Belytschko T. - Introduction (779635), страница 25

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Recall that F is given in Eq. (E3.5.4)), from which we obtainππ˙F =  A,t c − 2 As −B, t s − 2 Bc ,ππ A,t s+ 2 Ac B,t c − 2 Bs L = ˙F ⋅F− 1 =21 BacAB + Abs2 cs( Ba− Ab)F−1 =1 AB − AsBcBs Accs( Ba − Ab )  π  0+ Bas2 + Abc 2  2 1−1 0 (E3.5.9)(E3.5.10)The first term on the RHS is the rate-of-deformation since it is the ymmetric partof the velocity gradient, while the second term is the spin, which is skewsymmetric. The rate-of-deformation at t = 1 is given byAb 0 b + ab1 1D = AB 0 Ba = 1+ a + b + ab  00 a + ab (E3.5.11)Thus, while in the intermediate stages, the shear velocity-strains are nonzero, inthe configuration at t = 1 only the elongational velocity-strains are nonzero.

Forcomparison, the rate of the Green strain at t = 1 is given by3-27T. Belytschko, Continuum Mechanics, December 16, 19982˙ =  Aa 0  = a + aE 0 Bb  00 b + b2 28(E3.5.12)Example 3.6 An element is rotated by an angleθ about the origin. Evaluatethe infinitesimal strain (often called the linear strain).For a pure rotation, the motion is given by (3.2.20), x = R ⋅ X, where thetranslation has been dropped and R is given in Eq.(3.2.25), so x  cosθ =  y  sin θ− sinθ  X  cosθ  Y  ux  cos θ − 1 −sin θ  X u  =  cos θ −1 Y  y   sinθ(E3.6.1)In the definition of the linear strain tensor, the spatial coordinates with respect towhich the derivatives are taken are not specified.

We take them with respect tothe material coordinates (the result is the same if we choose the spatialcoordinates). The infinitesimal strains are then given byεx =∂ux= cos θ − 1,∂Xεy =∂uy=cos θ − 1,∂Y2ε xy =∂ux ∂u y+=0∂Y ∂X(E3.6.2)Thus, if θ is large, the extensional strains do not vanish.

Therefore, the linearstrain tensor cannot be used for large deformation problems, i.e. in geometricallynonlinear problems.A question that often arises is how large the rotations can be before anonlinear analysis is required. The previous example provides some guidance tothis choice. The magnitude of the strains predicted in (E3.6.2) are an indication ofthe error due to the small strain assumption. To get a better handle on this error,we expand cos θ in a Taylor’s series and substitute into (E3.6.2), which givesθ2θ24ε x = cosθ − 1=1 −+O θ −1≈ −22( )(3.3.23)This shows that the error in the linear strain is second order in the rotation.

Theadequacy of a linear analysis then hinges on how large an error can be toleratedand the magnitudes of the strains of interest. If the strains of interest are of order10 −2 , and 1% error is acceptable (it almost always is) then the rotations can be oforder 10 −2 , since the error due to the small strain assumption is of order 10 −4 . Ifthe strains of interest are smaller, the acceptable rotations are smaller: for strainsof order 10 −4 , the rotations should be of order 10 −3 for 1% error.

Theseguidelines assume that the equilibrium solution is stable, i.e. that buckling is notpossible. When buckling is possible, measures which can properly account forlarge deformations should be used or a stability analysis as described in Chapter 6should be performed.3-28T. Belytschko, Continuum Mechanics, December 16, 1998a129b132451Fig. 3.7. An element which is sheared, followed by an extension in the y-direction and thensubjected to deformations so that it is returned to its initial configuration.Example 3.7 An element is deformed through the stages shown in Fig.

3.7.The deformations between these stages are linear functions of time. Evaluate therate-of-deformation tensor D in each of these stages and obtain the time integralof the rate-of-deformation for the complete cycle of deformation ending in theundeformed configuration.Each stage of the deformation is assumed to occur over a unit timeinterval, so for stage n, t = n −1 . The time scaling is irrelevant to the results, andwe adopt this particular scaling to simplify the algebra.

The results would beidentical with any other scaling. The deformation function that takes state 1 tostate 2 isx ( X,t ) = X + atY,y( X,t ) = Y 0 ≤ t ≤1(E3.7.1)To determine the rate-of-deformation, we will use Eq. (3.3.18), L = ˙F ⋅F− 1 so we˙ and F −1 . These arefirst have to determine F, F1 at  ˙ 0F=, F=0 1 0a−1  1 −at ,F=0 0 1 (E3.7.2)The velocity gradient and rate of deformation are then given by (3.3.10):0L = ˙F ⋅F− 1 = 0a  10 0−at  0=1  0a0 aT11,D=L+L=22 a 0  (E3.7.3)0()Thus the rate-of-deformation is a pure shear, for both elongational componentsvanish.

The Green strain is obtained by Eq. (3.3.5), its rate by taking the timederivativeat  ˙ 1  00E = 12 F T ⋅F− I = 21 2 2 , E=2 at a t a()a 2a2t (E3.7.4)The Green strain and its rate include an elongational component, E22 which isabsent in the rate-of-deformation tensor. This component is small when theconstant a, and hence the magnitude of the shear, is small.3-29T. Belytschko, Continuum Mechanics, December 16, 199830For the subsequent stages of deformation, we only give the motion, thedeformation gradient, its inverse and rate and the rate-of-deformation and Greenstrain tensors.configuration 2 to configuration 3x ( X,t ) = X + aY,y ( X, t) = (1 +bt )Y, 1≤ t ≤ 2, t = t −1a  ˙ 0 0 1+bt1−11 F=,F=,F=1+ bt  0 0 b 0 1+bt 01L = ˙F ⋅F− 1 = 1+bt0(E3.7.5a)−a 1 (E3.7.5b)0 0 01, D = 12 (L +LT ) = 1+btb0 b(E3.7.5c)a00 ˙0E = 12 (F T ⋅F− I) = 21 , E = 21 2 a a + bt (bt + 2 ) 0 2b( bt +1) (E3.7.5d)configuration 3 to configuration 4:x ( X,t ) = X + a(1− t )Y,1F=0y( X, t) = (1+ b)Y , 2 ≤ t ≤ 3, t = t −2a(1− t )  ˙ 0 −a1+ b a(t −1)1, F =, F −1 = 1+b1+ b 1 0 0  0(E3.7.6b) 0 −a  0 −a 1L = ˙F ⋅F− 1 = 1+1 b , D = 21 L + LT = 2 (1+b)0 0 −a 0 ((E3.7.6a))(E3.7.6c)configuration 4 to configuration 5:x ( X,t ) = X,y ( X,t ) = (1+ b − bt )Y , 3 ≤ t ≤ 4, t = t −3(E3.7.7a)01 ˙ 0 0 1+ b − bt 0 −11F=,F=,F=1+ b− bt 0 −b10 1+b − bt  0(E3.7.7b)01L = ˙F ⋅F− 1 = 1+ b−bt 0(E3.7.7c)0, D= L−bThe Green strain in configuration 5 vanishes, since at t = 4 the deformationgradient is the unit tensor, F = I .

The time integral of the rate-of-deformation isgiven by4∫ D(t )dt =0120 0 a  0 0 −a   01 a 0  + 0 ln (1 + b ) + 2 (1+ b) − a 0  +  00−ln ( 1+b )(E3.7.8a)3-30T. Belytschko, Continuum Mechanics, December 16, 19980ab = 2(1+b ) 11031(E3.7.8b)Thus the integral of the rate-of-deformation over a cycle ending in theinitial configuration does not vanish. In other words, while the final configurationin this problem is the undeformed configuration so that a measure of strain shouldvanish, the integral of the rate-of-deformation is nonzero. This has significantrepercussions on the range of applicability of hypoelastic formulations to bedescribed in Sections 5? and 5?. It also means that the integral of the rate-ofdeformation is not a good measure of total strain.

It should be noted the integralover the cycle is close enough to zero for engineering purposes whenever a or bare small. The error in the strain is second order in the deformation, which meansit is negligible as long as the strains are of order 10-2. The integral of the Greenstrain rate, on the other hand, will vanish in this cycle, since it is the timederivative of the Green strain E, which vanishes in the final undeformed state.3 .4 STRESS MEASURES3.4.1 Definitions of Stresses.

In nonlinear problems, various stressmeasures can be defined. We will consider three measures of stress:1. the Cauchy stress σ ,2. the nominal stress tensor P;3. the second Piola-Kirchhoff (PK2) stress tensor S.The definitions of the first three stress tensors are given in Box 3.1.3-31T. Belytschko, Continuum Mechanics, December 16, 199832Box 3.1Definition of Stress Measuresnn0dfdΓF -1 dfdfdΓοΩοΩreferenceconfigurationcurrentconfigurationCauchy stress: n⋅σdΓ = df = tdΓ(3.4.1)Nominal stress: n0 ⋅PdΓ0 = df = t 0dΓ0(3.4.2)2nd Piola-Kirchhoff stress: n 0 ⋅SdΓ0 = F −1 ⋅df = F −1 ⋅t0 dΓ0(3.4.3)df = tdΓ = t 0 dΓ 0(3.4.4)The expression for the traction in terms of the Cauchy stress, Eq.

(3.4.1),is called Cauchy’s law or sometimes the Cauchy hypothesis. It involves thenormal to the current surface and the traction (force/unit area) on the currentsurface. For this reason, the Cauchy stress is often called the physical stress ortrue stress. For example, the trace of the Cauchy stress, trace(σ) = −pI , gives thetrue pressure p commonly used in fluid mechanics. The traces of the stressmeasures P and S do not give the true pressure because they are referred to theundeformed area. We will use the convention that the normal components of theCauchy stress are positive in tension. The Cauchy stress tensor is symmetric, i.e.σ T = σ , which we shall see follows from the conservation of angular momentum.The definition of the nominal stress P is similar to that of the Cauchystress except that it is expressed in terms of the area and normal of the referencesurface, i.e. the underformed surface.

It will be shown in Section 3.6.3 that thenominal stress is not symmetric. The transpose of the nominal stress is called thefirst Piola-Kirchhoff stress. (The nomenclature used by different authors fornominal stress and first Piola-Kirchhoff stress is contradictory; Truesdell and Noll(1965), Ogden (1984), Marsden and Hughes (1983) use the definition given here,Malvern (1969) calls P the first Piola-Kirchhoff stress.) Since P is notsymmetric, it is important to note that in the definition given in Eq. (3.4.2), thenormal is to the left of the tensor P.3-32T. Belytschko, Continuum Mechanics, December 16, 199833The second Piola-Kirchhoff stress is defined by Eq.

(3.4.3). It differs fromP in that the force is shifted by F −1 . This shift has a definite purpose: it makesthe second Piola-Kirchhoff stress symmetric and as we shall see, conjugate to therate of the Green strain in the sense of power. This stress measure is widely usedfor path-independent materials such as rubber. We will use the abbreviations PK1and PK2 stress for the first and second Piola-Kirchhoff stress, respectively.3.4.2 Transformation Between Stresses.The different stress tensors areinterrelated by functions of the deformation. The relations between the stressesare given in Box 3.2.

These relations can be obtained by using Eqs. (1-3) alongwith Nanson’s relation (p.169, Malvern(1969)) which relates the current normal tothe reference normal byndΓ = Jn 0 ⋅ F−1dΓ0nidΓ = Jn 0j Fji−1dΓ0(3.4.5)Note that the nought is placed wherever it is convenient: “0” and “e” haveinvariant meaning in this book and can appear as subscripts or superscripts!To illustrate how the transformations between different stress measures areobtained, we will develop an expression for the nominal stress in terms of theCauchy stress.

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