1

解析事例の紹介(Introduction of numerical examples)

機械系(Department of mechanical engineering)

倉橋 貴彦(Takahiko Kurahashi)

1

2

Example of numerical simulationsExample of numerical simulations

有限要素法 （ Finite Element Method ）

Example1 : 浅水波の伝播問題（ Propagation problems of shallow water wave)

Example2 : 物体に対する伝熱の問題(Heat transfer problems)

Example3 : 非圧縮粘性流体の流れ場の計算(Computation of fluid field for incompressible viscose flow)

Example 4 : 構造内における応力分布の評価(Evaluation of stress distribution in structures)

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2x

1x3x

2x

1x

3x m4.0

mL 0.10

mt 1.0)( 0 mt 1.0)( 0

mB 8.0

mh 0.10

mx 2.0

y

mh 0.10

1.1 Surge problem

Surge phenomenon :

Discontinuous wave is propagated from high wave height to low wave height by Tsumami etc.

Problem : Find time history of wave form variation.

Waveform at initial time

Numerical example

44

Computational result by FEM

55

Example of forward analysis

Flow analysis in Tokyo Bay

State equation

(Shallow water equation)

f

f

f

tttiny

v

x

uh

t

tttiny

gt

v

tttinx

gt

u

,0

,0

,0

0

0

0

v

u Velocity component for x direction

Velocity component for y direction

Water elevation

h

),( vux

yz

Tokyo Image diagram of

shallow water flow

h

g Gravity acceleration

Mean water depth

Finite element approximation for shallow water equation

n

yx

y

x

n

yx

y

x

v

u

MSthSth

StgM

StgM

v

u

MSthSth

StgM

StgM

11

10

10

0

01

1.2 Shallow water flow analysis for Tokyo bay

6

X (m)

Y(m

)

10000 20000 30000 40000 50000 60000

10000

20000

30000

40000

50000

10000 20000 30000 40000 50000 60000

E-W (m)

10000

20000

30000

40000

50000

N-S (m)

2

1

; Land Boundary

(Slip B.C.)

; Inflow Boundary

1

4

1

on2

sin

ii

i

i T

tat

Inflow B.C.

(Main four tidal components)

Water depth distribution

Computational conditions(“Finite element mesh” and “boundary condition” and “water depth distribution”)

Total number of nodes : 17,160

Total number of elements : 33,433

6

77

Animation of water elevation Animation of velocity vector

Computational result of water elevation and velocity vector in Tokyo bay

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1.3 Application of shallow water flow analysis to actual water quality purification problem　

（ Example ： Kita-Chiba Water conduction project at lake Teganuma in Chiba prefecture ）

手賀沼導水管

導水地点坂川

年平均値

環境基準値 5mg/l

1970 1975 1980 1985 1990 19950

10

20

30

40

mg/l

COD

Purpose of this project

Reduction of COD concentration to environmental standard 5mg/l.

Investigation

Relationship of COD concentration between at conduction point and at target point

Problem

Find appropriate COD concentration at conduction point

so as to be close to environmental standard of COD concentration at target point.

⇒ Inverse problem ( optimal control theory )

Fig. Variation of COD concentration for each year

Lake Teganuma

8.86m3/s

2.68m3/s

Target point (Inside of this lake)

Water conduction points

Environmental standard 5mg/l

Average value in each year

9

Computational result based on optimal control theory

( COD concentration at conduction points )

11 U

D1

21 U

2

2

2

Conduction point

COD concentration on Γ1U-2

COD concentration on Γ1U-1

COD 2mg/l

COD 18mg/l

10

1 day later 2 days later

3 days later 4 days later

Distribution of COD concentration at optimal control of COD concentration at conduction points (1/2)

11

5 days later 6 days later

7 days later

Distribution of COD concentration at optimal control of COD concentration at conduction points (2/2)

12

Time history of COD concentration at target point

目的点

11 U

21 U

Target concentration 　(5mg/l)

Result at optimal control of COD concentration at conduction points

Result at without control of COD concentration at conduction points

Target point

1313

2.1 Heat transfer problems

1m

1m 2m

2m

Thermal diffusivity

Distribution of initial temperature

Problem : Find time history of temperature variation

Boundary condition

　 Temperature on outside boundary is set to 0 degree.

0.001m2/s

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5.00

5.01)2cos(5.0

yxr

r

rrT

x

y

1414

Computational result by FEM

15

2.2 Heat transfer analysis considering movement of heat source point

Point A

Point B

Milling machine Endmill

Aluminum plate after milling

16

Numerical experiment 　（ Machining problem of aluminum plate）

100mm

100mm

D=32mm Material ：　 Aluminum

Rotation speed of endmill ： 1,750rpm

Transferred speed of plate ： 120mm/min (=2mm/s)

（ Actual experiment ： thickness 　 8mm ）

Initial temperature （ room temperature ）　 21.975℃

Density(kg/m3) 2700

Specific heat (J/(kg ・℃ ) 899

Thermal conductivity(J/(m ・ s ・ K) 203

Thermal diffusivity (m2/s×10-4) 0.836

Tab. Thermal properties of Aluminum

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Fig ; Temperature distribution

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2.3 Nondestructive testing of reinforcement corrosion shape

based on FEM and adjoint equation method

Fig, System for deterioration diagnosis in reinforced concrete(Oshita et. al. (2008))

Reinforced concrete

Infrared sensor

Coil for electromagnetic induction heating

Observation system of temperature on concrete surface

by electromagnetic induction heating

Heat image on concrete surface is obtained by this system.

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[1st. step]

Heat reinforcement bars

by electromagnetic induction.

[2nd. step]

After the heating,

except the coil and take heat image

by infrared sensor.

Experimental process

(Heat image on concrete surface by electromagnetic induction heating)

Reinforcement bars

Coil

Heat image

Infrared sensor

Machine for electromagnetic induction heating

Heated reinforcement bars

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Reinforced concrete

Cavity

Region of reinforcement corrosion

Fig. Examples of heat image on concrete surface

Reinforcement bar

(a) 　 Cavity (b) 　 Reinforcement corrosion

Problem;

Find shape of reinforcement corrosion using observed temperature on concrete surface.

Examples of heat image

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Computational conditions (2) （ Physical and computational conditions ）　

Real time (sec.) 600

Convection coefficient (W/m2 )℃ 10.0

Ambient temperature ( )℃ 21.3

Time of heating to reinforcement bar (sec.) 240

Heat up ratio on surface of reinforcement bar ( /sec.)℃

0.081

Initial temperature in concrete ( )℃ 19.5

Total number of nodes 242,000

Total number of elements 1,140,480

Time increment (sec.) 5.0

Time steps 120

Maximum movement value at l=0 (mm) 0.10

Convergence criterion ε 10-6

Tab. Physical constants *

Tab. Computational conditions *

【 Example of measurement 】

Cover depth 30mm

Diameter 　 D16 ( D=16mm)

Region of reinforcement corrosion 　

Length of corrosion 100mm

Width of corrosion 1.0mm

Concrete Reinforcement bar

（ without corrosion ）

Reinforcement bar （ with corrosion ）

Density ρ (kg/m33) 2.40×103 7.85×103 5.30×103

Specific heat c (J/kg)℃

1.15×103 4.70×102 1.20×103

Heat conductivity κ (W/m )℃

2.70 5.13×10 6.97×10-2

Computational domain（ * Unite for numerical conditions are changed to

“mm”.)

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Final shape

Width ： h=1.26mm ，　 Volume ：V=4,980mm3

Computational condition : Initial corrosion length : 100mm

（ Correct volume ： V=4,708mm3)

（ Correct solution ： Width h=1mm ， Length l=100mm)

Computational results (shape of reinforcement corrosion)

Initial shape

Volume ： V=9,391mm3

(Width h=2mm ， Length l=100mm)

The obtained shape is quite close to the target shape.

2323

3.1 Fluid analysis around body based on FEM

L

Characteristic inflow velocity ： U

Characteristic length ：L

Reynolds number

Re=(UL)/ν

ν ： kinematic viscosity coefficientL : Characteristic lengthU : Characteristic inflow velocity

Vortex street occurred by difference of Reynolds number

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Fluid analysis around circular cylinder based on FEM

Reynolds numner （ Re=UL/ν=112 ）

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3.2 Finite element analysis using fictitious domain method

Advantage Application for moving body problem → It is not necessary to do re-meshing.

Background mesh(whole domain Ω)

Difference points between present method and traditional FEM・ Two type of domain (overlap domain)・ Connectivity of physical value between two domains is carried out by interpolation

Foreground mesh(sub domain ω)

Magnified figure of overlapped region

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12.0

8.0

1.0

Center of circle(X,Y) = (0,0)

p=0(vx,vy)=(1.0,0.0)

Fictitious dimain FEM(vx,vy)=(0.0,0.0) in domain

Conventional FEM(vx,vy)=(0.0,0.0) on boundary

tx=0, vy=0

tx=0, vy=0

Comparison between Fictitious domain FEM and conventional FEM

for flow analysis around circular cylinder

Reynolds number Re = 250, Δt=0.001

Computational model

Nodes Elements

Foreground mesh

331 600

Background mesh

1,907 3,716

Nodes Elements

Mesh 1,636 3,116

Fictitious domain FEM

Conventional FEM

T=5 T=10 T=15 T=20

Numerical results

Pressure distribution and velocity vector at each time

Fictitious Domain FEM

Conventional FEM

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Fluid analysis around circular cylinder based on FEM using Fictitious Domain Method

Reynolds numner （ Re=250, Δt=0.001 ）

Measurement areaQ2

Q1

3.3 Application of FEM to two phase flow problem in micro-channel

Q1 (Ethyl acetate : 20℃) Q2 (Pure water: 20℃)

10 μl/min 50 μl/min

Number of nodes : 11,991

Number of elements : 23,200

Number of nodes : 11,991

Number of elements : 23,200

Q2

Q1

Non-Slip B.C. on wallNon-Slip B.C. on wall

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Experimental condition

Ex: → on interface

Continuum surface force (CSF) model

(Model of surface tension)

Surface tension coefficient

Interface curvature

Density

Coefficient for density ratio

iii

i

i

ii

ii

V

i

xxx

xgx

xxf

][

][

,

,

12

212

1

)(xg

Fluid 2

interface

Fluid 1

2

1Interface curvature

iii

i ss

x ,,

1s

ss

21

1)( xg

Reference, J.U. Brackbill, D.B.Kothe and C.ZemachContinuum method for modeling surface tension,Journal of computational physics. 100, pp.335-354, 1992.

Reference, J.U. Brackbill, D.B.Kothe and C.ZemachContinuum method for modeling surface tension,Journal of computational physics. 100, pp.335-354, 1992.

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Case3 : Pure water 50 μl/min, Ethyl acetate 10 μl/min

b1/H = 0.243 b1/H =0.296

　【 Magnification 】 ×540, H=100μm

H b1

- 0.2

- 0.15

- 0.1

- 0.05

0

0.05

0.1

0.15

0.2

- 0.4 - 0.2 0 0.2 0.4 0.6 0.8 1 1.2

Numerical rsultExperimental resultPure water

Ethyl acetate

Comparison of interface line

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Numerical and experimental results

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rrij ,

Stress distribution on interface

Fig ; Image diagram of stress distribution

at vertex on interface

Material 2

Material 1

Singular point

σ

r

O

Interface

xy

r

M.L.Williams, The stress around a fault or crack in dissimilar media, bulletin of the Seismological Society of America, Vol.49, No.2, (1959), 199-204.

σyy

Purpose of this study ：

Evaluation of stress singularity field near interface edge of bonded structure based on FEM using singular element

4.1 Evaluation of stress singularity field based on FEM

Interpolation function

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Linear tetrahedron element

Akin Singular Element 　（ in case of λ=0.5 ） ．

N1 N2 N3

SN1 SN2 SN3

N4

SN4

1

2

3

4

1

2

3

4

1

2

3

4

1

2

3

4

12

3

4

12

3

4

12

3

4

12

3

4

Singular point

11N 2N 3N

),,(

),,(11 1

1

R

NSN

),,(

),,(22

R

NSN

),,(

),,(33

R

NSN

4N

),,(

),,(44

R

NSN

Singular point Singular point Singular point

44332211

44332211

uNuNuNuNu

uSNuSNuSNuSNu

・・・ In case of elements included singular

point・・・ In the other elements

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Material 1(Mild steel)

Material 2(Aluminium)

3mm

3mm

1mm 1mm

σzz=10MPa

Characteristic minimum mesh length

Δhmin≒

Case1 : 1μm ，

Case2 : 1.5μm

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Singular point

Interface edge

1mm

1mm

0.3mm 0.3mm

Uniform mesh division area by minimum mesh size

ΔxminΔymin

Δzmin

Mesh division by tetrahedron element

3minmin Vh

最小メッシュ寸法

Material 1 (Mild steel)

Material 2 (Aluminium)

Region of Singular Elements

Stress singularity point

Tab. Minimum mesh size and nodes and elements

NodesElements

Case1 (Δhmin = 0.001145mm) 1μm

51,303234,000

Case2(Δhmin = 0.01430mm) 1.5μm

27,881126,144

Young’s modulus (GPa) Poisson’s ratio

Mild steel 216.00 0.30

Aluminium 69.09 0.33

Tab. Material properties4.2 Application of singular element for 3D model

Minimum mesh length

rKr yyzz 45,901,,

Case1 Δhmin≒8μm

Case2 Δhmin≒15μm

,45,901 yyK

r

φ=45°

Interface

Comparison of stress distribution for each minimum element mesh size （ θ=90° ， φ=45° ）

Stress distribution around singular point

are obtained by least square method

It is found that gradient of stress distribution is close to correct order of singularity λ=0.121.

z

x

y

b

1.0[mm]

6.0[mm]

6.0[mm]

Al

Fe

Material propaties

z

o

1/8model

Background cell

Minimum nodal distance≒3.9[μm]

4.3 Application of mesh free method for evaluation ofIntensity of stress singularity for 3D bonded structures

Computation conditions Tensile stress z ： 10[MPa] Width of model b ： 0.25, 0.5, 1.0, 2.0, 4.0, 6.0, 8.0[mm]

Young’s Modulus E[GPa]

Poisson ratio ν

Fe 216 0.30

Al 69.09 0.33

x

y

Advantage of mesh free methodMesh division is not needed.

(It is not necessary to satisfy the connectivity condition of domain of integration.)

8.0

7.8

7.6

7.4

Inte

nsit

y of

str

ess

sing

ular

ity

K 1

,MP

amm

5 6 7 8 9

12

Width b , mm

Al-Fe interface (=90deg) MFM BEM

20

18

16

14

12

Stre

ss

,M

Pa

3 4 5 6 7 8 90.01

2 3

Distance from origin r , mm

Al-Fe interface (=90deg)[mm]

MFM_b=2.0 BEM_b=2.0 MFM_b=1.0 BEM_b=1.0 MFM_b=0.5 BEM_b=0.5 Curve fitting

Intensity of stress singularity Kobtained by MFM is close to that obtained by BEM. But It is seen that difference of Kbetween MFM and BEM increases with increasing width “b”.

Comparison of results obtained by mesh free method and Boundary element method

K1 r vertex K2

＊ ＊ vertexvertex ＝＝0.1210.121

KK11 ：： Intensity of stress singularity for Intensity of stress singularity for

distance r from singular point distance r from singular point vertexvertex : Order of singularity at vertex on interface : Order of singularity at vertex on interface

b 1.0[mm]

°

°r

o

Stress distribution for radius direction Variation of Kwith respect to “Width” b