Belytschko T. - Introduction (779635), страница 68

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

Текст из файла (страница 68)

Belytschko & B. Moran, Solution Methods, December 16, 1998Thus the problem with one unknown is augmented by a second equation, whichleads to two nonlinear algeraic equations in two unknowns. The load need notincrease in the step, and may in fact decrease.

It is only necessary for thearclength parameter to increase, which is a perfectly natural way of tracing thebranch.To describe the method in a more general case, we consider the problem with nγload parameters γ a . For each load parameter, a parametrization equation must beadded:( )pa d,γ b = ∆sa2or( )p d,γ = ∆s(6.5.8)where[p = p1 ...]pnγ ,∆s = ∆s12... ∆s 2  , γ = γ 1nγ ... γnγThe resulting augmented equations for the equilibrium path are then( ) =  0 ( ) ∆s f d, γ p d,γ(6.5.9)Thus for a system with nDOF degrees of freedom, we obtain an augmented systemof nDOF + nγ equations in the same number of unknowns.The resulting equations can be solved by the standard Newton methods we havedescribed.

The linearized equations for the Newton method are given by ∂f ∂d ∂f ∂γ ∆d  0   =  ∂p ∂d ∂p ∂γ ∆γ  ∆s intext∂p ∂γ  ∆d  0 K − γ a Ka = ∂p ∂γ  ∆γ  ∆s ∂p ∂d(6.5.10)where the Jacobians of the nodal forces have been expressed in terms of theinternal tangent stiffness and the load stiffness on the LHS. A subscript has beenadded to the load stiffness because the Jacobian for each group of external loadsmust be considered separately.

At times the internal tangent stiffness must also besubdivided into terms associated with different parameters, as when thetemperature changes and causes buckling.6-57T. Belytschko & B. Moran, Solution Methods, December 16, 1998The parametrization equations need not be arclength equations: anyparametrization which leads to a regular set of Newton equations is a candidate.A major difficulty in all branch continuation techniques is setting theincrement size and in the scaling of the parametrization equations. If theincrement size is too small, considerable effort is wasted in determiningunnecessary equilibrium points.

On the other hand, if the increment size is toolarge, the convergence of the Newton procedure can fail or too many iterations areneeded. The selection of an appropriate stepsize can be aided by an estimate ofthe location of the next turning point or bifurcation point.

The step size can thenbe set so that a reasonable fraction of that distance is covered in the nextincrement. It is stressed however that bifurcation paths can appear out ofnowhere, so if a good knowledge of the branches is essential, branchcontinuations should be repeated with different stepsizes.Scaling of Arclength Equation. The arclength equation, when posed interms of force parameters and displacement increments is often poorly scaled.We summarize in the following some of the scaling methods which have beenproposed that appear to be effective in structural mechanics problems.The simplest method is to introduce a scaling factor between theincrements in load and the increments in displacements.

The parametrization thenis()p(d, γ ) =∆ γ 2 ∆f0T ∆f0 +α∆d T ∆d =∆ s2(6.5.11)where α is a scaling factor. A candidate for a scaling factor is the square of theaverage diagonal of the initial stiffness matrix.Bifurcations. We consider first equilibrium bifurcations, i.e. we ignore Hopfbifurcations. The bifurcation then consists of the intersection of two equilibriumbranches. If we are tracing a given equilibrium branch, such as AB in Fig. ?, thenit is very easy to miss the intersecting branch and end up on an unstable branch.The objective of this Section is to describe some methods for detecting thecrossing of a bifurcation point and anticipating when a bifurcation point will comeup along thhe branch.The classical method for detecting bifurcations in structural mechanics isan eigenvalue analysis. In an eigenvalue approach, we exploit the fact that thelinearized equations for the increment, Eqs.

() are singular at the bifurcation point.LinearStability. In Example 1, we have employed a technique which isfrequently used to examine the stability of an equilibrium solution: a dynamicsolution to a perturbation of the equilibrium solution. The dynamic equations arelinear because the perturbations are small, so this is called a linearized model. Ifthe dynamic solution grows, then it is said that the equilibrium solution is linearunstable. Otherwise, it is linear stable.

In the following, we develop a generalprocedure for examining the linear stability of an equilibrium solution byexamining the characteristics of the Jacobian matrix.6-58T. Belytschko & B. Moran, Solution Methods, December 16, 1998We consider an equilibrium point deq associated with a parametrized load, λf extof a rate-independent system. A Taylor series expansion of the residual about theequilibrium solution gives() ( )f deq + d = f deq +∂f eqd d + higher order terms∂d( )(6.5.12)The first term on the RHS vanishes because deq is an equilibrium solution. FromEq. () we can see that the second term can be linearized as follows:∂f eqd = K ext d eq − Kint d eq = −A d eq∂d( )( )( )( )(6.5.13)We now add the inertial forces to the system.

Since the mass matrix does notchange with displacements, we can then write the equations of motion for a smallperturbations about the equilibrium point asMd 2d+ Ad = 0dt 2(6.5.14)Note, that in contrast to Section ??, we do not include the mass matrix in theJacobian matrix A . The above is a set of linear ordinary differential equations ind . Since the solutions to such linear ordinary differential equations areexponential, we take solutions of the formd = yeµtdi = yae µt(6.5.15)Substituting the above into Eq.

(6.5.14) gives(A + µ 2M)ye µt = 0(6.5.16)The characteristic values µi of this system can be obtained from the eigenvalueproblemAyi = −λi Myi , λ i = µ i2(6.5.17)where λi , i =1 to n are the n eigenvalues and yi are the n eigenvectors. Thelinear stability of the system then revolves around the character and magnitudes ofthe eigenvalues µi .

The eigenvalues will generally be complex. If the real part ofthe eigenvalue is positive the solution will grow, i.e. if( )if for any i, Real µi > 0, the equilibrium point is linearly unstable(6.5.18)6-59T. Belytschko & B. Moran, Solution Methods, December 16, 1998Here µi is the complex conjugate of µi . On the other hand, if the real parts of alleigenvalues are negative, then the linearized solutions about the equilibrium pointdo not grow and we can say that( )if for all i, Real µi ≤ 0, equilibrium point is linearly stable(6.5.19)When the linearized equations are symmetric, then the eigenvalues mustbe real. We can then see that if the matrix A is positive definite, then theeigenvalues must be negative, and consequently the parameters µi are strictlycomplex.

Therefore, when µ are complexthe perturbated solutions are harmonicand of the same magnitude as the perturbation and the equilibrium points is linearstable.This result has important ramifications for many structural systems. If thesystem has a potential, i.e. if the system is conservative, then the matrix A issymmetric and corresponds to the Hessian of the potential energy, i.e.,Aab = ∂ 2 W ∂da∂db by Eq. ???. Recall that an equilibrium solution is a stationarypoint of the potential. Since A is the matrix of second derivatives, the positivedefiniteness of A implies that the stationary point is a local minimum.

Thus anequilibrium point is linear stable if and only if the potential at the equilibriumpoint is a local minimum, which implies that the Jacobian and Hessian matricesare positive definite. In other words, if∆da∂ 2 W  deq ∂d a∂db ∆db = Aab d eq  ∆da ∆db = ∆dT A∆d > 0∀ ∆d (6.5.20)then the equilibrium point deq must be linear stable. On the other hand, if thereexists a ∆d for which the above inequality is violated, then the stationary pointmust be a saddle point, and the equilibrium solution is not linear stable.To summarize, an equilibrium solution for a conservative system is linearstable if it corresponds to a local minimum of the potential energy, which requiresthe positive definiteness of the Hessian or Jacobian matrices (they are the same inthat case).

If the equilibrium solution is a saddle point, then the equilibriumsolution is unstable.For nonconservative systems, an equilibrium solution is also linear stableif the Jacobian matrix is symmetric and positive definite. If the Jacobian is notpositive definite, it is not linear stable. Any system is linear stable if all real partsof the eigenvalues of the system (6.5.17) are negative.The information provided by a linear stability analysis is not conclusivefrom an engineering viewpoint.

Since linear stability analysis assumes thelinearity of the response in the vicinity of the equilibrium point, perturbationsmust be small enough so that the response can be predicted by a linear model.Linear stability of an equilibrium point does not preclude the possibility that aphysically realistic perturbation will grow. If the system is highly nonlinear in the6-60T.

Belytschko & B. Moran, Solution Methods, December 16, 1998neighborhood of the equilibrium point, moderate perturbations of the system maylead to unstable growth. A linear stability analysis only reveals how a systemwith properties obtained by a linearization of the system about the equilibriumpoint behaves. Nevertheless, it gives information which is useful in engineeringand scientific analysis of systems.Estimates of Bifurcation Points. It is often desirable to determine thelocation of bifurcation points as the equilibrium path is generated.

Bothbifurcation points which may have been passed or which are upcoming are ofinterest. Whether a bifuraction point has been passed can be determined bychecking when the determinant of the Jacobian changes sign. A change of sign inthe determinant of the Jacobian is an indication of the change of sign of aneigenvalue. The determinant of the Jacobian vanishes at a critical point and willoften change sign when the critical point is passed.

Характеристики

Список файлов книги