CH.3. COMPATIBILITY EQUATIONS. Multimedia Course on Continuum Mechanics

CH.3. COMPATIBILITY EQUATIONS Multimedia Course on Continuum Mechanics

Overview Introduction Lecture 1 Compatibility Conditions Lecture Compatibility Equations of a Potential Vector Field Lecture 3 Compatibility Conditions for Infinitesimal Strains Lecture 4 Integration of the Infinitesimal Strain Tensor Lecture 5 Integration of the Deformation Rate Tensor Lecture 6

3.1 Compatibility Conditions Ch.3. Compatibility Equations 3

Introduction Given a displacement field, the corresponding strain field is found: U( X,t) u( x,t) E ε ij ij 1 U U i j Uk U k = + + i, j {1,, 3} X j Xi Xi X j 1 u u i j = + i, j {1,, 3} j i Is the inverse possible? ε ( x,t) u( x,t) 4

Compatibility Conditions Given an (arbitrary) symmetric second order tensor field, ε x,t, s a displacement field, u( x,t), fulfilling ux (,) t = ε( x, t) cannot always be obtained: 1 u u i j εij = + i, j {1,, 3} 6 PDEs OVERDETERMINED j 3 unknowns SYSTEM i For ε( x,t) to match a symmetric strain tensor: It must be integrable. There must exist a displacement field from which it comes from. COMPATIBILITY CONDITIONS must be satisfied REMARK Given u x,t, there will always exist an associated strain tensor, ε( x,t), obtainable through differentiation, which will automatically satisfy the compatibility conditions. 5

Compatibility Conditions The compatibility conditions are the conditions a symmetric nd order tensor must satisfy in order to be a strain tensor and, thus, exist a displacement field which satisfies: 1 u u i j εij = + i, j {1,, 3} j i They guarantee the continuity of the continuous medium during the deformation process. E( X,t) Incompatible strain field 6

3. Compatibility Equations of a Potential Vector Field Ch.3. Compatibility Equations 7

Preliminary example: Potential Vector Field v( x,t) A vector field will be a potential vector field if there exists a scalar function φ( x,t) (named potential function) such that: (, t) = φ( x, t) φ( x, t) ( x t) v x { } v i, = i 1,,3 Given a continuous scalar function potential vector field v( x,t). Is the inverse true? v( x,t) (,t) i φ( x,t) there will always exist a φ x such that φ ( x, t) = v( x, t) 8

Potential Field v( x,t) (,t) φ x such that φ ( x, t) = v( x, t) In component form, ( x t) ( x t) φ, φ, v i( x, t) = v i( x, t) = 0 i 1,,3 i i { } 3 eqns. 1 unknown OVERDETERMINED Differentiating once these expressions with respect to : SYSTEM x j φ ( x, t) vi = i, j 1,,3 j j i { } 9 eqns. 9

Schwartz Theorem The Schwartz Theorem about symmetry of second partial derivatives guarantees that, given a continuous function Φ ( x1, x,..., xn ) with continuous derivatives, the following holds true: Φ Φ = i, j i j j i 10

Compatibility Equations Considering the Schwartz Theorem, vx φ vx φ vx φ = = = y xy z xz vy φ vy φ vy φ = = = yx y y z yz vz φ vz φ vz φ = = = z x y z y z z In this system of 9 equations, only 6 different nd derivatives of the unknown φ( x,t) appear: φ φ φ φ φ φ,,,, and y z y z y z They can be eliminated and the following identities are obtained: v vy v v vy v = = = y z z y x x z z 11

Compatibility Equations A scalar function φ x,t which satisfies φ x, t = v x, t will exist if the vector field v x,t verifies: v y v z v y INTEGRABILITY (COMPATIBILITY) EQUATIONS of a potential vector field x z def vx = 0 = S y def v z = 0 = S v def y = 0 = S z z y x where v = 0 v v i j = 0 i, j 1,,3 xj x i eˆ1 eˆ ˆ e3 S x S S y v y z S z v v v { } x y z REMARK A functional relation can be established between these three equations. ( v) 0 = 1

3.3 Compatibility Conditions for Infinitesimal Strains Ch.3. Compatibility Equations 13

Infinitesimal strains case The infinitesimal strain field can be written as: u 1 x u u x y 1 ux u z + + y x z εxx εxy ε xz uy 1 uy u z ε = εxy εyy εyz = + y z y εxz εyz ε zz u z symmetrical z 6 PDEs 3 unknowns 14

Infinitesimal strains case The infinitesimal strain field can be written as: ε ε ε xx yy zz u 1 x u u x y = 0 εxy + = 0 y x u y 1 ux u z = 0 εxz + = 0 y z uz 1 u y u z = 0 εyz + = 0 z z y 6 PDEs 3 unknowns The system will have a solution only if certain compatibility conditions are satisfied. 15

Compatibility Conditions The compatibility conditions for the infinitesimal strain field are obtained through double differentiation (single differentiation is not enough). u x ε xx,,,,, x y z xy xz yz 1 u y u z ε yz + z y =, y, z, y, z, yz = 6 equations 6 equations 6x6=36 equations 16

Compatibility Conditions 17 The compatibility conditions for the infinitesimal strain field are 3 obtained through: 3 3 ε xx u ε yz 1 u x y u z = = + 3 zx yx 18 equations for εxy, εxz, εyz 18 equations for εxx, εyy, εzz 3 3 3 ε xx u ε 1 yz u x y u z = = + 3 y xy y zy y 3 3 3 ε xx u ε x yz 1 uy u z =... = + 3 z xz z z yz 3 3 ε xx u ε x yz 1 3 uy u z = = + xy y xy zxy y x 3 3 3 ε xx u ε x yz 1 uy u z = = + xz z xz z x yxz 3 3 3 ε xx u ε x yz 1 uy u z = = + yz xyz yz z y y z

Compatibility Conditions All the third derivatives of ux, uy and uz appear in the equations: 3 ux 3 3 3, y, z, y, y x, y z, z, z x, z y, yz = 10 derivatives 3 y 3 3 3 u, y, z, y, y x, y z, z, z x, z y, yz 3 uz 3 3 3, y, z, y, y x, y z, z, z x, z y, yz = = 10 derivatives 10 derivatives which constitute 30 of the unknowns in the system of 36 equations: f n 3 u ε i ij, = 0 n 1,,...,36 jkl kl 30 { } 18

Compatibility Equations 19 Eliminating the 30 unknowns, 3 ui j k l only strain derivatives) are obtained: def εyy ε ε zz yz Sxx = + = 0 z y yz def ε zz εxx εxz S yy = + = 0 z xz def ε ε xx yy εxy Szz = + = 0 y xy def ε ε zz yz ε ε xz xy Sxy = + + = 0 xy z x y z def εyy εyz ε ε xz xy Sxz = + + = 0 xz y y z def ε ε xx yz ε ε xz xy S yz = + + + = 0 yz y z, 6 equations (involving COMPATIBILITY EQUATIONS for the infinitesimal strain tensor S = ( ε ) = 0

Compatibility Equations The six equations are not functionally independent. They satisfy the equation, In indicial notation: S = ε = 0 S S xx xy Sxz + + = 0 y z Sxy Syy Syz + + = 0 y z S S xz yz Szz + + = 0 y z 0

Compatibility Equations The compatibility equations can be expressed in terms of the permutation operator,. e ijk S = e e ε, = 0 ml, 1,,3 ml mjq lir ij qr Or, alternatively: { } εij, kl + εkl, ij εik, jl ε jl, ik = 0 i, jkl,, 1,, 3 REMARK Any linear strain tensor (1 st order polynomial) with respect to the spatial variables will be compatible and, thus, integrable. 1

3.4 Integration of the Infinitesimal Strain Tensor Ch.3. Compatibility Equations

Preliminary Equations Rotation tensor Ω x,t : Rotation vector θ x,t : 1 Ω= skew( u ) = ( u u) 1 u u i j Ω ij = i, j {1,, 3} j i θ1 Ω3 Ω yz 0 θ3 θ 1 θ = u = θ = Ω = Ω [ Ω ] = θ 0 θ θ 31 zx 3 1 θ3 Ω 1 Ω xy θ θ1 0 3

Preliminary Equations Differentiating Ω x,t with respect to : 1 u u i j Ωij 1 u u i j Ω ij = = xj x i k k j i x k Adding and subtracting the term k : 1 u Ωij 1 u u i j 1 uk 1 uk = + = k k j i ij ij 1 ui u k 1 u j u k ε ε ik = + + = j xk xi xi xk x j j i = ε ik = ε jk i j jk 4

Preliminary Equations Using the previous results, the derivative of θ θ Ω 1 yz ε ε θ Ω ε ε xz xy = = = = x x z x x x y z θ Ω εxy ε zx θ Ω ε ε θ = = y y z x y y y z θ Ω ε ε θ Ω 1 yz ε ε zz zy = = = = z z z x z z y z θ Ω 3 xy ε xy εxx = = x x x y θ Ω 3 xy εyy εxy θ3 = = y y x y θ Ω 3 xy ε yz εxz = = z z x y 1 yz yz yy 1 = = θ x,t is obtained: zx xx xz yz zx xz zz 5

Preliminary Equations 6 Considering the displacement gradient tensor J x,t, (, t) u x J = = ε+ Ω = ε ij ui 1 u u i j 1 u u i j Jij = = + + = εij +Ωij i, j 1,,3 j j i j i Introducing the definition of are rewritten: i = 1: i = : i = 3: = Ω ij, the components of { } θ( x,t) J( x,t) j = 1 j = j = 3 ux ux ux = εxx = εxy θ3 = εxz + θ y z uy uy uy = εxy + θ3 = εyy = εyz θ1 y z uz uz uz = εxz θ = εyz + θ1 = εzz y z

Integration of the Strain Field The integration of the strain field steps: ε( x,t) is performed in two 1. Integration of derivative of θ x,t using the1 st order PDE system derived for θ, θ and θ. The solution will be of the type: 1 3 ( xyzt,,, ) c( t) i { 1,, 3} θ = θ + i i i The integration constants ci t can be obtained knowing the value of the rotation vector in some points of the medium (boundary conditions). ε( x ). Known,t and θ x,t, u is integrated using the 1 st order PDE system derived for u REMARK. The solution will be: If the compatibility equations ui = u i( xyzt,,, ) + c i ( t) i { 1,, 3} The integration constants c i ( t) can be obtained knowing the value of are satisfied, the displacements in some point of space (boundary conditions) these equations will be integrable. 7

Integration of the Strain Field The integration constants that appear imply that an integrable strain tensor ε( x,t) will determine the movement in any instant of not not time except for a rotation c() t = θˆ () t and a translation c () t = uˆ () t : ˆ θ x, t = θ x, t + θ t ε x, t u( x, t) = u ( x, t) + uˆ ( t) A displacement field can be constructed from this uniform rotation and translation: u ( x,) t = Ωˆ ( ˆ θ ()) t x+ uˆ () t u =Ωˆ S * 1 T 1 ˆ ˆ T ( u ) = ( u + ( u ) ) = ( Ω+Ω ) = 0 This corresponds to a rigid solid movement. 8

3.5 Integration of the Deformation Rate Tensor Ch.3. Compatibility Equations 9

Compatibility Equations in a Deformation Rate Field There is a correspondence between The concept of compatibility conditions can be extended to deformation rate tensor. 1 1 1 j i ij j i j i ij j i u u x x u u x x ε = + Ω = = u u u ε θ v v 1 v v 1 w 1 j i ij j i j i ij j i d x x x x = + = = v dv v ω d v 30

Example Obtain the velocity field corresponding to the deformation rate tensor: such that In point ( 1, 1, 1) ty 0 te 0 ty d( x, t) = te 0 0 tz 0 0 te the following conditions is fulfilled: t v( x, t) = e ω( x t) x= ( 1,1,1) e e t t 0 1, = v= 0 x= ( 1,1,1) t te 31

Example - Solution ty 0 te 0 ty d( x, t) = te 0 0 tz 0 0 te Consider the correspondence: u ε( u) 1 θ= u v dv 1 ω= v Take the expressions derived for θ1, θ θ3 substitute,t with and x,t with d x,t : ε ω 1 d d xz xy = = 0 0 x y z ω d d 0 0 y y z ω d 1 d zz zy = = 0 0 z y z 1 yz yy ω 1 = = and θ( x ) ω( x,t) ( t) C ( t) ω = 1 1 3

Example - Solution ty 0 te 0 ty d( x, t) = te 0 0 tz 0 0 te ω dxx dxz = = 0 0 x z x ω d d 0 0 y z x ω d xz d zz = = 0 0 z z x xy yz ω = = ω d 3 xy d xx = = 0 0 x x y ω d d y x y ω d 3 yz d xz = = 0 0 z x y 3 yy xy ty ω 3 = = 0 te ( t) C ( t) ω = ty ty ( y, t) t e dy te C ( t) ω = = + 3 3 33

Example - Solution 1 1 ω =C t ω =C t ty 3 3 ω = te + C t So, For point 1, 1, 1 : 0 1 { ω( x, t) } = v= 0 t te 1 1 ω = 0 = C t ω = 0 = C t t ty ω 3 = te = te + C3 t Therefore, for any point, x = { ω( x t) } ( 1,1,1) 0, = 0 ty te C C C 1 3 ( t) ( t) ( t) = 0 = 0 = 0 34

Example - Solution { ω( x t) } 0, = 0 ; ty te ty 0 te 0 ty d( x, t) = te 0 0 tz 0 0 te Taking the expressions i = 1: i = : The components of the velocities can be obtained: v v y x x v z x = d = 0 xx 3 = d ω = te te = te xy = d +ω = 0+ 0 xz j = 1 j = j = 3 v v v = d = d ω = d +ω y z x x x xx xy 3 xz v v v = d +ω = d = d ω y z y y y xy 3 yy yz 1 v v v i = 3: = d ω = d +ω = d y z ty ty ty z z z xz yz 1 zz ty ty = = + v, x y t te dy e C1 t 35

Example - Solution { ω( x t) } 0, = 0 ; ty te ty 0 te 0 ty d( x, t) = te 0 0 tz 0 0 te The components of the velocities can be obtained: v v v = d +ω = te + te = 0 y ty ty xy 3 y y y z = d = 0 yy = d ω = 0 0 yz 1 = v y t C t v v y z z v z z = d ω = 0 0 xz = d +ω = 0+ 0 yz zz 1 = d = te tz tz tz = = + v, z z t te dz e C3 t 36

Example - Solution v tz z = e + C 3 t v ty x = e + C 1 t v y = C t For point 1, 1, 1 : { v( x, t) } t e t = e t e So, v = = + t v y = e = C ( t) v Therefore, for any point, t ty x e e C1 t = = + t tz z e e C3 t x = x = ( 1,1,1) ( 1,1,1) { v( x, t) } ty e t = e tz e 1 3 ( t) ( t) C = 0 C t = e C = 0 t 37

Chapter 3 Compatibility Equations 3.1 Introduction Given a sufficiently regular displacement field U(X, t), it is always possible to find the corresponding strain field (for example, the Green-Lagrange strain field) by differentiating this strain field with respect to its coordinates (in this case, the material ones) 1, E ij = 1 ( Ui + U j + U ) k U k not = 1 Ui, j +U j,i +U k,i U k, j X j X i X i X j (3.1) i, j {1,,3}. In the infinitesimal strain case, given a displacement field u(x, t), the strain field ε ij = 1 ( ui + u ) j not = 1 j i (u i, j + u j,i ) i, j {1,,3} (3.) is obtained. The question can be formulated in reverse, that is, given a strain field ε (x,t), is it possible to find a displacement field u(x,t) such that ε (x,t) is its infinitesimal strain tensor? This is not always possible and the answer provides the socalled compatibility equations. Expression (3.) constitutes a system of 6 (due to symmetry) partial differential equations (PDEs) with 3 unknowns: u 1 (x,t), u (x,t), u 3 (x,t). This system is overdetermined because there exist more conditions than unknowns, and it may not have a solution. Therefore, for a second-order symmetric tensor ε (x, t) to correspond to a strain tensor (and, thus, be integrable and there exist a displacement field from which it comes) it is necessary that this tensor verifies certain conditions. These conditions are denominated compatibility conditions or equations and guarantee Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 1 Here, the simplified notation U i / X j not = U i, j is used. 109

110 CHAPTER 3. COMPATIBILITY EQUATIONS Figure 3.1: Non-compatible strain field. the continuity of the continuous medium during the deformation process (see Figure 3.1). Definition 3.1. The compatibility conditions are conditions that a second-order tensor must satisfy in order to be a strain tensor and, therefore, for there to exist a displacement field from which it comes. Remark 3.1. Note that, to define a strain tensor, the 6 components of a symmetric tensor cannot be written arbitrarily. These must satisfy the compatibility conditions. Remark 3.. Given a displacement field, one can always obtain, through differentiation, an associated strain field that automatically satisfies the compatibility conditions. Therefore, in this case, there is no sense in verifying that the compatibility conditions are satisfied. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 3. Preliminary Example: Compatibility Equations of a Potential Vector Field A given vector field v(x,t) is a potential field if there exists a scalar function φ (x,t) (named potential function) such that its gradient is v(x,t), v(x,t)= φ (x,t), v i (x,t)= φ(x,t) i i {1,,3}. (3.3) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

Preliminary Example: Compatibility Equations of a Potential Vector Field 111 Therefore, given a scalar (continuous) function φ (x, t), it is always possible to define a potential vector field v(x,t) such that the scalar function is its potential, as defined in (3.3). Now, the reverse question is posed: given a vector field v(x,t), does there exist a scalar function φ (x,t) such that φ (x,t) =v(x,t)? This is written in component form as v x = φ v y = φ y v z = φ z = v x φ = 0, = v y φ y = 0, = v z φ z = 0, (3.4) which corresponds to a system of PDEs with 3 equations and 1 unknown (φ (x,t)), thus, the system is overdetermined and may not have a solution. Differentiating once (3.4) with respect to (x,y,z) yields v x = φ, v y = φ y, v z = φ z, v x y = φ y, v y y = φ y, v z y = φ z y, v x z = φ z, v y z = φ y z, v z z = φ z, (3.5) which represents a system of 9 equations. Considering the equality of mixed partial derivatives, it is observed that 6 different functions (second derivatives) of the unknown φ are involved in these 9 equations, φ, φ y, φ z, φ y, ϕ z and φ y z. (3.6) So, they can be removed from the original system (3.5) and 3 relations, named compatibility conditions, can be established between the first partial derivatives of the components of v(x,t). Hence, for there to exist a scalar function φ (x,t) such that φ (x,t)=v(x,t), the given vector field v(x, t) must satisfy the following compatibility conditions. v y v x y = 0 def = S z ê 1 ê ê 3 v x z v z = 0 def = S y v z y v y z = 0 def = S x Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar where S not S x S y S z y z v x v y v z not rot v not = v (3.7) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

11 CHAPTER 3. COMPATIBILITY EQUATIONS In consequence, from (3.7), the compatibility equations can be written as Compatibility equations of a potential vector field v = 0 v i v j = 0 i, j {1,,3} j i (3.8) Remark 3.3. The 3 compatibility equations (3.7) or(3.8) are not independent of one another and a functional relation can be established between them. Indeed, applying the condition that the divergence of the rotational of a vector field is null, ( v)=0. 3.3 Compatibility Conditions for Infinitesimal Strains Consider the infinitesimal strain field ε (x,t) with components ε ij = 1 ( ui + u ) j not = 1 j i (u i, j + u j,i ) i, j {1,,3}, (3.9) which may be written in matrix form as ( u x 1 ux ε xx ε xy ε xz y + u ) ( y 1 ux z + u ) z ( [ε]= ε xy ε yy ε yz = u y 1 uy ε xz ε yz ε zz y z + u ) z. y u z (symm) z (3.10) Due to the symmetry in (3.10), only 6 different equations are obtained, Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar ε xx u x = 0, ε xy 1 ( ux y + u ) y = 0, ε yy u y y = 0, ε xz 1 ( ux z + u ) z = 0, ε zz u z z = 0, ε yz 1 ( uy z + u ) z = 0. y (3.11) A theorem of differential geometry states that the divergence of the rotational of any field is null, [ ( )] = 0. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

Compatibility Conditions for Infinitesimal Strains 113 Equation (3.11) is a system of 6 PDEs with 3 unknowns, which are the components of the displacement vector u(x,t) not [u x, u y, u z ] T. In general, this problem will not have a solution unless certain compatibility conditions are satisfied. To obtain these conditions, the equations in (3.11) are differentiated twice with respect to their spatial coordinates, ( ε xx u ) x, y, z, y, z, yz = 6 equations providing a total of 36 equations, (.. ε yz 1 ( uy z + u )) z y, y, z, y, z, yz = 6 equations, ε xx = 3 u x 3 ε xx y = 3 u x y ε xx z = 3 u x z ε xx y = 3 u x y ε xx z = 3 u x z ε xx y z = 3 u x y z }{{} (18 eqns for ε xx, ε yy, ε zz ) ε yz = 1 ( 3 u y z + 3 ) u z y ε yz y = 1 ( 3 u y z y + 3 ) u z y 3 ε yz z = 1 ( 3 u y z 3 + 3 ) u z y z ε yz y = 1 ( 3 u y z y + 3 ) u z y ε yz z = 1 ( 3 u y z + 3 ) u z y z ε yz y z = 1 ( 3 u y z y + 3 ) u z y z }{{} (18 eqns for ε xy, ε xz, ε yz ) (3.1) (3.13) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar All the possible third derivatives of each component of the displacements u x, u y and u z are involved in these 36 equations. Thus, there are 30 different derivatives, 3 u x 3, y, z, y 3, y x, y z, z 3, z x, z = 10 derivatives, y, yz 3 u y 3, y, z, y 3, y x, y z, z 3, z x, z = 10 derivatives, y, yz 3 u z 3, y, z, y 3, y x, y z, z 3, z x, z = 10 derivatives, y, yz (3.14) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

114 CHAPTER 3. COMPATIBILITY EQUATIONS which constitute the 30 unknowns in the system of 36 equations ( ) 3 u i ε ij f n, n {1,... 36} (3.15) j k l k l } {{ } 30 defined in (3.13). Therefore, the 30 unknowns, which are the displacement derivatives 3 u i /( j k l ), can be eliminated from this system and 6 equations are obtained. In these equations, the third derivatives mentioned above do not appear, but there will be 1 second derivatives of the strain tensor ε ij /( k l ). After the corresponding algebraic operations, the resulting equations are Compatibility equations S xx def S yy def S zz def S xy def = ε yy z + ε zz y ε yz y z = 0 = ε zz + ε xx z ε xz z = 0 = ε xx y + ε yy ε xy y = 0 = ε zz y + ( εyz z + ε xz y ε ) xy = 0 z = ε yy z + ( εyz y ε xz y + ε ) xy = 0 z = ε xx y z + ( ε yz + ε xz y + ε ) xy = 0 z S xz def S yz def (3.16) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar which constitute the compatibility equations for the infinitesimal strain tensor ε. The compact expression corresponding to the 6 equations in (3.16) is Compatibility equations for the infinitesimal strain tensor { S = (ε )=0 (3.17) Another way of expressing the compatibility conditions (3.16) is in terms of the three-index operator named permutation operator ( e ijk ). In this case, the compatibility equations can be written as S mn = e mjq e nir ε ij,qr = 0. (3.18) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

Compatibility Conditions for Infinitesimal Strains 115 Remark 3.4. The 6 equations (3.16) are not functionally independent and, taking again into account the fact that the divergence of the rotational of a field is intrinsically null, the following functional relations can be established between them. S = ( (ε )) = 0 = S xx + S xy y + S xz z = 0 S xy + S yy y + S yz z = 0 S xz + S yz y + S zz z = 0 Remark 3.5. The three-index operator denominated permutation operator is given by 0 if an index is repeated, i = j or i = k or j = k 1 positive (clockwise) direction of the indexes, e ijk = i, j,k {13,31,31} 1 negative (counterclockwise) direction of the indexes, i, j,k {13,31,13} This definition is summarized in graphic form in Figure 3.. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar Figure 3.: Definition of the permutation operator, e ijk. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

116 CHAPTER 3. COMPATIBILITY EQUATIONS Finally, another possible expression of the compatibility conditions is ε ij,kl + ε kl,ij ε ik, jl ε jl,ik = 0 i, j,k,l {1,,3}. (3.19) Remark 3.6. Since the compatibility equations (3.16) only involve the second spatial derivatives of the components of the strain tensor ε (x,t), every strain tensor that is linear (first-order polynomial) with respect to the spatial variables will be compatible and, therefore, integrable. As a particular case, every uniform strain tensor ε (t) is integrable. 3.4 Integration of the Infinitesimal Strain Field 3.4.1 Preliminary Equations Consider the rotation tensor Ω(x, t) for the infinitesimal strain case (see Chapter, Section.11.6), Ω = 1 (u u), Ω ij = 1 ( ui u ) (3.0) j i, j {1,,3}. j i and the infinitesimal rotation vector θ (x, t), associated with said rotation tensor, defined as 3 θ = 1 rot u = 1 θ 1 Ω 3 Ω yz u not = =. (3.1) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar θ θ 3 Ω 31 Ω 1 Ω zx Ω xy Differentiating the infinitesimal rotation tensor in (3.0) with respect to a coordinate x k yields Ω ij = 1 ( ui u ) j = Ω ij = 1 ( ui u ) j. (3.) j i k k j i 3 The tensor Ω is skew-symmetric, i.e., Ω not 0 Ω 1 Ω 31. Ω 1 0 Ω 3 Ω 31 Ω 3 0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

Integration of the Infinitesimal Strain Field 117 Adding and subtracting in (3.) the term u k /( i j ) and rearranging the expression obtained results in Ω ij = 1 ( ui u ) j + 1 u k 1 u k = k k j i i j i j = ( 1 ui + u ) k ( 1 u j + u ) k = ε ik ε jk (3.3). j k i i k j j i }{{}}{{} ε ik ε jk This expression can now be used to calculate the Cartesian derivatives of the components of the infinitesimal rotation vector, θ (x, t), given in (3.1), as follows. θ 1 = Ω yz = ε xz y ε xy z θ 1 θ 1 y = Ω yz y = ε yz y ε yy z θ 1 z = Ω yz = ε zz z y ε zy z θ = Ω zx = ε xx z ε xz θ θ y = Ω zx y = ε xy z ε yz θ z = Ω zx = ε xz z z ε zz (3.4) (3.5) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar θ 3 θ 3 = Ω xy = ε xy ε xx y θ 3 y = Ω xy y = ε yy ε xy y (3.6) θ 3 z = Ω xy = ε yz z ε xz y Assume the value of the infinitesimal rotation vector θ (x,t) is known and, through it by means of (3.1), the value of the infinitesimal rotation tensor X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

118 CHAPTER 3. COMPATIBILITY EQUATIONS Ω(x,t) is also known. Then, the displacement gradient tensor J(x,t) (see Chapter, Section.11.6) becomes J = u(x,t) = ε + Ω J ij = u i = 1 ( ui + u ) j + 1 ( ui u ) j = ε ij + Ω ij j j i j i }{{}}{{} ε ij Ω ij i, j {1,,3}. (3.7) Finally, writing in explicit form the different components in (3.7) and taking into account (3.1), the following is obtained 4. i = 1: i = : i = 3: j = 1 j = j = 3 u x = ε u x xx y = ε u x xy θ 3 z = ε xz + θ u y = ε u y xy + θ 3 y = ε u y yy z = ε yz θ 1 u z = ε u z xz θ y = ε u z yz + θ 1 z = ε zz (3.8) 3.4. Integration of the Strain Field Consider ε (x,t) is the infinitesimal strain field one wants to integrate. This operation is performed in two steps: 1) Using (3.4) through (3.6), the infinitesimal rotation vector θ (x, t) is integrated. The integration, with respect to space, of the infinitesimal rotation vector in (3.4) through (3.6) leads to a solution of the type θ i = θ i (x,y,z,t)+c i (t) i {1,,3}, (3.9) Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar where the integration constants c i (t), which, in general, may be a function of time, can be determined if the value (or the evolution along time) of the infinitesimal rotation vector at some point of the medium is known. ) Once the infinitesimal strain tensor ε (x, t) and the infinitesimal rotation vector θ (x,t) are known, the displacement field u(x,t) is integrated. The system of first-order PDEs defined in (3.8) is used, resulting in 4 According to (3.1), Ω not u i = ũ i (x,y,z,t)+c i (t) i {1,,3}. (3.30) 0 Ω 1 Ω 31 Ω 1 0 Ω 3 = Ω 31 Ω 3 0 0 θ 3 θ θ 3 0 θ 1 θ θ 1 0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961.

Integration of the Infinitesimal Strain Field 119 Again, the integration constants c i (t) that appear, which, in general, will be a function of time, are determined when the value (or the evolution along time) of the displacements at some point of space is known. Remark 3.7. The integration processes in steps 1) and ) involve integrating systems of first-order PDEs. If the compatibility equations in (3.16) are satisfied, these systems will be integrable (without leading to contradictions in their integration process) and will finally allow obtaining the displacement field. Remark 3.8. The presence of the integration constants in (3.9) and (3.30) shows that an integrable strain tensor, ε (x, t), determines the motion of each instant of time except for a rotation c(t) not = ˆθ (t) and a translation c (t) not = û(t). { θ (x,t)= ε (x,t) θ (x,t)+ ˆθ (t) u(x,t)=ũ(x,t)+û(t) From these uniform rotation ˆθ (t) and translation û(t) the displacement field u (x,t)= ˆΩ(t)x + û(t) = u = ˆΩ can be defined, which corresponds to a rigid body motion 5. Indeed, the strain associated with this displacement is null, ε (x,t)= s u = 1 (u + u )= 1 ( ˆΩ + }{{} ˆΩ T ) = 0, ˆΩ as corresponds to the concept of rigid body (without deformation). Consequently, it is concluded that every compatible strain field determines the displacements of the continuous medium except for a rigid body motion, which must be determined by means of the appropriate boundary conditions. Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar 5 The rigid body rotation tensor ˆΩ(t) (antisymmetric) is defined based on the rotation vector 0 ˆΩ 1 ˆΩ 31 0 ˆθ 3 θ not ˆθ (t) as ˆΩ ˆΩ 1 0 ˆΩ 3 = θ 3 0 θ 1. ˆΩ 31 ˆΩ 3 0 θ θ 1 0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

10 CHAPTER 3. COMPATIBILITY EQUATIONS Example 3.1 A certain motion is defined by the infinitesimal strain tensor 8x y 3 x z ε (x,t) not y x 0 3. x z 0 x 3 Obtain the corresponding displacement vector u(x, t) and the infinitesimal not rotation tensor Ω(x,t) taking into account that u(x,t) x=[0,0,0] T [3t, 0, 0] T and Ω(x,t) x=[0,0,0] T = 0. Solution Infinitesimal rotation vector Posing the systems of equations defined in (3.4) through (3.6) results in θ 1 = 0 ; θ 1 y = 0; θ 1 z = 0 θ 1 = C 1 (t), θ = 3xz ; θ y = 0; θ z = 3 x θ = 3 x z +C (t), θ 3 = 0 ; θ 3 y = 3 ; θ 3 z = 0 θ 3 = 3 y +C 3 (t). The integration constants C i (t) are determined by imposing that Ω(x,t) x=(0,0,0) T = 0 (and, therefore, the infinitesimal rotation vector θ (x,t) x=(0,0,0) T = 0), that is, [ C 1 (t)=c (t)=c 3 (t)=0 = θ (x) not 0, 3 x z, 3 y Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar and the infinitesimal rotation tensor is 0 θ 3 θ Ω(x) not θ 3 0 θ 1 = θ θ 1 0 0 3 y 3 x z 3 y 0 0. 3 x z 0 0 ] T X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

Compatibility Equations and Integration of the Strain Rate Field 11 Displacement vector Posing, and integrating, the systems of equations in (3.8) produces u 1 = 8x ; u 1 y = y ; u 1 z = 0 u 1 = 4x y +C 1 (t), u = y ; u y = x ; u z = 0 u = xy +C (t), u 3 = u 3 3x z ; y = 0 ; u 3 z = x3 u 3 = x 3 z +C 3 (t). and imposing that u(x,t) x=(0,0,0) T [3t, 0, 0] T yields not C 1 (t)=3t ; C (t)=c 3 (t)=0 = u(x) not [ 4x y + 3t, xy, x 3 z ] T. 3.5 Compatibility Equations and Integration of the Strain Rate Field Given the definitions of the infinitesimal strain tensor ε, the infinitesimal rotation tensor Ω and the infinitesimal rotation vector θ, there exists a clear correspondence between these magnitudes and a) the strain rate tensor d, b) the rotation rate (or spin) tensor w and c) the spin vector ω given in Chapter. These correspondences can be established in the following manner: u ε (u) ε ij = 1 ( ui + u ) j j i Ω ij = 1 ( ui u ) j j i Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar d ij = 1 w ij = 1 v d(v) ( vi j + v j i ( vi j v j i ) ) (3.31) θ = 1 u ω = 1 v Then, it is obvious that the concept of compatibility of a strain field ε introduced in Section 3.1 can be extended, by virtue of the correspondence with (3.31), to the compatibility of a strain rate field d(x,t). X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

1 CHAPTER 3. COMPATIBILITY EQUATIONS To integrate this field, the same procedure as that seen in Section 3.4. can be used, replacing ε by d, u by v, Ω by w and θ by ω. Certainly, this integration can only be performed if the compatibility equations in (3.16) are satisfied for the components of d(x,t). Remark 3.9. The resulting compatibility equations and the integration process of the strain rate vector d(x,t) are not, in this case, restricted to the infinitesimal strain case. Continuum Mechanics for Engineers Theory and Problems 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...581.0961

Problems and Exercises 13 PROBLEMS Problem 3.1 Determine the spatial description of the velocity field that corresponds to the strain rate tensor te tx 0 0 d(x,t) not 0 0 te y + 1. 0 te y + 1 0 For x = 0, ω 0 not [t 1, 0, 0] T and v 0 not [t, 0, t] T for t is satisfied. Solution The problem is solved by integrating the corresponding differential equations, taking into account the existent parallelism between the variables: u v ε d θ ω Angular velocity of the rotation vector ω 1 = 0; ω 1 y = ω 1 tey ; z = 0 ω 1 = C 1 (t)+te y, ω = 0; ω y = 0; ω z = 0 ω = C (t), ω 3 = 0; ω 3 y = 0; ω 3 z = 0 ω 3 = C 3 (t). The boundary conditions are imposed for x = 0, t 1 t +C 1 C 1 = 1 not ω 0 0 = C = C = 0 0 C 3 = 0, Continuum Mechanics for Engineers Theory and Problems X. Oliver and C. Agelet de Saracibar C 3 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/rg...581.0961

14 CHAPTER 3. COMPATIBILITY EQUATIONS and the final result is Velocity vector ω (x,t) not te y 1 0 0. v 1 = v 1 tetx ; y = 0; v 1 z = 0 v 1 = C 1 (t)+etx, v = 0; v y = 0; v z = v = C (t)+z, v 3 = 0; v 3 y = v 3 tey ; z = 0 v 3 = C 3 (t)+tey. The boundary conditions are imposed for x = 0, t 1 +C 1 C 1 not v 0 0 = C = C t t +C 3 C 3 = t, and the spatial description of the velocity field is e tx +t 1 v(x) not z. te y t Continuum Mechanics for Engineers Theory and Problems 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...581.0961

Problems and Exercises 15 EXERCISES 3.1 Deduce the displacement field that corresponds to the infinitesimal strain tensor 0 te ty 0 ε (x,t) not te ty 0 0. 0 0 te tz At point (1,1,1), u not [e t, e t, e t ] T and θ not [0, 0, te t ] T is verified. 3. Determine the spatial description of the velocity field that corresponds to the strain rate tensor 0 0 te tz d(x,t) not 0 te ty 0. te tz 0 0 The following is known: { for z = 0: v x = v z = 0, t, x,y for y = 1: v y = 0, t, x,z Continuum Mechanics for Engineers Theory and Problems 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...581.0961