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Composite Plate Bending Analysis With Matlab Code

  • March 25, 2012
  • Jared Brown

Composite Plate Bending Analysis With Matlab Code

For simplicity, we first analyze or quasi-isotropic plates where ( B_{ij} = 0 ) and ( D_{16} = D_{26} = 0 ). Then the equation simplifies to:

For a simply supported cross-ply laminate, this simplifies to ( w = 0 ) and ( \partial^2 w / \partial n^2 = 0 ) on edges. Composite Plate Bending Analysis With Matlab Code

% Build finite difference matrix N_total = nx * ny; A_mat = sparse(N_total, N_total); F = zeros(N_total,1); For simplicity, we first analyze or quasi-isotropic plates

N = length(layup); z = cumsum([-sum(thicknesses)/2, thicknesses]); % interfaces ABD = zeros(6,6); for k = 1:N theta = layup{k} * pi/180; m = cos(theta); n = sin(theta); T = [m^2, n^2, 2 m n; n^2, m^2, -2 m n; -m n, m n, m^2-n^2]; Qbar = T \ Q * T; % transformed stiffness hk = z(k+1) - z(k); ABD(1:3,1:3) = ABD(1:3,1:3) + Qbar * hk; ABD(1:3,4:6) = ABD(1:3,4:6) + Qbar * (z(k+1)^2 - z(k)^2)/2; ABD(4:6,1:3) = ABD(4:6,1:3) + Qbar * (z(k+1)^2 - z(k)^2)/2; ABD(4:6,4:6) = ABD(4:6,4:6) + Qbar * (z(k+1)^3 - z(k)^3)/3; end A = ABD(1:3,1:3); B = ABD(1:3,4:6); D = ABD(4:6,4:6); Compare to analytical formula:

[ A_{11} \frac{\partial u_0}{\partial x} + \dots - B_{11} \frac{\partial^2 w}{\partial x^2} - \dots = 0 ] [ D_{11} \frac{\partial^4 w}{\partial x^4} + \dots - B_{11} \frac{\partial^3 u_0}{\partial x^3} - \dots = q ]

% Maximum deflection max_def = max(w(:)) * 1e3; fprintf('Maximum deflection: %.3f mm\n', max_def); For validation, replace the laminate with a single isotropic layer (set E1=E2, G12=E/(2(1+nu)), same thickness). Compare to analytical formula:

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For simplicity, we first analyze or quasi-isotropic plates where ( B_{ij} = 0 ) and ( D_{16} = D_{26} = 0 ). Then the equation simplifies to:

For a simply supported cross-ply laminate, this simplifies to ( w = 0 ) and ( \partial^2 w / \partial n^2 = 0 ) on edges.

% Build finite difference matrix N_total = nx * ny; A_mat = sparse(N_total, N_total); F = zeros(N_total,1);

N = length(layup); z = cumsum([-sum(thicknesses)/2, thicknesses]); % interfaces ABD = zeros(6,6); for k = 1:N theta = layup{k} * pi/180; m = cos(theta); n = sin(theta); T = [m^2, n^2, 2 m n; n^2, m^2, -2 m n; -m n, m n, m^2-n^2]; Qbar = T \ Q * T; % transformed stiffness hk = z(k+1) - z(k); ABD(1:3,1:3) = ABD(1:3,1:3) + Qbar * hk; ABD(1:3,4:6) = ABD(1:3,4:6) + Qbar * (z(k+1)^2 - z(k)^2)/2; ABD(4:6,1:3) = ABD(4:6,1:3) + Qbar * (z(k+1)^2 - z(k)^2)/2; ABD(4:6,4:6) = ABD(4:6,4:6) + Qbar * (z(k+1)^3 - z(k)^3)/3; end A = ABD(1:3,1:3); B = ABD(1:3,4:6); D = ABD(4:6,4:6);

[ A_{11} \frac{\partial u_0}{\partial x} + \dots - B_{11} \frac{\partial^2 w}{\partial x^2} - \dots = 0 ] [ D_{11} \frac{\partial^4 w}{\partial x^4} + \dots - B_{11} \frac{\partial^3 u_0}{\partial x^3} - \dots = q ]

% Maximum deflection max_def = max(w(:)) * 1e3; fprintf('Maximum deflection: %.3f mm\n', max_def); For validation, replace the laminate with a single isotropic layer (set E1=E2, G12=E/(2(1+nu)), same thickness). Compare to analytical formula:

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