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Version Aster	Author (s) Organization (s)	Description of changes
6.3	P. MASSIN EDF -R&D/ AMA	Initial text

Transverse shear correction factors for orthotropic or eccentric stratified plates

The matrix \({H}_{\text{ct}}\) is defined so that the surface density of transverse shear energy obtained in the case of the three-dimensional distribution of stresses resulting from the resolution of the equilibrium is equal to that of the plate model based on Reissner’s hypotheses, for simple flexural behavior. We should therefore find \({H}_{\text{ct}}\) such as:

\(\frac{1}{2}\underset{\mathrm{-}h\mathrm{/}2}{\overset{+h\mathrm{/}2}{\mathrm{\int }}}\tau {H}_{g}^{\mathrm{-}1}\tau \mathrm{=}\frac{1}{2}{\text{TH}}_{\text{ct}}^{\mathrm{-}1}T\mathrm{=}\frac{1}{2}\gamma {H}_{\text{ct}}\gamma\) with \(\tau \mathrm{=}(\begin{array}{c}{\sigma }_{\text{xz}}\\ {\sigma }_{\text{yz}}\end{array})\) and \(T\mathrm{=}\underset{\mathrm{-}h\mathrm{/}2}{\overset{+h\mathrm{/}2}{\mathrm{\int }}}\tau \text{dz}\mathrm{=}{H}_{\text{ct}}\gamma\).

To obtain \({H}_{\text{ct}}\) we use the distribution of \(\tau\) following \(z\) obtained from the resolution of the 3D equilibrium equations without external torques:

\({\sigma }_{\text{xz}}\mathrm{=}\mathrm{-}\underset{\mathrm{-}h\mathrm{/}2}{\overset{z}{\mathrm{\int }}}({\sigma }_{\text{xx},x}+{\sigma }_{\text{xy},y})d\zeta ;{s}_{\text{yz}}\mathrm{=}\mathrm{-}\underset{\mathrm{-}h\mathrm{/}2}{\overset{z}{\mathrm{\int }}}({\sigma }_{\text{xy},x}+{\sigma }_{\text{yy},y})d\zeta\) with \({\sigma }_{\text{xz}}\mathrm{=}{\sigma }_{\text{yz}}\mathrm{=}0\) for \(z\mathrm{=}\mathrm{\pm }h\mathrm{/}2\).

In the case where there is no membrane flexure coupling (symmetry with respect to \(z\mathrm{=}0\)), the stresses in the plane of the element \({\sigma }_{\text{xx}},{\sigma }_{\text{yy}},{\sigma }_{\text{xy}}\) are expressed in the case of pure flexure behavior:

\(\sigma \mathrm{=}\mathit{zA}(z)M\) with \(A(z)\mathrm{=}H(z){H}_{f}^{\mathrm{-}1}\).

If \(H(z)\) and \({H}_{f}\) do not depend on \(x\) and \(y\) we can determine \({H}_{\text{ct}}\). In fact:

\(\tau (z)\mathrm{=}{D}_{1}(z)T+{D}_{2}(z)\lambda\) where \(T\mathrm{=}(\begin{array}{c}{T}_{x}\\ {T}_{y}\end{array})\mathrm{=}(\begin{array}{c}{M}_{\text{xx},x}+{M}_{\text{xy},y}\\ {M}_{\text{xy},x}+{M}_{\text{yy},y}\end{array})\) and \(\lambda \mathrm{=}(\begin{array}{c}{M}_{\text{xx},x}\mathrm{-}{M}_{\text{xy},y}\\ {M}_{\text{xy},x}\mathrm{-}{M}_{\text{yy},y}\\ {M}_{\text{yy},x}\\ {M}_{\text{xx},y}\end{array})\)

as well as:

\({\mathrm{D}}_{1}\mathrm{=}\mathrm{-}\underset{\mathrm{-}h\mathrm{/}2}{\overset{z}{\mathrm{\int }}}\frac{z}{2}(\begin{array}{cc}{A}_{\text{11}}+{A}_{\text{33}}& {A}_{\text{13}}+{A}_{\text{32}}\\ {A}_{\text{31}}+{A}_{\text{23}}& {A}_{\text{22}}+{A}_{\text{33}}\end{array})\mathit{dz}\),

\({\mathrm{D}}_{2}\mathrm{=}\mathrm{-}\underset{\mathrm{-}h\mathrm{/}2}{\overset{z}{\mathrm{\int }}}\frac{z}{2}(\begin{array}{cccc}{A}_{\text{11}}\mathrm{-}{A}_{\text{33}}& {A}_{\text{13}}\mathrm{-}{A}_{\text{32}}& {\mathrm{2A}}_{\text{12}}& {\mathrm{2A}}_{\text{31}}\\ {A}_{\text{31}}\mathrm{-}{A}_{\text{23}}& {A}_{\text{33}}\mathrm{-}{A}_{\text{22}}& {\mathrm{2A}}_{\text{32}}& {\mathrm{2A}}_{\text{21}}\end{array})\mathit{dz}\).

As a result, \(\frac{1}{2}\underset{\mathrm{-}h\mathrm{/}2}{\overset{+h\mathrm{/}2}{\mathrm{\int }}}t{\mathrm{H}}_{g}^{\mathrm{-}1}t\mathrm{=}\frac{1}{2}(\begin{array}{c}\mathrm{T}\\ \lambda \end{array})(\begin{array}{cc}{\mathrm{C}}_{\text{11}}& {\mathrm{C}}_{\text{12}}\\ {\mathrm{C}}_{\text{12}}^{T}& {\mathrm{C}}_{\text{22}}\end{array})(\begin{array}{c}\mathrm{T}\\ \lambda \end{array})\) with: \(\begin{array}{}{C}_{\text{11}}=\underset{-h/2}{\overset{+h/2}{\int }}{D}_{1}^{T}{H}_{g}^{-1}{D}_{1}\text{dz};\\ {C}_{\text{12}}=\underset{-h/2}{\overset{+h/2}{\int }}{D}_{1}^{T}{H}_{g}^{-1}{D}_{2}\text{dz};\\ {C}_{\text{22}}=\underset{-h/2}{\overset{+h/2}{\int }}{D}_{2}^{T}{H}_{g}^{-1}{D}_{2}\text{dz}\end{array}\)

As in addition \(\frac{1}{2}\underset{\mathrm{-}h\mathrm{/}2}{\overset{+h\mathrm{/}2}{\mathrm{\int }}}t{H}_{g}^{\mathrm{-}1}t\mathrm{=}\frac{1}{2}{\text{TH}}_{\text{ct}}^{\mathrm{-}1}T\), we propose to take \({H}_{\text{ct}}\mathrm{=}{C}_{\text{11}}^{\mathrm{-}1}\) to best satisfy the two equations regardless of \(T\) and \(\lambda\).

By comparing \({H}_{\text{ct}}\) calculated in this way with \({\stackrel{ˉ}{H}}_{\text{ct}}\mathrm{=}\underset{\mathrm{-}h\mathrm{/}2}{\overset{+h\mathrm{/}2}{\mathrm{\int }}}{H}_{g}\text{dz}\), the following transverse shear correction coefficients are shown: \({k}_{1}\mathrm{=}{H}_{\text{ct}}^{\text{11}}\mathrm{/}{\stackrel{ˉ}{H}}_{\text{ct}}^{\text{11}};{k}_{\text{12}}\mathrm{=}{H}_{\text{ct}}^{\text{12}}\mathrm{/}{\stackrel{ˉ}{H}}_{\text{ct}}^{\text{12}};{k}_{2}\mathrm{=}{H}_{\text{ct}}^{\text{22}}\mathrm{/}{\stackrel{ˉ}{H}}_{\text{ct}}^{\text{22}}\).

For a homogeneous, isotropic or anisotropic plate, we thus find: \({\mathrm{H}}_{\text{ct}}\mathrm{=}\mathit{kh}{\mathrm{H}}_{g}\) with \(k\mathrm{=}5\mathrm{/}6\).

Notes:

This method is only valid when the composite plate is symmetric with respect to z=0.

• For a multilayer material, it is established that:

\(\begin{array}{c}{\mathrm{C}}_{\text{11}}\mathrm{=}\mathrm{\sum }_{i\mathrm{=}1}^{N}\frac{{h}_{i}}{4}(\mathrm{\sum }_{p\mathrm{=}1}^{i\mathrm{-}1}{h}_{p}{h}_{p}{\mathrm{A}}_{p}^{T}\mathrm{-}\frac{1}{2}{z}_{i}^{2}{\mathrm{A}}_{i}^{T}){\mathrm{H}}_{g}^{\mathrm{-}1}(\mathrm{\sum }_{p\mathrm{=}1}^{i\mathrm{-}1}{h}_{p}{h}_{p}{\mathrm{A}}_{p}\mathrm{-}\frac{1}{2}{z}_{i}^{2}{\mathrm{A}}_{i})+\\ \frac{1}{\text{24}}({z}_{i+1}^{3}\mathrm{-}{z}_{i}^{3})\mathrm{[}{\mathrm{A}}_{i}^{T}{\mathrm{H}}_{g}^{\mathrm{-}1}(\mathrm{\sum }_{p\mathrm{=}1}^{i\mathrm{-}1}{h}_{p}{h}_{p}{\mathrm{A}}_{p}\mathrm{-}\frac{1}{2}{z}_{i}^{2}{\mathrm{A}}_{i})+(\mathrm{\sum }_{p\mathrm{=}1}^{i\mathrm{-}1}{h}_{p}{h}_{p}{\mathrm{A}}_{p}^{T}\mathrm{-}\frac{1}{2}{z}_{i}^{2}{\mathrm{A}}_{i}^{T}){\mathrm{H}}_{g}^{\mathrm{-}1}{\mathrm{A}}_{i}\mathrm{]}\\ +\frac{1}{\text{80}}({z}_{i+1}^{5}\mathrm{-}{z}_{i}^{5}){\mathrm{A}}_{i}^{T}{\mathrm{H}}_{g}^{\mathrm{-}1}{\mathrm{A}}_{i}\end{array}\)

where: \({h}_{i}={z}_{i+1}-{z}_{i},{h}_{i}=\frac{1}{2}({z}_{i+1}+{z}_{i})\) and \({A}_{i}\) represent the \((\begin{array}{cc}{A}_{\text{11}}+{A}_{\text{33}}& {A}_{\text{13}}+{A}_{\text{32}}\\ {A}_{\text{31}}+{A}_{\text{23}}& {A}_{\text{22}}+{A}_{\text{33}}\end{array})\) matrix for layer i.

• The validity of choice \({H}_{\text{ct}}={C}_{\text{11}}^{-1}\) can be examined a posteriori when we have an estimate of the solution (fields of displacement and plane constraints, in particular). We can then estimate the difference between the two energy estimates. A two-step calculation approach for multilayer plates and shells (with \({H}_{\text{ct}}\) diagonal and two coefficients \({k}_{1}\) and \({k}_{2}\)) has also been developed by Noor and Burton [bib10] [bib11].

• In the case of an isotropic or anisotropic homogeneous plate, the equality between the two energies is satisfied in the strict sense since \({D}_{2}\mathrm{=}0\). The choice made above is then valid and no subsequent examination is necessary.

Shear stress calculation

The term \(\mathit{d1iel}(z)\), which makes it possible to calculate the stress and shear deformation in the presence of eccentricity, is explained here.

In fact, according to the R3.07.03 documentation, the shear stress — shear forces relationships are:

\({\sigma }_{\mathit{xz}}={T}_{x}\ast \mathit{d1iel}(z)\) \({\sigma }_{\mathit{yz}}={T}_{y}\ast \mathit{d1iel}(z)\)

In classic cases without eccentricity, we have: \(\mathit{d1iel}(z)=3/\mathrm{2h}\ast (1-4{z}^{2}/{h}^{2})\).

In more general cases (in the presence of eccentricity for example) \(\mathit{d1iel}(z)\) should be modified. To correctly approximate the shear stress, the choice is made to apply a general quadratic form for \(\mathit{d1iel}(z)=a\ast {z}^{2}+b\ast z+c\) such that the following conditions are met:

• \(\underset{-h/2}{\overset{h/2}{\int }}\mathit{d1iel}(z+d)\mathit{dz}=1\) shear effort-constraints relationship

• \(\mathit{d1iel}(z+d=-h/2)=0;\mathit{d1iel}(z+d=h/2)=0\) free edge condition

With an eccentricity of \(d\), the coefficients of \(\mathit{d1iel}(z)\) are:

\(a=-6/\mathit{quotient}\), \(b=(6\ast (\mathit{zmin}+\mathit{zmax}))/\mathit{quotient}\), \(b=(6\ast \mathit{zmin}\ast \mathit{zmax})/\mathit{quotient}\)

\(\mathit{quotient}=({\mathit{zmax}}^{3}-3\ast {\mathit{zmax}}^{2}\ast \mathit{zmin}+3\ast \mathit{zmax}\ast {\mathit{zmin}}^{2}-{\mathit{zmin}}^{3}),\mathit{zmin}=-h/2+d,\mathit{zmax}=h/2+d\)