K y k c yt lecture notes 51 061606 1225 pm mechanical

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Unformatted text preview: e F FT = TR ⋅ F = TR ⋅ Feq = 2.52 ⋅ 438.7 N = 1106 N Notice that due to the characteristics of the system, the force transmitted to the foundation is approximately 2.5 times the force applied to the system. (b) 600 rpm Again solving for the dynamic amplitude X using the rearranged 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 ω = 2π 600rpm = 20π rad s = 62.83 rad s s 60 min Lecture Notes -54- 06/16/06 12:25 PM Mechanical Vibrations I Feq = meeω 2 = 10kg ⋅ 0.1m ⋅ (62.83 rad s ) 2 = 3,948 N c = 2ζ km = 2 ⋅ 0.2 ⋅ 200,000 N m ⋅ 500kg = 4000 N ⋅s m Evaluating the dynamic amplitude yields, X = Feq k − mω + jω c 2 = 3,948 N 200,000 N m − 500kg (62.83 rad s )2 + j (62.83 rad s )(4000 N ⋅s m) X = 3,948 N = 0.0022m or 2.2mm ( −1.774 ×106 + j 251.3 ×103 ) kg s2 Evaluating the transmissibility yields, TR = k + jω c 200,000 N m + j (20.94 rad s )(4000 N ⋅s m) = 2 k − mω + jω c 200,000 N m − 500kg (62.83 rad s ) 2 + j (62.83 rad s )(4000 N ⋅s m) TR = ( −1.774 ×10 ( 200,000 + j 251.3 ×10 ) 3 6 N 3 + j 251.3 ×10 ) m N m = 0.179 From the definition of transmissibility, namely TR = foundation can be determined. FT , the force transmitted to the F FT = TR ⋅ F = TR ⋅ Feq = 0.179 ⋅ 3,948 N = 708 N Notice that although the equivalent applied force is 900% larger than for the first case, the force actually transmitted to the foundation is about 30% smaller. Lecture Notes -55- 06/16/06 12:25 PM Mechanical Vibrations I #11 - Multiple Degree of Freedom Systems MDOF Example Starting with the simple two degree-of-freedom system shown, the objective will be to determine the equations of motion and the natural frequency and mode shapes. The equations of motion can be found by either of two techniques, Newton’s Method or Lagrange’s Method. k1 m1 x1 (t ) f1 (t ) k2 Equations of Motion – Newton’s Method m2 x2 (t ) f 2 (t ) As has been shown, Newton’s Method involves identifying all forces, both internal and external, acting on each body and then expressing those forces in terms of the given coordinates. Therefore, the first action must be the development of the free body diagrams for each body. The given two degree-of-freedom system has the two free body diagrams, show at right. Evaluating Newton’s equation, ∑ fi = ma , for each body yields a set i f1 k1 x1 m1 = m1 body #1 of equations describing the motion of the system. f1 − k1 x1 − k2 ( x1 − x2 ) + m1 g = m1 x1 f 2 − k2 ( x2 − x1 ) + m2 g = m2 x2 m1 g k2 ( x1 − x2 ) k2 ( x2 − x1 ) m2 = f2 m1 x1 body #2 m2 m2 x2 Collecting and rearranging the terms...
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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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