An exact theoretical analysis is very challenging mathematically if not impossible. In
the following, we make a few approximations to break Eqs. (4.63), (4.64), and (4.65)
into three subproblems that are one-way coupled and solve them sequentially.
We begin the approximations by neglecting the piezoelectric coupling in the
bending equation in Eq. (4.63) and obtain the following initial-boundary-value
problem for the bending of an elastic beam:
‐c 33 Iv ,3333 ¼ ρA€ v,
v ¼ 0, v ,3 ¼ 0, x 3 ¼ 0,
M ¼ 0, Q ¼ f y , x 3 ¼ L,
v ¼ 0, _
v ¼ 0, t ¼ 0:
ð4:68Þ
To make the shear force boundary condition in Eq. (4.68) homogeneous, we make a
change of the unknown variable from v to b v through
v x 3 , t
ð
Þ ¼b v x 3 , t
ð
ÞÀ
f y
6c 33 I
x
3
3 þ
f y L
2c 33 I
x
2
3 :
ð4:69Þ
Then Eq. (4.68) becomes
Àc 33 Ib v ,3333 ¼ ρA €
b v,
b v ¼ 0, b v ,3 ¼ 0, x 3 ¼ 0,
M ¼ 0, Q ¼ 0, x 3 ¼ L,
b v ¼
f y
6c 33 I
x
3
3 À
f y L
2c 33 I
x
2
3 , _
b v ¼ 0, t ¼ 0:
ð4:70Þ
Equation (4.70) can be solved using the method of separation of variables without
any challenge. We skip the procedure and summarize the solution below:
v x 3 , t
ð
Þ ¼
X 1
m¼1
A
m
ð Þ Y
m
ð Þ x 3
ð Þcos ω
m
ð Þ t
À
f y
6c 33 I
x
3
3 þ
f y L
2c 33 I
x
2
3 ,
ð4:71Þ
where
Y
m
ð Þ x 3
ð Þ ¼ cos β
m
ð Þ x 3 À cosh β
m
ð Þ x 3
þξ
m
ð Þ sin β
m
ð Þ x 3 À sinh β
m
ð Þ x 3
,
ξ
m
ð Þ
¼ À
cos β
m
ð Þ L þ cosh β
m
ð Þ L
sin β
m
ð Þ L þ sinh β
m
ð Þ L
,
ð4:72Þ
are the free vibration modes. β
(m) is the roots of
104
4 Bending of Beams
the following, we make a few approximations to break Eqs. (4.63), (4.64), and (4.65)
into three subproblems that are one-way coupled and solve them sequentially.
We begin the approximations by neglecting the piezoelectric coupling in the
bending equation in Eq. (4.63) and obtain the following initial-boundary-value
problem for the bending of an elastic beam:
‐c 33 Iv ,3333 ¼ ρA€ v,
v ¼ 0, v ,3 ¼ 0, x 3 ¼ 0,
M ¼ 0, Q ¼ f y , x 3 ¼ L,
v ¼ 0, _
v ¼ 0, t ¼ 0:
ð4:68Þ
To make the shear force boundary condition in Eq. (4.68) homogeneous, we make a
change of the unknown variable from v to b v through
v x 3 , t
ð
Þ ¼b v x 3 , t
ð
ÞÀ
f y
6c 33 I
x
3
3 þ
f y L
2c 33 I
x
2
3 :
ð4:69Þ
Then Eq. (4.68) becomes
Àc 33 Ib v ,3333 ¼ ρA €
b v,
b v ¼ 0, b v ,3 ¼ 0, x 3 ¼ 0,
M ¼ 0, Q ¼ 0, x 3 ¼ L,
b v ¼
f y
6c 33 I
x
3
3 À
f y L
2c 33 I
x
2
3 , _
b v ¼ 0, t ¼ 0:
ð4:70Þ
Equation (4.70) can be solved using the method of separation of variables without
any challenge. We skip the procedure and summarize the solution below:
v x 3 , t
ð
Þ ¼
X 1
m¼1
A
m
ð Þ Y
m
ð Þ x 3
ð Þcos ω
m
ð Þ t
À
f y
6c 33 I
x
3
3 þ
f y L
2c 33 I
x
2
3 ,
ð4:71Þ
where
Y
m
ð Þ x 3
ð Þ ¼ cos β
m
ð Þ x 3 À cosh β
m
ð Þ x 3
þξ
m
ð Þ sin β
m
ð Þ x 3 À sinh β
m
ð Þ x 3
,
ξ
m
ð Þ
¼ À
cos β
m
ð Þ L þ cosh β
m
ð Þ L
sin β
m
ð Þ L þ sinh β
m
ð Þ L
,
ð4:72Þ
are the free vibration modes. β
(m) is the roots of
104
4 Bending of Beams