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Unformatted text preview: ycles. Which simplifies to: 1 x(t ) δ = ln 0 ≈ 2πζ r n x(tn ) -2606/16/06 12:25 PM If the damping ratio is “small” (eg. ζ < 0.1 -or- 10%), then the above expression can be simplified to: δ ≈ 2πζ r
(Error less than 1% when ζ < 0.1 )
Lecture Notes Mechanical Vibrations I Log Decrement Example
Using the figure given above, estimate both the damped natural frequency and the damping ratio. First, select two convenient time points to estimate the period ( τ ). Between t = 0 seconds and t = 9 seconds, there are 11 cycles. τ= 9 sec = 0.8182 sec cycle 11cycles Therefore the damped natural frequency is ωr = 11cycles = 1.222 Hz -or- 7.679 rad s 9 sec Second, select two convenient time points to estimate the log decrement ( δ ). Between t ≈ 1.8 seconds and t ≈ 6.7 seconds, there are 6 cycles with x(t ≈ 1.8) ≈ 3 and x(t ≈ 6.7) ≈ 2 . δ = ln 1 n x(t0 ) 1 x(t0 ≈ 1.8) 1 3 = ln = ln = 0.0676 x(tn ) 6 x(tn ≈ 6.7) 6 2 Since the log decrement ( δ ) is so “small”, first try the simplified approximation for damping ratio ( ζ ). ζ ≈ δ 2π = 0.0676 2π = 0.0108 -or- 1.08%
(Error check: ζ < 0.1 so the approximation is adequate.) Just for completeness, since ζ < 0.1 , 1 − ζ 2 ≈ 1 , therefore Ωr = ωr
2 r ≈ ω r = 7.679 rad s -or- 1.222Hz σ r = −ζ r Ωr = −0.0108 ⋅ 7.679 rad s = −0.0829 rad s -or- −0.0132Hz Lecture Notes -27- 06/16/06 12:25 PM Mechanical Vibrations I #6 - Lagrange’s Method
d ∂T ∂T ∂U ∂D + + = Fi − dt ∂qi ∂qi ∂qi ∂qi Where: • T = Kinetic Energy • U = Potential Energy • D = Dissipative Energy • Fi = Externally applied force/moment •
qi = Generalized coordinate Generalized Coordinates
q = x, y, z ,θ x ,θ y ,θ z Kinetic Energy
Nm j 1 1 T = ∑ M i x 2 + ∑ J iθ 2 i =1 2 i =1 2 N Potential Energy
Nk Nm 1 U = ∑ Ki x 2 + ∑ ± ( M i g ) x i =1 2 i =1 Dissipative Energy
Nc 1 D = ∑ Ci x 2 i =1 2 Lecture Notes -28- 06/16/06 12:25 PM Mechanical Vibrations I Lagrange’s Method
Typical problems with Lagrange formulations: • Be sure to establish number of degrees of freedom first and formulate all energy terms in only those variables. Clearly identify which degrees of freedom are relative coordinates versus absolute coordinates. Watch out for rotational/translational problems. • For kinetic energy terms, be sure to formulate absolute velocities before taking derivatives. Watch out for 2-D and 3-D vector motions. • For potential energy terms, be sure that the actual deflection, described by relative and/or absolute coordinates, in spring elements is described. Watch out for 2-D and 3-D vector motions. • There should be only one total kinetic energy equation, one total potential energy and one total dissipative energy equation for the system. The kinetic, potential and dissipative energy equations should involve only the N generalized coordinates and the constants (mass, damping, stiffness) of the system. • Apply the Lagra...
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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.
- Fall '11