Introduction to materials modelling Lecture 10 - Anisotropic - - PowerPoint PPT Presentation

Introduction to materials modelling

Lecture 10 - Anisotropic plasticity, continuum damage mechanics Reijo Kouhia

Tampere University, Structural Mechanics

November 13, 2019

• R. Kouhia (Tampere University, Structural Mechanics)

Introduction to materials modelling November 13, 2019 1 / 7

Anisotropic yield conditions

1

Transverse isotropy: f(trσ, trσ2, trσ3, tr(σM ), tr(σ2M )) = 0 where M = m ⊗ m is the structural tensor for transverse isotropy.

2

Orthotropy: f(tr(σM 1), tr(σM 2), tr(σM 3), tr(σ2M 1), tr(σ2M 2), tr(σ2M 3), trσ3) = 0 where M i = mi ⊗ mi are the structural tensors for orthotropy.

• R. Kouhia (Tampere University, Structural Mechanics)

Introduction to materials modelling November 13, 2019 2 / 7

Transversely isotropic yield condition

The general form f(trσ, trσ2, trσ3, tr(σM ), tr(σ2M )) = 0 can also be written in terms of the deviatoric stress s f(trσ, trs2, trs3, tr(sM ), tr(s2M )) = 0 If the yield is independent of hydrostatic stress then f(trs2, trs3, tr(sM ), tr(s2M )) = 0 Defining J2 = 1

2trs2,

J4 = tr(sM ), J5 = tr(s2M ), JT = J2 + 1

4J4 − J5,

JL = J5 − J2

4

then the maximum shear stresses in the transverse isotropy plane τT and in a plane containing the longitudinal axis τL are τT =

• JT,

τL =

• JL
• R. Kouhia (Tampere University, Structural Mechanics)

Introduction to materials modelling November 13, 2019 3 / 7

Orthotropic yield condition

Orthotropic von Mises-Hill type yield condition which does not depend on hydrostatic stress f(tr(sM 1), tr(sM 2), tr(sM 3), tr(s2M 1), tr(s2M 2), tr(s2M 3)) = 0 A classic formulation (Hill 1948, 1950) is based on the quadratic form (coordinate axes coincide to the

• rthotropy directions)

˜ sT P˜ s − 1 = 0 where ˜ s =         s11 s22 s33 s12 s13 s23         , P =         F + G −F −G −F F + H −H −G −H G + H 2L 2M 2N         i.e. F(s11 − s22)2 + G(s11 − s33)2 + H(s22 − s33)2 + 2Ls2

12 + 2Ms2 13 + 2Ns2 23 − 1 = 0

• R. Kouhia (Tampere University, Structural Mechanics)

Introduction to materials modelling November 13, 2019 4 / 7

Orthotropic yield condition (cont’d)

The six parameters F, G, H, L, M and N can be solved from the uniaxial yield strengths in the

• rthotropy directions and from the three shear strengths

F = 1

2(σ−2 y1 + σ−2 y2 − σ−2 y3 ),

G = 1

2(σ−2 y1 + σ−2 y3 − σ−2 y2 ),

H = 1

2(σ−2 y2 + σ−2 y3 − σ−2 y1 )

and L = 1

2τ −2 y12,

M = 1

2τ −2 y13,

N = 1

2τ −2 y23

• R. Kouhia (Tampere University, Structural Mechanics)

Introduction to materials modelling November 13, 2019 5 / 7

Continuum damage mechanics - introduction

In 1958 Kachanov introduced a model to describe continuous degradation of a material. He introcuded a single variable - damage index or integrity - which continuosly reduces the elastic properties σ = φC eεe For the evolution of the integrity φ, he proposed the following kinetic law (modified by Rabotnov 1959) ˙ φ = − A φp σ φ n , σ = φEε For an undamaged material φ = 1 and at fully damaged state φ = 0. Often used with the damage variable D = 1 − φ, which at undamaged state D = 0 and at fully damaged state D = 1.

• R. Kouhia (Tampere University, Structural Mechanics)

Introduction to materials modelling November 13, 2019 6 / 7

Continuum damage mechanics (cont’d)

The notion of an effective stress which is related to the surface that effectively resists the load (A − AD) ˜ σ = F A − AD Definition of the damage variable D = AD A Thus ˜ σ = F A(1 − AD/A) = σ 1 − D Strain equivalence principle (J. Lemaitre 1971): “Any strain constitutive equation for a damaged material may be derived in the same way as for a virgin material except that the usual stress is replaced by the effective stress”

• R. Kouhia (Tampere University, Structural Mechanics)

Introduction to materials modelling November 13, 2019 7 / 7