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

CH.6. LINEAR ELASTICITY

Continuum Mechanics Course (MMC) - ETSECCPB - UPC

Overview

 Hypothesis of the Linear Elasticity Theory  Linear Elastic Constitutive Equation

 Generalized Hooke’s Law  Elastic Potential

 Isotropic Linear Elasticity

 Isotropic Constitutive Elastic Constants Tensor  Lamé Parameters  Isotropic Linear Elastic Constitutive Equation  Young’s Modulus and Poisson’s Ratio  Inverse Isotropic Linear Elastic Constitutive Equation  Spherical and Deviator Parts of Hooke’s Law  Limits in Elastic Properties

2

Overview (cont’d)

 The Linear Elastic Problem

 Governing Equations  Boundary Conditions  The Quasi-Static Problem  Solution

 Displacement Formulation  Stress Formulation

 Uniqueness of the solution  Saint-Venant’s Principle

 Linear Thermoelasticity

 Hypothesis of the Linear Elasticity Theory  Linear Thermoelastic Constitutive Equation  Inverse Constitutive Equation  Thermal Stress and Strain

3

Overview (cont’d)

 Thermal Analogies

 Solution to the linear thermoelastic problem  1st Thermal Analogy  2nd Thermal Analogy

 Superposition Principle in Linear Thermoelasticity  Hooke’s Law in Voigt Notation

4

5

Ch.6. Linear Elasticity

6.1 Hypothesis of the Linear Elasticity Theory

 Elasticity: Property of solid materials to deform under the

application of an external force and to regain their original shape after the force is removed.

 Stress: external force applied on a specified area.  Strain: amount of deformation.

 The (general) Theory of Elasticity links the strain

experienced in any volume element to the forces acting on the macroscopic body.



Introduction

Linear Elasticity is a simplification of the more general (nonlinear) Theory of Elasticity.

6

Hypothesis of the Linear Elastic Model

 The simplifying hypothesis of the Theory of Linear 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.

• 3. Isothermal, isentropic and adiabatic processes



Isothermal - temperature remains constant



Isentropic - entropy of the system remains constant



Adiabatic - occurs without heat transfer

       

, , t t     x x x x    

7

Hypothesis of the Linear Elastic Model

 The simplifying hypothesis of the Theory of Linear Elasticity are:

• 1. ‘Infinitesimal strains and deformation’ framework
• 2. Existence of an unstrained and unstressed reference state
• 3. Isothermal, isentropic and adiabatic processes

8

Hypothesis of the Linear Elastic Model

• 1. ‘Infinitesimal strains and deformation’ framework

the displacements are infinitesimal:



material and spatial configurations or coordinates are the same



material and spatial descriptions of a property & material and spatial differential operators are the same:



the deformation gradient , so the current spatial density is approximated by the density at the reference configuration. Thus, density is not an unknown variable in linear elastic problems.

 x X

       

, , , , t t t t        x X X x

   

       X x

    x F 1 X

t t

     F

 x X    

    

 F 1

  x X u

0

9

Hypothesis of the Linear Elastic Model

• 1. ‘Infinitesimal strains and deformation’ framework

the displacement gradients are infinitesimal:



The strain tensors in material and spatial configurations collapse into the infinitesimal strain tensor.

     

, , , t t t   E X e x x 

10

Hypothesis of the Linear Elastic Model

2. Existence of an unstrained and unstressed reference state



It is assumed that there exists a reference unstrained and unstressed neutral state, such that,



The reference state is usually assumed to correspond to the reference configuration.

11

       

, , t t     x x x x    

Hypothesis of the Linear Elastic Model

3. Isothermal and adiabatic (=isentropic) processes



In an isothermal process the temperature remains constant.



In an isentropic process the entropy of the system remains constant.



In an adiabatic process the net heat transfer entering into the body is zero.

     

, , t t       x x x x ( , ) ( ) ds s t s dt    X X

e

Q

V V

r dV dS V V 



     

  q n

r t       q x  

   0

s  

heat conduction from the exterior internal fonts

12

13

Ch.6. Linear Elasticity

6.2 Linear Elastic Constitutive Equation

 R. Hooke observed in 1660 that, for relatively small deformations

• f an object, the displacement or size of the deformation is

directly proportional to the deforming force or load.

 Hooke’s Law (for 1D problems) states that in an elastic

material strain is directly proportional to stress through the elasticity modulus.

Hooke’s Law

E   





E

F

l

F k l  

F l E A l  

14

 This proportionality is generalized for the multi-dimensional case

in the Theory of Linear Elasticity.

 It constitutes the constitutive equation of a linear elastic material.  The 4th order tensor is the constitutive elastic constants tensor:

 Has 34=81 components.  Has the following symmetries, reducing the tensor to 21 independent

components:

Generalized Hooke’s Law

     

, ( ) , , 1,2,3

ij ijkl kl

t t i j           x x : x   C C

Generalized Hooke’s Law



ijkl jikl ijkl ijlk

  C C C C

minor symmetries

ijkl klij

 C C

major symmetries

REMARK

The current stress at a point depends only on the current strain at the point, and not on the past history of strain states at the point.

15

 The internal energy balance equation for the (adiabatic) linear

elastic model is

Where:



is the specific internal energy (energy per unit mass).



is the specific heat generated by the internal sources.



is the heat conduction flux vector per unit surface.

Elastic Potential

 

: d u r dt      q   

global form local form 

 

V V V V

d d u u dV dV dV r dV dt dt   



    

   

: d q   

q

r u

REMARK

The deformation rate tensor is related to the material derivative of the material strain tensor through: In this case, and .

T

   E F d F   E     F 1 stress power heat variation infinitesimal strains

 

V V t t  

internal energy

16

 The stress power per unit of volume is an exact differential of the

internal energy density, , or internal energy per unit of volume:

 Operating in indicial notation:

Elastic Potential



ˆ( , ) ˆ ( ) :

ˆ

d du t u u dt dt

u

    x    

ˆ u







 

)

:

: :

ˆ 1 ( ) 2 1 1 ( ) 2 2 1 2

ij ij ij ijkl kl ij ijkl kl ij ijkl kl ij ijkl kl kl klij ij ij ijkl kl ij ijkl kl ij ijkl kl ijkl kl jkl ij ijkl kl

i

i k j l d dt

du dt d dt







   

                             :                     



 

 

C

C

C C

C

C C C C C C C C

REMARK

The symmetries of the

constitutive elastic constants tensor are used:

ijkl jikl ijkl ijlk

  C C C C

minor symmetries

ijkl klij

 C C

major symmetries

 

ˆ 1 2 du d dt dt  : : C  

17



Consequences:

1. Consider the time derivative of the internal energy in the whole volume:

 In elastic materials we talk about deformation energy because the stress

power is an exact differential.

Elastic Potential

 

ˆ 1 : : : 2 du d dt dt    C          

ˆ ˆ ˆ , , , :

V V V

d d d u t dV u t dV t dV dt dt dt   

  

x x x    U

stress power

REMARK

Internal energy in elastic materials is an exact differential.

18

 Consequences:

2. Integrating the time derivative of the internal energy density, and assuming that the density of the internal energy vanishes at the neutral reference state, :

 Due to thermodynamic reasons the elastic energy is assumed

always positive

Elastic Potential

     

1 , , ( ) 2 t t a a     x : : x x x x C  

       

1 ˆ , , , 2 u t t t a   x x : : x x C  

 

ˆ , u t   x x

 0

 

1 ˆ : : 2 u    0 C     

 



1 1 ˆ : : ( ) : 2 2 u   C      



19

 

ˆ 1 : : : 2 du d dt dt    C    

 The internal energy density defines a potential for the stress

tensor, and is thus, named elastic potential. The stress tensor can be computed as

 The constitutive elastic constants tensor can be obtained as the

second derivative of the internal energy density with respect to the strain tensor field,

Elastic Potential

 

2 ˆ ijkl ij kl

u      C

 

ˆ u      

   

2 ˆ

: ( ) u           C C        

ˆ( ( , )) 1 1 1 1 ( : : : : ( ) 2 2 2 2 u t           x           C C C C C

 

T

 

 

20

21

Ch.6. Linear Elasticity

6.3 Isotropic Linear Elasticity

 An isotropic elastic material must have the same elastic

properties (contained in ) in all directions.

 All the components of must be independent of the orientation of

the chosen (Cartesian) system must be a (mathematically) isotropic tensor. Where:



is the 4th order unit tensor defined as



and are scalar constants known as Lamé parameters or coefficients.

Isotropic Constitutive Elastic Constants Tensor

 

 

2 , , , 1,2,3

ijkl ij kl ik jl il jk

i j k l                      I C C 1 1

C C C

REMARK

The isotropy condition reduces the number of independent elastic constants from 21 to 2.

 

1 2

ik jl il jk ijkl

          I

I 



22

 Introducing the isotropic constitutive elastic constants tensor

into the generalized Hooke’s Law ,

(in indicial notation)

 The resulting constitutive equation is,

Isotropic Linear Elastic Constitutive Equation

 

 

1 1 2 ( ) 2 2 2

ij ijkl kl ij kl ik jl il jk kl ij kl kl ik jl kl il jk kl ij ij

Tr                                          C

2      I C 1 1

 : C      

2 2 , 1,2,3

ij ij ll ij

Tr i j                     1

Isotropic linear elastic constitutive equation. Hooke’s Law

( )

ll

Tr  

 

ji ij

 

 

ij

 

1 2 1 2

i j ij j i



 

  

23

 If the constitutive equation is,

Then, the internal energy density can be reduced to:

Elastic Potential

   

2 2 , 1,2,3

ij ij ll ij

Tr i j                1

Isotropic linear elastic constitutive equation. Hooke’s Law

   

 

 

 



 

2

1 1 ˆ : 2 : 2 2 1 1 : 2 2 2 1 2

Tr

u Tr Tr Tr           

σ

     



                  1 1

REMARK

The internal energy density is an elastic potential of the stress tensor as:

     

ˆ 2 u Tr              1

24

1. is isolated from the expression derived for Hooke’s Law 2. The trace of is obtained: 3. The trace of is easily isolated: 4. The expression in 3. is introduced into the one obtained in 1.

Inversion of the Constitutive Equation

   

 

         

2 2 3 2 Tr Tr Tr Tr Tr Tr Tr                   1 1



   

1 3 2 Tr Tr      

 

 

2 Tr        1

 

 

1 2 Tr        1

 

1 1 2 3 2 Tr                 1

   

1 2 3 2 2 Tr             1

=3

25

 The Lamé parameters in terms of and :  So the inverse const. eq. is re-written:

Inverse Isotropic Linear Elastic Constitutive Equation

 

 

 

   

 

1 1 1 1 1 1

x x y z xy xy y y x z xz xz z z x y yz yz

E G E G E G                                 

   

1 1 , 1,2,3

ij ll ij ij

Tr E E i j E E                            1

 E

Inverse isotropic linear elastic constitutive equation. Inverse Hooke’s Law.

   

3 2 2 E              

    

1 1 2 2 1 E E G            

In engineering notation:

26

 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.

 Poisson's ratio is the ratio, when a solid is uniaxially stretched,

• f the transverse strain (perpendicular to the applied stress), to the

axial strain (in the direction of the applied stress).

Young’s Modulus and Poisson’s Ratio

 E

 

3 2 E        

 

2      

27

Example

Consider an uniaxial traction test of an isotropic linear elastic material such that: Obtain the strains (in engineering notation) and comment on the results

• btained for a Poisson’s ratio of and .

x



x



x y z

x



x



x y z xy xz yz

           

E,    0.5  

28

Solution

For : For :

 

1

x x xy y x xz z x yz

E E E                    1

x x xy y xz z yz

E             

0.5  

1 0.5 0.5

x x xy y x xz z x yz

E E E                  1 1 2 1 2

x x xy y x xz z x yz

E E E                 

There is no Poisson’s effect and the transversal normal strains are zero. The volumetric deformation is zero, , the material is incompressible and the volume is preserved.

tr

x y z

       

29

 

 

 

   

 

1 1 1 1 1 1

x x y z xy xy y y x z xz xz z z x y yz yz

E G E G E G                                 

x y z xy xz yz

           

 The stress tensor can be split into a spherical, or volumetric, part

and a deviatoric part:

 Similarly for the strain tensor:

Spherical and deviatoric parts of Hooke’s Law

 

1 : 3

sph m

Tr      1 1 dev

m

    1    

m

      1

1 ´ 3 e     1 1 1 Tr ( 3 3

sph

e      1 1

1 dev 3e       1

30

 Operating on the volumetric strain:  The spherical parts of the stress and strain tensor are directly

related:

Spherical and deviatoric parts of Hooke’s Law

 

Tr e  

 

1 Tr E E          1

     

1 Tr Tr Tr e E E         1

 

3 1 2

m

e E    

3 

3

m

 

 

3 1 2

m

E e    

: volumetric deformation modulus

K

2 3 3(1 2 )

def

E K       

m

K e  

31

 Introducing into :

Taking into account that :

 The deviatoric parts of the stress and strain tensor are related

component by component:

Spherical and Deviator Parts of Hooke’s Law

m

      1

 

3 1 2

m

E e    

 

1 Tr E E          1

       

1 1 1 1 3 1

m m m m m

Tr E E Tr Tr E E E E E E E                                                  1 1 1 1 1 1 1 1

3 

 

1 2 1 1 1 1 3 1 2 3 E e e E E E                        1 1



2 2 , {1,2,3}

ij ij

G G i j           

1 ´ 3 e     1

Comparing this with the expression 1 ´ E      

1 1 1 E 2 2G     

32

 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:

Spherical and deviatoric parts of Hooke’s Law

2

ij ij

G     

m

K e  

33

 The internal energy density defines a potential for the stress

tensor and is, thus, an elastic potential:

 Plotting vs. :

Elastic Potential

 

1 ˆ : : 2 u  C   

 

ˆ : u         

 

ˆ u 

   

2 2

ˆ ˆ u u

  

         C C C C

  

     

 

ˆ u 

    

ˆ = : u

 

   C

 

   There is a minimum for :

  REMARK

The constitutive elastic constants

tensor is positive definite due to thermodynamic considerations.

C

34

 The elastic potential can be written as a function of the spherical

and deviatoric parts of the strain tensor:

Elastic Potential

 

1 1 ˆ 2 2 : : : u        C

   

2

1 1 : : : 2 2 Tr Tr                1  

: : : : :         C C C C

 

1 2 2 Tr         :     1

 

 

Tr

e







e





 



2 2

3

1 1 ´ ´ 3 3 1 2 ´ ´ ´ 9 3 1 ´ ´ 3

Tr

e e e e e

 

                      : : : : :



       1 1 1 1 1

 

2 2 2

1 1 1 2 ˆ ´: ´ ´: ´ 2 3 2 3 u e e μ e μ                     

K

 

2

1 ˆ ´: ´ 0 2 u K e       

Elastic potential in terms of the spherical and deviatoric parts

• f the strains.

35

 The derived expression must hold true for any deformation

process:

 Consider now the following particular cases of isotropic linear

elastic material:

 Pure spherical deformation process  Pure deviatoric deformation process

Limits in the Elastic Properties

 

2

1 ˆ : : ´ 0 2 u K e        ´

   

1 1

1 3 e    0   1

 

1 2

1 ˆ 2 u K e   K 

   

2 2

e    0  

 

2

ˆ ´ ´ 0 u    :    

:

ij ij

       

REMARK

volumetric deformation modulus Lamé’s second parameter

36



and are related to and through:

 Poisson’s ratio has a non-negative value,  Therefore,

Limits in the Elastic Properties

K  E 

 

3 1 2 E K    

 

2 1 E G      

 

2 1 E     

E  Young’s modulus

 

3 1 2 E E    

1 2    Poisson’s ratio

REMARK

In rare cases, a material can have a negative Poisson’s ratio. Such materials are named auxetic materials.

37

38

Ch.6. Linear Elasticity

6.4 The Linear Elastic Problem

 The linear elastic solid is subjected to body forces and prescribed

tractions:

 The Linear Elastic problem is the set of equations that allow

• btaining the evolution through time of the corresponding

displacements , strains and stresses .

Introduction

t 

   

,0 ,0 b x t x

   

, , t t b x t x

Initial actions: Actions through time:

 

,t u x

 

,t x 

 

,t x 

39

 The Linear Elastic Problem is governed by the equations:

1. Cauchy’s Equation of Motion.

Linear Momentum Balance Equation.

2. Constitutive Equation.

Isotropic Linear Elastic Constitutive Equation.

3. Geometrical Equation.

Kinematic Compatibility.

Governing Equations

     

2 2

, , , t t t t        u x x b x  

   

, 2 t Tr     x    1

     

1 , , 2

S

t t      x u x u u    

This is a PDE system of 15 eqns -15 unknowns: Which must be solved in the space.

 

,t u x

 

,t x 

 

,t x 

3 unknowns 6 unknowns 6 unknowns

3 

 R R

40

 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 :  Prescribed tractions on :  Prescribed displacements and stresses on :

Boundary Conditions



{0}

u u u u u u

V

     

                     

u







u



     

* *

( , ) ( , ) , , 1,2,3

u i i

t t t u t u t i            u x u x x x x

     

* *

( , ) ( , ) , , 1,2,3

ij j j

t t t t n t t i



              x n x x x x  t

         

 

* *

, , , , 1,2,3 , ,

i i u jk k j

u t u t i j k i j t t n t t



              x x x x x

41

Boundary Conditions

42

 Boundary conditions in time. INTIAL CONDITIONS.  Affect the time argument of the unknowns.  Generally, they are the known values at :

 Initial displacements:  Initial velocity:

Boundary Conditions

t 

 

,0 V    u x x

     

, ,0

not t

t V t



      u x u x v x x 

43

 Find the displacements , strains and stresses

such that

The Linear Elastic Problem

 

,t u x

 

,t x 

 

,t x 

     

2 2

, , , t t t t        u x x b x  

   

, 2 t Tr     x    1

     

1 , , 2

S

t t      x u x u u    

Cauchy’s Equation of Motion Constitutive Equation Geometric Equation

* *

: :

u 

     u u t n 

Boundary Conditions in space

   

,0 ,0   u x u x v 

Initial Conditions (Boundary Conditions in time)

44

 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.

 Generally, actions and responses depend on time. In these cases, the

problem is a dynamic problem, integrated in .

 In certain cases, the integration space is reduced to . The problem is

termed quasi-static.

Actions and Responses

       

* *

, , , t t t b x t x u x v x

     

, , , t t t u x x x  

 

not

,t  x A

ACTIONS

 

not

,t  x R

RESPONSES Mathematical model EDPs+BCs

3 

 R R

3

R

45

 A problem is said to be quasi-static if the acceleration term can

be considered to be negligible.

 This hypothesis is acceptable if actions are applied slowly. Then,

The Quasi-Static Problem

2 2

( , ) t t     u x a

2 2

/ t    0 A

2 2

/ t    0 R

2 2

( , ) t t    u x

46

 Find the displacements , strains and stresses

such that

 

2 2

,t t     u x

The Quasi-Static Problem

 

,t u x

 

,t x 

 

,t x 

   

, , t t     x b x  

   

, 2 t Tr     x    1

     

1 , , 2

S

t t      x u x u u    

* *

: :

u 

     u u t n 

   

,0 ,0   u x u x v 

Equilibrium Equation Constitutive Equation Geometric Equation Boundary Conditions in Space Initial Conditions

47

 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.

 For each value of the actions -characterized by a fixed value - a

response is obtained.

 Varying , a family of actions and its corresponding family of responses is

• btained.

The Quasi-Static Problem

     

* *

, , ,    b x t x u x

     

, , ,    u x x x  

 

not

,  x A

ACTIONS

 

not

,  x R

RESPONSES Mathematical model EDPs+BCs

*



 

*

, x A

 

*

, x R

*



48

Consider the typical material strength problem where a cantilever beam is subjected to a force at it’s tip. For a quasi-static problem, The response is , so for every time instant, it only depends on the corresponding value of .

Example

 

F t

   

 

t t    

 

t 

49

 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 generates a system of 3 eqns. for the 3 unknown components of .

 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 generates a system of 6 eqns. for the 6 unknown components of .

 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.

Solution of the Linear Elastic Problem

 

,t u x

 

,t x 

 

,t x 

 

,t u x

 

,t x 

 

,t x 

50

The aim is to reduce this system to a system with as the only unknowns. Once these are obtained, and will be found through substitution.

Displacement formulation

     

2 2

, , , t t t t        u x x b x  

   

, 2 t Tr     x    1

     

1 , , 2

S

t t      x u x u u    

Cauchy’s Equation of Motion Constitutive Equation Geometric Equation

* *

: :

u 

     u u t n 

Boundary Conditions in Space

   

,0 ,0   u x u x v 

Initial Conditions  

,t u x

 

,t x 

 

,t x 

51

 Introduce the Constitutive Equation into Cauchy’s Equation of

motion:

 Consider the following identities:

Displacement formulation

     

2 2

, , , t t t t        u x x b x  

   

, 2 t Tr     x    1

 

2 2

( ) 2 Tr t            u b     1  

 

     

11

( ) 1,2,3

k k ij ij i j j k i k i i

u u Tr x x x x x x i                                              1 u u     





u 

 

 

u

i



  

 

 

 

Tr   u     1

52

 Introduce the Constitutive Equation into Cauchy’s Equation of

motion:

 Consider the following identities:

Displacement formulation

     

2 2

, , , t t t t        u x x b x  

   

, 2 t Tr     x    1

 

2 2

( ) 2 Tr t            u b     1

 

 

     

2 2 2

1 1 1 1 1 2 2 2 2 2 1 1 1,2,3 2 2

ij j j i i i j j j i j j i j i i i

u u u u x x x x x x x x x i                                                              

u

u u u



     

 

2 i



u  



u 

 

 i





u  

2

1 1 ( ) 2 2     u u     

53

 Introduce the Constitutive Equation into Cauchy’s Equation of

Movement:

 Replacing the identities:  Then,

 The Navier Equations

are obtained:

Displacement formulation

     

2 2

, , , t t t t        u x x b x  

   

, 2 t Tr     x    1

 

2 2

( ) 2 Tr t            u b     1

 

 

 

Tr   u     1

2

1 1 ( ) 2 2     u u     

 

2 2 2

1 1 2 ( ) 2 2 t                   u u u u b     

       

2 2 2 , ,

1,2,3

j ji i jj i i

t u u b u i                            u u u b    

2nd order PDE system

54

 The boundary conditions are also rewritten in terms of :  The BCs are now:

Displacement formulation

 

,t u x

   

, 2 t Tr     x    1

* 

 t n 

 

 

*

2 Tr      t n n  





u 

 

1 2

S

    

u

u u



 

   

*

         t u n u u n   

 

* *

1,2,3

i i

u u i         u u

u



• n





   

 

 

* * . , ,

1,2,3

k k i i j j j i j i

u n u n u n t i                     u n u u n t   

• n

REMARK

The initial conditions remain the same.

55

 Navier equations in a cylindrical coordinate system:

Where:

Displacement formulation

cos ( , , ) sin x r r z y r z z             x

dV r d dr dz  

 

2 2

2 2 2

z r r

u e G G G b r r z t



                  

 

2 2

1 2 2 2

r z

u e G G G b r z r t

 

                  

   

2 2

2 2 2

r z z

u e G G G r b z r r r t



                   1 1 2

z r z

u u r z

 

                1 2

r z zr

u u z r



              

 

1 1 1 2

r z r

ru u r r r

 

               

 

1 1

z r

u u e ru r r r z



         

56

 Navier equations in a spherical coordinate system:

Where:

Displacement formulation

 

sin cos , , sin sin cos x r r y r z r                  x x

2 sin

dV r dr d d    

 

 

2 2

2 2 2 sin sin sin

r r

u e G G G b r r r t

 

                      

 

 

2 2

2 2 2 sin sin sin

r

G u e G G r b r r r r t

  

                      

   

2 2

2 2 2 sin

r

u G e G G r b r r r r t

  

                    

 

1 1 1 ω sinθ 2 sinθ θ sinθ φ

r

u u r r

  

             

 

1 1 1 ω 2 sinθ φ

r r

ru u r r r

  

               

 

1 1 1 ω 2 θ

r r

u ru r r r

  

             

 

 

 

2 2

1 sin sin sin

r

e r u ru ru r r

 

                    57

The aim is to reduce this system to a system with as the only unknowns. Once these are obtained, will be found through substitution and by integrating the geometric equations.

Stress formulation

   

, , t t     x b x  

   

1 ,t Tr E E       x    1

     

1 , , 2

S

t t      x u x u u    

Equilibrium Equation (Quasi-static problem) Inverse Constitutive Equation Geometric Equation

* *

: :

u 

     u u t n 

Boundary Conditions in Space

 

,t x 

REMARK

For the quasi-static problem, the time variable plays the role of a loading factor.

 

,t u x

 

,t x 

58

 Taking the geometric equation and, through successive derivations,

the displacements are eliminated:

 Introducing the inverse constitutive equation into the compatibility

equations and using the equilibrium equation:

 The Beltrami-Michell Equations are obtained:

Stress formulation

Compatibility Equations (seen in Ch.3.)

 

2 2 2 2

, , , 1,2,3

ij jl kl ik k l i j j l i k

i j k l x x x x x x x x                     

1

ij pp ij ij

E E          

ij j i

b x      

   

 

 

2 2

1 , 1,2,3 1 1

j k i kk ij ij i j k j i

b b b i j x x x x x                            2nd order PDE system

59

 The boundary conditions are:  Equilibrium Equations:

This is a 1st order PDE system, so they can act as boundary conditions of the (2nd order PDE system of the) Beltrami-Michell Equations

 Prescribed stresses on :

Stress formulation

    b  





* 

   n t 

• n

60

 Once the stress field is known, the strain field is found by substitution.  The calculation, after, of the displacement field requires that the

geometric equations be integrated with the prescribed displacements

• n :

Stress formulation

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.  

*

1 ( ) ( ) ( ) 2 ( ) ( )

u

V         x u x u x x u x u x x   

u



   

1 ,t Tr E E       x    1

61

 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.”

 Expressed in another way:

“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.”

Saint-Venant’s Principle

REMARK

This principle does not have a rigorous mathematical proof.            

(I) (II) (I) (II) (I) (II)

, , , , , ,

P P P P P P

t t t t t t    u x u x x x x x    

| 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

 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

isotropic problems but also for orthotropic and anisotropic ones.

Uniqueness of the solution

C

64

65

Ch.6. Linear Elasticity

6.5 Linear Thermoelasticity

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.

• 3. Isentropic and adiabatic processes – no longer isothermal!!!



Isentropic - entropy of the system remains constant



Adiabatic - occurs without heat transfer

       

, , t t     x x x x    

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: We will assume the temperature field is known.



But the process is still isentropic and adiabatic:

 

s t cnt 

e

Q

V V

r dV dS V V 



     

  q n

r t       q x   s  

heat conduction from the exterior internal fonts

       

, ,0 , ,

not

t t t t            x x x x 

67

Generalized Hooke’s Law

 The Generalized Hooke’s Law becomes:

Where



is the elastic constitutive tensor.



is the absolute temperature field.



is the temperature at the reference state.



is the tensor of thermal properties or constitutive thermal constants tensor.

 It is a semi-positive definite symmetric second-order tensor.

           

, : , : , , 1,2,3

ij ijkl kl ij

t t t i j                        x x x C C C C      C

Generalized Hooke’s Law for linear thermoelastic problems

 

,t  x

 

,t    x



68

 An isotropic thermoelastic material must have the same elastic

and thermal properties in all directions:



must be a (mathematically) isotropic 4th order tensor: Where:



is the 4th order symmetric unit tensor defined as



and are the Lamé parameters or coefficients.



is a (mathematically) isotropic 2nd order tensor:

Where:



is a scalar thermal constant parameter.

Isotropic Constitutive Constants Tensors

 

 

2 , , . 1,2,3

ijkl ij kl ik jl il jk

i j k l                      I C 1 1 C



 

1 2

ik jl il jk ijkl

          I

I



 

 

, 1,2,3

ij ij i j

             1



69

 Introducing the isotropic constitutive constants tensors and

into the generalized Hooke’s Law,

(in indicial notation)

 The resulting constitutive equation is,

Isotropic Linear Thermoelastic Constitutive Equation

 

 

 

   

1 1 2 2 2

ij ijkl kl ij ij kl ik jl il jk kl ij ij kl kl ik jl kl il jk kl ij

                                                   C

2      I C 1 1

   

2 2 , 1,2,3

ij ij ll ij ij

Tr i j                                   1 1

Isotropic linear thermoelastic constitutive equation.

ll





ji ij

 

 

ij





ij





   1

 

:      C   

  

70

1. is isolated from the Generalized Hooke’s Law for linear thermoelastic problems: 2. The thermal expansion coefficients tensor is defined as: 3. The inverse constitutive equation is obtained:

Inversion of the Constitutive Equation





:        C

1 1

: : 

 

   C C C C   

1 def 

 : C  

It is a 2nd order symmetric tensor which involves 6 thermal expansion coefficients

1 :





   C   

71

 For the isotropic case:  The inverse const. eq. is re-written:

 Where is a scalar thermal expansion coefficient related to the

scalar thermal constant parameter through:

Inverse Isotropic Linear Thermoelastic Constitutive Equation

   

1 1 , 1,2,3

ij ll ij ij ij

Tr E E i j E E                                     1 1

Inverse isotropic linear thermo elastic constitutive equation.

 

1 2 E     

72

 

 

1 1

1 1 2 ( 1 , , . 1,2,3

ijkl ij kl ik jl il jk

E E E i j k l E E             

 

                     I : C C 1 1 1) C  

 Comparing the constitutive equations,

the decomposition is made: Where:



is the non-thermal stress: the stress produced if there is no thermal phenomena.



is the thermal stress: the “corrector” stress due to the temperature increment.

Thermal Stress

 

2 Tr            1 1

 

2 Tr        1

Isotropic linear thermoelastic constitutive equation. Isotropic linear elastic constitutive equation.

nt

 

t

 

nt t

    

nt



t



73

 Similarly, by comparing the inverse constitutive equations,

the decomposition is made: Where:



is the non-thermal strain: the strain produced if there is no thermal phenomena.



is the thermal strain: the “corrector” strain due to the temperature increment.

Thermal Strain

 

1 Tr E E              1 1

 

1 Tr E E          1

Inverse isotropic linear thermoelastic constitutive eq. Inverse isotropic linear elastic constitutive eq.

nt

 

t

 

nt t

    

nt



t



74

 The thermal components appear when thermal processes are considered.

TOTAL NON-THERMAL COMPONENT THERMAL COMPONENT

Thermal Stress and Strain

nt t

    

nt t

    

nt 

: C  

1 nt 

 : C  

t

    

t

    

( ) 2

nt

Tr        1

t

     1 1 ( )

nt

Tr E E          1

t

     1

Isotropic material: Isotropic material: Isotropic material: Isotropic material: : :

nt t

       C C C C    

1 1 nt t  

       : : C C C C     These are the equations used in FEM codes.

75

Thermal Stress and Strain

REMARK 1

In thermoelastic problems, a state of zero strain in a body does not necessarily imply zero stress.

REMARK 2

In thermoelastic problems, a state of zero stress in a body does not necessarily imply zero strain.

nt

    

t

        0   1

nt

    

t

      0   1

76

77

Ch.6. Linear Elasticity

6.6 Thermal Analogies

 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:

 1st thermal analogy – Duhamel-Neumann analogy.  2nd thermal analogy

Solution to the Linear Thermoelastic Problem

78

1st Thermal Analogy

 The governing eqns. of the quasi-static isotropic linear thermoelastic

problem are:

   

, , t t     x b x  

   

, , t t      x : x C   1

     

1 , , 2

S

t t      x u x u u     Equilibrium Equation Constitutive Equation Geometric Equation

* *

: :

u 

     u u t n 

Boundary Conditions in Space

79

1st Thermal Analogy

 The actions and responses of the problem are:

       

* *

, , , , t t t t   b x t x u x x

     

, , , t t t u x x x  

  



not

,

I

t  x A

ACTIONS

  



not

,

I

t  x R

RESPONSES Mathematical model EDPs+BCs

REMARK

is known a priori, i.e., it is independent of the mechanical response. This is an uncoupled thermoelastic problem.  

,t   x

80

1st 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

equations:

 Momentum equations

nt t

    



 

nt t nt t nt

 

 

       



                1

 

ˆ 1 ˆ

nt

         b b b    

 

1

nt

                  b b       ˆ

not

 b

81

1st Thermal Analogy

 Boundary equations:

ANALOGOUS PROBLEM – A linear elastic problem can be solved as:

0ˆ nt

    b   ( ) 2

nt

Tr      : C     1

     

1 , , 2

S

t t      x u x u u     Equilibrium Equation Constitutive Equation Geometric Equation

1 ˆ ( )       b b  with

* *

: ˆ :

u nt 

     u u n t 

Boundary Conditions in Space

* *

ˆ      t t n

with



* * *

ˆ

( )

nt t

 

 

      

 

t

n t n t n

n

       1

* * *

ˆ ˆ ( )

nt

       n t t t n 



 

* nt t

    n t    

* nt t

    n n t  

82

1st Thermal Analogy

 The actions and responses of the ANALOGOUS NON-THERMAL PROBLEM are:

     

* *

ˆ , ˆ , , t t t b x t x u x

     

, , ,

nt

t t t u x x x  

  



not

,

II

t  x A

ACTIONS

  



not

,

II

t  x R

RESPONSES Mathematical model EDPs+BCs ANALOGOUS (ELASTIC) PROBLEM (II) ORIGINAL PROBLEM (I)

83

Responses are proven to be the solution of a thermoelastic problem under actions

 If the actions and responses of the original and analogous problems are

compared:

1st Thermal Analogy

     

( , ) ( , )

def I II III nt nt

t t                                                    u u x x R R R R R       1

RESPONSES ACTIONS

   

   

 

* * * *

1 ˆ ˆ ( , ) ( , ) ˆ ˆ

def I II III

t t        

 

                                                                          b b b b u u x x t t t t n A A A A A 

t

 



 b

*

 t



84

1st Thermal Analogy

 The solution of the ORIGINAL PROBLEM is:

ANALOGOUS PROBLEM (II) ORIGINAL PROBLEM (I) TRIVIAL PROBLEM (III)

           

I II I II I II

       u u     1

85

 

t

         u    1

1st Thermal Analogy

ANALOGOUS ELASTIC PROBLEM (II) THERMOELASTIC ORIGINAL PROBLEM (I) THERMOELASTIC TRIVIAL PROBLEM (III)

       

* *

, , , , t t t t   b x t x u x x

     

, , , t t t u x x x  

  



,

I

t x A

  



,

I

t x R

  



,

II

t x A

  



,

II

t x R

  



,

III

t x A

  



,

III

t x R

         

* * *

1 ˆ , ˆ , , t t t               b x b t x t n u x 

     

, , ,

nt

t t t u x x x  

     

*

1 ,t      



       b t n u x    

86

2nd Thermal Analogy

 The governing eqns. of the quasi-static isotropic linear thermoelastic

problem are:

   

, , t t     x b x  

   

• 1

, , t t      x : x C   1

     

1 , , 2

S

t t      x u x u u     Equilibrium Equation Inverse Constitutive Equation Geometric Equation

* *

: :

u 

     u u t n 

Boundary Conditions in Space

87

2nd Thermal Analogy

 The actions and responses of the problem are:

       

* *

, , , , t t t t   b x t x u x x

     

, , , t t t u x x x  

  



not

,

I

t  x A

ACTIONS

  



not

,

I

t  x R

RESPONSES Mathematical model EDPs+BCs

REMARK

is known a priori, i.e., it is independent of the mechanical response. This is an uncoupled thermoelastic problem.  

,t   x

88

 If the strain field is integrable, there exists a field of thermal

displacements which satisfies:

 Then, the total displacement field is decomposed as:

2nd Thermal Analogy

 

,

t

t u x

         

1 , 2 1 , 1,2,3 2

t S t t t t t j i t ij ij j i

t u u i j x x                               x u u u     1

The hypothesis is made that and are such that the strain field is integrable (satisfies the compatibility equations).

( , ) t   x ( )  x

( , ) ( , ) ( , )

def nt t

t t t   u x u x u x

REMARK

The solution is determined except for a rigid body movement characterized by a rotation tensor and a displacement vector . The family of admissible solutions is . This movement can be arbitrarily chosen.

   

*

, ,

t

t t



    u x u x x c  

 

,

t

t u x





*

c

nt t

  u u u

89

2nd 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

and replaced into the governing equations:

 Geometric equations:  Boundary equations:

nt t

    

nt t

  u u u



t t

( )

S S nt t S nt S t S nt t nt

         u u u u u u          



* nt t

  u u u

nt S nt

 u  

* *

:

nt t u

    u u u u u

* nt t

   u u u u u

90

2nd Thermal Analogy

ANALOGOUS PROBLEM – A linear elastic problem can be solved as:

    b  

• 1

nt 

: C  

nt S nt

 u   Equilibrium Equation Constitutive Equation Geometric Equation

* *

: :

nt t u 

      u u u n t 

Boundary Conditions in space

91

 The actions and responses of the ANALOGOUS PROBLEM are:

       

* *

, , , ,

nt t

t t t t   b x t x u u x u x

2nd Thermal Analogy

     

, , ,

nt nt

t t t u x x x  

  



not

,

II

t  x A

ACTIONS

  



not

,

II

t  x R

RESPONSES Mathematical model EDPs+BCs ANALOGOUS PROBLEM (II) ORIGINAL PROBLEM (I)

92

 If the actions and responses of the original and analogous problems are

compared:

2nd Thermal Analogy

     

( , ) ( , )

nt t t def I II III nt t

t t                                                  u u u u x x R R R R R      1

RESPONSES ACTIONS

     

* *

( , ) ( , )

t t def I II III

t t  

 

                                                  b b u u u u x x t t A A A A A

Responses are proven to be the solution of a thermo-elastic problem under actions

93

2nd Thermal Analogy

 The solution of the ORIGINAL PROBLEM is:

ANALOGOUS PROBLEM (II) ORIGINAL PROBLEM (I) TRIVIAL PROBLEM (III)

           

I II t I II I II

        u u u     1

94

   

,

t t

t        u u x    1

2nd Thermal Analogy

ANALOGOUS ELASTIC PROBLEM (II) THERMOELASTIC ORIGINAL PROBLEM (I) THERMOELASTIC TRIVIAL PROBLEM (III)

       

* *

, , , , t t t t   b x t x u x x

     

, , , t t t u x x x  

  



,

I

t x A

  



,

I

t x R

  



,

II

t x A

  



,

II

t x R

  



,

III

t x A

  



,

III

t x R

     

, , ,

nt nt

t t t u x x x  

   

*

, ,

t

t t 



    b u u x t x   

* t





    b u u u t

95

2nd Analogy in structural analysis

:

t x x t x x t u x x x l

u u x u u l        



         



*

:

nt x x nt x x t u x x x x l

u u u u u l    



       

96

 Although the 2nd analogy is more common, the 1st analogy

requires less corrections.

 The 2nd analogy can only be applied if the thermal strain field

is integrable.

 It is also recommended that the integration be simple.  The particular case

 Homogeneous material:  Lineal thermal increment:

is of special interest because the thermal strains are: and trivially satisfy the compatibility conditions (involving second

• rder derivatives).

Thermal Analogies

( ) cnt     x ax by cz d      

t

     1 linear polinomial

97

 In the particular case

 Homogeneous material:  Constant thermal increment:

the integration of the strain field has a trivial solution because the thermal strains are constant , therefore: The thermal displacement is:

Thermal Analogies

( ) cnt     x ( ) cnt       x

t

cnt      

 

,

t

t  

 

     u x x x c 

rigid body movement (can be chosen arbitrarily)

 

,

t

t     u x x

 

1

t

           x u x x x

HOMOTHECY (free thermal expansion)

98

99

Ch.6. Linear Elasticity

6.7 Superposition Principle

Linear Thermoelastic Problem

 The governing eqns. of the isotropic linear thermoelastic problem are:

   

, , t t     x b x  

   

, , t t      x : x C   1

     

1 , , 2

S

t t      x u x u u     Equilibrium Equation Constitutive Equation Geometric Equation

* *

: :

u 

     u u t n 

Boundary Conditions in space

   

,0 ,0   u x u x v 

Initial Conditions

100

Linear Thermoelastic Problem

 Consider two possible systems of actions:

and their responses :

  



  



  



  



   

1 1 * 1 * 1 1

, , , , t t t t   b x t x u x x v x

  



  



  



1 1 1

, , , t t t u x x x  

  



1

,t  x A

  



1

,t  x R

  



  



  



  



   

2 2 * 2 * 2 2

, , , , t t t t   b x t x u x x v x

  



2

,t  x A

  



  



  



2 2 2

, , , t t t u x x x  

  



2

,t  x R

101

 The solution to the system of actions where

and are two given scalar values, is .

 This can be proven by simple substitution of the linear

combination of actions and responses into the governing equations and boundary conditions, as shown in pp. 210-211 of the course book.

 When dealing with non-linear problems (plasticity, finite

deformations, etc), this principle is no longer valid.

The net response to the lineal thermoelastic problem caused by two or more groups of actions is the lineal combination of the responses which would have been caused by each action individually.

Superposition Principle

         

3 1 1 2 2

λ λ   A A A A A

         

3 1 1 2 2

λ λ   R R R R R

 

1

λ

 

2

λ

102

103

Ch.6. Linear Elasticity

6.8 Hooke’s Law in Voigt Notation

 Taking into account the symmetry of the stress and strain tensors,

these can be written in vector form:

Stress and Strain Vectors

x xy xz xy y yz xz yz z

                    

.

1 1 2 2 1 1 2 2 1 1 2 2

x xy xz x xy xz not xy y yz xy y yz xz yz z xz yz z

                                                

 

6 x y def z xy xz yz

                              R 

 

6 x y def z xy xz yz

                              R 

REMARK

The double contraction is transformed into the scalar (dot) product :

 

  

   

 

  

   

      

ij ij i i

    

2nd order tensors vectors

VOIGT NOTATION

104

 The inverse constitutive equation is rewritten:

Where is an elastic constants inverse matrix and is a thermal strain vector:

Inverse Constitutive Equation

 

1 Tr E E              1 1

     

1

ˆ

t 

   C   

1

1 1 1 ˆ 1 1 1 E E E E E E E E E G G G      



                                         C

1

ˆ  C

 

t



t

                    

 

t

                            

105

 By inverting the inverse constitutive equation, Hooke’s Law in terms of the

stress and strain vectors is obtained: Where is an elastic constants matrix :

Hooke’s Law

ˆ C

     

 

ˆ

t

   C   

          

1 1 1 1 1 1 1 1 1 1 ˆ 1 2 1 1 2 2 1 1 2 2 1 1 2 2 1 E                                                                          C

106

107

Ch.6. Linear Elasticity

Summary

 The simplifying hypothesis of the Theory of Linear Elasticity are:

1. Infinitesimal strains and deformation framework 2. Existence of an unstrained and unstressed reference state 3. Isothermal, isentropic and adiabatic processes

 Constitutive eq. of a linear elastic material - Generalized Hooke’s Law:  The internal energy density defines a potential for the stress tensor, it is an

elastic potential.

Summary

   

, , t t  x : x C  

ijkl jikl ijkl ijlk

  C C C C

minor symmetries

ijkl klij

 C C

major symmetries

The 4th order constitutive elastic constants tensor has 34=81 components.  

ˆ 1 2 du d dt dt   : : :  C    

 

ˆ u      

108

 Isotropic elastic material:

Where:



is the 4th order unit tensor defined as



and are the Lamé parameters.

 Constitutive eq. of an isotropic linear elastic material:

Summary (cont’d)

2      I C 1 1  

1 2

ik jl il jk ijkl

          I

I 



 

2 Tr        1

 

 

 

   

 

1 1 1 1 1 1

x x y z xy xy y y x z xz xz z z x y yz yz

E G E G E G                                 

 

1 Tr E E          1

Inverse equation:

 

3 2 E        

 

2      

Young’s Modulus Poisson’s ratio In engineering notation:

109

 The stress and strain tensors can be split into a spherical or volumetric

part and a deviator part:

 Limits in the elastic properties for isotropic linear elastic materials:

Summary (cont’d)

m

      1

1 ´ 3 e     1

2 3 3(1 2 )

def

E K       

m

K e  

2G     

 

2

1 ˆ ´ ´ 0 2 u K e     :   

K   

volumetric deformation modulus Lamé’s second parameter E  Young’s modulus 1 2    Poisson’s ratio

110

 The Linear Elastic Problem:

 Dynamic problem - actions and responses depend on time.  Quasi-static problem - if actions are applied slowly, the acceleration can be

considered to be negligible. The time variable disappears.

Summary (cont’d)

       

* *

, , , t t t b x t x u x v x

     

, , , t t t u x x x  

 

not

,t  x A ACTIONS

 

not

,t  x R

RESPONSES Mathematical model EDPs+BCs

111

 The Isotropic Linear Elastic Problem – displacement formulation:

The aim is to reduce this system to a system with as the only unknowns. Once these are obtained, and will be found through substitution.

Summary (cont’d)

     

2 2

, , , t t t t        u x x b x  

   

, 2 t Tr     x    1

     

1 , , 2

S

t t      x u x u u    

Cauchy’s Equation of Motion Constitutive Equation Geometric Equation

* *

: :

u 

     u u t n 

Boundary Conditions in Space

   

,0 ,0   u x u x v 

Initial Conditions

 

,t u x

 

,t x 

 

,t x 

   

2 2 2

t             u u u b   

*

 u u

u



• n





   

*

         u n u u n t   

• n

Navier Equations BCs

112

 The Isotropic Linear Elastic Problem – stress formulation:

The aim is to reduce this system to a system with as the only unknowns. Once these are obtained, will be found through substitution and by integrating the geometric equations.

Summary (cont’d)

   

, , t t     x b x  

   

1 ,t Tr E E       x    1

     

1 , , 2

S

t t      x u x u u    

Equilibrium Equation Inverse Constitutive Equation Geometric Equation

 

,t u x

 

,t x 

 

,t x  Boundary Conditions in Space

* *

: :

u 

     u u t n 

Beltrami-Michell Equations BCs    

 

2 2

1 1 1

j k i kk ij ij i j k j i

b b b x x x x x                          

    b  

* 

   n t 

• n

113

 The solution of the lineal elastic problem is unique. This can be proven by

Reductio ad absurdum.

 Saint Venant’s Principle:

“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.”

Summary (cont’d)

           

(I) (II) (I) (II) (I) (II)

, , , , , ,

P P P P P P

t t t t t t    u x u x x x x x    

| P    

114

 In the Theory of Linear Thermoelasticity the process is no longer

isothermal:

 The Generalized Hooke’s Law becomes:

Where



is the temperature field.



is the temperature distribution in the reference state.



is the tensor of thermal properties or constitutive thermal constants tensor.

 Isotropic thermoelastic material:

Summary (cont’d)

       

, , ,0 ,

not

t t t t            x x x x 

     

 

, , , t t t      x : x x C   

 

,t  x

 

,t    x



2      I C 1 1    1

115

 Constitutive eq. of an isotropic linear thermoelastic material:  Thermal stress and strain:

Summary (cont’d)

 

2 Tr            1 1

Inverse equation: Lamé parameters

 

1 Tr E E              1 1     

1 1 2 2 1 E E G            

With the thermal expansion coefficient:

1 2 E     

TOTAL NON-THERMAL COMPONENT THERMAL COMPONENT

nt t

    

nt t

     :

nt  C

 

1 : nt 

 C  

t

    

t

    

( ) 2

nt

Tr        1

t

     1

1 ( )

nt

Tr E E          1

t

     1

Isotropic material: Isotropic material: Isotropic material: Isotropic material:

116

 To solve the isotropic linear thermoelastic problem 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 and Stress Formulations).

 Although the 2nd analogy is more common, the 1st analogy requires less

corrections.

 The 2nd analogy can only be applied if the strain field is integrable.

Summary (cont’d)

117

1st Thermal Analogy - Duhamel-Neumann analogy

Summary (cont’d)

ANALOGOUS ELASTIC PROBLEM (II) THERMOELASTIC ORIGINAL PROBLEM (I) THERMOELASTIC TRIVIAL PROBLEM (III)

       

* *

, , , , t t t t   b x t x u x x

     

, , , t t t u x x x  

  



,

I

t x A

  



,

I

t x R

  



,

II

t x A

  



,

II

t x R

  



,

III

t x A

  



,

III

t x R

         

* * *

1 ˆ , ˆ , , t t t               b x b t x t n u x 

     

, , ,

nt

t t t u x x x  

     

*

1 ,t      



       b u t n x    

 

       u   1

118

ANALOGOUS ELASTIC PROBLEM (II) THERMOELASTIC TRIVIAL PROBLEM (III)

Summary (cont’d)

THERMOELASTIC ORIGINAL PROBLEM (I)

       

* *

, , , , t t t t   b x t x u x x

     

, , , t t t u x x x  

  



,

I

t x A

  



,

I

t x R

  



,

II

t x A

  



,

II

t x R

  



,

III

t x A

  



,

III

t x R

     

, , ,

nt nt

t t t u x x x  

   

*

, ,

t

t t 



    b u u x t x   

   

,

t

t        u u x   1

 

*

,

t

t 



   b u u u t x

2nd Thermal Analogy

119

 Superposition Principle:

“The net response to the lineal thermoelastic problem caused by two or more groups of actions is the lineal combination of the responses which would have been caused by each action individually.”

 Hooke’s Law in Voigt Notation:

Summary (cont’d)

 

6 x y def z xy xz yz

                              R 

 

6 x y def z xy xz yz

                              R 

     

1

ˆ

t 

   C   

     

 

ˆ

t

   C   

Inverse equation:

120