Belytschko T. - Introduction (779635), страница 23

Просмтор этого файла доступен только зарегистрированным пользователям. Но у нас супер быстрая регистрация: достаточно только электронной почты!

Текст из файла (страница 23)

As we shall see later, in many casesit is convenient to rotate the coordinate at each point of the material with thematerial. The rate-of-deformation is then expressed in terms of its corotationalˆ , which can be obtained from the global components by (3.2.30).components DijThese components can be obtained directly from the velocity field byˆ = 1  ∂vˆ i + ∂ˆvj  ≡ sym ∂vˆ i  ≡ vDˆi ,ˆjij2  ∂xˆ j ∂xˆ i  ∂xˆ j (3.2.39)where vˆ i ≡ vˆi are the components of the velocity field in the corotational system.the corotational system can be obtained from the polar decomposition theorem tobe described later or by other techniques; see section 4.6.Example3.1 RotationandStretchofTriangularElement.Consider the 3-node triangular finite element shown in Fig.

3.4. Let the motion ofthe nodes be given by3-15T. Belytschko, Continuum Mechanics, December 16, 199816x1(t ) = y1(t ) = 0πt,2πtx3 (t ) =− (1+ bt ) sin ,2x2 ( t) = 2(1+ at )cosπt2πty 3 (t ) = (1+ bt )cos2y2( t) = 2(1+ at ) sin(E3.1.1)Find the deformation function and the Jacobian determinant as a function of timeand find the values of a(t) and b(t) such that the Jacobian determinant remainsconstant.y2y,Yπω =232(1+a)13212x,X1+bx1Fig. 3.4.

Motion descrived by Eq. (E3.1.1) with the initial configuration at the left and thedeformed configuration at t=1 shown at the right.In terms of the triangular element coordinates ξI , the configuration of atriangular 3-node, linear displacement element at any time can be written as (seeAppendix A if you are not familiar with triangular coordinates)x (ξ,t ) = ∑ x I (t )ξI = x1(t )ξ1 + x2( t)ξ 2 + x 3( t)ξ3Iy(ξ, t) = ∑ yI ( t)ξ I = y1(t )ξ1 + y2 (t )ξ2 + y3 (t )ξ3(E3.1.2)IIn the initial configuration, i.e. at t=0:X = x ( ξ, 0) = X1ξ1 + X2 ξ2 + X3ξ3(E3.1.3)Y = y (ξ, 0) = Y1ξ1 + Y2ξ2 + Y3ξ 3Substituting the coordinates of the nodes in the undeformed configuration into theabove, X1 = X3 = 0, X2 = 2, Y1 = Y2 = 0, Y3 = 1 yieldsX = 2ξ2 , Y = ξ3(E3.1.4)3-16T.

Belytschko, Continuum Mechanics, December 16, 199817In this case, the relations between the triangular coordinates and the materialcoordinates can be inverted by inspection to give1ξ2 = 2 X,ξ3 = Y(E3.1.5)Substituting (E3.1.1) and (E3.1.5) into (E3.1.2) gives the following expression forthe motionπtπt−Y (1+ bt ) sin22πtπty( X,t ) = X (1+ at) sin +Y (1+ bt )cos22x ( X,t ) = X (1+ at)cos(E3.1.6)The deformation gradient is given by Eq.(3.2.16): ∂xF = ∂X∂y ∂X∂x   (1+ at )cos πt∂Y  = 2∂y  πt∂Y  (1+ at ) sin 2πt 2πt (1+bt ) cos 2−(1+ bt ) sin(E3.1.7)The deformation gradient is a function of time only and at any time constant in theelement because the displacement in this element is a linear function of thematerial coordinates.

The determinant of the deformation gradient is given byπtπt J =det ( F) = (1 +at )(1+ bt ) cos 2 +sin 2 22(E3.1.8)When a=b=0 the Jacobian determinant remains constant, J=1. This is a rotationwithout deformation. As expected, the Jacobian determinant remains constantsince the volume (or area in two dimensions) of anypart of a body does notchange in a rigid body motion. The second case in which the Jacobiandeterminant J remains constant is when b =− a / (1+ at ) , which corresponds to adeformation in which the area of the element remains constant. This is the type ofdeformation is called an isochoric deformation; the deformation of incompressiblematerials is isochoric.Example 3.2 Consider an element which is rotating at a constant angularvelocity ω about the origin.

Obtain the accelerations using both the material andspatial descriptions. Fine the deformation gradient F and its rate.The motion for a pure rotation about the origin is obtained from Eq.(3.2.20) using the rotation matrix in two dimensions (3.2.25): x cosωt −sin ωt  X x (t ) = R(t )X ⇒   =   y  sin ωt cos ωt   Y (E3.2.1)where we have used θ = ωt to express the motion is a function of time; ω is theangular velocity of the body. The velocity is obtained by taking the derivative ofthis motion with respect to time, which gives3-17T. Belytschko, Continuum Mechanics, December 16, 199818v x  ˙x  −sin ωt − cosωt  X v  = ˙  = ω   cos ωt −sin ωt  Y  y  y (E3.2.2)The acceleration in the material description is obtained by taking time derivativesof the velocitiesa x  v˙ x 2  −cos ωt a  = v˙  = ω − sin ωt y  ysin ωt   X  −cos ωt  Y (E3.2.3)To obtain a spatial description for the velocity, the material coordinates X and Y in(E3.2.2) are first expressed in terms of the spatial coordinates x and y by inverting(E3.2.1):v x  −sin ωt −cosωt  cos ωt sin ωt  x  =ω   cos ωt − sinωt − sin ωt cos ωt y  vy 0=ω1(E3.2.4)−1  x −y =ω0  y  xThe material time derivative the velocity field in the spatial description,Eq.(E3.2.4), is obtained via Eq.(3.2.11):∂v x ∂t   ∂vx ∂x ∂vx ∂y vx Dv ∂v=+ v⋅∇ v = +  Dt ∂t∂v y ∂t   ∂v y ∂x ∂vy ∂y vy (E3.2.5) 0 −ω  v x  −v y = 0+ =ωω 0   vy  vx If we then express the velocity field in (E3.2.5) in terms of the spatial coordinatesx and y by Eq.(E3.2.4), we havea x 0 −1  0  =ωω 1 0  1a y −1  x−1 0   x x   =ω 2   = −ω 2  0 y  0 −1 y y (E3.2.6)This is the well known result for the centrifugal acceleration: the acceleration(vector points toward the center of rotation and its magnitude is ω 2 x 2 + y2)12.To compare the above with the material form of the acceleration (E3.2.3)we use (E3.2.1) to express the spatial coordinates in (E3.2.6) in terms of thematerial coordinates, which gives0  cosωt −sinωt  X sin ωt  X v˙ x 2 −12  −cos ωt=ωv˙  = ω   0 −1 sin ωt cosωt  Y − sinωt −cosωt   Y  ywhich agrees with Eq.

(E3.2.3).3-18(E3.2.7)T. Belytschko, Continuum Mechanics, December 16, 199819The deformation gradient in obtained from its defintion (3.2.14) and(E3.2.1)F=sinωt cos ωt −sin ωt ∂x−1  cosωt= R =.F=∂X sinωt cos ωt −sin ωt cos ωt (E3.2.8)Example 3.3 Consider a square 4-node element, with 3 of the nodes fixed asshown in Fig.

3.5. Find the locus of positions of node 3 which results in avanishing Jacobian.y4y34J>03x12x12J<0J=0Figure 3.5. Original configuration of a square element and the locus of points for whichdeformed configuration with J < 0 is also shown.J = 0; aThe displacement field for the rectangular element with all nodes but node3 fixed is given by the bilinear fieldux ( X,Y ) = u3 x XY,uy ( X, Y ) = u3 y XY(E3.3.1)Since this element is a square, an isoparametric mapping is not needed.

Thisdisplacement field vanishes along the two shaded edges. The motion is given byx = X + ux = X + u3 x XYy = Y + uy = Y +u3y XY(E3.3.2)The deformation gradient is obtained from the above and Eq. (3.2.14):u3x X 1+ u3 x YF= u3 y Y 1+ u3 y X (E3.3.3)The Jacobian determinant is thenJ = det(F ) = 1+u3 xY + u3 y X(E3.3.4)3-19T. Belytschko, Continuum Mechanics, December 16, 199820We now examine when the Jacobian determinant will vanish. We need onlyconsider the Jacobian determinant for material particles in the undeformedconfiguration of the element, i.e.

the unit square X ∈[ 0,1] , Y ∈[ 0,1] . From the Eq.(E3.3.4), it is apparent that J is minimum when u3x < 0 and u3y < 0 . Then theminimum value of J occurs at X=Y=1, soJ ≥ 0 ⇒ 1+ u3xY + u3y X ≥ 0 ⇒ 1+u3 x + u3 y ≥ 0(E3.3.5)The locus of points along which J=0 is given by a linear function of the nodaldisplacements shown in Fig. 3.5, which also shows one deformed configuration ofthe element for which J < 0 . As can be seen, the Jacobian becomes negativewhen node 3 crosses the diagonal of the undeformed element.Example 3.4.

The displacement field around a growing crack is given byθθux = kf (r ) a + 2sin 2  cos22(E3.4.1)θθuy = kf (r) b − 2cos 2  sin22r 2 = ( X −ct )2 + Y 2,θ =tan −1(Y X )(E3.4.2)where a,b, c, and k are parameters which would be determined by the solution ofthe governing equations. This displacement field corresponds to a crack openingalong the X-axis at a velocity c; the configuration of the body at two times isshown in Fig.

3.6.yy,YyrΩ0ct1θx,Xct2Ω(t1)xΩ(t2 )Figure 3.6. The initial uncracked configuration and two subsequent configurations for a crackgrowing along x-axis.Find the discontinuity in the displacement along the line Y=0, X ≤ 0. Doesthis displacement field conform with the requirements on the motion given inSection 3.2.7?The motion is x = X + ux , y = Y +uy . The discontinuity in the displacementfield is found by finding the difference in (E3.4.1) for θ = π − and θ = π + , whichgives3-20xT.

Belytschko, Continuum Mechanics, December 16, 1998θ = π − ⇒ u x = 0, uy = kf (r )b21(E3.4.3)so the jumps, or discontinuities, in the displacement areux = 0, uy = 2kf (r)b(E3.4.4)Everywhere else the displacement field is continuous.This deformation function meets the criteria given in Section 3.3.6because the discontinuity occurs along only a line, which is a set of measure zeroin a two dimensional problem. From Fig. 3.6 it can be seen that in thisdeformation, the line behind the crack tip splits into two lines. It is also possibleto devise deformations where the line does not separate but a discontinuity occursin the tangential displacement field.

Both types of deformations are now commonin nonlinear finite element analysis.3.3 STRAIN MEASURESIn contrast to linear elasticity, many different measures of strain and strainrate are used in nonlinear continuum mechanics. Only two of these measures areconsidered here:1. the Green (Green-Lagrange) strain E2. the rate-of-deformation tensor D , also known as the velocity strain orrate-of-strain.In the following, these measures are defined and some key properties are given.Many other measures of strain and strain rate appear in the continuum mechanicsliterature; however, the above are the most widely used in finite element methods.It is sometimes advantageous to use other measures in describing constitutiveequations as discussed in Chapter 5, and these other strain measures will beintroduced as needed.A strain measure must vanish in any rigid body motion, and in particularin rigid body rotation.

If a strain measure fails to meet this requirement, thisstrain measure will predict the developnet of nonzero strains, and in turn nonzerostresses, in an initially unstressed body due to rigid body rotation. The key reasonwhy the usual linear strain displacement equations are abandoned in nonlineartheory is that they fail this test. This will be shown in Example 3.6. It will beshown in the following that E and D vanish in rigid body motion. A strainmeasure should satisfy other criteria, i.e.

Характеристики

Список файлов книги