Solution of Coupled Thermoelasticity Problem In Rotating DisksSolution of Coupled Thermoelasticity Problem In Rotating Disks by Ayoob Entezari ... Ceramic matrix composites (CMC) Functionally

Doctoral Dissertation on

Solution of Coupled Thermoelasticity Problem In Rotating Disks

byAyoob Entezari

26 September, 2017

Supervisors:Prof. M. A. Kouchakzadeh¹ and Prof. Erasmo Carrera²

¹Sharif University of TechnologyDepartment of Aerospace Engineering, Tehran, Iran

²Polytechnic University of Turin, Department of Mechanical and Aerospace Engineering, Italy

²MUL2 research group, Polytechnic University of Turin, Italy

Advisor:Dr. Matteo Filippi²

Cotutelle Doctoral Program

1. Introduction to rotating disks

2. Fundamentals of Linear Thermoelasticity

3. Literature review & present work

4. Analytical approach

5. Numerical approach

6. Conclusion

Outlines






6. Conclusion

Outlines

4

Aerospace (aero-engines, turbo-pumps, turbo-chargers, etc.)

Mechanical (spindles, flywheel, brake disks, etc.)

Naval

Power plant (steam and gas turbines, turbo-generators, )

Chemical plant

Electronics (electrical machines)

Applications

Introduction to rotating disks

5

Introduction to rotating disksConfigurations

6

Transient thermal load

In some of applications, the disks may be exposed to sudden temperature changes in short periods of time (for Ex. start and stop cycles)

These sudden changes in temperature can cause time dependent thermal stresses.

Thermal stresses due to large temperature gradients are higher than the steady-state stresses.

In such conditions, the disk should be designed with consideration of transient effects.

start and stop cyclesI) start up,

II) shut down

Introduction to rotating disksOperating conditions

Main Loads

Centrifugal forces

Thermal loads.

7

Effective properties of FGMs

ceramic-metal FGM

Metals: steels, super alloys

Ceramic matrix composites (CMC)

Functionally graded materials (FGMs)

Introduction to rotating disksDisk materials

Peff= VmPm +Vc Pc =Vm (Pm−Pc)+Pc

ceramic-metal FGM

Pm and Pc : properties of metal and ceramicVm and Vc : volume fractions of metal and ceramicVm = ( , , )

8

Introduction to rotating disksFGM disk

power gradation law for metal volume fraction along the radius

Vm = −−

met

al v

olum

e fra

ctio

n

radius

Effective properties of FGMs

Peff= VmPm +Vc Pc =Vm (Pm−Pc)+Pc

1. Introduction to rotating disk





6. Conclusion

Outlines

10

Fundamentals of Linear Thermoelasticity

Inertia effects• static problems• dynamic problems

displacement and temperature fields interaction• uncoupled problems• coupled problems

Classification of thermoelastic problems

11

( , ), −( ), + = 0equation of motion

energy equation( , ), =

Fundamentals of Linear ThermoelasticityClassification of thermoelastic problems

static steady-state problems

• → temperature change• → displacements• → elastic coefficients• → body forces• → thermoelastic moduli• → thermal conductivity• → internal heat source

12


Under axisymmetric & plane stress assumptions

( ) = + ( ⁄ ) ln( ⁄ )

ℎ − ℎ + ℎ = 0

static steady-state problems

equation of motion

energy equation

13

− , , =


Quasi-static problems

( , ), −( ), + = 0equation of motion

energy equation

• → density• → specific heat

14

( , ), −( ), + =− , , =


equation of motion

energy equation

Dynamic uncoupled problems

15

( , ), −( ), + = +− , , =


equation of motion

energy equation

Dynamic uncoupled problems

Considering mechanical damping

• → mechanical damping coefficient of material

16

Fundamentals of Linear Thermoelasticity

Coupled thermoelasticity

the time rate of strain is taken into account in the energy equation

elasticity and energy equations are coupled.

these coupled equations must be solved simultaneously.

Classification of thermoelastic problems

equation of motion

energy equation

Mechanical and thermal BCs and ICs

( , ), ( , )

17

( , ), −( ), + =− , , + , =


equation of motion

energy equation

Classical coupled problems

• → reference temperature

18

( , ), −( ), + =− , , + , =


equation of motion

energy equation

Classical coupled problems

→ reference temperature

infinite propagation speed for the thermal disturbances !!!

19


in the classical thermoelasticity

heat conduction equation is of a parabolictype.

Predicting infinite speed for heat propagation

The prediction is not physically acceptable.

thermal wave disturbances are not detectable.

generalized theories of thermoelasticity

non-classical theories with the finite speed of the thermal wave.

20

( , ), −( ), + =+ − ( , ),+ , + , = +

( , ), −( ), −( ), + =+ − 2 , − ( , ),+ , =

( , ), −( ), + =− ∗ , , + , =

without energy dissipation


• → LS relaxation time • , → GL relaxation times• → GL material constants

• ∗→ GN material constants






6. Conclusion

Outlines

22

Coupled thermoelasticity problems are still topics of active research.

Analytical solution of the these problems are mathematically difficult.

Number of papers on analytical solutions is limited.

Numerical methods are often used to solve these problems.

Numerical solutions of these problems have been presented in many articles.

Finite element method is still applied as a powerful numerical tool in such problems.

The major presented solutions are related to the basic problems (infinite medium, half-space, layer and axisymmetric problems).

Analytical and numerical solution of rotating disk problems has never before been presented.

Literature review & present work

Conclusion of the literature review

23

Literature review & present work

Present work

Study of coupled thermoelastic behavior in disks subjected to thermal shock loads

based on the generalized and classic theories

Disks with constant and variable thickness

Made of FGM

• Main purpose

• Implementation

Analytical approach

Numerical approach

24





Solution method

Numerical evaluation


6. Conclusion

Outlines

25

Consider• An annular rotating disk with constant thickness, • made of isotropic & homogeneous material, • Under axisymmetric thermal and mechanical shock loads.

Analytical approach - Solution method

Governing equations

( , ), −( ), + =+ − ( , ),+ , + , = +

+ 1 − 1 +− + 1 + + 1 = 0

( + 2 ) + 1 − 1 − − = −

Based on LS generalized coupled theory

Eq. of motion

energy Eq.

= 2+ 2 = 2+ 2 3 + 2 • & → Lame constants• → coefficient of linear thermal expansion

I

II

26


Governing equations

+ − 1 +

− + + + = 0( + 2 ) + 1 − 1 − − = −

Inner radius of the disk

Outer radius of the disk

- time dependent known functions

constant parameters − known functions of , 0 = ,( , 0) = ( )

, 0 = ,( , 0) = ( )

| + , =| + ( , ) = ( )

| + ( , ) = ( )| + ( , ) = ( )

thermal BCs. & ICs Mechanical BCs. & ICs

Coupled System Of Equations

I

II

27

propagation speed of elastic longitudinal wave


Governing equations in Non-dimensional form

Non-dimensional parameters

unit length

= + 2 )⁄= ⁄

= , = , == , = , == + 2 ) , =

28

Thermoelastic damping or coupling parameter

where


Governing equations in Non-dimensional form


I

II

+ 1 − 1 − − = − + 1 − 1 + − + 1 + + 1 = 0

= ( + 2• Non-dimensional propagation speed of thermal wave → = 1 ⁄• Non-dimensional propagation speed of elastic longitudinal wave → = 1

29


Solution of non-dimensional equations


I

II

+ 1 − 1 − − = − + 1 − 1 + − + 1 + + 1 = 0

| + ( , ) = ( ) | + ( , ) = ( ) | + ( , ) = ( ) | + ( , ) = ( )( , 0) = ( ), ( , 0) = ( )( , 0) = ( ), ( , 0) = ( )

Thermal and mechanical BCs. & ICs

30


+ 1 − 1 + − + 1 + + 1 = 0Solution of non-dimensional equations

energy Eq.

+ 1 − − = 0| + ( , ) = ( ) | + ( , ) = ( )( , 0) = 0 , ( , 0) = 0

+ 1 − − = , + + , +| + ( , ) = 0 | + ( , ) = 0( , 0) = ( ) , ( , 0) = ( )

( , ) = ( , ) + ( , )principle of superposition

decomposition

31



principle of superposition

decomposition

( , ) = ( , ) + ( , )

+ 1 − 1 − − = − Eq. of motion

+ 1 − − = 0| + ( , ) = ( ) | + ( , ) = ( )( , 0) = 0 , ( , 0) = 0

+ 1 − − = , −| + ( , ) = 0 | + ( , ) = 0( , 0) = ( ) , ( , 0) = ( )

32



Bessel equation and can beseparately solved using finite Hankel transform

decomposition

+ 1 − 1 − − = − Eq. of motion

+ 1 − − = 0| + ( , ) = ( ) | + ( , ) = ( )( , 0) = 0 , ( , 0) = 0

+ 1 − − = , −| + ( , ) = 0 | + ( , ) = 0( , 0) = ( ) , ( , 0) = ( )

33

and are positive roots of the following equations


Finite Hankel transform


kernel functions

ℋ[ ( , )] = ( , ) = ( , ) ( , )ℋ[ ( , )] = ( , ) = ( , ) ( , )

( , ) = ( ) ( ) | + ( − ( ) ( ) | + (( , ) = ( ) ( ) | + ( − ( ) ( ) | + (

( ) | + ( ( ) | + ( − ( ) | + ( ( ) | + (= 0( ) | + ( ( ) | + ( − ( ) | + ( ( ) | + ( = 0

34

2 20 1 1 1 2 1

1

2( ) ( )m

dt T T T f t f t

dx

p

æ ö÷ç ÷+ + = -ç ÷ç ÷è ø 2 4

1 1 4 33

2( ) ( )n

du u f t f t

dh

p

æ ö÷ç ÷+ = -ç ÷ç ÷è ø

21 1 1

12 2

131 32 1 3

141 42 1 4

1 1

10

( , ) ( )

( , ) ( )

( , 0) 0 , ( , 0) 0

r a

r b

u u uu

r rr r

uk k u a t f t

ru

k k u b t f tr

u r u r

=

=

¶ ¶+ - - =

¶¶

¶+ =

¶¶

+ =¶

= =

21 1

1 0 12

111 12 1 1

121 22 1 2

1 1

10

( , ) ( )

( , ) ( )

( , 0 ) 0 , ( , 0 ) 0

r a

r b

T TT t T

r rr

Tk k T a t f t

rT

k k T b t f tr

T r T r

=

=

¶ ¶+ - - =

¶¶

¶+ =

¶¶

+ =¶

= =

Taking the finite Hankel transform


Uncoupled sub-IBVPs (Bessel equations)


Solving ODEs

1( , )nu t h 1( , )mT t x

35

2 20 1 1 1 2 1

1

2( ) ( )m

dt T T T f t f t

dx

p

æ ö÷ç ÷+ + = -ç ÷ç ÷è ø 2 4

1 1 4 33

2( ) ( )n

du u f t f t

dh

p

æ ö÷ç ÷+ = -ç ÷ç ÷è ø


Uncoupled sub-IBVPs (Bessel equations)


Solving ODEs

1( , )nu t h 1( , )mT t x

Inverse finite Hankel transforms

= 1‖ ( , )‖ , = 1‖ ( , )‖

( , ) = ( , ) ( , ( , ) = ( , ) ( , )

36



( , ) = ( ) ( , ) , ( , ) = ( ) ( , )

decomposition

+ 1 − 1 − − = − Eq. of motion

+ 1 − − = 0| + ( , ) = ( ) | + ( , ) = ( )( , 0) = 0 , ( , 0) = 0

+ 1 − − = , −| + ( , ) = 0 | + ( , ) = 0( , 0) = ( ) , ( , 0) = ( )

37




I

II

+ 1 − 1 − − = − + 1 − 1 + − + 1 + + 1 = 0

( , ) = ( , ) ( , ) + ( ) ( , )( , ) = ( , ) ( , + ( ) ( , )





Solution method

Numerical evaluation


6. Conclusion

Outlines

39

Analytical approach - Numerical evaluation

material properties

geometry

Specifications of numerical example

= 1= 2= 40.4 GPa= 27 GPa= 23 × 10 K= 2707 kg/m3= 204 W/m ⋅ K= 903 J/kg ⋅ K

( ) = 0 01 0

Boundary conditions

at = → − = ( )= 0at = → = 0= 0

40

Time history of the non-dimensional solution at mid-radius


Based on classical theory of coupled thermoelasticity

Temperature Radial displacement

mid-radius

Nondimensional Time (t)

Non

dim

ensio

nalT

empe

ratu

re(T

)

0 2 4 6 8 10 12 14

0.08

0.16

0.24

0.32

0.4

0.48

Numerical Solution (Bagri & Eslami, 2004)Exact Solution

C =0.02, =0, r =1.5


Non

dim

ensio

nalR

adia

lDisp

lace

men

t(u)

0 2 4 6 8 10 12 14

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Exact SolutionNumerical Solution (Bagri & Eslami, 2004)

C =0.02, =0, r =1.5

Validation

41

Time history of the non-dimensional solution at mid-radius



Non

dim

ensio

nalT

empe

ratu

re(T

)

0 2 4 6 8 10 12 140

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Exact SolutionNumerical (Bagri & Eslami, 2004)

t0 =0.64, C =0.02, =0, r =1.5


Non

dim

ensio

nalR

adia

lDisp

lace

men

t(u)

0 2 4 6 8 10 12 14

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Exact SolutionNumerical (Bagri & Eslami, 2004)

t0 =0.64, C =0.02, =0, r =1.5

Temperature Radial displacement

mid-radius

Based on LS generalized theory of coupled thermoelasticityValidation

42


Nondimensional Radius(r)

Non

dim

ensio

nalT

empe

ratu

re(T

)

1 1.2 1.4 1.6 1.8 2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

steady state

T0=293 K, t0=0.64, =0.01

1.25

t=0.25

0.5

0.75

1

+

+

+

+

+

+

+

+

+

+

++

++

++

+ + + + + +

Nondimensional Radius (r)

Non

dim

ensio

nalR

adia

lDisp

lace

men

t(u)

1 1.2 1.4 1.6 1.8 2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

steady statet =0.25

1

T0=293 K, t0=0.64, =0.01

1.25

t =0.50.75

10


Non

dim

ensio

nalT

ange

ntia

lStre

ss(

)

1 1.2 1.4 1.6 1.8 2

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

steady statet = 0.25t = 0.5t = 0.75t = 1t = 1.25

T0=293 K, t0=0.64, =0.01


Non

dim

ensio

nalR

adia

lStre

ss(

rr)

1.2 1.4 1.6 1.8 2

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

steady statet = 0.25t = 0.5t = 0.75t = 1t = 1.25

T0=293 K, t0=0.64, =0.01

temperature change radial stress circumferential stressradial displacement

radius radius radius radius

Based on LS generalized theory of coupled thermoelasticity

Radial distribution for different values of the time.

Results and discussion






6. Conclusion

Outlines






Motivation

Development of method

Evaluations and results

6. Conclusion

Outlines

45

Analytical solutions are limited to those of a disk with simple geometry and boundary conditions.

FE method is more widely used for this class of problems.

1D and 2D FE models are not able to provide all the desired information.

3D FE modeling techniques may be required for a detailed coupled thermoelastic analysis.

3D FE models still impose large computational costs, specially, in a time-consuming transient solution.

There is a growing interest in the development of refined FE models with lower computational efforts.

A refined FE approach was developed by Prof. Carrera et al.

They formulated the FE methods on the basis of a class of theories of structures.

Numerical approach

Motivations

46

Numerical approachMain characteristics of FE models refined by Carrera

3D capabilities

lower computational costs

ability to analyze multi-field problems and multi-layered structuresMUL2 research group,Polytechnic University,

Turin, Italywww.mul2.polito.it






Motivation



6. Conclusion

Outlines

48

Approaches to FE modeling

Numerical approach - Development of method

Variational approach

Weighted residual methods

Weighted residual method based on Galerkin technique

Efficient, high rate of convergence

most common method to obtain a weak formulation of the problem

49

Governing equations

, , ,, ,Equation of motion Energy equation


)Hooke’s law

For anisotropic and nonhomogeneous materials.

Including LS, GL and classical theories of thermoelasticity.

Considering mechanical damping effect.

= = = = 0 → classical theory

= = = 0→ LS theory

= 0→ GL theory.

50

( , , , ) = ( , , ) ( )( , , , ) = ( , , ) ( )• = 1,⋯ ,• = number of nodal points in a element


FE formulation through Galerkin technique

• In 3D conventional FE method

51

, + − − = 0 ( + ) + − 2 , − , ,+ , + , − − = 0Equation of motion energy equation

Weighting function



( , , )

52

( ) + ( ) + ( T σ)= ( ) + ( )

( βT ) + ) + )+ ( βT ) + ) − (2 T∇ )

+ (∇T κ∇ ) = ( T ) + ( ) + )

Eq. of motion

energy Eq.



53


Refined 1D FE model through Carrera unified formulation

= ( )= ( )

1D FE3D beam-type structures

• = 1,⋯ ,• = number of bar nodes

54

• = 1,⋯ ,• = number of terms of the expansion.

Carrera unified formulation (CUF)

( , ) = ( , ) ( )( , ) = ( , )Θ ( )


Refined 1D FE model through Carrera unified formulation

• = 1,⋯ ,• = number of bar nodes

= ( )= ( )

1D FE3D beam-type structures

55

= ( )= ( )1D FE CUF( , ) = ( , ) ( )( , ) = ( , )Θ ( ) ( , , , ) = ( , , ) ( )( , , , ) = ( , , ) Θ ( )

1D FE-CUF

3D 8-nodes elementrefined 1D 2-nodes element


Refined 1D FE model through CUF

weighting function in 1D FE-CUF → ( , , )= ( ) ( , )

56



( , , , ) = ( ) ( , ) ( )( , , , ) = ( ) ( , ) Θ ( )1D FE-CUF

1D FE modeling elements and shape functions in 1D FE modeling

B2 linear

element

B2 quadratic

B4 cubic

57

selection of ( , ) and ( = 1,⋯ , ) is arbitrary.

various kinds of basic functions such as polynomials, harmonics and exponentials of any-order.

For instance, different classes of polynomials such as Taylor, Legendre and Lagrange polynomials.

In Carrera unified formulation



( , , , ) = ( ) ( , ) ( )( , , , ) = ( ) ( , ) Θ ( )1D FE-CUF

58

( , ) → bi-dimensional Lagrange functions

cross-sections can be discretized using Lagrange elements

• linear three-point (L3)• quadratic six-point (L6) • bilinear four-point (L4)



( , , , ) = ( ) ( , ) ( )( , , , ) = ( ) ( , ) Θ ( )1D FE-CUF

• biquadratic nine-point (L9) • bi-cubic sixteen-point (L16)

59

Substituting

into the weak forms of equation of motion and energy equation gives

lm s ls lm s ls lm s ls mt t t t+ =+M G K p d d d

• , and → 4×4 fundamental nuclei (FNs)of the mass, damping, and stiffness matrices

• → 4×1 FN of the load vector

• δ → 4×1 FN of the unknowns vector


FE equations in CUF form

( , , , ) = ( ) ( , ) ( )( , , , ) = ( ) ( , ) Θ ( )1D FE-CUF ( , , )= ( ) ( , )weighting function

60

Rayleigh damping model

→ structural damping effect

or


0 + + 0 =

= +

lm s ls lm s ls lm s ls mt t t t+ =+M G K p d d d

FE equations in CUF form

Different theories of thermoelasticity through the 1D FE-CUF

61

+ + =M G K P D D Dfor whole structure

assembly procedure of FNsfor each elementlm s ls lm s ls lm s ls mt t t t=+ +M G K pd d d

2 L4

3 B4


total degrees of freedom

DOF = 4 ×a model with 3 B4 / 2 L4, DOF=240

Assembly procedure via Fundamental Nuclei

62

taking Laplace

Transfinite element technique

2[ ]lm seq

lm s lm s lm s ls ms st

t t t t* *+ + =

K

M G pK d

numerical inversion


Time history analysis

lm s ls lm s ls lm s ls mt t t t=+ +M G K pd d d

eq* *=K PD

Assembling & ∗ for whole structure

solution in Laplace domain Δ∗solve

solution in time domain (Δ( ))

• →the Laplace variable• → FN of the equivalent stiffness matrix• ∗ denotes Laplace transform of the terms.

63

diffusivity velocity of elastic longitudinal wave unit length


Non-dimensional Equation for isotropic FGMs

Non-dimensional parameters

m

m

m

m m

m m0 0

m m m

m m m m

m m

m m m

ˆˆ ; ;

( 2 )ˆ ˆˆ; ;

1 1 1ˆˆ ˆ; ;

ˆ ˆ;( 2 )

eii

ei i

d d

n ni i ij ij i i

d e d d

i id m d

Vxx t t

l l

VTT u u t t

T l T l

q q t tc T V T T

l DX X R R

T c T

l mb

s sr b b

b l m

= =

+= = =

= = =

= =+

m m m m( 2 )/eV l m r= +mm m / el D V= m m m m/D ck r=

64


δ ∗ = ∗or ∗∗∗∗

=∗∗∗∗

Non-dimensional FNs for isotropic FGMs based on LS theory

Transfinite element equation2

11 22 , ,

, ,66 44 , ,

, ,12 66 , 23 ,

13 44 , , 21 , ,

14 ,

41

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ

y y

y y

slm ml mls L x s x L

m l mls L z s z L

m l mlslms x L x s L

slm ml mlz s x L x s z L

slm mlx s L

s

K s C F F I C F F I

C F F I C F F I

K C F F I C F F I

K C F F I C F F I

K C F F I

K

tr t t

t t

tt t

tt t

tb t

t

= +

+ +

= +

= +

=-

20 ,

,242 0

243 0 ,

244 0

, ,, ,

, ,

ˆˆ( )

ˆˆ( )

ˆˆ( )

ˆ ˆˆ( )

ˆ ˆ

ˆ

y

y y

lm mls x L

mlslms L

slm mls z L

slm mlc s L

m lmlx s x L s L

mlz s z L

C s t s C F F I

K C s t s C F F I

K C s t s C F F I

K s t s C C F F I

C F F I C F F I

C F F I

b t

tb t

tb t

tr t

k t k t

k t

= +

= +

= +

= + +

+ +

+

( ) ( )

( ) ( )

( ) ( )

( ) ( )

*1

*2

*3

* *4 0

ˆt

ˆt

ˆt

ˆˆ ˆ[( 1) ] ( )

e e

e e

e e

e e

m nx m x m

S Vm n

y m y m

S Vm n

z m z m

S Vm

m i i m

V S

p F N dS X F N dV

p F N dS X F N dV

p F N dS X F N dV

p t s R F N dV q n F N dS

tt t

tt t

tt t

tt t

* *

* *

* *

*

= +

= +

= +

= + +

ò ò

ò ò

ò ò

ò ò

65


Non-dimensional FNs for isotropic FGMs based on LS theory

11 22 , ,

, ,66 44 , ,

, ,12 66 , 23 ,

2

13 44 , , 21 , ,

14 ,

41

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ

y y

y y

slm ml mls L x s x L

m l mls L z s z L

m l mlslms x L x s L

slm ml mlz s x L x s z L

slm mlx s L

s

K C F F I C F F I

C F F I C F F I

K C F F I C F F I

K C F F I C F F I

K C F I

s

F

K

tr t t

t t

tt t

tt t

tb t

t

= +

+ +

= +

= +

=-

0 ,

,42 0

43 0 ,

44 0

, ,, ,

,

2

2

,

2

2

ˆˆ( )

ˆˆ( )

ˆˆ( )

ˆ ˆˆ( )

ˆ ˆ

ˆ

y

y y

lm mls x L

mlslms L

slm mls z L

slm mlc s L

m lmlx s x L s L

mlz s z L

s t C F F I

K t C F F I

K t C F F I

K t C C F F I

C F F I C F F I

C F F I

C

C

C

s

s s

s s

s s

b t

tb t

tb t

tr t

k t k t

k t

= +

= +

= +

= + +

+ +

+

( )( )

eAdA= ò

( )( ) , , , ,

, , , ,y y y y

e y y y y

m l ml m lmlL L L L m l m l m l m lLI I I I N N N N N N N N dy=ò

( )11 22 33

m m

44 55 66m m

12 13 23m m

2ˆ ˆ ˆ( 2 )

ˆ ˆ ˆ( 2 )

ˆ ˆ ˆ( 2 )

C C C

C C C

C C C

m l

l mm

l ml

l m

+= = =

+

= = =+

= = =+

m m m m

ˆ ˆ ˆ ˆ, , , cc

C C C Ccr b k

r b kr b k

= = = =

20 m

m m m m( )

T

cC

br l m

=+

thermoelastic coupling parameter →






Motivation



6. Conclusion

Outlines

1. Introduction to rotating disk2. Fundamentals of Linear Thermoelasticity3. Literature review & present work4. Analytical approach5. Numerical approach Motivation Development of method Evaluations and results

o Static structural analysisExample 1. Rotating variable thickness diskExample 2. Rotating variable thickness disk subjected thermal loadExample 3. Complex rotor

o Static structural-thermal analysis – Example 4. simple beam

o Quasi-static structural-thermal analysis – Example 5. simple beam

o Dynamic coupled structural-thermal analysisExample 6. Constant thickness disk made of isotropic homogeneous materialsExample 7. Constant thickness disk made of isotropic FGMs Example 8. variable thickness disk made of isotropic FGMs

6. Conclusion

Outlines





o Dynamic coupled structural-thermal analysis

Example 6. Constant thickness disk made of isotropic homogeneous materialsExample 7. Constant thickness disk made of isotropic FGMs Example 8. variable thickness disk made of isotropic FGMs

6. Conclusion

Outlines

69

Example 1. Rotating variable thickness disk

Young’s modulus 207GPaPoisson’s ratio 0.28density ( ) 7860kg/m

• = 2000rad/s• hub is assumed to be fully fixed = 0.05m= 0.2m ℎ = 0.06m

ℎ = 0.0134 . ℎ = 0.03m

Material properties

Static structural analysis

Numerical approach - Evaluations and results

annular disk with hyperbolic profile

70


Static structural analysisExample 1. Rotating variable thickness disk

DOFDiscretizingModel Over the corss sectionsAlong the axis6240 (2/6/8) × 32L48 B2, 3 CS*(1)5472 (2/4/6/8) × 32L48 B2, 4 CS (2)7200 (2/4/6/8) × 32L410 B2, 4 CS(3)7584 (1/2/4/6/8) × 32L412 B2, 5 CS(4)8352 (1/2/3/4/6/8) × 32L414 B2, 6 CS(5)9504 (1/2/3/4/5/6/8) × 32L416 B2, 7 CS(6)11040(1/2/3/4/5/6/7/8) × 32L418 B2, 8 CS(7)14496(1/2/3/4/5/6/7/8) × 32L422 B2, 8 CS(8)

* 3 types of cross section (CS) with different radii

1D FE-CUF modeling

Different 1D FE-CUF models of the disk

discretization along the axis

71

XX

XX

XXXX

XX

XXXX

XX

XXXX

XX

XXXX

XX

XXXX

XX

XXXXXX

XXXX XX XX

Radius (m)

Rad

ialD

ispla

cem

ent,

u r(

m)

0.05 0.075 0.1 0.125 0.15 0.175 0.20

15

30

45

60

75

90

105

120

135

150

165

Analytical Solutionmodel (1): DOF=6240model (2): DOF=5472model (3): DOF=7200model (4): DOF=7584model (5): DOF=8352model (6): DOF=9504model (7): DOF=11040model (8): DOF=14496ANSYS: DOF=14400

X

Radial displacement (μm)DOFModel At outer radiusAt mid-radius157.57119.011Analytical

1D CUF- FE(1.00)156.00(1.10)120.326240 (1)(0.36)157.00(0.54)118.365472 (2)(0.73)156.42(0.54)118.367200 (3)(0.27)157.15(0.22)118.757584 (4)(0.32)158.08(0.41)119.508352 (5)(0.23)157.93(0.43)118.509504 (6)(1.68)154.92(1.47)117.2611040(7)(1.63)155.00(1.64)117.0614496(8)(0.30)157.10(0.01)119.00144003D ANSYS

( ): % difference with respect to the analytical solution.

Radial displacement


Static structural analysisExample 1. Rotating variable thickness disk verification of results

72

Radial displacement (μm)DOFModel At outer radiusAt mid-radius157.57119.011Analytical

1D CUF- FE(1.00)156.00(1.10)120.326240 (1)(0.36)157.00(0.54)118.365472 (2)(0.73)156.42(0.54)118.367200 (3)(0.27)157.15(0.22)118.757584 (4)(0.32)158.08(0.41)119.508352 (5)(0.23)157.93(0.43)118.509504 (6)(1.68)154.92(1.47)117.2611040(7)(1.63)155.00(1.64)117.0614496(8)(0.30)157.10(0.01)119.00144003D ANSYS

( ): % difference with respect to the analytical solution.

Radial displacement



Model (2)

Error < 0.6%

2.6 times less DOFs of the 3D ANSYS model !!

73

Mesh refinement over the cross-sections


Static structural analysisExample 1. Rotating variable thickness disk 1D FE-CUF modeling

DOF1D FE-CUF Modelmodel31688B2, (1/2/3/4) × 32L4154728B2, (2/4/6/8) × 32L4289288B2, (5/7/9/14) × 32L43100808B2, (4/8/12/16) × 32L44135368B2, (10/12/14/20) × 32L45

74

XX

XX

XXXX

XX

XXXX

XX

XXXX

XX

XXXX

XX

XXXX

XX

XXXX

XXXXXX XX

Radius (m)

Rad

ialD

ispla

cem

ent,

u r(

m)

0.05 0.075 0.1 0.125 0.15 0.175 0.20

15

30

45

60

75

90

105

120

135

150

165

Analytical Solution8 B2, (1/2/3/4)32 L4, DOF=31688 B2, (2/4/6/8)32 L4, DOF=54728 B2, (5/7/9/14)32 L4, DOF=89288 B2, (4/8/12/16)32 L4, DOF=100808 B2, (10/12/14/20)32 L4, DOF=13536ANSYS, DOF= 14400

X

Radial displacement



effect of enriching the radial discretization Radial displacement (μm)DOFModel

At outer radiusAt mid-radius157.57119.011Analytical

1D CUF FE(2.27)154.00(3.79)114.503168 8B2, (1/2/3/4) × 32L4(0.36)157.00(0.54)118.365472 8B2, (2/4/6/8) × 32L4(0.36)157.00(0.01)119.008928 8B2, (5/7/9/14) × 32L4(0.27)158.00(0.01)119.00100808B2, (4/8/12/16) × 32L4(0.36)157.00(0.01)119.00135368B2, (10/12/14/20) × 32L4(0.30)157.10(0.01)119.00144003D FE (ANSYS)

( ) Absolute percentage difference with respect to the analytical solution.

Converged solution

with 1.6 times less DOFs of the 3D ANSYS model !!







6. Conclusion

Outlines

76

Radius (m)

Tem

prat

ure

chan

ge(

C)

0.05 0.075 0.1 0.125 0.15 0.175 0.2500

510

520

530

540

550

560

570

580

590

600

radial steady-state temperature distribution

• The disk is subjected to radial temperature gradient.• hub is assumed to be axially fixed.



Example 2. Rotating variable thickness disk subjected thermal load

77

radial displacement Radial and circumferential stresses


0.5 1 1.5 2

ur(mm)

Max=1.94

Min=0.55

50 150 250

rr(MPa)

Max=238

Min=0

150 350 550

(MPa)

Max=541

Min=0

radius radius


Example 2. Rotating variable thickness disk subjected thermal load

radialstress

circumferentialstress







6. Conclusion

Outlines

79

3D model of a complex rotor

The profile hyperbolic for the turbine disk

web-type profile for the compressor disks

Both ends of the shaft are fully fixed.



Example 3. Complex rotor

80

y (m)

z(m

)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.2

-0.1

0

0.1

0.2

28 Beam Elements

y (m)

z(m

)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.2

-0.1

0

0.1

0.2

32 Beam Elements

y (m)

z(m

)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.2

-0.1

0

0.1

0.2

40 Beam Elements

28 B2

uniform 12 × 32 L4

Lagrange mesh over the cross-section with the largest radius

refined 17 × 32 L4

32 B2 40 B2

discretizing along the axis



Example 3. Complex rotor 1D FE-CUF modeling

81

y (m)

z(m

)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7-0.2

-0.1

0

0.1

0.2

32 Beam Elements

refined 17 × 32L432 B2 along the axis

Converged model

computational model, DOF=27072



Example 3. Complex rotor 1D FE-CUF modeling

82



Example 3. Complex rotor verification of results

0 25 50 75 100 125 150 175 200ur(m)

1D FE-CUF solutionDOFs= 27,072

3D FE ANSYSDOFs=44,280

xx x

xx

xx

xx

xx

xx

xx

xx

x xx

x x x x x x x x x x

Radius (m)

Rad

ialD

ispl

acem

ent(m

)

0.05 0.075 0.1 0.125 0.15 0.175 0.20

20

40

60

80

100

120

140

160

180

200

Turbine disk in the rotor (3D ANSYS result)Turbine disk in the rotor (1D CUF result)Comp. Disk 1 in the rotorComp. Disk 2 in the rotor

x

1D FE-CUF solution

Radial displacement

with 1.6 times less DOFs

of the 3D ANSYS model !!

83

Radius (m)

Stre

ss(M

Pa)

0.05 0.075 0.1 0.125 0.15 0.175 0.20

50

100

150

200

250

300

350

400

450

500

550

Turbine disk in the rotorSingle Turbine disk with rigid hub

rr

rr


50 100 150 200 250 300 350 400 450 500 550 600 650 700rr(MPa)

50 100 150 200 250 300 350 400 450 500 550(MPa)


Example 3. Complex rotor verification of results

Radial and circumferential stresses

Circumferential stress

Radial stress







6. Conclusion

Outlines

85

Static structural-thermal analysis


Example 4. simple beam

Lagrange elements over the cross-section

discretizing along the axis

1D FE-CUF modeling

86




Temperature change

Axial displacement

results

1 L4 over the cross-section

87



Example 4. simple beam results



88

Yes!!

Does heat conduction equation satisfy?



Example 4. simple beam verification of results

89

At = 0.1→ = (0.1 )(23.1 × 10 ) . . = 0.219 mm

At = 0.5→ = (0.5 )(23.1 × 10 ) . = 0.6092 mm

Elongation =



Example 4. simple beam verification of results

Check free thermal expansion !







6. Conclusion

Outlines

91



Quasi-static structural-thermal analysis

Transient heat flux (thermal load)

92



Quasi-static structural-thermal analysis

results

Temperature changeAxial displacement

10B4/1L4 model

Axial displacement Temperature change

93





o Dynamic coupled structural-thermal analysisExample 6. Constant thickness disk made of isotropic homogeneous materialsExample 7. Constant thickness disk made of isotropic FGMs Example 8. variable thickness disk made of isotropic FGMs

6. Conclusion

Outlines

94

Lame’constant λ 40.4GPaLame’ constant μ 27GPacoefficient of linear thermal expansion (α) 23 × 10 Kdensity (ρ) 2707kg/mthermal conductivity (κ) 204W/m ∙ Kspecific heat (c) 903J/kg ∙ K

Material properties

Dynamic coupled structural-thermal analysis


Example 6. Constant thickness disk made of isotropic homogeneous materials

geometry= 1= 2Thickness = 0.1

Boundary conditions

= → − = ( )= 0 = → = 0= 0

where ( ) = 0 01 0

95



Example 6. Constant thickness disk made of isotropic homogeneous materials 1D FE-CUF modeling

DOFDiscretizingModel corss sectionsAlong the axis1680(6 × 30)L41 B2(1)25201 B3(2)33601 B4(3)

16803 × 15 L91 B2

(4) 2 × 10 L16(5)37446 × 18 L9(6)

Lagrange mesh over the cross-section with the largest radius

discretizing along the axis Different 1D FE-CUF models for the constant thickness disk

96


Non

dim

ensio

nalR

adia

lDisp

lace

men

t(u)

0 2 4 6 8 10 12 14

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Exact Solution (Entezari, 2016)Axisymetric FE (Bagri, 2004)CUF: 1 B2/(618) L9

t0 =0.64, C =0.02, r =1.5


Non

dim

ensio

nalT

empe

ratu

re(T

)

0 2 4 6 8 10 12 140

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Exact Solution (Entezari, 2016)Axisymetric FE (Bagri, 2004)CUF: 1 B2/(618) L9

t0 =0.64, C =0.02, r =1.5



Example 6. Constant thickness disk made of isotropic homogeneous materials verification of results

Based on the LS theoryof thermoelasticity

Time history of solution at mid-radius of the disk.

Temperature change Radial displacement

1B2 / 6 × 18 L9

97



Example 6. Constant thickness disk made of isotropic homogeneous materials verification of results

Based on the LS theoryof thermoelasticity

Time history of solution at mid-radius of the disk.

1B2 / 6 × 18 L9 Nondimensional Time (t)

Non

dim

ensio

nalT

ange

ntia

lStre

ss(

)

0 2 4 6 8 10 12 14

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Exact Solution (Entezari, 2016)CUF: 1 B2/(618) L9

t0 =0.64, C =0.02, r =1.5


Non

dim

ensio

nalR

adia

lStre

ss(

rr)

0 2 4 6 8 10 12 14

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Exact Solution (Entezari, 2016)CUF: 1 B2/(618) L9

t0 =0.64, C =0.02, r =1.5

Circumferential stressRadial stress





o Dynamic coupled structural-thermal analysisExample 6. Constant thickness disk made of isotropic homogeneous materialsExample 7. Constant thickness disk made of isotropic FGM Example 8. variable thickness disk made of isotropic FGMs

6. Conclusion

Outlines

99

Metal: Aluminum Ceramic: AluminaLame’constant λ 40.4 GPa 219.2 GPashear modulus μ 27.0 GPa 146.2 GPadensity (ρ) 2707 kg/m3 3800 kg/m3

coefficient of linear thermal expansion (α) 23.0106 K1 7.4106 K1

thermal conductivity (κ) 204 W/mK 28.0 W/mKspecific heat (c) 903 J/kgK 760 J/kgKdimensionless relaxation time ( ) 0.64 1.5625



Example 7. Constant thickness disk made of isotropic FGM

Material properties Metal-Ceramic FGM

geometry= 1= 2Thickness = 0.1

Geometry and material

effective properties

P=VmPm +Vc Pc =Vm (Pm−Pc)+Pc

metal volume fraction

Vm = − −

100

z

xFixed

Free

y

z

Adiabatic

T(t)Non-dimensional form

= 293 K, =0.01

at = 0 → = = = = 0



Example 7. Constant thickness disk made of isotropic FGM Operational, boundary & initial conditions

( ) = (1 − [1 + 100 mm] m m⁄ )

( ) = 1 − (1 + 100 )

101

Time history



Example 7. Constant thickness disk made of isotropic FGM results

the material properties linearly change through the radius ( = 1)

Based on the LS theory of thermoelasticity

Temperature change

Radial displacement

102

Axial deformation


Deformations of disk profile



Radial displacement radial deformation

Time history



103



Example 7. Constant thickness disk made of isotropic FGM results Speed range of the thermal wave



104



Example 7. Constant thickness disk made of isotropic FGM results Speed range of the thermal wave



wave reflection ≃ 1/1.5 = 0.6

105

1 1.25⁄ 1 0.27⁄0.8 3.7




c = c m⁄ 1 c⁄ = 0.27m = 1 m⁄ = 1.25

Thermal wave propagation

m, , − m m

+ − , + ,+ + = 0Non-dimensional form of energy equation



Speed range of the thermal wave 0.27 ,FGM 1.25wave reflection ≃ 1/1.5 = 0.6

106



Example 7. Constant thickness disk made of isotropic FGM results Elastic wave propagation



wave reflection

107



Example 7. Constant thickness disk made of isotropic FGM results Elastic wave propagation



wave reflectionNon-dimensional form of equation of motion

m + 2 m, + +

m + 2 m, + 1

m + 2 m, ,+ 1

m + 2 m, ( , + , ) −

m− 1

m, + , + = 0 c = 1.96

m = 1 →1 FGM 1.96 0.51 1Speed range of elastic wave

108

X

X

X

X

X

X

X

X

X

X

XX X

X XX

X X XX X X X X X X X X X X X X X X X X X

Nondimensional time(t)

Non

dim

ensio

nalt

empe

ratu

rech

ange

(T)

0 2 4 6 8 10 12 14-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

n = 0n = 1n = 2n = 5X

at mid radius

X X X X X X X X XX

X X XX X X

XX X

XX X

XX X

XX

X XX

X XX

X X XX X

Nondimensional time(t)N

ondi

men

siona

lrad

iald

ispla

cem

ent(

u r)0 2 4 6 8 10 12 14

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

n = 0n = 1n = 2n = 5X

at mid radius

Time history based on the LS theory at mid-radius of the disk




Temperature change Radial displacement

mid-radius

effects of power law index ( )


Vm = − −

109

Time history based on the LS theory at mid-radius of the disk




Radial stress Circumferential stress

mid-radius

effects of power law index ( )


Vm = − −X X

XX

XX

X XX

XX X

XX

X X

X

XX

X

X

X

X

X X

X

X

X

X

X XX

X

XX

XX X

X

X XX

XX X

XX

X

XX

X

X

XX

XX

X

X

XX

X X

X

X

X XX X

XX

XX

XX

X


Non

dim

ensio

nalr

adia

lstre

ss(

rr)

0 2 4 6 8 10 12 14-1.6

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

n = 0n = 1n = 2n = 5X

at mid radius

XX

XX

XX

X XX

X

X X XX

X

X

X

XX

XX

XX

X

X X

X

XX X

X X

X

X X X

X

X

X

XX

XX

XX

XX X

X XX

X

XX X X X

X

XX

X X X

X

XX

X XX

X

X

XX X

X


ondi

men

siona

lcirc

umfe

rent

ials

tress

(

)0 2 4 6 8 10 12 14

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

n = 0n = 1n = 2n = 5X

at mid radius

110

Time history based on the LS theory at mid-radius of the disk (n =1)




mid-radius

effects of reference temperature ( )

X

X

X

X X

XX

X XX

XX X

XX X X

X X X X X X X X X X X X X X X X X X X X


Non

dim

ensio

nalt

empe

ratu

rech

ange

(T)

0 2 4 6 8 10 12 14-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

T0 = 0T0 = 293T0 = 800

X

at mid radius

X X X XX X X

XX

X

X

X

X

X

X

X

X

X

XX

X

X

X

X

X X X

X

X

X

X X X

XX

X

X

X

X X

XX

X

X

X

XX

X

X XX

X

XX

X

X

XX

X

X X

X

X

X

XX

X XX

X

X

X

XX

X


ondi

men

siona

lrad

iald

ispla

cem

ent(

u r)0 2 4 6 8 10 12 14

0

0.1

0.2

0.3

0.4

0.5

0.6

T0 = 0T0 = 293T0 = 800

X

at mid radius

Radial displacementTemperature change = mm m( m + m)

coupling parameter

111

Time history based on the LS theory at mid-radius of the disk (n =1)




mid-radius

effects of reference temperature ( )

= mm m( m + m)

coupling parameter

X XX X X X

XX

X

X XX

X

X X

X

X

X

X

X

X

X

X

X

X

X

XX

X

X

X X

X

X

X

X

X

XX

X

X

X

X

X

X

X

X X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

XX

X

XX

X

X

X

X


Non

dim

ensio

nalr

adia

lstre

ss(

rr)

0 2 4 6 8 10 12 14-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

T0 = 0T0 = 293T0 = 800

X at mid radius

X XX

XX

X

X X X X

X

X

X

X

X

X

X

XX

X

X

X

X

X

X

X

X

X

X

X

X X

X

X

X

X

X

X

X

X XX

X

X

X

X

X

X X

X

X X

X

X

XX

X

X

X

X

X

XX X X

X

X X

X X

XX

X

X

X


ondi

men

siona

lcirc

umfe

rent

ials

tress

(

)0 2 4 6 8 10 12 14

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

T0 = 0T0 = 293T0 = 800

X at mid radius

Radial stress Circumferential stress

112





o Dynamic coupled structural-thermal analysisExample 6. Constant thickness disk made of isotropic homogeneous materialsExample 7. Constant thickness disk made of isotropic FGMExample 8. variable thickness disk made of isotropic FGM

6. Conclusion

Outlines

113

0.3

2

0.5

0.6

h = 0.42 r-0.5

r



Example 8. variable thickness disk made of isotropic FGM Geometry and material

Metal: Aluminum Ceramic: AluminaLame’constant λ 40.4 GPa 219.2 GPashear modulus μ 27.0 GPa 146.2 GPadensity (ρ) 2707 kg/m3 3800 kg/m3

coefficient of linear thermal expansion (α) 23.0106 K1 7.4106 K1

thermal conductivity (κ) 204 W/mK 28.0 W/mKspecific heat (c) 903 J/kgK 760 J/kgKdimensionless relaxation time ( ) 0.64 1.5625

Material properties Metal-Ceramic FGM

geometry = = 0.5 = = 2ℎ = 0.6ℎ = 0.3effective properties

P=VmPm +Vc Pc =Vm (Pm−Pc)+Pc


Vm = − −

114

= 293 K, =0.05

at = 0 → = = = = 0



Operational, boundary & initial conditionsExample 8. variable thickness disk made of isotropic FGM

Non-dimensional form

( ) = (1 − m⁄ )

( ) = 1 −

115



Example 8. variable thickness disk made of isotropic FGM Results for = 0

Nondimensional time

Non

dim

ensi

onal

tem

pera

ture

chan

ge

0 2 4 6 8 10 12 14

-0.2

0

0.2

0.4

0.6

0.8

1

Classic theoryLS theory

Temperature change

Nondimensional time

Non

dim

ensi

onal

radi

alst

ress

0 2 4 6 8 10 12 14

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4


Radial stress

Nondimensional time

Non

dim

ensi

onal

von

mis

eseq

uiva

lent

stre

ss

0 2 4 6 8 10 12 140

0.2

0.4

0.6

0.8

1


Von mises stress

Nondimensional time

Non

dim

ensi

onal

circ

umfe

rent

ials

tress

0 2 4 6 8 10 12 14

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4


Circumferential stress

Nondimensional time

Non

dim

ensi

onal

radi

aldi

spla

cem

ent

0 2 4 6 8 10 12 14

-0.2

0

0.2

0.4

0.6

0.8

1


Radial displacement

116



Example 8. variable thickness disk made of isotropic FGM Results for = 0 based LS theory

117



Example 8. variable thickness disk made of isotropic FGM

Rad

iald

ispl

acem

ent

Axi

al d

ispl

acem

ent

Results for = 0 based LS theory

118

Outlines






6. Conclusion

119

Some results obtained from coupled thermoelasticity solution

Conclusion - Summary of results

Transient deformations and stresses may be higher than those of a steady-state condition.

Time history of temperature is damped faster than time history of displacements.

Deformations and stresses oscillate along the time in a harmonic form.

Under the propagating longitudinal elastic waves along the radius, thickness of the disk also expands and

contracts, due to the Poisson effect.

When the coupling parameter takes a greater value, the amplitudes of oscillations of temperature

increase.

Lord–Shulman generalized coupled thermoelasticity predicts larger temperature and stresses compared

to the classical theories.

A functionally graded disk may be used as thermal barrier to reduce the thermal shock effects.

120

Conclusion - Summary of results

Some general points on the 1D FE-CUF modeling of disks The 1D FE method refined by the CUF can be effectively employed to analyze disks reduce the

computational cost of 3D FE analysis without affecting the accuracy.

the models provides a unified formulation that can easily consider different higher-order theories wherelarge bending loads are involved in the problem.

Increasing 1D elements along the axis of disks may not have significant effect on accuracy of resultsand only leads to more DOFs.

A proper distribution of the Lagrange elements and type of element used over the cross sections maylead to a reduction in computational costs and the convergence of results.

Making use of higher-order Lagrange elements (like L9 and L16) can reduce DOFs, while preservingthe accuracy.

increase of number of elements along the radial direction, compared to circumferential direction, ismore effective in improving the results.

121

Conclusion - Future works

Nonlinear thermoelasticity problems

Dynamic analysis of rotors subjected to transient thermal pre-stresses.

Study of thermoelastic damping effect on dynamic behaviors of rotors.

It is of interests to extend the study to

122

1. Entezari A, Filippi M, Carrera E., Kouchakzadeh M A, 3D Dynamic Coupled Thermoelastic Solution For ConstantThickness Disks Using Refined 1D Finite Element Models. European Journal of Mechanics - A/Solids. (Underreview).

2. Entezari A, Filippi M, Carrera E. Unified finite element approach for generalized coupled thermoelastic analysisof 3D beam-type structures, part 1: Equations and formulation. Journal of Thermal Stresses. 2017:1-16.

3. Filippi M, Entezari A, Carrera E. Unified finite element approach for generalized coupled thermoelastic analysisof 3D beam-type structures, part 2: Numerical evaluations. Journal of Thermal Stresses. 2017:1-15.

4. Entezari A, Filippi M, Carrera E. On dynamic analysis of variable thickness disks and complex rotors subjectedto thermal and mechanical prestresses. Journal of Sound and Vibration. 2017;405:68-85.

5. Kouchakzadeh MA, Entezari A, Carrera E. Exact Solutions for Dynamic and Quasi-Static ThermoelasticityProblems in Rotating Disks. Aerotecnica Missili & Spazio. 2016;95:3-12.

6. Entezari A, Kouchakzadeh MA, Carrera E, Filippi M. A refined finite element method for stress analysis of rotorsand rotating disks with variable thickness. Acta Mechanica. 2016:1-20.

7. Entezari A, Kouchakzadeh MA. Analytical solution of generalized coupled thermoelasticity problem in a rotatingdisk subjected to thermal and mechanical shock loads. Journal of Thermal Stresses. 2016:1-22.

8. Carrera E, Entezari A, Filippi M, Kouchakzadeh MA. 3D thermoelastic analysis of rotating disks having arbitraryprofile based on a variable kinematic 1D finite element method. Journal of Thermal Stresses. 2016:1-16.

9. Kouchakzadeh MA, Entezari A. Analytical Solution of Classic Coupled Thermoelasticity Problem in a RotatingDisk. Journal of Thermal Stresses. 2015;38:1269-91.

Publications in international Journals

Acknowledgements

Professor M. A. Kouchakzadeh and Erasmo Carrera, my supervisors

Dr. Matteo Filippi, my advisor and colleague in Italy.

Professors Hassan Haddadpour and Ali Hosseini Kordkheili, my Iranian committee members

Professors Maria Cinefra and Elvio Bonisoli, my Italian committee members

Doctoral Dissertation on

Solution of Coupled Thermoelasticity Problem In Rotating Disks

byAyoob Entezari

26 September, 2017

Supervisors:Prof. M. A. Kouchakzadeh¹ and Prof. Erasmo Carrera²

¹Sharif University of TechnologyDepartment of Aerospace Engineering, Tehran, Iran

²Polytechnic University of Turin, Department of Mechanical and Aerospace Engineering, Italy

²MUL2 research group, Polytechnic University of Turin, Italy

Cotutelle Doctoral Program

Thank you for your attention!

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