Belytschko T. - Introduction (779635), страница 69
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It would appear at first thatthe sign of the Jacobiann determinant would always change when passing abifurcation but things are not that simple: sometimes, two eigenvalues changesign at a bifurcation point and then the Jacobian determinant does not changesign, so the determinant test is not conclusive. Thus tracking the determinantprovides some guidance in finding bifurcation points, but it is not foolproof.Bifurcation points can also be estimated by tracking the eigenvalues of thesystem. The estimation of eigenvalues is simplified in solid mechanics becausethe stress appears linearly in the geometric stiffness and varies approximately inproportion with the load.
As we have seen from Example 6.?, the stability of anequilibrium path chamges when the lowest eigenvalue of the system changes sign.Thus the critical points can be located by an eigenvalue problem. There areseveral ways to do this:1. interpolate the Jacobian matrix of the system by a linear relationship.2. assume that the geometric and load stiffness are linearly proportional tothe load parameter in the neighborhood under consideration;Both methods can be applied with only a single load parameter. In the firstmethod, we assume that the Jacobian A, is a linear function of the load parameterγ .
The Jacobian matrix in the vicinity of the state n can then be written in termsof the states around n-1 and n byA(d,γ ) = ( 1−ξ)A(d n −1 ,γ n −1 ) + ξA(dn ,γ n ) ≡ (1− ξ )A− + ξA0( )γ = 1− ξ γ n −1 + ξγ n(6.5.21)(6.5.21b)where the last term in the above defines a more compact notation we will use inthe following. At the critical point, the determinant of the Jacobian matrix Avanishes, so()det A d, γ crit = 0(6.5.22)From (6.5.21) and the fact that a ystem with a zero determinant has a nontrivialhomogenrous soloution, we deduce that there exists a ξ such that6-61T.
Belytschko & B. Moran, Solution Methods, December 16, 1998(1−ξ )A− y +ξA 0y = 0(6.5.23)This can be put in the standard form of the generalized eigenvalue problem by thefollowing rearrangement of terms:()A0 y = ξ A 0 − A− y(6.5.24)The solution of this eigenvalue problem then gives an estimate of the critical loadby (6.5.21b).The lowest eigenvalues of (6.5.24) can be either positive or negative.
Negativeeigenvalues indicate critical points which have been passed and are known about,or they may indicate critical values which have inadvertently been passed. In thelatter case, state n may no longer be on a stable equilibrium path.For many structural problems, the eigenvalue problem may be simplified bytaking advantage of the following:1. the material stiffness in a linear material will not change significantly if thedisplacements prior to the critical point are small;2.
the geometric stiffness depends linearly on the load parameter, since it dependsalmost linearly on the stresses if the displacements are small (see the geometricstiffness for the bending and axial response in Eqs. ());3. the load is independent of the displacements, so the load stiffness vanishes.Since the geometric stiffness varies linearly with the load, if the above threeconditions are met we can then write( )( )A0 = K mat + K geo λ0 , A− = Kmat +K geo λ−(6.5.25)where K geo is the geometric stiffness for a unit value of the load parameter.Substituting into () then gives(( )( ))K mat y = ξ Kgeo λ0 − Kgeo − λ− y(6.5.26)The critical load is then given by(λcrit = ξ λ0 − λ−)(6.5.27)The procedure of determining the location of a nearly critical point then consistsof storing the following and they geometric stiffness is saved from the last step,and using the current geometric and material stiffness, the eigenvalues areobtained.
The eigenvalue which leads to the smallest critical load is the one ofinterest. When the parameter 0 ≤ξ ≤ 1, the critical point has been passed in thelast step. When ξ >1, the critical point is estimated to be further ahead in thebranch.6-62T. Belytschko & B. Moran, Solution Methods, December 16, 1998In analyzing structures, it is often desirable to estimate the first bifurcationpoint along the fundamental equilibrium path after a single load increment λ1 hasbeen applied.
An initial estimate of the bifurcation point can be found in terms ofthe geometric stiffness computed after one load increment. We assume thatK geo λ− = 0 since the first step is stress-free. Then Eq () gives( )( )K mat y = ξKgeo λ0 y(6.5.28)The critical value of the load parameter is then λcrit = ξλ0 , where λ0 is the loadparameter for the first load increment. this is the formula commonly cited inmatrix structural texts for obtaining the buckling load of a structure. Note that itassumes that the geometry of the structure changes so little with increasing loadthat the first estimate of the geometric stiffness suffices for extrapolating thecritical point.
It is effective primarily for bifurcation points. Prior to reaching alimit point, the geometric stiffness changes significantly, so an estimate based on() is quite erroneous.The study of systems stability has in the past two decades become a richscience known as dynamical systems theory. It includes topics such as chaos,fractals, attractors, repellors. These topics are beyond the scope of this book;some good references are Argyris and Melenk (), Moon () and Temam ().Example 6.4. A simple example of a problem with stable and unstable pathsconnected by a turning point is the shallow truss shown in Fig.
6.11. The initialcross sectional areas of the elements are A0 and the initial lengths of the twoelements are l 0 , which is given by l 20 = a 2 + b2 . A vertical load f is applied asshown, and since this is the only load we consider f to be the load parameter.
Thematerial is governed by a Kirchhoff law, (see Eq. ())S = CE x(E6.1.1)where C; as pointed out in Section ??, for a small strain, large-displacementproblem such as this, this constitutive equation is almost to a small-strain elastic,linear law. We will determine the equilibrium path as a function of the load anddetermine which branches are stable.The deformation of the truss in is described by the variable y, the currentvertical coordinate of the centerpoint, which leads to simpler equations than usingthe displacement.
Since this material is path-independent, we can use the theoremof minimum potential energy to develop the discrete equations. The potentialenergy, Eq.(), in this case is given byintW=W−Wext,Wint=212∑ ∫Ωe =1e0ˆ 2 dΩ, W ext = f ( b − y)CExx(E6.1.2)where the Green strain is uniaxial with a only component along the axes of thebars contributing to the internal energy. The Green strain in for both elements ismost easily evaluated by Eq.(), which gives6-63T. Belytschko & B.
Moran, Solution Methods, December 16, 1998() () 12 ( y 2 − b 2)ˆ = 1 l 2 − l2 = 1 a2 + y 2 − a 2 − b2 =Exx022(E6.1.3)so the internal energy is given by(W int = k y 2 − b2)2where k = 14 CA0l0(E6.1.4)Combining the above with the potential of the external forces gives the totalpotential(W = k y2 − b2)2− f ( b − y)(E6.1.5)The equilibrium equation is now obtained by applying the theorem of minimumpotential energy, which states that the equilibrium equation is given by thestationary points of the potential W given above, so the equilibrium equation is()220 = dWdy = 4k y − b y + f(E6.1.6)As can be seen from the above, the force is a cubic function of the verticalposition of the centerpoint, which is shown in Fig.
?. The equilibrium path hastwo turning points, usually called limit points in structural mechanics, and threebranches, denoted by AB, BC and CD in Fig. ?.We will now examine the stability of the branches of the equilibrium path. Thedynamic response is examined at a position y0 subject to a perturbation. Asolution to the linearized equations is considered, soy = y 0 + y where y = εeµt(E6.1.7)where ε is a small parameter. The equations of motion for this problem are givenby()d2yM 2 = f ext − f int = f0 −4k y2 − b2 ydt(E6.1.8)where M is the mass of the node.
Substituting () into () gives()f0 − 4k ( y0 + y ) − b2 ( y0 + y ) = M2d 2ydt2(E6.1.9)Expanding the above and dropping all terms which are higher order than linear iny , gives (it is expressed in terms of[ () ()]f0 − 4k y0 y02 − b 2 + y 3y02 − b2 = Md 2ydt2(E6.1.10)The load cancels the first term in the brackets, so the equations of motion become6-64T.
Belytschko & B. Moran, Solution Methods, December 16, 1998M()(d 2y22222 + 4ky 3y0 − b = 0 → α = ±i 3y0 − bdt)12(E6.1.11)where the (b)follows from substituting () into (a). It can then be seen that theperturbation solution () is real with one positive exponential whenever3y 20 −b 2 < 0. So the branch defined by−b3 < y0 < b3 isunstable(E6.1.12)The results of the above stability analysis can be obtained directly be examiningthe second derivative of the potential energy function, which from () is given by()d 2Wd 2Wd2W22=4k3y−b,<0when−b<3y<b,> 0 otherwisedy2dy 2dy 2(E6.1.13)So in the unstable equilibrium branches of a conservative system, the secondderivative of the potential energy changes sign.
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