CH.6. LINEAR ELASTICITY Continuum Mechanics Course (MMC) - ETSECCPB - PowerPoint PPT Presentation

Elastic Potential ˆ du 1 d         : : :  C dt 2 dt  Consequences: 2. Integrating the time derivative of the internal energy density, 1             ˆ u x , t x , t : : x , t a x C 2 and assuming that the density of the internal energy vanishes at the   neutral reference state, :   ˆ u x , t 0 x 0  0 1 1 1                 ˆ       u : : ( ) : x , t : : x , t a x a ( ) x 0 x C C  0 0 2 2 2   Due to thermodynamic reasons the elastic energy is assumed always positive 1          0  ˆ u : : 0 C 2 19

Elastic Potential  The internal energy density defines a potential for the stress tensor, and is thus, named elastic potential . The stress tensor can be computed as    T      ˆ u    u ˆ( ( , )) x t 1 1 1 1                ( : : : : ( ) C C C C C       2 2 2 2    The constitutive elastic constants tensor can be obtained as the second derivative of the internal energy density with respect to the strain tensor field, 2 ˆ 2 ˆ                u u : ( ) C       C C ijkl         ij kl 20

6.3 Isotropic Linear Elasticity Ch.6. Linear Elasticity 21

Isotropic Constitutive Elastic Constants Tensor  An isotropic elastic material must have the same elastic properties (contained in ) in all directions. C  All the components of must be independent of the orientation of C the chosen (Cartesian) system must be a (mathematically) C isotropic tensor.       2 I C 1 1                  i j k l , , , 1,2,3 C   ijkl ij kl ik jl il jk Where: 1   is the 4 th order unit tensor defined as         I I    ik jl il jk ijkl 2   and are scalar constants known as Lamé parameters or coefficients.  REMARK The isotropy condition reduces the number of independent elastic constants from 21 to 2. 22

Isotropic Linear Elastic Constitutive Equation  Introducing the isotropic constitutive elastic constants tensor         : 2 I into the generalized Hooke’s Law , C C 1 1 (in indicial notation)                    C ij ijkl kl ij kl ik jl il jk kl    ij  1 1                     2 Tr ( ) 2   ij kl kl ik jl kl il jk kl ij ij  2 2      ( )     Tr ll ji ij 1 1       ij i j i j 2 2  The resulting constitutive equation is,   Isotropic linear elastic         Tr 2  1 constitutive equation.            2 i j , 1,2,3   Hooke’s Law ij ij ll ij 23

Elastic Potential  If the constitutive equation is,   Isotropic linear elastic        Tr 2 1 constitutive equation.           2 i j , 1,2,3 Hooke’s Law ij ij ll ij Then, the internal energy density can be reduced to: 1 1 REMARK                   ˆ u : Tr 2 : 1 The internal energy density is an 2 2       σ elastic potential of the stress tensor 1 1 as:             Tr : 2 1    2 2   ˆ u    Tr              Tr 2 1   1           2 Tr 2 24

Inversion of the Constitutive Equation  1. is isolated from the expression derived for Hooke’s Law 1              Tr 2       1 Tr 1  2  2. The trace of is obtained:                                   Tr Tr Tr 2 Tr Tr 2 Tr 3 2 Tr 1 1 =3  3. The trace of is easily isolated: 1        Tr Tr    3 2 4. The expression in 3. is introduced into the one obtained in 1.    1 1 1           1       Tr Tr   1            2 3 2 2 3 2 2   25

Inverse Isotropic Linear Elastic Constitutive Equation  E  The Lamé parameters in terms of and :        E 3 2    E        1 1 2     E      G          2 2 1  So the inverse const. eq. is re-written:       1       Tr Inverse isotropic linear 1   E E  elastic constitutive equation.    1            i j , 1,2,3 Inverse Hooke’s Law.   ij ll ij ij E E   1 1              x x y z xy xy E G 1 1                In engineering notation: y y x z xz xz E G   1 1              z z x y yz yz E G 26

Young’s Modulus and Poisson’s Ratio E  Young's modulus is a measure of the stiffness of an elastic material. It is given by the ratio of the uniaxial stress over the uniaxial strain.       3 2  E      Poisson's ratio is the ratio, when a solid is uniaxially stretched, of the transverse strain (perpendicular to the applied stress), to the axial strain (in the direction of the applied stress).         2 27

Example Consider an uniaxial traction test of an isotropic linear elastic material such that:   0 E,  x           0 y y z xy xz yz   x x   x x x z Obtain the strains (in engineering notation) and comment on the results obtained for a Poisson’s ratio of and .     0.5 0 28

  1 1              x x y z xy xy E G   0 Solution x 1 1                          0 y y x z xz xz E G y z xy xz yz   1 1              z z x y yz yz E G For :   0 1      0 1 There is no Poisson’s effect      0 x x xy E x x xy E and the transversal normal            0 0 0 strains are zero. y x xz y xz E     0 0  z yz       0 z x yz E For :   0.5 The volumetric deformation is 1 1      0      0         zero, , the tr 0 x x xy E x x xy E x y z 1 material is incompressible 0.5       0       0 y x xz 2 E and the volume is preserved. y x xz E 1 0.5       0       0 z x yz 2 E z x yz E 29

Spherical and deviatoric parts of Hooke’s Law  The stress tensor can be split into a spherical, or volumetric, part and a deviatoric part: 1        : Tr 1 1 sph m 3       1 m         dev 1 m  Similarly for the strain tensor: 1 1 Tr (      e 1 1 1 sph 3 3     3 e ´ 1 1        dev 3 e 1 30

Spherical and deviatoric parts of Hooke’s Law  Operating on the volumetric strain:   e   Tr    1         Tr 1 E E    1            e Tr Tr Tr 1 E E   3  3 m : volumetric K      E 3 1 2 deformation modulus   e  e   m   3 1 2 2 E m def E      K   3 3(1 2 )  The spherical parts of the stress and strain tensor are directly related:   K e m 31

Spherical and Deviator Parts of Hooke’s Law    1    Introducing into :             Tr 1 1 m E E    1                 Tr 1 1 1 m m E E  0              1 1 1 3 1                      Tr Tr 1 1 1 1   1 m m m E E E E  E E  E  3 E Taking into account that :   e   m   3 1 2 Comparing this         1 2 1 E 1 1 1 with the expression          e e   1 1     1  E  3 1 2 E 3 E     3 e ´ 1    1 1 1   1      ´  E 2 2G E  The deviatoric parts of the stress and strain tensor are related component by component:            2 G 2 G i j , {1,2,3} ij ij 32

Spherical and deviatoric parts of Hooke’s Law  The spherical and deviatoric parts of the strain tensor are directly proportional to the spherical and deviatoric parts (component by component) respectively, of the stress tensor:        2 G K e ij ij m 33

Elastic Potential   u   The internal energy density defines a potential for the stress ˆ tensor and is, thus, an elastic potential : REMARK     1 u ˆ       ˆ u : : The constitutive elastic constants      C : 2   tensor is positive definite due C to thermodynamic considerations.   u   ˆ  Plotting vs. :   0 There is a minimum for :     u ˆ     = : 0 C     0   0         2 ˆ 2 ˆ u u    C C C C           0     0 0 34

Elastic Potential  The elastic potential can be written as a function of the spherical and deviatoric parts of the strain tensor:           : : : : : C C C C 1 1 1                      ˆ u : : : Tr 2  : 1 C  2 2 2       1 1       e ´ : e ´  1   1   e      e Tr  3   3  1 1                    2 Tr : : Tr : 1 1 2 2 2        2 e : e : ´ ´ ´ : 1 1 1   9 3 3      Tr 0 K   1 1 1 2 1                      ˆ 2 2 2 2 u e e μ ´: ´ e μ ´: ´ e ´ ´ :   2 3 2  3  3 Elastic potential in terms of the 1   spherical and deviatoric parts        2 ˆ u K e ´: ´ 0 2 of the strains. 35

Limits in the Elastic Properties  The derived expression must hold true for any deformation process: 1     2      ˆ u : K e : ´ 0 ´ 2  Consider now the following particular cases of isotropic linear elastic material:  Pure spherical deformation process 1    1  3 e volumetric 1 1   1   K  ˆ 2 u K e 0 0 deformation modulus 2     1  0  Pure deviatoric deformation process     2   Lamé’s second     0 2      ˆ u ´ ´ 0 : parameter   2  0 e REMARK         : 0 ij ij 36

Limits in the Elastic Properties   K and are related to and through: E  E E       K 0 G 0         3 1 2 2 1 REMARK In rare cases, a material can  Poisson’s ratio has a non-negative value, have a negative Poisson’s ratio. E Such materials are named  0   Young’s auxetic materials .   2 1 E  0 modulus   0  Therefore, E  0 1     3 1 2    0 Poisson’s ratio 2  E 0 37

6.4 The Linear Elastic Problem Ch.6. Linear Elasticity 38

Introduction  The linear elastic solid is subjected to body forces and prescribed tractions: Initial actions:   b x ,0 t  0   t x ,0   b x , t Actions   t x , t through time:  The Linear Elastic problem is the set of equations that allow obtaining the evolution through time of the corresponding       displacements , strains and stresses .   u x , t x , t x , t 39

Governing Equations  The Linear Elastic Problem is governed by the equations: 1. Cauchy’s Equation of Motion. Linear Momentum Balance Equation.     2 u x , t           x , t b x , t 0 0  2 t 2. Constitutive Equation. Isotropic Linear Elastic Constitutive Equation. This is a PDE system of     15 eqns -15 unknowns:        x , t Tr 2 1   u x , t 3 unknowns    x , t 6 unknowns 3. Geometrical Equation.    Kinematic Compatibility. x , t 6 unknowns 1 Which must be solved in                S x , t u x , t u u  3 2 the space. R R  40

Boundary Conditions  Boundary conditions in space  Affect the spatial arguments of the unknowns   Are applied on the contour of the solid, which is divided into three parts:   Prescribed displacements on : u   * u x ( , ) t u x ( , ) t     x t          u * u x , t u x , t i 1,2,3   i i         V  Prescribed tractions on :      u u               * ( , ) x t n ( , ) x t {0}     t        u u u u x t             * x , t n t x , t i 1,2,3   ij j j   Prescribed displacements and stresses on : u        * u x , t u x , t      i i      i j k , , 1,2,3 i j x t          u * x , t n t x , t   jk k j 41

Boundary Conditions 42

Boundary Conditions  Boundary conditions in time. INTIAL CONDITIONS.  Affect the time argument of the unknowns. t   Generally, they are the known values at : 0  Initial displacements:      u x ,0 0 x V  Initial velocity:    u x , t not         u x ,0 v x x V  0  t  t 0 43

The Linear Elastic Problem        Find the displacements , strains and stresses   u x , t x , t x , t such that     2 u x , t           x , t b x , t Cauchy’s Equation of Motion 0 0  2 t            x , t Tr 2 Constitutive Equation 1 1       Geometric Equation    S       x , t u x , t u u 2   * : u u u Boundary Conditions in space     * : t n     u x ,0 0 Initial Conditions (Boundary Conditions in time)    u x ,0 v  0 44

Actions and Responses  The linear elastic problem can be viewed as a system of actions or data inserted into a mathematical model made up of the EDP’s and boundary conditions seen which gives a series of responses or solution in displacements, strains and stresses.   b x , t   u x , t   Mathematical * t x , t    x , t model   * u x , t   EDPs+BCs  x , t   v x 0 not    x , t RESPONSES R not    x , t ACTIONS A  Generally, actions and responses depend on time. In these cases, the problem is a dynamic problem , integrated in .  3 R R   In certain cases, the integration space is reduced to . The problem is 3 R termed quasi-static . 45

The Quasi-Static Problem  A problem is said to be quasi-static if the acceleration term can be considered to be negligible.  2 u x ( , ) t   a 0  2 t  This hypothesis is acceptable if actions are applied slowly. Then,  2 u x ( , ) t    0    0  2 2 2 2 / t / t 0 A R  2 t 46

The Quasi-Static Problem        Find the displacements , strains and stresses   u x , t x , t x , t such that     2 u x , t  0 0  2 t           x , t b x , t 0 Equilibrium Equation 0            x , t Tr 2 Constitutive Equation 1 1       Geometric Equation          S x , t u x , t u u 2   * : u u u Boundary Conditions in Space     * : t n     u x ,0 0 Initial Conditions    u x ,0 v  0 47

The Quasi-Static Problem  The quasi-static linear elastic problem does not involve time derivatives .  Now the time variable plays the role of a loading factor : it describes the evolution of the actions.      b x ,  u x , Mathematical      * t x ,   x , model     EDPs+BCs  * u x ,   x , not not      ,   ,  x x RESPONSES ACTIONS A R    *  For each value of the actions -characterized by a fixed value - a ,  * x A   ,  * response is obtained. x R  *  Varying , a family of actions and its corresponding family of responses is obtained. 48

Example Consider the typical material strength problem where a cantilever beam is   subjected to a force at it’s tip. F t For a quasi-static problem,           t t The response is , so for every time instant, it only depends on   the corresponding value of .  t 49

Solution of the Linear Elastic Problem  To solve the isotropic linear elastic problem posed, two approaches can be used:  Displacement formulation - Navier Equations     Eliminate and from the general system of equations. This   x , t x , t   generates a system of 3 eqns. for the 3 unknown components of . u x , t  Useful with displacement BCs.  Avoids compatibility equations.  Mostly used in 3D problems.  Basis of most of the numerical methods.  Stress formulation - Beltrami-Michell Equations.      Eliminate and from the general system of equations. This x , t u x , t   generates a system of 6 eqns. for the 6 unknown components of .  x , t  Effective with boundary conditions given in stresses.  Must work with compatibility equations.  Mostly used in 2D problems.  Can only be used in the quasi-static problem. 50

Displacement formulation     2 u x , t           x , t b x , t Cauchy’s Equation of Motion 0 0  2 t            , t Tr 2 x Constitutive Equation 1 1       Geometric Equation          S x , t u x , t u u 2   * : u u u Boundary Conditions in Space     * : t n     u x ,0 0 Initial Conditions    u x ,0 v  0   The aim is to reduce this system to a system with as the only unknowns. u x , t       Once these are obtained, and will be found through substitution. x , t x , t 51

Displacement formulation  Introduce the Constitutive Equation into Cauchy’s Equation of motion:            x , t Tr 2 1   2 u               Tr ( ) 2 b 1 0 0    2 t   2 u x , t           x , t b x , t 0 0  2 t  Consider the following identities:         u i           u u                      Tr ( ) 1 k k u       11 ij ij       x x x x x x i     j j k i k i    u            u i 1,2,3   i              u Tr 1 52

Displacement formulation  Introduce the Constitutive Equation into Cauchy’s Equation of motion:            x , t Tr 2 1   2 u               Tr ( ) 2 b 1 0 0    2 t   2 u x , t           x , t b x , t 0 0  2 t  Consider the following identities:     i     2     u u i               u   u  2 1 u 1 u 1 1 1         ij   j   j         2  i  i   u u              i x x 2 x x 2 x x 2 x x 2 2 x         i j j j i j j i j i    u   1 1       2      u u i 1,2,3    2 2  i 1 1          2 ( u ) u 2 2 53

Displacement formulation  Introduce the Constitutive Equation into Cauchy’s Equation of Movement:            x , t Tr 2 1   2 u               Tr ( ) 2 b 1 0 0  2   t   2 u x , t           x , t b x , t 0 0  2 t  Replacing the identities:   1 1            u Tr          2 1 ( u ) u 2 2 2 nd order   2   1 1 u                 2 u 2 ( u ) u b  Then,   PDE system 0 0  2  2 2  t   2 u                  2 u u b   The Navier Equations 0 0  2 t                are obtained: u u b u i 1,2,3   j ji , i jj , 0 i 0 i  54

Displacement formulation    The boundary conditions are also rewritten in terms of : u x , t        u        x , t Tr 2 1            * t Tr n 2 n 1 *           S  u u   u t n 2     *             t u n u u n  The BCs are now:   * u u   on  u     * u u i 1,2,3   i i     REMARK              * u n u u n t   on  The initial conditions            * u n u n u n t i 1,2,3   k k . i i j , j j i , j i remain the same. 55

Displacement formulation  Navier equations in a cylindrical coordinate system :       2 e 2 G u      x r cos          2 G z 2 G b r  r      2 r r z t     x ( , , ) r z y r sin          2 1 e u   z z           2 G 2 G r 2 G z b       2 r z r t      2 e 2 G 2 G u              2 G r r b z      z  2 z r r r t     1 1 u u       z    r z    2  r z    dV r d dr dz     1 u u Where:      r z    zr   2 z r         ru u 1 1 1    u 1 1 u         r       e ru z  z r    2 r r r r       r r r z 56

Displacement formulation  Navier equations in a spherical coordinate system :      2 e 2 G 2 G u                2 G sin b r  r            2  x r sin cos r r sin r sin t           x x r , , y r sin sin       2 G     2 u e 2 G 2 G              r r sin b  z r cos            2 r r sin r sin r t      2     u 2 G e 2 G 2 G           2 sin r r b     dV r dr d d          2 r sin r r r t     1 1 1 u        ω u sin θ     r   2 r sin θ θ r sin θ φ     Where:    ru  1 1 u 1        ω r     r   2 r sin θ φ r r       1 1 1 u       ω ru r      r   2  r r r θ       1        2     e r u sin ru sin ru     r 2       r sin r   57

Stress formulation Equilibrium Equation           x , t b x , t 0 0 (Quasi-static problem)    1           Inverse Constitutive Equation x , t Tr 1 E E 1       Geometric Equation          S x , t u x , t u u 2   * : u u u Boundary Conditions in Space     * : t n     x , t The aim is to reduce this system to a system with as the only unknowns.      x , t Once these are obtained, will be found through substitution and by u x , t integrating the geometric equations. REMARK For the quasi-static problem, the time variable plays the role of a loading factor. 58

Stress formulation  Taking the geometric equation and, through successive derivations, the displacements are eliminated: Compatibility Equations  2   2      2 2   ij    jl   kl ik 0 i j k l , , , 1,2,3 (seen in Ch.3.)         x x x x x x x x k l i j j l i k  Introducing the inverse constitutive equation into the compatibility equations and using the equilibrium equation:    1        ij pp ij ij E E   ij    b 0  0 j x i 2 nd order  The Beltrami-Michell Equations are obtained: PDE system             b   2  b b 1   0 j          2 0 k 0 i kk i j , 1,2,3 ij       ij    1 x x 1 x x x i j k j i 59

Stress formulation  The boundary conditions are:        Equilibrium Equations: b 0 0 This is a 1 st order PDE system, so they can act as boundary conditions of the (2 nd order PDE system of the) Beltrami-Michell Equations      *  Prescribed stresses on : n t on   60

Stress formulation  Once the stress field is known, the strain field is found by substitution.    1           x , t Tr 1 E E  The calculation, after, of the displacement field requires that the geometric equations be integrated with the prescribed displacements  on : u 1           ( ) x u x ( ) u x ( ) x V 2  *   u x u x x ( ) ( ) u REMARK This need to integrate the second system is a considerable disadvantage with respect to the displacement formulation when using numerical methods to solve the lineal elastic problem. 61

Saint-Venant’s Principle  From A. E. H. Love's Treatise on the mathematical theory of elasticity : “According to the principle, the strains that are produced in a body by the application, to a small part of its surface, of a system of forces statically equivalent to zero force and zero couple, are of negligible magnitude at distances which are large compared with the linear dimensions of the part.” REMARK This principle does not have a  Expressed in another way: rigorous mathematical proof. “The difference between the stresses caused by statically equivalent load systems is insignificant at distances greater than the largest dimension of the area over which the loads are acting.”      (I) (II) u x , t u x , t P P            (I) (II) x , t x , t P | P P        (I) (II) x , t x , t P P 62

Saint-Venant’s Principle  Saint Venant’s Principle is often used in strength of materials.  It is useful to introduce the concept of stress: The exact solution of this problem is very complicated. This load system is statically equivalent to load system (I). The solution of this problem is very simple. Saint Venant’s Principle allows approximating solution (I) by solution (II) at a far enough distance from the ends of the beam. 63

Uniqueness of the solution  The solution of the lineal elastic problem is unique:  It is unique in strains and stresses .  It is unique in displacements assuming that appropriate boundary conditions hold in order to avoid rigid body motions.  This can be proven by Reductio ad absurdum ("reduction to the absurd"), as shown in pp. 189-193 of the course book.  This demonstration is valid for lineal elasticity in small deformations.  The constitutive tensor is used, so the demonstration is not only valid for C isotropic problems but also for orthotropic and anisotropic ones. 64

6.5 Linear Thermoelasticity Ch.6. Linear Elasticity 65

Hypothesis of the Linear Thermo-elastic Model  The simplifying hypothesis of the Theory of Linear Thermo- elasticity are: 1. Infinitesimal strains and deformation framework Both the displacements and their gradients are infinitesimal.  2. Existence of an unstrained and unstressed reference state The reference state is usually assumed to correspond to the reference      configuration.     x x , t 0 0 0         x x , t 0 0 0 3. Isentropic and adiabatic processes – no longer isothermal!!! Isentropic - entropy of the system remains constant  Adiabatic - occurs without heat transfer  66

Hypothesis of the Linear Thermo-Elastic Model 3. (Hypothesis of isothermal process is removed) The process is no longer isothermal so the temperature changes  throughout time: not          x , t x ,0 0     x , t       x , t 0  t We will assume the temperature field is known. But the process is still isentropic and adiabatic :     s t cnt s  0    q n        Q r dV dS 0 V V e  V V internal heat conduction fonts from the exterior         r q 0 x t 67

Generalized Hooke’s Law  The Generalized Hooke’s Law becomes:                        x , t : x , t : x , t  C C C C Generalized Hooke’s Law for 0      linear thermoelastic problems          i j , 1,2,3  C  ij ijkl kl ij 0 Where is the elastic constitutive tensor.     x , t is the absolute temperature field .       x , t is the temperature at the reference state .  0 0  is the tensor of thermal properties or constitutive thermal  constants tensor .  It is a semi-positive definite symmetric second-order tensor. 68

Isotropic Constitutive Constants Tensors  An isotropic thermoelastic material must have the same elastic and thermal properties in all directions: must be a (mathematically) isotropic 4 th order tensor:         2 I 1 1 C                  i j k l , , . 1,2,3  C  ijkl ij kl ik jl il jk Where: 1   is the 4 th order symmetric unit tensor defined as         I I    ik jl il jk ijkl 2   and are the Lamé parameters or coefficients.  is a (mathematically) isotropic 2 nd order tensor:        1         ij i j , 1,2,3   ij Where:  is a scalar thermal constant parameter.  69

Isotropic Linear Thermoelastic Constitutive Equation     Introducing the isotropic constitutive constants tensors and 1                into the generalized Hooke’s Law, : 2 I C 1 1 C 0 (in indicial notation)                                   C ij ijkl kl ij 0 ij kl ik jl il jk kl 0 ij   ij   1 1                      2   ij kl kl ik jl kl il jk kl 0 ij  2 2           ll ji ij   ij  The resulting constitutive equation is,                Tr 2  1 1  Isotropic linear thermoelastic     constitutive equation.              2 i j , 1,2,3     ij ij ll ij ij 70

Inversion of the Constitutive Equation  1. is isolated from the Generalized Hooke’s Law for linear thermoelastic problems:            1     1  : : : C C C C C  2. The thermal expansion coefficients tensor is defined as: def     1 : C It is a 2 nd order symmetric tensor which involves 6 thermal expansion coefficients 3. The inverse constitutive equation is obtained: 1 :         C 71

Inverse Isotropic Linear Thermoelastic Constitutive Equation  For the isotropic case:     1      1 I C 1 1     1 2 E E        1 : (  C 1)    1 E                 i j k l , , . 1,2,3 C   ijkl ij kl ik jl il jk E E  The inverse const. eq. is re-written:       1           Tr 1 1  Inverse isotropic linear thermo  E E  elastic constitutive equation.    1                 i j , 1,2,3  ij ll ij ij ij  E E   Where is a scalar thermal expansion coefficient related to the  scalar thermal constant parameter through:   1 2    E 72

Thermal Stress  Comparing the constitutive equations,          Tr 2 Isotropic linear elastic constitutive equation. 1              Isotropic linear thermoelastic constitutive equation. Tr 2 1 1 t   nt   the decomposition is made:      nt t Where:  nt is the non-thermal stress : the stress produced if there is no  thermal phenomena.  t is the thermal stress : the “corrector” stress due to the  temperature increment. 73

Thermal Strain  Similarly, by comparing the inverse constitutive equations,    1 Inverse isotropic linear elastic         Tr 1 constitutive eq. E E    1             Tr Inverse isotropic linear thermoelastic 1 1 E E   t constitutive eq.   nt the decomposition is made:      nt t Where:  nt is the non-thermal strain : the strain produced if there is no  thermal phenomena.  t is the thermal strain : the “corrector” strain due to the  temperature increment. 74

Thermal Stress and Strain  The thermal components appear when thermal processes are considered. NON-THERMAL THERMAL TOTAL COMPONENT COMPONENT nt     t     : C      nt t Isotropic material: Isotropic material:             nt t Tr ( ) 2 1 1          nt 1 t : C      nt t Isotropic material: Isotropic material:    1  nt      Tr ( )      1 t 1 E E These are the equations used          nt t : : C C C C in FEM codes.              1 nt 1 t : : C C C C   75

Thermal Stress and Strain REMARK 1 In thermoelastic problems, a state of zero strain in a body does not necessarily imply zero stress.      nt 0 0           0 t 1 REMARK 2 In thermoelastic problems, a state of zero stress in a body does not necessarily imply zero strain.      nt 0 0         0 t 1 76

6.6 Thermal Analogies Ch.6. Linear Elasticity 77

Solution to the Linear Thermoelastic Problem  To solve the isotropic linear thermoelastic problem posed thermal analogies are used.  The thermoelastic problem is solved like an elastic problem and then, the results are “corrected” to account for the temperature effects.  They use the same strategies and methodologies seen in solving isotropic linear elastic problems:  Displacement Formulation - Navier Equations.  Stress Formulation - Beltrami-Michell Equations.  Two basic analogies for solving quasi-static isotropic linear thermoelastic problems are presented:  1 st thermal analogy – Duhamel-Neumann analogy.  2 nd thermal analogy 78

1 st Thermal Analogy  The governing eqns. of the quasi-static isotropic linear thermoelastic problem are:           x , t b x , t 0 Equilibrium Equation 0            x , t : x , t 1 C Constitutive Equation 1                S x , t u x , t u u Geometric Equation 2   * : u u u Boundary Conditions in Space  *    : t n  79

1 st Thermal Analogy  The actions and responses of the problem are:       not not    I x , t  I ACTIONS A x , t RESPONSES R   b x , t   u x , t Mathematical   * t x , t    x , t model   * u x , t    x , t EDPs+BCs     x , t REMARK     is known a priori , i.e., it is x , t independent of the mechanical response. This is an uncoupled thermoelastic problem. 80

1 st Thermal Analogy  To solve the problem following the methods used in linear elastic problems, the thermal term must be removed.  The stress tensor is split into and replaced into the governing      nt t equations:  Momentum equations      nt   t        nt      t     nt          1       b 0 0 ˆ    nt    b 0 1   0      nt         b 0   1 0    ˆ         b b 0  not 0 ˆ  b 81

1 st Thermal Analogy  Boundary equations:    nt   t  nt    t   * n n t ˆ nt *    n t    * n t    ˆ *  *     t t ( ) n  nt   *   t   *     n t n t ( ) n           n 1 ˆ * t ANALOGOUS PROBLEM – A linear elastic problem can be solved as: 1 0 ˆ ˆ       nt       Equilibrium Equation b 0 b b ( ) with  0          nt : Tr ( ) 2 Constitutive Equation 1 C 1                S x , t u x , t u u Geometric Equation 2   * : u u u Boundary Conditions in Space ˆ ˆ          nt * * * : n t with t t n  82

1 st Thermal Analogy  The actions and responses of the ANALOGOUS NON-THERMAL PROBLEM are:     not not      II x , t  II ACTIONS A x , t RESPONSES R     u x , t ˆ Mathematical b x , t    x , t   ˆ model * t x , t    nt   x , t EDPs+BCs * u x , t ORIGINAL PROBLEM (I) ANALOGOUS (ELASTIC) PROBLEM (II) 83

1 st Thermal Analogy  If the actions and responses of the original and analogous problems are compared:   b   1        ˆ   ˆ     b b b b   ACTIONS          0  u          u 0 def       I  II      III ( , ) x t ( , ) x t 0         A A A A A ˆ ˆ * t  * * * t t t             t *     n                0          RESPONSES         u u 0 0         def       I  II        III ( , ) x t ( , ) x t 0 0         R R R R R                  nt nt        1  t   Responses are proven to be the solution of a thermoelastic problem under actions 84

1 st Thermal Analogy  The solution of the ORIGINAL PROBLEM is:     I  II u u      I   II      I   II     1 ORIGINAL PROBLEM (I) TRIVIAL PROBLEM (III) ANALOGOUS PROBLEM (II) 85

1 st Thermal Analogy THERMOELASTIC ANALOGOUS THERMOELASTIC ORIGINAL ELASTIC TRIVIAL PROBLEM (I) PROBLEM (II) PROBLEM (III) 1 1     ˆ           b x , t b       b x , t b            0 * t x , t 0          ˆ   I II      III * * t x , t t n x , t x , t x , t       t n A A A *   * u x , t   * u x , t  u   0     x , t        0 x , t      u x , t u 0 u x , t               I II      x , t x , t  0 x , t x , t III R R x , t R                  nt x , t x , t t 1 86

2 nd Thermal Analogy  The governing eqns. of the quasi-static isotropic linear thermoelastic problem are:           x , t b x , t 0 Equilibrium Equation 0            -1 x , t : x , t C 1 Inverse Constitutive Equation 1                S x , t u x , t u u Geometric Equation 2   * : u u u Boundary Conditions in Space  *    : t n  87

2 nd Thermal Analogy  The actions and responses of the problem are:       not not    I x , t  I ACTIONS A x , t RESPONSES R   b x , t   u x , t Mathematical   * t x , t    x , t model   * u x , t    x , t EDPs+BCs     x , t REMARK     is known a priori , i.e., it is x , t independent of the mechanical response. This is an uncoupled thermoelastic problem. 88

2 nd Thermal Analogy  x   The hypothesis is made that and are such that the ( ) ( , ) x t strain field is integrable (satisfies the compatibility equations).  If the strain field is integrable, there exists a field of thermal   displacements which satisfies: u x , t t 1                    x , t u u u t S t t t 1 2     u t 1 u t             j   i  i j , 1,2,3 t   ij ij   2 x x   j i REMARK   u x , t The solution is determined except for a rigid body movement t   c * characterized by a rotation tensor and a displacement vector .          u x , t u x , t  x c t  * The family of admissible solutions is . This movement can be arbitrarily chosen.  Then, the total displacement field is decomposed as: def     nt t nt t u ( , ) x t u x ( , ) t u x ( , ) t u u u 89

2 nd Thermal Analogy  To solve the problem following the methods used in linear elastic problems, the thermal terms must be removed.  The strain tensor and the displacement vector splits, and       nt  t nt t u u u and replaced into the governing equations:  Geometric equations:    S   S nt  t   S nt   S t   S nt   t u ( u u ) u u u  nt S nt     u t nt t       Boundary equations:  * u u * nt * t     nt  t  * : u u u u u u u u u  nt  t u u u 90

2 nd Thermal Analogy ANALOGOUS PROBLEM – A linear elastic problem can be solved as:       b 0 Equilibrium Equation 0 nt    -1 : Constitutive Equation C  nt   S nt Geometric Equation u    nt * t : u u u u Boundary Conditions in space     * : n t  91

2 nd Thermal Analogy  The actions and responses of the ANALOGOUS PROBLEM are:     not not      II x , t  II ACTIONS A x , t RESPONSES R   nt   u x , t Mathematical b x , t    nt x , t   model * t x , t    x , t EDPs+BCs       nt * t u u x , t u x , t ORIGINAL PROBLEM (I) ANALOGOUS PROBLEM (II) 92

2 nd Thermal Analogy  If the actions and responses of the original and analogous problems are compared:       b b 0 ACTIONS          t t u u u u       def       I  II     III ( , ) x t ( , ) x t       A A A A A * * t t 0                 0       RESPONSES         nt t t u u u u         def       I II III      nt   t      ( , ) x t ( , ) x t         R R R R 1 R           0 0         Responses are proven to be the solution of a thermo-elastic problem under actions 93

2 nd Thermal Analogy  The solution of the ORIGINAL PROBLEM is:     I  II  t u u u     I II        1     I II    ORIGINAL PROBLEM (I) ANALOGOUS PROBLEM (II) TRIVIAL PROBLEM (III) 94

2 nd Thermal Analogy THERMOELASTIC ANALOGOUS THERMOELASTIC ORIGINAL ELASTIC TRIVIAL PROBLEM (I) PROBLEM (II) PROBLEM (III)     b x , t b b 0            *   t x , t  u u  u u   u x , t t t      I II III x , t x , t x , t A A A     * u x , t t t 0 * *          0 x , t   x , t       u x , t nt u x , t  u u x , t t                 I II   x , t x , t  nt  x , t x , t        III t R R 1 x , t R       x , t   x , t 0 95

2 nd Analogy in structural analysis nt  u u  t     u u x x x x x    nt t        x x x x   *  t      : u u u l   t     : u u l u x x x x l   u x x x l  0 96

Thermal Analogies  Although the 2 nd analogy is more common, the 1 st analogy requires less corrections.  The 2 nd analogy can only be applied if the thermal strain field is integrable.  It is also recommended that the integration be simple.  The particular case     ( ) x cnt  Homogeneous material:       ax by cz d  Lineal thermal increment: is of special interest because the thermal strains are:      linear polinomial t 1 and trivially satisfy the compatibility conditions (involving second order derivatives). 97

Thermal Analogies  In the particular case     ( ) x cnt  Homogeneous material:       ( ) x cnt  Constant thermal increment: the integration of the strain field has a trivial solution because       cnt the thermal strains are constant , therefore: t rigid body movement (can be chosen           u x , t x  x c  t arbitrarily) The thermal displacement is:                    u x , t x x u x x 1 x t t HOMOTHECY (free thermal expansion) 98

6.7 Superposition Principle Ch.6. Linear Elasticity 99

Linear Thermoelastic Problem  The governing eqns. of the isotropic linear thermoelastic problem are:           x , t b x , t 0 Equilibrium Equation 0            x , t : x , t 1 C Constitutive Equation 1                S x , t u x , t u u Geometric Equation 2   * : u u u Boundary Conditions in space     * : t n     u x ,0 0 Initial Conditions    u x ,0 v  0 100