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Rectangular Plates

Plate Under Uniform Pressure Load

<<"MaterialMinds`rectangularPlate`"

Define material properties and a stacking sequence

prop = {{21 10^6, 1.3 10^6, .85 10^6, .35}} ;

transverseShear = {{0.4  10^6, 0.4 10^6}} ;

Use the parseLaminate functions and notation to build a stacking sequence

s = {45, -45, 0, 90}//Repeat[2]//Mat[1]//PlyThickness[.0055]//Sym ;

ReportLaminate[s]

Angle material Ply Thick
1 45 1 0.0055
2 -45 1 0.0055
3 0 1 0.0055
4 90 1 0.0055
5 45 1 0.0055
6 -45 1 0.0055
7 0 1 0.0055
8 90 1 0.0055
9 90 1 0.0055
10 0 1 0.0055
11 -45 1 0.0055
12 45 1 0.0055
13 90 1 0.0055
14 0 1 0.0055
15 -45 1 0.0055
16 45 1 0.0055

Simple supported plate, with dimensions x = 10 and y = 5. Use 8 solution terms in both the x and y directions. A unit pressure is applied.

ans = pressureSolution[prop, transverseShear, s, 10, 5, 8, 8,  {W[X], W[-X], W[Y], W[-Y]}, 1]

--rectangularPlateSoltn--

Find the displacement at the center of the plate

MMDisplacement[ans][3][{0, 0}]

0.0132101

Create a contour plot of w displacement

plateContourPlot[MMDisplacement[ans][3], 10, 5]

[Graphics:HTMLFiles/index_11.gif]

⁃Graphics⁃

Find another solution with clamped conditions

ans2 = pressureSolution[prop, transverseShear, s, 10, 5, 8, 8,  {W[X], W[-X], W[Y], W[-Y], TX[X], TX[-X], TY[Y], TY[-Y]}, 1] ;

MMMoment[ans2][3][{0, 0}]

0.0982647

Contour plot of the x-moment for a clamped plate

plateContourPlot[MMMoment[ans2][1], 10, 5]

[Graphics:HTMLFiles/index_17.gif]

⁃Graphics⁃

Y-component of transverse shear load

plateContourPlot[MMTransverseShear[ans2][2], 10, 5, PlotPoints40]

[Graphics:HTMLFiles/index_20.gif]

⁃Graphics⁃

Plate Under Non-Uniform Pressure Load

A non-uniform pressure load can be defined using a pure function. In this case (1 + #1)& is the pure function. It is equivalent to (1+Overscript[x, _]), where Overscript[x, _]is the normalized x coordinate that runs from -1 to +1.

c = pressureSolution[prop, transverseShear, s, 10, 5, 8, 8,  {W[X], W[-X], W[Y], W[-Y]}, (1 + #1) &]

--rectangularPlateSoltn--

The resulting displacement plot shows the influence of the variable pressure load

plateContourPlot[MMMoment[c][1], 10, 5, PlotPoints35]

[Graphics:HTMLFiles/index_27.gif]

⁃Graphics⁃

Anisotropic Plate

Look at a highly anisotropic laminate constructed by using all +45 degree plies in place of a balanced construction

sa = symStack @ homoStackDeg[{45, 45, 0, 90, 45, 45, 0, 90}, .0055] ;

c = pressureSolution[prop, transverseShear, sa, 10, 5, 8, 8,  {W[X], W[-X], W[Y], W[-Y]}, 1]

--rectangularPlateSoltn--

The resulting contour plot shows that the displacements are skewed because of the anisotropic nature of the laminate

plateContourPlot[MMDisplacement[c][3], 10, 5]

[Graphics:HTMLFiles/index_33.gif]

⁃Graphics⁃

Buckling

Define material properties and a stacking sequence

prop = {{21 10^6, 1.3 10^6, .85 10^6, .35}} ;

transverseShear = {{0.4  10^6, 0.4 10^6}} ;

Use the parseLaminate functions and notation to build a stacking sequence

s = {45, -45, 0, 90}//Repeat[2]//Mat[1]//PlyThickness[.0055]//Sym ;

ReportLaminate[s]

Angle material Ply Thick
1 45 1 0.0055
2 -45 1 0.0055
3 0 1 0.0055
4 90 1 0.0055
5 45 1 0.0055
6 -45 1 0.0055
7 0 1 0.0055
8 90 1 0.0055
9 90 1 0.0055
10 0 1 0.0055
11 -45 1 0.0055
12 45 1 0.0055
13 90 1 0.0055
14 0 1 0.0055
15 -45 1 0.0055
16 45 1 0.0055

Buckling solution for a simply-supported plate. The load vector is a unit x-direction compression.

{buck, c} = eigenSolution[Buckling, prop, transverseShear, s, 10, 5, 8, 8, {W[X], W[-X], W[Y], W[-Y]}, {-1, 0, 0}]

{831.992, --rectangularPlateSoltn--}

The first element of the list is the buckling load factor. It multiples in the input load list to yield a buckling load. The solution object (the second item returned in the list) can be used to generate buckling mode plots. Note that the number of terms in the solution series must be large enough to give a good representation of the buckling mode. For very large or small aspect ratio plates, there may not be enough terms to capture all of the buckle mode half-waves.

plateContourPlot[MMDisplacement[c][3], 10, 5, plateLegendFalse]

[Graphics:HTMLFiles/index_42.gif]

⁃ContourGraphics⁃

Report the solution with a unit shear load (third component of the load vector)

{buck, c} = eigenSolution[Buckling, prop, transverseShear, s, 15, 5, 8, 8,  {W[X], W[-X], W[Y], W[-Y]}, {0, 0, 1}]

{1028.63, --rectangularPlateSoltn--}

The contour plot shows a typical shear buckling pattern

plateContourPlot[MMDisplacement[c][3], 15, 5, plateLegendFalse]

[Graphics:HTMLFiles/index_47.gif]

⁃ContourGraphics⁃


Created by Mathematica  (March 7, 2004)