ENGN 2290: Plasticity Computational plasticity in Abaqus

Transcription

1 ENGN 229: Plasticity Computational plasticity in Abaqus The purpose of these exercises is to build a familiarity with using user-material subroutines (UMATs) in Abaqus/Standard. Abaqus/Standard is a finite-element program which implicitly solves the equations of equilibrium. Briefly, consider the strong form of the boundary-value problem on a domain : σ ij +b i = i in, σ ij n j = t i on t, u i = û i on u, where t and u are complementary subsurfaces of the surface ( t u =, t u = ), b i is the externally-applied body force, t i is the externally-applied surface traction on t, and û i are the prescribed displacements on u. We recast the strong form in the weak form by introducing a vectorial test function w i and writing ( σij +b i ) w i dv = for all test functions w i. Following standard arguments, we have ( ) ( ) σij (σij w i ) w i +b i w i dv = σ ij +b i w i dv (product rule) w i = σ ij w i n j da σ ij dv+ b i w i dv (divergence theorem) = =, t i w i da σ ij w i dv + b i w i dv (traction condition) which gives the weak form of the boundary-value problem w i σ ij dv = t i w i da+ b i w i dv. (1) In the finite-element method, the domain is approximated using finite elements, = e. The nodal solution variables are the displacements, which are interpolated inside each element by u i = u A i NA, (2) with the index A = 1,2,... denoting the nodes of the element, u A i the nodal displacements, and N A the shape functions. In a Galerkin approach, the test functions are interpolated by the same shape functions w i = w A i NA. () 1

2 Using (2) and () in (1), we obtain the following element-level system of equations: e N A σ ij dv = t i N A da+ b i N A dv. e e Abaqus/Standard solves this system of equations interactively using a Newton-Raphson procedure by defining the following residuals, e Ru A N A i = σ ij dv+ t i N A da+ b i N A dv, e e and using the corresponding tangents, K AB u i u k = RA u i u B k = e N A σ ij ǫ kl N B x l dv, where we have assumed that the traction and body force are fixed. The task of a UMAT is to calculate the stress σ ij and material tangent σ ij / ǫ ij at a given time so that Abaqus/ Standard may perform its Newton-Raphson iteration. For the time-discrete version of elastoplasticity, this amounts to calculating the stress σ n+1, updating internal variables, and calculating the algorithmically-consistent tangent C alg = σ n+1 / ǫ n+1 at a time t n+1. We undertook these tasks on the first exam for three-dimensional, small-deformation, rate-independent plasticity with isotropic hardening and a summary of these algorithms are included at the end of this document. Posted on Canvas, you will find a UMAT corresponding to this model with Vocé-type isotropic hardening, umat iso plasticity.for. The UMAT is written for three-dimensional and plane-strain problems; it is not intended for plane stress problems. There are also five input files, which you may use to familiarize yourself with the capability of the code. It is easiest to submit jobs using subroutines using the Abaqus Command interface. The command for submitting an Abaqus job is abaqus double inp=input filename.inp user=umat filename.for j=jobname inter 1. The first exercise is for simple shear of a single element, one shear.inp. The input file is set up to perform five load/unload cycles. Using this input file and the hardening parameters Y = 1 MPa, H = 125 MPa, and Y s = 25 MPa, calculate and plot the shear stress/strain relation, σ 2 versus γ 2. Your result should look like the following: 15 1 σ2 (MPa) γ 2 2

3 2. It is always good practice to exercise your code using a multi-element calculation involving inhomogeneous deformation. Such a calculation often exposes bugs not evident in a single element simulation. For this purpose, see the input file CurvedBar.inp. A schematic of this problem is shown below: Undeformed mesh Deformed mesh Moved face Fixed face The input file is set up to displace the surface denoted as the moved face above in the 2-direction through three cycles. Such a loading involves bending and torsion in different parts of the bar. The calculation result of the load applied to the face versus the displacement of the face in the 2-direction is shown below: 5 x 14 Force (N) Displacement (m). The next exercise involves tension/compression of a single element in the 1-direction, one ten.inp. In this input file, you may also specify a lateral normal stress σ 22. Consider a perfectly-plastic material, Y = 1 MPa, H = MPa, and Y s = 1 MPa. You may specify a value for the lateral normal stress σ 22 in the range from Y to Y and subsequently deform the element in tension or compression in the 1-direction. The value of σ 11 obtained from the calculation result then defines a point (σ 11,σ 22 ) in a plane-stress projection of stress space. Following this procedure, map out the plane-stress projection of the Mises yield surface. Your result should look like the following:

4 1.5 σ2/y σ 1 /Y 4. Next, consider the problem of plane-strain pure bending of a beam of thickness t, beam.inp. The input file is set up to impose a constant curvature deformation to the beam. Generate a plot of the the applied moment versus the bending angle. (The bending angle is simply the curvature times the length of the beam.) Your result should look like the following: 25 Moment (N-m) Angle (radians) The fully-plastic moment for a perfectly-plastic material in plane-strain is M p = 1 2 t2 Y, denoted as the dotted line above. Indeed, the calculated fully-plastic moment matches the analytical expression. 5. Finally, let us once again consider the problem of Hill indentation, HillIndent.inp. This input file is set up to perform plane-strain flat punch indentation of a perfectlyplastic material. (a) Verify that the calculated collapse load matches the analytical result of slip line theory P = kw(2+π). (b) In the *solid section command, change the material from the user material mat1 to built-in Abaqus perfect plasticity mat2, and verify that the result is the same. Open the.sta files for both jobs and compare the respective performances of the two calculations. 4

5 A plot of normalized indenter force P/kW versus the indent depth for both cases is given below: P/kW 2 UMAT plasticity 1 Abaqus plasticity Analytical solution Indent depth (m) The result of the UMAT is indistinguishable from the result of built-in Abaqus plasticity. Both calculation results asymptote to the analytical solution. 5

6 Summary of the constitutive equations for three-dimensional, small-deformation rate-independent plasticity with isotropic hardening 1. Kinematics: For small deformations, the strain is additively decomposed into elastic and plastic parts ǫ = ǫ e +ǫ p, with trǫ p = due to plastic incompressibility. 2. Elastic stress-strain relationship: The stress is given by σ = C[ǫ ǫ p ], where the fourth-order elasticity tensor, C, for an isotropic material given by C ijkl = G(δ ik δ jl +δ il δ jk )+ (K 2 ) G δ ij δ kl, with G and K denoting the shear and bulk moduli, respectively. The equivalent tensile stress (Mises stress) is then σ = 2 σ.. Flow rule: The evolution of ǫ p is given by ǫ p = 2 ǫ p N p, N p = σ 2 σ, ǫ p = 2 ǫp. 4. Flow strength: We introduce a stress-dimensioned flow strength with Y = Y( ǫ p ), Y() = Y, ǫ p (t) = t ǫ p (ζ)dζ denoting the equivalent tensile accumulated plastic strain. 5. Yield function: We introduce the yield function, defined as f = σ Y( ǫ p ). 6. Kuhn-Tucker complementary conditions: The values of the yield function f and ǫ p are constrained by f, ǫ p, f ǫ p =. 7. Consistency condition: The consistency condition, ǫ p f = when f =, determines the magnitude of ǫ p whenever it is non-zero. 6

7 Summary of the implicit time-integration procedure for three-dimensional, rate-independent plasticity with isotropic hardening Problem: Given: {ǫ n,ǫ p n, ǫ p n} at time t n, as well as ǫ n+1 at time t n+1 = t n + t Calculate: {σ n+1,ǫ p n+1, ǫp n+1 } 1. Calculate trial quantities σ tr = C[ǫ n+1 ǫ p n ], σ tr = 2 σ tr,, N p σ tr, tr =, 2 σ tr f tr = σ tr Y( ǫ p n). 2. Determine whether the step is elastic or plastic. if f tr then elastic step else plastic step end if. Elastic step: Plastic step: Solve σ n+1 = σ tr, ǫ p n+1 = ǫ p n, and ǫ p n+1 = ǫ p n, g( ǫ p ) = σ tr G ǫ p Y( ǫ p n + ǫp ) = for ǫ p and update σ n+1 = σ tr 6G ǫ p N p tr, ǫ p n+1 = ǫ p n + /2 ǫ p N p tr, ǫ p n+1 = ǫ p n + ǫp. 7

8 Algorithmically consistent tangent if f tr then elastic step C alg = C else plastic step end if C alg = C where 6G 2 G+H( ǫ p n+1) Np tr N p tr ǫ p6g2 σ tr (I N p tr N p tr) H( ǫ p n+1) = dy( ǫp ) d ǫ p ǫ p = ǫ p n+1 8