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 6.1 Hypothesis of the Linear Elasticity Theory Ch.6. Linear Elasticity

5 Introduction

 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.

 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. 00xx0  ,t     00xx0 ,t  3. Isothermal, isentropic and adiabatic processes  Isothermal - temperature remains constant  Isentropic - entropy of the system remains constant  Adiabatic - occurs without heat transfer

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  0 xXu xX  material and spatial descriptions of a property & material and spatial differential operators are the same: xX xXXx,,,,tttt         Xx  F1 x  the deformation gradient  X F1  , so the current spatial density is approximated by the density at the reference configuration.

0 ttF Thus, density is not an unknown variable in linear elastic problems.

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.

EX ,,,ttt ex   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,

00xx0  ,t 

00xx0 , t  

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

11 Hypothesis of the Linear Elastic Model

3. Isothermal and adiabatic (=isentropic) processes  In an isothermal process the temperature remains constant.

xx,,tt  00  x x   0

 In an isentropic process the entropy of the system remains constant. ds sts(,)XX () 0 s  0 dt  In an adiabatic process the net heat transfer entering into the body is zero.

Q  rdV qn dS 0 V V e0 VV internal heat conduction fonts from the exterior

0 rt  qx0 

12 6.2 Linear Elastic Constitutive Equation Ch.6. Linear Elasticity

13 Hooke’s Law

 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.  Fkl Fl  E E l A l F   E

 Hooke’s Law (for 1D problems) states that in an elastic material strain is directly proportional to stress through the elasticity modulus.

14 Generalized Hooke’s Law

 This proportionality is generalized for the multi-dimensional case in the Theory of Linear Elasticity. xx:x,(),tt   C Generalized   ijC ijkl kl ij,1,2,3Hooke’s Law   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: REMARK minor CCijkl jikl symmetries The current stress at a point CCijkl ijlk depends only on the current strain at the point, and not on the past major  symmetries CCijkl klij history of strain states at the point.

15 Elastic Potential

 The internal energy balance equation for the (adiabatic) linear elastic model is  global form stress power heat variation ddu udV 0 dV :d dV r q dV dt0 dt  0 VVVV  internal energy infinitesimal strains REMARK VVtt    local form The deformation rate tensor is related to d the material derivative of the material T 0 ur  :  q dt strain tensor through: EFdF   In this case, E     and F  1 . Where:  u is the specific internal energy (energy per unit mass).  r is the specific heat generated by the internal sources.  q is the heat conduction flux vector per unit surface.

16 Elastic Potential

 The stress power per unit of volume is an exact differential of the internal energy density, uˆ , or internal energy per unit of volume: ˆ ddut(,)x REMARK ()0 uuˆ  : dt dt uˆ The symmetries of the constitutive elastic  Operating in indicial notation: constants tensor are used: duˆ 1  : ()minor CCijkl jikl   ij ij ijCCC ijkl kl  ij ijkl kl ij ijkl kl dt :  2  symmetries  C Cijkl kl ik CCijkl ijlk jl major 11  symmetries CCijkl klij ()ijCC ijkl  kl kl klij  ij  ij CC ijkl  kl ij ijkl kl 22  d Cijkl ) dt ijCijkl kl 1 d duˆ d 1 ijC ijkl kl   ::C  2 dt   dt dt 2 ::C 17 Elastic Potential

duˆ 1 d  ::: C dt2 dt

 Consequences: 1. Consider the time derivative of the internal energy in the whole volume: dd dˆ utdVutdVˆˆxxx,,,: U  t  dVstress VVdt dt dt Vpower

 In elastic materials we talk about deformation energy because the stress power is an exact differential.

REMARK Internal energy in elastic materials is an exact differential.

18 Elastic Potential

duˆ 1 d  ::: C dt2 dt

 Consequences: 2. Integrating the time derivative of the internal energy density, 1 utˆ xx::xx,, tC , ta   2 and assuming that the density of the internal energy vanishes at the

neutral reference state, ut ˆ  xx ,0 0    :

1  0 11 x::x,,ttaa00C  x ()0 x x uˆ  ::C (): 2 22  Due to thermodynamic reasons the elastic energy is assumed always positive 1 uˆ  ::C  0 0 2

19 Elastic Potential

 The internal energy density defines a potential for the stress tensor, and is thus, named elastic potential. The stress tensor can be computed as  T   utˆ((,)) x 1 1 1 1 uˆ  (:: CC :   : C (   )    2222   

 The constitutive elastic constants tensor can be obtained as the second derivative of the internal energy density with respect to the strain tensor field,

2 2 () uˆ  C :   uˆ   C Cijkl   ij kl

20 6.3 Isotropic Linear Elasticity

Ch.6. Linear Elasticity

21 Isotropic Constitutive Elastic Constants Tensor

 An isotropic elastic material must have the same elastic properties (contained in C ) in all directions.  All the components of C must be independent of the orientation of the chosen (Cartesian) system C must be a (mathematically) isotropic tensor.

C 112 I   ijkl,,,  1,2,3 Cijkl ij   kl    ik jl il jk 

Where: 1  is the 4th order unit tensor defined as I   I ijkl 2  ik jl il jk    and  are scalar constants known as Lamé parameters or coefficients. REMARK The isotropy condition reduces the number of independent elastic constants from 21 to 2.

22 Isotropic Linear Elastic Constitutive Equation

 Introducing the isotropic constitutive elastic constants tensor C  112 I into the generalized Hooke’s Law   C :  , (in indicial notation)

ijC ijkl kl ij kl  ik jl  il jk  kl   11ij  ij  kl kl2()2    ik  jl  kl il jk  kl Tr  ij   ij 22 ll Tr()  jii j 1 1    2 ij 2 ij ij  The resulting constitutive equation is, Isotropic linear elastic Tr  1 2 constitutive equation.     ij ij ll2,1,2,3 ij ij  Hooke’s Law

23 Elastic Potential

 If the constitutive equation is, Tr  1 2 Isotropic linear elastic constitutive equation.   ij ij ll2,1,2,3 ij ij  Hooke’s Law

Then, the internal energy density can be reduced to:

11 REMARK uTrˆ  :2:1 22   The internal energy density is an σ elastic potential of the stress tensor 11as: Tr 1:2 22 uˆ   Tr  Tr  1 2    1 2  Tr  2

24 Inversion of the Constitutive Equation

1.  is isolated from the expression derived for Hooke’s Law 1  Tr  1 2   Tr   1 2 2. The trace of  is obtained: Tr Tr Tr 112 Tr  Tr   2  Tr  32 Tr  =3 3. The trace of  is easily isolated: 1 Tr Tr  32 4. The expression in 3. is introduced into the one obtained in 1. 11 1   Tr   1  Tr  1   232 23  2  2

25 Inverse Isotropic Linear Elastic Constitutive Equation

 The Lamé parameters in terms of E and  :  32  E E      112  E    G 2 21  So the inverse const. eq. is re-written:  1 Tr 1  Inverse isotropic linear  EE  elastic constitutive equation. 1   ij,1,2,3   Inverse Hooke’s Law.  ijEE ll ij ij 11  x xyz    xyxy  EG 11      In engineering notation: yEG y x z xz xz 11  z z x y   yz  yz E G

26 Young’s Modulus and Poisson’s Ratio

 Young's modulus E is a measure of the stiffness of an elastic material. It is given by the ratio of the uniaxial stress over the uniaxial strain.

 32  E  

 Poisson's ratio  is the ratio, when a solid is uniaxially stretched, of the transverse strain (perpendicular to the applied stress), to the axial strain (in the direction of the applied stress).

   2  

27 Example

Consider an uniaxial traction test of an isotropic linear elastic material such that:

 x  0 E,  y y z xy  xz  yz 0  x  x  x  x x

z

Obtain the strains (in engineering notation) and comment on the results obtained for a Poisson’s ratio of   0 and   0.5 .

28 11  x xyz    xyxy    0 EG x 11  y y x z   xz  xz y z xy  xz  yz 0 EG Solution 11  z z x y   yz  yz E G

For   0 : 1 xx  xy 0 1 There is no Poisson’s effect E xx  xy 0 E and the transversal normal yxxz 0 yxz 00 E strains are zero.  00 zyz  0 zxyzE 

For   0.5 : 1 1 The volumetric deformation is   0 xx  xy 0 xxE xy E zero, tr    xyz   0 , the 0.5 1 material is incompressible yxxz 0 yxxz 0 E 2E and the volume is preserved. 0.5 1 zxyz 0 zxyz 0  E 2E

29 Spherical and deviatoric parts of Hooke’s Law

 The stress tensor can be split into a spherical, or volumetric, part and a deviatoric part: 1 : 11Tr  sph m 3    m1     dev m1

 Similarly for the strain tensor: 11 sph e11Tr (  1 33  e1  ´ 1  dev  e1 3 3

30 Spherical and deviatoric parts of Hooke’s Law

 Operating on the volumetric strain: e  Tr

 1 Tr 1   EE

 1 e Tr Tr1   Tr EE  3  3 m K : volumetric 31 2 E deformation modulus e    m  e m 31 2 def 2 E E K   33(12)   The spherical parts of the stress and strain tensor are directly related:  m  K e

31 Spherical and Deviator Parts of Hooke’s Law

 1  Introducing    1    into   Tr   1  : m EE  1  Tr mm11   1  EE    0 11 13  1  mmTr11  Tr   1  1    m 1  EE 3 EEEEE E Taking into account that  m  e : 31 2 Comparing this 12 1E  1 1 1  ee11    with the expression  1 EEE312 3   e1  ´ 3 111 1  ´   E22G E  The deviatoric parts of the stress and strain tensor are related component by component:    2GGij ij 2 ij , {1, 2, 3}

32 Spherical and deviatoric parts of Hooke’s Law

 The spherical and deviatoric parts of the strain tensor are directly proportional to the spherical and deviatoric parts (component by component) respectively, of the stress tensor:

   2G  m  K e ij ij

33 Elastic Potential

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

There is a minimum for   0 :

uˆ  =:C   0  0 0

22uuˆˆ    CC    0 00

34 Elastic Potential

 The elastic potential can be written as a function of the spherical and deviatoric parts of the strain tensor:  ::CC  :: : 111 uˆ ::C :Tr  1 2 : 222    11 ee11´´:   Tr  e  e 33 112 Tr1:: Tr    : 12 22 ee2 11::: 1´´´ 933 K Tr 0 1122 12 2 1 2 ueeˆ  μ ´: ´e μ ´: ´ e ´´: 23 23 3 Elastic potential in terms of the 1 uKeˆ  2  ´: ´ 0 spherical and deviatoric parts 2 of the strains.

35 Limits in the Elastic Properties

 The derived expression must hold true for any deformation process: 1 ´ uKeˆ ::´02  2  Consider now the following particular cases of isotropic linear elastic material:  Pure spherical deformation process 1 1  e 1 1 volumetric 3 uKeˆ1  2  0 K  0 2 deformation modulus 1  0  Pure deviatoric deformation process

2   2 Lamé’s second  uˆ   ´´0:    0 e 2  0 parameter REMARK :0 ij ij  36 Limits in the Elastic Properties

 K and  are related to E and  through:

E E K   0   G 0 31 2 21  REMARK  In rare cases, a material can  Poisson’s ratio has a non-negative value, have a negative Poisson’s ratio. E Such materials are named  0 21 E  0 Young’s auxetic materials. modulus   0

 Therefore, E  0 1 31 2 0   Poisson’s ratio E  0 2

37 6.4 The Linear Elastic Problem

Ch.6. Linear Elasticity

38 Introduction

 The linear elastic solid is subjected to body forces and prescribed tractions: Initial actions: bx ,0 t  0 tx,0

bx ,t Actions through time: tx,t

 The Linear Elastic problem is the set of equations that allow obtaining the evolution through time of the corresponding displacements ux  , t  , strains   x , t  and stresses   x , t  .

39 Governing Equations

 The Linear Elastic Problem is governed by the equations: 1. Cauchy’s Equation of Motion. Linear Momentum Balance Equation. 2ux ,t xbx,,tt   00 t 2 2. Constitutive Equation. Isotropic Linear Elastic Constitutive Equation. This is a PDE system of  x,2tTr 1  15 eqns -15 unknowns: ux ,t 3 unknowns 3. Geometrical Equation. x,t 6 unknowns Kinematic Compatibility.  x,t 6 unknowns S 1 Which must be solved in xuxu,,tt   u 3 2 the RR   space.

40 Boundary Conditions

 Boundary conditions in space  Affect the spatial arguments of the unknowns  Are applied on the contour  of the solid, which is divided into three parts:

 Prescribed displacements on  u : ux(,)tt u* (,) x  * xu t ututiiixx,,1,2,3   

 Prescribed tractions on   : uu  V  (,)xnttt* (,) x     {0}  uuuu   * x  t  ijxx,,1,2,3tn j t j t  i   

 Prescribed displacements and stresses on  u  : * ututiixx,,  i,,j ki 1,2,3 j x t  *  u  jkxx,,tn k t j t

41 Boundary Conditions

42 Boundary Conditions

 Boundary conditions in time. INTIAL CONDITIONS.  Affect the time argument of the unknowns.  Generally, they are the known values at t  0 :  Initial displacements: ux ,0 0 x V

 Initial velocity: ux ,t not  ux ,0  v0 x  x V t t0

43 The Linear Elastic Problem

 Find the displacements ux  , t  , strains   x , t  and stresses   x , t  such that

2ux ,t xbx,,tt00   Cauchy’s Equation of Motion  t 2  x,2tTr 1  Constitutive Equation 1 xuxuu,,ttS   Geometric Equation 2

* u : uu * Boundary Conditions in space  : tn

ux ,0  0 Initial Conditions (Boundary Conditions in time) ux ,0  v0

44 Actions and Responses

 The linear elastic problem can be viewed as a system of actions or data inserted into a mathematical model made up of the EDP’s and boundary conditions seen which gives a series of responses or solution in displacements, strains and stresses. bx,t * ux ,t tx,t Mathematical  *  model  x,t ux,t   EDPs+BCs  x,t vx0 not not RESPONSES  R x,t ACTIONS  A x,t  Generally, actions and responses depend on time. In these cases, the 3 problem is a dynamic problem, integrated in RR   . 3  In certain cases, the integration space is reduced to R . The problem is termed quasi-static.

45 The Quasi-Static Problem

 A problem is said to be quasi-static if the acceleration term can be considered to be negligible. 2ux(,)t a0  t 2

 This hypothesis is acceptable if actions are applied slowly. Then, 2ux(,)t 22/ t 0 22/ t 0  0 A R  t 2

46 The Quasi-Static Problem

 Find the displacements ux  , t  , strains   x , t  and stresses   x , t  such that 2ux ,t   0 0  t 2

xbx0,,tt0    Equilibrium Equation

 x,2tTr 1  Constitutive Equation 1 xuxu,,ttS   u Geometric Equation 2

* u : uu * Boundary Conditions in Space  : tn

ux ,0 0 Initial Conditions ux ,0  v0

47 The Quasi-Static Problem

 The quasi-static linear elastic problem does not involve time derivatives.  Now the time variable plays the role of a loading factor: it describes the evolution of the actions.

bx,  Mathematical ux , * tx, model  x, *   ux, EDPs+BCs  x, not not ACTIONS  A x, RESPONSES  R x,

* *  For each value of the actions A  x ,   -characterized by a fixed value  - a * response R  x ,   is obtained. *  Varying  , a family of actions and its corresponding family of responses is obtained.

48 Example

Consider the typical material strength problem where a cantilever beam is subjected to a force Ft   at it’s tip. For a quasi-static problem,

The response is   tt       , so for every time instant, it only depends on the corresponding value of   t  .

49 Solution of the Linear Elastic Problem

 To solve the isotropic linear elastic problem posed, two approaches can be used:  Displacement formulation - Navier Equations Eliminate   x , t  and   x , t  from the general system of equations. This generates a system of 3 eqns. for the 3 unknown components of ux  , t  .  Useful with displacement BCs.  Avoids compatibility equations.  Mostly used in 3D problems.  Basis of most of the numerical methods.

 Stress formulation - Beltrami-Michell Equations. Eliminate ux  , t  and   x , t  from the general system of equations. This generates a system of 6 eqns. for the 6 unknown components of   x , t  .  Effective with boundary conditions given in stresses.  Must work with compatibility equations.  Mostly used in 2D problems.  Can only be used in the quasi-static problem.

50 Displacement formulation

2ux ,t xbx,,tt00   Cauchy’s Equation of Motion  t 2  x,2tTr 1  Constitutive Equation 1 xuxu,,ttS   u Geometric Equation 2

* u : uu * Boundary Conditions in Space  : tn

ux,0  0  Initial Conditions ux ,0  v0

The aim is to reduce this system to a system with ux  , t  as the only unknowns. Once these are obtained,   x , t  and   x , t  will be found through substitution.

51 Displacement formulation

 Introduce the Constitutive Equation into Cauchy’s Equation of motion:  x,2tTr 1  2u Tr()1 2 00b 2 2ux ,t  t xbx,,tt   00 t 2  Consider the following identities:   u  i uu Tr()1u  kk   i 11 ij ij  xxxxxxjjkiki    u   u i 1, 2, 3 i

Tr 1 u

52 Displacement formulation

 Introduce the Constitutive Equation into Cauchy’s Equation of motion:  x,2tTr 1  2u Tr()1 2 00b 2 2ux ,t  t xbx,,tt   00 t 2  Consider the following identities:  2 u   u    i i  11111uuuu2    ijii  j   j  2 u   u i     xx22222  xx  xxxx i  x jj ji jjij  i

112    u uui 1, 2,  3 22 i 11  () uu  2 22

53 Displacement formulation

 Introduce the Constitutive Equation into Cauchy’s Equation of Movement:  x,2tTr 1  2u Tr()1 2 00b 2 2ux ,t  t xbx,,tt   00 t 2  Replacing the identities: 11 Tr  1 u  () uu  2 22 2 nd 112  u 2 order  Then, uuub2()     00  2 22  t PDE system  2u    uub 2  The Navier Equations  002  t are obtained: uubui jji,,00   ijj i i 1, 2, 3  

54 Displacement formulation

 The boundary conditions are also rewritten in terms of ux  , t  :

 x,2tTr 1     u tnn*  Tr  2   1 *  S u uu tn   2

tunuun*    

 The BCs are now:

* uu on u  * uuiii1, 2, 3

* REMARK un u  u n t   on  The initial conditions un un un ti* 1, 2, 3  kk.,, i i j j ji j i remain the same.

55 Displacement formulation

  Navier equations in a cylindrical coordinate system: eG2   2u 22GGbzr   xr cos rr  zr  t2  x(,rz , ) yr sin 1 e  2u  222GGGb rz  zz rzrt 2 2 eG22 Grzu 2Gr   bz 2 zrr r   t 

11uz u rz   2 rz 

1 uurz dV r d dr dz Where:  zr   2 zr  11ru  1ur 11 u u      z zrerur 2 rr r rr r z

56 Displacement formulation 

 Navier equations   in a spherical coordinate system: eG22 G 2u 2sinGb  r rrsin r sin r  t2 xr sin cos  xxryr,,  sin sin   2G eG22 G 2u  r   rbsin  zr cos rrsin rr sin  t2    2   2G eG22 G   u  rbr 2 rrrrsin    t2 dV rsin dr d d      11 1u ωr  u sin θ  2sinrrθθ sin θφ Where:  11u 1 ru  ω  r  r 2sinrrrθφ 

11 1ur ωr  ru  2 rr rθ  1  2  erururu2  r sin sin  rrsin  57 Stress formulation

Equilibrium Equation xbx,,0tt    0 (Quasi-static problem)  1 x,tTr1  Inverse Constitutive Equation EE 1 xuxuu,,ttS   Geometric Equation 2

* u : uu * Boundary Conditions in Space  : tn

The aim is to reduce this system to a system with   x , t  as the only unknowns. Once these are obtained,   x , t  will be found through substitution and ux  , t  by integrating the geometric equations. REMARK For the quasi-static problem, the time variable plays the role of a loading factor.

58 Stress formulation

 Taking the geometric equation and, through successive derivations, the displacements are eliminated: 2222 Compatibility Equations ijkl ik jl 0,,,1,2,3ijkl   (seen in Ch.3.) xkl  x  xx i j  x jl  x  xx ik

 Introducing the inverse constitutive equation into the compatibility equations and using the equilibrium equation:   1    ij EEpp ij ij  ij  0bj 0 xi   2nd order  The Beltrami-Michell Equations are obtained: PDE system 2   b 2 1  kk 00bbki 0 j  ij  ij  ij,1,2,3  11xxij   x k  x j  x i

59 Stress formulation

 The boundary conditions are:

 Equilibrium Equations:   0b 0 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 :  nt on 

60 Stress formulation

 Once the stress field is known, the strain field is found by substitution.  1 x,tTr1  EE

 The calculation, after, of the displacement field requires that the geometric equations be integrated with the prescribed displacements  on u : 1 ()xux  () uxx () V 2 * ux() u () x x u

REMARK This need to integrate the second system is a considerable disadvantage with respect to the displacement formulation when using numerical methods to solve the lineal elastic problem.

61 Saint-Venant’s Principle

 From A. E. H. Love's Treatise on the mathematical theory of elasticity: “According to the principle, the strains that are produced in a body by the application, to a small part of its surface, of a system of forces statically equivalent to zero force and zero couple, are of negligible magnitude at distances which are large compared with the linear dimensions of the part.” REMARK This principle does not have a  Expressed in another way: rigorous mathematical proof. “The difference between the stresses caused by statically equivalent load systems is insignificant at distances greater than the largest dimension of the area over which the loads are acting.”

(I) (II) ux PP,,tt u x (I)xx,,tt (II) PP P |    (I) (II)  xxPP,,tt

62 Saint-Venant’s Principle

 Saint Venant’s Principle is often used in strength of materials.  It is useful to introduce the concept of stress: The exact solution of this problem is very complicated.

This load system is statically equivalent to load system (I). The solution of this problem is very simple.

Saint Venant’s Principle allows approximating solution (I) by solution (II) at a far enough distance from the ends of the beam.

63 Uniqueness of the solution

 The solution of the lineal elastic problem is unique:  It is unique in strains and stresses.  It is unique in displacements assuming that appropriate boundary conditions hold in order to avoid rigid body motions.

 This can be proven by Reductio ad absurdum ("reduction to the absurd"), as shown in pp. 189-193 of the course book.  This demonstration is valid for lineal elasticity in small deformations.  The constitutive tensor C is used, so the demonstration is not only valid for isotropic problems but also for orthotropic and anisotropic ones.

64 6.5 Linear Thermoelasticity

Ch.6. Linear Elasticity

65 Hypothesis of the Linear Thermo-elastic Model

 The simplifying hypothesis of the Theory of Linear Thermo- elasticity are:

1. Infinitesimal strains and deformation framework  Both the displacements and their gradients are infinitesimal.

2. Existence of an unstrained and unstressed reference state  The reference state is usually assumed to correspond to the reference configuration. 00xx0  ,t 

00xx0 , t   3. Isentropic and adiabatic processes – no longer isothermal!!!  Isentropic - entropy of the system remains constant  Adiabatic - occurs without heat transfer

66 Hypothesis of the Linear Thermo-Elastic Model

3. (Hypothesis of isothermal process is removed)  The process is no longer isothermal so the temperature changes throughout time: not  xx,,0t    0    x,t  x,0t    t We will assume the temperature field is known.

 But the process is still isentropic and adiabatic: stcnt s  0

Q  rdV qn dS 0 V V e  VV internal heat conduction fonts from the exterior  rt  qx0 

67 Generalized Hooke’s Law

The Generalized Hooke’s Law becomes:

xx,:,ttCC 0  :, x t Generalized Hooke’s Law for   linear thermoelastic problems  ijC ijkl kl ij 0 ij,1,2,3 

Where  is the elastic constitutive tensor.   x,tis the absolute temperature field.

 00 x,t  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.

68 Isotropic Constitutive Constants Tensors

 An isotropic thermoelastic material must have the same elastic and thermal properties in all directions:   must be a (mathematically) isotropic 4th order tensor: C 112 I   ijkl,,.  1,2,3 Cijkl ij   kl    ik jl il jk  Where: 1 th  I is the 4 order symmetric unit tensor defined as I ik jl il jk  ijkl 2     and  are the Lamé parameters or coefficients.

  is a (mathematically) isotropic 2nd order tensor:    1    ij ij ij,1,2,3 Where:   is a scalar thermal constant parameter.

69 Isotropic Linear Thermoelastic Constitutive Equation

 Introducing the isotropic constitutive constants tensors    1 and

C 112 I into the generalized Hooke’s Law,   C :  0  (in indicial notation)

ij C ijkl kl ij 00  ij kl  ik jl  il jk kl   ij    11ij ij kl kl2  ik jl kl  il jk kl 0 ij               22  ll  jii j    ij  The resulting constitutive equation is,

Tr 112  Isotropic linear thermoelastic    constitutive equation. ij ij ll2,1,2,3 ij ij ij 

70 Inversion of the Constitutive Equation

1.  is isolated from the Generalized Hooke’s Law for linear thermoelastic problems:

11 C :   CC:: 

2. The thermal expansion coefficients tensor  is defined as: def 1  C : It is a 2nd order symmetric tensor which involves 6 thermal expansion coefficients

3. The inverse constitutive equation is obtained:

1 C :  

71  Inverse   Isotropic Linear Thermoelastic Constitutive Equation

 For the isotropic case:

 1 1 C 11   I  EE 1 12   C : ( 1)   1 E Cijkl ij kl  ik jl  il jk ijkl,,.  1,2,3  EE  The inverse const. eq. is re-written:  1 Tr 11     EE Inverse isotropic linear thermo  1 elastic constitutive equation.      ij,1,2,3   ijEE ll ij ij ij

 Where  is a scalar thermal expansion coefficient related to the scalar thermal constant parameter  through: 12     E 72 Thermal Stress

 Comparing the constitutive equations,

  Tr 1 2 Isotropic linear elastic constitutive equation.

 Tr 112 Isotropic linear thermoelastic constitutive equation. t nt  

the decomposition is made: nt  t Where: nt   is the non-thermal stress: the stress produced if there is no thermal phenomena.  t is the thermal stress: the “corrector” stress due to the temperature increment.

73 Thermal Strain

 Similarly, by comparing the inverse constitutive equations,

 1 Inverse isotropic linear elastic  Tr 1  EE constitutive eq.  1 Tr 11    Inverse isotropic linear thermoelastic t EEnt constitutive eq.     the decomposition is made: nt  t Where: nt   is the non-thermal strain: the strain produced if there is no thermal phenomena.  t is the thermal strain: the “corrector” strain due to the temperature increment.

74 Thermal Stress and Strain

 The thermal components appear when thermal processes are considered.

NON-THERMAL THERMAL TOTAL COMPONENT COMPONENT

nt t   C :     nt t   Isotropic material: Isotropic material: nt Tr()1 2 t   1

nt 1 t   C :     nt  t Isotropic material: Isotropic material: nt  1  Tr()1  t   1 EE These are the equations used nt t in FEM codes. CC::  11nt t CC::  

75 Thermal Stress and Strain

REMARK 1 In thermoelastic problems, a state of zero strain in a body does not necessarily imply zero stress.  00nt

 t 1 0

REMARK 2 In thermoelastic problems, a state of zero stress in a body does not necessarily imply zero strain. 00nt 

 t 1 0

76 6.6 Thermal Analogies

Ch.6. Linear Elasticity

77 Solution to the Linear Thermoelastic Problem

 To solve the isotropic linear thermoelastic problem posed thermal analogies are used.

 The thermoelastic problem is solved like an elastic problem and then, the results are “corrected” to account for the temperature effects.

 They use the same strategies and methodologies seen in solving isotropic linear elastic problems:  Displacement Formulation - Navier Equations.  Stress Formulation - Beltrami-Michell Equations.

 Two basic analogies for solving quasi-static isotropic linear thermoelastic problems are presented:  1st thermal analogy – Duhamel-Neumann analogy.  2nd thermal analogy

78 1st Thermal Analogy

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

xbx0,,tt0    Equilibrium Equation

x:x,,ttC     1 Constitutive Equation 1 xuxuu,,ttS  Geometric Equation 2

* u : uu * Boundary Conditions in Space  : tn

79 1st Thermal Analogy

 The actions and responses of the problem are:

not not I I ACTIONS  A x,t RESPONSES  R x,t bx ,t ux ,t * Mathematical tx,t model x,t *  ux,t  x,t  EDPs+BCs  x,t

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

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    nt  t and replaced into the governing equations:  Momentum equations  nt  t     nt    t   nt     1

0b0  nt ˆ  0 b0 nt 1 0 b0   1 0 bbˆ     not 0  bˆ

81 1st Thermal Analogy

 Boundary equations:   nt t nt nnt t * nt * *  ntˆ  nt   ˆ** ntnt**  t  n  t ()  n tt()  n   1 n tˆ* ANALOGOUS PROBLEM –A linear elastic problem can be solved as:

nt ˆ ˆ 1  0b0 with bb ()  Equilibrium Equation 0 nt  C : Tr()1  2  Constitutive Equation 1 xuxuu,,ttS   Geometric Equation 2

* u : uu nt ˆ* ˆ** Boundary Conditions in Space  :  nt with tt  n

82 1st Thermal Analogy

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

not not II II ACTIONS  A x,t RESPONSES  R x,t

bxˆ  ,t Mathematical ux ,t txˆ* ,t model x,t nt  ux* ,t EDPs+BCs  x,t

ORIGINAL PROBLEM (I) ANALOGOUS (ELASTIC) PROBLEM (II) 83 1st Thermal Analogy

 If the actions and responses of the original and analogous problems are compared:  b 1 b bbbˆˆ    ACTIONS   0   def IIIu u0    III (,)xxtt (,)   0  AA* ***    A t tttˆˆ  *    t n   0         RESPONSES uu    0   0  def III          III  RR(,)xxtt (,)     0   0   R  nt   nt          1   t

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

84 1st Thermal Analogy

 The solution of the ORIGINAL PROBLEM is:

uuIII       III    III  1 ORIGINAL PROBLEM (I)

ANALOGOUS PROBLEM (II) TRIVIAL PROBLEM (III)

85 1st Thermal Analogy

THERMOELASTIC ANALOGOUS THERMOELASTIC ORIGINAL ELASTIC TRIVIAL PROBLEM (I) PROBLEM (II) PROBLEM (III) ˆ 1 1 bx ,t bx,t  b   b     * 0 0 tx,t **   I  II  txˆ ,t  t n III  *  x,t * x,t x,t tn   A   ux,t A   A   *    ux,t u0   x,t  0  x,t

ux ,t ux ,t u0 I  II  x,t x,t x,t III    0 R   R x,t R x,t  x,t nt x,t   t   1

86 2nd Thermal Analogy

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

xbx0,,tt0    Equilibrium Equation

-1 x:x,,ttC     1 Inverse Constitutive Equation 1 xuxuu,,ttS  Geometric Equation 2

* u : uu * Boundary Conditions in Space  : tn

87 2nd Thermal Analogy

 The actions and responses of the problem are:

not not I I ACTIONS  A x,t RESPONSES  R x,t bx ,t ux ,t * Mathematical tx,t model x,t *  ux,t  x,t  EDPs+BCs  x,t

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

88 2nd Thermal Analogy

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

 Then, the total displacement field is decomposed as: def uxnt(,)ttt ux (,) ux t (,) uu nt u t 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,    nt  t and uu  nt  u t and replaced into the governing equations:  Geometric equations:

SSnttSntStSntuuuu ()     uu    t t nt Su nt nt t

 Boundary equations:

uu * uuunt t  *  : uu** unt u u t uunt u t u

90 2nd Thermal Analogy

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

0b0  Equilibrium Equation nt -1  C : Constitutive Equation nt Su nt Geometric Equation

nt* t u : uuu * Boundary Conditions in space  :  nt

91 2nd Thermal Analogy

 The actions and responses of the ANALOGOUS PROBLEM are:

not not II II ACTIONS  A x,t RESPONSES  R x,t

nt bx ,t Mathematical ux ,t nt x,t tx*,t model    x,t uuxuxnt* ,,tt t EDPs+BCs

ORIGINAL PROBLEM (I) ANALOGOUS PROBLEM (II)

92 2nd Thermal Analogy

 If the actions and responses of the original and analogous problems are compared: bb 0 ACTIONS   tt def III  uuuu  III   AA(,)xxtt (,) **  A tt  0 0 

RESPONSES nt t t uu u  u  def I II   nt t   III   RR(,)xxtt (,)       1  R        00  

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:

uuuIII  t    III  1    III ORIGINAL PROBLEM (I)

ANALOGOUS PROBLEM (II) TRIVIAL PROBLEM (III)

94 2nd Thermal Analogy

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

bx ,t b b0  tx* ,t  t  t I   II  uu  u III  uux  ,t x,t * x,t x,t A   ux,t A   t* A   t0*     x,t  0  x,t

nt ux ,t ux ,t uux t  ,t I  II  nt x,t x,t  x,t III  t   1 R   R x,t R x,t  x,t x,t   0

95 2nd Analogy in structural analysis

nt t uuxx uuxx x nt t  xx xx t  : uuu* t   l  : uu  l ux xxxl ux xxl   0

96 Thermal Analogies

 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: ()x  cnt    Lineal thermal increment:  ax by cz d is of special interest because the thermal strains are:  t  1  linear polinomial and trivially satisfy the compatibility conditions (involving second order derivatives).

97 Thermal Analogies

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

HOMOTHECY (free thermal expansion)

98 6.7 Superposition Principle

Ch.6. Linear Elasticity

99 Linear Thermoelastic Problem

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

xbx0,,tt0    Equilibrium Equation

x:x,,ttC     1 Constitutive Equation 1 xuxuu,,ttS  Geometric Equation 2

* u : uu * Boundary Conditions in space  : tn

ux,0  0  Initial Conditions ux ,0  v0

100 Linear Thermoelastic Problem

 Consider two possible systems of actions:

bx1  ,t bx2  ,t tx*1 ,t tx*2 ,t 1 2 A x,t ux*1 ,t A x,t ux*2 ,t  1 x,t  2 x,t

1 2 vx0  vx0 

and their responses :

ux1  ,t ux2  ,t 1 1 2 2 R x,t  x,t R x,t  x,t 1 x,t 2 x,t

101 Superposition Principle

31122          The solution to the system of actions AAA  λλ  where 1 2 31122         λ and λ are two given scalar values, is RRR  λλ  .

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.

 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.

102 6.8 Hooke’s Law in Voigt Notation

Ch.6. Linear Elasticity

103 Stress and Strain Vectors

 Taking into account the symmetry of the stress and strain tensors,   these can be written in vector form:  x  11  y  xxyxz   22 xxyxz   def  not. z  11  R6   xyyyz   xyy yz   22 xy    xz yz z 11    xz xz yz z   22  yz REMARK VOIGT   The double contraction     is NOTATION x  transformed into the scalar (dot) y xxyxz  product     :   def  z 6   xy y yz   vectors   R   xy  xz yz z      ij ij i i   xz 2nd order  tensors yz 

104 Inverse Constitutive Equation

 The inverse constitutive equation is rewritten:

 1 t ˆ 1 Tr 11     C     EE

t ˆ 1 Where C is an elastic constants inverse matrix and   is a thermal strain vector:  00 1  t   000   00 EE E   00  1   000 EE E 1 000 EEE ˆ 1 C     1   00 0 00  G      1 t 00 0 0 0     G  0  1   00 0 00 0 G    0 

105 Hooke’s Law

 By inverting the inverse constitutive equation, Hooke’s Law in terms of the stress and strain vectors is obtained:

ˆ t  C         1 000  11  ˆ    Where C is an elastic   1 000 constants matrix : 11      1000 11  ˆ E 1 C   12  112 000 0 0  21   12   000 0 0  21     12   000 0 0   21      106 Summary

Ch.6. Linear Elasticity

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 state 3. Isothermal, isentropic and adiabatic processes  Constitutive eq. of a linear elastic material - Generalized Hooke’s Law:  x:x,,tt C   The 4th order minor CCijkl jikl symmetries constitutive elastic CCijkl ijlk constants tensor has major CCijkl klij 34=81 components. symmetries

 The internal energy density defines a potential for the stress tensor, it is an elastic potential. duˆ 1 d uˆ   ::: C   dt2 dt 

108 Summary (cont’d)

 Isotropic elastic material: C  112 I Where: th 1  I is the 4 order unit tensor defined as I   ijkl 2  ik jl il jk    and  are the Lamé parameters.

 Constitutive eq. of an isotropic linear elastic material: Inverse  1   Tr 1 2 equation:  Tr 1  EE  32  Young’s E  In engineering notation: Modulus   11   x xyz   xyxy  Poisson’s   EG ratio 2   11      yyxzxzxzEG 11  zzxyyzyz    E G

109 Summary (cont’d)

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

  1   def 2 E m   Ke   2G  K   1 m 33(12)   e1 ´ 3

 Limits in the elastic properties for isotropic linear elastic materials: 1 K  0 volumetric deformation modulus uKeˆ  2  ´´0: 2   0 Lamé’s second parameter E  0 Young’s modulus 1 0   Poisson’s ratio 2

110 Summary (cont’d)

 The Linear Elastic Problem: not RESPONSES  R x,t bx ,t Mathematical ux ,t tx* ,t  model x,t ux* ,t   EDPs+BCs  x,t vx0 not ACTIONS  A x,t

 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.

111 Summary (cont’d)

 The Isotropic Linear Elastic Problem – displacement formulation: 2ux ,t xbx,,tt00   2 Cauchy’s Equation of Motion  t  x,2tTr 1  Constitutive Equation

1 xuxuu,,ttS   Geometric Equation 2

* u : uu * Boundary Conditions in Space  : tn

ux,0  0  Initial Conditions ux ,0  v0 The aim is to reduce this system to a system with ux  , t  as the only unknowns. Once these are obtained,   x , t  and   x , t  will be found through substitution. Navier Equations * 2 BCs uu on u 2  u    uub  002 * t un u u n ton 

112 Summary (cont’d)

 The Isotropic Linear Elastic Problem – stress formulation:

 xbx,,0tt0   Equilibrium Equation  1 x,tTr1  Inverse Constitutive Equation EE 1 xuxuu,,ttS   Geometric Equation 2

: uu* u Boundary Conditions in Space *  : tn

The aim is to reduce this system to a system with   x , t  as the only unknowns. Once these are obtained,   x , t  will be found through substitution and ux  , t  by integrating the geometric equations.  Beltrami-Michell Equations BCs    b 0 2   b 0 2 1  00bbki0 j kk    * ij ij  nton 11xxij   x k  x j  x i 

113 Summary (cont’d)

 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.”

(I) (II) ux PP,,tt u x (I)xx,,tt (II) PP P |    (I) (II)  xxPP,,tt

114 Summary (cont’d)   

 In the Theory of Linear Thermoelasticity the process is no longer isothermal: not  x,t xx,,0tt   x ,  0 0 t

 The Generalized Hooke’s Law becomes:

x:xx,,,tttC      0  Where   x,t is the temperature field.

 00 x,t is the temperature distribution in the reference state.   is the tensor of thermal properties or constitutive thermal constants tensor.

 Isotropic thermoelastic material: C 112 I    1

115 Summary (cont’d)

 Constitutive eq. of an isotropic linear thermoelastic material: Inverse  1  Tr 112 Tr 11    equation: EE

 E With the thermal   Lamé 112  expansion coefficient:  12  parameters E    G E 21  Thermal stress and strain: TOTAL NON-THERMAL COMPONENT THERMAL COMPONENT

nt t   C :     nt t   Isotropic material: Isotropic material: nt Tr()1 2 t   1

nt 1 t   C :     nt t   Isotropic material:  1 Isotropic material: nt Tr()1  t   1 EE 116 Summary (cont’d)

 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.

117 Summary (cont’d)

1st Thermal Analogy - Duhamel-Neumann analogy

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

1 1 bx ,t bxˆ ,t  b   b     * tx,t 0  0 **    I  II  txˆ ,t  t n III  u0 ux* ,t    A x,t A x,t *  A x,t  ux,t tn*    x,t  0  x,t

ux ,t ux ,t u0 I II  R x,t x,t R x,t  x,t III    0  R x,t  x,t nt x,t    1

118 Summary (cont’d)

2nd Thermal Analogy

THERMOELASTIC ANALOGOUS THERMOELASTIC ORIGINAL ELASTIC TRIVIAL PROBLEM (I) PROBLEM (II) PROBLEM (III) bx ,t b b0  * tx,t uu   ut uux  t ,t I II III   *     A x,t ux,t A x,t t* A x,t t0*     x,t  x,t  x,t

ux ,t uxnt  ,t uux t  ,t I II  x,t x,t x,t nt x,t III  1 R  R    R x,t     x,t x,t   0

119 Summary (cont’d)

 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:

t ˆ 1 Inverse ˆ t C    equation:  C     

 x  x   y y def  def  z z   R6  R6 xy  xy    xz  xz  yz yz  

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