Homogenization and Frequency Analysis
119
For calculation of the component C 44 , C 66 = C 55 , the following strain is applied to
the RVE
γ
0
4 = ε
0
23 + ε
0
31 = 1, γ
0
6 = ε
0
12 + ε
0
21 = 1.
(4)
The coefficient C 44 , C 66 is calculated as
C 44 = ¯
σ 4 =
1
V
σ
V
4 =
1
V
V
σ 4 dV C 66 = ¯
σ 6 =
1
V
σ
V
6 =
1
V
V
σ 6 dV .
(5)
E 1 = C 11 −
2EC 2
12
C 22 + C 23
,
(6)
E 2 =
2C 11 C 22 + 2C 11 C 23 − 4C 2
12
(C 22 − C 23 + 2C 44 )
3C 11 C 22 + C 11 C 23 + 2C 11 C 44 − 4C 2
12
,
(7)
G 12 = G 13 = C 66 ,
(8)
ν 12 = ν 13 =
C 12
C 22 + C 23
,
(9)
ν 23 =
C 11 C 22 + 3C 11 C 23 − 2C 11 C 44 − 4C 2
12
3C 11 C 22 + C 11 C 23 + 2C 11 C 44 − 4C 2
12
.
(10)
The transverse shear modulus G 23 can be written in terms of additional constants
such as:
G 23 =
C 22
4
−
C 23
4
+
C 44
2
=
E 2
2(1 + ν 23 )
.
(11)
3 Free Vibration of Sandwich Plates
The equations to determine the natural frequencies of the symmetric sandwich panel are
used [1]:
D 11
∂ 2 Φ x
∂x 2 + D 66
∂ 2 Φ x
∂y 2 + (D 12 + D 66 )
∂ 2 Φ y
∂x∂y
− k
s A 55
Φ x +
∂w 0
∂x
− I 2
∂ 2 Φ x
∂t 2 = 0,
(D 12 + D 66 )
∂ 2 Φ x
∂x∂y
+ D 66
∂ 2 Φ y
∂x 2 + D 22
∂ 2 Φ y
∂y 2 − k
s A 44
Φ y +
∂w 0
∂y
− I 2
∂ 2 Φ y
∂t 2 = 0,
k
s A 55
∂Φ x
∂x
+
∂ 2 w
∂x 2
+ k
s A 44
∂Φ y
∂y
+
∂ 2 w 0
∂y 2
− ρ m h
∂ 2 w 0
∂t 2 = 0,
(12)
where k s is the transverse shear deformation factor given by value 5/6.
ρ m =
1
h
N
k=1
ρ k
z
(k)
− z
(k−1)
,
(13)