lectureacm_1

8/13/2019 LectureACM_1

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Mechanics

PGS.TS. Nguyn Xun Hng

Universit of Science

School of Math & Computer Science


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Contents

1. Introduction

.

3. Overview of Partial Differential Equations (PDE)

4. Introduction to Variational Principles

5. Overview of some advanced variational forms

6. Advanced numerical methods in deformable solids

7. Advanced numerical methods in plates and shells structure

8. Practical a lications


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References

1. Zienkewicz and Taylor, The finite element method, Vol., , .

2. J.N. Reddy, Principles of Continuum mechanics,

3. Boo Cheon Khoo et al. Numerical Methods for PartialDifferential Equations, Lecture Notes, MIT (2008)


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1. Introduction


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Introduction

Applied

Mechanics

Experimental


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Introduction

Nano/Micro

Comput.ec an cs

Methodology


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Introduction


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Introduction

Nanomechanics the molecular and atomic levels particle physics and chemistry.

Mt nano chnhbng 1/80.000 ng knh si tc


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Introduction

Micromechanics the crystallo-graphic and granular levels the design and fabrication of materials and micro-devices.

Mt tbo mu ln 2,5 micromt = 2,500 nanomt

1103 mm 10.0001 m =


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Introduction

Continuum mechanics studies bodies at the macroscopiclevel, usin continuum models in which the microstructure

is homogenized by phenomenological averages- Solids and Structures- Fluids

- Multi-physics

Our interest in this course !

Marco

Scale in meter, , millimeter


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Introduction

System : mechanical objects such as airplanes, buildings,bridges, engines, cars, microchips,

Biomechanics objects such as a whale, amoeba, inner

Ecological, astronomical and cosmological entities also


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Advanced numerical methods

Discretization Methodology:

Finite Difference (FDM)Finite Element (FEM)

Discontinuous Galerkin FEM

Boundary Element (BEM)Spectral methodsMeshfree methodsIso eometric anal sis

Multiscale- FE analysis


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Introduction

See Carlos Felippa, IFEM


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Introduction

Displacement

HybridMixed

Pure equilibriumMixed/hybrid

gy

StiffnessMixed/hybride

ner

formulation:FlexibilityMixed

DOFs


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Introduction

Total potential energy

ar at ona pr nc p e

Hellinger-Reissner (HR)Hybrid principlesVeubeke-Hu-Washizu (VHW)

Displacements (one field)

Approximate field:Stresses (one field)Displacements & stresses (two fields)

, ,(three fields)


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Some applications

Truss bridge

model

Derived from COMSOL

Deformation under self weight


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Some applications

3D trussPlate structures

compositepressureFrame


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Composite structures

Source: Apatech & Vinashin Vietnam


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analysis of composite plate0.5

y

q0

0.1

0.2

0.3

0.4

.

icknessz/h

a/h=10HSDT

a/h=4HSDT

a/h=10FSDT

a/h=4FSDTSupported

plate

x

z

h a/

2

a/

090

900

h

0.4

0.3

0.2

0.1

0

Nomalizedthi

x0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.80.5 Stress, x(a/2,b/2)

0.4

0.5

a/h=10HSDTa/h=4HSDT

a/h=10FSDT

0.4

0.5

0.4

0.5 a/h=10HSDT

a/h=4HSDT

a/h=10FSDT

0.1

0

0.1

0.2

0.3

malizedthicknessz/h

a/h=4FSDT

0.1

0

0.1

0.2

0.3

malize

ic

nessz

a/h=10HSDT

a/h=4HSDT

a/h=10FSDT

a/h=4FSDT 0.1

0

0.1

0.2

0.3

malizedthicknessz/h

a/h=4FSDT

0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.80.5

0.4

0.3

0.2

Stress, y(a/2,b/2)

No

0 0.1 0.2 0.3 0.4 0.5 0.6

0.5

0.4

0.3

0.2

Stress, xz

(0,b/2)

o

0 0.05 0.1 0.15 0.2 0.25

0.5

0.4

0.3

0.2

Stress, yz

(a/2,0)

No

y xz yzSource: H. Nguyen-Xuan et al. (2012), MAMS, accepted


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Sandwich plates

layer1 Temperature

decreases

layer2

layer3 Step 1

150C

Axialstressx

Temperature

decreases

150C to20C

displacement


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0.3

0.4

0.5

z

/h

0.1

0

0.1

.

alizedthickness

a/h=20

a/h=10

a/h=4

1.5 1 0.5 0 0.5 1 1.50.5

0.4

0.3

.

No

,x ,

0.3

0.4

0.5

z/h

0.3

0.4

0.5

z/h

0.3

0.4

0.5

/h

0.2

0.1

0

0.1

.

omalizedthickness

a/h=20

a/h=10

a/h=4

0.2

0.1

0

0.1

.

Nomalizedthickness

a/h=20

a/h=10

a/h=4

0.2

0.1

0

0.1

.

Nomalizedthickness

a/h=20

a/h=10

a/h=4

0.4 0.3 0.2 0.1 0 0.1 0.2 0.30.5

0.4

0.3

Stress, y(a/2,b/2)

0.2 0.15 0.1 0.05 0 0.05 0.1 0.15

0.5

0.4

0.3

Stress, xy

(0,b/2)

0.2 0 0.2 0.4 0.6 0.8 1 1.20.5

0.4

0.3

Stress, xz

(0,b/2)

Stresses distributions in (0/core/0) supported sandwich composite plates

Source: H. Nguyen-Xuan et al. (2012), MAMS, accepted


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Material propertyTc=300C

Numerical computation

( ) 0.5 /c m c m

n

c

z z

V z z h

Tm=20C

0.2

0.3

0.4

0.5

z/h

n=5

n=10

d dT

Thermal distribution through thickness

0.4

0.5

0.1

0

0.1

.

dimentionalthicknes

n=3

n=1

n=0.5

d dz z

0

0.1

0.2

.

nalthicknes

z/h

n=0

n=0.1 50 100 150 200 250 300 350 4000.5

0.4

0.3

0.2

Non-

n=0.1

n=0.3

0.3

0.2

0.1

Non-dime

nti n=0.3

n=0.5

n=1

n=3

n=5

n=10

Effective modulus Eef f

Effective modulus of Al/Al2O3 FGM

0 50 100 150 200 250 3000.5

0.4

Temperature o

C

n=100

Source: H. Nguyen-Xuan et al. (2011), Composite & Structures, 93: 3019-3039


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0.3

0.4

0.5

ss

ceramic0.51

-0.1

0.0

-0.1

0.0

0.1

0.2

tionalthickn

metal

-0.4

-0.3

-0.2

entionaldeflectio

ceramic0.5

-0.4

-0.3

-0.2

.

Non-d

ime

through thickness

-0.7

-0.6

-0.5

N

on-di 1

2

metalDeflectionunder mechanical load

-0.5

-0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3

Non-dimentional axial stress

-14 -12 -10 -8 -6 -4 -2 0

Load parameter

0.2

0.3

ceramic

0.50

= = = =

-

0.0

0.1

tionaldeflection

.12metal

0.40

0.45

tress

.

-0.3

-0.2

.

Non-dim

en

Deflection under- 0.25

0.30

.

Central

-14 -12 -10 -8 -6 -4 -2 0-0.4

Load parameter

Source: H. Nguyen-Xuan et al. (2012), Thin-Walled Structures 54:118

1 10 100 1000 10000 100000 10000000.20

Ratio L/t


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2000

n=125

30

1000

1500

gthermalT

cr

uniformlinear15

20

lingthermal

CCCC (non-uniform)SSSS (uniform)SSSS (non-uniform)

500

Criticalb

uckli -

5

10

Critica

lbuc

20 40 60 80 100

0

a/h

0 2 4 6 8 100

volume fraction exponent n

Source: H. Nguyen-Xuan et al., Journal of Thermal stresses (submitted, 2012)

The first four buckling mode shapes of FG circular plate


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Piezoelectric structure

Simulating the linear tilt angle of the reflected light through amirror of a MEMs device

BimorphMEMs device

20

25

structures 15Tiltangle(0)

Tilt angle ofmirror in thebimorph MEMs

5 PCM

T3

Q4

EST3

Source:Nguyen Xuan et al., SMS (2009)

0 5 10 15 20 25 30 35 40 45 500

Applied voltage (V)


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1

10-5

10 V

ricplate 0V (s)

5V (s)

10V (s)

-

-2

-1

5 V

/45]piezoelect a

5V (a)

10V (a)

-6

-5

-4

0 V

lectionof

[p/-4

-0.50.0

0.5

1.0

1.5

10 V

tricplaete

0V ()

5V ()

10V()

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Def

=

-3.5

-3.0

-2.5

-2.0

-1.5

-1.0

5 V

]apiezoele 0V ()

5V ()

10V()

0V ()

5V ()

Piezoelectric

late

-6.5

-6.0

-5.5

-5.0

-4.5

-4.0

0 Vctionof[p/-

Shape control

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

-7.5

-7.0

Defl

ydirection (x=a/2)Source: Phan et al., IJCM (2011), in press


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Vibration Control of satellite arms

A scale model to testactive vibration controlschemes of satellite

Hubbard, 1985),(Lammering, 1991).The accelerometer ismodeled by a tip massM.


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Other applications: Vibration analysis

ree v ra on o a mac ne par

Source: Nguyen-Thanh, JCAM, 233 (2010) 2112-2135


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Composite circular plates

Mode 2

Mode 1o e

Mode 4 Mode 5 Mode 6


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Buckling analysis

Plates under axial andbiaxial compression


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Mode shape of buckling behavior

SSSS SSFF SSCC

SSSC SSFC SSFS

Source: H. Nguyen-Xuan et al. (2010), CMAME, Vol. 199 (9-12), 471-489


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Static structures shell analysis

deformedconfiguration

ScordelisLoroof

Pinchedcylinder

nc e em sp ere

Source: our publication in CMAME, Vol. 200,, p. 3410-3424, 2011


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Pinched composite hemispherical shell with 18hole

h = 0.08 inE11 = 20.4610

6 psiE22 = 4.09210

6 psi= = 612 13 .

G23 = 1.26852106 psi

12 = 0.313

Inward deflection Outward deflection

Source: Vu Duy, Proceedings of Structural Engineering, 2010


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Free vibration in macro/microtubes

Source: our publication in KSCE, Vol. 12, No. 2, pp. 347-361, 2011


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Heterogeneous media

www.scorec.rpi.edu


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Multiscale approach to homogenization problems

The steady state heat conduction of microprocessor made of a layered material

processor core

Heat distribution in amicroprocessor

in cellphones orautomotive electronicsHeatsink


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in ,a u fe e- = W

1 1Find ( )such that forall ( )D Du H v H e

e e e e

W W

Weak form:Strong form:

( )on ,

on .

D D

N N

u g

n a u g

e

e e

= W = W

,

: ( )N

N Dfvdx g vdx a g vdx l ve e

W

W W W= + - = 1 1where : : 0onH v H v W = W = W

Micro solver: Under some constrains

(sample domain size, periodic or

Dirichlet boundary condition v.v), we

solve on each sampling domain .l

h

Kv d lKd

Post processing:

1( , ) ( ) .

l

l llH l

LKH H h h

K KKK T l

w

B v w a x v w dx K d dd

e

d ==

deg=2l 3 04


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104

deg=3

deg=4

deg=5

slope=3.04

slope=2.81

slope=2.73

103

CPU-t

ime(s)

slope=3.11

101

102

103

102

slope=2.03 102

101

number of macro elements per dimensionIsogeometric approach

105

104

uHH1

0

H1

slope=3.09

105

104

103

L2

2

slope=3.07

107

106

u0

deg=2

deg=3

slope=5.36

slope=4.17

107

106

u0

uH

u

0

L

slope=4.98

s ope= .

101

108

1/Hmax

deg=4

deg=5

10110

9

108

1/hmax

eg=

deg=3

deg=4

deg=5

slope=6.26

Nguyen-Xuan and Hoang, J. Com Physics, submitted, 2012


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Simulation of failure cracked structures

Plate is a rectangle of 4 8 ,

E = 1e9,E = 1e8, = =0.3 and = 1.

0.5 8performed for two cases, namely:

(1) hard inclusion (R = 0.1) and (2)2so t nc us on =

matrixE

2ncE fiber

rac n compos e ma er a


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Finite Element Mesh

Crack rowth in

compositematerial

Enrichment of singularfield around crack, bi-

material interface


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Crack path

8Comparison of Crack Growth Paths in Presence of Soft and Hard Inclusion

Ref

XFEM soft

6

7

Ref

XFEM (hard)

4

5

2

3

1

3 2 1 0 1 2 3 4 5 6 70


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Simulation of failure crack in materials

Concrete

Collaboration with: Prof T. Rabczuk and from Dr M. Duflot et al

Simulation of crack growth in materials


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Simulation of crack growth in materials

Crack growth in concrete at mesoscale Crack growth in polycrystals at mesoscale

Multi-delamination research:

Application: laminated composite

Source: Nguyen Vinh et al, IMM (2012), 3(4):1-42

s ruc ures, n ms


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Some other applications

Fuel Cell Bipolar Plate

The thermal and structuralanalysis in a bipolarp a e n a pro on exc angemembrane fuel cell

Source: Comsol software


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Some applications


ModelSource: Comsol software


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Some applications


Temperature distribution in the plate



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Some applications

Pinched Hemispherical Shell structure

Model



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Some applications

Shell Diffusion

Model

Electric potential distribution across the

surface (V).Source: Comsol software


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Some applications

Power Transistor

Model geometry and position of transistor chip

The power transistor is mounted on the circuit board usingthrough-hole technology. The solder in the holes give mechanical

suppor an e ec ron c con ac e ween e copper rou es an e

transistor pinsSource: Comsol software

S li i


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Some applications

Power Transistor

Temperature distributionSource: Comsol software

S li ti


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Some applications

Radar Cross Section

-cut plane lies above the water surface


S li ti


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Some applications

The total field norm for a 30 degree angle of incidence.The arrow represents the propagation direction of the

incident background field. Source: Comsol software


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2. Vectors and Tensors

Source: Thanks anonymous author contributed to this part

V t


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Vectors

zz

The Cartesian coordinates is considered !

Az

y

y

Ax

Vector is denoted by A or

Magnitude


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Magnitude

z

Ay

AxUnit vector:

x

2 2 2A Ae

x y z

Vector Addition


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Vector Addition

x x xA B C

( )

y y

z z zA B C A + B = C

Number axes


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Number axes

1x x

2

x z

1,2,3i i i

A B C i or simply:

i i i

This is Cartesian Tensor or indicial notation.

This is called 1st tensor Rank

A

Remark


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Remark

1

2 vectoriA A A

3A

Unless otherwise stated, any repeated indices within a

term require a sum on these indices from 1 to 3 or from

to or two- mens ona wor .


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Examples:

1 1 2 2 3 3i iB A B A B A B

31 2k

f ff f

1 2 2kx x x x

, 1,1 2,2 3,3k k k

kf f f fx


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Examples:

i i k k B A B A B

jj ii kkC C C

, , ,k k i i j jf f f

Look at Gibbs notation


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A

A

x y z, , .

, , can e use n t e terature.

1 2 3, ,

numbered coordinate directions.


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x y zA A A A j ki

or

x x y y z z

or

1 1 2 2 3 3 A A A A e e e

3

A A

1

n n

n

Matrix This is called 2nd


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11 12 13

21 22 23=a a aa a a

A

ensor an

31 32 33

a a a

21 22 2 1 2

.....

....

n n

n na a a a

11 21 1 1 1

= ... ... .... ... ...

.....

ij

m m m n m na a a a

1 2 1....m m mn mna a a a ,

The unit matrix


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1 0 0

= 0 1 0

I

0 0 1 Kronecker delta or the unit tensor:

0 if i j

i j

Second uniti j

tensor


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11 12 13 1 0 0

21 22 23 0 1 0 31 32 33 0 0 1

Recall Summation Convention

1 1 2 2 3 3ij i j j jA A A A ij i j

Recall Summation Convention


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3

and

ij jk ik

Strain and Stress:

11 22 33ii

Permutation constant


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ermutaioevenin3-2-1inki1e

permutaionoddin1)-2-3(inkj,i,1e kjikiork,j,iif0e kji

lso called the alternating tensor or the

Levi-Civita density.

Permutation constant Alternative definition


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e 2i j k j k k i i j no sum on i,j, and k

i,j,k 1,2,3


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1 2 3i i i

1 2 2

ei j k j j j

1 2 3k k k

2 2 2ei j k i j k

3 3 3i j k


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31 32 33

312 11 12 12

e

21 22 23

0 0 1312e 1 0 0 1

0 0 1

Determinants


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11 12 13a a a21 22 23 ija a a a a

( 1) ij ,

Determinants


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1 2 3e

ijk i j k a a a a

1 2 3eijk i j k a a a a

Vector Multiplication


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Scalar Product (Dot Product)

Tensor Product

Scalar Product


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B

cosA B A B

Scalar Product


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(1,0,0)i (0,1,0)j (0,0,1)kzz

y zA A A i j k Az

i y

Ay. . .

. 0i j . 0jk . 0i k

x

x

. ( ).( )

=

x y z x y zA A A B B B A B i j k i j k

x x y y z z

We can rename unit vectors as:

1 (1,0,0)e 2 (0,1,0)e 3 (0,0,1)e

Scalar Product


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(scalar)A B

x x y y z zA B A B A B 3

i iA B 1i


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Scalar Product With

i iA B

Vector Product


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C = A B

( )or C A B

C

Bn

A

Vector Product


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(vector)C A B

sinC A B n

sinC A B

is unit vector to lane defined b A B

n

Vector Product


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is area of parallelopipedC

with adjacent sides andA BC

B

Example: Moment vector of load


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( )or M r F M r F

Translation


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2

1

Quantities Transformed


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Scalars

Tensors

Scalar


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A scalar is a quantity that does not change its

coordinate system to another.

Vector


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22

1

1

3


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1 1 1 1 2 1 2 3 1 3cos , cos , cos ,V V x x V x x V x x

2 1 2 1 2 2 2 3 2 3cos , cos , cos ,V V x x V x x V x x 2 1 3 1 2 3 2 3 3 3cos , cos , cos ,V V x x V x x V x x

i ij jV a V


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1 1 1 1cos , cos ,x x x x


2 1 2 1

2 2 2 2

cos , cos ,

cos , cos ,

x x x x

x x x x


3 1 3 1

3 2 3 2

, ,

cos , cos ,x x x x


T f ti M t i [A]


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Transformation Matrix [A]

i ij j

i ji j

Tensor Product


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(tensor) orAB CA B=C

C A B

1 2 3 1 2 3C A A A B B B

S d O d T


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Second-Order Tensors

s mmetric tensorC Canti symmetric tensori iC C

Tensors Identity


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Any tensor symmetric tensor anti symmetric tensor

1 1

2 2

Tensors Operations


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additionB C

ij ij ij

scalar multiplicationB A

ij ijB A

Tensor Fields


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y

depend on spatial location and on time.

Examples are the displacement vector (1st

, .

, ,ij ij k T T x t T T x t , ,i i j jv v x t v v x t

Derivative


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Take partial derivative w.r.t. time of a vector

e :

31 2 vv v vt t t t

dvdv

kdt dt e

Unit Vectors in cylindrical coordinates


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Cylindrical components:

, , , , , ,

r rv r z t v r z t v e e

, , ,z z

r rde e

d

e

re

Take partial deri ati e r t time of a ector


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Take partial derivative w.r.t. time of a vector

r r

r r

dvd dv

dt dt dt

ev

e

z z

dv dvdv

dt dt dt

e

e e


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r

e

r

e

e

The gradient operator


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A directed rate of change of a tensor field

w.r.t. the coordinate directions may be

obtained by the del operator.

k

kx

( , , )x y z

kComponents of this operator:

y zx y z

Scalar Field


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gradient

k

grad

x y z

x y z

e e e

In polar coordinates:

r z

r r z

e e e

Gradient

Gradient vector is normal to surfaces of constant


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Gradient vector is normal to surfaces of constant. To et the rate of chan e in a articular

direction:

n

n

e

Plasticity field

f

Vector Fields


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xv

grad v v

v

v e eix


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Polar Coordinates

1


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1 v v v

v e e e e e e

1

r z r r z z

r r z

r r z

vv v v

vr r r z

r zvv v

1 1 1

r r r r z

r r zr r z

r r r

v vv v v

e e e e e e e e e e

r zz r z z z

r r r r r

vv v

e e e e e e

Curl e kv u v u


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eijk iux

v u

e mj

ijk im

TT S S

j

Polar Coordinates

r zT

r r z

e e e

[

r rr r r r r rz z

T T T

T T T

e e e e e e

e e e e e e

]z zr r z z z zz zT T T e e e e e e

Gradient and divergence theorems

LetF(x,y,z) and G(x,y,z) be functions of class C1().


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gra

n

(Gradient theorem)

( )d d ddiv

G G n G

vergence eoremwhere n=(nx, ny, nz) is the normal vector of the surface .Assume that , G are scalar functions. We have

d d dG F F G FG n2 d . d d

FG F F G G

n

Curl theorem

d d F n F


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d d F n F

Homework:

Find Gradient and divergence, Curl theorems

,

Example

1 2 0 1 5


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=, ,

0 3 2 1 4

Find : Det (K), Ku

KX = FX such that:

Example

Assumed that the components of the stress dyadic (tng cp)i i f i di i b


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at a certain point of a continuous medium are given by

200 400 300 400 0 0 psi

300 0 100

Find the stress vectort and its normal and tangential components at

the point on the plane, (x1, x2) x1 + 2x2 + 2x3 = constant, passing

through the point.

Example

Solution:1


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1

1 2 3

3

16001

400 ps3

100

t nSurface stress vector

Example


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(11/02/1847 -18/10/1931 )

"Ti khm ph ra 10.000 cht khng th sdn lm d tc bn n tc sau 10.000 lntht bi trong vic chto dy tc bngn)", "Ttcnhng ngi tht biu c mtim chung : ng n n ra r ng c c m n n

cng chnh l lc htbn lc cui cng ca" ",

li ca ngy mai"


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. verv ew o ar a eren a

Equations (PDEs)

, ,Einstein

Overview of Partial Differential Equations (PDEs)

Elliptic, Parabolic and Hyperbolic Equations

Th i ti2 22ax bx c dx e


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The conic e uation 2ax bx c dx e

2 0 : Ellipseb ac

: ara o aac 2


Elliptic, Parabolic and Hyperbolic Equations2 2 2u u u u u


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u u u u u 2 2x x y y x y

20 Laplace equation

, , , , , ,

, , Poisson equationxx yy f

2) 0 : Para o c , , Heat equat onxx tac u u

2 , ,xx tt

PDE Elliptic form

1. Poisson equation in one dimension (1D) (spring):d du


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d duq

dx dx

where u = u(x) is unknown, q = q(x), k> 0 are given functions

tea y orm

-An elastic bar (k = EA) subjected to body force or the

Some real phenomena

distributed load (q)

- Stead -State Linear Heat Conduction: k = the

thermal conductivity coefficient, q QA isdistribution heat production,uT- temperature)

- Steady potential flow: (k=-mass density,q is source field,

u- potential function)

Quantities corresponding to the variables of the spring


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Overview of Partial Differential Equations (PDE)

- Steady-State electric field: (k = the dielectric material

coefficient e is the source field u electric


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coefficient e is the source field u - electricpotential field)

Homework: 1) Establish model and corresponding

differential equations of above physical models

s a s mo e an correspon ngdifferential equations of Magnetostatics, and

Saint-Venant torsion problem (see Chapter 2

in AFEM by Prof. Carlos A. Felippa)

Example 1D

Ex1: Derive the governing equation of a rod subjected tothe forces inside the member and at the ends


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the forces inside the member and at the ends

Solution: Due to the very small cross-sectional dimensioncom ared to its len th it is assumed that the stress isuniform at any section and all other stresses are zero.

Example

Consider an element of length x


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x x x x x

Leading( ) 0x x x f x

x

or ( ) ( ) 0d

A f x d du

Hookes law: duE E dx dx

x

Example

( ( ) ) ( ) 0d du

EA x f x Governing equation


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( ( ) ) ( )fg qx x

Boundary conditions:

0sP A F at x L

where Fs is the compressive force in the spring

Thermal conduction

Ex2: Derive the governing equation for heat conductionthrough a wall


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g

( ) ( )q x A x ( ) ( )q x x A x x

x

)s x

A: section area

s: heat source

q: heat flux

Thermal conduction

Energy balance in volume of the wall( ) ( ) ( ) ( ) 0s x q x A x q x x A x x


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( ) ( ) ( ) ( )q qheat generated

heat flowin heat flowout

Rearran e

( ) ( ) ( ) ( )q x x A x x q x A xs

or

( )d qAs

dx

According to Fouriers law

dTq k dx 0

d dT

Ak sdx dx

Governing equation

Thermal conduction

Governing equation


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0d dT

Ak sdx dx

0,on x l

and the boundary condition

dTq k q

dx atx= 0

T T atx=

Example

Homework :

A bridge is supported by several concrete piers, and


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the below figure. The load 20 103 N/m2 represents

the weight of the bridge and an assumed distribution ofthe traffic on the bridge. The concrete weighsapproximately = 25 103N/m3 and its modulus isE = 28 109 N/m2.

Determine the axial displacement,,

a one-dimensional model.

Example

Ex3: Consider the bending of a straight beam accordingto the classical (EulerBernoulli) beam theory.


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The beam is subjected to distributed axial force f (x)

and transverse load q(x). derive the equationsgovern ng e equ r um o e eam.

Example

Solution:


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beam acted by area-integrated forces and moments

, ,xx xz xxA A

N dx V dx M zdx Axial force Shear force Bending moment

Example

Equilibrium equations:

x


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( ) 0

x

dN f x

0 : ( ) ( ) 0z

x

F V V dV q x dx ( ) 0

dVq x

dx

0 : ( ) ( ) 0y

M Vdx M M dM q x dxdx

0V

dx

Example

Hence: 2d M

2


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2dx

2 2

erw se, rom u er ernou ypo es s

2 2xxA AE zdx E z dA EIdx dx

ence, one o ta ns t e u er ernou equa on

2 2d d w 2 2

- ,dx dx


2. Poisson equation in two and three dimension (2D &3D): u


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u u

x x

x x y y

( ) ( ) ( ) 0k k k qx x y y z z 3D

Generalized Poisson equation

where

( , , ),x y z

.( , , )

u u uu

x y z


3. Poissons equation reduces to Laplaces equation:

s t e constant + t e source q = 0


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s t e constant + t e source q = 0

2 2

2 2.( ) 0 or 0,where =k u u

x y

Having solutions is calledharmonic functions, whichhave been widely studied over the last times

Homework: 1) Establish partial differential equations of

2) Find a general solution of Laplaces equation

Example

2 0u Laplaces equation:


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Note: this equation no dependence on time, just on the

, .situations such as:

Stead state tem erature distributions

Steady state stress distribution

Steady state flows, for example in a cylinder, around acorner,

Example

Ex3: Let consider the temperature distribution in anisolated wire with constant temperature T1 and T2 at twoend


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end.

Solution: At the steady-state without heat source, the governingequa on n x s rewr en as:

2

0d T

k with BC 10T T at x

x 2

2 1T Tur so u on s 1

l

-

solution is basically just straight line

Example

0,b

y u=0 Ex3: Let consider this problem

t i bl


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u=0u=0

two variables

(0,0) (a,0)

x

u= x

, ,xx yy

The prescribed value along edges(0, ) 0 0u y for y b In general, suppose that (2)

( , ) 0 0

( , ) 0 0

u a y for y b

u x b for y a

1( ) sinn

n

n x

f x b a

(3)

(4)

( ,0) ( ) 0u x f x for y a (5)

Solution: separate solution into two independent variables

u x X x Y


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0 " " 0u X Y Y X

" "X Y

" 0

"

X X

(1)

Eq. (1) combine with BC Eq. (2) combine with BC

sinn n xX ( )

sinh n b y

aY

a sinh n b

a

Using linear combination the solution of u is)(n b y

( ) ( sin)a n x

u x y b u y bx


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1 1

( , ) ( , sin)sinh

nn n abn

a

n n

u x y b u y bx

4

6

4

5

Apply for a=b=

0

2

3 1

4( ) sinn

x nxn

0

1

4

2

1

2

3

4 01

23

0


4. Convection diffusion- reaction equations:*

( )d k u u ru q


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.( )ad k u u ru qt

= , ...

Temperature -- Heat transfer

Pollutant concentration Coastal engineering

Probability distribution Statistical mechanics

......

The heat flow equation is parabolic (as is the convectiondiffusion equation)(*)


Ifu Independent time and no source (q = 0), and no......reac on, convec on - us on s a onary equa ons


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y

. inu k u

This belongs to a class of elliptic problems

Heat transfer in a cooling fin

Example : No convection diffusion- no reactione uations:

( )d k u q


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.( )ad k u qt


.


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.( ) inad k u qt

ad C C is the heat capacityk is thermal conductivity

This belongs to a class of parabolic problems


6. Wave equations (No diffusion term) : .( ) 0k u


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u at

This belongs to a class of hyperbolic problems



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=.

2 0 ink u u Find (u,) such that

0 onu


8. Wave equation:

inu

e k u q


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2inae k u q

t

9. Navier-Stokes equations (fluid mechanics)

T u

0

t

u


10. Structural Mechanics:


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View in 2D


Strain-DisplacementRelationship

u v w

The symmetric strain

tensor

11 22 33, ,x y zx y z

x xy xz


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1xy

y

u v

xy y yz

23

2 2

1( )

y

yz

yz

y x

v w

x

13

1( )

2 2

xzxz

u w

z x

y

z

y

z

yz

xz

Stress-Strain relationship

D xy

yz

xz

Vector form


Isotropic material

1 0 0 0

1 0 0 0

1 0 0 0


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1 20 0 0 0 0E

D

1 1 21 2

0 0 0 0 02

1 2

0 0 0 0 02

The equilibrium equations2

2t

u

f = Transient problem(please write more detail this equation ???)


Plane StressPlane Strain

Axial

symmetry


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The components of the The components of the Solving the equations for stress tensor in z-directionare assumed to be zero strain tensor in z-directionare assumed to be zero the global displacement (u,w) in therandzdirections

for three above cases ?


Boundary initial value problems

Boundary value problem

0 1( ) = for

xd duk f


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( ) for

(0,1)

k f

dx dx

0 1 0(0) |xdu

u u k g dx

2

+d u

Initial value problem

2

dt

du0 0

dt

What are you looking for ?

,nh hai pht. Khi bn ngi trn mt ci l nng la,


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nh hai pht. Khi bn ngi trn mt ci l nng la,

hai pht tng nh hai gi. l s tngi.

Albert Einstein

lectureacm_1

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