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1
11/6/2006 Finite Shaft Element YK-1
FEM Formulation of Shaft element
Y.A Khulief, PhD, PEProfessor of Mechanical EngineeringKFUPM
11/6/2006 Finite Shaft Element YK-2
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
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11/6/2006 Finite Shaft Element YK-3
Shaft Coordinates:Consider the following shaft element
p
11/6/2006 Finite Shaft Element YK-4
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
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11/6/2006 Finite Shaft Element YK-5
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
11/6/2006 Finite Shaft Element YK-6
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
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11/6/2006 Finite Shaft Element YK-7
Dropping the index i
1- rotate by an angle about the X axis
( )φΩ+
2- Then by an angle about the new y-axis
yθ
1y
3- Then by an angle about the new z-axis
zθ Reference Shaft rotation
11/6/2006 Finite Shaft Element YK-8
Now, let us express the instantaneous angular velocity vector
Rotational Vector:
Ω
( ) ( ) ( )1 2ˆˆ ˆ
y zI j kω φ θ θ= Ω+ + + (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
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11/6/2006 Finite Shaft Element YK-9
Now, let us express the instantaneous angular velocity vector
Rotational Vector:
(4)( ) ( ) ( )[ ]
( ) ( ) ( ) ( ) ( )[ ]KJI
KJI
yyyz
y
ˆcoscosˆcossinˆsin
ˆsinˆcosˆ
φθθφθθ
φφθφω
+Ω++Ω−−+
+Ω++Ω++Ω=
In the linear theory of elasticity, small deformations are assumed, and hence small angles approximations are invoked in rewriting Eq.4 as
11/6/2006 Finite Shaft Element YK-10
Now, let us express the instantaneous angular velocity vector
Rotational Vector:
(5)
( ) ( )( ) ( )
ˆ ˆ ˆ( ) [cos sin ]ˆ ˆ ˆ[ sin cos ]
ˆ ˆ( ) [ cos( ) sin( )]ˆ[ sin( ) cos( )]
y
z y
z y y z
y z
I J K
I J K
I J
K
ω φ θ φ φ
θ θ φ φ
φ θ θ θ φ θ φ
θ λ φ θ φ
= Ω+ + Ω+ + Ω+
+ − − Ω+ + Ω+
= Ω+ − + Ω+ − Ω+
+ + + Ω+
( ) ( )( ) ( )
cos sinsin cos
x z y
y y z
z y z
ω φ θ θω ω θ φ θ φ
ω θ φ θ φ
⎧ ⎫Ω+ −⎧ ⎫⎪ ⎪⎪ ⎪= = Ω+ − Ω+⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪Ω+ + Ω+⎩ ⎭ ⎩ ⎭
Or, in matrix for as
(6)
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11/6/2006 Finite Shaft Element YK-11
Now, let us differentiate Eq.1 with respects to time
Velocity Vector:
(7)
where
(8)
[ ] pp p p p
drr r r r
dtω ω= + × = +
[ ]0
00
z y
z x
y x
ω ωω ω ω
ω ω
⎡ ⎤−⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦
11/6/2006 Finite Shaft Element YK-12
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
[ ] [ ] [ ]pv p v
p
edrN e r N
rdtω ω
⎧ ⎫= + = ⎨ ⎬
⎩ ⎭and e is the vector of nodal coordinates
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11/6/2006 Finite Shaft Element YK-13
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
ρ
ρ ωω
⎧ ⎫ ⎧ ⎫= ⎨ ⎬ ⎨ ⎬
⎩ ⎭ ⎩ ⎭⎧ ⎫ ⎧ ⎫⎪ ⎪⎡ ⎤= ⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎪ ⎪ ⎩ ⎭⎩ ⎭
∫
∫
11/6/2006 Finite Shaft Element YK-14
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 ρ
ω
ω
ω ω
⎡= ⎣
+
+
⎤+ ⎦
∫
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11/6/2006 Finite Shaft Element YK-15
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
11/6/2006 Finite Shaft Element YK-16
Kinetic Energy:To this end, one may utilize the following expression:
[ ] [ ]⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+−−−+−−−+
=22
22
22
~~
xyzyzx
zyxzyx
xzyxyzT
ωωωωωωωωωωωωωωωωωω
ωω (13)
[ ] [ ] ( )2 2 2
0
1 12 2
lT T
p p x x y y z zV
r r dV I I I dxρ ω ω ρ ω ω ω= = + +∫ ∫
The last term =
(14)
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11/6/2006 Finite Shaft Element YK-17
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
ρ ω ω ρ φ θ θ
θ φ θ φ
θ φ θ φ
= Ω+ −
+ Ω+ − Ω+
+ Ω+ + Ω+
∫ ∫
which can be further simplified as
11/6/2006 Finite Shaft Element YK-18
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
ρ ω ω φ φ
φ θ θ θ θ
= Ω + + Ω
− Ω+ + +
∫ ∫ ∫
∫ ∫
( )
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
φ φ φ
θ θφ θ θ
θ θ
= Ω + + Ω
⎧ ⎫ ⎧ ⎫− Ω+ + ⎨ ⎬ ⎨ ⎬
⎩ ⎭ ⎩ ⎭
∫ ∫ ∫
∫ ∫(17)
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11/6/2006 Finite Shaft Element YK-19
Kinetic Energy:Note that
(18)
y y DI I Iρ ρ= = x pI Iρ =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
φ φ
θ θ θ θ φ
θ θ
θ θ
ρ ω ω φ⎡ ⎤ ⎡ ⎤= Ω + + Ω⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤− Ω − ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪+ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
∫ ∫ ∫ ∫
∫ ∫
∫
Using FEM notations, Eq.17 becomes
11/6/2006 Finite Shaft Element YK-20
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 dxφΩ∫Ω
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11/6/2006 Finite Shaft Element YK-21
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
θ
φ φ φ
θ
θ
θ θ
φ
θ
θ θ
⎡ ⎤⎡ ⎤ ⎡
=
⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤⎡ ⎤ =⎣ ⎦ ⎣ ⎦
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ =⎨ ⎬ ⎨ ⎬⎪
⎤ =⎣ ⎦
⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
⎣ ⎦ ⎣ ⎦∫
∫
∫
∫
∫
11/6/2006 Finite Shaft Element YK-22
Kinetic Energy:
(20)
Now, Eq.18 reduces to
[ ] [ ] [ ] [ ]
[ ] [ ] eMeeMe
eGeeMeCdVrr
rT
eT
TT
Vp
TTp
21
21
21~~
21
12
1
+−
Ω−+Ω=∫ φωωµ
Note that is the inertia coupling between torsional and transverse vibrations which is time dependent
[ ]eM
12
11/6/2006 Finite Shaft Element YK-23
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
Ω−Ω+=
+−
Ω−+Ω+= φ
[ ] [ ] [ ] [ ]2t r eM M M M Mφ⎡ ⎤= + + −⎣ ⎦where
[ ] 12
Te M e=
11/6/2006 Finite Shaft Element YK-24
Kinetic Energy:The KE is finally expressed as
∫=l
vT
vt dxNANM0
][][][ µ
∫=l
DT
r dxNINM0
][][][ θθ
∫=l
pT dxNINM
0
][][][ ϕϕϕ
[ ] [ ] [ ]( )0
lT T
e p z y y zM I N N e N N N e N dxϕ θ θ ϕ θ θ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦∫
translational
rotational
torsional
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11/6/2006 Finite Shaft Element YK-25
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 dxθ θ⎡ ⎤⎡ ⎤= ⎣ ⎦ ⎣ ⎦∫
[ ] [ ]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)
11/6/2006 Finite Shaft Element YK-26
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 θθ , ( )φ
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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11/6/2006 Finite Shaft Element YK-27
The elastic rotations are related to bending deformations by
Strain Energy: (Bending & Shear)
( , )( , )
( , )( , )
by
bz
w x tx tx
v x tx tx
θ
θ
∂= −
∂∂
=∂
(24)
The strain due to bending is given by
2 2
2
*
2
*b bv wy z
x xε ∂ ∂= − −
∂ ∂(25)
11/6/2006 Finite Shaft Element YK-28
(26)
The shear strains are given by
(27)
Strain Energy: (Bending & Shear)
**
**
bxz
bxy
wwx x
vvx x
γ
γ
∂∂= −∂ ∂
∂∂= −∂ ∂
112
T
VU E dVε ε= ∫
Strain Energy due to bending
15
11/6/2006 Finite Shaft Element YK-29
(28)
Strain Energy: (Bending & Shear)
∫=A
z dAyI 2y 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 dAzI 2
Now defining
(29)
11/6/2006 Finite Shaft Element YK-30
(30)
Strain Energy: (Bending & Shear)
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 dVτ γ τ γ= +∫
16
11/6/2006 Finite Shaft Element YK-31
(32)
Recalling the constitutive relationships
(33)
Strain Energy: (Bending & Shear)
, '2(1 )
EG Poission s ratioνν
= ≡+
xy xy xz xzG and Gτ κ γ τ κ γ= =
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
νκν
ν µκν µ ν µ
+=
++ +
=+ + + +
(34)
Shear modulus
Shear factor
/i oR Rµ =and
11/6/2006 Finite Shaft Element YK-32
(35)
Strain Energy: (Bending & Shear)
(36)
2
2 2* ** *
0
1 ( )2
1 ( )2
xy xzV
lb b
U G dV
v wv wGA x dxx x x x
κ γ γ
κ
= +
⎧ ⎫⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂∂ ∂⎪ ⎪= − + −⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎢ ⎥⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭
∫
∫Expression strain energy in terms of v and w components of displacements, using
*
*
cos sinsin cos
v v ww v w
θ θθ θ
= −
= +
We can express Equations (30) and (35) as
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11/6/2006 Finite Shaft Element YK-33
(35)
Strain Energy: (Bending & Shear)
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
θ θ θ θ
θ θ
⎛ ⎞⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎜ ⎟= − + +⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞∂ ∂⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠⎛ ⎞∂⎛ ⎞ ∂⎛ ⎞⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠⎝ ⎠
∫
∫
∫
11/6/2006 Finite Shaft Element YK-34
(36)
Strain Energy: (Bending & Shear)2 2
20
1 ( )2
ls sv wU GA x dx
x xκ⎧ ⎫⎡ ⎤∂ ∂⎪ ⎪⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭∫
Strain Energy due to torsion
2
30
12
l
pU GI dxxφ∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠∫ (37)
18
11/6/2006 Finite Shaft Element YK-35
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
ε⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞= − + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
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.
11/6/2006 Finite Shaft Element YK-36
Strain Energy: (axial)
(39)
The strain energy is obtained by the following relationship:
24
0
1 12 2
l
V
U dV EA dxεσ ε= =∫ ∫
Substituting the strain expression from Eq.38 into Eq.39, results, upon some algebraic manipulations, in the following:
19
11/6/2006 Finite Shaft Element YK-37
Strain Energy: (axial)
(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
⎛ ⎞⎜ ⎟⎝ ⎠
⎫⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎪⎭
11/6/2006 Finite Shaft Element YK-38
Strain Energy: (axial)
(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
11/6/2006 Finite Shaft Element YK-39
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
θ θ
φκ
⎧ ⎫⎡ ⎤∂⎛ ⎞ ∂⎪ ⎪⎛ ⎞= +⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎢ ⎥⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭⎧ ⎫⎡ ⎤∂ ∂ ∂⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎢ ⎥⎨ ⎬ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
⎧⎪ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞+ − − −⎨⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∫
∫ ∫
0
L
dx⎫⎪⎬
⎪ ⎪⎩ ⎭∫
11/6/2006 Finite Shaft Element YK-40
(43)
Assumed displacement field
)(00000000000000000000000000
),(),(),(
4321
4321
21
teNNNN
NNNNNN
txwtxvtxu
vvvv
vvvv
uu
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
)(0000000000000000
4321
4321 teNNNN
NNNN
z
y⎥⎦
⎤⎢⎣
⎡−−
=⎭⎬⎫
⎩⎨⎧
θθθθ
θθθθ
θθ
-
(44)
(45) 1 2
( , ) 0 0 0 0 0 0 0 0 0 0 ( )x t N N e tφ φ φ⎡ ⎤= ⎢ ⎥⎣ ⎦
(FEM expressions)
21
11/6/2006 Finite Shaft Element YK-41
[ ] ( , )( , ) ( ) ( ) ( )( , )
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 2
T
y z y ze t u v w u v wθ θ φ θ θ φ= (46)Equations (43), (44) and (45) can be written as
(47)
( ) ( ) ( )yz
yy
zz
Ne t N x e t
Nθ
θθ
θθ
⎡ ⎤⎧ ⎫ ⎡ ⎤= =⎨ ⎬ ⎢ ⎥ ⎣ ⎦⎩ ⎭ ⎣ ⎦
where the nodal coordinate vector is given by
( , ) ( )x t N e tφφ ⎡ ⎤= ⎣ ⎦
(48)
(49)
(FEM expressions)
11/6/2006 Finite Shaft Element YK-42
[ ] [ ] [ ], , , , ,y zu v wN N N N N Nθ θ ϕ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦⎣ ⎦⎣ ⎦
yθ
where
are the shape functions associated with axial u, bending v and w, elastic rotations and , and torsional deformations , respectively.
zθ φ
(FEM expressions)
22
11/6/2006 Finite Shaft Element YK-43
[ ] [ ]
[ ] [ ]
[ ] [ ]
,
,
,
,
,
,
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
φ φ
θ θ
θ θ
φ φ
θ θ
θ θ
= =
= =
= =
⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦
(FEM expressions)
11/6/2006 Finite Shaft Element YK-44
1
2 31 [1 3 2 (1 )]1vN ξ ξ ξ= − + +Φ −+Φ
2
2 3 21 [ 2 ( )]1 2vN ξ ξ ξ ξ ξΦ
= − + + −+Φ
3
2 31 [3 2 ]1vN ξ ξ ξ= − +Φ+Φ
4
2 3 21 [ ( )]1 2vN ξ ξ ξ ξΦ
= − + + − ++Φ
(FEM shape functions)
23
11/6/2006 Finite Shaft Element YK-45
1
26 [ ](1 )
Nlθ ξ ξ= − +
+Φ
2
2 31 [1 4 3 (1 )]1
Nθ ξ ξ ξ ξ= − + + +Φ −+Φ
3
26 [ ](1 )
Nlθ ξ ξ= −
+Φ
4
31 [ 2 3 ]1
Nθ ξ ξ ξ= − + +Φ+Φ
(FEM shape functions)
11/6/2006 Finite Shaft Element YK-46
11Nφ ξ= −
2Nφ ξ=
11uN ξ= −
2uN ξ=
where ( / )x lξ =
(FEM shape functions)
24
11/6/2006 Finite Shaft Element YK-47
where the strain energy expression
[ ] eKeU T
21
=
Strain Energy:
(50)
11/6/2006 Finite Shaft Element YK-48
Stiffness Matrices:
(54)
[ ] [ ] [ ] [ ]a e sK k k k kφ⎡ ⎤= + + + ⎣ ⎦
[ ] [ ] [ ] [ ] [ ] ≡−−−= 4321 kkkkk a
where the matrix is the augmented stiffness matrix given by [ ]K
[ ] [ ] [ ] ≡= ∫l
Te dxBEIBk
0θθ
[ ] [ ] [ ] ≡= ∫l
pT dxBGIBk
0φφφ
[ ] [ ] [ ]0
l
s ssTk GA BB dxκ= ≡∫
Axial stiffness matrix
Elastic stiffness matrix
Shear stiffness matrix
Torsional stiffness matrix
(51)
(52)
(53)
25
11/6/2006 Finite Shaft Element YK-49
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
θθθθ
θθθθ
and
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.
11/6/2006 Finite Shaft Element YK-50
Equation of Motion:
(59)QqL
qL
dtd
=∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
Using the Lagrangean approach
L=T-U : Lagrangean function
,TTq e= Ω : Generalized coordinates
Q : vector of generalized forcesT : total kinetic energyU : total strain energy
26
11/6/2006 Finite Shaft Element YK-51
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 =+Ω+
Ω
P.S. This is the elemental equation of motion, which can be assembled using the standard finite element assembly procedure.
11/6/2006 Finite Shaft Element YK-52
Modal Characteristics:Solving the eigenvalue problem to obtain the rotor’s 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