Involves the application of classical laminate theory (CLT). This question might

Question

Involves the application of classical laminate theory (CLT). This question might be used ABD Matrix to define the values by using the following steps.

1. Set Nx = 25 kN/m, Ny = Nxy = Mx = My = Mxy = 0, and determine the strains and curvatures that result from the prescribed loading condition for the [90, 45,-45, 0] material.

2. Calculate and illustrate graphically the through thickness stress distribution in the [90, 45,-45, 0] material in the laminate (x, y) coordinate system.

3. Calculate and illustrate graphically the through thickness stress distribution in the [90, 45,-45, 0] material in the material (1, 2) coordinate system.

4. Using the material properties in the table below determine the ply or plies that fail first and the value of Nx for the [90, 45,-45, 0] material. Use the following failure criteria:

a. Maximum stress

b. Maximum strain

c. Tsai-Wu, where the interaction parameter F12* is specified as F12 * = -0.5

5. Consider a rectangular sandwich plate with composite face sheets that is simply supported along all four edges. The top and bottom face sheets are identical and manufactured using a GFRP material system with the elastic properties quoted in the table in the beginning of this document. The sandwich core material is a cross linked PVC material with shear moduli Gc,xz=Gc,yz=38 MPa, and Young’s modulus Ec=110 MPa (isotropic core). The sandwich plate dimensions are: Length a (x-direction); Width b (y-direction), Core thickness tc. The sandwich plate is subjected to uniform in-plane compressive loading in the x-direction. Use the results of the sandwich plate theory (based on CLT modified to account for transverse shear deformations), the design formula for local face sheet instability (wrinkling), and the CLT analyses presented in the lecture notes to answer the following:

a. Consider the sandwich plate discussed above and with GFRP laminate face sheets composed of 4 plies with a stacking sequence of [0,90]S, a=b=750 mm, and tc=25 mm. Calculate the critical load for global buckling (expressed as critical value of the compressive normal stress resultant) and determine the corresponding buckling mode (number of half waves in the x and y directions).

b. Consider the sandwich plate problem defined in question a. Calculate the critical load for local face sheet buckling or wrinkling (expressed as the critical value of the compressive normal stress resultant). Discuss the result obtained in the context of the answer to question a above, and in the context of what the load correspond to ‘material’ failure (compression) would be.

c. Now consider the sandwich plate configuration defined in question a, but with a=250 mm and a=2250 mm, b=750 mm (i.e. two cases a/b=1/3 and a/b=3). Calculate the critical loads (expressed as critical value of compressive normal stress resultant) and determine the corresponding buckling modes (number of half waves in the x and y directions). Discuss the results obtained.

d. For the problem defined in question a, change the length to a=1500 mm, replace the core with a higher density foam material with Gc,xz=Gc,yz=76 MPa, and Young’s modulus Ec=220 MPa (isotropic core). Calculate the critical load for global buckling (critical value of compressive normal stress resultant), determine the corresponding buckling mode (number of half waves in the x and y directions), and calculate the critical load for local face sheet buckling or wrinkling (critical value of the compressive normal stress resultant). Discuss the results obtained.

Please help me writing Matlab code and give me a short explanation for each answer.

This is the first part :

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Failure stresses and strains data sheet values: Value 1040 35 570 114 72 0.021 0.002 0.011 0.0064 0.038 Prope Xt (MPa) Y (MPA) Xc (MPa) Yc (MPa 62c X and Y are the stresses to failure (strengths) in the directions parallel and transverse to the fibres and S is the shear strength. The subscriptst and c represent tension and compression

Explanation / Answer

%%Program for Calculation of A, B and D matrices for a Uniform laminate. clc; close all; clear all; digits(32); format short e; disp('Please Select the Material') disp('1 NCT 301') disp('2 New Material') comm=input('Enter the number correspond to your choice>> '); n=input('Enter Number of Layers (Total Number)>> '); disp('----------------------------------------------------------') disp('please enter layer stacking, one example is provided below'); disp('If the Layer Stacking(LS) is [+-30 0]s enter " [30 -30 0 0 -30 30] " Do Not Forget to Use Brackets [] !!!') disp('---------------------------------------------------------------------------------------------------------') l=input('Enter LS >> '); clc %Mechanical Properties of NCT 301 if comm==1 E1=155E+9; E2=12.1E+9; E3=E2; n21=0.248; n31=n21; n23=0.458; G12=4.4E+09; G13=G12; G23=3.20E+09; else disp('Please enter the material properties') disp('Enter the values for extensional and shear modulus in GPa') mm=1e+9; E1=input('E1 = '); E1=E1*mm; E2= input('E2 = '); E2=E2*mm; E3= input('E3 = '); E3=E3*mm; n21= input('n21 = '); n31= input('n31 = '); n23= input('n23 = '); G12= input('G12 = '); G12=G12*mm; G13= input('G13 = '); G13=G13*mm; G23= input('G23 = '); G23=G23*mm; end t = 125e-6; all_ang = l; k = {}; N=length(all_ang); h=zeros(N+1); h(N+1)=N*t/2; lstf=zeros(3*N,3); for k = 1 : N %%Compliance Matrix elements S = [1/E1 -n21/E1 -n31/E1 0 0 0; -n21/E1 1/E2 -n23/E2 0 0 0; -n31/E1 -n23/E2 1/E3 0 0 0; 0 0 0 1/G23 0 0; 0 0 0 0 1/G13 0; 0 0 0 0 0 1/G12]; %%Stiffness Matrix elements C = inv(S); %Reduced Stiffness Q = zeros(3,3); Q(1,1) = C(1,1) - (C(1,3))^2 / C(3,3); Q(2,2) = C(2,2) - (C(2,3))^2 / C(3,3); Q(1,2) = C(1,2) - C(1,3)*C(2,3) / C(3,3); Q(2,1) = Q(1,2); Q(3,3) = C(6,6); %Transformatioin Matrix m = cosd (all_ang(k)); n = sind (all_ang(k)); T = zeros(3,3); T(1,1)=m^2; T(1,2)=n^2; T(1,3)=2*m*n; T(2,1)=n^2; T(2,2)=m^2; T(2,3)=-2*m*n; T(3,1)=-m*n; T(3,2)=m*n; T(3,3)=m^2-n^2; %Transformed Stiffness q=zeros(3,3); q([1:2],[1:2])=Q([1:2],[1:2]); q(3,3)=2*Q(3,3); q=inv(T)*q*T; for i = 1:3 q(i,3)=q(i,3)/2; end lstf([3*k-2:3*k],[1:3])=q([1:3],[1:3]); h(k)=(k-N/2-1)*t; end % To Calculate A B and D Matrices for Uniform Laminate A=zeros(3,3); B=zeros(3,3); D=zeros(3,3); for i=1:3 for j=1:3 q([1:3],[1:3])=lstf([1:3],[1:3]); A(i,j) = q(i,j) * (h(2) - h(1)); B(i,j) = 1/2*(q(i,j) * (h(2)^2 - h(1)^2)); D(i,j) = 1/3*(q(i,j) * (h(2)^3 - h(1)^3)); for k = 2 : N q([1:3],[1:3]) = lstf( [3*k-2:3*k] , [1:3] ); A(i,j) = q(i,j) * (h(k+1) - h(k)) + A(i,j); B(i,j) = 1/2*(q(i,j) * (h(k+1)^2 - h(k)^2)) + B(i,j); D(i,j) = 1/3*(q(i,j) * (h(k+1)^3 - h(k)^3)) + D(i,j); end end end LamnStf=zeros(6,6); a=zeros(3,3); b=zeros(3,3); d=zeros(3,3); LamnStf([1:3],[1:3])=A([1:3],[1:3]); LamnStf([4:6],[4:6])=D([1:3],[1:3]); LamnStf([1:3],[4:6])=B([1:3],[1:3]); LamnStf([4:6],[1:3])=B([1:3],[1:3]); LamnCmp=inv(LamnStf); a([1:3],[1:3])=LamnCmp([1:3],[1:3]); b([1:3],[1:3])=LamnCmp([1:3],[4:6]); d([1:3],[1:3])=LamnCmp([4:6],[4:6]); if rem(length(all_ang),2)==0 A= A.*[1 1 0 ; 1 1 0 ; 0 0 1]; B= B.*[0 0 0 ; 0 0 0 ; 0 0 0]; D= D.*[1 1 1 ; 1 1 1 ; 1 1 1]; a= a.*[1 1 0 ; 1 1 0 ; 0 0 1]; b= b.*[0 0 0 ; 0 0 0 ; 0 0 0]; d= d.*[1 1 1 ; 1 1 1 ; 1 1 1]; end;

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