Belytschko T. - Introduction (779635), страница 22

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that F be invertible.When the deformation gradient F is regular, the Jacobian determinant J must benonzero, since the inverse of F exists if and only if its determinant J ≠ 0. Thusthe second and third conditions are related. We have stated a stronger conditionthat J be positive rather than just nonzero, which will be seen in Section 3.5.4 tofollow from mass conservation.3.2.7 Rigid Body Rotation and Coordinate Transformations.Rigid body rotation plays a crucial role in the theory of nonlinear continuummechanics. Many of the complexities which permeate the field arise from rigidbody rotation. Furthermore, the decision as to whether linear or nonlinearsoftware is appropriate for a particular linear material problem hinges on themagnitude of rigid body rotations. When the rigid body rotations are largeenough to render a linear strain measure invalid, nonlinear software must be used.A rigid body motion consisting of a translation x T (t ) and a rotation aboutthe origin is written asx ( X, t) = R(t ) ⋅X + xT ( t)xi (X,t ) = Rij( t) X j + xTi (t )(3.2.20)where R(t ) is the rotation tensor, also called a rotation matrix.

Because rigidbody rotation preserves length, and noting that dx T = 0 in rigid body motion, wehavedx ⋅ dx = dX ⋅  RT ⋅R ⋅ dX()dx idxi = RijdX j Rik dXk = dX j RTji Rik dXkSince the length must stay unchanged in rigid body motion, it follows thatRT ⋅ R = I(3.2.20b)and its inverse is given by its transpose:R−1 = R TRij−1 = RijT = Rji(3.2.21)The rotation tensor R is therefore said to be an orthogonal matrix and anytransformation by this matrix, such as x = RX, is called an orthogonaltransformation. Rotation is an example of an orthogoanl transformation.A rigid body rotation of a Lagrangian mesh of rectangular elements isshown in Fig. 3.2.

As can be seen, in the rigid body rotation, the element edges3-10T. Belytschko, Continuum Mechanics, December 16, 199811are rotated but the angles between the edges remain right angles. The elementedges are lines of constant X and Y, so when viewed in the deformedconfiguration, the material coordinates are rotated when the body is rotated asshown in Fig. 3.2.Specific expressions for the rotation matrix can be obtained in variousways.

We obtain it here by relating the components of the vector in r twodifferent coordinate systems with orthogonal base vectors e i and ˆe i ; a twodimensional example is shown in Fig. 3.3. The components in the rotatedcoordinate system are shown in Fig. 3.3. Since the vector r is independent of thecoordinate systemr = ri e i = rˆi ˆei(3.2.22)Yy, YXΩ0Ωφ ( X, t)x, XFig. 3.2. A rigid body rotation of a Lagrangian mesh showing the material coordinates whenviewed in the reference (initial, undeformed) configuration and the current configuration.3-11T.

Belytschko, Continuum Mechanics, December 16, 1998y^12yrx^e y^eye x^θexxFig. 3.3. Nomenclature for rotation transformation in two dimensions.Taking the scalar product of the above with e j givesri e i ⋅e j = ˆriˆe i ⋅e j → riδij = rˆi eˆ i ⋅e j → rj = R jirˆi , R ji = e j ⋅ ˆei(3.2.23)The second equation follows from the orthogonality of the base vectors, (3.2.21).The above shows that the elements of the rotation matrix are given by the scalarproducts of the corresponding base vectors; thus R12 = e1 ⋅ˆe2 .So thetransformation formulas for the components of a vector areri = Rijˆrj ≡ Riˆjrˆj , ˆrj = RjiT ri = Riˆjri(3.2.24)where the equation on the right follows from (3.3.20b).

In the second term of theindicial forms of the equations we have put the hat on the component associatedwith the hatted coordinates, but later it is often omitted. Note that the hatted indexis always the second index of the rotation matrix; this convention helps inremembering the form of the transformation eqaution. In matrix form the aboveare written asr = Rˆr,ˆr = RT rThe above is a matrix expression, as indicated by the absence of dots between theterms. The column matrices of components r and ˆr differ, but they pertain tothe same tensor.

In many works, this distinction is clarified by using differentsymbols for matrices and tensors, but the notation we have chosen does not pemitthis distinction.3-12T. Belytschko, Continuum Mechanics, December 16, 199813Writing out the rotation transformation in two dimensions givesRxˆy  ˆrx  e x ⋅ e ˆx =Ryˆy ˆry  ey ⋅e ˆxrx   Rxˆx = ry   Ryˆxex ⋅ e ˆy rˆx   cos θ -sin θ rˆx  = ey ⋅ eˆy  ˆry  sin θ cos θ ˆry (3.2.25)In the above, it can be seen that the subscripts of the rotation matrix correspond tothe vector components which are related by that term; for example, in theexpression for the x component in row 1, the Rxˆy is the coefficient of the ˆycomponent of r . The last form of the transformation in the above is obtained byevaluating the scalar products from Fig.

3.3 by inspection.The rotation of a vector is obtained by a similar relation. If the vector w isobtained by a rotation of the vector v, the two are related byw = R⋅v, wi = Rij v j(3.2.26)The first of the above can be written as()()w = R⋅ v j e j = v j R⋅e j = v jˆe j(3.2.27)where we have used the fact that the base vectors transform exactly like thecomponents; this can easily be derived by using (3.2.23). Taking the innerproduct of the first and last expressions of the above with the rotated base vectorˆe i gives()wˆ i = eˆ i ⋅w = v j eˆ i ⋅ˆe j = v jδij = vi(3.2.28)This shows that the components of the rotated vector w in the rotated coordinatesystem are identical to the components of the vector v in the unrotated coordinatesystem.The components of a second order tensor D are transformed betweendifferent coordinate systems byeˆ RTD = RDˆ RTDij = Rik Dkl lj(3.2.30a)The inverse of the above is obtained by premultiplying by RT , postmultiplying byR and using the orthogonality of R , (3.2.20b):ˆ = R T DRDˆ RTDij = Rik Dkl lj(3.2.30b)Note that the above are matrix expressions which relate the components of thesame tensor in two different coordinate systems.The velocity for a rigid body motion can be obtained by taking the timederivative of Eq.

(3.2.20). This gives3-13T. Belytschko, Continuum Mechanics, December 16, 1998˙ ( t) ⋅X + ˙x ( t )˙x ( X,t ) = RTor˙xi ( X, t) = R˙ ij (t ) Xj + ˙xTi (t )14(3.2.31)The structure of rigid body rotation can be clarified by expressing the materialcoordinates in (3.2.31) in terms of the spatial coordinates via (3.2.20), giving˙ ⋅RT ⋅ ( x − x ) + ˙xv ≡ ˙x = RTT(3.2.32)The tensor˙ ⋅R TΩ =R(3.2.33)is called the angular velocity tensor or angular velocity matrix, Dienes(1979, p221).

It is a skew symmetric tensor, skew symmetric tensors are also calledantisymmetric tensors. To demonstrate the skew symmetry of the angularvelocity tensor, we take the time derivative of (3.2.21) which givesDDt˙ ⋅RT + R ⋅R˙ T = 0 → Ω = −Ω T= 0→R(R⋅ R T ) = DIDt(3.2.34)Any skew symmetric tensor can be expressed in terms of the components of avector, cakked the axial vector, and the corresponding action of that matrix on avector can be replicated by a cross product, so if ω if the axial vector of Ω , thenΩr = ω ×rorΩ ijrj = eijkω j rk(3.2.34b)for any r andeijk1 foran even permutationof ijk= -1for an odd permutationof ijk 0 if anyindex is repeated(3.2.36)The tensor eijk is called the alternator tensor or permutation symbol.ω areThe relations between the skew symmetric tensor Ω and its axial vectorωi = 21 eijkΩ jk , Ω ij = eijkω k(3.2.35)which can be obtained by enforcing (3.2.34b) for all r .In two dimensions, a skew symmetric tensor has a single independentcomponent and its axial vector is perpendicular to the two dimensional plane ofthe model, so 0Ω =−Ω 12Ω12 0=0  ω3−ω 3 0 (3.2.37a)3-14T.

Belytschko, Continuum Mechanics, December 16, 199815In three dimensions, a skew symmetric tensor has three independent componentsand which are related to the three components of its axial vector by (3.2.25)giving 0Ω =  −Ω12−Ω 13Ω120−Ω23Ω13  0ω3 −ω 2  Ω23  = −ω 30ω1 0   ω2 −ω 10 (3.2.37b).When Eq.

(3.3.32) is expressed in terms of the angular velocity vector, wehave()vi ≡ x˙ i = Ωij x j − x Tj + vTi= eijkω j ( x k − x Tk ) + vTior v ≡ ˙x = ω× ( x − x T ) + vT(3.2.38)where we have exchanged k and j in the second line and used ekij = eijk . Thesecond equation is the well known equation for rigid body motion as given indynamics texts. The first term on the left hand side is velocity due to the rotationabout the point x T and the second term is the translational velocity of the pointx T . Any rigid body velocity can be expressed by (3.2.28).This concludes the formal discussion of rotation in this Chapter.However, the topic of rotation will reappear in many other parts of this Chapterand this book. Rotation, especially when combined with deformation, isfundamental to nonlinear continuum mechanics, and it should be thoroughlyunderstood by a student of this field.CorotationalRate-of-Deformation.

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