5 Critical Points NFEM Ch 5 Slide
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
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
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
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
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
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
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
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
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
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
The Circle Game Example Nonlinear FEM L Control parameter λ 0.5 0 0.5 T 2 B 2 S 2 R S B T Total residual: L 2 0.5 0 0.5 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
9 Special Points in Circle Game Example L Control parameter λ 0.5 0 0.5 T 2 B 2 S 2 R S B T L 2 0.5 0 0.5 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
Incremental Flow for Circle Game Example Nonlinear FEM Control parameter λ 0.5 0 0.5 2.00.00 0 0.50 0.50 0.30 0.5 0.30 0.30 0.5 0.50 0.5 0.08 0.04 0 0.04 0.08 0.5 0.30 0.50.00 2.00 0.5 0 0.5 State parameter µ NFEM Ch 5 Slide 4
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
Stiffness Sign Regions for Circle Game Example Nonlinear FEM Control parameter λ 0.5 0 0.5 K<0 Unstable T 2 B 2 L K >0 Stable R L 2 0.5 0 0.5 State parameter µ B T K = 0 Neutral NFEM Ch 5 Slide 6
Incremental Load Sign Regions for Circle Game Example Nonlinear FEM Control parameter λ 0.5 0 0.5 T 2 q< 0 L B R B 2 L 2 q > 0 q = 0 0.5 0 0.5 State parameter µ T NFEM Ch 5 Slide 7
Incremental Velocity Sign Regions for Circle Game Example Nonlinear FEM Control parameter λ 0.5 0 0.5 v=q/k=0/0 S T 2 B 2 L B R L 2 T S 2 0.5 0 0.5 State parameter µ q=0 & v=0 K = 0 NFEM Ch 5 Slide 8
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
Perfect PRC Column: Response Control parameter λ.4.2 0.8 0.6 Unstable 0.4 Unstable 0.2 T 2 R T 0.5 0 0.5 State parameter µ B Unstable Stable λ cr Stiffness coefficient K 0.75 0.5 0.25 0 0.25 0.5 0.75 0.5 0 0.5 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
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
Imperfect PRC Column: Response Control parameter λ.4.0 0.75.2 0.5 0.5.0 0.2 0.05 0.05 0. 0.0 0.0 0. 0.2 0.5 0.5 0.5 0.02 0.02 0.2 0.2 0.05 0.05 0.00.0 0.25 0. 0. 0.02 0.02 0.8 0.02 0.02 0.0 0.0 0 0.05 0.05 0.0 0.0 0.6 0. 0. 0.02 0.02 0.25 0.05 0.05 0.2 0.4 0.2 0.5 0. 0. 0.2 0.5 0.5 0.75 0.2 0.2 0 0.5 0 0.5 0.5 0 0.5 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
Imperfect PRC Column: Response (cont'd) Control parameter λ.4.2 0.8 0.6 0.4 0.2 λ(µ)=( µ 2 3/2 ) Unstable Stable Unstable 0 0 0.5 0 0.5 0.5 0 0.5 State parameter µ Imperfection parameter ε Critical load parameter λ.4.2 0.8 0.6 0.4 0.2 λ(ε)=( ε 2/3 ) 3/2 Critical point locus as stableunstable region separator Imperfection sensitivity diagram NFEM Ch 5 Slide 23