Composite Plate Bending Analysis With Matlab Code [upd] 〈Cross-Platform〉

% If symmetric laminate (B=0), we can solve simply for curvatures: % M = D * k => k = D_inv * M

function [Nw, dN] = shape_functions(xi, eta) % Shape functions and derivatives for w (12-term polynomial) % xi, eta in [-1,1] for master element (size 2a x 2b) % Returns Nw (1x4) for nodal w, dN (2x4) for slopes? Actually we need 12 DOF. % Here simplified: we return shape functions for w only. % For full [B] matrix, we need derivatives of w wrt x,y.

% For a fully functional version, please contact author or % implement shape functions from "Analysis of Laminated Composite Plates" by Reddy.

= -z * κ , where κ = ∂²w/∂x² , ∂²w/∂y² , 2∂²w/∂x∂y ^T

w(x,y)=∑m=1∞∑n=1∞Wmnsin(mπxa)sin(nπyb)w open paren x comma y close paren equals sum from m equals 1 to infinity of sum from n equals 1 to infinity of cap W sub m n end-sub sine open paren the fraction with numerator m pi x and denominator a end-fraction close paren sine open paren the fraction with numerator n pi y and denominator b end-fraction close paren For a uniform distributed load , the load coefficients Qmncap Q sub m n end-sub Composite Plate Bending Analysis With Matlab Code

% For symmetric laminates (B=0), deflection depends on D matrix w_max = q0 / (D( ); fprintf( 'Maximum Deflection: %e m\n' Use code with caution. Copied to clipboard

Learn how to perform structural analysis using the finite element method with Partial Differential Equation ToolboxTM in MATLAB®.

If the fibers are oriented at an angle $\theta$ relative to the plate axis ($x-y$), we transform the stiffness matrix: $$ [\barQ] = [T]^-1 [Q] [T]^-T $$ (This is handled via transformation matrices involving $\sin\theta$ and $\cos\theta$).

The following MATLAB script automates this analysis. It calculates the ABD matrix, solves for the center deflection of a simply supported cross-ply laminate under a uniform load, and plots the deflected surface shape. Use code with caution. Interpreting the Analysis % If symmetric laminate (B=0), we can solve

[ \mathbfB_m = \beginbmatrix \frac\partial N_i\partial x & 0 & 0 & 0 & 0 \ 0 & \frac\partial N_i\partial y & 0 & 0 & 0 \ \frac\partial N_i\partial y & \frac\partial N_i\partial x & 0 & 0 & 0 \endbmatrix, ]

f_e = ∫_-1^1∫_-1^1 p * [N_w]^T * det(J) * (a*b) dξ dη

[ (A_ij, B_ij, D_ij) = \sum_k=1^N \int_z_k-1^z_k \barQ_ij^(k) (1, z, z^2) , dz, \quad i,j=1,2,6, ]

Navier’s method solves this differential equation for simply supported boundary conditions using double Fourier series expansions for both the load and the displacement: % For full [B] matrix, we need derivatives of w wrt x,y

zk1 = z(k+1); zk = z(k);

function Ke = element_stiffness(xy, ABD, As) % xy: 4x2 matrix of element nodal coordinates (x,y) % ABD: 6x6 matrix [A B; B D] % As: 2x2 shear stiffness matrix % Returns 20x20 element stiffness matrix

function [ABD, As] = laminate_ABD(plies, z_coords, mat_props) % Compute laminate ABD and shear stiffness matrices % plies: cell array with fiber angles (deg) for each layer % z_coords: vector of z-coordinates of layer interfaces (from bottom to top) % mat_props: structure with E1, E2, G12, G23, G13, nu12

% At each node i, shape function for w gives 1 at node i, 0 at others. % Using bilinear shape functions for w alone would cause incompatibility. % For a working element, we use the ACM element (12 DOF). Simplified here:

Dij=13∑k=1N(Q̄ij)k(zk3−zk−13)cap D sub i j end-sub equals one-third sum from k equals 1 to cap N of open paren cap Q bar sub i j end-sub close paren sub k open paren z sub k cubed minus z sub k minus 1 end-sub cubed close paren Transverse Shear Stiffness Matrix (A_s) For FSDT formulations, an additional matrix Ascap A sub s accounts for transverse shear stiffness: