Nonlinear FEM. Critical Points. NFEM Ch 5 Slide 1

Transcription

1 5 Critical Points NFEM Ch 5 Slide

2 Assumptions for this Chapter System is conservative: total residual is the gradient of a total potential energy function r(u,λ) = (u,λ) u Consequence: the tangent stiffness matrix (Hessian of Π) K = r(u,λ) u 2 = (u,λ) u u is real symmetric NFEM Ch 5 Slide 2

3 Spectral Properties of Tangent Stiffness Matrix Nonlinear FEM The algebraic eigenproblem for K has the form Kz i = κ i z i, i =, 2,...N Since K is real symmetric, it enjoys two important spectral properties:. All eigenvalues of K are real. K has a complete set of real eigenvectors, which can be orthonormalized so that z T i z j = δ ij in which δ ij denotes the Kronecker delta NFEM Ch 5 Slide 3

4 Regular Versus Critical Regular point: K is nonsingular Critical point: K is singular (also called singular or nonregular points) Useful criterion for small systems with small # of DOF: The determinant of K vanishes at a critical point NFEM Ch 5 Slide 4

5 Why Are Critical Points Important? Along a static equilibrium path of a conservative system, the transition from stability to instability (or vice-versa) always occurs at a critical point Notes: () This does not mean that the transition will happen (2) Property does not extend to nonconservative systems Consequence: in designing against instability, the analyst is primarily interested in locating critical points NFEM Ch 5 Slide 5

6 Critical Point Classification as per Number of Zero Eigenvalues (= Rank Deficiency) Nonlinear FEM Isolated critical point: K has one zero eigenvalue (= rank deficiency of K is one) Multiple critical point: K has two or more zero eigenvalues (= rank deficiency of K is two or more) We will focus on the isolated type for now, since multiple critical points are more difficult to deal with NFEM Ch 5 Slide 6

7 Classification of Isolated Critical Points Nonlinear FEM Limit point: path continues with no branching, but tangent is normal to λ (control parameter) axis Bifurcation point: two or more paths cross and there is no unique tangent A limit point may be a maximum, minimum or inflexion point. If a maximum or minimum, its occurrence is also called snap-through or snap-buckling by structural engineers NFEM Ch 5 Slide 7

8 Limit or Bifurcation? Denote the values of state vector, staging control parameter, tangent stiffness matrix and incremental load vector at the critical point as u λ K q cr cr Denote the null eigenvector of K by z so K z = 0 T Compute the indicator z cr q cr (a scalar). Then (proven in the Notes) cr cr cr cr cr cr zcr T q cr 0 : limit point z T q cr cr = 0 : bifurcation point NFEM Ch 5 Slide 8

9 Response Example (2 DOF, pictured in 2D control-state space) λ L B λ L B L 2 L 2 B 2 u Limit point before bifurcation u Bifurcation before limit point NFEM Ch 5 Slide 9

10 Response Example (2 DOF, pictured in 3D control-state space) λ L B L 2 B 2 u 2 u Bifurcation before limit point NFEM Ch 5 Slide 0

11 A Simple Example (from Notes) Nonlinear FEM Find the critical points of the scalar residual equation 2 3 r = u 2 u + u λ = 0 To be worked out on whiteboard NFEM Ch 5 Slide

12 The Circle Game Example Nonlinear FEM L Control parameter λ T 2 B 2 S 2 R S B T Total residual: L State parameter µ r µ, λ = λ µ λ 2 + µ 2 = 0 (Artificial, not associated with a real structure) NFEM Ch 5 Slide 2

13 9 Special Points in Circle Game Example L Control parameter λ T 2 B 2 S 2 R S B T L State parameter µ One reference point R, atλ = µ = 0 Two bifurcation points B and B 2,atλ =±/ 2, µ =±/ 2 Two limit points L and L 2,atλ =±, µ = 0 Two turning points T and T 2,atλ = 0, µ =± Two non-equilibrium stationary points ("vortex points") S and S 2,atλ =±/ 6, µ = / 6 NFEM Ch 5 Slide 3

14 Incremental Flow for Circle Game Example Nonlinear FEM Control parameter λ State parameter µ NFEM Ch 5 Slide 4

15 Tangent Stiffness, Incremental Load, and Incremental Velocity for Circle Game Example For DOF these matrix and vector quantities reduce to scalars: K = r µ = λ2 + 2λµ 3µ 2 q = r λ = 3λ2 + 2λµ µ 2 v = q/k NFEM Ch 5 Slide 5

16 Stiffness Sign Regions for Circle Game Example Nonlinear FEM Control parameter λ K<0 Unstable T 2 B 2 L K >0 Stable R L State parameter µ B T K = 0 Neutral NFEM Ch 5 Slide 6

17 Incremental Load Sign Regions for Circle Game Example Nonlinear FEM Control parameter λ T 2 q< 0 L B R B 2 L 2 q > 0 q = State parameter µ T NFEM Ch 5 Slide 7

18 Incremental Velocity Sign Regions for Circle Game Example Nonlinear FEM Control parameter λ v=q/k=0/0 S T 2 B 2 L B R L 2 T S State parameter µ q=0 & v=0 K = 0 NFEM Ch 5 Slide 8

19 Propped Rigid Cantilevered (PRC) Column - Perfect Structure Nonlinear FEM ; C k ; A rigid L ;; B ;; C' ;; u A= L sin θ A θ spring stays A' horizontal as column tilts L L cos θ ;; B P = λ kl NFEM Ch 5 Slide 9

20 Perfect PRC Column: Response Control parameter λ Unstable 0.4 Unstable 0.2 T 2 R T State parameter µ B Unstable Stable λ cr Stiffness coefficient K State parameter µ B Control versus state parameter response for perfect column Tangent stiffness versus state parameter response for perfect column NFEM Ch 5 Slide 20

21 Propped Rigid Cantilevered (PRC) Column - Imperfect Structure Nonlinear FEM ; k C ; Initial Imperfection εl A θ 0 A 0 L ; C' ; spring stays horizontal as column tilts u A= L sin θ A θ 0 θ A 0 L P = λ kl A' L cos θ rigid B B NFEM Ch 5 Slide 2

22 Imperfect PRC Column: Response Control parameter λ State parameter µ State parameter µ Stiffness coefficient K Control versus state parameter response for varying imperfection Tangent stiffness versus state parameter response for varying imperfection NFEM Ch 5 Slide 22

23 Imperfect PRC Column: Response (cont'd) Control parameter λ λ(µ)=( µ 2 3/2 ) Unstable Stable Unstable State parameter µ Imperfection parameter ε Critical load parameter λ λ(ε)=( ε 2/3 ) 3/2 Critical point locus as stableunstable region separator Imperfection sensitivity diagram NFEM Ch 5 Slide 23