Clamped, Square Isotropic Plate With a Uniform Pressure Load
This example shows how to calculate the deflection of a structural plate acted on by a pressure loading using the Partial Differential Equation Toolbox™.
PDE and Boundary Conditions For A Thin Plate
The partial differential equation for a thin, isotropic plate with a pressure loading is
where is the bending stiffness of the plate given by
and is the modulus of elasticity, is Poisson's ratio, and is the plate thickness. The transverse deflection of the plate is and is the pressure load.
The boundary conditions for the clamped boundaries are and where is the derivative of in a direction normal to the boundary.
The Partial Differential Equation Toolbox™ cannot directly solve the fourth order plate equation shown above but this can be converted to the following two second order partial differential equations.
where is a new dependent variable. However, it is not obvious how to specify boundary conditions for this second order system. We cannot directly specify boundary conditions for both and . Instead, we directly prescribe to be zero and use the following technique to define in such a way to insure that also equals zero on the boundary. Stiff "springs" that apply a transverse shear force to the plate edge are distributed along the boundary. The shear force along the boundary due to these springs can be written where is the normal to the boundary and is the stiffness of the springs. The value of must be large enough that is approximately zero at all points on the boundary but not so large that numerical errors result because the stiffness matrix is ill-conditioned. This expression is a generalized Neumann boundary condition supported by Partial Differential Equation Toolbox™
In the Partial Differential Equation Toolbox™ definition for an elliptic system, the and dependent variables are u(1) and u(2). The two second order partial differential equations can be rewritten as
which is the form supported by the toolbox. The input corresponding to this formulation is shown in the sections below.
Create the PDE Model
Create a pde model for a PDE with two dependent variables.
numberOfPDE = 2; model = createpde(numberOfPDE);
E = 1.0e6; % modulus of elasticity nu = .3; % Poisson's ratio thick = .1; % plate thickness len = 10.0; % side length for the square plate hmax = len/20; % mesh size parameter D = E*thick^3/(12*(1 - nu^2)); pres = 2; % external pressure
For a single square, the geometry and mesh are easily defined as shown below.
gdm = [3 4 0 len len 0 0 0 len len]'; g = decsg(gdm,'S1',('S1')');
Create a geometry entity.
Plot the geometry and display the edge labels for use in the boundary condition definition.
figure; pdegplot(model,'EdgeLabels','on'); ylim([-1,11]) axis equal title 'Geometry With Edge Labels Displayed';
The documentation on PDE coefficients shows the required formats for the a and c matrices. The most convenient form for c in this example is from the table where is the number of differential equations. In this example . The tensor, in the form of an matrix of submatrices is shown below.
The six-row by one-column c matrix is defined below. The entries in the full a matrix and the f vector follow directly from the definition of the two-equation system shown above.
c = [1 0 1 D 0 D]'; a = [0 0 1 0]'; f = [0 pres]'; specifyCoefficients(model,'m',0,'d',0,'c',c,'a',a,'f',f);
k = 1e7; % spring stiffness
Define distributed springs on all four edges.
bOuter = applyBoundaryCondition(model,'neumann','Edge',(1:4),... 'g',[0 0],'q',[0 0; k 0]);
generateMesh(model, 'Hmax', hmax);
Finite Element and Analytical Solutions
The solution is calculated using the solvepde function and the transverse deflection is plotted using the pdeplot function. For comparison, the transverse deflection at the plate center is also calculated using an analytical solution to this problem.
res = solvepde(model); u = res.NodalSolution; numNodes = size(model.Mesh.Nodes,2); figure pdeplot(model,'XYData',u(1:numNodes),'Contour','on'); title 'Transverse Deflection' numNodes = size(model.Mesh.Nodes,2); fprintf('Transverse deflection at plate center(PDE Toolbox) = %12.4e\n',... min(u(1:numNodes,1)));
Transverse deflection at plate center(PDE Toolbox) = -2.7632e-01
Ccompute analytical solution.
wMax = -.0138*pres*len^4/(E*thick^3); fprintf('Transverse deflection at plate center(analytical) = %12.4e\n', wMax);
Transverse deflection at plate center(analytical) = -2.7600e-01