Generalized Hooke's

Multimedia Course on Continuum Mechanics Overview

 Hypothesis of the Linear Elasticity Theory Lecture 1  Linear Elastic Constitutive Equation Lecture 2  Generalized Hooke’s Law  Elastic Potential Lecture 3  Isotropic Linear Elasticity  Isotropic Constitutive Elastic Constants Tensor Lecture 4  Lamé Parameters  Isotropic Linear Elastic Constitutive Equation  Young’s Modulus and Poisson’s Ratio Lecture 5  Inverse Isotropic Linear Elastic Constitutive Equation  Spherical and Deviator Parts of Hooke’s Law Lecture 6  Limits in Elastic Properties Lecture 7

2 Overview (cont’d)

 The Linear Elastic Problem  Governing Equations Lecture 8  Boundary Conditions  The Quasi-Static Problem Lecture 9  Solution Lecture 10  Displacement Formulation  Stress Formulation Lecture 11  Saint-Venant’s Principle Lecture 12  Uniqueness of the solution  Linear Thermoelasticity  Hypothesis of the Linear Elasticity Theory Lecture 13  Linear Thermoelastic Constitutive Equation  Inverse Constitutive Equation Lecture 14  Thermal Stress and Strain

3 Overview (cont’d)

 Thermal Analogies  Solution to the linear thermoelastic problem Lecture 15  1st Thermal Analogy  2nd Thermal Analogy Lecture 16 Lecture 17

 Superposition Principle in Linear Thermoelasticity Lecture 18

 Hooke’s Law in Voigt Notation Lecture 19

4 6.1 Hypothesis of the Linear Elasticity Theory Ch.6. Linear Elasticity

5 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 x= Xu + xX≈  material and spatial descriptions of a property & material and spatial differential operators are the same: xX= γγ()()()()xXXx,t= , t =Γ=Γ ,, tt ∂•() ∂•() = ∇∇()•=() • ∂∂Xx ∂x  the deformation gradientF1 = ≈ F1 ≈ , so the current spatial ∂X density is approximated by the density at the reference configuration.

ρρ0 =ttF ≈ ρ 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≈= exε 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,

εε00()()xx0=,t =

σσ00()()x= x0,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.

θθ()()()xx,,tt≡≡00 θ x ∀ x

 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.

= ρ − ⋅ = ∀∆ ⊂ Qe0 ∫∫ r dV qn dS 0 V V VV∂ heat conduction internal REMARK from the exterior sources An isentropic process is an

ρ0 rt−⋅=∇ qx0 ∀∀ idealized thermodynamic process that is adiabatic, isothermal and reversible. 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. σ F= kl ∆ Fl∆ = E E Al F ε σ=E ε

 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. = σε(x,tt) () x:( x ,)  C Generalized  σεij= C ijkl kl ij,∈{ 1, 2, 3} Hooke’s Law   It constitutes the constitutive equation of a linear elastic material. th  The 4 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 transfer rate d duρ ρρ u dV = 0 dV=σ :d dV +() r −∇⋅q dV dt ∫00 ∫dt ∫∫ V V VV= ε internal energy infinitesimal strains REMARK ()Vt ≡∀ Vt local form The rate of strain tensor is related to the d material derivative of the material strain T ()ρρ0 ur=σ : ε + −∇⋅q  = ⋅⋅ dt tensor through: E F dF In this case, E  = ε  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: d duˆ(,)x t  REMARK ()ρ0 uu= =ˆ = σε: dt  dt The symmetries of the uˆ constitutive elastic  Operating in indicial notation: constants tensor are used: duˆ 1 = =σε:  =εσ=εε=εε+εε=C()  CC  minor CCijkl jikl  ij ij ij ijkl kl ij ijkl kl ij ijkl kl dt C:ε 2     symmetries = Cijklε kl ik↔ CCijkl ijlk jl↔ major 11 CCijkl= klij =εε+ε=εε+εε=()ijC ijkl klε kl C klij ij ( ij CC ijkl kl ij ijkl kl ) symmetries 22   C d ijkl (εεC ) dt ij ijkl kl 1 d duˆ d 1 =εεijC ijkl kl = ()εε::C 2 dt    dt dt 2 εε::C 17 Elastic Potential

duˆ 1 d =σε: = () ε ::C ε dt2 dt

 Consequences: 1. Consider the time derivative of the internal energy in the whole volume: dd d ∫∫uˆˆ()xx,, t dV= u() t dV=Uˆ() t = ∫σε : dV stress VVdt dt dt Vpower

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

REMARK The stress power, in elastic materials is an exact differential of the internal energy U ˆ . Then, in elastic processes, we can talk of the elastic energy U ˆ () t . 18 Elastic Potential

duˆ 1 d =σε: = () ε ::C ε dt2 dt

 Consequences: 2. Integrating the time derivative of the internal energy density, 1 utˆ ()x,,= εε()()() x t ::xC , ta+ x 2 and assuming that the density of the internal energy vanishes at the

neutral reference state, ut ˆ () xx ,0 0 = ∀ :

1 = 0 11 εε+==∀ ˆ ε= ε ε = σε ε ()()()x,t00 ::xC , ta x a () x 0 x u () :C : ( ): 2 22σ  Due to thermodynamic reasons the internal energy is assumed always positive 1 uˆ ()ε= εε::C > 0∀≠ ε0 2

19 Elastic Potential σε()()x,tt= C () x: x ,

 The internal energy density defines a potential for the stress tensor, and is thus, named elastic potential. The stress tensor can be computed as =σσT = ∂∂utˆ(ε (x , )) 1 1 1 1 ∂uˆ ()ε =(ε::CC ε) = : ε + ε : C = ( σ += σ ) σ σ = ∂∂εε ∂ε 2 222= σ

 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 ∂σε() ∂∂2uˆ ()εε()C : ∂ uˆ ()ε = = = C Cijkl = ∂ε ∂ εε ⊗∂ ∂ ε ∂εij ∂ε 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 =λµ11 ⊗+2 I  =++λδ δ µ δ δ δ δ ∈ Cijkl ij kl( ik jl il jk ) i, jkl , ,{} 1, 2, 3

Where: 1  is the 4th order unit tensor defined as []I =δδ + δδ I ijkl 2 ik jl il jk  λ and µ are scalar constants known as Lamé‘s 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 = λµ 11 ⊗+ 2 I into the generalized Hooke’s Law σε = C : , in index notation:

σij=C ijkl ε kl =( λδ ij δ kl + µ( δ ik δ jl + δ il δ jk) ) ε= kl = ε 11ij =λδδij kl ε+ kl 2 µδδik jl ε+ kl δδ il jk ε kl = λTr()ε δµij + 2 ε ij 22 =εε = =εll = Tr(ε) ji ij 1 1 = ε+ εε= 2 ij 2 ij ij  And the resulting constitutive equation is, Isotropic linear elastic σεε=λµTr ()1 + 2  constitutive equation. σ=+∈ λδ ε µ ε  ij ij ll2 ij ij ,{} 1, 2, 3 Hooke’s Law

23 Elastic Potential

 If the constitutive equation is, σεε=λµTr ()1 + 2 Isotropic linear elastic constitutive equation. σij=+∈ λδ ij ε ll2 µ ε ij ij ,{} 1, 2, 3 Hooke’s Law

Then, the internal energy density can be reduced to:

11 REMARK uˆ ()ε= σ : ε=(λµTr () ε1 + 2 ε) : ε= 22 The internal energy density is an =σ elastic potential of the stress tensor 11 as: =λµ ε1 ε+ ε:ε= Tr () :2 22ε ∂uˆ ()ε Tr( ) =σε()() =λµTr εε1 + 2 ∂ε 1 =λµTr2 ()ε + ε:ε 2

24 Inversion of the Constitutive Equation

1. ε is isolated from the expression derived for Hooke’s Law 1 σεε=λµTr ()1 + 2 ε=( σ−λ Tr () ε 1) 2µ 2. The trace of σ is obtained: Tr ()()σ= Tr(λ Tr ε11 +2 µ ε=) λ Tr()()()()() ε Tr+=+ 2 µ Tr ε 3 λµ 2 Tr ε =3 3. The trace of ε is easily isolated: 1 Tr ()εσ= Tr () 32λµ+ 4. The expression in 3. is introduced into the one obtained in 1. 11λ 1 ε= σ−λ Tr ()σ 1 ε=− Tr ()σ1 + σ 2µ 32 λµ+ 23µλ()+ 2 µ 2 µ

25 Inverse Isotropic Linear Elastic λ 1 ε=− Tr ()σ 1 + σ Constitutive Equation 23µλ()+ 2 µ 2 µ

 The Lamé parameters in terms of E and ν : µλ()32+ µ ν E E = λ = λµ+ ()()1+−νν 12 λ E ν = µ =G = 2()λµ+ 21()+ν  So the inverse const. eq. is re-written:  νν1+ εσσ=−+Tr () 1 Inverse isotropic linear  EE  elastic constitutive equation. νν1+ ε=−+∈ σδ σ ij,{} 1, 2, 3 Inverse Hooke’s Law.  ij EEll ij ij 11 εx =( σx −+ νσ() y σ z ) γxy = τ xy EG 11 εy =( σy −+ νσ()x σ z ) γxz = τ xz In engineering notation: EG 11 εz =( σz −+ νσ()x σ y ) γyz = τ yz EG

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.

µλ()32+ µ E = λµ+

 Poisson's ratio ν 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).

λ ν = 2()λµ+

27 Example

Consider an uniaxial traction test of an isotropic linear elastic material such that: σ > x 0 E, ν y σστy= z = xy = τ xz = τ yz = 0 σ x σ x

σ x σ x x

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

28 11 εx =( σx −+ νσ() y σ z ) γxy = τ xy σ > 0 EG x 11 εy =( σy −+ νσ() x σ z ) γxz = τ xz σστy= z = xy = τ xz = τ yz = 0 EG Solution 11 εz =( σz −+ νσ() x σ y ) γyz = τ yz EG

For ν = 0 : 1 εσx= x γxy = 0 1 There is no Poisson’s effect E εσx= x γxy = 0 E ν and the transversal normal ε=−= σγ0 εγ= 00= y E x xz y xz strains are zero. εγ= 00= ν z yz ε=−= σγ0 z E x yz

For ν = 0.5 : 1 1 The volumetric deformation is εσx= x γxy = 0 εσx= x γxy = 0 ε =++=εεε E E zero, tr xyz 0 , the 1 0.5 ε=−= σγ material is incompressible ε=−= σγ0 y x xz 0 y E x xz 2E and the volume is preserved. 1 0.5 ε=−= σγ ε=−= σγ0 z x yz 0 z E x yz 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 σσ:=σ 11 = Tr () sph m 3 σσ=σ m1 + ′ σ′ =dev σ = σ−σ m1

 Similarly for the strain tensor: 11 εε=11 = ) sph e Tr ( 1 33 εε= +e1 ´ 1 ε′ = dev εε= − e1 3 3

30 Spherical and deviatoric parts of Hooke’s Law

 Operating on the volumetric strain: e = Tr()ε

νν1+ εσ=−+Tr () 1 σ EE

νν1+ e =−+Tr()()σσ Tr1 Tr() EE = 3 = 3σ m K : bulk modulus 31()− 2ν E (volumetric strain modulus) e = σ σ m = e E m 31()− 2ν def 2 E K =+=λµ 3 3(1− 2ν )  The spherical parts of the stress and strain tensor are directly related: σ m = Ke

31 Spherical and Deviator Parts of Hooke’s Law

νν1+  Introducing σσ = σ 1 + ′ into εσσ =−+ Tr () 1 : m EE νν1+ ε =−+++Tr ()σσ11σσ′′() 1 EEmm ν ν = 0 ++ν ν + νν + ν ′′1 1 13 1 ′ =−−σσmmTr()11 Tr ()σσ 1 + 1 +=−+ σm1 σ E= 3 E E E EE E E Taking into account that σ m = e : 31()− 2ν Comparing this 12−ν 1E 1 ++ νν 1 1 ε = ee11+=+σσ′′with the expression − ν 1 E312() EE3 εε= e1 ´+ 3 1+ν 11 1+ν = = ε´= σ ′ E 2µ 2G E  The deviatoric parts of the stress and strain tensor are related component by component: σ ′′= 2Gε σεij ′′=2G ij 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:

σε′′= 2G σ m = Ke ij ij

33 Elastic Potential

 The internal energy density uˆ () ε defines a potential for the stress tensor and is, thus, an elastic potential: REMARK 1 ∂ ε ˆ ε= εε uˆ ( ) The constitutive elastic constants u () ::C σε= =  : 2 ∂ε tensor C is positive definite due to thermodynamic considerations.  Plotting u ˆ () ε vs. ε :

There is a minimum for ε = 0 :

∂uˆ ()ε =:()C ε = 0 ∂ε ε=0 ε=0

∂∂22uuˆˆ()εε() = =CC = ∂εε ⊗∂ ∂ εε ⊗∂ ε=0 εε=00=

34 Elastic Potential

 The elastic potential can be written as a function of the spherical and deviatoric parts of the strain tensor: ε::CC ε= ( :: ε) ε=σ : ε 111 uˆ (ε) = ε::C ε = σε := λµTr (ε) 1 + 2 ε: ε= 222 = σ 11 =ee11 +εε´´: += = Tr (ε ) = e = e2 33 112 =λTr (ε) 1:: ε += µλ εεTr ( ε) + µ εε: 12 22=ee2 11: + 1 ::ε´ += εε ´´ 93  3 Tr(ε′)=0 K 122 1 122 1 2 uˆ (ε) =λ ee ++=+ µμεε´: ´  λµeμ +εε´: ´ =e + εε´´: 2 3 23 3 Elastic potential in terms of the 1 uˆ (ε) = Ke2 +≥µ εε´: ´ 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 uˆ ()ε=:Ke2 +≥µ εε´ :´ 0 2  Consider now the following particular cases of isotropic linear elastic material:  Pure spherical deformation process ()1 1 ε = e 1 1 3 uˆ()1 = Ke2 ≥ 0 K > 0 bulk modulus 1 2 ε′() = 0  Pure deviatoric deformation process εε()2 = ′ Lamé’s second uˆ()2 =µ εε´´0: ≥ µ > 0 e()2 = 0 parameter REMARK ′′ εε:0=εεij ij ≥ 36 Limits in the Elastic Properties

 K and µ are related to E and ν through:

E E K = > 0 µ =G = > 0 31()− 2ν 21()+ν 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()+ν E ≥ 0 Young’s auxetic materials. modulus ν ≥ 0

 Therefore, E > 0 − ν 1 31() 2 0 ≤≤ν 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 ε () x , t and stresses σ () 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 ∇σ ⋅+()()x,,ttρρ bx = 00∂ t 2 2. Constitutive Equation. Isotropic Linear Elastic Constitutive Equation. This is a PDE system of σ ()()x,2t=λµ Tr εε1 + 15 eqns -15 unknowns: ux(),t 3 unknowns 3. Geometrical Equation. ε()x,t 6 unknowns Kinematic Compatibility. σ ()x,t 6 unknowns S 1 Which must be solved in ε()()x,,tt= ∇ ux =( u ⊗+⊗∇∇ u) 3 2 the RR × + space.

40 Boundary Conditions

 Boundary conditions in space  Affect the spatial arguments of the unknowns  Are applied on the boundary Γ of the solid, which is divided into three parts:  Prescribed displacements on Γ u : ux(,)tt= u* (,) x  * ∀x ∈Γu ∀t uii()(){}xx, tu= , ti ∈ 1, 2, 3

 Prescribed tractions on Γ σ : Γuu Γσσ Γ =Γ≡∂V σ ⋅= *  (,)xnttt (,) x Γu Γ=Γσ uu Γ σ =Γ u σσ  Γ=/{0}  ∀x ∈Γσ ∀t σ ⋅= * ∈  ij ()(){}xx, tnj t j , t i 1, 2, 3

 Prescribed displacements and stresses on Γ u σ : * uii()()xx,, tu= t  (i, jk ,∈{} 1, 2, 3 i ≠ j) ∀x ∈Γσ ∀t σ ⋅=* u  jk ()()xx,,tnk 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 = v x ∀∈ x V ∂ 0 t t=0

43 The Linear Elastic Problem

 Find the displacements ux () , t , strains ε () x , t and stresses σ () x , t such that

∂2ux(),t ∇σ⋅+()() x,,ttρρ00 bx = Cauchy’s Equation of Motion ∂ t 2 σ()()x,2t=λµ Tr εε1 + Constitutive Equation 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) Geometric Equation 2

* Γ=u : uu * Boundary conditions in space Γ=⋅σ : tnσ

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 , which gives a response (or solution) in displacements, strains and stresses. bx(),t * ux(),t tx(),t Mathematical * model ε()x,t ux(),t EDPs+BCs σ ()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 × + .  In certain cases, the integration space is reduced to R 3 . 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 22 ≈ ∂A/ ∂≈t 0 ∂R/ ∂≈t 0 2 0 ∂ t

46 The Quasi-Static Problem

 Find the displacements ux () , t , strains ε () x , t and stresses σ () x , t such that ∂2ux(),t ρ ≈ 0 0 ∂ t 2

∇σ ⋅+()()x,,ttρ0 bx = 0 Equilibrium Equation

σ()()x,2t=λµ Tr εε1 + Constitutive Equation 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) Geometric Equation 2

* Γ=u : uu * Boundary Conditions in Space Γ=⋅σ : tnσ

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 descriptor: it describes the evolution of the actions.

λ bx(), Mathematical ux(),λ * tx(),λ model ε()x,λ * ux(),λ EDPs+BCs σ ()x,λ not not ACTIONS = A ()x,λ RESPONSES = R ()x,λ

*  For each value of the actions A () x , λ * -characterized by a fixed value λ - a response R () x , λ * is obtained.  Varying λ * , a family of actions and its corresponding family of responses are 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 δ () tt = δλ (( )) , so for every time instant, it only depends on the corresponding value λ () 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 σ () x , t and ε () 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. Eliminates ux () , t and ε () x , t from the general system of equations. This generates a system of 6 eqns. for the 6 unknown components of σ () 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 ∇σ⋅+()()x,,ttρρ00 bx = Cauchy’s Equation of Motion ∂ t 2 σ()()x,2t=λµ Tr εε1 + Constitutive Equation 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) Geometric Equation 2

* Γ=u : uu * Boundary Conditions in Space Γ=⋅σ : tnσ

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, ε () x , t and σ () x , t will be found through substitution.

51 Displacement formulation

 Introduce the Constitutive Equation into Cauchy’s Equation of motion: σ()()x,2t=λµ Tr εε1 + ∂2u λ∇⋅ ε1 + µ ∇ε ⋅+ ρρ = []Tr() 2 00b 2 ∂2ux(),t ∂ t ∇σ⋅+()()x,,ttρρ bx = 00∂ t 2  Consider the following identities: = ∇∇⋅u ()() i ∂ ∂∂uu  ∂∂  ∂ ∇ε⋅()Tr()1u =εδ =kk δ = =()∇ ⋅= i ()11 ij ij  ∂xj ∂∂ xxjk  ∂∂ xx ik  ∂ x i = ∇ ⋅ u =⋅∈∇∇(){}u i 1, 2, 3 i

∇⋅=⋅()Tr () ε1 ∇∇()u

52 Displacement formulation

 Introduce the Constitutive Equation into Cauchy’s Equation of motion: σ()()x,2t=λµ Tr εε1 + ∂2u λ∇⋅ ε1 + µ ∇ε ⋅+ ρρ = []Tr() 2 00b 2 ∂2ux(),t ∂ t ∇σ⋅+()()x,,ttρρ bx = 00∂ t 2  Consider the following identities: = ∇2 u = ()∇∇()⋅u () i i ∂∂ε ∂1∂∂uuuu  112 ∂∂ ∂ 11 ()∇ε⋅ =ij =ii +j  = +j =∇2 u +()∇ ⋅=u i   () ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ i ∂ xj x j2 x j x i  22 xxjj x i x j 22xi = ∇ ⋅ 112 u =∇uu +⋅ ∇∇(){}i ∈1, 2, 3 22 i 11 ∇⋅= ε ∇∇() ⋅uu + ∇2 22

53 Displacement formulation

 Introduce the Constitutive Equation into Cauchy’s Equation of Movement: σ()()x,2t=λµ Tr εε1 + ∂2u λ∇⋅ ε1 + µ ∇ε ⋅+ ρρ = []Tr() 2 00b 2 ∂2ux(),t ∂ t ∇σ⋅+()()x,,ttρρ bx = 00∂ t 2  Replacing the identities: 11 ∇⋅=⋅()Tr () ε1 ∇∇()u ∇⋅= ε ∇∇() ⋅uu + ∇2 22 2 nd 112 ∂ u 2 order  Then, λ∇() ∇⋅+u2 µ ∇∇ () ⋅+ u ∇ ub +ρρ00 = 22 ∂ t 2 PDE system  ∂2u ()()λµ+∇∇ ⋅+u µ ∇2 ub + ρ = ρ  The Navier Equations  002  ∂t are obtained: ()λµ+uj, ji + µ u i ,0 jj += ρ bu i ρ 0 i i ∈{}1, 2, 3 

54 Displacement formulation

 The boundary conditions are also rewritten in terms of ux () , t : σ()()x,2 t=λµ Tr εε1 + = ∇ ⋅ u t* =λµTr ()εε nn +⋅2 () 1 * =∇ S u =(uu ⊗+⊗∇∇ ) tn=σ ⋅ 2 t* =λµ()(∇ ⋅ un + u ⊗+⊗ ∇∇ u) ⋅ n  The BCs are now:

uu= * on Γu  * uii= ui ∈{}1, 2, 3

* REMARK λµ()(∇⋅un + u ⊗ ∇∇ + ⊗ u) ⋅= n t  on Γσ The initial conditions λµun+ un +=∈ un t* i{}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 ∂∂ω ∂ω 2u λ+ −zr +θ +=ρρ xr= cosθ ()22G Gbr 2  ∂∂rrθ ∂ z ∂ t x(r ,θθ , z )≡= yr sin 1 ∂e ∂∂ωω ∂2u  ()λ+2G − 22 GGrz + +=ρρ b θ zz= r∂∂∂θ zrθ ∂ t2 ∂∂eG22 G∂∂ω 2u () λ+2Gr −()ω +rz += ρρ b ∂∂z rrθ r ∂θ z ∂t2

11∂uz ∂uθ ωrz= −Ωθ = − 2 rz∂∂θ

1 ∂∂uurz dV= r dθ dr dz Where: ωθ = −Ωzr = − 2 ∂∂zr

11∂ ()ruθ 1∂u r 11∂ ∂uθ ∂uz ωzr= −Ωθ = − e=() ru ++ 2 rr∂∂ rθ rr∂r r ∂∂θ z

56 Displacement formulation

 Navier equations in a spherical coordinate system:

2 ∂∂eG22G∂ωθ ∂ ur ()λ+2Gb − ()ωθϕ sin + +=ρr ρ ∂∂rrsinθθ rsinθ ∂ ϕ ∂t2 xr= sinθϕ cos  =θϕ ≡= θ ϕ λ + 2G ∂∂∂ω ∂2 xx()r, , yr sin sin () eG22r G uθ  − + ()rbωϕθsin θρ+= ρ zr= cosθ r ∂∂∂θ rsin θϕ rr sin θ ∂t2  λ + ∂2 () 2G ∂∂eG22 G∂ωr uϕ −ω + += ρρ 2 ()rbθϕ2 = θ θϕ rsinθϕ∂∂ rr r ∂θ ∂t dV rsin dr d d

11∂ 1∂uθ ωr = −Ωθϕ = ()uϕ sin θ − 2rr sin θ∂∂ θ sin θ φ Where:  11∂u 1∂ ()ruϕ ω = −Ω = r − θϕr  2r sin θφ∂∂ rr

11∂ 1∂ur ωϕθ= −Ωr = ()ruθ− 2 rr∂∂ r θ  1 ∂ ∂∂ =2 θθ ++ e 2 ()r ur sin ()ruθϕsin ()ru rrsinθ∂∂∂ θϕ 57 Stress formulation

Equilibrium Equation ∇σ⋅+x,ttρ bx ,0 = ()()0 (Quasi-static problem) νν1+ ε()x,t=−+ Tr () σσ1 Inverse Constitutive Equation EE 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) Geometric Equation 2

* Γ=u : uu * Boundary Conditions in Space Γ=⋅σ : tnσ

The aim is to reduce this system to a system with σ () x , t as the only unknowns. Once these are obtained, ε () 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: 2222 Compatibility Equations ∂∂εεij ∂∂εε jl +−−=kl ik 0i , jkl , , ∈{} 1, 2, 3 ∂x ∂ x ∂∂ xx ∂ x ∂ x ∂∂ xx (seen in Ch.3.) kl i j jl ik  Introducing the inverse constitutive equation into the compatibility equations and using the equilibrium equation: νν1+ ε=−σδ + σ ij EEpp ij ij ∂σ ij +=ρ0bj 0 ∂xi 2nd order  Beltrami-Michell Equations The are obtained: PDE system 2 ∂∂ρρ∂ ρ b 2 1 ∂ σ kk ν ()00bbki() ()0 j ∇+σδij =− ij − − ij,∈{} 1, 2, 3 11+νν ∂∂xxij − ∂xk ∂ xj ∂ x i

59 Stress formulation

 The boundary conditions are:  Equilibrium Equations: ∇σ⋅+ρ0b =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 : Γ σ σ ⋅=nt on Γσ

60 Stress formulation

 Once the stress field is known, the strain field is found by substitution. νν1+ ε()x,t=−+ Tr () σσ1 EE  The calculation, after, of the displacement field requires that the geometric equations be integrated with the prescribed displacements on Γ : u 1 ε()x=( ux () ⊗+⊗∇∇ ux ()) x ∈V 2 * ux()= u () x ∀ x ∈Γu

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) (II) εε()()xxPP,,tt≈ ∀P | δ >>  (I) (II) σσ()()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 proof is valid for lineal elasticity in infinitesimal strains.  The constitutive tensor C is used, so proof 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. εε00()()xx0=,t =

σσ00()()x= x0,t = 3. Isentropic and adiabatic processes – no longer isothermal !!!  Isentropic: entropy of the system remains constant  Adiabatic: deformation 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 θθ()()xx,t ≠= ,0 θ0 ∂θ ()x,t θ()x,0t = ≠ ∂t We will assume the temperature field is known.

 But the process is still isentropic and adiabatic: s() t≡ cnt s = 0

=ρ − ⋅ = ∀∆ ⊂ Qe ∫∫ r dV qn dS 0 V V VV∂ internal heat conduction sources from the exterior ρ rt−⋅=∇ qx0 ∀∀

67 Generalized Hooke’s Law

 The Generalized Hooke’s Law becomes:

σ()()()()xx,tt=CC :, εβ −θθ −0 =:, εβ x t −∆θ Generalized Hooke’s Law for  σ=C ε −− βθθ(){}ij, ∈ 1, 2, 3 linear thermoelastic problems  ij ijkl kl ij 0 Where  is the elastic constitutive tensor.  θ () x , t is the absolute temperature field.

 θθ 00 = () x , t is the temperature at the reference state.  β is the tensor of thermal properties or constitutive thermal constants tensor.  It is a positive semi-definite symmetric second-order tensor. REMARK A symmetric second-order tensor A is positive semi-definite when zT·A·z > 0 for every non-zero column vector z. 68 Isotropic Constitutive Constants Tensors

 An isotropic thermoelastic material must have the same elastic and thermal properties in all directions: th   must be a (mathematically) isotropic 4 order tensor: C =λµ11 ⊗+2 I  =++λδ δ µ δ δ δ δ ∈ Cijkl ij kl( ik jl il jk ) i, j , kl .{} 1, 2, 3 Where: 1 th   I is the 4 order symmetric unit tensor defined as []I =δδik jl + δδ il jk ijkl 2   λ and µ are the Lamé parameters or coefficients.

 β is a (mathematically) isotropic 2nd order tensor: β = β 1  β=βδ ∈  ij ij ij,{} 1, 2, 3 Where:  β is a scalar thermal constant parameter.

69 Isotropic Linear Thermoelastic Constitutive Equation

 Introducing the isotropic constitutive constants tensors β = β 1 and

C = λµ 11 ⊗+ 2 I into the generalized Hooke’s Law, σ = C : εβ −−()θθ0 (in indicial notation)

σij=C ijkl ε kl − βθθ ij () −=00( λδδij kl + µδδ( ik jl + δδ il jk) ) ε kl − βθθδ() −ij = = ε 11ij = λδδεij kl kl +2 µ δδεik jl kl + δδε il jk kl −− βθθδ()0 ij 22 = εll =εεji = ij = ∆θ = εij  The resulting constitutive equation is,

σεε=λTr ()11 +2 µ −∆ βθ Isotropic linear thermoelastic  σ= λδ ε + µ ε −∆ β θ δ ∈ constitutive equation. ij ij ll2 ij ij ij,{} 1, 2, 3

70 Inversion of the Constitutive Equation

1. ε is isolated from the Generalized Hooke’s Law for linear thermoelastic problems: εσ=−−11:: +∆θ β σε=: −∆β θ CC C α 2. The thermal expansion coefficients tensor α is defined as: def − αβ= C 1 : It is a 2nd order symmetric tensor which involves 6 thermal expansion coefficients

3. The inverse constitutive equation is obtained: − εσ=C 1 : +∆θ α

71 Inverse Isotropic Linear Thermoelastic Constitutive Equation

 For the isotropic case:

 − νν1+ C 1 =−11 ⊗+ I  12− ν EE →=α=−1 ββ11  C : ()  −1 νν1+ E Cijkl =−++δδij kl ( δδik jl δδ il jk ) i, j , kl . ∈{} 1, 2, 3  EE  The inverse const. eq. is re-written:  νν1+ εσσ=−Tr () 11 + +∆αθ  EE Inverse isotropic linear thermo  νν1+ elastic constitutive equation.  ε=− σ δ + σ +∆ α θδ ij, ∈ 1, 2, 3  ij ll ij ij ij {}  EE  Where α is a scalar thermal expansion coefficient related to the scalar thermal constant parameter β through: 12− ν αβ= E 72 Thermal Stress

 Comparing the constitutive equations,

σεε=λµ Tr ()1 + 2 Isotropic linear elastic constitutive equation. σεε=λ Tr ()11 +2 µ −∆ βθ Isotropic linear thermoelastic constitutive equation. t = σnt = σ

the decomposition is made: σσ=nt − σ t Where:  σ nt is the non-thermal stress: the stress produced if there is no temperature increment.  σ t is the thermal stress: the “corrector” stress due to the temperature increment.

73 Thermal Strain

 Similarly, by comparing the inverse constitutive equations,

νν1+ Inverse isotropic linear elastic εσσ=−+ Tr () 1 EE constitutive eq. νν1+ εσσ=− Tr () 11 + +∆αθ Inverse isotropic linear thermoelastic EE t constitutive eq. = εnt = ε the decomposition is made: εε=nt + ε t Where:  ε nt is the non-thermal strain: the strain produced if there is no temperature increment.  ε 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

σεnt = C : σβt = ∆θ σσ=nt − σ t Isotropic material: Isotropic material: σnt =λµTr() εε1 + 2 σt =βθ ∆ 1

− εσnt = C 1 : εαt = ∆θ εε=nt + ε t Isotropic material: Isotropic material: nt νν1+ ε=−+Tr() σσ1 εt =αθ ∆ 1 EE These are the equations used nt t in FEM codes. σ=CC:: ε = εε − −−11nt t ε=CC:: σ = σσ +

75 Thermal Stress and Strain

REMARK 1 In thermoelastic problems, a state of zero strain in a body does not necessarily imply zero stress. εσ=→=00nt

σσ=−t =−∆βθ1 ≠0

REMARK 2 In thermoelastic problems, a state of zero stress in a body does not necessarily imply zero strain. σε=→=00nt

εε==∆≠t αθ1 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:

∇σ⋅+()()x,,ttρ0 bx = 0 Equilibrium Equation

σε()()x,,tt=C :x −∆βθ1 Constitutive Equation 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) Geometric Equation 2

* Γ=u : uu * Boundary Conditions in Space Γ=⋅σ : tnσ

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 tx* (),t Elastic model ε()x,t * ux(),t EDPs+BCs σ ()x,t ∆θ ()x,t

REMARK ∆ θ () 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 σσ = nt − σ t and replaced into the governing equations:  Momentum equations σ= σnt − σ t ∇ ⋅= σ ∇⋅σnt − ∇⋅ σ t = ∇⋅σnt − ∇()βθ ∆  βθ∆ 1

∇σ⋅+ρ0b0 = ∇⋅σnt +=ρ ˆ 1 0 b0 ∇ nt ⋅σ+ρ −∇ βθ ∆= 0 b0() 1 ρ0 bbˆ =−∆∇ ()βθ not ρ0 = bˆ

81 1st Thermal Analogy

 Boundary equations: σ=σnt − σ t σσnt⋅−n t ⋅= nt* nt * * σ ⋅=ntˆ σ ⋅=nt Γσ : ˆ** σσnt ⋅=nt** +t ⋅ n = t +()βθ ∆ n tt=+∆()βθ n       βθ∆⋅1 n tˆ* ANALOGOUS PROBLEM – A linear elastic problem can be solved as:

nt ˆ ˆ 1 ∇σ⋅+ρ0b0 = with bb=−∆∇()βθ Equilibrium Equation ρ0 σεεnt =C : =λµTr()1 + 2 ε Constitutive Equation 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) Geometric Equation 2

* Γ=u : uu nt ˆ* ** Boundary Conditions in Space Γσ : σ ⋅=nt with ttˆ = +∆βθ 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 ux(),t Elastic model txˆ* (),t ε()x,t EDPs+BCs nt ux* (),t σ ()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 ˆˆ−  ∇ ()βθ∆ ACTIONS b bb ρ ∗   0 u u0∗   def ()I (,)tt−()II (,) = −= =0 =()III AAxx* * **   A t tˆˆ tt− *   ≡ t −∆βθn     ()  ∆θ 0 ∆θ  ∆θ

RESPONSES uu    0   0  def ()I ()II        ()III RR(,)xxtt− (,) =−=εε   0  =  0  =R  nt  nt    σ  σ   σσ−   −∆βθ1  = −σt

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

84 1st Thermal Analogy

THERMOELASTIC ANALOGOUS THERMOELASTIC ORIGINAL ELASTIC (TRIVIAL) PROBLEM (I) PROBLEM (II) PROBLEM (III) 1 1 bx(),t bxˆ (),t =−∆ b ∇ ()βθ  =∇ βθ ∆ ρ b () * 0 ρ0 tx(),t ** ()I ()II txˆ ()(),t =+∆ tβθ n ()III * x,t * x,t x,t tn=−∆()βθ A () ux(),t A () A () ux* (),t u0 ∗ = ∆θ ()x,t ∆=θ 0 ∆θ ()x,t

ux(),t ux(),t u0= ()I ()II ε()x,t ε()x,t ()III ε = 0 R ()x,t R ()x,t R ()x,t σ ()x,t σnt ()x,t σ =−=−∆σ t ()βθ1

86 2nd Thermal Analogy

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

∇σ⋅+()()x,,ttρ0 bx = 0 Equilibrium Equation

εσ()()x,,tt=C-1 :x +∆αθ1 Inverse Constitutive Equation 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) Geometric Equation 2

* Γ=u : uu * Boundary Conditions in Space Γ=⋅σ : tnσ

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 tx* (),t Elastic model ε()x,t * ux(),t EDPs+BCs σ ()x,t ∆θ ()x,t

REMARK ∆ θ () 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 assumption is made that ∆ θ (,) x t and α () x are such that the thermal t =αθ ∆ strain field ε1 is integrable (satisfies the compatibility equations).  If the thermal strain field is integrable, there exists a field of thermal displacements, ux t () , t , which satisfies: 1 εt()()x,t =αθ ∆1 =∇Stuu =( t ⊗+⊗ ∇∇ ut) 2 1 ∂ ut ∂ ut εt =∆=+() α θδ i j ij, ∈{} 1, 2, 3 ij ij ∂∂ 2 xxji REMARK The solution ux t () , t is determined except for a rigid body motion characterized by a rotation tensor Ω ∗ and a displacement vector c * . The family of admissible solutions is u t ()() x ,, tt = ux + Ω ∗ ⋅+ x c * . This movement can be arbitrarily chosen (at convenience).  Then, the total displacement field is decomposed by defining: def unt (,) xtt= ux (,) − ut (,) x 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, εε = nt + ε t and uu = nt + u t are replaced into the governing equations:  Geometric equations:

S S nt t S nt S t S nt t ε∇=u = ∇() uu += ∇ u + ∇ u = ∇ u + ε εt εnt εnt= S∇u nt εε=nt + εt

 Boundary equations:

uu= * unt+= uu t * Γ=: uu** unt =− u ut uu=nt + u t u

90 2nd Thermal Analogy

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

∇σ⋅+ρ0b0 = Equilibrium Equation εσnt = C-1 : Inverse constitutive Equation ε∇nt= Su nt Geometric Equation

Γ=−nt * t u : u uu * Boundary Conditions in space Γσ : σ ⋅=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 ux(),t Elastic model nt * ε ()x,t tx() ,t EDPs+BCs σ ()x,t unt = ux* ()(),,tt − uxt

ORIGINAL PROBLEM (I) ANALOGOUS PROBLEM (II)

92 2nd Thermal Analogy

 If the actions and responses of the original and analogous problems are compared:   ACTIONS bb 0 ∗∗tt def ()I ()II u uu−  u ()III AA(,)xxtt−=−== (,) ** A tt  0 ∆∆θθ0 

RESPONSES uunt  u t  ut  def ()I ()II      ()III RR(,)xxtt− (,) =−εε nt  =  ε t  =∆= αθ1  R        σσ 00  

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

93 2nd Thermal Analogy

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

bx(),t b b0 = * ∗ t ∗ t ()I tx(),t ()II uu= − u ()III u = ux(),t x,t * x,t x,t A () ux(),t A () t* A () t0* = ∆θ ()x,t ∆=θ 0 ∆θ ()x,t

ux(),t uxnt (),t u= uxt (),t ()I ()II nt ε()x,t ε ()x,t ()III εε=t =()αθ ∆ 1 R ()x,t R ()x,t R ()x,t σ ()x,t σ ()x,t σ = 0

95 2nd Analogy in structural analysis

nt t uuxx= uuxx= =αθ ∆ x nt t εεxx= εxx= ε = αθ ∆ * t t Γ: uuu = − =−∆αθ Γ=: uu =∆αθ ux x xx=  ux xx=     0

96 Thermal Analogies

 Although the 2nd analogy is more commonly used , 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: αα()x =const . =  Lineal thermal increment: ∆=θ ax + by + cz + d is of special interest because the thermal strains are: ε t =α∆θ 1 =linear polinomial and trivially satisfy the compatibility conditions (involving second order derivatives).

97 Thermal Analogies

 In the particular case  Homogeneous material: αα()x =const . =  Constant thermal increment: ∆=θθ()x const . =∆ the integration of the strain field has a trivial solution because the thermal strains are constant ε= t = ∆ θα 1 const . , therefore: rigid body motion (can be chosen arbitrarily: t ∗∗ ux(),t =αθ ∆ x +Ω ⋅+ xc at convenience) The thermal displacement is: uxt (),t =αθ ∆ x xu+t =+∆ xαθ x =+∆()1 αθ 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:

∇σ⋅+()()x,,ttρ0 bx = 0 Equilibrium Equation

σε()()x,,tt=C :x −∆βθ1 Constitutive Equation 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) Geometric Equation 2

* Γ=u : uu * Boundary Conditions in space Γ=⋅σ : tnσ

ux(),0 = 0 Initial Conditions ux (),0 = v0

100 Linear Thermoelastic Problem

 Consider two possible systems of actions:

()1 2 bx(),t bx() (),t ()1 2 tx* (),t tx*() (),t ()1 ()2 ()1 2 A ()x,t ≡ ux* (),t A ()x,t ≡ ux*() (),t ()1 2 ∆θ ()x,t ∆θ () ()x,t ()1 ()2 vx0 () vx0 ()

and their responses :

1 ()2 ux() (),t ux(),t ()1 1 ()2 ()2 R ()x,t ≡ ε() ()x,t R ()x,t ≡ ε ()x,t 1 ()2 σ() ()x,t σ ()x,t

101 Superposition Principle

()()()()()3 11 2 2  The solution to the system of actions AAA = λλ + where ()1 ()2 ()()()()()3 11 2 2 λ and λ are two given scalar values, is RRR = λλ + .

The response to the lineal thermoelastic problem caused by two or more groups of actions is the lineal combination of the responses 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.

 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: ε x 11  ε γγ ε y x xy xz  εε ε 22 def x xy xz ε z not. 11 6 ε = ε εε=  γ ε γ {}ε = ∈ R xy y yz xy y yz γ 22 xy εε ε  xz yz z γ 11 xz γxz γε yz z  22 γ yz REMARK VOIGT σ The double contraction () σ:ε is NOTATION x σ transformed into the scalar (dot) y στx xy τ xz product () {}{} σε ⋅ : def σ z 6 σ ≡ τxy στ y yz {}σ = ∈ R vectors  τ  xy ττσxz yz z σ:ε={}{} σ ⋅ ε σij ε ij= σε i i   τ xz 2nd order  τ tensors yz

104 Inverse Constitutive Equation

 The inverse constitutive equation is rewritten:

νν1+ t ε=−Tr () σσ11 + +∆αθ {}{}{}ε=⋅+Cˆ −1 σε EE

t Where Cˆ − 1 is an elastic constants inverse matrix and {} ε is a thermal strain vector: αθ∆ 00  1 −−νν εt ≡∆αθ 0 00 00 EE E 00αθ∆ −−νν1 0 00 EE E  −−νν1 0 00 E EE Cˆ −1 =  αθ∆ 1  0 0 0 00 αθ∆ G   αθ∆ 1 t 00 0 0 0 {}ε =  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:

t {}{}{}σ=⋅−Cˆ () εε νν 1 000 −−νν ˆ 11 Where C is an elastic νν 1 000 constants matrix : 11−−νν  νν 1000 −ν 11−−νν ˆ E ()1 C = 12− ν ()()1+−νν 12 000 0 0 21()−ν  12− ν 000 0 0 21()−ν  12− ν 000 0 0 21()−ν

106 Chapter 6 Linear Elasticity

6.1 Hypothesis of the Linear Theory of Elasticity The linear theory of elasticity can be considered a simplification of the general theory of elasticity, but a close enough approximation for most engineering ap- plications. The simplifying hypotheses of the linear theory of elasticity are a) Infinitesimal strains. The displacements and its gradients are small, see Chapter 2. • Small displacements. The material configuration (corresponding to the reference time t0) is indistinguishable from the spatial one (corresponding to the present time t) and, consequently, the spatial and material coordinates cannot be distinguished from each other either, see Figure 6.1.

= + =⇒ ≈ x X u x X (6.1) Theory≈ 0 and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 6.1: Small displacements are considered in the linear theory of elasticity.

263 264 CHAPTER 6. LINEAR ELASTICITY

From (6.1), one can write ∂x F = = 1 =⇒ |F| ≈ 1 . (6.2) ∂X

Remark 6.1. As a consequence of (6.1), there is no difference between the spatial and material descriptions of a property,

x = X =⇒ γ (x,t)=γ (X,t)=Γ (X,t)=Γ (x,t) , and all references to the spatial and material descriptions (in addition to any associated concepts such as local derivative, material derivative, etc.) no longer make sense in inﬁnitesimal elasticity. Likewise, the spatial Nabla differential operator (∇) is indistinguishable from the material one ∇¯ , ∂ (•) ∂ (•) = =⇒ ∇(•)=∇¯ (•) . ∂X ∂x

Remark 6.2. As a consequence of (6.2) and the principle of conser- vation of mass, the density in the present configuration ρt ≡ ρ (X,t) coincides with the one in the reference configuration ρ0 ≡ ρ (X,0) (which is assumed to be known), ρ = ρ |F| ≈ ρ , Theory0 t and Problemst and, therefore, the density is not an unknown in linear elasticity problems. Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar • Small displacement gradients. As a consequence, no distinction is made between the material strain tensor E(X,t) and the spatial strain tensor e(x,t), which collapse into the infinitesimal strain tensor ε (x,t).

E(X,t) ≈ e(x,t)=ε (x,t) ⎧ ⎪ 1 ⎨⎪ ε = ∇Su = (u ⊗ ∇ + ∇ ⊗ u) 2 (6.3) ∂ ⎪ 1 ∂ui u j ⎩ εij = + i, j ∈{1,2,3} 2 ∂x j ∂xi

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Linear Elastic Constitutive Equation. Generalized Hooke’s Law 265

b) Existence of a neutral state. The existence of a neutral state in which the strains and stresses are null is accepted. Usually, the neutral state is under- stood to occur in the reference conﬁguration. ε (x,t )=0 0 (6.4) σ (x,t0)=0

c) The deformation process is considered (in principle) to be isothermal1 and adiabatic.

Deﬁnition 6.1. Isothermal processes are those that take place at a temperature θ (x,t) that is constant along time,

θ (x,t) ≡ θ (x) . Adiabatic processes are those that take place without heat generation at any point and instant of time.

Heat generated inside a domain V per unit of time:

Qe = ρrdV− q · n dS = 0 ∀ΔV ⊂ V V ∂V =⇒ ρr − ∇ · q = 0 ∀x ∀t Slow deformation processes are commonly considered to be adiabatic.

Theory and Problems 6.2 Linear Elastic Constitutive Equation. Generalized Hooke’sContinuum Law Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar Hooke’s law for unidimensional problems establishes the proportionality between the stress, σ, and the strain, ε, by means of the constant named elastic modulus, E, σ = Eε . (6.5) In the theory of elasticity, this proportionality is generalized to the multidimen- sional case by assuming the linearity of the relation between the components of the stress tensor σ and those of the strain tensor ε in the expression known as generalized Hooke’s law,

1 The restriction to isothermal processes disappears in the linear theory of thermoelasticity, which will be addressed in Section 6.6.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 266 CHAPTER 6. LINEAR ELASTICITY

⎧ ⎨ σ (x,t)=C : ε (x,t) Generalized (6.6) Hooke’s law ⎩ σij = Cijklεkl i, j ∈{1,2,3} which constitutes the constitutive equation of a linear elastic material. The fourth-order tensor C (denoted as tensor of elastic constants) has 34 = 81 components. However, due to the symmetry of the tensors σ and ε, it must exhibit certain symmetries in relation to the exchange of its indexes. These are: C = C ijkl jikl → C = C minor symmetries ijkl ijlk (6.7)

Cijkl = Ckli j → major symmetries

Consequently, the number of different constants in the tensor of elastic constants C is reduced to 21.

Remark 6.3. An essential characteristic of the elastic behavior (which is veriﬁed in (6.5)) is that the stresses at a certain point and time, σ (x,t), depend (only) on the strains at said point and time, ε (x,t), and not on the history of previous strains.

6.2.1 Elastic Potential Consider the speciﬁc internal energy u(x,t) (internal energy per unit of mass) and the density of internalTheory energy uˆ(x and,t) (internal Problems energy per unit of volume), which related through uˆ(x,t)=ρ u(x,t) , Continuum Mechanics0 for Engineers © X. Oliver and C.uˆ Agelet de Saracibar (6.8) (ρ ) ρ du ≈ ρ du = d 0 u = duˆ , dt 0 dt dt dt where ρ0 ≈ ρ (see Remark 6.2) has been taken into account. Consider now the energy equation in its local form2, du duˆ . duˆ . ρ = = σ : d + ρ r − ∇ · q = σ : ε =⇒ = σ : ε , (6.9) 0 dt dt 0 dt

. 2 The identity d = ε, characteristic of the inﬁnitesimal strain case, is considered here.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Linear Elastic Constitutive Equation. Generalized Hooke’s Law 267

where the adiabatic nature of the deformation process (ρ0 r − ∇ · q = 0) has been considered. Then, the global (integral) form of the energy equation in (6.9)is obtained by integrating over the material volume V.

Global form of the energy equation in linear elasticity dU d duˆ . = udVˆ = dV = σ : ε dV dt dt dt ≡ (6.10) Vt V V V with U (t)= uˆ(x,t) dV V

Here, U (t) is the internal energy of the material volume considered.

Remark 6.4. The stress power (in the case of linear elasticity) is an exact differential, . dU stress power = σ : ε dV = . V dt

Replacing now (6.6)in(6.9), i ↔ k ↔ j l duˆ not . . . . 1 . . = uˆ = σ : ε = ε σ = ε C ε = ε C ε + ε C ε = dt ij ij ij ijkl kl 2 ij ijkl kl ij ijkl kl = 1 ε. C Theoryε + ε. C andε = Problems1 ε. C ε + ε C ε. = 2 ij ijkl kl kl kli j ij 2 ij ijkl kl ij ijkl kl = 1 d ε C ε = 1 d (ε C ε) , Continuumij ijkl Mechanicskl : : for Engineers 2 dt © X. Oliver2 dt and C. Agelet de Saracibar (6.11) where the symmetries in (6.7) have been taken into account. Integrating the expression obtained and imposing the condition that the density of internal energy 3 uˆ(x,t0) in the neutral state be null (for t = t0 ⇒ ε (x,t0)=0) produces the density of internal energy.

3 The conditionu ˆ(x,t0)=0 can be introduced without loss of generality.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 268 CHAPTER 6. LINEAR ELASTICITY ⎫ 1 ⎪ uˆ(x,t)= (ε (x,t) : C : ε (x,t)) + a(x) ⎬ 2 =⇒ ⎪ uˆ(x,t )=0 ∀x ⎭ 0 (6.12) 1 =⇒ ε (x,t ) : C : ε (x,t )+a(x)=a(x)=0 ∀x 2 0 0 = 0 Density of 1 1 uˆ(ε)= (ε : C : ε)= ε C ε (6.13) internal energy 2 2 ij ijkl kl

Now, (6.13) is differentiated with respect to ε, considering once more the symmetries in (6.7). ⎧ ⎪ ⎪ ∂uˆ(ε) 1 1 1 1 ⎨⎪ = C : ε + ε : C = C : ε + C : ε = C : ε = σ ∂εε 2 2 2 2 ⎪ ⎪ ∂uˆ(ε) 1 1 1 1 ⎩ = Cijklεkl + εkl Ckli j = Cijklεkl + Cijklεkl = Cijklεkl = σij ∂εij 2 2 2 2 ⎧ (6.14) ⎪ ∂ (ε) ⎨⎪ uˆ = σ =⇒ ∂εε ⎪ (6.15) ⎪ ∂uˆ(ε) ⎩ = σij i, j ∈{1,2,3} ∂εij Equation (6.15) qualiﬁes the density of internal energyu ˆ(ε) as a potential for the stresses (which are obtained by differentiation of this potential), named elastic potential. Theory and Problems ⎧ ⎪ ⎪ 1 1 ⎨⎪ uˆ(ε)= ε : C : ε = σ : ε Continuum Mechanics2 for Engineers2 © X. Oliver and C. Agelet= σ de Saracibar Elastic potential ⎪ (6.16) ⎪ ∂ (ε) ⎩⎪ uˆ = σ ∂εε

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Isotropy. Lame’s´ Constants. Hooke’s Law for Isotropic Linear Elasticity 269

6.3 Isotropy. Lame’s´ Constants. Hooke’s Law for Isotropic Linear Elasticity

Deﬁnition 6.2. An isotropic material is that which has the same properties in all directions.

The elastic properties of a linear elastic material are contained in the tensor of elastic constants C deﬁned in (6.6) and (6.7). Consequently, the components of this tensor must be independent of the orientation of the Cartesian system used4.

Consider, for example, the systems {x1,x2,x3} and {x1 ,x2 ,x3 } in Figure 6.2, the constitutive equation for these two systems is written as

{x1,x2,x3} =⇒ [σ ]=[C] : [ε] (6.17)

{x1 ,x2 ,x3 } =⇒ [σ ] =[C] : [ε] and, for the case of an isotropic material, the components of C in both systems must be the same ( [C]=[C] ). Therefore, the aforementioned deﬁnition of isotropy, which has a physical character, translates into the isotropic character, in the mathematical sense, of the tensor of elastic constants C. ⎧ ⎪ ⎨⎪C = λ1 ⊗ 1 + 2μI Tensor of elastic C = λδ δ + μ δ δ + δ δ (6.18) constants ⎪ ijkl ij kl ik jl il jk ⎩⎪Theory and Problems i, j,k,l ∈{1,2,3} Continuum Mechanics for Engineers Here, λ and μ are known© X. as OliverLame’s´ constantsand C. Agelet, whichde characterize Saracibar the elastic behavior of the material and must be obtained experimentally.

Remark 6.5. The isotropy condition reduces the number of elastic constants of the material from 21 to 2.

4 A tensor is isotropic if it maintains its components in any Cartesian coordinate system. The most general expression of a fourth-order isotropic tensor is C = λ1⊗1+2μI , ∀λ, μ. Here, the fourth-order symmetric (isotropic) unit tensor I is deﬁned by means of its components as [ ] = δ δ + δ δ / I ijkl ik jl il jk 2.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 270 CHAPTER 6. LINEAR ELASTICITY

Figure 6.2: Representation of the Cartesian systems {x1,x2,x3} and {x1 ,x2 ,x3 }.

Introducing (6.18)in(6.6) results in the isotropic linear elastic constitutive equation, 1 1 σ = C ε = λδ δ ε +2μ δ δ ε + δ δ ε . (6.19) ij ijkl kl ijklkl 2 ik jl kl 2 il jk kl ε ε ε = ε ll ij ji ij εij ⎧ ⎨ Constitutive eqn. for a σ = λ Tr (ε)1 + 2μεε linear elastic material. ⎩ (6.20) Hooke’s law. σij = λδijεll + 2μεij i, j ∈{1,2,3}

6.3.1 Inversion of Hooke’s Law. Young’s Modulus. Poisson’s Ratio The constitutive equation (6.20) provides the stresses in terms of the strains. To obtain its inverse expression,Theory the following and procedure Problems is followed. a) The trace of (6.20) is obtained, ⎫ ⎪ TrContinuum(σ )=λ Tr (ε)Tr (1 Mechanics)+2μ Tr (ε)=(3λ for+ 2μ Engineers)Tr (ε) ⎪ © X. Oliver and C. Agelet de Saracibar⎬⎪ 3 =⇒ ⎪ ( = )=⇒ σ = λε δ + με =( λ + μ)ε ⎪ i j ii ll ii 2 ii 3 2 ll ⎭⎪ (6.21) 3 1 =⇒ Tr (ε)= Tr (σ ) . (3λ + 2μ) b) ε is isolated from (6.20) and introduced in (6.21), 1 1 λ 1 ε = − λ Tr (ε)1 + σ = − Tr (σ )1 + σ . (6.22) 2μ 2μ 2μ (3λ + 2μ) 2μ

The new elastic properties E (Young’s modulus) and ν (Poisson’s ratio) are de- ﬁned as follows. ⎫ ⎪ Young’s modulus or μ (3λ + 2μ) ⎪ E = ⎬ tensile (elastic) modulus λ + μ =⇒ ⎪ λ ⎪ Poisson’s ratio ν = ⎭ 2(λ + μ) ⎧ (6.23) ⎪ ⎪ ν ⎨⎪ λ = E ( + ν)( − ν) =⇒ 1 1 2 ⎪ ⎪ E ⎩ μ = = G shear (elastic) modulus 2(1 + ν)

Equation (6.22) can be expressed in terms of E and ν, resulting in the inverse Hooke’s law. ⎧ ⎪ ⎪ ν 1 + ν ⎪ ε = − Tr (σ )1 + σ Inverse constitutive ⎨ E E ν + ν (6.24) equation for an isotropic ⎪ ε = − σ δ + 1 σ ⎪ ij ll ij ij linear elastic material ⎪ E E ⎩ i, j ∈{1,2,3}

Finally, (6.24) is rewritten, using engineering notation for the components of the strain and stress tensors. Theory and Problems 1 1 εx = (σx − ν (σy + σz)) γxy = τxy ContinuumE Mechanics for EngineersG ε = ©1 ( X.σ − Oliverν (σ + andσ )) C. Ageletγ = 1 deτ Saracibar (6.25) y E y x z xz G xz ε = 1 (σ − ν (σ + σ )) γ = 1 τ z E z x y yz G yz

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 272 CHAPTER 6. LINEAR ELASTICITY

Example 6.1 – Consider an uniaxial tensile test of a rectangular cuboid composed of an isotropic linear elastic material with Young’s modulus E and shear modulus G, such that its uniform stress state results in

σx = 0 and σy = σz = τxy = τxz = τyz = 0 . Obtain the strains in engineering notation.

Solution

From (6.25) one⎧ obtains ⎧ ⎪ σ ⎪ τxy ⎪ ε = x ⎪ γ = = 0 ⎨⎪ x ⎨⎪ xy E σ τG σ = σ = =⇒ ε = −ν x τ = τ = τ = =⇒ γ = xz = y z 0 ⎪ y xy xz yz 0 ⎪ xz 0 ⎪ E ⎪ G ⎩⎪ σx ⎩⎪ τyz ε = −ν γ = = 0 z E yz G Therefore, due to these strains, the rectangular cuboid subjected to an uniaxial tensile test, shown in the ﬁgure below, stretches in the x-direction and contracts in the y- and z-directions.

Theory and Problems 6.4 Hooke’s Law in Spherical and Deviatoric Components ConsiderContinuum the decomposition Mechanics of the stress tensor forσ and Engineers the deformation tensor ε in their spherical and deviatoric© X. Oliver parts, and C. Agelet de Saracibar

1 σ = Tr (σ )1 + σ = σ 1 + σ , (6.26) 3 m

1 1 ε = Tr (ε)1 + ε = e1 + ε . (6.27) 3 3

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Hooke’s Law in Spherical and Deviatoric Components 273

The volumetric strain e = Tr (ε) is obtained by computing the trace of (6.24).

ν 1 + ν 1 − 2ν e = Tr (ε)=− Tr (σ )Tr (1)+ Tr (σ )= Tr (σ ) = E E E 3 3σm (6.28) ( − ν) = 3 1 2 σ E m ⎧ ⎪ ⎪ E ⎨ σm = e = Ke =⇒ 3(1 − 2ν) ⎪ (6.29) ⎪ def 2 E ⎩ K = λ + μ = = bulk modulus 3 3(1 − 2ν) Introducing (6.26), (6.27) and (6.29)in(6.24), results in

ν 1 + ν 1 − 2ν 1 + ν ε = − 3σ 1 + (σ 1 + σ )= σ 1 + σ = E m E m E m E E e 3(1 − 2ν) 1 1 + ν 1 1 1 + ν (6.30) = e1 + σ =⇒ ε = e1 + ε = e1 + σ 3 E 3 3 E

+ ν =⇒ ε = 1 σ = 1 σ = 1 σ . E 2μ 2G Equations (6.29) and (6.30) relate the spherical part (characterized by the mean σ σ ε stress m and the volumetricTheory strain e) and and the Problems deviatoric part ( and ) of the stress and strain tensors as follows.

Continuumσ = Mechanics for Engineers m Ke © X. Oliver and C.Spherical Agelet de part Saracibar (6.31) σ = 2Gε Deviatoric part σ ij = 2Gε ij i, j ∈{1,2,3}

Remark 6.6. Note the proportionality between σm and e as well as

between σ ij and ε ij (component to component), see Figure 6.3.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 274 CHAPTER 6. LINEAR ELASTICITY

Figure 6.3: Hooke’s law in spherical and deviatoric components.

6.5 Limits in the Values of the Elastic Properties Thermodynamic considerations allow proving that the tensor of elastic constants C is positive-deﬁnite5, and, thus, ε : C : ε > 0 ∀ε = 0 . (6.32)

Remark 6.7. As a consequence of (6.32), the elastic potential is always null or positive, 1 uˆ(ε)= ε : C : ε ≥ 0 . 2

Remark 6.8. The elastic potential has a minimum at the neutral state, that is, for ε = 0 (see Figure 6.4). In effect, from (6.15), 1 Theory∂uˆ(ε and) Problems∂ 2uˆ(ε) uˆ(ε)= ε : C : ε , σ = = C : ε and = C . 2 ∂εε ∂εε ⊗ ∂εε ε = Then,Continuum for 0, Mechanics for Engineers ∂uˆ(ε) © X. Oliver( andε) C. Agelet de Saracibar = 0 =⇒ uˆ has an extreme ∂εε (maximum-minimum) at ε = 0. ε= 0 ∂ 2 (ε) uˆ = C =⇒ ∂εε ⊗ ∂εε The extreme is a minimum. ε=0 positive- deﬁnite

5 A fourth-order symmetric tensor A is defined positive-definite if for all second-order tensor x = 0 the expression x : A : x = xij Aijkl xkl > 0 is satisfied and, in addition, x : A : x = 0 ⇔ x = 0.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Limits in the Values of the Elastic Properties 275

Figure 6.4: Elastic potential.

Consider the expression of the elastic potential (6.16) and the constitutive equation (6.20), then,

1 1 1 uˆ(ε)= ε : C : ε = σ : ε = λ Tr (ε)1 + 2μεε : ε = 2 2 2 1 1 (6.33) = λ Tr (ε) 1 : ε +μεε : ε = λ Tr2 (ε)+μεε : ε . 2 2 Tr (ε)

Expression (6.33) can also be written in terms of the spherical and deviatoric components of strain6,

1 2 1 uˆ(ε)= λ Tr (ε) + μεε : ε = λe2 + μεε : ε . (6.34) 2 2 e Theory and Problems Here, the double contraction of the inﬁnitesimal strain tensor is

1 1 1 2 ε : ε =Continuume1 + ε : Mechanicse1 + ε = e2 1 : for1 + e Engineers1 : ε + ε : ε = 3 ©3 X. Oliver and9 C. Agelet3 de Saracibar 3 Tr (ε )=0 (6.35) 1 = e2 + ε : ε . 3 Replacing (6.35)in(6.34),

1 1 1 2 uˆ(ε)= λe2 + μe2 + μεε : ε = λ + μ e2 + μεε : ε . (6.36) 2 3 2 3 K

6 The trace of a deviatoric tensor is always null, Tr(ε )=0.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 276 CHAPTER 6. LINEAR ELASTICITY

1 uˆ(ε)= Ke2 + μεε : ε ≥ 0 (6.37) 2

Consider now an isotropic linear elastic material, characterized by a certain value of its elastic properties. Equation (6.37) must be satisﬁed for any deformation process. Consider two particular types: a) A pure spherical deformation process ⎫ ⎪ ε (1) = 1 ⎬ e 1 ( ) 1 3 =⇒ uˆ 1 = Ke2 ≥ 0 =⇒ K > 0 (6.38) ( ) ⎪ ε 1 = 0 ⎭ 2

b) A pure deviatoric deformation process7 (2) ε = ε ( ) =⇒ uˆ 2 = μεε : ε ≥ 0 =⇒ μ > 0 (6.39) e(2) = 0 Equations (6.38) and (6.39) lead to E E K = > 0 and μ = G = > 0 (6.40) 3(1 − 2ν) 2(1 + ν) which are the limits in the values of the elastic constants K and G. Experience proves that the Poisson’s ratio ν is always non-negative8 and, therefore ⎫ E > ⎬⎪ ( + ν) 0 2 1 =⇒ E > 0 , ⎭⎪ νTheory≥ 0 and Problems ⎫ (6.41) E > ⎬⎪ Continuum( − Mechanicsν) 0 for Engineers1 3 1© X.2 Oliver and=⇒ C. Agelet0 ≤ ν ≤ de. Saracibar ⎭⎪ 2 E ≥ 0

7 The double contraction or double dot product of a tensor by itself is always equal or greater than zero: ε : ε = ε ijε ij ≥ 0. 8 In rare cases, a material can have a negative Poisson’s ratio. Such materials are named auxetic materials.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 The Linear Elastic Problem 277

Initial actions: b(x,0) t = 0 → t(x,0)

Actions along time t: b(x,t) t(x,t)

Figure 6.5: Linear elastic problem.

6.6 The Linear Elastic Problem Consider the linear elastic solid9 in Figure 6.5, which is subjected to certain actions characterized by the vector of body forces b(x,t) in the interior of the volume V and the traction vector t(x,t) on the boundary ∂V. The set of equations that allow determining the evolution along time of the displacements u(x,t), strains ε (x,t) and stresses σ (x,t) is named linear elastic problem.

6.6.1 Governing Equations The linear elastic problem is governed by the following equations: a) Cauchy’s equation (balance of linear momentum)

∂ 2u(x,t) ∇ · σ (x,t)+ρ b(x,t)=ρ 0 0 ∂ 2 t (3 equations) (6.42) ∂σ Theory∂ 2 and Problems ij + ρ = ρ u j ∈{ , , } 0 b j 0 2 j 1 2 3 ∂xi ∂t Continuum Mechanics for Engineers b) Constitutive equation©(isotropic X. Oliver linear and elastic) C. Agelet10 de Saracibar

σ (x,t)=λ Tr (ε (x,t))1 + 2μεε (x,t) (6 equations) (6.43) σij = λδijεll + 2μεij i, j ∈{1,2,3}

9 Here, linear elastic solid refers to a continuous medium constituted by a material that obeys the linear elastic constitutive equation. 10 The symmetry of the stress and strain tensors entails that only six of the nine equations are different from one another. In addition, when listing the unknowns, only the different components of these tensors will be considered.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 278 CHAPTER 6. LINEAR ELASTICITY

c) Geometric equation (compatibility relation between inﬁnitesimal strains and displacements)

1 ε (x,t)=∇Su(x,t)= (u ⊗ ∇ + ∇ ⊗ u) 2 (6 equations) (6.44) 1 ∂ui ∂u j εij = + i, j ∈{1,2,3} 2 ∂x j ∂xi

These equations involve the following unknowns: • u(x,t) (3 unknowns) • ε (x,t) (6 unknowns) (6.45) • σ (x,t) (6 unknowns) and constitute a system of partial differential equations (PDEs). The system is composed of 15 differential equations with the 15 unknowns listed in (6.45). These are of the type (•)(x,y,z,t), and, thus, must be solved in the R3 × R+ space. The problem is well deﬁned when the adequate boundary conditions are provided.

6.6.2 Boundary Conditions 6.6.2.1 Boundary Conditions in Space

Consider the boundary Γ ≡ ∂V of the solid is divided into three parts, Γu, Γσ and Γuσ , characterized by (see Figure 6.6) Γu Γσ Γuσ = Γ ≡ ∂V , (6.46) Γu Γσ = Γu Γuσ = Γuσ Γσ = {/0} . Theory and Problems These allow deﬁning the boundary conditions in space, that is, those conditions that affect the spatial arguments (x,y,z) of the unknowns (6.45) of the problem.

• BoundaryContinuumΓu: prescribed Mechanics displacements for Engineers © X. Oliver and C. Agelet de Saracibar u(x,t)=u∗ (x,t) ∀x ∈ Γ ∀t (6.47) ( , )= ∗ ( , ) ∈{ , , } u ui x t ui x t i 1 2 3

• Boundary Γσ : prescribed tractions σ (x,t) · n = t∗ (x,t) ∀x ∈ Γσ ∀t (6.48) σ ( , ) · = ∗ ( , ) , ∈{ , , } ij x t n j ti x t i j 1 2 3

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 The Linear Elastic Problem 279

Figure 6.6: Boundary conditions in space.

• Boundary Γ σ : prescribed displacements and tractions11 u ( , )= ∗ ( , ) ui x t ui x t i, j,k ∈{1,2,3}, i = j ∀x ∈ Γ σ ∀t (6.49) σ ( , ) · = ∗ ( , ) u jk x t nk t j x t

Example 6.2 – Exempliﬁcation of the boundary conditions in space.

Solution The different types of boundary conditions in space are illustrated in the following ﬁgure of a beam.

Theory and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

11 In Γuσ certain components (components i ) have prescribed displacements while the others (components j ) have the traction vector prescribed.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 280 CHAPTER 6. LINEAR ELASTICITY

6.6.2.2 Boundary Conditions in Time: Initial Conditions In general, at the initial or reference time t = 0 the displacements and velocities are known. ⎫ ( , )= ⎬⎪ u x 0 0 ∂ ( , ) ∀x ∈ V (6.50) u x t not= . ( , )= ( ) ⎭⎪ ∂ u x 0 v0 x t t=0

6.6.3 Quasi-Static Problem The system of equations (6.42)to(6.50) can be visualized, from a mechanical point of view, as a system of actions or data (the body forces b(x,t), the traction vector t∗ (x,t), the prescribed displacements u∗ (x,t) and the initial velocities v0 (x)) that, introduced into a mathematical model composed of the differential equations given in Section 6.6.1 and the boundary conditions described in Section 6.6.2, provides the response or solution in the form of the displacement field u(x,t), the deformation field ε (x,t) and the stress field σ (x,t).

⎫ b(x,t) ⎪ ⎧ ⎬⎪ ⎨ ( , ) ∗ ( , ) MATHEMATICAL u x t u x t ⇒ ⇒ ε ( , ) ∗ ( , ) ⎪ MODEL : ⎩ x t t x t ⎭⎪ PDEs + BCs σ ( , ) (6.51) ( ) x t v0 x not not Responses = R (x,t) Actions = A (x,t)

In the most general caseTheory12, both the and actions Problems and the responses depend on time (see Figure 6.7) and the system of PDEs must be integrated over both the space Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 6.7: Evolution of the response along time.

12 In this case (general problem), the problem is named dynamic problem.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 The Linear Elastic Problem 281

and the time variables R3 × R+ . However, in certain cases, the integration space can be reduced in one dimension, the one corresponding to time. This is the case for the so-called quasi-static problems.

Deﬁnition 6.3. A quasi-static linear elastic problem is a linear elastic problem in which the acceleration is considered to be negligible,

∂ 2u(x,t) a = ≈ 0 . ∂t2 This hypothesis is acceptable when the actions are applied slowly. In such case, the variation of the actions A along time is slow ∂ 2A/∂t2 ≈ 0 and, due to the continuous dependency of the results on the data, the variation of the response R along time is also small ∂ 2R/∂t2 ≈ 0 . Consequently, the second derivative of the response is considered negligible and, in particular,

∂ 2u(x,t) ≈ 0 . ∂t2

The governing differential equations are reduced to the following in the case of a quasi-static problem: a) Cauchy’s equation, also known as equilibrium equation.

∂ 2u(x,t) ∇ · σ (x,t)+ρ b(x,t)=ρ = 0 (6.52) Theory0 and0 Problems∂t2

Continuum Mechanics for Engineers b) Constitutive equation© X. Oliver and C. Agelet de Saracibar

σ (x,t)=λ Tr (ε (x,t))1 + 2μεε (x,t) (6.53)

c) Geometric equation, which no longer involves any time derivative.

1 ε (x,t)=∇Su(x,t)= (u ⊗ ∇ + ∇ ⊗ u) (6.54) 2

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 282 CHAPTER 6. LINEAR ELASTICITY

The system of differential equations only needs to be integrated in space (solved in R3 ) with the boundary conditions in space of Section 6.6.2.1. More- over, time merely serves as a parameter describing the evolution of the actions, which are usually described in terms of the load factor or pseudo-time λ (t).

⎫ ⎧ b(x,λ) ⎬ MATHEMATICAL ⎨ u(x,λ) ∗ ( ,λ) ⇒ ⇒ ε ( ,λ) u x ⎭ MODEL : ⎩ x t∗ (x,λ) PDEs + BCs σ (x,λ) (6.55) Actions not= A (x,λ) Responses not= R (x,λ)

In other words, for each value of the actions (characterized by a ﬁxed value of λ ∗), A (xλ ∗), a response R (x,λ ∗) is obtained. Varying the value of λ ∗ produces a family of actions and its corresponding family of responses.

Example 6.3 – Application to a typical problem of strength of materials.

Solution Consider a cantilever beam subjected to a force F (t) at its free end. Under the quasi-static problem hypothesis, and considering a parametrized action of the type λF∗, the response (deﬂection at its free end) can be computed as

∗ 3 δ (λ)=λ F l . 3EI This is the classical solution obtained in strength of materials. Now, if the evolution along time of λ (t) can take any form, the value of δ (t)=δ (λ (t)) correspondingTheory to andeach instant Problems of time only depends on the corresponding value of λ. Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Solution to the Linear Elastic Problem 283

6.7 Solution to the Linear Elastic Problem The linear elastic problem can be typically solved following two different approaches: a) Displacement formulation b) Stress formulation Their names are directly related to which is the main unknown being considered in each formulation (displacements or stresses, respectively).

Remark 6.9. At present, the displacement formulation has greater application because most numerical methods used to solve the linear elastic problem are based on this approach. Theory and Problems

6.7.1 DisplacementContinuum Formulation: Mechanics Navier’s for Equation Engineers © X. Oliver and C. Agelet de Saracibar Consider the equations that constitute the linear elastic problem:

∂ 2u ∇ · σ + ρ b = ρ Cauchy’s equation 0 0 ∂t2 σ = λ Tr (ε)1 + 2μεε Constitutive equation (6.56) 1 ε = ∇Su = (u ⊗ ∇ + ∇ ⊗ u) Geometric equation 2

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 284 CHAPTER 6. LINEAR ELASTICITY

∗ Γu : u = u ∗ Boundary conditions in space (6.57) Γσ : t = σ · n u(x,0)=0 . Initial conditions (6.58) u(x,0)=v0

The aim is to pose a reduced system in which only the displacement ﬁeld u(x,t) intervenes as an unknown. The ﬁrst step consists in replacing the constitutive equation in the Cauchy’s equation, both given in (6.56). ∂ 2u ∇ · σ + ρ b = ∇ · λ Tr (ε)1 + 2μεε + ρ b = ρ 0 0 0 ∂t2 (6.59) ∂ 2u =⇒ λ∇ · Tr (ε)1 + 2μ∇ · ε + ρ b = ρ 0 0 ∂t2 Consider the following identities13.

∂ε ∂ ∂ ∂ ∂ 2 ∂ ∂ ∇ · ε = ij = 1 ui + u j = 1 ui + 1 u j = i ∂x j ∂x j 2 ∂x j ∂xi 2 ∂x j∂x j 2 ∂xi ∂x j ! " 1 1 ∂ 1 1 = ∇2u + (∇ · u)= ∇2u + ∇(∇ · u) i ∈ {1,2,3} i ∂ 2 2 xi 2 2 i 1 1 ∇ · ε = ∇(∇ · u)+ ∇2u 2 2 Theory and Problems (6.60)

∂ ∂ ∂ ∂ ∂ ∇ · Continuum(ε) = (ε Mechanicsδ )= ul δ for= Engineersul = Tr 1 i © X.ll ij Oliver and C.ij Agelet de Saracibar ∂x j ∂x j ∂xl ∂xi ∂xl ∂ = (∇ · )= ∇(∇ · ) ∈ { , , } u u i i 1 2 3 ∂xi ∇ · Tr (ε)1 = ∇(∇ · u) (6.61)

13 ∇2 def= ∂ 2 /(∂ ∂ ) The Laplace operator of a vector v is deﬁned as v i vi x j x j .

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Solution to the Linear Elastic Problem 285

Equation (6.59) can be rewritten by replacing the expressions in the identities (6.60) and (6.61), resulting in ⎧ ⎨ ∂ 2u Navier’s (λ + μ)∇(∇ · u)+μ∇2u + ρ b = ρ 0 0 ∂ 2 (6.62) equation ⎩ t (λ + μ)u j, ji + μ ui, jj+ ρ0 bi = ρ0 u¨i i ∈ {1,2,3} which constitutes a system of second-order PDEs in displacements u(x,t) (that must be, thus, integrated in R3 × R+), and receives the name of Navier’s equation. The boundary conditions can also be written in terms of the displacements as follows. Replacing the constitutive equation of (6.56) in the boundary conditions in Γσ of (6.57) results in t∗ = σ · n = λ Tr (ε)1 + 2μεε · n = λ (Tr (ε))n + 2μεε · n = (6.63) = λ (∇ · u)n + 2μ ∇S · u · n = λ (∇ · u)n + μ (u ⊗ ∇ + ∇ ⊗ u) · n and the boundary conditions in space (6.57) expressed in terms of the displacements are obtained. = ∗ u u Γ = ∗ ∈ { , , } in u ui ui i 1 2 3 λ (∇ · ) + μ ( ⊗ ∇ + ∇ ⊗ ) · = ∗ (6.64) u n u u n t Γ λ + μ ( + )= ∗ , ∈ { , , } in σ ul,l ni ui, j n j u j,i n j ti i j 1 2 3

The initial conditions (6.58) remain unchanged. Integrating the system (6.62) yields the displacement field u(x,t). Differentiation of this field and substitution in the geometric equationTheory of (6.56) produces and the Problems strain field ε (x,t), and, finally, replacing the strain in the constitutive equation results in the stress field σ (x,t). Continuum Mechanics for Engineers 6.7.1.1 Navier’s Equation© in X. Cylindrical Oliver and and SphericalC. Agelet Coordinates de Saracibar Navier’s equation (6.62) is expressed in compact or index notation and is independent of the coordinate system considered. The components of this equation are expressed as follows in the cylindrical and spherical coordinate systems (see section 2.15).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 286 CHAPTER 6. LINEAR ELASTICITY

Cylindrical coordinates

∂e 2μ ∂ω ∂ωθ ∂ 2u (λ + 2μ) − z + 2μ + ρb = ρ r ∂r r ∂θ ∂z r ∂t2 2 1 ∂e ∂ωr ∂ωz ∂ uθ (λ + 2μ) − 2μ + 2μ + ρbθ = ρ (6.65) r ∂θ ∂z ∂r ∂t2

∂e 2μ ∂ (rωθ ) 2μ ∂ω ∂ 2u (λ + 2μ) − + r + ρb = ρ z ∂z r ∂r r ∂θ z ∂t2 where

1 1 ∂uz ∂uθ ω = −Ωθ = − r z 2 r ∂θ ∂z

1 ∂ur ∂uz ωθ = −Ω = − zr 2 ∂z ∂r

1 1 ∂ (ruθ ) 1 ∂ur ω = −Ω θ = − z r 2 r ∂r r ∂θ

1 ∂ (ru ) 1 ∂uθ ∂u e = r + + z r ∂r r ∂θ ∂z

⎡ ⎤ x = r cosθ x(r,θ,z) not≡ ⎣ y = r sinθ ⎦ z = z Theory and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 6.8: Cylindrical coordinates.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Solution to the Linear Elastic Problem 287

Spherical coordinates

∂e 2μ ∂ ωφ sinθ 2μ ∂ωθ ∂ 2u (λ + 2μ) − + + ρb = ρ r ∂r r sinθ ∂θ r sinθ ∂φ r ∂t2 ∂ ω θ 2 1 ∂e 2μ ∂ωr 2μ r φ sin ∂ uθ (λ + 2μ) − + + ρbθ = ρ (6.66) r ∂θ r sinθ ∂φ r sinθ ∂r ∂t2 2 1 ∂e 2μ ∂ (rωθ ) 2μ ∂ωr ∂ uφ (λ + 2μ) − + + ρbφ = ρ r sinθ ∂φ r ∂r r ∂θ ∂t2 where ' ( 1 1 ∂ uφ sinθ 1 ∂uθ ω = −Ωθφ = − r 2 r sinθ ∂θ r sinθ ∂φ ' ( ∂ 1 1 ∂ur 1 ruφ ωθ = −Ωφ = − r 2 r sinθ ∂φ r ∂r

1 1 ∂ (ruθ ) 1 ∂ur ω = −Ω θ = − z r 2 r ∂r r ∂θ ' ( 2 1 ∂ r ur sinθ ∂ (ruθ sinθ) ∂ ruφ e = + + r2 sinθ ∂r ∂θ ∂φ

⎡ ⎤ x = r sinθ cosφ x(r,θ,φ) not≡ ⎣ y = r sinθ sinφ ⎦ Theory and Problemsz = z cosθ

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 6.9: Spherical coordinates.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 288 CHAPTER 6. LINEAR ELASTICITY

6.7.2 Stress Formulation: Beltrami-Michell Equation This formulation is solely valid for the quasi-static case discussed in Sec- tion 6.6.3. Consider, thus, the equations that constitute the quasi-static linear elastic problem:

∇ · σ + ρ0b = 0 Equilibrium equation ν 1 + ν ε = − Tr (σ )1 + σ Inverse constitutive equation (6.67) E E 1 ε = ∇Su = (u ⊗ ∇ + ∇ ⊗ u) Geometric equation 2 ∗ Γu : u = u ∗ Boundary conditions in space (6.68) Γσ : t = σ · n where the inverse constitutive (6.24) (strains in terms of stresses) has been considered in (6.67). The starting point of the stress formulation is the geometric equation of (6.67) from which, by means of successive differentiation, the displacements are eliminated and the compatibility equations14 are obtained,

εij,kl + εkl,ij− εik, jl − ε jl,ik = 0 i, j,k,l ∈ {1,2,3} . (6.69) Then, the equations of the problem are deduced in the following manner: a) The constitutive equation of (6.67) is replaced in the compatibility equations (6.69). b) The resulting expression is introduced in the equilibrium equation of (6.67). This results in the equationTheory and Problems

Beltrami-Michell equation Continuum Mechanicsν for Engineers 2 1 © X. Oliver and C. Agelet de Saracibar ∇ σ + σ , = − δ (ρ b ), − (ρ b ), − (ρ b ) (6.70) ij 1 + ν ll ij 1 − ν ij 0 l l 0 i j 0 j ,i i, j ∈ {1,2,3} which receives the name of Beltrami-Michell equation and constitutes a system of second-order PDEs in stresses σ (x) that must be solved in R3. The boundary conditions of this system are the equilibrium equation of (6.67), which, being a system of ﬁrst-order PDEs, acts as the boundary conditions of the second-order system in (6.70), and the boundary conditions in Γσ .

14 The deduction of the compatibility equations has been studied in Chapter 3, Section 3.3.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Unicity of the Solution to the Linear Elastic Problem 289

∇ · σ + ρ0b = 0 Equilibrium equation (6.71)

∗ σ · n = t in Γσ Boundary conditions in Γσ (6.72)

The integration of the system in (6.70) yields the stress ﬁeld σ (x). Substi- tution of the stresses in the inverse constitutive equation of (6.67) results in the strains ε (x). However, to obtain the displacement ﬁeld u(x), the geometric equations must be integrated, taking into account the boundary conditions 15 in Γu . ⎧ ⎪ 1 ⎨ ε (x)= u(x) ⊗ ∇ + ∇ ⊗ u(x) x ∈ V 2 (6.73) ⎩⎪ ∗ u(x)=u (x) ∀x ∈ Γu

Thus, the system of second-order PDEs must be integrated in R3.

Remark 6.10. The need to integrate the second system (6.73) (when the stress formulation is followed) is a disadvantage (with respect to the displacement formulation described in Section 6.7.1) when numerical methods are used to solve the linear elastic problem.

6.8 Unicity of the Solution to the Linear Elastic Problem Theory and Problems Theorem 6.1. The solution ⎡ ⎤ u(x,t) Continuum MechanicsR ( , ) not≡ ⎣ ε ( , ) for⎦ Engineers © X. Oliverx t andx C.t Agelet de Saracibar σ (x,t) to the linear elastic problem posed in (6.42) to (6.44) is unique.

Proof A not ∗ ∗ T Consider the actions deﬁned by A (x,t) ≡ [b(x,t), u (x,t), t (x,t), v0 (x)] , in the domains V, Γu, Γσ and V, respectively, (satisfying Γσ Γu = ∂V and

15 An analytical procedure to integrate these geometric equations was provided in Chapter 3, Section 3.4.2

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 290 CHAPTER 6. LINEAR ELASTICITY

Figure 6.10: Linear elastic problem.

Γσ Γu = 0)/ act on the linear elastic problem schematically represented in Fig- ure 6.10. The possible solutions R (x,t) not≡ [u(x,t), ε (x,t), σ (x,t)]T to the linear elastic problem must satisfy the equations: ∂ 2u ∇ · σ + ρ b = ρ Cauchy’s equation 0 0 ∂t2 σ = λ Tr (ε)1 + 2μεε Constitutive equation (6.74) 1 ε = ∇Su = (u ⊗ ∇ + ∇ ⊗ u) Geometric equation 2 ∗ Γu : u = u ∗ Boundary conditions in space (6.75) Γσ : t = σ · n u(x,0)=0 . TheoryInitial and conditions Problems (6.76) u(x,0)=v0 Continuum Mechanics for Engineers The unicity of the solution© X. Oliver is proven and as follows. C. Agelet Suppose de Saracibar the solution is not unique, that is, there exist two different solutions to the problem, ⎡ ⎤ ⎡ ⎤ u(1) (x,t) u(2) (x,t) ( ) not ⎢ ⎥ ( ) not ⎢ ⎥ R 1 (x,t) ≡ ⎣ ε (1) (x,t) ⎦ and R 2 (x,t) ≡ ⎣ ε (2) (x,t) ⎦ σ (1) (x,t) σ (2) (x,t) (6.77) such that R(1) = R(2) , which, therefore, must satisfy equations (6.74)to(6.76) and are the elastic re- not ∗ ∗ T sponses to the same action A (x,t) ≡ [b(x,t), u (x,t), t (x,t), v0 (x)] . Con-

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Unicity of the Solution to the Linear Elastic Problem 291 sider now a possible response constituted by the difference R(2) −R(1), ⎡ ⎤ ⎡ ⎤ u(2) (x,t) − u(1) (x,t) u(x,t) def ( ) ( ) not ⎢ ( ) ( ) ⎥ def ⎢ ⎥ R (x,t) = R 2 −R 1 ≡ ⎣ ε 2 (x,t) − ε 1 (x,t) ⎦ = ⎣ ε(x,t) ⎦ . (6.78) σ (2) (x,t) − σ (1) (x,t) σ (x,t) Note how the answer R satisﬁes the following equations: • Cauchy’s equation with b = 0 16 ∇ · σ (x,t)=∇ · σ (2) (x,t) − σ (1) (x,t) = ∇ · σ (2) − ∇ · σ (1) = ' ( ' ( ∂ 2u(2) ∂ 2u(1) = −ρ b + ρ − −ρ b + ρ = (6.79) 0 0 ∂t2 0 0 ∂t2

∂ 2 (2) ∂ 2 (1) ∂ 2 = ρ u − ρ u = ρ u 0 ∂t2 0 ∂t2 0 ∂t2 • Constitutive equation17

σ (x,t)=σ (2) (x,t) − σ (1) (x,t)=C : ε (2) −C : ε (1) = (6.80) = C : ε (2) − ε (1) = C : ε

• Geometric equation

ε(x,t)=ε (2) (x,t) − ε (1) (x,t)=∇Su(2) − ∇Su(1) = Theory and Problems (6.81) = ∇S u(2) − u(1) = ∇Su Continuum Mechanics for Engineers © X. Oliver and∗ C. Agelet de Saracibar • Boundary conditions in Γu with u = 0 ⎧ ⎪ ⎨ u(x,t)=u(2) (x,t) − u(1) (x,t)=u∗ − u∗ = 0 ∀t =⇒ Γ → u ⎪ (6.82) ⎩⎪ ∂u(x,t) . =⇒ = u(x,t)=0 ∂t 16 The fact that the Nabla operator (∇ ∗ (•)) is a linear operator is used advantageously here, that is, ∇ ∗ (a + b)=∇ ∗ a + ∇ ∗ b, where ∗ symbolizes any type of differential operation. Likewise, the operator ∂ 2 (•,t)/∂t2 is also a linear operator. 17 The property that the operator C : is a linear operator is applied here, that is, C : (a + b)=C : a +C : b.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 292 CHAPTER 6. LINEAR ELASTICITY

∗ • Boundary conditions in Γσ with t = 0 ( ) ( ) ( ) ( ) Γσ → σ (x,t) · n = σ 2 (x,t) − σ 1 (x,t) · n = σ 2 · n − σ 1 · n = (6.83) = t∗ − t∗ = 0

• Initial conditions with v0 = 0 ⎧ ⎪ ⎨ u(x,0)=u(2) (x,0) − u(1) (x,0)=0 − 0 = 0 ⎪ (6.84) ⎩⎪ ∂u(x,0) . . ( ) . ( ) = u(x,0)=u 2 (x,0) − u 1 (x,0)=v − v = 0 ∂t 0 0 Consider now the calculation of the integral = Γ 0in σ Divergence . . Theorem . n· σ · u dS = n · σ · u dS = ∇· σ · u dV = 0 , (6.85) ∂V Γ Γσ V u = 0inΓu where the conditions (6.82) and (6.83) have been applied. Operating on (6.85) results⎧ in ⎪ 2 ⎪ . . . ∂ u . . T ⎨ ∇ · σ · u = ∇ · σ · u + σ : ∇u = ρ · u + σ : ∇u 0 ∂t2 ⎪ . . ⎪ ∂ . ∂σ . ∂ ∂ 2 . ∂ ⎩⎪ σ = ij + σ u j = ρ u j + σ u j , ∈ { , , } iju j u j ij 0 2 u j ji i j 1 2 3 ∂xi ∂xi Theory∂xi and∂t Problems∂xi (6.86) where the condition (6.79) has been considered. On the other hand18, Continuum Mechanics for Engineers . . T . 1©. X. Oliver and. 1 C.. Agelet de Saracibar. . ∇u = u ⊗ ∇ = u ⊗ ∇ + ∇ ⊗ u + u ⊗ ∇ − ∇ ⊗ u = ε + Ω =⇒ 2 2 . . . S . ε = ∇ u Ω = ∇au . T . . . T . σ ∇ = σ ε + σ Ω =⇒ σ ∇ = σ ε . : u : : : u : = 0 (6.87) . 18 σ Ω The fact that is. a symmetric. tensor and is an antisymmetric one is considered here, which leads to σ :Ω = σijΩij = 0.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Unicity of the Solution to the Linear Elastic Problem 293

In addition19, v2 . . . ∂ · ∂ 2 u . ∂ 2 u . 1 u u 1 ∂ v ·v ρ0 · u = ρ0 · u = ρ0 = ρ0 = ∂t2 ∂t2 2 ∂t 2 ∂t (6.88)

d 1 ∂ 2 u . d 1 = ρ v2 =⇒ ρ · u = ρ v2 . 0 dt 2 0 ∂t2 0 dt 2

Replacing (6.88) and (6.87)in(6.86), and the resulting expression in (6.85), and taking into account the deﬁnition of internal energy U givenin(6.10) produces . d 1 . ∇ · σ · u dV = ρ v2 dV + σ : ε dV = 0 =⇒ 0 dt 2 V V V d 1 . ρ v2dV + σ : ε dV = 0 =⇒ (6.89) dt 2 0 V V dK/dt dU/dt

dK dU d + = K + U = 0 ∀t ≥ 0 . (6.90) dt dt dt

Note, though, that at the initial time t = 0 the following is satisﬁed (see (6.10), (6.13) and (6.84)) ⎫ ⎪ K = 1ρ 2 = 1ρ · = ⎪ 0v dV 0 v0 v0 dV 0 ⎪ t=0 2 t=0 2 ⎪ V V . ⎬ = u0 0 ⇒ K + U = 0 Theory and Problems⎪ = U = ( , ) = 1 ε C ε = ⎪ t 0 uˆ x t dV : : dV 0 ⎪ t=0 t=0 2 t=0 t=0 ⎪ V V ⎭ Continuum Mechanics= 0 for Engineers © X. Oliver and C. Agelet de Saracibar (6.91) and the integration of (6.90) with the initial condition (6.91) leads to

K + U = 0 ∀t ≥ 0 , (6.92) where 1 K = ρ v2 dV ≥ 0 ∀t ≥ 0 . (6.93) 2 0 V ≥ 0

def 19 Here, the deﬁnition |v| = v is used.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 294 CHAPTER 6. LINEAR ELASTICITY

Comparing (6.92) and (6.93) necessarily leads to the conclusion K + U = 0 ∀ ≥ =⇒ U = 1ε C ε ≤ ∀ ≥ . t 0 : : dV 0 t 0 (6.94) K ≥ 0 2 V On the other hand, since the tensor of elastic constants C is positive-deﬁnite (see (6.32)) ,

ε(x,t) : C : ε(x,t) ≥ 0 ∀x ∈ V ∀t ≥ 0 =⇒ 1 (6.95) U = ε : C : ε dV ≥ 0 ∀t ≥ 0 . 2 V Then, comparing (6.94) and (6.95) necessarily leads to U ( ) ≤ t 0 ∀ ≥ =⇒ U ( )= 1ε C ε = ∀ ≥ . t 0 t : : dV 0 t 0 (6.96) U (t) ≥ 0 2 V

Considering once more the positive-deﬁnite condition of tensor C 20, 1 U = ε : C : ε dV = 0 ∀t ≥ 0 =⇒ ε : C : ε = 0 ∀x, ∀t ≥ 0 (6.97) 2 V ≥ 0 and, necessarily, from the positive-deﬁnite condition of C it is deduced that ε : C : ε = 0 ⇐⇒ ε(x,t)=0 ∀x, ∀t ≥ 0 (6.98)

( ) ( ) ( ) ( ) ε(x,t)=ε 2Theory− ε 1 = 0 and=⇒ Problemsε 2 = ε 1 . (6.99) In addition, replacing (6.99)in(6.81) results in

Continuum Mechanics1 ∂u ∂u for Engineers ε(x,t)=∇S · u = 0 ©=⇒ X. Oliveri and+ C.j Agelet= 0 i, dej ∈ Saracibar{1,2,3} , (6.100) 2 ∂x j ∂xi

20 The following theorem of integral calculus is applied here: If φ (x) ≥ 0 and φ (x) dΩ = 0 =⇒ φ (x)=0 ∀x ∈ Ω . Ω

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Saint-Venant’s Principle 295 which is a system of six homogeneous and ﬁrst-order PDEs. Its integration leads to the solution21 ( , )= Ω · + u x t x c with rotation translation

⎡ ⎤ ⎡ ⎤ (6.101) 0 −θ θ c ⎢ 3 2 ⎥ ⎢ 1 ⎥ Ω not≡ not≡ , ⎣ θ3 0 −θ1 ⎦ and c ⎣ c2 ⎦ −θ2 θ1 0 c3 where Ω is an antisymmetric tensor (rotation tensor dependent on three constants θ1,θ2 and θ3) and c is a constant vector equivalent to a translation. Ulti- mately, the solution (6.100) to the system (6.101) are the displacements u(x,t) compatible with a null strain ε(x,t)=0, which correspond to a rigid body motion. The integration constants in Ω and c are determined by imposing the boundary conditions (6.82) (u(x,t)=0 ∀x ∈ Γu), therefore, if the rigid body motion is impeded through the restrictions in Γu, one obtains Ω = 0 and c = 0. In conclusion, ( , )=Ω · + u x t x c =⇒ ( , )= (2) − (1) = =⇒ (2) = (1) . u x t u u 0 u u Ω ≡ 0 ; c ≡ 0 (6.102) Finally, replacing (6.99)in(6.80) yields

( ) ( ) ( ) ( ) σ (x,t)=C : ε = 0 = σ 2 − σ 1 =⇒ σ 2 = σ 1 . (6.103) Then, observing (6.99),(Theory6.102) and (6.103 and) leads Problems to the conclusion ⎫ u(2) = u(1) ⎬ Continuumε (2) = Mechanicsε (1) =⇒ R for(2) = R Engineers(1) . © X. Oliver⎭ and C. Agelet de Saracibar(6.104) σ (2) = σ (1)

Therefore, the solution is unique (QED).

6.9 Saint-Venant’s Principle Saint-Venant’s principle is an empirical principle that does not have a rigorous proof. Consider a solid Ω that is subjected to a system of forces on its

21 This solution can be obtained applying the methodology used in Chapter 3, Section 3.4.2 to integrate the strain ﬁeld.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 296 CHAPTER 6. LINEAR ELASTICITY

Figure 6.11: Saint-Venant’s principle. boundary characterized by the traction vector t∗ (see Figure 6.11). These actions will lead to a solution or response in displacements, strains and stresses, ( ) not ( ) ( ) ( ) T R I (x,t) ≡ u I (x,t), ε I (x,t), σ I (x,t) . Consider now a part Γˆ of the boundary Γσ Γˆ ⊂ Γσ of said medium, whose typical dimension is , and re- place the system of actions applied on the boundary, t(I), by another system, t(II), that is statically equivalent to t(I) 22, without modifying the actions on the Γ rest of σ . Modifying the actions in this way will presumably result in the new responses R(II) (x,t) not≡ u(II) (x,t), ε (II) (x,t), σ (II) (x,t) T . Saint-Venant’s principle states that, for the points belonging to the domain Ω that are sufﬁciently far from the boundary Γˆ, the solution in both cases is prac- tically the same, that is, for a point P of the interior of Ω, Theory and⎫ Problems (I) ( , ) ≈ (II) ( , ) ⎪ u xp t u xp t ⎬ ( ) ( ) ε I (x ,t) ≈ ε II (x ,t) ∀P δ . (6.105) Continuump Mechanicsp ⎭⎪ for Engineers (I) © X. Oliver(II) and C. Agelet de Saracibar σ (xp,t) ≈ σ (xp,t)

In other words, if the distance δ between the point being considered and the part of the boundary in which the actions have been modiﬁed is large in com- parison with the dimension of the modiﬁed zone, the response in said point is equivalent in both cases.

22 Two systems of forces t(I) and t(II) are said to be statically equivalent if the resultant (forces and moments) of both systems is the same.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Linear Thermoelasticity. Thermal Stresses and Strains 297

Example 6.4 – Description of Saint-Venant’s principle in strength of materials and how it relates to the concept of stress.

Solution Consider a beam (or prismatic piece) with a cross-section A subjected to a tensile point force F in its ends, as shown in the ﬁgure below. The exact solution to the original elastic problem (system (I)) is extremely complicated, especially in the vicinity of the points of application of the point forces. If the forces F are now replaced by a statically equivalent system of uniformly distributed tensile loads in the end sections σ = F/A (system (II)), the elastic solution to the corresponding problem is extremely simple and coincides (for a Poisson’s ratio of ν = 0) with the axial stress solution provided by strength of materials (uniformly distributed stresses in all the piece, σx = F/A). At a far enough distance from the beam’s ends (once or twice the edge), Saint- Venant’s principle allows approximating solution (I) with solution (II), and also allows dimensioning the strength characteristics of the piece for practical purposes.

Theory and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

6.10 Linear Thermoelasticity. Thermal Stresses and Strains The main difference of linear thermoelasticity with respect to the linear elasticity studied up to this point is that the deformation process is no longer assumed to be isothermal (see Section 6.1). Now, the thermal effects are included and the

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 298 CHAPTER 6. LINEAR ELASTICITY temperature θ (x,t) is considered to evolve along time, that is,

not θ (x,t) = θ (x,0) = θ0 , . ∂θ(x,t) (6.106) θ (x,t)= = 0 . ∂t Nevertheless, the hypothesis that the processes are adiabatic (slow) is maintained and, thus, ρ0 r − ∇ · q ≈ 0 . (6.107)

6.10.1 Linear Thermoelastic Constitutive Equation Hooke’s law (6.6) in this case is generalized to

σ = C : ε − β (θ − θ ) 0 , (6.108) σij = Cijklεkl − βij(θ − θ0) i, j ∈ {1,2,3}

Here, C is the tensor of elastic constants defined in (6.7), θ (x,t) is the temperature field, θ0 (x)=θ (x,0) is the distribution of temperatures in the neutral state (reference configuration) and β is the (symmetric) tensor of thermal properties. T Tensor of thermal β = β (6.109) properties βij = β ji i, j ∈{1,2,3}

In the case of an isotropic material, tensor C must be a fourth-order isotropic tensor and β , a second-order isotropic one23, that is, Theory and Problems C = λ ⊗ + μ 1 1 2 I Cijkl = λδijδkl + μ δikδ jl + δilδ jk i, j,k,l ∈ {1,2,3} Continuum Mechanics for Engineers (6.110) β = β1 © X. Oliver and C. Agelet de Saracibar

βij = βδij i, j ∈ {1,2,3} where now a single thermal property β appears in addition to the elastic constants λ and μ. Replacing (6.110) in the constitutive equation (6.108) and deﬁn- not ing (θ − θ0) = Δθ, yields

23 The most general expression of a second-order isotropic tensor is β = β1 ∀β.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Linear Thermoelasticity. Thermal Stresses and Strains 299

Constitutive equation of an isotropic linear thermoelastic material (6.111) σ = λ Tr (ε)1 + 2μεε − βΔθ1

σij = λεllδij+ 2μεij− βΔθδij i, j ∈ {1,2,3}

6.10.2 Inverse Constitutive Equation Equation (6.111) can be inverted as follows. ⎧ ⎪ σ = C ε − Δθ β ⇒ ε = C−1 σ + Δθ C−1 β = C−1 σ + Δθ α ⎨ : : : : α ⎪ α ⎩ def α = C−1 : β → Tensor of thermal expansion coefﬁcients (6.112) where α is a second-order (symmetric) tensor involving six thermal properties named coefﬁcients of thermal expansion. For an isotropic case, in agreement with (6.111) and (6.24), and after certain algebraic manipulation, one obtains

Inverse constitutive equation of an isotropic linear thermoelastic material ν 1 + ν ε = − Tr (σ )1 + σ + αΔθ 1 (6.113) E E ν 1 + ν εij = − σll δij+ σij+ αΔθ δij i, j ∈ {1,2,3} E TheoryE and Problems

Here, α is a scalar denoted as coefﬁcient of thermal expansion, related to the Continuumβ Mechanics for Engineers thermal property in (6.111© X.) Oliverby means and of C. Agelet de Saracibar 1 − 2ν Thermal expansion → α = β . (6.114) coefﬁcient E

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 300 CHAPTER 6. LINEAR ELASTICITY

6.10.3 Thermal Stresses and Strains Comparing the linear elastic constitutive equation (6.20) and the linear thermoelastic one (6.111) suggests the following decomposition.

σ = λ (ε) + μεε −βΔθ = σ nt − σ t Tr 1 2 1 σ nt σ t ⎧ (6.115) ⎪ def ⎨ Non-thermal stress → σ nt = λ Tr (ε)1 + 2μεε ⎩⎪ def Thermal stress → σ t = βΔθ1

Here, σ nt represents the stress produced if there do not exist any thermal phe- nomena and σ t is named thermal stress and acts as the “correcting” stress due to the thermal increment. A similar operation can be performed on the inverse constitutive equations for the linear elastic and linear thermoelastic cases of (6.24) and (6.113), respectively, resulting in ν 1 + ν ε = − Tr (σ )1 + σ +αΔθ1 = ε nt + ε t E E t ε nt ε ⎧ (6.116) ⎪ def ν 1 + ν ⎨ Non-thermal strain → ε nt = − Tr (σ )1 + σ E E ⎩⎪ def Thermal strain → ε t = αΔθ1

In conclusion, the stress and strain tensors in linear thermoelasticity can be decomposed into Theory and Problems

Non-thermal Thermal TotalContinuumcomponent Mechanics forcomponent Engineers © X. Oliver and C. Agelet de Saracibar σ nt = C : ε σ t = Δθββ σ = σ nt − σ t Isotropic material: Isotropic material: (6.117) σ nt = λ Tr (ε)1 + 2μεε σ t = βΔθ1

ε nt = C−1 : σ ε t = Δθαα ε = ε nt + ε t Isotropic material: Isotropic material: (6.118) ν 1 + ν ε t = αΔθ ε nt = − Tr (σ )1 + σ 1 E E

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Linear Thermoelasticity. Thermal Stresses and Strains 301 where the thermal components appear due to the thermal processes being taken into account. The following expressions result from (6.117) and (6.118). − ε nt = C 1 : σ =⇒ σ = C : ε nt = C : ε − ε t (6.119) − − σ nt = C : ε =⇒ ε = C 1 : σ nt = C 1 : σ + σ t (6.120)

Remark 6.11. Unlike what occurs in elasticity, in the thermoelastic case a state of null strain in a point of a medium does not imply a state of null stress in said point. In effect, for ε = 0 in (6.117),

ε = 0 =⇒ σ nt = 0 =⇒ σ = −σ t = −βΔθ1 = 0 .

Δθ = 0 σ = −σ t = −βΔθ1 ε = 0

Remark 6.12. Analogously, in thermoelasticity a state of null stress in a point of a mediumTheorydoes not and imply a Problems state of null strain in said point since (6.118) with σ = 0 yields Continuumσ = 0 =⇒ ε Mechanicsnt = 0 =⇒ ε = ε fort = αΔθ Engineers1 = 0 . © X. Oliver and C. Agelet de Saracibar

Δθ = 0 ε = ε t = αΔθ1 σ = 0

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 302 CHAPTER 6. LINEAR ELASTICITY

Figure 6.12: Actions on a continuous medium.

6.11 Thermal Analogies The thermal analogies arise from the search of procedures to solve the linear thermoelastic problem using the strategies and methodologies developed in Sec- tion 6.7 for the linear elastic problem (without considering thermal effects). Two analogies are presented in this section which, for the sake of simplicity, are restricted to the isotropic quasi-static problem, although they can be directly extrapolated to the general anisotropic dynamic problem.

6.11.1 First Thermal Analogy (Duhamel-Newman Analogy) Consider the continuous medium in Figure 6.12 on which the body forces b(x,t) and an increment of temperature Δθ(x,t) are acting, and on whose boundaries Γ Γ ∗ ( , ) ∗ ( , ) u and σ act the prescribedTheory displacements andu Problemsx t and a traction vector t x t , respectively. The equations of the (isotropic quasi-static) linear thermoelastic problem are Continuum⎧ Mechanics for Engineers ⎪ © X. Oliver and C. Agelet de Saracibar ⎪ ∇ · σ + ρ = ⎨⎪ 0b 0 Equilibrium equation Governing σ = C : ε − βΔθ1 Constitutive equation equations ⎪ ⎪ ⎩ ε = ∇Su Geometric equation (6.121) Γ = ∗ Boundary u : u u ∗ conditions Γσ : σ · n = t

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Thermal Analogies 303 which compose the actions (data) A (x,t) and responses (unknowns) R (x,t) of the problem24.

⎫ b(x,t) ⎪ ⎧ ⎬⎪ ⎨ ( , ) ∗ ( , ) MATHEMATICAL u x t u x t ⇒ ⇒ ε ( , ) ∗ ( , ) ⎪ MODEL : ⎩ x t t x t ⎭⎪ PDEs + BCs σ ( , ) (6.122) Δθ( , ) x t x t not ( ) not ( ) Responses = R I (x,t) Actions = A I (x,t)

To be able to apply the resolution methods typical of the liner elastic problem developed in Section 6.7, the thermal term in the equations of the thermoelastic problem (6.121) must be eliminated (at least, in appearance). To this aim, the decomposition of the stress tensor σ = σ nt −σ t is replaced in (6.121) as follows. a) Equilibrium equation

σ = σ nt − σ t =⇒ ∇ · σ = ∇ · σ nt − ∇ · σ t = ∇ · σ nt − ∇(βΔθ) (6.123) βΔθ1 ∇ · σ + ρ = =⇒ ∇ · σ nt + ρ − 1 ∇(βΔθ) = 0b 0 0 b ρ 0 0 (6.124) not= bˆ =⇒ ∇ · σ nt + bˆ = 0 which constitutes the equilibriumTheory equation and Problems of the medium subjected to the pseudo-body forces bˆ (x,t) deﬁned by ⎧ Continuum⎪ Mechanics for Engineers ⎪ © X. Oliver1 and C. Agelet de Saracibar ⎨ bˆ (x,t)=b(x,t) − ∇(βΔθ) ρ0 ⎪ (6.125) ⎪ 1 ∂ (βΔθ) ⎩ bˆi (x,t)=bi (x,t) − i ∈ {1,2,3} ρ0 ∂xi b) Constitutive equation

σ nt = C : ε = λ Tr (ε)1 + 2μεε (6.126)

24 The ﬁeld of thermal increments Δθ(x,t) is assumed to be known a priori and, therefore, independent of the mechanical response of the problem. This situation is known as the uncoupled thermomechanical problem.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 304 CHAPTER 6. LINEAR ELASTICITY

c) Geometric equation (remains unchanged)

ε = ∇2u (6.127)

d) Boundary condition in Γu ∗ Γu : u = u (6.128)

e) Boundary condition in Γσ σ = σ nt − σ t =⇒ σ nt · n − σ t · n = t∗ =⇒ σ · n = t∗ (6.129) σ nt · = ∗ + σ t · = ∗ + βΔθ =⇒ Γ σ nt · = ˆ∗ n t n t n σ : n t ∗ βΔθ1 · n tˆ where tˆ∗ (x,t) is a pseudo-traction vector deﬁned by

∗ ∗ tˆ = t + βΔθn . (6.130)

Equations (6.123)to(6.130) allow rewriting the original problem (6.121)as ⎧ ∇ · σ nt + ρ ˆ = ⎪ 0b 0 ⎪ Equilibrium ⎪ ˆ = − 1 ∇(βΔθ) ⎪ with b b equation ⎨ ρ0 Governing ⎪ σ nt TheoryC ε andε Problemsε Constitutive equations ⎪ σ = C : ε = λ Tr (ε)1 + 2μεε ⎪ equation ⎪ (6.131) ⎩⎪ Geometric ε = ∇Su Continuum Mechanics for Engineersequation © X. Oliver and C. Agelet de Saracibar Γ = ∗ Boundary u : u u ∗ conditions Γσ : σ nt · n = tˆ with tˆ = t + βΔθn which constitutes the so-called analogous problem, a linear elastic problem that can be solved with the methodology indicated for this type of problems in Sec- tion 6.7 and characterized by the following actions and responses.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Thermal Analogies 305

⎫ ⎧ bˆ (x,t) ⎬ MATHEMATICAL ⎨ u(x,t) u∗ (x,t) ⇒ MODEL : ⇒ ε (x,t) ⎭ ⎩ (6.132) tˆ∗ (x,t) PDEs + BCs σ nt (x,t) ( ) ( ) Actions not= A II (x,t) Responses not= R II (x,t)

Comparing the actions and responses of the original problem (6.122) with those of the analogous problem (6.132), reveals the difference between them to be ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 ˆ − ˆ ∇(βΔθ) b b b b ⎢ ρ ⎥ ⎢ ∗ ⎥ ⎢ ∗ ⎥ ⎢ ⎥ ⎢ 0 ⎥ A(I) −A(II) not≡ ⎢ u ⎥ − ⎢ u ⎥ = ⎢ 0 ⎥ = ⎢ 0 ⎥ def= A(III) ( , ) ⎣ ∗ ⎦ ⎣ ˆ∗ ⎦ ⎣ ∗ − ˆ∗ ⎦ ⎢ ⎥ x t t t t t ⎣ −βΔθn ⎦ Δθ 0 Δθ Δθ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ u u ⎢ 0 ⎥ 0 ( ) ( ) not ⎣ ⎦ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ def ( ) R I −R II ≡ ε − ε = ⎣ 0 ⎦ = ⎣ 0 ⎦ = R III (x,t) σ σ nt σ − σ nt −βΔθ 1 −σ t (6.133) where (6.130) and (6.117) have been taken into account.

Remark 6.13. It can be directly veriﬁed that, in (6.133), R(III) is the response corresponding to the system of actions A(III) in the thermoelastic problem (Theory6.121). and Problems

Continuum Mechanics for Engineers Equation (6.133) suggests© X. that Oliver the original and C. problem Agelet(I de) may Saracibar be interpreted as the sum (superposition) of two problems or states:

STATE (II) (to be solved): analogous elastic state in which the temperature does not intervene and that can be solved by means of elastic procedures. + STATE (III) (trivial): trivial thermoelastic state in which the responses R(III) (x) givenin(6.133) are known without the need of any calculations.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 306 CHAPTER 6. LINEAR ELASTICITY

Once STATE (II) is computed, the solution to the original thermoelastic problem of STATE (I) is obtained as ⎧ (I) (II) Solution to the ⎨ u = u ε (I) = ε (II) original thermoelastic ⎩ (6.134) problem σ (I) = σ (II) − βΔθ1

The procedure to solve the thermoelastic problem based on the ﬁrst thermal analogy is summarized as a superposition of states in Figure 6.13.

STATE ACTION RESPONSE

⎡ ⎤ b(x,t) ⎡ ⎤ ⎢ ⎥ u(x,t) ⎢ u∗ (x,t) ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ε (x,t) ⎦ ⎣ t∗ (x,t) ⎦ σ (x,t) Δθ(x,t)

(I) Thermoelastic (original)

⎡ ⎤ ˆ = − 1 ∇(βΔθ) ⎢ b b ρ ⎥ ⎡ ⎤ ⎢ 0 ⎥ ⎢ ⎥ u(x,t) ⎢ ∗ ( , ) ⎥ ⎢ ⎥ ⎢ u x t ⎥ ⎣ ε ( , ) ⎦ ⎢ ⎥ x t ⎢ ˆ∗ = ∗ + βΔθ ⎥ ⎣ t t n ⎦ σ nt (x,t) Theory and ProblemsΔθ = 0 (II) Elastic (analogous) Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar ⎡ ⎤ 1 ⎡ ⎤ ⎢ b = ∇(βΔθ) ⎥ ⎢ ρ0 ⎥ u = 0 ⎢ ∗ ⎥ ⎢ ⎥ ⎢ u = 0 ⎥ ⎣ ε = ⎦ ⎢ ⎥ 0 ∗ = −βΔθ ⎣ t n ⎦ σ = −βΔθ1 Δθ(x,t)

(III) Thermoelastic (trivial)

Figure 6.13: First thermal analogy.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Thermal Analogies 307

6.11.2 Second Thermal Analogy The second thermal analogy is based on expressing the equations that constitute the problem in terms of the thermal strains ε t deﬁned in (6.118). Consider the equations of the original thermoelastic problem, with the constitutive equation in its inverse form ⎧ ⎪ ⎪ ∇ · σ + ρ = ⎨⎪ 0b 0 Equilibrium equation Governing ε = C−1 σ + αΔθ Inverse constitutive ⎪ : 1 equations ⎪ equation ⎩⎪ ε = ∇Su Geometric equation (6.135) Γ = ∗ Boundary u : u u ∗ conditions Γσ : σ · n = t which constitute the actions (data) A (x,t) and responses (unknowns) R (x,t) of the problem.

⎫ b(x,t) ⎪ ⎧ ⎬⎪ ⎨ ( , ) ∗ ( , ) MATHEMATICAL u x t u x t ⇒ ⇒ ε ( , ) ∗ ( , ) ⎪ MODEL : ⎩ x t t x t ⎭⎪ PDEs + BCs σ ( , ) (6.136) Δθ( , ) x t x t not ( ) not ( ) Responses = R I (x,t) Actions = A I (x,t)

Hypothesis 6.1. AssumeTheory that the and coefficient Problems of thermal expansion α (x) and the thermal increment Δθ(x,t) are such that the thermal strain field Continuumt Mechanics for Engineers ©ε X.(x Oliver,t)=α and(x)Δθ C.(x Agelet,t)1 de Saracibar is integrable (satisfies the compatibility conditions).

Consequently, there exists a thermal displacement ﬁeld ut (x,t) that satisﬁes ⎧ ⎪ 1 ⎨⎪ ε t (x,t)=αΔθ1 = ∇Sut = (ut ⊗ ∇ + ∇ ⊗ ut) ' 2 ( ∂ t ∂ut (6.137) ⎪ εt = αΔθδ = 1 ui + j , ∈{ , , } ⎩ ij ij i j 1 2 3 2 ∂x j ∂xi

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 308 CHAPTER 6. LINEAR ELASTICITY

Remark 6.14. The solution ut (x,t) to the system of differential equations (6.137) exists if and only if the strain ﬁeld ε t (x,t) satisﬁes the compatibility conditions (see Chapter 3). In addition, this solution is determined except for a rigid body motion characterized by a rota- ∗ tion tensor Ω and a displacement vector c∗ (both constant). That is, there exists a family of admissible solutions of the form t ( , )= ( , )+ Ω∗ · + ∗ . u x t u x t x c rotation translation rigid body motion

The rigid body motion may be chosen arbitrarily (in the form which is most convenient for the resolution process).

Once the thermal displacements have been deﬁned, a decomposition of the total displacements into their thermal and non-thermal parts can be performed as follows.

def unt (x,t) = u(x,t) − ut (x,t)=⇒ u = unt + ut (6.138) To eliminate the thermal term in the equations that constitute the thermoelastic problem (6.135), the decompositions of the displacements and strains (u = unt + ut and ε = ε nt + ε t) is introduced in the equations of (6.135), which result in a) Equilibrium equationTheory(remains unchanged) and Problems ∇ · σ + ρ0b = 0 (6.139) Continuum Mechanics for Engineers b) Inverse constitutive© equation X. Oliver and C. Agelet de Saracibar

− ν 1 + ν ε nt = C 1 : σ = − Tr (σ )1 + σ (6.140) E E

c) Geometric equation ⎫ ε = ∇S = ∇S ( nt + t)=∇S nt + ∇S t = ∇S nt + ε t u u u u u u ⎬ =⇒ ε nt = ∇S nt ε t ⎭ u ε = ε nt + ε t (6.141)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Thermal Analogies 309

d) Boundary condition in Γu ∗ u = u ∗ =⇒ Γ : unt = u − ut (6.142) u = unt + ut u

e) Boundary condition in Γσ (remains unchanged)

∗ Γσ : σ · n = t (6.143)

Equations (6.139)to(6.143) allow rewriting the original problem (6.135)as ⎧ ⎪ ⎪ ∇ · σ + ρ = Equilibrium ⎪ 0b 0 ⎪ equation ⎪ ⎨ − Governing ε nt = C 1 : σ = ν + ν Constitutive equations ⎪ 1 ⎪ = − Tr (σ )1 + σ equation ⎪ E E (6.144) ⎪ ⎩⎪ Geometric ε nt = ∇Sunt equation Γ = ∗ − t Boundary u : u u u ∗ conditions Γσ : σ · n = t which constitutes the so-called analogous problem, a linear elastic problem characterized by the following actions and responses

⎫Theory and Problems⎧ bˆ (x,t) ⎬ MATHEM. ⎨ unt (x,t) u∗ (x,t) − ut (x,t) ⇒ MODEL : ⇒ ε nt (x,t) ⎭ ⎩ (6.145) tˆ∗ (Continuumx,t) MechanicsPDEs + BCs for Engineersσ (x,t) © X. Oliver and C. Agelet de Saracibar ( ) ( ) Actions not= A II (x,t) Responses not= R II (x,t)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 310 CHAPTER 6. LINEAR ELASTICITY

Comparing the actions and responses of the original problem (6.136) and the analogous problem (6.145), reveals the difference between them to be ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ b b 0 ⎢ ∗ ⎥ ⎢ ∗ − t ⎥ ⎢ t ⎥ A(I) A(II) not ⎢ u ⎥ ⎢ u u ⎥ ⎢ u ⎥ def A(III) A −A ≡ ⎣ ∗ ⎦ − ⎣ ∗ ⎦ = ⎣ ⎦ = A (x,t) t t 0 Δθ 0 Δθ (6.146) ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ u unt ut ut ( ) ( ) not ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎢ ⎥ def ( ) R I −R II ≡ ε − ε nt = ε t = ⎣ αΔθ1 ⎦ = R III (x,t) σ σ σ σ 0 0 where equations (6.138) and (6.118) have been taken into account.

Remark 6.15. It can be directly veriﬁed that, in (6.146), R(III) is the response corresponding to the system of actions A(III) in the thermoelastic problem (6.135).

Therefore, the original problem (I) can be interpreted as the sum (superposition) of two problems or states:

STATE (II) (to be solved): analogous elastic state in which the temperature does not intervene and that can be solved by means of elastic procedures. + STATE (III) (trivial): trivial thermoelastic state in which the responses R(III) (x) givenin(6.146Theory) are known and without Problems the need of any calculations.

OnceContinuumSTATE (II) is computed, Mechanics the solution to for the original Engineers thermoelastic problem of STATE (I) is obtained© X. as Oliver and C. Agelet de Saracibar ⎧ ⎪ (I) = (II) + t Solution to the ⎨ u u u original thermoelastic ε (I) = ε (II) + αΔθ1 (6.147) ⎩⎪ problem σ (I) = σ (II) where ut is known from the integration process of the thermal strain ﬁeld in (6.137). The procedure to solve the thermoelastic problem based on the second thermal analogy is summarized as a superposition of states in Figure 6.14.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Thermal Analogies 311

STATE ACTION RESPONSE

⎡ ⎤ b(x,t) ⎡ ⎤ ⎢ ⎥ u(x,t) ⎢ u∗ (x,t) ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ε (x,t) ⎦ ⎣ t∗ (x,t) ⎦ σ (x,t) Δθ(x,t)

(I) Thermoelastic (original)

⎡ ⎤ b(x,t) ⎡ ⎤ ⎢ ⎥ unt (x,t) ⎢ u∗ − ut ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ε nt (x,t) ⎦ ⎣ t∗ (x,t) ⎦ σ (x,t) Δθ = 0

(II) Elastic (analogous)

Theory and⎡ Problems⎤ b = 0 ⎡ ⎤ ⎢ ⎥ u = ut (x,t) ⎢ u∗ = ut ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ε = αΔθ1 ⎦ Continuum Mechanics⎣ fort∗ = 0 Engineers⎦ © X. Oliver and C. Agelet de Saracibarσ = 0 Δθ(x,t)

(III) Thermoelastic (trivial)

Figure 6.14: Second thermal analogy.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 312 CHAPTER 6. LINEAR ELASTICITY

Example 6.5 – Solve the problem of a beam fully-ﬁxed at its ends and subjected to a constant thermal increment Δθ using the second thermal analogy.

Solution The classic procedure followed in strength of materials to solve this problem consists in the superposition (sum) of the following situations: 1) The structure is initially considered to be hyperstatic; 2) the right end is freed to allow for thermal expansion, which takes place with null stresses (since it is an iso- static structure); and 3) the displacement of the beam’s right end is recovered until it is brought again to zero. This procedure coincides exactly with the application of the second thermal analogy in which the thermal displacement ﬁeld ut is deﬁned by the thermal expansion of the piece with its right end freed (state III). Said expansion produces a displacement in the right end of value u|x= = αΔθ and, when recovering the displacement at this end, the boundary condition Γ = ∗ − t = − t , u : u u u u 0 which corresponds exactly to state II of Figure 6.14, is being implicitly applied.

Theory and Problems

Remark 6.16. The application of the second thermal analogy essen- tially resides in the integration of the thermal strain field ε t (x,t) to obtain the thermal displacement field ut (x,t) (see Remark 6.14). If the thermal strains are not integrable, the analogy cannot be applied. Comparing its advantages and disadvantages with respect to the first thermal analogy, it is also recommended that the integration of the thermal strains be, in addition to possible, simple to perform.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Thermal Analogies 313

Remark 6.17. The case involving • a homogeneous material (α (x)=const. = α) • a linear thermal increment (Δθ = ax + by + cz + d) is of particular interest. In this case, the product Δθαα is a linear polynomial and the thermal strains ε t = Δθαα automatically satisfy the compatibility conditions (6.69) (which are equations that only contain second-order derivatives) and, therefore, the thermal strain ﬁeld is guaranteed to be integrable.

Remark 6.18. In the case involving • a homogeneous material (α (x)=const. = α) • a constant thermal increment (Δθ = const.) the integration of the thermal strain ﬁeld ε t = Δθα1 = const. is trivial, resulting in t ( , )=αΔθ +Ω∗ · + ∗ , u x t x x c rigid body motion

where the rigid body motion can be chosen arbitrarily (see Re- mark 6.14). If this motion is considered to be null, the solution to the thermal displacement ﬁeld is ut (x,t)=αΔθxTheory=⇒ x + andut = x + ProblemsαΔθx =(1 + αΔθ)x , which means that STATE (III) in the second thermal analogy (see FigureContinuum6.15) is an homothecy, Mechanics with respect for to the Engineers origin of coordinates, of value (1 +©αΔθ X. Oliver). This homothecy and C. Agelet is known de as Saracibarfree thermal expansion (see Figure 6.15). The value of the thermal displacement (associated with the free thermal expansion) in the boundary Γu can be trivially determined in this case without need of formally integrating the thermal strains.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 314 CHAPTER 6. LINEAR ELASTICITY

Figure 6.15: Free thermal expansion in a homogeneous material subjected to a constant thermal increment.

6.12 Superposition Principle in Linear Thermoelasticity Consider the linear thermoelastic problem in Figure 6.16 and its corresponding governing equations

∂ 2u ∇ · σ + ρ b = ρ Cauchy’s equation 0 0 ∂t2 σ = λ (ε) + μεε −βΔθ Tr 1 2 1 Constitutive equation (6.148) C : ε 1 ε = ∇Su = (u ⊗ ∇ + ∇ ⊗ u) Geometric equation 2

∗ Γu : u = u ∗ TheoryBoundary and conditions Problems in space (6.149) Γσ : t = σ · n Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 6.16: Linear thermoelastic problem.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Superposition Principle in Linear Thermoelasticity 315

u(x,0)=0 . Initial conditions (6.150) u(x,0)=v0 which deﬁne the generic set of actions and responses ⎫ ˆ ( , ) ⎪ b x t ⎪ ⎧ ∗ ⎪ u (x,t) ⎬ MATHEMATICAL ⎨ u(x,t) t∗ (x,t) ⇒ MODEL : ⇒ ε (x,t) ⎪ ⎩ Δθ(x,t) ⎪ PDEs + BCs σ (x,t) (6.151) ⎭⎪ v (x) 0 Responses not= R (x,t) Actions not= A (x,t)

Remark 6.19. The different (scalar, vector, tensor and differential) operators that intervene in the governing equations of the problem (6.148)to(6.150) are linear, that is, given any two scalars a and b,

∇ · (•) → linear =⇒ ∇ · (ax + by)=a ∇ · x + b ∇ · y ,

C : (•) → linear =⇒ C : (ax + by)=a C : x + b C : y ,

∇S (•) → linear =⇒ ∇S (ax + by)=a ∇Sx + b ∇Sy , ∂ 2 ∂ 2 (ax + by) ∂ 2x ∂ 2y (•) → linearTheory=⇒ and Problems= a + b . ∂t2 ∂t2 ∂t2 ∂t2 Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar Consider now two possible systems of actions A(1) and A(2), ⎡ ⎤ ⎡ ⎤ b(1) (x,t) b(2) (x,t) ⎢ ⎥ ⎢ ⎥ ⎢ ∗(1) ( , ) ⎥ ⎢ ∗(2) ( , ) ⎥ ⎢ u x t ⎥ ⎢ u x t ⎥ ( ) not ⎢ ∗( ) ⎥ ( ) not ⎢ ∗( ) ⎥ A 1 (x,t) ≡ ⎢ t 1 (x,t) ⎥ and A 2 (x,t) ≡ ⎢ t 2 (x,t) ⎥ , (6.152) ⎢ ( ) ⎥ ⎢ ( ) ⎥ ⎣ Δθ 1 (x,t) ⎦ ⎣ Δθ 2 (x,t) ⎦ (1) ( ) (2) ( ) v0 x v0 x

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 316 CHAPTER 6. LINEAR ELASTICITY and their corresponding responses, R(1) and R(2), ⎡ ⎤ ⎡ ⎤ u(1) (x,t) u(2) (x,t) ( ) not ⎢ ⎥ ( ) not ⎢ ⎥ R 1 (x,t) ≡ ⎣ ε (1) (x,t) ⎦ and R 2 (x,t) ≡ ⎣ ε (2) (x,t) ⎦ . (6.153) σ (1) (x,t) σ (2) (x,t)

Theorem 6.2. Superposition principle. The solution (response) to the system of actions A(3) = λ (1) A(1) + λ (2) A(2)

(where λ (1) and λ (2) are any two scalars) is R(3) = λ (1) R(1) + λ (2) R(2)

In other words, the solution to the linear thermoelastic problem when considering a linear combination of different systems of actions is the same linear combination of the individual solutions to each of these systems of actions.

Proof Replacing the actions A(3) = λ (1) A(1) + λ (2) A(2) and the responses R(3) = λ (1) R(1) + λ (2) R(2) in the equations of the problem, and taking into account the linearity of theTheory different operators and Problems (see Remark 6.19) yields a) Cauchy’s equation σ (3) (3) (1) σ (1) (1) (2) σ (2) (2) ∇ · σContinuum+ ρ0 b = λ Mechanics∇ · σ + ρ0b for+λ Engineers∇ · σ + ρ0b = © X. Oliver and C. Agelet de Saracibar ∂ 2 (1) ∂ 2 (2) ρ u ρ u 0 ∂t2 0 ∂t2 ( ) ( ) ( ) ( ) ∂ 2 λ 1 u 1 + λ 2 u 2 ∂ 2 (3) = ρ = ρ u 0 ∂t2 0 ∂t2

2 (3) ( ) ( ) ∂ u ∇ · σ 3 + ρ b 3 = ρ 0 0 ∂t2 (6.154)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Superposition Principle in Linear Thermoelasticity 317

b) Constitutive equation ( ) ( ) ( ) σ (3) − C : ε (3) − βΔθ(3)1 = λ (1) σ 1 − C : ε 1 − βΔθ 1 1 + = 0 ( ) ( ) ( ) λ (2) σ 2 − C : ε 2 − βΔθ 2 1 = 0 = 0

( ) ( ) ( ) σ 3 = C : ε 3 − βΔθ 3 1 (6.155)

c) Geometric equation ( ) ( ) ( ) ( ) ε (3) − ∇Su(3) = λ (1) ε 1 − ∇Su 1 +λ (2) ε 2 − ∇Su 2 = 0 = 0 = 0 (6.156) ( ) ( ) ε 3 = ∇Su 3

d) Boundary condition in Γu ( ) ∗( ) ( ) ∗( ) u(3) − u∗(3) = λ (1) u 1 − u 1 +λ (2) u 2 − u 2 = 0 = 0 = 0 (6.157) (3) ∗(3) Γu : u = u

e) Boundary condition in Γσ Theory and Problems ( ) ∗( ) ( ) ∗( ) σ (3) · n − t∗(3) = λ (1) σ 1 · n − t 1 +λ (2) σ 2 · n − t 2 = 0 Continuum Mechanics for Engineers © X. Oliver= 0 and C. Agelet= de0 Saracibar(6.158) (3) ∗(3) Γσ : σ · n = t

f) Initial conditions . ( ) (3) ( ) . ( ) (1) ( ) . ( ) (2) u 3 (x,0) − v = λ 1 u 1 (x,0) − v +λ 2 u 2 (x,0) − v = 0 0 0 0 = 0 = 0 . (3) ( , )= (3) u x 0 v0 (6.159)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 318 CHAPTER 6. LINEAR ELASTICITY

Consequently, R(3) = λ (1) R(1) + λ (2) R(2) not≡ u(3), ε (3), σ (3) T is the solution to the thermoelastic problem subjected to the actions: A(3) = λ (1) A(1) + λ (2) A(2) (QED).

6.13 Hooke’s Law in terms of the Stress and Strain “Vectors” The symmetry of the stress and the strain tensors, σ and ε, means that only six of its nine components in a certain Cartesian system are different. Therefore, and to “economize” in writing, only these six different components are used in engineering, and they are expressed in the form of the stress and strain “vectors”. These are constructed in R6, systematically arranging the elements of the upper triangle of the matrix of components of the corresponding tensor in the following manner25. ⎡ ⎤ σ ⎢ x ⎥ ⎡ ⎤ ⎢ σ ⎥ ⎢ y ⎥ σx τxy τxz ⎢ ⎥ ⎢ ⎥ def ⎢ σ ⎥ σ not≡ ⎣ τ σ τ ⎦ → {σ } = ⎢ z ⎥ (6.160) xy y yz ⎢ τ ⎥ ⎢ xy ⎥ τxz τyz σz ⎢ ⎥ ⎣ τxz ⎦

τyz

The same arrangement is followed in the case of the strains, with the particularity that the strain vector {ε} is constructed using the angular strains γxy = 2εxy, γxz = 2εxz and γyz = 2εyz (see Chapter 2, Section 2.11.4). ⎡ ⎤ Theory⎡ and⎤ Problems εx 1 1 ⎢ ⎥ ⎡ ⎤ ⎢ ε γ γ ⎥ ⎢ ε ⎥ ε ε ε ⎢ x 2 xy 2 xz ⎥ ⎢ y ⎥ Continuumx xy xz ⎢ Mechanics⎥ for Engineers⎢ ε ⎥ not ⎢ ⎥ not ⎢ ⎥ def ⎢ z ⎥ ε ≡ ⎣ ε ε ε ⎦ =©⎢ X.1γ Oliverε and1γ C.⎥ Agelet→ { deε} Saracibar= ⎢ ⎥ (6.161) xy y yz ⎢ xy y yz ⎥ ⎢ γ ⎥ ε ε ε ⎢ 2 2 ⎥ ⎢ xy ⎥ xz yz z ⎣ ⎦ ⎢ γ ⎥ 1γ 1γ ε ⎣ xz ⎦ 2 xz 2 yz z γyz

25 The notation {x} is used to denote the vector in R6 constructed from the symmetric tensor x.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Hooke’s Law in terms of the Stress and Strain “Vectors” 319

Remark 6.20. An interesting property of this construction is that the double contraction of the stress and strain tensors is transformed into the dot product (in R6) of the stress and strain vectors, σ ε = {σ } · {ε} ⇐⇒ σ ε = σ ε : ij ij i i second-order vectors tensors which can be veriﬁed by performing said operations, using the deﬁ- nitions in (6.160) and (6.161).

The inverse constitutive equation (6.113), ν 1 + ν ε = − Tr (σ )1 + σ + αΔθ 1 , (6.162) E E can now be rewritten in terms of the stress and strain vectors as − {ε} = Cˆ 1 · {σ } + {ε}t , (6.163)

− where Cˆ 1 is the inverse matrix of elastic constants, ⎡ ⎤ −ν −ν ⎢ 1 ⎥ ⎢ 000⎥ ⎢ E E E ⎥ ⎢ ⎥ ⎢ −ν 1 −ν ⎥ ⎢ 000⎥ ⎢ E E E ⎥ ⎢ ⎥ ⎢ −ν Theory−ν 1 and Problems ⎥ ⎢ 000⎥ ⎢ ⎥ Cˆ −1 not≡ ⎢ E E E ⎥ ⎢ ⎥ (6.164) ⎢ ⎥ Continuum⎢ Mechanics1 for Engineers⎥ ⎢ 000© X. Oliver and C. Agelet00 de Saracibar⎥ ⎢ G ⎥ ⎢ ⎥ ⎢ 1 ⎥ ⎢ 000 0 0 ⎥ ⎢ G ⎥ ⎣ ⎦ 1 000 00 G and {ε}t is a thermal strain vector deﬁned by means of an adequate translation of the thermal strain tensor ε t = αΔθ1,

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 320 CHAPTER 6. LINEAR ELASTICITY

⎡ ⎤ αΔθ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ αΔθ ⎢ αΔθ⎥ ⎢ 00⎥ ⎢ ⎥ ⎢ ⎥ ⎢ αΔθ⎥ ε t not≡ ⎢ αΔθ ⎥ → {ε}t def= ⎢ ⎥ . ⎢ 0 0 ⎥ ⎢ ⎥ (6.165) ⎣ ⎦ ⎢ 0 ⎥ ⎢ ⎥ 00αΔθ ⎣ 0 ⎦ 0

Finally, the inversion of equation (6.163) provides Hooke’s law in terms of the stress and strain vectors,

Hooke’s law in terms of the {σ } = Cˆ · {ε} − {ε}t (6.166) stress and strain vectors where Cˆ is the matrix of elastic constants. ⎡ ⎤ ν ν ⎢ ⎥ ⎢ 1 000⎥ ⎢ 1 − ν 1 − ν ⎥ ⎢ ⎥ ⎢ ν ν ⎥ ⎢ 1 000⎥ ⎢ 1 − ν 1 − ν ⎥ ⎢ ⎥ ⎢ ν ν ⎥ ⎢ 1 000⎥ ( − ν) ⎢ 1 − ν 1 − ν ⎥ not E 1 ⎢ ⎥ Cˆ ≡ ⎢ ⎥ (6.167) (1 + ν)(1 − 2ν) ⎢ − ν ⎥ ⎢ 1 2 ⎥ ⎢ 000 00⎥ ⎢ 2(1 − ν) ⎥ ⎢ ⎥ ⎢ 1 − 2ν ⎥ ⎢ ⎥ ⎢ 000 0 ( − ν) 0 ⎥ ⎢ 2 1 ⎥ ⎣ Theory and Problems 1 − 2ν ⎦ 000 00 2(1 − ν) Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

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PROBLEMS

Problem 6.1 – Justify whether the following statements are true or false. a) The terms isentropic and adiabatic are equivalent when dealing with a thermoelastic material. b) The second thermal analogy is always applicable to linear thermoelastic materials.

Solution a) According to the second law of thermodynamics (5.114), ρ θ .i = ρ θ .− (ρ − ∇ · ) ≥ . 0 sloc 0 s 0 r q 0 All processes are reversible in the case of a thermoelastic material and, thus, the inequality becomes an equality,

ρ θ .i = ρ θ .− (ρ − ∇ · )= 0 sloc 0 s 0 r q 0 . [1]

. An isentropic process (entropy remains constant) is characterized by s = 0. On the other hand, an adiabatoc process (variation of heat is null) satisfies ρ − ∇ · = . Theory0 r andq 0 Problems Therefore, if an isentropic process is assumed, and its mathematical expression is introduced in [1], the definition of an adiabatic process is obtained, Continuum Mechanics for Engineers ρ θ .−(ρ© X.− ∇ Oliver· )= and=⇒ C. Ageletρ − ∇ de· Saracibar= . 0 s 0 r q 0 0 r q 0 = 0 Conversely, if an adiabatic process is assumed, and its mathematical expression is introduced in [1], the definition of an isentropic process is obtained, ρ θ .− (ρ − ∇ · ) = =⇒ . = . 0 s 0 r q 0 s 0 = 0 In conclusion, the statement is true.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 322 CHAPTER 6. LINEAR ELASTICITY b) The second thermal analogy is not always applicable. The condition that the thermal strain field be integrable must be verified, that is, the thermal strain field ε t (x,t) must satisfy the compatibility conditions (3.19),

εij,kl + εkl,ij− εik, jl − ε jl,ik = 0 i, j,k,l ∈ {1,2,3} . Given that these involve second-order derivatives of the components of the strain tensor with respect to x, y and z, they will be automatically satisﬁed if α = const. and Δθ = const.,orifαΔθ is linear in x, y and z (which is the deﬁnition of a linear thermoelastic material). Therefore, the statement is true.

Problem 6.2 – An isotropic linear elastic solid is subjected to a constant pressure of value p on all of its external boundary, in addition to a thermal increment of Δθ = θ (x,y,z) in its interior. Both actions cancel each other out such that no displacements are observed in the solid. Obtain the value of Δθ in each point of the solid.

Solution The ﬁrst thermal analogy described in Section 6.11.1 will be applied. To this aim, the original problem I is decomposed into the sum of problems II and III as described in Figure 6.13.

PROBLEM I ⎧ ⎪ b = 0 ⎧ ⎪ ⎪ ⎨ ∗ ⎨ u t = −pnTheoryin Γσ and Problems Actions: Responses: ε ⎪ ∗ = Γ ⎪ ⎪ u 0 in u ⎩ ⎩⎪ σ ContinuumΔθ = Δθ Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar PROBLEM III This problem is solved ﬁrst since its solution is trivial. ⎧ ⎪ = 1 β∇(Δθ) ⎧ ⎪ bIII ρ ⎨⎪ ⎨⎪ uIII = 0 ∗ t = −βΔθn in Γσ Actions: III Responses: ε III = 0 ⎪ ∗ ⎩⎪ ⎪ u = 0 in Γu σ = −βΔθ ⎩⎪ III III 1 ΔθIII = Δθ

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 323

PROBLEM II ⎧ ⎪ = 1 β∇(Δθ) ⎪ bII ⎨⎪ ρ ∗ =(− + βΔθ) Γ Actions: tII p n in σ ⎪ ∗ ∗ ⎪ u = u = 0 in Γu ⎩⎪ II ΔθII = 0 To solve problem II, Navier’s equation (6.62) is taken into account, together with the fact that uII = 0.

2 (λ + μ)∇(∇ · uII)+μ∇ uII + ρbII = 0 =⇒

bII = 0 =⇒ β∇(Δθ)=0 =⇒ Δθ is uniform

In addition, uII = 0 also results in

1 ε = (u ⊗ ∇ + ∇ ⊗ u )=0 , II 2 II II

σ II = λ (∇ · uII)1 + μ (uII ⊗ ∇ + ∇ ⊗ uII)=0 .

∗ σ Since the traction vector tII is deﬁned in terms of the stress tensor II, σ · = ∗ =(− + βΔθ) = ∀ =⇒−+ βΔθ= , II n tII p n 0 n p 0 and the value of the thermal increment is ﬁnally obtained,

Problem 6.3 – A cylindrical shell of height h, internal radius R and external radius 2R is placed inside an inﬁnitely rigid cylindrical cavity of height h and radius 2R + a, with a << R. Assume the cylindrical shell is subjected to a uniform temperature ﬁeld Δθ. a) Determine the value of Δθ∗ required for the external lateral walls of the cylindrical shell and the rigid walls of the cavity to come into contact.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 324 CHAPTER 6. LINEAR ELASTICITY

b) Plot, indicating the most signiﬁcant values, the curve δ − Δθ, where δ is the lengthening of the internal radius of the cylindrical shell. Deter- mine the value of Δθ such that this radius recovers its initial value. c) Plot, indicating the most signiﬁcant values, the curves σrr −Δθ, σθθ − Δθ and σzz − Δθ, in points A and B.

Hypotheses: 1) Young’s modulus: E 2) Poisson’s coefﬁcient: ν = 0 3) Thermal expansion coefﬁcient: α 4) Isotropic linear elastic material 5) Weights can be neglected 6) The friction betweenTheory the walls and is negligible Problems

Continuum Mechanics for Engineers Solution © X. Oliver and C. Agelet de Saracibar a) Two distinct phases can be identiﬁed in this problem:

First phase The cylindrical shell has not come into contact with the rigid walls of the cavity. The boundary condition on the lateral walls, both internal and external, will be null radial stress. The two cylinders will come into contact when

ur (r = 2R)=a .

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 325

Second phase The cylindrical shell and the rigid walls of the cavity are in contact and, therefore, the boundary condition on the external lateral wall is different than that of the first phase. In this case, a null radial displacement will be imposed. Nonetheless, the internal wall will retain the same boundary condition as in the previous phase. A positive Δθ will reduce the internal radius since the external radius cannot increase because it is limited by the infinitely rigid walls of the cavity. Then, the only possibility is that the cylindrical shell continues expanding inwards. There will be a point in which the internal radius, which had increased in the first phase, will recover its initial value.

The ﬁrst thermal analogy (see Section 6.11.1) and the superposition principle (see Section 6.12) will be applied. To this aim, the original problem (problem I) is decomposed into the sum of problems II and II as described in Figure 6.13. PROBLEM III The actions in problem III, the trivial problem, are

= 1 ∇ · (ββΔθ) . bIII ρ

In this case, however, Δθ is uniform and β is a spherical and constant tensor (β = β1). Therefore, bIII = 0 . The boundary conditions are

1) Prescribed displacements in Γu : uIII = 0. 2) Prescribed stresses in Γσ : t = −ββΔθn = −βΔθn. The solution to this problemTheory is known and to be Problems

PROBLEM II The actions in problem II, the analogous problem, are

= − 1 ∇ · (ββΔθ) . bII b ρ

Here, b = 0 because the weight of the cylinder is assumed to be negligible and the second term is zero, as seen in problem III. Therefore,

bII = 0 .

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 326 CHAPTER 6. LINEAR ELASTICITY

The boundary conditions are ∗ ∗ 1) Prescribed displacements in Γu : uII = u , where u is the displacement imposed in problem I. ∗ ∗ 2) Prescribed stresses in Γσ : tII = σ II · n = t + ββΔθn = t + βΔθn, where t∗ is the traction vector imposed in problem I. The analogous problem will now be solved assuming an inﬁnitesimal strains hypothesis, since a << R and the strains are due to Δθ, which are generally inﬁnitesimal. Due to cylindrical symmetry, the displacement vector u is known to be of the form not T uII (r,z) ≡ [ur (r) , 0, uz (z)] .

In addition, uz (z)=0 will be imposed in all points since no information on the top and bottom surfaces of the cylindrical shell is given. Boundary conditions in displacements cannot be imposed for these surfaces because there is no way to determine the integration constants of uz that would appear if uz = 0 were considered. Therefore, the displacement vector

not T uII (r,z) ≡ [ur (r) , 0, 0 ] is adopted. Navier’s equation (6.62) will be used to solve this problem,

∂ 2u (λ + μ)∇(∇ · u )+μ∇2u + ρ b = ρ II = 0 . II II 0 II 0 ∂t2 Note that the problem requires working in cylindrical coordinates and, thus, the equation must be adapted to this system of coordinates. Given the simpliﬁcations introduced into the problem, only the radial component of the equation will result in a non-trivial solution,

∂e 2G ∂ω ∂ωθ ∂ 2u (λ + 2G) −Theoryz + 2G and+ Problemsρb = ρ r , [2] ∂r r ∂θ ∂z r ∂t2 where bContinuumr is the radial component Mechanics of bII and with for Engineers © X. Oliver and C. Agelet de Saracibar

1 ∂ur ∂uz ωθ = + = 0 , 2 ∂z ∂r

1 1 ∂ (ruθ ) 1 ∂u ω = − r = 0 , z 2 r ∂r r ∂θ

1 ∂ (ru ) 1 ∂uθ ∂u 1 ∂ (ru ) e = r + + z = r . r ∂r r ∂θ ∂z r ∂r

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 327

The values of the parameters λ, G and β that intervene in Navier’s equation must also be determined from the known parameters (E , α , ν = 0) as follows. λ ν = = 0 =⇒ λ = 0 2(λ + μ) E E μ = =⇒ μ = G = [3] 2(1 + ν) 2 E β = α =⇒ β = Eα 1 − 2ν

The problem can be considered to be a quasi-static and, taking into account bII = 0 and the relations derived in [3], the Navier’s stokes equation [2] is reduced to ∂e ∂ 1 ∂ (λ + 2G) = 0 =⇒ E (ru) = 0 . ∂r ∂r r ∂r r Integrating this last expression leads to 1 ∂ ∂ (ru)=2A =⇒ (ru)=2Ar =⇒ ru = Ar2 + B r ∂r r ∂r r r ! " [4] B B T =⇒ u = Ar + =⇒ u (r) not≡ Ar + , 0, 0 , r r II r where A and B are the integration constants. The strain tensor corresponding to this displacement vector is easily obtained by means of the geometric equation (6.3), ⎡ ⎤ − B ⎢ A 00⎥ Theory⎢ r2 and Problems⎥ ε ( ) not≡ ⎢ B ⎥ II r ⎣ 0A+ 0 ⎦ . [5] r2 Continuum Mechanics000 for Engineers © X. Oliver and C. Agelet de Saracibar Finally, the stress tensor is obtained through the constitutive equation of an isotropic linear elastic material (6.20), particularized with the expressions in [3],

σ II = λ (Tr (ε II))1 + 2μεε II =⇒ σ II = Eε II . [6]

First phase The integration constants A and B must be determined by means of the boundary conditions. Stresses can be imposed in both lateral walls of the cylindrical shell as follows.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 328 CHAPTER 6. LINEAR ELASTICITY

BOUNDARY CONDITION AT r = 2R If r = 2R and according to the boundary conditions in Γσ of the analogous problem, ∗ ∗ tII = σ II · n = t + ββΔθn = t + βΔθn . Here, the following is known: n =[1, 0, 0]T : outward unit normal vector. t∗ = 0, since, for this phase, problem I has no loading on the lateral walls. σ II is given by [5] and [6]. Therefore, the boundary condition is reduced to

σrr (r = 2R)=βΔθ, which, replacing the value of the radial stress from [6] and, considering [3], results in

B A − = αΔθ . [7] 4R2

BOUNDARY CONDITION AT r = R If r = R and according to the boundary conditions in Γσ of the analogous problem, ∗ ∗ tII = σ II · n = t + ββΔθn = t + βΔθn . Here, the following is known: n =[−1, 0, 0]T : outward unit normal vector. t∗ = 0, since, for thisTheory phase, problem and I has Problems no loading on the lateral walls. σ II is given by [5] and [6]. Therefore,Continuum the boundary Mechanics condition is reduced for to Engineers © X. Oliver and C. Agelet de Saracibar σrr (r = R)=βΔθ, which, replacing the value of the radial stress from [6] and, considering [3], results in

B A − = αΔθ . [8] R2 From [7] and [8], the values

A = αΔθ and B = 0 [9]

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 329

are obtained. Now, replacing [9] in [4], [5] and [6] results in the displacements, strains and stresses of the analogous problem. not T uII ≡ αΔθr , 0, 0 ⎡ ⎤ αΔθ 00 not ⎢ ⎥ ε II ≡ ⎣ 0 αΔθ 0 ⎦ 000 [10] ⎡ ⎤ EαΔθ 00 not ⎢ ⎥ σ II ≡ ⎣ 0 EαΔθ 0 ⎦ 000

Taking into account the superposition principle (see Section 6.12), and expressions [1], [3] and [10], the original problem is solved for the ﬁrst phase. u not≡ αΔθr , 0, 0 T ⎡ ⎤ αΔθ 00 not ⎢ ⎥ ε ≡ ⎣ 0 αΔθ 0 ⎦ 000 [11] ⎡ ⎤ 00 0 not ⎢ ⎥ σ ≡ ⎣ 00 0⎦ 00−EαΔθ Theory and Problems To obtain the value of Δθ∗ for which the external lateral walls of the cylindrical shell and the rigid walls of the cavity come into contact, it is enough to impose that Continuum Mechanics for∗ Engineers ur (r©= X.2R)= Olivera = and⇒ C.αΔθ Agelet2R = dea . Saracibar Then, the temperature ﬁeld required for the external lateral walls of the cylindrical shell and the rigid walls of the cavity to come into contact is

∗ a Δθ = . [12] 2αR b) First, the value Δθ∗∗ for which the internal radius recovers its initial position will be determined. To this aim, the same geometry as in the initial problem will be used, but now there will exist contact between the cylindrical shell and the

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 330 CHAPTER 6. LINEAR ELASTICITY rigid walls of the cavity, which corresponds to the second phase deﬁned in the previous section. So, a new problem must be solved, with the same geometry as before but considering different boundary conditions.

Second phase The ﬁrst phase will be obviated in this section, but one must bear in mind that the solid now starts from a state that results from the previous phase, that is, it has already suffered certain displacements, strains, stresses and thermal increments. The variable Δθ will be used. As before, the ﬁrst thermal analogy will be applied. Problem III remains unchanged and, thus, so does its result [1]. Therefore, problem II must be solved with the same expressions [4], [5] and [6]. The integration constants A and B must be determined by means of the boundary conditions. Stresses can be imposed on the internal lateral wall of the cylindrical shell and displacements, on its external lateral wall.

BOUNDARY CONDITION AT r = 2R

If r = 2R and according to the boundary conditions in Γu of the analogous problem, ur (r = 2R)=0 . Therefore, the following condition is obtained, B A2R+ = 0 . [13] 2R

BOUNDARY CONDITION AT r = R If r = R and accordingTheory to the boundary and conditions Problems in Γσ of the analogous problem, ∗ ∗ tII = σ II · n = t + β Δθn = t + β Δθn . Continuum Mechanics for Engineers Here, the following is© known: X. Oliver and C. Agelet de Saracibar n not≡ [−1, 0, 0]T : outward unit normal vector. t∗ = 0, since, for this phase, problem I has no loading on the lateral walls. σ II is given by [5] and [6]. Therefore, the boundary condition is reduced to

σrr (r = R)=β Δθ , which, replacing the value of the radial stress from [6], and considering [3], results in B A − = α Δθ . [14] R2

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 331

From [13] and [14], the values

1 4 A = α Δθ and B = − α ΔθR2 [15] 5 5 are obtained. Introducing now [15] in [4], [5] and [6] results in the displacements, strains and stresses of the analogous problem.

! "T 1 4R2 u not≡ α Δθ r − , 0, 0 II 5 r ⎡ ⎤ 1 4R2 ⎢ α Δθ 1 + 00⎥ ⎢ 5 r2 ⎥ ⎢ 2 ⎥ ε not≡ ⎢ 1 4R ⎥ II ⎢ 0 α Δθ 1 − 0 ⎥ ⎣ 5 r2 ⎦ 000[16] ⎡ ⎤ 1 4R2 ⎢ Eα Δθ 1 + 00⎥ ⎢ 5 r2 ⎥ ⎢ 2 ⎥ σ not≡ ⎢ 1 4R ⎥ II ⎢ 0 Eα Δθ 1 − 0 ⎥ ⎣ 5 r2 ⎦ 000 Taking into account the superposition principle (see Section 6.12), and expressions [1], [3] and [16], the original problem is solved for the second phase. ! "T 1 4R2 u not≡Theoryα Δθ r and− Problems, 0, 0 5 r ⎡ ⎤ 2 1α Δθ + 4R Continuum⎢ Mechanics1 for00 Engineers⎥ [17a] ⎢ 5 © X. Oliverr2 and C. Agelet de Saracibar⎥ ⎢ 2 ⎥ ε not≡ ⎢ 1 4R ⎥ ⎢ 0 α Δθ 1 − 0 ⎥ ⎣ 5 r2 ⎦ 000

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 332 CHAPTER 6. LINEAR ELASTICITY

⎡ ⎤ 4 R2 ⎢ Eα Δθ −1 + 00⎥ ⎢ 5 r2 ⎥ ⎢ 2 ⎥ σ not≡ ⎢ 4 R ⎥ ⎢ 0 Eα Δθ −1 − 0 ⎥ [17b] ⎣ 5 r2 ⎦ 00−Eα Δθ Note that, up to this point, the second phase has been solved assuming an initial neutral state. In reality, this phase starts from the ﬁnal state of the ﬁrst phase, which has the displacements, strains, stresses and thermal increments corresponding to Δθ = Δθ∗, ∗ uinitial = u first phase (Δθ = Δθ ) , ε ε ∗ ε initial = ε first phase (Δθ = Δθ ) , [18] σ σ ∗ σ initial = σ first phase (Δθ = Δθ ) .

In fact, the variable Δθ in [17] is not a total thermal increment but the difference in temperature at the moment corresponding to Δθ∗, that is,

Δθ = Δθ − Δθ∗ . [19]

Then, considering [17], [18] and [19], the actual displacements, strains and stresses during the second phase are obtained, = + Δθ , usecondphase uinitial u ε = ε + ε Δθ , [20] secondphase initial σ σ σ σ secondphase = σ initial + σ Δθ . Therefore, to determineTheoryΔθ∗∗, it is enough andProblems to impose that the displacement, according to the ﬁrst phase, of the internal radius be equal but of opposite sign to that of the second phase. In this way, the total displacement will be null. Continuum Mechanics for∗ Engineers First phase: displacement© X. for Oliverr = R and andΔθ C.= AgeletΔθ . From de [11] Saracibar and [12], a δ = u (r = R, Δθ = Δθ∗)=αΔθ∗ R = . [21] 1 r 2

Second phase: displacement for r = R and Δθ = Δθ∗∗. From [17],

3 δ = u r = R, Δθ = Δθ∗∗ = − α Δθ∗∗ R . [22] 2 r 5 Then, 5a δ = −δ =⇒ Δθ∗∗ = . [23] 1 2 6αR

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 333

Finally, from [19] the total thermal increment is obtained,

Δθ∗∗ = Δθ∗∗ + Δθ∗ = 5a + a =⇒ Δθ∗∗ = 4a 6αR 2αR 3αR

Now, the curve δ −Δθ can be plotted, where δ is the displacement of the internal radius of the cylindrical shell.

c) Expressions [11] and [17] must be used to plot the curves σrr − Δθ, σθθ − Δθ and σzz − Δθ for points B(r = R) and A(r = 2R).

Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 334 CHAPTER 6. LINEAR ELASTICITY

Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 335

EXERCISES

6.1 – A cylinder composed of an isotropic linear elastic material stands on a rigid base. At a very small distance “a” (a h) of its top face there is another rigid surface. A uniform pressure p acts on all the lateral surface of the cylinder. Plot, indicating the most signiﬁcant values, the following curves: a) Curve p − δ, where δ is the shortening of the radius of the cylinder, R.

b) Curve p − σA, where σA is the stress normal to the bottom contact surface at point A.

Additional hypotheses: 1) Weights can be neglected. 2) Lame’s´ constants: λ = μ 3) The problem is assumed to be quasi-static.

Theory and Problems

Continuum Mechanics for Engineers 6.2 – The solid sphere A©with X. external Oliver radius and C.R1 Ageletand the de solid Saracibar spherical B, with external radius R2 are composed of the same material. The external surface of A and the internal surface of B are separated by a very small distance “a” (a R1 and a R2). a) Determine what value of the uniform normal pressure p shown in the ﬁgure is required for the two surfaces to be in contact. b) Plot, indicating the most signiﬁcant values, the curve p − δ, where δ is the shortening of R2.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 336 CHAPTER 6. LINEAR ELASTICITY

Additional hypotheses: 1) Young’s modulus: E 2) Lame’s´ constants: λ = μ

3) R1 = R 4) R2 = 2R

6.3 – Two solid cylinders composed of different elastic materials are vertically superimposed and confined between two infinitely rigid walls. The cylinders are subjected to the external pressures p and α p (p > 0, α > 0) as shown in the figure. a) Determine the displacement field of the two cylinders in terms of the integration constants (justify the assumptions used). b) Indicate the boundary conditions that need to be applied for the different boundaries of the problem. c) Assuming a constant value α such that the contact surface between the two cylinders does not have a vertical displacement, calculate the integration constants and the value of α. Theory and Problems

Continuum MechanicsAdditional for Engineers hypotheses: © X. Oliver and C. Agelet de Saracibar 1) Top cylinder: λ1 = μ1 2) Bottom cylinder: λ2 = μ2 3) The friction between the cylinders and between the cylinders and the walls is assumed to be null. 4) Weights can be neglected.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 337

6.4 – A cylinder with radius Ri is placed in the interior of a cylindrical shell with internal radius Ri + 2e and external radius Re. There is an elastic gasket between the cylinder and the cylindrical shell which has an internal radius Ri and a thickness “e”. The cylindrical shell is subjected to an external pressure p. a) Determine the displacement, strain and stress ﬁelds of the cylinder and the cylindrical shell.

b) Plot the curves Ur − p, where Ur is the radial displacement, and σrr − p, where σrr is the radial stress at points A, B and C of the ﬁgure.

Data: Ri = 1 Re = 2 ν = 0 E (Young’s modulus)

Additional hypotheses: Theory and Problems 1) The constitutive law of the elastic gasket is p∗ = K δ ∗, where p∗ is the pres- δ ∗ sureContinuum acting on the gasket, Mechanicsis the shortening for of Engineers its thickness and K is its elastic modulus. © X. Oliver and C. Agelet de Saracibar 2) e Ri 3) A plane strain behavior in an inﬁnitesimal strain framework may be assumed.

6.5 – The ﬁgure below schematizes the layout of a railway rail composed of straight rails of length “L”, separated by an elastic gasket with elastic modulus K. Due to symmetry and construction considerations, it can be assumed that the section x = 0 suffers no longitudinal displacements and the inferior part of the rail suffers no vertical displacements. A constant thermal increment Δθ is imposed in all points of the rail.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 338 CHAPTER 6. LINEAR ELASTICITY

a) Obtain the displacement, strain and stress ﬁelds in terms of the corresponding integration constants. b) Indicate the boundary conditions that must be applied to determine the integration constants. c) Determine the integration constants and obtain the corresponding displacement, strain and stress ﬁelds. d) Particularize these results for the cases K = 0 (open junction) and K → ∞ (continuous rail).

Additional hypotheses: 1) Assume the displacements are of the form u =[u(x) , v(y) , w(z)]T . 2) Linear elastic material 3) λ = μ 4) The weight of the rail can be neglected.

6.6 – A solid cylinder with radius R and height h is placed between two infinitely rigid walls, fitting perfectly between them without producing any stress. A thermal increment Δθ > 0 is applied on the cylinder. Determine: a) The displacement field in terms of the corresponding integration constants. Theory and Problems b) The integration constants. c)Continuum The stress state. Plot Mechanics its variation along for the radius. Engineers © X. Oliver and C. Agelet de Saracibar

Additional hypotheses: 1) Material properties: λ = μ and α = α (r)=α0 + α1r 2) The friction between the cylinder and the walls is negligible. 3) Weights can be neglected.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961