% Shape functions for w and slopes (σ = -dw/dx, τ = dw/dy) % Node 1 (xi=-1, eta=-1) N(1) = 1/8 * (1-xi) (1-eta) ( (1+xi)^2*(1+eta)^2 - (1+xi)*(1+eta) - (1+xi)^2 - (1+eta)^2 + 2 ); % Similar for others – too lengthy. Instead, we use a simplified approach: % For demonstration and educational clarity, we assume a reduced integration % and approximate B using bilinear w + constant slopes. Full derivation is long.
% Jacobian for rectangular element detJ = a_elem * b_elem; Composite Plate Bending Analysis With Matlab Code
% 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 % Shape functions for w and slopes (σ
% Element connectivity elements = zeros(Nx_elem * Ny_elem, 4); elem_id = 0; for iy = 1:Ny_elem for ix = 1:Nx_elem elem_id = elem_id + 1; n1 = (iy-1)*nx + ix; n2 = n1 + 1; n3 = n2 + nx; n4 = n3 - 1; elements(elem_id, :) = [n1, n2, n3, n4]; end end % Jacobian for rectangular element detJ = a_elem
% Transformation matrix for stresses (3x3) T = [m^2, n^2, 2*m*n; n^2, m^2, -2*m*n; -m*n, m*n, m^2-n^2];