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Matlab code of chapter 4
%matlab code of composite structure
%chapter four code
%michromechanics composite material
%calculation of material properties
%inputs: Em,Ef,Vm,Vf,Gm,Gf,alpha m, alpha f, neu
f, neu m
%_______________________________________________
_____________
Ef=230;
Em=4.6;
Vf=0.4;
Vm=0.6;
Gm=1.769;
Gf=92;
alpham=30*10^-6;
alphaf=15*10^-6;
neuf=0.25;
neum=0.3;
E1=(Ef*Vf)+(Em*Vm)
E2=(Em*Ef)/((Vf*Em)+(Vm*Ef))
alpha1=((Ef*Vf*alphaf)+(Em*Vm*alpham))/E1
alpha2=(alphaf*Vf)+(alpham*Vm)
G12=(Gm*Gf)/((Gf*Vm)+(Gm*Vf))
neu12=(neuf*Vf)+(neum*Vm)
Verification the results
We will use the data of the problem 4.5 in chapter 4
The specification of this problem is :
Ef=230 Gpa
Em= 4.6 Gpa
Gm= 1.769 GPA
Gf= 92 Gpa
alpham=30*10^-6
alphaf=15*10^-6
analytical results :
G12 2.911 Gpa
E1 94.76
E2 7.565
υ 12 0.28
Alpha 1 1.5437e-05
Alpha 2 2.4000e-05
MAtlab code results :
E1 94.7600
E2 7.5658
G12 2.9110
υ 12 0.2800
Alpha 1
1.5437e-05
Alpha 2 2.4000e-05
As we see when we compare the results we found
that the results is the same and the error don’t
exceeds .01 %
So the code is work correctly

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Matlab code of chapter 4

  • 1. Matlab code of chapter 4 %matlab code of composite structure %chapter four code %michromechanics composite material %calculation of material properties %inputs: Em,Ef,Vm,Vf,Gm,Gf,alpha m, alpha f, neu f, neu m %_______________________________________________ _____________ Ef=230; Em=4.6; Vf=0.4; Vm=0.6; Gm=1.769; Gf=92; alpham=30*10^-6; alphaf=15*10^-6; neuf=0.25; neum=0.3; E1=(Ef*Vf)+(Em*Vm) E2=(Em*Ef)/((Vf*Em)+(Vm*Ef)) alpha1=((Ef*Vf*alphaf)+(Em*Vm*alpham))/E1 alpha2=(alphaf*Vf)+(alpham*Vm) G12=(Gm*Gf)/((Gf*Vm)+(Gm*Vf)) neu12=(neuf*Vf)+(neum*Vm)
  • 2. Verification the results We will use the data of the problem 4.5 in chapter 4 The specification of this problem is : Ef=230 Gpa Em= 4.6 Gpa Gm= 1.769 GPA Gf= 92 Gpa alpham=30*10^-6 alphaf=15*10^-6 analytical results : G12 2.911 Gpa E1 94.76 E2 7.565 υ 12 0.28 Alpha 1 1.5437e-05 Alpha 2 2.4000e-05
  • 3. MAtlab code results : E1 94.7600 E2 7.5658 G12 2.9110 υ 12 0.2800 Alpha 1 1.5437e-05 Alpha 2 2.4000e-05 As we see when we compare the results we found that the results is the same and the error don’t exceeds .01 % So the code is work correctly