ºÝºÝߣ

ºÝºÝߣShare a Scribd company logo
Home work (3):
¦¤L= ¦Å L0 + B¦¤m+¦Á¦¤T
Hook¡¯s law:
1-according to load (¦Ò):
¦Ò = E ¦Åm
2-according to temp change ¦¤T:
¦Åt = ¦Á . ¦¤T
Total strain is ¦Å =¦Åm+¦Åt =¦Ò/E + ¦Á . ¦¤T
For direction (1):
¦Å1=¦Åf=¦Åm
¦Åf=¦Òf/Ef+¦Áf.¦¤T ¦Ò1
¦Åm=¦Òm/Em +¦Ám.¦¤T
¦Òf=Ef(¦Åf-¦Áf T)
¦Òm=Em(¦Åm-¦Ám.T)
¦Ò1=Vf ¦Òf + Vm ¦Òm
¦Ò1=E1(¦Å1-¦Á1 ¦¤T)
¦Ò1=Vf Ef (¦Åf-¦Áf¦¤ T)+Vm Em (¦Åm-¦Ám.¦¤T)
by sub with ¦Ò1
E1( ¦Å1-¦Á1 ¦¤T)= Vf Ef (¦Åf-¦Áf¦¤ T)+Vm Em (¦Åm-¦Ám.¦¤T)
We have: ¦Å1=¦Åf=¦Åm
We can get:
¦Á1 E1 = Vf Ef ¦Áf + Vm Em ¦Ám
for direction 2: ¦Ò2
¦Ò2=¦Òm=¦Òf
¦Å2=¦Åf Vf+¦Åm Vm
¦Å= ¦Ò/E + ¦Á . ¦¤T
¦Ò2/E2+¦Á2 ¦¤T= (¦Òf/Ef+¦Áf ¦¤T)Vf+(¦Òm/Em+¦Ám ¦¤T) Vm
by solve the last eguation:
¦Á2= ¦Áf Vf +¦Ám Vm

More Related Content

Home work 3

  • 1. Home work (3): ¦¤L= ¦Å L0 + B¦¤m+¦Á¦¤T Hook¡¯s law: 1-according to load (¦Ò): ¦Ò = E ¦Åm 2-according to temp change ¦¤T: ¦Åt = ¦Á . ¦¤T Total strain is ¦Å =¦Åm+¦Åt =¦Ò/E + ¦Á . ¦¤T For direction (1): ¦Å1=¦Åf=¦Åm ¦Åf=¦Òf/Ef+¦Áf.¦¤T ¦Ò1 ¦Åm=¦Òm/Em +¦Ám.¦¤T ¦Òf=Ef(¦Åf-¦Áf T) ¦Òm=Em(¦Åm-¦Ám.T) ¦Ò1=Vf ¦Òf + Vm ¦Òm ¦Ò1=E1(¦Å1-¦Á1 ¦¤T) ¦Ò1=Vf Ef (¦Åf-¦Áf¦¤ T)+Vm Em (¦Åm-¦Ám.¦¤T) by sub with ¦Ò1 E1( ¦Å1-¦Á1 ¦¤T)= Vf Ef (¦Åf-¦Áf¦¤ T)+Vm Em (¦Åm-¦Ám.¦¤T) We have: ¦Å1=¦Åf=¦Åm We can get: ¦Á1 E1 = Vf Ef ¦Áf + Vm Em ¦Ám
  • 2. for direction 2: ¦Ò2 ¦Ò2=¦Òm=¦Òf ¦Å2=¦Åf Vf+¦Åm Vm ¦Å= ¦Ò/E + ¦Á . ¦¤T ¦Ò2/E2+¦Á2 ¦¤T= (¦Òf/Ef+¦Áf ¦¤T)Vf+(¦Òm/Em+¦Ám ¦¤T) Vm by solve the last eguation: ¦Á2= ¦Áf Vf +¦Ám Vm