Consider static deformation of a free beam. There are no mechanical loads. f 2 ¼ 0,
f 3 ¼ 0, and m 1 ¼ 0. Equations (7.118), (7.120), and (7.121) reduce to
b
T ,3 ¼ 0,
Q ,3 ¼ M ,33 ¼ 0,
ð7:122Þ
b
D 3,3 ¼ q Δp À Δn
ð
Þ A
2
ð Þ ,
ð7:123Þ
0 ¼ ÀJ
p
3,3 ,
0 ¼ J
n
3,3 :
ð7:124Þ
For the mechanically free and electrically isolated beam we are considering, the
boundary conditions are
b
T 0
ð Þ ¼ b
T L
ð Þ ¼ 0, M 0
ð Þ ¼ M L
ð Þ ¼ 0,
Q 0
ð Þ ¼ Q L
ð Þ ¼ 0, b
D 0
ð Þ ¼ b
D L
ð Þ ¼ 0,
J
p
3 0
ð Þ ¼ J
p
3 L
ð Þ ¼ 0, J
n
3 0
ð Þ ¼ J
n
3 L
ð Þ ¼ 0:
ð7:125Þ
Δp and Δn must satisfy the following global charge neutrality conditions:
Z L
0
Δpdx 3 ¼ 0,
Z L
0
Δndx 3 ¼ 0:
ð7:126Þ
Only one of Eq. (7.126) is independent. To determine the mechanical displacements
and the electric potential uniquely, we set
v 0
ð Þ ¼ 0, v L
ð Þ ¼ 0, w L=2
ð
Þ ¼ 0,
ð7:127Þ
φ L=2
ð
Þ ¼ 0:
ð7:128Þ
Equations (7.122), (7.123), and (7.124) can be written as a fourth-order equation
mainly for v, a second-order equation mainly for w, and second-order equations
mainly for φ, Δp, and Δn, respectively, with couplings among them. The four
equations can be further manipulated into
φ ,33 À k
2
φ ¼ À
C 1 þ C 2
ð
Þ A
2
ð Þ
e ε
,
v ,333 ¼ αφ ,33 ,
w ,33 ¼ βφ ,33 ,
ð7:129Þ
7.5 Extension and Bending of Composite Beams
207