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07_Frequency-Domain Analysis(SDOF)

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Harmonic Excitation of Undamped Systems
? Consider the usual spring mass damper system with applied force F(t)=F0cosωt ω ? ω is the driving frequency ? F0 is the magnitude of the applied force ? We take c = 0 to start with
Displacement x F=F0cosω t

M

k

Equations of motion
? Solution is the sum of homogenous and particular solution ? The particular solution assumes form of forcing function (physically the ?? mx(t ) = ?kx(t ) + F0 cos(ωt ) input wins):
2 ??(t ) + ωn x(t ) = f 0 cos(ωt ) x

where f 0 =

F0

m

,

ωn = k m

Substitute particular solution into the equation of motion:
??p x
2 ωn x p

2 ?ω 2 X cos ωt + ωn X cos ωt = f 0 cos ωt

f0 solving yields: X = 2 ωn ? ω 2
Thus the particular solution has the form:

f0 x p (t) = 2 cos(ω t) 2 ωn ? ω

Add particular and homogeneous solutions to get general solution:
x(t ) =
particular

f0 A1 sin ωnt + A2 cos ωn t + 2 cos ωt 2 ωn ? ω
homogeneous

A1 and A2 are constants of integration.

Apply the initial conditions to evaluate the constants
f0 f0 x(0) = A1 sin 0 + A2 cos 0 + 2 cos 0 = A2 + 2 = x0 2 2 ωn ? ω ωn ? ω ? A2 = x0 ? ? x(0) = ωn ( A1 cos 0 ? A2 sin 0) ? f0 2 ωn ? ω 2

f0 sin 0 = ωn A1 = v0 2 2 ωn ? ω

? A1 = x(t ) = v0

ωn

v0

?

? ? f f sin ωnt + ? x0 ? 2 0 2 ? cos ωnt + 2 0 2 cos ωt ωn ? ω ? ωn ? ω ωn ?

Comparison of free and forced response
? Sum of two harmonic terms of different frequency ? Free response has amplitude and phase effected by forcing function ? Our solution is not defined for ωn = ω because it produces division by 0. ? If forcing frequency is close to natural frequency the amplitude of particular solution is very large

Response for m=100 kg, k=1000 N/m, F=100 N, ω = ωn +5 v0=0.1m/s and x0= -0.02 m.
0.05 Displacement (x)

0

-0.05

0

2

4 6 Time (sec)

8

10

Note the obvious presence of two harmonic signals

What happens when ω is near ωn?
x(t ) = 2 f0 ? ω ? ω ? ? ωn + ω ? t ? sin ? t? sin ? n 2 2 ωn ? ω ? 2 ? ? 2 ? (2.13)
2 f0 ? ω ?ω ? sin ? n t? 2 ωn ? ω 2 2 ? ?
When the drive frequency and natural frequency are close a beating phenomena occurs

1 Displacement (x) 0.5 0

Larger amplitude
-0.5 -1

0

5

10

15 Time (sec)

20

25

30

x p (t) = tX sin(ω t) f X= 0 2ω

What happens when ω is ωn?
grows with out bound

substitute into eq. and solve for X

f0 x(t ) = A1 sin ωt + A2 cos ωt + t sin(ωt ) 2ω
5

When the drive frequency and natural frequency are the same the amplitude of the vibration grows without bounds. This is known as a resonance condition. The most important concept in Chapter 2!

Displacement (x)

0

-5 0 5 10 15 Time (sec) 20 25 30

Harmonic Excitation of Damped Systems

? Extending resonance and response calculation to damped systems

Harmonic excitation of damped systems
?? ? mx(t ) + cx(t ) + kx(t ) = F0 cos ωt ??(t ) + 2ζωn x(t ) + ω x(t ) = f 0 cos ωt ? x
2 n

x p (t ) = X cos (ωt + θ )
now includes a phase shift

Displacement x F=F0cosωt

k

M
c

Let xp have the form:

x p (t ) = As cos ωt + Bs sin ωt ? Bs ? X = As + Bs , θ = tan ? ? ? As ? ? x p = ?ω As sin ωt + ω Bs cos ωt
2 2 ?1

??p = ?ω As cos ωt ? ω Bs sin ωt x
2 2
Note that we are using the rectangular form, but we could use one of the other forms of the solution.

Substitute into the equations of motion
(?ω As + 2ζω nωBs + ω As ? f 0 ) cos ωt
2

+ ? ω Bs + 2ζω nωAs + ω Bs sin ωt = 0
2 2 n

for all time. Specifically for t = 0,2π / ω ? (ω ? ω ) As + (2ζω nω ) Bs = f 0
2 n 2

(

2 n

)

(?2ζω nω ) As + (ω ? ω ) Bs = 0
2 n 2

Write as a matrix equation:
2 ?(ωn ? ω 2 ) 2ζωnω ? ? As ? ? f 0 ? ? ?=?0? 2 2 ?? ? ?2ζωnω (ωn ? ω ) ? ? Bs ? ? ?

Solving for As and Bs:

(ω ? ω ) f 0 As = 2 2 (ωn ? ω ) + (2ζωnω )
2 n 2 2 2

2ζω nω f0 Bs = 2 2 2 2 (ω n ? ω ) + (2ζω nω )

Substitute the values of As and Bs into xp:
? 2ζωnω ? x p (t ) = cos(ωt ? tan ? 2 ) 2 ? 2 (ωn ? ω 2 ) 2 + (2ζωnω ) 2 ? ωn ? ω ? f0
X ?1

θ

Add homogeneous and particular to get total solution:

x(t ) = Ae ?ζωnt sin(ωd t + φ ) + X cos(ωt ? θ )
homogeneous or transient solution particular or steady state solution

Note: that A and φ will not have the same values as in Ch 1, for the free response. Also as t gets large, transient dies out.

Things to notice about damped forced response
? If ζ = 0, undamped equations result ? Steady state solution prevails for large t ? Often we ignore the transient term (how large is ζ, how long is t?) ? Coefficients of transient terms (constants of integration) are effected by the initial conditions AND the forcing function ? For underdamped systems at resonance the, amplitude is finite.

Proceeding with ignoring the transient
? Always check to make sure the transient is not significant ? For example, transients are very important in earthquakes ? However, in many machine applications transients may be ignored

Proceeding with ignoring the transient
Magnitude:

X=

f0
2 (ωn ? ω 2 ) 2 + (2ζωnω ) 2

Frequency ratio:

ω r= ωn
2 1 Xk X ωn = = F0 f0 (1 ? r 2 )2 + (2ζ r ) 2

Non dimensional Form:

Phase:

θ = tan ?1 2ζ r

(

1 ? r2

)

Magnitude plot
? Resonance is close to r = 1 ? For ζ = 0, r =1 defines resonance ? As ζ grows resonance moves r <1, and X decreases ? The exact value of r, can be found from differentiating the magnitude

X=
40 30 20 X (dB) 10 0 -10 -20 0

1 (1 ? r 2 )2 + (2ζ r)2
ζ =0.01 ζ =0.1 ζ =0.3 ζ =0.5 ζ =1

0.5

1 r

1.5

2

Phase plot
? Resonance occurs at φ = π/2 ? The phase changes more rapidly when the damping is small ? From low to high values of r the phase always changes by 1800 or π radians

θ = tan ?1 ? 2ζr ?
?
3.5 3 2.5

? ? 1? r ?
2

Phase (rad)

2 1.5 1 0.5 0 0

ζ =0.01 ζ =0.1 ζ =0.3 ζ =0.5 ζ =1

0.5

1

1.5

2

r

Base Excitation
? Important class of vibration analysis – Preventing excitations from passing from a vibrating base through its mount into a structure ? Vibration isolation – Vibrations in your car – Satellite operation – Disk drives, etc.

FBD of SDOF Base Excitation
System Sketch
x(t)

System FBD
m

m
k c
? ? k ( x ? y ) c(x ? y)

y(t)

base

? ? ?? ∑ F =-k (x-y)-c(x-y)=mx ?? ? ? mx +cx + kx = cy + ky (2.61)

SDOF Base Excitation (cont)
Assume: y(t ) = Y sin(ωt ) and plug into Equation(2.61) ?? ? mx+cx + kx = cωY cos(ωt ) + kY sin(ωt ) (2.63)
For a car,
ω=


harmonic forcing functions
= 2πV

τ

λ

The steady-state solution is just the superposition of the two individual particular solutions (system is linear).
f0 c f0 s 2 2 ??+2ζωn x + ωn x = 2ζωnωY cos(ωt ) + ωn Y sin(ωt ) ? x

(2.64)

Particular Solution (sine term)
With a sine for the forcing function,
2 ??+2ζωn x + ωn x =f0 s sin ωt ? x

x ps = As cos ωt + Bs sin ωt = X s sin(ωt ? φs ) where As = Bs = (ω ? ω ) + ( 2ζωnω )
2 n 2 2

?2ζωnω f0 s

Use rectangular form to make it easier to add the cos term
2

(ω ? ω ) + ( 2ζωnω )
2 n 2 2

2 (ωn ? ω 2 ) f0 s

2

Particular Solution (cos term)
With a cosine for the forcing function, we showed
2 ??+2ζωn x + ωn x =f0c cos ωt ? x

x pc = Ac cos ωt + Bc sin ωt = X c cos(ωt ? φc ) where Ac = Bc = (ω ? ω ) + ( 2ζωnω )
2 n 2 2 2 (ωn ? ω 2 ) f0c 2

(ω ? ω ) + ( 2ζωnω )
2 n 2 2

2ζωnω f0c

2

Magnitude X/Y
Now add the sin and cos terms to get the magnitude of the full particular solution
2 (2ζω )2 + ω n 2 2

X =

(ω ? ω ) + (2ζω nω )
2 n 2 2

2 2 f0c + f0s

2

= ω nY

(ω ? ω ) + (2ζω nω )
2 n

2

2 where f0c = 2ζω nωY and f0s = ω nY

if we define r = ω ωn this becomes

X =Y

(1? r ) + (2ζ r )
2 2

1+ (2ζ r)2

2

(2.70)

X 1 + (2ζ r)2 = 2 2 2 Y (1 ? r ) + (2ζ r )

(2.71)

The relative magnitude plot of X/Y versus frequency ratio: Called the Displacement Transmissibility
40 30 20 X/Y (dB) 10 0 -10 -20 0

ζ =0.01 ζ =0.1 ζ =0.3 ζ =0.7

0.5

1

1.5 2 Frequency ratio r

2.5

3

From the plot of relative Displacement Transmissibility observe that:
? X/Y is called Displacement Transmissibility Ratio ? Potentially severe amplification at resonance ? Attenuation for r > sqrt(2) Isolation Zone ? If r< sqrt(2) transmissibility decreases with damping ratio Amplification Zone ? If r >> 1 then transmissibility increases with damping ratio Xp~2Yζ/r

Next examine the Force Transmitted to the mass as a function of the frequency ratio

? ? ?? FT = ? k ( x ? y ) ? c( x ? y ) = mx At steady state, x(t ) = X cos(ωt ? φ ), so ??=-ω X cos(ωt ? φ ) x
2
x(t)

From FBD

m
k c FT

FT = mω X = k r X
2 2
y(t)

base

Plot of Force Transmissibility (in dB) versus frequency ratio
40 30 20 F/kY (dB) 10 0 -10 -20

ζ =0.01 ζ =0.1 ζ =0.3 ζ =0.7

0

0.5

1

1.5 2 Frequency ratio r

2.5

3

Comparison between force and displacement transmissibility
Force Transmissibility

Displacement Transmissibility

Example: Effect of speed on the
amplitude of car vibration

Model the road as a sinusoidal input to base motion of the car model
Approximation of road surface:
y(t) = (0.01 m)sin ω bt 1 ? ? ? hour ? ? 2π rad ? ω b = v(km/hr) ? ?? ?? ? = 0.2909v rad/s ? 0.006 km ? ? 3600 s ? ? cycle ?

ω b (20km/hr) = 5.818 rad/s

From the data give, determine the frequency and damping ratio of the car suspension:
ωn =
k = m 4 × 10 4 N/m = 6.303 rad/s ( ≈ 1 Hz) 1007 kg

c ζ= = 2 km 2

(4 × 10

2000 Ns/m
4

N/m (1007 kg )

)

= 0.158

From the input frequency, input amplitude, natural frequency and damping ratio use equation (2.70) to compute the amplitude of the response:
ω b 5.818 r= = ω 6.303

1 + (2ζ r)2 X =Y (1 ? r 2 )2 + (2ζ r)2 = (0.01 m )

(1 ? (0.923) ) + (2 (0.158)(0.923))
2 2

1 + [2(0.158)(0.923)]

2 2

= 0.0319 m

What happens as the car goes faster?

Example: Compute the force transmitted to
a machine through base motion at resonance
From (2.77) at r =1:
FT ? 1 + (2ζ ) ? =? kY ? (2ζ )2 ? ?
2 1/2

? FT =

kY 1 + 4ζ 2 2ζ

From given m, c, and k: ζ =

c 900 = ? 0.04 2 km 2 40, 000i 3000

From measured excitation Y = 0.001 m:
kY (40, 000 N/m)(0.001 m) 2 FT = 1 + 4ζ = 1 + 4(0.04) 2 = 501.6 N 2ζ 2(0.04)

Rotating Unbalance
? ? ? ? Gyros Cryo-coolers Tires Washing machines
Machine of total mass m i.e. m0 included in m

m0
e

ω?t
k c

e = eccentricity mo = mass unbalance ω? = rotation frequency

Rotating Unbalance (cont)
Rx

ω?
e

m0

What force is imparted on the structure? Note it rotates with x component:
Ry

θ

xr = e sin ω r t ? a x = ??r = ?eω r2 sin ω r t x

From sophomore dynamics,
Rx = m0 a x = ? mo eωr2 sin θ = ? mo eωr2 sin ωr t R y = m0 a y = ? mo eωr2 cos θ = ? mo eωr2 cos ωr t

Rotating Unbalance (cont)
The problem is now just like any other SDOF system with a harmonic excitation
m0eω?2sin(ω?t)
x(t)

?? ? mx + cx + kx = mo eωr2 sin ωr t
2 n

(2.82)

m
k c

mo 2 ? or ?? + 2ζωn x + ω x = x eωr sin ωr t m
Note the influences on the forcing function (we are assuming that
the mass m is held in place in the y direction as indicated in Figure 2.18)

Rotating Unbalance (cont)
? Just another SDOF oscillator with a harmonic forcing function ? Expressed in terms of frequency ratio r

x p (t ) = X sin(ωr t ? φ ) moe r X= m (1 ? r 2 )2 + ( 2ζ r )2 ? 2ζ r ? φ = tan ? 2 ? ? 1? r ?
?1 2

(2.83) (2.84)

(2.85)

Displacement magnitude vs frequency caused by rotating unbalance

Example:Helicopter rotor unbalance
Given Fig 2.21

k = 1 × 10 5 N/m mtail = 60 kg mrot = 20 kg m0 = 0.5 kg ζ = 0.01

Fig 2.22

Compute the deflection at 1500 rpm and find the rotor speed at which the deflection is maximum

Solution
The rotating mass is 20 + 0.5 or 20.5. The stiffness is provided by the Tail section and the corresponding mass is that determined in Example 1.4.4. So the system natural frequency is
k 105 N/m ωn = = = 46.69 rad/s mtail 60 kg 20.5 + m+ 3 3 The frequency of rotation is

rev min 2π rad ωr = 1500 rpm = 1500 = 157 rad/s min 60 s rev 157 rad/s ? r= = 3.16 49.49 rad/s

Now compute the deflection at r = 3.16 and ζ =0.01 using eq (2.84)
m0 e r2 X= m (1 ? r 2 ) 2 + (2ζ r ) 2

( 0.5 kg )( 0.15 m ) =
20.5 kg

( 3.16 )
2 2

2

(1 ? (3.16) ) ? ( 2(0.01)(3.16) )

2

= 0.004 m

At around r = 1, the max deflection occurs:

At r = 1:

rad rev 60 s r = 1 ? ω r = 49.69 rad/s = 49.69 = 474.5 rpm s 2π rad min

( 0.5 kg )( 0.15 m ) X=
20.5 kg

1 = 0.183 m or 18.3 cm 2(0.01)

Measurement Devices
? A basic transducer used in vibration measurement is the accelerometer. ? This device can be modeled using the base equations developed in the previous section

? ? ?? ∑ F =-k (x-y)-c(x-y)=mx ?? ? ? ? mx = -c( x ? y ) - k ( x ? y ) (2.86) and (2.61)
Here, y(t) is the measured response of the structure

Base motion applied to measurement devices
Let z (t ) = x(t ) ? y (t ) (2.87) : ? ?? ? mz + cz (t ) + kz (t ) = mωb2Y cos ωbt (2.88) Z r2 ? = Y (1 ? r 2 ) 2 + (2ζ r ) 2 and (2.90)
Accelerometer

θ = tan ?1 ?

? 2ζ r ? 2 ? ? 1? r ?

(2.91)

These equations should be familiar from base motion. Here they describe measurement!

Strain Gauge

Magnitude and sensitivity plots for accelerometers.
Effect of damping on proportionality constant

Fig 2.27

Fig 2.26
Magnitude plot showing Regions of measurement In the accel region, output voltage is nearly proportional to displacement




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