Dynamics of Machines - Part II - IFS.pdf
Dynamics of Machines - Part II - IFS.pdf Dynamics of Machines - Part II - IFS.pdf
equations of motion can really describe the movement of the physical system. After creating the mechanical model for the physical system, the next step is to derive the equation of motion based on the mechanical model. The mechanical model is built by lumped masses m1, m2, m1 (assumption !!!), springs with equivalent stiffness coefficient (calculated using Beam Theory) and dampers with equivalent viscous coefficient (obtained experimentally). While creating the mechanical model and assuming that the mass is a particle, the equation of motion can be derived using Newton’s or Lagrange axioms. For the 3 D.O.F system one can write: M¨y(t) + D˙y(t) + Ky(t) = f(t) (75) or ⎡ ⎣ m11 m12 m13 m21 m21 m23 m31 m32 m33 The mass coefficients ⎫ m11 = m1 + m2 m12 = 0 m13 = 0 m21 = 0 m22 = m3 + m4 m23 = 0 m31 = 0 m32 = 0 m33 = m5 + m6 ⎤⎧ ⎨ ¨y1 ⎦ ¨y2 ⎩ ¨y3 ⎪⎬ ⎪⎭ ⎫ ⎬ ⎭ + ⎡ ⎣ d11 d12 d13 d21 d22 d23 d31 d32 d33 ⎡ + ⎣ ⎤⎧ ⎨ ⎦ ⎩ ˙y1 ˙y2 ˙y3 k11 k12 k13 k21 k22 k23 k31 k32 k33 ⎫ ⎬ ⎭ + ⎤⎧ ⎨ ⎦ ⎩ y1 y2 y3 ⎫ ⎬ ⎭ = ⎧ ⎨ ⎩ f1 f2 f3 ⎫ ⎬ ⎭ ejωt can easily be achieved either by measuring the masses or by having the material density and mass dimensions. The equivalent damping coefficients can be approximated by d11 = 2ξ k11m11 d12 = 0 d13 = 0 d21 = 0 d22 = 2ξ k22m22 d23 = 0 d31 = 0 d32 = 0 d33 = 2ξ ⎫ ⎪⎬ (Approximation!!!) (78) ⎪⎭ k33m33 or by assuming, for example, proportional damping D = αM + βK. The coefficients α and β can be chosen, so that the damping factor ξ of the first resonance is of the same order as the damping factor achieved in the previous section. Please, note that this is just an approximation 50 (76) (77)
which could be verified using Modal Analysis Techniques. Another way of approximating the damping coefficients is given by eq.(78), which should also be verified through experiments! The stiffness coefficients k11 = 3EIL 3 2 (L2 − 4L3) L 3 1 (L1 − L2) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3) k12 = −3EI(−3L2(L2 − 2L3)L3 + L1(L 2 2 − 2L2L3 − 2L 2 3 )) L1(L1 − L2) 2 (L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3) k13 = −9EIL 2 2 L1(L1 − L2)(L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3) k21 = −3EI(−3L2(L2 − 2L3)L3 + L1(L 2 2 − 2L2L3 − 2L 2 3 )) L1(L1 − L2) 2 (L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3) k22 = k23 = k31 = k32 = k33 = 3EI(L1 − 4L3)(L1 − L3) 2 (L1 − L2) 2 (L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3) −3EI(L 2 1 − 2L1L2 − 2L 2 2 − 3L1L3 + 6L2L3) (L1 − L2)(L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3) −9EIL2 2 L1(L1 − L2)(L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3) −3EI(L 2 1 − 2L1L2 − 2L 2 2 − 3L1L3 + 6L2L3) (L1 − L2)(L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3) 3EI(L1 − 4L2) (L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3) can be calculated using the geometry of the beam (I, L1 , L2, L3) and its material properties (E), according to section 1.3. You can try to get these coefficients using the information presented in section 1.5. Differential Equations 2nd order → 1st order M D ¨y 0 M ˙y + 0 K ˙y −M 0 y = ¯f 0 A˙z(t) + Bz(t) = fe jωt z(t) = ˙y(t) y(t) ⎧ ⎪⎨ = ⎪⎩ ˙y1(t) ˙y2(t) ˙y3(t) y1(t) y2(t) y3(t) ⎫ ⎪⎬ ⎪⎭ → velocity → velocity → velocity → displacement → displacement → displacement 51 ⎫ ⎪⎬ ⎪⎭ e jωt (79) (80) (81)
- 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 and 20: 1.6.6 Homogeneous Solution or Free-
- Page 21 and 22: (a) Amplitude [m/s 2 ] x Signal 10
- 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: 1.8 Mechanical Systems with 3 D.O.F
- 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
- Page 71 and 72: acc [m/s 2 ] acc [m/s 2 ] 0.8 0.6 0
which could be verified using Modal Analysis Techniques. Another way <strong>of</strong> approximating the<br />
damping coefficients is given by eq.(78), which should also be verified through experiments!<br />
The stiffness coefficients<br />
k11 =<br />
3EIL 3 2 (L2 − 4L3)<br />
L 3 1 (L1 − L2) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
k12 = −3EI(−3L2(L2 − 2L3)L3 + L1(L 2 2 − 2L2L3 − 2L 2 3 ))<br />
L1(L1 − L2) 2 (L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
k13 =<br />
−9EIL 2 2<br />
L1(L1 − L2)(L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
k21 = −3EI(−3L2(L2 − 2L3)L3 + L1(L 2 2 − 2L2L3 − 2L 2 3 ))<br />
L1(L1 − L2) 2 (L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
k22 =<br />
k23 =<br />
k31 =<br />
k32 =<br />
k33 =<br />
3EI(L1 − 4L3)(L1 − L3) 2<br />
(L1 − L2) 2 (L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
−3EI(L 2 1 − 2L1L2 − 2L 2 2 − 3L1L3 + 6L2L3)<br />
(L1 − L2)(L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
−9EIL2 2<br />
L1(L1 − L2)(L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
−3EI(L 2 1 − 2L1L2 − 2L 2 2 − 3L1L3 + 6L2L3)<br />
(L1 − L2)(L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
3EI(L1 − 4L2)<br />
(L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)<br />
can be calculated using the geometry <strong>of</strong> the beam (I, L1 , L2, L3) and its material properties (E),<br />
according to section 1.3. You can try to get these coefficients using the information presented<br />
in section 1.5.<br />
Differential Equations 2nd order → 1st order<br />
<br />
M D ¨y<br />
0 M ˙y<br />
+<br />
<br />
0 K ˙y<br />
−M 0 y<br />
=<br />
<br />
¯f<br />
0<br />
A˙z(t) + Bz(t) = fe jωt<br />
z(t) =<br />
˙y(t)<br />
y(t)<br />
⎧<br />
⎪⎨<br />
=<br />
⎪⎩<br />
˙y1(t)<br />
˙y2(t)<br />
˙y3(t)<br />
y1(t)<br />
y2(t)<br />
y3(t)<br />
⎫<br />
⎪⎬<br />
⎪⎭<br />
→ velocity<br />
→ velocity<br />
→ velocity<br />
→ displacement<br />
→ displacement<br />
→ displacement<br />
51<br />
<br />
⎫<br />
⎪⎬<br />
⎪⎭<br />
e jωt<br />
(79)<br />
(80)<br />
(81)
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