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WU/CCM-93/3
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Solution
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Problems
of the
Finite
Method
Stefan
i--
b-_
-J
(,9 _<. <]
O_u_C)
c") "_
t,,.'%
Stress Analysis
by the p-Version
Barna
to
_'_ m
Element
A. Szab6,
v
-J
"3
q_
1993
_J
"9
CENTER FOR
COMPUTATIONAL
-MECHANICS
•
L. Actis
_,
I[
I I.."_-1]
_
m
Ricardo
M. Holzer
November,
I
of Elastic-Plastic
WASI-fiNGTON UNIVERSITY
CAMPUS BOX 1129
ST. LOUIS, MO 63130
Prepared for
Lyndon B. Johnson
National Aeronautics
Space
and
Space Administration
Houston,
Texas 77058
Center
and
,#
j <.
Center
for
Computational
Washington
St.
Louis,
Report
SOLUTION
BY
OF
THE
Missouri
OF
P. and
Blanche
STRESS
THE
Barna
Albert
FINITE
Y. Greensfelder
Senior
Post-Doctoral
ELEMENT
PROBLEMS
METHOD
Professor
of Mechanics
L. Actis
Research
Stefan
ANALYSIS
A. Szab5
Ricardo
Associate
M. Holzer
Research
November,
Prepared
National
63130
WU/CCM-98/S
ELASTIC-PLASTIC
F-VERSION
Mechanics
University
Fellow
1993
for:
Lyndon
B. Johnson
Space Center
Aeronautics
and Space Administration
Houston,
TX 77058
TABLE
Abstract
ii
........................
Introduction
...........................
Formulation
of the mathematical
of the solution
Examples
algorithm
problem
matrix
matrix
matrix
..............
in the case
in the case
in the case
3
4
of of plane stress
. . 6
of of plane
strain
. . 7
of of axial symmetry
9
.................
10
...........................
Example
Example
Example
Example
References
ii
I
Assumptions
.........................
The elastic-plastic
compliance
The elastic-plastic
compliance
The elastic-plastic
compliance
Summary
CONTENTS
............................
Acknowledgement
Outline
OF
I:
2:
3:
4:
and
11
Plane strain
...................
Plane stress
...................
An axisymmetric
problem
.............
Limit load in the case of plane strain
conclusions
....................
...........................
........
II
15
17
20
23
24
-i-
ABSTRACT
The solution of small-strain elastic-plastic stress analysis problems by the pversion of the finite element method is discussed. The formulation is based on
the deformation theory of plasticity and the displacement
method.
Practical
realization
of controlling
discretization
errors for elastic-plastic
problems
focus of the paper.
Numerical
examples,
which include
comparisons
deformation
and incremental
theories
of plasticity
under tight control
tion errors, are presented.
is the main
between
the
of discretiza-
ACKNOWLEDGEMENT
This
National
work has been supported
by Lyndon
Aeronautics
and Space Administration
KEY
Plasticity,
estimation.
deformation
theory,
B. Johnson
Space
Center
under Grant NAG 9-622.
of the
WORDS
numerical
-ii-
analysis,
finite
element
method,
error
INTRODUCTION
This paper is concernedwith application of the p-version of the finite element
method to elastic-plastic stressanalysis problems with emphasison the deformation theory of plasticity. Our interest in
this subject
is motivated
by the following
considerations:
(1)
The effects
practical
by mathematical
models
is illustrated
the
plasticity
The
the
loads
ple,
(4)
of ductile
can
deformation
be well
theory
materials
represented
plasticity
[1], [2].
materials
is based
on
is generally
the
correlated
deformation
theory
of
susceptible
to Poisson
In the
known
ratio
conventional
is is used.
as mixed
locking
(h-version)
For this
methods,
must
and
hence
locking
reason
correct
occurs
alternative
be employed.
when
formula-
See,
for exam-
[5]).
tional
mathematical
initial
cost.
The
nonlinear
problems
errors
may
tion
errors:
without
with
not
finite
in the
be
element
refinement.
of linear
schemes,
this
important
the
because
solution
for controlling
can be increased
the
gains
multiplied
by
hence
which
uti-
discretizasubstantially
to use
in performance,
the
of
initial
process,
is well
because
case
the
p-version,
advantageous
well.
in the
The
suited
a capability
purpose
of iteration.
of freedom
are
have
at a low computa-
problems
throughout
spaces,
cycles
serves
is more
It is particularly
in nonlinear
unadapted
case
must
of nonlinearities
errors
course
of degrees
systems
of plasticity
adequate
in the
The number
mesh
p-distributions
than
accumulate
hierarchic
theory
of discretization
may
lizes
of real physical
for the effects
deformation
control
discretization
models
estimates
Adaptive
steps.
J-integral
formulation
generally
Realistic
pared
on the
effects
in strain-hardening
are obtained.
to provide
(5)
Such
made
examples.
The
is not
displacement
tious,
on structures
[3], [4].
p-version
limit
based
of cracks
J-integral.
event
importance.
by four
The propagation
with
(3)
overload
are of substantial
This
(2)
of a single
number
adaptive
as com-
of iteration
(6)
Prom
the
the point
the
deformation
The
p-version
1980's.
are much
of the
are
therein.
of implementation,
theory
Its theoretical
characteristics
listed
of view
the
(see,
for example,
been
to Linear problems,
smaller
than
fnite
element
basis
is now thoroughly
extensively
With
the data
method
documented.
exception
[7]), virtually
and
for the
well
developed
We
refer
requirements
incremental
became
of adaptive
to
and
to problems
during
its performance
[6] and
the
references
in fluid
applications
for
theory.
established
hp-extensions
all documented
particularly
storage
of the
dynamics
p-version
belonging
to the
have
following
two categories:
Category
A: The
exact
solution
is analytic
on the
entire
solution
domain
and
its
B: The
exact
solution
is analytic
on the
entire
solution
domain
and
its
boundaries.
Category
boundaries,
finite
are
with
number
called
The
nential
the
of points
singular
and
rates
in engineering
Certain
nonlinear
class
the
the
and
solved
smoothness
of the
case
behavior
The
of linear
presented
the
of the
deformation
by the
exact
(in three
solution
dimensions
is not
analytic
in this
paper.
compliance
axisymmetric
problems;
the
matrices
an outline
-2-
mesh
with
A and
theory
of plane
and
procedure
with
even
a broad
do not
perturb
Therefore
the
p-
the performance
the
B. This
same
and
four of
is based;
plane
as
expected
problems,
of plasticity
stress,
to _sx.
associated
In fact,
to a set of benchmark
cases
respect
to be substantially
in categories
normally
of the underlying
of plasticity,
problems.
expo-
meshes.
significantly.
of the algorithmic
paper.
of accuracy
of plasticity,
theory
deformation
for the
element
B because
nonlinearities
be expected
respect
range
element
such
belonging
A and
on the smoothness
solution
problems
with
finite
finite
incremental
confirmed
the
theory
for solving
on which
within
as material
can
assumptions
in this
the
in Categories
depends
p-version
elastic-plastic
presented
of points
where
simple
such
method
of the
has been
are
design
underlying
is an effective
in the
which
and
elasticity
characteristics
with
of nonlinearities,
of problems
version
points
can be achieved,
of the p-version
_sx
types
The
number
for problems
practice,
The effectiveness
solution
lines).
is effective
convergence
exact
of a finite
points.
p-version
expected
exception
the
strain
and
examples
are
FORMULATION
The cases
of plane
OF THE
stress,
plane
MATHEMATICAL
strain
and
two-dimensional
formulations,
are considered
are based
displacement
method.
on the
PROBLEM
axially
symmetric
problems,
in the
following.
The
that
is,
formulations
Notation.
The
components
of the displacement
uy = uy(z, y) The components
of the
vector
small
def
_ are denoted
strain
tensor,
by u. -- u.(z,y)
by definition,
are:
a t_
" = a'-'z"
(1,,)
,,.,a,,,,
(lb)
ey = a"'ydef
_ Ul
" = a'-;-
(10
(ld)
"" =_ \ ay +T= / "
In addition, '_-vd,_ _-e., will be used
of strain. The
e (resp.p). The
and
to represent
the
usual
engineering
definition
elastic(resp. plastic)strains will be indicated by the superscript
three principal strains are denoted by ex,e2,es. The
equivalent
elasticstrain is defined by:
e d,J V_ _/(,I - _)_ + (_ - _,_)_
+ (e3- ,;)_
(2,,)
2(1 + p)
where
v is Poisson's
ratio.
The equivalent
plastic
strain
is defined
by
_,da=V_
-]- _/(_ - 4) 2 + (4 -- 4) 2+ (4 --_)2
and
the
total
equivalent
strain
(_b)
is, by definition,
_' e + e.
The uniaxial
The
cipal
strain
stress
stress
at the
tensor
onset
of yielding
components
components
are
denoted
(2c)
is denoted
axe denoted
by _,,
by (71, _2, _s.
by er.
_y, #,,
r-u.
The equivalent
The
stress
three
prin-
is defined
by:
_. de|
= T _/('_ - "_)_+ (`'_- "s)_+ ('_ - '')_
-3-
(3)
The components of the stress deviator tensor are denoted by
_.,
_y, _,,
_.,.
By
definition,
1
(4a)
I
(4b)
1
=,. _sy
The
second
defined
invariant
+-.)
(4c)
(4d)
----1"zy
of the
stress
deviator
tensor
is denoted
by
J2 and
is
by:
J2 de_
1
=
In the
z instead
case
of axial
of =, y, z.
subscripts
are
symmetry
In the
"2
-2
+ ,;,.
the independent
one-dimensional
(5)
variables
case
(i.e.,
are denoted
uniaxial
stress
by r, O,
state)
the
omitted.
Assumptions.
The
based
assumptions
are
described
Assumption
The
and
its
in the
the
formulation
components
are much
boundary,
and
deformations
written
the
for the
equations
Assumption
undeformed
written
smaller
than
are
small
configuration
for the
deformed
total
strain
to Fig.
is the sum
1, in the
of the elastic
case
is
unity
on the
solution
in the
sense
that
are essentially
the
domain
equilibrium
same
as the
configuration.
E0 is the
secant
by Hooke's
modulus.
strain
of uniaxial
('+e=
stress
problem
2:
Referring
the
mathematical
1:
equilibrium
where
of the
following.
strain
equations
The
on which
Since
and
stress
the
state
plastic
strain.
the stress-strain
law is:
E,
the
law:
E
--4-
elastic
part
of the
strain
is related
to
where B
is the
modulus
of elasticity,
we have:
,--
f
1
/
Assumption
Ep
Fig.
1. Typical
values
of the
uniaxiai
stress-strain
curve.
3:
The
the
(6)
absolute
stress
tensor
components
stress
remain
tensor
components
in a fixed
are
proportion
non-decreasing
as
the
and
deformation
progresses.
Assumption
4:
The
plastic
Assumptions
experimental
uniaxial
strain
3 and
information
stress
state
tensor
is proportional
4 allow
generalization
is available
_ = 2v/3 and
hence
_'=]
In two-dimensional
to two
to the
stress
deviator
of the uniaxiai
and
three
dimensions.
eq. (6) can be written
_
stress
tensor.
state
for which
In the
case
of
as:
(Ta)
_.
problems:
_
=_
' .
-5-
'.
(Tb)
Rexnark:
In the
incremental
the following
tensor
respect
derivatives
_. and
assumption,
components
J2 with
theory
of plasticity
analogous
of the plastic
to the
of J2 with
strain
_=., respectively.
_,,
Thus,
one-dimensional
4, is made:
components
_,
yon Mises
are proportional
of the
yield
criterion
Increments
to the
first
stress
_, and r=, can be shown
of the
derivatives
tensor.
The
to be equal
of
first
to _,
_y,
for example;
de = d__
In the
on the
to Assumption
corresponding
respect
based
= d_,_..
case:
2
de" = de = dA # = .dA
3
hence:
3 dVp .
d_._= _ -g-_,,,.
Analogous
strain
The
relationships
hold
for the
plastic
(s)
increments
of each
component
of the
tensor.
elutie-plastie
Using
between
material
the
the
definition
plastic
compliance
of the
strain
and
matrix
stress
in the
deviator,
deviatoric
ease
given
stress
of plane
by (3),
stress.
and
the relationship
(Tb), we have:
{}/31 /r/31/3
il{}
"l_y
Using
and
(E} = {_'}+(eP}
the stress
a relationship
is obtained
r=y
between
the total
strain
components
tensor:
{i:
_/ffiy
0
oI [1-1/2}
0
0
E-E.
+ E.-'-T-
2(1+ v)
-6-
-1/2
" 1
_rU
0
0
,._
(9a)
The matrix
is readily
in the
brackets
invertible
is the elastic-plastic
to obtain
the
material
material
stiffness
compliance
matrix
which
matrix.
Remark:
Referring
equivalent
to Fig.
stress,
1 and
the
it can be easily
definitions
shown
of Eq.
(8),
in the
_
a
incremental
equivalent
plastic
strain
and
that
(E-E0)
EE,
In view
for the
_'
theory
of plasticity
the
equation
analogous
to (ga) is:
d%
= _
d'y. v
0
The elastic-plastic
In the
:
1
case
material
of plane
&ry
2(1 + _,)
drffiv
compliance
matrix
strain
ilia/1
+ T
%
0
0
in the case
rffiy
of plane
strain.
we have
_.=_:+e.=o
where
1
3{1
E)
3(_
})
(}_.
1
_1
)
Therefore:
1C1-},1-_,],..+..,
and
1
_. = a. - ] (a, + ay + a.)
=
=
_1
[1
_
1 [1_
__(1_
2v)] (a. + %) )
] [11_. 1
+6T1E. (I - 2_,1 _ffi-7--
g_-
(1 - 2L,) cry.
"
(_)
Similarly:
@, = --
The
elastic
strain
_ _-(1
components
v
e_------_ v (3
the
plastic
rE..1
of the
2v)
- _E (1--2v)
+ _-_-(I
stress
-
components
) v(_;
or=-
) _=-I-_ 1(
-
(1--2v)
1-_+_-_-{1v
rE,.
are:
)
(10a)
2_))_y
(10b)
_u
2(1+ v)
E
r_u
_:Y=
and
_= +
in terms
(_-_+_l-
_'-=_ 1
- 2v)
strain
(10c)
components
E-E.
[_
in terms
1E,
E - E.
of the
]
stress
E-E.
1 E. (1 - 2_) .=
components
[3
are:
1E,(1_2v)]o.
E_
_-_(I
(11a)
-
E-E.
Combining
ance
equations
matrix
can
be
(10a,b,c)
written
and
in
the
(11a,b,c),
the
elastic-plastic
material
form:
[c]_ c_2 c.
Oll
0
C12
0
o
0 ]
Css
where1
(
v
rE..1
2v)
)
0,2 = ---E (_ - 2-_(1- 2v))
+
+
-2v)
,-E.E_E..:
[_L._4E'IE"
(1- 2v)]
oss = 2(I+_) +sB -E •
E
To
obtain
the
elastic-plastic
EE.
material
stiffness
-8-
matrix,
[(7] is inverted.
]
compli-
The
elastic-plastic
In the
and
axial
by Hooke's
material
axisymmetric
strain
compliance
case
components
matrix
the
elastic
are related
in the cue
part
of the
to the
of axial
the
symmetry.
radial,
corresponding
circumferential
stress
components
law:
I
(_r
--
Vo'#
--
vo'z)
1
_ ffi
--/sO" r -_-O' 0 --
//O's)
1
_: =
The plastic
strain
components
are
--VO. r --
related
VO"e -_- O's).
to the
stress
by:
f2/31/31{r}
_._-_(_._)/-,_ .. -1_
/ _..
_
The elastic-plastic
L-1/s
compliance
matrix
[e]def=
C12
(711
-1/3
is of the
L_'
Cn
C12
C12
0
2/3.1
o.,
form:
Cz2
C11
C12
0
X
0 ]
C44
where:
Since
the elastic-plastic
be readily
computed
compliance
to obtain
the
matrix
has a special
elastic-plastic
-9-
stiffness
structure,
matrix.
its inverse
can
OUTLINE
Following
lems
The
based
is an outline
on the
procedure
tice
solution.
the
3.
Using
used
iteration.
the
5 percent,
in solving
the elastic-plastic
described
The
equivalent
iteration
from
the
In each
Gauss
stiffness
matrix.
> Ey in one
the
stress
in the
number
prob-
next
section.
is represented
strain
by observing
modulus
point
_ > er
is
prac-
or computing
the
Gauss
determine
matrices
points,
obtain
and
point
and
let
(#)(I) =
to each
Gauss
curve.
the stiffness
Gauss
norm
It is good
E_,h) corresponding
stress-strain
for which
in energy
1 percent.
r' in each
one-dimensional
or more
under
error
field.
secant
Recompute
relative
discretization
elastic
the
that
preferably
of the
in the
(_)(k), compute
point
ALGORITHM
of plasticity
Ensure
quality
of continuity
Compute
theory
as direct
under
to check
2.
SOLUTION
in brackets.
certainly
degree
4.
deformation
a linear
small,
THE
of the procedure
is known
by a superscript
1. Obtain
OF
the
elastic-plastic
for those
a new
material
elements
finite
for which
element
solution
_;+I).
5.
Using
E,(k) and _+i),
Gauss
point,
plastic
material
and
pute
the
each
part
plastic
Compute
Gauss
the
total
stress
strain
matrix.
of the
strain
the
total
stiffness
the elastic
corresponding
6.
using
compute
computed
Determine
material
the
components
from
the
stiffness
by subtracting
strain
tensor
elastic
matrix,
elastic
_+_)
and
strains
i.e.,
strain
{_(k+l))
in each
the
elastic-
from
Hooke's
components
{=(k+1))
law.
Com-
from
the
components.
equivalent
strains
(v')(h+l),
point
the following
(ff){k+_)
criterion
and
_{k+l) from
(2.a,b,c).
If in
is met:
[_k+1_ _ _h_I
where
_, is a pre-specified
E,(k+_}, increment/c
_/c+
tolerance,
1 and
return
-10-
then
to step
stop,
3.
else
using
z{h+l),
compute
EXAMPLES
The solutions
of representative
obtained
by application
of the
p-version
are
with
theory
compared
of plasticity
and
The boundary
tial)
traction
vector
Example
In
this
models
models
based
on the deformation
of plasticity
Model
2 was
program
obtained
solution
the
have
is I000,
domain
material
and
using
the
product
space
of degree
ment,
Inc.,
yield
tangen-
tangential)
element
Clayton
2 is based
criterion.
The
discretization
er-
Model
yield
I is
criterion,
on the
im-
by the
incremental
numerical
program,
mesh
are
element
shown
are applied,
stress
obtained
p is the span
Road,
alternative
was obtained
Model
relationship:
mesh
is a trademark
of the
solution
to be elastic-perfectly
points
to
are compared:
computer
conditions
were
attributable
control
The numerical
(v) is 0.3 the yield
quadrilateral
data
T, = 30, Tt = 0 and along
six-element
quadrature
7750
(reap.
solution
called
for
FEASIBLE
capabilities.
finite
CD
solutions
method
t PEGASY$
by the
incremental
(resp.
and the yon Mises
PEGASYSt.
the stress-strain
ratio
numerical
of Ganssian
the normal
response
of plasticity
boundary
characterize
on the standard
of the
of the normal
and
tight
material
yon Mises
is assumed
Poisson's
The
_)
computed
under
p-extension
BC T_ = 24, Tt = 0. Along
rameters
in terms
by an experimental
AB and DE symmetry
The
in
examined
theory
and
programs
The
solution
Tt).
in this paper.
analysis
theory
[8]. Both
and numerical
Results
h- and p-extensions.
u, (resp.
of elastic-plastic
as described
element
following.
by applications
described
T_ (reap.
are
Two
finite
are
differences
rors.
plemented
obtained
in the
Strain.
example
elastic-plastic
theory
obtained
component
component
1: Plane
results
solutions
vector
are solved
deformation
conditions
displacement
examples
shown
of the
that
is, _
= Tt = 0.
plastic.
Therefore
The modulus
(_r)
is 20.
in Fig.
2 and
the
set of monomials
of elasticity
of the
pa(E)
is unity.
Finite element
product
_',
Along
three
The thickness
by the p-version
2.
= Tt = 0. Along
EAT,
([_[ _< I, [7[ < I). For both
was fixed
in Figure
space.
The
i,y = 0, I,_,...
models
the number
at 14 × 14 for all p-levels.
o/" Eng/neer/ng
St. Louis,
-11-
Software
A_O 63117.
Research
and
Develop-
,p
C
E
---_X
10 10
Fig.
2. Solution
The number
of degrees
the finite
element
are given
in Table
1 for the
0.01
Table
and 6-element
of freedom
(IIrs)
and
linear
1. Estimated
for the
errors
due
to the
behavior.
In Model
energy
relative
It is seen
that
relative
error
linear
solution.
alternative
IIrs
error
1.
computed
from
in energy
norm
the numerical
error
at
7.13
1.80
0.44
0.12
0.03
0.01
0.O0
0.O0
CO
--40.1562351166
0.O0
in the
iterative
be considered
1 r_ ffi 0.001 was
used
-12-
solutions
negligible.
mathematical
eq.
were
Therefore
models
(see
norm
(e_)8
-39.9521346350
-40.1432844703
-40.1554486036
-40.1561743285
--40.1562305887
--40.1562347043
-40.1562350709
-40.1562351098
errors
can
in energy
16
56
120
208
320
456
616
800
CO
variations
for Model
potential
the estimated
.,V
1
2
3
4
5
6
7
8
the approximation
the
mesh
percent.
p
tolerances
(N),
solution.
for the
The
B
6.0
domain
solutions
p = 8 is less than
6.0
set so small
the
of elastic-plastic
(12)).
In Model
results
that
show
material
2 the
error
tolerance was set
tolerance
was
stiffness
current
the
residuals
after
load
the
current
the
equilibrium
where
iteration.
6r d,_ [6K]z-
iteration,
z is the
2.
perimeter
Circumferential
of the
0
circular
Model
0
22.5
45
67.5
9O
along
the
strain
hole.
1
1 and
2 are
perimeter
in
Table
relative
2.
compared
of the
and
in physical
the
es
+
the
hole.
(V
at
the
vector
in
and
By
Model
in Table
_
2
the
basis
properties
p-Convergence
perimeter
p
N
2
3
4
5
56
120
208
320
of the
0=0
0.1383
0.1346
0.1329
0.1340
used
for
of which
0=22.5
0.1112
0.1258
0.1268
0.1269
are
Model
effi45
0.0927
0.1013
0.1004
0.1012
-13-
not
alternative
in
between
greater
yield
are
computing
differences
circumferential
hole.
°
the
results
than
criteria
determined.
of the
circular
The
reference,
that
of plasticity
are
strain
(et)
sin 28
x-axis.
It is seen
theories
on
3.
2.
circumferential
definition:
positive
2 were
p=8.
di .
of the
cos 20 -
the
space,
+1.7
+1.9
+4.0
+4.9
-2.2
basis
e. - _
from
incremental
material
Table
for
shown
experiments
requisite
change
Relative
0.1318
0.1248
0.0989
0.0576
0.0362
on
circular
measured
results
differences
deformation
and
angle
The
the
the
(_,)
2
2
s is the
is
solution
Product
Model
0.1340
0.1272
0.1029
0.0604
0.0354
def
where
r, [6K]
current
degrees
Models
Specifically,
vector.
Table
at
in the
_< 1.0 -4
_(6r)T6r/r_r
matrix
is the
on
°
strain
1, product
e=67.5
0.0610
0.0593
0.0606
0.0604
°
(_t)
space.
0=90
0.0407
0.0372
0.0372
0.0366
°
listed
the
the
the
errors
are
tested
r
Control
tive.
of the
Letting
which
results
as in the
case
a circular
to uniform
criterion.
with
Models
herein
the
are
opening,
very
primarily
solution
Models
with
differences
to be minor
separate
I and
major
Galin's
because
caused
and
in the
2 are shown
axis of 3.05/ro,
in Fig.
1.64/ro
and
medium,
conditions,
solutions
the
and
domain
the solutions
is much
problem
ro is the radius
for the
of the circular
Model 1
Model
2
alin's Solution
2.0--
1.0--
Fig.
3. Contours
I
I
I
1.0
2.0
4.0
separating
the yielded
-14-
and
unyielded
Gaiin
this contour
Legend
-- ....
are
conditions.
hole.
4.0-
larger
in boundary
yield
materials
3. For Galin's
yield
in the solution
Tresca
unyielded
Tresca
sub-
as compared
differences
and
where
[7], [8]. Gaiin
on the
differences
yon Mises
the yielded
plastic
by the
therefore
same
and iteration
methods
is based
boundary
between
the
3.
elastic-perfectly
Nevertheless,
by the differences
lines which
is an ellipse
comparison
substantially
classical
solution
effec-
of quadrature
of discretization
in Table
in the
to be very
number
to obtain
using
Galin's
differences
the
likely
The contour
and
by Gaiin
was found
for the
control
in an infinite
is not possible.
circular
conditions
caused
2, strict
possible
are shown
at infinity.
are
values
tight
results
opening
there
I and
it was
was solved
stresses
Because
presented
than
problem
The
by p-extension
default
of the extremely
above.
A similar
jected
the
are p-dependent,
described
considered
errors
r, = 0.01, and using
points,
errors
discretization
regions.
Example
2: Plane
In this
strip
Stress.
problem
we consider
of a strain-hardening
results
of the numerical
by Theocaris
and
the
elastic-plastic
material
analysis
Marketos
response
to loading
of a thin
by enforced
are compared
with
those
perforated
displacements.
obtained
The
experimentally
in [11].
C
1.75
7.0
\
'"
21 0
Fig.
The
strip
vantage
is shown
of symmetry,
discretized
using
symmetry
three
was stress
free
the
The
part
smallest
corresponding
was
are in inch
one fourth
as shown
prescribed
units.
of the
in Fig.
4.
CD
Along
and
ad-
which
was
strip
AB
(u,, = Tt = 0); along
= A, Tt = 0); along
Taking
EA
and
BC
the
DE
normal
boundary
typical
of the strip
(_),
ultimate
slope
the
strength
Poisson's
stress
in uniaxial
values
of an aluminum
_urs
ratio
with
yield
strength
= 40, 000 psi.
The
modulus
plastic
tangent
v = 0.30, and
conditions
tension,
alloy
were
shown
assumed.
in Fig.
of the
linear
part
stress
at the
end of the
of stress
(E),
(_2 = 35,500
the
the
5, was characterized
slope
linear
psi)
and
of the
slope
constant
(_
strain
= 2e, ooo
(_2 = 0.0055)
to F_.
For each value
edge
are
Plane
curve
parameters:
and
(t_
and
= 3.2 x 106 psi.
strain-hardening
psi),
imposed
Notation.
dimensions
domain
were
is E = 9.956 x 106 psi,
_
All
elements,
properties
The stress-strain
by five
finite
_r = 34, 500 psi,
of elasticity
modulus
solution
strip.
(T, = Tt = 0).
The material
in tension
4.
conditions
(A) was
"-
4. Perforated
in Fig.
the
boundary
displacement
/A
of the imposed
and the maximum
displacement
strain
CA), the resultant
_, at the point
-15-
force
of first yield
F along
(which
the
occurs
Stress (ksi)
4O
3O
2O
I0
0
I
0
Fig.
at the
5. Example
edge
of the
for polynomial
The
trunk
element
(I I < 1,
hole)
were
space
curve
computed
8 using
I
the
in tension
from
trunk
of degree
;
0.01
0.012
based
the finite
space
(also
p is defined
on
by the
monomials
_'n,
0.014
on five parameters.
element
solution
known
the
as the
standard
< 1) as the span of the set of monomiais
+ Y -< p, augmented
obtained
'serendipity'
quadrilateral
_n_', _,j = O,l, 2,. . . ,p,
_np for p _> 2 and
by the monomiai
_
1.
The
in energy
criterion
'
0.006
o.oos
Strain
0.004
2: Stress-strain
degree
space).
for p=
0.002
number
norm
for the
Table
of degrees
of the
of freedom
starting
nonlinear
(linear)
solution
4. Results
for the
0.0050
0.0100
0.0125
0.0150
0.0175
0.0200
0.0225
0.0250
0.0275
0.0300
The results
of the
analysis
was
was
211.
solution
The
was
estimated
relative
0.23 percent
The
error
stopping
set at r, = 0.001 (see eq. (12)).
perforated
0.217
0.433
0.541
0.645
0.744
0.836
0.917
0.981
1.027
1.055
for various
-16--
strip
shown
in Fig.
4.
0.469
0.948
1.251
1.686
2.188
2.659
3.273
4.064
5.269
6.862
values
of the imposed
displacement
are
presented
which
in Table
were
normalized
the
4 and in Fig.
extracted
stress,
yield
6. Figure
by reading
=Av/_Y,
the
is defined
6 also includes
values
from
the
as the
ratio
between
the
normalized
strain
strain
provided
the
in [10].
The
average
stress
and
strain
e= and
the
F
(7}-
yield
plots
results
strength:
_AV
and
the experimental
strain
(_y/E).
reaching
is defined
The plastic
a value
Amin
as the
region
ratio
is confined
of approximately
Normalized
O'y
between
the
up to the maximum
normalized
4.0.
Stress
1.2
[]
1.0
0.8
(3
o/
0.6
0.4
0.2
PEGASYS
Theocaris
1
0.0
0.0
f
1.0
2.0
I
3.0
_
4.0
5.0
Normalized
Fig.
6.
Perforated
It is seen
observed
in Fig.
strain.
explanation
involve
Example
some
6 that
Other
is that
an infinitesimal
strip:
investigators
gauge
length
length
=: An _etric
In this
spherical
The results
problem
pressure
of the
strain
of the numerical
of finite
I
6.0
7.0
vs. Maximal
8.0
normalized
is larger
reported
wheres
[
Strain
stress
the computed
in the case
gauge
Average
et al
similar
solution
experimentally
than
strain
the experimentally
discrepancies.
the strain
determined
A possible
is reported
strains
with
invariably
size.
problem.
we
vessel
analysis
consider
with
the
elastic-plastic
a cylindrical
for an elastic
nozzle
- perfectly
-17-
behaviour
under
uniform
plastic
of a thin-walled
internal
material
pressure.
are compared
with those obtained experimentally by Dinno and Gill in [12], and
Zienkiewicz
numerically
by
in [13].
The generating
section
are shown
in Fig.
are
of a steel
typical
of elasticity
7.
and
the finite
All dimensions
alloy
with
yield
element
are
mesh,
in inch
strength
E = 29.12 x 10e psi, Poiseon's
consisting
units.
The
in tension
ratio
of 14 elements,
material
properties
_r = 40, 540 psi, modulus
v = 0.30, and
zero
strain
hardening
(,_, =0).
-.--I
1
"ill"
m
"aiD.
.-Jim.
2.8125
0.125
CO
P_,j
O3
IP"-
,4
A
0.25
0.545
I
8.687
_1
=_
r
Fig.
7. Spherical
Uniform
vessel.
The
constraints
The
of point
uz=0
I
pressure
pressure
(T,, = -p,
external
surface
vessel.
objectives
of the
was
stress
analysis
A (= A) for a range
and
A sequence
The
to determine
of linear
estimated
in Fig.
section
imposed
free
were
(T,
and mesh.
on the
inner
ffi Tt = 0).
surface
The
7.
the
of pressure
values
which
cause
the
shape
of the
resulting
plastic
by p-extension
using
size and
error
was
obtained
in energy
-18-
of the
displacement
to determine
solutions
relative
Generating
Tt = 0) was
(u, ffi u, = 0) are indicated
extensively
space.
Ur
=1"
norm
of the
vertical
the
finite
displacement
vessel
element
to
yield
zone.
the
trunk
solution
Table 5. Results for the pressure vessel (Example
Pressure
Displ.
(psi)
u_
(in)
760
Internal
3).
7.02 × I0-s
900
8.51 x 10-s
I000
10.78 x I0 -s
1080
15.34 x I0 -s
1120
19.66 x 10 -s
1140
22.45
1160
27.91
x 10 -s
1180
35.65
× I0 -s
1200
51.44
× 10 -s
Pressure
x 10 -s
(psi)
1400
1200
m
' I
1000
800
600
400
/
PEGASYS
200
0
0.0
10.0
Vertical
Fig.
8. Example
3: Internal
r
20.0
[]
Dinno and Gill
"1-
Zienkiewicz
r
30.0
Deflection
pressure
I
40.0
60.0
60.0
of A (x 1000 in)
vs.
axial
displacement
=A.
at polynomial degree of 8 (trunk space), was 1.0 percent. There were 1056 degrees
of freedom. The nonlinear analysis was performed
at p-level 8 with the stopping
criterionr= = 0.01 (see eq. (12)). The results of the analysis for various values of
the internalpressure are presented in Table 5 and in Fig. 8 which also includes the
experimental and the finiteelements resultsgiven in [12].It is worth noting that
the plasticzone spreads over the entiresection of the nozzle-sphere intersectionfor
values of p ___
9oo psi (see Fig. 9). Good
agreement
with the experimental results
was obtained even for high values of pressure. The boundaries of the plastic zone
for various values of the applied pressure axe shown
in Fig. 9. These
substantiallythe same as those presented in reference [13].
-19-
results axe
1080
100o
900
760
760
Fig.
9.
Example
Example
3: Boundaries
4: Limit
load in the cue
In reference
lutions
the
based
fully
limit
on the
by substantial
plastic
based
on the
displacement
same
occurs.
This
point
p-extensions,
mulation
rate
is discussed
however
described
occurs
as the
[6], [14].
in this
The
paper
pressure
cases
constant
detail
following
too
have
values.
often
volume.
load
in
the
at all.
In h-extensions
constraints
of freedom,
in [6] also.
Locking
grow
hence
load
that
when
at
locking
does not occur
demonstrates
limit
so-
exceed
limit
of degrees
correct
element
stiff" response
no
volume
example
the
finite
solutions
at a constant
number
will give
that
much
element
the
in some
observed
in some
formulation
for various
exhibit
finite
and
deformation
or comparable
Rice
formulation
amounts
is because
zone
=train.
and
Consequently,
This
the
Parks
displacement
range.
plastic
of plane
[5] Nagtega_l,
plastic
load
of the
in
the for-
p-extension
is used.
The most
the
The
limit
challenging
load for a deep
example
1/gth
of the
The
follows:
example
double-edge-notch
is challenging
crack
solution
size,
0); on segment
BC
because
hence
domain
On segments
presented
AB
uniform
the
the
crack
is shown
and
EA
normal
in reference
(DEN)
crack
plane
is very
tip singularity
in Fig.
symmetry
10.
-2o-
strain
deep,
computation
tensile
the
of
specimen.
ligament
is only
is strong.
The
conditions
displacement
[5] is the
boundary
are
is imposed
conditions
prescribed
(=,
= 6/2,
are
(_
as
= _ =
_
= o);
D
C
Ligament
1.0
E
0.1
A
--,
3.0
Fig.
segments
CD
and DE
Poisson's
ratio
were
The finite
gular
point
checked
tions.
1.0 and
The smoothness
indicator
of mesh
space
at
final
load
degrees
case
With
used.
was
Net
space
across
of elasticity
was
toward
the
of the
mesh
of degrees
final
the
was
computa-
boundaries
load
of freedom
case,
was
sin-
quality
nonlinear
interelement
of the
and
1.0 also.
progression
course
exception
was used
modulus
In addition,
in the
number
4.
stress
in geometric
contours
the
The
the product
of freedom
contours
The
the yield
of 0.15.
of the stress
quality.
p = s was
factor
stress
Example
(T, = T, = 0).
was graded
grading
the
free
domain.
0.3 respectively,
mesh
the
by plotting
Solution
are stress
element
using
I0.
B
>
the
2143.
at p = s. The corresponding
is an
trunk
In the
number
of
4183.
Stress/Yield
Strength
3.5
3.0
2.5
2.0
1.5
1.0
PEGASYS
0.5
Theoretical
0.0
0
Limit
,
I
'
t
'
r
'
2
4
6
8
10
12
14
16
Normalized Displacement
Fig.
11. Example
For the
the
ligament
4. Normalized
yon Mises
is given
yield
average
condition
the
by _t_,, = (2 + ,r)_r/v_
-21-
stress
limit
vs. normalized
load
_ 2.97_y.
in terms
The
displacement.
of the
normalized
net stress
stress
on
(i.e.,
the average
stress
displacement
where
in Fig.
w is the
theoretical
on the ligament
limit
width
load
11.
The
of the
divided
normalized
strip
is reached
(in this
by _r is plotted
displacement
example
at high values
-22-
of the
against
is defined
tp = 2.0).
imposed
the normalized
by
E6/(_rw)
It is seen
displacement.
that
the
SUMMARY AND CONCLUSIONS
The deformation theory of
havior
under
certain
restrictive
On comparing
fully
are
controlled
smaller
potheses
than
examples
not
of the
deformation
theory
based
While
the
with
that
theory
are
Budiansky's
of the general
and
be said
within
soundness
physical
with
that
incremental
other
the differences
hy-
the
natural
vari-
the
plastic
zone
was smooth.
exper-
reference
results
from
a few
those
obtained
than
solutions
obtained
by means
reported
experiments
by other
theories
not
[2].
In fact,
the
of the
[1]
may
be
violation
differences
the
be
White
without
between
in any
and
not
of plasticity
loading
distinguish
of plasticity
would
by Hodge
proportional
theory"
could
theories
in a care-
numerical
of results
"deformation
of a plasticity
experiments
the
paper.
alternative
solution
of the
be-
theory.
consistent
paths
and
In the
accuracy
that
that
experiment
a few percent
conclusions
observation
of loading
the
were
general
results
used for a range
it can
theory
to test
conclusion:
2 to 4 the
on the incremental
of course
warranted,
small
Examples
nevertheless
designed
and the exact
same
in this
incremental
In this
zone
described
relationships
properties.
to the
were
of elastic-plastic
1, it was found
constitutive
point
model
the
experiments
by an elastic
under
with
Example
in physical
known,
investigators
theory
material
The
described
which
elastic-plastic
confined
was
and
errors
completely
iments
assumptions
experiment,
in elastic-plastic
other
is a useful
deformation
numerical
concerning
ations
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the
plasticity
are
so
deformation
examples
discussed
in this paper.
The formulation
on the
displacement
ceptible
to Poisson
It has
extensions
elastic-plastic
cremental
been
are
of the elastic-plastic
method.
ratio
locking.
demonstrated
effective
material
theory
This
even
for
problem
is possible
The correct
for the
in cases
limit
errors
Similar
where
because
deformation
controlling
behavior.
described
results
unloading
-23-
of
in this paper
p-extensions
loads
are
theory
resulted
been
not
sus-
obtained.
of plasticity
discretization
have
are
is based
that
associated
obtained
in reverse
for the
plasticity
pwith
in[15].
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