Shaft FEM.pdf

1

11/13/2006 Finite Shaft Element YK-1

FEM Formulation of Shaft element

Y.A Khulief, PhD, PEProfessor of Mechanical EngineeringKFUPM


Assumptions:This is a Lagrangean formulation with the following ASSUMPTIONS:

Material is elastic, homogeneous, isotropic Plane x-secions initially perpendicular to neutral axis

remain plane, but no longer perpendicular to neutral axis after bending deformation

Deflections of the rotor are produced by displacements of points on the centerline

Disks are treated as rigidMaterial damping and fluidelastic forces are neglected

2


Shaft Coordinates:Consider the following shaft element

p


Shaft Coordinates:

( )i i ix y z Element Coordinate deformed state

The following Coordinates are assigned:

X Y Z Fixed inertial frame

( )i i iX Y Z Element coordinate undeformed state

Consider an arbitrary point pi on the undeformed element, which is then transformed into point p in the deformed state of the element

3


Shaft Coordinates:The global position of point p is defined by vector

p = +r R r (1)

Or, simply as

urRrp ++= 0 (2)

where is the deformation vectoru


Shaft Coordinates:The element undergoes axial deformation u in the X direction and two bending deformations v and w in the Y and Z direction, respectively.

Now, let us describe the element x-section orientation after deformation; i.e. to establish the coordinate transformationfrom

i i i i i itoX Y Z x y z

See next figure for rotational angles

4


Dropping the index i

1- rotate by an angle about the X axis

( )+

2- Then by an angle about the new y-axis

y

1y

3- Then by an angle about the new z-axis

z Reference Shaft rotation


Now, let us express the instantaneous angular velocity vector

Rotational Vector:

( ) ( ) ( )1 2 y zI j k = + + + (3)The unit vectors directions are shown on previous figure.

Note that is the rotor angular velocity.

Transforming the velocity vector of Eq.3 into the global coordinate system , one obtains X Y Z

5



Rotational Vector:

(4)( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( )[ ]KJIKJI

yyyz

y

coscoscossinsin

sincos

++++

+++++=

In the linear theory of elasticity, small deformations are assumed, and hence small angles approximations are invoked in rewriting Eq.4 as



Rotational Vector:

(5)

( ) ( )( ) ( )

( ) [cos sin ] [ sin cos ]

( ) [ cos( ) sin( )][ sin( ) cos( )]

y

z y

z y y z

y z

I J K

I J K

I J

K

= + + + + +

+ + + +

= + + + +

+ + + +

( ) ( )( ) ( )

cos sinsin cos

x z y

y y z

z y z

+ = = + +

+ + +

Or, in matrix for as

(6)

6


Now, let us differentiate Eq.1 with respects to time

Velocity Vector:

(7)

where

(8)

[ ]{ }p p p p pdr

r r r rdt

= + = +

[ ]0

00

z y

z x

y x

=


Using the FEM notations, one can express the deformation vector in the form:

Velocity Vector:

(9)

where is the shape function matrix. Now Eq.7 can be expressed as

(10)

{ } [ ]{ }eNuu v==

[ ]vN

[ ]{ } [ ]{ } [ ]p v p vp

edrN e r N

rdt

= + =

and {e} is the vector of nodal coordinates

7


Kinetic Energy:

(11)

The kinetic energy of the element is obtained by integrating thekinetic energy of the infinitesimal volume at point p over the volume V

[ ]

12

12

Tp p

V

TTT v

p vTpV

dr drKE dV

dt dt

eNe r N dVr

=

=


Kinetic Energy:

(12)

Which can be written in the form

{ } [ ] [ ]{ }

{ } [ ] [ ]{ }{ } [ ] [ ]{ }{ } [ ] [ ]{ }

12

TTv v

V

TTv p

T Tp v

T Tp p

e N N e

e N r

r N e

r r dV

KE

=

+

+

+

8


Kinetic Energy:The first term in Equation ( 12) gives the kinetic energy due to translation; the second and third terms are identically zero if moments of inertia are calculated with respect to center of massof the element. The last term gives kinetic energy due to rotation that includes gyroscopic moments.

Now, let us evaluate the last term of Eq.12


Kinetic Energy:To this end, one may utilize the following expression:

[ ] [ ]

+++

=22

22

22

~~

xyzyzx

zyxzyx

xzyxyzT

(13)

{ } [ ] [ ]{ } ( )2 2 20

1 12 2

lT T

p p x x y y z zV

r r dV I I I dx = = + +

The last term =

(14)

9


Kinetic Energy:Substituting from Eq.6 into Eq.14, one gets

(15){ } [ ] [ ]{ } ( ){

( ) ( )( )( ) ( )( ) }

2

02

2

cos sin

sin cos

lT T

p p x z yV

y y z

z y z

r r dV I

I

I dx

= +

+ + +

+ + + +

which can be further simplified as


Kinetic Energy:

(16)

Or, simply as

{ } [ ] [ ]{ } ( ) ( )

( ) ( )

2 2

0 0

2 2

0 0

1 12 2

l lT T

p p p pV

l l

p z y D y z

r r dV I dx I dx

I dx I dx

= + +

+ + +

( )

2

0 0 0

0 0

12

l l lT

p p p

Tl lT y y

p z y Dz z

I dx I dx I dx

I dx I dx

= + +

+ +

(17)

10


Kinetic Energy:Note that

(18)

y y DI I I = = x pI I =and

{ } [ ] [ ]{ } { } { }

{ } { } { } { } { }

{ } { }

2

0 0 0

0 0

0

1 12 2

[ ] [ ]

[ ] [ ]

z y z y

y y

z z

l l lT TT T

p p p p pV

l lT TT T

p p

Tl

TD

r r dV I dx e N I N e dx I dx

e N I N e dx e N I N e N e dx

N Ne I e dx

N N

= + +

+

Using FEM notations, Eq.17 becomes


Kinetic Energy:The term gives the inertial coupling between rigid body coordinates and elastic coordinates. For constant this term has no contribution to the equation of motion of the drillstring, and can be neglected.

Now, let us introduce some matrix expressions to simplify the final form of the KE expression :

0

l

pI dx

11


Kinetic Energy:

(19)[ ]

{ } [ ]

[ ]

0

10

0

10

0

12

[ ] [ ]

[ ] [ ]

z

z y

y y

z z

y

lT

l

p

lT

p

lT

p

Tl

D

p

r

e

I dx C

N I N dx M

N I N d

I

x G

N NI dx

N N e N d

MN

x M

N

=

=

=

=

=


Kinetic Energy:

(20)

Now, Eq.18 reduces to

{ } [ ] [ ]{ } { } [ ]{ } { } [ ]{ }

{ } [ ]{ } { } [ ]{ }eMeeMe

eGeeMeCdVrr

rT

eT

TT

Vp

TTp

21

21

21~~

21

12

1

+

+=

Note that is the inertia coupling between torsional and transverse vibrations which is time dependent

[ ]eM

12


Kinetic Energy:

(21)

The KE is finally expressed as

{ } [ ]{ } { } [ ]{ } { } [ ]{ }

{ } [ ]{ } { } [ ]{ }

{ } [ ]{ } { } [ ]{ }eGeCeMe

eMeeMe

eGeeMeCeMeKE

TT

rT

eT

TTt

T

12

1

12

1

21

21

21

21

21

21

+=

+

++=

[ ] [ ] [ ] [ ]2t r eM M M M M = + + where

{ } [ ]{ }12

Te M e=


Kinetic Energy:The KE is finally expressed as

=l

vT

vt dxNANM0

][][][

=l

DT

r dxNINM0

][][][

=l

pT dxNINM

0

][][][

[ ] [ ]{ } { }[ ]( )0

lT T

e p z y y zM I N N e N N N e N dx =

translational

rotational

torsional

13


Kinetic Energy:The gyroscopic matrix [G] and can be represented by the following expression , where for constant angular speed

[ ]10

z y

lT

pG I N N dx =

[ ] [ ]1 1[ ]TG G G=

Next, is to carry out the integrations to arrive at explicit expressions of the non-zero entries of the aforementioned element coefficient matrices; see Appendix

(22)


The deformation of a typical cross-section of the drillstring may be expressed by three translations and three rotations. Two of the translations (v, w) are due to bending in the Y and Zdirections and the third one (u) is due to axial translation. The three rotations are due to bending and due to torsion .

Strain Energy:

( ),s sv w( ),b bv w

( )zy , ( )

The two translations (v, w) consist of contributions due to bending, and due to shear; that is

( , ) ( , ) ( , )( , ) ( , ) ( , )

b s

b s

v x t v x t v x tw x t w x t w x t

= += +

(23)

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The elastic rotations are related to bending deformations by

Strain Energy: (Bending & Shear)

( , )( , )

( , )( , )

by

bz

w x tx tx

v x tx tx

=

=

(24)

The strain due to bending is given by

2 2

2

*

2

*b bv wy z

x x =

(25)


(26)

The shear strains are given by

(27)


**

**

bxz

bxy

wwx x

vvx x

=

=

112

T

VU E dV =

Strain Energy due to bending

15


(28)


=A

z dAyI2

y zI I I= =

22 * 2 *

1 2 20

2 22 * 2 * 2 * 2 *2 2

2 2 2 20

2

22

lb b

A

lb b b b

A

v wEU y z dAdxx x

v v w wE y yz z dAdxx x x x

=

= + +

=A

y dAzI2

Now defining

(29)


(30)


2 22 * 2 *

1 2 202

lb b

z yv wEU I I dxx x

= +

(31)

Strain energy due to bending

Strain Energy due to shear

2 ( )xy xy xz xzVU dV = +

16


(32)

Recalling the constitutive relationships

(33)


, '2(1 )

EG Poission s ratio

= +

xy xy xz xzG and G = =

2 2

2 2 2

6(1 )7 6

6(1 )(1 )(7 6 )(1 ) (20 12 )

for solid circular cross section

for hollow circular cross section

+=

++ +

=+ + + +

(34)

Shear modulus

Shear factor

/i oR R =and


(35)


(36)

2

2 2* ** *

0

1 ( )2

1 ( )2

xy xzV

lb b

U G dV

v wv wGA x dxx x x x

= +

= +

Expression strain energy in terms of v and w components of displacements, using

*

*

cos sinsin cos

v v ww v w

=

= +

We can express Equations (30) and (35) as

17


(35)


Similarly, strain energy due to shear

2 22 2 2 2

1 2 2 2 20

2 22 2

2 20

2 2

0

cos sin cos sin2

( )2

( )2

lb b b b

z y

lb b

ly z

v v w wEU I I dxx x x x

v wE I x dxx x

E I x dxx x

= + + = + = +


(36)

Strain Energy: (Bending & Shear)2 2

20

1 ( )2

ls sv wU GA x dx

x x = +

Strain Energy due to torsion

2

30

12

l

pU GI dxx = (37)

18


Strain Energy: (axial & bending)

(38)

The axial displacement can be defined to account for the effect of bending large deflection on the axial movement.

Therefore, the strain in the axial direction can be defined fromEulerian strain tensor as [ See Continuum Mech. Ref]:

2 2212

b bdv dwdu dudx dx dx dx

= + +

The first term in Eq.36 is the linear term of axial strain and it will generate the linear terms in the stiffness matrix. The remainingterms are second order terms which are usually neglected in linear structural analysis.


Strain Energy: (axial)

(39)

The strain energy is obtained by the following relationship:

24

0

1 12 2

l

V

U dV EA dx = =

Substituting the strain expression from Eq.38 into Eq.39, results, upon some algebraic manipulations, in the following:

19



(40)

The strain energy is obtained by the following relationship: 22 22

40

2 22 3

0

24 2 2

1 12 2

12

1 1 14 2 2

Lb b

Lb b

b

dv dwdu duU EA dxdx dx dx dx

dv dwdu du du duEAdx dx dx dx dx dx

dv ddu du dudx dx dx dx

= + +

=

+ + +

2

4 4 2 21 1 14 4 2

b

b b b b

wdx

dv dw dv dw dxdx dx dx dx

+ + +



(41)

Neglecting higher order terms leads to : 2 22 3

40

12

Lb bdv dwdu du du duU EA dx

dx dx dx dx dx dx =

Now, total strain energy becomes

20


Strain Energy: (total)

(42)

Now the total strain energy is 1 2 3 4U U U U U= + + +

2 2

0

2 2 2

0 0

2 22 3

( )2

1 1( )2 2

12

ly z

l ls s

p

b b

EU I x dxx x

v wGA x dx GI dxx x x

dv dwdu du du duEAdx dx dx dx dx dx

= + + + +

+

0

L

dx


(43)

Assumed displacement field

{ })(00000000000000000000000000

),(),(),(

4321

4321

21

teNNNN

NNNNNN

txwtxvtxu

vvvv

vvvv

uu

=

{ })(0000000000000000

4321

4321 teNNNN

NNNN

z

y

=

-

(44)

(45){ }1 2

( , ) 0 0 0 0 0 0 0 0 0 0 ( )x t N N e t =

(FEM expressions)

21


{ } [ ]{ }( , )( , ) ( ) ( ) ( )( , )

u

v t

w

u x t Nv x t N e t N x e tw x t N

= =

( ){ } { }1 1 2 21 1 1 1 2 2 2 2T

y z y ze t u v w u v w = (46)Equations (43), (44) and (45) can be written as

(47)

{ } { }( ) ( ) ( )yz

yy

zz

Ne t N x e t

N

= =

where the nodal coordinate vector is given by

{ }( , ) ( )x t N e t =

(48)

(49)

(FEM expressions)


[ ] [ ] [ ], , , , ,y zu v w

N N N N N N

y

where

are the shape functions associated with axial u, bending v and w, elastic rotations and , and torsional deformations , respectively.

z

(FEM expressions)

22


[ ]{ } [ ]{ }

[ ]{ } [ ]{ }

[ ]{ } [ ]{ }

{ } { }

{ } { }

{ } { }

,

,

,

,

,

,

y y

z z

u u

v v

w w

y y

z z

duu N e B edxdvv N e B edxdww N e B edx

N e N e

N e N e

N e N e

= =

= =

= =

= =

= =

= =

(FEM expressions)


1

2 31 [1 3 2 (1 )]1v

N = + + +

2

2 3 21 [ 2 ( )]1 2v

N = + + +

3

2 31 [3 2 ]1v

N = ++

4

2 3 21 [ ( )]1 2v

N = + + ++

(FEM shape functions)

23


1

26 [ ](1 )

Nl

= ++

2

2 31 [1 4 3 (1 )]1

N = + + + +

3

26 [ ](1 )

Nl

= +

4

31 [ 2 3 ]1

N = + ++



11N =

2N =

11uN =

2uN =

where ( / )x l =


24


where the strain energy expression

{ } [ ]{ }eKeU T21

=

Strain Energy:

(50)


Stiffness Matrices:

(54)

[ ] [ ] [ ] [ ]a e sK k k k k = + + + [ ] [ ] [ ] [ ] [ ] = 4321 kkkkk a

where the matrix is the augmented stiffness matrix given by [ ]K

[ ] [ ] [ ] = l

Te dxBEIBk

0

[ ] [ ] [ ] = l

pT dxBGIBk

0

[ ] [ ] [ ]0

l

s ssTk GA BB dx=

Axial stiffness matrix

Elastic stiffness matrix

Shear stiffness matrix

Torsional stiffness matrix

(51)

(52)

(53)

25


Stiffness Matrices:

(55)

(56)

(57)

(58)

[ ] [ ] [ ]

[ ] [ ] [ ]{ }[ ]

[ ] [ ] [ ]{ }[ ] [ ] [ ]{ }[ ]

[ ] [ ] [ ]{ }[ ] [ ] [ ]{ }[ ]

+=

+=

=

=

L

uTT

u

L

uTT

u

L

uuT

u

L

uT

u

dxNeBNNeNBEAk

dxNeBNNeNBEAk

dxBeBBEAk

dxBEABk

zzzz

yyyy

04

03

02

01

21

21

23

and


Stiffness Matrices:P.S. All stiffness matrices are state independent and symmetric except which are state dependent and nonsymmetrical.

See details of FEM in Ref: Mohiuddin & Khulief, Modal characteristics of rotors using a conical shaft finite element, CMAME, Vol. 115, 1994, p.125-144.

[ ] [ ] [ ]2 3 3,k k and k

26


Equation of Motion:

(59)QqL

qL

dtd

=

Using the Lagrangean approach

L=T-U : Lagrangean function

{ }, TTq e= : Generalized coordinates Q : vector of generalized forcesT : total kinetic energyU : total strain energy


Equation of Motion:

(60)

Substituting the Lagrangean function into Eq.59, and carry out the required differentiation, one gets for a shaft rotating at aconstant angular speed

[ ]{ } [ ]{ } [ ]{ } { }QeKeGeM =++

P.S. This is the elemental equation of motion, which can be assembled using the standard finite element assembly procedure.

27


The DISK:

{ } { } { } { }2 11 1( ) 2 2T Td d d d d d d

disk pKE e M e I e G e = + (61)

The disk is assumed to be rigid and solely characterized by its kinetic energy. Using the same procedure followed for the shaft element leading to Eq.21, one can write the kinetic energy of the disk as

2d d d d dt r eM M M M M = + +

where


The DISK:0 0 0 0 0 0

0 0 0 00 0 00 0 0

0 00

d

ddt

mm

M

=

(62)

0 0 0 0 0 00 0 0 0 0

0 0 0 00 0

00

dr d

DdD

MI

I

=

(63)

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The DISK:0 0 0 0 0 0

0 0 0 0 00 0 0 0

0 0 00 0

d

dp

M

I

=

0 0 0 0 0 00 0 0 0 0

0 0 0 00 0

0 00

de

y

M

=

(64)

(65)


The DISK:

1

0 0 0 0 0 00 0 0 0 0

0 0 0 00 0

0 00

ddp

GI

=

(66)

(67){ } { } { }d d d d dM e G e Q + =

Now, applying the Lagrangean, one gets for the disk

29


The BEARINGS:

yy yz zz zy

yy yz zz zy

W K v v K w v K w w K v w

C v v C w v C w w C v w

=

(68)

Assuming that the bearing characteristics; i.e. stiffness and damping are know, on may utilize the principle of virtual work to include the bearing dynamics into equation of motion.

The virtual work done due to bearing forces acting on theshaft is given by

Or, simply as

v wW F v F v = +


The BEARINGS:

{ } { } { }b b b b bC e K e Q + =

yy yz yy yzv

zy zz zy zzw

K K C CF v vK K C CF w w

=

yy yz

zy zz

K KK K

where Fv and Fw are the components of the bearing generalized coordinates, which can written in matrix form as

yy yz

zy zz

C CC C

(69)

Or as

where

(70)

30


Modal Characteristics:Solving the eigenvalue problem to obtain the rotors modal characteristics.

Damped Eigenvalue Mode Shape Plot

-1.5-1

-0.50

0.51

1.5

0 0.5 1 1.5 2 2.5

Axial Location, meters

Re(x)

Im(x)

Re(y)

Im(y)

Boiler Feed Pump (BFP)On 2 Journal & 1 Thrust Bearings

Damped Eigenvalue Mode Shape PlotBoiler Feed Pump (BFP)

On 2 Journal & 1 Thrust Bearings


Modal Characteristics:Simple demonstration

31


Modal Characteristics:In order to obtain the eigenvalue solution associated with the homogenous equation of motion, one can represent Eq.60 in the following state-space form

[ ]{ } [ ]{ } { }A y B y Q+ =

{ }TT Ty e e =

[0] [ ][ ]

[ ] [ ]M

AM G

=

[ ] [0][ ]

[0] [ ]M

BK

=

{ } {0 }T TQ Q=

where

(61)


Modal Characteristics:Because of the nonlinearities introduced in the mass and stiffness matrices due to torsion-transverse and axial-transversecouplings, [A] and [B] are non symmetric state space matrices. In modal analysis, [A] and [B] will retrieve their symmetric properties as the nonlinear coupling vanishes. Therefore, [A] will be a skew symmetric matrix and [B] will be a symmetric matrix

The dimension of each of the matrices [M], [K] and [G] is (6n x6n) where n is the number of nodes, while the matrices [A] and [B] are of dimension (12n x 12n).

32


Eigenvalue Solution:In rotordynamics, the eigenvalue problem can be viewed from the prospective of the either of the following equations of motion:

(a) The eigenvalue associated with equation of motion as writtenwith respect to the inertial frame; i.e. Eq.61

(b) The eigenvalue associated with equation of motion as writtenwith respect to the rotating frame.


Eigenvalue Solution:

{ } { } ty y e=

(a) The eigenvalue associated with equation of motion as written with respect to the inertial frame; i.e. Eq.61

Since [A] and [B] are not necessarily symmetric, then the eigenvalues can be obtained by solving self-adjoint eigenvalue problem associated with

(62)

Assuming solution in the form

[ ]{ } [ ]{ } { }0A y B y+ =

(63)

33



[ ] [ ]( ){ } { }0T Ti iA B L + =

[ ] [ ]( ){ } { }0ti A B y e + = (64)Due to loss of symmetry of the matrices, right and left eigenvectors must be introduced, i.e.

(66)

Substituting in Eq.62, one gets

[ ] [ ]( ){ } { }0i iA B R + = (65)



[ ] [ ]( )i A B +

[ ] [ ] [ ] [ ] 0T Ti iA B A B + = + =

[ ] [ ]( )T Ti A B +and

If either [A] or [B] or both being not symmetric, then [R] does not equal [L], however for symmetric matrices, [R] = [L].

(67)

Since [R] and [L] are not zero, then

must be singular

The solution of Eq.67 yields 2n complex eigenvalues of the form

j = (68)

34



Where the imaginary part is the damped natural frequency.

The real part is related to modal damping as

000

>

= =

Critically damped mode

Unstable mode

Stable mode

How the real part is related to the logarithmic decrement ?


Modal Transformation:Now, one can introduce the modal transformation {y} = [R] {u}, where {u} is the vector of modal coordinates. In general, the modal matrices [R] and [L] are composed of a set of complex eigenvectors (mode shapes) that account for a selected set of significant modes.

Pre-multiplying both sides of Eq.62 by [L]T and substituting for {y} in terms of modal coordinates , the truncated modal form of the equations of motion can be written as

35


Modal Transformation:

[ ] [ ] [ ][ ]TrB L B R=

[ ] [ ] [ ]TrQ L Q=

[ ]{ } [ ]{ } { }r r rA u B u Q+ =This is the reduced-order modal form of the equation of motion, where the coefficient modal matrices are given by

(69)

[ ] [ ] [ ][ ]TrA L A R=

What is r ?



[ ]

cos sin 0 0sin cos 0 0

0 0 cos sin0 0 sin cos

t tt t

Tt tt t

=

(b) The eigenvalue associated with equation of motion as written with respect to the rotating frame.

Now let {p} be the transformed vector of the nodal coordinate vector {e}, such that {e} = [T] {p}

(70)

If is the angle between the two reference frames, thent

36



{ }{ } Ty zp v w =

[ ] [ ]{ } [ ]{ }e S p T p= +

(71)

where, after ignoring the axial and torsional deformations, the nodal coordinate vector becomes

Now, differentiating the coordinate equation twice

2[ ] [ ]({ } { }) 2 [ ][ ]{ }e T p p T S p = +



[ ] 1 [ ]S T

=

=

where,

Now, for a shaft rotating at a constant speed , define the following whirl ratio:

Proceed to write the equation of motion in the rotating coordinate system.

37


Eigenvalue Solution: where,

{ } ( ){ }( )( ){ } { }2

2 [ ]

[ ] [ ] 0

M p M G p

K M G p

+ +

+ + =

Limitations:Can only be solved for undamped isotropic bearings; It can only be solved for a specific whirl ratio

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Shaft FEM.pdf