For bending in the y-z plane, the main stress components are T 3 and T 4 . Therefore we
introduce the following stress relaxation for thin beams:
T 1 ¼ T 2 ¼ T 5 ¼ T 6 ffi 0:
ð4:8Þ
From the constitutive relations in Eq. (4.3), for the relevant strain and electric
displacement components, we have
S 3 ¼ s 33 T 3 þ d 33 E 3 ,
S 4 ¼ s 44 T 4 þ d 15 E 2 ,
D 2 ¼ d 15 T 4 þ ε 11 E 2 ,
D 3 ¼ d 33 T 3 þ ε 33 E 3 :
ð4:9Þ
We invert Eq. (4.9) 1,2 for expressions of stresses in terms of strains and substitute the
resulting expressions into Eq. (4.9) 3,4 . This results in
T 3 ¼ T 33 ¼ c 33 S 3 À e 33 E 3 ¼ c 33 w ,3 þ x 2 ψ ,3
À
Á þ e 33 ϕ
0
ð Þ
,3 þ x 2 ϕ
1
ð Þ
,3
,
T 4 ¼ T 32 ¼ c 44 S 4 À e 15 E 2 ¼ c 44 v ,3 þ ψ
ð
Þþe 15 ϕ
1
ð Þ ,
D 2 ¼ e 15 S 4 þ ε 11 E 2 ¼ e 15 v ,3 þ ψ
ð
ÞÀε 11 ϕ
1
ð Þ ,
D 3 ¼ e 33 S 3 þ ε 33 E 3 ¼ e 33 w ,3 þ x 2 ψ ,3
À
Á À ε 33 ϕ
0
ð Þ
,3 þ x 2 ϕ
1
ð Þ
,3
,
ð4:10Þ
where Eq. (4.7) has been used and the one-dimensional effective material constants
for thin beams are
c 33 ¼ 1=s 33 , c 44 ¼ 1=s 44 , e 33 ¼ d 33 =s 33 , e 15 ¼ d 15 =s 44 ,
ε 11 ¼ ε 11 À d
2
15 =s 44 , ε 33 ¼ ε 33 À d
2
33 =s 33 :
ð4:11Þ
The relevant constitutive relations for the currents are the following ones from
Eq. (4.4):
J
p
2 ¼ qp 0 μ
p
11 E 2 À qD
p
11 Δp
ð Þ ,2 ¼ Àqp 0 μ
p
11 ϕ
1
ð Þ
À qD
p
11 p
1
ð Þ
:
J
p
3 ¼ qp 0 μ
p
33 E 3 À qD
p
33 Δp
ð Þ ,3
¼ Àqp 0 μ
p
33 ϕ
0
ð Þ
,3 þ x 2 ϕ
1
ð Þ
,3
À qD
p
33 p
0
ð Þ
,3 þ x 2 p
1
ð Þ
,3
,
J
n
2 ¼ qn 0 μ
n
11 E 2 þ qD
n
11 Δn
ð Þ ,2 ¼ Àqn 0 μ
n
11 ϕ
1
ð Þ
þ qD
n
11 n
1
ð Þ ,
J
n
3 ¼ qn 0 μ
n
33 E 3 þ qD
n
33 Δn
ð Þ ,3
¼ Àqn 0 μ
n
33 ϕ
0
ð Þ
,3 þ x 2 ϕ
1
ð Þ
,3
þ qD
n
33 n
0
ð Þ
,3 þ x 2 n
1
ð Þ
,3
,
ð4:12Þ
4.1 One-Dimensional Equations for Bending
91