Engineering Vibration 3rd Edition Prentice Hall © D. J. Inman
[email protected], We make our living in dynamics, vibration and control
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Motivation The Millennium Bridge a Recent Vibration Problem (2000-1)
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OPENED AND CLOSED WITHIN A VIEW HOURS BECAUSE OF UNDESIRABLE VIBRATION (Show movies)
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Modeling and Degrees of Freedom The bridge has many degrees of freedom, we will start with one and work towards many.
Lack of consideration of dynamic loads and Vibration caused this to new bridge to vibrate wildly Virginia Tech © D. J. Inman
The goal of this course is to understand such 3/43 phenomenon and how to prevent it
Example 1.1.1 The Pendulum • Sketch the structure or part of interest • Write down all the forces and make a “free body diagram” • Use Newton’s Law and/or Euler’s Law to find the equations of motion
∑M
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0
= J 0α , J 0 = ml
2
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The problem is one dimensional, hence a scalar equation results J 0α (t ) = −mgl sin θ (t ) ⇒ m 2θ (t ) + mg sin θ (t ) = 0 restoring force
Here the over dots denote differentiation with respect to time t This is a second order, nonlinear ordinary differential equation We can linearize the equation by using the approximation sin θ ≈ θ
g
⇒ m θ (t ) + mg θ (t ) = 0 ⇒ θ (t ) + θ (t ) = 0 2
Requires knowledge of θ (0) and θ (0) Virginia Tech © D. J. Inman
the initial position and velocity.
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Next consider a spring mass system and perform a static experiment: • From strength of materials recall: FBD:
nonlinear
linear A plot of force versus displacement: Virginia Tech © D. J. Inman
experiment ⇒ fk = kx
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Free-body diagram and equation of motion
•Newton’s Law:
mx(t ) = −kx(t ) ⇒ mx(t ) + kx(t ) = 0 (1.2) x(0) = x0 , x(0) = v0 Virginia Tech © D. J. Inman
Again a 2nd order ordinary differential equation
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Stiffness and Mass Vibration is cause by the interaction of two different forces one related to position (stiffness) and one related to acceleration (mass). Proportional to displacement
Stiffness (k)
Displacement x
fk = −kx(t)
k
statics
m
Mass (m)
f m = ma (t ) = mx(t )
Mass
Spring
dynamics
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Proportional to acceleration
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Examples of Single-Degree-ofFreedom Systems Pendulum
Shaft and Disk
l =length θ Gravity g
Torsional Stiffness k
m θ
g θ (t) + θ (t) = 0 l Virginia Tech © D. J. Inman
Moment of inertia J
Jθ (t ) + kθ (t ) = 0 9/43
Solution to 2nd order DEs Lets assume a solution:
x(t)
x(t) = Asin(ω n t + φ )
t
Differentiating twice gives:
x(t ) = ωn A cos(ωn t + φ ) x(t ) = −ωn2 A sin(ωnt + φ ) = -ωn2 x(t ) Substituting back into the equations of motion gives:
−mω n2 Asin(ω nt + φ ) + kAsin(ω nt + φ ) = 0 −mω + k = 0 2 n
Virginia Tech © D. J. Inman
or
ωn =
k m
Natural frequency rad/s
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Summary of simple harmonic motion x(t)
Period
T=
x0
2π
ωn
Amplitude A
Slope here is v0
φ
t Maximum Velocity
ωn
ωn A
ω n rad/s ω n cycles ω n fn = = = Hz 2π rad/cycle 2π s 2π Virginia Tech © D. J. Inman
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Initial Conditions If a system is vibrating then we must assume that something must have (in the past) transferred energy into to the system and caused it to move. For example the mass could have been: •moved a distance x0 and then released at t = 0 (i.e. given Potential energy) or •given an initial velocity v0 (i.e. given Kinetic energy) or •Some combination of the two above cases From our earlier solution we know that:
x0 = x(0) = A sin(ωn 0 + φ ) = A sin(φ ) v0 = x(0) = ωn A cos(ωn 0 + φ ) = ωn A cos(φ ) Virginia Tech © D. J. Inman
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Initial Conditions Solving these equation gives:
A=
1
ωn
⎛ ωn x0 ⎞ ω x + v , φ = tan ⎜ ⎟ ⎝ v0 ⎠ 2 2 n 0
−1
2 0
Amplitude
Phase
Slope here is v0
x(t)
1
ωn
x0
ω n2 x02 + v02 x0
t
φ Virginia Tech © D. J. Inman
φ
ωn
v0
ωn
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The total solution for the spring mass system is ω n2 x02 + v02 ⎛ −1 ω n x0 ⎞ x(t) = sin ⎜ ω nt + tan ωn v0 ⎟⎠ ⎝
(1.10)
Called the solution to a simple harmonic oscillator and describes oscillatory motion, or simple harmonic motion. Note (Example 1.1.2)
ω n2 x02 + v02 x(0) = ωn Virginia Tech © D. J. Inman
ω n x0 ω x +v 2 2 n 0
as it should
2 0
= x0
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A note on arctangents •
•
Note that calculating arctangent from a calculator requires some attention. First, all machines work in radians. The argument atan(-/+) is in a different quadrant then atan(+/-), and usual machine calculations will return an arctangent in between -π/2 and +π/2, reading only the atan(-) for both of the above two cases. +
+ φ
-
+ -
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_ φ
+
In MATLAB, use the atan2(x,y) function to get 15/43 the correct phase.
Example 1.1.3 wheel, tire suspension m=30 kg, ωn= 10 hz, what is k? ⎛ cylce 2π rad ⎞ 5 i = × k = mω = ( 30 kg ) ⎜10 1.184 10 N/m ⎟ sec cylce ⎠ ⎝ 2 n
There are of course more complex models of suspension systems and these appear latter in the course as our tools develope
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Section 1.2 Harmonic Motion The period is the time elapsed to complete one complete cylce
2π rad 2π T= = s ω n rad/s ω n
(1.11)
The natural frequency in the commonly used units of hertz:
ωn ω n rad/s ω n cycles ω n fn = = = = Hz 2π 2π rad/cycle 2π 2π s For the pendulum:
ωn =
(1.12)
g l rad/s, T = 2π s l g
For the disk and shaft:
ωn = Virginia Tech © D. J. Inman
k J rad/s, T = 2π s J k
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Relationship between Displacement, Velocity and Acceleration 1
x(t) = Asin(ω nt + φ )
x
Displacement
0
-1 0 20
x(t) = ω n A cos(ω nt + φ )
v
Velocity
a
x(t) = −ω n2 Asin(ω nt + φ )
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1
0.2
0.3
0.4
0.5 0.6 Time (sec)
0.7
0.8
0.9
1
0
-20 0 200
Acceleration
0 -200 0
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A=1, ωn=12
Note how the relative magnitude of each increases for ωn>1
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Example 1.2.1Hardware store spring, bolt: m= 49.2x10
-3
kg, k=857.8 N/m and x0 =10 mm. Compute ωn and the max amplitude of vibration.
k 857.8 N/m ωn = = = 132 rad/s -3 m 49.2 ×10 kg
ωn = 21 Hz fn = 2π 2π 1 1 T= = = ωn fn 21 cyles x(t) max = A = Virginia Tech © D. J. Inman
1
ωn
Note: common Units are Hertz
To avoid Costly errors use fn when working in Hertz and ωn when in rad/s
0.0476 s sec 0
ω x + v = x0 = 10 mm 2 2 n 0
2 0
Units depend on system 19/43
Compute the solution and max velocity and acceleration
v(t)max = ωn A = 1320 mm/s = 1.32 m/s a(t) max = ω n A = 174.24 ×10 mm/s 2
3
2.92 mph
2
= 174.24 m/s ≈ 17.8g! g = 9.8 m/s π −1 ⎛ω n x0 ⎞ 90° φ = tan ⎜ ⎟ = rad ⎝ 0 ⎠ 2 x(t) = 10 sin(132t + π / 2) = 10 cos(132t) mm 2
2
~0.4 in max Virginia Tech © D. J. Inman
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Does gravity matter in spring problems? Let Δ be the deflection caused by hanging a mass on a spring (Δ = x1-x0 in the figure) Then from static equilibrium:
mg = kΔ
Next sum the forces in the vertical for some point x > x1 measured from Δ:
mx = −k ( x + Δ ) + mg = −kx + mg − k Δ =0
⇒ mx(t ) + kx(t ) = 0 So no, gravity does not have an effect on the vibration Virginia Tech © D. J. Inman
(note that this is not the case if the spring is nonlinear)
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Example 1.2.2 Pendulums and measuring g • A 2 m pendulum swings with a period of 2.893 s • What is the acceleration due to gravity at that location?
4π 2 4π 2 g= 2 l = 2m 2 2 T 2.893 s 2 ⇒ g = 9.434 m/s
This is g in Denver, CO USA, at 1638m and a latitude of 40° Virginia Tech © D. J. Inman
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Review of Complex Numbers and Complex Exponential (See Appendix A) A complex number can be written with a real and imaginary part or as a complex exponential
c = a + jb = Ae jθ Where
a = A cosθ ,b = Asin θ
ℑ a
Multiplying two complex numbers:
c1c2 = A1 A2 e
j (θ1 + θ 2 )
Dividing two complex numbers:
A
b
ℜ
c1 A1 j (θ1 −θ2 ) = e c2 A2 Virginia Tech © D. J. Inman
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Equivalent Solutions to 2nd order Differential Equations (see Window 1.4) All of the following solutions are equivalent:
x(t) = Asin(ω nt + φ )
Called the magnitude and phase form
x(t) = A1 sin ω nt + A2 cos ω nt x(t) = a1e jω n t + a2 e− jω n t
Sometimes called the Cartesian form
Called the polar form
The relationships between A and φ, A1 and A2, and a1 and a2 can be found in Window 1.4 of the course text, page 17.
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•Each is useful in different situations •Each represents the same information •Each solves the equation of motion 24/43
Derivation of the solution Substitute x(t ) = aeλt into mx + kx = 0 ⇒ mλ 2 aeλt + kaeλt = 0 ⇒ mλ + k = 0 ⇒ 2
k k λ = ± − =± j = ±ωn j ⇒ m m −ωn jt ωn jt ⇒ x(t ) = a1e and x(t ) = a2 e ωn jt
x(t ) = a1e
+ a2 e
−ωn jt
(1.18)
This approach will be used again for more complicated problems Virginia Tech © D. J. Inman
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Is frequency always positive? From the preceding analysis, λ = + ωn then
ω n jt
x(t) = a1e
+ a2 e
− ω n jt
Using the Euler relations for trigonometric functions, the above solution can be written as (recall Window 1.4)
x(t) = Asin (ω nt + φ )
(1.19)
It is in this form that we identify as the natural frequency ωn and this is positive, the + sign being used up in the transformation from exponentials to the sine function. Virginia Tech © D. J. Inman
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Calculating RMS May need to be limited due to physical constraints
A = peak value T
Not very useful since for a sine function the average value is zero
1 x = lim ∫ x(t)dt = average value T →∞ T 0 T
1 2 2 x = lim ∫ x (t)dt = mean-square value T →∞ T 0 xrms = x 2 = root mean square value
(1.21) Proportional to energy
Also useful when the vibration is random Virginia Tech © D. J. Inman
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The Decibel or dB scale It is often useful to use a logarithmic scale to plot vibration levels (or noise levels). One such scale is called the decibel or dB scale. The dB scale is always relative to some reference value x0. It is define as: 2
⎛ x⎞ ⎛ x⎞ dB = 10 log10 ⎜ ⎟ = 20 log10 ⎜ ⎟ ⎝ x0 ⎠ ⎝ x0 ⎠
(1.22)
For example: if an acceleration value was 19.6m/s2 then relative to 1g (or 9.8m/s2) the level would be 6dB, 2
⎛ 19.6 ⎞ 10 log10 ⎜ = 20 log10 (2 ) = 6dB ⎟ ⎝ 9.8 ⎠ Virginia Tech © D. J. Inman
Or for Example 1.2.1: The Acceleration Magnitude 28/43 is 20log10(17.8)=25dB relative to 1g.
1.3 Viscous Damping All real systems dissipate energy when they vibrate. To account for this we must consider damping. The most simple form of damping (from a mathematical point of view) is called viscous damping. A viscous damper (or dashpot) produces a force that is proportional to velocity. Mostly a mathematically motivated form, allowing a solution to the resulting equations of motion that predicts reasonable (observed) amounts of energy dissipation.
Damper (c)
f c = −cv(t ) = −cx(t )
x
fc Virginia Tech © D. J. Inman
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Differential Equation Including Damping For this damped single degree of freedom system the force acting on the mass is due to the spring and the dashpot i.e. fm= - fk - fc. Displacement x
mx(t ) = −kx(t ) − cx(t ) M or mx(t ) + cx(t ) + kx(t ) = 0 (1.25)
k
c
To solve this for of the equation it is useful to assume a solution of the form:
x(t) = ae
Virginia Tech © D. J. Inman
λt
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Solution to DE with damping included (dates to 1743 by Euler) The velocity and acceleration can then be calculated as: x(t) = λae λt x(t) = λ2 ae λt
If this is substituted into the equation of motion we get:
aeλt (mλ 2 + cλ + k) = 0 (1.26) Divide equation by m, substitute for natural frequency and assume a non-trivial solution
λt
ae ≠ 0 Virginia Tech © D. J. Inman
⇒
(λ + cλ 2
m
+ω ) = 0 2 n
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Solution to DE with Damping Included For convenience we will define a term known as the damping ratio as:
c ζ= 2 km
(1.30)
Lower case Greek zeta
The equation of motion then becomes:
(λ + 2ζωn λ + ω ) = 0 2
2 n
Solving for λ then gives,
λ1,2 = −ζω n ± ω n ζ 2 − 1 Virginia Tech © D. J. Inman
(1.31) 32/43
Possibility 1. Critically damped motion Critical damping occurs when ζ=1. The damping coefficient c in this case is given by:
ζ =1 ⇒ c = ccr = 2 km = 2mωn Solving for λ then gives,
definition of critical damping coefficient
λ1,2 = −1ωn ± ωn 1 −1 = −ωn 2
The solution then takes the form
A repeated, real root
x(t) = a1e−ω n t + a2te−ω n t Virginia Tech © D. J. Inman
Needs two independent solutions, hence the t in the second term 33/43
Critically damped motion a1 and a2 can be calculated from initial conditions (t=0),
x = (a1 + a2t)e−ω n t k=225N/m m=100kg and
⇒ a1 = x0 v0 = −ω n a1 + a2 ⇒ a2 = v0 + ω n x0 • •
No oscillation occurs Useful in door mechanisms, analog gauges
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0.6 x0=0.4mm v0=1mm/s
0.5 Displacement (mm)
v = (−ω n a1 − ω n a2t + a2 )e−ω n t
ζ=1
x0=0.4mm v0=0mm/s x0=0.4mm v0=-1mm/s
0.4 0.3 0.2 0.1 0 -0.1
0
1
2 Time (sec)
3
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4
Possibility 2: Overdamped motion An overdamped case occurs when ζ>1. Both of the roots of the equation are again real.
k=225N/m m=100kg andz=2
0.6
λ1,2 = −ζωn ± ωn ζ −1 2
ζ −1
+ a2 eω n t
ζ −1 2
a1 and a2 can again be calculated from initial conditions (t=0),
−v0 + ( − ζ + ζ 2 −1)ωn x0 a1 = 2ωn ζ 2 −1 v0 + (ζ + ζ −1)ωn x0 a2 = 2ωn ζ 2 −1 2
Virginia Tech © D. J. Inman
0.5
)
Displacement (mm)
x(t) = e−ζωn t (a1e−ωn t
2
x0=0.4mm v0=1mm/s x0=0.4mm v0=0mm/s x0=0.4mm v0=-1mm/s
0.4 0.3 0.2 0.1 0 -0.1 0
1
2 Time (sec)
Slower to respond than critically damped case
3
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4
Possibility 3: Underdamped motion An underdamped case occurs when ζ
