1. Home work (3):
ΔL= ε L0 + BΔm+αΔT
Hook’s law:
1-according to load (σ):
σ = E εm
2-according to temp change ΔT:
εt = α . ΔT
Total strain is ε =εm+εt =σ/E + α . ΔT
For direction (1):
ε1=εf=εm
εf=σf/Ef+αf.ΔT σ1
εm=σm/Em +αm.ΔT
σf=Ef(εf-αf T)
σm=Em(εm-αm.T)
σ1=Vf σf + Vm σm
σ1=E1(ε1-α1 ΔT)
σ1=Vf Ef (εf-αfΔ T)+Vm Em (εm-αm.ΔT)
by sub with σ1
E1( ε1-α1 ΔT)= Vf Ef (εf-αfΔ T)+Vm Em (εm-αm.ΔT)
We have: ε1=εf=εm
We can get:
α1 E1 = Vf Ef αf + Vm Em αm

2. for direction 2: σ2
σ2=σm=σf
ε2=εf Vf+εm Vm
ε= σ/E + α . ΔT
σ2/E2+α2 ΔT= (σf/Ef+αf ΔT)Vf+(σm/Em+αm ΔT) Vm
by solve the last eguation:
α2= αf Vf +αm Vm