Transverse vibration of a cantilever beam
The differential equation of motion that describe the transverse vibration of a beam is given as ∂ 2 y ∂ 4 y m 2 + EI 4 = f ( f (x, t) ∂t ∂x
(1)
where y , m, E , and I are the transverse deflection, mass per unit length, Young modulus of elasticity, and second moment of area of the cross-section of the beam, and f ( f (x, t) is the loading per unit length of the beam. Free vibration of the beam: In the case of free vibration, f ( f (x, t) = 0, and the equation reduces to 4 ∂ 2 y 2 ∂ y +a =0 ∂t 2 ∂x 4
(2)
where a2 =
EI . m
(3)
Using the method of separation of variables, we assume a solution of the form y (x, t) = φ(x)q (t)
(4)
where φ(x) is a function which depends only on space, and q (t) only on time. time. Sub Sub-stituting Eq. (4) into Eq. (2), and simplifying, we obtain 1 d4 φ 1 d2 q = = ω2 a 4 2 φ dx q dt
−
2
(5)
where ω is a constant to be determined. Equation (5) can be rewritten as d4 φ dx4 and
where
4
− β φ = 0
(6)
d2 q + ω 2 q = q = 0 2 dt
(7)
ω2 mω2 β = 2 = . a EI
(8)
4
The general solutions of Eqs. (6) and (7) are given as φ(x) = c1 sin(βx) + c2 cos(βx) + c3 sinh(βx) + c4 cosh(βx)
(9)
q (t) = c5 sin(ωt) + c6 cos(ωt)
(10)
and
Free vibration of a cantilever beam :
The above solutions apply to free transverse vibration of any beam. In this section, we will derive the solution specifically for a cantilever beam shown in Fig. 1. For this beam, the boundary conditions are 1. The displacement and the slope of the beam at the fixed end are zero, and 2. The moment and the shear force on the beam at the free end are zero. These conditions can be written in mathematical form as ∂y ∂ 2 y ∂ 3 y (0, t) = 2 (L, t) = 3 (L, t) = 0 y(0, t) = ∂x ∂x ∂x
(11)
where L is the length of the beam. Equation (11) implies that dφ d2 φ d3 φ (0, t) = 2 (L, t) = 3 (L, t) = 0 dx dx dx
φ(0, t) =
(12)
Differentiating Eq. (9) three times with respect to x, we obtain dφ = β (c1 cos(βx) dx
−c
d2 φ = β 2 ( c1 sin(βx) 2 dx
sin(βx) + c3 cosh(βx) + c4 sinh(βx)),
(13)
cos(βx) + c3 sinh(βx) + c4 cosh(βx)),
(14)
d3 φ = β 3 ( c1 cos(βx) + c2 sin(βx) + c3 cosh(βx) + c4 sinh(βx)). 3 dx
(15)
−
and
2
−c
2
−
Substituting Eq. (12) into Eqs. (9), (13), (14) and (15), we obtain
−c
1
sin(βL)
−c
2
c2 + c4 = 0,
(16)
c1 + c3 = 0,
(17)
cos(βL)
−c
1
2
sinh(βL)
−c
2
cosh(βx)) = 0,
(18)
Piezoceramics
Point of load
a x0 Figure 1. Cantilever Beam
and
−c
1
cos(βL) + c2 sin(βL)
−c
1
cosh(βL)
−c
2
sinh(βL) = 0.
(19)
Equations (18) and (19) can be written in matrix form as
a
11
a21
a12 a22
c 0 1
c2
=
(20)
0
where a11 , a12 , a21 , and a22 are given as a11 = sin(βL) + sinh(βL),
(21)
a12 = cos(βL) + cosh(βL)),
(22)
a21 = cos(βL) + cosh(βL),
(23)
a22 =
− sin(βL) + sinh(βL).
(24)
For nontrivial solution, the determinant of the matrix in Eq. (19) must be 0. This leads to cos(βL) cosh(βL) + 1 = 0.
(25)
This is a nonlinear equation which has infinite roots, the first 4 roots of which are given as βL = 1.8751, 4.6941, 7.8548, 10.9955.
(26)
For a rectangular beam, we have I = bh3 /12 and m = ρbh, where b and h are the width and the thickness of the beam, and ρ is the density of the beam material. Using Eq. (8) and the expressions for I and m, we get 3 2 EI 4 Ebh 4 Eh = β = β ω = β 12ρbh 12ρ m 2
4
or
β 2 E (βL)2 h E = ω= h ρ 12 12 L2 ρ Using Eqs. (26) and (27), the first 4 frequencies of the beam are given as
√
h ω = 1.0150 2 L
E
h , 6.3608 2 ρ L
√
E
h , 17.8105 2 ρ L
E
h , 34.9014 2 ρ L
(27)
E ρ
.
(28)
Since βL has infinite values, ω also has infinite values. In other words, the system has infinite frequencies. This is because the system is continuous, and all continuous systems have infinite number of frequencies. 3
Using Eq. (18), the ratio of c2 upon c1 can be written as sin(βL) + sinh(βL) − cos(βL) + cosh(βL))
c2 = c1 Substituting c1 =
−c
3
and c4 =
φ(x) = c1 (sin(βx)
−c
2
(29)
(see Eqs. 16 and 17) into Eq. (9), we get
− sinh(βx) + cc (cos(βx) − cosh(βx)) 2
(30)
1
where c2 /c1 is given by Eq. (29). Note that for different values of β , we get different φ(x)s, known as the mode shapes , and different q (t)s, known as the generalized coordinates . Properties of the mode shapes:
Let us represent different mode shapes as φ j (x), j = 1, mode shapes satisfy Eq. (6). Thus, we have
··· , ∞.
Note that these
d4 φ j (x) = β j4 φ j (x) 4 dx
(31)
Multiplying both sides by φk (x) and integrating from 0 to L, we get 4
2
2
β φ (x)φ (x)dx = d φ (x) φ (x)dx = d φ (x) d φ (x) dx+ dx dx dx d φ (x) dφ (x) d φ (x) φ (x) − dx dx dx Using the boundary conditions (Eq. (12)), Eq. (31) reduces to β φ (x)φ (x)dx = d φ (x) d φ (x) dx L
0
L
4
j
j
k
k
4
0
3
L
j
L
k
3
2
0
2
j
j
0
k
2
L
j
k
2
(32)
0
L
j
0
L
4
j
k
2
2
j
dx
2
0
k
dx2
(33)
Switching the role of φ j (x) and φk (x), we obtain 2
2
β φ (x)φ (x)dx = d φ (x) d φ (x) dx L
0
L
4
k
k
j
0
k
dx2
j
dx2
(34)
Equations (33) and (34) lead to L
φ (x)φ (x)dx = 0 j
0
and
2
k
2
d φ (x) d φ (x) dx = 0 L
0
k
dx
j
dx2
2
4
(35)
(36)
Let us select c1 such that L
(φ (x)) dx = 1, 2
(37)
j
0
then from Eq. (33), it follows that 2
d φ (x) dx = β 2
L
j
4
dx
0
(38)
j
2
Equation (35) is known as the orthogonality condition . In other words, the mode shapes φ j (x), j = 1,
··· , ∞ are orthogonal.
The mode shapes are also called the
eigenfunctions . When Eq. (37) is satisfied, the mode shapes (eigenfunctions) are
called the normal mode shapes (normalized eigenfunctions ) , and Eq. (35) is called the orthonormality condition. Forced vibration:
Let us now consider force vibration of the cantilever beam. For this case, 4 1 ∂ 2 y 2 ∂ y + = a f (x, t) ∂t 2 ∂x 4 m
(39)
Using the method of separation of variables, the displacement y can be written as ∞
φ (x)q (t) y(x, t) = j
(40)
j
j =1
where φ j (x) are the normal mode shapes determined above. Substituting this into Eq. (39), we obtain 2
4
φ (x) ∂ q + a ∂ φ (x) q (t) = 1 f (x, t) ∞
j
j =1
∞
j
2
∂t 2
j
∂x 4
j =1
j
m
(41)
Multiplying Eq. (41) both sides by φk (x) and integrating from 0 to L, we obtain (after some simplification) 1 ∂ 2 q j 2 + (t) = ω q j j ∂t 2 m
L
φ (x)f (x, t), 0
j
j = 1,
·· · , ∞
(42)
Note that for a given value of j, Eq. (42) represents force vibration of a single degree-of-freedom spring-mass system. Since j varies from 1 to
∞, a cantilever beam
can be thought of as a system consisting of infinite single degree-of-freedom spring-
mass system. Consequently, the response of a vibrating cantilever beam may have infinite frequency. Further, force f (x, t) would cause all generalized coordinates q j (t), 5
j = 1,
··· , ∞ to change with the following exception.
If the initial conditions of a
generalized coordinate are zero and the associated mode shape leads the right hand side of Eq. (42) to be zero, then this generalized coordinate would not be excited. To get a better understanding of the system and the response of the system, assume that initially the beam is in rest position and f (x, t) is a point force acting at a point x0 . In this case, the initial conditions and the force f (x, t) can be written as y(x, 0) =
∂y(x, 0) =0 ∂t
(43)
and f (x, t) = δ (x where δ (x
− x )f (t)
(44)
0
− x ) is the delta function. Since, Eq. (43) must be satisfied at all points 0
of the beam, we obtain, using Eqs. (40) and (43) that q j (0) = q ˙ j (0) = 0
(45)
Substituting Eq. (44) into Eq. (42), yields 1 ∂ 2 q j 2 + (t) = ω q φ j (x0 )f (t), j j ∂t 2 m
j = 1,
(46)
··· , ∞
Note that if for certain j = J , φJ (x0 ) = 0, then q J (t) = 0, i.e. q J (t) will not be excited. Let us now consider that the force is an impulsive force. In this case, force f (t) can be written as f (t) = F 0 δ (t), and Eq. (46) reduces to 1 ∂ 2 q j 2 + (t) = ω q φ j (x0 )F 0 δ (t), j j ∂t 2 m
j = 1,
Here, F 0 is the amplitude of the force. Integrating Eq. (47) from 0 0+
0−
∂ 2 q j dt + ∂t 2
0+
0−
0+
ω q (t)dt = 2
j j
0−
(47)
··· , ∞.
1 φ j (x0 )F 0 δ (t)dt, m
−
to 0+ , we obtain
j = 1,
·· · , ∞,
or q ˙(0+ )
− q ˙(0
−
) = q ˙(0+ ) =
1 φ j (x0 )F 0 , m
j = 1,
·· · , ∞,
(48)
Here we have used the following identities: 0+
0−
q j (t)dt = 0, 6
(49)
0+
0−
δ (t)dt = 1,
(50)
and q ˙(0 ) = 0.
(51)
−
Equation (49) follows because q j (t) is continuous, Eq. (50) is a property of the delta function, and Eq. (51) is the second part of Eq. (45). For t > 0, Eq. (47) is given as ∂ 2 q j + ω j2 q j (t) = 0 2 ∂t
(52)
The solution of this equation is given as q j (t) = A j sin(ω j t) + B j cos(ω j t)
(53)
Since, q j (0) = 0 and q ˙ j (0+ ) = φ j (x0 )F 0 /m, it follows that A j =
φ j (x0 )F 0 , mω j
and
B j = 0
(54)
Using Eqs. (53) and (54), we obtain q j (t) =
φ j (x0 )F 0 sin(ω j t) mω j
(55)
Substituting Eq. (55) into Eq. (40), we obtain the Vertical displacement as F 0 φ j (x0 ) sin(ω j t) y(x, t) = φ j (x) m j =1 ω j ∞
(56)
Note that the contribution of a mode to the overall signal depends on the location where the load is applied, the frequency of the mode, and the point where the signal is measured. Measurement of vibration using a piezoceramic :
Modeling of a piezoceramic is difficult. This is because piezoceramic patches have finite dimension, and the strain over that patch may vary. For simplicity, we will assume that the strain over the piezoceramic is uniform, and it is equal to the strain at the center of the patch. Further, many other factors such as bonding effect, Poisson effect, error in orientation, etc. are neglected. With these in mind, the strain at the piezoceramic patch can be given as h1 h d2 y(a, t) F 0 h d2 φ j (x) = = = 2 rc 2 dx2 2m j =1 dx2 ∞
7
|
x=a
φ j (x0 ) sin(ω j t) ω j
(57)
where rc is the radius of curvature of the centerline of the beam, and a is the location of the center of the patch. Since, the voltage generated in the piezoceramic patch is proportional to the strain in the patch, the voltage is given as kF 0 h d2 φ j (x) V = k = m j =1 dx2 ∞
|
x=a
φ j (x0 ) sin(ω j t) ω j
(58)
where k is a proportionality constant. Here, a factor 2 has been applied with the assumption that we have two piezoceramic patches in series such that the voltage add.
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