H.W #3 solution
1
1- Derivation of thermal coefficient for composite material having a stress in the fiber direction ∈ f=¿ ∈ m=¿ ∈ 1 ¿ ¿ A f ∗σ f + A m ∗σ m = A total ∗σ 1 F f +F m =F total ∆x∗A f ∗σ f + ∆x∗A m ∗σ m = A total ∗σ 1 ÷ ∆x∗A total V f ∗σ f + V m ∗σ m =σ 1 ∈= σ E + α∗T↔σ =E∗( ∈−α∗T) V f ∗E f ( ∈ f −α f ∗T ) + V m ∗E m ( ∈ m −α m ∗T ) =σ 1 ∈ 1 ( V f ∗E f + V m ∗E m ) −T ( V f ∗E f ∗α f + V m ∗E m ∗α m ) =σ 1 σ 1 =E 1 ( ∈ 1 −α 1 ∗T ) ∴E 1 =V f ∗E f + V m ∗E m ∴α 1 = V f ∗E f ∗α f + V m ∗E m ∗α m V f ∗E f + V m ∗E m 2- Derivation of thermal coefficient for composite material having a stress perpendicular to the fiber direction σ f =σ m =σ 2 ∈ f ∗l f +∈ m ∗l m =∆l=∈ 2 ∗( l ¿¿ f +l m ) ¿ ÷ ( l ¿¿ f+l m ) ¿ ∈ 2 =∈ f ∗V f + ∈ m ∗V m ∈ 2 =V f ( σ f E f +α f ∗T ) + V m ( σ m E m + α m ∗T ) ∈ 2 =σ 2 ( V f ∗E m +V m ∗E f E f ∗E m ) +T ( V f ∗α f + V m ∗α m ) ∈ 2 = σ 2 E 2 +α 2 ∗T∴E 2 = E f ∗E m V f ∗E m + V m ∗E f ∴α 2 = V f ∗α f +V m ∗α m
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Transcript of H.W #3 solution
1- Derivation of thermal coefficient for composite material having a stress in the fiber direction
∈f =¿∈m=¿∈1 ¿¿
A f∗σ f+Am∗σm=A total∗σ 1F f+Fm=F total∆ x∗A f∗σ f+∆ x∗Am∗σm=A total∗σ 1 ÷∆ x∗A total
V f∗σ f+V m∗σ m=σ 1∈= σE
+α∗T↔σ=E∗(∈−α∗T )
V f∗E f (∈f−α f∗T )+V m∗Em (∈m−αm∗T )=σ 1 ∈1 (V f∗E f+V m∗Em )−T (V f∗Ef∗α f+V m∗Em∗αm )=σ1 σ 1=E1 (∈1−α1∗T )∴E1=V f∗E f+V m∗Em
∴α1=V f∗E f∗α f+V m∗Em∗αm
V f∗E f+V m∗Em
2- Derivation of thermal coefficient for composite material having a stress perpendicular to the fiber direction
σ f=σm=σ2∈f∗lf+∈m∗lm=∆l=∈2∗(l¿¿ f +lm)¿ ÷(l¿¿ f+ lm)¿ ∈2=∈f∗V f+∈m∗V m
∈2=V f ( σ fEf +α f∗T )+V m( σmEm+αm∗T )∈2=σ 2(V f∗Em+V m∗E f
E f∗Em )+T (V f∗α f+V m∗αm )
∈2=σ2E2
+α2∗T ∴E2=Ef∗Em
V f∗Em+V m∗E f∴α2=V f∗α f+V m∗αm
