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Vibration of Multi-Degree-of-Freedom (MDOF) Systems - Newtons

Method

H.P. LEEDepartment of Mechanical Engineering

EA-05-20Email: mpeleehp@nus.edu.sg

Semester 2 2014/2015

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Degrees of freedom The degrees of freedom (DOF) of a

rigid body is defined as the number of independent movements it has.

The figure shows a rigid body in a plane.

The bar (rigid body) can be translatedalong the x axis, translated along the y axis, and rotated about its centroid (centre of mass). (or change of the orientation of the bar)

Therefore, for this rigid body in planar motion, there are 3 DOF.

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Example

A door has one degree of freedom

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Some are not that obvious

The following mechanisms only have one

degree of freedom.

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Example

A three DOF robotic arm

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Example

A slider crank in an engine

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Example Windshield wiper mechanism

There are two four bar mechanisms links 1234 and links 1456. 1234 is a crank rocker mechanism. 1456 is a double rocker mechanism,, also a parallel mechanism.

It only has one degree of freedom.

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Example of spring mass damper system

A one degree of freedom system

A two degree of freedom system

Some of these simple models are used for analyzing complex motions

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Another example

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Review on Mechanical Components for a Vibrating System

Spring Force

k is the spring constant

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Mechanical Components

Damping force

c is the damping coefficient

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Mechanical Components

Inertia force

(or kinetic force)

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One degree of freedom systems

A spring mass system

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One degree of freedom system

A spring mass damper system

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Degrees of freedom back to something more fundamental

In general, the number of degrees of freedom of a dynamical system is the number of independent parameters (or coordinates) required to describe the motion of the dynamical system.

For a spring mass system, it has one degree of freedom. The variable x is the independent coordinate required to describe the motion of the mass.

k

x0

m

x

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Example : A Torsional Vibration System

An elastic shaft and a rigid

rotor.

A one degree of freedom

system.

J is the moment of inertia of

the rotor.

K is the torsional stiffness of

the shaft.

J

KShaft

Disc

1 DOF

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A two-degree-of-freedom Torsional Vibration System

The following is an example of a 2 DOF vibration

system.

1J1

2J2

K1

K2

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Example for a two DOF spring-mass-damper system

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Equation of motion The equation of motion for

the 2DOF system can be derived either by the Newtons method (a vector approach) or the Lagranges equation (a scalar approach).

The Newtons method is to construct the force diagram and then apply the Newtons second law.

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Newtons Law (Revision)

For a mass m acted upon by the resultant F of

external forces, the acceleration a is

F = m a

F m a=

In graphical form, or so called the force diagram, or

the free body diagram:

ma is known as the kinetic force. It is a vector quantity.

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DAlemberts Principle (Revision)

The kinetic force can be moved to the

left hand side of the equation by

changing the sign: (changing the

direction)

Fm a

= 0

If the kinetic force is moved to the

left hand side, it is known as the

inertia force. The inertia force

and the external force are in

static equilibrium.

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Force Diagram

For the 2 DOF rotor system.

Governing equations

Newtons method

. Force equilibrium diagram

1

2

Disc-1

Disc-2

)(K 122

22J

11J11K

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Equation of Motion From force equilibrium diagram of disc 1.

From force equilibrium diagram of disc 2.

0)(KKJ 1221111

0K)K(KJ 2212111

0)(KJ 12222

0KKJ 221222

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Example spring mass systems

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2 DOF spring mass systems

Free body force diagram

x1 x2

m1 m2k1 x1

k2(x2 -x1)

m1 a2a1 m2

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2 DOF spring mass systems

1 1 1 1 2 2 1

2 2 2 2 1

1 1 1 2 1 2 2

2 2 2 1 2 2

( ) ( ) ( ) ( ) (4.1)

( ) ( ) ( )

Rearranging terms:

( ) ( ) ( ) ( ) 0 (4.2)

( ) ( ) ( ) 0

m x t k x t k x t x t

m x t k x t x t

m x t k k x t k x t

m x t k x t k x t

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General observations

A 2 Degree-of-Freedom system has Two natural frequencies (we will discuss this later).

Two equations of motion.

Free vibrations, so the forms are homogeneous equations.

Equations are coupled: Both have x1 and x2.

If only one mass moves, the other follows

In this case the coupling is due to k2. Mathematically and Physically.

If k2 = 0, no coupling occurs and can be solved as two independent SDOF systems.

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Initial Conditions For a 2 DOF systems, the two equations are

linear second order differential equations with constant coefficients.

The two equations will result in four unknown constants from the integration process.

Four initial conditions are therefore required for the solutions.

The initial conditions are typically in terms of initial positions and velocities.

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Initial conditions The initial conditions for the previous examples are

1 10 1 10 2 20 2 20(0) , (0) , (0) , (0)x x x x x x x x

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Matrix form The two equations can be written in the form of a single

matrix equation

1 1 1

2 2 2

1 1 2 2

2 2 2

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

0

0

x t x t x tt , t , t

x t x t x t

m k k kM , K

m k k

M K

x x x

x x 0

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Matrix form The matrix equation, if expanded, is the same as the

system of two ordinary differential equations

0)()()(

0)()()()(

221222

2212111

txktxktxm

txktxkktxm

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Initial conditions The initial conditions in vector form are

10 10

20 20

(0) , and (0)x x

x x

x x

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Solution process

2

2

Let ( )

1, , , unknown

-

-

j t

j t

t e

j

M K e

M K

x u

u 0 u

u 0

u 0

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Observation This process changes the system of two ordinary

differential equations into an algebraic vector equation.

2

1

2

- (4.17)

This is two algebraic equation in 3 uknowns

( 1 vector of two elements and 1 scalar):

= , and

M K

u

u

u 0

u

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Solution For a non trivial solution, the inverse cannot exist

2

12

2

If the inv - exists : which is the

static equilibrium position. For motion to occur

- does not exist

or det - (4.19)

M K

M K

M K

u 0

u 0

0

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Solution characteristic equation The determinant results in 1 equation in one unknown

known as the characteristic equation

det -2M K 0

det2m1 k1 k2 k2

k2 2m2 k2

0

m1m24 (m1k2 m2k1 m2k2 )

2 k1k2 0

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Solution characteristic equation The equation is quadratic and will result in two solutions

2 2

1 2 1 2 and and

2

1 1

2

2 2

( ) (4.22)

and

( ) (4.23)

M K

M K

u 0

u 0

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Solution

Each of these matrix equations represents 2

equations in the 2 unknown components of the

vector, but the coefficient matrix is singular so

each matrix equation results in only 1

independent equation

The following examples will clarify this

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Example

m1=9 kg,m2=1kg, k1=24 N/m and k2=3 N/m

The characteristic equation becomes4-62+8=(2-2)(2-4)=0

2 = 2 and 2 =4 or

1,3 2 rad/s, 2,4 2 rad/s

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Solution

112

1 1

12

2

1 1

11

12

11 12 11 12

For =2, denote then we have

(- )

27 9(2) 3 0

3 3 (2) 0

9 3 0 and 3 0

u

u

M K

u

u

u u u u

u

u 0

2 equations, 2 unknowns but DEPENDENT!(the 2nd equation is -3 times the first)

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Solution Only the direction of vectors u can be determined, not

the magnitude as it remains arbitrary

1111 12

12

2

1

1 1 results from both equations:

3 3

only the direction, not the magnitude can be determined!

This is because: det( ) 0.

The magnitude of the vector is arbitrary. To see this suppose

t

uu u

u

M K

1

2

1 1 1

2 2

1 1 1 1

hat satisfies

( ) , so does , arbitrary. So

( ) ( )

M K a a

M K a M K

u

u 0 u

u 0 u 0

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Solution

212

2 2

22

2

1

21

22

21 22 21 22

For = 4, let then we have

(- )

27 9(4) 3 0

3 3 (4) 0

19 3 0 or

3

u

u

M K

u

u

u u u u

u

u 0

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Solution How about the magnitudes? We can only determine the

relative magnitudes at this stage

u12 1 u1 1

3

1

u22 1 u2 1

3

1

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Mode shapes u1 and u2 are known as the mode shapes

1,3 2, has mode shape u1 1

3

1

2,4 2, has mode shape u2 1

3

1

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The complete solution As the system is linear, the solution is

1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

1 2

1 1 1 1 2 2 2 2

1 2 1 2

( ) , , ,

( )

( )

sin( ) sin( )

where , , , and are const

j t j t j t j t

j t j t j t j t

j t j t j t j t

t e e e e

t a e b e c e d e

t ae be ce de

A t A t

A A

x u u u u

x u u u u

x u u

u u

ants of integration

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Physical interpretation of the results

Each of the TWO masses is oscillating at TWO

natural frequencies 1 and 2

The relative magnitude of each sine term, and

hence of the magnitude of oscillation of m1 and

m2 is determined by the value of A1u1 and A2u2

The vectors u1 and u2 are called mode shapes

because the describe the relative magnitude of

oscillation between the two masses

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What is a mode shape? First note that A1, A2, 1 and 2 are determined by the

initial conditions

Choose them so that A2 = 1 = 2 =0

Thus each mass oscillates at (one) (the same)

frequency 1 with magnitudes proportional to u1 the 1st

mode shape

x(t) x1(t)

x2(t)

A1

u11

u12

sin1t A1u1 sin1t

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Graphical representation of mode shapes

Mode 1:

k1

m1

x1

m2

x2k2

Mode 2:

k1

m1

x1

m2

x2k2

x2=A

x2=Ax1=A/3

x1=-A/3

u1 1

3

1

u2 1

3

1

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Time response with given initial conditions

To compute the time response with the given initial

conditions

2211

22

11

2

1

2211

22

11

2

1

2cos22cos2

2cos23

2cos23

)(

)(

2sin2sin

2sin3

2sin3

)(

)(

0

0)0( mm,

0

1=(0)consider

tAtA

tA

tA

tx

tx

tAtA

tA

tA

tx

tx

xx

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Time response At t = 0

2211

22

11

2211

22

11

cos2cos2

cos3

2cos23

0

0

sinsin

sin3

sin3

0

mm 1

AA

AA

AA

AA

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Time response There are four equations with four unknowns

3 A1 sin 1 A2 sin 2

0 A1 sin 1 A2 sin 2

0 A1 2 cos 1 A2 2cos 2

0 A1 2 cos 1 A2 2cos 2

A1 1.5 mm,A2 1.5 mm,1 2

2 rad

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Time response The final solution is

x1(t) 0.5cos 2t 0.5cos2t

x2 (t) 1.5cos 2t 1.5cos2t

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Graphical solutions

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The final solution It is a combination of the two mode shapes

x(t) a1u1cos1t a2u2 cos2t

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Conclusion A two degree-of-freedom system will have two natural

frequencies.

These two frequencies are present in the general response.

Frequencies are not those of two component systems

There are simpler approach for solving the same problem.

1 2 k1m1

1.63,2 2 k2m2

1.732