ME 242 Selected Answers Chapter 9

ME 242 Selected Answers Chapter 9 - 9.1 Undamped, Free...

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Unformatted text preview: 9.1 Undamped, Free Response for Single Degree of Freedom Systems x = 0.556 m, v = 6.50 m/s, a = -5.00 m/s mB = 34.1 kg n = 20 rad/s n = IG n n n n 2kL2 m1 L2 2 +m L 2 3 = 1.1310-2 slgft2 6g = 5L = 5.82 rad/s = 9.18 rad/s = 2.80 rad/s 3g 2L 2 9.2.12 9.2.14 9.2.16 9.2.18 m = 8g xeq = 0.327 mm Fsp = (15 N) sin(t) xeq = 0.049 mm 9.1.2 9.1.6 9.1.8 9.1.10 9.1.12 9.1.14 9.1.16 9.1.18 9.1.20 9.1.22 9.1.24 9.1.26 9.1.28 9.1.30 9.1.32 9.1.34 9.1.36 9.1.38 9.1.40 9.1.42 9.1.44 9.1.46 9.1.48 9.1.50 9.3 Damped, Free Response for Single Degree of Freedom Systems t = 1.02 s c = 25.5 Ns/m t1 = 0.140 s, t2 = 0.137 s x(t) = v + x (x0 cos d t + 0 n 0 sin d t)e-n t d 9.3.2 9.3.4 9.3.6 9.3.8 n = n = g 6r y(t) = [1.5 - cos((4.63 s-1 )t)] ft 2g n = l k1 k2 k3 n = m(k1 k2 + k2 k3 + k3 k1 ) gwater n = lwood n = 3.05 rad/s g n = (r( - 2) g n = r n = 18.6 rad/s 2g n = l k1 + k2 n = m 2g n = 3 r2 - r 1 n = n = 2 (k1 + k2 )l3 2 2 m1 l1 + m2 l2 9.4 Damped, Sinusoidally Forced Response for Single Degree of Freedom Systems x(t) = (0.156 m) sin[(8 rad/s)t - 2.47 rad] x(t) = -(3.6110-2 m/s) sin[(100 rad/s)t -0.109 rad] peak = 4.8 rad/s, C = 0.051 m 30 m = 150 m/s v = 0.2 s 3.1% = 0.025 9.4.2 9.4.4 9.4.6 9.4.8 9.4.10 9.4.12 (9mA + 6mB )g 2h(6mA + mB ) 9.2 Undamped, Sinusoidally Forced Response for Single Degree of Freedom Systems h = 0.269 m x0 = -6. 610-2 m x = 4.9110-2 mm x = 6.2310-2 ft v = 48 mph 66 9.2.2 9.2.4 9.2.6 9.2.8 9.2.10 ...
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This note was uploaded on 09/19/2009 for the course ME 242 taught by Professor Kam during the Spring '06 term at Nevada.

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