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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
4.6*0.6+230*0.4
E1=94.76
E2=(Em*Ef)/((Em*Vf)+(Ef*Vm))
E2=7.565
υ 12 = υf*Vf+ υm*Vm
υ 12= 0.25*0.4+0.3*0.6
υ 12 =0.28
problem 4.11 :
E1>30 Gpa
E1/E2<3.5
E1= E1=Ef*Vf+Em*Vm
E2=(Em*Ef)/((Em*Vf)+(Ef*Vm))
3.5*E2 > Ef*Vf+Em*Vm > 30
3.5E2-Em*Vm > Ef*Vf > 30-Em*Vm/Ef
(3.5E2-Em*Vm)/Ef > Vf > 30-Em*Vm/Ef
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

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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
  • 2. 4.6*0.6+230*0.4 E1=94.76 E2=(Em*Ef)/((Em*Vf)+(Ef*Vm)) E2=7.565 υ 12 = υf*Vf+ υm*Vm υ 12= 0.25*0.4+0.3*0.6 υ 12 =0.28 problem 4.11 : E1>30 Gpa E1/E2<3.5 E1= E1=Ef*Vf+Em*Vm E2=(Em*Ef)/((Em*Vf)+(Ef*Vm)) 3.5*E2 > Ef*Vf+Em*Vm > 30 3.5E2-Em*Vm > Ef*Vf > 30-Em*Vm/Ef (3.5E2-Em*Vm)/Ef > Vf > 30-Em*Vm/Ef
  • 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