1225 pm lecture notes mechanical vibrations i 9

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Unformatted text preview: ting Example A A machine of 100 kg mass has a 20 kg rotor with 5 mm eccentricity. The mounting springs have k = 85 kN m and the damping is negligible. The operating speed is 600 rpm and the unit is constrained to move vertically. (a) Determine the dynamic amplitude of the machine. (b) Redesign the mounting so that the dynamic amplitude is reduced to one half of the original value, but maintaining the same natural frequency. Solution Given the following parameters: m = 100kg , me = 20kg , e = 5mm , k = 85 kN m , c 0 N ⋅s m & ω = 600rpm (a) Starting with the frequency response function previously developed, ω m e me x(t) k c H (ω ) = X ω2 (ω ) = mee −ω 2 m + jω c + k Solve for the dynamic amplitude X by rearranging the expression and evaluating its magnitude, mee ω 2 X (ω ) = −ω 2 m + jω c + k The additional parameters needed are the unbalance and the rotational frequency in mee = 20kg ⋅ 5 × 10−3 m = 0.1kg ⋅ m 600rpm ω = 2π = 20π rad s = 62.83 rad s 60 s min rad s . Evaluating the dynamic amplitude with c = 0 yields, X = mee ω 2 (0.1kg ⋅ m)(20π rad s ) 2 = k − mω 2 85,000 N m − 100kg (20π rad s ) 2 X = 394.78 kg⋅m s2 = −0.001274m or 1.274mm −309,784 kg s2 Lecture Notes -47- 06/16/06 12:25 PM Mechanical Vibrations I (b) There are two options for reducing the dynamic amplitude by half and yet keeping the same natural frequency, increasing mass and increasing damping. X Ω= reduced = 1 2 X original = 0.6372mm k 85,000 N m = = 29.15 rad s 100kg m Option 1: increasing mass. Manipulating the expression for dynamic amplitude mee ω 2 mee ω 2 mee ω 2 X= = = k k − mω 2 m( m − ω 2 ) m(Ω 2 − ω 2 ) yields an expression for the increased mass. m= mee ω 2 (0.1kg ⋅ m)(20π rad s ) 2 = = 200kg X (Ω 2 − ω 2 ) −0.6372mm((29.15 rad s ) 2 − (20π rad s ) 2 ) However, keeping the same Ω implies a corresponding increase in k . Therefore, k = m Ω 2 = 200kg (29.15 rad s ) 2 = 170,000 N m or 170 kN m Notice how the solution makes sense intuitively. The operating speed is above the natural frequency of the system. In that region, the frequency response function magnitude 1 X ( ) approaches , so it would be expected that to reduce the dynamic amplitude by m mee half would require approximately doubling the mass. Option 2: increasing damping. Again, manipulating the expression for dynamic amplitude X = mee ω 2 − mω 2 + jω c + k X −mω 2 + jω c + k = mee ω 2 −mω + jω c + k = 2 mee ω 2 X Lecture Notes -48- 06/16/06 12:25 PM Mechanical Vibrations I ( k − mω ) (ω c) = 2 2 2 + (ω c) 2 = 2 mee ω 2 X 2 2 mee ω 2 X 2 − ( k − mω 2 ) 2 yields an expression for the necessary damping mee ω 2 c= X rad 2 2 − ( k − mω 2 ) 2 ω2 s 2 ( 0.1kg ⋅ m(20π c= 2 2 )2 ) 2 (0.006372m) − ( 85,000 N m − 100kg (20π rad s ) 2 ) 2 (20π rad s ) 2 155.85 × 103 kg ⋅m s4 2 2 2 − ( −309,784 kg s2 ) −9 2 383.87 ×109 kg s4 − 95.966 × 109 kg s4 406.0 × 10 m = = 8,540 kg s c= 2 2 3,947....
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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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