Cc 2 km cc c 8540 n s m 1465 or 1465 2

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Unformatted text preview: rms, we get the desired isolation relationship. X cs + k ( s) = 2 Y ms + cs + k Or expressed in the Frequency Domain as: X jω c + k (ω ) = 2 −ω m + jω c + k Y k c y(t) Lecture Notes -51- 06/16/06 12:25 PM Mechanical Vibrations I Conclusion Comparing the expression for transmissibility to the expression for isolation, we find that they are identical. This interesting result implies that the process of determining the force transmitted to the base is equivalent to determining the base motion transmitted to the 10 9 10 8 M agnitude 6 4 2 8 7 6 M agnitude 5 4 3 0 10 5 0 -5 Om ega [rad/s ec] -10 -1 -0.5 S igm a [rad/s ec ] 0.5 0 1 2 1 0 -10 -8 -6 -4 -2 0 2 Om ega [rad/s ec ] 4 6 8 10 system. 4 3 4 2 2 1 P hase [rad] 0 P has e [rad] 0 -2 -1 -4 10 5 0 -5 Om ega [rad/s ec] -10 -1 -0.5 S igm a [rad/s ec ] 0.5 0 1 -2 -3 -4 -10 -8 -6 -4 -2 0 2 Om ega [rad/s ec ] 4 6 8 10 Lecture Notes -52- 06/16/06 12:25 PM Mechanical Vibrations I Transmissibility Example A A vertical single-cylinder diesel engine of 500 kg mass is mounted on springs with k = 200 kN m and dampers with ζ = 0.2 . The rotating parts are well balanced. The mass of the equivalent reciprocating parts is 10 kg and the stroke is 200 mm. Find the dynamic amplitude of the vertical motion, the transmissibility, and the force transmitted to the foundation, if the engine is operated at (a) 200 rpm; (b) 600 rpm. Solution Given the following parameters: stroke m = 500kg , mR = 10kg , stroke = 200mm , k = 200 kN m & ζ = 0.2 (a) 200 rpm x(t) m Starting with the single degree-of-freedom frequency response function previously developed, H (ω ) = X 1 (ω ) = 2 −ω m + jω c + k F k c Solve for the dynamic amplitude X by rearranging the expression and evaluating its magnitude, X (ω ) = F (ω ) −ω m + jω c + k 2 The additional parameters needed are the equivalent force, the damping and the operating frequency expressed in rad s . From the mechanical unbalance development, the equivalent force can be expressed as Feq = meeω 2 . Recognizing that the reciprocating mass is the effective mass, me = mR , and that the effective eccentricity is half the stroke, e = stroke , yields, 2 200rpm 20 rad = π s = 20.94 rad s 60 s min 3 2 Feq = meeω = 10kg ⋅ 0.1m ⋅ (20.94 rad s ) 2 = 438.7 N ω = 2π c = 2ζ km = 2 ⋅ 0.2 ⋅ 200,000 N m ⋅ 500kg = 4000 N ⋅s m Lecture Notes -53- 06/16/06 12:25 PM Mechanical Vibrations I Evaluating the dynamic amplitude yields, X = Feq k − mω + jω c 2 = 438.7 N 200,000 N m − 500kg (20.94 rad s ) 2 + j (20.94 rad s )(4000 N ⋅s m) X = 438.7 N = 0.0051m or 5.1mm ( −19,324 + j83,776 ) kg s2 Evaluating the transmissibility yields, TR = k + jω c 200,000 N m + j (20.94 rad s )(4000 N ⋅s m) = 2 200,000 N m − 500kg (20.94 rad s ) 2 + j (20.94 rad s )(4000 N ⋅s m) k − mω + jω c TR = ( 200,000 + j83,776 ) N m ( −19,324 + j83,776 ) N m = 2.52 From the definition of transmissibility, namely TR = foundation can be determined. FT , the force transmitted to th...
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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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