Energy methods for damped systems

17,020 views
16,599 views

Published on

0 Comments
9 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
17,020
On SlideShare
0
From Embeds
0
Number of Embeds
55
Actions
Shares
0
Downloads
317
Comments
0
Likes
9
Embeds 0
No embeds

No notes for slide

Energy methods for damped systems

  1. 1. Section 1.4 Modeling and Energy Methods • Provides an alternative way to determine the equation of motion, and an alternative way to calculate the natural frequency of a system • Useful if the forces or torques acting on the object or mechanical part are difficult to determine • Very useful for more complicated systems to be discussed later (MDOF and distributed mass systems)College of Engineering @ProfAdhikari, #EG260 1/53© Eng. Vib, 3rd Ed.
  2. 2. Potential and Kinetic EnergyThe potential energy of mechanicalsystems U is often stored in “springs” x=0 x0(remember that for a spring F = kx) k x0 x0 M 1 2 U spring = ∫ F dx = ∫ kx dx = kx0 0 0 2 Mass Spring The kinetic energy of mechanical systems T is due to the motion of the “mass” in the system 1 2 1 2  Ttrans = mx , Trot = Jθ 2 2 College of Engineering College of Engineering 2/53 © Eng. Vib, 3rd Ed.
  3. 3. Conservation of Energy For a simple, conservative (i.e. no damper), mass spring system the energy must be conserved: T + U = constant d or (T + U ) = 0 dt At two different times t1 and t2 the increase in potential energy must be equal to a decrease in kinetic energy (or visa- versa). U1 − U 2 = T2 − T1 and U max = TmaxCollege of Engineering College of Engineering 3/53© Eng. Vib, 3rd Ed.
  4. 4. Deriving the equation of motion from the energy x=0 x k M Mass Spring d d !1 2 1 2$  (T +U) = # mx + kx & = 0 dt dt " 2 2 % ⇒ x ( m + kx ) = 0  x  Since x cannot be zero for all time, then m + kx = 0 xCollege of Engineering College of Engineering 4/53© Eng. Vib, 3rd Ed.
  5. 5. Determining the Natural frequency directly from the energyIf the solution is given by x(t)= Asin(ωt+ϕ) then the maximumpotential and kinetic energies can be used to calculate the naturalfrequency of the system 1 2 1 U max = kA Tmax = m(ωn A)2 2 2 Since these two values must be equal 1 2 1 2 kA = m(ωn A) 2 2 2 k ⇒ k = mωn ⇒ ωn = m College of Engineering College of Engineering 5/53 © Eng. Vib, 3rd Ed.
  6. 6. Example 1.4.1 Compute the natural frequency of this roller fixed in place by a spring. Assume it is a conservative system (i.e. no losses) and rolls with out slipping." 1 2 1 2  Trot = Jθ and Ttrans = mx 2 2College of Engineering College of Engineering 6/53© Eng. Vib, 3rd Ed.
  7. 7. Solution continued  x = rθ ⇒ x = rθ⇒T = J  1 x2 Rot 2 r2 The max value of T happens at vmax = ω n A 1 (ω n A)2 1 1" J% 2 ⇒ Tmax = J + m(ω n A)2 = $ m + 2 ω n A 2 2 r2 2 2# r & The max value of U happens at xmax = A 1 ⇒ U max = kA 2 Thus Tmax = U max ⇒ 2 1" J% 2 2 1 2 k $ m + 2 ω n A = kA ⇒ ω n = 2# r & 2 " J% $m + 2 # r & Effective mass College of Engineering College of Engineering 7/53 © Eng. Vib, 3rd Ed.
  8. 8. Example 1.4.2 Determine the equation ofmotion of the pendulum using energy l θ 2 m! J = ml mg College of Engineering College of Engineering 8/53 © Eng. Vib, 3rd Ed.
  9. 9. Now write down the energy 1 2 1 2 2 T = J 0θ = m θ 2 2 U = mg(1− cosθ ), the change in elevation is (1− cosθ ) d d " 1 2 2 % (T +U) = $ m θ + mg(1− cosθ ) = 0 dt dt # 2 &College of Engineering College of Engineering 9/53© Eng. Vib, 3rd Ed.
  10. 10.  2  m θθ + mg(sin θ )θ = 0 ⇒θ  m 2θ + mg(sin θ ) = 0  ( )  2 ⇒ m θ + mg(sin θ ) = 0  g ⇒ θ (t) + sin θ (t) = 0   g g ⇒ θ (t) + θ (t) = 0 ⇒ ωn =  College of Engineering College of Engineering 10/53© Eng. Vib, 3rd Ed.
  11. 11. Example 1.4.4 The effect of including the mass ofthe spring on the value of the frequency. y y +dy m s, k l m x(t) College of Engineering College of Engineering 11/53 © Eng. Vib, 3rd Ed.
  12. 12. m !mass of element dy : s dy #  # " assumptions yvelocity of element dy: vdy = x (t), #  #  $  2 1 ms & y )Tspring =  ∫  (  x+ dy (adds up the KE of each element) 2 0 * 1 , ms / 2 = . 1x  2- 3 0 1 2 & 1 , ms / 1 ) 2 1, ms / 2 2  Tmass = mx ⇒ Ttot = ( . 1 + m+ x ⇒ Tmax = . m + 1ω n A 2 2 - 3 0 2 * 2- 30 1 2U max = kA 2 k ⇒ ωn = • This provides some m m+ s simple design and 3 modeling guides"College of Engineering College of Engineering 12/53© Eng. Vib, 3rd Ed.
  13. 13. What about gravity? kΔ" mg − kΔ = 0, from FBD, and static equilibrium m!k +x(t)" 0" mg 1 U spring = k(Δ + x)2 m! Δ 2 +x(t) U grav = −mgx 1 2  T = mx 2 College of Engineering College of Engineering 13/53 © Eng. Vib, 3rd Ed.
  14. 14. d Now use (T +U) = 0 dt d $1 2 1 2  ⇒ & mx − mgx + k(Δ + x) ) = 0 dt % 2 2 ( ⇒ mx − mgx + k(Δ + x) x x   ⇒ x (m + kx) + x (kΔ − mg) = 0  x     0 from static equilibiurm ⇒ m + kx = 0 x • Gravity does not effect the equation of motion or the natural frequency of the system for a linear system as shown previously with a force balance.College of Engineering College of Engineering© Eng. Vib, 3rd Ed. 14/53
  15. 15. Lagrange’s Method for deriving equations of motion.Again consider a conservative system and its energy.It can be shown that if the Lagrangian L is defined as L = T −UThen the equations of motion can be calculated from d " ∂L % ∂L $ − =0 (1.63)  dt # ∂q & ∂q Which becomes d " ∂T % ∂T ∂U $ − + =0 (1.64)  dt # ∂q & ∂q ∂q College of Engineering College of Engineering 15/53 © Eng. Vib, 3rd Ed. Here q is a generalized coordinate
  16. 16. Example 1.4.7 Derive the equation of motionof a spring mass system via the Lagrangian 1 2 1  and U = kx 2 T = mx 2 2 Here q = x, and and the Lagrangian becomes 1 1  L = T −U = mx 2 − kx 2 2 2 Equation (1.64) becomes d " ∂T % ∂T ∂U d $ − +  = ( mx ) − 0 + kx = 0  dt # ∂x & ∂x ∂x dt ⇒ m + kx = 0 x College of Engineering College of Engineering 16/53 © Eng. Vib, 3rd Ed.
  17. 17. Example l x = sin θ 2  k 2 k  θ 2 h = l (1 − cos θ ) m 1 2 1 2 U = kx + kx + mgl (1 − cos θ ) 2 2 kl 2 = sin θ + mgl (1 − cos θ ) 2 College of Engineering © Eng. Vib, 3rd Ed. 4 College of Engineering 17/53
  18. 18. 1 2 1 2 2The Kinetic energy term is: T = J 0θ = m θ 2 2 Compute the terms in Lagrange’s equation: d " ∂T % d   $ = dt # ∂θ & dt ( m 2θ = m 2θ ) ∂T =0 ∂θ ∂U ∂ " k 2 2 % k 2 = $ sin θ + mg(1− cosθ ) = sin θ cosθ + mgsin θ ∂θ ∂θ # 4 & 2 Lagrange’s equation (1.64) yields: d " ∂T % ∂T ∂U  k 2 $ − + = m 2θ + sin θ cosθ + mgsin θ = 0  dt # ∂q & ∂q ∂q 2College of Engineering College of Engineering 18/53© Eng. Vib, 3rd Ed.
  19. 19. Does it make sense: 2 2  k m θ + sin θ cosθ + mgsin θ = 0 2     0 if k=0Linearize to get small angle case:  k 2 m 2θ + θ + mgθ = 0 2  + " k + 2mg %θ = 0 ⇒θ $ # 2m & k + 2mg ⇒ ωn = 2m What happens College of Engineering College of Engineering if you linearize first? 19/53 © Eng. Vib, 3rd Ed.
  20. 20. Follow me in twitter @TheSandy36 Hashtag EG-260College of Engineering College of Engineering 20/44© Eng. Vib, 3rd Ed.
  21. 21. 1.5 More on springs and stiffness •  Longitudinal motion •  A is the cross sectional area (m2) l k= EA l •  E is the elastic modulus (Pa=N/m2) m" •  l is the length (m) •  k is the stiffness (N/m) x(t)"College of Engineering College of Engineering 21/53© Eng. Vib, 3rd Ed.
  22. 22. Figure 1.21 Torsional Stiffness •  Jp is the polar moment of inertia of the rod GJ p •  J is the mass Jp k= moment of inertia of l the disk 0 •  G is the shear J! θ(t) modulus, l is the lengthCollege of Engineering College of Engineering 22/53© Eng. Vib, 3rd Ed.
  23. 23. Example 1.5.1 compute the frequency of a shaft/mass system {J = 0.5 kg m2} From Equation (1.50)   ∑ M = Jθ ⇒ Jθ (t) + kθ (t) = 0 (t) + k θ (t) = 0 ⇒θ Figure 1.22 J k GJ p πd4 ⇒ ωn = = , Jp = J J 32 For a 2 m steel shaft, diameter of 0.5 cm ⇒ GJ p (8 ×1010 N/m 2 )[π (0.5 ×10 −2 m)4 / 32] ωn = = J (2 m)(0.5kg ⋅ m 2 ) = 2.2 rad/s College of Engineering College of Engineering 23/53 © Eng. Vib, 3rd Ed.
  24. 24. Fig. 1.22 Helical Spring d = diameter of wire 2R= diameter of turns 2R " n = number of turns x(t)= end deflection x(t) " G= shear modulus of spring material" " 4 GdAllows the design of springsto have specific stiffness k= 3 College of Engineering 64nR College of Engineering 24/53 © Eng. Vib, 3rd Ed.
  25. 25. Fig 1.23 Transverse beam stiffness •  Strength of materials f and experiments yield: m 3EI k= 3 l x With a mass at the tip: 3EI ωn =College of Engineering College of Engineering ml 3 25/53© Eng. Vib, 3rd Ed.
  26. 26. Example for a Heavy BeamConsider again the beam of Figure 1.23 and whathappens if the mass of the beam is considered. P = applied static load Much like example 1.4.4 M = mass of beam where the mass of a spring m was considered, the procedure is to calculate the kinetic energy of the beam y x(t) itself, by looking at a differential element of theFrom strength of materials the static beam and integrating overdeflection of a cantilever beam of the beam lengthlength l is: Py 2 x ( y) = ( 3l − y ) 6 EI Pl 3 Which has maximum value of (at x =l ): x max = College of Engineering College of Engineering 3EI © Eng. Vib, 3rd Ed.
  27. 27. Next integrate along the beam to compute the beam’s kinetic energy contribution 1 Tmax = ∫ (mass of differential element)i(velocity of differential)2 2 0 1 " 2 M 2 1 M xmax   % = ∫ # x ( y)$ dy = 6 ∫0 ( 3y 2 − y 3 ) dy 2  2  4  0 Mass of element dy 1 33 * 2 = )  M , xmax 2 ( 140 + 33 Thus the equivalent mass of the beam is: M eq = M 140 And the equivalent mass of the beam- mass system is: 33 msystem = M +m 140 College of Engineering College of Engineering 27/53 © Eng. Vib, 3rd Ed.
  28. 28. With the equivalent mass known thefrequency adjustment for including the mass of the beam becomes 3EI k l3 ωn = = meq 33 m+ M 140 3EI = rad/s ⎛ 33 ⎞ l 3 ⎜ m + M ⎟ ⎝ 140 ⎠College of Engineering College of Engineering 28/53© Eng. Vib, 3rd Ed.
  29. 29. Samples of Vibrating Systems•  Deflection of continuum (beams, plates, bars, etc) such as airplane wings, truck chassis, disc drives, circuit boards…•  Shaft rotation•  Rolling ships•  See the book for more examples.College of Engineering College of Engineering 29/53© Eng. Vib, 3rd Ed.
  30. 30. Example 1.5.2 Effect of fuel on frequency of an airplane wing •  Model wing as transverse beam •  Model fuel as tip mass •  Ignore the mass of the wing and see how the E, I m! frequency of the system changes as the fuel is l used up x(t) "College of Engineering College of Engineering 30/53© Eng. Vib, 3rd Ed.
  31. 31. Mass of pod 10 kg empty 1000 kg full  = 5.2x10-5 m4, E =6.9x109 N/m,  = 2 m•  Hence the natural 3EI 3(6.9 × 10 9 )(5.2 × 10 −5 ) ωfull = = frequency ml 3 1000 ⋅ 23 changes by an = 11.6 rad/s=1.8 Hz order of 3EI 3(6.9 × 10 9 )(5.2 × 10 −5 ) magnitude ωempty = = while it ml 3 10 ⋅ 23 empties out = 115 rad/s=18.5 Hz fuel. This ignores the mass of the wing College of Engineering College of Engineering 31/53 © Eng. Vib, 3rd Ed.
  32. 32. Example 1.5.3 Rolling motion of a ship  Jθ (t) = −W GZ = −Whsin θ (t) For small angles this becomes  Jθ (t) + Whθ (t) = 0 hW ⇒ ωn = JCollege of Engineering College of Engineering 32/53© Eng. Vib, 3rd Ed.
  33. 33. Combining Springs: Springs are usually onlyavailable in limited stiffness values. Combing them allows other values to be obtainedA k1 B k2 C •  Equivalent Spring 1 series: k AC = k1 1 1 a b + k1 k2 k2 parallel: kab = k1 + k2 This is identical to the combination of capacitors in electrical circuits College of Engineering College of Engineering 33/53 © Eng. Vib, 3rd Ed.
  34. 34. Use these to design from available parts •  Discrete springs available in standard values •  Dynamic requirements require specific frequencies •  Mass is often fixed or + small amount •  Use spring combinations to adjust ωn •  Check static deflection College of Engineering College of Engineering 34/53 © Eng. Vib, 3rd Ed.
  35. 35. Example 1.5.5 Design of a spring mass system using available springs: series vs parallel k2 •  Let m = 10 kg k1 •  Compare a series and m parallel combination •  a) k1 =1000 N/m, k2 = 3000 k3 N/m, k3 = k4 =0 •  b) k3 =1000 N/m, k4 = 3000 k4 N/m, k1 = k2 =0College of Engineering College of Engineering 35/53© Eng. Vib, 3rd Ed.
  36. 36. Case a) parallel connection: k3 = k4 = 0, keq = k1 + k2 = 1000 + 3000 = 4000 N/m keg 4000 ⇒ ω parallel = = = 20 rad/s m 10 Case b) series connection: 1 3000 k1 = k2 = 0, keq = = = 750 N/m (1 k3 ) + (1 k4 ) 3 + 1 keg 750 ⇒ ωseries = = = 8.66 rad/s m 10Same physical components, very different frequency"Allows some design flexibility in using off the shelf components" College of Engineering College of Engineering 36/53 © Eng. Vib, 3rd Ed.
  37. 37. Example: Find the equivalent stiffness k of thefollowing system (Fig 1.26, page 47) k 3 k4 k1+k2+k5 k1 + k2 + k5 + k3 + k 4 k1 k2 m m m k3 k3 1 k3 k4 = = k5 k4 1 1 k3 + k4 + k3 k4 k4 k1k3 + k2 k3 + k5 k3 + k1k4 + k2 k4 + k5 k4 + k3 k4College of Engineering ωn = 37/53 m ( k3 + k 4 )© Eng. Vib, 3rd Ed.
  38. 38. Example 1.5.5 Compare the naturalfrequency of two springs connected to a mass inparallel with two in seriesA series connect of k1 =1000 N/m and k2 =3000 N/m with m = 10 kgyields: 1 750 N/m keq = = 750 N/m ⇒ ω series = = 8.66 rad/s 1 / 1000 + 1 / 3000 10 kgA parallel connect of k1 =1000 N/m and k2 =3000 N/m with m = 10 kgyields: 4000 N/m keg = 1000 N/m + 3000 N/m = 4000 N/m ⇒ ω par = = 20 rad/s 10 kg Same components, very different frequency College of Engineering College of Engineering 38/53 © Eng. Vib, 3rd Ed.
  39. 39. Static DeflectionAnother important consideration in designing with springs is thestatic deflection mg Δk = mg ⇒ Δ = k This determines how much a spring compresses or sags due to the static mass (you can see this when you jack your car up) The other concern is “rattle space” which is the maximum deflection A College of Engineering College of Engineering 39/53 © Eng. Vib, 3rd Ed.
  40. 40. Section 1.6 Measurement•  Mass: usually pretty easy to measure using a balance- a static experiment•  Stiffness: again can be measured statically by using a simple displacement measurement and knowing the applied force•  Damping: can only be measured dynamicallyCollege of Engineering College of Engineering 40/53© Eng. Vib, 3rd Ed.
  41. 41. Measuring moments of inertiausing a Trifilar suspension system gT 2r02 ( m0 + m ) J= − J0 4π l 2 T is the measured period g is the acceleration due to gravity College of Engineering College of Engineering 41/53 © Eng. Vib, 3rd Ed.
  42. 42. Stiffness MeasurementsFrom Static Deflection: Force or stress Linear Nonlinear F = k x or σ = E ε F ⇒k= x Deflection or strain From Dynamic Frequency: k 2 ωn = ⇒ k = m ωn m College of Engineering College of Engineering 42/53 © Eng. Vib, 3rd Ed.
  43. 43. Example 1.6.1 Use the beam stiffnessequation to compute the modulus of a materialFigure 1.24  = 1 m, m = 6 kg, I = 10-9 m4 , and measured T = 0.62 s ml 3 T = 2π = 0.62 s 3EI 3 4π ml 2 3 4π 2 ( 6 kg )(1 m ) ⇒E= = = 2.05 × 1011 N/m 2 3T 2 I ( 3(0.62 s)2 10 −9 m 4 ) College of Engineering College of Engineering 43/53 © Eng. Vib, 3rd Ed.
  44. 44. Damping Measurement (Dynamic only)Define the Logarithmic Decrement: x(t) δ = ln (1.71) x(t + T ) Ae−ζω n t sin(ω d t + φ ) δ = ln −ζω n (t +T ) Ae sin(ω d t + ω dT ) + φ ) (1.72) δ = ζω nT c δ δ ζ= = = 2 2 ccr ωnT 4π + δ (1.75)College of Engineering College of Engineering 44/53© Eng. Vib, 3rd Ed.
  45. 45. Section 1.7: Design Considerations Using the analysis so far to guide the selection of components.College of Engineering College of Engineering 45/53© Eng. Vib, 3rd Ed.
  46. 46. Example 1.7.1 •  Mass 2 kg < m < 3 kg and k > 200 N/m •  For a possible frequency range of 8.16 rad/s < ωn < 10 rad/s •  For initial conditions: x0 = 0, v0 < 300 mm/s •  Choose a c so response is always < 25 mm College of Engineering College of Engineering 46/53 © Eng. Vib, 3rd Ed.
  47. 47. Solution:•  Write down x(t) for 0 initial displacement•  Look for max 1 amplitude 0.5•  Occurs at time of first Amplitude peak (Tmax) 0•  Compute the amplitude at Tmax -0.5•  Compute ζ for A(Tmax)=0.025 -1 0 0.5 1 1.5 2 Time(sec) College of Engineering College of Engineering 47/53 © Eng. Vib, 3rd Ed.
  48. 48. v0 −ζωntx(t) = e sin(ω d t) ωd     Amplitude⇒ worst case happens at smallest ω d ⇒ ω n = 8.16 rad/s⇒ worst case happens at max v0 = 300 mm/sWith ω n and v0 fixed at these values, investigate how varies with ζFirst peak is highest and occurs atd ( x(t)) = 0 ⇒ ω d e−ζωnt cos(ω d t) − ζω n e−ζωnt sin(ω d t) = 0dt 1 1 # 1− ζ 2 & −1 ω d −1Solve for t = Tmax ⇒ Tm = tan ( )= tan % % ζ ( ( ωd ζω n ω d $ ζ 1−ζ 2 v0 − tan −1 ( ζ ) # 1− ζ 2 & 1−ζ 2 −1Sub Tmax into x(t) : Am (ζ ) = x(Tm ) = e sin(tan % % ζ () ( 2 ω n 1− ζ $ ζ 1−ζ 2 − tan −1 ( ) v 1−ζ 2 ζ Am (ζ ) = 0 e ωn College of Engineering College of Engineering 48/53 © Eng. Vib, 3rd Ed.
  49. 49. To keep the max value less then 0.025 m solve Amax (ζ ) = 0.025 ⇒ ζ = 0.281 Using the upper limit on the mass (m = 3 kg) yields c = 2mω nζ = 2 ⋅ 3⋅ 8.16 ⋅ 0.281= 14.15 kg/s v0 FYI, ζ = 0 yields Amax = = 37 mm ωn College of Engineering College of Engineering 49/53 © Eng. Vib, 3rd Ed.€
  50. 50. Example 1.7.3 What happens to a good design whensome one changes the parameters? (Car suspension system). Howdoes ζ change with mass? Given ζ =1, m=1361 kg, Δ=0.05 m, compute c, k . k 2 mg ωn = ⇒ k = 1361ω n , mg = kΔ ⇒ k = m Δ mg 9.81 ⇒ ωn = = = 14 rad/s ⇒ mΔ 0.05 k = 1361(14)2 = 2.668 × 10 5 N/m ζ =1 ⇒ c = 2mω n = 2(1361)(14) = 3.81 × 10 4 kg/s College of Engineering College of Engineering 50/53 © Eng. Vib, 3rd Ed.
  51. 51. Now add 290 kg of passengers and luggage. What happens? m = 1361 + 290 = 1651 kg mg 1651⋅ 9.8 ⇒Δ= = 5 ≈ 0.06 m k 2.668 × 10 g 9.8 ⇒ ωn = = = 12.7 rad/s Δ 0.06 c 3.81 × 10 4 So some oscillation" ζ= = = 0.9 results at a lower" ccr 2mω n frequency."College of Engineering College of Engineering 51/53© Eng. Vib, 3rd Ed.
  52. 52. Section 1.8 StabilityStability is defined for the solution of freeresponse case:Stable: x(t) < M, ∀ t > 0Asymptotically Stable: lim x(t) = 0  t→∞Unstable:if it is not stable or asymptotically stable College of Engineering College of Engineering 52/53 © Eng. Vib, 3rd Ed.
  53. 53. Examples of the types of stability Stable Asymptotically Stable x(t) x(t) t t x(t) x(t) t t Divergent instability Flutter instability College of Engineering College of Engineering 53/53 © Eng. Vib, 3rd Ed.
  54. 54. Example: 1.8.1: For what values of the spring constant will the response be stable? Figure 1.37 ! k 2 $ k 2  m 2θ + #  sin θ & cosθ − mgsin θ = 0 ⇒ m 2θ + θ − mgθ = 0 " 2 % 2  ⇒ 2mθ + ( k − 2mg)θ = 0 (for small θ ) ⇒ k l > 2mg for a stable response College of Engineering College of Engineering 54/53 © Eng. Vib, 3rd Ed.

×