Lecture in Nonlinear FEM on. the Building- and Civil Engineering sectors 8.th. semester for

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1 Lecture in Nonlinear FEM on the Building- and Civil Engineering sectors 8.th. semester for the Building- and Civil Engineering, B8k, and Mechanical Engineering, B8m AALBORG UNIVERSITY ESBJERG, DENMARK ***************** Theme: Design of marine constructions. 1

2 Outline: Updated: 16. februar Introduction Notes 2. Geometrical nonlinearity - strain measures Cook 17.1, Geometrical nonlinearity - appl. in buckling analysis Cook Stress stiffness Cook Buckling Cook Material nonlinearity - introduction Cook Material nonlinearity - solution methods Cook 17.6, Contact nonlinearity Cook Nonlinear dynamic problems Cook Nonlinear dynamic problems Cook Literature: Noter A. Kristensen: Cook Cook, R. D. 2002: Concepts and applications of finite element analysis. John Wiley & Sons 2

3 3. Geometrical nonlinearity - appl. in buckling analysis Programme: Last time 4 Solution of nonlinear equations 5 The Newton-Raphson procedure 6 The Modified Newton-Raphson procedure 8 The BFGS procedure 10 Example 1: Torsional spring on a clamped beam 12 Example 2: Torsional spring on a clamped column 15 Example 3: Simple rod 17 Examples Assignments 3

4 Last time The nonlinear equations of equilibrium in the residual formulation is given by: The load is applied in load-steps n = 1,2,... The tangent-stiffnessmatrix [K T ] is given as r(d, f ) = p(d) f = 0 (1) [K T ] = [K 0 ] + [K L ({d})] + [K σ ({σ})] (2) 4

5 Solution of nonlinear equations Geometric nonlinearity behaviour relates to that during loading changes in geometry (deformations, displacements) will have an effect on the loaddisplacement response. Geometric nonlinear FEA can be categorized into three groups: 1. large displacements/large rotations/large strains 2. large displacements/large rotations/small strains 3. large displacements/small rotations/small strains The Newton-Raphson solution procedure is described generally (1D), but there exist various solution schemes, which improve the classical Newton- Raphson procedure, e.g. the procedure outlined by Broyden-Fletcher-Goldfarb- Shanno (BFGS). 5

6 The Newton-Raphson procedure The Newton-Raphson procedure for a typical increment: load-step n = 1,2,... f n = f n 1 + f n d0 n = dn 1 iterations i = 0,1,... compute ri n = p(di n) f n stop iterations when ri n < ε stop f n compute K T (di n)δdn i (factorization with time) (forward-backward substitu- solve the equation of equilibrium δdi n = K T (di n) 1 ri n tion) update the displacement d n i+1 = dn i + δdn i for number of load-step where ε stop = 0,01 0,001. 6

7 The Newton-Raphson procedure NAFEMS p. 248: Figure

8 The Modified Newton-Raphson procedure The Modified Newton-Raphson procedure for a typical increment: load-step n = 1,2,... f n = f n 1 + f n d0 n = dn 1 compute K T (di n)δdn i (factorization with time) iterations i = 0,1,... compute ri n = p(di n) f n stop iterations when ri n < ε stop f n (forward-backward substitu- solve the equation of equilibrium δdi n = K T (di n) 1 ri n tion) update the displacement d n i+1 = dn i + δdn i for number of load-step where ε stop = 0,01 0,001. 8

9 The Modified Newton-Raphson procedure NAFEMS p. 248: Figure

10 The BFGS procedure The BFGS procedure is a Quasi-Newton procedure, which is based on secant methods, where K T (d n i ) 1 is updated. NAFEMS p. 252: Figure

11 The BFGS procedure The BFGS Quasi-Newton procedure for a typical increment: load-step n = 1,2,... f n = f n 1 + f n d0 n = dn 1 compute K T (di n)δdn i (factorization with time) iterations i = 0,1,... compute ri n = p(di n) f n stop iterations when ri n < ε stop f n if i > 0, update K T (di n) 1 by BFGS inverse update scheme (forward-backward substitu- solve the equation of equilibrium δdi n = K T (di n) 1 ri n tion) update the displacement d n i+1 = dn i + δdn i for number of load-step where ε stop = 0,01 0,

12 Example 1: Torsional spring on a clamped beam Example on large displacements/large rotations/small strains NAFEMS p. 17: Figure

13 Example 1: Torsional spring on a clamped beam Moment equilibrium about the hinge (in the deformed state): PLcosθ = M P = Kθ Lcosθ where K is the torsional stiffness of the spring and M = Kθ. If the angle θ is small, then cosθ 1 and the linear equation of equilibrium is obtained: NAFEMS p. 18: Figure P = Kθ L 13

14 Example 1: Torsional spring on a clamped beam 14

15 Example 2: Torsional spring on a clamped column Example on large displacements/large rotations/small strains NAFEMS p. 46: Figure

16 Example 2: Torsional spring on a clamped column Moment equilibrium about the hinge (in the deformed state): PLsinθ = M P = Kθ Lsinθ This equation of equilibrium is interesting as it has two solutions PL θ = 0 for arbitrary P and K = θ sinθ P = Kθ L These are shown on figure 2.5. The intersection point between the two curves of equilibrium is termed the bifurcation point. Linearization of the term: θ 0 so sinθ θ and the equation of equilibrium becomes: (K PL)θ = 0 Equilibrium is obtained for each value of P if θ = 0, but if P = K/L then the column is in equilibrium for each value of θ. This equation of equilibrium represents, in the simplest form, a linearized buckling problem (instability problem, Euler buckling). The critical force P c = K/L is termed the elastic critical buckling load. 16

17 Example 3: Simple rod Example on large displacements/small rotations/small strains. This simple example provide a good illustration of geometric nonlinear analysis, especially a discussion of strain measures, discussion of stress, equilibrium and the notation of the tangent- stiffness - NAFEMS p. 48: Figure

18 Example 3: Simple rod There is an initial tensile force N 0 and the assumption small strain imply that the cross-sectional area A of the rod is assumed constant. The equation of equilibrium is derived for the deformed configuration: Nsinα P = N u + h L P = 0 (3) Introducing the constitutive conditions, i.e. it is assumed a linear elastic material with elasticity modulus E, provide that: where ε is the strain. (N N 0 ) = EAε 18

19 Example 3: Simple rod Engineering strain ε E is the traditional definition used in theory based on infinitesimal strains: Engineering stress: σ E = P A 0 = E ε E where A 0 is the cross-sectional at beginning of the test. Engineering strain: ε E = L where L is the current length and is the length on the tensile rod at the beginning of the test. Poisson s ratio: ν = ε r ε a where ε r is the strain in radial direction and ε a is the strain in the axial direction. 19

20 Example 3: Simple rod True stress: σ = P A where A is the current cross-sectional area of the tensile rod. True strain: ( ) L ε = ln Stress-strain curve for true strain can be expressed by: σ = K ε n where K is the coefficient and n is deformation hardening exponent. Since nonlinear problems often encounter large displacements and small strains it is imperative to apply a strain definition which yield zero strain for arbitrary rigid body movements. It is reasonable to employ L 2 instead of L as basis for the strain definition. 20

21 Example 3: Simple rod An advantage to employ L 2 is that this strain definition is relatively easy to generalize in 2D and 3D continuum. Applying the engineering strain definition and multiplying by L + provide ε E = (L )(L + ) = (L2 L0 2) (L + ) L + L0 2 = (L2 L ( 0 2) ) L L (L 2 L0 2 = ( ) L0 2 L0 + L ) = (L2 L0 2) L (2 + ε E) For small strains is ε E 0 and can be removed from the numerator. Thereby the Green-Lagrangian strain definition ε G is obtained: ε G = L2 L 2 0 2L

22 Example 3: Simple rod For this example the engineering strain definition is used: ε E = L where = ( ) 2 h a 2 + h 2 = a 1 + a and L = ( h + u 2 ) a 2 + (h + u) 2 = a 1 + a The expression for the strain ε E becomes complex. This problem was avoided by George Green by introducing another strain measure termed Green s strain: This yield: ε G = ( h ε G = L2 L 2 0 2L 2 0 )( ) u + 1 ( ) 2 u (4) 2 22

23 Example 3: Simple rod The relation between the two main strain measures is: ε G ε E ε E = 1 2 ε E ε G = ε E ε2 E It is assumed that N 0 = 0 to obtain a simplified expression and the stress is determined by σ = N A = Eε E As ε G ε E = ε E this expression can be written σ = N A = Eε E = E ( ( )) 1 1 L + 1 ε G (5) 2 For small strains it is given that L/ 1. This is illustrated by NAFEMS p. 72: Figure

24 Example 3: Simple rod The figure illustrates, the 2. Piola-Kirchhoff spænding [S] which is similar to the rotated Cauchy stress [σ ]. 24

25 Example 3: Simple rod Generally it is given: The Cauchy stress σ (the current force per unit deformed area) is work consistent with the engineering strain ε E The 2. Piola-Kirchhoff stress S (transformed forces per unit un-deformed area) is work consistent with Green-Lagrangian strain ε G In this example with small strains the difference between ε E and ε G can neglected and thereby the constitutive equation (N N 0 ) = EAε G (6) 25

26 Example 3: Simple rod Substitution of the constitutive equation 6 in the equation of equilibrium 3 and inserting the Green-Lagrangian strain 4 under the assumption that L/ 1 due to small strains yields ( ) u + h P = N = EA L 3 0 (hu + 12 ) u2 (h + u) }{{}}{{} b) a) ( ) u + h +N 0 It is shown that these two types of nonlinearities is included in the equation of equilibrium: a): a nonlinear strain definition is applied b): equilibrium is derived for the deformed state (7) 26

27 Example 3: Simple rod Notice the equation of equilibrium 7 is derived for the deformed situation, now referring to the original configuration, i.e. the variables A, h and - NAFEMS p. 50: Figure

28 Example 3: Simple rod The equation of equilibrium 1 is rewritten to ( ) u + h r(d, f ) = p(d) f = 0 r = N P = 0 r = p(u) P = 0 The incremental equation of equilibrium has the form K T δu = r as the indices is removed. It can be seen that ( ) u + h r = p(u) P = N P 28

29 Example 3: Simple rod The tangent-stiffness matrix K T can now be determined by K T = d p(u) = d ( ) u + h N(u) du du = d ( ) ( ) u + h u + h dn N + du du = d ( ) ( ) u + h u + h d N + du du (EAε G + N 0 ) = d ( ) ( ) ( ( )) u + h u + h d hu N + EA du du L0 2 + u2 2L0 2 = N ( ) ( ) u + h EA u + h + where dn/du has been determined from the constitutive equation 6 and the expression for the Green-Lagrangian strain, i.e. equation 4. 29

30 Example 3: Simple rod The expression for the tangent-stiffness can be rewritten providing: where K 0 = EA K T = K 0 + K L + K σ ( ) 2 h is the linear stiffness, K L = EA [ 2 u h + ( u h) 2 ]( h ) 2 is the initial shear stiffness and K σ = N is the initial stress stiffness. K 0 is known from static analysis (small displacements). K L reflects the effects on the stiffness due to changes in the displacement. K σ is the effect on the stiffness due to, e.g. membrane forces. K 0 and K σ is used for linear stability analysis. Now the incremental equation of equilibrium can be solved by: ( ) u + h K T δu = r where r = N P This example should illustrate the effects, which should be included in a geometric nonlinear analysis. 30