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

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