Dynamics of Machines - Part II - IFS.pdf
Dynamics of Machines - Part II - IFS.pdf Dynamics of Machines - Part II - IFS.pdf
Using the information of figure 11 (signal in time domain), where yo = 3.0 · 10 −5 [m/s 2 ] (first peak of signal in time domain, figure 11) yN = y6 = 2.0 · 10 −5 [m/s 2 ] (after 6 peaks) N = 6, one gets • Damping Factor of the system ”A” (ξ) ξ = • Log Dec 1 + 1 2π·6 ln 1 2π·6 ln 3.0·10 −5 2.0·10 −5 3.0·10 −5 2.0·10 −5 β = 2π ξ = 2π 0.010 = 0.068 • Equivalent Viscous Damping (d) = 2 0.010755 ≈ 0.010 1.000058 d = 2 · ξ · ωn · m = 2 · 0.010 · (0.87 · 2 · π) · 0.382 ≈ 0.016 [N · s/m] (a) Amplitude [m/s 2 ] (b) Amplitude [m/s 2 ] 2 0 −2 x Signal 10−5 4 (a) in Time Domain − (b) in Frequency Domain −4 0 5 10 15 time [s] 20 25 30 0.8 0.6 0.4 0.2 x 10−5 1 0 0 5 10 15 20 25 frequency [Hz] Figure 11: Transient Vibration – Acceleration of the clamped-free flexible beam when two concentrated masses m = m1 +m2 = 0.382 Kg are attached at its free end (L = 0.610 m) – Natural frequency of the mass-spring system ”A”: 0.87 Hz. 20
(a) Amplitude [m/s 2 ] x Signal 10−5 5 0 (a) in Time Domain − (b) in Frequency Domain −5 0 5 10 15 time [s] 20 25 30 (b) Amplitude [m/s 2 ] x 10−5 2 1 0 0 5 10 15 20 25 frequency [Hz] Figure 12: Transient Vibration – Acceleration of the clamped-free flexible beam when two concentrated masses m = m1 +m2 = 0.382 Kg are attached at its free end (L = 0.310 m) – Natural frequency of the mass-spring system ”B”: 1.75 Hz. • Damping Factor of the system ”B” – From fig.12, one gets: yo = 4.6 · 10 −5 [m/s 2 ], yN = y54 = 1.0 · 10 −5 [m/s 2 ] and N = 54: ξ = 1 + 1 2π·54 ln 1 2π·54 ln 4.6·10 −5 1.0·10 −5 4.6·10 −5 1.0·10 −5 • Equivalent Viscous Damping (d) = 2 0.004498 ≈ 0.005 (46) 1.000010 d = 2 · ξ · ωn · m = 2 · 0.005 · (1.75 · 2 · π) · 0.382 ≈ 0.04 [N · s/m] 21
- Page 1 and 2: DYNAMICS OF MACHINES 41614 PART I -
- Page 3 and 4: 1 Introduction to Dynamical Modelli
- Page 5 and 6: 1.3 Data of the Mechanical System
- Page 7 and 8: 1.5 Calculating Stiffness Matrices
- Page 9 and 10: 1.6 Mechanical Systems with 1 D.O.F
- Page 11 and 12: Demanding (λ 2 + 2ξωnλ + ω 2 n
- Page 13 and 14: 1 yini − A det λ1 vini − A C2
- Page 15 and 16: 1.6.4 Analytical and Numerical Solu
- Page 17 and 18: (a) y(t) [m] (b) y(t) [m] (c) y(t)
- Page 19: 1.6.6 Homogeneous Solution or Free-
- Page 23 and 24: (a) Amplitude [m/s 2 ] (b) Amplitud
- Page 25 and 26: Imag(A(ω)) [m/N] 0 −1 −2 −3
- Page 27 and 28: 1.6.11 Superposition of Transient a
- Page 29 and 30: (a) Amplitude [m/s 2 ] (b) Amplitud
- Page 31 and 32: 1.7 Mechanical Systems with 2 D.O.F
- Page 33 and 34: ⎧ ⎫ ⎪⎨ ˙y1(t) ⎪⎬ ˙y
- Page 35 and 36: zini = U c + A ⇒ c = U −1 {(zin
- Page 37 and 38: 1.7.4 Modal Analysis using Matlab e
- Page 39 and 40: 0.7 0.6 0.5 0.4 0.3 0.2 0.1 First M
- Page 41 and 42: %__________________________________
- Page 43 and 44: ||y 1 (ω)|| [m/N] Phase [ o] Excit
- Page 45 and 46: ||y i (ω)|| [m/N] Phase [ o] 0.8 0
- Page 47 and 48: Imag(y i (ω)/f 1 (ω)) (i=1,2) [m/
- Page 49 and 50: 1.8 Mechanical Systems with 3 D.O.F
- Page 51 and 52: which could be verified using Modal
- Page 53 and 54: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
- Page 55 and 56: 1.8.4 Theoretical Frequency Respons
- Page 57 and 58: (a) Amplitude [m/s 2 ] (b) Amplitud
- Page 59 and 60: 4. Vary the number of masses attach
- Page 61 and 62: 1.10 Project 0 - Identification of
- Page 63 and 64: (a) (b) REAL(Acc/force) [(m/s 2 )/N
- Page 65 and 66: 6. Model application - As explained
- Page 67 and 68: changeable unbalanced mass for simu
- Page 69 and 70: %Modal Matrix u with mode shapes %M
Using the information <strong>of</strong> figure 11 (signal in time domain), where<br />
yo = 3.0 · 10 −5 [m/s 2 ] (first peak <strong>of</strong> signal in time domain, figure 11)<br />
yN = y6 = 2.0 · 10 −5 [m/s 2 ] (after 6 peaks)<br />
N = 6,<br />
one gets<br />
• Damping Factor <strong>of</strong> the system ”A” (ξ)<br />
ξ = <br />
• Log Dec<br />
1 +<br />
1<br />
2π·6 ln<br />
1<br />
2π·6 ln<br />
3.0·10 −5<br />
2.0·10 −5<br />
<br />
3.0·10 −5<br />
2.0·10 −5<br />
β = 2π ξ = 2π 0.010 = 0.068<br />
• Equivalent Viscous Damping (d)<br />
<br />
=<br />
2 0.010755<br />
≈ 0.010<br />
1.000058<br />
d = 2 · ξ · ωn · m = 2 · 0.010 · (0.87 · 2 · π) · 0.382 ≈ 0.016 [N · s/m]<br />
(a) Amplitude [m/s 2 ]<br />
(b) Amplitude [m/s 2 ]<br />
2<br />
0<br />
−2<br />
x Signal 10−5<br />
4<br />
(a) in Time Domain − (b) in Frequency Domain<br />
−4<br />
0 5 10 15<br />
time [s]<br />
20 25 30<br />
0.8<br />
0.6<br />
0.4<br />
0.2<br />
x 10−5<br />
1<br />
0<br />
0 5 10 15 20 25<br />
frequency [Hz]<br />
Figure 11: Transient Vibration – Acceleration <strong>of</strong> the clamped-free flexible beam when two concentrated<br />
masses m = m1 +m2 = 0.382 Kg are attached at its free end (L = 0.610 m) – Natural<br />
frequency <strong>of</strong> the mass-spring system ”A”: 0.87 Hz.<br />
20
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