CH.6. LINEAR ELASTICITYmmc.rmee.upc.edu/documents/Slides/Ch6_v17.pdfin the Theory of Linear Elasticity. It constitutes the constitutive equation of a linear elastic material. The 4th

CH.6. LINEAR ELASTICITYContinuum 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

0 0

0 0

,

,

t

t

x x 0

x x 0

7


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


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

xF 1X

0 t t F

x X

F 1

x X u0

9


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


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

0 0

0 0

,

,

t

t

x x 0

x x 0


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.

0 0, ,t t x x x x

( , ) ( ) 0dss t sdt

X X

e 0Q 0V V

r dV dS V V

q n

0 0r t q x

0

0s

heat conduction from the exterior

internal fonts

12

13


6.2 Linear Elastic Constitutive Equation

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

Fl

F k l F lEA 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, 2,3ij ijkl kl

t t

i j

x x : x C

CGeneralized Hooke’s Law

ijkl jikl

ijkl ijlk

C CC C

minor symmetries

ijkl klijC Cmajor 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

0 :d u rdt

q

global form

local form

00 0

V V V V

d d uu dV dV dV r dVdt dt

: d q

qru

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 tt

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

0ˆ( , ) ˆ( ) :

ˆ

d du tu udt dt

u

x

u

)

:

: :

ˆ 1 ( )2

1 1( )2 2

12

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

jklij ijkl kl

i

i kj l

ddt

dudt

ddt

:

C

C

C

CC

C C C

C C C C

C

REMARK The symmetries of the constitutive elastic constants tensor are used:

ijkl jikl

ijkl ijlk

C CC C

minor symmetries


ˆ 12

du ddt 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 ddt dt

C

ˆˆ ˆ, , , :V V V

d d du t dV u t dV t dVdt 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

0 01 , , ( ) 02

t t a a x : : x x x xC

1ˆ , , ,2

u t t t a x x : : x xC

0ˆ , 0u t x x

0

1ˆ : : 02

u 0C

1 1ˆ : : ( ) :2 2

u C

19

ˆ 1: : :2

du ddt 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 îjkl

ij kl

u

C

u

2 ˆ :( ) u

C

C

ˆ( ( , )) 1 1 1 1( : : : : ( )2 2 2 2

u t

x

C C C

T

20

21


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, 2,3ijkl ij kl ik jl il jk i j k l

IC

C

1 1

CC

C

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

12 ik jl il jkijkl II

22

Introducing the isotropic constitutive elastic constants tensorinto the generalized Hooke’s Law ,

(in indicial notation)

The resulting constitutive equation is,

Isotropic Linear Elastic Constitutive Equation

1 12 ( ) 22 2

ij ijkl kl ij kl ik jl il jk kl

ij kl kl ik jl kl il jk kl ij ijTr

C

2 IC 1 1 :C

2

2 , 1, 2,3ij ij ll ij

Tr

i j

1 Isotropic linear elastic constitutive equation.

Hooke’s Law

( )ll Tr ji ij

ij

12

12 i jij j i

23

If the constitutive equation is,

Then, the internal energy density can be reduced to:

Elastic Potential

2

2 , 1, 2,3ij ij ll ij

Tr

i j

1 Isotropic linear elastic constitutive equation.

Hooke’s Law

2

1 1ˆ : 2 :2 2

1 1: 22 2

12

Tr

u Tr

Tr

Tr

σ

1

1

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

ˆ

2u

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 2Tr Tr Tr Tr Tr Tr Tr 1 1

13 2

Tr Tr

2Tr 1 12

Tr

1

1 12 3 2

Tr

1 1

2 3 2 2Tr

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 , 1,2,3ij ll ij ij

TrE E

i jE E

1

E

Inverse isotropic linear elastic constitutive equation.

Inverse Hooke’s Law.

3 2

2

E

1 1 2

2 1

E

EG

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, of 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 2E

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 obtained for a Poisson’s ratio of and .

x x

x

y

z

xx

00

x

y z xy xz yz

E,

0 0.5

28

Solution

For :

For :

0 1 0

0

0

x x xy

y x xz

z x yz

E

E

E

1 0

0 00 0

x x xy

y xz

z yz

E

0.5

1 0

0.5 0

0.5 0

x x xy

y x xz

z x yz

E

E

E

1 0

1 021 0

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 0x 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

00

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

1:3sph m Tr 1 1

dev m 1 m 1

1 ´3

e 1

1 1 Tr (3 3sph e 1 1

1dev3

e 1

30

Operating on the volumetric strain:

The spherical parts of the stress and strain tensor are directly related:


Tre

1TrE E

1

1Tr Tr TreE E

1

3 1 2me

E

3 3 m

3 1 2mE e

: volumetric deformation modulusK

23 3(1 2 )

def EK

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 2mE e

1TrE E

1

1

1 1 1 3 1

m m

m m m

TrE E

Tr TrE E E E E E E

1 1 1

1 1 1 1 13

1 2 1 1 1 1

3 1 2 3E e e

E E E

1 1

0

2 2 , {1,2,3}ij ijG G i j

1 ´3

e 1

Comparing this with the expression

1É

1 1 1E 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:


2ij ijG 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

1ˆ : :2

u C ˆ:

u

u

2 2

00 0

ˆ û u

C C

u

0

0

ˆ= : 0

u

C

There is a minimum for : 0

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 1ˆ2 2: : :u C

21 1: : :2 2

Tr Tr 1

: : : : : C C

1 22

Tr : 1

Tr e e

2

2

03

1 1´ ´3 31 2 ´ ´ ´9 3

1 ´ ´3

Tr

e e

e e

e

:

: : :

:

1 1

1 1 1

2 2 21 1 1 2ˆ : ´ : ´2 3 2 3

u e e μ e μ

K

21ˆ : ´ 02

u K e Elastic potential in terms of the spherical and deviatoric parts of 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

21ˆ : : ´ 02

u K e ´

1

1

13

e

0

1 1 21ˆ 02

u K e 0K

2

2e

0

2ˆ ´ ´ 0u : 0

: 0ij 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

0

3 1 2EK

02 1

EG

0

2 10

E

0E Young’s

modulus

0

3 1 20

E

E

102

Poisson’s ratio

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

37

38


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 obtaining the evolution through time of the corresponding displacements , strains and stresses .

Introduction

0t

,0

,0

b x

t x

,

,

t

t

b x

t x

Initial actions:

Actions through time:

, tu x , tx , tx

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

0 0 2

,, ,

tt t

t

u x

x b x

, 2t Tr x 1

1, ,2

St t x u x u u

This is a PDE system of 15 eqns -15 unknowns:

Which must be solved in the space.

, tu x , tx

, tx

3 unknowns

6 unknowns

6 unknowns

3R 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 tt

u t u t i

u x u xx

x x

*

*

( , ) ( , ), , 1,2,3ij j j

t tt

t n t t i

x n xx

x x t

*

*

, ,, , 1, 2,3

, ,i i

ujk k j

u t u ti j k i j t

t n t t

x xx

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

0t

,0 V u x 0 x

00

,,0

not

t

tV

t

u x

u x v x x

43

Find the displacements , strains and stresses such that

The Linear Elastic Problem

, tu x , tx , tx

2

0 0 2

,, ,

tt t

t

u x

x b x

, 2t Tr x 1

1, ,2

St 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

, 0

u x 0

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

*

*

0

,,,

ttt

b xt xu xv x

,,,

ttt

u xxx

not

, t xAACTIONS

not, t xRRESPONSES

Mathematical model

EDPs+BCs

3R R

3R

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,


2

2

( , )tt

u xa 0

2 2/ t 0A 2 2/ t 0R2

2

( , )tt

u x 0

46

Find the displacements , strains and stresses such that 2

0 2

, tt

u x

0


, tu x , tx , tx

0, ,t t x b x 0

, 2t Tr x 1

1, ,2

St t x u x u u

*

*

:

:u

u u

t n

0

,0

,0

u x 0

u x v

Equilibrium 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 obtained.


*

*

,,,

b xt xu x

,,,

u xxx

not

, xAACTIONS not

, xRRESPONSES

Mathematical model

EDPs+BCs

* *,xA *,xR

*

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

, tu x , tx , tx

, tu x , tx

, tx

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

0 0 2

,, ,

tt t

t

u x

x b x

, 2t Tr x 1

1, ,2

St t x u x u u



Geometric Equation

*

*

:

:u

u u

t n Boundary Conditions in Space

0

,0

,0

u x 0

u x vInitial Conditions

, tu x , tx , tx

51

Introduce the Constitutive Equation into Cauchy’s Equation of motion:

Consider the following identities:


2

0 0 2

,, ,

tt t

t

u x

x b x

, 2t Tr x 1

2

0 0 2( ) 2Trt

ub 1

11( )

1,2,3

k kij iji

j j k i k i

i

u uTrx 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:


2

0 0 2

,, ,

tt t

t

u x

x b x

, 2t Tr x 1

2

0 0 2( ) 2Trt

ub 1

2

2

21 1 1 1 12 2 2 2 2

1 1 1, 2,32 2

ij j ji ii

j j j i j j i j i

i

i

u uu ux x x x x x x x x

i

u u

u u

2

i u

u

i u

21 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:


2

0 0 2

,, ,

tt t

t

u x

x b x

, 2t Tr x 1

2

0 0 2( ) 2Trt

ub 1

Tr u 1 21 1( )2 2

u u

2

20 0 2

1 12 ( )2 2 t

uu u u b

22

0 0 2

, , 0 0 1, 2,3j ji i jj i i

tu u b u i

uu u b

2nd order PDE system

54

The boundary conditions are also rewritten in terms of :

The BCs are now:


, tu x

, 2t Tr x 1

* t n * 2Tr t n n

u

12

S u u u

* t u n u u n

*

* 1, 2,3i iu u i

u uuon

*

*. , , 1, 2,3k k i i j j j i j iu n u n u n t i

u n u u n t on

REMARK The initial conditions remain the same.

55

Navier equations in a cylindrical coordinate system:

Where:


cos ( , , ) sin

x rr z y r

z z

x

dV r d dr dz

2

2

22 2z rr

ue GG G br r z t

2

2

12 2 2r z ueG G G br z r t

2

2

2 22 r zz

ue G GG r bz r r r t

1 12

zr z

uur z

12

r zzr

u uz r

1 1 12

rz r

ru ur r r

1 1 z

ru ue ru

r r r z

56

Navier equations in a spherical coordinate system:

Where:


sin cos

, , sin sin cos

x rr y r

z r

x x

2 sin dV r dr d d

2

2

2 22 sinsin sin

rr

ue G GG br r r t

2

2

2 2 2 sinsin sin

rG ue G G r br r r r t

2

2

2 2 2sin

r uG e G Gr br r r r t

1 1 1ω sinθ2 sinθ θ sinθ φr

uur r

1 1 1ω2 sinθ φ

rr

ruur r r

1 1 1ω2 θ

rr

urur r r

22

1 sin sinsin re 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

0, , 0t t x b x

1, t TrE E

x 1

1, ,2

St t x u x u u

Equilibrium Equation(Quasi-static problem)


Geometric Equation

*

*

:

:u

u u


, tx

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

, tu x , tx

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 22 2

0 , , , 1, 2,3ij jlkl ik

k l i j j l i k

i j k lx x x x x x x x

1ij pp ij ijE E

0 0ijj

i

bx

2

00 02 1 , 1, 2,31 1

jk ikkij ij

i j k j i

bb bi 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

0 0 b

*

n t on

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 on :

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 TrE 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.”


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 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


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

0 0

0 0

,

,

t

t

x x 0

x x 0

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

eQ 0V V

r dV dS V V

q n

0r t q x

0s

heat conduction from the exterior

internal fonts

0, ,0

,, 0

nott

tt

t

x x

xx

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.

0

0

, : , : ,

, 1, 2,3ij ijkl kl ij

t t t

i j

x x xC C

CGeneralized Hooke’s Law for linear thermoelastic problems

, t x 0 0, 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, 2,3ijkl ij kl ik jl il jk i j k l

IC 1 1

C


, 1, 2,3ij ij i j

1

69

Introducing the isotropic constitutive constants tensors andinto the generalized Hooke’s Law,

(in indicial notation)

The resulting constitutive equation is,

Isotropic Linear Thermoelastic Constitutive Equation

0 0

01 122 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 IC 1 1

2

2 , 1,2,3ij ij ll ij ij

Tr

i j

1 1 Isotropic linear thermoelasticconstitutive equation.

ll ji ij

ij

ij

1 0: 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

1def

: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 , 1,2,3ij ll ij ij ij

TrE E

i jE E

1 1 Inverse isotropic linear thermo elastic constitutive equation.

1 2E

72

1

1

11 2(

1 , , . 1,2,3ijkl ij kl ik jl il jk

E EEi j k l

E E

I:

CC

1 11)

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

2Tr 1 1

2Tr 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

1TrE E

1 1

1TrE E

1

Inverse isotropic linear thermoelasticconstitutive eq.

Inverse isotropic linear elasticconstitutive eq.

nt t

nt t

nt

t

74

The thermal components appear when thermal processes are considered.

TOTALNON-THERMAL COMPONENT

THERMAL COMPONENT


nt t

nt t

nt :C

1nt :C

t

t

( ) 2nt Tr 1 t 1

1( )nt TrE E

1 t 1

Isotropic material: Isotropic material:


: :nt t C C 1 1nt t : :C C

These are the equations used in FEM codes.

75


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 0 0

t 0 1

nt 0 0

t 0 1

76

77


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 ThermoelasticProblem

78

1st Thermal Analogy

The governing eqns. of the quasi-static isotropic linear thermoelastic problem are:

0, ,t t x b x 0

, ,t t x : xC 1

1, ,2

St t x u x u u



Geometric Equation

*

*

:

:u

u u


79

1st Thermal Analogy

The actions and responses of the problem are:

*

*

,,,

,

ttt

t

b xt xu x

x

,,,

ttt

u xxx

not

,I t xAACTIONS not

,I t xRRESPONSES

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

0

0

ˆ

1ˆ

nt

b 0

b b

0

00

1nt

b 0

b 0

ˆnot b

81

1st Thermal Analogy

Boundary equations:

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

0ˆnt b 0

( ) 2nt Tr :C 1

1, ,2

St t x u x u u



Geometric Equation

0

1ˆ ( )

b b with

*

*

:ˆ:

u

nt

u u

n tBoundary Conditions in Space* *ˆ t t nwith

* *

*ˆ( )nt t

t

n t n t nn

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 ,

,

ttt

b xt xu x

,,,nt

ttt

u xxx

not

,II t xAACTIONS not

,II t xRRESPONSES

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 0 0x x 0 0R R R

1

RESPONSES

ACTIONS

0

* * * *

1ˆ ˆ

( , ) ( , ) ˆ ˆ

0

defI II IIIt t

b b b bu u 0 0x xt t t t n

A A A

t

b

* t

84

1st Thermal Analogy

The solution of the ORIGINAL PROBLEM is:

ANALOGOUS PROBLEM (II)


TRIVIAL PROBLEM (III)

I II

I II

I II

u u

1

85

t

u 00

1

1st Thermal Analogy

ANALOGOUSELASTIC

PROBLEM (II)

THERMOELASTIC ORIGINAL

PROBLEM (I)

THERMOELASTIC TRIVIAL

PROBLEM (III)

*

*

,,,

,

ttt

t

b xt xu x

x

,,,

ttt

u xxx

,I txA

,I txR

,II txA

,II txR

,III txA

,III txR

0* *

*

1ˆ ,

ˆ ,,0

t

tt

b x b

t x t nu x

,,

,nt

tt

t

u xxx

0

*

1

, t

b

t nu 0

x

86

2nd Thermal Analogy

The governing eqns. of the quasi-static isotropic linear thermoelastic problem are:

0, ,t t x b x 0

-1, ,t t x : xC 1

1, ,2

St t x u x u u



Geometric Equation

*

*

:

:u

u u


87

2nd Thermal Analogy

The actions and responses of the problem are:

*

*

,,,

,

ttt

t

b xt xu x

x

,,,

ttt

u xxx

not

,I t xAACTIONS not

,I t xRRESPONSES

Mathematical model

EDPs+BCs

REMARKis 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 tu x

1,2

1 , 1, 2,32

t S t t t

ttjit

ij ijj i

t

uu i jx 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 tt 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 tu x *c

nt t u u u89

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 tu u u u u u

*

nt t

u uu u u

90

2nd Thermal Analogy

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

0 b 0 -1nt :C

nt S nt u



Geometric Equation

*

*

:

:

nt tu

u u u

n tBoundary Conditions in space

91

The actions and responses of the ANALOGOUS PROBLEM are:

*

*

,,

, ,nt t

tt

t t

b xt xu u x u x

2nd Thermal Analogy

,,

,

nt

nt

tt

t

u xxx

not

,II t xAACTIONS not

,II t xRRESPONSES

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

defI II IIInt tt t

u u u ux x

0 0R R R

1

RESPONSES

ACTIONS

* *( , ) ( , )

0

t t defI II IIIt t

b b 0u u u u

x xt t 0

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)


TRIVIAL PROBLEM (III)

I II t

I II

I II

u u u

1

94

,t

t

t

u u x

0

1

2nd Thermal Analogy

ANALOGOUSELASTIC

PROBLEM (II)


PROBLEM (I)


PROBLEM (III)

*

*

,,,

,

ttt

t

b xt xu x

x

,,,

ttt

u xxx

,I txA

,I txR

,II txA

,II txR

,III txA

,III txR

,,

,

nt

nt

tt

t

u xxx

*

,

,

t t

t

b 0u u xt 0

x

*

0

t

bu u ut

95

2nd Analogy in structural analysis

:

tx x

tx x

tu x x x l

u u x

u u l

*

0

:

ntx x

ntx x

tu 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 order derivatives).

Thermal Analogies

( ) cnt xax 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 1t x u x x x

HOMOTHECY(free thermal expansion)

98

99


6.7 Superposition Principle

Linear Thermoelastic Problem

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

0, ,t t x b x 0

, ,t t x : xC 1

1, ,2

St t x u x u u



Geometric Equation

*

*

:

:u

u u

t n Boundary Conditions in space

0

,0

,0

u x 0

u x v Initial Conditions

100

Linear Thermoelastic Problem

Consider two possible systems of actions:

and their responses :

1

1*

1*

1

10

,

,

,

,

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 xA

1 , t xR

2

2*

2*

2

20

,

,

,

,

t

t

t

t

b x

t x

u x

x

v x

2 , t xA

2

2

2

,

,

,

t

t

t

u x

x

x

2 , t xR

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 2λ λ A A A 3 1 1 2 2λ λ R R R 1λ 2λ

102

103


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 12 2

1 12 21 12 2

x xy xz

x xy xz not

xy y yz xy y yz

xz yz z

xz yz z

6

x

y

defz

xy

xz

yz

R

6

x

y

defz

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:


1TrE E

1 1 1ˆ t C

1

1 0 0 0

1 0 0 0

1 0 0 0ˆ

10 0 0 0 0

10 0 0 0 0

10 0 0 0 0

E E E

E E E

E E E

G

G

G

C

1ˆ C t

0 00 00 0

t

000

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

C

ˆ t C

1 0 0 01 1

1 0 0 01 1

1 0 0 01 11ˆ 1 21 1 2 0 0 0 0 0

2 11 20 0 0 0 0

2 11 20 0 0 0 0

2 1

E

C

106

107


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 state3. 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 tx : xC ijkl jikl

ijkl ijlk

C CC C

minor symmetries


The 4th order constitutive elastic

constants tensor has 34=81 components.

ˆ 12

du ddt dt

: : : C 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 IC 1 1


2Tr 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

1TrE E

1Inverse

equation:

3 2E

2

Young’s ModulusPoisson’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 11 ´3

e 1

23 3(1 2 )

def EK

m K e 2G

21ˆ ´ ´ 02

u K e : 0K 0

volumetric deformation modulusLamé’s second parameter

0E Young’s modulus102

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)

*

*

0

,,,

ttt

b xt xu xv x

,,,

ttt

u xxx

not

, t xAACTIONS

not

, t xRRESPONSES

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

0 0 2

,, ,

tt t

t

u x

x b x

, 2t Tr x 1

1, ,2

St t x u x u u



Geometric Equation

*

*

:

:u

u u

t nBoundary Conditions in Space

0

,0

,0

u x 0

u x vInitial Conditions

, tu x , tx , tx

2

20 0 2t

uu u b

*u u uon

* u n u u n t on

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)

0, , 0t t x b x

1, t TrE E

x 1

1, ,2

St t x u x u u



Geometric Equation

, tu x , tx , tx

Boundary Conditions in Space*

*

:

:u

u u

t n

Beltrami-Michell Equations BCs 2

00 02 11 1

jk ikkij ij

i j k j i

bb bx x x x x

0 0 b *

n t on

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

,, ,0 , 0

not tt t

t

x

x x x

0, , ,t t t x : x xC

, t x 0 0, t x

2 IC 1 1 1

115

Constitutive eq. of an isotropic linear thermoelastic material:

Thermal stress and strain:

Summary (cont’d)

2Tr 1 1 Inverse equation:

Laméparameters

1TrE E

1 1

1 1 2

2 1

E

EG

With the thermal expansion coefficient:

1 2E

TOTAL NON-THERMAL COMPONENT THERMAL COMPONENT

nt t

nt t :nt C

1 :nt C

t

t

( ) 2nt Tr 1 t 1

1( )nt TrE E

1 t 1



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)

ANALOGOUSELASTIC

PROBLEM (II)


PROBLEM (I)


PROBLEM (III)

*

*

,,,

,

ttt

t

b xt xu x

x

,,,

ttt

u xxx

,I txA

,I txR

,II txA

,II txR

,III txA

,III txR

0* *

*

1ˆ ,

ˆ ,,0

t

tt

b x b

t x t nu x

,,

,nt

tt

t

u xxx

0

*

1

, t

b

u 0t n

x

u 00

1

118

ANALOGOUSELASTIC

PROBLEM (II)


PROBLEM (III)

Summary (cont’d)


PROBLEM (I)

*

*

,,,

,

ttt

t

b xt xu x

x

,,,

ttt

u xxx

,I txA

,I txR

,II txA

,II txR

,III txA

,III txR

,,

,

nt

nt

tt

t

u xxx

*

,

,

t t

t

b 0u u xt 0

x

,t t

u u x

0

1

*

,

t

t

bu u ut

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

defz

xy

xz

yz

R 6

x

y

defz

xy

xz

yz

R

1ˆ t C ˆ t C Inverse equation:

120

Documents

CH.6. LINEAR ELASTICITYmmc.rmee.upc.edu/documents/Slides/Ch6_v17.pdfin the Theory of Linear Elasticity. It constitutes the constitutive equation of a linear elastic material. The 4th