shells_details

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GeometryEquilibriumDivergence theoremsPointwise form of the equilibrium equationsEquilibrium in the reference configuration and component representationVirtual workUnshearable shellsUnshearable hyperelastic shellsKoiter shellsSmall displacements and rotationsSpecial case: stress-free reference configurationGeneral case: shells with initial stressAppendixGradient and divergence on a surfaceThe effective Lamé constants for a linearly elastic shell

$\def\bP{\boldsymbol P}$ $\def\bF{\boldsymbol F}$ $\def\bN{\boldsymbol N}$ $\def\bG{\boldsymbol G}$ $\def\bM{\boldsymbol M}$ $\def\bK{\boldsymbol K}$

Geometry

${\mathcal S}$ , which we use as label set to identify the particles that comprise the shell. The label set should not be confused with the configuration of the shell, the latter being specified by

placement $\boldsymbol y:{\mathcal S}\to\mathcal E$

and

director field $\boldsymbol d:{\mathcal S}\to \textrm{Unit sphere}=\{\boldsymbol v\in T\mathcal E:|\boldsymbol v|=1\}$ .

Equilibrium

¹ $\boldsymbol S$ $\boldsymbol M$ $\mathcal P\subset {\mathcal S}$ are given by

\boldsymbol r(\mathcal P)=\int_{x\in\mathcal P}\boldsymbol b+\int_{x\in\partial\mathcal P}\boldsymbol S\boldsymbol\nu,\qquad \boldsymbol m(\mathcal P)=\int_{x\in\mathcal P}\big(\boldsymbol y\times\boldsymbol b+\boldsymbol d\times\boldsymbol c\bigr)+\int_{x\in\partial\mathcal P}\bigl(\boldsymbol y\times\boldsymbol S\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 $x$ $\boldsymbol\nu(x)\in T_x{\mathcal S}$ $\mathcal S$ $\partial\mathcal P$ $\mathcal P$ $x$ $\boldsymbol S(x)\boldsymbol \nu(x)$ $\boldsymbol d(x)\times\boldsymbol M(x)\boldsymbol \nu(x)$ represent, respectively, the linear densities of contact forcecontact couple $\mathcal P$ $x$ .

Remarks:

$\boldsymbol d$ encodes the assumption that a shell cannot sustain and transmit couples parallel to the director.
$\boldsymbol c$ $\boldsymbol M\boldsymbol\nu$ $\boldsymbol d$ .
$\boldsymbol S(x)$ $T_x \mathcal S$ $T\mathcal E$ $\boldsymbol M(x)$ $T_x\mathcal S$ $\boldsymbol d(x)$ :

\boldsymbol M(x):T_x\mathcal S\to\operatorname{span}\{\boldsymbol d(x)\}^\perp,\qquad\operatorname{span}\{\boldsymbol d\}^\perp=\{\boldsymbol v\in T\mathcal E:\boldsymbol v\cdot\boldsymbol d=0\}.

$\boldsymbol S$ $\boldsymbol M$ are the homologous of the Piola (AKA 1st Piola-Kirchhoff) stress.
$\mathcal S$ $\mathcal S$ $\mathcal S$ $\mathcal S$ to be a surface makes the resulting equations more compact when we develop a small-displacement, small-strain theory.
$x$ $T_x\mathcal S$ .
tangential $T_x\mathcal S$ .

Divergence theorems

fluxes $\mathcal S$ $\partial\mathcal P$ $\mathcal P\subset\mathcal S$ $\mathcal P$ $\boldsymbol v$ $\mathcal S$ gradient $\boldsymbol v$ $\mathcal S$ by setting

\nabla\boldsymbol v=\boldsymbol v_{,\alpha}\otimes \mathbf a^\alpha,

$\mathbf a^\alpha$ $\mathcal S$ $\mathbf a_\alpha=x_{,\alpha}$ $\mathbf P(x)=\boldsymbol I-{\boldsymbol n}(x)\otimes{\boldsymbol n}(x)$ $T_x\mathcal S$ $\mathcal S$ and work with differential forms (as far as balance laws are concerned).

$\mathbf v$ is a tangential vector field, then the following divergence theorem holds:

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

$\boldsymbol B=\boldsymbol b\otimes\mathbf v$ $\boldsymbol b$ $\mathbf v$ is a tangential tensor field. Then

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

$\{\boldsymbol e_i,i=1,2,3\}$ $\mathcal E$ $\boldsymbol T$ $\boldsymbol T=\boldsymbol e_i\otimes\boldsymbol v^{(i)}$ $\boldsymbol v^{(i)}$ $\eqref{eq:divergence2}$ $\boldsymbol T$ . Thus, we can conveniently take

\operatorname{div}\boldsymbol T=\boldsymbol T_{,\alpha} \mathbf a^\alpha=\nabla\boldsymbol T(\cdot,\mathbf a^\alpha)=\nabla\boldsymbol T:\mathbf P.

$\boldsymbol T$ . Here double dots denote contraction with respect to the last two indices.

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\\ \boxed{\operatorname{div}\boldsymbol S+\boldsymbol b=\boldsymbol 0\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

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

$\mathcal P$ is equivalent to

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

Then same procedure as before 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)+\boldsymbol y\otimes(\operatorname{div}&\boldsymbol S+\boldsymbol b)+\nabla\boldsymbol y\,\boldsymbol S^T+\nabla\boldsymbol d\,\boldsymbol M^T\bigr)=\boldsymbol 0\quad \text{in }\mathcal S,\label{eq:pw1} \end{aligned}

On introducing the deformation gradient and the director gradient

\boxed{\boldsymbol F=\nabla\boldsymbol y,\qquad \boldsymbol G=\nabla\boldsymbol d,}

$\operatorname{div}\boldsymbol S+\boldsymbol b=\boldsymbol 0$ $\operatorname{skw}\boldsymbol A=-(\operatorname{skw}\boldsymbol A)^T=-\operatorname{skw}(\boldsymbol A^T)$ $\boldsymbol A$ , the last equation yields

\operatorname{skw}\bigl(\boldsymbol d\otimes(\operatorname{div}\boldsymbol M+\boldsymbol c)-\boldsymbol S \boldsymbol F^T-\boldsymbol M\boldsymbol G^T\bigr)=\boldsymbol 0\quad \text{in }\mathcal S\label{eq:eq1}

Note that this equation coincides with the symmetry condition (5.18) found in [^1]

We now define

\boldsymbol P=\boldsymbol I-\boldsymbol d\otimes\boldsymbol d.

Warning $\boldsymbol P$ $\mathbf P=\boldsymbol I-\boldsymbol n\otimes\boldsymbol n$ $T\mathcal E$ $T\mathcal E$ . Upright boldface symbol denote tangential vector fields or tangential tensor fields.

Thanks to the above decomposition we now can write:

\boldsymbol S=(\boldsymbol d\otimes\boldsymbol d)\boldsymbol S+\boldsymbol P\boldsymbol S=\boldsymbol d\otimes\boldsymbol S^T\boldsymbol d+\boldsymbol P\boldsymbol S,

$\eqref{eq:eq1}$ can be written as

\operatorname{skw}\bigl(\boldsymbol d\otimes(\operatorname{div}\boldsymbol M+\boldsymbol c)-\underbrace{(\boldsymbol d\otimes\boldsymbol d)\boldsymbol S\boldsymbol F^T}_{\displaystyle\boldsymbol d\otimes \boldsymbol F\boldsymbol S^T\boldsymbol d}-\boldsymbol P\boldsymbol S \boldsymbol F^T-\boldsymbol M\boldsymbol G^T\bigr)=\boldsymbol 0\quad \text{in }\mathcal S\label{eq:eq2b}

Whence,

\operatorname{skw}\boldsymbol A=\boldsymbol 0, \qquad\textrm{where}\quad \boldsymbol A=\boldsymbol d\otimes(\operatorname{div}\boldsymbol M+\boldsymbol c-\boldsymbol F\boldsymbol S^T\boldsymbol d)-\boldsymbol P\boldsymbol S \boldsymbol F^T-\boldsymbol M\boldsymbol G^T\label{eq:eq3}

$\boldsymbol P(\operatorname{skw}\boldsymbol A)\boldsymbol d=\boldsymbol 0$ $\boldsymbol P\boldsymbol c=\boldsymbol 0$ ):

\begin{align} &\boxed{\boldsymbol P(\operatorname{div}\boldsymbol M-(\boldsymbol F\, \boldsymbol S^T+\boldsymbol S\boldsymbol F^T)\boldsymbol d)+\boldsymbol c=\boldsymbol 0.} \end{align}

$\boldsymbol P(\operatorname{skw}\boldsymbol A)\boldsymbol P=\boldsymbol 0$ , which gives:

\boxed{\operatorname{skw}(\boldsymbol P\boldsymbol S\boldsymbol F^T+\boldsymbol M\boldsymbol G^T)=\boldsymbol 0.}\label{eq:0318151000}

The aforementioned calculation is based on the splitting

\boldsymbol S=\boldsymbol N+\boldsymbol d\otimes\mathbf q

where

\boldsymbol N=\boldsymbol P\boldsymbol S,\qquad \mathbf q=\boldsymbol F^T\boldsymbol d.

The equilibrium equation for the forces becomes

\operatorname{div}\boldsymbol N+\boldsymbol G\boldsymbol q+(\operatorname{div}\mathbf q)\boldsymbol d+\boldsymbol b=\boldsymbol 0

The above equilibrium equations can be written as

\begin{align} &\boxed{\boldsymbol P(\operatorname{div}\boldsymbol M-\boldsymbol F\mathbf q)\, +\boldsymbol N\boldsymbol F^T\boldsymbol d+\boldsymbol c=\boldsymbol 0,}\\ &\boxed{\operatorname{skw}(\boldsymbol N\boldsymbol F^T+\boldsymbol M\boldsymbol G^T)=\boldsymbol 0.}\label{eq:03181510001} \end{align}

$\mathbf q=\boldsymbol S^T\boldsymbol d$ shear-force flux $\mathbf q\cdot\boldsymbol\nu$ $\boldsymbol d$ $\boldsymbol \nu$ .
$\boldsymbol N$ membrane-force flux $\boldsymbol N$ not a tangential field $\boldsymbol d$ $\boldsymbol n$ (the referential unit normal).

Equilibrium in the reference configuration and component representation

$\boldsymbol y(\boldsymbol x)=\boldsymbol x$ $\boldsymbol d={\boldsymbol n}$ . Then, the equilibrium equations become

\begin{align} &\boxed{{\mathbf P}(\operatorname{div}{\mathbf N}+\boldsymbol b)+\mathbf G\mathbf q=\boldsymbol 0}\\ &\boxed{\operatorname{div}\mathbf q-\mathbf N\cdot\mathbf G+\boldsymbol b\cdot\boldsymbol n=\boldsymbol 0}\\ &\boxed{{\mathbf P}\operatorname{div}{\mathbf M}-{\mathbf q}+{\boldsymbol c}=\boldsymbol 0,}\\ &\boxed{\operatorname{skw}({\mathbf N}+{\mathbf M}{\mathbf G}^T)=\boldsymbol 0.}\label{eq:0318151001} \end{align}

$x$ $T_x\mathcal S\to T_x\mathcal S$ $\mathbf P(\boldsymbol x)$ $\boldsymbol x$ .

Most references dealing with thin shells contain equilibrium equations written using contravariant components. To compare our result with theirs, we introduce the representation

\mathbf N=N^{\alpha\beta}\mathbf a_\alpha\otimes \mathbf a_\beta,\qquad \mathbf q=q^\alpha \mathbf a_\alpha,\qquad\boldsymbol M=M^{\alpha\beta}\mathbf a_\alpha\otimes \mathbf a_\beta,

$\mathbf a_\alpha=x_{,\alpha}$ is the covariant basis. Furthermore, we write

\mathbf G=\nabla\boldsymbol n=\boldsymbol n_{,\alpha}\otimes \mathbf a^\alpha=(\boldsymbol n_{,\beta}\cdot \mathbf a^\alpha) \mathbf a_\alpha\otimes \mathbf a^\beta=-B_{\cdot\beta}^{\alpha} \mathbf a_\alpha\otimes \mathbf a^\beta=-B_{\alpha\beta} \mathbf a^\alpha\otimes\mathbf a^\beta.

We then have

\boldsymbol P\operatorname{div}\mathbf N=\boldsymbol N_{,\beta}\mathbf a^\beta=N^{\alpha\beta}|_{\beta}\mathbf a_\alpha,\qquad \mathbf G\mathbf q=-B^\alpha_{\cdot\beta}q^\beta\mathbf a_\alpha,\qquad \mathbf P\boldsymbol b=b^\alpha\mathbf a_\alpha.

Thus, the first equilibrium equation becomes

\boxed{N^{\alpha\beta}|_\beta-B^\alpha_{\cdot\beta}q^\beta+b^\alpha,\qquad \alpha=1,2.}

Furthermore,

\operatorname{div}\boldsymbol q=\boldsymbol q_{,\alpha}\cdot\boldsymbol a^\alpha=q^\alpha|_{\alpha},\qquad \mathbf N\cdot\mathbf G=-N^{\alpha\beta}B_{\alpha\beta}

Hence the second equilibrium equation becomes

\boxed{q^\alpha|_\beta+B_{\alpha\beta} N^{\alpha\beta}+b^3=0.}

Analogous calculations performed on the equilibrium equations for the moments yield standard expressions.

Virtual work

We define the external work as:

\delta W=\int_{\mathcal P}\boldsymbol b\cdot\delta\boldsymbol y+\boldsymbol c\cdot\delta\boldsymbol d+\int_{\partial\mathcal P}\boldsymbol S\boldsymbol \nu\cdot\delta\boldsymbol y+\boldsymbol M\boldsymbol\nu\cdot\delta\boldsymbol d.

On using the divergence theorem

\delta W=\int_{\mathcal P}\underbrace{(\operatorname{div}\boldsymbol S+\boldsymbol b)}_{\displaystyle 0}\cdot\delta\boldsymbol y+\underbrace{(\boldsymbol P\operatorname{div}\boldsymbol M-\boldsymbol F\,\mathbf q+\boldsymbol c)}_{\displaystyle \boldsymbol 0}\cdot\delta\boldsymbol n+\underbrace{\boldsymbol F\,\mathbf q\cdot\delta\boldsymbol d}_{\displaystyle\mathbf q\cdot \boldsymbol F^T\delta\boldsymbol d}+\boldsymbol S\cdot\delta\boldsymbol F+{\boldsymbol M\cdot\delta\boldsymbol G}

We next write

\boldsymbol S\cdot\delta\boldsymbol F=\boldsymbol N\cdot\boldsymbol F+\boldsymbol d\otimes\mathbf q\cdot\delta\boldsymbol F=\boldsymbol N\cdot\delta\boldsymbol F+\mathbf q\cdot\delta\boldsymbol F^T\boldsymbol d.

We define the shear strain:

\mathbf g=\boldsymbol F^T\boldsymbol d.

We arrive at

\delta W=\int_{\mathcal P} \bN\cdot\delta\boldsymbol F+\mathbf q\cdot\delta \mathbf g+\boldsymbol M\cdot\delta\boldsymbol G.

$\mathbf g$ $\boldsymbol F\mapsto\boldsymbol{\mathsf Q}\boldsymbol F$ $\boldsymbol d\mapsto\boldsymbol{\mathsf Q}\boldsymbol d$ $\boldsymbol{\mathsf Q}:T\mathcal E\to T\mathcal E$ $\mathbf g$ $\mathbf q$ .

We would like to express virtual work in terms of variations of the following strain measures:

\mathbf C=\bF^\top\bF,\qquad \mathbf K=\bF^\top\bG.

$\mathbf C$ $\mathbf K$ $T_x\mathcal S$ $T_x\mathcal S$ into itself.

$\mathbf a_\alpha=\boldsymbol x_{,\alpha}$ $\boldsymbol a_\alpha=\boldsymbol y_{,\alpha}$ , (again, keep in mind the difference between upright and slanted fonts!) respectively, the covariant basis in the reference and deformed configuration, then

\begin{aligned} &&\mathbf C=\boldsymbol F^T\boldsymbol F&=(\mathbf a^\alpha\otimes \boldsymbol a_\alpha)(\boldsymbol a_\beta\otimes \mathbf a^\beta)=(\boldsymbol a_{\alpha}\cdot \boldsymbol a_{\beta})\mathbf a^\alpha\otimes \mathbf a^\beta\\ &&&=a_{\alpha\beta}\mathbf a^\alpha\otimes \mathbf a^\beta \\ &&\mathbf K=\boldsymbol F^{\top} \boldsymbol G &=(\mathbf a^{\alpha} \otimes \boldsymbol a_{\alpha} )(\boldsymbol d_{,\beta} \otimes \mathbf a^{\beta})=\boldsymbol a_{\alpha} \cdot \boldsymbol d_{,\beta} \mathbf a^{\alpha} \otimes \mathbf a^{\beta} \\ &&&=\kappa_{\alpha \beta} \mathbf a^{\alpha} \otimes \mathbf a^{\beta} \end{aligned}

$\boldsymbol F=\boldsymbol a_\alpha\otimes \mathbf a^\alpha$ $T_x \mathcal S$ $T_y \mathcal S'$ $\mathcal S'$ $\mathcal S$ $\boldsymbol F^{-1}$ is well defined. We now introduce the referential stress measures

\mathbf N=\boldsymbol F^{-1}\boldsymbol N,\qquad \mathbf M=\boldsymbol F^{-1}\boldsymbol M.

Then we can compute

\begin{aligned} \bN \cdot \delta{\bF}+\bM \cdot \delta{\bG} &=\bN \cdot \dot{\bF}+{\bF}^{-1} \bM \cdot \bF^{T} \delta{\bG} \\ &=\bN \cdot \delta{\bF}+\mathbf M \cdot\left[{\delta(\bF^{T} \bG)} -\delta{\boldsymbol F}^{T} \bG\right] \\ &=\bN \cdot \delta{\bF}+\mathbf M \cdot \delta{\mathbf{K}}-\bG \mathbf M^T \cdot \delta{\bF} \\ &=\left(\bN-\boldsymbol G \mathbf M^{T} \right) \cdot \delta{\bF}+\bF^{-1} \bM \cdot \delta{\bK} \\ &=\left(\bF^{-1} \bN-\bF^{-1} \bG \mathbf M^{T} \right) \cdot \bF^{\top} \delta{\bF}+\mathbf{M} \cdot \delta{\mathbf K} \\ &=\left(\mathbf{N}-\widetilde{\mathbf{K}} \mathbf{M}^{T}\right) \cdot \frac{\delta{\mathbf C}}{2}+\mathbf{M} \cdot \delta{\mathbf K} \end{aligned}

$\widetilde{\mathbf K}=\boldsymbol F^{-1}\boldsymbol G$ $\mathbf N-\widetilde{\mathbf K}\mathbf M^T$ $\eqref{eq:0318151000}$ $\boldsymbol N\boldsymbol F^T+\boldsymbol M\boldsymbol G^T\in\operatorname{Sym}$ $\boldsymbol F^{-1}(\boldsymbol N\boldsymbol F^T+\boldsymbol M\boldsymbol G^T)\boldsymbol F^{-T}\in\operatorname{Sym}$ $\mathbf N+\mathbf M\widetilde{\mathbf K}^T\in\operatorname{Sym}$ $\mathbf N-(\mathbf M\widetilde{\mathbf K}^T)^T\in\operatorname{Sym}$ . Note that:

\widetilde{\mathbf K}=\boldsymbol F^{-1}\boldsymbol G=\kappa^{\alpha}_{\cdot\beta}\mathbf a^\alpha\otimes \mathbf a^\beta,\qquad \kappa^\alpha_{\cdot\beta}=\boldsymbol a^\alpha\cdot\boldsymbol d_{,\beta}

We introduce the Green-Lagrange strains

\mathbf E=\frac 12(\mathbf C-\mathring{\mathbf C}),\qquad \mathbf L=\mathbf K-\mathring{\mathbf K}.

$\mathring{\mathbf C}=\nabla\boldsymbol x={\mathbf P}$ $\mathring{\mathbf K}=\nabla{\boldsymbol n}$ $\mathbf C$ $\mathbf K$ .

Then the above equation becomes

\begin{aligned} \delta W(\mathcal P)=\int_{\mathcal P}\left(\mathbf{N}-\widetilde{\mathbf{K}} \mathbf{M}^{T}\right) \cdot{\delta{\mathbf E}}+\mathbf q\cdot\delta\mathbf g+\mathbf{M} \cdot \delta{\mathbf L} \end{aligned}

Remarks

the book by Taylor, Zienkiewicz and Fox obtains the same expression of the internal power by starting from three-dimensional elasticity. We work out the expression of the stress working and our frame-indifference requirements without invoking a parent three-dimensional theory.
$\mathbf M$ $\mathbf N$ are the homologous of the Cosserat (AKA 2nd Piola-Kirchhoff) stress in nonlinear elasticity
$\boldsymbol C$ is the analogue of the right Green strain.
Our derivation is completely component- and derivative-free. In fact, a definition of gradient of scalar and tensor field can be given, based on the notion of tangent space and Gateaux derivative can be given, which does not rely on coordinates (although in differential geometry the very notion of manifold requires the introduction of chart, and hence a coordinate system.)

Unshearable shells

From this point on, we shall restrict our attention to shells satisfying the constraint:

\mathbf g=\boldsymbol F^T\boldsymbol d=\boldsymbol 0.

$\boldsymbol F^T\boldsymbol d=\boldsymbol F[\boldsymbol d,\cdot]=\boldsymbol 0$ $\boldsymbol F^T\boldsymbol G+\nabla\boldsymbol F[\boldsymbol d,\cdot,\cdot]$ $\nabla\boldsymbol F[\boldsymbol n,\cdot,\cdot]=0$ $\boldsymbol d$ . ²

$\boldsymbol F:T_{\boldsymbol x} \mathcal S\to T_{\boldsymbol y(\boldsymbol x)}\mathcal S'$ is invertible, whence

\boldsymbol G=\boldsymbol F^{-T}\nabla\boldsymbol F[\boldsymbol d,\cdot,\cdot].

Observe that in this case we have

\boldsymbol P\boldsymbol F=\boldsymbol F-(\boldsymbol d\otimes\boldsymbol d)\boldsymbol F=\boldsymbol F.

Thus the equilibrium equations become

\begin{align} &\boxed{\boldsymbol P\operatorname{div}\boldsymbol M-\boldsymbol F\boldsymbol q+\boldsymbol c=\boldsymbol 0,}\label{eq:aa}\\ &\boxed{\operatorname{skw}(\boldsymbol N\boldsymbol F^T+\boldsymbol M\boldsymbol G^T)=\boldsymbol 0.} \end{align}\label{eq:29}

$\boldsymbol S=\boldsymbol N+\boldsymbol n\otimes\boldsymbol q$ $\boldsymbol N^T\boldsymbol n=\boldsymbol 0$ , we can write the equilibrium equation for the forces as

\operatorname{div}\boldsymbol N+\boldsymbol G \boldsymbol q+(\operatorname{div}\boldsymbol q)\boldsymbol n+\boldsymbol b=\boldsymbol 0.\label{30}

$\boldsymbol n\cdot\operatorname{div}\boldsymbol N=\operatorname{div}(\boldsymbol N^T\boldsymbol n)-\boldsymbol N\cdot \nabla\boldsymbol n$ $\boldsymbol N$ $\boldsymbol n$ $\boldsymbol n\cdot\operatorname{div}\boldsymbol N=-\boldsymbol N\cdot\nabla\boldsymbol n=-\boldsymbol N\cdot\boldsymbol G$ $\boldsymbol G^T\boldsymbol n=\boldsymbol 0$ $\eqref{30}$ $\boldsymbol n$ , we obtain

\operatorname{div}\boldsymbol q=\boldsymbol N\cdot\boldsymbol G-\boldsymbol n\cdot\boldsymbol b.

$\eqref{eq:aa}$ $\boldsymbol F^{-1}$ $\boldsymbol F^{-1}$ $T_{\boldsymbol y(\boldsymbol x)}\mathcal S'\to T_{\boldsymbol x}\mathcal S$ :

\boldsymbol q=\boldsymbol F^{-1}(\boldsymbol P\operatorname{div}\boldsymbol M+\boldsymbol c).

$\operatorname{div}\boldsymbol q$ . The result is:

\boxed{\operatorname{div}(\boldsymbol F^{-1}(\boldsymbol P\operatorname{div}\boldsymbol M+\boldsymbol c))-\boldsymbol N\cdot\boldsymbol G-\boldsymbol n\cdot\boldsymbol b=\boldsymbol 0.}\label{eq:351143}

$\eqref{30}$ :

\boxed{\boldsymbol P(\operatorname{div}\boldsymbol N+\boldsymbol b)+\boldsymbol G\boldsymbol F^{-1}(\boldsymbol P\operatorname{div}\boldsymbol M+\boldsymbol c)=\boldsymbol 0.} \label{eq:361144}

Unshearable hyperelastic shells

Assume that the Helmholtz free energy per unit referential area is

\psi=\widehat\psi(\mathbf E,\mathbf L).

$\mathcal P\subset\mathcal S$ , the time derivative of the free energy be not greater than the external power:

\frac{\rm d}{{\rm d}t}\int_\mathcal P\psi\le\int_{\mathcal P}\boldsymbol b\cdot\dot{\boldsymbol y}+\boldsymbol c\cdot\dot{\boldsymbol d}+\int_{\partial\mathcal P}\boldsymbol S\boldsymbol\nu\cdot\dot{\boldsymbol y}+\boldsymbol M\boldsymbol\nu\cdot\dot{\boldsymbol d}\qquad\forall \mathcal P\subset\mathcal S.

By making use of the principle of virtual powers, the above inequality is equivalent to

\frac{\rm d}{{\rm d}t}\int_\mathcal P\psi\le\int_{\mathcal P}\left(\mathbf{N}-\widetilde{\mathbf{K}} \mathbf{M}^{T}\right) \cdot {\dot{\mathbf E}}+\mathbf{M} \cdot \dot{\mathbf L}

A standard argument yields, for a hyperelastic material, the constitutive equations for the Cosserat-like stresses are

\begin{aligned} &\mathbf N=\partial_{\mathbf E}\widehat\psi(\mathbf E,\mathbf L)+\widetilde{\mathbf K}\mathbf M^T\qquad\text{and}\quad &\mathbf M=\partial_{\mathbf L}\widehat\psi(\mathbf E,\mathbf L). \end{aligned}

As a result, the constitutive equation for the Piola-like stresses are

\begin{aligned} &\boldsymbol N=\boldsymbol F\mathbf N=\boldsymbol F\partial_{\mathbf E}\widehat\psi\left(\frac 12 (\boldsymbol F^T\boldsymbol F-{\mathbf P}),\boldsymbol F^T\boldsymbol G-\mathring{\mathbf K}\right)+\boldsymbol{G}\widehat{\boldsymbol M}^T(\boldsymbol F,\boldsymbol G)\boldsymbol F^{-T}=:\widehat{\boldsymbol N}(\boldsymbol F,\boldsymbol G)\\ \\ &\boldsymbol M=\boldsymbol F\mathbf M=\boldsymbol F\partial_{\mathbf L}\widehat\psi\left(\frac 12 (\boldsymbol F^T\boldsymbol F-{\mathbf P}),\boldsymbol F^T\boldsymbol G-\mathring{\mathbf K}\right)=:\widehat{\boldsymbol M}(\boldsymbol F,\boldsymbol G) \end{aligned}

Remark $\mathbf E$ $\mathbf L$ $\mathbf N$ $\mathbf M$ , they do not determine them constitutively.

Koiter shells

$h$ a thickness parameter, we assume that the strain energy is separable, and has the form

\psi=\widehat\psi(\mathbf E,\mathbf L)=\frac h2\mathbb C\left[\mathbf E-\mathbf E_R\right]\cdot \mathbf E+\frac {h^3}{24}\mathbb C[\mathbf L-\mathbf L_R]\cdot(\mathbf L-\mathbf L_R).

$\mathbf E_R$ $\mathbf L_R$ $\mathbb C$ is the fourth-order tensor defined by

\mathbb C[\mathbf E]=2\widetilde\mu\operatorname{sym}\mathbf E+\widetilde\lambda\operatorname{tr}(\mathbf E)\mathring{\mathbf P},

$\widetilde\mu$ $\widetilde\lambda$ are the effective Lamé constants. We compute

\mathbf N=h\mathbb C[\mathbf E-\mathbf E_R]+\widetilde{\mathbf K}\mathbf M^T\in\operatorname{Sym},\qquad \mathbf M=\mathbf M_0+\frac{h^3}{12}\mathbb C[\mathbf K]\in\operatorname{Sym},

We write

\widetilde{\mathbf L}=\widetilde{\mathbf K}-\mathring{\widetilde{\mathbf K}}=\widetilde{\mathbf K}-\mathring{\mathbf K}=\boldsymbol F^{-1}\boldsymbol G-\mathring{\mathbf K}.

Then the constitutive equations for hyperelastic shells specialize to

\begin{aligned} &\mathbf N=h\mathbb C[\mathbf E-\mathbf E_R]+\widetilde{\mathbf K}\mathbf M^T=\mathring{\mathbf N}+h\mathbb C[\mathbf E]+\widetilde{\mathbf L}{\mathbf M}+\frac {h^3}{12}\mathring{\mathbf K}\mathbb C[\mathbf L],\\ &\mathbf M=\frac{h^3}{12}\mathbb C[\mathbf L-\mathbf L_R]=\mathring{\mathbf M}+\frac {h^3}{12}\mathbb C[\mathbf L], \end{aligned}

$\mathring{\mathbf M}=\mathring{\mathbf M}^T=-\frac {h^3}{12} \mathbb C[\mathbf L_R]$ $\mathring{\mathbf N}=-h\mathbb C[\mathbf E_R]+\mathring{\mathbf K}\mathring{\mathbf M}$ .

$\boldsymbol N$ $\boldsymbol M$ $\boldsymbol F$ $\boldsymbol G$ are:

\boxed{ \begin{aligned} &\boldsymbol N=\boldsymbol F\mathbf N=\boldsymbol F\mathring{\mathbf N}+h\boldsymbol F\mathbb C\left[\frac 12 (\boldsymbol F^T\boldsymbol F-\mathbf I)\right]+(\boldsymbol G-\boldsymbol F\mathring{\mathbf K})\mathring{\mathbf M}+\frac {h^3}{12}\boldsymbol G\mathbb C[\boldsymbol F^T\boldsymbol G-\mathring{\mathbf K}] \\ &\boldsymbol M=\boldsymbol F\mathring{\mathbf M}+\frac {h^3}{12}\boldsymbol F\mathbb C[\boldsymbol F^T\mathbf G-\mathring{\mathbf K}] \end{aligned}}

Small displacements and rotations

$\boldsymbol y(\boldsymbol x)=\boldsymbol x+\boldsymbol u(\boldsymbol x)$ $\boldsymbol d(\boldsymbol x)={\boldsymbol n}(\boldsymbol x)+\boldsymbol\varphi(\boldsymbol x)$ $\boldsymbol u$ $\boldsymbol\varphi$ , respectively, as the displacementrotation $\boldsymbol n$ i.e. $\boldsymbol\varphi\cdot\boldsymbol n=0$ .

$\mathbf P$ $\boldsymbol P$ $\mathbf P$ $\boldsymbol P$ in the reference configuration)

\begin{aligned} &\boldsymbol F={\mathbf P}+\boldsymbol H,\qquad \mathbf P=\nabla\boldsymbol x=\mathbf I-\boldsymbol n\otimes\boldsymbol n,\quad\boldsymbol H=\nabla\boldsymbol u,\\ &\boldsymbol G=\mathring{\mathbf K}+\boldsymbol\Phi,\qquad \boldsymbol\Phi=\nabla\boldsymbol\varphi. \end{aligned}

We assume small displacements and small rotations:

|\boldsymbol H|<<1,\qquad |\boldsymbol\Phi|<<1.

\boldsymbol{S}=\boldsymbol{\mathring S}+\widetilde{\boldsymbol S},

Special case: stress-free reference configuration

$\boldsymbol{\mathring S}=\boldsymbol 0$ $\boldsymbol{\mathring M}=0$ $\eqref{eq:351143}$ $\eqref{eq:361144}$ yields

\begin{aligned} &{\mathbf P}(\operatorname{div}\boldsymbol {N}+\boldsymbol {b})+\mathring{\mathbf K}(\operatorname{div}{\boldsymbol M}+{\boldsymbol c})=\boldsymbol 0, \\ &\mathbf P\operatorname{div}({\boldsymbol M}+\widetilde{\boldsymbol c})-{\boldsymbol N}\cdot\mathring{\mathbf K}-{\boldsymbol n}\cdot{\boldsymbol b}=0, \end{aligned}

Moreover, the incremental constitutive equations take the form

\begin{aligned} &\boldsymbol N=h\mathbb C\left[\frac 1 2({\mathbf P}\boldsymbol H+\boldsymbol H^T{\mathbf P})\right]+\frac {h^3}{12}\mathring{\mathbf K}\mathbb C[\boldsymbol H^T\mathring{\mathbf K}+\boldsymbol\Phi]\\ &\boldsymbol M=\frac{h^3}{12}\mathbb C[\boldsymbol H^T\mathring{\mathbf K}+\boldsymbol\Phi] \end{aligned}

General case: shells with initial stress

\begin{aligned}&\boldsymbol N=\mathring{\mathbf N}+\boldsymbol H\mathring{\mathbf N}+h\mathbb C\left[\frac 1 2(\mathring{\mathbf P}\boldsymbol H+\boldsymbol H^T\mathring{\mathbf P})\right]+(\boldsymbol\Phi-\boldsymbol H\mathring{\mathbf K})\mathring{\mathbf M}+\frac {h^3}{12}\mathring{\mathbf K}\mathbb C[\boldsymbol H^T\mathring{\mathbf K}+\boldsymbol\Phi]\\&\boldsymbol M=\mathring{\mathbf M}+\boldsymbol H\mathring{\mathbf M}+\frac{h^3}{12}\mathbb C[\boldsymbol H^T\mathring{\mathbf K}+\boldsymbol\Phi]\end{aligned}

Remark: $\operatorname{div}\boldsymbol S+\boldsymbol b=\mathbf 0$ , and only then perform the relevant projections of the resulting equations.

Appendix

Gradient and divergence on a surface

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

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

covariant derivative $\boldsymbol v$ is

\nabla_s\boldsymbol v=\boldsymbol P\nabla\boldsymbol v.

$T_x$ $T_x$ .

$\boldsymbol a$ $\operatorname{div}\boldsymbol a=\boldsymbol P\cdot\nabla\boldsymbol a=\boldsymbol P\nabla_s\boldsymbol a$ $\boldsymbol P(x)$ $V$ $T_x$ .

$\boldsymbol A$ $T_x$ into some linear space is to consider first tensors of the form

\boldsymbol A(x)=\boldsymbol v\otimes\boldsymbol a(x),

$\boldsymbol v$ is a constant vector. In this case, we have

\operatorname{div}\boldsymbol A=(\operatorname{div}\boldsymbol a)\boldsymbol v.

$\boldsymbol b$ is another tangential tensor field, then

\operatorname{\boldsymbol A^T\boldsymbol a}=\operatorname{div}\boldsymbol a \boldsymbol a\cdot\operatorname{div}\boldsymbol A+\nabla\boldsymbol a\cdot\boldsymbol A.

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

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

The divergence of a tensor field can be written as

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

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

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

We obtain

\operatorname{div}(\boldsymbol y\otimes\boldsymbol S)=\boldsymbol y\otimes\operatorname{div}\boldsymbol S+\nabla\boldsymbol y\,\boldsymbol S^T,\qquad \operatorname{div}(\boldsymbol d\otimes\boldsymbol M)=\boldsymbol d\otimes\operatorname{div}\boldsymbol M+\nabla\boldsymbol d\,\boldsymbol M^T.

Hence,

\int_{\partial\mathcal P}(\boldsymbol y\otimes\boldsymbol S\boldsymbol\nu+\boldsymbol n\otimes\boldsymbol M\boldsymbol\nu)=\int_{\mathcal P}(\boldsymbol y\otimes\operatorname{div}\boldsymbol S+\nabla\boldsymbol y\,\boldsymbol S^T+\boldsymbol n\otimes\operatorname{div}\boldsymbol M+\nabla\boldsymbol d\,\boldsymbol M^T).

On introducing the decomposition

\boldsymbol S=\boldsymbol d\otimes\boldsymbol S^T\boldsymbol d+(\boldsymbol I-\boldsymbol d\otimes\boldsymbol d)\boldsymbol S=\boldsymbol d\otimes\boldsymbol q+\boldsymbol N,

we can write

\operatorname{skw}\int_{\partial\mathcal P}(\boldsymbol y\otimes\boldsymbol S\boldsymbol\nu+\boldsymbol d\otimes\boldsymbol M\boldsymbol\nu)=\operatorname{skw}\int_{\mathcal P}\Big[\boldsymbol y\otimes\operatorname{div}\boldsymbol S+\boldsymbol d\otimes(\operatorname{div}\boldsymbol M-\nabla\boldsymbol y^T\boldsymbol q)-(\boldsymbol N\nabla\boldsymbol y^T+\boldsymbol\nabla\boldsymbol d^T)\Big].

An incomplete list of 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] G. Del Piero: On the method of virtual powers in continuum mechanics. J. Mech. Mat. Struct. 4 (2004), 281-292.

[5] A. Di Carlo, P. Podio-Guidugli, and W.O. Williams, Shells with thickness distension. International Journal of Solids and Structures 38(6-7):1201-1225 · February 2001

1 Details to obtain the pointwise form of the balance equations↩

2 To prove this fact, one can use a coordinate system. Indeed

$\boldsymbol F=\boldsymbol y_{,\alpha}\otimes\boldsymbol a^\alpha$ . This yields

$\nabla\boldsymbol F=\boldsymbol y_{,\alpha\beta}\otimes\boldsymbol a^\alpha\otimes\boldsymbol{a}^\beta+\boldsymbol y_{,\alpha}\otimes\boldsymbol a^\alpha_{,\beta}\otimes\boldsymbol a^\beta$ . Since

$\boldsymbol y_{,\alpha}\cdot\boldsymbol n=0$ , we have

$\nabla\boldsymbol F[\boldsymbol n,\cdot,\cdot]=\boldsymbol n\cdot\boldsymbol y_{,\alpha\beta}\boldsymbol a^\alpha\otimes\boldsymbol a^\beta$ , which is a tangential tensor.↩