Solution of Coupled Thermoelasticity Problem In Rotating Disks

Cotutelle Doctoral Program Doctoral Dissertation on Solution of Coupled Thermoelasticity Problem In Rotating Disks by Ayoob Entezari Supervisors: Prof. M. A. Kouchakzadeh¹ and Prof. Erasmo Carrera² Advisor: Dr. Matteo Filippi² ¹Sharif University of Technology Department of Aerospace Engineering, Tehran, Iran ²MUL2 research group, Polytechnic University of Turin, Italy 26 September, 2017 ²Polytechnic University of Turin, Department of Mechanical and Aerospace Engineering, Italy

Outlines 1. Introduction to rotating disks 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach 6. Conclusion

Introduction to rotating disks Applications 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) 4

Introduction to rotating disks Configurations 5

Introduction to rotating disks Operating conditions Main Loads Centrifugal forces Thermal loads. start and stop cycles I) start up, II) shut down 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. 6

Introduction to rotating disks Disk materials Metals: steels, super alloys Ceramic matrix composites (CMC) Functionally graded materials (FGMs) ceramic-metal FGM Effective properties of FGMs P eff = VmPm +Vc Pc =Vm (Pm Pc)+Pc ceramic-metal FGM Pm and Pc : properties of metal and ceramic Vm and Vc : volume fractions of metal and ceramic Vm =(,,) 7

Introduction to rotating disks FGM disk power gradation law for metal volume fraction along the radius Vm = metal volume fraction Effective properties of FGMs P eff = VmPm +Vc Pc =Vm (Pm Pc)+Pc 8 radius

Outlines 1. Introduction to rotating disk 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach 6. Conclusion

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems Inertia effects static problems dynamic problems displacement and temperature fields interaction uncoupled problems coupled problems 10

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems static steady-state problems equation of motion (, ), ( ), + =0 energy equation (, ), = temperature change displacements elastic coefficients body forces thermoelastic moduli thermal conductivity internal heat source 11

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems static steady-state problems Under axisymmetric & plane stress assumptions equation of motion h h + h =0 energy equation () = + ln( ( ) ) 12

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems Quasi-static problems equation of motion (, ), ( ), + =0 energy equation,, = density specific heat 13

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems Dynamic uncoupled problems equation of motion (, ), ( ), + = energy equation,, = 14

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems Dynamic uncoupled problems Considering mechanical damping equation of motion (, ), ( ), + = + energy equation,, = mechanical damping coefficient of material 15

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems 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. equation of motion energy equation Mechanical and thermal BCs and ICs (,), (,) 16

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems Classical coupled problems equation of motion (, ), ( ), + =, energy equation, +, = reference temperature 17

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems Classical coupled problems equation of motion (, ), ( ), + =, energy equation, +, = reference temperature infinite propagation speed for the thermal disturbances!!! 18

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems in the classical thermoelasticity heat conduction equation is of a parabolic type. 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. 19

Fundamentals of Linear Thermoelasticity Classification of thermoelastic problems without energy dissipation (, ), ( ), + = (, ), ( ), ( ), + = (, ), ( ), + = + (, ), +, +, =+ + 2, (, ), +, =,, +, = 20 LS relaxation time, GL relaxation times GL material constants GN material constants

Outlines 1. Introduction to rotating disk 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach 6. Conclusion

Literature review & present work Conclusion of the literature review 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. 22

Literature review & present work Present work Main purpose 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 Implementation Analytical approach Numerical approach 23

Outlines 1. Introduction to rotating disk 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach Solution method Numerical evaluation 5. Numerical approach 6. Conclusion 24

Analytical approach - Solution method Governing equations Consider An annular rotating disk with constant thickness, made of isotropic & homogeneous material, Under axisymmetric thermal and mechanical shock loads. Based on LS generalized coupled theory Eq. of motion (, ), ( ), + = + 1 + 1 1+ + + 1 =0 I energy Eq. + (, ), +, +, =+ ( +2) + 1 1 = II 25 = 2 +2 = 2 +2 3 + 2 & Lame constants coefficient of linear thermal expansion

Analytical approach - Solution method Governing equations Coupled System Of Equations I + 1+ + + + =0 II ( +2) + 1 1 = thermal BCs. & ICs Mechanical BCs. & ICs + (,)= () + (,)= () +, = + (,)= () - Inner radius of the disk Outer radius of the disk time dependent known functions 26, 0 =, (, 0) = (), 0 =, (, 0) = () constant parameters known functions of

Analytical approach - Solution method Governing equations in Non-dimensional form Non-dimensional parameters =, =, = =, = +2), =, = = propagation speed of elastic longitudinal wave = +2) unit length = 27

Analytical approach - Solution method Governing equations in Non-dimensional form Coupled System Of Equations I + 1 1 = II + 1 1+ + 1 + + 1 =0 where = ( +2 Thermoelastic damping or coupling parameter Non-dimensional propagation speed of thermal wave = 1 28 Non-dimensional propagation speed of elastic longitudinal wave =1

Analytical approach - Solution method Solution of non-dimensional equations Coupled System Of Equations I + 1 1 = II + 1 1+ + 1 + + 1 =0 Thermal and mechanical BCs. & ICs 29 + (, ) = ( ) + (, ) = ( ) + (,) = ( ) + (, ) = ( ) (,0)= ( ), (,0)= ( ) (,0)= ( ), (,0)= ( )

Analytical approach - Solution method Solution of non-dimensional equations + 1 1+ + 1 + + 1 =0 energy Eq. decomposition + 1 =0 + 1 =, + +, + + (, ) = () + (, ) = () (, 0) = 0, (, 0) = 0 + (, ) = 0 + (, ) = 0 (, 0) = (), (, 0) = () 30 principle of superposition (, ) = (, ) + (, )

Analytical approach - Solution method Solution of non-dimensional equations + 1 1 decomposition = Eq. of motion + 1 =0 + (, ) = () + (, ) = () (, 0) = 0, (, 0) = 0 + 1 =, + (, ) = 0 + (, ) = 0 (, 0) = (), (, 0) = () 31 principle of superposition (, ) = (, ) + (, )

Analytical approach - Solution method Solution of non-dimensional equations Finite Hankel transform kernel functions (, )= ( ) ( ) (, )= ( ) ( ) H[ (,)]= (, )= (, ) and are positive roots of the following equations (, ) H[ (, )] = (, )= (, ) (, ) + ( ( ) ( ) + ( ( ) ( ) + ( + ( 33 ( ) ( ) + ( =0 ( ) + ( ( ) + ( ( ) + ( ( ) ( ) + ( + ( ( ) + ( + ( = 0

Analytical approach - Solution method Solution of non-dimensional equations Uncoupled sub-ibvps (Bessel equations) 2 u1 1 u1 u1 + - - u 2 2 1 = r r r r 0 2 T1 1 T1 + - T 2 1 - t0t 1 = r r r 0 u k + k u ( a, t) = f ( t) 1 31 32 1 3 r r = a u k + k u (, b t) = f () t 1 41 42 1 4 r r = b u ( r, 0) = 0, u ( r, 0) = 0 1 1 T k + k T ( a, t) = f ( t) 1 11 12 1 1 r r = a T k + k T ( b, t) = f ( t) 1 21 22 1 2 r r = b T ( r, 0) = 0, T ( r, 0) = 0 1 1 Taking the finite Hankel transform 2 æ d ö u u f () t f () t ç ø 2 4 1 + hn 1 = 4-3 p è d3 Solving ODEs 2 æ d ö tt T T f() t f() t ç ø 2 2 0 1+ 1+ xm 1= 2-1 p è d1 34 u1 (, t hn ) T1 (, t x ) m

Analytical approach - Solution method Solution of non-dimensional equations Uncoupled sub-ibvps (Bessel equations) 2 æ d ö u u f () t f () t ç ø 2 4 1 + hn 1 = 4-3 p è d3 Solving ODEs 2 æ d ö tt T T f() t f() t ç ø 2 2 0 1+ 1+ xm 1= 2-1 p è d1 u1 (, t hn ) T1 (, t x ) m Inverse finite Hankel transforms (, ) = (, ) (, (, ) = (, ) (, ) 35 = 1 (, ), = 1 (, )

Analytical approach - Solution method Solution of non-dimensional equations Coupled System Of Equations I + 1 1 = II + 1 1+ + 1 + + 1 =0 37 (, ) = (, ) (, ) + () (, ) (, ) = (, ) (, + () (, )

Analytical approach - Numerical evaluation Specifications of numerical example geometry Boundary conditions =1 =2 material properties =40.4 GPa =27 GPa =23 10 K = 2707 kg/m 3 = 204 W/m K = 903 J/kg K at = = () =0 at = =0 =0 () = 0 0 1 0 39

Analytical approach - Numerical evaluation Validation Based on classical theory of coupled thermoelasticity Temperature Radial displacement Nondimensional Temperature (T) 0.48 0.4 0.32 0.24 0.16 0.08 C =0.02, =0, r =1.5 Numerical Solution (Bagri & Eslami, 2004) Exact Solution Nondimensional Radial Displacement (u) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 C =0.02, =0, r =1.5 Exact Solution NumericalSolution(Bagri&Eslami,2004) mid-radius 0 2 4 6 8 10 12 14 Nondimensional Time (t) 0 2 4 6 8 10 12 14 Nondimensional Time (t) 40 Time history of the non-dimensional solution at mid-radius

Analytical approach - Numerical evaluation Validation Based on LS generalized theory of coupled thermoelasticity Temperature Radial displacement 0.5 1 Nondimensional Temperature (T) 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 t 0 =0.64, C =0.02, =0, r =1.5 Exact Solution Numerical (Bagri & Eslami, 2004) Nondimensional Radial Displacement (u) 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 t 0 =0.64, C =0.02, =0, r =1.5 Exact Solution Numerical(Bagri&Eslami,2004) 0 0 2 4 6 8 10 12 14-0.1 mid-radius 0 0 2 4 6 8 10 12 14 Nondimensional Time (t) -0.2 Nondimensional Time (t) 41 Time history of the non-dimensional solution at mid-radius

Analytical approach - Numerical evaluation Results and discussion Based on LS generalized theory of coupled thermoelasticity temperature change radial displacement radial stress circumferential stress Nondimensional Temperature (T) 0.8 0.7 0.6 0.5 0.4 0.75 0.3 0.5 0.2 t=0.25 1.25 1 T 0 =293 K, t 0 =0.64, =0.01 steady state 0.1 0 1 1.2 1.4 1.6 1.8 2-0.1 Nondimensional Radius(r) Nondimensional Radial Displacement (u) 0.7 0.6 0.5 0.4 0.3 0.2 T 0 =293 K, t 0 =0.64, =0.01 + + + + + + + + + steady state t =0.25 + + + + -0.1 Nondimensional Radius (r) + + + + + + + 1.25 0.1 1 0.75 t =0.5 0 + 1 1.2 1.4 1.6 1.8 2 10+ Nondimensional Radial Stress ( rr ) 0.1 0-0.1-0.2-0.3-0.4-0.5-0.6-0.7 1.2 1.4 1.6 1.8 2 steady state t =0.25 t =0.5 t =0.75 t =1 t =1.25 Nondimensional Radius (r) T 0 =293 K, t 0 =0.64, =0.01 Nondimensional Tangential Stress ( ) 0.2 0.1 T 0 =293 K, t 0 =0.64, =0.01 0 1 1.2 1.4 1.6 1.8 2-0.1-0.2-0.3 steady state -0.4 t =0.25 t =0.5-0.5 t =0.75 t =1-0.6 t =1.25-0.7 Nondimensional Radius (r) radius radius radius radius Radial distribution for different values of the time. 42

Outlines 1. Introduction to rotating disk 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach 6. Conclusion

Numerical approach Motivations 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. 45

Numerical approach Main characteristics of FE models refined by Carrera 3D capabilities lower computational costs ability to analyze multi-field problems and multi-layered structures MUL2 research group, Polytechnic University, Turin, Italy www.mul2.polito.it 46

Numerical approach - Development of method Approaches to FE modeling 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 48

Numerical approach - Development of method Governing equations For anisotropic and nonhomogeneous materials. Including LS, GL and classical theories of thermoelasticity. Considering mechanical damping effect. Equation of motion Energy equation,,,,, Hooke s law ) = = = =0 classical theory = = =0 LS theory =0 GL theory. 49

Numerical approach - Development of method FE formulation through Galerkin technique In 3D conventional FE method (,,,)= (,,) () (,,,)= (,,) () =1,, = number of nodal points in a element 50

Numerical approach - Development of method FE formulation through Galerkin technique Weighting function (,,) Equation of motion energy equation, + = 0 ( + ) + 2,, +, +, = 0, 51

Numerical approach - Development of method FE formulation through Galerkin technique ( ) + ( ) + ( T σ) Eq. of motion = ( ) + ( ) ( β T ) + ) + ) energy Eq. + ( β T ) + ) (2 T ) + ( T κ ) = ( T ) + ( ) + ) 52

Numerical approach - Development of method Refined 1D FE model through Carrera unified formulation 3D beam-type structures 1D FE = () = () 53 =1,, = number of bar nodes

Numerical approach - Development of method Refined 1D FE model through Carrera unified formulation 3D beam-type structures 1D FE Carrera unified formulation (CUF) = () = () (, ) = (, ) () (, ) = (, )Θ () 54 =1,, = number of bar nodes =1,, = number of terms of the expansion.

Numerical approach - Development of method Refined 1D FE model through CUF 1D FE CUF 1D FE-CUF = () = () (, ) = (, ) () (, ) = (, )Θ () (,,, ) = (,,) () (,,, ) = (,, ) Θ () weighting function in 1D FE-CUF (,, )= () (, ) 55 refined 1D 2-nodes element 3D 8-nodes element

Numerical approach - Development of method Refined 1D FE model through CUF 1D FE-CUF (,,, ) = () (, ) () (,,, ) = () (, ) Θ () 1D FE modeling elements and shape functions in 1D FE modeling element B2 linear B2 quadratic 56 B4 cubic

Numerical approach - Development of method Refined 1D FE model through CUF 1D FE-CUF (,,, ) = () (, ) () (,,, ) = () (, ) Θ () In Carrera unified formulation 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. 57

Numerical approach - Development of method Refined 1D FE model through CUF 1D FE-CUF (,,, ) = () (, ) () (,,, ) = () (, ) Θ () (, ) bi-dimensional Lagrange functions cross-sections can be discretized using Lagrange elements linear three-point (L3) quadratic six-point (L6) bilinear four-point (L4) biquadratic nine-point (L9) bi-cubic sixteen-point (L16) 58

Numerical approach - Development of method FE equations in CUF form Substituting 1D FE-CUF (,,, ) = () (, ) () (,,, ) = () (, ) Θ () weighting function (,, )= () (, ) into the weak forms of equation of motion and energy equation gives M d + G d + K d = p lmts ls lmts ls lmts ls mt, 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 59

Numerical approach - Development of method FE equations in CUF form 0 M d + G d + K d = p lmts ls lmts ls lmts ls mt + or + 0 = Different theories of thermoelasticity through the 1D FE-CUF structural damping effect Rayleigh damping model = + 60

Numerical approach - Development of method Assembly procedure via Fundamental Nuclei for each element M d + G d + K d = p lmts ls lmts ls lmts ls mt assembly procedure of FNs 3 B4 for whole structure MD + GD + KD = P total degrees of freedom DOF = 4 a model with 3 B4 / 2 L4, DOF=240 2 L4 61

Numerical approach - Development of method Time history analysis M d + G d + K d = p lmts ls lmts ls lmts ls mt Transfinite element technique taking Laplace lmts 2 lmts lmts ls mt [ M s + G s + K ] d = p K lmts eq * * the Laplace variable FN of the equivalent stiffness matrix denotes Laplace transform of the terms. Assembling & for whole structure K eq D * * = P solve 62 solution in Laplace domain Δ numerical inversion solution in time domain (Δ())

Numerical approach - Development of method Non-dimensional Equation for isotropic FGMs Non-dimensional parameters m ˆ i x = ; t = t ; i x l m ˆ m T u u t t m e m ˆ T ( l + 2 m ) V = ; ˆi = ; = T l T l d V l m m e ˆ i 0 0 mbm d m 1 1 ˆ 1 qˆ q t t n n i = i ; sˆ ij = sij ; i = i cmtdrmve bmtd Tdbm ˆ l ˆ D i = R = R T ( 2 ) m m i ; dbm cmtd lm + mm velocity of elastic longitudinal wave unit length diffusivity 63 V = ( l + 2 m )/ r e m m m m lm D m / Ve m = k / c r = m m m m D

Numerical approach - Development of method Non-dimensional FNs for isotropic FGMs based on LS theory 64 Transfinite element equation δ = or * * p ˆt F N ds ˆ F N dv ( e ) ( e ) = m t n * 1 = x t m + x t m ( e ) ( e ) S V * * m t n ˆ * 2 = ò ˆt y t m + ò y t m ( e ) ( e ) S V * * m t n ˆ * 3 = ò ˆt z t m + ò z t m ( e ) ( e ) S V * m t ˆ ˆ * * 4 = [( 0 + 1) ] t m + ( ˆi i ) t m V ò p F N ds F N dv p F N ds F N dv ò p t s R F N dv q n F N ds ò ò S K = s Cˆ F F I + Cˆ F F I tslm 2 ml ml 11 r t s L 22 t, x s, x L + Cˆ F F I + Cˆ F F I m, yl, y ml 66 t s L 44 t, z s, z L K Cˆ F F I Cˆ F F I tslm m, yl ml, = + 12 66 t sx, L 23 t, x s L K = Cˆ F F I + Cˆ F F I tslm ml ml 13 44 t, z s, x L 21 t, x s, z L K =-Cˆ F F I K tslm ml 14 b t, x s L tslm 41 = C s tˆ + s Cˆ F F I tslm 2 42 0 2 ( ) 0 b t sx, K = C( s tˆ + s) Cˆ F F I K = C( s tˆ + s) Cˆ F F I tslm 2 43 0 b t sz, tslm 2 44 0 ml, L ( ˆ ) ˆ ˆ ml K = s t + s C C F F I + + Cˆ F F I + Cˆ F F I ml m, yl, k t, x s, x L k t s L + Cˆ F F I k r ml t, z s, z L b t s c t s L ml L y ml L y y

Numerical approach - Development of method Non-dimensional FNs for isotropic FGMs based on LS theory = ò =ò A ( ) da ( e ) ( ) ml m, yl ml, y m, yl, y L L L L L m l m l m l m l I I I I N N N N N N N N dy ( e), y, y, y, y ˆ r ˆ b,, ˆ k C = C = C =, Cˆ c = r b k rm bm km cm c K = s Cˆ F F I + Cˆ F F I tslm 2 ml ml 11 r t s L 22 t, x s, x L + Cˆ F F I + Cˆ F F I m, yl, y ml 66 t s L 44 t, z s, z L K Cˆ F F I Cˆ F F I tslm m, yl ml, = + 12 66 t sx, L 23 t, x s L K = Cˆ F F I + Cˆ F F I tslm ml ml 13 44 t, z s, x L 21 t, x s, z L K =-Cˆ F F I tslm ml 14 b t, x s L y 65 ( 2m+ l) Cˆ 11 = Cˆ 22 = Cˆ 33 = ( lm + 2 mm) ˆ ˆ ˆ m C44 = C55 = C66 = ( lm + 2 mm) Cˆ ˆ ˆ l 12 = C13 = C23 = ( l + 2 m ) thermoelastic coupling parameter C = m m c T 2 0bm r ( l + m ) m m m m K tslm 41 = C s tˆ + Cˆ F F I tslm 2 42 0 tslm 2 43 0 b t sz, tslm 2 44 0 2 ( s ) 0 b t sx, K = C ( s tˆ + s) Cˆ F F I K = C ( s tˆ + s) Cˆ F F I ml, L ( ˆ ) ˆ ˆ ml K = s t + s C C F F I + + Cˆ F F I + Cˆ F F I ml m, yl, k t, x s, x L k t s L + Cˆ F F I k r ml t, z s, z L b t s c t s L ml L y ml L y

Outlines 1. Introduction to rotating disk 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach Motivation Development of method Evaluations and results 6. Conclusion o Static structural analysis Example 1. Rotating variable thickness disk Example 2. Rotating variable thickness disk subjected thermal load Example 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 analysis Example 6. Constant thickness disk made of isotropic homogeneous materials Example 7. Constant thickness disk made of isotropic FGMs Example 8. variable thickness disk made of isotropic FGMs

Numerical approach - Evaluations and results Static structural analysis Example 1. Rotating variable thickness disk Material properties Young s modulus 207 GPa Poisson s ratio 0.28 density () 7860 kg/m annular disk with hyperbolic profile =0.05 m h =0.06 m =0.2 m h =0.03 m = 2000 rad/s hub is assumed to be fully fixed h = 0.0134. 69

Numerical approach - Evaluations and results Static structural analysis Example 1. Rotating variable thickness disk 1D FE-CUF modeling Different 1D FE-CUF models of the disk Model Discretizing Along the axis Over the corss sections DOF (1) 8 B2, 3 CS * (2/6/8) 32 L4 6240 (2) 8 B2, 4 CS (2/4/6/8) 32 L4 5472 (3) 10 B2, 4 CS (2/4/6/8) 32 L4 7200 (4) 12 B2, 5 CS (1/2/4/6/8) 32 L4 7584 (5) 14 B2, 6 CS (1/2/3/4/6/8) 32 L4 8352 (6) 16 B2, 7 CS (1/2/3/4/5/6/8) 32 L4 9504 (7) 18 B2, 8 CS (1/2/3/4/5/6/7/8) 32 L4 11040 (8) 22 B2, 8 CS (1/2/3/4/5/6/7/8) 32 L4 14496 * 3 types of cross section (CS) with different radii 70 discretization along the axis

Numerical approach - Evaluations and results Static structural analysis Example 1. Rotating variable thickness disk verification of results Radial displacement 71 Model DOF Radial displacement (μm) At mid-radius At outer radius Analytical 1D CUF- FE 1 119.01 157.57 (1) 6240 120.32 (1.10) 156.00 (1.00) (2) 5472 118.36 (0.54) 157.00 (0.36) (3) 7200 118.36 (0.54) 156.42 (0.73) (4) 7584 118.75 (0.22) 157.15 (0.27) (5) 8352 119.50 (0.41) 158.08 (0.32) (6) 9504 118.50 (0.43) 157.93 (0.23) (7) 11040 117.26 (1.47) 154.92 (1.68) (8) 14496 117.06 (1.64) 155.00 (1.63) 3D ANSYS 14400 119.00 (0.01) 157.10 (0.30) ( ) : % difference with respect to the analytical solution. Radial Displacement, u r (m) 165 150 135 120 105 90 75 60 45 30 15 Radius (m) Analytical Solution model (1): DOF=6240 model (2): DOF=5472 model (3): DOF=7200 model (4): DOF=7584 model (5): DOF=8352 model (6): DOF=9504 model (7): DOF=11040 model (8): DOF=14496 ANSYS: DOF=14400 0 0.05 0.075 0.1 0.125 0.15 0.175 0.2

Numerical approach - Evaluations and results Static structural analysis Example 1. Rotating variable thickness disk verification of results Radial displacement 72 Model DOF Radial displacement (μm) At mid-radius At outer radius Analytical 1D CUF- FE 1 119.01 157.57 (1) 6240 120.32 (1.10) 156.00 (1.00) (2) 5472 118.36 (0.54) 157.00 (0.36) (3) 7200 118.36 (0.54) 156.42 (0.73) (4) 7584 118.75 (0.22) 157.15 (0.27) (5) 8352 119.50 (0.41) 158.08 (0.32) (6) 9504 118.50 (0.43) 157.93 (0.23) (7) 11040 117.26 (1.47) 154.92 (1.68) (8) 14496 117.06 (1.64) 155.00 (1.63) 3D ANSYS 14400 119.00 (0.01) 157.10 (0.30) ( ) : % difference with respect to the analytical solution. Error < 0.6% Model (2) 2.6 times less DOFs of the 3D ANSYS model!!

Numerical approach - Evaluations and results Static structural analysis Example 1. Rotating variable thickness disk 1D FE-CUF modeling Mesh refinement over the cross-sections model 1D FE-CUF Model DOF 1 8 B2, (1/2/3/4) 32 L4 3168 2 8 B2, (2/4/6/8) 32 L4 5472 3 8 B2, (5/7/9/14) 32 L4 8928 4 8 B2, (4/8/12/16) 32 L4 10080 5 8 B2, (10/12/14/20) 32 L4 13536 73

Numerical approach - Evaluations and results Static structural analysis Example 1. Rotating variable thickness disk verification of results Radial displacement 74 Radial Displacement, u r (m) 165 150 135 120 105 90 75 60 45 30 15 effect of enriching the radial discretization 0 0.05 0.075 0.1 0.125 0.15 0.175 0.2 Radius (m) Analytical Solution 8 B2, (1/2/3/4)32 L4, DOF=3168 8 B2, (2/4/6/8)32 L4, DOF=5472 8 B2, (5/7/9/14)32 L4, DOF=8928 8 B2, (4/8/12/16)32 L4, DOF=10080 8 B2, (10/12/14/20)32 L4, DOF=13536 ANSYS, DOF= 14400 Model DOF Radial displacement (μm) At mid-radius At outer radius Analytical 1D CUF FE 1 119.01 157.57 8 B2, (1/2/3/4) 32 L4 3168 114.50 (3.79) 154.00 (2.27) 8 B2, (2/4/6/8) 32 L4 5472 118.36 (0.54) 157.00 (0.36) 8 B2, (5/7/9/14) 32 L4 8928 119.00 (0.01) 157.00 (0.36) 8 B2, (4/8/12/16) 32 L4 10080 119.00 (0.01) 158.00 (0.27) 8 B2, (10/12/14/20) 32 L4 13536 119.00 (0.01) 157.00 (0.36) 3D FE (ANSYS) 14400 119.00 (0.01) 157.10 (0.30) ( ) Absolute percentage difference with respect to the analytical solution. Converged solution with 1.6 times less DOFs of the 3D ANSYS model!!

Numerical approach - Evaluations and results Static structural analysis Example 2. Rotating variable thickness disk subjected thermal load The disk is subjected to radial temperature gradient. hub is assumed to be axially fixed. Temprature change (C) 600 590 580 570 560 550 540 530 520 510 500 0.05 0.075 0.1 0.125 0.15 0.175 0.2 Radius (m) radial steady-state temperature distribution 76

Numerical approach - Evaluations and results Static structural analysis Example 2. Rotating variable thickness disk subjected thermal load radial displacement Radial and circumferential stresses u r (mm) rr (MPa) (MPa) 0.5 1 1.5 2 Max=1.94 Min=0 50 150 250 Min=0 150 350 550 Max=238 Min=0.55 Max=541 radius radial stress circumferential stress radius 77

Numerical approach - Evaluations and results Static structural analysis Example 3. Complex rotor The profile hyperbolic for the turbine disk web-type profile for the compressor disks Both ends of the shaft are fully fixed. 3D model of a complex rotor 79

Numerical approach - Evaluations and results Static structural analysis Example 3. Complex rotor 1D FE-CUF modeling discretizing along the axis 0.2 0.1 28 B2 0.2 0.1 32 B2 40 B2 0.2 0.1 z (m) 0 z (m) 0 z (m) 0-0.1 28 Beam Elements -0.1 32 Beam Elements -0.1 40 Beam Elements -0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y (m) -0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y (m) -0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y (m) Lagrange mesh over the cross-section with the largest radius 80 uniform 12 32 L4 refined 17 32 L4

Numerical approach - Evaluations and results Static structural analysis Example 3. Complex rotor 1D FE-CUF modeling Converged model 0.2 0.1 z (m) 0-0.1 32 Beam Elements -0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y (m) 81 32 B2 along the axis refined 17 32 L4 computational model, DOF=27072

Numerical approach - Evaluations and results Static structural analysis Example 3. Complex rotor verification of results Radial displacement 200 1D FE-CUF solution with 1.6 times less DOFs of the 3D ANSYS model!! u r (m) 0 25 50 75 100 125 150 175 200 1D FE-CUF solution DOFs= 27,072 3D FE ANSYS DOFs=44,280 Radial Displacement (m) 180 160 140 120 100 80 60 40 20 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 x x x x Turbine disk in the rotor (3D ANSYS result) Turbine disk in the rotor (1D CUF result) Comp. Disk 1 in the rotor Comp. Disk 2 in the rotor 82 0 0.05 0.075 0.1 0.125 0.15 0.175 0.2 Radius (m)

Numerical approach - Evaluations and results Static structural analysis Example 3. Complex rotor verification of results Radial and circumferential stresses 550 rr (MPa) 50 100 150 200 250 300 350 400 450 500 550 600 650 700 Radial stress 500 450 rr Turbine disk in the rotor Single Turbine disk with rigid hub 400 350 (MPa) 50 100 150 200 250 300 350 400 450 500 550 Stress (MPa) 300 250 200 rr Circumferential stress 150 100 50 83 0 0.05 0.075 0.1 0.125 0.15 0.175 0.2 Radius (m)

Numerical approach - Evaluations and results Static structural-thermal analysis Example 4. simple beam 1D FE-CUF modeling discretizing along the axis Lagrange elements over the cross-section 85

Numerical approach - Evaluations and results Static structural-thermal analysis Example 4. simple beam results Temperature change Axial displacement 86 1 L4 over the cross-section

Numerical approach - Evaluations and results Static structural-thermal analysis Example 4. simple beam results 1 L9 over the cross-section 87 1 L16 over the cross-section

Numerical approach - Evaluations and results Static structural-thermal analysis Example 4. simple beam verification of results Does heat conduction equation satisfy? Yes!! 88

Numerical approach - Evaluations and results Static structural-thermal analysis Example 4. simple beam verification of results Check free thermal expansion! Elongation = At =0.1 =(0.1 )(23.1 10 ).. = 0.219 mm At =0.5 =(0.5 )(23.1 10 ). = 0.6092 mm 89

Numerical approach - Evaluations and results Quasi-static structural-thermal analysis Example 5. simple beam Transient heat flux (thermal load) 91

Numerical approach - Evaluations and results Quasi-static structural-thermal analysis Example 5. simple beam results 10B4/1L4 model Axial displacement Temperature change Axial displacement Temperature change 92

Outlines 1. Introduction to rotating disk 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach Motivation Development of method Evaluations and results o Static structural analysis Example 1. Rotating variable thickness disk Example 2. Rotating variable thickness disk subjected thermal load Example 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 analysis Example 6. Constant thickness disk made of isotropic homogeneous materials Example 7. Constant thickness disk made of isotropic FGMs Example 8. variable thickness disk made of isotropic FGMs 93 6. Conclusion

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 6. Constant thickness disk made of isotropic homogeneous materials Boundary conditions Material properties Lame constant λ Lame constant μ coefficient of linear thermal expansion (α) density (ρ) thermal conductivity (κ) specific heat (c) 40.4 GPa 27 GPa 23 10 K 2707 kg/m 204 W/m K 903 J/kg K = = () =0 = =0 =0 where () = 0 0 1 0 geometry =1 =2 Thickness = 0.1 94

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 6. Constant thickness disk made of isotropic homogeneous materials 1D FE-CUF modeling discretizing along the axis Different 1D FE-CUF models for the constant thickness disk Model Discretizing Along the axis corss sections DOF (1) 1 B2 1680 (2) 1 B3 (6 30) L4 2520 (3) 1 B4 3360 (4) 3 15 L9 (5) 1 B2 2 10 L16 1680 (6) 6 18 L9 3744 Lagrange mesh over the cross-section with the largest radius 95

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 6. Constant thickness disk made of isotropic homogeneous materials verification of results Based on the LS theory of thermoelasticity 0.5 Temperature change 1 Radial displacement 0.45 t 0 =0.64, C =0.02, r =1.5 0.9 t 0 =0.64, C =0.02, r =1.5 Nondimensional Temperature (T) 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 Exact Solution (Entezari, 2016) Axisymetric FE (Bagri, 2004) CUF: 1 B2/(618) L9 Nondimensional Radial Displacement (u) 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1-0.1 Exact Solution (Entezari, 2016) Axisymetric FE (Bagri, 2004) CUF: 1 B2/(618) L9 0 0 2 4 6 8 10 12 14 1B2 / 6 18 L9 0 0 2 4 6 8 10 12 14 Nondimensional Time (t) -0.2 Nondimensional Time (t) 96 Time history of solution at mid-radius of the disk.

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 6. Constant thickness disk made of isotropic homogeneous materials verification of results Based on the LS theory of thermoelasticity 1 Radial stress 0.8 Circumferential stress Nondimensional Radial Stress ( rr ) 0.8 0.6 0.4 0.2-0.2-0.4 t 0 =0.64, C =0.02, r =1.5 0 0 2 4 6 8 10 12 14 Exact Solution (Entezari, 2016) CUF: 1 B2/(618) L9 Nondimensional Tangential Stress ( ) 0.6 0.4 0.2-0.2 t 0 =0.64, C =0.02, r =1.5 Exact Solution (Entezari, 2016) CUF: 1 B2/(618) L9 0 0 2 4 6 8 10 12 14-0.6-0.4 1B2 / 6 18 L9-0.8 Nondimensional Time (t) -0.6 Nondimensional Time (t) 97 Time history of solution at mid-radius of the disk.

Outlines 1. Introduction to rotating disk 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach Motivation Development of method Evaluations and results 6. Conclusion o Static structural analysis Example 1. Rotating variable thickness disk Example 2. Rotating variable thickness disk subjected thermal load Example 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 analysis Example 6. Constant thickness disk made of isotropic homogeneous materials Example 7. Constant thickness disk made of isotropic FGM Example 8. variable thickness disk made of isotropic FGMs

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM Geometry and material Material properties Metal-Ceramic FGM Metal: Aluminum Ceramic: Alumina Lame constant λ 40.4 GPa 219.2 GPa shear modulus μ 27.0 GPa 146.2 GPa density (ρ) 2707 kg/m 3 3800 kg/m 3 coefficient of linear thermal expansion (α) 23.010 6 K 1 7.410 6 K 1 thermal conductivity (κ) 204 W/mK 28.0 W/mK specific heat (c) 903 J/kgK 760 J/kgK dimensionless relaxation time ( ) 0.64 1.5625 effective properties P=VmPm +Vc Pc =Vm (Pm Pc)+Pc metal volume fraction Vm = geometry =1 =2 Thickness = 0.1 99

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM Operational, boundary & initial conditions = 293 K, =0.01 at =0 = = ==0 Free T(t) () = (1 [1 + 100 m m ] m Non-dimensional form ( ) = 1 (1 + 100 ) m ) Adiabatic Fixed z x z y 100

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM results Time history the material properties linearly change through the radius ( =1) Based on the LS theory of thermoelasticity Temperature change Radial displacement 101

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM results Speed range of the thermal wave the material properties linearly change through the radius ( =1) Based on the LS theory of thermoelasticity 103

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM results Thermal wave propagation the material properties linearly change through the radius ( =1) Based on the LS theory of thermoelasticity Non-dimensional form of energy equation m,, mm + + + =0, +, c = c m 1 c =0.27 wave reflection 1/1.5=0.6 105 m = 1 m =1.25 Speed range of the thermal wave 0.27,FGM 1.25 1 1.25 1 0.27 0.8 3.7

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM results Elastic wave propagation the material properties linearly change through the radius ( =1) Based on the LS theory of thermoelasticity Non-dimensional form of equation of motion wave reflection + m +2m +, + m +2m 1, + m +2m,, 1 m +2m,(, +, ) m 1 m, +, + =0 Speed range of elastic wave c =1.96 m =1 1 FGM 1.96 0.51 1 107

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM results effects of power law index () Nondimensional temperature change (T) 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Temperature change -0.2 0 2 4 6 8 10 12 14 Nondimensional time(t) at mid radius n =0 n =1 n =2 n =5 Nondimensional radial displacement (u r ) 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0-0.2 Radial displacement n =0 n =1 n =2 n =5-0.4 0 2 4 6 8 10 12 14 Nondimensional time(t) at mid radius metal volume fraction Vm = mid-radius Time history based on the LS theory at mid-radius of the disk 108

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM results effects of power law index () Nondimensional radial stress ( rr ) 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 Radial stress -1.6 0 2 4 6 8 10 12 14 Nondimensional time(t) n =0 n =1 n =2 n =5 at mid radius Circumferential stress Time history based on the LS theory at mid-radius of the disk Nondimensional circumferential stress ( ) 0.8 0.6 0.4 0.2 0-0.2-0.4-0.6-0.8-1 0 2 4 6 8 10 12 14 Nondimensional time(t) n =0 n =1 n =2 n =5 at mid radius metal volume fraction Vm = mid-radius 109

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM results effects of reference temperature ( ) Nondimensional temperature change (T) 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Temperature change -0.2 0 2 4 6 8 10 12 14 Nondimensional time(t) at mid radius T 0 =0 T 0 =293 T 0 =800 Nondimensional radial displacement (u r ) 0.6 0.5 0.4 0.3 0.2 0.1 0 Radial displacement 0 2 4 6 8 10 12 14 Nondimensional time(t) at mid radius T 0 =0 T 0 =293 T 0 =800 coupling parameter = m mm(m +m) Time history based on the LS theory at mid-radius of the disk (n =1) mid-radius 110

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 7. Constant thickness disk made of isotropic FGM results effects of reference temperature ( ) Nondimensional radial stress ( rr ) 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0-0.2-0.4-0.6-0.8 Radial stress T 0 =0 T 0 = 293 T 0 = 800 at mid radius Nondimensional circumferential stress ( ) 0.8 0.6 0.4 0.2 0-0.2-0.4-0.6-0.8 Circumferential stress T 0 =0 T 0 =293 T 0 =800 at mid radius coupling parameter = m mm(m +m) -1 0 2 4 6 8 10 12 14 Nondimensional time(t) -1 0 2 4 6 8 10 12 14 Nondimensional time(t) Time history based on the LS theory at mid-radius of the disk (n =1) mid-radius 111

Outlines 1. Introduction to rotating disk 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach Motivation Development of method Evaluations and results o Static structural analysis Example 1. Rotating variable thickness disk Example 2. Rotating variable thickness disk subjected thermal load Example 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 analysis Example 6. Constant thickness disk made of isotropic homogeneous materials Example 7. Constant thickness disk made of isotropic FGM Example 8. variable thickness disk made of isotropic FGM 112 6. Conclusion

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 8. variable thickness disk made of isotropic FGM Geometry and material 0.3 Material properties Metal-Ceramic FGM Metal: Aluminum Ceramic: Alumina Lame constant λ 40.4 GPa 219.2 GPa shear modulus μ 27.0 GPa 146.2 GPa density (ρ) 2707 kg/m 3 3800 kg/m 3 coefficient of linear thermal expansion (α) 23.010 6 K 1 7.410 6 K 1 thermal conductivity (κ) 204 W/mK 28.0 W/mK specific heat (c) 903 J/kgK 760 J/kgK dimensionless relaxation time ( ) 0.64 1.5625 2 0.5 0.6 h =0.42r -0.5 r 113 effective properties P=VmPm +Vc Pc =Vm (Pm Pc)+Pc metal volume fraction Vm = geometry ==0.5 = =2 h =0.6 h =0.3

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 8. variable thickness disk made of isotropic FGM = 293 K, =0.05 at =0 = = ==0 Operational, boundary & initial conditions () = (1 Non-dimensional form m) ( ) =1 114

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 8. variable thickness disk made of isotropic FGM Results for =0 1 Temperature change 1 Radial displacement Nondimensional temperature change 0.8 0.6 0.4 0.2 Classic theory LS theory 0 0 2 4 6 8 10 12 14 Nondimensional radial displacement 0.8 0.6 0.4 0.2 Classic theory LS theory 0 0 2 4 6 8 10 12 14-0.2-0.2 Nondimensional time Nondimensional time 115 Nondimensional radial stress 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 14-0.2-0.4-0.6-0.8-1 Radial stress Nondimensional time Classic theory LS theory Nondimensional circumferential stress 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 14-0.2-0.4-0.6-0.8-1 Circumferential stress Nondimensional time Classic theory LS theory Nondimensional von mises equivalent stress 1 0.8 0.6 0.4 0.2 Von mises stress Classic theory LS theory 0 0 2 4 6 8 10 12 14 Nondimensional time

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 8. variable thickness disk made of isotropic FGM Results for =0based LS theory 116

Numerical approach - Evaluations and results Dynamic coupled structural-thermal analysis Example 8. variable thickness disk made of isotropic FGM Results for =0based LS theory Axial displacement Radial displacement 117

Outlines 1. Introduction to rotating disks 2. Fundamentals of Linear Thermoelasticity 3. Literature review & present work 4. Analytical approach 5. Numerical approach 6. Conclusion 118

Conclusion - Summary of results Some results obtained from coupled thermoelasticity solution 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 alongthetimeinaharmonic 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. 119 A functionally graded disk may be used as thermal barrier to reduce the thermal shock effects.

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 where large bending loads are involved in the problem. Increasing 1D elements along the axis of disks may not have significant effect on accuracy of results and only leads to more DOFs. A proper distribution of the Lagrange elements and type of element used over the cross sections may lead 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 preserving the accuracy. increase of number of elements along the radial direction, compared to circumferential direction, is more effective in improving the results. 120

Conclusion - Future works It is of interests to extend the study to Nonlinear thermoelasticity problems Dynamic analysis of rotors subjected to transient thermal pre-stresses. Study of thermoelastic damping effect on dynamic behaviors of rotors. 121

Publications in international Journals 1. Entezari A, Filippi M, Carrera E., Kouchakzadeh M A, 3D Dynamic Coupled Thermoelastic Solution For Constant Thickness Disks Using Refined 1D Finite Element Models. European Journal of Mechanics - A/Solids. (Under review). 2. Entezari A, Filippi M, Carrera E. Unified finite element approach for generalized coupled thermoelastic analysis of 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 analysis of 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 subjected to 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 Thermoelasticity Problems 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 rotors and rotating disks with variable thickness. Acta Mechanica. 2016:1-20. 7. Entezari A, Kouchakzadeh MA. Analytical solution of generalized coupled thermoelasticity problem in a rotating disk 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 arbitrary profile based on a variable kinematic 1D finite element method. Journal of Thermal Stresses. 2016:1-16. 122 9. Kouchakzadeh MA, Entezari A. Analytical Solution of Classic Coupled Thermoelasticity Problem in a Rotating Disk. Journal of Thermal Stresses. 2015;38:1269-91.

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

Cotutelle Doctoral Program Doctoral Dissertation on Solution of Coupled Thermoelasticity Problem In Rotating Disks by Ayoob Entezari Supervisors: Prof. M. A. Kouchakzadeh¹ and Prof. Erasmo Carrera² Thank you for your attention! ¹Sharif University of Technology Department of Aerospace Engineering, Tehran, Iran ²MUL2 research group, Polytechnic University of Turin, Italy 26 September, 2017 ²Polytechnic University of Turin, Department of Mechanical and Aerospace Engineering, Italy