The change of phase through resonance the peak in the

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 68 Hz. Lecture Notes -65- 06/16/06 12:25 PM Mechanical Vibrations I Ω3 = 9.068Hz The peak FRF magnitude equals 0.0204 m/N, therefore the half-power magnitude would be 0.0144 m/N. So the half-power points are estimated as 9.0383 Hz and 9.0982 Hz, which result in an estimate of damping ratio of 0.0033. ζ3 = ∆f 9.0982 Hz − 9.0383Hz = = 0.0033 -or- 0.33% 2 fc 2 ( 9.068 Hz ) The peak imaginary values, for each output (response) location for the first input (force) location, are estimated as [-0.0164 0.0204 –0.0091] m/N. ℑ{ H11} = -0.0164 m N , ℑ{ H 21} = 0.0204 m N & ℑ{ H 31} = -0.0091 m N Therefore the mode shape is estimated as, −0.80 1.00 −0.0164 {ψ }3 = 0.0204 -or- 1.00 -or- −1.24 −0.45 0.55 −0.0091 So the modal parameter estimate for the third mode is: 1.00 Ω3 = 9.068Hz , ζ 3 = 0.33% & {ψ }3 = −1.24 0.55 NOTE: By convention, modes are numbered from lowest frequency to highest frequency. Lecture Notes -66- 06/16/06 12:25 PM Mechanical Vibrations I #13 – Transient Solution of Equation of Motion The last piece of the vibration problem is to solve for the transient solution of the differential equation of motion. In general, since the excitation (forcing) function can be literally anything, a closed form solution is not practical and numerical techniques are used. (The numerical methods will be discussed in the next section.) However, a study of the general solution for certain forcing functions is important to the basic understanding of the complete solution for the equation of motion. Impulse Response The impulse response is the response of the system to an impulsive excitation. An impulse is frequently a large magnitude, short duration force. f (t ) ˆ The impulse ( F ) of a force is defined as ∫ f (t )dt . ˆ F ˆ F ε t ξ ε ∫ f (t )dt ˆ If the impulse is unity ( F = 1 ) and the duration of the pulse ( ε ) is allowed to approach zero ( ε → 0 ) then the force ( f (t ) ) is referred to as a unit impulse. The unit impulse is equivalent to the mathematical expression called a unit delta function ( δ (t ) ). The delta function is defined by: δ (t − ξ ) = 0 ∞ when t ≠ ξ when 0 < ξ < ∞ ∫ δ (t − ξ )dt = 1 0 Further, the delta function has a number of interesting and useful properties. Of particular interest is the result of multiplication by another function. ∞ ∫ f (t )δ (t − ξ )dt = f (ξ ) 0 Recalling the impulse/momentum relationship ( f (t )dt = m dv ) and integrating both sides shows that an impulse acting upon a mass results in a change in velocity ( ∆v ) without a significant change in position. Therefore, Lecture Notes -67- 06/16/06 12:25 PM Mechanical Vibrations I ˆ ∆v = F m The free response of a single degree-of-freedom system to initial conditions is given by: x(0) − σ r x(0) sin(ω r t ) + x(0) cos(ω r t ) x(t ) = eσ rt ωr Assuming the system is initially at rest, then combining the free response with the change in momentum yields, x(t ) = ˆ F σ...
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