EG260 Solutions-example_3

EG260 Solutions-example_3 - EG-260 EG-260 Dynamics I -...

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Unformatted text preview: EG-260 EG-260 Dynamics I - Example Sheet 3 Answers Damped single degree of freedom systems 1. k = 400 N / m c = 9.03kg / s mg = 2 N so ccr = 2mn = 2 km = 2 400 * 2 / g = 40 b) n = k 400 g g = = 20 m 2 2 = 44.3rad / s ( 7.05Hz ) c) = c 9.03 1 = = cr 18.06 2 2 = 18.06kg / s g so that d = n 1 - 2 = n 3 4 = 44.3 3 = 38.4 rad / s 4 (6.11 Hz) @Avril Slone 2007 EG-260 2. Mg = 5N k = 100 N / m a. 1 = ln 0.202 0.98 b. = nT = = c. n = n * 2 n 1 - 2 = 3.18.10 -3 4 2 + 2 100 = 20 g = 14rad / s 5/ g d n = 14 rad / s = 2.23 Hz d. c = 2n m c = 2 * 3.18 * 10 -3 *14 * 5 / g = 0.0459kg / s @Avril Slone 2007 EG-260 3. Take moments about the pivot & ma 2& = sum of moments = - a (ac& ) - b(bk ) so & ma 2& + a 2 c& + b 2 k = 0 Finding the eigenvalues 1,2 = - a 2c (a c ) 2 2 - 4 ma 2b 2 k <1 2 ma 2 = - n jd so 4 ma 2b 2 k - (a 2c ) d = 2 ma 2 2 2 k b c = - m a 2m For the damping k b ccr - =0 m a 2m ccr = 2 b km a 2 2 2 @Avril Slone 2007 ...
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This note was uploaded on 05/18/2010 for the course ENGINEERIN EG 260 taught by Professor Stone during the Spring '10 term at Swansea UK.

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EG260 Solutions-example_3 - EG-260 EG-260 Dynamics I -...

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