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

A multiaxial high-cycle transversely isotropic fatigue model

Reijo Kouhia1, Sami Holopainen1, Timo Saksala1 and Andrew Roiko2

1Tampere University of Technology

Department of Mechanical Engineering and Industrial Systems

2VTT Technical Research Centre of Finland

Partially funded by TEKES - the National Technology Agency of Finland Project SCarFace, number 40205/12 14th European Mechanics of Materials Conference August 27–29, 2014, Gothenburg, Sweden

Outline

• Motivation and background
• Isotropic model
• Transversely isotropic model
• Parameter estimation
• Results
• Conclusions and future

developments

20 40 60 80 −200 200 400 600 800

stress [MPa]

80 20 40 60 80 0.005 0.01 0.015 0.02

Damage

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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.

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Continuum approach

Proposed by Ottosen, Stenstr¨

• m and Ristinmaa in 2008.

Endurance surface postulated as β = 1 σoe (¯ σ + AI1 − σoe), where ¯ σ =

• 3J2(s − α) =
• 3

2(s − α) : (s − α),

I1 = tr σ. Back stress and damage evolution eqs. ˙ α = C(s − α) ˙ β, ˙ D = g(β, D) ˙ β = K exp(Lβ) ˙ β.

I1 s β > 0 β < 0 β = β = α = 0

σ1 σ2 σ3 α’ α dα’ dα A ds B β < 0 β > 0

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Transversely isotropic model

The stress is decomposed as σ = σL + σT, where σT = 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 I1 = tr σ, I2 = 1

2tr (σ2),

I3 = 1

3tr (σ3),

I4 = tr (σB), I5 = tr (σ2B).

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Endurance surface

Present transversely isotropic formulation β =

• ¯

σ + ALIL1 + ATIT 1 −

• (1 − ζ)ST + ζSL
• /ST = 0,

where ¯ σ =

• 3J2(s − α),

IL1 = tr σL = I4, IT 1 = tr σT = I1 − I4, and ζ = σL : σL σ : σ n = 2I5 − I2

4

2I2 n . In uniaxial loading σ = σn ⊗ n the ζ-factor has the form ζ = (2 cos2 ψ − cos4 ψ)n, where ψ is the angle between n and b.

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Shape in the π-plane and ζ-factor

, σ1 σ2 σ3

n = 2 n = 1 n = 0.5 ψ ζ π/8 π/4 3π/8 π/2 1 0.8 0.6 0.4 0.2

SL/ST = 1 dotted black line, 1.5 dashed blue line, 2 red line AL = 0.225, AT = 0.275 b = (0, 0, 1)T

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Evolution equations for α and D

Damage and the back-stress evolves only when moving away from the endurance surface ˙ D = K 1 − D exp(Lβ) ˙ β, ˙ α = C(s − α) ˙ β.

time

∽

1 2 3 4 5 σ1 σ2 σ3 σ4 α1 -

• α2 = α3
• α4 = α5

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Estimation of the parameters

Five material parameters in the endurance surface SL, ST, AL, AT 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. SL = 447 MPa, ST = 360 MPa, AL = 0.225, AT = 0.300, C = 33.6, K = 12.8 · 10−5, L = 4.0

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Fatigue strengths σm = 0

300 400 500 600 104 105 106 107 σa [MPa]

!" !" !" !" !" !" !" !"

10

2

10

3

10

4

10

5

10

6

0.2 0.4 0.6 0.8 1 Damage N N

△ denotes experimental results, • model predictions

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Effect of mean stress

σx = σxm + σxa sin(ωt) σy = σym + σya sin(ωt) longitudinal transverse

0.5 0.6 0.7 0.8 0.9 1.0 0.5 1.0 1.5 2.0 σxm/σxa σxa(σxm)/σxa(0)

!" !" !"

0.5 0.6 0.7 0.8 0.9 1.0 0.5 1.0 1.5 2.0 σym/σya σya(σym)/σya(0)

!" !"

△ denotes experimental results from McDiarmid 1985 (34CrNiMo6), • model predictions

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Effect of mean shear stress

τxy = τxym + τxya sin(ωt)

!" !" !" !" !"

0.7 0.8 0.9 1.0 0.5 1.0 1.5 2.0 τxym/τxya τxya(τxym)/τxya(0)

100 1000 10000 50000 0.2 0.4 0.6 0.8 1 .0 1.0

N Damage

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Effect of phase shift

σx = σxm + σxa sin(ωt) σx = σxa sin(ωt) σy = σxm + σxa sin(ωt − φy) τxy = 1

2σxa sin(ωt − φxy)

(a) (b) 0.6 0.7 0.8 0.9 1.0 1.1 45 90 135 180 φy σxa(φy)/σxa(0) 0.9 1.0 1.1 1.2 1.3 30 60 90 φxy σxa(φxy)/σxa(0)

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Effect of frequency difference

model based on isotropic AISI SAE 4340 transversely isotropic 34CrMo6

• exp. results shown

25CrMo4 (Liu & Zenner) 34CrNiMo6 (McDiarmid) σx = σxa sin(ωxt) σx = σxa sin(ωxt) τxy = 1

2σxa sin(ωxyt)

σy = σxa sin(ωyt)

0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 10−1 100 101 ωxy/ωx σxa(ωxy)/σxa(1)

! ! ! !

0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1 2 3 4 5 6 7 8 ωy/ωx σxa(φy)/σxa(0)

"# "# "#

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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!

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Deviatoric invariants

Deviatoric invariants and max shear in the longitudinal and in the isotropy plane J2 = 1

2tr (s2),

J4 = tr (sB), J5 = tr (s2B). τmax(σT) =

• J2 + 1

4J2 4 − J5,

τmax(σL) =

• J5 − J2

4.

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