LECT06 - MDOF Part 2 [Compatibility Mode]

8/10/2019 LECT06 - MDOF Part 2 [Compatibility Mode]

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MULTI DEGREE OFFREEDOM (M-DOF)


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Outline-Part 2

1 Using Newtons Second Law to derive

Equations of Motion

2 Influence Coefficients

3 Eigenvalue problem


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Using Newtons 2nd Law to deriveEquations of Motion

Step 1: Set up suitable coordinates to describe

positions of the various masses in the system.

Step 2: Measure displacements of the massesfrom their static equilibrium positions

Step 3: Draw free body diagram and indicateforces acting on each mass

Step 4: Apply Newtons 2nd

Law to each mass


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Example 1

Derive the equations of motion of the spring-

mass damper system shown below.


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Solution

Free-body diagram is as shown:

Applying Newtons 2nd Law gives:

Set i=1 with x0=0and i=nwith xn+1=0:

( ) ( ) ( ) ( ) 1,...,3,2,111111

=+++=++++

niFxxcxxcxxkxxkxm iiiiiiiiiiiiiii &&&&&&

( ) ( )

( ) ( ) nnnnnnnnnnnnn Fxkxkkxcxccxm

Fxkxkkxcxccxm

=++++

=++++

++ 1111

1221212212111

&&&&

&&&&


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Solution

Notes:

The equations of motion canbe expressed in matrix form:

[m] is the mass matrix

[c] is the damping matrix [k] is the stiffness matrix

[ ] [ ] [ ] Fxkxcxmrr&r&&r

=++

[ ]

=

nm

m

m

m

0...0

0

00

0...0

2

1

OM

M

[ ] [ ] ( )( )

( )

( )( )

( )

=

=

+

+

+

+

=

+

+

+

=

+

+

tF

tFtF

F

tx

txtx

x

kkk

kkkkkkk

kkk

k

ccc

cccc

ccc

c

nnn

nnn

3

2

1

3

2

1

1

433

3322

221

1

3322

221

,,

000

0

0000

00

,000

00

rr

OO

L

MM

M

M


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Influence Coefficients

One set of influence coefficients is

associated with each matrix involved in theequations of motion

Equation of Motion Influence coefficient

Stiffness matrix Stiffness influencecoefficient

Mass matrix Inertia influence coefficient

Inverse stiffnessmatrix

Flexibility influencecoefficient

Inverse mass matrix Inverse inertia coefficient


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Stiffness Influence Coefficient, kij

kij is the force at point idue to unit

displacement at pointjand zero displacementat all other points.

Total force at point i,

Matrix form:

=

==

n

j

jiji nixkF1

,...,2,1,

[ ] [ ]

==

nnnn

n

n

kkk

kkk

kkk

kxkF

...

...

...

where

21

22221

11211

MM

rr


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Stiffness Influence Coefficient, kij

Note:

kij= kji

kij for torsional systems is defined as thetorque at point idue to unit angular

displacement at pointjand zero angulardisplacements at all other points.


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Example 2

Find the stiffness influence coefficients of the

system shown below.


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Solution

Let x1, x2and x3be the displacements of m1, m2 and

m3 respectively.

Set x1=1 and x2=x3=0.

Horizontal equilibrium of forces:

Mass m1: k1= -k2+ k11 (E1)

Mass m2: k21= -k2 (E2)

Mass m3: k31 = 0 (E3)

Solving E1 to E3 gives k11=k1+k2

k21= -k2

k31

= 0


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Solution

Next set x2=1 and x1=x3=0.

Horizontal equilibrium of forces: Mass m1: k12 + k2 = 0 (E4)

Mass m2: k22 -k3 = k2 (E5)

Mass m3: k32 = -k3 (E6) Solving E4 to E6 gives

k22 = k2+k3

k12 = -k2k32 = -k3


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Solution

Finally set x3=1 and x1=x2=0.

Horizontal equilibrium of forces: Mass m1: k13 = 0 (E7)

Mass m2: k23 + k3 = 0 (E8)

Mass m3: k33 = k3 (E9)

Solving E7 to E9 givesk33 = k3k13 = 0

k23 = -k3

Thus the stiffness matrix is:[ ]

+

+

=

33

3322

221

0

0

kk

kkkk

kkk

k


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Flexibility Influence Coefficient,aij

Have to solve n sets of linear equations to

obtain all the kijs in an n DOF system

Generating aijs is simpler.

aij is defined as the deflection at point i due tounit load at pointj,

xij= aijFj, where xij is the displacement at

point i due to external force Fj

Matrix form:

==

===

n

j

jij

n

j

iji niFaxx11

,...,2,1,

[ ]Faxrr

=


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Flexibility Influence Coefficient,aij

Note:

Stiffness and flexibility matrices are theinverse of each other.

aij= aji

aijfor torsional systems is defined as theangular deflection of point idue to unit torque

at pointj.

[ ] [ ] [ ] [ ] 11, == kaak


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Example 3

Find the flexibility influence coefficients of the

system shown below.


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Solution

Let x1, x2and x3be the displacements of m1,

m2 and m3 respectively.

Set F1=1 and F2=F3=0.


Mass m1: k1a11= k2(a21 a11) + 1 (E1)

Mass m2: k2(a21 a11) = k3(a31 a21) (E2) Mass m3: k3(a31 a21) = 0 (E3)

Solving E1 to E3 gives a11 = a21 = a31 = 1/k1


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Solution

Next set F2=1 and F1=F3=0.


Mass m1: k1a12 = k2(a22 a12) (E4) Mass m2: k2(a22 a12) = k3(a32 a22) +1 (E5)

Mass m3: k3(a32 a22) = 0 (E6)

Solving E4 to E6 gives a12 = 1/k1

a22 = a32 = 1/k1 + 1/k2


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Solution

Next set F3=1 and F1=F2=0.


Mass m1: k1a13 = k2(a23 a13) (E7) Mass m2: k2(a23 a13) = k3(a33 a23) (E8)

Mass m3: k3(a33 a23) = 1 (E9)

Solving E7 to E9 gives a13 = 1/k1

a23 = 1/k1 + 1/k2

a33 = 1/k1 + 1/k2 + 1/k3


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Eigenvalue Problem

(Eigenvalue problem)

For a non-trivial solution, determinant ofthe coefficient matrix must be zero.

i.e. = |kij-2mij| = |[k]- 2[m]| =0(Characteristic equation)

2 is the eigenvalue

[ ] [ ][ ] 02rr

= Xmk


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Solution of the Eigenvalue Problem

Eigenvalue problem:

Multiplying by [k]-1:

[I] is identity matrix [D]=[k]-1[m] is dynamical matrix.

For a non-trivial solution of

characteristic determinant =|[I]-[D]|=0

Use numerical methods to solve if DOF ofsystem is large

[ ] [ ][ ]2

10

== whereXmkrr

[ ] [ ][ ] [ ] [ ]XDXXDIrrrr

== Ior0

,Xr


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Example 4

Find the natural frequencies and mode

shapes of the system shown below fork1=k2=k3=k and m1=m2=m3=m


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Solution

Dynamical matrix [D]=[k]-1[m] [a][m]

Flexibility matrix

Mass matrix

Thus

[ ]

=

321

221

1111

ka

[ ]

=

100

010

001

mm

[ ]

=

321

221111

k

mD


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Solution

Frequency equation:

=|[I]-[D]|=

Dividing throughout by ,

2

1where0

321

221

111

00

00

00

==

k

m

km

km

2

23 where0165

31

2

2

21

1

===+=


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Solution

Once the natural freq are known, the

eigenvectors can be calculated using

m

k

k

m8025.1,2490.3 3

2

33 ===

m

k

k

m2471.1,5553.1 2

2

22 ===

m

k

k

m44504.0,19806.0 1

2

11 ===

[ ] [ ][ ] ( ) ( )

( )

( )

( )

(E1)where3,2,1,0

3

2

1

===

i

i

i

ii

i

X

X

X

XiXDIrrr


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Solution

1st mode: Substitute 1=5.0489 into (E1):

3 unknowns X1(1),X2(1), X3(1) in 3 equations

Can express any 2 unknowns in terms of the

remaining one.

( )

( )

( )

=

0

00

321

221111

100

010001

0489.51

3

1

2

1

1

X

XX

k

m

k

m

k

m

( )

( )

( )

=

0

0

0

0489.221

20489.31

110489.4

..1

3

1

2

11

X

X

X

ei


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Solution

X2(1) + X3(1) = 4.0489 X1(1)

3.0489X2(1) 2X3(1) = X1(1)

Solving the above, we get

X2(1)

=1.8019X1(1)

and X3(1)

=2.2470X1(1)

Thus first mode shape

where X1(1) can be chosen arbitrarily.

( ) ( )

=

2470.2

8019.1

11

1

1XX

r


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Solution

2nd mode: Substitute 2=0.6430 into (E1):


Can express any 2 unknowns in terms ofthe remaining one.

k

m

( )

( )

( )

=

0

0

0

321

221

111

100

010

001

6430.02

3

22

2

1

X

X

X

km

km

( )

( )

( )

=

0

0

0

3570.221

23570.11

113570.0

..2

3

22

2

1

X

X

X

ei


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Solution

X2(2) X3(2) = 0.3570X1(2)

-1.3570X2(2) 2X3(2) = X1(2)


X2(2)

=0.4450X1(2)

and X3(2)

=-0.8020X1(2)

Thus 2nd mode shape


( ) ( )

=

8020.0

4450.0

12

1

2XX

r


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Solution

3rd mode: Substitute 3=0.3078 into (E1):


Can express any 2 unknowns in terms of the

remaining one.

( )

( )

( )

=

0

0

0

321

221

111

100

010

001

3780.03

3

3

2

3

1

X

X

X

k

m

k

m

( )

( )

( )

=

0

0

0

6922.221

26922.11

116922.0

..3

3

3

2

3

1

X

X

X

ei

k

m


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Solution

-X2(3) - X3(3) = 0.6922X1(3)

-1.6922X2(3) 2X3(3) = X1(3)


X2(3)

=-1.2468X1(3)

and X3(3)

=0.5544X1(3)

Thus 3rd mode shape


( ) ( )

=

5544.0

2468.1

13

1

3XX

r


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Solution

When X1(1) = X1(2) =

X1(3) =1, the modeshapes are as follows:

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LECT06 - MDOF Part 2 [Compatibility Mode]

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