Solution of Coupled Thermoelasticity Problem In Rotating Disks by - PowerPoint PPT Presentation

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 � � �� � + 1 �� − �� � � � � �� 1 + � � � � �� I (� ���� � �,� ) ,� −(� �� �) ,� +� � = ��� � Eq. of motion � � �� � + � � � � ���� � + 1 ���� + 1 � �� � � � −� �� � = 0 � � ��� � �� + ���� − (� �� �, � ) ,� � � � � − � � � �� � + 1 �� − 1 � � �� � + 2�) �� = −��� � (� �� � � − � +� � � � � �� �� �,� + � � � �� �� �,� = � + � � �� II energy Eq. � 2� 2� � & � → Lame constants • � = � = � � + 2� � � � + 2� 3� + 2� � � → coefficient of linear thermal expansion • 25

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

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

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

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

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

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

Analytical approach - Solution method Solution of non-dimensional equations � � � �̂ � − � � ��̂ � + 1 ��̂ − 1 � � − �� � � ��̂ � � ��̂ = −�̂� Eq. of motion �̂ decomposition � � � � � � � � �� � + 1 �� � �� − � � �� � + 1 �� � �� − � � ,� − �� � � � − �� � = � � � − �� � = 0 � � �� � �� � � �� �� | ��� + � �� � � (�, �) = 0 � �� �� | ��� + � �� � � (�, �) = � � (�) �� � �� � � �� ��̂ | ��� + � �� � � (�, �) = 0 � �� �� | ��� + � �� � � (�, �) = � � (�) � � (�, 0) = � � (�) , �� � (�, 0) = � � (�) � � (�, 0) = 0 , �� � (�, 0) = 0 Bessel equation and can be separately solved using finite Hankel transform 32

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

Analytical approach - Solution method Solution of non-dimensional equations Uncoupled sub-IBVPs (Bessel equations) 2 ¶ ¶ u u u 2 1 ¶ ¶ T T 1   1 + 1 - 1 -  = u 0 1 1 + - T - t T = 0 1 1 0 1 2 r ¶ r 2 ¶ 2 ¶ r r r r ¶ r ¶ ¶ u T 1 1 k + k u ( , ) a t = f ( ) t + = k k T ( , ) a t f ( ) t 31 32 1 3 11 12 1 1 ¶ ¶ r r r = a r = a ¶ ¶ T u 1 + = 1 + = k k T ( , ) b t f ( ) t k k u ( , ) b t f ( ) t 21 22 1 2 41 42 1 4 ¶ ¶ r r = = r b r b  = = T ( , 0) r 0 , T ( , 0) r 0 u ( , 0) r = 0 , u  ( , 0) r = 0 1 1 1 1 Taking the finite Hankel transform æ ö d 2 æ ö ÷ ç d 2 2 ÷  + h = - 4 ÷ ç u u f t ( ) f t ( ) ç   2 + + x = - 2 ÷ ÷ t T T T f t ( ) f t ( ) ç ç 1 n 1 4 3 ÷ ÷ p è d ø ç 0 1 1 m 1 2 1 ÷ p è d ø 3 1 Solving ODEs t h t x u 1 ( , ) T 1 ( , ) m n 34

Analytical approach - Solution method Solution of non-dimensional equations Uncoupled sub-IBVPs (Bessel equations) æ ö d 2 æ ö ÷ ç d 2 2 + h = - 4 ÷ ÷ u  u f t ( ) f t ( ) ç   ç 2 2 + + x = - ÷ ÷ t T T T f t ( ) f t ( ) ç ç ÷ 1 n 1 4 3 ÷ p è d ø ç ÷ 0 1 1 m 1 2 1 p è d ø 3 1 Solving ODEs t h t x u 1 ( , ) T 1 ( , ) m n Inverse finite Hankel transforms � � � � � � � (�, � � )� � (�, � � ) � � (�, �) = �� � � (�, � � )� � (�, � � � � � (�, �) = � � � � � ��� ��� 1 1 � � = � � � = ‖� � (�, � � )‖ � , � ‖� � (�, � � )‖ � 35

Analytical approach - Solution method Solution of non-dimensional equations � � � �̂ � − � � ��̂ � + 1 ��̂ − 1 � � − �� � � ��̂ � � ��̂ = −�̂� Eq. of motion �̂ decomposition � � � � � � � � �� � + 1 �� � �� − � � �� � + 1 �� � �� − � � ,� − �� � � � − �� � = � � � − �� � = 0 � � �� � �� � � �� �� | ��� + � �� � � (�, �) = 0 � �� �� | ��� + � �� � � (�, �) = � � (�) �� � �� � � �� ��̂ | ��� + � �� � � (�, �) = 0 � �� �� | ��� + � �� � � (�, �) = � � (�) � � (�, 0) = � � (�) , �� � (�, 0) = � � (�) � � (�, 0) = 0 , �� � (�, 0) = 0 � � � � � � (�, �) = � �� �� (�)� � (�, � � ) , � � (�, �) = � �� �� (�)� � (�, � � ) ��� ��� 36 ��� ���

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

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

Analytical approach - Numerical evaluation Specifications of numerical example geometry Boundary conditions � at �̂ = � → �− �� � = 1 ��̂ = � � �� (�) � = 2 � � = 0 material properties � = 0 at �̂ = � → � � � � �� = 0 � = 40.4 GPa � = 27 GPa � = 23 × 10 �� K �� � �� (�) = �0 �̂ � 0 � � = 2707 kg/m3 1 �̂ � 0 � = 204 W/m ⋅ K � = 903 J/kg ⋅ K 39

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

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

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

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

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 MUL2 research group, Polytechnic University,  ability to analyze multi-field problems and multi-layered structures Turin, Italy www.mul2.polito.it 46

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

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 classica l theories of thermoelasticity.  Considering mechanical damping effect . Energy equation Equation of motion � � � ,� �� � ,� ��,� � � � � � �� �,� � �� �,� � 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 � � (�, �, �) energy equation Equation of motion � ���(� � + � � )�� + ���� − 2�̃ � �� ,� − � �� �, � ,� � � ��,� + � � − ��� � − ��� � � � �� = 0 � � � � +� � � � � �� �� �,� + � � � �� �� �,� − � − � � ���� � �� = 0 51

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

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

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

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 → � � (�, �, � )= � � (�)� � (�, �) refined 1D 2-nodes element 3D 8-nodes element 55

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 linear B2 quadratic B2 B4 cubic 56

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) • biquadratic nine-point (L9) • quadratic six-point (L6) • bi-cubic sixteen-point (L16) • bilinear four-point (L4) 58

Numerical approach - Development of method FE equations in CUF form Substituting 1D FE-CUF weighting function � (�, �) � �� (�) �(�, �, �, �) = � � (�)� � � (�, �, � )= � � (�)� � (�, �) � (�, �) Θ �� (�) �(�, �, �, �) = � � (�)� into the weak forms of equation of motion and energy equation gives   t t t t lm s ls lm s ls lm s ls m d + d + d = M G K p • � ���� , � ���� 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   t t t t lm s ls lm s ls lm s ls m d + d + d = M G K p or ���� ���� ���� ���� ���� � �� = � �� � �� 0 �� �� + � �� � �� �� �� + � �� � �� � �� � �� � �� � � � �� ���� ���� ���� ���� ���� � �� � �� � �� � �� 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 assembly procedure of FNs for each element   t t t t lm s ls lm s ls lm s ls m d + d + d = M G K p 3 B4 for whole structure   D + D + D = M G K 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 Transfinite element technique   lm s ls t lm s ls t lm s ls t m t d + d + d = M G K p �̃ → the Laplace variable taking Laplace • ���� → FN of the equivalent stiffness matrix � �� • ∗ denotes Laplace transform of the terms. • * * lm s t 2 lm s t lm s t ls m t + + d = [ M s  G s  K ] p   t lm s K eq ���� & � �� ∗ for whole structure Assembling � �� * * D K = P eq solve numerical solution in Laplace domain Δ ∗ solution in time domain ( Δ (�) ) inversion 62

Numerical approach - Development of method Non-dimensional Equation for isotropic FGMs Non-dimensional parameters V x e ˆ i = = x ˆ ; t t ; m i l l m m V l + m ( 2 ) T e ˆ ˆ m m T = u ˆ = u t = t ; ; m i i 0 0 b T l T l d m m d m 1 1 1 ˆ n n ˆ = s ˆ = s = q q ; ; t t i i ij ij i i r b b c T V T T m d m e m d d m m l D ˆ ˆ m m = = X X ; R R i i b l + m T c T ( 2 ) d m m d m m velocity of elastic longitudinal wave unit length diffusivity = k r V = ( l + 2 m )/ r D / c = l D m / V e m m m m e m m m m m m 63

Numerical approach - Development of method Non-dimensional FNs for isotropic FGMs based on LS theory ˆ ˆ t slm =   2 ml + ml K s C F F  I  C F F  I Transfinite element equation r t s L t x s x L 11 22 , , ���� δ �� ∗ = � �� ∗ ˆ ˆ m , l , + +  C F F  I  C F F  I ml � �� y y t t 66 s L 44 , z s z , L or ˆ ˆ m , l ml , t = + K slm  C F F  I  C F F  I y y �� ∗ 12 66 t s x , L 23 t , x s L �� ∗ ���� ���� ���� ���� � � � �� � �� � �� � �� � � ˆ ˆ t = + K slm  C F F  I ml  C F F  I ml �� ∗ �� ∗ ���� ���� ���� ���� � �� � �� � �� � �� � � � � t t 13 44 , z s x , L 21 , x s z , L = ˆ �� ∗ ���� ���� ���� ���� t = - K slm  C F F  I ml � �� � �� � �� � �� �� ∗ � � � � b t 14 , x s L �� ∗ ���� ���� ���� ���� � �� ∗ � �� � �� � �� � �� � � ˆ ˆ t = + K s lm C s t (  2 s   ) C F F  I ml b t 41 0 s x , L ˆ * ˆ * t ò ò m n * = + p t F N d S X F N d V ˆ ˆ ml , t = + K slm C s t (  2 s   ) C F F  I 1 x t m x t m y 42 0 b t s L ( e ) ( e ) S V ˆ * * ˆ ˆ ò ˆ ò m t n * t = + K slm C s t (  2 s   ) C F F  I ml p = F N d S + X F N d V t t t 2 y m y m 43 0 b t s z , L ( e ) ( e ) S V ˆ ˆ ˆ t = + + K slm ( s t  2 s   ) C C F F  I ml * * ˆ t ò ˆ ò m n * = + p t F N d S X F N d V r t 44 0 c s L t t 3 z m z m ˆ ˆ m , l , + +  C F F  I ml  C F F  I ( e ) ( e ) S V y y k t , x s x , L k t s L * ˆ ˆ t ò ò m * * = + + p [( t s 1) R ] F N d V ( q n ˆ ) F N d S 4 0 t m i i t m ˆ +  C F F  I ml ( e ) ( e ) k t , z s z , L V S 64

Numerical approach - Development of method Non-dimensional FNs for isotropic FGMs based on LS theory = ò ˆ ˆ t slm =  2 ml + ml K s  C F F  I  C F F  I    ) (  ) d A r t s L t x s x L 11 22 , , ( e A ˆ ˆ m , l , + +  C F F  I  C F F  I ml y y t t 66 s L 44 , z s z , L ( ) m l , ml , m l , , = ò ml ˆ ˆ I I I I N N N N N N N N dy y y y y m , l ml , t = + K slm  C F F  I  C F F  I y y L L L L m l m l m l m l 12 66 t s x , L 23 t , x s L ( ) e L , y , y , y , y ˆ ˆ t = + K slm  C F F  I ml  C F F  I ml t t 13 44 , z s x , L 21 , x s z , L r b k c ˆ ˆ ˆ ˆ ˆ = = = = C , C , C , C t = - K slm  C F F  I ml r b k c b t 14 , x s L r b k c m m m m ( ) m + l 2 ˆ ˆ t = + K s lm C ( s t  2 s  )  C F F  I ml ˆ ˆ ˆ C = C = C = b t 41 0 s x , L 11 22 33 l + m ( 2 ) m m ˆ ˆ ml , t = + K slm C ( s  2 t s  )  C F F  I y m ˆ ˆ ˆ 42 0 b t s L = = = C C C 44 55 66 l + m ( 2 ) ˆ ˆ t = + K slm C ( s  2 t s  )  C F F  I ml m m 43 0 b t s z , L l ˆ ˆ ˆ = = = C C C ˆ ˆ ˆ t = + + 12 13 23 K slm ( s  2 t s  )  C C F F  I ml l + m ( 2 ) r t 44 0 c s L m m ˆ ˆ m , l , +  C F F  I ml +  C F F  I y y k t , x s x , L k t s L 2 b T 0 m = C ˆ thermoelastic coupling parameter → +  C F F  I ml r l + m c ( ) k t , z s z , L 65 m m m m

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

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 6. Conclusion

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 6. Conclusion

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 7860 kg/m � density ( � ) • � = 2000 rad/s annular disk with hyperbolic profile • hub is assumed to be fully fixed ℎ �� = 0.06 m � �� = 0.05 m � � = 0.2 m ℎ � = 0.03 m 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 Discretizing Model DOF Along the axis Over the corss sections (2/6/8) × 32 L4 (1) 8 B2, 3 CS * 6240 (2/4/6/8) × 32 L4 (2) 8 B2, 4 CS 5472 (2/4/6/8) × 32 L4 (3) 10 B2, 4 CS 7200 (1/2/4/6/8) × 32 L4 (4) 12 B2, 5 CS 7584 (1/2/3/4/6/8) × 32 L4 (5) 14 B2, 6 CS 8352 (1/2/3/4/5/6/8) × 32 L4 (6) 16 B2, 7 CS 9504 (1/2/3/4/5/6/7/8) × 32 L4 (7) 18 B2, 8 CS 11040 (1/2/3/4/5/6/7/8) × 32 L4 (8) 22 B2, 8 CS 14496 * 3 types of cross section (CS) with different radii discretization along the axis 70

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

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

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 8 B2, (1/2/3/4) × 32 L4 1 3168 8 B2, (2/4/6/8) × 32 L4 2 5472 8 B2, (5/7/9/14) × 32 L4 3 8928 8 B2, (4/8/12/16) × 32 L4 4 10080 8 B2, (10/12/14/20) × 32 L4 5 13536 73

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

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 6. Conclusion

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. 600 590 Temprature change (  C) 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  rr (MPa)   (MPa) u r (mm) 0.5 1 1.5 2 50 150 250 150 350 550 Max=1.94 Min=0 Min=0 Max=238 Max=541 Min=0.55 radial circumferential stress stress radius radius 77

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 6. Conclusion

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 0.2 0.2 28 B2 32 B2 40 B2 0.1 0.1 0.1 z (m) z (m) z (m) 0 0 0 -0.1 -0.1 -0.1 28 Beam Elements 32 Beam Elements 40 Beam Elements -0.2 -0.2 -0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y (m) y (m) y (m) Lagrange mesh over the cross-section with the largest radius uniform 12 × 32 L4 refined 17 × 32 L4 80

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 BeamElements -0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y (m) refined 17 × 32 L4 32 B2 along the axis computational model, DOF=27072 81

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

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

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 6. Conclusion

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 � � 1 L4 over the cross-section � 86

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

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

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 6. Conclusion

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 � � = 0 40.4 GPa Lame’constant λ 27 GPa Lame’ constant μ � = 0 �̂ = � → � � 23 × 10 �� K �� coe ffi cient of linear thermal expansion ( α ) � � �� = 0 2707 kg/m � density ( ρ ) 204 W/m ∙ K geometry thermal conductivity ( κ ) 903 J/kg ∙ K specific heat ( c ) � = 1 where � �� (�) = �0 �̂ � 0 � = 2 � 1 �̂ � 0 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 Different 1D FE-CUF models for the constant thickness disk discretizing along the axis Discretizing Model DOF Along the axis corss sections (1) 1 B2 1680 (6 × 30) L4 (2) 1 B3 2520 (3) 1 B4 3360 3 × 15 L9 (4) 1680 2 × 10 L16 (5) 1 B2 6 × 18 L9 3744 (6) 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 Temperature change Radial displacement of thermoelasticity 0.5 1   t 0 =0.64, C =0.02, r =1.5   0.9 0.45 t 0 =0.64, C =0.02, r =1.5 Nondimensional Radial Displacement ( u ) 0.8 Exact Solution (Entezari, 2016)  0.4 Axisymetric FE (Bagri, 2004)  Nondimensional Temperature ( T ) 0.7 CUF: 1 B2/(6  18) L9 0.35 0.6 0.3 0.5 0.25 0.4 0.3 0.2 Exact Solution (Entezari, 2016) Axisymetric FE (Bagri, 2004) 0.2 0.15 CUF: 1 B2/(6  18) L9 0.1 0.1 0 0 2 4 6 8 10 12 14 0.05 -0.1 0 -0.2 0 2 4 6 8 10 12 14   1B2 / 6 × 18 L9 Nondimensional Time ( t ) 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 Radial stress Circumferential stress of thermoelasticity 0.8 1   t 0 =0.64, C =0.02, r =1.5   t 0 =0.64, C =0.02, r =1.5 0.8 0.6 Exact Solution (Entezari, 2016) Nondimensional Tangential Stress (   ) Exact Solution (Entezari, 2016) CUF: 1 B2/(6  18) L9  Nondimensional Radial Stress (  rr ) CUF: 1 B2/(6  18) L9 0.6  0.4 0.4 0.2 0.2 0 0 0 2 4 6 8 10 12 14 0 2 4 6 8 10 12 14 -0.2 -0.2 -0.4 -0.4 -0.6 -0.8 -0.6   Nondimensional Time ( t ) Nondimensional Time ( t ) 1B2 / 6 × 18 L9 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 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 6. Conclusion

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 23.0  10  6 K  1 7.4  10  6 K  1 coe ffi cient of linear thermal expansion ( α ) 204 W/m  K 28.0 W/m  K thermal conductivity ( κ ) 903 J/kg  K 760 J/kg  K specific heat ( c ) dimensionless relaxation time ( �̂ � ) 0.64 1.5625 metal volume fraction geometry effective properties � � = 1 � − �̂ P=VmPm +Vc Pc =Vm (Pm − Pc)+Pc Vm = � = 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 � �� � ⁄ � m ]� ������ � m � m �(�) = � � (1 − [1 + 100 � m ) � = �� = � = 0 at � = 0 → � = � Non-dimensional form T ( t ) Free � �(�̂) = 1 − (1 + 100�̂)� ����� � Adiabatic z z Fixed x y 100