two dof system - mechanical vibration

8/17/2019 Two Dof System - MECHANICAL VIBRATION

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TWO DEGREE OF FREEDOM SYSTEMS


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DANIEL Bernoulli (1700-1782):

Swiss mathematician who proposed

principal of super position


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• Requires two independent coordinates to describe their motion

• Assumption:

• Mass is considered as rigid body with two possible motions.• Vibration in vertical plane

• Idealised as a bar of mass, m and MI, J0 supported on two springs of sand k2

• Displacements given by x(t) and θ(t)

• Displacements can also be given by x1 (t) and x2 (t)



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• Packaging of instrument of mass , m

supported on springs from all 4 sides

• Here there is one point mass but two

displacements x and y and two DOF

• DOF = (No. of masses in the system) X

( No. of possible types of motion of each

mass)

• If a 2 DOF system is vibrating at one of thenatural frequency, the amplitudes of the 2

DOFs (coordinates) are related in a specific

manner & the configuration is called normal

/ principal/natural mode of vibration

INTRODUCTION



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• If an arbitrary initial excitation is given to the system, the re

vibration will be a superposition of two normal modes of vibration.• Generalized co-ordinates: The configuration of the system can b

by a set of independent coordinates such as displacement, ang

other physical parameters.

• The EOMs are generally coupled but we can find some

coordinates which will give uncoupled equations.



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Equation of Motion for Forced Vibration

By Newton’s Law

ሷ+ (+ ) ሶ - ሶ 1 2 1 22 = 1 ሷ+ (+ ) ሶ - ሶ 2 3 2 21 = 2

(1) Coupled 2nd Orde


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Equation (1) can be written in matrix form as

[m] ሷ (t) + [c] ሶ (t) + [k] = F (t) (2)

[m] = 00

= {()()

}

[c] = 1 2 2 3 F = {()()

}

[k] =1 2


Equation of Motion for Forced Vibration

[m],[c] and [k] are symmetric matr

[m]T =[m], [c]T =[c], [k]T =[k]

Equation (1) becomes uncoupled w

and k2 are zero


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Free Vibration of an undamped system

For undamped free vibrationsBy Newton’s Law

ሷ+ 1 2 1 22 = 0 (3)

ሷ 2 3 2 21 = 0 (4)

We are interested in knowing whether m1 and m2 can oscillate harmonicafrequency and phase angle but with different amplitudes.

Take the solution as

1 = 1 ( ) (5)2 = 2 ( )


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Substituting (5) in (3 & 4)

[{-m1 2 + 1 2 }1 2 2 ] ( ) = 0[-2 1+ {-m2 2 + 2 3 }2] ( ) = 0 (6)Equation (6) must be satisfied for all values of time t,

∴ {-m1 2 + 1 2 } 2 = 0-2 1+ {-m2 2 + 2 3 } = 0 (7)

Trivial solution is 1= 2 = 0

For Non-trivial solution, determinants of coefficients must be zero

4-{ 1 2 + 2 3 }

+{ 1 2 2 3 - } = 0

This is quadratic equation in with solutions and


No Vibration


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Determination of X1 and X2

• Depends of and

• and

are values of X1 and X2 for =

• and

are values of X1 and X2 for = • Since equation (7) is homogeneous, only ratios

= ൗ

and = ൗ

can be found

• Substituting 2 = and 2 =

equation (7) gives

= ൗ

=

−m1 + +

=

−m2 + +

= ൗ

=

−m1 + +

=

−m2

+ +(


{-m1 2 + 1 -2 1+ {-m2


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The two ratios given for each r i (i = 1,2) in equation (9) are identical.

The normal modes of vibration corresponding to and can be eas

=

=

1

=

=

2 (10)

and

denotes normal modes of vibration and are called moda

the system



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Free vibration solution / motion in time

()1 = {(t) }(t) T = { cos(t + 1) r 1 cos(t + 1) }T Fi

()2 = {

(t) }(t) T = {

cos(t + 2) r 2 cos(t + 2) }T Se

,

, 1 , 2 are constants to be determined from initial conditions.

Initial Conditions:

• At time t =0, x1 (t=0) and ሶ1(t=0) , x2 (t=0) and ሶ2(t=0) ???• For any general initial conditions both modes will get excited.

• Resulting motion

• ()

= c1 () c2 ()

(12)


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() ()involves unknown constants

,

(see 11) ,

choose c1 = c2 =1 with no loss of generality

1 = (t) +

(t) = cos(t + 1) +

cos(t + 2)

2 = (t) + (t) = 1 cos(t + 1) + r 2 cos(t + 2)

Constants ,

, 1,, 2 are to be determined from initial conditions

x1 (t=0)= x1(0) , ሶ1(t=0)= ሶ1(0)

x2 (t=0) = x2(0) , ሶ2(t=0)= ሶ2(0)



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Substituting (14) in (13)

x1(0)= cos 1 +

cos 2 ሶ1(0) = -

sin 1 - sin 2

x2(0) = r 1 cos 1 + r 2

cos 2 ሶ2(0) = -r 1

sin 1 - r 2 sin 2 (

These four equations in four unknowns can be solved.



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Find the natural frequencies and

mode shapes of spring – masssystem shown in figure which is

constrained to move in the vertical

direction only. Take n = 1.

Problem 1


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Mode Shapes

In phase mode180 degree out of phase mod


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The EOM of the system is

J1 ሷ= -kt1 1 + kt2(2 - 1) + Mt1J2 ሷ= -kt2(2 - 1) - kt32+ Mt2

Rearranging

J1 ሷ (kt1 + kt2 )1 - kt22 = Mt1J2 ሷ -kt21 + (kt2 +kt3 ) 2 = Mt2 (16)

Torsional Systems


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Find the natural frequency and mode

shapes for the torsional systemshown in Figure for J1 = J0, J2 = 2J0and kt1= kt2 = kt

problem

Coordinate Coupling and Principal Coordinates


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• There can be n different coordinate systems defining coniguration

system called generalised coordinates.

• Lathe bed-rigid body (mass,inertia), headstock & tailstock-lumped




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Assumptions:

• Lathe bed-rigid body (mass,inertia), headstock & tailstock-lumped m

• Bed supported on springs• (x1, x2), (x,), (x1,), (y,)- generalised co-ordinates




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Equation of Motion using (x(t),())

• Positive values of motion as per figure.

• σ = 0m ሷ = - k1(x – l1 ) – k2(x + l2 ) (17)

• σ = 0

J0 ሷ = k1(x – l1 ) l1 – k2(x + l2 ) l2 (18)

00 J0

ሷ ሷ +

(k1 k2) (k1l1 k2 l2)(k1l1 k2 l2) (k1l12 k2l2 2)

x = 0

(19)

These equations become independent when

(k1l1 k2 l2) = 0 Static /elastic Coupling


C di t C li d P i i l C di t


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Equation of Motion using (y(t),())• Positive values of motion as per figure.

• σ = 0m ሷ = - k1(y – l’1 ) – k2(x + l’2 ) - me ሷ (20)

• σ = 0Jp ሷ = k1(y – l’1 ) l’1 – k2(y + l’2 ) l’2- me ሷ (21)

J p

ሷ ሷ

+(k1 k2) (k1l’1 k2 l’2)

(k1l’1 k2 l’2) (k1l’12

k2l’2 2

)

y

= 0

Static /Elastic and Dynamic / Mass Coupling

(22)


C di t C li d P i i l C di t


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Equation of motion of viscously damped 2 dof system

11 1221 22

ሷ ሷ

+ 11 1221 22 ሶ ሶ

11 1221 22

=

• System vibrates in its own natural way regardless of coordinates us

choice of coordinates are for convenience.

• (19) and (22) nature of coupling depends on choice of coordin

• Principal coordinates: No static/dynamic coupling


Nondiagonal stiffness matrix Static / elastic Coupling

Nondiagonal mass matrix Dynamic / mass Coupling

Nondiagonal damping matrix Velocity / damping Coupling

Dynami

coupling


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Determine the principal coordinates for the spring mass system show

Problem 3


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Determine the pitch (angular motion) and

bounce ( up and down motion) frequencies

and the location of oscillation centers(nodes) of an automobile with following data:

Mass (m) =1000 kg

Radius of gyration ( r) =0.9m

Distance between front axle and CG (l1 ) = 1 m

Distance between rear axle and CG (l2 ) = 2 mFront spring stiffness (kf ) = 18 kN/m

Rear spring stiffness (kr ) = 22 kN/m

Problem 4


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MODE SHAPES


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Equation of motion of viscously damped 2 dof system under external

11 1221 22

ሷ ሷ

+ 11 1221 22 ሶ ሶ

11 1221 22

=

Assume F j(t) = F j0 , j = 1,2

Assume steady state solution as

x j

(t) = X j

, j = 1,2X1and X2,in general complex quantities depends on and system par

Substituting (24) and (25) in (23)

(211 11 11) (212 12 12)(212 12 12) (222 22 22)

=

Forced Vibration Analysis



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Let Zrs() = (2 ), r,s = 1,2 Mechanical Impeden

Substituting (27) in (26)

[Z()] X = F0

[Z()]= (211 11 11) (212 12 12)(212 12 12) (222 22 22)

IMPEDEN

X =

, F0

=

, [Z()

] =11 1221 22

Solving

X = [Z()]-1F0

Substituting (29) in (25) we get solution x1(t) and x2(t)



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Find the steady state response of the system

shown in Figure. Plot the frequency response

curve

Problem 4

At a particular frequency

ω,vibration of mass m1 isbasis of vibration absorb

Vibration Absorber


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Use: If a machine is operating a frequency near its natural frequen

can be reduced by another spring mass system which acts

neutralizer / dynamic vibration absorber.

Principal: Natural frequency of resulting system is away fro

frequency.

Assumption: Machine to be single degree of freedom system.

Vibration Absorber


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Equation of Motion

ሷ+ 1 1 2(1 2) =F0 sin ωt

ሷ 2(1 2) = 0 (30)Assume harmonic solutionx j(t) = X j sin ωt , j = 1,2 (31)

Steady state solution is

X1 =−

+− − −

(32)

X2 =

+− − − (33)


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To make X1 = 0 (aim of absorber) ,

ω = 0

ω = (34)

If the machine , before addition of the dynamic vibration absorber,

operates near its resonance, ω ≅ ω=

If ω =

=

(35)

Amplitude of vibration of machine, while operating at its original resofrequency, will be zero.

∴ ω =

(36)

=F0

(37)


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Equation (32), (33) becomes

(39)

(38)

• Two peaks for two natural frequencies.

• X1 = 0 at ω = ω

• X2 = -

= - F0

(40)

Force exerted by the aux

opposite to impressed fo

and neutralizes it making


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Size of dynamic vibration absorber can be found

from (40) and (35):

X2 = - F0= ω X2 (41)Two new resonant frequencies Ω1 and Ω2 are

introduced in the system at which amplitude is

infinite.

Hence operating frequency ω should be away from

Ω1and

Ω2.

Setting denominator of (38) to zero, leads

+1 = 0


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The values of Ω1 and Ω2 are

(42)

• Machine must pass through Ω1 duringstartup and stopping resulting in large

amplitude.

• As dynamic absorber is tuned to one

excitation frequency ω , the steady stateamplitude of the machine will be zero only

at that frequency.

The difference betweincreases with increas

P bl 5


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A diesel engine, weighing 3000 N, is supported on a pedestal mbeen observed that the engine induces vibration into surrou

through its pedestal mount at an operating speed of 6000 rpm. Deparameters of the vibration absorber that will reduce the vibrmounted on pedestal. The magnitude of the exciting force is 250amplitude of motion of the auxiliary mass is limited to 2 mm.

Problem 5

Problem 6


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A motor generator set shown in figure is designed to operate in the speed ra

4000 rpm. However, the set is found to vibrate violently at a speed of 3000

slight unbalance in the rotor. It is proposed to attach a cantilever mounted

absorber system to eliminate the problem. When the cantilever carrying a triatuned to 3000 rpm is attached to the set, the resulting natural frequencies of t

found to be 2500 rpm and 3500 rpm. Design the absorber to be attached by

mass and stiffness so that the natural frequencies of the total system fa

operating speed range of the motor generator set.

Problem 6

two dof system - mechanical vibration

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