# Clamped Square Isotropic Plate with Uniform Pressure Load

This example shows how to calculate the deflection of a structural plate under a pressure loading.

The partial differential equation for a thin isotropic plate with a pressure loading is

$${\nabla}^{2}(D{\nabla}^{2}w)=-p,$$

where $$D$$ is the bending stiffness of the plate given by

$$D=\frac{E{h}^{3}}{12(1-{\nu}^{2})},$$

and $$E$$ is the modulus of elasticity, $$\nu $$ is Poisson's ratio, $$h$$ is the plate thickness, $$w$$ is the transverse deflection of the plate, and $$p$$ is the pressure load.

The boundary conditions for the clamped boundaries are $$w=0$$ and $${w}^{\prime}=0$$, where $${w}^{\prime}$$ is the derivative of $$w$$ in a direction normal to the boundary.

Partial Differential Equation Toolbox™ cannot directly solve this fourth-order plate equation. Convert the fourth-order equation to these two second-order partial differential equations, where $$v$$ is the new dependent variable.

$${\nabla}^{2}w=v$$

$$D{\nabla}^{2}v=-p$$

You cannot directly specify boundary conditions for both $$w$$ and $${w}^{\prime}$$ in this second-order system. Instead, specify that $${w}^{\prime}$$ is 0, and define $${v}^{\prime}$$ so that $$w$$ also equals 0 on the boundary. To specify these conditions, use stiff "springs" distributed along the boundary. The springs apply a transverse shear force to the plate edge. Define the shear force along the boundary due to these springs as $$n\cdot D\nabla v=-kw$$, where $$n$$ is the normal to the boundary, and $$k$$ is the stiffness of the springs. This expression is a generalized Neumann boundary condition supported by the toolbox. The value of $$k$$ must be large enough so that $$w$$ is approximately 0 at all points on the boundary. It also must be small enough to avoid numerical errors due to an ill-conditioned stiffness matrix.

The toolbox uses the dependent variables ${\mathit{u}}_{1}$ and ${\mathit{u}}_{2}$ instead of $$w$$ and $$v$$. Rewrite the two second-order partial differential equations using variables ${\mathit{u}}_{1}$ and ${\mathit{u}}_{2}$:

$$-{\nabla}^{2}{u}_{1}+{u}_{2}=0$$

$$-D{\nabla}^{2}{u}_{2}=p$$

Create a PDE model for a system of two equations.

model = createpde(2);

Create a square geometry and include it in the model.

len = 10; gdm = [3 4 0 len len 0 0 0 len len]'; g = decsg(gdm,'S1',('S1')'); geometryFromEdges(model,g);

Plot the geometry with the edge labels.

figure pdegplot(model,EdgeLabels="on") xlim([-1,11]) ylim([-1,11]) title("Geometry With Edge Labels Displayed")

PDE coefficients must be specified using the format required by the toolbox. For details, see

The c coefficient in this example is a tensor, which can be represented as a 2-by-2 matrix of 2-by-2 blocks:

$$\left[\begin{array}{cccc}c(1)& c(2)& \cdot & \cdot \\ \cdot & c(3)& \cdot & \cdot \\ \cdot & \cdot & c(4)& c(5)\\ \cdot & \cdot & \cdot & c(6)\end{array}\right]$$

This matrix is further flattened into a column vector of six elements. The entries in the full 2-by-2 matrix (defining the coefficient a) and the 2-by-1 vector (defining the coefficient f) follow directly from the definition of the two-equation system.

E = 1.0e6; % Modulus of elasticity nu = 0.3; % Poisson's ratio thick = 0.1; % Plate thickness pres = 2; % External pressure D = E*thick^3/(12*(1 - nu^2)); 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);

To define boundary conditions, first specify spring stiffness.

k = 1e7;

Define distributed springs on all four edges.

bOuter = applyBoundaryCondition(model,"neumann",Edge=(1:4),... g=[0 0],q=[0 0; k 0]);

Generate a mesh.

generateMesh(model);

Solve the model.

res = solvepde(model);

Access the solution at the nodal locations.

u = res.NodalSolution;

Plot the transverse deflection.

pdeplot(model,XYData=u(:,1),Contour="on") title("Transverse Deflection")

Find the transverse deflection at the plate center.

numNodes = size(model.Mesh.Nodes,2); wMax = min(u(1:numNodes,1))

wMax = -0.2762

Compare the result with the deflection at the plate center computed analytically.

wMax_exact = -.0138*pres*len^4/(E*thick^3)

wMax_exact = -0.2760