Shaft FEM-1.pdf

110/30/2006 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


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


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


Dropping the index i

1- rotate by an angle about the X axis

( )+

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

y1y

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



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)


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


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

dt dt

eNe r N dVr

= =


Kinetic Energy:

(12)

Which can be written in the form

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

12

TTv v

VTT

v p

T Tp v

T Tp p

e N N e

e N r

r N e

r r dV

KE

= ++

+


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)


Kinetic Energy:Substituting from Eq.6 into eq.14, on 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


Kinetic Energy:The KE is finally expressed as

= l vTvt dxNANM0

][][][

= l DTr dxNINM0

][][][

= l pT dxNINM0

][][][

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