# Vibs1_review_notes

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PM Mechanical Vibrations I The transformation to the Laplace Domain can be easily accomplished by using an assumed solution approach. x ( t ) = Xe st & f ( t ) = Fe st Taking derivatives of the assumed solutions yields, x ( t ) = sXe st & x ( t ) = s 2 Xe st Recognize that the vector form of the assumed solution is simply, { x} = { X } e st & { f } = {F } e st { x} = s { X } e st & { x} = s 2 { X } e st Substituting into the equation of motion yields, m1 0 2 X 1 st k1 + k2 0 m s X e + −k 2 2 2 −k2 X 1 st F1 st e = F e k2 X 2 2 Because e st ≠ 0 for all time, the solution reduces to, m1 0 2 X 1 k1 + k2 0 m s X + −k 2 2 2 − k2 X 1 F1 = k2 X 2 F2 Since the objective is to calculate natural frequency and mode shape, it is necessary to solve the homogeneous portion of the solution. Collecting terms and setting the force side of the equation to zero yields, m1 0 2 k1 + k2 s + −k 2 0 m2 − k 2 X 1 0 = k 2 X 2 0 The solution is now in one of the eigenvalue/eigenvector forms. The simplest solution approach is to recognize that in order for the equation to be valid, either of two conditions must hold. First, the vector { X } is identically zero (the trivial solution) or second, the determinant of the coefficient matrix is zero (i.e. the coefficient matrix is singular for certain values of s .) The second is the only useful solution and will be pursued here. Setting the determinant of the coefficient matrix to zero yields, m1 0 2 k1 + k2 0 m s + −k 2 2 − k2 =0 k2 Lecture Notes -59- 06/16/06 12:25 PM Mechanical Vibrations I This determinant is called the characteristic polynomial and the solution effectively becomes the problem of finding the roots of the characteristic polynomial. m1s 2 0 0 k1 + k2 + m2 s 2 − k2 − k2 =0 k2 (m s 1 2 + k1 + k2 )( m2 s 2 + k2 ) − ( − k2 )( − k2 ) = 0 m1s 2 + k1 + k2 − k2 =0 m2 s + k2 − k2 2 ( m1m2 ) s 4 + ( m1k2 + m2k1 + m2k2 ) s 2 + ( k1k2 ) = 0 In general, the characteristic polynomial is complete (has all terms), but because this example problem does not include damping, all the odd powers of s are eliminated. This makes finding the roots ( λr ) of the equation somewhat easier. λ = 2 r − ( m1k2 + m2 k1 + m2 k2 ) ± ( m1k2 + m2k1 + m2k2 ) 2 ( m1m2 ) 2 − 4 ( m1m2 )( k1k2 ) At this point, continued manipulation of the arbitrary variables does not contribute to the understanding of the solution, but only proves pure obstinacy. Returning to the eigenvalue/eigenvector problem above and substituting some numerical values should prove more interesting. Let m1 = 2 kg , m2 = 1 kg , k1 = 3 N m and k2 = 2 N m . 2 0 2 3 + 2 −2 X 1 0 = s + −2 2 X 2 0 0 1 Evaluating the characteristic polynomial yields, ( 2s Which has the solution, 2 + 5 )( s 2 + 2 ) − ( −2 )( −2 ) = 0 ( 2s 4 + 9s 2 + 6 ) = 0 λ = 2 r −9 ± 92 − 4 ( 2 )( 6 ) 2 ( 2) = −9 ± 33 4 λr2 ≈ −9 ± 5.745 2 ≈ −0.814, −3.686 rad s 2 4 Lecture Notes -60- 06/16/06 12:25 PM Mechanical Vibrations I Finally, ta...
View Full Document

## 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.

Ask a homework question - tutors are online