1. Homework no.5
Ply mechanics

2. Problem (5-2)
%% Problem 5.2%%
%%used material carbon epoxy(AS4/3501-6)%%
%for this material F1t=1830, F2t=57 , F6=71%
clear all;
clc;
sigmax=10;
sigmay=5;
sigmaxy=2.5;
m=cosd(60);
n=sind(60);
T=[m^2 n^2 2*n*m; n^2 m^2 -2*m*n; -m*n m*n (m^2)-(n^2)];
sigma=[sigmax;sigmay;sigmaxy];
sigmadash=T*sigma
RESULT:
sigmadash =
8.4151
6.5849
-3.4151
F1t=1830>8.41 F2t=57>6.584, F6=71>3.415
Then  F.s=allowable strength/applied load=57/6.584=8.6

3. Problem (5-3)
clear all;
clc;
E1=35;
E2=3.5;
neu12=0.3;
neu21=(E2*neu12)/E1;
G12=1.75;
delta=1-(neu12*neu21);
Q=[E1/delta (E2*neu12)/delta 0; neu12*E2/delta E2/delta 0; 0 0
G12]
m=cosd(45);
n=sind(45);
T=[m^2 n^2 2*n*m; n^2 m^2 -2*m*n; -m*n m*n (m^2)-(n^2)];
Tinverse=inv(T);
Qbar=Tinverse*Q*T
Results
Q =
35.3179 1.0595 0
1.0595 3.5318 0
0 0 1.7500
Qbar =
11.1172 9.3672 15.8930
9.3672 11.1172 15.8930
7.9465 7.9465 18.3653

4. Problem (5-5)
%problem (5-5)
clear all;
clc;
%geometry
w=10;
t=0.01;
L=150;
E1=35;
A1=w*t;
A2=L*t;
A12=L*w;
%material property
E2=3.5;
neu12=0.3;
neu21=(E2*neu12)/E1;
G12=1.75;
delta=1-(neu12*neu21);
Q=[E1/delta (E2*neu12)/delta 0; neu12*E2/delta E2/delta 0; 0 0
G12]
%transformation
m=cosd(10);
n=sind(10);
T=[m^2 n^2 2*n*m; n^2 m^2 -2*m*n; -m*n m*n (m^2)-(n^2)];
Tinverse=inv(T);
Qbar=Tinverse*Q*T
%applied forces and stresses
Faxial=200;
sigma1=Faxial/A1;
sigma2=Faxial/A2;
sigma12=Faxial/A12;
sigma=[sigma1; sigma2 ;sigma12];
%(a)stress strain relation in local coordinate
epsilon=inv(Q)*sigma
%(b)strain in global coordinate

5. epsilondash=Tinverse*epsilon
Results
Q =
35.3179 1.0595 0
1.0595 3.5318 0
0 0 1.7500
Qbar =
33.3876 2.0313 10.7758
2.0313 3.5184 0.0957
5.3879 0.0478 3.6936
epsilon =
56.0000
20.9524
0.0762
epsilondash =
54.9171
22.0353
6.0651