Finite Element Method (FEM) for Vibration

(1)

1652103-433 Introduction to Mechanical Vibration orn University, THAILANDhanical Engineering Department

CHAPTER 7

FINITE ELEMENT METHOD (FEM) FOR VIBRA TION

2103-433 Introduction to Mechanical Vibration Chulalongkorn University, THAILANDMechanical Engineering Department

Finite Element Method (FEM) f or Vibration

Outlines:

•

Introduction

•

and of bar elements

•

and of beam elements

•

Lumped mass matrices

•

Trusses

•

Model reduction KMKM

(2)

Vibration of Fle xib le Structures

so called “Distributed-Parameter Systems Suspension headsin hard disk drives

Airplane wingstructure

Flexible robot arm

Model of Fle xib le Structures

Flexible structures can be modeled as bar, beam, plate, shell, and etc.

Two ways to model:

1. Analytical model

(beyond scope of this class)

•

use the knowledge of strength of material and dynamics to derive EOM.

•

EOMs are in form of Partial Differential Equations (PDE).

•

Use discretization technique to solve PDE.

2. Finite Element Model (FEM)

(3)

FEM: Appr o ximation method

Procedure:

•

Divide (discretize) the structure into a number of small simple elements. The elements are connected to each others by nodes.

•

All node displacement are chosen as DOFs, so called nodal DOF.

•

Displacement function within the element are approximated by a linear combination of low-order polynomials.

•

For vibration application, we construct the

matrices and w.r.t. all nodal DOF using energy method. KM d1d2 d4d3

di is nodal degree of freedom d1d2Mdn Element nodes

Bar Elements

From solid mechanics, static displacement of the bar is governed by

Step I: divide the bar into elements.

3 elements, 4 nodes

, , , and are nodal displacements.

Step II: a displacement function

Consider any element, e.g., the 1 st element.Assume a displacement function within the element as EAx 2 2d du0=

u1t()u2t()u3t()u4t()

uxt,()c1t()xc2t()+= u(x,t)x

E, ρ, L

u1(t)u2(t)u3(t)u4(t)

123

1234

112

(4)

B.C.s:

,

Hence and

Rewrite a displacement function in matrix form:

Differentiate u w.r.t.x: x0=uxt,()u1t()=

xl=uxt,()u2t()=

c1t() u2t()u1t()–l --- =c2t()u1t()=

uxt,()1 xl --,– xl -- u1t()u2t() Nut()==

x∂ ∂uxt,()xd dNut()1l ---,– 1l --- u1t()u2t() But()=== Shape functions1 1-x/lx/l

xl xl u2 u1 u(x,t)

Step III: Construct local

and

Str ain energy:

Strain energy of the bar element V(t) is

or

where is a local stiffness matrix defined as KM

Vt() 12 ---EAx∂ ∂uxt,() 2xd0 l

∫

=

Vt() 12 ---EAu TB TBu[]xd0 l

∫

=

Vt()12 ---u Tt()EAl --- 11–1–1 ut()12 ---u Tt()kut()==

k

kEAl --- 11–1–1 =

(5)

Kinetic energy:

Kinetic energy of the bar element T(t) is

where is the bar density.

Since

Therefore

or

where is a local mass matrix defined as Tt() 12 ---Aρx()t∂ ∂uxt,() 2xd0 l

∫

=ρx()

t∂ ∂uxt,()Nt∂ ∂ut()Nu˙t()==

Tt() 12 ---Aρx()u˙ TN TNu˙[]xd0 l

∫

=

Tt()12 ---u˙ Tt()ρAl6 --- 2112 u˙t()12 ---u˙ Tt()mu˙t()==

m

mρAl6 --- 2112 =

Step IV : Assemb ly

Let’s consider a three-element bar belows.

For any one element,

In this case ,

Str ain energy:

1-st element:

2-nd element: kEAl --- 11–1–1 =

lL3 ---=k3EAL --- 11–1–1 =

V1t()12 ---u Tku3EA2L --- u1u2

  

T11–1–1 u1u2

  

==

V2t()3EA2L --- u2u3

  

T11–1–1 u2u3

  

= lll

L u1(t)u2(t)u3(t)u4(t)

(6)

3-rd element:

Total strain energy is

or

With boundary conditions;, then V3t()3EA2L --- u3u4

  

T11–1–1 u3u4

  

=

VV1V2V3++=

Vt()3EA2L --- u1u2u3u4 T11–001–21–001–21–001–1 u1u2u3u4 =

u0t,()u1t()0==

Vt()3EA2L --- 0u2u3u4 T11–001–21–001–21–001–1 0u2u3u4 =

2103-433 Introduction to Mechanical Vibration Chulalongkorn University, THAILANDMechanical Engineering DepartmentV(t) can be reduced towhere K is a global stiffness matrix.

, for the 3-element bar Vt()3EA2L --- u2u3u4 T21–01–21–01–1 u2u3u4 12 ---u TKu==

K3EAL --- 21–01–21–01–1 =

(7)

Again, consider the three-element bar:

For any one element,

Kinetic energy:

1-st element:

2-nd element:

3-rd element:

Total kinetic energy is

or mρAl6 --- 2112 ρAL18 --- 2112 ==

T1t()12 ---u˙ Tmu˙ρAL36 --- u˙1u˙2 T2112 u˙1u˙2 ==

T2t()ρAL36 --- u˙2u˙3 T2112 u˙2u˙3 =

T3t()ρAL36 --- u˙3u˙4 T2112 u˙3u˙4 =

Tt()T1T2T3++=

2103-433 Introduction to Mechanical Vibration Chulalongkorn University, THAILANDMechanical Engineering DepartmentWith boundary conditions;, thenV(t) can be reduced towhere M is a global mass matrix.

, for the 3-element bar TρAL36 --- u˙1u˙2u˙3u˙4 T2100141001410012 u˙1u˙2u˙3u˙4 =

u˙0t,()u˙1t()0==

TρAL36 --- 0u˙2u˙3u˙4 T2100141001410012 0u˙2u˙3u˙4 =

TρAL36 --- u˙2u˙3u˙4 T410141012 u˙2u˙3u˙4 12 ---u˙ TMu˙==

MρAL18 --- 410141012 =

(8)

Notes:

•

To analyze vibration of the structure, we

use the global mass matrix and global

stiffness matrix to construct the matrix equation

Questions:

•

Can we determine and , if the element size is not uniform?

•

How accurate is the FEM result? MK

Mu˙˙t()Kut()+0=

KM

Comparison of Natural Frequencies

For L = 1m, ρ = 2700 kg/m 3, and

N/m 2. E710 7×=

Exact ωnωn from FEM%difference

1-element model:

ωn1 = 8,819 rad/sωn1 = 7,998 rad/s

3-element model:

ωn1 = 8,092 rad/s ωn2 = 26,458 rad/s

ωn3 = 47,997 rad/s ωn1 = 7,998 rad/s

ωn2= 23,994 rad/s

ωn3= 39,900 rad/s 1.18%10.3%20.3%

Rule of thumb: At least twice as manyelements must be used than number of accurate frequencies required. 10.3%

(9)

Beam Elements

From solid mechanics, static displacement of the beam is governed by

Step I: divide the beam into elements.

1 element, 2 nodes

and are nodal linear displacement.

and are nodal angular displacement.

Step II: a displacement function

Assume a displacement function within the element as x 2 2

d dEIx 2 2d dw0=

w1t()w2t()φ1t()φ2t()

wxt,()c1t()x 3c2t()x 2c3t()xc4t()+++= w(x,t)x

EA, ρ, I, L

w1(t) w2(t)

φ1(t)φ2(t) l

2103-433 Introduction to Mechanical Vibration Chulalongkorn University, THAILANDMechanical Engineering DepartmentB.C.s:

, and

Hence

and

Rewrite a displacement function in matrix form:

where x0=w0t,()w1t()=x∂ ∂w0t,()φ1t()=

xl=wlt,()w2t()=x∂ ∂wlt,()φ2t()=

c1t()1l 3 ---2w1w2–()lφ1φ2+()+[]=

c2t()1l 2 ---3w2w1–()l2φ1φ2+()–[]=

c3t()φ1t()=c4t()w1t()=

wxt,()N1N2N3N4 w1t()φ1t()

w2t()φ2t() Ndt()==

Shape functions

(10)

1832103-433 Introduction to Mechanical Vibration orn University, THAILANDhanical Engineering Department₀₀

.10.20.30.40.50.60.70.80.91 0 0.2 0.4 0.6 0.8 1

00.10.20.30.40.50.60.70.80.91 5 5 0 0.1 N1x()13x 2l 2 ---–2x 3l 3 ---+=N2x() x2x 2l --- –x 3l 2 ---- +=

N3x()3x 2

l 2 --- 2x 3l 3 --- –=N4x() x 2l ----–x 3l 2 ----+=

N1(x)N3(x)

N2(x)

N4(x) 1 1 x/l

x/l

Step III: Construct local

and

Strain ener gy

Strain energy of the beam element V(t) is

where is the flexural rigidity of the beam.

Plug into and integrate , we get

where is a local stiffness matrix defined as KM

Vt() 12 ---εσ 12 ---EIx 2 2

∂ ∂wxt,() 2

xd0 l

∫

==

EIwxt,()Nx()dt()=Vt()Vt()

Vt()12 ---d Tt()kdt()=

k

kEI

l 3 --- 126l12–6l

6l4l 26l–2l 2

12–6l–126l–

6l2l 26l–4l 2 =

(11)

Kinetic ener gy

Kinetic energy of the beam element T(t) is

where is the beam density.

Since

Therefore

or

where is a local mass matrix defined as Tt() 12 ---Aρx()t∂ ∂wxt,() 2xd0 l

∫

=ρx()

t∂ ∂wxt,()Nt∂ ∂dt()Nd˙t()==

Tt() 12 ---Aρx()d˙ TN TNd˙[]xd0 l

∫

=

Tt()12 ---d˙ Tt()md˙t()=

mmρAl420 --- 15622l5413l–

22l4l 213l3l 2–5413l15622–l

13l–3l 2–22l–4l 2 =

Step IV : Assemb ly

Let’s consider a two-element beam belows.

For one element,

For 2-element , then kEIl 3 --- 126l12–6l

6l4l 26l–2l 2

12–6l–126l–

6l2l 26l–4l 2 =

mρAl420 --- 15622l5413l–

22l4l 213l3l 2–5413l15622–l

13l–3l 2–22l–4l 2 =

lL2 --- = w1(t)

φ1(t)φ2(t)φ3(t) w2(t)w3(t)ll

(12)

Strain ener gy

1-st element: k8EIL 3 --- 123L12–3L

3LL 23L–0.5L 2

12–3L–123L–

3L0.5L 23L–L 2 =

mρAL840 --- 15611L546.5L–

11LL 26.5L0.75L 2–546.5L15611L–

6.5L–0.75L 2–11L–L 2 =

V1t()12 ---d1 Tkd1 w1φ1w2φ2

      

Tk w1φ1w2φ2

      

==

2103-433 Introduction to Mechanical Vibration Chulalongkorn University, THAILANDMechanical Engineering Department2-nd element:Total strain energy is

Extended matrix form of V(t) is

With fixed-free B.C.s,

and

,

V(t) is then V2t()12 ---d2 Tkd2 12 --- w2φ2w3φ3

      

Tk w2φ2w3φ3

      

==

VV1V2+=

Vt()12 --- w1φ1w2φ2w3φ3 Tw1φ1w2φ2w3φ3 = k

k

w0t,()w1t()0==

x∂ ∂w0t,()φ1t()0==

(13)

V(t) can be reduced to

where K is a global stiffness matrix.

,

for the 2-element beam Vt()12 --- 00w2φ2w3φ3 T00w2φ2w3φ3 = k

k

Vt()8EI

2L 3 ---- w2φ2w3φ3 T1212+()3L–3L+()12–3L

3L–3L+()L 2L 2+()3L–0.5L 2

12–3L–123L–

3L0.5L 23L–L 2 w2φ2w3φ3 12 ---d TKd==

K8EI

L 3 --- 24012–3L

02L 23L–0.5L 2

12–3L–123L–

3L0.5L 23L–L 2 =

Kinetic ener gy

1-st element:

2-nd element:

Total kinetic energy is T1t()12 ---d˙1 Tmd1 ˙ w˙1φ˙1w˙2φ˙2

      

Tm w˙1φ˙1w˙2

φ˙2

      

==

T2t()12 ---d˙2 Tmd2 ˙ w˙2φ˙2w˙3

φ˙3

      

Tm w˙2φ˙2w˙3

φ˙3

      

==

TT1T2+=

(14)

Extened matrix form of T(t) is

With fixed-free B.C.s,

and

,

T(t) is then Tt()12 --- w˙1φ˙1w˙2φ˙2w˙3φ˙3 Tw˙1φ˙1w˙2φ˙2w˙3φ˙3 = m

m

w˙0t,()w˙1t()0==

x∂ ∂w˙0t,()φ˙1t()0==

Tt()12 --- 00w˙2φ˙2w˙3

φ˙3 T00w˙2φ˙2w˙3φ˙3 = m

m

2103-433 Introduction to Mechanical Vibration Chulalongkorn University, THAILANDMechanical Engineering DepartmentT(t) can be reduced towhere M is a global mass matrix.

,

for the 2-element beam Tt()12 ---ρAL840 ---- w˙2φ˙2w˙3φ˙3 T156156+()11L–11L+()546.5L–

11L–11L+()L 2L 2+()6.5L0.75L 2–546.5L15611L–

6.5L–0.75L 2–11L–L 2 w˙2φ˙2w˙3

φ˙3 12 ---d˙TMd˙==

MρAL840 --- 3120546.5L–

02L 26.5L0.75L 2–546.5L15611L–

6.5L–0.75L 2–11L–L 2 =

(15)

Lumped-Mass Matrices

•

constructed from kinetic energy are called “consistent-mass matrix” and they are a full matrix.

•

It is difficult to calculate of the full matrix.

•

An alternative method is to construct as a “lumped-mass matrix”.

Consider a bar element:

Consider a beam element: M

M 1–

M

ρ, A, L mρAL=

m/2m/2 M ρAL2 --- 1001 =

ρ, A, L mρAL=

I 13 ---m2 ---- 

L2 --- 

2= m/2m/2M ρAL2 --- 10000 L 212 ---00

0010

000 L 2

12 --- =

T russ Structure

Let’s consider element 2:

Rewrite in a matrix form

or u3t()U3θU4θsin+cos=

u4t()U5θU6θsin+cos=

u3t()u4t() θcosθsin0000θcosθsin U3U4U5U6 = 1

2

3 u1 U2U1

U5 U6

u2

U5 U6

u4

u3 U3 U4

U3 θ θ l l

l 1

2 1

2

u3U4

(16)

where:

is the local coordinates of

element 2,

is the global

coordinates of element 2.

in global coor dinates

Element 2:

where is the global stiffness matrix of element 2.

or u2t()ΓU2t()=

u2u3u4 T=

U2U3U4U5U6 T=

KV2t()12 ---u2 TKeu2 12 ---U2 TΓ TKeΓU2 12 ---U2 TK2()U2===

K2()

K2()EAl ---- θcos0θsin00θcos0θsin 11–1–1 θcosθsin0000θcosθsin = K2()EAl ---- θcos() 2θθcossinθcos() 2–θθcossin–

θθcossinθsin() 2θθcossin–θsin() 2–

θcos() 2–θθcossin–θcos() 2θθcossin θθcossin–θsin() 2–θθcossinθsin() 2 =

2103-433 Introduction to Mechanical Vibration Chulalongkorn University, THAILANDMechanical Engineering DepartmentElement 1:where:

is the global

coordinates of element 1,

for this example.

Assemb ly:

where:

is the full

global coordinates ,

is the e xpanded global

stiffness matr ix associated

with .

V1t()12 ---U1 TK1()U1=

U1U1U2U5U6 T=

K1()K2()=

Vt()V1t()V2t()+=

Vt()12 ---U1 TK1()U1 12 ---U2 TK2()U2+12 ---U TKU==

UU1U2U3U4U5U6 T=

K

U

(17)

in global coor dinates

The full mass matrix associated with

can be determined the same way as . M

MU˙

K

CHAPTER 8

VIBRA TION TESTING