fprintf('D Matrix (N.m):\n'); disp(D);
[ 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) ] Composite Plate Bending Analysis With Matlab Code
Boundary conditions (simply supported): [ w = 0,\quad M_{xx}=0 \Rightarrow \frac{\partial^2 w}{\partial x^2}=0 \text{ on } x=0,a ] (same for y-direction) fprintf('D Matrix (N
% Calculate stresses in each ply at top and bottom of ply fprintf('\nStress recovery at center (x=%.3f, y=%.3f):\n', x(i_center), y(j_center)); for k = 1:num_plies theta_k = theta(k) pi/180; m = cos(theta_k); n = sin(theta_k); T = [m^2, n^2, 2 m n; n^2, m^2, -2 m n; -m n, m n, m^2-n^2]; Q = [Q11, Q12, 0; Q12, Q22, 0; 0, 0, Q66]; z_top = z(k+1); z_bot = z(k); % Stress at top of ply (global coordinates) sigma_global_top = z_top * (D(1:3,1:3) \ kappa); % M = D kappa, sigma = M z/I?? Actually sigma_global = Q_bar * kappa * z % Correct method: curvatures -> strains = z kappa, then stress = Q_bar * strain strain_global = [kxx; kyy; 2*kxy] * z_top; stress_global_top = Q_bar * strain_global; stress_local_top = T \ stress_global_top; % transform to material coordinates (1,2,6) fprintf('D Matrix (N.m):\n')
% Load (uniform pressure) F(n) = 1000; % Pa end end
% 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;
