Belytschko T. - Introduction (779635), страница 26

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To begin, we equate df written in terms of the Cauchy stress andthe nominal stress, Eqs. (3.4.2) and (3.4.3), givingdf = n ⋅σdΓ= n0 ⋅PdΓ0(3.4.6)Substituting the expression for normal n given by Nanson’s relation, (3.4.5) into(3.4.6) givesJn0 ⋅ F−1 ⋅σdΓ0 = n 0 ⋅PdΓ0(3.4.7)Since the above holds for all n0 , it follows thatP = JF−1 ⋅ σ or Pij = JFik−1σ kj or Pij = J∂Xiσ∂x k kjJσ = F ⋅P or Jσ ij = Fik Pkj(3.4.8a)(3.4.8b)It can be seen immediately from (3.4.8a) that P ≠ P T , i.e.

the nominal stresstensor is not symmetric. The balance of angular momentum, which gives theCauchy stress tensor to be symmetric, σ = σ T , is expressed asF ⋅P = PT ⋅ FT(3.4.9)The nominal stress can be related to the PK2 stress by multiplying Eq.(3.4.3) by F giving()df = F ⋅ (n0 ⋅S)dΓ0 = F⋅ ST ⋅n0 dΓ0 = F ⋅ST ⋅n 0dΓ03-33(3.4.10)T. Belytschko, Continuum Mechanics, December 16, 199834The above is somewhat confusing in tensor notation, so it is rewritten below inindicial notation()dfi = Fik n0j S jk dΓ0 = Fik SkjT n0j dΓ0(3.4.11)The force df in the above is now written in terms of the nominal stress using(3.4.2):df = n 0 ⋅PdΓ0 = PT ⋅ n 0dΓ0 = F⋅S T ⋅n 0 dΓ0(3.4.12)where the last equality is Eq.

(3.4.10) repeated. Since the above holds for all n0 ,we haveP = S⋅ F TorPij = Sik FkjT = Sik F jk(3.4.13)Taking the inverse transformation of (3.4.8a) and substituting into (3.4.13) givesσ = J −1F⋅S⋅ FTorσ ij = J −1 Fik Skl FljT(3.4.14a)The above relation can be inverted to express the PK2 stress in terms of theCauchy stress:S = JF −1 ⋅σ⋅F − TorSij = JFik−1σ klFlj−T(3.4.14b)The above relations between the PK2 stress and the Cauchy stress, like(3.4.8), depend only on the deformation gradient F and the Jacobian determinantJ = det( F) .

Thus, if the deformation is known, the state of stress can always beexpressed in terms of either the Cauchy stress σ , the nominal stress P or the PK2stress S. It can be seen from (3.4.14b) that if the Cauchy stress is symmetric, thenS is also symmetric: S = ST . The inverse relationships to (3.4.8) and (3.4.14) areeasily obtained by matrix manipulations.3.4.3.CorotationalStressandRate-of-Deformation. In someelements, particularly structural elements such as beams and shells, it isconvenient to use the Cauchy stress and rate-of-deformation in corotational form,in which all components are expressed in a coordinate system that rotates with theˆ , is also called the rotatedmaterial. The corotational Cauchy stress, denoted by σstress tensor (Dill p.

245). We will defer the details of how the rotation and therotation matrix R is obtained until we consider specific elements in Chapters 4and 9. For the present, we assume that we can somehow find a coordinate systemthat rotates with the material.The corotational components of the Cauchy stress and the corotationalrate-of-deformation are obtained by the standard transformation rule for secondorder tensors, Eq.(3.2.30):ˆ = R T ⋅ σ⋅R or σˆ ij = RikTσ kl Rljσ3-34(3.4.15a)T. Belytschko, Continuum Mechanics, December 16, 1998ˆ = R T ⋅D ⋅R or Dˆ = RT D RDijik kl lj35(3.4.15b)The corotational Cauchy stress tensor is the same tensor as the Cauchy stress, butit is expressed in terms of components in a coordinate system that rotates with thematerial.

Strictly speaking, from a theoretical viewpoint, a tensor is independentof the coordinate system in which its components are expressed. However, such afundamentasl view can get quite confusing in an introductory text, so we willsuperpose hats on the tensor whenever we are referring to its corotationalcomponents. The corotational rate-of-deformation is similarly related to the rateof-deformation.By expressing these tensors in a coordinate system that rotates with thematerial, it is easier to deal with structural elements and anisotropic materials.The corotational stress is sometimes called the unrotated stress, which seems likea contradictory name: the difference arises as to whether you consider the hattedcoordinate system to be moving with the material (or element) or whether youconsider it to be a fixed independent entity.

Both viewpoints are valid and thechoice is just a matter of preference. We prefer the corotational viewpointbecause it is easier to picture, see Example 4.?.Box 3.2Transformations of StressesCauchy StressσσPSˆσNominal StressP2nd PiolaKirchhoffStress SJ −1F ⋅PJ −1F ⋅S⋅ FTS⋅F TJF−1 ⋅σJF−1 ⋅σ ⋅F− TP ⋅F− TRT ⋅ σ⋅RJ −1U ⋅P⋅RNote: dx = F ⋅dX = R⋅U ⋅dX in deformation,U is the strectch tensor, see Sec.5?dx = R ⋅ dX = R⋅dxˆ in rotationJ −1U ⋅S⋅UCorotationalCauchyˆStress σˆ ⋅R TR⋅ σˆ ⋅RTJU −1 ⋅ σˆ ⋅U−1JU −1 ⋅ σExample 3.8 Consider the deformation given in Example 3.2, Eq.

(E3.2.1).Let the Cauchy stress in the initial state be given byσ 0σ (t = 0) =  x 00σy0 (E3.8.1)Consider the stress to be frozen into the material, so as the body rotates, the initialstress rotates also, as shown in Fig. 3.8.3-35T. Belytschko, Continuum Mechanics, December 16, 199836yyσ 0yσ 0xΩ0σ 0xσ 0yxΩxσ 0yσ 0xFigure 3.8.

Prestressed body rotated by 90˚.This corresponds to the behavior of an initial state of stress in a rotating solid,which will be explored further in Section 3.6 Evaluate the PK2 stress, thenominal stress and the corotational stress in the initial configuration and theconfiguration at t = π 2ω .In the initial state, F = I , soσ 0xˆS= P= σ =σ = 00σ 0y (E3.8.2)In the deformed configuration at t =π, the deformation gradient is given by2ωcosπ 2 −sinπ 2   0 −1F==, sinπ 2 cosπ 2  1 0 J =det( F) = 1(E3.8.3)Since the stress is considered frozen in the material, the stress state in the rotatedconfiguration is given byσ y0 0 σ=0 0 σx(E3.8.4)The nominal stress in the configuration is given by Box 3.2: 0 1 σ 0yP = JF σ = −1 0 0−1σ 0x 0   0= 0σ 0x  −σ y 0 3-36(E3.8.5)T.

Belytschko, Continuum Mechanics, December 16, 199837Note that the nominal stress is not symmetric. The 2nd Piola-Kirchhoff stress canbe expressed in terms of the nominal stress P by Box 3.2 as follows: 0σ x0  0S = P⋅F − T =  0 −σ y 0  1−1 σ 0x 0 =0   0 σ 0y (E3.8.6)Since the mapping in this case is a pure rotation, R = F , so when t =π2ωˆ =S., σThis example used the notion that an initial state of stress can beconsidered in a solid is frozen into the material and rotates with the solid. Itshowed that in a pure rotation, the PK2 stress is unchanged; thus the PK2 stressbehaves as if it were frozen into the material.

This can also be explained bynoting that the material coordinates rotate with the material and the componentsof the PK2 stress are related to the orientation of the material coordiantes. Thusin the previous example, the component S11 , which is associated with Xcomponents, corresponds to the σ 22 components of physical stress in the finalconfiguration and the components σ11 in the initial configuration. Theˆ are also unchanged by thecorotational components of the Cauchy stress σrotation of the material, and in the absence of deformation equal the componentsof the PK2 stress. If the motion were not a pure rotation, the corotational Cauchystress components would differ from the components of the PK2 stress in the finalconfiguration.The nominal stress at t = 1 is more difficult to interpret physically.

Thisstress is kind of an expatriate, living partially in the current configuration andpartially in the reference configuration. For this reason, it is often described as atwo-point tensor, with a leg in each configuration, the reference configuration andthe current configuration. The left leg is associated with the normal in thereference configuration, the right leg with a force on a surface element in thecurrent configuration, as seen from in its defintion, Eq.

(3.4.2). For this reasonand the lack of symmetry of the nominal stress P , it is seldom used in constitutiveequations. Its attractiveness lies in the simplicity of the momentum and finiteelement equations when expressed in terms of P .Example 3.9 Uniaxial Stress.Z,zyzY,yab0Ω0l0ba0X,xΩlFigure 3.9. Undeformed and current configurations of a body in a uniaxial state of stress.3-37xT.

Belytschko, Continuum Mechanics, December 16, 199838Consider a bar in a state of uniaxial stress as shown in Fig. 3.9. Relate thenominal stress and the PK2 stress to the uniaxial Cauchy stress. The initialdimensions (the dimensions of the bar in the reference configuration) are l0 , a0 andb0 , and the current dimensions are l, a sox=laX, y = Y,l0a0z=bZb0(E3.9.1)Therefore∂x ∂X ∂x ∂YF = ∂y ∂X ∂y ∂Y ∂z ∂X ∂z ∂YF0a a000 0 b b0 (E3.9.2)abla0b0 l0(E3.9.3) l0 l00 = 0a0 a0 0b0 b  0(E3.9.4)J =det ( F) =−1∂x ∂Z   l l0∂y ∂Z  =  0∂z ∂Z   0The state of stress is uniaxial with the x-component the only nonzero component,soσ x 0σ=0 0 0 0000(E3.9.5)Evaluating P as given by Box 3.2 using Eqs.

(E3.9.3-E3.9.5) then givesl 0 labl P=0a0b0 l0  00a0 a00 σ x0  0b0 b  0 abσ x0 0  a b0 00 0 =  00 0   00 00 00 0(E3.9.6)Thus the only nonzero component of the nominal stress isP11 =abAσ xσx =a0 b0A0(E3.9.7)where the last equality is based on the formulas for the cross-sectional area, A=aband A0 = a0 b0 ; Eq. (E3.9.7) agrees with Eq. (2.2.7). Thus, in a state of uniaxialstress, P11 corresponds to the engineering stress.3-38T. Belytschko, Continuum Mechanics, December 16, 199839The relationship between the PK2 stress and Cauchy stress for a uniaxialstate of stress is obtained by using Eqs.

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