Belytschko T. - Introduction (779635), страница 24

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it should increase as the deformationincreases, etc. (Hill, ). However, the ability to represent rigid body motion iscrucial and indicates when geometrically nonlinear theory must be used.3.3.1 Green strain tensor. The Green strain tensor E is defined byds 2 − dS 2 = 2dX ⋅ E⋅ dXordxi dxi − dX idX i = 2dXi Eij dX j(3.3.1)so it gives the change in the square of the length of the material vector dX.Recall the vector dX pertains to the undeformed configuration. Therefore, theGreen strain measures the difference of the square of the length of an infinitesimalsegment in the current (deformed) configuration and the reference (undeformed)3-21T. Belytschko, Continuum Mechanics, December 16, 199822configuration.

To evaluate the Green strain tensor, we use (3.2.15) to rewrite theLHS of (3.3.1) as()dx ⋅dx = ( dX⋅F ) ⋅ ( F ⋅dX ) = dX ⋅ FT ⋅F ⋅dX(3.3.2)The above are clearer in indicial notation()dx ⋅dx = dxi dxi = Fij dX j FikdX k = dX j FjiT Fik dXk = dX ⋅ F T ⋅ F ⋅dXUsing the above with (3.3.1) and dX ⋅ dX = dX ⋅I ⋅ dX givesdX⋅ FT ⋅ F⋅ dX− dX ⋅ I⋅dX − dX⋅ 2E⋅ dX = 0(3.3.3)Factoring out the common terms then yields()dX⋅ FT ⋅ F − I − 2E ⋅ dX = 0(3.3.4)Since the above must hold for all dX, it follows thatE=()1 TF ⋅F − I2or Eij =(1 TF F −δ2 ik kj ij)(3.3.5)The Green strain tensor can also be expressed in terms of displacement gradientsbyE=11  ∂u ∂u ∂u ∂u (∇X u)T +∇ X u + (∇ Xu )T ⋅∇ X u , Eij =  i + j + k k  (3.3.6)22  ∂X j ∂Xi ∂Xi ∂X j ()This expression is derived as follows. We first evaluate F T ⋅F in terms of thedisplacements using indicial notation.FikT Fkj = Fki Fkj =∂xk ∂xk∂X i ∂Xj(definition of transpose and Eq.

(3.2.14)) ∂u ∂X   ∂u∂X =  k + k   k + k  (by Eq. (3.2.7)) ∂Xi ∂X i   ∂X j ∂X j  ∂u  ∂u=  k +δ ki   k +δ kj  ∂Xi  ∂X j ∂u ∂u ∂u ∂u=  i + j + k k +δ ij  ∂Xj ∂Xi ∂X i ∂X jSubstituting the above into (3.3.5) gives (3.3.6).3-22T. Belytschko, Continuum Mechanics, December 16, 199823To show that the Green strain vanishes in rigid body motion, we considerthe deformation function for a general rigid body motion described in Eq.(3.2.20): x = R ⋅ X+ x T . The deformation gradient F according to Eq (3.2.14) isthen given by F = R .

Using the expression for the Green strain, Eq. (3.3.5). gives()E = 12 RT ⋅R − I = 21 (I − I) = 0where the second equality follows from the orthogonality of the rotation tensor,Eq.(3.2.21). This demonstrates that the Green strain will vanish in any rigid bodymotion, so it meets an important requirement of a strain measure.3.3.2 Rate-of-deformation.The second measure of strain to be consideredhere is the rate-of-deformation D . It is also called the velocity strain and thestretching tensor.

In contrast to the Green strain tensor, it is a rate measure ofstrain.In order to develop an expression for the rate-of-deformation, we firstdefine the velocity gradient L byL=∂v= (∇v)T = (grad v)T∂xor Lij =∂vi,∂x j(3.3.7)dv = L⋅ dx or dvi = Lij dx jWe have shown several tensor forms of the definition which are frequently seen,but we will primarily use the first or the indicial form. In the above, the symbol∇ or the abbreviation “grad” preceding the function denotes the spatial gradientof the function, i.e., the derivatives are taken with respect to the spatialcoordinates. The symbol ∇ always specifies the spatial gradient unless adifferent coordinate is appended as a subscript, as in ∇ X , which denotes thematerial gradient.The velocity gradient tensor can be decomposed into symmetric and skewsymmetric parts byL=() (11L + LT + L −LT22) or Lij=() (11Lij + L ji + Lij − L ji22)(3.3.8)This is a standard decomposition of a second order tensor or square matrix: anysecond order tensor can be expressed as the sum of its symmetric and skewsymmetric parts in the above manner; skew symmetry is also known asantisymmetry.The rate-of-deformation D is defined as the symmetric part of L, i.e.

thefirst term on the RHS of (3.3.8) and the spin W is the skew symmetric part of L,i.e. the second term on the RHS of (3.3.8). Using these definitions, we can writeL = (∇v ) = D + WTorLij = vi , j = Dij + Wij3-23(3.3.9)T. Belytschko, Continuum Mechanics, December 16, 1998D=(1L +LT2W=()1L − LT2or)1  ∂v ∂v j Dij =  i +2  ∂x j ∂xi orWij =1  ∂vi ∂v j −2  ∂x j ∂xi 24(3.3.10)(3.3.11)The rate-of-deformation is a measure of the rate of change of the square ofthe length of infinitesimal material line segments.

The definition is∂∂ds2 = ( dx⋅ dx) = 2dx⋅ D⋅ dx∂t∂t( )∀dx(3.3.12)The equivalence of (3.3.10) and (3.3.12) is shown as follows. The expression forthe rate-of-deformation is obtained from the above as follows:2dx⋅D⋅ dx =∂( dx(X, t) ⋅dx (X,t )) = 2dx⋅dv (using(3.2.8))∂t= 2dx⋅∂v⋅dx by chain rule∂x= 2dx⋅ L⋅dx(using (3.3.7))(= dx⋅ ( L + LT ) ⋅dx)= dx⋅ L + LT +L − LT ⋅dx(3.3.13)by antisymmetry of L − LT ; (3.3.10) follows from the last line in (3.3.13) due tothe arbitrariness of dx .In the absence of deformation, the spin tensor and angular velocity tensorare equal, W = Ω .

This is shown as follows. In rigid body motion D = 0 , soL = W and by integrating Eq. (3.3.7b) we havev = W ⋅( x − xT ) + v T(3.3.14)where x T and v T are constants of integration. Comparison with Eq. (3.2.32) thenshows that the spin and angular velocity tensors are identical in rigid bodyrotation. When the body undergoes deformation in addition to rotation, the spintensor generally differs from the angular velocity tensor. This has importantimplications on the character of objective stress rates, which are discussed inSection 3.7.3.3.3. Rate-of-deformation in terms of rate of Green strain. Therate-of-deformation can be related to the rate of the Green strain tensor.

To obtain3-24T. Belytschko, Continuum Mechanics, December 16, 199825this relation, we first obtain the material gradient of the velocity field, defined inEq. (3.3.7b), in terms of the spatial gradient by the chain rule:L=∂v ∂v ∂X=⋅,∂x ∂X ∂xLij =∂vi∂v ∂X k= i∂x j ∂Xk ∂x j(3.3.15)The definition of the deformation gradient is now recalled, Eq. (3.3.10),Fij = ∂xi ∂X j . Taking the material time derivative of the deformation gradientgives˙ = ∂v ,F∂X˙ = ∂viFij∂X j(3.3.16)By the chain rule∂xi ∂Xk∂X∂X= δij → Fik k =δij → Fkj−1 = k ,∂Xk ∂xj∂xj∂xjF −1 =∂X∂x(3.3.17)Using the above two equations, (3.3.15) can be rewritten asL = ˙F ⋅F− 1,Lij = F˙ ik Fkj−1(3.3.18)When the deformation gradient is known, this equation can be used to obtain therate-of-deformation and the Green strain rate. To obtain a single expressionrelating these two measures of strain rate, we note that from (3.3.10) and (3.3.18)we haveD=12(L +L ) = (F˙ ⋅FT12−1˙T+ F −T ⋅ F)(3.3.19)Taking the time derivative of the expression for the Green strain, (3.3.5) gives˙ =Ε12() ()D T˙ + ˙FT ⋅ FF ⋅F− I = 21 F T ⋅ FDt(3.3.20)Premultiplying Eq.

(3.3.19) by F T F and postmultiplying by F givesF T ⋅D⋅ F =12(FT ⋅ F˙ + ˙FT ⋅F) → E˙ = F T ⋅D⋅For E˙ ij = FikT DklFlj (3.3.21)where the last equality follows from Eq. (3.3.20). The above can easily beinverted to yield˙ ⋅F −1D = F− T ⋅ Eor˙ F −1Dij = Fik−T Ekl lj(3.3.22)As we shall see in Chapter 5, (3.3.22) is an example of a push forward operation,(3.3.21) of the pullback operation. The two measures are two ways of viewing thesame tensor: the rate of Green strain is expresses in the reference configurationwhat the rate-of-deformation expresses in the current configuration. However, the3-25T. Belytschko, Continuum Mechanics, December 16, 199826properties of the two forms are somewhat different. For instance, in Example 3.7we shall see that the integral of the Green strain rate in time is path independent,whereas the integral of the rate-of-deformation is not path independent.These formulas could be obtained more easily by starting from thedefinitions of the Green strain tensor and the rate-of-deformation, Eqs.

(3.3.1) and(3.3.9), respectively. However, Eq. (3.3.18), which is very useful, would then beskipped. Therefore the other derivation is left as an exercise, Problem ?.Example 3.5.Strain Measures in CombinedRotation. Consider the motion of a body given byStretchandx ( X,t ) = (1+ at ) Xcos π2 t − (1+ bt )Y sin π2 t(E3.5.1)y( X,t ) = (1+ at )X sin π2 t + (1+ bt )Y cos π2 t(E3.5.2)where a and b are positive constants. Evaluate the deformation gradient F , theGreen strain E and rate-of-deformation tensor as functions of time and examinefor t = 0 and t = 1.For convenience, we defineA( t) ≡ (1 +at ), B(t ) ≡ (1 +bt ) , c ≡cos π2 t , s ≡ sin π2 t(E3.5.3)The deformation gradient F is evaluated by Eq.(3.2.10) using (E3.5.1): ∂xF = ∂X∂y ∂X∂x ∂Y  =  Ac −Bs ∂y As Bc ∂Y(E3.5.4)The above deformation consists of the simultaneous stretching of thematerial lines along the X and Y axes and the rotation of the element.

Thedeformation gradient is constant in the element at any time, and the othermeasures of strain will also be constant at any time. The Green strain tensor isobtained from (3.3.5), with F given by (E3.5.4), which givesE==1 T1   Ac As Ac −Bs   1 0 F ⋅F − I =  −22  −Bs Bc  As Bc  0 1 (1 A2 )200   1 0  1  2at + a2 t2 =−B 2  0 1  2 002bt + b2t 2 (E3.5.5)It can be seen that the values of the Green strain tensor correspond to what wouldbe expected from its definition: the line segments which are in the X and Ydirections are extended by at and bt, respectively, so E11 and E22 are nonzero.The strain E11 = EXX is positive when a is positive because the line segment alongthe X axis is lengthened.

The magnitudes of the components of the Green strain3-26T. Belytschko, Continuum Mechanics, December 16, 199827correspond to the engineering measures of strain if the quadratic terms in a and bare negligible. The constants are restricted so that at >−1and bt >−1, forotherwise the Jacobian of the deformation becomes negative. When t = 0, x = Xand E = 0 .For the purpose of evaluating the rate-of-deformation, we first obtain thevelocity, which is the material time derivative of (E3.5.1):v x = (ac − π2 As) X − (bs +vy = ( as+π2π2Bc)Y(E3.5.6)Ac) X + (bc − π2 Bs)YThe velocity gradient is given by (3.3.7b),L = (∇v )T ∂v∂xx=  ∂vy ∂x  ac − ωAs −bs − ωBc =  as +ωAc bc− ωBs ∂y ∂vx∂y∂vy(E3.5.7)Since at t = 0, x = X, y = Y , c=1, s=0, A = B =1, so the velocity gradient at t = 0is given byL = (∇v )T a= π 20 −1− π2 a 0 π ,W=→ D= 2 10 b 0 b (E3.5.8)To determine the time history of the rate-of-deformation, we first evaluate thetime derivative of the deformation tensor and the inverse of the deformationtensor.

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