This document is a report submitted to Dr. Mohammad Tawfik containing solutions to homework problems involving analyzing stresses and strains in composite materials. It includes the material properties of carbon-epoxy used, calculations of local stresses and strains using MATLAB code in the appendices, and results showing the local and global stresses and strains from applying a tensile load. The safety factor of the material is also calculated based on the maximum stresses.

1. Smart Materials 4th
year
Report: Solution to HW (V)
Report No: V Date: 18/4/2013
Submitted to: Dr. Mohammad Tawfik
Name
 Mohammad Tawfik Eraky
‫عراقي‬ ‫أحمد‬ ‫توفيق‬ ‫محمد‬
2013/2014

2. Pb.1
Solution
Carbon –epoxy (AS4/3501-6)
E1 (longitudinal Modulus E1) 45 Gpa
E2 (Transverse Modulus) 12 Gpa
𝜈12 (Poisson’s Ratio) 0.19
F1t (longitudinal tensile strength) 1830 Mpa
F2t (transverse tensile strength ) 57 Mpa
F6 ( Inplane shear strength) 71Mpa
F1c(longitudinal compressive strength) 1096Mpa
F1c(transverse compressive strength) 228Mpa
Using MATLAB code (appendix A) we got the following local stresses
𝜎𝑥′𝑥′ 8.4151 Mpa
𝜎 𝑦′𝑦′ 6.5849 Mpa
𝜏 𝑥′𝑦′ -3.4151 Mpa

3. Using the maximum stress criterion
8.4151 Mpa <1096 Mpa<1830Mpa
6.5849 Mpa <57 Mpa<228Mpa
-3.4151 Mpa<71Mpa
The safety factor (minimum’’ R’’) =
57
6.5849
= 8.656
2b.P
Solution
Using MATLAB code (appendix B) we got the following results
matrixotationR
𝑇 =
0.5 0.5 1
0.5 0.5 −1
−0.5 0.5 0

4. 𝑄 =
35.3179 1.0595 0
1.0595 3.5318 0
0 0 3.500
𝐺𝑝𝑎
𝑄̅ =
11.9922 8.4922 15.8930
8.4922 11.9922 15.8930
7.9445 7.9445 18.3453
𝐺𝑝𝑎
Pb.3
Solution
Using MATLAB code (appendix c) we got the following results
(a) Strains in local axes
ϵx′x′
εy′y′
τx′y′
=
0.2745
0.003
−0.4884
x 𝟏𝟎−𝟑

5. (b) Strains in global axes
ϵxx
εyy
τxy
=
0.4334
−0.1559
−0.4127
x 𝟏𝟎−𝟑
(c)
As we see the results stains is non zero value due to Poisson
ratio when tensile load applied in x-direction there is a strain
too in transverse direction

6. Appendix A
clc;clear all;close all;
%calculating(local)stresses
ceta=45*pi/180;
c=ceta;
T=[cos(c)^2 sin(c)^2 2*sin(c)*cos(c);
sin(c)^2 cos(c)^2 -2*sin(c)*cos(c);
-sin(c)*cos(c) sin(c)*cos(c) cos(c)^2-sin(c)^2];
sigma_global=[10 ; 5 ; 2.5 ];
sigma_local_Mpa=T*sigma_global;
display(sigma_local_Mpa);

7. Appendix B
clc;clear all;close all;
%calculating(local)stresses
ceta=45*pi/180;
c=ceta;
e_1=35;
e_2=3.5;
new_12=0.3;
g_12=1.75;
g_23=0.35;
%outputs
t=[cos(c)^2 sin(c)^2 2*sin(c)*cos(c);
sin(c)^2 cos(c)^2 -2*sin(c)*cos(c);
-sin(c)*cos(c) sin(c)*cos(c) cos(c)^2-sin(c)^2];
new_21=new_12*e_2/e_1;
delta=1-new_12*new_21;
q=[e_1/delta new_12*e_2/delta 0;
new_12*e_2/delta e_2/delta 0;
0 0 2*g_12];
t_inv=inv(t);
q_bar=(t_inv)*q*t;
sigma_global=[10 ; 5 ; 2.5 ];
sigma_local_Mpa=t*sigma_global;
%display
display (q_bar);

8. Appendix C
clc;clear all;close all;
%calculating(local)stresses
ceta=10*pi/180;
c=ceta;
e_1=35;
e_2=3.5;
new_12=0.3;
g_12=1.75;
g_23=0.35;
%Transformations Matrixes
t=[cos(c)^2 sin(c)^2 2*sin(c)*cos(c);
sin(c)^2 cos(c)^2 -2*sin(c)*cos(c);
-sin(c)*cos(c) sin(c)*cos(c) cos(c)^2-sin(c)^2];
new_21=new_12*e_2/e_1;
delta=1-new_12*new_21;
q=[e_1/delta new_12*e_2/delta 0;
new_12*e_2/delta e_2/delta 0;
0 0 2*g_12];
t_inv=inv(t);
q_bar=(t_inv)*q*t;
% stresses in global axes
sigma_global=[200/20 ; 0 ; 0];
%stresses in local axes
sigma_local=t*sigma_global;
%strains in local axes
strain_local=10^-3*(inv(q))*sigma_local;
%strains in Global Axes
strain_global=(t_inv)*strain_local;
%display ouputs
display (q_bar);
display(sigma_local);
display(strain_local);
display(strain_global);