Unformatted text preview: fore, since the system matrix becomes rank deficient for certain values of s, the problem reduces to a polynomial root solver. (The determinant of the system matrix is zero.) Lecture Notes 19 06/16/06 12:25 PM Mechanical Vibrations I [ M ] s 2 + [C ] s + [ K ] = 0
This determinant results in a polynomial whose roots are the poles of the system. This is known as the characteristic polynomial. Often the determinant notation is dropped (but still written equal to the scalar zero implying a determinant) and the matrix characteristic polynomial is expressed as, [ M ] s 2 + [C ] s + [ K ] = 0
Returning to the scalar (1 DOF) problem as initially presented, we have: ms 2 + cs + k = 0 Since the determinant of a scalar is simply the scalar itself, the characteristic polynomial is simply: ms 2 + cs + k = 0
Since this is a quadratic equation, there are two roots, namely: λ1,2 = −c ± c 2 − 4mk 2m This results in the homogeneous (transient) solutions form of: x(t ) = A1eλ1t + A2eλ2t The characteristic behavior of this system solution is determined by the relationship of m, c, & k. c 2 > 4mk c 2 = 4mk
c 2 < 4mk Two distinct real roots. Two equal real roots. One pair of complex conjugate roots. For the condition of c 2 > 4mk , no oscillation occurs. This is called overdamped. Since no oscillation occurs, this is not particularly interesting from a vibration point of view and we will not investigate it any further. For the condition of c 2 = 4mk , this is the boundary between oscillatory and nonoscillatory behavior. This is called criticallydamped. We will return to this later. Lecture Notes 20 06/16/06 12:25 PM Mechanical Vibrations I The interesting oscillatory behavior occurs when c 2 < 4mk . This is called underdamped. When a system is underdamped, the system will respond to initial conditions with harmonic oscillatory behavior. First, we recognize the roots of the characteristic equation as the system poles ( λr ). For the given SDOF system, the roots have the following form: λr = λ1,2 = −c k c ±j − 2m m 2m 2 Examining the two parts (real and imaginary) of the poles, we can write: λr = σ r ± jω r
Where, −c is the damping for mode r (for single degreeoffreedom systems, it is 2m often written as σ d ) and σr = k c ωr = ± − is the damped natural frequency for mode r (for single m 2m degreeoffreedom system, it is often written as ω d ) 2 Since λ1 and λ2 are complex conjugates, the homogeneous solution for the SDOF problem can be written as:
x(t ) = Xe λt + X *e λ t
* k . This is called the undamped natural frequency, written m as Ω r (for single degreeoffreedom system, it is often written as ω n .) This is the frequency at which the system will oscillate if there is no damping in the system. If we let c → 0 , then ω r → Returning to the critically damped condition, where c 2 = 4mk , we can now define a few more vibration parameters. Defining the critical damping as:
cc = 2 mk Allows us to define the damping ra...
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This note was uploaded on 09/29/2013 for the course MECHANICAL ME taught by Professor Regalla during the Fall '11 term at Birla Institute of Technology & Science, Pilani  Hyderabad.
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