%% 5. Loop Through Layers to Build ABD for k = 1:n_plies theta = layers(k) * (pi/180); % Convert to radians
% ========================================================================= % Composite Plate Bending Analysis using Classical Laminate Plate Theory % Boundary Conditions: Simply Supported on all edges (Navier's Method) % ========================================================================= clear; clc; close all; %% 1. Geometry and Load Data a = 0.5; % Plate length along x-axis (m) b = 0.5; % Plate width along y-axis (m) q0 = -10000; % Uniform transverse load (Pa) %% 2. Material Properties (E1, E2, nu12, G12) E1 = 140e9; % Longitudinal elastic modulus (Pa) E2 = 10e9; % Transverse elastic modulus (Pa) nu12 = 0.3; % Major Poisson's ratio G12 = 5e9; % In-plane shear modulus (Pa) nu21 = nu12 * E2 / E1; %% 3. Layup Configuration % Define ply orientations (degrees) and thicknesses (m) theta = [0, 90, 90, 0]; ply_t = 0.00125 * ones(1, length(theta)); % 1.25 mm per ply total_h = sum(ply_t); % Calculate Z-coordinates for each ply interface relative to mid-plane z = zeros(1, length(theta) + 1); z(1) = -total_h / 2; for k = 1:length(theta) z(k+1) = z(k) + ply_t(k); end %% 4. Calculate Laminate Bending Stiffness Matrix [D] Q = zeros(3,3); Q(1,1) = E1 / (1 - nu12*nu21); Q(1,2) = nu21 * E1 / (1 - nu12*nu21); Q(2,1) = Q(1,2); Q(2,2) = E2 / (1 - nu12*nu21); Q(3,3) = G12; D = zeros(3,3); for k = 1:length(theta) t = deg2rad(theta(k)); % Transformation matrix components m = cos(t); n = sin(t); T = [m^2, n^2, 2*m*n; n^2, m^2, -2*m*n; -m*n, m*n, m^2-n^2]; % Transformed reduced stiffness matrix Qbar = T \ Q * (T^-1).'; % Accumulate into bending stiffness matrix D D = D + (1/3) * Qbar * (z(k+1)^3 - z(k)^3); end %% 5. Navier's Solution for Deflection Nx = 50; Ny = 50; % Grid resolution for plotting x = linspace(0, a, Nx); y = linspace(0, b, Ny); [X, Y] = meshgrid(x, y); W = zeros(Nx, Ny); % Initialize deflection matrix M_max = 30; N_max = 30; % Fourier series truncation limits for m = 1:2:M_max for n = 1:2:N_max % Load coefficient for uniform distribution Qmn = (16 * q0) / ((pi^2) * m * n); % Denominator based on governing equation denom = (pi^4) * (D(1,1)*(m/a)^4 + 2*(D(1,2) + 2*D(3,3))*(m/a)^2*(n/b)^2 + D(2,2)*(n/b)^4); % Deflection amplitude Wmn = Qmn / denom; % Construct displacement field W = W + Wmn * sin(m * pi * X / a) .* sin(n * pi * Y / b); end end %% 6. Visualization figure('Color', 'w'); surf(X, Y, W*1000, 'EdgeColor', 'interp'); colormap jet; colorbar; title('Transverse Deflection of Composite Plate'); xlabel('Length a (m)'); ylabel('Width b (m)'); zlabel('Deflection w (mm)'); view(-37.5, 30); grid on; % Display max deflection in command window fprintf('Maximum central deflection: %.4f mm\n', min(W(:))*1000); Use code with caution. Result Analysis and Interpretation
%% 8. Stress Analysis at Top and Bottom of Plies disp('--- Ply Stresses ---'); z_coords = []; sig_global = []; for k = 1:n_plies % Get z-coordinates for top and bottom of current ply z_bot_k = z(k); z_top_k = z(k+1); Composite Plate Bending Analysis With Matlab Code
This article provides a step-by-step approach to implementing a for composite plate bending using MATLAB . We will use Classical Laminated Plate Theory (CLPT) and a 4-node rectangular element with 12 degrees of freedom per element (w, θx, θy at each node). A complete working code is provided, along with validation against an analytical solution.
% Analytical solution (Navier, simply supported, symmetric laminate) % For square plate a=b, load p=p0, D11, D22, D12, D66, D16=D26=0 if symmetric balanced % Assume D16=0, D26=0 for cross-ply [0/90]s D11 = D(1,1); D12 = D(1,2); D22 = D(2,2); D66 = D(3,3); w_analytical = 0; for m = 1:2:25 for n = 1:2:25 denom = pi^4 * ( D11*(m/a)^4 + 2*(D12+2 D66) (m/a)^2*(n/b)^2 + D22*(n/b)^4 ); w_analytical = w_analytical + (16 p0/(m n pi^2)) / denom; end end w_analytical = w_analytical * sin(m pi/2) sin(n pi/2); % at center Material Properties (E1, E2, nu12, G12) E1 =
that is (SS-1 boundary conditions), Naviers' solution utilizes a double Fourier series to solve the governing differential equations. Load Representation A uniform distributed load is expressed as:
[ \mathbfd^e = [ \mathbfd 1 \ \mathbfd 2 \ \mathbfd 3 \ \mathbfd 4 ]^T, \quad \mathbfd i = [u 0i \ v 0i \ w 0i \ \phi xi \ \phi yi]^T. ] Navier's Solution for Deflection Nx = 50; Ny
%% 1. Material Properties (Example: Carbon/Epoxy) E1 = 181e9; % Longitudinal Modulus (Pa) E2 = 10.3e9; % Transverse Modulus (Pa) G12 = 7.17e9; % Shear Modulus (Pa) nu12 = 0.28; % Poisson's Ratio
%% 3. Calculate Reduced Stiffness Matrix [Q] for 0-degree ply % Using Plane Stress assumption Q11 = E1 / (1 - nu12*nu21); Q22 = E2 / (1 - nu12*nu21); Q12 = (nu12 * E2) / (1 - nu12*nu21); Q66 = G12;
Computes the stiffness matrices A, B, D by iterating through layers, transforming the Q-matrix, and integrating through the thickness.
w0=∑i=1num_nodesNiwi,ϕx=∑i=1num_nodesNiϕxi,ϕy=∑i=1num_nodesNiϕyiw sub 0 equals sum from i equals 1 to n u m _ n o d e s of cap N sub i w sub i comma space phi sub x equals sum from i equals 1 to n u m _ n o d e s of cap N sub i phi sub x i end-sub comma space phi sub y equals sum from i equals 1 to n u m _ n o d e s of cap N sub i phi sub y i end-sub Stiffness Matrix Integration The element stiffness matrix is split into bending-extension stiffness and transverse shear stiffness