Page 118 - The Mechatronics Handbook

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8.3 Vibration and Modal Analysis

As mentioned earlier, the time response of a continuum structure requires the solution of Eqs. (8.10)
with the acceleration terms present. For linear systems this solution can be represented by an infinite
superposition of characteristic functions (modes). Associated with each such mode is also a characteristic
number (eigenvalue) determining the time response of the mode. The analysis of these modes is called
modal analysis and has a central role in the design of resonant cantilever sensors, flapping wings for
micro-air-vehicles (MAVs) and micromirrors, used in laser scanners and projection systems. In the case
of a cantilever beam, the flexural displacements are described by a fourth-order differential equation

(
IE ∂ wx, t) ∂ wx, t( )
4
2
------- ----------------------- + ----------------------- = 0 (8.31)
rA ∂x 4 ∂t 2
where I is the moment of inertia, E is the Young’s modulus, ρ is the density, and A is the area of the cross
section. When the thickness of the cantilever is much smaller than the width, E should be replaced by
2
the reduced Young’s modulus E 1 = E/(1 − ν ). For a rectangular cross section, (8.31) is reduced to

Eh ∂ wx, t( ) ∂ wx, t)
(
4
2
2
--------- ----------------------- + ----------------------- = 0 (8.32)
12r ∂x 4 ∂t 2
where h is the thickness of the beam. The solution of (8.32) can be written in terms of an infinite series
of characteristic functions representing the individual vibration modes
∞
w = ∑ Φ i x()sin ( w i t + d i ) (8.33)
i=1

where the characteristic functions Φ i are expressed with the four Rayleigh functions S, T, U, and V:

(
(
(
(
Φ i = a i S l i x) + b i T l i x) + c i U l i x) + d i V l i x)
Sx() = 1 -- cosh( x + cos x), Tx() = 1 x + sin x)
-- sinh(
2 2 (8.34)
4
Ux() = 1 -- cosh( x – cos x), Vx() = 1 x – sin x), l i = w i 2 rA
-- sinh(
-------
2 2 IE
The coefficients a i , b i , c i , d i , ω i , and δ i are determined from the boundary and initial conditions of (8.34).
For a cantilever beam with a fixed end at x = 0 and a free end at x = L, the boundary conditions are
2
----------------------- =
w 0, t( ) = 0, ∂ wL, t( ) 0
∂x 2
(8.35)
(
(
∂w 0, t) 0, ∂ wL, t) 0
3
-------------------- =
----------------------- =
∂x ∂x 3
Since (8.35) are to be satisfied by each of the functions Φ i , it follows that a i = 0, b i = 0 and
cosh(λ L)cos(λ L) = −1 (8.36)
i
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