Anisotropic plasticity, anisotropic failure and their application to the simulation of aluminium extrusions under crash loads

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1 Anisotropic plasticity, anisotropic failure and their application to the simulation of aluminium extrusions under crash loads Paul Du Bois, Consulting engineer Oasys LS-DYNA Users Meeting 28th January 2016 The Arup Campus, UK

2 Anisotropy from a metallurgical point of view : texture index Sample BLM 45 BLM 45 BLM 46 BLM Texture Index Predominant Texture Type Transverse Basal Transverse Transverse Transverse Basal Basal Basal, Transverse Degrees off-center Texture index denotes degree of anisotropy Index ranges from 1 (isotropic) to (single crystal) Aluminum foil (highly textured) has an index of 4.14 Anisotropy depends upon the manufacturing process Distinguish between materials with strong texture vs. materials with weak texture

3 Anisotropy from a mathematical point of view : symmetry Fully anisotropic : no symmetry Monoclinic : 1 symmetry plane f xx, yy, zz, xy, yz, zx f xx, yy, zz, xy, yz, zx Orthotropic : 2 symmetry planes f xx, yy, zz, xy, yz, zx f xx, yy, zz, xy, yz, zx f xx, yy, zz, xy, yz, zx f xx, yy, zz, xy, yz, zx Transversely isotropic : axisymmetric isotropic Most numerical models assume orthotropic response Properties in 45 degree and 135 degrees to rolling direction are assumed to be the same

4 Anisotropy from a CAE engineering point of view : process Levels of increasing complexity of simulations involving shell elements : isotropic Transversely isotropic (R<1 or R>1 ) Orthotropic with known material direction ( stamping, extrusions ) Orthotropic with material directions to be mapped ( stamped or cold formed parts ) May be a blessing in disguise as results may become less random

5 Anisotropy from an industrial point of view : formability Science of metalforming started in the UK and the US in the 1950 s Hosford, Swift, Hill, Lankford and others identified the Rparameter(s) as a measure of formability : high R is good formability as it means little thinning occurs during deformation The R-value for any angle is determined as the ratio of transversal to thickness plastic strain rate in uniaxial tension : R 22p p 33 sin 2 f f f sin 2 cos 2 xx xy yy f f xx yy R is equal to unity for an isotropic metal So formability is directly related to the transverse anisotropy and/or inplane anisotropy of the metal sheet

6 When is a material isotropic? Example for plane stress isotropic in the elastic range y00 y45 y90 Isotropic yield Isotropic flow E 00 E R 00 R 45 R 90 1 Isotropic failure pf pf pf

7 Traditional crash analysis based on MAT_024 : isotropic multi-linear plasticity Why? Because for mild steel and HSLA we can get away with it Because it is difficult to determine the rolling direction of the sheet in every IP after stamping The approach still seems to work for materials such as DP, CP, TRIP and UHSS

8 Traditional metalforming analysis Rolled sheet is orthotropic Before forming the rolling direction is known R-values are a measure of formability Thus considering orthotropy in the simulation is both possible and important First orthotropic law was Hill-48 Associated flow : anisotropic yield and anisotropic flow are related Mainly interested in small deformations ( below necking strain ) Consequently off-axis yield curves are considered scaled versions of the yield curve in the rolling direction : call this isotropic or homologuous hardening

9 Hill 48 in plane stress Quadratic orthotropic 4 parameter law (00=rolling direction ) G 002 F 902 H N Traditional way of working : input the yield curve in rolling directions and 3 R-values, compute the yield values in the off-axis directions yyp H R00 p zz G xxp H R90 p zz F p 135 N 1 R45 p zz F G F H [( F G 2 N )]

10 Typical R-values for steel and aluminium sheet

11 Higher order anisotropic material laws At Alcoa Frederic Barlat refined the anisotropic material laws with the purpose of simulating aluminium sheet Associated flow Homologuous hardening ( single yield curve ) Higher order yield surface ( order 6 or 8 ) Barlat-89 is implemented in MAT_036, 4 parameter law, reduces to Hill 48 if m=2, popular workhorse in the stamping community Barlat-2000 is implemented as MAT_133, 8 parameter law Aretz model was implemented as MAT_135 by SimLab Trondheim, 8 parameter model

12 [kn] Aluminium sheet with R= Force 4 2 Reference : Increasing rupture predictability for aluminum, influence of anisotropy Daniel Riemensperger, LS-DYNA forum 2015, Bamberg displacement 30 [mm]

13 Aluminium extrusions Lightweight material Fracture is an issue Increasing use in automotive industry Yield seems mildly anisotropic or even isotropic Flow is strongly anisotropic, indicating non-associated flow R-values can be very accurately measured using DIC Values are highly directionally dependent in extrusions

14 Aluminium extrusions For crash simulations the prime concern is and remains capturing the correct stress levels However the R-values ( anisotropic flow ) will influence the failure This leads to a consideration of anisotropy in the crash simulation Possible as the extrusion direction is known Interest goes up to failure far beyond necking, this has lead to the introduction of distortional hardening ( multiple yield curves )

15 Aluminium extrusion XXXX and Hill 48 / Barlat 89 The example below illustrates the failure of 4 parameter laws to accurately model an aluminium extrusion H G 1 H G 0.71 R G H 0.29 H R F 0.71 F N 1 N R F G 2 s N F G 2N F H R00 R45 R s test Hill

16 Aluminium extrusion XXX and Hill 48 / Barlat 89 The example below illustrates the failure of 4 parameter laws to accurately model an aluminium extrusion Comparison of MAT_024 and MAT_036 with R00=R90=0.4 and R45=1.2 R00 R45 R90 The surfaces under 45 Degree differ from 00 degree

17 The Fleisher-Borrvall model : a first shot at distortional hardening It seems desirable to match yield curves and variable R-values as the shape of yield curves in different directions may vary : y00 y p y00 E y90 p45 45 y45 E p The input of 3 yield curves is expected in the option HR=7 y90 E

18 Distortional hardening The yield curves are internally converted in function of equivalent plastic strain : y00 y45 y90 y00 y45 p45 00 p y00 y90 90 p The conversion is based on work hardening The 45 and 90 degree curves are then scaled so that all 3 curves have the same initial yield point

19 Distortional hardening : The yield function is rewritten ( unconventionally ) as : f eff 00 y00 p 45 y45 p 90 y90 p With the parameters a b c a 1 4 b 2 : q 4q12 1 q d bq12 1 b 1 q c d 1 q 45 1 c q c 4 c c 1 d 1 q c 4

20 Distortional hardening The Fleisher-Borrvall model was implemented in MAT_036 as the HR=7 option, 6 load curve model For loading in 00, 45 and 90 degrees we get resp. : f eff y00 p f eff y45 p f eff y90 p And for other loading angles we get a convex combination of the hardening curves in each direction A non-associated flow rule allows to respect R-values as specified by the user in 3 directions

21 Aluminium extrusion YYYY Material has isotropic yield

22 Alumimium extrusion YYYY Reference shows R values as: R00 = 0.48, R45 = 0.29, R90 = 1.76 Bumper Beam Longitudinal System Subjected to Offset Impact Loading Kokkula (PhD Thesis) AA-6060 T1 Aluminum R values for AW-6060 T66 are: R00 = 0.49, R45 = 0.27, R90 = 1.69

23 Aluminium extrusion YYYY

24 Aluminium extrusion YYYY Reference : Testing and Validation of a New Damage Model for Application in Automotive Crashworthiness Simulations Sean Haight, 2012 ( internal report )

25 Further enhancements of MAT_036 HR=7 Reference : Development of an Anisotropic Material Model for the Simulation of Extruded Aluminum under Transient Dynamic Loads, Anthony Smith, LS-DYNA conference Wuerzburg, 2015 Orthotropic plasticity with distortional hardening seemed the way to go for aluminium extrusions Many questions remain at this point How good is the response under shear/biaxial loads? Does the material law have sufficient free parameters?

26 6000 series extruded aluminum Some anisotropy in flow stress High anisotropy in R-Values Need to investigate anisotropic models in LS-Dyna YS0 = 180 MPa R0 = 0.56 YS45 = MPa YS90 = MPa R45 = R90 = 1.70 Tensile Test Schematic courtesy of: Structural Impact Laboratory Norwegian University of Science and Technology Engineering Strain R-Value Engineering Stress 6000 series extruded aluminium Effective Plastic Strain 26

27 MAT_036, HR=7 for pure shear 00p 1p R p p degree 00 degree 1 2 Precise definition of material system needed 00p 90p 1p 2p 2p 1 R

28 MAT_036, HR=7 for equibiaxial tension 00p 1p R p p degree 00 degree 1 2 Isotropic : independent of material system 00p 90p 1p 2p 2p 1 R

29 Enhanced formulation of MAT_036 with HR=7 The yield function is rewritten as : f eff 00 y00 p 45 y45 p 90 y90 p s ys p b yb p With the parameters b max(0,1 4 d2 ) s max(0,4 d2 1) d : c 00 1 c d 1 q 1 b s 4 c 45 1 c q 1 b s 2 c 90 1 c 1 d 1 q 1 b s 4

30 MAT_036 with HR=7 is now a 10 load curve model Enhancements to Mat RSHR RBIA 36 HR=7 X RBIAX.GT.0: Constant biaxial R-value RBIAX.LT.0: Biaxial R-value as function of (plastic) strain RBIAX = deps11/deps22 RSHR.GT.0: Constant shear R-value RSHR.LT.0: Shear R-value defined as function of (plastic) strain RSHR = - deps11/deps22 30

31 New Mat 36 HR=7 Yield Surface 90 Biaxial 45 0 Biaxial Biaxial Shear Shear 0, 45, 90 Material direction; yield surface at plastic strain = 0.0 YS0 = YS45 = YS90 = YSShear = YSBiaxial= Mat 36 HR 3 Mat 36 HR 7 Old Mat 36 HR 7 - New 31

32 Enhanced MAT_036 HR=7 model for extruded aluminium Input Plastic Strain Hardening Curves True Stress vs. Plastic Strain 0 Yield Surface 0 Material direction, yield surface evolution from plastic strain = 0.00 to 0.50 This material behavior is a good example of distortional hardening; note how the yield surface expands and distorts with increasing plastic strain. Similar behavior is observed in the 45 and 90 material directions. 32

33 Uniaxial Tension 0 Force vs. Displacement Uniaxial Tension 45 Force vs. Displacement Uniaxial Tension 90 Force vs. Displacement 33

34 Shear butterfly test Cases Force vs. Displacement 34

35 Punch test to Test Cases Biaxial Punch Test Force vs. Displacement 35

36 Plane stress orthotropic failure : directional dependency upon the state of stress

37 MAT_ADD_GENERALIZED_DAMAGE or MAGD xx D11 yy D21 D xy 41 D12 D22 D42 D14 xxeff eff D24 yy D44 xyeff xx 1 d 0 0 xxeff eff 1 d 0 yy yy d xyeff xy 37

38 introduction Isotropic failure can be modeled in LS-DYNA using MAT_ADD_EROSION_GISSMO, this model takes into account the dependency of failure on the state of stress and the load path but has no directionality In the following we will generalize GISSMO to model orthotropic failure meaning the material has 3 symmetry planes and in particular failure strains under 45 degree and under 135 degree to the material x-axis are the same, the model will at first be limited to plane stress ( shell elements ) This approach will be realized with the versatile MAT_ADD_GENERALIZED_DAMAGE option in LS-DYNA We remain consistent with orthotropic plasticity models in plane stress where material properties are specified under 0, 45 and 90 degrees to the material x-axis. Consequently the model will require GISSMO type input based on experiments with the first principal stress direction under 0, 45 and 90 degrees to the material x-axis. Thus NHIS=3.

39 introduction The model will be orthotropic with respect to failure, different failure strains can be defined in different directions for the same state of stress The dependency of failure on state of stress and load path remains the same as in GISSMO Damage coupling is isotropic : a scalar function defines the relationship between true stress and effective stress, thus IFLG3=1 How fast damage accumulates depends upon the loading direction Since all data are defined in the material system, damage must be accumulated in the material system, thus IFLG2=1 Instead of using history variables from the elasto-plastic material law, the drivers for the 3 GISSMO models are internally computed according to the theory outlayed in these notes, thus IFLG1=2

40 introduction Damage coupling is isotropic : a scalar function defines the relationship between true stress and effective stress, thus IFLG3=1 Since all data are defined in the material system, damage must be accumulated in the material system, thus IFLG2=1 Instead of using history variables from the elastoplastic material law, the drivers for the 3 GISSMO models are internally computed, thus IFLG1=2

41 Experimental database Tensile, notched and shear tests needed in 3 directions direction of the shear tests is the first principal stress direction (tensile diagonal) Only 1 biaxial test needed

42 Input data preparation LCSDG-90 LCSDG-00 LCSDG-45

43 Input data generation A consistent input deck also requires the instability curves (ECRIT) to have coincident biaxial points Swift curves generated for material hardening curves in 3 directions will usually not have coincident biaxial points

44 the plastic strain rate tensor in plane stress We can compute an equivalent plastic strain rate for every material direction : xxp p xy 1 b p xyp 1 0 b b p p p yy 0 b b b 1 1 b 1 b b 1 1 b 13 1 b ep p 1 1 b 2 b 2 1p 2 cos sin b2 b 3 90ep 2 90p 1 1 b 2 b 2 1p sin 2 cos sin 3 ep p 1 1 b 2 b 2 1p cos 2 cos sin b2 b 3 2 b

45 The plastic strain rate tensor in plane stress Plot the coefficients in function of the angle theta for unit principal strain rate : Angle theta in degrees 00p 1p cos 2 cos sin 90p 1p sin 2 cos sin p 45p 135 1p 2 cos sin 0

46 Damage accumulation Damage can now be accumulated in the material system using the failure criteria that were determined from testing in each direction : Then the damage is computed as : 1 1 n ep 00 00f 1 1 n 90ep 90f d 90 d 90 dt 1 1 n ep 45 45f d 45 d 45 dt d 00 nd 00 d 90 nd 90 d 45 nd 45 d 00 d 00 dt No problems with the sign of the damage as equivalent strain rates are all positive

47 isotropic damage Propose a damage formulation for IFLG3=1 using only the strain rate components with positive coefficients Then, assuming a positive first principal strain rate : cos sin cos 0 d d 452 d d 452 d 002 p cos 2 sin 2 cos 4 2 cos 3 sin sin sin cos 0 d d 452 d d 452 d 902 p cos 2 sin 2 sin 4 2 cos sin \

48 isotropic damage This approach results in a monotonic model ( no new maxima or minima are generated between the failure surfaces defined for degrees )

49 isotropic damage This approach results in a monotonic model ( no new maxima or minima are generated between the failure surfaces defined for degrees )

50 isotropic damage This approach results in a monotonic model ( no new maxima or minima are generated between the failure surfaces defined for degrees )

51 Large scale validation

52 Uniaxial tensile test : MAT_036 + MAT_ADD_D_G 00 degree 45 degree 90 degree

53 Conclusions Due to pronounced anisotropic flow the prediction of failure in aluminium extrusions has been an elusive goal Progress was made thanks to extensive code development in both plasticity ( MAT_036 ) and anisotropic failure (M_A_G_D) Remarkable work done by Thomas Borrvall from Dynamore Nordic and Tobias Erhart from Dynamore The development was customer driven by Mercedes-Benz and Honda-NA The orthotropic failure model in M_A_G_D is to our knowledge the only failure model that allows for directional dependency of the failure strain upon the state of stress Procedures need to be developed for the data generation of aluminium extrusions to speed up the process Magnesium extrusions could be even more challenging

54 The (preliminary) end

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