Hw solution number 4
- 1. Hw solution #4
Pb:4-5:
Compute E1,E2,G12 and υ12
Giveb :
Ef=230 gpa Em=Ef/50
Gm=Em/2.6 Vf=0.4 Vm=0.6
υ f=0.25 υ m= 0.3
solution :
Em=230/50=4.6 Gpa
Gm=Ef/2.6=230/2.6=1.769 Gpa
Gf=250/2.6=92 GPA
G12=(Gm*Gf)/(GmVf+GfVm)
G12=(92*1.769)/((92*0.6)+(1.769*0.4))
G12=2.9111 Gpa
E1=Ef*Vf+Em*Vm
- 3. Assume that we have E2 and Vm and Ef,Em
We can get the other one Vf and we are sure that it satisfies the
condition :
E1>30 Gpa
E1/E2<3.5