Transcript Training

Decomposition and Visualization of
Fourth-Order Elastic-Plastic Tensors
Alisa G. NeemanSC, Rebecca BrannonU,
Boris JeremicD, Allen Van GelderSC,
and Alex PangSC
7/6/2015
1
Driving Problem
How to visualize fourth-order tensor fields
representing a solid’s time-varying stiffness
(such as soil during an earthquake).
Two Stages:
1.
Meaningful decomposition to reduce number of
components per field location:
3x3x3x3
2.
6x6
3x3
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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?
Order
Notation
Zero
Represented
by
Scalar
First
1D array
vi
Second
2D array
Aij
Fourth
4D array
Eijkl
c
Symmetry: Aij = Aji
Eijkl = Ejikl = Eijlk = Ejilk = Eklij
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Solid Mechanics Vocabulary



Elastic:
recoverable deformation
Elastic-Plastic:
recoverable + permanent
deformation
Localized failure
4
Basic Constitutive Equation
Hooke’s 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
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axes x, y, z
Stiffness Changes With
Increasing Stress
Material Behavior
Elastic
Elastic-plastic
Failure, localized
Eijkl Properties
Positive-definite,
fully symmetric
Eigenvalues get smaller,
non-singular, may lose
major symmetry
Non-positive-definite,
possibly singular
Non-trivial to decompose (or visualize!)
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Approach
1.
Unroll the 3x3x3x3 tensor into a 6x6 matrix.
»
2.
Polar decomposition on the matrix produces
»
3.
5.
two 6 x 6 matrices; the rotation part and the
symmetric stretch part.
Eigen-decomposition on the stretch yielding
»
4.
Numerical methods only exist for matrices
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:
E = QS
where Q is a pure rotation matrix (orthonormal, positive
determinant) and S is a stretch (symmetric,
positive-semi-definite).
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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




As the solid deforms,
stress changes induced at
Gauss integration points
z
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
node
y
Gauss
point
x
Finite Element
(8 node brick) 12
Physical Meaning
Stress Modes
Isotropic, 3 equal
eigenvalues
Two equal eigenvalues One zero eigenvalue,
+ distinct eigenvalue with 2 others equal but
more compressive
opposite in sign
than the others
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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





Pressuredependent
Non-associated
flow
Stage 1:
induced hardening
Stage 2:
softening and
singularity
Difficult to correlate
stress to soil’s
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(σ):
delineates stress that
causes elastic versus
elastic-plastic
deformation.
Yield Surface:
is a convex isosurface
in stress space where
F(σ)= 0.
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Associated vs. Non-Associated



Plastic Flow G(σ): 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 application
http://neesforge.nees.org/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 = √S2 = T√J T
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Rotation with one zero column
(find C!)


Define C = QT
B = QST= C√J



Cj = Bj/√Jjj for j = 2,…,n



QST = QT √JTTT
but TTT = T-1T = I, since T is orthogonal
Remove √J 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 √2E1112 √2E1123 √2E1131
E2211 E2222 E2233 √2E2212 √2E2223 √2E2231
E3311 E3322 E3333 √2E3312 √2E3323 √2E3331
√2E1211 √2E1222 √2E1233 √2E1212 √2E1223 √2E1231
√2E2311 √2E2322 √2E2333 √2E2312 √2E2323 √2E2331
√2E3111 √2E3122 √2E3133 √2E3112 √2E3123 √2E3131
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