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kappa = [kxx; kyy; 2*kxy]; % engineering curvatures
% 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)
%% Compute ABD Matrix A = zeros(3,3); B = zeros(3,3); D = zeros(3,3); for k = 1:num_plies theta_k = theta(k) * pi/180; m = cos(theta_k); n = sin(theta_k); % Transformation matrix T = [m^2, n^2, 2 m n; n^2, m^2, -2 m n; -m n, m n, m^2-n^2]; % Q_bar = T * Q * T_inv Q = [Q11, Q12, 0; Q12, Q22, 0; 0, 0, Q66]; Q_bar = T * Q * T'; % Integrate through thickness A = A + Q_bar * (z(k+1)-z(k)); B = B + Q_bar * 0.5 * (z(k+1)^2 - z(k)^2); D = D + Q_bar * (1/3) * (z(k+1)^3 - z(k)^3); end % For symmetric laminate, B should be zero (numerically small) B = zeros(3,3); % enforce symmetry Composite Plate Bending Analysis With Matlab 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) ]
% Interior points for i = 3:Nx-2 for j = 3:Ny-2 n = idx(i,j); % w_xxxx K(n, idx(i-2,j)) = K(n, idx(i-2,j)) + c1; K(n, idx(i-1,j)) = K(n, idx(i-1,j)) - 4 c1; K(n, idx(i,j)) = K(n, idx(i,j)) + 6 c1; K(n, idx(i+1,j)) = K(n, idx(i+1,j)) - 4 c1; K(n, idx(i+2,j)) = K(n, idx(i+2,j)) + c1; % w_yyyy K(n, idx(i,j-2)) = K(n, idx(i,j-2)) + c3; K(n, idx(i,j-1)) = K(n, idx(i,j-1)) - 4 c3; K(n, idx(i,j)) = K(n, idx(i,j)) + 6 c3; K(n, idx(i,j+1)) = K(n, idx(i,j+1)) - 4 c3; K(n, idx(i,j+2)) = K(n, idx(i,j+2)) + c3; % w_xxyy K(n, idx(i-1,j-1)) = K(n, idx(i-1,j-1)) + c2; K(n, idx(i-1,j)) = K(n, idx(i-1,j)) - 2 c2; K(n, idx(i-1,j+1)) = K(n, idx(i-1,j+1)) + c2; K(n, idx(i,j-1)) = K(n, idx(i,j-1)) - 2 c2; K(n, idx(i,j)) = K(n, idx(i,j)) + 4 c2; K(n, idx(i,j+1)) = K(n, idx(i,j+1)) - 2 c2; K(n, idx(i+1,j-1)) = K(n, idx(i+1,j-1)) + c2; K(n, idx(i+1,j)) = K(n, idx(i+1,j)) - 2*c2; K(n, idx(i+1,j+1)) = K(n, idx(i+1,j+1)) + c2; kappa = [kxx; kyy; 2*kxy]; % engineering curvatures
% 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;
% Reduced stiffness matrix (plane stress) Q11 = E1/(1-nu12 nu21); Q12 = nu12 E2/(1-nu12 nu21); Q22 = E2/(1-nu12 nu21); Q66 = G12; We discretize the plate into (N_x \times N_y) points
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);
% Apply simply supported boundary conditions: w=0 and Mxx=0 => w,xx=0 on x-edges % We'll set w=0 on all edges and use ghost points to enforce curvature=0 % For simplicity, we set w=0 on boundary nodes and eliminate their equations.
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:
kappa = [kxx; kyy; 2*kxy]; % engineering curvatures
% 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)
%% Compute ABD Matrix A = zeros(3,3); B = zeros(3,3); D = zeros(3,3); for k = 1:num_plies theta_k = theta(k) * pi/180; m = cos(theta_k); n = sin(theta_k); % Transformation matrix T = [m^2, n^2, 2 m n; n^2, m^2, -2 m n; -m n, m n, m^2-n^2]; % Q_bar = T * Q * T_inv Q = [Q11, Q12, 0; Q12, Q22, 0; 0, 0, Q66]; Q_bar = T * Q * T'; % Integrate through thickness A = A + Q_bar * (z(k+1)-z(k)); B = B + Q_bar * 0.5 * (z(k+1)^2 - z(k)^2); D = D + Q_bar * (1/3) * (z(k+1)^3 - z(k)^3); end % For symmetric laminate, B should be zero (numerically small) B = zeros(3,3); % enforce symmetry
[ 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) ]
% Interior points for i = 3:Nx-2 for j = 3:Ny-2 n = idx(i,j); % w_xxxx K(n, idx(i-2,j)) = K(n, idx(i-2,j)) + c1; K(n, idx(i-1,j)) = K(n, idx(i-1,j)) - 4 c1; K(n, idx(i,j)) = K(n, idx(i,j)) + 6 c1; K(n, idx(i+1,j)) = K(n, idx(i+1,j)) - 4 c1; K(n, idx(i+2,j)) = K(n, idx(i+2,j)) + c1; % w_yyyy K(n, idx(i,j-2)) = K(n, idx(i,j-2)) + c3; K(n, idx(i,j-1)) = K(n, idx(i,j-1)) - 4 c3; K(n, idx(i,j)) = K(n, idx(i,j)) + 6 c3; K(n, idx(i,j+1)) = K(n, idx(i,j+1)) - 4 c3; K(n, idx(i,j+2)) = K(n, idx(i,j+2)) + c3; % w_xxyy K(n, idx(i-1,j-1)) = K(n, idx(i-1,j-1)) + c2; K(n, idx(i-1,j)) = K(n, idx(i-1,j)) - 2 c2; K(n, idx(i-1,j+1)) = K(n, idx(i-1,j+1)) + c2; K(n, idx(i,j-1)) = K(n, idx(i,j-1)) - 2 c2; K(n, idx(i,j)) = K(n, idx(i,j)) + 4 c2; K(n, idx(i,j+1)) = K(n, idx(i,j+1)) - 2 c2; K(n, idx(i+1,j-1)) = K(n, idx(i+1,j-1)) + c2; K(n, idx(i+1,j)) = K(n, idx(i+1,j)) - 2*c2; K(n, idx(i+1,j+1)) = K(n, idx(i+1,j+1)) + c2;
% 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;
% Reduced stiffness matrix (plane stress) Q11 = E1/(1-nu12 nu21); Q12 = nu12 E2/(1-nu12 nu21); Q22 = E2/(1-nu12 nu21); Q66 = G12;
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);
% Apply simply supported boundary conditions: w=0 and Mxx=0 => w,xx=0 on x-edges % We'll set w=0 on all edges and use ghost points to enforce curvature=0 % For simplicity, we set w=0 on boundary nodes and eliminate their equations.
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:
