HCMUT 2004 Faculty of Applied Sciences Hochiminh City University of Technology The Finite Element...

HCMUT 2004

Faculty of Applied SciencesHochiminh City University of Technology

The Finite Element Method

PhD. TRUONG Tich ThienDepartment of Engineering Mechnics

The Finite Element MethodIntroduction

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Linear Structural Analyse

- Truss Structure

- Beam

- Shell

- 3-D Solid

Material nonlinear

- Plasticity (Structure with stresses above yield stress)

- Hyperelasticity (ν = 0.5, i.e. Rubber)

- Creep, Swelling

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Geometric nonlinear

- Large Deflection

- Stress Stiffening

Dynamics

- Natural Frequency

- Forced Vibration

- Random Vibration

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Stability

- Buckling

Field Analysis

- Heat Transfer

- Magnetics

- Fluid Flow

- Acoustics

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Evolution of the Finite Element Method

1941 HRENIKOFF, MC HENRY, NEWMARKApproximation of a continuum Problem through a Framework

1946 SOUTHWELLRelaxation Methods in theoretical Physics

1954 ARGYRIS, TURNEREnergy Theorems and Structural Analysis (general StructuralAnalysis for Aircraft structures)

1960 CLOUGHFEM in Plane Stress Analysis

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- Dividing a solid in Finite Elements- Compatibility between the Elements through a displacement function- Equilibrium condition through the principal of virtual work

FE = Finite Elementi, j, k = Nodal points (Nodes)

of an Element

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The stiffness relation:

[K] {d} = {F}

or K d = F

K = Total stiffness matrixd = Matrix of nodal displacementsF = Matrix of nodal forces

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K d = F

dT = [u1 v1 w1 . . . un vn wn]

FT = [Fx1 F y1 . . . F xn F yn F zn]

K is a n x n matrixK is a sparse matrix and symmetric

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K d = F

Solving the stiffness relation by:

- CHOLESKY – Method- WAVE – FRONT – Method

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1, 2 = NodesF1, F2 = Nodal forces

k = Spring rateu1, u2 = Nodal displacements

u1 u2

F1 F21 2k

F1 = k (u1 – u2)

F2 = k (u2 – u1)

Spring Element

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Element stiffness matrix

Fdk

2

1

2

1

F

F

u

u

kk

kk

kk

kkk

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Spring System

u2 u3

F1 F3

1 3k2

k1

2

u1

F2

Element stiffness matrices

11

111 kk

kkk

22

222 kk

kkk

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the stiffness relation by using superposition

FdK

3

2

1

3

2

1

22

2211

11

F

F

F

u

u

u

kk0

kkkk

0kk

22

2211

11

kk0

kkkk

0kk

K

Total stiffness matrix

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Truss Element

x

yu2

u1

F2

F1 1

2

A

AE

k

Element stiffness matrix c = cosα s = sinα

= lengthA = cross-sectional areaE = Young´s modulus

Spring rate of a truss element

22

22

22

22

kscsscs

csccsc

scsscs

csccsc

AE

k

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1 = 450 2 = 1350

y

x

Fx3

Fy3

1 2

3

A E A E

Element : Element :Node 1 1 Node 1 2Node 2 3 Node 2 3

3

3

3

3

2

2

1

1

0

0

0

0

200000

021000

001000

000100

000010

000001

2

y

x

F

F

v

u

v

u

v

u

AE

Stiffness relation

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Beam Element

EJM1 M2x

y

1 2Q1 Q2

x

y

v1 v2

1 212

Forces Displacements

A = Cross – sectional area E = Young’s modulusI = Moment of inertia = Length

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the stiffness relation

d

k

F

2

2

2

1

1

1

22

2323

22

2323

2

2y

2x

1

1y

1x

v

u

v

u

EI4EI60

EI2EI60

EI6EI120

EI6EI120

00EA

00EA

EI2EI60

EI4EI60

EI6EI120

EI6EI120

00EA

00EA

M

Q

Q

M

Q

Q

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Example for practical FEM application

Engineering system Possible finite element model

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Plane stress Triangular Element

1 2

3

x

y

u1

v1

u3

v3

u2

v2

Equilibrium condition: Principal of virtual workCompatibility condition: linear displacement function

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General displacements (Displacement function)u(x,y) = α1 + α2x + α2y

v(x,y) = α4 + α5x + α6y

Nodal displacementsu1= α1 + α2x1+ α3y1

v1= α4 + α5x1+ α6y1

similar for node 2 and node 3.

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u = N d General displacements to nodal displacements

ε = B d Strains to nodal displacements

σ = D ε Stresses to strains

σ = D B d Stresses to nodal displacements

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Other displacement functions

1

2

3

4

5

quadratic displacement function

u(x,y) = α1 + α2x + α3y+ α4x2 + α5y

2+α6xy

v(x,y) = α7 + α8x + α9y+ α10x2 + α11y

2+α12xy

Triangular element with 6 nodes6

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cubic displacement function

- stress field can be better approximated- more computing time- less numerical accuracy- geometry cannot be good approximated

1

2

45

6

7

10

Triangular element with 10 nodes

8

9

3

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σ = stress matrix p = force matrixε = strain matrix u = displacement matrix

V

T

V

dVδdVUδδU εσ

sA

T

V

Tm dAdVW upuf

Principal of Virtual Work

δU = virtual work done by the applied forceδW = stored strain energy

δU + δW = 0

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Element stiffness matrix

V

T dVBDBk

2

100

01

01

1

ED

2 D = Elasticity matrix

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b1 = y2 – y3 c1 = x3 – x2

b2 = y3 – y1 c2 = x1 – x3 AΔ = Area of element

b3 = y1 – y2 c3 = x2 – x1

332211

321

321

bcbcbc

c0c0c0

0b0b0b

A2

1B

linear displacement function yields :- linear displacement field- constant strain field- constant stress field

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Dynamics

k1 k2

c1 c2

m1 m2m0

u0

F0 F1 F2

u1 u2

Equation of motion

FdKdCdM

2

1

0

2

1

0

22

2211

11

2

1

0

22

2211

11

2

1

0

2

1

0

F

F

F

u

u

u

kk0

kkkk

0kk

u

u

u

cc0

cccc

0cc

u

u

u

m00

0m0

00m

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M = Mass matrixC = Damping matrixK = Stiffness matrixd = Nodal displacement matrix

= Nodal velocity matrix= Nodal acceleration matrix

FKddCdM

dd

or

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for a continuum

u = N dε = B d

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ρ = Mass densityμ = Viscosity matrix

V

Te dVNNC

V

Te dVB DBk

the element matrices

V

Te dVNNM

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o Modal analysiso Harmonic response analysis - Full harmonic - Reduced harmonico Transient dynamic analysis - Linear dynamic - Nonlinear dynamic

tFKddCdM

General Equation of Motion

Types of dynamic solution

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Purpose: To determine the natural frequencies and mode shapes for the structure

Assumptions: Linear structure (M, K, = constant)No Damping (c = 0 )Free Vibrations (F = 0)

0 KddM

Modal Analysis

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Transformation methods Iteration methods

JACOBI INVERSE POWERGIVENS INVERSE POWER WITH SHIFTSHOUSEHOLDER SUB – SPACE ITERATIONQ – R METHOD

for harmonic motion: d = d0 cos (ωt)

(-ω2M + K) d0 = 0

Eigenvalue extraction procedures

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Purpose: To determine the response of a linear system

Assumptions: Linear Structure (M, C, K = constant)Harmonic forcing function at known frequency

ti0e FKddCdM

Harmonic Response Analysis

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K

Forcing funktion F = F0e-iωt

Response will be harmonic at input frequency d = d0 e-iωt

(-ω2M – iωC + K) d = F0

is a complex matrixd will be complex (amplitude and phase angle)

00 FdK

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Limiting cases:

ω = 0 : K d = F0 Static solution

C = 0 : (-ω2M + K) d = F0 Response in phase

C = 0, ω = ωn : (-ωn2M + K) d = F0 infinite amplitudes

C = 0, ω = ωn : (-ωn2M - iωnC + K) d = F0 finite amplitudes,

large phase shifts

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Transient Dynamic Analysis

F(t) = arbitrary time history forcing function

tFKddCdM

periodic forcing function

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impulsive forcing function

Earthquake in El Centro,

California18.05.1940

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Two major types of integration:

- Modal superposition

- Direct numerical integration

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T0

Q0

T1

Q1

T2

Q2

1 2

1 2

1, A1 2, A2

, A

0

0ne-dimensional heat flow principle

, = conductivity elements = convection element0, 1, 2 = temperature elements

A = Cross-sectional area = Lengthλ = Conductivity Aα = Convection surface

T = Temperature Q = Heat flow C = Specific heat α = Coefficient of

thermal expansion

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Heath flow through a conduction element:

Heat stored in a temperature element:

cp = specific heat capacity

C = specific heat

Heat transition for a convection element: Q = A(T – T2)

12 TTA

Q

TCTVcQ p

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QT

K

TC

TAQ

Q

Q

T

T

T

AAA

0

AAAA

0AA

T

T

T

C00

0C0

00C

2

1

0

2

1

0

2

22

2

22

2

22

2

22

1

11

1

11

1

11

1

11

2

1

0

2

1

0

Heat balance

or QKTTC

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C = specific heat matrixK = conductivity matrixQ = heat flow matrix T = temperature matrix = time derivation of TT

For the stationary state with = 0

KT = Q

T

QKTTC

HCMUT 2004 Faculty of Applied Sciences Hochiminh City University of Technology The Finite Element...

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