,-
CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
WEIGHT OPTIMIZATION OF STEPPED BEAMS
A graduate project submitted in partial
satisfaction of the requirements for
the degree of Master of Science in
ENGINEERING
by
GEORGE WILLIAM ABSHER
/
June, 1977
r------------------·------- ·------------------------·---·-.-- --------··------ ·- ·- --·-- ·----------·------· ---------------]
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The Graduate Project of George William Absher
is ap roved:
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California State University, Northridge
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~---·---·--·
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----- ···- - -----·-·· ------·--
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---------~-------------------------------------------------
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---- ·-'- __________________ j
:·-------- -------·---------·-··------·-
---·-·-----·-·-------:-·---·------·-··-~----
·--· - -----·-------- --·-·-·--·-··-------1
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ACKNOWLEDGEMENT
To Professor Spunt for suggesting the
study area and Leslie Rosman for typing.
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-~--·---- ·-·----------------·-~--- --------~------------- --··
.iii
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------------------------ ··---------- -------~---------_.J
r--·---------------·------·--·--------------------------~-----·----------~-----------~-----··-----i
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TABLE OF CONTENTS
PAGE
ACKNOWLEDGEMENT
iii
ABSTRACT
v
INTRODUC'fiON
1
SINGLE POINT LOADED CANTILEVER BEAMS
3
UNIFORMLY LOADED CANTILEVER BEAMS
18
CENTER POINT LOADED SIMPLY SUPPORTED BEAMS
33
UNIFORMLY LOADED SIMPLY SUPPORTED BEAMS
48
COMPARISON BETWEEN TWO DIFFERENT TYPES OF
BRIDGE STRUCTURES
64
UNIFORMLY LOADED BEAMS SUPPORTED AT THREE Pw\CES
68
CONCLUSIONS
80
LIST OF REFERENCES
82
LIST OF SYMBOLS
83
APPENDIX A
LISTING OF OPTIMIZATION PROGRAM
84
APPENDIX B
LISTING OF NU}IERICAL INTEGRATION PROGRAM
89
i
L___ --------------------------------------------------··--- ------------------------------'
iv
,-------------------------------
ABSTRACT
WEIGHT OPTIMIZATION OF STEPPED BEAMS
by
George William Absher
Master of Science in Engineering
The purpose of this project is to investigate the
effects on the minimum'weight requirements of a beam when
different types of beam tapering are used.
The results
are then compared with the minimum weight requirement of a
uniform beam.
The optimization technique employed for the continuous
tapers was first described in Ref. (1), and later extended
3
to the stepped taper case by the same author. ( )
volves the use of a systems merit function.
This in-
An optimiza-
tion program based on the Monte Carlo method was set up to
handle many of the more complex merit functions.
Two groups of cantilever beam load.ing configurations
were investigated.
The first group consisted of. single
point loaded cantilevers, while the second group was com·posed of_uniformly loaded cantilevers.
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1(_
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__________________________________ _
_________________________ _j
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,-----------·---·-·---- · . - · - - - · - · - - - - - - - - - - · · - - · - - - --------·-
·-··--·--···-------------------·--·-------~
.
!
Two groups of simply supported beams were also investigated.
The first group
consi~ted
of simply supported
beams with a point load at the center of their span.
The
members of the second group were uniformly loaded.
The cantilever beams and the simply supported beams
had the lowest merit function, therefore the least weight,
when they were of the continuous taper configuration.
Small weight differences were found between step tapered
beams with equal length steps, and step tapered beams with
optimized length steps.
Weight comparisons were also done for two examples
of bridge beam structures.
The first involved two canti-
lever beams meeting in the center of the span compared
with a simply supported span which turned out to be the
lightest bridge structure.
A comparison was done between two different methods
of supporting a beam at three places.
It was shown that a
three point support beam structure which allows overhang
can be lighter than a beam supported at both ends and the
center.
.
....
)."-. ~
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:
~------·-·----------------------------
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----·------------------------------'
vi
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---------------------:---------------~
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INTRODUCTION
The primary interest of this study is in the weight
ratios between different types of beams.
For this purpose
I
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-.a merit function based on weight will be used.
This merit
function will be a function which is proportional to the
optimized weight of the beam.
To find this merit function we consider the following
things.
The minimum cross-sectinal area of a beam is
dependent upon the maximum bending moment, and is in fact
.
*
. proportional to M2/ 3 (l). A function which is proportional
2
to the weight of the beam is therefore M / 3 L, where L is
l
1
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l,i
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the length of the beam.
The merit function M2/ 3 L is for a uniform beam but
for a step tapered beam we use the lengths of the individ-
!.
ual steps (L') in place of (L), and we use the maximum
bending moment in each section (M.)
instead of the maximum
l.
moment in the beam (M).
Summing these individual terms we
have a merit function equal to ~ M~/ 3 L•.
. 1
].<=
l.
I
When dealing with-a continuously tapered and optimized!
beam the merit function is again different.
For this case
j
the maximum moment (M) is replaced with the maximum moment
!
(M(x)), on a differential length of beam (dx). The merit I
L 2/3
I
function over the length of the beam is therefore
(x)dx. i
I
. 0
!
[M
*
Supers~r~pts
refer to
refer~n~-~~----------
1
__________________________
2
~----·------------------------------------·--
i
!
------------------------------- ----------------·------------- --------6
The following four types of problems were evaluated
in this fashion.
1. Single point loaded cantilever beams.
2. Uniformly loaded cantilever beams.
3. Center point loaded, simply supported beams.
4. Uniformly loaded, simply supported beams.
--·-:
~
!
--------------- ----
---
-- ---------------. -------
----------- --- . - -- ·------
-
------------
----
.
------------- -----· --------
---~
CANTILEVER BEAMS
SINGLE POINT LOADED
The following cases of point loaded cantilever beams
were investigated.
1. Uniform beams.
2. Beams with continuously optimized taper.
3. Step tapered beams with equal length sections.
4. Step tapered beams with optimised length steps.
Comparisons were made between these four types of
beams to determine the ratio's between the minimum values
of the beam cross-sectional area and hence the weight
ratio's between the beams.
. - - - - - . .--J
3
4
~---------------------------;:~~.=~:--~~:----------------------------------1
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-
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SINGLE POINT LOAD
1
UNIFORM CROSS-SECTION
I
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p
I
.,
~~------- L--------~
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.I
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In the case of a uniform beam the maximum moment (M)
is equal to the product of the load (P) and the distance
of the load from the support (L), or, M =PL.
The merit
II
I
I.
I
function is therefore:(!)
~-
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------------------------------------- . . ---.. --- __· . . ___ _I
-------------- ·-·----------·---~-------------------- ---------:-----------c-·--..,----·-------------
.
-----------------1
CANTILEVER BEAM
..
i
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SINGLE POINT LOAD
I
CONTINUOUSLY OPTIMIZED TAPER
I
p
'~----
:
L _ _ _ ____.
-
~
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I!
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For the case of the point loaded cantilever beam with
a continuously optimized taper, the merit function assumes
the integral configuration and becomes (l)
Merit ~
LA
fM
{x)dx
0
Merit
f_ (Px)
L
Where A = 2/3
~
1
i
I
I
2/3
dx
0
Merit
=
.6P
2/3 5/31L
=
X
.
.6P
2/3 5/3
L
.0
Figure 1 shows graphically how the cross sectional area
changes over the length of the beam.
!
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L____ ---------------------------------------------------------------
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6
FIGURE 1
CANTILEVER BEAM
'
i-
SINGLE POINT LOAD
CONTINUOUSLY OPTIMIZED TAPER
RELATIVE CROSS-SECTIONAL AREA
vs
LOCATION ALONG BEAM
1.0
.9
~
~
.8
<
..:I
<
z .7
0
.....
-~
u~
rl.l
rl.l
rl.l
0
~
.6
.5
!
u
f;l:1
::> .4
.....
E-4
j
i
~
I
~
.3
I
I
.2
.1
I
I
.9L
I
L-------~------~------~--
l,l
x.:::L
LOCATION ALONG BEAM
--~--------------- ----·-·-·------ _____ j
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7
----~--
·--
~-- -·~---~----<---------------------------~----------~---------------------·----------,
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CANTILEVER BEAM
r'
i
SINGLE POINT LOADED
I
STEP TAPERED WITH EQUAL LENGTH SECTIONS
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The next beam considered was the. single point loaded
cantilever with equal length sections. The Merit function
(M 2/ 3 L) for this case is slightly differento The maximum
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.I
: mqment (M) becomes the maximum moment for a particular
section of beam (M 1 ).
The length (Li) is replaced with
the length of a section
I
~
(*), where (N) is the number of
sections the beam is divided into.
An example case for (N) equal to four follows:
p
~--~
,
,"',
,,
,.
,
11~~l"~i
+-r+-r-+l
J
t =
I
M:eri t
'"I"'
~
(~~ 2/3~ + (~) 2/3~
p2/3L5/3
4
[(i) 2/3 +
2 3
2/3
/ L
+ (3PL)
(PL)
I!
4 +
(if/3 +
m
.
2/3 +
i
(it
3
]
I
----· ·-- - .... -··-·-···---· ...... ··-- ---·-·------· --··----------·---·---------·--·-·-·-----J
8
---
------~--
-~-----------
-
----------------------- -----·- ---- - - - - - -
-------------,
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Merit
I
I
For the general case the merit function becomes:
p
~ NL
~
""
,. '
---i-+-- L
;'
N
-
"'"
LL
N
""
""
""
""
1
-------
r
---,
L
N
--'-------~
.................
--
""
.;'
,
"'
Merit ==
+ 2
A
+ ••• +
Table 2 gives the numerical values for this group of
merit functions.
Figure 2 gives a graphical representa-
tion of the result obtained by changing the number of beam
sections "Nn.
I:
.
.
.
[_ ___ ------------------------------------- --------------------------------------- --------------------
1
FIGURE 2
CANTILEVER BEAM
SINGLE POINT LOAD
1.0
-EQUAL LENGTH BEAM SECTIONS
MERIT FUNCTION
i
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~
vs
.9
1-1
I
I
NUMBER OF BEAM SECTIONS
E-i
u
z
I
~
E-i
1-1
~
I
i
.s
I
~
-~
I
E-4
6
I
1-1
~
;:=
I
.7
I
I
lI ~-+---l---t--+---+--+---t--+--~--~1-=-----:1:---~~~-~
I
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I
16 I
I
I
•6
I
1
2
3
4
5
6
7
I
8
9
10
12
14
I
NUMBER OF BEAM SECTIONS
L____________________ -------~----------------------------------.------------------------------------------- ------------·
j
i
---··--·---·-J
(J)
10
r-----~----·-
--------------·-----·· -----------------------------------·---------,
CANTILEVER BEAMS
SINGLE POINT LOADED
STEP TAPERED WITH OPTIMIZED LENGTH STEPS
When dealing with a step tapered beam with different
step lengths it is again necessary to modify the merit
2 3
function for the general beam (M / L). As in the case of
a beam with equal length steps, the maximum moment for each
individual section of beam (Mi) replaces (M).
The length
(L) is replaced by the length of each individual section
(K.L), where (K.) is a fractional multiplier that indicates
~
~
what fraction of the total length each section represents.
It is these fractional multipliers that are variables subject to optimization based on minimum weight.< 3 )
The merit function has now become; let A=2/3
N
=
I
(M. 2/3K.L)
~
~
i=l
For an example the case of N=3 will be shown.
Merit
p
....h-K1 L-K 2 L 1;"
'*"
K~LTK 1 L-
:
- I
i
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-I
~
,,
~
,.
_i
I
L-.----------·-----
i
11
r-·---------------------------------
!
Merit
-
------------------------------------------------~
~ .~(~K1 L) ~ (PK1 L)~1 L +
1~
I
[P(K LtK2 L>] AK 2 L
1
A
+ (PL) (L-K1 L-K 2L)
Merit
= (~L5 / 3 ) [~K 1+(K1+K 2 )AK 2+(1-K 1 -K 2>}
I
I
For the general case the merit function can be repre-·l
I
sented as:
------------------------t:
-j-<
t
:;
L£KiL
p
L
KiL
----------------------~
KiL
.,.
,.
1______
,.
,.
rKiL
~~~I____,
~
,..
,....- ,. ....
~
:.,....
_,.
,..
/
Merit
Merit
A
••• +(Kl+K2+•••+KN-2+KN-l)
•••
.
(KN-l)+(l-Kl-K2- •••
-~-2-K:r\f-1 ~
Table 2 gives the numerical results for the values of
the merit functions for this type of beam, and figure 3
shows a graphical representation of the results.
The
values obtained for the lengths of the optimized sections
are presented in tible
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l__________________ ·- --------- .
1.
. ______________ j
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~--··-·---------------------.-·-------~-----·-·-------·----·-------------------,
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TABLE 1
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CANTILEVER BEAM
I
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II
SINGLE POINT LOADING
I
OPTIMIZED LENGTH SECTIONS
_I
-
NmffiER OF
BEAM SECTIONS
MERIT
FUNCTIO~
K.
l.
N
K1
.4648
3
.74691
K1
K2
.2941
.3387
K1
K2
K3
.2117
.2438
.2644
• 71201
5
.69057
KJ.
K2
K3
K4
.1638
.1886
• 2046
.2166
6
.67605
K1
K2
K3
K4
K5
.1326
.1528
.1657
.1754
.1834
7
.66556
K1
K2
K3
K4
.1109
.1277
.1385
• 1467
.1533
.1590
K5
K6
I
Ki
.81410
,,,
-
VALUE OF
2
4
L---·----·--···-·----------~-----------------------
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---------·--
I
l
I
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I
13
,-----------------------------------------------------------------------------------------------------~
TABLE 1 CONTINUED
:t{UMBER OF
BEAM SECTIONS
N
g
MERIT
FUNCTION
K.
.65762
K1
l.
K4
K5
K6
K7
.65140
K1
K2
K3
K4
K5
K6
K7
Kg
10
.64640
K.
l.
K2
K3
9
VALUE OF
K1
K2
K3
K4
K5
K6
K7
Kg
Kg
.0949
.1093
.11g6
.1255
.1312
.1361
.1403
.og27
.0953
.1033
.1094
.1144
.11g6
.1223
.1256
.0731
.og42
.0913
.0967
.1011
.104g
.10g1
.1110
.113'6
.. . . .
;
L________________ ----------
---------- __I
~------
·--·····
i
FIGURE 3
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CANTILEVER BEAM
I
1.0
SINGLE POINT LOAD
I
I
OPTIMIZED LENGTH BEAM SECTION
MERIT FUNCTION
~
vs
.9
H
E-4
NUMBER OF BEAM SECTIONS
(.)
~
~
~
.8
·~
~
,.
H
fiz;1
i:t=
.7
~6~~~~1
2
3
4
5
6
7
8
9
10
NUMBER OF BEAM SECTIONS
12
14
16
i
I
I
-----· ----- ··-··-.J
.....
•
15
,----------------------------------------------------------------------------- ------ --·- ----------------------1
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TABLE 2
:
POINT LOADED CANTILEVER BEAMS
MERIT FUNCTION RESULTS
(1)
*Mer1t
.
for uniform cross section beam = 1.0
Merit for continuous taper beam
(2)
N
MERrr FUNCTIONS
{3)
OPTIMIZED
LENGTH
SECTIONS
(4)
EQUAL
LENGTH
SECTIONS
PERCENT
DIFFERENCE
(4)-(3)
(4)
0.6
MERIT RATIO
(4)
m
1.67
1.
1
=
2
.8141
. .8150
.11
1.36
3
• 7469
• 7480
.15
1.25
4
.7120
.7131
.15
1.19
i
5
.6906
.6916
.14
1.15
6
.6760
.6770
.15
1.13
7
.6656
.6665
.14
1.11
8
.6576
.6585
.14
1.098
9
.6514
.6523
.14
1.087
10
.6464
.• 6472
.12
1.079
20
*
.6241
i-
!
[
!
!
i
!
1.040
3 5 3
All merit value are q 21 L /
i
_____________________________ j
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--- --------···--···-·-·----------------------- ---------,
~---------
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FIGURE 4
CANTILEVER BEAM
i
!
SINGLE POINT LOADED
I
WEIGHT MERIT RATIO
1.7
vs
1.6
NUMBER OF BEAM SECTIONS
0
H
~
E-4
H
~
I
I
I
I
I
I
I
i
::a!
E-4
ffi
1.5
1.4
1.3
H
rii
IS:
1.2
1.1
1~0
T-----r-----r-~~r---~----~~---+----~-
1
3
5
7
9
13
11
15
I
L________
NUMBER OF BEAM SECTIONS "N"
.
----------------------------------·--------------------------j
~,.
17
r · - - - - - . - - - - : - · · · · · - · · - . ---------·---·------····----·-····----········-·· ---
·---~---·
.....
·-------··--·-·---~
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CANTILEVER BEAMS
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SINGLE POINT LOADED
SUMMARY
!I
!
Now that the merit functions for these four types of
end loaded cantilever beams have been determined, we can
compare their differences.
By looking at the results shown in table 2 it can be
I
seen that the maximum percentage difference between the
weight merit functions for beams with equal length sections and beams with optimized length sections is only
I!
0.15 percent.
I
It does not appear that the use of opti-.
:;::d0 ~e:::: :::t::::i:: ::::~:::::l:fa::::c::::r::: ::1:
I
0
I.
be used.
The ratio of the merit function for a fully tapered
beam to that of step tapered beams with equal length steps
is shown in figure 4, and can also be seen in table 2.
.,
The percentage weight difference ranges from 67% for a
I
uniform beam to four percent for a beam with 20 equal
length sections.
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IL--------~-..- · · - · - - - - - ·
.
I.
--··--
r---·~-----·-~----·----------·-~-----·-------------·------------------·
'
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CANTILEVER BEAMS
I
UNIFORMLY LOADED
Four cases of uniformly loaded cantilever beams were
compared to determine the ratios of their optimum weights.
The four types of beams considered are listed below.
1. Uniform beams.
2. Beams with a continuously optimized taper.
3. Step tapered beams with equal length sections.
4. Step tapered beams with optimized length steps.
I
!________________________- ___________________________j
18
1~
r-------------~---------------------- --------------------------------~
.
i
CANTILEVER BEAMS
UNIFORMLY LOADED
UNIFORM CROSS SECTION
For a uniformly loaded cantilever beam the maximum
moment is:
M=(qL) ~5L)
= • 5q (L) 2 •
Therefore the merit
function becomes:
Where (q) is the load per
unit length, and L is the length of the beam.
I
I
I
\----------~------------------~-------------------------------------- -- ------------~-----_!
20
r--------------------------------------------------------
CANTILEVER BEAMS
1
I
UNIFORMLY LOADED
l
CONTINUOUSLY OPTIMIZED TAPER
1
l
I -
J
I
I
I
I!
I:
i
The minimum cross-sectional area of the beam, being
2 3
proportional to {M / ) at each point along the beam can
I
lead us to a relation that is proportional to the weight
I
of the beam.
I
The moment (M) at any point on the beam is
2
M={qx)(.5x)=.5qx • To find the merit function we take
2 3
(M 1 ) over the length of the beam and get the relation:
L(, 2/3 {l)Lf
2 A
where A=2/3
Merit = .bM(x)dx= _ Jo(.5qx ) dx
q 2/3L Lfo · 4/3
Merit =
A
x dx=
2
q 2/3L
'
2
A
3
7x
7/3
f
b
2 3
3
Merit = .267 q / L7 /
; or letting W=qL we get:
Merit = .267
I.
I
I
I
I
I
I
w21 3 L5 / 3
I
Figure 5 gives the relationship between cross sectioned
I
I
area and beam length.
l ____________________ _
II
------------------
J
21
;-------------1
I
I
FIGURE 5
CANTILEVER BEAM
I
i
UNIFORMLY LOADED
CONTINUOUS TAPER
.I
RELATIVE CROSS-SECTIONAL AREA
I
vs
POSITION ALONG BEAM LENGTH
1.0
<
~
~
<
~
<
z
0
~
E-4
0
~
rll
I
rll
.s
.7
Ql
I.
~
I
I
~ .6
0
l>
~
8
.5
~
~
.4
j
I
.3
.2
.1
I
~---1---4-----+--+--1---t---;--t--r--r-l
.lL .2L .3L .4L .5L .6L .7L .SL .9L
I
POSITION ALONG BEAM LENGTH
x=L
[
--------------·--'------------·___ _ j
22
------
·--
--~-------- -~--
----·------------------
-------------
---~
I
CANTILEVER BEAMS
I
UNIFORMLY LOADED
STEP TAPERED WITH EQUAL LENGTH STEPS
The merit function for a step tapered beam with equal
length sections and a uniform load is:
N
Merit =
i~(Mi)A(i)
where A is equal to (2/3), and- (M.)
1
is the maximum moment in each individual section of beam.
For.a beam with three equal length sections this
turns out to be:
,"'
,
1
~
"'
"'
I
Merit
Merit
Merit
=
E2~L) <~>] \~)
(~~ .tf
l
[---------·--·-·----------
+
J;
+ (:)A
~qL) <~>]\~)
And when W =
(qL)
I
I
I
---------------·-___j
-----·----------------·-··---··--. ·---------------------------,
II
For the general case of N sections all of uniform
length we have:
I
I
I
..
:;
I
~
~
......
::..
i
lI
N
Merit
Merit
= i~(Mi)A(~);
=
W~L~~:
where A equals (2/3), and
w=
(qL)
[(l2)A+ ( 2 2)A+ •. •+ ((N-l) 2)A+(N2)AJ
2 N
..
.
.
.
.
I
I
~
I
I
Table 4 gives the merit function values for this type of
beam.
Figure 6 gives
a
graphical representation of the
effects on the merit function caused by using different
I
Ii
I
--I
number of sections in the beam.
I
I
I
I\_____________________
I
I
·---·-----~----_j
FIGURE 6
CANTILEVER BEAM
I
I
UNIFORM LOAD
EQUAL LENGTH SECTIONS
MERIT FUNCTION
vs
IZi
o.
NUMBER OF BEAM SECTIONS "N"
H
E-c
~
~
~
H
~
ril
=a
~
H
ril
iii=
I
I
I
I
!
I
1
2
3
4
I
l
L-------------~--------
5
6.
7
8
9
10
NUMBER OF BEAM SECTIONS "N"
!
I
I
_______ _ j
~
-~,_
~--------~--
--·----------
-
----------------~
1
l
CANTILEVER BEAMS
I
UNIFORMLY LOADED
Ij
STEP TAPERED WI'l'H OPTIMIZED LENGTH STEPS
I
.
.
1
I
I
This is again a case of step tapered beam with differ-!i
ent length steps.
Merit
=
N
Like before the merit function becomes:
A
(M.K.L)
2
1
1
i=l
For the case of N=3, we have:
.,
"~
I
,"'
~
3
, Merit =
) (M~K. L)
f=i
1 .
multiplier.
1
It
1
., ..
~s
Where A=2/3, and K.]. is a fractional
this fractional multiplier, K.,
that is
].
optimized to produce a minimum merit function.
:Merit -
[ (qK 1 L)(+)J \ K l L)+ [q(K Li-K 2 L) (Kl r.;K2L)J A (K L)
1
2
+ [(qL)
(~)]A (L-K 1 L-K2 L);
+
And when W equals (qL)
we get:
:
WAL5/3[ 2
Merit =
K K +
:
1_ 1
2A
i
'
This equation is solved using the optimization program listed in the appendix and the results are as follows:
l__________________________ -
I
1
_________________]
26
Merit
=
.38009WALS/ 3
Where W equals {qL)
Kl = .36767
K2 == .32647
For the general case of N sections of different
lengths we have:
n
fit+ f f t f t f l f f tIff l t t f tit
... ____ ___
.
It
~q
- -: [______
.....
~L£KiL-~~-KiL -~•-KiLL -~------ L
K.L
1
Table 3 gives the optimum length of the sections and the
results of the merit functions and compares them with the
.
results of the other three cases of uniformly loaded cantilever beams.
t
I
Figure 7 gives a graphical interpretation
of the results.
i
I
I
I
I
L.--------.-----~-------
I
I
___
- ________
· _______ _j
21-
,----------------------------------------------- ------------------------------------,
I
I
I-
I
1
TABLE 3
i
;
CANTILEVER BEAMS
UNIFORM LOAD
OPTIMIZED LENGTH BEAM SECTIONS
NUMBER OF
BEAM SECTIONS
N
MERIT
FUNCTION
K.
2
.43929
K1
.52968
3
.38009
K1
K2
.3677
.3265
4
.35144
K1
K2
K3
.2846
.2526
.2366
.33457
K1
K2
K3
K4
.2333
.
5
-
6
7
.32347
.31562
L _____________________________ _
1
VALUE OF
K.
1
.2074
.1942
.1858
K2
K3
K4
K5
.1989
.1764
.1653
.1581
.1527
K1
K2
K3
K4
K5
K6
.1736
.1541
.1444
.1380
.1334
.1298
K]_
---
-----------~
28
--------------- ---- ---------- -------1
TABLE 3 CONTINUED
NU:t.ffiER OF
BEAM SECTIONS
N
MERIT
FUNCTION
8
.30977
K.l.
VALUE OF
Ki
i
!
I
I
K3K4
K --5
K6
K7
.1544
.1370
.1284
.1228
.1187
.1154
.1128
Kl
K2
K3
K4
K5
K6
K7
K8
.1392
.1236
.1158
.1108
.1070
.1041
.1017
.0997
Kl
K2
K3
K4
K5
K6
K7
K8
Kg
.1270
.1127
.1056
.1011
.0976
.0949.
K1
K2
.30525
9
.
10
!
.30165
.,
.0928
.0910
.0894
'
'4
I
I
L__________________
- - _ _ _ _ _ _ ___j
r-----·-----------··-- .-··---------·- ·- ------------- ·-·-------------
FIGURE 7
CANTILEVER BEAM
.s
UNIFORMLY LOADED
OPTIMIZED LENGTH STEPS
.7
MERIT FUNCTION
vs
.6
~
1-1
NUMBER OF BEAM SECTIONS (N)
•5
1 .
E-4
C)
z
~
.4
E-4
1-1
~
::;g
.3
.2
.1
1
2
3
4
5
6
~tiMBER
7
8
9
10
OF BEAM SECTIONS
··-·--···-·-··--------------·-----·------------·------·
ty
~
30
~-----·--------------------------------------------------------,
TABLE 4
i
UNIFORM LOADED CANTILEVER BEAMS
I
MERIT FUNCTION RESULTS
I
(1) *Merit function for a uniform cross section beam=.63
(2)
MERIT FUNCTIONS
N
(3)
(4)
OPTIMIZED
LENGTH
SECTIONS
1
*
=. 2699
Merit function for a continuous taper beam
EQUAL
LENGTH
SECTIONS
PERCENT
DIFFERENCE
(4)-{3)
(4)
.63
MERIT RATIO
(4)
t2T
2.33
2
.4393
.4400
.16
1.63
3
.3801
.3808
.18
1.41
4
.3514
.3521
.20
1.30
5
.3346
.3352
.18
1.24
6
.3235
.3240
.15
1.20
7
.3156
.3161
.16
1.17
8
.3098
.3103
.16
1.15
9
.3053
.3057
.13
1.13
10
.3017
.3021
.13
1.12
.
.
All Mer1t funct1ons are q 2/3 L 5/3
I
II
I
l___
- ...... ---·----· -·- ----·- ·---· .•. -----------
,---------;------·--·-----------------------------------------·-··--·
I
II
FIGURE 8
CANTILEVER BEAM
I
I
I
-··-----'----·-·-··---~---~--~
I
UNIFORMLY LOADED
I
WEIGHT MERIT RATIO
2.4-
i
vs
2.2
NUMBER OF BEAM SECTIONS
i
i
5
7
9
11
13
15
I
I
NUMBER OF BEAM SECTIONS "N"
I
•
--------------------------·
______ j
w
~
32
------------------------------·------------
-·-------------------------,
I
I
i
CANTILEVER BEAMS
UNIFORMLY LOADED
SUMMARY
From table 4 we can see that a considerable weight
advantage can be obtained by tapering uniformly loaded.
cantilever beams.
We shall now compare the different
types of tapering.
Table 4 has a comparison between the merit functions
for uniformly loaded cantilever beams with step tapering.
Those beams using optimized length step tapering are shown
to have less than two tenths of a percent weight advantage
over those with equal length step tapering.
In normal
circumstances the added expense would not justify the advantage.
Figure 8 shows the ratio between the weight merit
functions of beams with equal length step tapering to that
of beam with continuous taper.
Even as the number of sec-
tions reaches ten for the step tapered beam, there is
still an 11.2 percent advantage for the continuously tapered beam.
'
IL _____________ _
[--------------·-----------~-------------------------:------------·-------·-----:
I
I
I
I
;
SIMPLY SUPPORTED BEAMS
ii
CENTER POINT LOADED
l
I
I
.
I
The following cases of center point loaded, simply
'
I
supported beams were investigated.
•
I
!
iI
'
I. Beams of uniform cross section.
I
2. Beams with continuously optimized taper.
3. Step tapered beams with equal length sections.
!
4. Step tapered beams with optimized length sections •.
The merit functions for these beams were calculated
and compared to determine the improvement over a uniform
cross section beam caused by different types of beam
tapering.
I!
I
I
I
I
I
!
•
,
i
.
I
I
I
.
.
I
L__________________________________________________________ _
33
34.
i
SIMPLY SUPPORTED BEAMS
CENTER POINT LOADED
UNIFORM CROSS SECTION
-!
I
I
p
I
l
i
L
I
i
:":•·
.A
I
..t
I
.5P
.5P
In this case of a simply supported beam with center
point loading, and a uniform cross section we can use the
merit function:
The maximum moment {M) is {.5P)(.5L)=.25PL.
There-
fore the merit function becomes:
'Merit= {.25PL)AL
!
i
L___ ··-- -
I
= .3969PAL5 / 3 ; Where A is equal to(2/3).
35
r-~-------~----·------------------.------------------~--------~-------~----~.
SIMPLY SUPPORTED BEAMS
CENTER POINT LOADED
CONTINUOUSLY OPTIMIZED TAPER
p
!
I
I
.5P
.5P
Tbis is another case of a continuously optimized
I
I
. I·
taper, therefore we again use the integral form of the
I
merit ffnction:
Merit =
!~(x)dx.
(1)
I
!
Because both the load and the beam are symmetric about
the beam center we can simplify the problem by integrating
from zerQ to (L/2) and double it.
.oL
Merit
= 2Jr<x)Adx
Merit =
Merit
=
is
Where A is equal to (2/3).
~ I[(.5P) (x)] Adx .2381PALS/ 3
The merit function
J
I
n""i
I
2(.5P)A(i) x 5/3 .SL
O
1
I
!
Figure 9 shows the relationship between cross sectional
l!
area required and the distance from the beam end.
I
IL_ ______________
i
I!
_j
,---------------· --------
·--····----····--------·------~
1
I
iI
i
FIGURE 9
SIMPLY SUPPORTED BEAM
I
II
CENTER POINT LOADED
i
I
I
I
Ii
CONTINUOUSLY OPTIMIZED TAPER
I
i
I
I
RELATIVE CROSS-SECTIONAL AREA
<
pq
~
<
vs
..:l
~
1-1
POSITION ALONG BEAM
.5
E-4
{.)
I
pq
Cf.l
I
Cf.l
Cf.l
0
~
u
.4
.3
pq
I>
1-1
E-4
:§
~
ex:
.2
.I
x=O
.IL
.2L
.3L
.4L
.5L
.6L
.7L
.SL
.9L
xaL
POSITION ALONG BEAM LENGTH
L:__ _ _ _ _
----------------------------------------------
·---------~--------
--------(.A)
c.
37
---------
r-·--
1
-·-----------·~-------------- -------~-----------1
SIMPLY SUPPORTED BEAMS
I
CENTER POINT LOADED
STEP TAPERED WITH EQUAL LENGTH SECTIONS
I!
For the case of a step tapered, simply supported beam I
with equal length sections the merit function takes on the
1,
· form:
N
Merit ==
2<M
M.
1
)A(L)
Where
is the maximum moment for
i=l i
R
1
each individual section arid (A) is equal to (2/3).
j
_
:
I
'
Because the beam loading is symetric about its center !!
I
point, we shall also make the beam sections symmetric
about the center point.
This enables us to take the sum-
i
i
mation over one half of the beam and double it to obtain
II
the merit function for the complete beam.
I
Taking N equals five the merit function becomes:
I
p
• 5T
). ~
~
.5P
I'
...
r~
~
5
+
~
5
)'I "If
I·
.lL..
5
.. ,.
.lL..
5
~
,; ~
..1..~
5
.5P
38
~-----------------------~----------------
.. ----,....----_-,--·
-------· --- ---------·---·-
For the general case of "N" sections the Merit function becomes:
N
.Merit = )' (M.) A (L)
:1=1 1
if
Mer 1. t
=
2~L
N
5/31 (1 )A+ (2 )A+ • • •
2N
+(
'2"N
.5(N-l)-l)A (·5(N-l).)A
2N
+
~~
+.
S( 2 r-)Al
• <:J
Table 6 gives the numerical results of the Merit functions.
Figure 10 gives a graphical representation of these merit
function.
r----..-- . ----··----- ····----
FIGURE 10
SIMPLY SUPPORTED BEAM
CENTER POINT LOADED
EQUAL LENGTH SECTIONS
I
.40
WEIGHT MERIT FUNCTION
vs
z
0
E-4
~
NUMBER OF BEAM SECTIONS "N"
1-4
.35
~
E-4
1-4
~
~
::g
E-4
:X:
0
30
I·
l
(!:)
1-4
~
I
iS=
.25
.225
4--------+--~----+-------~--------+--=~---+--------+--------+~---
1
3
5
7
11
9
13
15
NUMBER OF BEAM SECTIONS "N"
---------·· ..... ····-- - - - - · - - - - - - - - - - - · · · · - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - · ._ _________ j
w
t.D
:-.·::-;-~-
---:--
4Q
r - - - - - - - .- - - - · - - - - - · - .- - - - - .- - - - - - - - - - - - · · · - - - - - - - - - · - · - - - - - - - - ,
I
I
;
I
.
.
!
SIMPLY SUPPORTED BEAMS
!
!
CENTER POINT LOADED
TAPERED BEAM WITH OPTIMIZED LENGTH SECTIONS
The merit function used for the case of a simply supported, center point loaded, tapered beam with optimized
length sections is:
N
. 1
].<=
A
is the optimized length of
Merit = }.: (M.) (K. L); where K.L
l.
l.
l.
an individual section.
The solution can be simplified by
taking twice the summation over half the beam.
For a beam with five optimized length sections we
have:
p
5
Merit ...
Merit
=
i~ (Mi)A(Ki L) ; Where "A" is equal to (2/3).
J
t
21 [<;)(K 1 L) \K 1 L) + [<;) (K 1 Irt-K 2 L)
+
~;><~>f[L-2K1~-2K2LJ
Merit == 2pi\L 5/31
!!____________________
(~)A (K~ [K1;K2
I
r
(K 2 L) +
.
(K2) + ( • 5-K1 -K2) (. 25) A
I
I
I
- - - - - ·------"------··--·--··------------J
41
~--------------------~----------·
'
----------------~---------------
'
'
j
For the general case of "N" beam sections we have:
p
.5P
.5P
N
Merit
Merit
= 2:
A
i==l
=
.
(M.) K.L
1
2PA L
5/31
•••+
1
'
.,
.
.
A + [ .5(K -+K )] A K +•••
(.5K 1 ).K
1
1 2
2
A
[. 5 (Kl+K2+•••+K.5(N-!)-l+K~5(N-1))] K.5(N-l)
+ (. 25) A[l-2Kl-2K2-· •• -2K. 5
(N-~)1
I
These equations -for the optimimum length of the beam
sections, and the merit functions were solved using the
optimization program in appendix "A".
Table 5 gives the optimum section lengths for the
beams.
Table 6 ~ives the numerical values for the merit
functions.
Figure 11 gives a graphical representation of
the merit function for beams of different numbers of sec·tions.
I
1
I
.
I
I
I
I
~------~---------------------------------------------------------------------.---------------j
42
SIMPLY SUPPORTED BEAMS
CENTER POINT LOADED
OPTIMIZED LENGTH BEAM SECTIONS
NUMBER OF
BEAM SECTIONS
N
MERIT
FUNCTION
1
.3968
3
.3231
K1
.2324
5
.2964
K1
K2
.1470
.1693
7
.2826
K1
K2
K3
.1059
.1219
.1322
9
.2741
K1
K2
K3
K4
.0819
.0943
.1023
.1083
11
.2683
K1
K2
K3
K4
K5
.0663
.0764
.0828
.0877
.0917
13
.2641
K1
K2
K3
K4
Ks
K6
.0555
.0639
.0693
.0733
.0767
.0795
K.
1
VALUE OF
.
K.1
43
:--------·--·----------·- - - - - - - - · - · - - - - - - - - - -----------------
----------------~
TABLE 5 CONTINUED
NUMBER OF
BEAM SECTIONS
N
15
MERIT
FUNCTION
K.
.2610
K1
~
K2·
K3
K4
17
.25g5
K1
.0414
K2
K3
K4
.0476
.0517
.0547
.0572
• 0593
Kl
K2
K3
K4
K5
K6
K7
Kg
K9
L - __ _ _ _ _ _ _ _ _ _ _
.0475
.0547
.0593
.062g
.0656
.o6go
K6
K7
Kg
.2565
Ki
K5
K6
K7
K5
19
VALUE OF
i
I
II
i
I
.0701
.0611
.062g
.0366
.0421
.0457
.o4g4
.0505
.0524
.0540
.0555
.056g
I,.
i
r-··---i
FIGURE 11
I
•·.
SIMPLY SUPPORTED BEAM
CENTER POINT LOAD
OPTIMIZED LENGTH SECTIONS
.40
WEIGHT MERIT FUNCTION
vs
~
.....
t;z
~
i
.I
NUMBER OF BEAM SECTIONS "N"
I
.35
E-4
.....
~
·::a
E-4
• 30
~
.....
~
I:J:
.25
I
.225
I
1
I
I
3
5
1
9
11
NUMBER OF BEAM SECTIONS "N"
L----------------··-------· ------··-----------·· ----· ····-- ....
13
15
16 I
I
···-·-·--···--··------·----··----------------------· ____.J
~
~
-------- -----------------------------l
!
TABLE 6
CENTER POINT LOADED, SIMPLY SUPPORTED
MERIT FUNCTION RESULTS
(1)
*Merit
(2)
for uniform cross section beam
Merit for continuous taper beam
N
MERIT FUNCTIONS
(4)
(3)
OPTIMIZED
LENGTH
SECTIONS
1
.3969
.,. .2381
PERCENT
DIFFERENCE
( 4) -(3)
(4)
.;3969
MERIT RATIO
(4)
m
1.67
3
.3231
.3342
3.44
1.40
5
• 2964
.3023
1.99
1.27
7
.2826
.2863
1.31
1.20
9
.2741
.2766
.91
1.16
11
-.• 2683
.2702
.71
1.14
13
.2641
.2657
.61
1.12
15
.2610
.2622
.·47
1.10
17
.2585
.2596
.42
1.09
19
.2565
.2574
.34
1.08
* All
!
l
EQUAL
LENGTH
SECTIONS
=
.
.
mer1t funct1ons are W2/3 L 5/3
.
•
I
I
i[___________________________________________________________
_ _j
I
--- --- --------------- -----------------------'-----------------------------------'I
~------------'-----
I
FIGURE 12
1
SIMPLY SUPPORTED BEAM
I
!
CENTER POINT LOADED
I
WEIGHT MERIT RATIO
1.7
vs
NUMBER OF BEAM SECTIONS "N"
1.6
0
E-c
<
1-1
~
E-c
1-1
~
1 •5
-1.4
::a
E-4 -1 3
=
.
0
1-1
f;l;.1
Iii=
i
I
1.2
1.1
I
I
I
I
I
I
i
I
I
I
1
2
3
4
5
6
7
8'
9
10
11
13
15
NUMBER OF BEAM SECTIONS "N"
[___ -------------------
~
-en
r--------
------------------------------------------:-------_-----------~
I
-
I
i-
SIMPLY SUPPORTED BEAMS
.
I
CENTER POINT LOADED
SUMMARY
Looking at table 6 where the results for center point
loaded simply supported beams are tabulated we can make
the following observations.
The continuously tapered beam has a 67 percent weight
advantage over the beam with a uniform cross section.
The advantage of an optimized step length when using
step tapered beams is greater than that for the cantilever
environmnet, rising to 3.4 percent for the core of a beam
with three sections.
Figure 12 shows the ratio between the merit functions
for beams with equal length step taper and that of a beam
with continuous taper.
!·
l:
At the level of a step tapered
beam with 19 equal length sections there is still an eight
I
percent weight advantage for the continuously tapered beam. I
I
I
i
\
I
I
I
l
I
L-------~------------
I
- - - ------------ _____-_______________ _j
-------------------------------- ------------------------- -------------------l
I
I
SIMPLY SUPPORTED BEAMS
UNIFORMLY LOADED
The following cases of uniformly loaded, simply sup-
I
ship between their weight merit functions.
I
1. Beams of uniform cross_section.
!
I
2. Beams with continuously optimized taper.
I
sections. I
3. Step tapered beams with equal length sections.
4. Step tapered beams with optimized length
I
48,
t!9
SIMPLY SUPPORTED BEAMS
UNIFORMLY LOADED
UNIFORM CROSS SECTION
£+
I·
f f i i t i I l
i t
l.,q
~
L
.5qL
.5qL
The merit function for the case of a simply supported
beam, uniformly loaded and with a uniform cross section is:
Merit
=
2/3 (l)
M
L, where "M" is the maximum bending moment
found along the length of the beam.
The merit function
thus becomes:
Merit
r:
.
] 2/3L,
= l{. 5qL) (. 5L)- (. 5qL) (. 25L~
Merit= q 2 / 3 L7 / 3 (.125) 2 / 3 = .25W 2 /SLS/S
l------·-·------------·--·--------·------·--·-----------·.---··-· -----------
---·----~--------------------,
SIMPLY SUPPORTED BEAMS
-'UNIFORMLY LOADED
CONTINUOUSLY OPTIMIZED TAPER
II
I
I
I
jt t
t~
f f f
*
t i
f i
L
tJJ:
I
t
.5qL
.5qL
When we want to find the merit function of a simply
supported, uniformly loaded beam with a continuously optimized cross section we must use the integral form:
Merit
fL
=J
Merit=
M(x)
0
2/3
dx
(l)L[
2]2/3
.5qLx-.5qx
dx
0
=J
(.5q) 2/~(Lx-x 2 ) 213 dx
A computer program based on the trapazoidal rule was
written to integrate this merit function.
A listing of
this program appears in the appendix
I
I .
The solution of the merit function turns out to be
o
1864W 2 / 3 L S/ 3
Figure 13 graphically shows the relative cross sec-
·-------------
I
I
where "W" is equal to qL.
tional areas of the beam over its length.
I
I
I
----------- J
~---------·-----------------------------------·---
-··--
----:---------- ------ -----------------------1
..
I
I
I
FIGURE 13
SIMPLY SUPPORTED BEAM
I
UNIFORMLY LOADED
I
CONTINUOUSLY OPTIMIZED TAPER
i
I
I!
I
RELATIVE CROSS SECTIONAL AREA
vs
~
LOCATION ON BEAM
~
<:
t-=1
~
0
E-4
.4
1-1
C)
l't'l
CZl
I
CZl
CZl
8
.3
.2
C)
r.-.:l
>
.1
1-1
j
1"4
~
.1L
.2L
.3L
.4L
.SL
.6L
.7L
.BL
.9L .
x•L
x-=()
POSITION ALONG BEAM LENGTH
L______ -------------------------------------------------------------c.n
~
52
~'
UNIFORMLY LOADED
STEP TAPERED WITH EQUAL LENGTH SECTIONS
For a simply supported, uniformly loaded beam with
equal length step tapering, the merit function is of the
.
N
2 3
form:
Merit =
M. / L • To simplify the calculations
Ir=l 1
N
we shall take advantage of the beams symmetry and use
L
twice the varue of hal:f the. beam.
The merit function for a beam with five equal length
sections is calculated below as an ex:ample.
• t t i
~
J t I f i t 'I f f l t f J ~ J lq
I
~?r-
rc--..k.---:h-..k.
5
5
... I..
_.k
5
.. I~
L
_k
5
~-
~
+:f
.5qL
.5qL
5
Merit ==
L
(M.)
i=l
!
1
2 3
1
L
0
~ 21[(~)(~)- (if' )(tr)f (~)+
.
·
+[(*)(¥)-(2~L'(¥ff)f/3(~}+ [(~)(~)-(~)(i)Y/3(fu) j·
2 3
2 3
2 3 5
.
2W / L /~ ~(1 1 ) /
(2 4 ) /
2/3
1 .
Merit
+ 5 - 25
+ ( • 25)
( • 5)J
273 Vo'2'5
13
Merit
=
5(2)
I
Merit
=
.2216
·
.
w21 3 L5 / 3
IL______________________
·----------------.-----.----- J
'
/
53
-------·--------1
~----·----~
For the general case with "N" equal length beam sec-
I
!
tions the merit function becomes:
=
Merit
~
2 3
(M.) /
L
i=l . ~ .
N'
I
I
I
l
i
t
f
t
t
t t
i
I
k-----1:=~
f
t
-- --
t
t f
t i i
----- ~--
t t "t t f iq
I
-~=~--1-----lo
I!
I
I
I
I·
I
.5qL
Merit
.5qL
=
~
2W2/3L 5/3 (1
N ( ) 2/3
2
·: [ .
2/3
22)
1 )2/3 (2
N - i?
+ N - i?
(
• • •+ .5(N-l)- ~.5(n-l)
I
i
!
I
+•• •
)2]2/3+(:·~25) 2/3 (.5)}
~
I
I
I
Table 8 gives the numerical values for the merit function.
Figure 14 shows the effect of different numbers of section
on the merit function.
I
II
.
L------------··---·-------------------------------· -·-------·-----·--·-------
I
J
.. -
,--·--·-~------·---,·-·-
I
I
-----·--·-···--·-·----··---~
II
FIGURE 14
I
I
I
I
I
SIMPLY SUPPORTED BEAM
;
I
I
I
.26
i
•
EQUAL LENGTH STEPS
I;
I
.25
I
I
~
I e::
u
I z
I
I
I
I
I
I
l
I
UNIFORMLY LOADED
~
MERIT FUNCTION
I
i
vs
.• 24
I
NUMBER OF BEAM SECTIONS "N"
I
I
.23
·!
8
H
~
~
~
.22
8
=
s .. 21
I
I
"'-l
I.
Iii=
I
I
.20
I
I
I
I
I
I
I
.19
.18
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
NUMBER OF BEAM SECTIONS "N"
L----·----····-·--····------·-·---·-;'------·---------·-------·-- ··--- ··---·--·-··--
c.n
~
,<
55
,-----------~-~---------~---------------------~~----------------~-
;
.
--------------1
II
SIMPLY SUPPORTED BEAMS
!
at only half the beam.
For an example, the merit function for "N" equal to
,•
f~ve
will be calculated.
Merit =
Merit
where A=2/3
=
I
------------~-----------J
56
the merit function becomes:
Merit ==
N
2 3
'(M.) / K.L
~
l.
l.
i=l
------L--------
----~
.5qL
.5qL
Merit
2 W2/3L 5/3
=
:.-
2
2/3
{
[
2 2/3
2] 2/3
(Kl-Kl)
Kl+ (Kl+K2)-(Kl+K2)
K2+ ..
1
2
·+ [( K1+K2+ • • •+K. 5<N-1J- (Kl+K2+ • • •+K .5 (N-1)) ]AK. iHN-1):
2 3
+ ( .0625) / .(- 5-K1 -K2 !
• • • .-K •
.
J
5 ~N-1))
I
.
[_____________________________________ ··---·--·-·--------------·--_____ j
I
57
.-----------------·---~---·--
.
~~----------·-·--
··-
------------- --
·------------~-
Table 7 gives the values for the optimized
..
--------------·'
secti~n
I
lengths.i
I
Table 8 gives the values for the merit functions.
Figure
15 shows graphically the differences in merit function due
j
i
1
to different numbers of beam sections.
I
I
!
I
II
I
I
I
Il
I
I
I
I
I
I
I
I
I
I
I
i
I
l - - - - - - · · - - - - - - - - - - - · - - - - · - - · · - · - · - - · - · - - - - · - ·-·
·- - - - - - - -· ·- - - J
58
- - ------------------------------,
f
I
I.
J,
I
.
TABLE 7
!I
!
SIMPLY SUPPORTED BEAMS
I
UNIFORMLY LOADED
1
I
NUMBER OF
BEAM SECTIONS
N
MERIT
FUNCTION
1
.1864
3
K.].
VALUE OF
.2229
K1
.1538
5
.2119
.0955
.1292
7
.2060
K1
K2
K1
K2
K3
.06864
.08713
.11154
K1
K2
K3
K4
.05319
.06566
• 07878
.09872
K1
K2
K3
K4
K5
.04316
.05245
.06117
.07185
.08901
K1
K2
K3
K4
K5
K6
.03619
.04353
.04993
.05695
• 06610
.08135
.
9
11
13
.2022
.1995
.1976
K.].
L
•
I
I
I
.
L - - - - - - - - - - - - - - - - - · - - - - - - - - - - - - - - - - - · _______________ _ ]
·-------~-------~--,
r----~~---
:
l
I
I.
TABLE 7 CONTINUED
I
NUMBER OF
BEAM SECTIONS
MERIT
FUNCTION
K.
.1961
K1
K2
K3
K4
K5
K6
K7
.03102
.03715
.04197
.04761
.05299
.06183
Kl
K2
K3
K4
K5
K6
K7
K8
.02712
.03222
.03631
.04030
.04468
• 04999
.05734
.07001
K1
K2
K3
K4
K5
K6
K7
K8
Kg
.02401
.02842
.03186
.03510
.03850
.04236
.04717
.05393
.06567
l..
.1950
17
.
19
.1940
K.
l..
N
15
VALUE OF
I
I
.07507
.
i.
I!
'
.I
I!
i
i
I
I
I
!
I
i
I!
~--·--·----~----~-----------------
I
l
I
---·-------·---·--·------_j
~-------·--------------·-·····-----------------·-
·---·--·-·---- --·-·-----······-·· ---··-.
. .....
I
I
I
I
·-··········· -· ·-··· ··-·-·-----·
·- ·---·-··--·-·-·-···-----·-·----------]
FIGURE 15
I
SIMPLY SUPPORTED BEM!
I
I
I!
I
··-
UNIFORMLY LOADED
.26
OPTIMIZED LENGTH STEPS
I
MERIT FUNCTION
.25
z
vs
0
E:: .• 24
~
fi!
•
NUMBER OF BEAM SECTIONS "N"
.23
E-4
t-1
~
~
::a
'.22
E-4
~
t-1
.21
~
;:::::
.20
.19
.18 .
1
2
3
4
5
6
7
8
9
10
11
NUMBER OF BEAM SECTIONS "N"
I
L _ _ _ . · - - - - - - - - - - - · - - - · - - - - - - - · - · · · - - - - - - - - - - - - · · · · - - · · - - - - - - - ·---
12
13
14
15
16 i
i
---------·- _____ _j
en
0
61:
-------------------------------------------~
I.
TABLE 8
UNIFORMLY LOADED, SIMPLY SUPPORTED
MERIT FUNCTION RESULTS
(1) *Merit for uniform cross section beam (2)
Merit for continuous taper beam
N
MERIT FUNCTIONS
(3)
OPTIMIZED
LENGTH
SECTIONS
1
(4)
I
EQUAL
LENGTH
SECTIONS
= .1864
PERCENT
DIFFERENCE
{ 4)- (3)
(4)
.25
.25
MERI'l' RATIO
(4)
m
1.341
3
.2229
.2374
6.51
1.274
5
.2119
.2216
4.58
1.189
7
.2060
.2129
3.35
1.142
9
.2022
.2076
2.67
1.114
11
.1995
.2040
2.26
1.094
13
.1976
.2013
1.87
1 .. 080
15
.1961
.1993
1.63
1.069
17
.1950
.1978
1.44
1.061
19
.1940
.1965
1.29
1.054
-I
* All merit functions are
-
'
j
!
L-------~--- ~-
w21 3 L5 / 3
I
I
-··---·-- ··--·---·-
FIGURE 16
SIMPLY SUPPORTED BEAM
1.45
UNIFORMLY LOADED
I
1.40
WEIGHT MERIT RATIO
vs
1.35
NUMBER OF BEAM SECTIONS "N"
0
H
~
f-4
H
p::
1.30
1.25
~
:s
f-4
:::=
1.20
C!J
1-1
~
l:=
i
I
1.15
1.10
1.05
1
2
3
4
5
6
7
8
9
10
11
NUMBER OF BEAM SECTIONS "N"
12
13
14
15
16
i
i
I
-··-·--'
0)
w
63
SIMPLY SUPPORTED BEAMS
UNIFORMLY LOADED
SUMMARY
Table 8 gives the results of the merit function calculation for the four types of uniformly loaded, simply
supported beams.
From this it can be seen that the con-
tinuously tapered beam has a 34 percent weight advantage
over a beam with a uniform cross section.
The weight ad-
vantage for the step tapered beams with optimized length
steps over the step tapered beams with equal length steps
j,:s
JlOW
a maximum of six and one half percent for a three
section beam, and decreasing as the number of beam section
increases.
The ratio of the merit function of the step tapered
beams with equal length steps, to that of a continuously
tapered beam is shown in figure 16 and table 8.
This ratio
ranges from a high of 1.34 for a beam with one section (a
beam with a uniform cross section) to 1.05 for a beam with
19 equal length sections.
!
L______________________ ------
____________________________ _j
A COMPARISON BETWEEN TWO DIFFERENT
TYPES OF BRIDGE STRUCTURES
A weight merit function comparison between the cantilever type of bridge structure as used on the Forth Railway
Bridge( 2 ) which spans the Firth of Forth at Queensferry,
and a simply supported type of beam structure will now be
done.
First a look at the cantilever type.
The figure below
shows how the two cantilever beams meet in the center to
form a continuous span.
The loading -is uniform.
The two
cantilever sections are symmetrical to each other, therefore, the merit function for only one need be calculated.
j + t t t t t t f i t i ~q
t...-c',..__- ~ --~~li-4'1f-- f ----1~"'"1
_________________________________________________ __!
65
~---------.
!
---------:------------··-----------------------------------l
The moment
eq~ation
is M(x)
=
(qx) (.Sx)
.5qx 2 •
=
·1
.!
For a beam with continuous taper, the merit function is:
L
L
2/3
2 3
2
2 3 7 3
Merit = 2~~ / (x) dx ~ 2
5qx )
dx - • 2143 q 1 L 1
2 3 5 3
; where W = qL
Merit
.1071 w 1 L /
£<.
=
Figure 17 gives the change in the cross sectional area
along the beam.
The simply supported bridge structure is shown in the
figure below.
~ t f +t t t f f l t f f
A
t..
.5qL
X
t f iq
A
·I
L
•
.
.5qL
2
The moment equation is (.5qLx- .5qx ).
Putting this
I·
into the merit function for a continuously tapered beam
gives u~:
Merit
=
1M
213
L
L
=foe. 5qLx
-
2 213
.5qx )
dx.
Solving this by the trapazoid rule gives
2 3 5 3
Merit = .1864 w 1 L /
• Figure 18 gives the beam cross
sectional area required-along its length.
A comparison between the weight merit functions of the
two beams shows that when considering only these factors
the supported type beam is preferable to the can;tilever
types.
_________ _j
·--·-···---- ----·-·-··-·
-·---- -- ---··---------------------------------------·-1
I
I
I
FIGURE 17
i
I
CANTILEVER CONFIGURATION
I
I
RELATIVE CROSS-SECTIONAL AREA REQUIRED
I
vs
I
POSITION ALONG BEAM LENGTH
i
.7
I
I
i
I
I
.6·
i
I
.5
I
I
.4
i
.3
I
.2
.1
x•O
.lL
.2L
.3L
.4L
.5L
.6L
.7L
.SL
.9L
i
x•L
POSITION ALONG BEAM LENGTH
------->-----'--'------;-------1-------··-_.______________________
I
I
_____ __j
--;--~------
Q')
m
- - - · - · -----------·--·--------------·--·--------------1
·--~
FIGURE 18
I
SIMPLY SUPPORTED CONFIGURATION
I
RELATIVE CROSS-SECTIONAL AREA
I
vs
I
<
(:;r:l
~
POSITION ALONG BEAM
~
<
<
~
1-1
:
.,_':~
E-4
u(:;r:l
Cl.l
I
Cl.l
Cl.l
g
.3
(:;r:l
.2
u
·I>
1-1
E-4
j
~
.1
~
.1L
.2L
.3L
.4L
.SL
.6L
.7L
x=O
.8L
.9L
x=L
i
I
POSITION ALONG BEAM
i
I
I
·------------------------------
_j
0)
...:t
-~----~----~
---~--. ---------~
I
I!
UNIFORMLY LOADED BEAM
SUPPORTED AT THREE PLACES
In this example, two uniformly loaded beams supported
at three points were investigated.
Both beams are sup-
ported at three places by optimized reactions.
The differ-
ence in the two beams lies in the position of the supports.
The f1rst case deals with a beam with a support at
each end, and another at the center.
The weight merit
function for that beam will be compared with the weight
merit function for a beam with a support at the center and
two other supports set.in an optimum distance from the
ends.
In other words this second beam will have an over-
hang or cantilever at each end.
II
l
i
I' - - · · - · - - · · - - - -.
- - - ___________________________ j
f:)8
I
69
r
l
------------------------------------------~
i;
I
UNIFORMLY. LOADED BEAM
UNIFORM CROSS SECTION
I
SUPPORTED AT THREE PLACES
NO BEAM OVERHANG
I
.I
!
f t f f
~
t l
L
,.. X
2
'
.5(qL-R)
t f t f t l I t t
f
R
L
2
~ q
~t
.5(qL-R}
Shown above is the beam configuration and its approximate shear diagram.
For this type of beam we can describe
the reactions at each point by calling the center reaction
"R'' and then because of symmetry the outer reactions are
.5(qL-R),where "R" is equal to KqL.
The moment equation from zero to .5L can now be found
using the reaction forces.
Beacuse the beam is symmetric
we need calculate the merit function for only half the beam
and double it.
The moment equation from zero to .5L is
M=.5(qL-R)(X) - .5qX
2
= .5qLX- .5KqLX
.5qX 2
L__ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ - - - - - -
.
I
I
____________!
70
----------·--------cl
,-------1
The location of the maximum moment is found. by setting the first derivative equal to zero and solving for
I
: "x" •
i
dM
ax
= .5qL - .5KqL x = .5L(l-K) which
· figure.
qx = 0
I
.I
is point "a" on the preceding
By substituting "x" back into the moment equation
we get the equation for the maximum moment at point "a".
'
2
Ma = .5L(l-K) (.5qL-.5KqL) - .5q [.5L(l-K)]
Ma
= .25qL 2 ~1-K) 2 -.5(1-K) 2 ] =
2
2
.125qL (1-K)
The other maximum moment occurs at x=.5L which is
point "C" on the
figur~.
The moment at this point is ob-
tained by substituting x=.5L into the moment equation.
• 5L ( • 5qL-. 5KqL)
- • 5q ( • 5L)
2
Mc
=
Me
= .25qL 2 [1-K -.;] = .125qL 2 (1-2K)
For optimum conditions on a uniform cross section
beam we will want the magnitude of Ma= Me , which gives us:
2
.125qL (2K-l)
2
l-2K + K = 2K-l
2
K -4K
I
+ 2 = 0
L ___- - - - - - - - - - - - - - - - - - - - - .- - - - - - - - - - - -
71
~-~~~~---------'------~----~---~---------------------~
I
.
.
·
Solving by the quadratic equation we get K Therefore,R
i
= KqL =
.5858qL.
,
.5858 •
We can now substitute this
into the maximum moment equation and get
2
2
2
2
Ma = .125qL (1-K) = .125qL (1-.5858)
=
.077lqL
2
The merit function used to compare optimum weight of
2 3
beams is M / L therefore the merit function in this case
l
~
,
I
I
I
' is:
Merit
Figure 19 gives the relative cross sectional area required
with respect to the distance down the beam.
The merit function for a continuous taper beam with
these restraints would'be the solution to the integral:
- .5L
.
2
2 ]2/3
Merit = 2
Gl25qL (1-K)
dx. Solving this by the .
. trapazoidal rule gives a merit equal to 0.1097 w21 3 L5 1 3 •
L
This could be further improved upon by setting the areas
under the curves of figure 19 equal to each other for the
true solution for a continuously tapered beam.
J
I
l__________________
II
----
REACTIONS LOCATED AT BOTH ENDS AND CENTER
RELATIVE CROSS-SECTIONAL AREA
vs
.1
~
p:::
<tl
POSITION ALONG BEAM LENGTH
.075
..:I
<x:
z .05
0
H
E-4 .025
C)
~
rl.l
4
0
·\
rl.l
!g
: o;.025
i
!r:£l
1
~
I~
i,..:I
~
~~
I
.o5
.075··
~0
.lL
.2L
.4L
.3L
x=.5L
.6L
.SL
I
.9L
x•L
POSITION ALONG BEAM LENGTH
l
.7L
--.!-·-----'"---
--- ----;----
------·---- -------------
j
______ j
~
w
~
73
--------------------~
I
UNIFORMLY LOADED 'BEAM
--1
UNIFORM CROSS SECTION
SUPPORTED AT THREE PLACES
INCLUDES BEAM OVERHANG
·II
I
I
i
I
~t4
.5L
J--
' t
.5L
f t f 1 l f i f f f l t
tn~ti
K1qL
t
qL(l-2K 1 )
·t
f t l
I
q
f
K1qL
II
I
I
I
I
I.
I
v
For this case where we must also optimize the amount
I
of beam overhang we end up with two variables, K1 and K2 • · ·j
K describes the reaction forces and K2 describes the lo1
\
cation of the outer beam supports. The above figures show
the loading, beam reactions, and shear diagram for this
case.
The moment equation from x=O to x~K 2 is
2
i
-(qx)(.5x)= -.5qx • This gives us the moment at point!
M=
1
"A" ' which is one of our maximum moment points, as
.
'
:_MA __:__
2
2 2
I
I.
_::_~q (K?_~2--~--=-=-~~~-~~-~-----------~~~----------------~
74
-·---------···------··•------·-----~----··--·----C-~-~---------------•·-•·--1
The moment equation at the center point or point "C"
is
c = (.5qL)(.25L) + K1 qL(.5L-K 2L)
2
M = qL [-.125 + .5K -K K~]
(2)
1 1
c
M
-I
I
We can now calculate the moment between point "A" and "C" •
2 2
MA-C= -.5qX L + (K 1 qL)(XL-K 2 L)
MA-C = qL2
~<:lX
- .5X2 - KlK2]
The location of the maximum moment between
is found by setting
A
and
C
dMA-C = 0 , and solving for X.
dx
clMA-C
----=
dx
We can now find MB which is the maximum moment between points A and C by substituting X=K1 into the equation for MA-B
MB = qL 2 [Kl (Kl) - .5 K1 2 - KlK2]
2
2
M = qL (a5K
KlK2)
B
1
i
I
(3)
I
i
I
l~-----·~------·-
------- .. ------~-~--- - - - - - - - - - - - - - - - - - - ·
·-----~------J
75
,---T: require th-:-:mal~:s~ eros:-se~~i~= ~~~:-;or-:-~ni=l
form beam we want the magnitude of the maximum moments to
•. be equal.
Therefore,
JMA}
~om {MAJ
IMBI
==
==
==
[Mel
fMc)
we get
-.5qL 2K22 = qL 2[-.125
2
K2 -.25
K =
.
1
2K 2 -I
From
~A) ..==
[
MB l
(4)
we get
-.5qL 2 K~ =- -qL 2 (.sKi-K1 K2 ]
K~ = K~
- 2K 1 K2
(5)
Now substitute equation (4) into equation (5).
2
K2 -.25
K2 -.25
2
K2 ""' 2K -I - 2K2 2K2-r=
2
which becomes
3
2
4
.0625
7K 2 - 6K 2 + .5K 2 + .5K2
~2 J
Figure 20 shows a graph of the pertinent part of this
function.
K
2
I
L
=
For our purposes the only logical solution is
.1306 •
r·~·
···- ..
I
FIGURE 20
. GRAPH OF
7K~-6K~+. 5K~+. 5K 2 -1/16
• Y
.1
.05
y
0
.05
i
I
I
I
.1
I
i
I
I
o
-o1
0
•1
L_
•2
•3
•4
•5
.55:
i
II
i
•2
I
----------:---~~---------------------------------------~'
----·
...:!
0)
77
r--------· ------------------------------,---------------
----------------~--
-------1
Now substituting the value of K into equation (4) we
2
1,,
'
i
!
'can determine the value of K1 •
II
I
2
= (.1306)2-.25- 3153
K1 = K2-.25
2K 2 -1
2(.1306)-1
- •
The maximum moments along the beam are now found by
. substituting the values of K and K into any of the three
1
2
• moment equations.
2 2
=-.5qL K2
-
= -qL 2 (.5)(.1306) 2 = -.00853qL 2
2
2
2
qL [.SKi -K 1 K2 ] - qL [. 5 (. 3153) - (. 3153) (
=
.00853qL
.1306~
2
Mc = qL 2 r[.125+. 5~ 1 -K 1 K~-]
=
2
qL [-.125+. 5 (. 3153)- (. 3153) (.1306)]
-.00853qL
2
.
-
.
.
The merit function for this beam of uniform cross
section is:
The merit function for a continuous taper beam with
. these same restraints is by using the trapazoid rule
.0316
w21 3 L513 •
The required relative cross sectional area for each
·point along the beam is shown in figure 21.
_________________
]
---·-------..·----------------·- ----
..
"'
FIGURE 21
UNIFORMLY LOADED BEAM
THREE OPTIMIZED REACTIONS
LOCATION OF REACTIONS OPTIMIZED
RELATIVE CROSS-SECTIONAL AREA
~
IX:
<:
vs
~
LOCATION ALONG BEAM LENGTH
~
0
1-1
E-4
!.os f
C)
tf.l
~ .025
C)
J:il
>
1-1
e..
j
.025
rz:l
p:;
.·o5
x=O
.lL
.2L
.3L
.4L
x=.5L
.6L
.7L
.sL
I
I
LOCATION ALONG BEAM LENGTH
L__________.________._.:____________-. _____
• ------------·___·____
.9L
_'_j
...:a
co
79
,---------~----------------.
------------·-
i
THREE PLACE SUPPORT
SUMMARY
When comparing the merit functions for these two
beam configurations we see that the beam with an overhang
has a considerable advantage over the beam configuration
which is a
with no overhang. This difference is .1811
.04175
4.3 to 1 advantage.
L _____· ___
·--
r----------------··--------··-·--1
·--··-·-------------··---·------------,
.
CONCLUSIONS
l
I
From the comparisons of different types of beams done
in this study, it becomes apparent that there can be a
significant weight saving realized by using a tapered beam
instead of a beam with a uniform cross section.
The comparison of different types of beam tapers indicates that the continuous taper beam can be made lighter
than the other types of tapered beams.
The continuous
}.--···
taper beams would probably also be more expensive to use.
Therefore, there would be a trade-off between weight savings and increased
cos~.
The difference in the minimum weights between the·
step tapered beams with equal length steps, and the step
tapered beams with optimized length steps is very small.
This small weight difference would not make the use of the
lighter step tapered beams with optimized length steps
attractive enough to justify their higher cost.
We can gain some insight into the best probable number of steps to use on a beam by looking at the curves of
MERIT FUNCTION VS NUMBER OF BEAM SECTIONS N•
. 11, 15 show these curves for the beams with optimized
· length steps.
Figures 2, 6, 10, 14 show the same curves
for the beams with equal length steps.
In all of these
cases, a flattening out of the curves can be observed at
·about N - 4.
L__ _ _ - - - - - - - - - -
80
81
-------------------------------,
added section becomes small.
Additional beam steps,_which
I
usually mean increased cost, would yield a relatively
smaller weight savings per step.
Additional steps would
probably not be economically justifiable for the small
advantage gained.
L_____________________________ ---
I
_______. ______]
--------------------------------·--·---------·---~
I
I
LIST OF REFERENCES
I
.. (1)
Spunt, L., "OPTIMUM STRUCTURAL DESIGN", Prentice
Hall, New Jersey, 1971.
(2)
Beckett, D., "GREAT BUILDINGS OF THE WORLD-BRIDGES",
Hamlyn Publishing Group limited, London, 1969.
(3)
Spunt L., Lecture Notes Engr. 437, CSUN, Spring 1976. ,
i
'
I
I
l
i
I
_·I
i.
I
I·
I
I
I
I
i
II
!·
1
I
I
I
I
I
I
'
I
L
-----~_j
82
~-------~---~-----------------------------------
-----
LIST OF SYMBOLS
A
=
2/3
L = Beam total length, or load cantilevered length
L 1 = L/N
M ==
Bending moment
p
= Applied point load
X
= Distance along horizontal axis
N
= The number of sections a beam is divided· into
M.=
Maximum moment in section i
~
K.=
The fraction of total beam length
~
q == Beam load per unit. length
W == Total beam· load
R == KqL
V
=
=
qL
KW
= Shear
~----------
-----~------------------
83
-----------.-----_----------------------------------·-----------l
I
I
APPENDIX A
COMPUTER LISTINGS OF THE
MONTE CARLO BASED
OPTIMIZATION PROGRAM
I
I
\.
I
i
I
I
I
.
-
I
i
i
i
I
_____··_J
I
~!~------------·
84
·-·- ···----- - C 0 MM 0 N ··X ( 15)
·- ·· --··
· ···
G H1 ~: NS I 0 N STEP ( 15 ) , STEPS ( 15 ) , CE L <15
CIHENSION A{15),8(15),C(15) ,0(15> _
~
- ···· ---·-·· · ---------------
C
I~FUl
NUMGER CF VARIABLES, SAMPLE SIZE AND MAX VECTOR ATTEMPTS
C
l~FUl
LO~ER
---.g D 0 & R: E i1 G ( 6!) , Hll D ) · N ~- MAX S, MAX V ··-····- ·
··-- ---· --- ---·--·----- · --·1010 FO~MAT<3I10)
_
READ
A~O
10000030
1 0 Q0 0 Q50 -
"'~----------------------··--
10000060
UPPER BOUND AND INITIAL STEP SIZE FOR EACH VARIABLE
(60,1020)
FO"<Mi\1(.~Ei7.5)
----1020
00000i U
1 0 0 0 0 D2 0
···-------·-------~---1.
, i< ( 15 l , Xl ( 15 ) , XU ( 15 ) , S X ( 15 )
(XL(l),XUtl),OELCU,l=1.,N)
·
- ··
--· ···
-·
··········
10000070
UH10tH!iH1
----·-······-···· -----------------------10tlUilU90
IFCIFEOFC601.£Q.-Il GO TO 9500
C l~ITI~LIZE MERIT FUNCTION AT SOME LARGE NUMBER AND COUNTER AT 1
10000100
SAVE = 1000000.
10000110
------------COUNT= 1.
·· ···
···· · · · ------------'----------100ll01.ZD
WRITE (61,6000)
10000121
FOLLOWED BY RANOOM V10000122
1ECTOR
TERMINATES WHEN A MAXI10000123
-------2MUM NUt18EF: OF \JECiOR ATTEMPTS FAILS 10 CEERMINE A DIRECTION OF AC10000124
3CEPTA8LE CESIG~ I~PROVEMENT
,!,4X,125HP~OGR~M ~EOUCES INITIAL ST1000Ui25
bEP SIZE BY FACTOR OF 10 FCR SECOND STAGE ITERATIICN tNO HAS PROVIS1000D126
SION FCR ACGE~ERATlON OF STEP SIZE,/,4X~36HCUSABLE DIRECTION HAVING10JOG127
6000
FO~MATC1H1,25XK73HMONTE CARLO RANDOM SA~PLING
SEQUENTI~L ~OVEMENT ,1/,4X124HPROGRAM
6 B EE N F 0 U 1\ 0 i , I 14 6 h 4 0 H I NPUT f 0 i< C U .'< R E t-.; T F t-< 0 B L EM
WRIT f
(61 , 1 0 2 5)
N, MAX S, MAX 1/ 1 (I, Xl. (
I) ,
XU ( I}
F CL L 0 WS 8 E L 0 W -- )- ~-- 1 0 0 G0 1 2 B
1 0 t: L (I) , I= 1 , N)
1 0 0 Cl 0 13 0
1025 FORMAi(//99X,22HNLMaER OF VARIABLES = ,I2,~X,29HNUMBER OF RANDOM $10000140
1tMPLI~G3 = ,I6,5X,32HMAX NUMBER OF VECTOR ATTEMPTS = ,I10,///,32X,10000150
---------- 21 tHL 0 ?iE R EO UN 0, 1 uX 9 HHUPP E R. BGUNO ,- 10 X; 17HIN l'f·lA L---S-TEP--Sl-lE-tii+25X-, 10 0 0 0 16G ·
32HX{ ,I2,1H) dX7E12.5,9X,E12 .. S~12X,E12 .. Sn)
100!HJl7U
1090 OC 10:iiJ I:::l,tl
10G00150
YFL
= RANC:Jt'i(OJ
-------:R (-I ) =Y F L -- ·· · • · ----- --------------------1030 X(I)
XL<IH·fdillf(XU(!)-XL(I))
=
CALL CONSTtCOOE>
C CCCE = 1, SATISFIED ••••• CODE = O, VIOLAlEO
- - - - - - I F(C OOE- •J • S )·1 GL~ 0 9 10-lf 0 1 1 OS 0--------~-------. 1050 CALL MERITCFUNCT)
C
~AS
MERIT
IM~ROVEO
IF<FU~CT-SAVE)1060,1060,1040
-1-G-£:-U--S t VE -.::: F U !'-. C T
_1
i'\~ITE
h11,U
··- -- · --- · ·
FUNCT,COUNT, (I,Xti>,I=1,N)
FUNCT =,E12.7 1 5~,7HCOUNT
= ~~L2o5,!/,58),2HXCI2,4H) = tE1~.5))
----·GO -1 B8 D---I =b N---1H)
=
=
--------------------------1 0 0 0 0 2? 0 -1000()260
10000270
· ------···-------
FORMAT!~X,16H!MPRCVEO
SX<Il
X(I)
COUNT
COUNT + 1. ..
. ·..
If(COUNT ~ ~AXSl1090,1090,i095.
00 OQ 2Hl ··
10000220
10000230
10000240
-------'------~1
10000280
-10000 -29-G
=,F6.3,5X,2HXt,I2,4
1 0 0 -0 0 3 0 0 ·-·-
10000310
10000320
10000338
co
(}1
---1 G1.75---DG .. tl) 96 -·I= 1-,. N·------------------·------~-----·----------"
1096 XU) = SXU)
c
C ••••••••BEGIN VECTOR ~OVEHENT •••••••••
SIZE=l.
C INTIALIZE COUNTER AND FLAG/ACCEL ACCELERATIC~ MONITORING
4030 COUNT = 1 ..
-tr--£-5-lABt; ISH · ST f P--- S l·Z E:--MO N!TOR---------· · - - -
--------FLAG··= 0 ,. {} ----·--------·---- - -..
C
ACCEL =0.0
CCMFUTE STEP COHFCNENTS
3030 CO 2000 I=1,N
=- RAN CO H HH----- - - - - - R(Il=YFL
STEP (I l =0 EL <I> "'DCGS ( 3 .. 141 E.v-R (1))
SAVE STEF CO~PONENTS
--------Y F1:--
C
-2GOIJ-~TEPSH,
C
~----..:..
- - - 1 0 0 GO 340 ·
10000350
10000360
·--------,-··------------,-------........... ___________
:::STEP{!)·----·--·--------
CALL CONSTCCOOE)
--·2 (130 -CALl·
2040
ME~
2040,2040,2020
SAVE = FU~CT
WRITE (61,1) FUNCT,COUNT, <I,XU),I=l,N)
FE~FCRMANCE
. ~~~~
VIOLATED
I< (fliNG l l ···-···-·-· --·--· --- ---·-----·-·-..·---··-
IF(FUNCT -
- -.. -------·£. G T0 · 2 0 8 0
C
= o,
SAVE)
-----·-----------------
HAS NOT
I~PROVEO
~~If~~~IJ =~~~F(!)
,
......
10000480
1 QO 00 4g0
100005GO
OESIG~
C IF CCGE 1, SATISFIEO ••••••• COOE
IF(CCCE-.5) 2020,2030,2030
1oaoo~.tou
10000410
-------------------------------10UG0420 ·
10000430
10000440
10000450
VARIA8LES
2005 00 2010 I=l,N
£010 X(!) = X(I) -t STEP(!)
·--B----Crf.CK· CONSTRAINTS·· -- ·--·····---·-··- .. -------·----------..-----ACVA~CE
10000370
---·-.--·-1 0 0 0 0 ,31j 0 .
iOOOOJgo
----------------------------------~-----------:....:..-_.:___;_.
10000310
10000520
10000530
10000540
-10000 550 -
100G0560
10000570
10000580
1 0 U0 0 59 U· -·
10000600
10000610
1 0 0 0 0 £> 2 0 --
10000630
i8~~~t~8_
_______
00
0'1
---C- C HECI< I F ACC E L E RA T I 0 N H AS 8£ E N AP PL<t-ED--·---------··----------·-----~ACCEl)2060~2060,2070
IF(.5
2070 IF(COUNT - MAXV)3010,3010,3020
~010
---..
·
FLAG=O.n
COUNT :: COUNT
GOTO 3030
+ ·h
-·
10000690
·--·"·"·-·--..-----·--------···-..··--·--·-··--...............--.--·-------·------JLOGOulOO ·
10000710
2080 IF{FLAG -.5)3040,3040,3050
----·
10000720
3040 ACCEL =1.
==
D0 3 G50 I· 1 , N · ·· · · · · --··---·----------------·
3060 ;;:TEP<Il
2 • ..,.STEPU)
--------
= 1.
3050 COUNT
GOTO 200~
--·2 0 f: 0 f U• G ; L · · ··· --··---·-·· ·----
= o.a
ACCEL
1 N
==
STEPS(!)
CO 3070 I
3070 STEP(I)
-····· .. ---- G0 T 0 2 H 015 ·
· - ··
----------
----·
10000790
10000d1G
· .........................................- - - - - - - - -..------~----·--------..,..-----....... 1 0 !J 0 0 ij 2 0 ..
402~
=.
c
hRITE
(61,3080)
10000770
---1 G0 0 u7 3 0 ··
10000800
I= 1 N
4020 OELfl) =.1¥-0ELli>
------ -· s I zr-:
2. .. ...... ______ .......---------......--------GOTO 4030
L010
10u00!3D
--------1 0 0 0 0 7 4 0
10000750
10000760
3020 IFCSIZE - t.S) 4000,4000,4010
400G DO
----·1 0 G0 0 6 6 f.J ·
10000670
100u0650
SAVE
,<I,
. .
X(I),
!=l,N)
100008JO
10000840
10000350
-- 1 oooo g s n .
1000Ud70
1000ll83U
10000890
---3-(l-8-~ .. --f-0~ MATl////-~2-X...,-2 SH-LEA-S-T--VAtUE---CF--ME-R-l-+--I-S---,.£.t-2-.-5-~5-X....,...2-6H-HiE----OE-S-IG-1 00 0 0 9 J U ··
iN VARIABLES ARE
25))
,2HX(,I2,4H)
= ,E12.5/,(/,90X,2HX(,I2,4Hl
= ,E12.10000910
.
.
10000920
GO TO 9000
----9StH}- S T 0 P -------- ·· ............___. _
END
~~----·-
-----sueR-our-r-r-re--cottsi·tco-oE'l _____
---..---1'00 01010 ----
COMMON Xt15)
COOE=1.
RETURN
- - - - ' - - t:Nn--
..................__ ...
10000930
-----·-----'-;...;._;_._
10001020
100'01049
10001050
_;_;;_~---4-0002tl00·-···--
co
-.J
i
- : - - - 5 UB R 0 UT-I NE· - t1 E Rll·{ FU NC-1)----- -------------------------------------------~-----------------------·20 0 0 0-0 2 0--·-··"'-
.
COMMON
XC15)
20000030
A :: 2e/3,.
e::: 0<{1).Y.-¥-(5./3.l)'~'(.S-¥•A-t·1.) + X(2)"'{0<(1H·.5•X(2Jl.Y.•A +(X(U+X(
-+-----12)} 'I' ... tl)
. . ...
........ - ·········
..... ------·-------~-----------·-·
C
X(3)¥- ((X(1H-X(2) +.5•X(3))'HA + (X(t> +X(2) +X(3))••A) +X(i.t)-¥
2 ( ( X ( 1 I 't X ( 2 ) + X ( 3 ) + • 5 • X ( 4 ) ) • • A + ( X ( 1 ) + X ( 2 ) +X ( 3 ) +X ( 4 ) ) • • A )
i
0:: XC5)"'((X(1) +XC2l+ X(3)t- X(4)+ .5'~-X(5})••A +(X(llt X(2)+ X(3}
-'---3+-X(i;) +X{5))>~-'*A)
-------------- ---------····-···---------------·--·------··------ -----------!
E:: X{6)¥((XCU+X(2)+X(3)+X(4)+X(5)+,.5'~-X(6))••A + \X(1)+X(2)+X(,3)+
.
1X(4) +X(5) tX(6) )'«-'~<A)
=
F
=
X(l)lf-((X(U+X<2H-X{3)+X(4)+)((5ltX(6)+.5-'f-X(7))'~-""A
+ {X(U+X(2)+
------1X(.3}+X(4)+X{5>+X(6)+X(7»)••A)
-------------- - - ·-------- --------G:: X(8)¥((X(i)+X(2)+X(3)+X(4)+X(5)+X(o)+X(7)+.5•X(8) >••A+(X(i)+
I
1 X ( 2) +X ( 3) +X ( 4) +X ( 5) +X ( 6) +X ( 7) +X ( 8) ) ••A)
1
I
= X(9)•((X(1)+X(2)+X(3)+X(4)+X(5)+X(6)+X(7)+X(6)+.S¥X(q) )••A +
X ( 1 )+X ( 2) +X ( 3 )+X t 4} +X ( 5 H· X-( 6) +X( 7) +X ( 8) +X{ 9) -) • "'A)-------------------------.. ··-·----------------------------FUNCT = C.5-X(1)-X(2)-X(3)-X(4)-X(5)-X(E)-X(7)-X(8)•XC9) ) 4 ( t.S+X
~
-r------1 l
1
,
1(1)+XC2)+XI3)+X(4)+XC5)+X(6)+X(7)+X(8l+X(9) )••A+l.} + l2.••A)•(8+
1C+D+t+F+G+H)
t-----R ETU R N--------------------------------------1
END
2 0 0 0 0 -05 0-------
200000
ro
(J)
~---
1
I
I
I
I
I
I
APPENDIX B
~I
!
COMPUTER LISTING OF
NUMERICAL INTEGRATION
Il
I
!.
BY THE
TRAPAZOID RULE
I
I
II
I
I
I
I
i1
I
I
I
i
I
!
!
l
---------------
'--~------ ---~----··
89
90
----·-~-
~-~-----·
-------------------:-1
I
I
I
A=.446
B=.55
ERROR=.OOOl
MAX=500.
1
I
:~:
;3
H=(B-A)/Z
S=O.
,11
I
I
~A.
W=l.
Q=2./3.
]I
S=S+(-.5*X**2~.3153*X-.04118)**Q+
l
7
6
8
I
D=C
.9
Z=2.*Z
-
;
.
-
IF(Z-~~X)3,3,15
15
.C
17
20
CONTINUE
NO CONVERGENCE
WRITE(2,20)C
FORMAT(lOX,l4HTHE AREA IS= ,El2.4)
STOP
END
i
I
I!
i
I
1 (-.5*(X+H)**2+.3153*(X+H)-.04118)**Q
-~
X=X+H
W=W+l.
I
IF(W-Z)5,5,7
C=(S*H)/2.
I
WRITE(2,6)Z,C,W
i
FORMAT (lOX, 2HZ=, El2. 4, 5X, 2HC=, El2. 4, 5X, 2IDV=, E12. 4) !
R=(C-D)/C
IF(AB.S (R) -ERROR) 17, 17, 9
.
L~------
I
I