1. )4HW (
Example (1):
Givens:
= = 4.6 Gpa
= 230 GPa
= = 1.769 GPa
= = 92 GPa
Req:
, , ,
:Sol
= . + . = (0.4*230) + (0.6*4.6) = 94.7GPa
= = = 7.55GPa
= . + . = (0.3*0.6) + (0.25*0.4) = 0.26
= = = 2.9 GPa

2. Example (2):
Req:: , , , for the following material
combinations: = 0.45 and = 0.55
a))E-glass and polyester(isophthalic)
From table 2.1:
E-glass
= 72.345GPa
= 0.22
= 3.45 GPa
From table 2.4:
Polyester
= 3.4GPa
= 0.38
= 75.9 GPa
= . + . = (0.55*72.345) + (0.45*3.4) = 41.32
= = = 7.145GPa
= . + . = (0.38*0.55) + (0.22*0.45) = 0.308
= = = 6.047GPa

3. b))S-glass and epoxy9310:
From table 2.1:
S-glass
= 85GPa
= 0.22
= 4.8 GPa
From table 2.4:
Epoxy9310
= 3.12GPa
= 0.38
= 75.8 GPa
= . + . = (0.55*85) + (0.45*3.12) = 48.154
= = = 6.635GPa
= . + . = (0.38*0.55) + (0.22*0.45) = 0.308
= = = 8.297GPa
c))Carbon T300 and vinyl esler:
From table 2.1:
Carbon T300
= 230GPa
= 0.2
= 3.53 GPa

4. From table 2.4:
Vinyl esler
= 3.4GPa
= 0.38
= 82.7 GPa
= . + . = (0.55*230) + (0.45*3.4) = 128.03
= = = 7.421GPa
= . + . = (0.38*0.55) + (0.2*0.45) = 0.299
= = = 6.202GPa
Example (3)
Select a matrix, to obtain material with and
Using matlab for different fibers and matrices we find that to
get these conditions we use:
D-glass as fiber and Bismaleiraide as a matrix
Matlab code:
clear;clc;closeall;
Vm = 0.24;
Vf = 1-Vm;
Ef = 55;
Em = 3.583;
E1 = (Vf*Ef)+(Vm*Em)
E2 = (Em*Ef)/((Vf*Em)+(Vm*Ef))
F = E1/E2
Output:
E1 = 42.6599
E2 =12.3761
F =3.4470