Introduction to Mathematical Physics/Energy in continuous media/Generalized elasticity

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Introduction[edit]

In this section, the concept of elastic energy is presented. \index{elasticity} The notion of elastic energy allows to deduce easily "strains--deformations" relations.\index{strain--deformation relation} So, in modelization of matter by virtual powers method \index{virtual powers} a power P that is a functional of displacement is introduced. Consider in particular case of a mass m attached to a spring of constant k.Deformation of the system is referenced by the elongation x of the spring with respect to equilibrium. The virtual work \index{virtual work} associated to a displacement dx is

deltWfdx


\delta W=f.dx

Quantity f represents the constraint , here a force, and x is the deformation. If force f is conservative, then it is known that the elementary work (provided by the exterior) is the total differential of a potential energy function or internal energy U :

eqdeltaWdU


\delta W=-dU

In general, force f depends on the deformation. Relation f=f(x) is thus a constraint--deformation relation .


The most natural way to find the strain-deformation relation is the following. One looks for the expression of U as a function of the deformations using the physics of the the problem and symmetries. In the particular case of an oscillator, the internal energy has to depend only on the distance x to equilibrium position. If U admits an expansion at x=0, in the neighbourhood of the equilibrium position U can be approximated by:


U(x)=a_0+a_1x^1+a_2x^2+O(x^2)

As x=0 is an equilibrium position, we have dU=0 at x=0. That implies that a_1 is zero. Curve U(x) at the neighbourhood of equilibrium has thus a parabolic shape (see figure figparabe

figparabe

In the neighbourhood of a stable equilibrium position x_0, the intern energy function U, as a function of the difference to equilibrium presents a parabolic profile.

As


dU=-fdx

the strain--deformation relation becomes:


f=\frac{dU}{dx}

Oscillators chains[edit]

Consider a unidimensional chain of N oscillators coupled by springs of constant k_{ij}. this system is represented at figure figchaineosc. Each oscillator is referenced by its difference position x_i with respect to equilibrium position. A calculation using the Newton's law of motion implies:


U=\sum\frac{1}{2}k_{ij}(x_i-x_{j-1})^2

figchaineosc

A coupled oscillator chain is a toy example for studying elasticity.

A calculation using virtual powers principle would have consisted in affirming: The total elastic potential energy is in general a function U(x_1,\dots,
x_N)<math> of the differences x_i</math> to the equilibrium positions. This differential is total since force is conservative\footnote{ This assumption is the most difficult to prove in the theories on elasticity as it will be shown at next section} . So, at equilibrium: \index{equilibrium} :


dU=0

If U admits a Taylor expansion:

eqdevliUch


U(x_1=0,\dots x_N=0)=a+a_i x_i+a_{ij}x_ix_j+O(x^2)

In this last equation, repeated index summing convention as been used. Defining the differential of the intern energy as:


dU=f_i dx_i

one obtains


f_i=\frac{\partial U}{\partial x_i}.

Using expression of U provided by equation eqdevliUch yields to:


f_i=a_{ij}x_j.

But here, as the interaction occurs only between nearest neighbours, variables x_i are not the right thermodynamical variables. let us choose as thermodynamical variables the variables \epsilon_i defined by:


\epsilon_i=x_i-x_{i-1}.

Differential of U becomes:


dU=F_id\epsilon_i

Assuming that U admits a Taylor expansion around the equilibrium position:


U=b+b_i\epsilon_i+b_{ij}\epsilon_i\epsilon_j+O(\epsilon^2)

and that dU=0 at equilibrium, yields to:


F_i=b_{ij}\epsilon_j

As the interaction occurs only between nearest neighbours:


b_{ij}=0 \mbox{ si  } i\neq j\pm 1

so:


F_i=b_{ii}\epsilon_i+b_{ii+1}\epsilon_{i+1}

This does correspond to the expression of the force applied to mass i :


F_i=-k(x_{i}-x_{i-1})-k(x_{i+1}-x_{i})

if one sets k=-b_{ii}=-b_{ii+1}.

secmaterelast

Tridimensional elastic material[edit]

Consider a system S in a state S_X which is a deformation from the state S_0. Each particle position is referenced by a vector a in the state S_0 and by the vector x in the state S_X:


x=a+X

Vector X represents the deformation.

Remark:

Such a model allows to describe for instance fluids and solids.

Consider the case where X is always "small". Such an hypothesis is called small perturbations hypothesis (SPH). The intern energy is looked as a function U(X).

Definition: The deformation tensor SPH is the symmetric part of the tensor gradient of X.


\epsilon_{ij}=\frac{1}{2}(X_{i,j}-X_{j,i})

At section secpuisvirtu it has been seen that the power of the admissible intern strains for the problem considered here is:


P_i=\int K_{ij}^s u_{i,j}^s d\tau

with


dU=-P_idt

Tensor u_{i,j}^s is called rate of deformation tensor. It is the symmetric part of tensor u_{i,j}. It can be shown [ph:fluid:Germain80] that in the frame of SPH hypothesis, the rate of deformation tensor is simply the time derivative of SPH deformation tensor:


u_{i,j}^s=\frac{d\epsilon_{ij}}{dt}

Thus:

dukij


dU=-\int K_{ij}^s d\epsilon_{ij}d\tau.

Function U can thus be considered as a function U(\epsilon_{ij}). More precisely, one looks for U that can be written:


U=\int \rho e_l d\tau

where e_l is an internal energy density with\footnote{ Function U depends only on \epsilon_{ij}.} whose Taylor expansion around the equilibrium position is:

eqrhoel


\rho e_l=a+a_{ij}\epsilon_{ij}+a_{ijkl}\epsilon_{ij}\epsilon_{kl}

We have\footnote{

footdensi

Indeed:


\frac{d}{dt}U=\frac{d}{dt}\int \rho e_l d\tau

and from the properties of the particulaire derivative:


\frac{d}{dt}\int \rho e_l d\tau=\int\frac{d}{dt}( \rho e_l d\tau)

Now,


\frac{d}{dt} (\rho e_ld\tau)=e_l\frac{d}{dt} (\rho d\tau) + \rho
d\tau\frac{d}{dt}e_l

From the mass conservation law:


\frac{d}{dt} \rho=0

}

eqdudt

\frac{dU}{dt}=\int \rho (\frac{d}{dt}e_l) d\tau

Thus


dU=\int \rho de_l d\tau

Using expression eqrhoel of e_l and assuming that dU is zero at equilibrium, we have:


dU=\int \rho  [a_{ijkl}\epsilon_{ij}d\epsilon_{kl}+
a_{ijkl}d\epsilon_{ij}\epsilon_{kl}]d\tau

thus:


dU=\int \rho  b_{ijkl}\epsilon_{kl}d\epsilon_{ij}d\tau

with b_{ijkl}=a_{ijkl}+a_{klij}. Identification with equation dukij, yields to the following strain--deformation relation:


K_{ij}^s=b_{ijkl}\epsilon_{kl}

it is a generalized Hooke law\index{Hooke law}. The b_{ijkl}'s are the elasticity coefficients.

Remark: Calculation of the footnote footdensi show that calculations done at previous section secchampdslamat should deal with volumic energy densities.

secenernema

Nematic material[edit]

A nematic material\index{nematic} is a material [ph:liqcr:DeGennes74] whose state can be defined by vector field\footnote{ State of smectic materials can be defined by a function u(x,y). } n. This field is related to the orientation of the molecules in the material (see figure figchampnema)

figchampnema

Each molecule orientation in the nematic material can be described by a vector n. In a continuous model, this yields to a vector field n. Internal energy of the nematic is a function of the vector field n and its partial derivatives.

Let us look for an internal energy U that depends on the gradients of the n field:


U=\int u d\tau

with


u=u_1(\partial_in_j)+u_2(n_i\partial_jn_k)+ u_3(\partial_i
n_j\partial_kn_l)+ u_4(n_pn_q\partial_i n_j\partial_kn_l)+\dots

The most general form of u_1 for a linear dependence on the derivatives is:

eqsansder


u_1(\partial_in_j)= K_{ij} \partial_i n_j

where K_{ij} is a second order tensor depending on r. Let us consider how symmetries can simplify this last form.

  • Rotation invariance. Functional u_1 should be rotation invariant.

  u_1(\partial_i n_j)=u_1(R_{ik}R_{jl}\partial_k n_l)

where R_{mn} are orthogonal transformations (rotations). We thus have the condition:

  K_{ij}=R_{ik}R_{jl}K_{kl},

that is, tensor K_{ij} has to be isotrope. It is know that the only second order isotrope tensor in a three dimensional space is \delta_{ij}, that is the identity. So u_1 could always be written like:

  u_1=k_0\mbox{ div } n

  • Invariance under the transformation n maps to -n . The energy of distortion is independent on the sense of n, that is u_1(n)=u_1(-n). This implies that the constant k_0 in the previous equation is zero.

Thus, there is no possible energy that has the form given by equation eqsansder. This yields to consider next possible term u_2. general form for u_2 is:

u_2=L_{ijk}n_k\partial_in_j

Let us consider how symmetries can simplify this last form.

  • Invariance under the transformation n maps to -n . This invariance condition is well fulfilled by u_2.
  • Rotation invariance. The rotation invariance condition implies that:

  L_{ijk}=R_{il}R_{jm}R_{kn}L_{lmn}

It is known that there does not exist any third order isotrope tensor in R^3, but there exist a third order isotrope pseudo tensor: the signature pseudo tensor e_{jkl} (see appendix secformultens). This yields to the expression:

  u_2=k_1e_{ijk}n_k\partial_in_j=k_1n\mbox{ rot } n.

  • {\bf Invariance of the energy with respect to the axis transformation x\rightarrow -x, y\rightarrow -y, z\rightarrow -z.} The energy of nematic crystals has this invariance property\footnote{ Cholesteric crystal doesn't verify this condition.} . Since e_{ijk} is a pseudo-tensor it changes its signs for such transformation.

There are thus no term u_2 in the expression of the internal energy for a nematic crystal. Using similar argumentation, it can be shown that u_3 can always be written:

u_3=K_1 (\mbox{ div } n)^2

and u_4:

u_4=K_2 (n.\mbox{ rot } n)^2+K_3(n \wedge \mbox{ rot } n)^2

Limiting the development of the density energy u to second order partial derivatives of n yields thus to the expression:

u=K_1 (\mbox{ div } n)^2+K_2 (n.\mbox{ rot } n)^2+K_3(n \wedge \mbox{ rot } n)^2