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Analysis of the static response of cross-ply simply supported plates and shells based on a higher-order theory

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Mechanics of Composite Materials Aims and scope

The boundary-discontinuous double Fourier series-based solution methodology is used to solve the problem of higher-order shear deformation of cross-ply plates and doubly curved panels, which are characterized by a system of five highly coupled linear partial differential equations with mixed-type simply supported boundary conditions prescribed at all four their edges. The present solution is related to a number of unsolved boundary-value problems and can serve as a tool in particular for early design stages and for benchmark comparisons and verifications of numerical results. The analytical results obtained are compared with finite-element calculations, and a good agreement is found to exist between them.

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Acknowledgment

The first author was financed by the Portuguese Foundation of Science and Technology under the contract number SFRH/BPD/47687/2008.

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Authors and Affiliations

  1. Centre for Marine Technology and Engineering (CENTEC), Technical University of Lisbon, Instituto Superior Técnico, Portugal, Av. Rovisco Pais, 1049-001, Lisbon, Portugal

    A. S. Oktem & C. Guedes Soares

Authors
  1. A. S. Oktem
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  2. C. Guedes Soares
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Corresponding author

Correspondence to C. Guedes Soares.

Additional information

Russian translation published in Mekhanika Kompozitnykh Materialov, Vol. 48, No. 1, pp. 95-110, January-February, 2012.

 

 

$$ \begin{array}{*{20}{c}} {{a_1} = \frac{{{A_{11}}}}{{{R_1}}} + \frac{{{A_{12}}}}{{{R_2}}},\;{a_2} = {B_{11}} - \frac{{4{E_{11}}}}{{3{h^2}}},\;{a_3} = {B_{12}} - \frac{{4{E_{12}}}}{{3{h^2}}},\;{a_4} = \frac{{4{E_{11}}}}{{3{h^2}}},\;{a_5} = \frac{{4{E_{12}}}}{{3{h^2}}},\;{a_6} = \frac{{{A_{12}}}}{{{R_1}}} + \frac{{{A_{22}}}}{{{R_2}}},} \\ {{a_7} = {B_{22}} - \frac{{4{E_{22}}}}{{3{h^2}}},\;{a_8} = \frac{{4{E_{22}}}}{{3{h^2}}},\;{a_9} = {B_{66}} - \frac{{4{E_{66}}}}{{3{h^2}}},\;{a_{10}} = \frac{{8{E_{66}}}}{{3{h^2}}};} \\ \end{array} $$
$$ \begin{array}{*{20}{c}} {{b_1} = \frac{{{B_{11}}}}{{{R_1}}} + \frac{{{B_{12}}}}{{{R_2}}},\;{b_2} = {D_{11}} - \frac{{4{F_{11}}}}{{3{h^2}}},\;{b_3} = {D_{12}} - \frac{{4{F_{12}}}}{{3{h^2}}},\;{b_4} = \frac{{4{F_{11}}}}{{3{h^2}}},\;{b_5} = \frac{{4{F_{12}}}}{{3{h^2}}},\;{b_6} = \frac{{{B_{12}}}}{{{R_1}}} + \frac{{{B_{22}}}}{{{R_2}}},\;{b_7} = {D_{22}} - \frac{{4{F_{22}}}}{{3{h^2}}},} \\ {\;{b_8} = \frac{{4{F_{22}}}}{{3{h^2}}},\;{b_9} = {D_{66}} - \frac{{4{F_{66}}}}{{3{h^2}}},\;{b_{10}} = \frac{{8{F_{66}}}}{{3{h^2}}},\;{b_{11}} = \frac{{{E_{11}}}}{{{R_1}}} + \frac{{{E_{12}}}}{{{R_2}}},\;{b_{12}} = {F_{11}} - \frac{{4{H_{11}}}}{{3{h^2}}},\;{b_{13}} = {F_{12}} - \frac{{4{H_{12}}}}{{3{h^2}}},\;{b_{14}} = \frac{{4{H_{11}}}}{{3{h^2}}};} \\ {{b_{15}} = \frac{{4{H_{12}}}}{{3{h^2}}},\;{b_{16}} = {F_{22}} - \frac{{4{H_{22}}}}{{3{h^2}}},\;{b_{17}} = \frac{{4{H_{22}}}}{{3{h^2}}},\;{b_{18}} = \frac{{4{H_{66}}}}{{3{h^2}}},\;{b_{19}} = \frac{{8{H_{12}}}}{{3{h^2}}};} \\ \end{array} $$
$$ {d_1} = {A_{44}} - \frac{{4{D_{44}}}}{{{h^2}}},\;{d_2} = {A_{55}} - \frac{{4{D_{55}}}}{{{h^2}}},\;{d_3} = {D_{44}} - \frac{{4{F_{44}}}}{{{h^2}}},\;{d_4} = {D_{55}} - \frac{{4{F_{55}}}}{{{h^2}}}; $$
$$ \begin{array}{*{20}{c}} {{e_1} = {B_{12}} + {B_{66}} - \frac{{4{E_{12}}}}{{3{h^2}}} - \frac{{4{E_{66}}}}{{3{h^2}}},\;{e_2} = {b_1} - {d_2} + \frac{{4{D_4}}}{{{h^2}}} - \frac{{4{B_{11}}}}{{3{h^2}}},\;{e_3} = {b_2} - \frac{{4{b_{12}}}}{{3{h^2}}},\;{e_4} = {b_3} + {b_9} - \frac{{4{b_{13}}}}{{3{h^2}}} - \frac{{4{b_{18}}}}{{3{h^2}}},\;{e_5} = {b_9} - \frac{{4{b_{18}}}}{{3{h^2}}},} \\ {{e_6} = - {b_4} + \frac{{4{b_{14}}}}{{3{h^2}}},\;{e_7} = - {b_5} - {b_{10}} + \frac{{4{b_{15}}}}{{3{h^2}}} + \frac{{4{b_{19}}}}{{3{h^2}}},\;{e_8} = - {d_2} + \frac{{4{d_4}}}{{{h^2}}},\;{e_9} = {b_7} - \frac{{4{b_{16}}}}{{3{h^2}}},\;{e_{10}} = {b_8} + \frac{{4{b_{17}}}}{{3{h^2}}},} \\ {{e_{11}} = - {d_1} + \frac{{4{d_3}}}{{{h^2}}},\;{e_{12}} = {b_6} - {d_1} + \frac{{4{d_3}}}{{{h^2}}} - \frac{{4{b_{20}}}}{{3{h^2}}};} \\ \end{array} $$
$$ \begin{array}{*{20}{c}} {{f_1} = {A_{12}} + {A_{66}},\;{f_2} = {a_3} + {a_9},\;{f_3} = - {a_{10}} - {a_5},\;{f_4} = {d_2} - \frac{{4{d_4}}}{{{h^2}}} - \frac{{{a_2}}}{{{R_1}}} - \frac{{{a_3}}}{{{R_2}}},\;{f_5} = {d_1} - \frac{{4{d_3}}}{{{h^2}}} - \frac{{{a_3}}}{{{R_1}}} - \frac{{{a_7}}}{{{R_2}}},} \\ {{f_6} = {f_2} - \frac{{4{d_4}}}{{{h^2}}} + \frac{{4{b_{11}}}}{{3{h^2}}} + \frac{{{a_4}}}{{{R_1}}} + \frac{{{a_5}}}{{{R_2}}},\;{f_7} = {d_1} - \frac{{4{d_3}}}{{{h^2}}} + \frac{{4{b_{20}}}}{{3{h^2}}} + \frac{{{a_5}}}{{{R_1}}} + \frac{{{a_8}}}{{{R_2}}},\;{f_8} = \frac{{4{E_{12}}}}{{3{h^2}}} + \frac{{8{E_{66}}}}{{3{h^2}}},\;{f_9} = \frac{{4{b_{12}}}}{{3{h^2}}},\;{f_{10}} = \frac{{4{b_{13}}}}{{3{h^2}}} + \frac{{8{b_{18}}}}{{3{h^2}}},} \\ {{f_{11}} = \frac{{4{b_{14}}}}{{3{h^2}}},\;{f_{12}} = \frac{{8{b_{15}}}}{{3{h^2}}} - \frac{{8{b_{19}}}}{{3{h^2}}},\;{f_{13}} = \frac{{4{b_{16}}}}{{3{h^2}}},\;{f_{14}} = \frac{{4{b_{17}}}}{{3{h^2}}},\;{f_{15}} = \frac{{{a_1}}}{{{R_1}}} - \frac{{{a_6}}}{{{R_2}}}.} \\ \end{array} $$

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Oktem, A.S., Soares, C.G. Analysis of the static response of cross-ply simply supported plates and shells based on a higher-order theory. Mech Compos Mater 48, 65–76 (2012). https://doi.org/10.1007/s11029-012-9252-z

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