Lab 7 – Forced Vibration and Frequency Response Measurements

MEEN 364 V. Alladi, 2001. A. Bhattacharaya, 2002. A. Rynn 2002. J. Mlcak 2005 A. G. Parlos 2006 Last Update: 22 August 2006

Lab 7 – Forced Vibration and Frequency Response Measurements Introduction The response of a system subject to forced excitation will depend on the forcing function. The frequency response will in general consist of the frequency of the forcing function along with other harmonics. Objective 1. To measure the natural frequency and damping of the system subject to forced vibration using time domain measurements. 2. To observe the spectrum of the vibration signal. 3. To relate the spectrum of the vibration signal to the frequencies in the system. 4. To see the effect of sampling rate on the spectrum. PreLab 1) L

w (side view of the beam)

A beam of length L = 21.25cm, rectangular cross-sectional area of 3.125cm2 and a thickness of 0.635cm. Find the equivalent stiffness K of the beam. Also, determine the natural frequency of the beam. Assume the density of steel as 7850 Kg/m3. (Hint: Refer to Strength of materials book to find the stiffness and calculate the natural frequency from the K and m values. Be sure to use the equivalent mass for the beam.) 2)

An accelerometer with a sensitivity of 10mV/g is used to measure the vibration of a beam. It produces a harmonic signal with a frequency of 300 rad/s and a peak-peak amplitude of 51 mv. What is the peak amplitude of the beam vibration (i.e., displacement)? An accelerometer provides a voltage proportional to the magnitude of the acceleration.

Setup Description The system consists of a simple cantilever beam with a motor mounted at its free end. The DC motor is imbalanced to provide a periodic excitation force to the beam. It is powered by means

1

MEEN 364 V. Alladi, 2001. A. Bhattacharaya, 2002. A. Rynn 2002. J. Mlcak 2005 A. G. Parlos 2006 Last Update: 22 August 2006 of a variable power supply, in order to change the speed. A piezoelectric accelerometer mounted on the free end of the cantilever is used for measuring the acceleration of the tip of the beam. DC motor with imbalanced load d

Ω

0.25 in.

beam accelerometer 1.5 in.

10 in.

Figure 9-1 - Cantilever Beam System (drawn by Christopher Cullum) The system shown above can be modeled as a second order system. Feq(t) = mdω2sin(ωt)

Y(t)

Meq

Equation of Motion .. . MeqY(t) + C eqY(t) + K eqY(t) = Feq(t) Keq

Ceq

Figure 9-2 - Equivalent Second Order System (drawn by Christopher Cullum) Theory

In order to derive the equation of motion, one must first determine the mass center of the system and then apply Newton’s law.

M t a g = ΣF y

(1)

2

MEEN 364 V. Alladi, 2001. A. Bhattacharaya, 2002. A. Rynn 2002. J. Mlcak 2005 A. G. Parlos 2006 Last Update: 22 August 2006

d

ω

lm

ld

block lblk beam acc la Ceq

Keq

Figure 9-3 – Detailed Schematic of Equivalent Second Order System (drawn by Christopher Cullum)

Let (2)

M t = M a + M bl + M b + M m + M w where

Mb =

13 M beam 55

(3)

let d = offset of the wheels mass center from its geometric center. Mass center Yg is given by

Yg = (M bY + M blk (Y + l blk ) + M m (Y + l m ) + M a (Y − l a ) + M w (Y + l d + d sin(ω t )) ) / M t (4) Differentiating (4) twice gives ..

..

Y g = Y − mdω 2 sin(ω t )

(5)

where

m = Mw / Mt we know

(6)

..

M t Y g = ΣF y

(7) 3

MEEN 364 V. Alladi, 2001. A. Bhattacharaya, 2002. A. Rynn 2002. J. Mlcak 2005 A. G. Parlos 2006 Last Update: 22 August 2006 Also since we are modeling the system as a second order system, .

ΣFy = − K eq Y − C eq Y

(8)

Using (5) and (7) in (8) we get ..

.

M t Y + C eq Y + K eq Y = M w dω 2 sin(ω t )

(9)

The solution to the above differential equation gives the response of the system — the displacement of the tip of the beam as a function of time for a given frequency of the exciting function. The response of the system to the above forcing function will consist of a transient part and a steady state solution. We are interested in the steady state solution alone as the transient of the actual system dies down quickly, as you will see when you perform the experiment. To solve the above equation, we use the real part of − je jwt as the forcing function and find out the particular solution. To do that let ..

.

M t Y + C eq Y + K eq Y = M w dω 2 Re( − je jwt ) −

Y = Y e jwt and we are interested in the real part of the solution Using (10) and (11) −

( K eq − M tω 2 + jCeqω ) Y = − M w dω 2 j

(10) (11) (12)

Recall that −

where

Y = Y cos(ω t + φ )

(13)

φ = angle(Y )

(14) −

We just need to determine Y andφ

−

Y=

=

− M wdω 2 j

(15)

K eq − M tω 2 + jCeqω M w dω 2

(16)

( K eq − M tω 2 ) 2 + (Ceqω ) 2

Define 4

and

ω n 2 = K eq / M t β = ω /ωn M wω 2 M w M t =

K eq

Ceqω K eq

=

M t K eq

MEEN 364 V. Alladi, 2001. A. Bhattacharaya, 2002. A. Rynn 2002. J. Mlcak 2005 A. G. Parlos 2006 Last Update: 22 August 2006 (17) (18)

ω 2 = mβ 2

(19)

( 2 M tω nζ )ω = 2ζβ K eq

(20)

From (15), (16) and (17) mdβ 2 Y = ( 1 − β 2 )2 + ( 2ζβ )2

(21)

φ = ∠ − M w dω 2 j − ∠K eq − ω 2 M t + jC eqω C eqω ) = tan −1 (− mdω 2 / 0) − tan −1 ( K eq − ω 2 M t ) 2ζβ φ = 2700 − tan −1 ( ) 1− β 2 ..

The accelerometer measures a= Y = −ω 2 Y cos(ω t + φ ) md ω 2 β 2 a =ω2Y = ( 1 − β 2 )2 + ( 2ζβ )2

(22) (23) (24) (25) (26)

Do not forget to realize that this is the same response you obtain from the transfer function. Note that if the damping is small ζ → 0andβ → 1 ⇒ ω = ω n . In other words, the motor speed is the natural frequency of the system. For this case Y → ∞ and a → ∞ . This is called resonance. Rewriting equation (26) as 2 a md ω n = 4 β ( 1 − β 2 )2 + ( 2ζβ )2

(27)

The above system has a maximum at β = 1 or ω = ω n

(28)

Let us now try to see how we can find out our damping ratio ζ . Let

a ∗ = max(

a

β4

)

(29)

Combining (27),(28) and (29) we arrive at

a

β

=

4

β =1

md ω n

2

( 1 − β ) + ( 2ζβ ) 2

2

= md ω 2 / 2ζ

2

β =1

5

(30)

MEEN 364 V. Alladi, 2001. A. Bhattacharaya, 2002. A. Rynn 2002. J. Mlcak 2005 A. G. Parlos 2006 Last Update: 22 August 2006 Let us find 2 points, β1 and β2, such that a 2 = .707a ∗ = .707mdω n / 2ζ 4

(31)

β

=

mdω n

2

(32)

( 1 − β 2 )2 + ( 2ζβ )2

⇒ ( 1 − β 2 )2 + ( 2ζβ )2 = 8ζ 2

(33)

⇒ β 2 = ( 1 − 2ζ 2 ) ± 2ζ 1 + ζ 2

(34)

For small values of ζ , we have ζ 2