Analysis With Matlab Code | Composite Plate Bending

% Load (uniform pressure) F(n) = 1000; % Pa end end

dx2 = dx^2; dy2 = dy^2; kxx = (w(i_center-1,j_center) - 2 w(i_center,j_center) + w(i_center+1,j_center)) / dx2; kyy = (w(i_center,j_center-1) - 2 w(i_center,j_center) + w(i_center,j_center+1)) / dy2; kxy = (w(i_center-1,j_center-1) - w(i_center-1,j_center+1) - w(i_center+1,j_center-1) + w(i_center+1,j_center+1)) / (4 dx dy);

strain_global_bot = [kxx; kyy; 2*kxy] * z_bot; stress_global_bot = Q_bar * strain_global_bot; stress_local_bot = T \ stress_global_bot;

%% Stress Recovery % Compute curvatures at center element (using central diff) i_center = round(Nx/2); j_center = round(Ny/2); if mod(Nx,2)==0, i_center=i_center+1; end if mod(Ny,2)==0, j_center=j_center+1; end Composite Plate Bending Analysis With Matlab Code

boundary_nodes = []; for i = 1:Nx for j = [1, Ny] boundary_nodes = [boundary_nodes, idx(i,j)]; end end for j = 2:Ny-1 boundary_nodes = [boundary_nodes, idx(1,j), idx(Nx,j)]; end boundary_nodes = unique(boundary_nodes);

For interior node (i,j):

% Set rows for boundary to identity K(boundary_nodes, :) = 0; for n = boundary_nodes K(n,n) = 1; F(n) = 0; end % Load (uniform pressure) F(n) = 1000; %

% Central difference coefficients c1 = D(1,1)/dx^4; c2 = (2*(D(1,2)+2 D(3,3)))/(dx^2 dy^2); c3 = D(2,2)/dy^4;

We’ll solve for deflection and then compute stresses in each ply. We discretize the plate into (N_x \times N_y) points. The biharmonic operator is approximated using central differences:

We assemble a sparse linear system ( [K] {w} = {f} ) and solve. Below is the complete code. It computes deflections, curvatures, and then stresses in each ply at Gauss points. Below is the complete code

[ D_{11} \frac{\partial^4 w}{\partial x^4} + 2(D_{12}+2D_{66}) \frac{\partial^4 w}{\partial x^2 \partial y^2} + D_{22} \frac{\partial^4 w}{\partial y^4} = q(x,y) ]

fprintf('D Matrix (N.m):\n'); disp(D);

% Max deflection fprintf('Max deflection = %.2e m\n', max(w(:)));

% Ply stacking [0/90/90/0] (symmetric) theta = [0, 90, 90, 0]; % degrees z = linspace(-h/2, h/2, num_plies+1); % ply interfaces

% Reduced stiffness matrix (plane stress) Q11 = E1/(1-nu12 nu21); Q12 = nu12 E2/(1-nu12 nu21); Q22 = E2/(1-nu12 nu21); Q66 = G12;