3 Residual Force Equations NFEM Ch 3 Slide 1
Total Force Residual Equation Vector form r(u,λ) = 0 r = total force residual vector u = state vector with displacement DOF Λ = array of control parameters Indicial form r (u, Λ ) = 0 i j k NFEM Ch 3 Slide 2
Black Box Interpretation Control: Λ State u Residual evaluation Residual r NFEM Ch 3 Slide 3
Example Nonlinear FEM r 1 = 4 u 1 u 2 + u 2 u 3 6 1 = 0 r 2 = 6 u 2 u 1 + u 1 u 3 3 2 = 0 r 3 = 4 u 3 + u 1 u 2 3 2 = 0 Vector form: [ ] r1 (u, Λ) [ ] 4 u1 u 2 + u 2 u 3 6 1 r = r(u, Λ) = r 2 (u, Λ) r 3 (u, Λ) = 6 u 2 u 1 + u 1 u 3 3 2 4 u 3 + u 1 u 2 3 2 = 0 with u = [ u1 u 2 u 3 ] Λ = [ 1 2 ] NFEM Ch 3 Slide 4
Correlation with Linear FEM Master stiffness equations: K u = f Transforming to force residual: r = K u f = 0 Control parameters are not needed in linear FEM. NFEM Ch 3 Slide 5
Conservative System: Derivation From Energy total potential energy Π r = = 0 u Physical meaning: force equilibrium is associated with total potential energy Π being stationary (its gradient of r w.r.t. state vector u vanishes) NFEM Ch 3 Slide 6
Balanced Force Residual Form Split total residual as r = p f = 0 Move f to RHS p(u) = f(u,λ) internal force vector external force vector NFEM Ch 3 Slide 7
Balanced Force Residual Form: Black Box Control: Λ State u Ext force evaluation Int force evaluation f p Residual r NFEM Ch 3 Slide 8
Previous Example Nonlinear FEM r 1 = 4 u 1 u 2 + u 2 u 3 6 1 = 0 r 2 = 6 u 2 u 1 + u 1 u 3 3 2 = 0 r 3 = 4 u 3 + u 1 u 2 3 2 = 0 [ ] r1 (u, Λ) [ ] 4 u1 u 2 + u 2 u 3 6 1 r = r(u, Λ) = r 2 (u, Λ) r 3 (u, Λ) = 6 u 2 u 1 + u 1 u 3 3 2 4 u 3 + u 1 u 2 3 2 = 0 [ ] p1 (u) [ ] 4 u1 u 2 + u 2 u 3 [ ] f1 (Λ) [ ] 2 1 p = p 2 (u) p 3 (u) = 6 u 2 u 1 + u 1 u 3 4 u 3 + u 1 u 2, f = f 2 (Λ) f 3 (Λ) = 3 2 2 NFEM Ch 3 Slide 9
Conservative System: Derivation from Energy total potential energy Π = U W internal energy external work potential p = U u, f = W u, NFEM Ch 3 Slide 10
(Tangent) Stiffness and Control Matrices K = r u with entries K ij = r i u j Q = r Λ with entries Q ij = r i j An example next NFEM Ch 3 Slide 11
Residual Rate Forms Nonlinear FEM Define state and control in terms of pseudo time t : u = u(t) Λ = Λ(t) The first two derivatives of r with respect to t are, using indicial form ṙ i = r i u j u j + r i j j r i = r [ i 2 r i ü j + u j + u k + u j u k [ 2 r i u k + j u k 2 r ] i k u j + r i j u j k j 2 r i j k k ] j in which a superposed dot denotes derivative wrt t, as in real dynamics NFEM Ch 3 Slide 12
Residual Rate ODEs in Matrix Form Nonlinear FEM Recall that K = r u Q = r Λ Using these definitions the previous ODEs in indicial form can be rewritten in matrix form as ṙ = K u Q Λ first order rate form r = Kü + K u Q Λ Q Λ second order rate form Examples in next slide NFEM Ch 3 Slide 13
Residual Computation Example Nonlinear FEM A E,A constant B A C' P = λ EA B θ θ u C k = β EA L k L 2L L Only one DOF: vertical motion of midspan joint C Only one control parameter denoted by λ instead of Λ 1 NFEM Ch 3 Slide 14
Residual Computation Example (cnt'd) A C' P = λ EA B θ θ u k C' P = λ EA State parameter µ = u/l = tan θ Control parameter λ = P/EA F AC F s F BC NFEM Ch 3 Slide 15
Residual Computation Example (cnt'd) Nonlinear FEM r(µ, λ) = µ ( 2 + β ) 2 λ = 0 1 + µ 2 K = r µ = β + 2(1 + µ2 ) 3/2 1 (1 + µ 2 ) 3/2 NFEM Ch 3 Slide 16
Residual Computation Example (cnt'd) Load factor λ = P/EA 0.6 0.4 0.2 0 0.2 0.4 0.6 Response using engineering strain measure β = 1 β = 1/10 β = 1/100, 1/1000 (indistinguishable at plot scale) 1 0.5 0 0.5 1 Dimensionless displacement µ = u/l NFEM Ch 3 Slide 17
Residual Computation Example (cnt'd) Stiffness coefficient K = dr/dµ 2.5 2 1.5 1 0.5 0 Stiffness using engineering strain measure 1 0.5 0 0.5 1 Dimensionless displacement µ = u/l NFEM Ch 3 Slide 18
Staging Complicated nonlinear systems, such as structures, are analyzed in stages because the superposition principle no longer applies Staging reduces multiple control parameters to only one over each loading stage. This single parameter is called the staging parameter and will be usually denoted by λ NFEM Ch 3 Slide 19
Staging (cont'd) Nonlinear FEM Suppose that we have solved the residual equation for the control parameters Λ A corresponding to point A in the control space. We next want to advance the solution to Λ corresponding to point B Interpolate the control parameters linearly B Λ = (1 λ)λ A + λλ B where λ is the staging parameter that varies from 0 at A to 1 at B The residual equation to ber solve over the stage is with the I.C. u = u at λ = 0. A r (u,λ) = 0 This equation has only one control parameter. It is studied in Chapter 4. NFEM Ch 3 Slide 20
To Illustrate Staging We Will Look at a Suspension Bridge NFEM Ch 3 Slide 21