Models_of_linearly_elastic

This page is work in progress

Models of linearly elastic shells

The equations that govern the equilibrium of linearly elastic shells are

\begin{aligned} &\operatorname{div}\bm F+\bm b=\bm 0\\ &\operatorname{div}_s\bm M+\bm c+\bm q=0. \end{aligned}

Geometry

$\mathcal S$ $\mathcal E$ $\mathcal S$ $\boldsymbol n(x)$ $x$ .

$V$ $x\in\mathcal S$ tangent space $x$

T_x=\{\boldsymbol v\in V:\boldsymbol v\cdot\boldsymbol n(x)=0\}.

$\mathcal S$ as the label set to identify the particles that comprise the shell. The label set should not be confused with the configurationplacement $\widehat y:\mathcal S\to\mathcal E$ director field $\bm d:\mathcal S\to \partial B(0,1)$ $\widehat{y}$ $\mathcal S'$ in the Euclidean space, but does not embody complete information on its configuration.

Equilibrium

$\boldsymbol F$ $\boldsymbol M$ $\mathcal P$ are given by

\boldsymbol r(\mathcal P)=\int_{x\in\mathcal P}\boldsymbol b+\int_{x\in\partial\mathcal P}\boldsymbol F\boldsymbol\nu,\qquad \boldsymbol m(\mathcal P)=\int_{x\in\mathcal P}\big(\bm y\times\boldsymbol b+\boldsymbol d\times\boldsymbol c\bigr)+\int_{x\in\partial\mathcal P}\bigl(\bm y\times\boldsymbol F\boldsymbol\nu+\boldsymbol d\times\boldsymbol M\boldsymbol \nu\bigr).\label{eq:balance2}

$\boldsymbol b(x)$ $\boldsymbol d (x)\times \boldsymbol c(x)$ reference area $x$ $\bm y(x)$ $\widehat y(x)$ $\boldsymbol\nu(x)\in T_x$ $\mathcal S$ $\partial\mathcal P$ $\mathcal P$ $x$ $\boldsymbol F(x)\boldsymbol \nu(x)$ $\boldsymbol d(x)\times\boldsymbol M(x)\boldsymbol \nu(x)$ represent, respectively, the linear densities of contact forcecontact couple $\mathcal P$ $x$ .

Remarks:

$\bm d$ encodes the assumption that a shell cannot sustain couples parallel to the director.
$\bm c$ $\bm M\bm\nu$ $\bm d$ .
$\boldsymbol F(x)$ $T_x$ $V$ $\boldsymbol M(x)$ $T_x$ $\bm d(x)$ :

\bm M(x):T_x\to\operatorname{span}\{\bm d(x)\}^\perp,\qquad\operatorname{span}\{\bm d\}^\perp=\{\bm v\in V:\bm v\cdot\bm d=0\}.

$\bm F$ $\bm M$ are the homologous of the Piola stress.
$\mathcal S$ $\mathcal S$ .

Divergence theorems

$\partial\mathcal P$ $\mathcal P$ . To begin with, we record the following identity

\int_{\partial\mathcal P}\boldsymbol v\cdot\boldsymbol\nu=\int_{\mathcal P}\operatorname{div}\boldsymbol v,\quad \text{where}\quad \operatorname{div}\boldsymbol v=\boldsymbol v_{,\alpha}\cdot\boldsymbol a^\alpha.

$\mathbf A=\mathbf a\otimes\mathbf v$ $\mathbf a$ is a constant vector. Then

\int_{\mathcal P}\boldsymbol A\boldsymbol\nu=\boldsymbol a\otimes\int_{\partial\mathcal P}\boldsymbol v\cdot\boldsymbol\nu=\boldsymbol a\otimes\int_{\mathcal P}\operatorname{div}\boldsymbol v=\int_{\mathcal P}\boldsymbol A_{,\alpha}\boldsymbol a^\alpha.\label{eq:divergence2}

$\mathbf A$ $\boldsymbol A=\boldsymbol e_i\otimes\boldsymbol v^{(i)}$ $\eqref{eq:divergence2}$ $\boldsymbol A$ . Thus, we can conveniently take

\operatorname{div}\boldsymbol T=\boldsymbol T_{,\alpha}\boldsymbol a^\alpha.

$\boldsymbol T$ .

Pointwise form of the equilibrium equations

$\boldsymbol r(\mathcal P)$ $\boldsymbol m(\mathcal P)$ $\mathcal P \subset \mathcal S$ . It is not difficult, using the divergence theorem and a standard localization argument to deduce the equivalence:

\begin{aligned} \boldsymbol r(\mathcal P)=\boldsymbol 0&\quad\forall\mathcal P\subset\mathcal S\\ &\Updownarrow\\ \operatorname{div}\boldsymbol F+\boldsymbol b=\boldsymbol 0\color{black}&\quad\text{everywhere in }\mathcal S. \end{aligned}\label{eq:balance_pointwise1}

A similar argument can be applied to the equilibrium of moments. However, the result of this procedure is more transparent if we use the identity

(\bm a\times\bm b)\times\bm c=(\bm a\otimes\bm b-\bm b\otimes\bm a)\bm c=2\operatorname{skw}(\bm a\otimes\bm b)\bm c.

$\mathcal P$ is equivalent to

\boldsymbol 0=\operatorname{skw}\Bigl[\int_{\mathcal P}\bigl(\bm y\otimes\boldsymbol b+\boldsymbol d\otimes\boldsymbol c\bigr)+\int_{\partial\mathcal P}\bigl(\boldsymbol y\otimes\boldsymbol F\boldsymbol\nu+\boldsymbol d\otimes\boldsymbol M\boldsymbol \nu\bigr)\Bigr]\qquad\forall\mathcal P\subset\mathcal S,

Then a procedure similar as above yields the following equivalence

\begin{aligned} \boldsymbol m(\mathcal P)=\boldsymbol 0&\quad\forall\mathcal P\subset\mathcal S \\ &\Updownarrow \\ \operatorname{skw}\bigl(\boldsymbol d\otimes(\operatorname{div}\boldsymbol M+\boldsymbol c+\bm F^T\bm d)+\boldsymbol y\otimes(\operatorname{div}&\boldsymbol F+\boldsymbol b)+\nabla\bm y\,\bm F^T+\nabla\bm d\,\bm M^T\bigr)=\boldsymbol 0\quad \text{in }\mathcal S,\label{eq:pw1} \end{aligned}

From this fact it is not difficult to conclude that

\begin{aligned} &(\bm I-\bm d\otimes\bm d)\operatorname{div}\boldsymbol M+\boldsymbol c+\bm q=\boldsymbol 0,\\[0.2em] &\bm N\nabla\bm y^T +\bm M\nabla\bm d^T\in{\rm Sym}. \end{aligned}

where we have introduced the decomposition:

\boldsymbol F=\boldsymbol d\otimes\boldsymbol q+\boldsymbol N,\qquad \boldsymbol q=\boldsymbol F^T\boldsymbol d,\qquad \bm N=(\bm I-\bm d\otimes\bm d)\bm F.

Small displacements

$\bm y(\bm x)=\bm x$ $\bm d=\bm n$ $\nabla\bm y=\bm P$ $\nabla\bm d=-\bm W$ , where

\bm P=\nabla\bm x,\qquad \text{and}\qquad -\bm W=\nabla\bm n

$\bm P=\bm I-\bm n\otimes\bm n$ $\bm P$ $T_x$ .

We introduce the surface divergence

\operatorname{div}_s(\cdot)=\bm P\operatorname{div}(\cdot),

$\bm W$ is symmetric, then we recover the standard equilibrium equations used in the theory of linearly elastic shells:

\begin{aligned}&\operatorname{div}_s\bmsf M+\bmsf c+\bmsf q=\boldsymbol 0,\\[0.2em]&\bmsf N-\bmsf M\bmsf W\in{\rm Sym}.\end{aligned}

$\boldsymbol q$ shear vector $\boldsymbol q\cdot\boldsymbol\nu$ $\boldsymbol n$ $\boldsymbol \nu$ .

Remarks.

The geometrical structure of the shell emerges only at this stage, when we single out a stress-free configuration.

$\bm P$ $\bm W$ are superficial tensor fields, hence:

\bm P=\nabla_x\bm x,\qquad -\bm W=\nabla_s\bm n.

Virtual work

$\boldsymbol u$ $\boldsymbol\varphi$ $x$ $\boldsymbol y=\boldsymbol x+\boldsymbol u(x)$ $\boldsymbol d=\bm n+\boldsymbol\omega\times\boldsymbol n=\bm n+\bm\varphi$ . Then the work performed by the applied couples on a virtual rotation is:

\boldsymbol n\times\boldsymbol c\cdot\delta\boldsymbol\omega=\boldsymbol c\cdot\delta\boldsymbol\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 q\cdot\delta\boldsymbol\varphi+\boldsymbol M\cdot\nabla_s\delta\boldsymbol\varphi.

Using the previously introduced decomposition, we can write

\delta W(\mathcal P)=\int_{\mathcal P}\bm N\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.

Note that the vector

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

$\nabla_s\delta\bm u$ $\nabla_s\delta\bm\varphi$ are not. 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.

$\bm W=-\nabla\bm n$ $T_x$ 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.

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].

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}].

Appendix

Gradient and divergence on a surface

$\bm v$ $\mathcal S$ $x$ $T_x$ $V$ $\bm a$ $a:\mathbb R\to\mathcal S$ $a(0)=x$ $a'(0)=\bm a$ , and by letting

\nabla\boldsymbol v[\boldsymbol a]=(\bm v\circ a)'(0).

covariant derivative $\bm v$ $D\bm v=\bm P\nabla\bm v$ $T_x$ $T_x$

Invariance under a change of observer

A rigid displacement from the reference configuration has the form

\delta\bm u(x)=\delta\bm v+\delta\bm\psi\times\bm x(x),\qquad \delta\bm\varphi(x)=\delta\bm\psi\times\bm n(x).

$[\delta\bm\psi\times]$ $\delta\bm\psi$ , we can write

\nabla\delta\bm u=[\bm\psi\times]\bm P,

hence

\delta\bm\gamma=\delta\bm\varphi-\nabla\bm u^T\bm n=[\delta\bm\psi\times]\bm n+[\delta\bm\psi\times]^T\bm n=\bm 0.

$\delta\bm\gamma$ is invariant under a change of observer.

The effective Lamé constants for a linearly elastic shell

For a three-dimensional isotropic, linearly elastic material, the stress-strain relation has the form

\mathbf S=2\mu\mathbf E+\lambda(\operatorname{tr}\mathbf E)\mathbf I. \label{eq:resp}

$\mu$ $\lambda$ https://doi.org/10.1007/s10659-005-4738-8 $h$ $\eqref{eq:resp}$ is captured by a shell with Lamé constants (See the discussion about the convergence

\widetilde\mu=\mu,\qquad \widetilde\lambda=\frac{2\mu}{2\mu+\lambda}\lambda.

$\mathbf S\bm n\cdot\bm n=0$ , which leads to the system

\begin{aligned} \bmsf S=2\mu\bmsf E+\lambda({\rm tr}\bmsf E+\varepsilon)\bmsf P\\ (2\mu+\lambda)\varepsilon+\lambda{\rm tr}\bmsf E=0. \end{aligned}

Details to obtain the pointwise form of the balance equations

The divergence of a tensor field can be written as

\operatorname{div}\bm T=\bm T_{,\alpha}\bm a^\alpha.

$\bm v$ $\bm T$ is a tensor field, we have

\operatorname{div}(\bm v\otimes\bm T)=\bm v\otimes\operatorname{div}\bm T+\bm v_{,\alpha}\otimes(\bm T\bm a^\alpha)=\bm v\otimes\operatorname{div}\bm T+\bm (\bm v_{,\alpha}\otimes \bm a^\alpha)\bm T^T=\bm v\otimes\operatorname{div}\bm T+\nabla\bm v\ \bm T^T

$\nabla\bm x=\bm P$ $\nabla\bm n=-\bm W$ , we obtain

\operatorname{div}(\bm x\otimes\bm F)=\bm x\otimes\operatorname{div}\bm F+\bm P\bm F^T,\qquad \operatorname{div}(\bm n\otimes\bm M)=\bm n\otimes\operatorname{div}\bm M-\bm W\bm M^T.

$\bm F=\bm F\bm P$ $\bmsf P=\bmsf P^T$ $\bm 1=\bm P+\bm n\otimes\bm n$ ,

\bm P\bm F^T=\bm F^T=\bm F^T\bm P+\bm F^T\bm n\otimes\bm n.

Hence,

\int_{\partial\mathcal P}(\bm x\otimes\bm F\bm\nu+\bm n\otimes\bm M\bm\nu)=\int_{\mathcal P}(\bm x\otimes\operatorname{div}\bm F+\bm F^T\bmsf P+\bm n\otimes\operatorname{div}\bm M+\bm F^T\bm n\otimes\bm n-\bm W\bm M^T).

Linear Koiter energy

A finite deformation of the shell is described by an one-to-one smooth map

\bm y:\mathcal S\to\mathcal S'.

$\nabla\bm y(\bm x)$ $T_x$ $T'_{\bm y(\bm x)}$ $T'_{\bm x'}$ $\mathcal S'$ $\bm x'$ . We define the strain as

\bmsf D=\nabla\bm y^T\nabla\bm y/2.

$\bm x:\mathbb R^2\to\mathcal S$ $\mathcal S$ $\bm x'=\bm y\circ\bm x$ $\mathcal S'$ .

$\bm n':\mathcal S'\to V$ $\mathcal S'$ is

\bmsf W'=-\nabla'\bm n'=-\nabla(\bm n'\circ\bm y)(\nabla\bm y)^{-1}.

$\mathcal P$ is

E(\mathcal P)=\frac 12\int_{\mathcal P}h\mathbb C[\bmsf D]\cdot\bmsf D+\frac {h^3}{12}\mathbb C[\bmsf W'\circ\bm y^{-1}-\bmsf W]:(\bmsf W'\circ\bm y^{-1}-\bmsf W)

$\bmsf D$ is

d\bmsf D=\frac 12(\bmsf P\nabla\bm u+\nabla\bm u^T\bmsf P)=\frac 12(\bmsf P\nabla\bm v+\nabla\bm v^T\bmsf P)-w\bmsf W.

$\bmsf W$ is

d\bmsf W=-d\bm n_{,\alpha}\otimes\bm a^\alpha-\bm n_{,\alpha}\otimes d\bm a^\alpha=-\bm\varphi_{,\alpha}\otimes\bm a^\alpha+\bmsf W\bm a_\alpha\otimes d\bm a^\alpha=-\bm \Phi+\bm W\bm a_\alpha\otimes d\bm a^\alpha.

$d\bm x=\bm u$ $d\bm a_\alpha=d\bm x_{,\alpha}=\bm u_{,\alpha}$ . Thus,

\begin{aligned} -\bm n\otimes d\bm n-d\bm n\otimes\bm n&=d\bmsf P=\bm a_\alpha\otimes d\bm a^\alpha+d\bm a_\alpha\otimes \bm a^\alpha\\ &=\bm a_\alpha\otimes d\bm a^\alpha+\bm u_{,\alpha}\otimes \bm a^\alpha\\ &=\bm a_\alpha\otimes d\bm a^\alpha+\nabla\bm u\\ &=\bm a_\alpha\otimes d\bm a^\alpha+\nabla \bm v+\bm n\otimes\nabla w-w\bmsf W. \end{aligned}

$\bmsf P$ ,

\bm W\bm a_\alpha\otimes d\bm a^\alpha=-\bmsf Wd\bm n\otimes\bm n-\bm W\nabla\bm v+w\bmsf W^2

$d\bm n=\bm\varphi$ , hence

\bm W\bm a_\alpha\otimes d\bm a^\alpha=-\bmsf W\bm\varphi\otimes\bm n-\bm W\nabla\bm v+w\bmsf W^2

Thus,

d\bmsf W=-\bm\Phi-\bmsf W\nabla\bm v+w\bmsf W^2.

References:

[1] P. Podio-Guidugli: Lezioni sulla teoria lineare dei gusci elastici sottili, Masson, 1991

[2] C. Davini: Lezioni di Teoria dei Gusci (lecture notes in Italian).

[3] P.M. Naghdi: The Theory of Shells and Plates, Handbuch der Physik VIa/2 p.425-640, Springer-Verlag, 1972

[4] P.G. Ciarlet: An introduction to differential geometry with applications to elasticity, Springer, 2005.

\newcommand\bm[1]{\boldsymbol{#1}} \def\bmn{\bm n} \def\bma{\bm a} \def\bmv{\bm v} \newcommand\bmsf[1]{\boldsymbol{{#1}}}