ASE 365 - HW 4 Solutions

# ASE 365 - HW 4 Solutions - Solutions Homework Set 4 1...

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Unformatted text preview: Solutions: Homework Set 4 1. Problem 3.16. The paragraph before Eq. (3.52) explains why the amplitude of the vibration, for the rotating unbalanced mass excitation case, is obtained by replacing the constant A in Eq. (3.28) with ( m/M ) eω 2 /ω 2 n . The amplitude of the transmitted force found in Eq. (3.94) is for the case of harmonic force excitation, rather than rotating unbalanced mass excitation. When the vibration amplitude is given by Eq. (3.52) rather than Eq. (3.28), the transmitted force amplitude becomes: F tr = bracketleftBig m M e ( ω/ω n ) 2 bracketrightBig k bracketleftBig 1 + (2 ζ ( ω/ω n )) 2 bracketrightBig | G ( iω ) | = m M e ( ω/ω n ) 2 k 1 + [2 ζ ( ω/ω n )] 2 bracketleftBig 1 − ( ω/ω n ) 2 bracketrightBig 2 + [2 ζ ( ω/ω n )] 2 1 2 Considering that ( ω/ω n ) = 4, the quantity in {} ’s is ( 1 + (8 ζ ) 2 ) / ( 225 + (8 ζ ) 2 ) , which has a value of 1 / 225 when ζ = 0 and asymptotically approaches 1 as ζ increases. Inserting the given parameters and setting F tr equal to the limit of 250 N yields a maximum value for ζ : F tr = 5 80 × . 1 × 4 2 × 8000 × braceleftbigg 1 + (8 ζ ) 2 (1 − 4 2 ) 2 + (8 ζ ) 2 bracerightbigg 1 2 = 250 ⇒ ζ max = 0 . 6026 This corresponds to a dashpot coefficient of c max = 2 ζ max ω n M = 964 . 2 N · s m . Any value of c below c max (even c = 0) would keep F tr below the limit. However, some damping may be desirable for controlling the amplitude of vibration and transmitted force as the system accelerates through resonance to the operating speed. If vibration during the acceleration through resonance is a significant issue, a value of c close to c max should be chosen. If not, a lower level of damping would reduce the amount of heat produced in the damper at operating speed....
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ASE 365 - HW 4 Solutions - Solutions Homework Set 4 1...

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