CALIFORNIA STATE UNIVERSITY, NORTHRI:OOE
NONLINEAR ANALYSIS BY FINITE ELEMENT METHODS
A project submitted in partial satisfaction of the
requirements for the degree of Master of Science in
Engineering
by
Russell Howard England
May, 1975
The project of Russell Howard England is approved:
CALIFORT\JIA STATE UNIVERSITY, NORTHRIDGE
September, 19 74
TABLE OF CONTENTS
·Page
iv
ABSTRACT
INTRODUCTION
1
I.
NONLINEAR ANALYSIS TECHNIQUES
10
II.
EXAMPLE:
INCREMENTAL METHOD
14
III.
EXAMPLE:
ITERATIVE METHOD
24
IV.
COMPUTER
PROGRAM
34
A.
FLOW CHART FOR THE INCREMENTAL METHOD
34
B.
FLOW CHART FOR THE ITERATION METHOD
36
C.
MATRIX INVERSION BY DECOMPOSITION
38
D.
INPUT INSTRUCTIONS FOR THE INCRE-
44
MENTAL _METHOD
E.
INPUT INSTRUCTIONS FOR 'rHE ITERATIVE
47
METHOD
F.
COMPUTER LISTING FOR THE INCREMENTAL
50
METHOD
G.
COMPUTER LISTING FOR THE
ITERATI\~
57
METHOD
v.
BIBLIOGRAPHY
66
APPENDIX A - DERIVATION OF THE STIFFNESS MATRIX
67
FOR A THREE DIMENSIONAL TRUSS .
APPENDIX B - EXAMPLES USING THE INCREMENTAL
73
AND THE ITERATIVE COMPUTER PROGRAMS
I!
_________j
iii
ABSTRACT
NONLINEAR ANALYSIS BY FINITE ELEMENT METHODS
by
Russell Howard England
Master of Science in Engineering
September, 19 7 4
This project deals with the solution of nonlinear
structural problems by the use of the finite element
method.
The method is described for the general
structural problem.
Two computer programs are developed
using nonlinear analysis.
These computer programs solve
a three dimensional nonlinear truss by the use of the
incremental and the iterative methods.
The theory
behind these programs along with a user manual is
presented herein.
These results compare favorably with
those obtained by exact methods.
I
l
II
----·-----------·-----·-··----·-·--·----------·------····----------~---'
iv
·:-,
INTRODUCTION
The finite element method has evolved since the
development of the high speed digital computer.
The
finite element method requires the solution of a set of
many simultaneous equations.
With the use of the
computer the solution to a set of simultaneous
equations
can be found rapidly with modern numerical methodso
In the past, structural analysis has proceeded
as follows:
1.
A free body cut of a structural member is taken
and the six stress components are applied to
their respective locations.
2.
The six stress components are related through
· equilibriumo
3o
The stresses are related to strains by the use
of Hooke's law.
i:
4.
Strains are related to displacements.
5. From relating equilibrium a mathematical
expression, often a differential equation is
found and analysis is reduced to finding a
solution to the differential equation that
satisfies the boundary conditions.
Often it is not possible to determine a closed form
solution to the mathematical
mod~
so numerical methods
are used to provide approximate solutions.
1
The finite
2
•
-~
~
_ , , ___ •••, ••
,••-~
•v-·-··-·~··~-·"•>•• -----~···--- 4 ~-
-•
--~-
_,,L •·-~• ~ >•-•
·-
, • ,__,~,.~-~··"-- -·-··--'"·'•----'L~~---~~•-.••··--~ -~-~_,--
,_...,__ ..........,,_,,~~-------·-·--~n~-..........,._.,
I
element method divides the structure into a number of
discrete volumes by dividing it into an equivalent system
of smaller elements.
Instead of trying to find a closed
form solution for the entire body, solutions are found for
each small element with various boundary conditions.
A model of the structure is divided into a finite
number of elements.
' elements at nodeso
These elements are connected to other
.A displacement function is assumed to
. approximate the exact displacement for each element.
The
amplitudes of the displacement function are determined
from the displacements of the nodeso
The strains are
determined from the displacement functipn; then from
the potential energy of the element the stiffness matrix
1
is determined.
The potential energy of an elastic
structure is the sum of the internal energy (strain
energ~plus
the potential energy of the external loads.
If a structure is in a state of equilibrium then this
energy is a minimum.
The element stiffness matrix for a linear material
is derived as follows:
The selection of a displacement
is ehosen, generally a polynomial.
A polynomial is chosen
for a displacement function because polynomials are easy
to differentiate and intergrate.
The order of the
polynomial is equal to_ the number of degrees of freedom
needed to describe the displacements of all the nodeso
3
; A displacement function must be continuous within each
element and displacements must be compatible between
elementso
Compatible means each element must deform
without gaps or overlaps between elements.
A displacement
function must include rigid body displacements and also
include constant states of straino
A constant state of
strain is necessary because if an element approaches
infinitesimally small the strains in each element will
approach a constanto
Examples:
one dimensional
= c:w,
u~,
two dimensional
\Jl'4.~
::.
=
c;~, "\Ola.K
o~'l -4-~e"
'I (X.Y).
three dimensional
UC.X,'<,~)
=
"(~-<.l) ,.
~
'\-
a-~X'(-\
-'t.t .,..... ""cwS' '(\ ~
"'"•ct 'ICY t-Qito )Ca. + <:aiu '(l.
ot' '1
+.J\a. "(
~d,)'. +'o) '( ""'al"t~
o.~s .rc:~~t,x -+c:ot..,
'< +Qiac
w(;,c., '<t ~) -:: -'C\ ~ c:W.\o~ + C~~,, '< "'c~,'L t.
' In matrix format
l
l
u\ ,. \l ~~~~~
~~ ~ = l ~1{o~)
l~. 'l \
w
'(,
Where
displacement function
.
i
4
[$]
=
polynomials of the displacements
{C<'
=
the coefficients of the polynomials
Since the number of degrees of freedom is equal to the
number of coefficients, the coefficients can be determined
from matrix algebra.
The displacement function is
evaluated at the nodal points.
{ q'
=
u
u
u
u
(node
(node
(node
(node
.
1)
2)
3)
4)
u (node n)
(\>(node
(node
+(node
t,l) (node
~
=
1)
2)
3) {"'} =
4)
[A J(..,
.
; (node n)
From matrix algebra
t~\ =
-·lq~
[A]
Where
~\]
=
displacement transformation matrix
Substituting for
~gives
the displacement function as
a function of the nodal displacements.
5
' The strains are functions of the displacement function
and can be computed as direct function of the nodal
displacement.
£,. : ~U/J.'1
l,'(
=
£~ =
~V/~'1
~wf ~l:
' Where
f(\ . ,
vector of strains
From the generalized Hooke 1 s law equation for elastic
isotropic materials.
(.,.
•
1
1£
(.::r.,. - v ( 4'~ \ 11:))
. £.1 = 'l\i (. o-'1- vt~~ +cr1:']
£ ~ ::: 'lr;. l"1: - Y (cr,. "'4'"<' J
~
. .'( =. "'.._.'(I~
¥._.,. = rr-'f."'./ '
~-<' : If""(" I G::.
Where
E
= Young's
modulus
)) = Poissons ratio ·
G
= Shear
modulus
6
.
The stresses are functions of strains
· { cr } = [ C ]
{E }
Where
{cr} = vector of stress
~J = matrix of material constants
{cr} =
[c J
[ B)
{q}
The stresses and strains are now a function of the nodal
displacements {q}.
{E}
=
{cr} =
[B l
{q}
[c J
[B]
{q}
In order to determine a method for calculating the
element stiffness matrix it is necessary to determine the
internal strain energy.
'(
Consider a cube with sides
'
7
dx, dy, and dz inside an elastic body subject to an
external system of forces.
cr dydz,
X
The normal forces are
y dxdz,
y
The shear forces are
T
xy
dydz,
T
T
yz
yx
dzdx,
T
dzdx,
T
ZX
zy
dxdy
dxdy
Under the action of these forces the faces of the cube
will displace in the normal direction by amounts
£
X
dx,
sydy, s 2 dz and will distort by amounts Yxy' Yyz' and Yzx•
The total work done by these forces is the mechanical
energy stored in the differential element volume dxdydz.
This mechanical energy is called strain energy.
The
strain energy does not depend on the manner in which the
forces are applied but only on their final values.
The
relation between the normal force cr dydz and the displacex
ment
£
X
dx is linear.
8
The corresponding work is equal to the area under the
triangle.
a
dU =
X
£
X
dV
For the other normal forces and shear forces the same
procedure yields the following results.
y
xz
dV
The total strain energy stored in a elastic body of
volume V is
u =
~
v(a £ +a £ +a £ +T
x x
y y
z z
v-
xy'xy
+T
v
xz'xz
+T
v
yz'yz
)dV
The internal strain energy is equal to the work done
by the external loads {Q}:
The external work is equal
to one half the sum of the products of the intensities
of the external forces and the components of the displacements of their points of application in the directions
of the external forces.
N
=
~
E
i=l
Qi qi
9
From matrix algebra
[£
J=
t~'T:
cr =
[B] {11)
{\\
[c.] ~] {\)
e,
~=
I Jv {\\T r~l [c..] [s]
~~
zf·ir( ~ t"l rcJ [B]d") t'l.\
{\\ <Av
Clapeyron's theory states that the internal strain
energy is equal to the external energy.
= U:r T.
thfl~\ .. ! h\ {Jv [8] [t][l3j .iv) {1,'
UE
[~1:: (fv IBII<IIel dv)
['J.~
By definition of the stiffness matrix the external loads
are linear combination of the external displacements.
Therefore the stiffness matrix has been derived.
j v te]
(C] [B] dV
.
I
_j
Section I
i
'
NONLINEAR ANALYSIS TECHNIQUES
Often in structural analysis it is convenient and
accurate to use linear analysiso
In reality all materials
are nonlinear, some more than others and a linear analysis
' will often lead to results that ·are not accurateo
In this
case it is necessary to develop a nonlinear analysis
technique.
There are two types of problems in dealing
·with the finite element method.
One is the nonlinear
stress-strain relationship of a material and the other
is a change in the geometry that results after loading
of the structure.
geometry will be
Nonlinear problems
n~glected,
due to changes in
and nonlinear analysis will
be based on the nonlinear relationship of the stressstrain of the materials.
Geometry is assumed to remain
constant.
As in elastic analysis, nonlinear analysis uses the
---
-
------
basic condi t:Loris- of equirrtirllim -an.-a--compatTol1Tty~- Tn:e----- ----·--·---..- - -
major difference in the two analyses is the stress-strain
relationshiPo 'In the inelastic problem the strain is
not proportional to the stress since each strain component
contains a linear elastic component and a plastic
component not linearly related.
10
The solution of nonlinear
11
' problems by the finite element method can be solved by
various methods.
Two methods shall be describedo
One
is the incremental or stepwise procedure, and the other
is a step iterative or deformation theoryo
The incre-
mental theory relates a inrement of strain to a increment
of stress for a given
path is followedo
s~ress
stateo
Hence the true load
The step iterative theory establishes
a relationship between the stresses and the total
strains.
The path by which the stress distribution is
reached does not influence the strains.
true for proportional loading.
This is only
The ·two theories give
equivalent answers for the case of proportional loadings.
The incremental stiffness method is used to solve
nonlinear problems by dividing the load into several
increments.
The final stress and strain states are based
on the accumulation of the results of the individual
loading increments where the stiffness of the structure
is based on the current stress and strain states.
: Basically, the incremental procedure solves the nonlinear
problem by a series
of linear problems, or the analysis
.
is based on a piece-wise linear stress-strain curveo
If the loads are incremented smaller and smaller, the
results become more and more accurateo
If the structure
is a truss or beam, in the limit the results are exacto
12
•·-•· -• _,
~·-
,, •·· • •··•
•·•·~- --~~
••·~-~--~;~··-~-·---~~-~-~~"'-•·- --------~--~.--~~·---~~-••·---~------a•-...__•~-----~>
'
With the incremental method, the loads are first
divided into increments usually but not necessarily equalo
I~ "'L
The total load applied to the structure in any given
increment
.
~
is given by
~
~
.):
'
iA(Q\i
The incremental displacement due to a incremental load
can be determined from the stiffness matrix which is a
' function of the total displacements.
f•t,.~ ~K~\'\~)j-'A.Q
.a
"
h' =j~ lo.'t\i
The step iterative procedure differs
f~om
the
incremental procedure in that the structure is comletely
loaded during each iteration.
In the first iteration
Young 1 s modulus is taken as the tangent modulus for a
strain of zero.
The stiffness matrix is evaluated and
the displacement of the nodes are found.
From these
displacements the stresses are found from the stress-strain
relationship for each elemento
Since the stiffness matrix
' was approximated, the internal loads produced in the
structure will no\b equal the external loads applied to
the structure.
satisfied.
Therefore equilibrium will not be
After each iteration the load that is not in
13
balance is used for the loads to compute an additional
displacement.
This process is continued until equilibrium
is approximated to a desirable solution.
First the displacements are calculated based on
applying all the load.
The stiffness matrix is evaluated
for Young's modulus equal to the tangent modulus for
zero strain.
This displacement gives loads that do not satisfy
equilibrium.
These loads that do not satisfy equilibrium are used to
satisfied to a desired amount.
The stiffness matrix
is based on the secant modulus at the end of. the previous
iterative step.
L_____
!
--~-------------~
Section l i
EXAMPLE:
INCREMENTAL METHOD
In this section a nonlinear three bar
st~tically
indeterminate truss is solved using the incremental
method.
I
lOO
\
Member #
Area
_l,
i
Length
·1
1
200
2
1
100
3
1
200
The material properties are the same for all three
e1ement members.
I
~-----------------------~-J
14
15
The stress-strain curve for all members
40k
30k-
6
(j =
10
X
10 £
-6.25
X
8 2.
10 (
20k
lOk
.004
The load of 50,000 will be applied in five increments.
E for each cycle will be taken as the tangent modulus
for each members strain state.
The tangent modulus
can be defined as follows
=
For this problem
tr
=
6
10
X
-6.25
10 (.
X
8
10 {.l..
6
= 10
X
10
9
-1. 25
X
10 £.
For the first increment the tangent modulus is evaluated
at zero strain.
Since this problem has been divided
into five load increments, the first load applied will
II
II
_ _ _ _ _ _ _ _____..J
16
be 10,000.
The stiffness matrix is
= E.A.
l l /L.l
2
cos e,
~sine.cose]
.
. -2 e
[ -s1necose,
Sln.
• 0,
[ • o,
I
.
75,
.OJ
1•
.433]
.25
• 433,
For the first increment E. is taken as the tangent
l
modulus at zero strain.
E. = 10,000,000
l
The total stiffness matrix for the first increment
Kt
=
o.]
1. 5,
50,000
[ 0.'
2.5
The deflections for the first increment
l___
I
I
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ jJ
17
.666,
=
{:~L
=
1/50,000
[ .o,
c~)
The strain for each member
=
A(.:,
1/L \cose,
sine'
}6q~'
~q2
A~\ 'l:
=
112oo f.s66,
l
·~ ~ql\
~q2
b.f.l.':t
A(~"l
=
=
1/lOO
l/200
l·o,
l.J tql\
\-.866,
.~v·ql\
.. q2
boq2
For the first increment
=
.0002
Ai."tr =
.0008
C,;t =
.0002
.t:..E:er
u..
The change in the stress can be calculated
I
I
j
... = .
A (i.·
E~
l
~t.~
...J
18
~\1"'"'.
=
~ cr,.r
= 8 ' 000
~ <f"3::t
= 2 ' 000
2' 000
After the first increment the total stresses, strains,·
and displacements are equal to the first increment.
0'\'f
O'"l.T
<S~,.
= A <T', X
= ~ 6"'~ X:
= ""crJ :t
1•'t
=
A1er
11-T
=
~
1z.l.
For the second increment the tangent modulus for the
strains after the first increment are calculated as
E.
l
=
10
X
El = 9.75
E2
E3
=
=
10
X
9.00
X
9-75
X
6
-1.25
10 6
X
10 9£.
10 6
10 6
The stiffness matrix for the second increment
• 7312'
[ .o,
L _____
.o]
1.144
·--------~------
J
19
o.J
p..l37,
=
Lo.,
.8744
.000218600
[
A
cr, ll'
IJ..t:T,
or~,.
""d"'~
=
=
.000874320
.000218600
E. AUU
l.a.
2,131.155
= 7,868.5
II=
2,131.155
The total dis-placements, stresses, and strains after
the second increment
E:t-r
(1"'(.. T
~''t
=
=
A.
=
AtC:.t
b.
t.::t
o\ ~t~ 1T
6'\.'1.
+
~C'i. U'
-\- A\C:lf
l\T
= .000i.r1858
cr.,.. =
(aT
= .00167432
<Ji,- = 15,868.5
!~'f
= .00041858
CfaT = o.
1...,. = .167422
-----~--~----
(f~T
4,131.155
= 4,131.155
I
I
i
----------------~
20
For the third increment the tangent modulus is calculated
Elm.
= 9.4768
X
m
E3 nr
= 7.9071
X
= 9.4768
X
E2
10
6
10 6
6
10
For the third increment the stiffness matrix is
r-71076,
Lo.,
o~
1.0276J
= 10 _ 5 .[1.4069,
o.,
o.] { o \
.973113
10,000)
~(,:ur=
.00024328
•CT.m= 2,305.516
A{2.ln=
.00097311
"' a;.m =
at,m= . 0002 4 328
A.<fl\11
.,,IQ
A4
\'a.lU
=
7 , 69 4 • 5
= 2' 305.516
.o.
= .093133
The total displacements, stresses, and strains after the
third increment
+ Af.'.II +
(~T
=
Cfi.,,.
= a <t.:.'l +
'tC::r
=
~(;:l
~0"\.11
~,c:x+ ~1'-11
+
A((.
lU
AO'i.
m
+ ~~c:.:nr
I
l----~--------------------·-------------------------__j
21
e:,-r
= • 00066186
cr,~
=
6,436.671
(a.T = • 00264733
O"z_,.
=
2 3, 56 3 •
('JT
a.,, =
= .00066186
1\T
1'-T
=
o.
=
.260535
6, 4 36.6 71
For the fourth increment the tangent modulus is calculated
=
ElW
9.1727
X
E2m = 6.6908 X
10 6
10 6
E3W' = 9.1727 X 10
6
For the fourth increment the stiffness matrix is
=
\A q'm
=
{A q}m
=
10
5
r.
1
6 8 79 5 ,
0•
_o.,
.8984o_
[Kt]
10- 5
~Q1~
e·
0·J
4536,
o.,
1.1131
!1:,000,
.a.(,m= .00027827
Ali', a=
2,552.52
A("t.ti' = .00111309
b.<f;.'[2 =
7,447.48
.00027827
A G~,m=
2,552.52
a.t,.u
=
= o.
.u.\l.N
= .111309
22
The total displacements, stresses, and strains after the
fourth increment
~ ( (.~
(;:T
=
<:ftT
= Atr(.r
+ Af.tll + A.€.:.~ +
+
A. a'\. D:
cr, m +
+
&:~..
+
I)~(.
'l,i.T = A'&C: :t + ~~'-li
m. +
A
E. '-1St
J,)r. (J i.
A
.m:
1,{1!L
( \T
= • 0 0 0 9 40 14
CT,,..
~~T
= .00376042
<rz:t' = 31,010.48
(IT
= .00094014
cr')-r = 8, 9 89 .15
4( \T
~'Z.'T
= 8,989~15
= 0•
= • 371844
For the fifth increment the tangent modulus is calculated
6
ElY
= 8.8248 X 10
E2Y
= 5.2995
X
10
E3 :sf = 8.8248
X
10 6
6
For the fifth increment the stiffness matrix is
[Kt]~
~A
ql3[
\~q'~
=
10 5
. [.66186,
o.,
0"1
.75057
-\
=
=
[Kt] \AQ\"£
10-5 [ 1.51089'
o.,
o.] t:,oo~
1.33232
23
.a. q 1 "]t
o•
=
.133232
A q2'1['
ll-£'5[=
.00033308
~G"," =
A(a.~ =
• 001332 32
~cf1.'S(
Ai)SJ
= .00033308
A
2,939.366
= 7 , 0 60 • 6 34
G'Js:c = 2, 9 39. 366
The total displacements, stresses and strains after
the fifth increment give the approximate final results
for the nonlinear truss subject to the given loads.
C.i:r =
f"C:T =
1,-:r =
A(c.r
b.
<f'i.:t.
A~i.r
+ a.(c:u. + A(,m + Af:i.lSZ·
+ A.fi'c::n
+
~ (J.:_m
+
+A ~iii + A"\ "til +
b.
+ .a..(c: ~
<fc: .ISr +
b.
<rc: -sz.
u.1t a + ~'iY
cr,t' = 11,928.516
£\T
= .0012732
(z..T
= .0050924
()LT
()T
= .0012732
O'lT = 11,928.516
~tT
= O.
1tT
=
= 38,071.114
.505076
This problem was solved for five increments.
The actual
displacements, stresses, and strains will be slightly
different than the results calculated for five increments.
The more increments used the more accurate the results.
If the number of increments goes to infinity the solution
is exact.
'-----------~J
Section III
EXAMPLE:
ITERATIVE METHOD
In this section the same example solved in section
III with the incremental method is solved using the
iterative method.
L
.
ql
50,000
'
\00
Member #
Area
Length
l
l
200
2
l
100
3
l
2DO
The material properties are the same for all three
element members.
24
25
The stress-strain curve for all members
40k
30k
(]'=
10
6
10 (
X
-6.25
10
X
8
£.'
20k
lOk
t.
.oo4
.002
The load of 50,000 will be solved for five iterations.
E for each cycle will be taken as the secant modulus
for each members stress state.
[ Ki
Kl = E 1 /200
K3
J
[
= E.A./L.
l
l
[·75,
l
The stiffness matrix is
cose,
2
.
-s1necose,
-. 433]
-;.433,
.25
J
-s1necose
.
•
;!
s1n e .
K2 = E 2 /100
a.]
[0·'
o.,
1•
.433]
75'
- E 3/200 [..433,
.25
For the first increment E. is taken as the secant or
l
tangent modulus at zero strain.
l
----'
26
Ei = 10,000,000
The stiffness matrix for the first iteration
o.J
l.5,
r..
2.5
0.'
The deflection for the first iteration
{ ql\
q2 I
\:~\I
=
=
.666,
l/50,000
[ o.,.
0.]
.4
t:1
The strains for each member
c._
=
1/Li { cose, sine\
!, =
1/200
~ • 86 6'
{l.
=
1/100
l
e:3
=
l/200 (-.866,
o.'
.• 5 ~
1.\
.5J
l:~\
&:1
~~\
t::\
27
• For the first iteration
£,
= .001
£a.
= .004
£3
= o001
From the stress-strain curve the stresses due to the
strains for the first iteration
cr,
cr\.
G3
= 9,375o
= 30, OOOo
= 9,375o
The load in each member for the first iteration is
... h.~
p1 = 9,375.
p2 = 30,000o
p3 = 9,375o
(7·
The loads of all three members produce a load in the Q2
direction of
Q2
= 39,375.
If this load would have equal the applied load of 50,000
the solution would have been reachedo
Since the truss
is not in equilibrium a second iteration is needed with a
28
load equal to (50,000 - 39,375) or 10,625.
The load
applied for the second iteration is 10,625.
For the
second iteration the secant modulus is
El = 9.375 X 10
6
E2 = 7.50 X 10
E3 = 9.375 X 10
6
6
The stiffness matrix for the second iteration
=
.703125,
[
o.,
.984375
10-5 [ 1.4222,
o.,
o.J
0.]
1.01587
~
0
~
110,62~
The total deflection after the second iteration
q 2T
The total strains
=
~re
.507937
calculated from the total
displacements.
I
'
__________________________j
29
£,T = • 0012698
£'-T
=
.00507937
£~, = .0012698
From the stress-strain curve the stress after the second
iteration are
cr,,.. =
11,690.19
cr~.T = 34,668.73
O""T
=
11,690.19
These stresses produce a load in the Q direction of
2
46,358.9.
(50,000 -
A third iteration is solved with a load of
46,358.9) or 3,641.1.
For the third iteration
the secant modulus is
6
El = 9.2063 X 10
6
E2 = 6.8254 X 10
6
E3 = 9.2063 X 10
The stiffness matrix for the third iteration
o.]
.69047,
[
o.,
_,
.91269
10-5 [1.4482,
o.,
11
0
\
1.09565J h,64l.l)
I
I
_ _ _____j
30
The total deflection after the third iteration
q
\T"
=
o.
.547831
The total strains are calculated for the total
displacements
£. \T = • 0 0 13 6 9 58
!1.-r = .00547831
f~,.
=
.00136958
From the stress-strain curve the stress after the third
iteration are
cr,"
=
12,5-23.46
<'i.r
=
36,025.68
0',,
=
12,523.46
The stresses produce a load in the Q2 direction of
48,549.14.
A
fourth iteration is solved for a load of
(50,000 - 48,549.14) or 1,450.86.
iteration the secant modulus is
L
El = 9.144
X
E2 = 6.576
X
10 6
10.6
9.144
X
10
E~ =
.J
6
For the fourth
31
The stiffness matrix for the fourth iteration
o.]
• 6858,
[
_,.o.,
\Aq\~
=
\"q\u
10 5
=
.8862
[Kt])<>Q\N
[~:: ~.1284]
58
J
} 1:4 5 a.8 5
The total deflection after the fourth iteration
=
o.
The total strains are calculated for the total
displacements
c.T
=
(LT
= .005642
(')T
= • 001411
.ool4ll
From the stress-strain curve the stresses after the third
iteration are
( j\ \
l
= 12 , 8 6 5 • 6 7
32
~T"
<T')T
= 36,524o97
= 12,865o67
These stresses produce a load in the Q direction of
2
49,390 which is approximately equal to the applied
load of 50,000.
If more accuracy is desired, the
iteration process can be continued for more cycles.
this
prob~e~,
this result will suffice.
For
The final
results for the displacements, stresses, and strains are
= 12,865.67
= 36,524.97
= 12,865.67
= .001411
= .005642
= .001411
These results are approximately equal to the results
obtained for the incremental methodo
For the exact
results, see the same example in the appendixo
In
general, the iterative method is faster than the incremental methodo
33
The exact solution for strains. stress and ilef'lections
are
0\
=
Cia. =
Oj =
=
t:a. =
£) =
13,000
(,
37,000
13,000
~' =
12-
=
.00143
.00572
.00143
0
.572
The largest error occurs in the deflection at q • For
2
the incremental method, the percent error in deflection
is 11.9%.
For the iterative method, the error is 1.4%.
The results can be improved by taking more steps.
Section IV
COMPUTER PROGRAM
IV-A
FLOW CHART FOR THE INCREMENTAL METHOD
lRead data
I
Zero out
Total deflections
Total stresses
Total strains
.
Calculate elements
tangent modulus
based on the total
element strains
;
Assemble t.otal
svstem stiffness
matrix
AssembLe renuced
stiffness matrix
and reduced
incremental loads
Calculate reduced
incremental deflections
l
...
_j
34
35
Calculate total
system incremental
deflections
I
Calculate element
incremental stresses
and strains
Calculate the sum~of
all the incremental
stresses, strains,
deflections and loads
1
No
Does the sum of all
incremental: loads
equal the total
applied load
Yes
IPrint
~e~ult~
I
I End]
------------~------------___j
36
IV-B
FLOW CHART FOR THE ITERATION METHOD
IRead data
'
Ze-ro out
Total deflections
-- ----.-----------
Calculate element
secant modulus based
on the total element
strains
--- ---- -
Assemble total
system stiffness
matrix
Assemble reduced
stiffness matrix
and reduced resultant
.load
•..
-------
-------------- --- --
Calculate reduced
iterative deflections
based on the
resultant loads
I
,
Calcutate total
svstem iterative
deflection
37
Calculate total
deflection based on
the sum of all
iterative deflections
Calculate element
strains
Calculate element
stresses based on
.element strains
,
Calculate resultant
loads b~sed on the
applied loads minus
the loads produced
by element stresses
No
Are the resultant
loads approximately
equal to the applied
loads
IPrint results)
38
IV-C
MATRIX INVERSION BY DECOMPOSITION
The finite element method requires the inversion
of the stiffness matrix
[ K] •
Much of the computer
calculation time is involved in inverting this stiffness
matrix.
An efficient method for inverting the stiffness
matrix is desirable.
One fact about the stiffness
matrix is that it is symmetrical.
This can be proved
by Maxwell's reciprocity theorem.
1.
First a load P
1
is applied to a structure and
gives a deflection at P
1
equal to x
11
•
The external energy is equal to
2.
Now a load P
2
is applied to the structure and
produces an additional deflection at P
to x
22
and an additional deflection at P
eaual to x
L_______________
2
12
•
equal
1
39
~'1.'1.
The external energy for both loads applied
3.
Now a load P
2
is first applied and it produces
a deflection at point 2 equal to x
22
•
I
The external energy is equal to
4.
Now a load pl is applied and produces a
deflection at pl equal to x
and an additional
11
deflection at p2 equal to x21·
I
J
40
~-
Pz.
,
1=
The external energy for both loads applied
Since from superposition the external energy is independent of the order in which the loads are applied.
Since
Therefore
Similarly it can be proved that
k ..
l,l
L
=
k ..
Jl
41
The symmetrical stiffness matrix can be decomposed
into the product of three matrices
T
[uJ (n] [u J
Where
,[D] is a diagonal matrix and
[u
J is
a upper
triangular matrix
""
ID] =
[u]
T
=
0
Daa.
C>
~
'
..
Uaa.l
u... u,.. .
ju] =
' "•z. .. u,...
. . u...,
'
o.,.
I
0
.
\
From matrix algebra
.a
.. ,
T-t
[uJ [n] [uJ
The coefficients of
[u1 ,
(u]
and
[D
Jcan
be
determined from matrix multiplication
Kll
=
Dll
K22
=
2
u12 +Dll + D22
Kl2
=
Dllul2
K23
=
ul2ul3Dll + D22u23
Kl3
=
D u
11 11
K24
=
ul2ul4Dll + D22u24
Kln - Dlluln
K2n
=
ul2ulnDll + D22u2n
I
I
I
--'
42
From a more exhaust matrix multiplication, the recursion
formulas can be determined as
Dll
=
All
uii
=
l
ulj
=
Klj/Dll
D..
ll
=
K.. - s
u ..
=
1/D ..
l.l
for i=l,n
.i
C:-'
~
2
2
uk. Dkk
.,e: 1
l
ll
( K ..
ll
The inversion of
lJ
(D] •
-
i~
2
\.-•
·~
"··
[UJ ,
ukiukj Dkk)
[u]
and
i 'Z 2. jz. i+l
can also. be found
from matrix multiplication
_,
D..
ll
=
1/D ..
D..
=
0
l.l
ll
for i rl .i
(UJ= (r]
[uJ
_,
_, \J . l
u.
Uu
u _, .
2l.
•
u,..,
v-'\.to)
Un.
\
•
..
v,"'
Ot,.,
-
• •
~
0
.
.,
u.. ~
[l.]
0
IJ~o~N
Ii
l
------------~
43
From matrix multiplication
1
=
-1
ull
1
=
-1
u22
1
=
-1
uii
0
=
u12 + ul2
0
=
-1
-1
ul3 +, u23ul2 + ul3
0
=
-1
-1
-1
ul4 + u24ul2 + u34ul~ + ul4
=
for i
l,n
-1
From an exhausted matrix multiplication, the recursion
formulas can be determined as
-1
u l..l
=
1
for i
=
l,n
.)-1
u -1
..
lJ
,.-·
[u]
-1
= - w.~.u.k
l
~"'
=
for i< j
uki
-\.,.
[u]
Now the inverse of the stiffness matrix can be determined
from matrix multiplication
_,
=
_,
"T-\
[u] [D] [u)
44
IV-D
INPUT INSTRUCTIONS FOR THE INCREMENTAL METHOD
The capacity of the program depends on the size
of the dimension statements.
At present. the program is
dimensioned to handle a maximum of 10 nodes, 30 elements
and 10 different materials.
The input instructions are
listed as follows:
I
Control cards
There are two control cards.
The first card reads
the total number of nodes, total number of elements. and
the total number of materials.
The second card reads
the number of nodes with loads and the total number of
increments.
(Note:
If NLI=l. it is for the elastic
case.)
Read (60,501) NNT,NNE,NMT
501 Format (3I5)
Read (60.502) NL,NLI
502 Format (2I5)
II
Material cards
The total number of material cards depend upon the
total number of materials and the type of material.
The
first card reads the type of material, 0 for elastic
plastic and 1 for a third order polynomial tangent
modulus verses strain curve.
If the material tvpe if 0,
then the next card reads the Young's modulus· for the
45
elastic portion and the strain when the tangent modulus
is zero.
If the material type is 1. then the next 2
cards read 4 values of tangent modulus on the first card
and four values of corresponding strains on the second
card.
Do 10 I= l,NMT
Read (60,503) ITYPE(I)
503 Format (I5)
For ITYPE(I)=O
Read (60~505) EL(I),SP(I)
505 Format (2Fl0.5)
Go to 10
For ITYPE(I)=l
Read (60~504) El(I),E2(I),E~(I),E4(I)
Read (60,504) Sl(I),S2(I),S3(I),S4(I)
504 Format (4fl0.5)
10 Continue
III
Nodal cards
The total number of cards is equal to the total
number of nodes.
These cards read node number, degree
of freedom in the I, J, K direction, (0 for free, 1 for
fixed), and the X, Y. Z location for the particular node.
The nodes must be read in node order.
Read (60,506) NN,IDF(I),IDF(J),IDF(K),X,Y.Z
506 Format (4I5,~Fl0.5)
IV
Element cards
The total number of cards is equal to the total
number of elements.
These cards read the element number,
,_______________j
46
the two nodes located on the end of the element, the
material number, and the area of the element.
Read ( 60,507) NE,NNI (NE) ,NNJ (NE) ,MN(NE) ,AREA(1'E)
507 Format (4I5,F10.5)
V Load cards
The total number of cards is equal to two times the
total number of nodes with loadso
The first card in a
sequence reads the node being loaded and the second reads
the load in the X, Y, Z direction.
Read (60,508) Nl~
·
508 Format (I5)
·
Read (60,509) P(I),P(J),P(K)
509 Format (3F10.5)
47
IV-E
INPUT INSTRUCTIONS FOR THE ITERATIVE METHOD
The capacity of the program depends on the size
of the dimension statements.
At present, the program is
dimensioned to handle a maximum of 10 nodes, 10 elements
and 10 different materials.
The input instructions are
listed as follows:
I
Control cards
There are two control cards.
the total number of
nodes~
The first card reads
total number of elements, and
the total number of materials.
The second card reads
the number of nodes with loads and the total number of
iterations.
(Note:
If NLI=l, it is for the elastic
case.)
Read (60,501) NNT,NNE,NMT
501 Format (3I5)
Read (60,502) NL;NLI
502 Format (2I5)
II
Material cards
The total number of material cards depend upon the
total number of materials and the type of material.
The
first card reads the tvpe of material, 0 for elastic
plastic and 1 for a third order polynomial stress verses
strain curve.
If the material type is
o.
then the next
card reads the Young's modulus for the elastic portion
48
and the strain when the
tan~ent
modulus is zero.
If the
material type is 1, then the next 2 cards read 4 values
of stresses on the first card and four values of
corresponding strains on the second card.
Do 10 I= l,NMT
Read (60.503) ITYPE(I)
503 Format (I5)
For ITYPE(I)=O
Read (60,505) EL(I).SB(I)
505 Format (2Fl0.5)
·
Go to 10
0
For ITYPE(I)=l
Read (60,504) Sil(I).SI2(I),SI3(I),SI4(I)
Read (60.504) Sl(I),S2(I).S3(I),S4(I)
504 Format (4Fl0.5)
10 Continue
III
Nodal cards
The total number of cards are equal to the total
number of nodes.
These cards read node number,
de~ree
of freedom in the I, J, K direction. (0 for free, l for
fixed), and the X, Y, Z location for the particular node.
The nodes must be read in node order.
Read (60,506) NN,IDF(I),IDF(J).IDF(K),X,Y,Z
506 Format (4I5,3Fl0.5)
IV
Element cards
The total number of cards are equal to the total
number of elements.
These cards read the element number,
I
------·------·------~__j
49
the two nodes located on the end of the element, the
material number, and the area of the element.
Read (60, 507) NE ,NNI(NE) ,NNJ (NE) ,IVJN(NE) ,AREA(NE)
507 Format (4I5,F10.5)
.
·
:
i
V Load cards
The total number of cards are equal to two times the
total number of nodes with loads.
The first card in a
sequence reads the node being loaded and the second
reads the load in the X, Y, Z direction.
Read (60,508) NN
508 Format (I5)
Read (60,509) P(I),P(J),P(K)
509 Format (3F10.5)
.l
50
IV-F COMPUTER LISTING FOR THE INCREMENTAL METHOD
51
!------
·-----,---------------
LN ODC1
COMMON AC10J,S1C101,CI101,01101,STRAINI301,SPI1CI,EI3GI,ElC10J
LN 0002 ___________ COMMON S1110I,S21lOJ,S31HI,S411CI ,ITYPEI1Ql,HNI1QI,NET
lh OOCJ
CI~EI.SICN £11101 ,<:2 I 101, E 3 !101,£4 I 101, IGF 130 I ,x 13C I, T(3Q I ,z (JOJ ___ _
LN 0004
OIHENSICN Bl3u,&I,P1301,STRESSI3Ql,PRIJQJ
.
lh 0005
OIHEN3ION SI30,3CI,NNII3GI,NNJI3DI,AREAI30l
-LN 000&-- -------- DI!-:EI';SJCN XTI301 ,GK IJC ,JQI, SR 130, 3GI ,SRI 130, 3;j'J-;OXI30f - - - - - - - - LN 0007
OIHEN~IO~ OXRI3DI,OSTRAINI30I,OSTRESSI30I,TLENGTHI301
LN 0008
C R~AO NUMBER OF NODES, NUMBER CF ELEMENTS, NUHOER OF HATER!ALS
-LN 11009
.. --READ l&u,5G11 t<NT,NET,NMT . -.- -- - --------------LN 0010
501 FORMAT 13151
LN 0011
C READ Nyr.BER OF NCOES WITH LCAOS, AND NUMBE?. OF INCREMENTS TO DIVIDE LOADS.
LN 0012
C t.OTE IF I'll IS EQUAL TG 1 IT IS .FOR ELASTIC CASE
---------- - - - - - LN 0013
READ 1&0,5021 hL,NLI
LN 0014
502 FORMAT 12151
--lN-0015 ---- --- ·oc
10
:t:i~NHT
LN
LN
LN
LN
LN
001&
C RE~O ITYPEIII 0 FOR ELASTIC PLASTIC, 1 FOR lRD ORDER POLYNO~IAL
0017
READ 1&0,503, ITYPE IIJ
0!118
503-FCi'IHAT 1!5) -- -------'--'-~--- - - - 0019
IF IITYFEIII.EC.OI GO TO 20
0020
C READ 4 VALUES FOR F iNO CORRESPONDING STRAINS
--LN 0021
C TWC CARCS, E ON FIRST CARD STRAit1S ON SECOND
lN 0022
READ lo0,5041 Eliii,E21Il,EHU ,E4(J)
LN 0023
REAC I&C,5041 Sliii,SZIII ,SJI!I,S4111
---LN. 0024____504 FOR HAT lloFlO .51 --- -LN 0025
•tll=Ellll
LN 0026
B11II=IE21Il-AII11/IS21II-S1CIII
--LN-Oif27_______C II I= IE31 II-A I 11-01111 •c SJI I)~-Slt'"I'"I'I'J'/LN 0028
111S31Il-S11III•IS31Il-S21IIll
LN 0029
01Il=IE41Il-AIII-B11II•IS41II-S11!11-CIII•IS41II-S11III•
---01 -OOJQ ---------+1 S4 II 1-SZI I IIIII 154 111-Stii l I •1 S<+l II-S21I fl • l::><+ I li-S31Ii1T-lN 0031
GO TO 111
LN 0032
C RE~D
FOR ELASTIC PLASTIC MATERIAL E ELASTIC. AND STRAIN WHEN E=O
--LN0033______
20 READ 1&0,5051 El.IU -,SPH.i _______________________ _
lN 0034
505 FORMAT 12F10.5J
tN 0035
!0 COt-<TINUE
.
t:.rr oo36 --------- r=c ----
LN 0037
DO 30 J=t,NNT
lN 0038
C REAC NODE NLHOER DEGRtE GF FREEDOH IN !,J,K
--lN-0039____ cx.y,z NOTE HUST BE RE4DIN NCOE O~LJER·1~2,J--;i;_E_CT~----l~ 00'+0
READ 160.50&1 t-<N,lOFIIl,lOFII+11,IOFII+21,X!JI,YIJI,ZCJI
LN 00'+1 ______
506 FORMAT
1<+!5,3F10.51
-TN-004i
1=1+3- -----------_•__:_____ _ _ _ _ _ _ _ _ _ _ __
LN 0043
JC CONTINUE
LN 0044
DO 40 I=l,NET
--l:N-CQ45
C REA!J ELEI'!fNT tJUMBER,IHl_N_ODE -NUH6ER.JTH- 1';00£ NUMBER.-HATU<fAl--N,JMtlfR;--;l'~EAlN 0046
REID 1&0.5071 NE,NNliNEI,~NJIN~I,HNI~EI,A~EAINEI
LN 0047
507 FO~MAT 14I5,F1Go51
LN 0048
N!,NNIII\EI -. - - - - - - - - - - - - - - LN 0049
NJ=I\1\JII\EI
lN 0050
XLX=XlNJI-XCNII
-LN 0051 ·------- VLY=YINJI-YINI)__ _
lN 0052
ZlZ=liNJl-Zit-<II
LN 0053
1LENGTHINEI=SQRTIXLX••2+YLY••2+ZLZ••21
Ltr oas"·--
BlNE.ta=-xLx-
--- ----· --------- ----··-
-·- --·-------l
I!
52
-·=·-~--~~~-~·~~-~~~:--------:.:------•---'L.·-·~---------___....,_
ANSI FORTRANlZ.31/HASTER
LN 0055
LN 0056
-----7-L:::-N-0~:0
LN 0058
LN 00~9
OFF o 0 OPTION IS
9fNEo21=-YLY
eBlNE,Jl=-ZLZ
INE ,1,) :xix=-------------,--------------_;_-.,,1
si
---~L~N~O~O·&O
INTEGER WORD SIZE = 1 • • OPTION IS
______ ......__
Bl~E,SI=YLY
~0
81NEo61=ZLZ
NOF~N~T·J;==------------------------------------------------
LN 0061
C ZERO OUT DISPLACEHENTSoSTRAINS.STRESSESo LOADS
LN 0062_ _ _ _ _ 00 50 I=t,NDF
----:L"-:N-·oo_&_J
__ c::.:.._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _,,
ifTil.=:a~
LN 006~
LN 0065
----='L-:.:N-Oil!i6
LN 0067
----~LN 0068
di0069
LN 0070
LN 0071
50 Plli=O.
DO 60 I=1oNET
S Tf~-Afh C f i : o • - " . ' - ' - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1
60 STRESSIII=O•
ICOUNT=1
C REAO-NOO-E--LCA-05 lNODE NUMBER, AND LOAD IN X,Y,ZI
00 70 I=1,NL
READ 160,5061 NN
----L~,~N'-"00 7z
50 8 F !JRHA·r-·i I 51··-'--'-=-------------'---------------__;1
LN 0073
NJ:NN•3-3
LN 007~
70 READ 160,5091 PINJ+11,PIN3+21,PlN3+31
--~-'L,~N--00 7 5
5 09 FOR 11AT--CJF 10~5-,-------··=-=-=~~~~----·-----------LN 0076
C ASSEMBLING REDUCED P COLUHN HATRIX·
LN 0077
NB=._O
---~L~N,~0078
IP=1----------------------------------lN 0079
00 60 I=t,NDF
r---~l~N~OO~O
IF IIDFCII~·~N~E~·~O~Ic~G~O~T~0~8~0~------------------------------'
LN 0081
N~-;-Ni<i-1-·
LN 0082
PRIIPI=FIII
lN 0083
IP=IP+t
----:L~N~OD8~
eo co.ii'rTiiu;.:Ec-----------------------------'----LN 0085
WRITE 161,6001 NNT,NET,NHT
---~1,-"li--:0086
600 FCRf'44 T I• NUMBER OF ELEMENT_S __ _:•;I5,/I,
LN oo87
f•-NUi-1eEF<-·aF ELEHENt-s··-=·.rs,ll,
-'..:....:...!_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _- '
LN 0088
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LN 0119
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C CALCULATE YCUNG'S HOOULUS FOR EACH ELEMENT
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RETURti·-=------------------------------litO WRITE 161,'1011
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FCR~AT c• TRUSS.~I~S~l~N~S~TA~B~L~E•_I~---------------------------______
END
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30 Elll=ELIHI
10 C~N~T~I~N~U~E~----------------------------------------------I
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Gv
f)
20
ST~=:S31Il=AIMl+l11~1•15TR~I~IIJ-S1('1lJ+
1 C I '11 • I 5 T ~ ~ I 'l I I l - S 1 I :i J I • I S T~ l !11 I r l - S 2 1:11 J +
2Q(qJ•Ist:>l!qiii-~11~11•1st~&INIII•S2i4(l*IST~liNIII-,31q)l
GO T 0 1 J
OJO~
LN 000'3
L~l 0010
20 IF !STi'.!.I'Hil,LT,SPC~ll Gll TO 10
VI 0011
------nrr-rr~~--------~s~Jr.o~.e~~T·~~.~~··c~~,~,----------------~---------------------------OH2
LN
GO
LN 0113
ro 10
L"l !101'-
30
<;TO-':S~!Il=ELCMl•.>T><.AI!dfl
Lfl 0015
LN 0116
LN OH7
LN OJ 11
LN OH<:l
lU
(;J"lllliel::.
USASI
00 '+0 I=t,N-':T
40 E!II=STiE551II/ST?AINIII
~::.I!Ji!•l
"Ill)
FQRT~AN
)10
DIAGNOSTIC RESULTS FOR SIGMA
fRROR.)
'r--------------------r
--r-------------- -
------~----
!
---------------------·~-
I
________ j
Section V
BIBLIOGRAPHY
1.
Desai,C. s •• Introduction to the Finite ETerrient
Method, Van Nostrand Reinhold Company, New York,
New York, 1972.
2.
Przernieniecki,J. S., Theory of Matrix st·ru:ctural
Analysis, McGraw-Hill,Inc •• New York. New York. 1968 •
•
66
Appendix
A
DERIVATION OF THE STIFFNESS MATRIX FOR A
THREE DIMENSIONAL TRUSS
In this section the stiffness matrix is derived
for a three dimensional truss using the method described
in the introduction.
The stiffness matrix for other
types of structural elements. can be derived using this
method •
.Y
......-x
~----
The displacement function is
ut~'
=
--<'
+ Ot~ s
U(s) is the position of any point on the truss as a
function of s.
q
and
q
are the position of the ends
of the truss after the loads have been applied.
(u~ -
67
68
Where
l u\
is the displacement function,
tC~tJ
polynomials of s and
1<t 1 is the
is the coefficients of the
polynomials.
Where
{q
I
is the displacements of the nodes,
[A
J 'is
the value of the dependent variables at their particular
node.
From matrix algebra
Where
A
lOf} can
be determined.
is the displacement transformation matrix
~:j
r
1,
= l-1/L,
l
l
J
loa
69
=
[1 -s/L,
The strain can now be determined from the displacement
function.
[-1/L,
J t~l
1/L
The stress can be determined from the strain.
For a
truss
= E
The internal strain energy
L
Ux
=
f !G:\
[-1/L,
1/L] EA
C)
Where A is the cross sectional area
L
=~J~::'
- '=
1::)
I
.~
______ j
70
The external work, the work done by the external forces
{Q)
\Je.
=
-i\q\' tQ1
Uw::
=
t
rlf rlJ
q2
Q2
The external work must equal the internal work
=
~ b:Y
1,
EA/1
[ -1
.
,
=
Therefore the stiffness matrix is
(K]
=
1,
EA/L
[ -1,
Transformation from the s coordinate system into the
x,y,z coordinates
Qlx
=
Q2x
=
1 Q2
Qly
=m
Ql
Q2y
=
m Q2
Qlz
=
n Ql
Q2z
=
n Q2
1 Ql
71
Where 1, m, n are the direction cosines between the
x, y' z coordinates and the s axis respectively.
Qlx
1, 0
Qly
m, 0
Qlz
n, 0
o,
1
Q2y
o,
m
Q2z
0, n
Q2x
=
[ 1, -1 J tq1\
EA/L
-1,
1
q2
ql
=
1 qlx + m qly + n qlz
q2
=
1 q2x + m q2y + n q2z
qlx'
qly
{ :~
[1,
=
m, n,
o, o, o,
o, o,
0
1, m, n
]
qlz
q2x
q2y
q2z
Qlz
=
1, 0
qlx
m, 0
qly
n, 0
EA/L
Q2y
o,
o,
Q2z
0, n
Q2x
1
m
[ 1, -1] [1,
-1,
1
m, n,
o, o,
o, o, o,
1, m,
..,
1
qlz
q2x
q2y
q2z
II
I
J
72
1, 0
m·, 0
[K]
=
n, 0
o,
o,
o,
EA/L
1
[ l,
-1,
-1
1
1, m, n, 0,
o,
o, o, o,
1, m,
m
~
n
Therefore the stiffness matrix is
12
1m,
1n,
-12
-1m,
-1n
1m,
mm,
-1m,
-mm,
-mn
1n,
mn,
mn,
2
n,
-1n,
-mn,
-n
-1n,
12
1m,
1n
-mn,
2
-n,
1m,
m,
mn
1n,
mn,
n
'
[K]
L
=
EA/L
-1~
-1m,
-1m,
2
--m,
-1n,
-mn,
'
'
2
2
Appendix B
EXAMPLES FOR THE INCREMENTAL AND ITERATIVE METHOD
Two separate examples are shown
incremental and iterative method.
both the
fo~
The input and output
for the computer programs are shown in this section.
The
first example is a two dimensional 3 bar truss where all
elements are composed of one material.
I
I
The second example!
is a three dimensional truss with 5 bars.
There are two
types of materials in the second example.
For the first
material, stress is a function of strain and a third
order polynomial fit is used in the program.
material is elastic-plastic.
The second
The number of total incre-
ments for both examples was taken as 50.
total iterations ror both examples was 10.
The· number of
It was found
that much fewer cycles were necessary to achieve the same
accuracy with the iterative method.
~-----·-------------------------------·----J
73
74
p
Q
Node number
Q
Element number
C:::,.
Material number
Example 1
L_ _
75
Node number
-X
y
-z
IDF(I)
IDF(J)
IDF(K)
1
o.
1
1
173.2
1
1
1
3
346.3
1
1
1
4
173.2
100.
o.
o.
o.
o.
1
2
o.
o.
o.
0
0
1
Element number
Material number
Node(I)
Node(J)
Area
1
1
1
4
1.
2
1
2
4
1.
1
1
3
4
1.
The only node that has an applied load is node
load applied is 50,000.
4.
The
The material properties are as
follows
~
~1:,
= 10
t~a-
= -
d£
X
6
10 €, -6.25
= 10
X
10
6
X
10
-1.25
8 t
f.
X
10 9f.
For the iterative method, any four different values of
stress and corresponding values of. strain are input for
the material properties.
For the incremental method,
any four different values of tangent modulus and
corresponding values of strain are input for the material
properties.
76
COMPUTER OUTPUT FOR THE INCREMENTAL METHOD
EXAMPLE l
·
77
'
NUMBER OF ELEMENTS
-Nui18ER OF ELEMENTS
OF MATERIALS
~!1J!~R
a
..
=
3
..
1
1
1
2
J
1
1
1
1
It
0
0
1
1
1
1
.
1
DECREE OF FREEDOM
y
X
z
NODE
~UMBER
.
LOCATION OF COORDINATES
---y-
NODE POINT NUMBERS
ELEMENT
NUMBER
I
1
1
J
.3
z
z
z
X
o.ooooo
173.20000
Jlt&.;,.ooo·o
17J.ZOOOO
...
..
J
o.ooooo
o.ooooo
o.ooooo
100.00000
MATERIAL
NUHBER
1
1
1
o.oooao
o.ooooo
o;aooao
o.oooao
'
AREA
1.00000
1.croooo
t.ooooa
MATERIAL NUHBER =
TYPECJ'RO ORDER POL VNOMIAl.J
1
-ei= o. fOOY+oaEZ= Oo875E_i_Q7___n_::_o;7stnat___E4= Oo&25E+07.
sz,.
Slt::z O.JODE-02
O.DOOE+OD
O.tOOE-02
Sl= o.zooe-02
S1=
DEFLECTION OF NODE
NODE
NUHBER
1
2
y
X
OoliOOE+OD
o.oooe+oo
O.OOOE+OO
Oo985E-11
..,
3
ELEMENT NUHBER
o.oooe+oo
O.OOOE+OO
o.oaoE+OO
o.si2EXOO
2
STRAIN
o.1ItlE-D2
Oo572E-02
3
a.t~tle-oz
1
O.OOGE+OO
O.OOOE+OO
O.OOOE+OD
o .ciliof+oo
I
;
z
STRESS
Oo130E+05
o.J70e+os
o.13oi:+os
.
----------
-
.
I
I
l=______
-
--·---·----·
~
78
COMPUTER OUTPUT FOR THE ITERATIVE METHOD
EXAMPLE 1
79
..
NUMBER OF ELEMENTS
NUMBER OF ELEMENTS
a
J
NUMBER OF MATERIALS a
1
NOOE
I
DECREE OF FREEDOM
-NUMifE~X--Y---Z----""-'-'-----
lOCATION OF COORDINATES
ic
--- -, Y --------z-------------...;!
1
2
3
1
1
·1
1
1
1
1
1
1
o.ooooo
o.ooooo
113.2oooo
o.ooooo
3,6-.-,.-oooo,-------------co'-.'-iiooooo
..
0
0
1
173.20000
ELEMENT
NUIIBER
o.ooooo
o.oooJo
o.ifoifo--<u------------o.ooooo
100.00000
NODE POINT N~HBERS
I
J ·
1
1
..
MATERIAl
NUMBER
1
3
3
1t
1
AREA
1.00000
ooooo·---------------,
t.ooooo
-----~2-----zi------.,.c----------~1'----.-1-'-.
MATERIAL NUMBER
=
1
TYPEI3'RD ORDER POLYNOMIAll
--sTRISs-f:-(f ;-a ooE+0 0----STR Essz;-··a: 937E + o4------ STRE.SSJi -Q-;JOQ E-+Q_5___STR~·ss4-=~-t;7Ei05-STRAIN1= OoOOOE+OO
NODE
NUMBER
STRAIN2= O.lOOE-02
STRAIN3s O.ltQOE-02
DXEFLECTION OF NOYOE
---'---'--~1'=''----.o•.•o"'oo E+o o
2
3
OoOOOE+OO
O.OOOE+OO
---7,.'----o~.~34-oE-u
STRAINit= Oo5G6E-D2
l
i'_
o. a·•o•o-E'" +0·-o..-:------:o•.--:o"'o"'"oE~'+c'o"'o'"'---------------------7
O.OOOE+OO
O.OOOE+OO
O.OOOE+OO
O.OOGE+CO
o. 57 7e+-oo<-------.o<-'.-.;-o aoE+oo:---------------'--------
--.E"l"E"'M'"E"'N'"'T---.o-N~U:-cMB"'E"'R"------SrRifi"'N.-------,S~'T"'RE"SS~-----~------------
1
2
OolltltE-02
Q.577E-02
--,-------;li------~-io-=-.iltlfE-02
0.131E+05
. 0.36qE+05
Oo1J1E+-If!:>------------------
-~--------------------·------------------------·---------·
--~
' '
...
i
'
''
iI.~
j:
--
l
I
c-1
80
Q
Node number
<:}
Element number
/:::,.
Material number
Example 2.
81
Node number
-X
y
-z
IDF(I)
IDF(J)
IDF(K)
1
o.
1
1
1
2
100.
o.
o.
1
1
1
3
100 • . 100.
1
1
1
1
1
1
1
1
1
0
0
0
4
o.
100.
5
50.
50.
o.
o.
o.
o.
o.
6
50.
50.
200.
Element number
Material number
Node(I)
Node ( J)
Area
1
1
1
6
1.
2
1
2
6
1.
~
1
3
6
1.
4
1
4
6
1.
5
2
5
6
1.
The only node that a load is applied is node 6.
applied is 100,000.
The load
There are two types of materials.
The first has the following properties.
For the iterative method. any four different values of
stress and corresponding values of strain are input for
the material
properties~
For the incremental method,
j
__ _j
83
COMPUTER OUTPUT FOR THE INCREMENTAL METHOD
EXAMPLE 2
82
any four different values of tangent modulus and
corresponding values of strain are input for the material
properties.
The second material is elastic-plastic.
aso,ooo-
------~--------------~(
. 005'
The input for elastic-plastic materials are the same
for both the incremental and iterative methods.
E = 10.000.000
(p
= .005
84
.
-
-
--·
NUI1BEA OF ELEMENTS
:II
-.rutiOf:A OF ELEMENTS
"
5
=
2
NUHBEA OF HATEAIALS
6
!
'
'
LOCATION OF COORDINATES
DEGREE OF FREEDOM
NODE
--.rul1aER
l
2
,.3
)(
y
1
l
1
1
1
1
0
5
6
z
1
1
1
1
1
1
1
1
l
0
0
X
O.OGOOO
100.00000
10~00000
o.ooooo
so.ooooo
50.00000
I
1
1
2
3
2
,.3
,.
5
5
0.0·0000
Q.OOOJO
o.ooooo
tinr;ooooli
100.00000
50.00000
o.~ou.ro
o.ooooo
o.ooouo
50~00000
2011-~-~~0000
MATERIAL
NUH.BER
1
1
AREA
J
6
6
&
1
NODE POINT NUMBERS
ELEMENT
NUHBER
z
y
o.ooo~o
1
6
2
&
t.ooooo
loOOOCiO
1.00000
1.ooooo
1.ooooo
..
..
TYPEC3 1 RD ORDER POL YNOHIAU
HAT ERIAL NUHBER
1
Elt= 0.1JOE•07
EJ= 0.755E+IH
E2= o.79&E+07
El• o.sJJE+07
Sit= Oo10JE-01
SJz O.ZOOE-02
Sl= O.OQOE+OO
S2= 0.100E-02
--HATER IAL-NUHBER =
TYPE
I
ELASfiC-P-i.ASt
ICI
Oo100E+J3
E
z
=
PLASTIC STRAIN = "0 ..
DEFLECTION OF NODE
NODE
NUMBER
1
2
3
y
)(
OollOOE+OD
0 .ooiiE+OC
Q.ODOE+OO
O.OOOE+OO
o.ooo-Hoo
o.oooE+oo
,.
5
6
ELEMENT NUMBER
1
2
J
,.
5
O.OOOE+OO
o. aoof:+oo
O.OOOE+OO
O.OOOE+OO
o.oOOE+OO
O.lltOE-10
STfiA IN
a ._?_5\IJ- o 2
o.ss9E-oz
0.559E-C2
Oe559E-02
o.629E:..cz
l
OoGOOE+OO
o~oooE+ifo
O.OOOE+OO
O.OOOE+OO
o .oo-oe+oo
0.12&E+G1
STRESS
o.J9&E+05
o.J9&E"ios
j
O.J9&E+G5
o.J9&E+Os
I
C~SO&E+05
---
I;
I\'.... ,.,.,,.._
i
1
85
COMPUTER OUTPUT FOR THE ITERATIVE METHOD
EXAMPLE 2
86
I
I
NUH:Jr::~
OF ::LE:Mi:<H5
NOMJ.:.::C OF
NU.r-l~£1P
cl:.M~
'II S
"
&
=
5
OF 11Al!':PilLS =
------
.
I
I
2
'lEG~!'!':
NOur:
NUMJ'".>(
z
\
,.
.!
s
i
~·
'
ELEM::NT
t
LOCATION OF
FPEED'l'1
1
1
1
1
1
1
u
0
n.o~ooo
1
1
1
1
1
1
::;oJ~tJIN~
1
1
1
u
[
a.ooooo
1tlO.JOOaO
a.~o~ao
1•JU, UuUuU
o.oooao
liJU,UUU•JU
100.00000
~o.uooao
:;o.aoooJ
'>U• OUOJO
7U.OOOOO
1.
i
~.ooE
NU•I•lt.>(
P'JI'H NU'1'1E"'S
~
l
HAT=:RI4L
"<J
'""I'(
1
1
o. oa.ono
o. oooco
u.uuut.u
o.oaooo
o.aoooo
.210. OiJOu•J
.)
b
L
l.uuuiJ~
"5
"5
&
5
1
l
t.ooooa
1.00000
&
I
1.00000
1.00000
2
.}
1
1
A 'I.E A
1
2
'!
II
TE.5
~
[
"
1
')I'
I
i
-----
--
l
!
II
Mille. RIAL f,.QM:!t:.R
-
SHt:SS1=
S T I'( A1'11
o.ao~.-·oo
Hid-:.~IAL
NO 1"3:. ~·
1 YPt.. (3 '~tj ORO"-R PJL TNO:OlJ.l\l J
sr<.::SSJ= 0.!)00:.+05
STRESS2= o.JOOE+O<;
ST-(A~'H= o.Bon.:-cz
STR..\ItiZ= 0,400€-02
1 YP:- It (s.~~-;:,llt...-PLH::.Ilvl
2
t. = ll. 1 UOt. +:1 K
1
= O.QOOE+OO
j
-
l
NOC-.
l ::.FL::.CTtJN
NUH'ER
~
I
1
'I
2
o.non
l
o.~oo
I
I
l
I
I
o.ooo •GO
,.
o.ooo
u.iloo
o.ooo
&
OF
r1.;ut
z
y
O,OOOE+OO
o. 000"+ 00
•00
o.ooo::·o~
o. no~~· oo
+00
+00
o.uoo;:·u~
O.OOE+OO
O.OOE+OD
+00
•oo
o. oao:::• oo
1), i) 0 0 ~. 0 I)
-o. ·121.0-tJ
o.oon~·oo
o.tZT::+at
I
l
I
I
j
ELE'IENT 'lUH:J':"!
i
3 lE+Uj
o.s.,s•-oz
I
0 • ')r;l)~.,. •Ol
o. J•JL+!l'>
"5
o.c;,t;~::-oz
o.J'lllL+05
o. SOil.:.•h
il.6.Hu:-lll
!
1
'
Q,
2
I
i
!
sr~~ss
ST~AIN
0.5&3<::-02
-
PC A::> Ill. ;::dRil tl
~
I
a. ,I
II
I
i
SH::SS't= 0.550 +0'>
ST~~I~i'+" 0 • 12U -01
o.Jill·-+il:•
I
IBl\£1
FORTRAN Coding Form
1 ~\ C.RE.M e:NTA\..
PROGRAM
['..
co
( EXA.MP\..e. \)
\"'\ \:i T\-'c 0 D
I
PROGRAMMER
L..__.___.
________
I
I~0
112
3
DATE
I~I
STATEMENT
NUMBER
u
---------------------
.
PUNCH
FORTRAN STATEMENT
. 0
u
4
GRAPHIC
PUNCHING
INSTRUCTIONS
51617
8
9101112131415
16 17
18 19
20 21
22 23 24 25 26 27
28 29 30 31
32 33 34 35
36
37 38 39 40 41
42 43 44 45
46 47 48 49 50
"I
52 53 54 55 56 57 58 59 60 61
rI
1
I
1
1
I1
!
!
62
63 64
~~-+--+--+--r
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I
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L\
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I I
I
tJ
I
:
i
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Ti·of"'\o·'_
v
...
i
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l
!
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I •
1--+_j1___
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l6 2.1sdo:do.
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: ~ -~ " ' 0 D~ - i c. A !\?.. 't?~ i ~j_J___
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