Introduction to materials modelling Lecture 4 - Deformation, strain - PowerPoint PPT Presentation

Introduction to materials modelling Lecture 4 - Deformation, strain Reijo Kouhia Tampere University, Structural Mechanics October 4, 2019 R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 1 / 13

Motion of a continuum body Balance Force Stress Constitutive equations Displacements Strains Kinematics B ∗ σ = f equilibrium σ = Cε constitutive model ε = Bu kinematical relation R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 2 / 13

Description of motion A material point has coordinates X in the undeformed state. After deformation it is moved to the place x . A mapping χ is called the motion x = χ ( X , t ) = X + u ( X , t ) , x i = χ i ( X , t ) = X i + u i ( X , t ) , and u is the displacement vector. X are the material coordinates. Frequently used in solid mechanics. x are the spatial coordinates. Much used in fluid mechanics. R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 3 / 13

Deformation gradient Deformation gradient F gives the change of an infinitesimal line element at P F = ∂ χ d x = F d X , ∂ X , Figure from G.Holzapfel: Nonlinear solid mechanics , p. 70 R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 4 / 13

Deformation gradient - cont’d In indicial notation d x i = F ij d X j , F ij = ∂χ i = ∂X i + ∂u i = δ ij + ∂u i . ∂X j ∂X j ∂X j ∂X j If there is no deformation, then F = I . It contains both strains and rigid body rotation and can be decomposed as (the polar decomposition) F = RU = VR , where R is orthogonal rotation tensor and U and V are the symmetric and positive definite right and left stretch tensors. R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 5 / 13

Definition of strain √ Length of a line element PQ is d S = d X · d X √ In deformed state | pq | = d s = d x · d x q u Q Q dS d u p u P P 2 [(d s ) 2 − (d S ) 2 ] = 1 1 2 (d x · d x − d X · d X ) 2 d X · ( F T F − I )d X = d X · E d X = 1 where E is the Green-Lagrange strain tensor. R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 6 / 13

Green-Lagrange strain tensor 2 ( F T F − I ) = 1 E = 1 2 ( C − I ) , where C = F T F is the right Cauchy-Green deformation tensor. E = 0 for pure rigid body rotation. G-L in terms of displacement � ∂ u � ∂ u � � T � � T E = 1 ∂ u + ∂ u ∂ X + 2 ∂ X ∂ X ∂ X If ∂ u /∂ X ≪ 1 , then � � T � E ≈ ε = 1 ∂ u � ∂ u ∂ x + , 2 ∂ x where ε is the infinitesimal strain tensor - notice x ≈ X . R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 7 / 13

Other strain tensors A general strain definition can be stated as E ( m ) = 1 m ( U m − I ) . m = 2 corresponds to the G-L strain tensor. The Hencky or logarithic strain tensor is obtained when m → 0 + m → 0 + E ( m ) = ln U . lim The Biot strain tensor for m = 1 E (1) = U − I . R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 8 / 13

Infinitesimal strain tensor Also known as the small strain tensor ε = sym grad u in index notation � ∂u i + ∂u j � ε ij = 1 2 ∂x j ∂x i Von K´ arm´ an notation 1 1   ε x 2 γ xy 2 γ xz 1 1 ε = 2 γ xy ε y 2 γ yz   1 1 2 γ xz 2 γ yz ε z R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 9 / 13

Strain in arbitrary direction Strain in direction n ( | n | = 1 ) ε n = n · ε n . Change in the angle between orthonormal vectors n and m γ nm = 2 n · ε m . R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 10 / 13

Principal strains Eigenvalues of the strain tensor ε n = λ n ( ε − λ I ) n = 0 Non-trivial solution for n if det( ε − λ I ) = 0 Characteristic polynomial − λ 3 + I ε 1 λ 2 + I ε 2 λ + I ε 3 = 0 where I ε 1 = tr ε = ε kk = ε 11 + ε 22 + ε 33 I ε 2 = 1 2 [tr( ε 2 ) − (tr ε ) 2 ] I ε 3 = det ε are called the principal invariants of the infinitesimal strain tensor. R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 11 / 13

Volumetric - isochoric split The strain tensor can be split into volumetric and isochoric i.e. volume preserving parts ε = 1 3 (tr ε ) I + e where tr ε = ε vol is the volumetric strain ε vol = V − V 0 . V 0 R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 12 / 13

Note on dual stress measure R. Kouhia (Tampere University, Structural Mechanics) Introduction to materials modelling October 4, 2019 13 / 13