A multiaxial high-cycle transversely isotropicfatigue model
Reijo Kouhia1, Sami Holopainen1, Timo Saksala1 and Andrew Roiko2
1Tampere University of TechnologyDepartment of Mechanical Engineering and Industrial Systems
2VTT Technical Research Centre of Finland
Partially funded by TEKES - the National Technology Agency of FinlandProject SCarFace, number 40205/12
14th European Mechanics of Materials ConferenceAugust 27–29, 2014, Gothenburg, Sweden
Outline
• Motivation and background
• Isotropic model
• Transversely isotropic model
• Parameter estimation
• Results
• Conclusions and futuredevelopments
0 20 40 60 80−200
0
200
400
600
800
stress
[MPa]
80 0 20 40 60 800
0.005
0.01
0.015
0.02
Damage
EMMC-14, August 27–29, 2014 2
Motivation and background
Certain materials exhibit transversely isotropic symmetry
• unidirectionally reinforcedcomposites
• forged metals
– elasticity isotropic– fatigue properties transversely
isotropic
Figures from http://aciers.free.fr
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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, Grubisic and Simburger 1976, Papadopoulos 1997.
Cumulative damage theories.
A more fundamental approach using evolution equations.
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Continuum approach
Proposed by Ottosen, Stenstrom and Ristinmaa in 2008.
Endurance surface postulated as
β =1
σoe
(σ + AI1 − σoe),
where
σ =√
3J2(s − α) =√
32(s − α) : (s − α),
I1 = trσ.
Back stress and damage evolution eqs.
α = C(s − α)β,
D = g(β,D)β = K exp(Lβ)β.
I1
‖s‖
β > 0β < 0
β =0
β =0
α 6= 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 ,
whereσT = PσP , P = I −B,
where B = b ⊗ b is the structural tensor and b is the unit vector normal to thetransverse isotropy plane.
Integrity basis of a transversely isotropic solid
I1 = trσ, I2 =12tr (σ
2), I3 =13tr (σ
3), I4 = tr (σB), I5 = tr (σ2B).
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Endurance surfacePresent transversely isotropic formulation
β ={
σ +ALIL1
+ATIT1
−[
(1− ζ)ST+ ζS
L
]}
/ST= 0,
where
σ =√
3J2(s−α), IL1 = trσL = I4, IT1 = trσT = I1 − I4,
and
ζ =(
σL : σL
σ : σ
)n
=
(
2I5 − I242I2
)n
.
In uniaxial loading σ = σn⊗ n the ζ-factor has the form
ζ = (2 cos2ψ − cos4ψ)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 = 2n = 1
n = 0.5
ψ
ζ
0 π/8 π/4 3π/8 π/2
1
0.8
0.6
0.4
0.2
0
SL/ST = 1 dotted black line, 1.5 dashed blue line, 2 red lineAL = 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 theendurance surface
D =K
1−Dexp(Lβ)β, α = C(s−α)β.
time
∽
1
2
3
4
5σ1
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 anddamage C,K and L.
Data from tests with forged 34CrMo6 steel. Due to the lack of data in theintermediate directions we have chosen n = 1.
SL = 447MPa, ST = 360MPa, 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[M
Pa]
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102
103
104
105
106
0
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 0.5 1.0 1.5 2.0σxm/σxa
σxa(σ
xm)/σxa(0)
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!"
!"
0.5
0.6
0.7
0.8
0.9
1.0
0 0.5 1.0 1.5 2.0σym/σya
σya(σ
ym)/σya(0)
!"
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△ denotes experimental results from McDiarmid 1985 (34CrNiMo6), • model predictions
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Effect of mean shear stress
τxy = τxym + τxya sin(ωt)
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!"
!"
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!"
0.7
0.8
0.9
1.0
0 0.5 1.0 1.5 2.0τxym/τxya
τ xya(τ
xym)/τ x
ya(0)
100 1000 10000 500000
0.2
0.4
0.6
0.8
1 .0 1.0
NDamage
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Effect of phase shift
σx = σxm + σxa sin(ωt) σx = σxa sin(ωt)
σy = σxm + σxa sin(ωt − φy) τxy = 12σxa sin(ωt − φxy)
(a) (b)
0.6
0.7
0.8
0.9
1.0
1.1
0 45 90 135 180
φy
σxa(φ
y)/σxa(0)
0.9
1.0
1.1
1.2
1.3
0 30 60 90
φxy
σxa(φ
xy)/σxa(0)
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Effect of frequency differencemodel 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 = 12σxa 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(ω
xy)/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(φ
y)/σxa(0)
"#
"#
"#
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Conclusions and future developments
• Transversally isotropiccontinuum based HCF-model
• More tests needed
• Microstructurally basedanisotropic damage model
• Constitutive model with anisotropicdamage
• 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 =12tr (s
2), J4 = tr (sB), J5 = tr (s2B).
τmax(σT) =
√
J2 +14J
24 − J5, τmax(σL
) =√
J5 − J24 .
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