The document provides solutions to two problems (6.5 and 6.2) regarding composite material analysis. For problem 6.5, it calculates stiffness coefficients (Qij) for a two-layer composite with given material properties and orientations, and determines transformed stiffness coefficients and stress-strain relationships. For problem 6.2, it determines expressions for strains and curvatures on a middle surface with given displacement and rotation fields.

1. Hw#6 solution
Problem 6.5:
Given :
Nx=1
T1=t2= 1.27 E1=137.8 Gpa
E2=9.6 Gpa υ12=0.3
Solution :
Δ=1- υ12^2(E2/E1)
Δ=1-0.3^2*(9.6/137.8)=0.9937
Q11=138.673
Q12=Q21=2.8982
Q22=9.6608
Q66=5.2
For Θ1=30
(Q11)1’=78.0035+4.986+2.4152=85.4047
(Q12)1’=23.9125
(Q22)1’=8.667+4.986+5.4342=19.0872
(Q16)1’=40.716+0.3937

2. (Q26)1’=13.5722+1.1812=14.753
(Q66)1’=24.7757+3.25=28.0257
Θ2=-30
(Q11)2’=85.4
(Q12)2’=23.9125
(Q22)2’=19.0872
(Q16)2’=-41.10
(Q66)2’=28.0257
Aij== ∑ ( 𝑄𝑖𝑗) ∗ ( 𝑍𝑘 − 𝑍𝑘 − 1)𝑘
A11=2*85.4*1.27=216.916
A12=2*23.9125*1.27=60.7377
A22=2*19.0872*1.27=48.4814
A16=0
A26=0 A66=71.185
α11=A22/((A11*A22)-(A12))
=48.4814/(216.91*60.7-48.4814)=.0036
α22=.02074

3. α66=.01404
α12=-.01
problem:6.2
given :
middle surface strain and curvature
Uo=Acos(αx)sin(βy)
Vo=Bsin(αx)cos(βy)
Φx=Qcos(αx)sin(βy)
Φy=Rsin(αx)cos(βy)
Solution :
W=wo(x,y)
U=uo(x,y)-z Φx
V=vo(x,y)-z Φy
Uox’=-Aαsin(αx)sin(βy)
(Φx)x’=-Qβsin(βy)sin(αx)
ϵxx=-Aαsin(βy)sin(αx)+ZQβsin(βy)sin(αx)
=[ZQβ-Aα]sin(βy)sin(αx)

4. ϵyy= [ZRβ-Bβ]sin(βy)sin(αx)
ϵxy=ϵoxy+zKxy
ϵoxy=0.5(Uoy’+V0x’)
=0.5cos(βy)cos(αx)[Aβ+Bβ]
Kxy=0.5((Φx)y’+( Φy)x’)
=0.5cos(βy)cos(αx)[βQ+αR]
ϵxy=0.5cos(βy)cos(αx)[βQ+αR+ Aβ+Bβ]