Decomposition and Visualization of FourthOrder ElasticPlastic Tensors

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Decomposition and Visualization of Fourth-Order
Elastic-Plastic Tensors

• Alisa G. NeemanSC, Rebecca BrannonU, Boris
  JeremicD, Allen Van GelderSC, and Alex PangSC

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Driving Problem
How to visualize fourth-order tensor fields
representing a solids time-varying stiffness
(such as soil during an earthquake).

• Two Stages
• Meaningful decomposition to reduce number of
  components per field location3x3x3x3
  6x6 3x3 1
• Meaningful visualization to find locations of
  softening and mode of stress to which the
  material is vulnerable.

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What is a Fourth-Order Tensor?
Symmetry Aij Aji Eijkl Ejikl Eijlk
Ejilk Eklij
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Solid Mechanics Vocabulary

• Elastic recoverable deformation
• Elastic-Plasticrecoverable permanent
  deformation
• Localized failure

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Basic Constitutive Equation
Hookes Law generalized to 3D (what gets solved
by the simulator) stress
increment 4th-order elastic-plastic
tangent stiffness strain increment i,
j, k, l range from 1 to 3, and represent
orthogonal spatial axes x, y, z
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Stiffness Changes WithIncreasing Stress
Non-trivial to decompose (or visualize!)
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Approach

• Unroll the 3x3x3x3 tensor into a 6x6 matrix.
• Numerical methods only exist for matrices
• Polar decomposition on the matrix produces
• two 6 x 6 matrices the rotation part and
  thesymmetric stretch part.
• Eigen-decomposition on the stretch yielding
• 6 (6-element) eigenvectors and 6 real
  eigenvalues.
• Select a single eigenvector and compose it into a
  symmetric 3 x 3 tensor.
• Visualize the second-order eigentensor with a
  glyph that shows its structure and also reflects
  its eigenvalue.

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Application Constraints

• Small Deformation Theory
• Small displacement gradients
• Stiffness cast with respect to reference
  configuration changes associated with rigid
  material rotation eliminated from consideration
• Second-order tensors must be symmetric
• Constrains fourth-order tensors to exhibit minor
  symmetry, i.e. Eijkl Ejikl Ejilk Eijlk
• Natural materials exhibit minor symmetry

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Polar Decomposition
Uniquely separates a matrix into two
components where Q is a pure rotation matrix
(orthonormal, positive determinant) and S is a
stretch (symmetric, positive-semi-definite).
E QS
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Polar Stretch and Mode of Vulnerability

• Eigen-decomposition applied to stretch S yields
  6 second-order eigentensors and eigenvalues.
• A reduced eigenvalue means the solid is less
  stiff and the associated eigentensor is the mode
  of stress for which the solid has the most
  reduction in stiffness.

terminology may vary
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Eigentensor Glyphs
Eigentensor glyphs drawn by stretching a unit
sphere according to the formula
n
where n is a unit length direction vector from
the center of the sphere to a point on the surface
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Glyphs and Finite Elements
node
Gauss point

• As the solid deforms, stress changes induced at
  Gauss integration points
• Irregular layout on X, Y, and Z
• 1-2 orders of magnitude difference in element
  sizes
• Glyph scale factor distance between Gauss
  points in single element

y
x
z
Finite Element (8 node brick)
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Physical Meaning
Stress Modes
Two equal eigenvalues distinct eigenvalue more
compressive than the others
One zero eigenvalue, with 2 others equal but
opposite in sign
Isotropic, 3 equal eigenvalues
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Experiments

• Stage 1
• Soil self-weight compression (Z), 25 steps
• Induces deformation and may change stiffness
• Stage 2
• Two point loads, 100 steps
• Small Z component (0.9659 kN) and
• Large X component (1294 kN)

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Deformation From Experiments
Black arrows indicate point load locations
Variation due to self weight
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Material 1 Drucker-Prager Soil Model

• Fails under tension
• Non-hardening (no change with compression)
• Stage 1 No change
• Stage 2 Experienced compression in front of
  point loads and tension behind

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Material 2 Dafalias-Manzari Soil Model

• Pressure-dependent
• Non-associated flow
• Stage 1 induced hardening
• Stage 2softening and singularity
• Difficult to correlate stress to soils behavior

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What About the Polar Rotation?

• We need a little more solid mechanics vocabulary

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Material Model Vocabulary

• Yield Function F(s)delineates stress that
  causes elastic versus elastic-plastic
  deformation.
• Yield Surface is a convex isosurface in stress
  space where F(s) 0.

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Associated vs. Non-Associated

• Plastic Flow G(s) determines plastic strain
  increment vectors. If the material is associated,
  the plastic flow is along the normal to the yield
  surface.
• Non-associated the plastic flow can diverge from
  the yield surface normal.
• This is where the stiffness tensor loses
  symmetry!

In the absence of elastic-plastic coupling
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Polar Rotation and Elastic-Plastic Materials

• We conjecture that rotation Q quantifies the
  misalignment between yield surface normal and the
  plastic flow in non-associated materials.
• Early results look promising on simulated and
  measured data.

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Summary

• Original Goals
• Meaningful decomposition method to facilitate
  visualization of 4th-order tensors
• Meaningful visualization technique
• Results
• First application of polar decomposition to
  analyze stiffness
• Technique meaningful within subset of solid
  mechanics simulations

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Where to Get More Information

• Supplementary Materials
• How to unroll a 3x3x3x3 tensor into 6x6
• Algorithm for polar decomposition in the face of
  a near-zero eigenvalue
• NEESforge source code repository for VEES
  visualization applicationhttp//neesforge.nees.or
  g/projects/vees/

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Thanks!!

• This work funded by a GAANN fellowship and the
  UCSC/Los Alamos Institute for Scalable Scientific
  Data Management (ISSDM)
• Thanks for your attention!

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Plastic Flow Along the Normal to the Yield Surface

• Mathematically, it means that the plastic strain
  increment tensor is a positive scalar multiple of
  the yield surface normal
• where the yield surface normal is itself a
  positive scalar multiple of the derivative of the
  yield function with respect to stress
• holding internal variables constant.

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Polar Decomposition With Zero/Near-Zero
Eigenvalue
E QS for

• Square Matrix
• Non-negative determinant
• If one zero eigenvalue, decomposition is unique
  with specification that
• det(Q) 1

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Stretch

• Define M ETE STS S2
• M T J TT
• Eigen-decomposition
• J diagonal, ascending order
• S vS2 TvJ T

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Rotation with one zero column (find C!)

• Define C QT
• B QST CvJ
• QST QT vJTTT
• but TTT T-1T I, since T is orthogonal
• Cj Bj/vJjj for j 2,,n
• Remove vJ from non-zero columns
• C1 to calculate, use Gramm-Schidt completion of
  6-D orthonormal basis
• Q CTT
• TT T-1

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More accurate stretch

• S sym(Q-1 E)

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Unrolling with Mandel Constants

• E1111 E1122 E1133 v2E1112
  v2E1123 v2E1131
• E2211 E2222 E2233 v2E2212
  v2E2223 v2E2231
• E3311 E3322 E3333 v2E3312 v2E3323
  v2E3331
• v2E1211 v2E1222 v2E1233 v2E1212 v2E1223 v2E1231
• v2E2311 v2E2322 v2E2333 v2E2312 v2E2323 v2E2331
• v2E3111 v2E3122 v2E3133 v2E3112 v2E3123 v2E3131