.
'
·CALIFORNIA STATE UNIVERSI'rY, NORTHRIDGE
FINITE
ELE~ffi1~
ANALYSIS OF
NONLINEAR GEOI-fETRIC STRUC'rURES
A project submitted in partial satisfaction of the
requirements for the degree of Master of Science in
Engineering
by
Romeo Gene Suarez
June, 1979
The Project of Romeo Gene Suarez is approved:
J. f!/Jiiy~ ly c>
j
"S,. Gad6mski, C6mmittee Chairman
California State University, Northridge
ii
ACKNOWLEDGEHEl'rT
Sincere gratiude and appreciation is expressed to my
wife Josephine, for her patience, understanding and
moral support, not only during the course of this
research but also throughout the course of completing
all the requirements for the degree.
iii
TABLE OF CON'rENTS
vi
ABSTRACT
1.0
2.0
I1~RODUCTION
1.1
The Finite Element Method
1.2
Nonlinear Elasticity Problem
2
1.3
Scope Of Present Work
THE NONLINEAR GEOHETRIC SOLUTION PROCEDURE
3
5
The Incremental Stiffness Method
5
2.1
2.2 .Comparison Between Linear And The
Nonlinear Method Of Analysis
8
3.0 APPLICATION OF THE INCREMENTAL STIFFNESS
l~ETHOD
3.1
TO A SIMPLE TRUSS PROBLEH
Stability Analysis
3.2 Load-Deflection Analysis
4.0 THE COMPUTER PROGRAM
4.1
14
15
~?
I I
29
Flow Chart For The Incremental
Stiffness Solution Procedure .
29
4.2 Input Instructi0ns
31
4.3 Example Problems
33
5.0 CONCLUSIONS AND RECONNENDATIONS
44
6.0
46
BIBLIOGRAPHY
Page
APPENDIX A - DERIVATION OF A TRUSS
NONLINEAR
STIFFr~SS
ELE~~1ff
47
11ATRIX FOR
LOAD-DEFLECTION RESPONSE AND
STABILITY ANALYSIS
APPENDIX B - DERIVATION OF A BEAN ELEHEJ:iT
NONLil\TEAR STU'FNESS
61
~IATRIX
APPENDIX C - COMPUTER PROGRAH LISTING
66
ABSTRACT
FINITE ELEHEN':P ANAIJYSIS OF
NONLINEAR GEOHETRIC STRUCTURES
by
Romeo Gene Suarez
_Master of Science in Engineering
.For some structures, linear analysis may not be adequate,
and nonlinear analysis should be applied. Here, an
incremental stiffness procedure which employs the finite
element displacement technique, is presented for nonlinear analysis of planar trusses. Elastic and geometric
stiffness matrices are derived for application to large
deflection and structural stability problem. Several
exa~ples,
including a five-element truss, have been
solved. For comparison, both linear and nonlinear
analyses are included. The numerical results are in good
agreement with known analytical results. The computer
program written for the nonlinear analysis appears in
Appendix
c.
v:i
'
_1, 0
IN~~RODUCTION
The incremental stiffness procedure embodying the finite
element displacement method is presented for nonlinear
analysis of planar trusses,
1,1
The Finite Element Method
The prublem in structural mechanics is that of finding the
forces and displacements (or the stresses and strains) in
a continuous system, In the finite element method, the
system is discretized by a number of elements which ar2
joined at selected node points. Each element has a finite
number of force and/or displacement parameter at the
nodes, Relationships between nodal force and displacements
are obtained by applying variational principles, The
response of the system is obtained by an assemblage of the
individual elements by requiring displacement continuity
at the nodal points., The assemblage of elements results in
a system of equations (with nodal displacements as the
dependent
v~riables)
\'ihich govern the structural behavior
for T.he entire structure, After applying the necessary
boundary conditions to the equations, the system of
equations ia solved for the nodal displacements,
'
~2
Nonlinear Problems
Nonlinearity is introduced into the theory of structures
in three ways:
1.
Through the strain-displacement equations.
2.
Through the equilibrium equations.
3.
Through the stress-strain relations.
The first two sources are termed geometric nonlinearities
while the third is due to material behavior and is,
therefore, classifi0d as physical nonlinearity or plastic,
behavior.
Four general types of problems arise as a result of the
considerations listed above. These are:
1.
Problems which are geometrically and physically
linear.
2.
Problems which are geometrically linear and
physically nonlinear.
3•
Problems which are physically linear and
geometrically nonlinear.
4.
Problems which are physically and geometrically
nonlinear.
The analyses pres8nted are limited to physically linear
and geometrically nonlinear structur9s.
"·
··---------
3
A finite element method of analysj.s for structural
problems which are physically linear and geometrically nonlinear is presented and illustrated# Since
large strains are encountered only occasionally during plastic deformation of ordinarily materials and
in rubber-like materials which are seldom used in
engineering structures, only the important sub-class
of geometric nonlinearity in which the displacements
"
are large but the strains are small will be discussed.
Method of analysis is illustrated for load-deflectiou
response of a truss structure. The incremental
stiff~
ness solution procedure embodying the finite element
method is presented in Section 2. Comparison l;>etween
the linear and nonlinear analysis is presented along
with the effect of the number of incremental loading
used. in analysis. Section 3 prese::1ts indetail the
analysis to determine the critical load at which a
simple truss becomes unstable utilizing the nonlinear
stiffness matrix derived in Appendix
A~
Results of
this analysis are compared to that obtained by the
classical method for substantiation.
The same simple truss problem is also analyzed for
maximum deflections for the applied load of same magnitude as the critical buckling load. The computer
program developed to analyze nonlinear load-deflection
response in a planar truss system is presented in
Section 4. This computer program is
open~ended
in that
any element can be added into it without affecting
the existing computational procedure for load-deflection
response of a truss structure. The nonlinear stiffness
matrix derivation for a truss and beam element is presented in Appendix A and in Appendix R respectively.
Listing for the computer program is presented in
Appendix c..
~--·-----
---~
- -"·- - - --
-
2, 0
THE NONLTNEAR GEOHETRIC SOLUTION PROCEDURE
2,1
The Incremental Stiffness Method
The solution procedure presented is the incremental
stiffness solution procedure embodying the finite element displaceffient method. In this method, the load is
applied as a sequence of sufficiently small increments
so that the structure can be assumed to respond linearly.
during each incremental loading. For each increment of
load, increments of displacements and corresponding
increments of strain are computed, These incremental
quantities are used to compute various corrective
stiffness matrices which serve to take into account
the deformed geometry of the structures. A subsequent
increment of load is applied and the process continued
until the desired number of load
in~rements
have been
applied, The net effect is to solve a sequence of linear
prbblems wherein the stiffness properties are computed
based on the current geometry prior to each load increment. The solution procedure takes the following
mathematical form:
( 1)
5
- - _ _ ,>:_;_,__
-
--
6
Where
[KLJ is
the linear stiffness
matrix,~rC~)J is
an
increrrgntal stiffness matrix which corrects for the
deformed geometry and is based on displacements at the
previous load step i-1, A q is the increment of displacement due to the ith load increment, and AQ is the increment of load applied. After the application of the
ith load increment, the total displacement is given by
,.,.
q
i
=~Aq
~~t
j
..
(2)
The general incremental stiffness solution procedure is
presented symbollically_in·Table· 1.
Table t
The Incremental Stiffness Solution Procedure
Step
No.
1
2
3
Equations To Solve
([KJ+[K1(q0)J) [MJ =[AQJ
(eJ+ [KI (ql) 1) [aqz] =[A Q2J
([K~+[K/q2) J) p~qJ =[AQ3J
Aq
1
Total
Load
Q = AQ
q = Aq.
1
1
,
q = Aq + _q
bq
2
~q
a
2
2
1
·q=A,q+q.
'3.
2
3
1
Q=
2
l
'
'
l
t
\
l
J
where [
-
-
-
----
KI(qO)
-
---··
n
n
1
I
Q = f1Q + Q
2
3
3
t
[l.q
!
2
1
([\]+[KI (qn-1 ~[Aqn] =[A\J
I
1
D. Q + Q
'
n
...
Total
Dis pl.
I ncr.
:Oispl.
)'\.
q =.2:_ A~~
n .i:::J
Q=2 LlQ.
n
.
}-=I
.
J-
I~
= 0
-~------~--~--~~-·---------
----~--
---------~-
----~-----
------------~-
-------------·-~------~-~----
-.J
8
I
2.2
Linear And Nonlinear Analysis
Co~~~ween
Consider the single degree-of-freedom elementary truss
problem shown in Figure 1,
F
100~
Figure
A Simple Truss Problem
Due to the support conditions of the truss, the displacements q 1 ,q 2 and q are all zero, These boundary
3
conditions will reduce the matrix expression, Equations
(A26) and (A27) of Appendix A, into the simpler forms:
EA
[\] =
and
L
EA
-L
lK}
NL
2
(3)
sin?.(;
(2 -8- sin
·6 cos 0
+
cos
2
"'6').
(4)
'
-·--~---
"'·
-·---
----
-
-- -·- -
-
~
-
-
- -
-
-
"'-'
-- -
-
-
9
From Equation(A25):
e = 1/L (q4 sin "0)
-9-= 1/L (q4 cos 7))
(5)
c(= 1/L (q4 sino) + (3/2L 2 ) ( q~ cos 2
~) •
-..
By substituting Equation(5) into Equation(4), .the·
nonlinear stiffness matrix is obtained as:
[J
K
NL
EA
q1,
-
( 6)
q1 2.
?
= - (3-..:t sino cosco + 1.5(-21;L ) COS~?().
L
L
~
The system geometric matrix,
Eqn~(A28)
LINEAR SOLUTION
The linear solution is obtained by applying or incrementing the applied load only once.
The vertical deflection at node 2 is
·
----
~-
--- -
A,
-
---
-
- ·--- -
--
--
-
-·- -·
~
-
-·--
--
10
q
4
=
L
EA sin 2
Q
where:
Q = P =-·- 2.0 pounds;
E
= 107
L
= 100.005
sin 2 1)
psi;
=
inches;
1/L 2 ,
therefore,
q
4
- -. 20 inch.
NONLINEAR ANALYSIS
The computational procedure as presented in Section 2.1
and as tabulated in Table 1 is used for this purpose.
The r0sults are shown in Table 2 and in
T~ble
3. Table 2
shows the results for 4 equal incremental loadings
and Table 3 shows the results for 20 equal incremental
loadings. These results are plotted in Figure 2 alongside a plot for the linear analysis for convenience in
compa=ing the two methods of analysis.
l?
Table 2.
Result Of Incremental Stiffness Method Of Analysis To The Simple Truss Of
Figure 1 For FOlffi Incremental Loading
Incr.
No.
[KNLJ
tLJ
[KG]
[KG]
AQ
Q
Aq
q
I
I
i
I
I
1
9.99850
0
9.99850
• 10002
-.50
-.50
-.05001
-.05001
2
9.99850
-1.46236
8.53614
• 11705
-.50 -1.00
-.05857
-. 10858
I
I
'
3
9.99850
-3.07986
6.91864
• 1445L~
-.50 -1.50
-.07227
-. 18085
4
9.99850
-4.93367
5.06483
• 19744
-.50 -2.00
-.09872
-.27957
--~~--
-------------
-
---·
-~--------------
------~-----
-----------------
-
1
-·--------
_.
.....
Table 3
Result Of Incremental Stiffness Method Of Analysis To The Simple Truss Of
Figure 1 For TWENTY Incremental Loading.
I~:~· [KL1
(KNL1
L~<"1
\.KJ'
b.Q
G.
2
3
4
5
6
7
8
9
10
11
12
13
14
15
9.9985
9c9985
9.9985
9.9985
9.9985
9.9985
9.9985
9.9985
9.9985
9.9985
9.9985
9.9985
9.9985
9.9985
9. 9985
0
-.298470
-.602984
-.914023
-1.23214
-1.55796
-1.89223
-2.23579
-2.58973
-2.95521
-3.33374
-3.72726
-4.13800
-4.56912
-5.02466
9.99850
9.70003
9.39552
9.08448
8.76636
8.44054
8.10627
7.76271
7.40878
7.04329
6.66476
6.2712LJ5.86050
5.42938
4. 97384
.10002
.10309
.10643
.11088
.11407
.11848
.12336
.12882
.13497
.14198
.15004
.159'+6
.17063
.18418
• 20105
-.10
-.10
-.10
-.10
-.10
-.10
-.10
-.10
-.10
-.10
-.10
-.10
-.10
-.10
-. 10
-.10
-.20
-.30
-.40
-.50
-.60
-.70
-.80
-.90
i
· 1
i
l'
I
I
I
I
1
20
9.9985
-8.02481
1.97369
,
-~.o
-1.1
-1.2
-1.3
-1.4
-1.5
~~
-.01000
-.01031
-.01064
-.01101
-.01141
-.01185
-.01234
-.01288
-.01350
-.0142o
-.01500
-.01595
-.01706
-.01842
-. 02011
~
I
-.01000
-.02031
-.03095
-.04196
-.05337 I
-.06522
-.07755
-.090Lt-4
-.10393
-.11813 1
-.13314 l
-.14908 I
-.16614
-.18456 I
-. 204-67 !
i
I
I
, '
_t
I
I
I
I
.I
-.20
-2.0
I?
I
I
. I
.50667
I
I
I
·
l
l
-.05?_~-~~3~
_.
I\)
------
A,
---
13
II
/·{)
L/NE~~
AN'ALY51.S"
.6€{=,/0
(AI//1/Cf<= 20)
Nt" IlL INcA~
ANALYSIS
~------~~-·------r--------4--------~-
·10
DEFLECTION, inch
Figure 2
Comparison of load-displacement curves for linear
and nonlinear analysis along with e:ffect of the
number of incremental loading.
-.\
'
-'
- -:
-"· -- --------r-
3,.0
-------
,~.
----~-~--
APPLICM!ION OF THE INCREMEH'rAL STIFFNESS l-'IET:fiQ12
The solution procedure for stability and load-deflection
response of a planar truss structure is presented in this
section. Consider the simple truss shown in Figure 3.
p
y
100.00
Figure 3: A Simple Truss Problem
Member
.
1
2
L
100.50
10.00
·-o
.5.71°
90.00 0
sin
0
0.09950
11.00000
cos?{
0.9950
o.oooo
Area
1. 0
1.0
E
I
107,....
10(
The aimple truss is first analyzed to determine the critical buekling load,P, and using this load, the loaddeflection response of the structure is determined. From
Figure 3, displacements q 1 , q , q , and q are zero. ThesP.
2
4
3
boundary conditio~s along with the tabulated data from
Figure 3 above, will be used in the .stability and loaddeflection analysis to this problem.
14
15
~
Stability Analysis
Applying Equation(A34) and the boundary conditions to
the simple truss problem of Figure
cs
EA
L
3~
p
+-
cs
-sc
The element forces for members 1 and 2 are F
F2
= P.
Substituting data from Figure
1
= 0' ·and
3, the element
stiffness matrices are
• 0099
.0010].
.0010
.0001
and
oJ
~\
o
+·P
r,.
10
Lo
0
0
The stability matrix of the system is therefore
16
.0099
.0010]
b
.
. + p
.0010
.1001
0
0
.10
0
0
0
•
From Equation(A35) and (A36)
[KJ
.0099E
.0010E
.0010E
• 100 1E
=
41
)
from which
i
.0010E
.0099E + • 10
0
=
.0010E
The lowest root
of~
• 1001E
"
gives the critical buckling load)
P
= -9.8420
critical
X
10
-2
E '
A detailed classical solution of this sim:ple truss is
found from Reference 5, page 147. The critical value
is found to be
p
critical
= -9.8420
X
10- 2 Ee
This result is identical to that obtained by using Eqn.
(A34). This substantiatep the formulation of
Eq~ation(A34)~
~.
--~-------
~-----
--··------
--·
1?
3,2
Load-Defl~ction
Analysis
The load-deflection response of the simple truss of
Figure 1 js determined for an applied load equal to
the critical buckling load obtained in Section 3.1. The
linear solution is first determined followed by the
nonlinear solution utilizing the incremental stiffness
method discussed in Section 2.0.
Applying the boundary conditions of the problem to
Equation(A26), Appendix A, the linear stiffness matrix
of the system is
sin i cos {(
'
sin -t cos ({
and from Equation(A27) the nonlinear stiffness matrix
of the system is
K
NL
=.
EA
where
s
c
-&
e
.,zsc-& + s
2
c<
-s 2 -G- + c 2 -G-- sc <><:
L
= sin 1( .
= cos "t
)
<
)
= rotational strain ..)
= axial
o< ·= e
+
strain ·)
3I 2 -e-2
,
(symm)
2sc -eo+
l
c~ j
18
FromEquation(.A25)~
e -1-
L1
[<\-
q ) cos 'il+ (q - q )sini
2
l
1
6
J
J-
1
q )cos~+ (q - q )sin~
e =T
tq54
2.
2
6
2
3
1
.go-,- 11
1
.g.-2- 1 2
c:J:...= e
1
o<..=
Eq,-
)sin~+
q
5
[y
I
(q - q)cos7)]
2
\
6
q )sin-(+ (q 5
z..
6
q)cos1!'"~
4
z
2
+ 3/2...r;;;,.
1
1
e + 3/2 .f1.2
2
2
2
From Section 3.1, the applied load Pis 984.20 kips. The
number of incremental loading, NINCR:. for our prolDlem . ±s
5 increments.
~.
-
-.---
~
-
.
------- --
----
--
19
Linear Analysis
For the linear solution,
~NI,J
= 0 and the load P
is incremented only once. ·· ·
The linear stiffness
0099
.00101
[ .0010
.100~
,
The nodal deflections
q
p5
5
q
q
= [\]
p5
11 [01.605
6
q
X
y
-1.00~
5 =E
-1.000
10.000
6
{ .098420}
-
-.984200
inch ,
t-9~.201
p,
~----~
---------------
--------------
20
Nonlinear AnalysiB
For the j_ncremental load, L\Q
The linear stiffness matrix,lKL
r
.0010l
L.0010
• 1001
rK 1= E .0099
l
L
1
j
I
Note that the load matrix
A G,\ and the linear
stiffness matrix[KL"J are constant throughout the
whole iterational process.
For the FIRST INCREMENT
tNLl= l\ (qO~ =
O
The total system geometric matrix and its inverse
is
tG1
\ l
= l\ =
r1
~r
G
~
[98.5l85
r
9.85185
.01016
= l_:-.00010 ..
9. 85185]
1000.99
-.0001001
.00100~
•
'
- "· - - - - - - - -
-~----
·------
---
- - - -"•- - - - -
---
21
The total deflection after the first increment is
equal to the first incremental deflection
By substituting the first incremental deflections into
the
line~
and rotational strain equations (ref. page
18), the following results are obtained for the
SECOND INCREMENT.
~-
;e 1
-9-1
=0
and
e 2 = -.019684
= ~ = -.0019684
oe, = .000006
= -.. 0196782
.The nonlinear stiffness matrix
r
l
J .03879
1
k
\l: l
NL1 Jz.l:•19204
k NL2
rK
l
r-19. 6782
1
=
L 9684
1•
= r:.19. 639LI-1
LNL J2 L1· 77636
-. 192041
•
-.0381!?.1
1.9:~
•
1. 776361 ..
-.0381~
,e.,
-
------------
---
---
-
e,
-----
--·-
-·
22
The system geometric matrix and its inverse
11 .. 628~
1000.95J
-,000148l
.001001J
The nodal deflection for the second incremental
loading is
.02904
-.19699
The total deflection after the second increment
is
q
5
q
6
=
~
.048724
\-.393830
This process is repeated for all loading
increme11tation.
For the THIRD INCREHENT
e
=
• 92484 6x10 -Lt·
1
e
2
..g.
1
.g.
2
=--.393831x10 -1
= -.394 756x1 o- 2
= -.487241 x1 o- 2
=
oe.,
o<2
.115859x10-3
= -.393475x10- 1
The nonlinear stiffness illatrix
lk J -
-19.5615
1.39020
1.3902
- .... 10458
~k}~21 =
-39.3475
4.87240
NL 1 ~
'4~8724
0
-58.9090
6. 26260
6.2626
-.10458
o,
----~-
~---
------
--------·-·----- --
24
The system geometric matrix and its inverse
[KJ =tK1+lK l
G
1
3
tG13
lKGJ~
NL
~
39.6095
16.1145]
16.1145
1000.88
=
.0254130
-.000409
-.000409
.001006
=
The nodal deflection for the third incremental
loading is
(
.0805381
~-.1979630
=
The total deflection after the third increment
is
• 12926
=
-.59179
-~-~-------
-
"·
··---~--
------------------
-
·---
"·--
-
-
--
.
-
--
-·-
--
- - - --- -
25
For the FOURTH
=
e
TNCRm~IENT
.633888x10-3
1
e
= -.591795xl0
-1
2.
..a-
= -. 598734x1 o-
2
1
..e-
== -.
129262x1 0-1
2
o(
=
.747660x10-3
1
o(
2
= -.589288x10- 1 i
'
The nonlinear stiffness matrix is
= [-5.8. 7903
5.6'1127
rkNL 1
J
5.671270l
-. '148894
-58.9?.90
l 2Jf \.. 12. 92623
-117.719
18.5975
18.59750
-.14889
J
The system geometric matrix and its inverse is
-19.2006
=
28.4493
1000.84
-.0499768
.,00142062
• 1Li.22062
.00095878
The nodal deflection for the fourth incremental
loading is
-.279635
=
-.188727
· The total deflection after the fourth increment
is
q5) = \-· 15037
\
(-.78052
·-
-·---·-~--·--
--
----
_'2_, _ _ _ _ _ _ _ _ _ _
-
--
For the ElE..'tH ,I:NCREHENT
e
I'
1
e
= -.22 6 1o3x10-?-
2
-&
1
-e2
= -.780521x10-l
2
- -.757905x10-
= .150372x10-l
-2
.t., =----. 217547x10
_,
e><
2
= -.780521x10 ·
The nonlinear stiffness matrix is
-117.5720
17.8797
17.87.97~
J
-.51255
-77.71300
-15.03724
-195.285
0
?.84246
-.512553
J
~.
-
---
~--
-- -
----
~
-
---
- - -- ~
-- --
- -- -
"
-
-
-
--- --
- -
-
-
-
---
-
--
-
-
- -
·-
--
-
28
The system geometric matrix and its inverse .is
-96.7664
12.6943
12~6943
1000.47
=
The nodal deflection for the
:f1~4 incremental
loading is
-.0257674
-.196420
The total deflection after the fi~ft increment
is
(\
l
q
-.17614
l..n~"-
=
-.97694
-
---~~--------
-------
_______ ________
_;.
------------------
--~-------
4.0
THE COHPUTER PROGRAM
Jt.1
Flow Chart For The Hechanized Incremental Stiffness
Solution Procedure
Input Data
,,,
Generate Global Linear
Stiffness Hatrix
Generate Global Nonlinear Stiffness Matrix
Boundary
Conditions ·
'"
Assemble Reduced
Assemble Rerlu~ed
,~
Linear Stiffness ~-~-~----~----~~~ Nonlinear StiLfness
Matrix, [KL]
/ 1
Matrix, [KNL
l
Assemble Into Total
Nonlinear Stiffness
Hatrix, (KG j
1l l
[KG
=
KL
,,
.....
ll
+ [ KNJ]
-
A,
------------
-----
30
Compute Incremental Deflections
tM\ ~' [KGrtAQl
Compute Total Deflections After Each
Incremental Loading,n
t1 ~ t ~
q
q
n
n-1 +
t 1
Aq
n
Compute Sum Of All Incremental Loads,
:;E_
(AQ1.
1.= 1 (
)....
\
n
If,
~~Q1.~
1.=1
F
Applied Load F
-
,..
Print Input Data and
Nodal Deflections
~END
I
31
~-2
Input Instructions
The input data required to ob"Lain the deflections of
a truss structure using the program (Append:t.x C)
developed for this analysis are described below along
·with the input data format.
One or more truss element can be analyzed. The present
capacity of the program is for a maximum of 10 elements,
11 nodes and as many materials as there are elements in
the system.
Input Data Cards
Card
No ..
(l)
Columns
I
Format
Description
-1 -·5
6 - 10
11
15
16 - 20
-
15
15
15
15
( 2)
1 - 5
6 - 10
11 - 15
T,...
-/
15
15
- 251 F10;5
26 - 35 F10.5
16
Parameter Card
Total no. of nodes,
Total no. of elements.
Total no. of loads.
No. of increments to divide loads,
Node specification card( The no.
o£ cards equal to the no. of
nodes and must be inputed in
node order 1,2,3 •••• ,etc.).
Node number.
Degree of freedom in x and y
direction respectively. 0 = ·free
ar.d 1 = restrained.
Coordinate in x axis.
Coordinate in y axis.
32
lJ.lJitlt data. Cards ( cont' d}
t - - - ·-r------.....-----.----~--------------c;..,
Card
No.
Co1umns
Format
(3)
1
-,5
6 - 10
11 - 15
16 - 25
26- 40
I5
I5
I5
F10.5
F15.3
(4)
(4a)
(4b)
Description
Element Specific.:;,tion Card· (No.
of cards equal to the number of
elements and must be i.nputed in
element order 1,2,3 •••• ,etc ).
Element number.
The ith node number.
The jth node number.
P...rea of element.
Young's modulus of elasticity.
Load Data Cards ( Requires 2
cards for each load application
at a node. The first card is for
the node number and the second
card is the magnitude of the
load in 'the x and y direction.
3ign of the load is positive in
the +x and +y direction and
negative on the opposite
directions.).
1 -
5
15
16 - 30
1 -
I5
F15.3
F15.3
Node number.
Node load in x-direction~
Node load in y-direction.
I
:z3
•.1
4.3
Example Problems
Two nonlinear truss problems are analyzed utilizing
the mechanized analysis for the solution procedure as
presented in Section 4.1.
The first example problem is the same problem solved
by longhand method in Section 3.2. Choice of this
problem is to countercheck the results attained by
the longhand method. Further description of this problem
is presented in Section 3.0.
The second example problem is a 5 member planar truss
system of uniform material for each member but of
differ8nt cross-sectional areas and length. This
problem is solve for a single and for 20 loading
incrementation to obtain a linear and nonlinear analysis
result.
"·
~--------------~--
--·------------~~----
---------
Determine the load-deflection response of the simple
truss shown below for an applied load equal to its
critical load at buckling,P = 984.20 kips. Use 5 equal
incremental loading.
p
y
®~­
T
.
@
10·0
EAtz.
®
~
Notes:
1.) Numbers enclosed by() denotes
Node number.
2.) Numbers enclosed by
element number.
Q denotes
Compute.r.. Output For Prob.lem No, 1
NUMBER OF
NOJ~S=
NUMBER OF
~LEMENlS=
~UMBER
3 "
2
OF NOOES WITH LOAO=
1
NUMBER OF INCREMENTAL LOADING=
5
I!
I
NOOC:
DEGREE OF FRC:EOOM
LOCATICN OF
COORDINATE~
y
NUMBER
u
v
1
2
3
1
1
1
1
1DO.Jv000
0.00!.100
0
i~O.tvOOQ
10.0000~
I)
X
o.c;;oao
o.. oooac
I
I
'
I
1:
NODE POlhT NUMBERS
ELEMENT
NUMBER
I
1
2
AREA
YOUNGS H.OD
J
1
3
1.00000
U.1JOOOE•OS
2
3
1,CCCOO
0.1COtJCE+05
\..,.,..5'
\Jt
.
..
LOAD
I~CR
NODE
NUMt3ER
1
l
l
2
3
1
LO~J
NOOE
INCR NUHdER
2
2
2
LOAD
4
v
u.JOOJDE+OO
.J.DCCJOEH:C
J.OD;JJ~E+OD
0.4872'+E-01
-0.39383E+QO
1
2
3
NO-J£
i
2
3
LOAD
NO Jt.
INCR NUMt3lR
5
1
5
2
5
fEFLECTIGNS AT THE NODES
l.l
2
3
!NCR NUMBER
4
4
•0.19E34E+OC
O.!JfJOilDE+OC
tWJE
1NCR NUMt3f.R
3•
0.19E:84E-Ci
l
LOt.D
3
3
DEFLECTIONS AT THE N00ES
v
u.00!:0!1E+QO
J.OOCJDE+OO
,).0 OCOOE+O\l
G.OOQOOE+OO
li
3
i ?
CEFLECTIONS AT TrlE NODES
u
v
O.O~~!JOE+CO
O.JQLOOE+OC
IJ.12S26::HD
'
I
I!.OCOJ2E+Ol
o.uccotE+J:
•0.5917S::+OIJ
CEFLtCTIONS AT TH::: NODES
u
C. ect:DDE+GC
Oe 0 f.i\·CJE+GC
-0.15G37E+DQ
v
o• ar oo.; E +oo
c.cro;:o~t:+DG
- 'J.. 7 8 C3 2 E + J
~
CEFLECTIONS AT THE NOO~S
v
u
t;.OO(uiJE+Oil
O.OOCOOE+OO
~.ClJ[QOE+OO
•ve1/El4E+OO
O.L0[110E+30
-C.976q4E+CO
VJ
0\
37
Example Problem No. 2
Determine the load=deflection response of the planar
truss shown below, first by linear analysis and finally
by nonlinear analysis using 20 incremental loading.
y
II
240·0
(D
®
I '
·-
Computer Output For Problem No, 2
NODE
NUM3ER
1
OEGP.E£ CF FREEOOH
u
v
1
1
0
0
2
3
'+
Eli.MENT
NU~9F..R
4
5
Q,GiJliOO
o.oocco
zr.u.oocoo
18C. 0 0 0 0 0
i
1Bo.oooac
2Z.O.OCGOO
D.CGtJCO
NODE POINT NUHaERS
1
2
4
1
i.,
O. C OODO
0
0
J
2
3
2
3
3
Ic
y
X
1
I
i
2
3
LOCATION OF COORDINATES
AREA
1.33333
1..oo~ov
I
.:..66667
·1· ot>6f:7
1.33333
YOUNGS H::>O
Oe100£lJE:+08
Uc.10LOOE·"J8
u .1COUC ::-4·(18
G.lCCJC ~a 3
iJ. 1CC.Ou +HI
\.N
en
f
Problem No. 2, Part a} LINEAR ANALY$IS
,.
NUHB£R OF NODES=
NUMBE~
OF
E~EM~NTS=
NUMBER
l~
NOOES
~ITH
.5
1
LOAO=
NUH3ER JF INGPEMENTAL LOADING=
1
P = 30 kips
3X
P = 20 kips
3y
LOAD
INCR
1
l
1
1
NJOE
~UHBER
1
2
3
4
DEFLECTIONS AT THE
u
OoCOCOQE+GO
u. uiHH!OF.+OO
-0.356ttuE+OC
c.CCCI)OE+0\1
I
NOO~S
?
v
I
c.cc:0cJ E+oo
o.o:tiu~E+Cu
'•
-c .1152J F.+c o
O.IH.:;;LCE+fJJ
\.N
\.0
Problem No. 2, Part b) NONLINEAR ANALYSIS
NUMBER OF NODES WITH LOAD=
NUM3ER OF
IHCREM~NTAL
1
LOADING=
20
I
I'
....
NJO£
I1olCP. NUH BE.R
CEFLiCTIONS AT 'THE NODES
LOt.IJ
1
1
1
1
LOAO
INCR
2
2
2
z
1
2
3
4
u
u.ooooaE+oo
O. IJOC&jOE+OO
-c. 17B2uE-o 1
Q,{;Q(.uCE+UJ
fi,.OOOCGE+CO
. -0.57bCOE·02
2:
,..
u
...
~
3
3
.3
2
3
4
LO AO
NJIJE
INCR NU~3ER
r.
4
'•
4
1
2
3
4
Q,i.JGOOCEtCIJ
v
O.l.iOC~O[iiJO
O.GGJCu£-t-G!J
-C.35641E•.H
•C.d~521E-G1
o .. zz759E·o5
lJ.OOOIJOE+C!I
o.cccucE+ao
LOAD NJDE
Il-.CR NU•l GE.R
3
:.
DEFLECTIONS AT THE NUCES
WlOE
NU~dER
1
2
v
o.co~aiiE+oc
C .11451 E-t g
DEFLECTIONS AT THE NODES
.v
u
DGC!JOE+OO
OcCCtGCE+OJ
O• .JOJCGE+OO
•D,5346SE-Ol
-'3.172:84-E-\: 1
0.95C5d£-05
L.t.H.O~OE+C'U
Q,
o • 3u .1. 4 a c:- o 9
DEFLECTIONS AT THE NODES
u
o.ccocc!E+C3
C. Ll.iC ..djE+OO
-J.712'33£eQl
u.z~T71.,f--JL•
I}
0.0LOL:JE+CG
-0.17088£-09
-G, 23 043 E-G 1
C.uCJhlit-CC
+=0
LOAO
NODE
INC~
NU"1dER.
5
...
5
5
2
s
LOAD
IN C~
6
6
6
6
LOAD
I~~cq
7
7
7
.,
1
3
4
NOOE
t~ UM? E R
1
2
3
4
NJUE
NU'1 '3ER
6
8
u~DtiCOGE+OIJ
-~.89125E~j1
(J.
I :
-0.32615£-08
-G.28o113E·{i1
!.l.JOCCJE+C:Ci
4 7576f -~ "•
CEFLECT!ONS AT THE NCO=s
u
v
tl,OuOCCE+OO
O•OOCOCE+GO
O, OGCJOE+OO
•Da12702E-C7
•0•1C6g6Et-OQ
-G.34592E-(1
0.83318E-C4
iJ,GOOGGt+G!J
DEFLECTIONS AT THE
u
NOD~S
v
o. OOCGCE+OO
L·CCGC.QEt-00
"-
o.;,;tr.JOE+OO
•(. 3Lt55J E·t 7
-J.124ill£-<·OO
Cel331f3£-03
-0.4C372E-Cl
O..JGUCJE+OO
.,
3
4
N)Of
INCR NU'13Ei<
e
IJ
o.ooco~£+C~
1
LOAD
~
CEFLECTIONS AT THE NODES
u
o.oonooE+ao
1
2
3
4
LOA!J
NODE
INCR
t!U'1dE~
q
1
q
q
g
DEFLECTIONS
u
u.cc~un£+oo
At·
THE NODES
J
G.
~IL.JCQ
E+OiJ
Q.COIJLHE+OD
•C • 77763 E-C 7
-L.14267CHlO
-C.46l.5!)t.-C1
0.20C35E-03
O.OOJLOE+CD
:?
I
DEFLECTIONS AT THE NODES
u
v
O.OGJOuE+Gil
2
J
O.uOG:iOf_+OiJ
u.ccnouE+:JO
-G.!.5'+7,3E-t6
-~~loG53E+GO
-C.5lg5:.:..E-G1
4
0.28656E-G3
l.'eliCiJOJE-rCJ
i-
-)"'"'
DEFLECTIONS AT THE NODES
NlDE
INCR NU1BER
LOAD
10
10
1.:1
10
LOAD
IN{;R
11
11
11
.L1
~
....
2
3
1
2
3
-t:.1963CE+OO
(j,S268dE-03
4
LOAO
NODE
INCR
12
12
tlU~ ~E~
12
4
LOAD
INCR
13
13
13
13
LOAD
-0.6356~E-C1
Q,Q(,,J!.JOE+OO
o.occooE+oo
O,OUOOOE+UO
i.l.CCOQOE+OO
-0.77fi.%E·C6
-c. 2142(.:£+01
t.68610E-03
-U.693doE•C:1
O.OIJC!JGE+(;J
CEFlECTIONS AT THE NJDES
R
I
u
v
1
2
3
O.~COOUE+OG
Q,OO~LCE+IJO
C.OijGl'C£tOO
-0.23212E+Ou
•0.12065E·05
·0.75219E•Ol
4
Q,6748~E-tJ3
O.OuGGJE+CJ
NOQE
INCh:. NW13ER
14
14
14
14
-O.S7753E-t 1
O.ODJOUC:+IJO
DEFLECTIONS AT THE NUDES
u
v
NODE
NU·F~f
-0.2821Lt£•C6
v
DEFLECTIONS AT THE NODES
.;
u
o. cr.u en E+t o
o.oococE+oo
o.or0rcE+oo
-o.<+BlluE-t::c
NO fiE
NU~GER
12
~.CvnOOE+O!l
O.OOOCCE+OO
-0.17841E+OO
0.39456E-03
Lt
1
2
3
u
C.COCOOE+OO
1
,.
::;
DEFLECTIONS AT THE NODES
>J
u
O.OCGC.JE+GO
O. LL ~GLE+Ou
2
3
(1, CU:~.;L£+00
-f..18J52E-G5
-u. 251.h16E+IJO
-(!.81iJ65E-(;1
~
0·10958E-u2
O.O£iCGJE+CC
+r\..1
lOAD
·
N)
CEFLECTIO~S
~E
15
...1
15
15
2
3
15
4
LOAD
NJCE
IN CR.
~~U:-18ER
1&
16
16
16
LOAD
INCR
2
INC'-t
NU~3ER
13
16
18
1
2
3
4
11
t~O
DE
NU~2EP
19
1
1g
19
19
2
3
LOAD
INCR
-G. 8692'+E-C 1
o ••nacJE+co
DEFLECTIONS AT T H£ NODES
4
N)DE
NU!i8~f
20
1
~!)
2
20
20
3
4
v
u
-0.28598E+QO
(;.16451£-02
N3DE
INCR
-J.2&216£-C5
2
3
4
lOAO
LOAO
u.ooc~uE+JO
- u • 2 6 8C 1 E + C\l
u .13 516£-ij 2
·c.O(IOGLE+OC
3
4
v
O.OGCCJE+OJ
1
1
T~c·NoD::S
o.C01JC(;E+CO
U.OOIJOiJE+CG
- J • 37 112 E- ~
~.CCuui!E+OJ
~
-c. 92798 E-c 1
f.laOuCCC;E+CO
DEFLECTIONS AT
NODE
NU'iBEF
17
17
17
17
Al
u
l'4C~ NU~ BER
u
0 • C.fJ CGCE +il U
D·CCC:OOE+Ofj
THE
NODES
v
iJ.OC~OGE~t:lO
-t.3~398E+OG
-G.51389E-t.: S
-o. 986e7 E--1
C.19791E-02
~.CGOOOE+CO
i
. DEFlECTIONS AT THE NODES
v
u
o.otcc~E+to
O.CCCOLE+OO
c. CCtHH.E+OG
=0.32199E+GO
-0.69798E-O 5
-O.iC459E+C ~
0•23565£-02
a.u~ilCilE+CO
DEFLECT!Ot.S AT
u
ll. GC OC OE +(j 0
u•{OOiJOE+OO
-..; .. 3581CE+OO
0.32535£-02
v
-C.932t.2E:-t S
-C.11i.l52E+GG
-(; •· 34(0 3E +0 1l
D • 278C 3E-O 2
CEFL~CTIONS
NODES
u.oc;ocaE+\JJ
C.COCOCE+GC
O.OOOCGE+OC
u
THE
O.(t!J~JE+CO
AT
THE
r-coc~s
v
o.OtCOiiE+OO
-J.12259E-C 4
-0 • .11646C:+(l0
a.ououuE+t~
?.0
CONCLUSIONS AND RECONHENDATIONS
The incremental stiffness procedure embodying the finite
element displacement method is presented for nonlinear
analysis of planar trusses.
The method of analysis is convenient and easy to use in
that it only requires the specification
size. Refining the load step
of a load step
or stepping the number
s~ze
of load incrementation increases the degree of accuracy
of the solution. It is evident, as shown in Figure 2,
that the linear method of structural analysis does not
adequately describe the behavior of a structure under
loads. 'The deformation of a structure alters the
directions of the initial internal
forces~
The
nonlinee~
cu:rves in Figure 2 reflect the effect of geometry changes
as the truss deflects under load. Some structures are
more nonlinear than others as illustrated by the example
problems presented in Section 4.3. The first example
problem is more nonlinear than the second problem. It is
therefore necessary, to include nonlinear effects to
obtain a more accurate
resul~.
The solution procedure
prese~ted
for a trliss structure
can be easily extended for beam analysis. The nonlinear
stiffness matrix derived for a beam element (Appendix B)
LS
L-
can be easily included to extend the computer program
presented in Appendix
c.
Further refinement or improve-
ment of the computer program can also be accomplished,
to include in the output, the corrAsponding load and
deflection result for each incremental solution. The
incremental and total strains and stresses along with
elemenc internal loads should also be included.
BIBLIOGRAPHY
6.0
1.
Ural,o., Finite Element Method- Basic Concepts
and Applicptions, Intext Educational Publishers,
New York, 1973.
2.
Zienkiewicz, o.
c.,
The Finite Element Method in
Structural and Continuum Hechanics, NcGraw-Hill,
New York, 1967.
3.
Desai,
c.
and Abel, J., Introduction To The Finite
Element Method, Van Nostrand Reinhold Co.,NeY., 1972.
4.
Przemieniecki, J.
s.,
Theory of Matrix Structura.l
.Analysis, HcGraw-Hill, New York, 1968.
5.
Timoshenko,
s.
P. and
G~re,
J. M., Theory Of Elastic
Stability, 2nd Edition, McGraw-Hill, New York, 1961.
6.
Stricklin, J., Haisler,
,,
w.
and Von Riesemann,
w.,
Geometrica_llv Nonlinear Str:uct_ural An.a.l:vsis Bv
_Dire.ct Sti:f,fness Method·; Journal of the Structural
Division, ASCE, Vol. 97, No. ST9, Sept. 1971.
I
.!.,..
,~
o
APPENDIX A
.PEEIVA:J:ION OF THE NONLINEAR GEOHETRIC HATRIX FOR
A
TRUSS ELEHENT
When a uniform truss is loaded in simple tension, the
forces do a certain amount of work as the member
stretches. If the truss element shown in Figure A-1
is subjected to a normal stress
force F x
Ex
= · (}x
t)x
only, there exists a
dy dz which does work on an extension
dx. The relation between the force Fx and the exten-
sion &X dx during loading is shown as line OA in Figure
A-1(c). Tne work done during deformation in the
triangle OAB is
W = 1/2 Fx
t:x
(.A 1)
dx
Let W = ·du, then .
du
=·
1/2~cr-x dx dy dz
(A2)
Integrating Eqn. (A2) throughout the length of the
element
U = 1/2
where
55
dy dz
S5.) (f'x t:x
"
= area
dx dy dz
of section, therefore the
internal strain energy for the truss element is
(A3)
L.
u
= _fl../2
5
(}X
I)
or
EA
y
f
~X
( t.. )
u1 = - 2
X
dx
2
(A4)
dx
0
~~
(b)
Figure A-1
The nonlinear relation between strain, ex' and the
displacement components u and v in the direction of the
truss element axis is
...
t
(AS)
=
X
Since strain is small compared to unity, ex can be tal~en
as the . true physical strain and sine e ( 4 u.j a >C ) 2 is
small compared to
a u./a y. , therefore ( ~ l)..j~ y.. ) 2 can be
.,
omitted. The te·rm ( alf/;;y. )c.. is the contribution of
rotation to
~-r'
X
therefore this term is retained.
Equation(A5) can be written as;
fX
+ 1/2 (
0 v
··-
ax
2
)
(A6)
where the second term is nonlinear. Equation(A6)
determines the strain in the direction of the element
axis (Ref. 4).
Considur a pin jointed constant stress truss element
with a uniform cross sectional area A, modulus of elastici ty E, and length L as shown in Figure A-2,, Under the
action of an applied load, the truss is displaced from
its uriginal location AB to A 1 B'. The displacement in
the global x and y directions at ends A and B respectively are q , q and q , q • 'fhe local displacements at
1 2
3 4
ends A and B in the local coorcUnate system is x 1 and
y 1 respectively. For the
disp~acement
model, let u(x)
and v(x) be linear functions at a node (linear two term
polynomial) :
u ( x) = a.
v(x)
0
= b0
+ a x
1
+ b x ..
(A1)
1
The choice of the displacement model, Equation(A7)z
provides the necessary constant strain along the length
of the
memb,~r
and also at the same time provides as many
constants as there are nodal degree of freedome The
first condition in Equation(A7) is the linear expression
for strain and the second condition is necessary to
50
express the strain energy in terms of nodal displacement.
Writing u and v displacements at each node of element in
Figure A-2, gives
and
a0 = u 1
::: _e.g_
a,
*
.ax
b1 =
= (u
=
v1
b 0 :::
2
,
(A8)
- u 1 )/ L •
(v 2 - v 1 )/ L -
By substituting Equation(A8) ·into Equation(A7), the
following expressions are obtained:
= u1
u(x)
+ (-au/~x)/L J
v(x) = v 1 +
(A9)
(~v/'l'x)/L
'
The displacements u and v is therefore varying linearly
along the length L.
~quation(A9)
can then be expressed
as
f X )
~ ) q1 + q3\1
u = ( 1-
(A10)
v
= ( 1-
~ ) q2 + q4( ~ )
•
In matrix forr:
(u}
tv =
( 1-x/L)
0
x/L
0
( 1-::::/L)
0
The axial energy stored in a truss.with linear
\.
5
stress-strain law (Hookean elasticity) is
2-
_A!Lc.
2
<::: x
0
dx
(A12)
51
Substituting Equation(A6) into Equation(A12) yields,
U
=
5
1.
A"E
2
'f.:llL
[~x
+ - 1·-(_u-_)
2
2.
2
]
ax
dx
D
L
.
AE \ [.:a_u_
au av
= ~J. tax) +-ax <-ax-) +
2
·-
2
1
":lV
-(....L.!-)
4
i)X
4]
(A13)
dX
•
D
The first term inside the bracket in the right hand
side of Equation(A13) represent the strain energy,u 1 ,
obtained by linear theory.
U
L
AE
= ----
2
51.(
~)
0 x
2
dx
(A14)
0
The last two terms inside the bracket in the right
hand side of Equation(A13) represent the strain
energy, UNL' due to the nonlinear terms in the straindisplacement relations
(r::'U... (..LL /
-2Jl~x ax
u = A.E:
NL
4
+ _]_(_1L) ]
4
ax
dx
(A15)
()
The Linear Stiffness Matrix
The linear stiffness matrix is now obtained by
substituting Equation(A8) into Equation(A14), integrating
and then evaluating the second partial derivative of
the
(Af")
52
Integration of Equation(A16) yields
•
The partial derivatives of UL with respect to q. and q.
J.
are:
J
.
'
(A17)
AE
= - (Lq 3 -
q )
1
•
Therefore:
tQl=(~:L1
AE
=
L
-q~
q1
...,
-q 1
(A18)
'
q3
and
a2. UL - AEL
aq-oq'.
J.
J
0
-1
0
0
0
0
0
-1
0
0
0
0
0
0
=[KLJ
• (A19)
l
Equation(A19) is the linear stiffness matrix of the system in the local coordinates.
53
l:!mhhinea.r_.§~t i
ffness Hatrix
The nonlinear stiffness matrix is similarly obtained
by determining UNL as a function of the displacements (in
terms of strains), substituting af:sumed displacement
functions and evaluating the second derivatives. From
Equation(A15), let
au
ax
= e
a v
and
=....ff-
ax
then, Equation( A 15) can be \vritten as
u
NL
JLG ....z _:.___.:}x
+
)
or
TJ
NL
=
4
0
AEL (
2
e-G-
+
2
1
4)
-.e•
(A20)
4
The nonlinear stiffness matrix is therefore
(1:.21)
~q.
~
oq.
J
+ ag. ..LL
(>q.
~
~~
~Ci·
J
+ 3r;;;.2
a~q.
.Qo a -f1
~q.
~
J
J
Substituting partial derivatives into Equation(A.21),
the nonlinear stiffness matrix in the local coordinate
system is obtained as follows:
54
[\~~ -~
0
-9--
0
.g..
00:::..
-&
-
-'6~~
(A22)
0
--&
-fr
0
-e-
-e<::
0
c:<.
•
w1wre:
3
~= e + -..e-2
2
i
e --
;
55
J;Tong,lli:LQT Geometric Matrix
The local to global transformation matrix is
[Tm}
cos<(
sin C:
0
0
-sin '6
cos {(
0
0
o'
0
0
cos -6
sin
0
0
-sin "5
cos -t
f.~;;
t~
t~~>~~
.k:tL:
&.t~
j
(A23)
X,Y = Global System
x,y = Undeformed Element
System
y
~/'I
I
~<~
t{
b"l
....__-----------------~X
Figure A-2:
Truss Element
:i!'rom Figure A-2, the displacement at tha nodes a:r·e
u, = -q, cos¥+ q2 sin'{
v,
u2
v2
= -q1
= q3
= -q3
sin?)+ q2 cos/{
cos <5 +
qLI-
sin ( + q
4
sin¥
ccs ?{
(A24)
56
a:1d
1
e =
:t
1
-L
1
-L
(u - u )
1
2
sin --t)
(q coso+ q
3
4
+ q ) cos"( +
(q
3
-
(-q
q )sin-(
(q
2
4
1
cos"(+ q sin'/)
1
2
1
-e- = -
(v
I,
'\
- v
2
1
1
-L
(-q sin
- -
(q
i
3
L
+ q cos 0):- (-q sin if+ q coso)
4
1
q ) cos ( + ( q
-
4
e + 3/2
(A25}
J
2
-
1
2
q ) sin (
3
2
-fr
•
The global linear and nonlinear stiffness matrices are
therefore
[\] =
[Tmr rLJ1Tm1
c2
[KJ~~
cs 2
s2
(symm)
-s2
-cs
c2
-cs
-s 2
cs
•
s~
(A26)
p
'
-'5M(
e.nd
rLNLJl= rl rnJl r\"Nljlm
lrJ
K
T
T
{!22)
T
1 -
(symm)
2
EL c e-s2e-sccA
2sc6+c 2 o<..
.L
s2.e-c~+scoc
- 2s c-8+ s '-o<.
. ., e
-c.SC
2
2sce-s 2c::<
?
--
2
~c "<.
c e-s 2e-·sc<:~C. 2scf:1+c~
Lwhere the subscript 1 denotes local coordinate system
and
s
= sin o and
c
==
coso •
Combining Equation(A26) and Eqn.(A27), the global
nonlinear geometric matrix of the system is obtained
LKGj ~[ \] [KNLJ
+
=
LKLJ+ \yq)J
•
(A28)
I
58
.TrJ}.s§._ ~.J-e,rqent
StiJ.llLe..§§~ri~tagili t.Y
NQ.n)..ine@.:
jmaly~i.s
In determining the load at which the truss structure
will become unstable, the nonlinear stiffness matrix
fK .
L NL
1
must be derived to allow the inclusion of the
nonlinear term av/ox.in.Equation(A6).for the case of
incrementing the applied load once only.
From Equation(A13)
.l
u = ..&..
2
jL
-
.
r\ (.2.1L) 2 - + . ..£.JL..(~) 2
~X
()X
·'
ox
+ __J_( ov_)
4 .3 X
4~
dx
0
Substituting Equation(A17) into the above expression
and neglecting the higher order Cov/.:>x)4, since this will
\_KNL1 as
only make
a function of displacement, and
integrating yields
(A29)
AE
2
2
AE
2
2
U = · - ( u -2"..1 u +u ) + - ( u -u ) (v -2v v +v )
21
1
1 2
2
2L
2
2
1 2
2
Since
AE(u -u )/ L = AE£= The Applied Load, P ;
2
1
therefore,
U ·=
AS
-a< u
2
2
.
2.
2
-2u 1u 2 +u ) + P(vl -2v \ v2. + vz.)
1
2
•
(A30)
59
From Castigliano's first
theorem~
the stiffness coeffi-
cient is
Therefore, from Equation(A30)
'
2JL_ -
AE
-
(3~ 1
an
L
_p
=-y-
3Uz.
(u;
-
(v
- v 2)
1
u2)
(A31)
~u
AE
=
au3
<1U
L
=~
t
-au4
(-u 1 + u 2 )
(-v
1 + v2)
Equation(A31) can be written in matrix form as
.
[~
K
G
= -L
AE
0
-1
0
0
0
0
0
-1
0
0
0
0
[Kj [K*l
=
L
+
~
0
NL
__.
p
+L
~
0
0
0
0
1
0
0
0
0
0
0
-1
0
0
-·1- '
(A32)
(A33)
•
60
Note th0.t Equation(A32) i.s no longer a function of
displacement unlike Equation(A2?).
Transforming Equation(A32) into the global coordinate
system by use of the transformation equation,Eqn.(A23),
yields
(A31+)
--.
·--c2
s2
cs 2
[~}Etl:2
s2
(symm)
-cs
c2
-s2
cs
p
+L
s2
... (symm)
-sc
c2
2
sc
s2
-c2
-sc
-R
sc
c2
From Equation(A32), the classical. stability or eigenvalue
problem for buckling is formulated by considering a
single increment without external loading(Ref. 2).
Assigning a scaling factor
unit value of
~
A and evaluating)!~:L1f0r a
, Equation(A33) can be ·written as
= 0 •
(A35)
EquationCll-35) has nontrivial solutions only if the
determinant
=0
•
{A36)
APPENDIX B
DERIVATION OF NONLINEA."R GEOMETRIC MATRIX FOR A
BEPJ,1 ELEMENT
·Consider the beam element shown in Figure B-1
y
X
Figure B-1:
The Beam Element
The nodal displacements q 1 ~nd q'f- are axial displacements,
q 2 and q are transverse displacements, and q and q
3
5
6
are rotations. From Ref. 6, the strain energy,U, for
a beam is
u=
~ 5~-E E~dV
I)
and tha total strain £..,for moderate rotation is taken
as
61
(B1)
62
e=e
2
(B2)
- yx
+ 1/2-&
where
e = du/dx
-&· =
X=
dv/dx
(B3)
d~/dx 2
and u and v are the mid-axial and transverse displacements respectively within the beam, e is the linear
strain component,-& is the rotation of the element and
x is the linear curvature. It is assumed that moderate
rotations occur without exdessive
deformations(curvature
j
.
changes) so that only the linear component of the
curvature,x, is considered.
Substituting Equations(B2) and (B3) into Equation(B1)
and integrating over the cross-sectional area yields
~1
('2
J:
(L 2
4
dx·+ ~- j~ (e& + 1/4-% ) dx
D
0
where A is the cross sectional area and I is the moment
of inertia about the z-axis. The strain energy Equation(
B4), can be rewritten to seperate the strain energy into
a contribution due to linear terms, UL' and due to
nonlinear terms in the strain displacement relations,
U = UL + UNL
(B5)
where u1" represents the first two terms on the .right
hand s:i.de of Equation(B4) and the last te:r·m represents
u1\TL•
Therefore
.Lt
dx +
E~ ~
~
2
>.<:
(B6)
dx )
0
and
~
u
NL
= _M_
2
~2 +
1/L,-G-4)
~)L •
( 137)
()
From Equation(B6), the linear stiffness matrix,
is obtained in the local coordinate system as
[
K~~
i.
0
12
0
6L
4L 2
0
0
AL 2/I
EI
~
-Arl/r
(symm)
0
-·12
-6L
0
12
0
6L
2L2
0
-6L
4L 2
_....1.
where the subscript 1 denotes local coordinate system.
The nonlinear stiffness matrix, KNL
~
is obtained by
determining UNL' Equation(B7), as a function of
displacements(in terms of strains), substituting assumed
displacement functions and evaluating the second
derivatives in similar manner as for the truss
KNL
in
Appendix A.
0
-9-
<:::><::
(symm)
~NJ= £:
where e,
0
0
-'&
(B9)
...-e-
-~
-&
0
0
0
0
0
0
0
0
0
0
.rfT and cf... are
0
the same notations as presented i!l
Appemlix A derivation for the truss element- LKNL
The local
t~
J.
global transformation mat:r.ix for the linear
and nonlinear stiffness matrix is
c
s
0
0
0
0
-s
c
0
0
0
0
0
0
1
0
0
0
0
0
0
c
s
0
0
0
0
-s
c
0
0
0
0
0
0
(B10)
where
C
=
COS
-6 and
= sin (5
S
e
The local to global transformation for the linear and
nonlinear stiffness matrix is given by
(B11)
and the system geometric matrix is similarly obtained
as :. previously shown in Appendix A as
l
K \
Gj
= fK
\_
L
l
-\ + \ K
j
NL j
t
=L\}t/q0·
(B12)
C0Iv1PUTER PROGRAH LISTING
This computer program determines the load-deflection
characteristics of a truss structure utilizing the
incremental stiffness solution procedure presented
in Section 2.1. Refer to Section
tions.
66
4~2
for input instruc-
lSI
~ ORT~AN
LN llOC1
!..N tl:lC2
LN Q CD 3
Lf\
LN
CC S
L
GOC7
L.N !1111J8
i..N ~0L~'3
LN 00l0
UJ 1011
LN · n12
i..t\ 1.1013
UJ 0,.,14
0 ;l i 5
LN DfJ16
u:
LN 0017
u~
J~lR
LN GC19
LN !ie2G
LN JC21
LN a.~z2
LN OC23
\,j
""~ ?~
LN v ; 2
C READ lOlAL
Lf'.i OC25
J !J 27
LN 0 0 2 8
.
I =1
u136
J :137
•
OPTION IS
OFF , 0 OPTION !3
OF NJJES, TOTAL
~UMBER OF ~LEME~TS,
~ !J M3 E::?. CJ F I 1\ CR:: HEN T S
N~OD~tNELEM,NLOAD,~INGR
'f 0 DIV! il E: l 0 A0
.
J=i,t~NC;:E
C ~EAO NODE N~HOER,OEG~iE OF FREEDOM IN TH~ U V DrREGTio~l(J FO~ FREE,
C 1 FOR F!XEO) ,AND THE X,Y LOCATION FOR THE PARTIGULA~ NOuE.(NOTE, THE
C NODES HUST BE R~AJ IN ~ODE OROE~ 1,2,3, •••••• )
R E AlJ t f. Q , 6 2 0 J .'~ !'. , 1 D,. ( I ) , I DF ( l +1 > , X{ J l , Y ( J )
620 FlJRt;A't(3I5t2Fl!J.S)
···
I :;I+?
2C CJNI H;JE
C READ ELEMENT OAT~
· 00 Sd I=i,NE~EM
C P.EAO ELCME:NT tJUhBE!'..,ITH NODi: i.U:18EF:,JTH 'lODE NUMBER,
C ELEMENT CFJSS SECTION A~EA ANO YOUNGS HODULUS
P£AC!6G 7 ~31Jl
NEth~I(N~i d~NJ<NE.l
FOP.I''~-1'(3c5iF1Qe?tri5.3J
630
C C0MPU1E
.
LE~GTH
N:=~.~I<t,El
GF ELEMENTS
·
,AR.EAtNE'
,EO~E)
N ...l=tii-' J:. N t.)
XLX=~('LJ)c·X (N!}
YLY=Y HiJl-Y <Nl)
TLE~GTrl(~~)=SQRTtXLK++2tYLY••z>
g ',
LN
~~AC(6Ci61G)
DC 21.
U-i QJ31
Lt.;
3>
Ul
~ 3~
LN a1n ..
L~
N~MaEK
610 Fl•RMAT {4!5)
C READ t10ltS OATA
LN Q C2 9
Ui a J3G
LN OJ35
t
C T0 T AL Nll t~ 8 f R 0 F N0 CC: S WIT H ;.. 0 1\ DS At·' !J
._I
I
N
u~
=1
DIMENSION SL<20t2tH,Sm.. (ZG,ZOl ,SKG(20,2J),SL.R.(2J,2J>
DJ~ENSION E<~G),!OF(2J)~X(2J),V(2C) 1 B(20t~>~T.E~GTH(10)
O.~.Hft,~ICJN il (2J) ,Frd2G) ,~KGFi..( 21.., 20) ,uSP{2iJ), EDSP<20)
lME;t~SfON DI<'F'(2(t.,LJPq; ·.J<2U)tSNP(~J,2Q),_~KZ!2: 2f))·
B H' EN';, cN N N I t 2 u) , ~~ ~ • J 1~ c ) , t•;;;. r.. A { :ted , "'K <2;; 1 i w~ ,s tc ~ { z·.J , 2 o3 •
~CClt
t...~ 3~D6
INTEGER WORD SIZE
( 2. 3) /,. .~STER
fl HI E , 1 ) ::: - XL X
8P!E~2t=-YL"f
C
B (t~ E , 3 l =XL X
8 8 B < r\ E , •• l ::: Y !.,. Y
COMP~!F.-
NU~~F~_CF
i~ u~= Nt~:..i[, E "".::
DEGREE OF
FRE~OO~
0\
·-..,J
'·
l~
l
t~
~
11C38
0 039
ll04G
1)041
0042
c· ZERO OUT NODAL DISPLACEMENTS
00 64 I=1,NOF
ED§PUl=(l.
or
. . P<ll=o.
64
LN Ju43
C ZERO uUT lOADS
DO
56 !=1, NOF
LN IJC45
C
j..~{
LN Cui+'+
·J '+ 6
LN GC47
u~
~
LN
ll048
LN JU4'3
LN Jt::;o
LN )051
~..~~ aos2
Ul H53
56 p(!)::C.
Do 74 !=i,NLOAO
R. E AC ( 6 G, 6 4 0 t N N
6 4 0 F 0 H· !H ( I 5 t
tJi=tlN"'2·2
74 REACd6Q,650) P(N1+1),P(N1+2)
650 ~ORMAT(2F15.3t
C ASSEMBLl
Nk=C
- P=1
Ul J )54LN Otl55
6o
uJ ac 56
~o.N
LN 1C60
LN 0061
LN 0(;62
LN QG63
LN J866
LN
0!173
J074
J J 75
0076
78 I=i,NDF
MATRIX
GO TO 78
PFd:iP):;F(U
600
Wk1TE(61t600) NNODE,NELEM,NLOADtNINCR
FC~~AT t• NUI·18E~ OF NOOES=•,I5,//,
1• NLM9ER OF ELEnE~TS=•,Is,/1,
2+ ~!UMoC:F. OF tWoE~ WITH LOAiJ:+,IStllt
3'*' ~~Ut'BER uF !NC~d::ME"!TAL LOAJING=..,,I~,///1/)
\-l~·ITE <&1, 60 1)
601 Fcru1AT
(
1• NCDE
OEG~EE OF FPEEDOH
2"" NLJI-IBER
U
II
DO 7C8 I=1tNNODE
LN JC70
Ltl
L.N
LN
LN
CCLU~N
IP=IP•t
78 COHT!NJE
J~69
LN OG71
LN J072
P
NR=!'IR.+1
LN 3:n;:..
LN ac65
LN OC67
l t: 0 2 E: 8
~ECUCEO
IF<IDF<IL'4E.0)
tHS7
U-4 OJ58
Lt-! JG59
NODE NUMBER AND NOOE LOADS IN X,Y DIRECTIONS
~EAD
LOCATION
OF
X
COORDINATES~./,
y
¥)
N2=I+2-2
WVrT£(61 1 602) I,loF(N2+1liiJF(N2+2),X(!),Y(I)
6 0 2 F u~ r-. AT ( I? , 2 I 8 , 2 t BX, F 1 0 • 5)
Wr\ITE!61,603)
7GJ
6C3 FORMAT(/////,
i+
ELE~ENT
NUMBER
NOD~
POINT NUMBERS
AR::A
\
\
YOUNGS MOO"',/,
..•
0".
Oo
INTEGER
tSl i='ORTRANC2.:H /MI\STE~
'·
2•
00 710 1=1tNELEM
Ul GC77'
t..N Jt7fl
LN
LN
~'J79
oc~o
Ui l1 C81
LN ):::82
U/ ·c;.; K3
LN ::084
L..N ~C&?
L ~; J l' 6
L t; UG1:\ 7
c
u~
t.~:
~N
~N
uub a
OC8Y
~~gc
i.:J91
t...N
LN
~C92
~ Cg 3
u~
•4
..l-:
95
:t r- 0 r
L~
aC':J9
LN OD94
~
tN
oC97
LN JC9B
Ul :J1CC!
LN
Q1C1
U1
LN
Lt\
l.N
di
J 1C 2
u~
.J 11· B
LN
LN
~103
OlC4
01G5
J 1G 6
J1~17
l)l(.g
lN l11G
UJ J 111
LN ~112
Lf\ J113
u.1 a114
LN J115
LN 011
o
WJ~D
7.1.0 WRITE<o1,6U4)
SIZE
I
=1
, • JPTION IS
OFF ,
~
OPTIC~
IS
.1;11-,
J
ltNIIICI),NNJUltAfi.E~~U),EU>
604
Fq~MATJ~I1QtF16.5,Elb.5)
~,g
FORM~Tt//1//)
WF~TE(o.,o~~)
.
C Zc°C CUT LINEAR AND NONLINEAR. STIFFNESS HATRIX
nO 1f'l3 I=l,tWF
=
0 0 1 f.: 3 J 1 , NO F
GK·c: ,J) =i1•
St<IrJ,=O.
~Nrq,Jl=Q•
..:~Nl (.._ tJ) =b•
S LiZ < I ! J ) =D•
103 SIJU \~,Jl=C.
C CALCULATE Ll~€AR GL08A~ STIFFNESS OF THE SYSTEM
DC 189 K~i,NELEH
Ni=l\til\K)
N .J::: l·i ~·.1 J ( K)
DO 11C
110
!~1,4
co 11C J=1,4
GK(I,J>=BiK,J)¥8(K,t)~AREA(K)•E<K•/iTLENGTH(K)••3)
ou 18? 1=1,2
00 187 J=1,2
!H::: L I ._ 2 - 2 l- I
N2=!li"2-2+J
N3=~J·2-2+I
4 !\ J lO. 2 - 2 .. j
~ < ~11 , .~ 2 >
Loa~
~~
s
=
=s
N2> • GKo,
J,
SLl~3,N4)=SL(N3,N4)+GK(I+2,J+2)
SL(N1,N4)=SL(N1,N4)+GK(i,J+2)
t87 S~IN3,.N?)·=5L(N3,N2)+GK<r+2,J)
18'3 CONTH~~t
C ASSEMdLl ~EDUCED LINEAR STIFFNESS HATRIX OF
1 A=J
lS?=t
C 0 . 13 G I =1 , NO F
JSR=l
TH~
~YSTEM
r::::.:~=ISR+IA
I A=D
C'\
'-D
'-
Do !30 J=1,NnF
LN lii7
LtJ 01115
LN J 11 q
LH Ji2C
lF<IDF(Jl.NE:u) GO TO 130
IFtiDF(I)•NE.Q) GO TO 130
I A-1
LN J121
LN J122
i..N J123
l.N J124
LN
LN
0125
:J126
LN J1:::7
LN J 1 Z 8
L~
J12g
Lt" 0 1~e
U4 J 131
LN ~ 13 2
LN · 13 3
LN C134
Ui J 13 S
LN
L. N
u.:
~136
• 1 .~ 7
Jt3~
L f'l
G1.3 g
LN
~:14G
LN J1Lt1
LN jl42
dl Q i 43
LN
Lr~
G 14 4
Q lit 5
LN Ji49
LN 111+
LN )148
u. 0 !4 g
L I~ J 1 S )
i..N J151
LN J.15 2
S~R<ISR,JSR)=SL(! 9 J)
J.A<= JS~+l
130 CON11NUE
.., S!ZE OF RfOUCEJ ~ATRlX=NR 9Y NR
c G0HFUTE NONLI~EA~ STIFFNESS MAT~IX CF
!CCUNT=c
80 I COUI~T=ICOUr~T+1
00 15~ L=i,NEL£M
N I= tdl I ( l. )
N J=~·;N J( U
COS= CX < NJ ) ·X ( NI ) ) IT.,. ENG T H ( U
~
.
TH~
SYSTEM
SIN=(Y(NJ~·V(Nl))/TLENGTHCL)
~!L 1 =NP 2-1
~~ L 2
2-1
=ru•
U(N!»=JISP\NL!.)
U (NJ.I=)lSP(Nl2)
V<Nl)=DISPl~L1+1)
V OLJ l =DIS;:, ( NL 2 +U
.
·
·
EST~A!~=(U(NJ)+U(N!))/TLENGTHCL·)•COS+(VCNJ)•V(N!J)/T~ENGTK(L)~S!N
\~TF AI!~= (U ( NI}
NJ)) /TL~t>JGT Ht L )•SIN+ ( V (NJ) ·VtN!) >/'fL!::NG"fH (I..) •coS
t1 !.. Ph A::: E 3 T ;;;: AI h + { 3 • J I 2 • 0 ) ~ t P. 5 f RA I 1\'"' 4 2)
·
-u (
$NP(NL1,Nl1):(-?~+SlN•COS•F3TRAIN+SI~~·2•ALPHA
~ r; P { NLl tt~ L 1 + 1 ) =1- !.5! N' '"'2 t ) ~:;: S T K A! N + ( G0 S •"' 2 ) <\f. E.S T ;(.A IN-SIN+ C0 S • ALP H /1.
s r, F { N L 1 .r~ L 2 , = < - s"' p ( ~ L 1. 1\ L.l > '
=(-
+i. )
3 ~J t-' ! N !.. 1 t ~o~ L 1 +l ) )
St~P (NL1+1,:H 1l =S;<;P(~.JLl,NL1+1.)
SNPf~L1+1,~L1+ii=2~SIN+CJS•~ST~AIN+(~03
S ~~ P (11 L 1 t 1 , 'i L 2) ::- S :~P~ ~l 1•1 , i'L 1)
S N D ( t JL 1 • 1 • r~ i.. 2 t 1 I ::: \ - ;> r~ ~· ( ~ L 1 +1 , 1·1 L 1 + :1 ) }
S ~-.:: { NL 2 , 1\J ' !. )
l'< p ( ri ~ 1 , ~ l. 2 )
S /, P <i\ L 1 , ~~ l 2
k
t
4
•2l•A~PHA
=:
S NP ( NL 2 , r~ 1 +1 ) :..: 3 t ? l I~~ 1 , r~:.. 2 +1 )
s t, P o'l L 2 • r~ L 2 } =s N P t i'"- 1 , N'· i ,
--.]
0
INTEGER WORD SIZE = 1 , • OPiiON IS
lSI rORTR.ANiZ.3l/M.e.STER
LN· l153
t~ ~r~~
:.,N 0156
1..N J 15 7
LN G15d
lN J 159
LN tl16!)
d~ n.1.61
LN Jl7J
1&C
180
I A= C
I SP=1
CJ 13 6
JSF=1
J17~
LN 3175
IF<lOF(J).NE.Cl GO TO 136
IF<ICF<Il .NE.O) GO TC 136
J17g
J 1 ea
I
u186
J!.37
~= 1
S~~R<!§R,JSRl=SNLt!,J}
Olt:~1
!J•b5
r:: 1 , ND F
ISF=IS~+lA
ll17~
0182
J:i.A3
01tl4
GEOMETRIC
I A=··
00 I36 J:t,NCF
LN J 17 6
LN )177
u~
SNL(~3,~2)=SNL(~3,N2)+SNP{N3,N2}•EtLl•AREA(L)/T~E~GTH(~}
CU~l.~,NUC.
~EDUCED SYSTE~
HAT~IX
C ASSEMELE
1..N J173
LN
!,.N
LN
LN
LN
Ll\
LN
LN
LN
Ni=N!""2-2+.L
N2=i'H""2o2+J
N 3=t.J• 2-2 t·I
N <+=~; J•2- 2 t-,_1
S i~L Oli, N 2) SNL Oh , Nz l +SNP ( N 1. , N 2) • E '( L) • ARE A ( L) IT LENGTH(!..)
S :\ L 0; 3, N4 ) =S NL \ :~ 3 , ~It ) +S NF ( N3 , l.j r. )4 E ( L ) • 1\ R.E A ( L ) IT 1.. £ NGT H { L)
S I~ L « ~~ 1 , tJ 4 ) ::.: SNL.. H l1 , N1+ , + '5 ~J P ( N1. , ~~ 4 ) 4 E CL. ) 4 AR. E A { L >IT .. E!>. GT H( U
=
tN 0172
u~
0 OPTION
SNP(NL2+1,NL1,=SNP(Nt 1+1,NL2)
SNP~NL2+1tNLi+1)=S~Pl~L1+1,N~2+1)
~ NP H< L2 t1 , 'I L2 \ = S NP U L i ~ 1 , ht L1 )
::; h P ( N L2 i 1 , ~ L 2 t 1 ) ::: S 'H· ( N \... 1 ... 1 , ~ L 1 + 1)
C ASSEMBLE TJTAL SVST~M G~OMETRIC HATRIX
Do i6C I=1,2
DO 16r J=tt.2
LN ~163
LN )164
L~J 0165
dl .) 166
lf-;IJ171
~
SNP<NL2,NL2~1):SNP!NL1tNL1+1)
uj :; 162
i..t~ ~ 16'
LN 0168
LN iJ169
OFF
J.:, :::JSK+1
136 corHlNJE
C ASStHJLE JOTAL
=
17L
~YSTEM N~NLINEAR
0 0 1 7 I; I 1 ' N Ft
0•) 170 J=l,NI<
STIFFNESS MATRIX
SKGR(I,Jt=Sl~(!,J)+SNLRl!.Jl
"'J
_...
·.
LN J1.8ti
LN l 189
LN J 1C3J
L~
J1131
U; 0 1g2
LN ~ 1 g 3
LN
L ~-;
~1q4
a1 g s
u-1 o19o
!..h 3i97
Ll\ Ji9~
Lt~ ~ l ::; g
i.. ~ D 20 G
i.. N u 2C 1
LN nz~:z
LN J2C3
l. 1.,1 [} 2 f) 4
L ~' 0 2( 5
LN U 2t~ b
u
1
LN J2C8
Ll> C 2C q
u~
-~
LN U2l.G
Ui 0 211
LN J212
LN CJ1"t
U4 0 2J 4
i.N 021.5
lti 0216
Ll\
0~17
U; t:i2l,'!.
d! 0 21 g
L N 1 22 J
lt<' J221
o 22 2
u: a 22 3
u~
Lt\
022t~
c stzE oF -R.EoucEQ . .,.ArR.i:x~si(GiitNR..~RJ
C CALCULATE iHE .o.NVE~.SE 8F THE FJ:.ouCED ~ATR.!X,SKGR<N~,NR)
CALL !NVERTlN~ SKGR SKI)
·
C COMPUTE NGDAL INC;.EM~NTA~ D!SPLACEHENTS AT EACH I~C,. LOADS
00 175 I=1,1\R
C SP (I)= 1; ..
00 175 J=1tNF.
175 OSP(l)=DSP(ll•SKI<I,J)~P~(J)/NINCR
C CALCULATE T07A- DISPLACEMENTS
1JX:::1
QO 176 I-=1 NDF
.i.FfiCF<I,.~f.:H GO fO 177
E~SP (I) =DsP ODX)
IuX=IOX+l
GO TO 176
r 1 E J s P cr ,, : c •
i 7b
815
c·r·ii•.:U:.
od'.s15 1.= dJOF
Dl~P1iJ=ofsp(l)+EOS~<Il
Wr. I 1 f ( S 1 , 65 5)
6 5 5 . F C~- t-; AT C I ,
1 "" L C A0
I'· 0!) E
2• It,CR. I'.UMaER
uDEFLECTT- ONS
1 =1
AT rHE
~ooEs•,t,
v
.,
DO 7'15 J=i, ;moDE
662
Wk!7E(51,662,
ICCU~T,J,OISp<Il~O!SPCI+1)
FO~MAT(215,4X,2{4X,~1&.5))
I;;:I+-2
715 CONliNUE
C COMPUTE ElEMENT STRAINS AND STRESSES
1 F <1 C OU NT • E Q • ~liN CR)
DO "19 I=t,NOF
00"_79 J=1~ WF
a
G
T0 6 8 0
1
79 SrH- CI,J,=u•
GO TO %
680 !COUNT=NlNCR
ENLI
"-'
N
SUBkOUfiNE IN~ERTtNtXK,XKI)
DIMENSICN XK(20,2G)tXKil20t2D)
LN OOC1
Ll'-; 0 CC2
0 I MENS! 0 N 0 t 2 :3 J ~ U ( 2 u , ~ iJ )
l.N :l(C3
LN
•1r'f!4
Lt-. ~(,~5
i..N :n:c-6
Ll~
J(1~7
L.N qrJC8
u~
wl;~.;g
tt~
j['t:J
~~
c: 1 3
I)
-1~.1Lt
jJ15
u; JC i 6
Lt·· Oi:l7
u. 0 c11
U; OJ19
Lh '1t;2()
LN :l02i
l.~ Jl22
LN uJ23
Lt-.; '•')24
Ul ~ G2 5
L~~
u: ' [ 26
tl
L
u~
U(!tl)=i.
N K= 1-1
K=:!..
3U
60
Jfi=I t1
00 50 J::JN,N
K=:J.
IFt~~K .. LT.t<) GO TO 50
U CI , .J) U 0, .1) - U ( K 9!) • U ( K, J) _.. 0 t K)
=
K::: K + 1
DG 7C I;;:!.,N
~ ., t.~::- '3
DO 70
dl
u~31
tNN
uC3,1
UJ K0 ~?
U c. 3 4
c
Ji;::J.i-1
I.
IJ
lj
35
ul~
~
LN uu 1
L ri J (: .3 cl
i..N J03q
·- - .. --
GG TC 60
50 U<I,J):(U(I,Jl+XK(l,J))/0(!)
:-:~28
Ot;10
LN
K=Kti
GO TC 4C
Ul~l~Il=1.
LN
:..~l
4f! IFUlK.LT.K) GO TO 30
0 <I J =0 \1 t -U ( K1 I) • U ( K, I~ "'D ( K,
l- '- '
!J
; '
:
oc s:; r:::J.,N
0 n) =0 .; 1) +X K (1 ~!)
0. ". . . ,.
L ~·
·
DO 1C J=i•N
U (JtJt=C.
10 UIC ... ~J):::O.
J c11
U!
~J
I<= J-1
J=J~~,
N
K=i
90 IF fNK.Lf.KI GO TQ 70 4
u:{ ,J)=UIC!,J)-Ui<I,Kl U(K, J)
K=K+1
70
GO TO 9J
o"'l ;·
•j ··~: -!._
·r >' uI I c..
-
8000
~.)
~ f
1J0 I=1,N
1UC J=1 N
1UO UlO(I,J):Uf(I,Jl/O(J)
LN ]"4()
CC 112 1:=1,N
d~
X K! (! ·• J):: iJ •
DC 112 K=l,~~
~N ~~41
•J;J42
Lr-. OC 1-t3
LN
OC44
LN :J::-..5
LN JO:.o
U J ( 2 (., , 20 ) t U I 0 ( 2\l • 2 ~ )
0(!)=C.
U: OC12
i.. t~
t
00 :tO I-=itN
t)Ci 112
112
J=l~N
XK!(I,J~=X~I(t,Jl•UIOf!,Kl•UitJ,KJ
PEfUt.:N
E"'lJ
·-._J
\~...;