On analysis of viscoelastic structures. Kim, Ju-Eon. Calhoun: The NPS Institutional Archive 1987

On analysis of viscoelastic structures. Kim, Ju-Eon. Calhoun: The NPS Institutional Archive 1987
Calhoun: The NPS Institutional Archive
Theses and Dissertations
Thesis Collection
1987
On analysis of viscoelastic structures.
Kim, Ju-Eon.
http://hdl.handle.net/10945/22393
OHOOL
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NAVAL POSTGRADUATE SCHOOL
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THESIS
ON ANALYSIS OF VISCOELASTIC STRUCTURES
by
Kim
,
Ju-Eon
September 1987
The; sis
Advisor:
Ramesh Kolar
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Viscoelastic Structures Problems
ABSTRACT (Continue on
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if
neceuary and identify by bicxk
num^r)
The thesis begins with a comprehensive formulation of continuum-based finite
element theory.
The theoretical portion concludes with the details of both
spatial and temporal discretization, including a discussion of nonlinearity
This study is to understand the effects of nonlinear behavior of structures
using a finite element method.
The nonlinear behavior equations are derived from equations of motion and
constitutive equations.
The basic theory is the principle of virtual work.
in
particular, large displacement problems and visco-elastic problems remain a
challenging engineering problems today.
The viscoelastic problems depend on the
relaxation function which is the source of material nonlinearities
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On Analysis of Viscoelastic Structure
by
/-Kim, Ju-Eon
Captain, Republic of Korea Air Force
B.S., Korea Air Force Academy, Seoul, 1980
B.S., Seoul University, Seoul, 1983
Submitted in partial fulfillment of the
requirements for the degree of
MASTER OF SCIENCE IN AERONAUTICAL ENGINEERING
from the
NAVAL POSTGRADUATE SCHOOL
»
September 1987
.
.
ABSTRACT
This study
is to
understand the
effects of nonlinear
behavior of structures using a finite element method.
The
nonlinear
equations of
behavior
equations
derived
are
constitutive equations.
motion and
from
The basic
theory is the principle of virtual work.
The thesis begins with
continuum-based
finite
with
portion concludes
temporal
a
comprehensive
element
the
discretization,
The theoretical
theory.
details
both
of
including
formulation of
a
spatial and
discussion
of
nonlinearity
In particular,
elastic problems
today.
large displacement
remain a
The viscoelastic
problems and visco-
challenging engineering problems
problems depend
on the relaxation
function which is the source of material nonlinearities
...
.
.
TABLE OF CONTENTS
I
II
.
.
INTRODUCTION
A.
ANALYSIS OF STRUCTURES
7
B.
LITERATURE REVIEW
7
C.
THESIS OVERVIEW
OVERVIEW
12
B.
NONLINEAR CONTINUUM EQUATIONS
12
.
.
Principle of Virtual Work
12
One-Dimensional Finite Element
15
C.
CONSISTENT NODAL FORCES
17
D
SOLUTION METHODS
21
E.
IV
12
A.
2.
.
11
THEORETICAL FORMULATION
1
III
7
1
Incremental Method
21
2
Newton-Raphson Method
22
3.
Modified Newton-Raphson Method
24
ANALYTIC SOLUTIONS
24
1
Elastic Materials
26
2.
Visco-Elastic Materials
30
PROGRAM IMPLEMENTATION
35
A.
OVERVIEW
35
B.
SOLUTION PROCEDURES
36
1
Algorithm Implementation
36
2.
Flow Chart of Finite Element Program
38
NUMERICAL EXAMPLES
39
4
V.
A.
ELASTIC MATERIALS
39
B.
VISCO-ELASTIC MATERIALS
45
CONCLUSION
48
LIST OF REFERENCES
49
INITIAL DISTRIBUTION LIST
59
ACKNOWLEDGEMENTS
I
am
appreciative
provided to me
by
faculty members
I
would
of
Professor
the
assistance
Ramesh
Kolar
and
and
guidance
the other
at the Naval Postgraduate School.
also like
to thank
the R.O.K.
Air Force for
allowing the opportunity to study.
I
am
greatful
to
my wife, Mi-Seon and my daughters,
Myeon-Joa and Nam-Lyang with whom God has blessed me.
INTRODUCTION
I.
A.
ANALYSIS OF STRUCTURES
structures subjected to large displacements
Analysis of
requires the inclusion
nonlinear
of
terms
the strain
in
In the formulation of the problem,
displacement relations.
either total Lagrangian or Eulerian approach is adopted.
In
order to assure structural integrity for loads beyond linear
range,
nonlinear analysis is very much mandated.
Another class of problems
of structures
problems
-
vary with time at
where stresses
in the components
deformation state
constant
a
viscoelastic behavior
is the
or strains
vary with time at constant stress states.
problems
These
as
applied
simple one-dimensional
to
structures will be addressed in this research work.
follows is
end
the
a
brief survey of literature in this area.
this
of
In what
chapter,
overview
of
At
thesis
the
is
described.
B.
LITERATURE REVIEW
A method for nonlinear static and dynamic analysis using
cable and beam
Formulation
large
elements
allows
for
deformations.
comparison
of
finite
is
given
material
Gadala
element
and
by
Bergan
et
formulations
[2]
[1].
and very
nonlinearities
Oravas
al
present
based
a
on the
Lagrangian, updated Lagrangian and Eulerian formulations for
nonlinear mechanics problems.
Henriksen [3] applies
to
a
single integral constitutive law
viscoelastic
describe
Ghoneim et al [4] treat the
viscoelastic and
behavior
his
in
strain
total
formulation.
rate
Pinsky and
use variational approach for studying the nonlinear
Kim [5]
viscoelastic
shell as
behavior
shell
representing
by
multi-director field.
a
within individual layers, the
transverse
strain
normal
Saigel and Yang present
describe
the layered
The formulation includes,
effects
layer thickness coordinate [6,
and
sum of
viscoplastic contributions in implementing
the constitutive equations in their formulation,
and
as
to
transverse shear
of
arbitrary orders in the
7].
quadrilateral
a
implementation
shell element
elastic-viscoplatic
of
constitutive law for large deformation.
A large collection
of papers on the application of ADINA may be found in [8].
Response
nonlinear
of
mechanical
excitation is given by To [9].
with thermoviscoplastic
discussed
Allen
contacting
methods
bodies
are
given
deformations of
on
to
random
Thermomechanical response of
uniaxial bars
by
systems
[10].
Large
the
contact
consideration
composite beams
constitutive laws is
deformations
[11,
the
and solution
surfaces
in
of
12].
Large
subjected to axial dynamic
load is reported in [13].
Based
on
Eulerian-Lagrangian
formulation,
frictional
contact analysis is discussed by Haber et al [14].
Halleux and Casadei [6]
analysis
structures
of
Polar-decomposition
derived tangent
based
used
is
biquadratic
on
arriving
in
large strain
at
elements.
explicitly
stiffness matrices
for a
frame element by
method
based
on co-rotational
Kondoh and Atluri [15].
coordinate
present transient
where
system,
system attached to
A
element
each
cartesian coordinate
local
the
translate
and
rotate is
described by Mattiasson [16].
based
problems
Inelastic
equations is given by Nagtegaal
rate
during
assumed
is
an
incremental virtual work
on
A
[17].
increment
constant strain
in integrating the
constitutive equations.
curved
Planar
beams
undergoing
rotations
large
is
analyzed using Helinger-Reissner mixed variational principle
by
Noor
Steele
[18].
discusses
several nonlinearities
including buckling, creep, and nonlinear elastic problems in
[19].
plastic
The use
model
Hirashima et al
of Lagrangian
is
presented
[21]
formulation using
an elasto-
by Voyiadjis and Navaee [20].
present
an
approach
for
the snap-
through analysis of curved shell panels and determination of
initial stress matrix.
Importance
pellet
are
of
viscoelastic
investigated
by
relaxation
Basombrio
[22],
observed that the duration of the ramp input
response considerably.
effects
on
a
where it is
influences the
Viscoplastic constitutive models are
obtained explicitly, without resorting
directly implemented
to inversion,
to be
finite element formulation [23].
in a
An intrinsic time scale that can vary with time and position
together
with
means
to
control
integration in the approximation
of
constitutive
is
used,
the numerical
model by
Chambers and Becker [24].
discuss
applications of visco-
plastic models in the metal-forming.
Elasto-plastic models
Chandra et
are
used
al
26]
[25,
analyzing
in
unilateral
problems by Cheng et al [12].
contact
Chung
et
al
[27]
suggest
varying parameters in guiding solutions for the
methods for
simulation of plastic forming processes.
propose
friction
and
formulation
for
et
al
[28]
path-dependent materials based on
arbitrary Lagrangian Euler method
plastic wave
Liu
and
include
an elastic-
propagation problem.
Rheinboldt and Riks
describe
several
algorithms
based on
continuation method and mesh refinements for the solution of
nonlinear finite element
propose
an
approach
equations.
to
solve
Ryu
nonlinear
and
Arora [29]
element
finite
equations based on substructures concept.
Several
plastic
[10,
7,
papers
behavior
4,
30,
pertaining
and
their
31].
10
to
visco-elastic
,
visco-
implementation is reported in
C.
THESIS OVERVIEW
The objective of this
thesis is
to review
some of the
literature available in the analysis of large deformation of
structures,
develop
to
displacements and
structures,
the
the
of
and
incremental
load
method.
large
of one dimensional
to understand the physics of
nonlinearities
theoretical formulation based on
include
to
viscoelastic behavior
and in the process,
effects
theory
Chapter
.
principle of
II
presents
virtual work
Descretization based on a
simplex element is given.
Chapter III
procedure.
gives the
algorithms used
in the solution
Full Newton-Raphson and Modified Newton-Raphson
methods are described.
Chapter
IV
describes
some
test
cases
and numerical
results obtained.
Finally, Chapter V provides conclusions and some remarks
for future work in this area.
11
THEORETICAL FORMULATION
II.
OVERVIEW
A.
chapter
this
In
we
consider
a
nonlinear
continuum
This approach is based
equations for a rod element.
on the
The resulting equations are
principle of virtual work [48].
in convenient form for numerical solution via finite element
methods [49].
course,
of
discretized-solid
the
J T Oden et
.
This,
al
.
[50]
the conceptual basis of
is
finite
introduced
elements,
by
in the elastic structures and nonlinear
continua.
B.
NONLINEAR CONTINUUM EQUATIONS
Discuss a rod element with the
It
may
proved
be
that
external virtual work.
equilibrium if
is equal
and only
the
to
total
total Lagrangian Method.
internal virtual work is equal to
Namely
a
system
deformable
is in
if the total external virtual work
internal
virtual
work
for every
virtual displacement consistent with the constraints.
1.
Principl e of Virtual Work
In this section we prove that total internal virtual
work
is equal to total external virtual work.
concept of
is
minimum potential energy [51].
satisfied
potential with
to
equilibrium
state
and
respect to the displacement
12
Consider the
The condition
the
change
of
must
remain
stationary.
And
consider the load-
displacement and stress-strain relationship
of figures
shown below.
-^sukFigure 2.1 Load-Displacement Curve
6-
Figure 2.2 Stress-Strain Curve
In
General,
the
total
potential
energy
is equal to the
summation of strain energy and potential energy.
ie,
where,
=
7T P
Kp
w
U
+
W
(2.1)
Total potential energy
Strain energy (internal work)
Potential energy of load system (external
work)
13
,
From Figure (2.1),
W
=
-
=
-
where,
p 6u
J
2 Pi
(2.2)
ui
i th
load
deflection under
Pi
ui
i th
load
From Figure (2.2),
\
U
(\
o 6€ d(vol)
=
vol
o
5
a € d(vol)
J
(2.3)
vol
Substituting the equation (2.2) and (2.3) in equation (2.1),
we get
a € d(vol)
5
vol
p
Equation
can
(2.4)
-
be
2
p.
(2.4)
u.
represented
by
matrix
form,
the
following is the matrix formula.
tup
=
-5
J
v
where,
{€} d(vol)
{a}
-
{
u.
}
{
p.
}
2.5
x
a vector of stresses
a corresponding vector of strains
a vector of elemental nodal displacements
a vector of elemental nodel forces
{a}
{€}
(ui
}
{Pi}
Let us define [B]
,
a
matrix relating strains and
displacements
le
{€}
=
and define [D]
le,
{a}
=
[B]
,
a
[D]
(2.6)
{u}
matrix relating stresses and strains,
(2.7)
{€}
14
}
Substituting equation
(2.6) &
(2.7)
equation (2.5), we
in
get
T
Kp
{ui
-
T
T
{B}
}
[D]
d(vol)
[B]
{ui
-
}
{ui
}
{pi
}
(2.8)
vol
Now we consider the minimum potential
energy, the condition
is as follows:
dTTp
(2.9
=
d{ui
}
with respect
Taking derivatives
to {ui
}
in equation (2.8)
and applying to equation (2.9), we get
{pi
}
:J
CD]
[B]
d(vol) {ui
[B]
(2.10)
}
vol
relating
From
loads
displacements,
and
elemental stiffness matrix [k]
le,
[K]
=
J
[B]
[D]
[B]
we
can
get the
.
d(vol)
(2. 11
vol
Summing all
the elements
of the entire structure,
equation
(2.10) becomes
{P} =
where,
2
.
[K]
{P}
-
[K]
[u]
=
=
(2.12)
{u}
1 {Pi
2 [K]
2 {ui
}
One Dimensional Finite Element
Consider a shape function which is linearly function
with
geometric
boundary
condition
15
with
one
dimensional
We define that the shape function,
finite element.
\J/M
(x)
aN
=
bN
+
tn,
is
X
In order to derive the constants in the shape function for a
with
element
rod
displacement
linear
field,
use
the
geometric boundary condition.
We may obtain the constants as follows:
X2
Lo
i
b
(2.13)
=
Ln
1
XI
a
2
Lo
1
b
2
The relationship between displacement and shape function is
given below
M
u
=
N
uN
^(x)
=
Another
T
|\{/N
(N=l;
u x (t)
^.(x)
+
consideration
M=2,
....)
(2.14)
u 2 (t)
the viscoelastic materials.
is
viscoelastic materials, the
stresses
may
terms of relaxation functions as given by
't
s(t)=
J
G(t
-
£
)
s
(£)
16
d£
be
For
expressed in
]
=
S
I(t)
(t)
e
e
where,
(2.15)
S
(t)
S
=
[
r
G
I(t)
(
t
t -
(
elastic response
:
)
£
)
=
S*
(
£
)
d£
ae
relaxation function.
G(t):
Assume the
element is in equilibrium and quasistatic state.
The motion equation for a one dimensional finite
given by
J T
.
.
ODEN [ref.6],
S\^
N
Vo
where,
Vo
P
element is
:
)
dV
=
M
M, x
Ao Lo
=
u
*£/
1
(
x
,
P
(u
N
M
,t
undeformed volume
total generalized force at node N
:
N
On
substituting
for
above
functions
shape
and
their
derivatives, we may obtain
fN
(u M
t,
,
I)
=
Vob N
Se
[
-
I]
(1
+
b
k
)
-
P N (u M ,t)
(2.16)
=
C.
uk
CONSISTENT NODAL FORCES
Consider a rod element in two configurations: undeformed
and deformed.
Let
the
distributed
forces
configurations be q and
q*
load vector
both the
depends on
per unit length in these
respectively.
the deformed state.
17
The consistent
magnitude of the load and
.
One dimensional rod element
under distributed
load is
showed as following figures.
(x,V>
-*.
%*C*>*^
*
*.
V
2-
i
2,--
i
a
L.
where,
Lo
L
undeformed rod element length
:
deformed rod element length
:
q(x,t) is distributed load per unit undeformed length
q*(y,t) is distributed load per unit deformed length
Figure
One-Dimensional
2.3
Rod Element under Distributed
Load
Consider
the
relationship
undeformed coordinate.
between
the
deformed
It may be written as
y(x,t)-x+u(x,t)
where
and
(2.17)
x:
undeformed coordinate of
a
y:
deformed coordinate of
point
u:
displacement of
a
a
point
point within element
The consistent nodal forces vector is given by
(2.18)
In index nutation,
P
=
[
nodal forces may.be written as
Pp%dV
-
fjh*
dVo
(2.19)
with
N
where
F
=
body forces per unit mass deformed
:
Fo
body forces per unit mass undeformed
:
P
deformed density
:
Po
2
1,
undeformed density
:
Relating q*(y,t) and q(x,t) is given by
q(x,t)
where,
(2.20)
dy/dx
=
A.
q*(y,t)
\
=
stretch
:
undeformed body
Consider the
force per
unit mass, then we
may write
(2.21)
r°"
dm
P Ao
-
"
P.
A
Using equation (2.17)
y
x + \Jf
=
(2.22)
u
M
M
dy
=
(2.23)
u
+ \|/
1
dx
M
M, x
The undeformed body force per unit mass is then
A
P
q*
=
(
o
'
+^M,x
1
um)
(2.24)
Ao
The consistent nodal forces take the following form,
L
f
PN
=
J
q*
L °
t
Vf?x
f
+
J
o
q*>^
MiX u M\^ N dx
(2.25)
Substituting the shape function in equation (2.25), we may
get
19
PLo
=
5
N
+ b u
M M)
(1
q*(a N
J
+
bN
x)
dx
(2.26)
Now we consider some special cases of equation (2.26)
One case is the uniformly distributed loads.
V
/
*>*
S*
Figure 2.4 Rod Element under Uniformly Distributed Loads
The uniformly distributed loads are given
q*(y,t)
=
h(t)
qo
Substituting the loads in
equation (2.26),
we may
get the
consistent nodal forces.
qo
p /I
{
P„
>
=
>
=
h(t)
[Lo
X J
+
*-
u
ui]
U
(2.27)
du
where,
h(t
dx
Another case is the linearly distributed loads
Figure 2.5 Rod Element under Linearly Distributed Loads
20
.
Linearly distributed loads are given
q(x,t)
q (t) + [q (t)
=
Substituting
loads
the
-
(t)]
q
ey
equation (2.26) we may get the
in
consistent nodal forces.
(Lo
P
N
<
D.
-
+ u
u
a
l"
)
<
>
13
2
M
(2.28)
»
,<\
SOLUTION METHODS
In this section we
incremental
method,
derive the
equilibrium equations of
Newton-Raphson,
Full
and
Modified
Newton-Raphson Method.
It is based on
Lagrangian
total
Method
use the
and
Taylor Serier at a given equilibrium point.
element
The
stiffness
matrices
vectors are assembled to form global
prescribed
The
manner.
and
consistent load
matrices in
the usual
displacements are implemented by
replacing the diagonal element
of the
corresponding row by
unity and the rest of the element of the rows and columns by
zeros [49]
1
Incremental Method (INC)
.
Consider the incremental process, we can rewrite the
equation of load-displacement relationship,
[K]i
(A>i
where,
{
i
APc
}i
+
:
:
i
r
(APMi
+
{APc}i
the current configuration coordinates.
the force unbalance
21
(2.29)
{/\P}i+i
an increment in the external load
:
The method is described by the Figure 2.6
c
c
-A
,
H
Pb
....v*
/f
\
'
"ICj
A
1
1
As
p.
1
*s>
J
1
1
1
'
k
where,
-p.
«
1
1
1
JL
The algorithm of equation (2.29)
AEC
AB'
—^—4
\
*
C
:
The algorithm of purely incremental
process
Figure 2.6. The Algorithm of Incremental process
The approximate values of A,
To improve
E,
is computed from p -f(u).
C,
these values, take the iteration process at each
load level with {AP}i+i=
0,
as is Figure 2.7.
The iteration processes are
load
levels
with
many
corrective
another is many load levels
This is
sometimes called
two
with
ways,
one
is
few
iterations in each, or
few
iterations
in each.
'incremental with one-step Newton-
Raphson method"
2.
Full Newton-Raphsor: Method CFNR)
Consider the load versus
f(u).
Take the Taylor series of P
22
displacement function
=
f(u)
P
-
,
,
.
<#>.
J(U^
JCUA + A,)=
A,
(2.30)
dP
where
tangent stiffness at A
K
du
f(u
)
A
f
(u
A
+
)
=
P
A.such that
We look for
k
(A)
A
=
p
-
f(u
A
P
-
)
,
SO
B
(2.31)
P
B
I
+
B
The consequent steps are explained by the Figure 2.7
Figure 2.7. The Newton Raphson solution of the equation
P
f(u)
=
The above algorithms are below;
update displacements
use
.
ui
:
ui
to get stiffness
find next increment
Eventually
ua
+
A
=
l|A
and load
ki
^a from
+•
,
A
+
ki
A^
(
Ax
Pi
)
=
+
Pb
-
Pi
=
ub
to a close
approximation
Then,
the load can be
process can
be started
increased
again.
23
from
Pb
to
Pc
,
and the
This solution algorithm is
known as the Newton-Raphson Method.
3.
Modified Newton-Raphson Method (MNR)
Newton-Raphson
The
[k]
solving in every
iteration
An
economic.
that
tangent
alternative
is
process
This
cycle.
modified
tangent
same
iterations cycles.
stiffness
matrix
is
is not
Newton-Raphson
also called "constant stiffness iteration".
iteration,
the
requires
generated and reduced for equation
stiffness matrix
be
method
used
Here
for several
The MNR is described by the Figure 2.8
Figure 2.8 The MNR Iteration
but
As compared with FNR,
the MNR requires more iterations,
each
quickly because [K] is formed and
is
done
more
reduced in only the first iteration.
E.
ANALYTIC SOLUTIONS
At first we define the concept of stresses [52] Eulerian
stress, a
,
is defined as follows:
24
)
:
N(y,t)
(y,t)
a
(2.32)
=
A(y)
and is called True stress.
Lagrangian stress,
T,
is defined
as follows:
N(y,t)
T(x,t)
(2.33)
-
Ao (x)
and is called Engineering stress.
Kirchloff stress,
S(x,t)
defined as follows:
N(y,t)
—
=
is
S,
(2.34)
\(x,t) Ao(x)
The relationship between strain
r
,
and
stretch
is as
follows
(
\*
"
1
du
\
=
i
(2.36)
+
dx
where,
N
A
For the
Applied force
:
deformed area
:
special case
in cross-sectional
of small strain,
area are
small.
both
u,
In this
x and change
case,
we can
write
r * u
=
€
a * T » S
where, €
is
Thus in
engineering strain.
small deformation case,
and engineering
strain
can
€
25
be
the Lagrangian stress T
used
to
replace the
)
and
strain
engineering
can
€
be
used
Kirchhoff stress S and Lagrangian strain
r
to replace the
.
Elastic Materials
1.
Consider
cross- sectional
one-dimensional
a
area,
Ao
Ao(x),
=
with
rod
varying
in equilibrium under
distributed external loading.
(T+£*X)
AoOO
(A*+% *x)
(a)
where,
(b)
external distributed load per unit length
f
T
:
Lagrangian stress within rod
Figure 2.9. One-Dimensional Rod under Axial Loading
For static problems and elastic materials, both
f
and T are
independent of time,
£
—
-C
f
From figure
n«-J
.. \
2.9(b),
1*
-
rr'
i
• -
x
the static equilibrium of forces in the
x-direction is
£ F(x) = -TA + fax
(T
(dT/dx) ax
Expanding in Taylor series,
26
)
(A
(dA
(2. 37
/
dx) Ax
)
=
=
^c
For
a
in the limit,
continuous rod,
i
+
i^TA^
(2.38)
-
(2.39)
Substitution of equations (2.34),
and (2.36) in the
(2.35),
above
(2.40)
where,
u(x)
displacement w.r.t. x
:
u, x
first derivative of u(x)
:
u.xx
second derivative of u(x)
:
Note that
equation (2.40)
force
(per
f
unit
displacement field
values of distributed
gives the
length)
u(x) of
a
required
to
one-dimensional
produce
the
rod once the
boundary conditions, the stress-strain law, and varying area
function
Ao
(
x
constant area
f(x)
= -A
have been
)
Ao
,
u,xx
Consider the rod with
specified.
equation (2.40) becomes
1(1
Some special cases of
+ u, x
)
**2
(
dS* /dr
)
+ S«]
equation (2.41)
case is the specified displacements.
27
(2.41)
is considered.
One
x
\oo>
/-=*
//-
where,
X
J A«
u(x)
"
u, x
+
ao
-
a
+
2
i
=
u, x x
2
+
a, x
aa x 2
aa x
a»
Figure 2.10 Test Model of "Inverse Solution"
Apply the boundary contition u,x(x)
a
u x
(
)
a
=
,
x
—
=0
at x
=
Lo
(2.42)
x2
-
2Lo
If
specified, we may get the exact solution u(x) and
is
a
t
the corresponding required distributed load f(x).
Linear displacements field:
a.
u(x)
u, x
U,
xx
=
=
a,
a,
=
From equation (2.42), we may get
f(x)
Note that
=
a
linear
displacement
with distributed load.
28
field
cannot
be produced
r
Quadratic displacement field:
b.
u(x)
u, x
u
11,
-
a,x
=
=
a
-
-
{
-
(
(ai/2Lo)x 2
-
y
(
a
i
Lo
/
a | / Lo
)
x
)
Consider the linear materials,
Se
E
-
r
We may get the distributed forces from equation (2.41)
f(x)
-
-
EAuU.xx
-
[(1
+
u, x
)
+
+
(u, x
% u2,
x )]
(2.43
Consider the nonlinear materials,
-
Se
Ei r
=
E2
+
(quadratic form)
2
From equation (2.41), we may get
f(x)
AoU.xx
-
=
+
(u,
El
x
{
*-
(1
+
u, x
^u2,x)
)
+
[Ei
E2
E2
2
+
(u,x
+
(u,x
>2U 2
,x)]
% u2,x)]
+
}
(2.44)
Another case is the prescribed loads.
/
^p
/e
A
-*
3C
Figure 2.11 Test Model of Rod with Prescribed End Element
From equation (2.39),
since no distributed loads exist,
d
(
TAo
)
=
with constant area.
dx
We may get
G
=
constant
(2.45)
9Q
r
)
s
The applied load is given
P
-
Ao\
-
(2.46)
S
Linear material;
^e
Er
r
E(u,
=
+
x
*i
u^
,
x
)
From equation (2.46), we may get
P
-
EAo
=
(u,
+
x
u2, x
+
^ u3,
x
(2.47)
)
Nonlinear material;
Se
El r
=
E2
+
2
(quadratic form)
From equation (2.46) we may get
P
Ao
=
+ U,
(1
X
[El
)
(U,
x
y2 U= ,x
+
)
+
E2
(U, x
+
% U2
,
x
)2]
(2.48)
2
.
Quas i lin ear Viscoel a tic materials
The Viscoelastic materials has
than
problems
the
elastic
materials,
the more complicated
since the external
loads and displacements are time-dependent.
lU-t)
u Lo c*>
> Fu c*)
<
x
where,
Ao
^
Lo
,
X=Lo
uo(t)
—^1
u<
:
:
uLo(t)
Fo
(
t
FLo(t)
Underf ormed Area and Length
time-varying left-end displacement
time-varying right-end displacement
:
time-varying left-end force
time-varying right-end force
Figure 2.12 One-Dimensional Rod Subjected to
End Loads
30
Time-varying
.
For quasilinear viscoelastic materials,
Sit)
Jn
The Lagrangian
(2.15)
&,
constant throughout the rod since
stress is
the distributed loads is not applied.
From equation (2.39),
the stress is time variable only.
T(x,t)
g(t)
-
\(t)
=
(2.49)
S(t)
And the internal forces produced with the rod is
F(t)
=
TAo
Aog(t)
=
(2.50)
Substituting equation (2.15) in equation (2.49) yields
g(t)
\(t)
=
(1
S(t)
+ u.) S
-
e
(r)
(2.51)
tf
o
of equation
hand side
The right
(2.51)
is a time function
only
Therefore
must also be a time function.
g(t)
du
u,x(x,t)
h(t)
=
(2.52)
dx
Integrating w.r.t.
u=
u(x,t)
where,
C
at equation (2.52)
x
xh(t)
-
+
C(t)
integrating function to be determined by the
:
boundary condition at one end.
Assume that
u
(
o t
,
)
(2.53)
=
uo
(
t
)
=
C t
(
)
,
31
)
)
therefore
u(x,t)=xh(t)+uo(t),
u,x (x, t)
=
(2.54)
h(t)
Substituting equation (2.54) in equation (2.51) yields
;]
(2.55)
Se
Once the G(t) and
(
r
been specified,
have
)
the equation
be used with the boundary conditions to find the
(2.55) can
Consider the
exact solution in some cases.
with linear material property.
boundary conditions at
3
element solid
The prescribed displacements
ends
both
described
is
in Figure
2.13.
/
/
/
/
>
/£
x
-J
L,
u(o, t)
u
(
Lo
,
uo
=
t
)
=
(
u,
t)
(
-
(2.56)
t
Figure 2.13 Test Model of
a
Rod with Prescribed Time Varying
Displacement at the End.
The prescribed displacements are linear,
u(x, t)
U, X
(
X t
,
)
(x/Lo
u.
I'
(
o
u.
)
t
)
(t)
/ Lo
(2.57)
=
h(t
32
(2.58)
.
where,
t^
(t)
=
Loh(t)
Assume linear material property, S e
And for a
G(t)
where,
element rod, G(t)is defined below
3
=
-
(1
a
Er
=
a
P
,
=
(2.59)
constants for material properties.
;
Note that G(o)
e-Pt
+
)
1
at t
=
G(t)
,
=
-
1
a
at
(3-0
No relaxation occurs and the material exhibits elastic
response
Specifying the prescribed displacement function u
r
(t).
L o
Ramp Function
a.
At x
=
Lo
C.'S:u(o,t)=0
B.
,
u(Lo,t)= u u
r
(t)
o
=
at (a:
constant)
From equation (2.57) and 2.58), we may get
u(x,t)
u, x
(
x t
,
=
(a/Lo
)
=
(
)
a/Lo
x t
)
(2.60)
t
Substituting equations (2.57)
-
We may get the reaction force,
EAo
at
[1
+
F.
a a
a2 1 2
[{at+
]
Lo
in equation (2.55)
(2.59)
2Lo
Lo
a a
a
{!-(
}
-|3t
)}e
pLo
(3
a
{(Pt
-
1)
+
(p 2 t 2
2
(3
-
2(3
1 + 2)}]
(2.61)
pLo
Equation (2.61) describes the reaction force produced at the
fixed end for the prescribed displacement
u
,
(
t
)
-
at
33
)
Step Function
b.
At x
Lo
=
(Relaxation Test)
B.C.'S
,
u ( Lo
where,
1
(
t
)
,
t)
u o t
)
ui^l
t
(
:
=
,
(
=
unit step function
:
From equation (2.57) and (2.58), we may get
ULo
u(x,t)
x l(t)
=
(2.62)
Lo
ULo
u, x
i
(t)
(2.63)
Lo
Substituting equations
(2.57)
-
(2.59) and (2.62) in
equation (2.55), we may get the reaction force,
EAo
F(t)
=
(
ULo
ULo
ulo
)
+
(1
)
G(t)
[1
+
2Lo
Lo
F.
)
(
1
(t)]
Lo
(2.64)
Consider the Kirchhoff stresses.
From equation (2.34)
ULo
S
=
E
(
ULo
)
(
+
1
)
ULo
rL
=
prescribed
E ru
G(t)
(2.65)
ULo
(
Lo
a
=
2Lo
Lo
where,
G(t)
1
+
)
Lagrangian strain due to
2Lo
step displacement function
end.
34
u,
l(t) at the rod
.
.
.
III.
PROGRAM IMPLEMENTATION
OVERVIEW
A.
are
We
analysis,
(P)
:
{u}
constrained stiffness matrix
know
we
For static
nodal displacements vector
{u}
If
analysis.
external loads vector
where {P}
[k]
static
in
load-displacement relationship is as following:
[k]
=
interested
problems by
{P}
or
then
{u},
using elementary
we can analyze the static
elastic theory.
In general,
the static analysis steps are shown as follows:
Step
Enter geometry of material properties,
1
loads,
et al
Step
2
Calculate element stiffness matrices.
Step
3
Determine appropriate part of [k] and {P}.
Step
4
:
If
[k]
or {P} is appropriate,
through step
Step
5
:
If
[k]
3
or {P} is not appropriate,
constraints
Step
6
Step
7
Step
8
:
:
:
then repeat step
Determine {u}.
Regenerate elements.
Calculate element stresses.
35
then apply
2
B.
SOLUTION PROCEDURES
structural problems, to find the solutions
In nonlinear
we use the INC and FNR and MNR.
in advance
what increments
In fact,
of loads or iteration should be
good approximation
used to obtain a
solution in program application.
for elastic
structural problems.
solution.
The
strategy subroutines
The
or visco-elastic time-dependent
Here the initial state for incremental
was selected as a null displacement vector.
loading process
Next we consider finite
viscoelastic
and
to the
was used in determining the strategy for
interactive method
were developed
it is hard to know
element method
(FEMEV)
programs.
for elastic (FEMEL)
IN
both
programs,
equilibrium was checked at each stage for solutions obtained
from other INC or FNR and MNR.
1
.
Algorithm Implementation
Consider
based on
total
Algorithm is
Step
1.
the
algorithm
Lagrangian
of
Method.
Newton-Raphson
Method
Full Newton-Raphson
as follows:
Establish local coordinates X-Y for the element at
hand.
Step
2.
Generate the element stiffness matrix [k] in local
coordinates so that it operates on the local D.O.F.
Step
3.
Transform [k] to global coordinates so that it
operates on global D.O.F.
Step
4.
Repeat step
1
through
36
3
until all elements have been
treated.
Step
5.
Compute displacements.
Step
6.
Compute internal forces,
Step
7.
Solve [k] {Au}
where,
Step
8.
uN +
=
1
(A p >
=
uN +
[k]{u}.
A
uN
Test for convergence.
If not satisfied,
Where,
convergence
return to step
=
((ART AR)
0.001
37
1.
/
(AprAp))**0.5
<
2.
Flow Cha rt of the Finite
In
E lement
Program
put: geometry, mat. prop; loads
3L
PROBLEMTYPE:
ELASTA/ISCOELASTIC
F"
ELE. MATRICES
STIFF/LOAD:
GLOBASSEMBLY
SOLN:FORWARD
SOLUTION
ELIMINATION
MONITOR
-r-
SOLN:BACKWARD
DIAGNOSTICS
SUBSTITUTION
£
UPDATE DISPS
4/
COMPUTE RESIDUAL
FORCES
<-
NEW LOAD INC/UPDATE
DISPS
e
PRINT RESULTS
STOP
[Flowchart of the Finite element Program
I
j
NUMERICA L EX AM PLES
IV.
two
consider
We
numerical
examples
for
elastic and
viscoelastic materials.
A.
ELASTIC MATERIAL
1
•
Presc r ibed conce ntrated fo rces
.
Specify the rod element.
/
/
»
Pz
/
/
M
L
Pii
-
22,5000 lbs
A
=
0.25 [in:
L
=10
E
-
[in]
106
[lbs/in2]
Consider the stiffness matrix of rod element,
the
[k]
.
For
rod element,
Kn
EA
r
L
|_k2i
ki
2-
[K]
(4.1)
kc
2
Consider the boundary conditions,
ui
=
Pi
=
39
P2
22,5000 at
=
From loads
-
x
-
L
displacements relationship,
Solve for the displacements,
Hence,
u2
,
we may get
LP2
9.0
U2
(in)
EA
2
P res cr ibed di
.
splacem e nts
/
X
/
/
-I
L
Consider the same rod
the
element before,
quadratic displacements, u(x),
u(x)
=
0.02 x
-
0.001 x
The first derivative is
du
—
0.0 2
=
-
0.002 x
dx
The second derivative is
d'^u
=
0.0U2
dx2
40
2
and specified
From equation
2.41
(
)
with the constant area, we may
getthe distributed loads for any
du
d2 u
f
(
X
)
=
-
Ao
[
(
1
+
3.0 * 10-3
dx
X2
ds e
)
dx 2
=
x.
+
2
3.06 x
-
41
S*
]
dr
+
530
(4.2)
e
a
o
ts
-2w
6
fiu
-o
O
2
T
00009
T
00000
>
I
000*
O'OOOB
saoaoi
42
I
00003
I
0001
00
s
8
O
U
CO
>
\
1
lit
oro
tot
to o
too
oo o
ooo
wo
S1N3W33VTHSK1
43
coo coo
too oot
S3
|
Q
O
O
CO
CO
w
on
fiU
o'er
09i
0*21
oe
oo
S3SS3H1S iJOHHOHIM
44
O'C
00
.
B.
)
.
)
VISCOELASTIC MATERIAL
Consider the rod element below,
U.(*>
F.C*)
->
-ya
Boundary conditions; u(0,t)
u(Lo,t)
Lagrangian Stress; T(x,t)
Internal Force; F(t)
Kirchhoff stress;
=
S
=
g(t)
T Au
g(t)
=
(t)
=
u
1
L o
L o
-
A(t) S(t)
(
t
(4.3)
=
t)
,
Ac
(x)
element solids, the relaxation function is
3
G(t)
u
=
X(x
For
(t)
uo
=
=
-
(1
a)
OL
where,
t
a e-pt
and
(4.4)
constants
are
|3
of
the
material
properties
Assume unit step function for prescribed displacements.
the boundary condition is
At that time,
u(o,
u ( Lo
t)
t) =
,
where,
We
may
=
u.
uo
=
t)
u.
u o
1
(
t
unit step function
l(t):
get
(
displacements
the
function
with
unit
step
function
u
(
x
,
t
)
=
--
-
ul
o
1
(
(4.5)
t
Lo
4 5
.
du
—
ULo
l(t)
=
dx
(4.6)
Lo
And the Kirchhoff stress is specified
S
=
(4.7)
E r
Substituting equations (4.4)
EAo
F t
(
)
-
(4.7)
UL
UL o
Ulo
=
1
G t
+
(
)
2Lo
Lo
in equation
1
o
where,
o
+
1
(
t
)
S,
G(t)
(4.9)
ULo
ULo
rLo
(4.8)
Lo
Hence, we may get the Kirchhoff stress,
S = r rL
(4.3),
=
1
+
:
Lagrangian strain
2Lo
Lo
due to the prescribed step displacement function
u
1
(
t
)
at the rod end
Equation
(4.9)
describes
the relaxation cases.
equation (4.8), we may get the internal forces.
46
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CONCLUSION
V.
study
This
directed
is
towards
understanding
the
nonlinear behavior of structures.
The formulation
based
is
work.
We consider the one
based
rod
viscoelastic effects.
incremental
process
dimensional nonlinear continuumincludes
that
element
To
with
Newton-Raphson Method.
principle of virtual
on the
the
get
large displacements and
solution
we
use the
full Newton-Raphson and modified
Numerical solutions agree
well with
exact solutions.
Further
structures,
Consideration
panels at high
studies
such
of
may
viscoelastic
to two dimensional
composite
laminated
as
the
extended
be
effects
temperatures
is
another
study
to
verify
of
possible
plates.
composite
area of
study.
An
experimental
present study is also recommended.
48
the results of the
,
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