Shell_theory

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Shell theory

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\}.

Equilibrium

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

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

$\boldsymbol b$ $\boldsymbol n \times \boldsymbol c$ $\boldsymbol x$ $x$ $\boldsymbol n$ $\boldsymbol n$ $\boldsymbol c(x)\in T_x$ .

$\boldsymbol\nu(x)\in T_x$ $\mathcal S$ $\partial\mathcal P$ $\mathcal P$ $x$ $\boldsymbol F(x)\boldsymbol \nu(x)$ $\boldsymbol n(x)\times\boldsymbol M(x)\boldsymbol \nu(x)$ represent, respectively, the linear densities of contect forcecontact couple $\mathcal P$ $x$ $\boldsymbol F(x)$ $T_x$ $V$ $\boldsymbol M(x)$ $T_x$ $[\boldsymbol n\times]$ ).

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

$\mathcal S$ $\boldsymbol r(\mathcal P)$ $\boldsymbol m(\mathcal P)$ be null for every part $\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\\ \color{blue}\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 start from

\boldsymbol 0=\operatorname{skw}\Bigl[\int_{\mathcal P}\bigl(\boldsymbol x\otimes\boldsymbol b+\boldsymbol n\otimes\boldsymbol c\bigr)+\int_{\partial\mathcal P}\bigl(\boldsymbol x\otimes\boldsymbol F\boldsymbol\nu+\boldsymbol n\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 n\otimes(\operatorname{div}\boldsymbol M+\boldsymbol c+\bm F^T\bm n)+2(\bm F^T\bmsf P-&\bmsf W^T\bm M)+\boldsymbol x\otimes(\operatorname{div}\boldsymbol F+\boldsymbol b)\bigr)=\boldsymbol 0\quad \text{in }\mathcal S,\label{eq:pw1} \end{aligned}

$\boldsymbol F^T(x)$ $\boldsymbol F(x)$ $V$ $T_x$ $\boldsymbol 1$ $V$ $\bmsf P=\nabla\bm x=\bm 1-\bm n\otimes\bm n$ $V$ $T_x$ .

$\boldsymbol W=-\nabla\boldsymbol n$ Weingarten map $\boldsymbol N$ $\boldsymbol M\boldsymbol W$ $T_x$ $T_x$ $\boldsymbol n$ . From this fact it is not difficult to conclude that

\color{blue}\begin{aligned} &\bm P(\operatorname{div}\boldsymbol M)+\boldsymbol c+\bm F^T\bm n=\boldsymbol 0,\\[0.2em] &\bm P\bm F-\boldsymbol M\boldsymbol W\in{\rm Sym}. \end{aligned}

The result is also more transparent if we introduce the decomposition

\boldsymbol F=\boldsymbol n\otimes\boldsymbol q+\boldsymbol N,\qquad \boldsymbol q=\boldsymbol F^T\boldsymbol n,\qquad \boldsymbol N=\bmsf P\boldsymbol F,\qquad \bmsf P=\boldsymbol 1-\boldsymbol n\otimes\boldsymbol n.

We also introduce the notation

\nabla_s\bm T=\bm P\nabla\bm T,\qquad \operatorname{div}_s\bm T=\bm P\operatorname{div}\bm T.

Then we can write:

\color{blue}\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 $\boldsymbol q\cdot\boldsymbol\nu$ $\boldsymbol n$ $\boldsymbol \nu$ .

Virtual work

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

\boldsymbol n\times\boldsymbol c\cdot\delta\boldsymbol\omega=\boldsymbol n\times\boldsymbol c\cdot\boldsymbol n\times\boldsymbol\delta\boldsymbol\varphi=\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\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\delta\boldsymbol\varphi.

It is useful at this point to introduce the decomposition of the displacement

\boldsymbol u=\boldsymbol v+w\boldsymbol n,\qquad \boldsymbol v\cdot\boldsymbol n=0.

Then

\nabla\boldsymbol u=\nabla\boldsymbol v+\boldsymbol n\otimes\nabla w-\boldsymbol W w.

On introducing the virtual strains: these should be symmetric

\delta\bm E=\nabla\delta\bm v-\bm W\delta w,\qquad \delta\bm g=\delta\bm\varphi+\nabla\delta w+\bm W\delta\bm v,\qquad \delta\bm K=\nabla\delta\bm\varphi,

the expression of the virtual work can be written as

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

$\delta\bm E$ $\delta\bm g$ $\delta\bm K$ as the virtual stretching, shearingbending $\delta\bm n=-\nabla\delta w-\bm W\delta\bm v$ $\delta\bm g=\delta\bm\varphi-\delta\bm n$ $\bm d$ $\bm n$ $\delta\bm g=\bm 0$ for a Kirchhoff plate. Then ,

\delta\bm K=-\nabla\nabla\delta w-\nabla(\bm W\delta\bm v)

the expression of the virtual work is

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

Koiter shells

$\mathcal P$ 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\bm\Rho\cdot\bm\Rho,

where

\bm\Rho=\bm K-\bm E\bm W,

$\mathbb C$ , defined by

\mathbb C\bm{\mathsf A}=2\widetilde\mu\bm{\mathsf A}+\widetilde\lambda\operatorname{tr}(\bm{\mathsf A}){\bmsf I}

$\bmsf I$ $T_x$ $x$ into itself. The time derivative of the strain energy is

\dot E(\mathcal P)=\int_{\mathcal P}h\mathbb C[\bm E]\cdot\dot{\bm E}+\frac {h^3}{12}\mathbb C[\bm\Rho]\cdot\dot{\bm\Rho}.

Then the internal power is

\dot W(\mathcal P)=\int_{\mathcal P}(\bm N-\bm M\bm W)\cdot\dot{\bm E}+\bm M\cdot\dot{\bm \Rho}

This leads to the constitutive equations

\color{blue}\bm N-\bm M\bm W=h\mathbb C[\bm E],\qquad \bm M=\frac {h^3}{12}\mathbb C[\bm\Rho].

$\bm N-\bm M\bm K$ is a symmetric tensor

Remarks

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

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.

$\bm a$ $\bm A$ 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(\bm M+\bm F^T\bm n)-\bm W\bm M^T).

By localization we obtain

Kirchhoff-Love kinematics

$\boldsymbol d$ $\bmn$ $\bm\varphi={d}\bmn$ $\bm{\mathsf P}\bmn=\bm 0$ ${\rm d}\bmn$ is a tangential vector, we obtain

\begin{aligned} \bm\varphi&={d}\bmn=- ({d}\bm{\mathsf P})\bm n=(d\bma^\alpha\otimes\bma_\alpha+\bma^\alpha\otimes d\bma_\alpha)\bmn=-\bma^\alpha\otimes d\bma_\alpha\bmn\\ &=-\bma^\alpha\otimes \bm u_{,\alpha}\bmn=-\nabla\bm u^T\bm n=-\nabla(\bm u\cdot\bm n)+\nabla\bm n^T\bm u. \end{aligned}

whence

\bm\varphi=-(\nabla w+\bm W\bm v).

Balance equations

Whereas it is natural to project the balance equations Note that it is not natural to introduce a similar decomposition for the force balance equation at this stage.

Derivation of the 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.

Note that

\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-w\bmsf W. \end{aligned}

Hence

\bm W\bm a_\alpha\otimes d\bm a^\alpha=-\bm W\nabla\bm v+w\bms 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}}}