A multiaxial high-cycle transversely isotropic fatigue model Reijo - PowerPoint PPT Presentation

A multiaxial high-cycle transversely isotropic fatigue model Reijo Kouhia 1 , Sami Holopainen 1 , Timo Saksala 1 and Andrew Roiko 2 1 Tampere University of Technology Department of Mechanical Engineering and Industrial Systems 2 VTT Technical Research Centre of Finland Partially funded by TEKES - the National Technology Agency of Finland Project SCarFace, number 40205/12 14 th European Mechanics of Materials Conference August 27–29, 2014, Gothenburg, Sweden

Outline 800 600 • Motivation and background stress [MPa] 400 200 • Isotropic model 0 • Transversely isotropic model −200 0 20 40 60 80 • Parameter estimation 0.02 • Results 0.015 • Conclusions and future Damage 0.01 developments 0.005 0 80 0 20 40 60 80 EMMC-14, August 27–29, 2014 2

Motivation and background Certain materials exhibit transversely isotropic symmetry • unidirectionally reinforced composites • forged metals – elasticity isotropic – fatigue properties transversely isotropic Figures from http://aciers.free.fr EMMC-14, August 27–29, 2014 3

Background - Fatigue models Either stress, strain or energy based. Stress based criteria are commonly used in high-cycle fatigue • stress invariant criteria, Sines 1955, Crossland 1956, Fuchs 1979 • critical plane criteria, Findley 1959, Dang Van 1989, McDiarmid 1990 • average stress criteria, Gr¨ ubisic and Simburger 1976, Papadopoulos 1997. Cumulative damage theories. A more fundamental approach using evolution equations. EMMC-14, August 27–29, 2014 4

Continuum approach Proposed by Ottosen, Stenstr¨ om and Ristinmaa in 2008. � s � Endurance surface postulated as α � = 0 β = 0 1 β < 0 β > 0 β = (¯ σ + AI 1 − σ oe ) , 0 σ oe = β where I 1 � � 3 B σ = ¯ 3 J 2 ( s − α ) = 2 ( s − α ) : ( s − α ) , σ 1 β > 0 d s I 1 = tr σ . Back stress and damage evolution eqs. d α ’ β < 0 α ’ d α α = C ( s − α ) ˙ ˙ β, α D = g ( β, D ) ˙ β = K exp( Lβ ) ˙ ˙ β. σ 2 σ 3 A EMMC-14, August 27–29, 2014 5

Transversely isotropic model The stress is decomposed as σ = σ L + σ T , where σ T = P σP , P = I − B , where B = b ⊗ b is the structural tensor and b is the unit vector normal to the transverse isotropy plane. Integrity basis of a transversely isotropic solid I 2 = 1 2 tr ( σ 2 ) , I 3 = 1 3 tr ( σ 3 ) , I 5 = tr ( σ 2 B ) . I 1 = tr σ , I 4 = tr ( σB ) , EMMC-14, August 27–29, 2014 6

Endurance surface Present transversely isotropic formulation � � �� β = σ + A L I L 1 + A T I T 1 − ¯ (1 − ζ ) S T + ζS L /S T = 0 , where � σ = ¯ 3 J 2 ( s − α ) , I L 1 = tr σ L = I 4 , I T 1 = tr σ T = I 1 − I 4 , and � n � 2 I 5 − I 2 � σ L : σ L � n 4 ζ = = . σ : σ 2 I 2 In uniaxial loading σ = σ n ⊗ n the ζ -factor has the form ζ = (2 cos 2 ψ − cos 4 ψ ) n , where ψ is the angle between n and b . EMMC-14, August 27–29, 2014 7

Shape in the π -plane and ζ -factor σ 1 1 0.8 0.6 ζ 0.4 n = 0 . 5 0.2 n = 1 n = 2 σ 2 σ 3 0 π/ 8 π/ 4 3 π/ 8 π/ 2 0 ψ , S L /S T = 1 dotted black line, 1 . 5 dashed blue line, 2 red line A L = 0 . 225 , A T = 0 . 275 b = (0 , 0 , 1) T EMMC-14, August 27–29, 2014 8

Evolution equations for α and D Damage and the back-stress evolves only when moving away from the endurance surface K 1 − D exp( Lβ ) ˙ α = C ( s − α ) ˙ ˙ D = β, ˙ β. 2 σ 2 5 1 σ 1 - α 2 = α 3 α 1 - - α 4 = α 5 3 σ 3 4 σ 4 ∽ time EMMC-14, August 27–29, 2014 9

Estimation of the parameters Five material parameters in the endurance surface S L , S T , A L , A T and n . Three material parameters in the evolution equations for the back-stress and damage C, K and L . Data from tests with forged 34CrMo6 steel. Due to the lack of data in the intermediate directions we have chosen n = 1 . S L = 447 MPa , S T = 360 MPa , A L = 0 . 225 , A T = 0 . 300 , K = 12 . 8 · 10 − 5 , C = 33 . 6 , L = 4 . 0 EMMC-14, August 27–29, 2014 10

!" !" !" !" !" !" !" !" Fatigue strengths σ m = 0 1 600 0.8 500 σ a [MPa] 0.6 Damage 0.4 400 0.2 0 300 2 3 4 5 6 10 10 10 10 10 10 4 10 5 10 6 10 7 N N △ denotes experimental results, • model predictions EMMC-14, August 27–29, 2014 11

!" !" !" !" !" Effect of mean stress σ x = σ x m + σ x a sin( ωt ) σ y = σ y m + σ y a sin( ωt ) longitudinal transverse 1 . 0 1 . 0 σ x a ( σ x m ) /σ x a (0) 0 . 9 σ y a ( σ y m ) /σ y a (0) 0 . 9 0 . 8 0 . 8 0 . 7 0 . 7 0 . 6 0 . 6 0 . 5 0 . 5 0 0 . 5 1 . 0 1 . 5 2 . 0 0 0 . 5 1 . 0 1 . 5 2 . 0 σ x m /σ x a σ y m /σ y a △ denotes experimental results from McDiarmid 1985 (34CrNiMo6), • model predictions EMMC-14, August 27–29, 2014 12

!" !" !" !" !" Effect of mean shear stress τ xy = τ xy m + τ xy a sin( ωt ) 1 .0 1.0 1 . 0 0.8 τ xy a ( τ xy m ) /τ xy a (0) Damage 0.6 0 . 9 0.4 0 . 8 0.2 0 . 7 0 100 1000 10000 50000 0 0 . 5 1 . 0 1 . 5 2 . 0 N τ xy m /τ xy a EMMC-14, August 27–29, 2014 13

Effect of phase shift σ x = σ x m + σ x a sin( ωt ) σ x = σ x a sin( ωt ) τ xy = 1 σ y = σ x m + σ x a sin( ωt − φ y ) 2 σ x a sin( ωt − φ xy ) (a) (b) 1 . 1 1 . 3 1 . 0 1 . 2 σ x a ( φ xy ) /σ x a (0) σ x a ( φ y ) /σ x a (0) 0 . 9 1 . 1 0 . 8 1 . 0 0 . 7 0 . 6 0 . 9 0 45 90 135 180 0 30 60 90 φ y φ xy EMMC-14, August 27–29, 2014 14

! ! "# "# "# ! ! Effect of frequency difference model based on isotropic AISI SAE 4340 transversely isotropic 34CrMo6 exp. results shown 25CrMo4 (Liu & Zenner) 34CrNiMo6 (McDiarmid) σ x = σ x a sin( ω x t ) σ x = σ x a sin( ω x t ) τ xy = 1 2 σ x a sin( ω xy t ) σ y = σ x a sin( ω y t ) 1 . 3 1 . 3 1 . 2 1 . 2 1 . 1 1 . 1 σ x a ( ω xy ) /σ x a (1) σ x a ( φ y ) /σ x a (0) 1 . 0 1 . 0 0 . 9 0 . 9 0 . 8 0 . 8 0 . 7 0 . 7 0 . 6 0 . 6 10 − 1 10 0 10 1 1 2 3 4 5 6 7 8 ω xy /ω x ω y /ω x EMMC-14, August 27–29, 2014 15

Conclusions and future developments • Transversally isotropic continuum based HCF-model • More tests needed • Microstructurally based anisotropic damage model • Constitutive model with anisotropic damage • Implementation into a FE code Alexander Roslin: Lady with the veil, 1768 Thank you for your attention! EMMC-14, August 27–29, 2014 16

Deviatoric invariants Deviatoric invariants and max shear in the longitudinal and in the isotropy plane 2 tr ( s 2 ) , J 5 = tr ( s 2 B ) . J 2 = 1 J 4 = tr ( sB ) , � � J 2 + 1 4 J 2 J 5 − J 2 τ max ( σ T ) = 4 − J 5 , τ max ( σ L ) = 4 . EMMC-14, August 27–29, 2014 17