Composite Plate Bending Analysis With Matlab Code -

% Material properties for each layer (E1, E2, nu12, G12, G13, G23) % Example: Carbon/Epoxy E1 = 140e9; E2 = 10e9; nu12 = 0.3; G12 = 5e9; G13 = 5e9; G23 = 3.8e9;

where (u₀, v₀) are mid-plane in-plane displacements and w₀ is the transverse deflection.

%% 9. Visualization figure; plot(sig_global(1,:)/1e6, z_coords*1000, '-o', 'LineWidth', 2); grid on; xlabel('Bending Stress \sigma_x (MPa)'); ylabel('Thickness z (mm)'); title('Through-Thickness Stress Distribution'); Composite Plate Bending Analysis With Matlab Code

% Compute ABD matrix Q = [E1/(1-nu12 nu21), nu12 E2/(1-nu12 nu21), 0; nu12 E2/(1-nu12 nu21), E2/(1-nu12 nu21), 0; 0, 0, G12];

Here is a simplified script to calculate the bending stiffness (D matrix) of a symmetric laminate: % Material Properties (e.g., Carbon/Epoxy) ; v21 = v12 * E2 / E1; % Reduced Stiffness Matrix [Q] -v12*v21), v12*E2/( -v12*v21), ; v12*E2/( -v12*v21), E2/( -v12*v21), % Layup: [0/45/-45/90]s ]; t_ply = % thickness of each ply n = length(theta); h = n * t_ply; z = -h/ : t_ply : h/ % z-coordinates of interfaces D = zeros( % Material properties for each layer (E1, E2,

For complex loading (like a point load), you would wrap the solution in a for loop to sum the Fourier series (e.g., 5. Conclusion

% Laminate layup: symmetric [0/90/90/0] (4 layers) layup_angles = [0, 90, 90, 0]; % degrees n_layers = length(layup_angles); t_layer = h_total / n_layers; % each layer thickness % degrees n_layers = length(layup_angles)

%% 4. Mesh Generation nx = Nx_elem + 1; ny = Ny_elem + 1; x_nodes = linspace(0, a, nx); y_nodes = linspace(0, b, ny); [X, Y] = meshgrid(x_nodes, y_nodes);

The angles array defines the laminate [0/90/90/0] .

%% Geometry and Mesh Lx = 0.2; % plate length in x (m) Ly = 0.2; % width in y (m) nx = 20; % elements along x ny = 20; % elements along y h_total = 0.005; % total thickness (m)