Course in. Nonlinear FEM

Transcription

1 Course in Dynamics

2 Outline Lecture 1 Introduction Lecture Geometric nonlinearity Lecture 3 Material nonlinearity Lecture 4 Material nonlinearity continued Lecture 5 Geometric nonlinearity revisited Lecture 6 Issues in nonlinear FEA Lecture 7 Contact nonlinearity Lecture 8 Contact nonlinearity continued Lecture 9 Dynamics Lecture 1 Dynamics continued Dynamics

3 Lecture 1 Introduction, Cook [17.1]: Types of nonlinear problems Definitions Lecture Geometric nonlinearity, Cook [17.1, ]: Linear buckling or eigen buckling Prestress and stress stiffening Nonlinear buckling and imperfections Solution methods Lecture 3 Material nonlinearity, Cook [17.3, 17.4]: Plasticity systems Yield criteria Lecture 4 Material nonlinearity revisited, Cook [17.6, 17.]: Flow rules Hardening rules Tangent stiffness Dynamics 3

4 Lecture 5 Geometric nonlinearity revisited, Cook [17.9, ]: - The incremental equation of equilibrium - The nonlinear strain-displacement matrix - The tangent-stiffness matrix - Strain measures Lecture 6 Issues in nonlinear FEA, Cook [17., ]: Solution methods and strategies Convergence and stop criteria Postprocessing/Results Troubleshooting Dynamics 4

5 Elements for Mass-Spring-Damper Systems

6 MASS1: Structural Mass Element Name Nodes Degrees of Freedom Real Constants Material Properties Surface Loads Body Loads Special Features KEYOPT() KEYOPT(3) MASS1 I UX, UY, UZ, ROTX, ROTY, ROTZ if KEYOPT(3) = UX, UY, UZ if KEYOPT(3) = UX, UY, ROTZ if KEYOPT(3) = 3 UX, UY if KEYOPT(3) = 4 MASSX, MASSY, MASSZ, IXX, IYY, IZZ if KEYOPT(3) = MASS if KEYOPT(3) = MASS, IZZ if KEYOPT(3) = 3 MASS if KEYOPT(3) = 4 None None None Large deflection Key for element coordinate system - 3-D mass with rotary inertia - 3-D mass without rotary inertia 3 - -D mass with rotary inertia 4 - -D mass without rotary inertia Dynamics 6

7 COMBIN14: Spring/Damper Element Name Nodes Degrees of Freedom Real Constants Material Properties Surface Loads Body Loads COMBIN14 I, J UX, UY, UZ if KEYOPT(3) = ROTX, ROTY, ROTZ if KEYOPT(3) = 1 UX, UY if KEYOPT(3) = etc. K, CV1, CV None None None Special Features KEYOPT(1) KEYOPT(3) Nonlinear (if CV is not zero), Stress stiffening, Large deflection, etc. - Linear Solution (default) 1 - Nonlinear solution (required if CV is non-zero) - 3-D longitudinal spring-damper 1-3-D torsional spring-damper - -D longitudinal spring-damper (-D elements must lie in an X-Y plane) Dynamics 7

8 Equivalent Mass/Spring/Damper Translational Systems Rotational Systems Kinetic Energy KE = 1 M eqy KE = 1 J eq θ Potential Energy PE = 1 K eq y PE = 1 K eq θ Dissipated Energy dde dt = D eqy dde = dt 1 D eq θ Dynamics 8

9 Design of Spring/Damper in a Recoil Landing System

10 Problem Description Vehicle mass M Piston Cylinder Spring K Damper D Rod Foot Dynamics 1

11 Modeling Considerations M y K My + Dy y() =, + Ky = y() = V D Dynamics 11

12 Equivalent Spring/Damper ΔL θ y PE = K eq = 1 3 K θ ( y cos ) 3K ( cosθ ) K dde dt = ( cos ) 3 D y θ ( cos ) D eq = 3D θ Dynamics 1

13 ANSYS Procedure (1/) 1 FINISH /CLEAR 3 /TITLE, Recoil Landing System (SI) 4 /PREP7 5 6 K = 7 7 M = 8 D = 4! Critically damped 9 V = 5! 18 km/hr 1 11 ET, 1, COMBIN14,,, 1 ET,, MASS1,,, 4 13 R, 1, K, D 14 R,, M N, 1,! Ground 17 N,, 1! Vehicle 18 TYPE, 1 $ REAL, 1 $ E, 1, 19 TYPE, $ REAL, $ E, FINISH Dynamics 13

14 ANSYS Procedure (/) /SOLU ANTYPE, TRANS TRNOPT, FULL D, ALL, UY, D, 1, UX, IC,, UX,, -V DELTIM,.1 TIME, OUTRES, NSOL, ALL SOLVE FINISH /POST6 NSOL,,, U, X, DISP /GRID, 1 /AXLAB, Y, DISPLACEMENT PLVAR, Dynamics 14

15 Design of Damper in a Spring Scale

16 Problem Description Item placed on platform mass m Plateform mass M p Spring K mass M k R Calibrated dial J c Gear ratio n Geared magnifier J g Rack mass M r Damper D mass M d Dynamics 16

17 Modeling Considerations Item m Scale M ( m + ) M y + Dy + Ky y( ) =, y() = = mg y Damper D Spring K M = M M k g c p + Mr + Md R R J J n Dynamics 17

18 Lumped Masses (1/4) Lumped Mass for a Spring L Equivalent lumped mass M eq V o x dm V V = Vo x L o 1 L 1 Vo x Mkdx KE = V dm = = L L L M kv 6 M = eq M 3 k Dynamics 18

19 Lumped Masses (/4) Lumped Mass for a Gear-and-Rack Set M r V o R θ J G Rolls without slipping KE 1 1 = MrVo + J Gθ = 1 MrV o + 1 J G Vo R = 1 M r + J R G V o Meq = Mr + J R G Dynamics 19

20 Dynamics Lumped Masses (3/4) Lumped Mass for a Geared Shafts Set J 1 J N 1 N J eq Gears Shafts g c g c g c g g g g g c c g g N N J J N N J J J J KE θ θ θ θ θ + = + = + = + = c g c g eq N N J J J

21 Dynamics 1 Lumped Masses (4/4) Lumped Mass for the Spring Scale System R n J J M M c g r eq + + = 3 R n J R J M M M M M c g k d r p = Gear set

22 ANSYS Procedure (1/) 1 FINISH /CLEAR 3 /TITLE, Spring Scale (cgs) 4 /PREP7 5 6 m = 15! G 7 Meq = 5! G 8 K = 3.E6! dyn/cm 9 D = 6.4E4! dyn-s/cm 1 11 ET, 1, COMBIN14,,, 1 ET,, MASS1,,, 4 13 R, 1, K, D 14 R,, m+meq N, 1, 17 N,, 1 18 TYPE, 1 $ REAL, 1 $ E, 1, 19 TYPE, $ REAL, $ E, FINISH Dynamics

23 ANSYS Procedure (/) /SOLU ANTYPE, TRANS TRNOPT, FULL DELTIM,. TIME, KBC, 1 D, ALL, UY, D, 1, UX, F,, FX, -m*981 OUTRES, NSOL, ALL SOLVE FINISH /POST6 NSOL,,, U, X, DISP /GRID, 1 PLVAR, Dynamics 3

24 Quenching of a Shaft

25 Problem Description Shaft T() m s =.69 lb Bath T b () m b = 1 lb T(t) T b (t) Before quenching After quenching Dynamics 5

26 Modeling Considerations m s, C s, T s h, A m b, C b, T b Dynamics 6

27 ANSYS Procedure (1/) FINISH /CLEAR /TITLE, Unit: lb(mass)-ft-(btu)-hr-f /PREP PI = 4*ATAN(1) 18 ET, 1, MASS71 19 R, 1, PI*(1/4/1)**/4*(5/1) MP, DENS, 1, MP, C, 1,.11 3 ET,, LINK34 4 R,,.8 5 MP, HF,, ET, 3, MASS71,,, 1 R, 3, 1/6 MP, DENS, 3, 6 MP, C, 3, 1. N, 1, N,, 1 TYPE, 1 $ REAL, 1 $ MAT, 1 E, 1 TYPE, $ REAL, $ MAT, E, 1, TYPE, 3 $ REAL, 3 $ MAT, 3 E, FINISH Dynamics 7

28 ANSYS Procedure (/) 3 /SOLU ANTYPE, TRANS 35 TRNOPT, FULL 36 DELTIM,.1 37 KBC, 1 38 TIME,.1 39 IC, 1, TEMP, 13 4 IC,, TEMP, OUTRES, NSOL, ALL 43 SOLVE 44 FINISH /POST NSOL,, 1, TEMP,, SHAFT 49 NSOL, 3,, TEMP,, WATER 5 /GRID, 1 51 PLVAR,, 3 Dynamics 8

29 Elements Related to Lumped-Mass Systems

30 Masses Dynamics 3

31 Springs/Dampers Dynamics 31

32 Thermal Links Dynamics 3

33 Circuit Element Dynamics 33

34 Dynamic Effects KD = F M D + CD + KD = F Inertia force Damping force Elastic Force External force Dynamic Effects Dynamics 34

35 Transient Dynamic Analysis M D + CD + KD = F Dynamics 35

36 Modal Analysis (1/3) M D + CD + KD = Dynamics 36

37 Modal Analysis (/3) M D + CD + KD = M D + KD = f d = f u 1 ξ πξ R = e Dynamics 37

38 Modal Analysis (3/3) Avoid resonance Exploit resonance Assess structural stiffness Structural modal degrees of freedom Further dynamic analyses etc. Dynamics 38

39 Harmonic Response Analysis MD + CD + KD = F sin ωt ( + Φ) Dynamics 39

40 Solution Methods

41 Solution Methods Solution Methods for Equation of Motion Direct Integration Mode Superposition Implicit Explicit Full Reduce Full Reduce Dynamics 41

42 Solution methods Dynamics 4

43 Direct Integration Implicit method (ANSYS) Explicit method (LS-DYNA) Dynamics 43

44 Implicit vs. Explicit Methods Implicit method (..., D, D ) D, t + Δt = f t Δt t Dt + Δt t + Δt = Explicit method ( D, D D ) D f,..., t Δt t Δt t Dynamics 44

45 Linear vs. nonlinear dynamics Dynamics or equation of motion: The causal relation between the present state and the next state in the future. It is a deterministic rule which tells us what happens in the next time step. In the case of a continuous time, the time step is infinitesimally small. Thus, the equation of motion is a differential equation or a system of differential equations: du/dt = F(u), where u is the state and t is the time variable. An example is the equation of motion of an undriven and undamped pendulum. In the case of a discrete time, the time steps are nonzero and the dynamics is a map: u n+1 = F(u n ), with the discrete time n. An example is the baker map. Note, that the corresponding physical time points t n do not necessarily occur equidistantly. Only the order has to be the same. That is, n < m implies t n < t m. The dynamics is linear if the causal relation between the present state and the next state is linear. Otherwise it is nonlinear. Dynamics 45

46 Implicit vs. Explicit What is the difference between implicit and explicit dynamics? (Difference between regular ANSYS and ANSYS/LS-DYNA?) For computers, matrix multiplication is easy. Matrix inversion is the more computationally expensive operation. The equations we solve in nonlinear, dynamic analyses in ANSYS and in LS-DYNA are: [M]{a} + [C]{v} + [K]{x} = {F} Hence, in ANSYS, we need to invert the [K] matrix when using direct solvers (frontal, sparse). Iterative solvers use a different technique from direct solvers, however, the inversion of [K] is the CPU-intensive operation for any 'regular' ANSYS solver, direct or iterative. We then can solve for displacements {x}. Of course, with nonlinearities, [K(x)] is also a function of {x}, so we need to use Newton-Raphson method to solve for [K] as well (material nonlinearities and contact get thrown into [K(x)]) Dynamics 46

47 Implicit vs. explicit In LS-DYNA, on the other hand, we solve for accelerations {a} first. It is assumed that the mass matrix is lumped. This basically forces the use of lower-order elements, i.e. for all explicit dynamics codes (ANSYS/LS-DYNA, MSC.Dytran, ABAQUS/Explicit), lower-order elements are used. The benefit of doing lumped mass is, if we solve for {a}, then [M], if lumped, is a diagonal mass matrix. This means that inversion of [M] is trivial (diagonal terms only) Another way to view it, is that we now have N set of *uncoupled* equations. Hence, we just have to do matrix multiplication, which is less CPU-intensive. Also, [K] does not need to be inverted, and accounting for material nonlinearties and contact is easier. Dynamics 47

48 Implicit vs. explicit The terms 'implicit' and 'explicit' refer to time integration For example backward Euler method, that is an example of an implicit time integration scheme central difference or forward Euler are examples of explicit time integration schemes It relates to when you calculate the quantities - either based on current or previous time step. In any case, this is a very simplified explanation, and the main point is that implicit time integration is unconditionally stable, whereas explicit time integration is not (there is a critical timestep the timestep delta(t) needs to be smaller than). Implicit, e.g. 'regular' ANSYS allows for much larger time steps Explicit, e.g. LS-DYNA requires much smaller time steps. Also, LS-DYNA requires very tiny steps, i.e. good for impact/short-duration events, not usually things like maybe creep where the model's time scale may be on the order of hours or more. Dynamics 48

49 Implicit vs. explicit 'Regular' ANSYS uses implicit time integration. This means that {x} is solved for, but we need to invert [K], which means that each iteration is computationally expensive. However, because we solve for {x}, it is implicit, and we don't need very tiny timesteps (i.e., each iteration is expensive, but we usually don't need too many iterations total). The overall timescale doesn't affect us much (although there are considerations of small enough timesteps for proper momentum transfer, capturing dynamic response). ANSYS/LS-DYNA uses explicit time integration. This means that {a} is solved for, and inverting [M] is trivial -- each iteration is very efficient. However, because we solve for {a}, then determine {x}, it is explicit, and we need very small timesteps (many, many iterations) to ensure stability of solution since we get {x} by calculating {a} first. (i.e., each iteration is cheap, but we usually need many, many iterations total) Dynamics 49

50 Mode Superposition Method D = C M + C M + C M C n M n Dynamics 5

51 Reduced Method KD = F K K mm sm K K ms ss D D m s = F F m s D s KD = K 1 ss m = F ( F K D ) s sm m where K = K mm F = F m K K ms ms K K 1 ss 1 ss K F s sm Dynamics 51

52 Methods for Nonlinear Dynamic Analysis For nonlinear analysis, the only methods applicable is DIRECT INTEGRATION method. Reduced method can not be used for nonlinear analysis. Either implicit or explicit methods can be used. Dynamics 5

53 Mass and Damping

54 Dynamics 54 Consistent vs. Lumped Mass Matrices x x x x x x x x x x x x x x x x x x x x x x x x x x ROTZ UY UX ROTZ UY UX j j j i i i Consistent mass matrix Lumped mass matrix

55 Damping Damping effects is the total of all energy dissipation mechanisms Hysteresis (solid damping) Viscous damping Dry-friction (Coulomb damping) Dynamics 55

56 Idealization of Structural Damping Structural dampings are usually small (%-7%). Equivalent viscous damping is assumed in ANSYS, i.e., F D = CD Dynamics 56

57 How ANSYS Forms Damping Matrix? N = c j j j j = 1 Ω k= 1 m N [ ] [ ] ( )[ ] [ ] e m ξ C α M + β + β K + β + β K + [ C ] + [ C ] Alpha damping Beta damping Material dependent beta damping Element damping matrices Frequency-dependent damping matrix k ξ Dynamics 57

58 Copper Cylinder Impacting on a Rigid Wall

59 Problem Description Initial Velocity V o y L D x Dynamics 59

60 Modeling Consideration Material: bilinear plastic model. VISCO16 (D viscoplastic solid) is used. Use axisymmetric model. Dynamics 6

61 ANSYS Procedure (1/4) 1 FINISH /CLEAR 3 /TITLE, UNITS: SI 4 /PREP7 5 6 ET, 1, VISCO16,,, 1 7 MP, EX, 1, 117E9 8 MP, NUXY, 1,.35 9 MP, DENS, 1, TB, BISO, 1 1 TBDATA,, 4E6, 1E6 13 TBPLOT, BISO, RECTNG,,.3,, LESIZE, 1,,, 4 17 LESIZE,,,, 18 MSHAPE,, D 19 MSHKEY, 1 AMESH, ALL 1 FINISH Dynamics 61

62 ANSYS Procedure (/4) /SOLU ANTYPE, TRANS TRNOPT, FULL NLGEOM, ON IC, ALL, UY,, -7 NSEL, S, LOC, X, D, ALL, UX, NSEL, S, LOC, Y, D, ALL, UY, NSEL, ALL /PBC, U,, ON EPLOT TIME, 8E-6 DELTIM,.4E-6 KBC, 1 OUTRES, ALL, 4 SOLVE FINISH Dynamics 6

63 ANSYS Procedure (3/4) /POST6 TOPNODE = NODE(,.34,) NSOL,, TOPNODE, U, Y, DISP DERIV, 3,, 1,, VELO /GRID, 1 /AXLAB, X, TIME s /AXLAB, Y, DISPLACEMENT m PLVAR, /AXLAB, Y, VELOCITY m/s PLVAR, 3 FINISH Dynamics 63

64 ANSYS Procedure (4/4) /POST1 SET, LAST PLDISP, PLNSOL, EPTO, EQV ANTIME, 3 Dynamics 64

65 Dynamic Loads

66 Dynamic Loads: An Example 1 /SOLU... 3 F,...!.5 at the nodes 4 TIME,.5! Ending time DELTIM,... KBC, AUTOTS, ON! Integration step! Ramped loading! Option Force (N).5 8 OUTRES,...! Option 9 SOLVE! Load step F,...! 1 at the nodes TIME, 1! Ending time Time (s) 13 SOLVE! Load step FDELE,...! Zero the force 16 TIME, 1.5! Ending time 17 KBC, 1! Stepped loading 18 SOLVE! Load step 3 Dynamics 66

67 Initial Conditions

68 Example: An Stationary Plate Subjected to an Impulse Load This is the default initial condition. No input is needed. Dynamics 68

69 Example: Initial Velocity on a Golf Club Head This simple initial condition can be specified by using IC command. NSEL, ALL IC, ALL, UY,, V Dynamics 69

70 Example: Plucking a Cantilever Beam /SOLU ANTYPE, TRANS... TIMINT, OFF! Transient effects off TIME,.1! Small time interval D,...! Apply displacement at desired nodes KBC, 1! Stepped loads NSUBST,! To avoid non-zero velocity SOLVE TIMINT, ON! Transient effects on TIME,...! Actual time at end of load DDELE,...! Delete the applied displacement SOLVE Dynamics 7

71 Example: Dropping an Object from Rest /SOLU TIMINT, OFF! Transient effects off TIME,.1! Small time interval NSEL,...! Select all nodes on the object D, ALL, ALL,! Temporarily fix them NSEL, ALL ACEL,...! Apply acceleration KBC, 1! Stepped loads NSUBST,! To avoid non-zero velocity SOLVE! Load step 1 TIMINT, ON! Transient effects on TIME,...! Actual time at end of load NSEL,...! Select all nodes on the object DDELE, ALL, ALL! Release them NSEL, ALL SOLVE! Load step Dynamics 71

72 Integration time Steps

73 Response Frequency Response Minimum response time Time Δt p Dynamics 73

74 Abrupt Changes in Loading Force (N) Time (s) Dynamics 74

75 Contact Frequency Δt T 3 Dynamics 75

76 Wave Propagation Δt Δx 3c Dynamics 76

77 Exercise: Rocket Flight y 3 Thrust 14 in. 1 lb 1 1 sec. Time Dynamics 77