Belytschko T. - Introduction (779635), страница 31

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Rigid body translation does not appear in this equation because dxand dX are differential line segments in the current and reference configurations,respectively, and the differential line segments are not affected by translation. IfEq. (3.7.3) were integrated to obtain the deformation function, x = φ ( X,t ) , thenthe rigid body translation would appear as a constant of integration.

In atranslation, F = I , and dx=dX.The polar decomposition theorem is proven in the following. To simplifythe proof, we treat the tensors as matrices. Premultiplying both sides of Eq.(3.7.1) by its transpose gives3-58T. Belytschko, Continuum Mechanics, December 16, 1998FT ⋅F = ( RU ) ( RU) = UT R T RU = UT U = UUT59(3.7.4)where (3.7.2) is used to obtain the third and fourth equalities.

The last term is thesquare of the U matrix. It follows that(U = FT ⋅F)12(3.7.5)The fractional power of a matrix is defined in terms of its spectralrepresentation, see e.g. Chandrasekharaiah and Debnath (1994, p96). It iscomputed by first transforming the matrix to its principal coordinates, where thematrix becomes a diagonal matrix with the eigenvalues on the diagonal. Thefractional power is then applied to all of the diagonal terms, and the matrix istransformed back. This is illustrated in the following examples. The matrixFT ⋅F is positive definite, so all of its eigenvalues are positive.

Consequently thematrix U is always real.The rotation part of the deformation, R, can then be found by applying Eq.(3.7.1), which givesR = F⋅ U−1(3.7.6)The existence of the inverse of U follows from the fact that all of its eigenvaluesare always positive, since the right hand side of Eq. (3.7.5) is always a positivematrix.The matrix U is closely related to an engineering definition of strain. Itsprincipal values represent the elongations of line segments in the principaldirections of U. Therefore, many researchers have found this tensor to beappealing for developing constitutive equations. The tensor U − I is called theBiot strain tensor.A deformation can also be decomposed in terms of a left stretch tensor anda rotation according toF =V⋅R(3.7.7)This form of the polar decomposition is used less frequently and we only note it inpassing here.

It will play a role in discussions of material symmetry for elasticmaterials at finite strain. The polar decomposition theorem, which is usuallyapplied to the deformation tensor, applies to any invertible square matrix: anysquare matrix can be multiplicatively decomposed into a rotation matrix and asymmetric matrix, see Chandrasekharaiah and Debnath (1994, p97).It is emphasized that the rotations of different line segments at the samepoint depend on the orientation of the line segment. In a three dimensional body,only three line segments are rotated exactly by R(X,t) at any point X. These arethe line segments corresponding to the principal directions of the stretch tensor U.It can be shown that these are also the principal directions of the Green straintensor.

The rotations of line segments which are oriented in directions other thanthe principal directions of E are not given by R.3-59T. Belytschko, Continuum Mechanics, December 16, 199860Example 3.10 Consider the motion of a triangular element in which the nodalcoordinates x I ( t ) and yI ( t) are given byx1( t) = a + 2aty1( t) = 2atx2 ( t) = 2atx3( t) = 3aty2 ( t ) = 2a − 2aty3( t ) = 0(E3.10.1)Find the rigid body rotation and the stretch tensors by the polar decompositiontheorem at t=1.0 and at t=0.5.The motion of a triangular domain can most easily be expressed by usingthe shape functions for triangular elements, i.e.

the area coordinates. In terms ofthe triangular coordinates, the motion is given byx (ξ,t ) = x1( t )ξ1 + x2( t)ξ2 + x 3( t)ξ 3(E3.10.2)y(ξ,t ) = y1( t)ξ1 + y2( t )ξ2 + y3 (t)ξ 3(E3.10.3)where ξ I are the triangular, or area, coordinates; see Appendix A; the materialcoordinates appear implicitly in the RHS of the above through the relationshipbetween the area coordinates and the coordinates at time t=0. To extract thoserelationships we write the above at time t=0, which givesx ( ξ,0 ) = X = X1ξ1 + X2 ξ2 + X3ξ3 = aξ1(E3.10.4)y( ξ,0) = Y = Y1ξ1 +Y2ξ2 +Y3 ξ3 = 2aξ2(E3.10.5)In this case, the relations between the triangular coordinates are particularlysimple because most of the nodal coordinates vanish in the initial configuration,so the relations developed above could be obtained by inspection.Using Eq. (E3.10.5) to express the triangular coordinates in terms of thematerial coordinates, Eq (E3.10.1) can be writtenx( X,1) = 3aξ1 + 2aξ2 + 3aξ3 X YY= 3X + Y + 3a 1− −  = 3a −a 2a 2y( X,1) = 2aξ1 + 0ξ 2 + 0ξ3(E3.10.6)(E3.10.7)=2XThe deformation gradient is then obtained by evaluating the derivatives of theabove motion using Eq.

(3.2.16)3-60T. Belytschko, Continuum Mechanics, December 16, 1998 ∂xF =  ∂∂Xy∂ X∂x ∂Y =∂y ∂Y 02−0. 50 61(E3.10.8)The stretch tensor U is then evaluated by Eq. (3.7.5):TU = ( F F)1/ 24=00  1/ 2  2 0 =0. 25 0 0. 5(E3.10.9)In this case the U matrix is diagonal, so the principal values are simply thediagonal terms.

The positive square roots are chosen in evaluating the square rootof the matrix because the principal stretches must be positive; otherwise theJacobian determinant would be negative since according to Eq. (3.7.1),J =det ( R) det (U) and det (R) = 1, so det (U) < 0 implies J < 0 . The rotationmatrix R is then given by Eq. (3.7.6): 0 −0.

5  0. 5 0  0R = FU−1 = =0   0 2  12−10 (E3.10.10)Comparing the above rotation matrix R and Eq. (3.2.25), it can be seen that therotation is a counterclockwise 90 degree rotation. This is also readily apparentfrom Fig. 3.9. The deformation consists of an elongation of the line segmentbetween nodes 1 and 3, i.e. dX, by a factor of 2, (see U 11 in Eq. (E3.10.9)) and acontraction of the line segment between nodes 3 and 2, i.e. dY, by a factor of 0.5,(see U 22 in Eq. (E3.10.9)), followed by a translation of 3a in the x-direction and a90 degree rotation.

Since the original line segments along the x and y directionscorrespond to the principal directions, or eigenvectors, of U, the rotations of theseline segments correspond to the rotation of the body in the polar decompositiontheorem.The configuration at t=0.5 is given by evaluating Eq. (E3.10.1) at that time,giving:x (X,0.5) = 2aξ1 + aξ2 +1.5aξ3= 2a(E3.10.11a)XYX Y+a+1.5a 1− −  = 1.5a + 0.5X −0.25Ya2aa 2a y( X,0.5) = aξ1 + aξ2 +0ξ3XY= a +a= X + 0.5Ya2a(E3.10.11b)The deformation gradient F is then given by ∂xF =  ∂∂Xy∂ X∂x ∂Y =∂y ∂Y 0.

5 −0. 25 10. 5 (E3.10.12)3-61T. Belytschko, Continuum Mechanics, December 16, 199862and the stretch tensor U is given by Eq. (3.7.6):TU = ( F F)1/ 20. 375  1/ 2 1. 0932 0. 2343 1. 25== 0. 375 0. 31250. 2343 0. 5076(E3.10.13)The last matrix in the above is obtained by finding the eigenvalues λi of F T F ,taking their positive square roots, and placing them on a diagonal matrix calledH = diag λ1 , λ2 . The matrix H transformed back to the global components()by U = AT HA where A is the matrix whose columns are the eigenvectors of F T F .These matrices are:−0.9436 0.3310 A=−0.3310 −0.94361.38150 H= 0.1810 0(E3.10.14)The rotation matrix R is then found byR = FU−1−1 0. 5 −0.

25  1. 0932 0. 2343 0. 6247 −0. 7809 ==0. 5   0. 2343 0. 50761 0. 7809 0. 6247 (E3.10.15)Example 3.11 Consider the deformation for which the deformation gradient isc − as ac− sF=s + ac as +c c =cos θ,s = sin θ(E3.11.1)1where a is a constant. Find the stretch tensor and the rotation matrix when a= 2 ,πθ = 2.For the particular values given− 12F=1−11 2 1.25 1 C = FT ⋅F =  1 1.25(E3.11.2)The eigenvalues and corresponding eigenvectors of C areλ1 = 0.25y1T =12[1−1]λ2 = 2.25yT2 =12[1 1](E3.11.3)3-62T.

Belytschko, Continuum Mechanics, December 16, 199863The diagonal form of C, diag(C), consists of these eigenvalues and the squareroot of diag(C) is obtained by taking the positive square roots of theseeigenvalues 14diag(C) = 0 12 0 01/ 29  ⇒ diag(C ) = 30 2 4(E3.11.4)The U matrix is then obtained by transforming diag(C) back to the x-y coordinatesystemU = Y⋅diag(C1/ 2)⋅Y =T121 1 1  2 0 −1 1 0 3 2121 −1 1 21 1  = 2 11(E3.11.5)2The rotation matrix is obtained by Eq.

(3.7.6):− 12R = FU = 1−1−1 2  2 −1  0 −11 3  = 1 0 −122 (E3.11.6)3.7.2 Objective Rates in Constitutive Equations.To explain whyobjective rates are needed for the Cauchy stress tensor, we consider the rod shownin Fig. 3.10. Suppose the simplest example of a rate constitutive equation is used,known as a hypoelastic law, where the stress rate is linearly related to the rate-ofdeformation:DσijDt= Cijkl DklDσ= C :DDtory(3.7.8)X, yσ0σ0xxYσ x = σ0 , σy = 0σ x = 0, σ y = σ 03-63T. Belytschko, Continuum Mechanics, December 16, 199864Fig. 3.10. Rotation of a bar under initial stress showing the change of Cauchy stress which occurswithout any deformation.The question posed here is: are the above valid constitutive equations?The answer is negative, and can be explained as follows.

Consider a solid,such as the bar in Fig. 3.10, which is stressed in its initial configuration withσ x = σ 0 . Now assume that the bar rotates as shown at constant length, so there isno deformation, i.e. D=0. Recall that in rigid body motion a state of initial stress(or prestress) is frozen in the body in a solid, i.e. since the deformation does notchange in a rigid body rotation, the stress as viewed by an observer riding with thebody should not change. Therefore the Cauchy stress expressed in a fixedcoordinate system will change during the rotation, so the material derivative of thestress must be nonzero.

However, in a pure rigid body rotation, the right handside of Eq. (3.7.8) will vanish throughout the motion, for we have already shownthat the rate-of-deformation vanishes in rigid body motion. Therefore, somethingmust be missing in Eq. (3.7.8).The situation explained in the previous paragraph is not just hypothetical;it is representative of what happens in real situations and simulations. A bodymay be in a state of stress due to thermal stresses or prestressing; an example isthe stress in prestressed reinforcement bars.

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