Models_of_linearly_elastic

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Equilibrium

$\mathcal S$ in the three-dimensional Euclidean points space are:

\begin{align} &\operatorname{div}\bm F+\bm b=\bm 0,\label{eq:111}\\ &\operatorname{div}_s\bm M-\bm F^T\bm n+\bm c=\bm 0.\label{eq:112} \end{align}

$\bm F$ $\bm M$ $\bm b$ $\bm c$ $\bm n$ $\bm P=\bm I-\bm n\otimes\bm n$ $\operatorname{div}\bm F$ $\bm F$ $\mathbb R^2\ni(\zeta^1,\zeta^2)\mapsto \bm x(\zeta^1,\zeta^2)\in\mathcal S$ by

\operatorname{div}\bm F=\bm F_{,\alpha}\bm a^\alpha,

$\{\bm a^\alpha\}_{\alpha=1,2}$ surface divergence $\bm M$ is

\operatorname{div}_s\bm M=\bm P\operatorname{div}\bm M.

$x\in\mathcal S$ $\bm b(x)$ $\bm n(x)\times\bm c(x)$ $\mathcal S$ $x$ $\bm n$ $\bm F(x)$ $T_x\mathcal S$ $x$ $\bm M(x)$ $T_x\mathcal S$ $\bm\nu(x)$ $x \in\mathcal S$ $\bm F(x)\bm\nu(x)$ $\bm n(x)\times\bm M\bm\nu(x)$ $\bm\nu(x)$ .

$\mathcal P\subset\mathcal S$ $\mathcal P$ must vanish:

\bm 0=\int_{x\in\mathcal P}\boldsymbol b+\int_{x\in\partial\mathcal P}\boldsymbol F\boldsymbol\nu,\qquad \boldsymbol 0=\int_{x\in\mathcal P}\big(\bm x\times\boldsymbol b+\boldsymbol n\times\boldsymbol c\bigr)+\int_{x\in\partial\mathcal P}\bigl(\bm x\times\boldsymbol F\boldsymbol\nu+\boldsymbol n\times\boldsymbol M\boldsymbol \nu\bigr).\label{eq:1111}

$\bm x=x-0$ is the position with respect to the origin.) Such consequence can be drawn by the application of the divergence identity

\int_{\partial\mathcal P}\bm A\bm \nu=\int_{\mathcal P}\operatorname{div}\bm A,

$\eqref{eq:1111}$ $\bm F$ $\bm M$ must obey

\bm P\bm F-\bm M\bm W\in{\rm Sym}.\label{eq:113}

$\bm a_3$ $\bm n$ , and using the component representation, we can write

\bm F=F^{i\beta}\bm a_i\otimes\bm a_\beta,\qquad \bm M=M^{\alpha\beta}\bm a_\alpha\otimes\mathbf a_\beta,\qquad \bm b=b^i\bm a_i,\qquad \bm c=c^\alpha\bm a_\alpha,

$\eqref{eq:111}-\eqref{eq:112}$ can be written as follows

\begin{align} &F^{i\beta}|_{\beta}+b^i=0,\qquad i=1,2,3,\\ &M^{\alpha\beta}|_{\beta}-F^{3\alpha}+c^\alpha=0,\qquad\alpha=1,2. \end{align}

$\eqref{eq:113}$ reads

\varepsilon_{\alpha\beta}(F^{\alpha\beta}-W^\alpha_\gamma M^{\gamma\beta})=0,

$\varepsilon_{\alpha\beta}$ $W^\alpha_\gamma=\bm W\cdot(\bm a^\alpha\otimes\bm a_\gamma)=\bm n_{,\gamma}\cdot\bm a^\alpha$ are the mixed components of the Weingarten tensor.

Principle of virtual work

$\bm u$ $\bm\varphi$ $\bm n$ $x$ $y(x)=x+\bm u(x)$ $\bm d(x)=\bm n(x)+\bm \varphi(x)$ .

$\delta\bm u$ $\delta\bm\varphi$ $\mathcal P$ is:

\delta W(\mathcal P)=\int_{\mathcal P}\bigl(\boldsymbol b\cdot\delta\boldsymbol u+\boldsymbol c\cdot\boldsymbol\delta\bm\varphi\bigl)+\int_{\mathcal P}\boldsymbol F\boldsymbol\nu\cdot\delta\boldsymbol u+\boldsymbol M\boldsymbol\nu\cdot\delta\boldsymbol\varphi.

Integration by parts by making use of the divergence theorem and of the equilibrium equations shows that

\delta W(\mathcal P)=\int_{\mathcal P}\boldsymbol F\cdot\nabla\delta\boldsymbol u+\boldsymbol F^T\bm n\cdot\delta\boldsymbol\varphi+\boldsymbol M\cdot\nabla\delta\boldsymbol\varphi.

Introducing the decomposition

\bm F=\bm N+\bm n\otimes\bm q,\qquad \bm N=\bm P\bm F,\qquad \bm q=\bm F^T\bm n,

$\bm M$ is a tangential tensor, we can write

\delta W(\mathcal P)=\int_{\mathcal P}\bm N\cdot\nabla_s\delta\bm u+\bm q\cdot(\delta\bm\varphi+\nabla\delta\bm u^T\bm n)+\bm M\cdot\nabla_s\delta\bm\varphi,

$\nabla_s=\bm P\nabla$ is the covariant derivative.

Rigid displacements have the form

\bm u(x)=\bm a+\bm\psi\times\bm x(x),\qquad \bm\varphi(x)=\bm\psi\times\bm n(x),

$\bm a$ $\bm\psi$ is a rotation vector. For a rigid displacement,

\nabla\bm u=[\bm\psi\times],\qquad \nabla\bm\varphi=-[\bm\psi\times]\bm W,

$[\bm\psi\times]$ $\bm\psi$ . Note that the shear vector

\bm g=\delta\bm\varphi+\nabla\delta\bm u^T\bm n

$\nabla_s\delta\bm u$ $\nabla_s\delta\bm\varphi$ although being tangential, do not have the latter property. Thus, we cannot take them as strain measures. However, we can rewrite the internal work as

\delta W(\mathcal P)=\int_{\mathcal P}(\bm N-\bm M\bm W)\cdot\nabla_s\delta\bm u+\bm q\cdot(\delta\bm\varphi+\bm n\cdot\nabla\delta\bm u^T)+\bm M\cdot(\nabla_s\delta\bm\varphi+\nabla_s\delta\bm u\, \bm W).

$\delta\bm\varphi(x)=\bm\psi\times\bm n(x)$ $\nabla_s\delta\bm\varphi=-[\bm\psi\times]\bm W$ $\nabla_s\delta\bm u=[\bm\psi\times]$ $\nabla_s\delta\bm\varphi+\nabla_s\delta\bm u\bm W$ $\bm N-\bm M\bm W$ is symmetric, then

\delta W(\mathcal P)=\int_{\mathcal P}(\bm N-\bm M\bm W)\cdot\operatorname{sym}\nabla_s\delta\bm u+\bm q\cdot(\delta\bm\varphi+\bm n\cdot\nabla\delta\bm u^T)+\bm M\cdot(\nabla_s\delta\bm\varphi+\nabla_s\delta\bm u\, \bm W)

$\operatorname{sym}\nabla_s\delta\bm u$ $\delta\bm u$ is an infinitesimal rigid displacement. This and the previous results suggest that we adopt

\bm E=\operatorname{sym}\nabla_s\bm u,\qquad \bm K=\nabla_s\bm\varphi+\nabla_s\bm u\bm W

as additional strain measures, so that the virtual work reads

\delta W(\mathcal P)=\int_{\mathcal P}(\bm N-\bm M\bm W)\cdot\delta\bm E+\bm q\cdot\delta\bm g+\bm M\cdot\delta\bm K.

The statement that

\int_{\mathcal P}(\bm N-\bm M\bm W)\cdot\delta\bm E+\bm q\cdot\delta\bm g+\bm M\cdot\delta\bm K=\int_{\mathcal P}\bigl(\boldsymbol b\cdot\delta\boldsymbol u+\boldsymbol c\cdot\boldsymbol\delta\bm\varphi\bigl)+\int_{\mathcal P}\boldsymbol F\boldsymbol\nu\cdot\delta\boldsymbol u+\boldsymbol M\boldsymbol\nu\cdot\delta\boldsymbol\varphi.

$\mathcal P$ Principle of Virtual Work $\mathcal P$ $\mathcal S$ , then

\int_{\mathcal P}(\bm N-\bm M\bm W)\cdot\delta\bm E+\bm q\cdot\delta\bm g+\bm M\cdot\delta\bm K=\int_{\mathcal P}\bigl(\boldsymbol b\cdot\delta\boldsymbol u+\boldsymbol c\cdot\boldsymbol\delta\bm\varphi\bigl)+\int_{\mathcal P}\bm f\cdot\delta\boldsymbol u+\bm m\cdot\delta\boldsymbol\varphi,

$\bm f$ $\bm m$ are the forces and moments imposed on the boundary of the shell.

$\bm W=-\nabla\bm n$ $T_x$ into itself, we have

\nabla\bm u^T\bm n=\nabla(\bm u\cdot\bm n)-\nabla\bm n^T\bm u=\nabla w+\bm W\bm v \label{eq:strain1}

where we have introduced the decomposition of the displacement into its tangential and normal components

\bm u=\bm v+w\bm n.

$\nabla\bm u=\nabla\bm v+\bm n\otimes\nabla w-w\bm W$ , we have

\nabla_s\bm u=\nabla_s\bm v-w\bm W.

$\bm W$ is symmetric,

\bm E=\operatorname{sym}\nabla_s\bm v-w\bm W,\qquad \bm g=\delta\bm\varphi+\nabla w+\bm W\bm v,\qquad \bm K=\nabla\bm\varphi+\nabla_s\bm v\bm W-w\bm W^2.

Unshearable shells

$\bm g=\bm 0$ , which yields

\bm\varphi=-\nabla\bm u^T\bm n=-\nabla w-\bm W\bm v.

In this case,

\bm K=-\nabla\bm u^T\bm n+\nabla_s\bm u\bm W=-\nabla_s\nabla w+\nabla_s(\bm W\bm v)+\nabla_s\bm v\bm W-w\bm W^2.

The internal work reduces to

\delta W(\mathcal P)=\int_{\mathcal P}(\bm N-\bm M\bm W)\cdot\delta\bm E+\bm M\cdot\delta\bm K.

$\bm q=\bm F^T\bm n$ $\bm b$ into its tangential and normal part:

\bm b=\bm b_t+b_n\bm n,

and we write the equilibrium equations as:

\begin{align} &\operatorname{div}_s\bm N+\bm b_t=\bm 0,\label{eq:11122}\\ &\operatorname{div}_s\bm q+b_n=0\\ &\operatorname{div}_s\bm M-\bm q+\bm c=\bm 0.\label{eq:1121} \end{align}

$\operatorname{div}\bm q=\bm q_{,\alpha}\cdot\bm a^\alpha$ $\bm q$ . Then we combine the last two equations to obtain a pair of equations in pure form:

\begin{align} &\operatorname{div}_s\bm N+\bm b_t=\bm 0,\\ &\operatorname{div}\operatorname{div}_s\bm M+\operatorname{div}\bm c-b_n=0. \end{align}

Linear Kirchhoff-Love shells

The expression of the internal work suggest a constitutive equation for the strain energy that is quadratic and separable, in the sense that it has the form

E(\mathcal P)=\frac h 2 \int_{\mathcal P}\mathbb C[\bm E]\cdot\bm E+\frac {h^3}{24}\int_{\mathcal P}\mathbb C[\bm K]\cdot\bm K,

$\mathbb C$ is an isotropic fourth-order tensor:

\mathbb C\bm A=2\widetilde\mu\bm A+\widetilde\lambda(\operatorname{tr}\bm A){\bm P}.

The time derivative of the strain energy is

\dot E(\mathcal P)=\int_{\mathcal P}\mathbb N[\bm E]\cdot\dot{\bm E}+\mathbb M[\bm K]\cdot\dot{\bm K},

where

\mathbb N=h\mathbb C,\qquad \mathbb M=\frac {h^3}{12}\mathbb C.

On taking into account the expression for the internal work, the dissipation principle writes:

\int_{\mathcal P}(\bm N-\bm M\bm W-\mathbb{N}[\bm E])\cdot\dot{\bm E}+(\bm M-\mathbb M[\bm K])\cdot\dot{\bm K}\ge 0,

This suggests the constitutive equations

\bm N=\mathbb N[\bm E]+\mathbb M[\bm K]\bm W,\qquad \bm M=\mathbb M[\bm K].

$\bm q=\bm F^T\bm n$ is constitutively undetermined. Hence, to obtain equilibrium equations in pure form we must combine

\begin{aligned} &\operatorname{div}_s\bm N-\bm W(\operatorname{div}_s\bm M)+\bm p=\bm 0,\\ &\operatorname{div}_s\operatorname{div}_s\bm M-q=0. \end{aligned}

Linear Koiter shells

$\mathcal P$ of a linear Koiter shell is

E(\mathcal P)=\frac h 2 \int_{\mathcal P} \mathbb C[\bm E]\cdot\bm E+\frac {h^3}{24}\int_{\mathcal P}\mathbb C[\widetilde{\bm K}]\cdot \widetilde{\bm K},

where

\widetilde{\bm K}=\bm K-\bm W\nabla\bm u-\nabla_s\bm u\bm W=\nabla\bm\varphi-\bm W\nabla\bm u.

The time derivative of the strain energy is

\dot E(\mathcal P)=\int_{\mathcal P}(\widetilde{\bm N}-\bm W\mathbb M[\widetilde{\bm K}]-\mathbb M[\widetilde{\bm K}]\bm W)\cdot\dot{\bm E}+\widetilde{\bm M}\cdot\dot{\bm K},

The dissipation principle yields

\int_{\mathcal P}(\bm N-\mathbb N[\bm E]+\bm W\mathbb M[\widetilde{\bm K}])\cdot\dot{\bm E}+(\bm M-\mathbb M[\widetilde{\bm K}])\cdot\dot{\bm K}\ge 0.

In particular, we have the constitutive equations

\bm N=\mathbb N[\bm E]-\bm W\mathbb M[\widetilde{\bm K}],\qquad \bm M=\mathbb M[\widetilde{\bm K}].

Additional details and references can be found here).

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