Belytschko T. - Introduction (779635), страница 12

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

Текст из файла (страница 12)

n The internal nodal forces (2.5.14) are then givenfei nt =∫ B0 PdΩ 0TΩ e0orf I,inte = ∫ B0 I PdΩ0(2.5.16)Ω e0where we have used dΩ 0 = A0 dX and Ω e0 is the initial domain of the element. In thisnotation the deformation gradient F and the one-dimensional strain are given by2-26T. Belytschko, Chapter 2, December 16, 1998ε = B0ue(2.5.17)Box 2.2. Discrete Equations in Total Lagrangian Formulationu( X,t ) = N( X )ue (t ) = Σ N I( X )u Ie (t )I(B2.2.1)in each elementε =∑I∂NI eu = B0 ue∂X I(B2.2.2)evaluate the nominal stress P by constitutive equation∂Nfei nt =∫ ∂X PdΩ0 = ∫ B0 PdΩ 0Ω e0TorΩ e0feIint =∂N IPdΩ0∂XΩe∫(B2.2.3)0feext = ∫Ω e ρ0N TbdΩ 0 +(N T A0 t x ) Γ te(B2.2.4)ρ 0 NT NdΩ0(B2.2.5)00Me = ∫Ω0e˙˙ + f int = f extMu(B2.2.6)Example 2.5.1.Two-Node, Linear Displacement Element.

Consider a twonode element shown in Fig. 3. The element shown is initially of length l 0 and constantcross-sectional area A 0 . At any subsequent time t , the length is l( t ) and the crosssectional area is A(t); the dependence l and A on time t will not be explicitly notedhenceforth. The cross-sectional area of the element is taken to be constant, i.e. independentof X.2-27T.

Belytschko, Chapter 2, December 16, 1998A , A o = constanttALu1Aou212X,xLoFig. 2.3. Two node element in one dimension for total Lagrangian formulation showing the initial,undeformed (reference) configuration and the deformed (current) configuration.Displacement field, strain, and B 0 matrix. The displacement field is given by the linearLagrange interpolant expressed in terms of the material coordinateu( X, t ) =1[X − X ,l0 2u1 (t ) X − X1 ]u2 (t )(2.5.18)where l 0 = X2 – X1 . The strain measure is evaluated in terms of the nodal displacementsby using Eq.

(2.5.18) with (B2.2.2):ε( X, t ) = u, X = l1 [–10u1( t) +1] u2 (t )(2.5.19)The above defines the B 0 matrix to beB0 = l1 [–10+ 1](2.5.20)Nodal Internal Forces. The internal nodal forces are then given by (2.5.16):X2T1  −1PA dXf inte = ∫Ω e B0 PdΩ X = ∫X l 0010  +1(2.5.21a)If we assume that the cross-sectional area and the nominal stress P is constant, theintegrand in (2.5.21a) is then constant, so the integral can be evaluated by taking theproduct of the integrand and the initial length of the element l 0 , which givesf inteint f1 –1=   = A0 P   f 2 e+1(2.5.21b)From the above, we can see that the nodal internal forces are equal and opposite, so theelement internal nodal forces are in equilibrium, even in a dynamic problem.

Thischaracteristic of element nodal forces will apply to all elements for which translation results2-28T. Belytschko, Chapter 2, December 16, 1998in no deformation; it does not apply to axisymmetric elements. Since P = T/A 0, (seeEq. (2.1.1)) the nodal forces are equal to the load T carried by the element.Nodal External Forces. The external nodal forces arising from the body force are givenby (B2.2.3)X 2 ρ  X2 – X feext = ∫Ω e ρ 0NT bA0 dX = ∫ X l 0  bA0 dX010 X – X 1(2.5.22a)If we approximate the body forces b(X,t) by a linear Lagrange interpolantX –XX–Xb( X,t ) = b1 (t )  2l  + b2 ( t )  l 1 00(2.5.22b)and taking A0 to be constant, the evaluation of the integral in (2.5.22a) givesf exte =ρ 0 A0 l 0 2b1 + b2 6 b1 + 2b2 (2.5.22c)The evaluation of the external nodal forces is facilitated by expressing the integral in termsof a parent element coordinateξ = ( X – X1 ) / l0 , ξ =∈[ 0,1](2.5.23)Element Mass Matrix.

The element mass matrix is given by (B2.2.5):1M e = ∫ e ρ0 N TNdΩ0 = ∫ ρ0 NT NA0l 0dξΩ01−ξ = ∫ ρ 0[ 1−ξ0 ξ 10ρ A l  2 1ξ ] A0 l 0 dξ = 0 6 0 0  1 2(2.5.24a)It can be seen from the above that the mass matrix is independent of time, since it dependsonly on the initial density, cross-sectional area and length.The diagonal mass matrix as obtained by the row-sum technique (2.4.26) isMe =ρ 0 A0l 0  1 0  ρ0 A0 l0= 2 I20 1 (2.5.24b)As can be seen from the above, in the diagonal mass matrix for this element, half of themass of the element is ascribed to each of the nodes. For this reason, it is often called thelumped mass matrix.Example 2.5.2.Example of Assembled Equations.

Consider a mesh of twoelements as shown in Fig. 4. The body force b(x) is constant, b. We will develop thegoverning equations for this mesh; the equation for the center node is of particular interestsince it represents the typical equation for the interior node of any one-dimensional mesh.2-29T. Belytschko, Chapter 2, December 16, 19981122L (1)3L (2)Fig. 4The connectivity matrices Le for this mesh are1 0 0 L (1) =  0 1 0 (2.5.25a)0 1 0 L (2) =  0 0 1 (2.5.25b)The global internal force matrix by Eq. (2.5.5) is given in terms of the element internalforces byint 0  f 1 =  f 2  +  f1  0 f  2 (2)(1)intTintf int = LT(1) f int(1) + L (2 )f (2)(2.5.26)which from (2.5.21b) givesfint–1 0 (2 )= A0 P(1 ) +1 + A0 P( 2 )  –1 0 +1( 1)(2.5.27)Similarlyext 0  f 1 =  f 2  +  f1  0 f  2 (2)(1)extextf ext = LT(1) f ext(1) + L(2) f (2)(2.5.28)and using (2.5.22c) with constant body force givesfextbρ0(1) A0(1) l (1)ρ (2) A(2) l (2)00=b + 0 02200bb(2.5.29)The global, assembled mass matrix is given by (2.5.9)M = LT(1) M(1) L(1) + LT(2) M (2) L(2)(2.5.30)2-30T.

Belytschko, Chapter 2, December 16, 1998and by (2.5.24a)M = LT(1)(1) (1)ρ(1)0 A 0 l0621(2) (2) (2)1T ρ0 A0 l0L+L(2)62 (1)2 1  1 2  L(2)(2.5.31)To simplify the form of the assembled equations, we now consider a uniform mesh with(2)(1)(2)(1)(2)constant initial properties, so ρ (1)0 = ρ 0 = ρ 0 , A0 = A0 = A0 , l 0 = l 0 = l 0 and we()(2) (2) (2)(1)define m1 = ( ρ0(1) A(1)0 l 0 ) / 6 , m2 = ρ 0 A0 l 0 / 6 so the assembled mass matrix is2m1m10 M =  m1 2(m1 + m2 ) m2  0m22m2 (2.5.32)Writing out the second equation of motion for this system (which is obtained from thesecond row of M, f ext and f int ) gives16()(1) (1) ˙˙(1) (1) (1)(2) (2) (2) ˙ ˙1˙˙ρ (1)u2 + 16 ρ0(2) A0(2) l (2)0 A0 l0 u1 + 3 ρ 0 A0 l0 + ρ 0 A0 l 00 u3–A P +A P(1)(1)(2)(2)=b2(ρ(1)0A l +ρ A l(1) (1)00(2)0(2) (2)00)(2.5.33)Using uniform properties as before and dividing by A0 l0 , we obtain the following equationof motion at node 2:(2)(1)ρ 0 ( 16 u˙˙1 + 23 u˙˙2 + 16 ˙u˙3 ) + P l– P = ρ 0 b0(2.5.34)If the mass matrix is lumped, the corresponding expression is(2)(1)ρ 0u˙˙2 + P l– P = ρ 0 b0(2.5.35)The above equation is equivalent to a finite difference expression for the momentumequation (2.2.4) with A 0 constant: it is only necessary to use the central differenceexpression P, X ( X2 ) = P (2 ) – P( 1) / l0 to reveal the identity.

Thus the finite elementprocedure appears to be a circuitous way of obtaining what follows simply and directlyfrom a finite difference approximation. The advantage of a finite element approach is that itgives a consistent procedure for obtaining semidiscrete equations when the element lengths,cross-sectional area, and density vary. Furthermore, for linear problems, a finite elementsolution can be shown to provide the best approximation in the sense that the error isminimized in the energy norm (see Strang and Fix); finite difference approximations forirregular grids and varying areas and densities, on the other hand, are difficult to construct.The finite element method also gives the means of obtaining consistent mass matrices andhigher order elements, which are more accurate. But the main advantage of finite elementmethods, which undoubtedly has been the driving force behind its popularity, is the easewith which it can model complex geometries.

This of course is masked in one dimensionalproblems, but it will become apparent when we study multi-dimensional problems.()2-31T. Belytschko, Chapter 2, December 16, 1998Example 2.5.3. Three-node quadratic displacement element. A 3-node elementof length L0 and cross-sectional area A0 is shown in Fig. 4. Node 2 is placed betweennodes 1 and 3; although in this analysis we do not assume it to be midway between thenodes, it is recommended that it be placed midway between the nodes in most models.

Themapping between the material coordinates X and the referential coordinate ξ is given by1X( ξ) = N( ξ)Xe =  ξ(ξ −1) 1– ξ 22 X1  ξ (ξ +1)  X2  2 X3 1(2.5.36)where N(ξ) is the matrix of Lagrange interpolants, or shape functions, and ξ is the elementcoordinate. The displacement field is given by the same interpolantsu(ξ , t ) = N(ξ )ue (t ) =[12ξ( ξ –1)1– ξ21ξ2u1 (t )(ξ + 1) u2 (t ) u (t ) 3 ](2.5.37)By the chain ruleε = F –1= u, X = u,ξξ, X = u,ξ X,ξ−1 =1[ 2ξ –1 – 4ξ2 X,ξ2ξ + 1] ue(2.5.38)We have used the fact that in one dimension, ξ ,x = X,−1ξ . We can write the above asε = B0ue where B0 =1[2ξ–1 – 4ξ2X,ξ2ξ + 1] ue(2.5.39)The internal nodal forces are given by Eq.

Характеристики

Список файлов книги