Belytschko T. - Introduction (779635), страница 65
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(6.4.92) is written aswhere h is the plastic modulus, pij =epCijkl= λδ ijδ kl + µ(δ ikδ jl + δ ilδ jk ) −4µ2p ph + 3µ ij kl(E6.1.11)2µ, p = p , p = p , p = p and( h +3µ ) 1 11 2 22 3 12τ 1 = τ11 , τ 2 = τ22 , τ 3 = τ12 , the tangent stiffness matrix is obtained asUsing Voigt notation and letting γ = λ + 2µ[ C ] = 0 λtanab p12−2µγ p2 p1 p3 p1GeometricStiffnessEq. (6.3.55), i.e.,λλ + 2µ0p1 p2p22p3 p200 µp1 p3 4τ 1 02τ 3 1p2 p3 − 0 4τ 22τ3 22τ 3 2τ 3 τ 1 +τ 2 p32 (E6.1.12)Matrix. The geometric stiffness matrix is given by6-42T.
Belytschko & B. Moran, Solution Methods, December 16, 1998TK geoIJ = I2 × 2 ∫A B I σB J dA = I2× 2HIJ(E6.1.13)0From Eq. (E4.1.18)B =1 y232A x32y12 x21 y31x13(E6.1.15)Substituting Eq. (E6.1.15) into Eq. (E6.1.16) gives y 231 H=y2A 31 y12x32 σ xx σ xy 1 y23x13 σ xy σ yy 2 A x32x21 y31x13y12 x21 (E6.1.18)Assuming the integrand to be constant, the geometric stiffness matrix is obtainedby multiplying the integrand in Eq.
(E6.1.13) by A to giveK geoIJ =Kgeo14AHIJ I 2× 2H 11 01 H21=4A 0 H31 00H120H13H1100H22H1200H23H2100H32H 2200H33H310H 3200 H13 0 H23 0 H33 (E6.1.19)The geometric stiffness matrix is independent of material response and as can beseen from Eqs. (E6.1.18 - E6.1.19) depends only on the current stress rate and thegeometry of the element. The load stiffness matrix is given by the same equationas described for the rod.Example 6.2. Two-Node Rod Element.We now consider the two-node rod element under a state of uniaxialstress. The rod is assumed to lie along the xˆ −axis .
The only non-zero Cauchystress component is σˆ 11 ≡ σˆ x . The tangent stiffness and the external load matricesare derived in the updated Lagrangian form, i.e. in the current configuration. Wefirst reconsider the constitutive relation for the special case of uniaxial stress. Thesuperscript hats are dropped in the following for convenience.The Truesdell rate of the Cauchy stress is assumed to be given by Eq.(6.3.??)σJσ ij∇T = CijklDkl(E6.2.1)6-43T. Belytschko & B.
Moran, Solution Methods, December 16, 1998For the case of uniaxial stress, the only non-zero components of the rate ofdeformation tensor are D11 , D22 , and D33 .The uniaxial stress rate is therefore given byσJσJσJσ 11∇J = C1111D11 + C1122D22 + C1133D33(E6.2.2)The traction-free condition on the surface of the rod can be stated asσJσJJσ 22∇c = C2211D11 + C2222D22 +Cσ2233D33 = 0σJσJσJσ 33∇c = C3311D11 + C3333D22 + C3333D33 = 0(E6.2.3)If the rod is initially transversely isotropic (with the axis of symmetry coincidentσJσJJσJwith the x1 -axis) the tangent moduli are related by C1133= C1122and Cσ2222= C3333.Furthermore, uniaxial stressing in the direction of the axis of isotropy preservesthe transverse isotropy and these relations hold throughout the deformation.Solving Eq.
(6.2.3), with these assumptions we obtain SD22 = D33 ,D22 = −σJC2211σJJ D11C2222+ Cσ2233(E6.2.4)Using Eq. (6.2.4) for D22 and D23 in Eq. (E6.2.3) gives the uniaxial relationσ 11∇J = E σT D11or[ C ] = [E ]στστ(E6.2.5)where Etan is the tangent modulus associated with the state of uniaxial stress andis given byσTEσ T = C1111−σTσT2C2211C1122TTCσ2233+ Cσ2222(E6.2.6)Material Tangent Stiffness Matrix. The tangent stiffness matrix for arate-independent material is given by Eq. (6.4.18) in the current configurationwhich we write in the local coordinate system asT σTˆ mat = BK∫ ˆ Cˆ Bˆ dΩ(E6.2.7)ΩUsing the B matrix from Eq.
(E4.6.3) and Cσ J as given by Eq. (E6.2.5), weobtainˆ matK −1 10 σT 1= ∫ [ E ] [−1 0l1 l0 0 1+1 0]Aldξ6-44(E6.2.8)T. Belytschko & B. Moran, Solution Methods, December 16, 1998Here, the B matrix has been expanded to a 4 ×1 matrix by adding zeros to reflectthat the xˆ -component of the rate-of-deformation is independent of the transversevelocities. If we assume Eσ T is constant in the element, thenσTˆ mat = AEKl+10 −1 00 −1 00 0 00 +1 00 0 0(E6.2.9)This is identical to the linear stiffness matrix for a rod if Eσ T is replaced byYoung's modulus E. The global stiffness is given by Eq.
(4.5.42):ˆ matTK mat = TT K(E6.2.10a)where T is given by cos θ sin θ− sin θ cos θT =000 000 00 cos θ sin θ−sin θ cos θ (E6.2.10b) cos 2 θcosθ sinθsin 2 θsymmetric−cos 2 θ−cos θsin θ −cos θ sin θ−sin 2 θ (E6.2.11)cos 2 θcos θ sin θ sin 2 θ soK mat =AEσ Tlwhere the material constant Eσ T relates the Truesdell rate of the Cauchy stress tothe rate-of-deformation in a uniaxial state of stress.GeometricStiffnessMatrix.
The geometric stiffness is developed in acoordinate system that at time t coincides with the axis of the bar but is fixed intime. Note that since the coordinate system is fixed in the orientation shown inFig. ??, it is not a true corotational coordinate system, so the rotation correctionsof an objective rate must be considered. We will use the Truesdell rate. We couldalso consider the xˆ , ˆy coordinate system corotational and derive the geometricstiffness by accounting for the channge of the transformation matrix T in(E4.6.11) .
Many such derivations are given in Crisfield. The result should beidentical, since the same mechanical effect is represented, but the derivation isgenerally more difficult. The geometric stiffness matrix is given by Eq. (6.4.18):.ˆ =Hˆ IKIJIJTˆ = BH∫ ˆ σBˆ dΩ(E6.2.12)Ω6-45T. Belytschko & B. Moran, Solution Methods, December 16, 1998where the geometric stiffness has been expressed in the local coordinate systemfor simplicity.
Using the B matrix from Eq. (4.6.3), it follows that1 −11H = ∫ [σˆ x ] [−1 +1]dΩl +1lΩ(E6.2.13)Assuming that the stress is constant givesˆ x A +1 −1ˆ =σHl −1 +1(E6.2.14)Expanding the above, we obtain the geometric stiffness as given by Eq. (E6.2.12)ˆ geoK +1 0 −1 0 Aσˆ x 0 +1 0 −1=l −1 0 +1 0 0 −1 0 +1(E6.2.15)Use of the transformation formula, Eq. (4.5.42), shows that the geometric stiffnessis independent of the orientation of the beam.ˆ geo T = Kˆ geoK geo = TT K(E6.2.16)The total tangent stiffness is then given by the sum of the material and geometricstiffnessesK int = K mat + Kgeo(E6.2.17)The matrix is symmetric, which is a consequence of choosing a constitutiveequation in terms of the Truesdell rate of the convected rate of the Kirchhoffstress.
The matrix is positive definite as long as the initial stress is smallcompared to the tangent modulus. THE STIFFNESS FOR JAUMANN RATEAND THE EIGENVALUES OF K ARE LEFT AS EXERCIZES.LoadStiffness. The load stiffness for the rod is given by Eq. (??). We writeonly the nonzero terms noting N, iη = 0 and that x,η = y,η = 0 , since the shapefunction is only a function of ξ . For simplicity, we first evaluate it in thecorotational system, which givesˆ = 1 pN N 0KIJ ∫0I J , ξ − z, ηz,η dξ0 (E6.2.18)In the above, z,η can be taken to the width of the element a.
Using (???) givesˆ = 1 pN N l 0 1 adξKIJ∫0 I J , ξ −1 0(E6.2.19)6-46T. Belytschko & B. Moran, Solution Methods, December 16, 1998Let1− ξ 1HIJ = ∫ N I NJ , ξ dξ = ∫ [ −1 +1]dξξ l001=11 −1 1 2l −1 1(E6.2.20)(E6.2.21)Thenˆ ext = plaHKIJIJ(E6.2.22)Taking the shape functions (???) and substituting into the above givesˆ extK0 −1 0 1 1 0 −1 0pa =2 0 −1 0 1 1 0 −1 0(E6.2.23)The above matrix is also invariant with rotation, i.e.,ˆ ext T = Kˆ extK ext = T TK(E6.2.24)for forces and velocities expressed in any other Cartesian coordinate system.Material Tangent Stiffness Matrix in Total Lagrangian Form.The material tangent stiffness matrix for a rate-independent material is given byEq.
(6.4.18) in the reference configurationK mat =∫BTΩ0CSEBdΩ 0(E6.2.25)Using the B matrix from Eq. (E4.???) and CSE as given by Eq. (E6.2.5), weobtain−cos θ 1 −sin θ SE 1matK =∫ E[ −cos θl cosθ l00 0 sinθ 1[ ]− sin θ cos θ sin θ ] A0l 0dξIf we assume E SE is constant in the element, then6-47(E6.2.26)T. Belytschko & B. Moran, Solution Methods, December 16, 1998K mat =A0 ESEl0 cos 2 θcos θ sinθsin 2 θsymmetric−cos 2 θ−cos θsin θcos 2 θ−cos θ sin θ −sin 2 θ (E6.2.27)cosθ sinθ sin 2 θ where the material constant E SE relates the Truesdell rate of the Cauchy stress tothe rate-of-deformation in a uniaxial state of stress.