k
2
¼
qp 0 μ
p
11
D
p
11
þ
qn 0 μ
n
11
D
n
11
A
2
ð Þ
e ε
,
e ε ¼
e 15 A
1
ð Þ
À
Á 2
b c b
A
þ b ε b
A:
ð6:129Þ
The general solution of Eq. (6.128) is
φ ¼ kC 3 sinh kx 1 þ kC 4 cosh kx 1 þ
C 1 þ C 2
ð
Þ A
2
ð Þ
k
2
e ε
,
ð6:130Þ
where C 3 and C 4 are undetermined constants. From Eqs. (6.114), (6.117) 2 , and the
mechanical boundary condition on the shear force at x 3 ¼ L in Eq. (6.121), the
moment equation takes the following form:
Dψ ,11 ¼ F,
ð6:131Þ
which implies that
ψ ¼
1
2
F
D
x
2
1 þ C 5 x 1 þ C 6 ,
ð6:132Þ
where C 5 and C 6 are undetermined constants. From Q ¼ F and Eq. (6.113), we
obtain
w ¼
Fx 1 À e 15 A
1
ð Þ
φ þ C 7
b c b
A
À
Z
ψdx 1 ,
ð6:133Þ
where C 7 is an integration constant. To determine C 1 through C 7 , we need to use the
remaining ones of Eq. (6.121), one of Eq. (6.122) and Eq. (6.123).
For a numerical example consider a beam of PZT4 and silicon. We choose the
following geometric parameters that L ¼ 600 nm, h ¼ 10 nm, c ¼ 15 nm, and
b ¼ 50 nm. p 0 ¼ n 0 ¼ 10
23 m
À3 . The basic fields of interest are shown in Fig. 6.14
for different values of the end shear force F. Although the shear force Q ¼ F is a
constant, the shear strain in (a) varies a little along the beam, especially near the two
ends. This is related to the fact that the electric potential and field in the beam change
near the two ends and sometimes significantly as shown in (b) because of the
hyperbolic functions in Eq. (6.130). Since the axial electric field E 1 is negative,
the holes are driven to the left and the electrons to the right in (c) and (d). This is
similar to the composite beam in the previous section but different from the
homogeneous beam in Sect. 4.3 where the charges are driven to the top and bottom
of the beam. For all fields, a larger F produces stronger fields as expected. Similar to
the previous section, there also exists an optimal thickness ratio between the two
types of layers for strong coupling between bending and mobile charge
redistribution.
6.4 Bending of Beams with e 15
167
2
¼
qp 0 μ
p
11
D
p
11
þ
qn 0 μ
n
11
D
n
11
A
2
ð Þ
e ε
,
e ε ¼
e 15 A
1
ð Þ
À
Á 2
b c b
A
þ b ε b
A:
ð6:129Þ
The general solution of Eq. (6.128) is
φ ¼ kC 3 sinh kx 1 þ kC 4 cosh kx 1 þ
C 1 þ C 2
ð
Þ A
2
ð Þ
k
2
e ε
,
ð6:130Þ
where C 3 and C 4 are undetermined constants. From Eqs. (6.114), (6.117) 2 , and the
mechanical boundary condition on the shear force at x 3 ¼ L in Eq. (6.121), the
moment equation takes the following form:
Dψ ,11 ¼ F,
ð6:131Þ
which implies that
ψ ¼
1
2
F
D
x
2
1 þ C 5 x 1 þ C 6 ,
ð6:132Þ
where C 5 and C 6 are undetermined constants. From Q ¼ F and Eq. (6.113), we
obtain
w ¼
Fx 1 À e 15 A
1
ð Þ
φ þ C 7
b c b
A
À
Z
ψdx 1 ,
ð6:133Þ
where C 7 is an integration constant. To determine C 1 through C 7 , we need to use the
remaining ones of Eq. (6.121), one of Eq. (6.122) and Eq. (6.123).
For a numerical example consider a beam of PZT4 and silicon. We choose the
following geometric parameters that L ¼ 600 nm, h ¼ 10 nm, c ¼ 15 nm, and
b ¼ 50 nm. p 0 ¼ n 0 ¼ 10
23 m
À3 . The basic fields of interest are shown in Fig. 6.14
for different values of the end shear force F. Although the shear force Q ¼ F is a
constant, the shear strain in (a) varies a little along the beam, especially near the two
ends. This is related to the fact that the electric potential and field in the beam change
near the two ends and sometimes significantly as shown in (b) because of the
hyperbolic functions in Eq. (6.130). Since the axial electric field E 1 is negative,
the holes are driven to the left and the electrons to the right in (c) and (d). This is
similar to the composite beam in the previous section but different from the
homogeneous beam in Sect. 4.3 where the charges are driven to the top and bottom
of the beam. For all fields, a larger F produces stronger fields as expected. Similar to
the previous section, there also exists an optimal thickness ratio between the two
types of layers for strong coupling between bending and mobile charge
redistribution.
6.4 Bending of Beams with e 15
167