasticity in three-dimensions, is simply stated without derivation as

\begin{equation} K = \rho \cdot\frac{\partial P}{\partial \rho} \quad, \end{equation}

with $\rho$ as mass density and $P$ as pressure.

I was wondering how the equation for $K$ above, for a spherical, isotropic and homogenous body can be derived and if it can be directly linked or expressed in terms of the one-dimensional Young modulus $E$?

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Once you're tired of the 1D limits of simple Hooke's Law $$\varepsilon=\frac{\sigma}{E},$$

you may wish to begin working with generalized Hooke's Law

$$\varepsilon_{ij}=\frac{1+\nu}{E}\sigma_{ij}-\frac{\nu}{E}\sigma_{kk}\delta_{ij}$$

(derivation here, for example), which is capable of describing all 3D linear elastic deformation (of an isotropic material).

This law couples all the isotropic elastic moduli that we can think of.

To link Young's modulus $E$ to the bulk modulus $K$, for instance, recall that the bulk modulus links volumetric strain to hydrostatic pressure (also known as equitriaxial compressive stress, where $\sigma_{xx}=\sigma_{yy}=\sigma_{yy}=-P$ and all other stresses are zero):

$$K\equiv -V\left(\frac{\partial P}{\partial V}\right)_T=\frac{P}{-\Delta V/V},$$

where the last term is for the standard linear elastic conditions that are the focus here. (You can obtain the astrophysics equation you mention by using the chain rule to write $K=-V\left(\frac{\partial P}{\partial \rho}\right)_T\left(\frac{\partial \rho}{\partial V}\right)_T$ and then plugging in the definition $\rho=\frac{m}{V}$.)

We can link the relative decrease in volume to the strain by

$$ -\frac{\Delta V}{V}=-[(1+\varepsilon)^3-1]\approx-3\varepsilon,$$

where the last term assumes the small strains that are also the focus here. From generalized Hooke's Law, setting $j=i$ (so that the Kronecker delta $\delta=1$) and from $\sigma_{kk}= \sigma_{11}+ \sigma_{22} + \sigma_{33}$ (Einstein summation), we have

$$\varepsilon_{ii}=\frac{1-2\nu}{E}\sigma_{ii}.$$

Combining these equations, we obtain

$$K=\frac{E}{3(1-2\nu)}$$

and a much better sense of how the elastic moduli can be linked via generalized Hooke's Law.

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  • $\begingroup$ Thank you very much for your very educative and enlightening answer!! I seems, I am very lucky that you found my question here, as you devoted yourself to the topic of the elastic moduli yourself quite a bit and summarized your thoughts on your homepage! Thank you very much for sharing this very instructive piece with very well conceived animations. :) I think I was able to fully follow your argumentation, and only one question remains after studying your answer: $\endgroup$ Aug 28, 2020 at 10:12
  • $\begingroup$ Am I correct that we apply the generalized Hooke's law in 3 dimensions we get, $$ \epsilon_{ii} = \frac{1+\nu}{E} \sigma_{ii} - \frac{\nu}{E} \sigma_{ii} \delta_{ii} = \frac{1-2\nu}{E} \sigma_{ii} $$ because $\delta_{ii} = 3$ in three dimensions? $\endgroup$ Aug 28, 2020 at 10:16
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    $\begingroup$ Not quite; $\delta_{ii}=1$ but $\sigma_{kk}=\sigma_{11}+ \sigma_{22}+ \sigma_{33}$ because $k$ doesn’t appear elsewhere. This is Einstein summation convention. $\endgroup$ Aug 28, 2020 at 14:40
  • $\begingroup$ Wait, are you sure? Because in three dimensions and also through the E.S., I recall that one gets, $$\delta_{ii} = \delta^{i}_{\; i} = \delta_{11} + \delta_{22} + \delta_{33} = 1 + 1 + 1 \equiv 3 \quad.$$ Just for reference, e.g. math.stackexchange.com/questions/2858877/… . $\endgroup$ Aug 28, 2020 at 14:48
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    $\begingroup$ $\delta_{ii}=3$ when $i$ is a dummy variable, but $i$ isn’t a dummy variable in generalized Hooke’s Law, it’s an index variable (used throughout the equation). $k$ is a dummy variable, repeated within one term and not otherwise defined. $\endgroup$ Aug 28, 2020 at 14:58
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The Bulk modulus in terms of the Lame parameters is $$ \kappa =\lambda +\frac 23 \mu. $$ It is not simply proportional to Youngs modulus. In stretching a wire we do not only make it longer, but we also allow its thickness $W$ to change. Young's modulus therefore needs to know Poisson's ratio, which is defined by $$ \frac{dW}{W}= -\sigma_{\rm Poisson} \frac{dL}{L} $$ and is given by $$ \sigma_{\rm Poisson}= \frac 12 \frac{\lambda}{\lambda+\mu}. $$ The bulk modulus would be relevent to stretching, only if we prevent the wire from getting thinner.

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