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Energy methods for damped systems
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.
Potential and Kinetic
Energy The potential energy of mechanical systems 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.
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 = Tmax College of Engineering College of Engineering 3/53 © Eng. Vib, 3rd Ed.
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 x College of Engineering College of Engineering 4/53 © Eng. Vib, 3rd Ed.
5.
Determining the Natural
frequency directly from the energy If the solution is given by x(t)= Asin(ωt+ϕ) then the maximum potential and kinetic energies can be used to calculate the natural frequency 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.
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 2 College of Engineering College of Engineering 6/53 © Eng. Vib, 3rd Ed.
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.
Example 1.4.2 Determine
the equation of motion 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.
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.
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.
Example 1.4.4
The effect of including the mass of the 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.
m
! mass of element dy : s dy # # " assumptions y velocity 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 2 U 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.
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.
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.
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 −U Then 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.
Example 1.4.7 Derive
the equation of motion of 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.
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.
1 2 1
2 2 The 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 + mgsin θ ∂θ ∂θ # 4 & 2 Lagrange’s equation (1.64) yields: d " ∂T % ∂T ∂U k 2 $ '− + = m 2θ + sin θ cosθ + mgsin θ = 0 dt # ∂q & ∂q ∂q 2 College of Engineering College of Engineering 18/53 © Eng. Vib, 3rd Ed.
19.
Does it make
sense: 2 2 k m θ + sin θ cosθ + mgsin θ = 0 2 0 if k=0 Linearize to get small angle case: k 2 m 2θ + θ + mgθ = 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.
Follow me in
twitter @TheSandy36 Hashtag EG-260 College of Engineering College of Engineering 20/44 © Eng. Vib, 3rd Ed.
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.
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 length College of Engineering College of Engineering 22/53 © Eng. Vib, 3rd Ed.
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.
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 Gd Allows the design of springs to have specific stiffness k= 3 College of Engineering 64nR College of Engineering 24/53 © Eng. Vib, 3rd Ed.
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.
Example for a
Heavy Beam Consider again the beam of Figure 1.23 and what happens 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 the From strength of materials the static beam and integrating over deflection of a cantilever beam of the beam length length 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.
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 ( 3y 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.
With the equivalent
mass known the frequency 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.
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.
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.
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.
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 = J College of Engineering College of Engineering 32/53 © Eng. Vib, 3rd Ed.
33.
Combining Springs: Springs
are usually only available in limited stiffness values. Combing them allows other values to be obtained A 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.
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.
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 =0 College of Engineering College of Engineering 35/53 © Eng. Vib, 3rd Ed.
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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 10 Same 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.
Example: Find the
equivalent stiffness k of the following 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 k4 College of Engineering ωn = 37/53 m ( k3 + k 4 ) © Eng. Vib, 3rd Ed.
38.
Example 1.5.5
Compare the natural frequency of two springs connected to a mass in parallel with two in series A series connect of k1 =1000 N/m and k2 =3000 N/m with m = 10 kg yields: 1 750 N/m keq = = 750 N/m ⇒ ω series = = 8.66 rad/s 1 / 1000 + 1 / 3000 10 kg A parallel connect of k1 =1000 N/m and k2 =3000 N/m with m = 10 kg yields: 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.
Static Deflection Another important
consideration in designing with springs is the static 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.
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 dynamically College of Engineering College of Engineering 40/53 © Eng. Vib, 3rd Ed.
41.
Measuring moments of
inertia using 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.
Stiffness Measurements From 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.
Example 1.6.1 Use
the beam stiffness equation to compute the modulus of a material Figure 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.
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.
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.
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.
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.
v0 −ζωnt x(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/s With ω n and v0 fixed at these values, investigate how varies with ζ First peak is highest and occurs at d ( x(t)) = 0 ⇒ ω d e−ζωnt cos(ω d t) − ζω n e−ζωnt sin(ω d t) = 0 dt 1 1 # 1− ζ 2 & −1 ω d −1 Solve 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 −1 Sub 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.
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.
Example 1.7.3 What
happens to a good design when some one changes the parameters? (Car suspension system). How does 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.
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.
Section 1.8 Stability Stability
is defined for the solution of free response case: Stable: x(t) < M, ∀ t > 0 Asymptotically 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.
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.
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 − mgsin θ = 0 ⇒ m 2θ + θ − mgθ = 0 " 2 % 2 ⇒ 2mθ + ( 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.
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