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Unformatted text preview: 8 rad s2 3,947.8 rad s2 or more familiarly, c = 8,540 N ⋅s m . This can be expressed as a fraction of critical damping c ( ζ = ) by recalling the expression for the critical damping value, cc = 2 km . cc ζ = c 8,540 N ⋅s m = = 1.465 or 146.5% 2 km 2 85,000 N m ⋅100kg Notice that to reduce the dynamic response of this system utilizing only damping requires an over-damped condition of nearly 150%. For this problem, the mass addition is most likely to be the easiest and cheapest practical solution. Lecture Notes -49- 06/16/06 12:25 PM Mechanical Vibrations I #10 - Steady-State Applications
In this section, two different but related steady-state applications will be developed. The first is the force transmissibility relationship. For transmissibility, the interest is in determining the ratio of the force transmitted to the base ( FT (ω ) ) to the applied force ( F (ω ) ) as a function of frequency. The second application is vibration isolation. For isolation, the interest is in determining the ratio of the response of the system ( X (ω ) ) to the motion of the base ( Y (ω ) ) as a function of frequency. (This is sometimes referred to as response ratio.) Just as in the earlier development of the Frequency Response Function (FRF), for both applications, the focus will be upon harmonic excitation. Transmissibility
For transmissibility, we want to determine the FT (ω ) relationship. We begin with the F equation of motion for our representative single degree-of -freedom system. (NOTE: we assume negligible motion of the base.) mx(t ) + cx(t ) + k x(t ) = f (t ) In addition, we need the restraining force at the other end of the spring/damper.
cx(t ) + k x(t ) = fT (t ) m x(t) f(t) Applying the same transformations developed earlier, namely that the forcing functions and responses can be express as complex exponentials, and collecting terms, we get: k c fT(t) ( ms
and 2 + cs + k ) Xe st = Fe st ( cs + k ) Xe st = FT e st Finally, by taking the ratio of the second equation to the first, we get the desired transmissibility relationship. FT cs + k ( s) = 2 F ms + cs + k Lecture Notes -50- 06/16/06 12:25 PM Mechanical Vibrations I Or expressed in the Frequency Domain as: FT jω c + k (ω ) = 2 −ω m + jω c + k F Isolation
For isolation, we want to determine the X (ω ) relationship. We begin with the equation Y of motion for our representative single degree-of –freedom system. (NOTE: this time the base motion is NOT negligible, however there is no explicitly applied external force.) Because there is motion for both the structure and the base, we need to use the difference in motion across the spring/damper. m x(t) mx(t ) + c( x(t ) − y (t ) ) + k ( x(t ) − y (t ) ) = 0 Transforming the equation as before, we get: ms 2 Xe st + c( sXe st − sYe st ) + k ( Xe st − Ye st ) = 0 Collecting terms, and moving the base motion to the other side of the equation yields, (ms 2 + cs + k ) Xe st = (cs + k )Ye st Finally, by cross dividing the relevant te...
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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.
- Fall '11