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G E N E R A L M O T O R S C O R P O R A T I O N
AEROElRBcE OPERATIONS DEmRTMENT
https://ntrs.nasa.gov/search.jsp?R=19650015244 2020-03-26T07:55:30+00:00Z
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G E N E R A L M O T O R S C O R P O R A T I O N
STRUCTURES I
THE RESPONSE OF BEAMS AND RINGS TO HIGH-INTENSITY, SHORT-DURATION LOADING
Sponsored by
N A S A H O U S T O N Contract No. N A S O - 3 0 8 1
R . E. SENNETT D. E. SKAAR
GM DEFENSE RESEARCH LABORATORIES SANTA BARBARA, CALIFORNIA
AEROSPACE OPERATIONS DEPARTMENT
T R 6 5 - 0 8 FEBRUARY 1 9 6 5
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N I
ABSTRACT
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Computer programs are given for determining the response of beams and rings
to impulsive loads, with their use described in detail.
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Two sample problems illustrate the input and output data formats, as well as
various output options.
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CONTENTS
Abstract
Introduction
Theory
Program Description and Use
Beam Input Data
Determination of Input Data for Sample Beam Problem
Input Data Format
Beam Problem Output
Sample Ring Problem
Ring Program Output
References
Appendix
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ILLUSTRATIONS
Figure 1
Figure 2
Figure 3
Figure 4
Figure 5
Figure 6
Figure 7
Figure 8
Coordinate System and Internal Forces
Idealized Thickness Model
Beam for Sample Problem
Lumped- Mass Beam Approximation
Sample Beam Input Format
Suggested Program Control (Columns 1, 15, 29, 43, and 57 are blank)
Sample Ring Input Format
Ring Geometry
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INTRODUCTION
In recent years, many investigators (1-4)* have attempted to determine responses
of various structural elements to blast or impulsive loadings. Symonds, (5) for
example, has developed a "rigid-plastic" theory of deformation in which it is assumed that no elastic deformation takes place, so that all the energy imparted
to the system is channeled into plastic deformation of the structure. This type of
analysis has two major disadvantages:
1) The energy input to the system must be much greater than the
elastic strain-energy that can be stored by the structure. Only permanent deformations a re obtained; i. e. , no time-history
of the response can be found. 2)
The most recent technique which is not subject to these limitations was devel-
oped by Witmer, et al, (6) a t the M. I. T. Aeroelastic and Structures Research
Laboratory. This technique is now described briefly.
THEORY
The dynamic equilibrium equations for the structural element shown in Figure 1
can be written \
i 3 3 - (N cos 9 ) -as (Q sin 8) + Fy - mi; = 0 iS
a a - (N sin 8) + - (Q cos 0) + F as as Z
- mw = 0
Q = O aM as - -
* Raised numbers in parentheses indicate references at the end of this report.
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where
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m = mass per unit length of structure
8 = slope of structural element
N, Q, M = normal force, shear force, and bending moment at a given cross section
V, w = accelerations in the horizontal and vertical directions
F , F = forces per unit a r ea in the horizontal and vertical directions
In the derivation of Eqs. (l), the effects of shear deformation and rotatory
inertia have been neglected. In the case of impulsive loading, F and FZ are zero, and the beam is considered to have an initial velocity.
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Equation (1) may be phrased in finite difference form and interpreted as describing a lumped-parameter model consisting of masses connected by
weightless, straight links. (These equations, as well as a complete description
of the model, can be found in Reference 6. )
Equations (1) and the corresponding strain-displacement equations have been
programed for an IBM 7040 digital computer using Fortran IV language. The
remainder of this report describes the use of this program.
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PROGRAM DESCRIPTION AND USE
Two program listings will be found in the appendix. The first , "Blast-Loaded
Beams, " calculates the response of a clamped-ended beam to an impulsive
load. The beam may be pre-tensioned, if desired. Sample output data a r e also presented.
The second, "Blast-Loaded Rings, " calculate:; the response of a circular ring
to impulsive loading. The ring need not be complete.
Boiii tile beam arid the ring are uf reciaiiguiar- cruss seciiuii, arid iiie cruss-
sectional area is distributed among six "flanges, " as shown in Figure 2.
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Definitions of the quantities contained i n the programs are given a t the beginning
of the listings.
BEAM INPUT DATA
CMT classification data (up to 72 characters)
This must be smaller than the time required for a longitudinal wave to traverse the distance between mass-points; otherwise the solution will be unstable.
Young's modulus for the beam material (lb/sq in. )
strain-hardening modulus (lb/sq in. )
yield s t ress of material (lb/sq in. )
DT Time interval (sec)
E
E P
S l G l
TI readout time (sec) width of beam (in. )
thickness of beam (in. )
WIN input displacement (in. ) The method for computing WIN is described in the sample problem. mass of a single mass-point (lb-sec /in. )
starting time (generally initialized a t zero)
time of termination of solution (sec)
length of beam (in. ) maximum and minimum limits of transverse displacement (in. ) pre-tension s t ress (lb/sq in. )
total number of plots (less than or equal to 10)
the number of the first mass-point that has non-zero input displacement the number of the last mass-point that has non-zero input i displacement
B
H
2 SM
T
T F
LN
XMAX, XMIN
SIGO N P L
K1
K 2
fixed point
TP(J) plotting time (sec)
DETERMINATION OF INPUT DATA FOR SAMPLE BEAM PROBLEM
Consider the response of a beam 20 inches long, 1/2 inch wide, and 1/16 inch
thick, made of 7075-T6 aluminum alloy pre-tensioned to 50% of the yield s t ress
and subjected a t time a 1-inch length a t the center of the beam. The beam is shown in Figure (3).
T = 0 to a uniformly distributed impulsive load over
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I It is desired to obtain the response of this beam for a total time of 200p sec.
Thus, T F = 200 x lom6.
The yield s t ress for 7075-T6 aluminum is approximately 71,400 pounds per
square inch. Therefore, SlGl = 71,400. Since this aluminum is not a strain-
hardening material , EP = 0 .
Because the beam is pre-tensioned to 50% of the yield s t ress ,
A s specified, LN = 20, B = 1/2, H = 1/16. It is desired to read out the
required data at intervals of 20psec; thus TI = 20 x 10
SIGO = 35,700.
-6 .
Since there are 60 mass points representing the total mass of the beam, then
(B) (H) ( L N P 60 SM =
where
SM = 0.27 x
32, and 33 will be loaded by the impulse. Thus,
p is the mass density of the beam material. In this example, . Referring to Figure 4, we notice that mass points 30, 31,
K1 = 30, and K2 = 33.
A time interval DT of 0.5 x seconds is found to satisfy the necessary
criterion for a convergent solution.
We assume that the velocities a t which the loaded mass-points begin to move
have been computed from impulse- momentum relations. Suppose these velocities
are constant and equal to 1,030 inches per second. The input displacement
WIN
for the f i rs t time increment DT
I is therefore 1,030 (DT) = 0.515 x l oq2 inches. That is, we assume that
, the loaded masses move without restraint.
I Since mass-points 30 - 33 received the initial load, the greatest strains will
occur in this vicinity. Therefore, the output codes L l , L2, L3, and L4 h
are assigned the values 29, 31, 32, and 34. A non-zero integer must always
appear for these output codes.
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Plots of the beam profile are desired at 50-usec intervals from zero to 200
psec. Thus, N P = 5 and TP(1) = 0, TP(2) = 0.5 x , TP(3) = 0 . 1 x
TP(4) = 0. 15 x , and TP(5) = 0.2 x . Because no more than five
plots are required, TP(6) through TP(10) need not be specified. The
quantity
vs time.
IP is set equal to 1 to obtaina plot of the center displacement
INPUT DATA FORMAT
With the exception of cards 1 and 5, all cards consist of five fields, 14 columns
wide. Card 1 uses the first 72 columns for data identification, and card 5 has
eight fields, two columns wide for fixed-point constants. Figure 5 shows the
layout of the data fields on a data sheet. If N P = 0, cards 6 and 7 must be
omitted and if N P is less than six, only card 7 must be omitted.
Possible sources of e r r o r in the input deck are the odd field widths on all
cards except 1 and 5. To minimize the occurence of these e r r o r s and to facili- I
tate the preparation of input decks, the program control feature of card-punching
equipment is strongly recommended. This program control card (Fig. 6) will
enable the user to prepare data without using the numeric shift key. U s e of
the shift key anywhere on the card wil l cause it to advance to the f i rs t column of the next numeric field. To avoid skipping, the ALPH shift key should be
depressed whenever a minus sign is to be punched. I I
i BEAM PROBLEM OUTPUT
A listing of the input data appears at the beginning of the output for each set of
data, followed by a set of header lines which identify the output in three basic
groups: Line 1 contains the time, centerline displacement, and the transverse
acceleration at the center; line 2 has the strain at flange 6 of the links specified
by the output codes; line 3 contains the strains in the links at flange 1. These
t three nutput grniups are printed heginning With the s t a r t i ~ g tim-g T , zfid the printout interval T F .
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Supplementary data can be obtained at the printout time by using the sense
switch options. This information is printed on tape unit No. 4. If sense switch
3 is on, the following data is printed for mass points 1 - 31: acceleration in
the horizontal direction, acceleration i n the transverse direction, total trans-
verse displacement of the mass points, and horizontal coordinates of the mass
points. When sense switch 4 is on, the strain in flange J (J = 1, . . . 6) of link
I (I = 1, . . . 32) is printed. If the sense switches a r e left on for an entire run,
the total run time is increased by 50%.
SAMPLE RING PROBLEM
Input Data
DT
E
E P
S lGl TC
B
H
ALFA
SM T
R TP
TF
fixed 1: point 1 K2
time increment (sec)
Young's modulus (lb/sq in. ) strain-hardening modulus (lb/sq in. )
yield s t r e s s (lb/sq in.) readout time (sec)
width of ring cross section (in. )
thickness of ring (in. )
defined in Figure 8 (rad) mass of a single mass-point (lb-sec /in. )
starting time (generally initialized a t zero)
radius of the ring (in.)
plotting time interval (sec)
time of termination of solution (sec) ring profile plot code (non-zero for plots)
the number of the first mass-point which has non-zero input displacement
the number of the last mass-point which has non-zero input displacement
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Calculation of the input data for the ring proceeds in the same manner as that
for the beam; only the calculations which differ are described here.
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The angle of the clamped supports from the vertical is defined by ALFA
The quantity RIN is analogous to WIN in the beam program. A displace-
ment toward the center of curvature is a positive input displacement.
The radius of the ring in inches is R . In the sample problem a complete
ring is uniformly loaded radially over a central angle of 120 degrees by an
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the loaded mass-points. Mass-points 22 (Kl) to 41(K2) are loaded.
The ring material, 6061-T6 aluminum alloy, is considered to be elastic-
perfectly-plastic. Thus, E = 10. 4 x lo6 , E P = 0, and S lGl = 41,400. I To complete the specification of the physical properties of the ring, R = 3 in.,
H = 1/8 in., and B = 1 in.
-3 Since output is desired every 100 psec, TC = 0. 1 x and T P = 0.1 x 10 . This produces a plot of the ring deformation every 100 p sec. The final time
T F is 300. 5 cc sec to insure output data a t 300 p sec.
RING PROGRAM OUTPUT
A listing of the input data appears at the beginning of the output for each data
set. Beginning with the starting time T and continuing a t printout intervals
TC , the following data is presented: horizontal and vertical coordinates for
mass-point I; s t ra in in link I f o r flanges 1 and 6; (I = 1, . . . , 31).
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Z
V -..
I
I I I
+ Y
Figure 1 Coordinate System and Internal Forces
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ACTUAL CROSS SECTION IDEALIZED CROSS SECTION
Figure 2 Idealized Thickness Model
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Figure 3 Beam for Sample Problem
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R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
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60 61
Figure 4 Lumped- Mass Beam Approximation
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*
c C
E E
. C z E
C
E E
I
i c
c z E
U
2 k
E $ m
cd cn
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1 I I I i I
I I 1
I I I I I t
o + a +
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+ +
g $
I
+ + + 2
1 m
+
i + 'q + + I t
% I i I+ + * $ * +
+ + + +
I t
t L
rl
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0 R
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(POSITIVE DIRECTION AS SHOWN)
------- '1 0
V -
Figure 8 Ring Geometry
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REFERENCES
1. H. H. Bleich and M. G. Salvadori, "Impulsive Motion of Elasto-Plastic
Beams, " Transactions of the American Society of Civil Engineers,
Sep 1953
2. M. M. Chen, P. T. Hsu, and T. H.H. Pian, "Impulsive Loading of
Rigid- Plastic Curved Beams, It Fourth U. S. National Congress of Applied Mechanics, 1962
3. M. F. Conroy, "The Plastic Deformation of Built-In Beams Due to
Distributed Dynamic Loading" (paper presented at the Summer
Conference of Applied Mechanics Division, American Society of
Mechanical Engineers, Jun 9 - 11, 1964, Boulder, Colorado)
4. P. E. Duwez, D. S. Clark, and H. F. Bohnonblust, "The Behavior of
Long Beams Under Impact Loading, I' J. Appl. Mech., Mar 1950
5. E. H. Lee, and P. S. Symonds, "Large Plastic Deformations of Beams
Under Transverse Impact, ' I J. Appl. Mech. Vol. 19, 1952
6. E.A. Witmer, H. A. Balmer, J. W. Leech, and T. H. H. Pian, "Large
Dynamic Deformations of Beams, Circular Rings, Circular Plates,
and Shells" (paper presented at the AIAA Launch and Space Vehicle
Shell Structures Conference, Apr 1 - 3, 1963, Palm Springs, Calif.)
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APPENDIX
1. Beam
a) Listing
b) Output
2. Ring
a) Listing
b) Output
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S I B F T C B E A M L I S T * R E F C C I N P U T D A T A C C
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B W I D T H O F B E A M CMT D A T A 1 D E N T IF I C A T I ON D T T I M E I N T E R V A L
1 : C C E YOUNG'S MODULUS C EP S T 2 A I N - H A R D E N I N G MODULUS C E P O S T R A I N A S S O C I A T E 0 W I T H S I G O C E P S T ( I 9 J ) T O T A L S T R A I N I N F L A N G E J O F L I N K I C H T H I C K N E S S O F B E A M C I P D I S P L A C E M E N T VSe T I M E P L O T CODE
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C K 1 T H E NUMBER OF T H E F I R S T M A S S - P O I N T W H I C H H A S NON-ZERO I C I N P U T 0 I S P L A C E M E N T I C K 2 T H E NUMBER O F T H E L A S T M A S S - P O I N T WHICH H A S W N - Z E R O
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C I N P U T D I S P L A C E M E N T C LN L E N G T H O F B E A M C NPL T O T A L NUMBER O F P L O T S C Q SHEAR F O R C E C Sl C1 Y I E L D S T R E S S O F M A T E R I A L
C S I G ( 1 r J ) S T 2 E S S IN F L A N G E J O F L I N K I C SM M A S S O F A S I N G L E M A S S - P O I N T
I C SIC0 P R E T E N S I ON S T R E S S
I C T S T A R T I NG T I M E I C T F T I M E OF T E R M I N A T I O N OF S O L U T I O N
C T I READOUT T I M E C T P P L O T T I N G T I M E C WDD A C C E L E R A T I O N I N T e A N S V E R S E D I R E C T I O N C WIN INPUT D I S P L A C E M E N T C XMAX M A X I M U M L I M I T O F TRANSVERSE D I S P L A C E M E N T C X M I N M I N I M U M L I M I T OF TRANSVERSE D I S P L A C E M E N T C C DEFI N I T IONS C C C BMOM BEND I N G MOMENT C DS I D I S T A N C E B E T W E E N MASS P O I N T S C E l A X I A L S T R A I N C E2 B E N D I N G S T R A I N C €PO S T R A I N A S S O C I A T E D W I T H S l G O C F N O R NORMAL F O R C E
C O O R D I N A T E S O F THE F L A N G E S PREiEiu'Siei\i
C P S I 2 SlGO C VDD A C C E L E R A T I O N I N THE H O R I Z O N T A L D I R E C T I O N C V I T ( 1 ) H O R I Z O N T A L C O O R D I N A T E O F M A S S I
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C vo I N I T I A L L O N G I T U D I N A L C O O R D I N A T E S C W I T ( I ) T 3 T A L T 3 A N S V E R S E D I S P L A C E M E N T OF M A S S I C wo I N I T I A L T R A N S V E R S E C O O R D I N A T E S
R E A L LN D I M E N S I O N D A T A ( B O 0 ) 0 1 MENS I O N T I MX (500 D I M E N S I O N S 1 G 0 ~ 6 1 ~ 6 ) r E P 0 ~ 6 1 ~ 6 ~ r P S 1 o r V 0 ~ 6 ) ~ V 0 ~ 6 2 ) ~ W 0 ~ 6 2 ~ * V 1 ~ 6 2 ~ ~ W 1 (62 ) *
C L D I S ( 500 1
1 2 WDD ( 6 1 1 * V 1 T 62 1 3 W I T ( 6 2 ) , E P S f ( 6 l r b ) r D E P S ( 6 1 r 6 l r S l G T ( 6 l r 6 ) r S l G C l ( 6 1 * 6 ) * 4
V O P ( 6 2 ) * O S I ( 6 l ) r S ( 6 1 I r C ( 6 1 ) * E l (61 ) r D T I ( 6 1 ) r E 2 ( 6 2 ) * E P S ( 6 1 * 6 ) * S 1 G ( 61 9 6 ) r F N O R (61 ) *BMOM (6 1 9 Q ( 61 )
S1 GC2 (61 6 ) * CMT ( 12 1
VDO (6 1
X (32 1 Y (32 ) T P ( 10 1 500 FORMAT (5E14.8) 501 FORMAT ( 1 2 A 6 ) 502 FORMAT (812) 5510 FORMAT (49H1 D E F O R M A T I O N P R O F I L E OF A N I M P A C T L O A D E D B E A M 1 4 2 X q
5 1 1 FORMAT ( l H O * 8 X 1 4 H T I M E INTERVAL=rE14.8r4X16HYOUNG-S MODULUSme9Xo 1 A 2 * 1 H/ * A 2 1 H / * A 2 r d X 5 H P A G E ! 4 / 5 X 1 2 A d 1
1 E14 0 8 9 4 X 1 3 H Y IELD S T R E S S = * E l 4. 8 / 9 X A 4HIN1 T I A L T I M E = t E 1 4 0 8 , 4 X * 2 2 5 H S T R A I N - H A R D E N I N G MODULUS=tE14.8*4X6HWIDTH=r7XE14.8/9X 3 1 4 H F I N A L T I M E = * E 1 4 . 8 r l X 2 5 H I N I T I A L D I S P L A C E M E N T = *E1408* 4 4x1 3 H T H I C K N E S S = r E 1 4 . 8 1 9 X 1 4 H R E A D - O U T T 1 M E = r E 1 4 0 8 * 4 X 5 2 5 H M A S S OF A M A S S - P O I N T S rE14.8,4X7HLENGTH=*6XE14.8)
512 FORMAT (7HOLINE ~ ~ ~ X ~ H T I M E I ~ S X ~ H C L D I S P ~ * ~ ~ X ~ H W D D / ~ H LINE 2 * S X 1 r4(4HEPS(rI2r3H,6)rllX)/*7H LINE 3t5X14(4HEPS(~I2r3H*l)*llX))
513 FORMAT (6HO LINE/SH 1 e 4 X E 1 5
I 5H 2 4x4 ( E 15. 8 5 X ) /EiH 3 * 4 X 4 ( € 1 5 0 8 * 5 X ) ) 514 FORMAT ( 5 H O 1 * 4 X 3 ( E 1 5 0 a r 5 X ) / 5 H 294x4 ( E 1 5 0 8 * 5 X ) / 5 H 3,
1 4 X 4 ( E 1 5 0 8 * 5 X ) ) 515 FORMAT (6HOTIME=,E12.5/11X3HVDD114X3HWD0,14X3HWDD~l4X3HWIT~l4X3HVZT~) 516 FORMAT ( 1 O X e 4 E 1 7 . 8 ) 517 FORMAT ( ~ H O T I M E = ~ E ~ E O ~ / ~ X ~ H I ~ ~ ( ~ X ~ H E P S ( I ~ ~ ~ ~ ~ ~ H ) ~ ~ X ) ) 518 FORMAT ( 8 X * 1 2 * 6 E 1 7 . 8 ) 519 FORMAT ( 9 X * 4 5 H I N I T I A L D I S P L A C E M E N T E X T E N D S FROM M A S S - P O I N T 0 1 2 r l X 1
1 1 4 H T O M A S S - P O I N T 9 1 2 ) C A L L P L O T S ( D A T A 800 1 C A L L P L O T (0.0*000*-3) C A L L F P T ( o T R U E O O O ) C A L L D A T E ( X D l r 1 0 2 ~ 1 D 3 ) N P G N - 1 NPCA = 1 INP=O
1 R E A D (5*501) ( C M T ) Y ~ I ~ E : ~ ; s : Q ) !E! ~ ! D ~ ~ ! D ~ ~ N P G N I C M T R E A D (5,500) D T , E ~ E P I S ~ G ~ ~ T I I ~ * H , W I N I S M , T ~ ~ F , L N I X M A X ~ X M I N ~ S I G O R E A D ( 5 9 S 0 2 ) NPL,KfrK2*Ll,L2*t3*L4*IP WRITE ( 6 9 5 1 1 ) D T I E I S ~ G ~ ~ T I E P * B * T F , W I N I H I ~ I I S M I L N W R I T E ( 6 r S 1 9 ) K l * Y 2
19
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
2 4
80
3
20
81 02
86
21 22 23 24
25 26
27
271
20
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N I I '
21
I G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
54 EPST( 1 9 J)=E1 ( I )-PSI (J)*E2( I + l )+EPO( I * J ) DO 541 Jtlr6
DO 58 I = l r 6 1 DO 58 J=l*6
I
54 1 EPST (61 J )=EPST (60 * J )
55 DEPS(IIJ)=EPST(I~J)-EPS(I*J) 56 EPS(ItJ)+EPST(IrJ)
SI GT ( I 9 J )=SI G( I J )+E*DEPS ( 1 t J 1 57 SlG(I*J)=SIGT(I*J)
IF(DEPS( 1. J) 1 58 CONTINUE
I
61 r 6 0 * 5 9
T=T+DT GO TO (150t151)*1PLOT
150 IF ~ T o G E o T P ~ J P ~ o A N D o J P o L E ~ ~ ~ L ~ GO TO 66 151 IF(ToGEoTC)GO TO 72 70 IF(TFoGToT)CO TO 68
IF (IPoNEoO) CALL GRHl (TIMX*CLDIS*INP) 69 GO T O 1 72 IF (LCoGE.52) GO TO 87 88 WRITE (6r514) T~WIT(3I)tWDD(31)*EPS(Ll~6)t~PS(~2*6)*E~S(L3*6)~
1 EPS(L4*6)rEPS(Llrl )*EPS(L2*1)*EPS(L3*1 )*EPS(L4*l 1 LC=LC+4 CALL SSWTCH (3rJSS) CALL SSWTCH ( 4 , ISS IF (JSSoEQ.2 ) GO TO 90 WRITE (4t510) ID1 *ID2*103tNPGA*CMT
1
NPGA=NPGA+ 1 WRITE (4*!515) T WRITE (4r516) ( V D D ( 1 )rWDD(S 1 ,WIT( I )*VIT( I Is1 9 3 1 1
90 I F (ISSoEQo2) GO T O 89 WRITE (41510) IDI*ID2*103*NPGAtCMT NPGA=NPGA+ 1 WRITE (4,537) T*(J*J=1,6) WR 1 TE ( 4 9518 ) ( 1 9 (EPS ( I J) * J O l e 6 1 * I = 1 * 32 1
I 89 INP=INP+I TI M X ( INP 1 =T CLDIS( INP)=WIT(31 1 TC=TC+T I GO TO 70
IF ( S 1 G ( f 9 J 1 -S I GC I ( I * J ) 133 * 339 34
GO TO 29
GO TO 29
GO TO 29
30 SlGCl(f~J)=SIGl+EP*EPS(I*J)
33 SlG(IrJ)~SIG(I*J)
34 SlG(1tJ) =SlCCI(I*J)
31 SlG(ltJ)=SIG(I*J)
22
I G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
c
32 S 1 G C 2 ~ I ~ J ~ ~ - S l G l + E P * E P S ~ I , J ~ IF ( S 1 G ( I t J 1-S 1 GC2 ( I 9 J ) 135 * 36- 36
35 S ~ G ( I I J ) = S ~ G C ~ ( I ~ J ) GO T O 29
36 SIC( I r J l = S I G ( I t J ) GO T O 29 S1 G C I ( 11 J )=S1 Gl+EP*EPS( I J ) I F ( S 1 G ( I t J I - S I G C l ( I ~ J l ) 6 2 r 6 3
6 2 S l G ( Z * J ) p S l C ( I t J ) GO TO 58
63 S l G ( I * J ) = S l G C l ( 1 1 J) GO TO 58
60 S 1 G ( Z t J = S I C ( I t J 1 GO TO 58
61 S l G C 2 ( I 1 J 1 o - S l C l +EP*EPS( I IF(SIG(l~J)-SIGC2(I~J))64*65t65
64 s1 G ( I t J ) = S I G C 2 ( I 1 J 1 GO TO 58
65 S1 G ( I t J) = S I G ( 11 J ) GO TO 58
W R I T E (6.512) L1 rL2rL3*L4*L l * L 2 r L 3 ~ L 4
59
J 1
1 87 WRITE (6.510) I D ~ , I D ~ I I D ~ ~ N P G N * C M T
NPGN=NPGN+ 1 L C = 6 GO TO 88
66 JP=JP+1 Y ( 1 )=0.0 Y ( 2 1 zF/20 Y ( 32 )= 10.0 DX= ( XMAX-XMI N 1 / 7 0 0 X ( 1 l = ( W I T ( l ) - X M I N ) / D X X (2 1 = ( W I T ( 2 1-XM I N ) / O X X ( 3 2 ) r ( W I T ( 3 2 ) - X M I N ) / D X DO 67 1 ~ 3 1 3 1 Y ( I ) = Y ( I - I ) + F
C A L L A X I S ( 0 0 0 ~ 0 0 0 1 2 8 H P O S I T 1 O N C O O R D I N A T E ( I N C H E S ) ~ + 2 8 t l O o O ~ 9 O o O ~
C A L L SYMBOL ( 0 0 2 ~ 1 0 0 0 5 1 0 0 1 4 ~ 2 3 H B E A M P R O F I L E A T T I M E = ,000123) T I M E = T + I O o * + 6 C A L L NUMBER C A L L PMU (3~8011000510~14) C A L L SYMBOL ( 3 0 9 2 r ~ 0 0 0 5 r 0 ~ ~ ~ * 4 H S € C 0 ~ 0 0 0 ~ 4 ) C A L L L I N E i X . Y i J 2 ; ! ! C A L L A X I S ( C O O I O O ~ ~ ~ ~ H D I S P L A C E M E N T ( I N C H E S ) ~ - ~ ~ ~ ~ O O ~ O ~ O O X M I N ~ O X )
67 X ( I ) = ( W I f ( I ) - X M I N ) / D X
1 0009 1.0)
(2 .96 10 0051 00 1 4 * T I ME t 0009 1 1
C A L L P L O T (805tOaOe-31 GO T O 151
23
G M D E F E N S E R E S E A R C H L A B O R A T O R i E S @ G E N E R A L M O T O R S C O R P O R A T I O N
END
SUBROUTINE G R H l (TIDIN) D I M E N S I O N T ( 1 ) * D ( I ) C A L L S C A L E ( T I N I ~ ~ O * X M I N * D X * I ) C A L L S C A L E ( D * N * ~ o O I Y M I N * D Y * ~ 1 C A L L P L O T ( 1 0 0 ~ l 0 0 ~ - 3 )
B l B F T C G R H l L I S T *REF
C A L L A X I S ~ 0 0 0 ~ 0 0 0 ~ 4 H T I M E ~ ~ 4 * 6 0 0 ~ 0 ~ 0 ~ X ~ 1 N ~ D X ~ C A L L A X I S ~ O ~ O ~ O ~ O ~ 1 ~ H ~ ~ S P L A C E M ~ N T ~ l 2 ~ 8 ~ O ~ ~ ~ ~ ~ Y M I N ~ D Y ~ C A L L L INE ( T I D I N * ! 1 C A L L P L O T (7.59-109-3) R E T U R N END
24
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N 1
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e-- 0 0 0 I l l w ?II w O O r n m N ? p e n eo.- @ N O * * e 0 m m 9 m m . . . n o m
d N m
d d 0 0 I I
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0 0 . .
m c r d 0 0 0
I 1 w w w PN.0 w e - m m .3 O m . * + m a l m N U ? 090 4 h '99
0 3 0 . . . I
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- N m
TR65-08
25
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
TR65-08
- 4 d d
0 0 0 0
m N P7N
00 00 . . . .
c o d - . m - 4 0 0 0 C O C
4-
0 0 I 1 L u u W O m N -0 n m
u.* -9 m N
00
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lL-4
c o o w 111 Id
C N u . c u m - NIP0 m r n m L n f l c .?a* V I 6 9
0 0 3
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. . . 1
0 - - 0 0 0
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0 0 0
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r - w . . .
m - c l 000 I l l
Id w u h m o c r C Q p m m * A * o o m m m w - m m 000
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0 0 I 1
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00 . .
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0 0 0 0 - u O N Q h m - l n 3 m 8-454
0 0 0
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m m m . . . 0 0 0
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s m m 6 9 4 6 - m *In9 0 6 N c-f. - r I n I n
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w w
N - 9 - I n - mcc . D m rCIN
00
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26
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
TR65-08
BEAM PROGRAM
Memory Required
Object Program (30552)8
1/0 Buffers (1530)8
Time for Execution (sample problem)
11 min
Routines Required (those not included with IBM’s IBSYS)
BEAM PMUZ
GRHI DATE
FPTMOD
27
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
-. 1000 .a999 .3000 5000 7000 .goo0 1 1000 1.3000 D I SPLRCEMENT ( INCHES I
28
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
1
TR65 -08
- * 00000 .05000 10000 .I5000 .20000 -25000 -30000 TIME ( ~ 1 0 - 3 1
29
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
S I B F T C R I N G L I ST *REF COMMON P S I ~ ~ ) r V 0 ~ 6 2 ~ r W 0 ~ 6 2 ~ ~ V 1 ~ 6 2 ~ ~ W l ~ 6 2 ~ r B E T A ~ 3 l ~ ~ C ~ E T ~ ~ 3 l ~ r S 6 € T ~
1 r 2 s(61 ) r C ( 6 1 ) r D S I W I 6 1 ) * S O ( 6 1 ~ * C O ~ 6 1 ~ ~ D ~ l O ~ 6 l ~ ~ E l ~ 6 l ~ ~ D T I ~ 6 l ~ r 3 €2(62)rEPS(61*6)rFNOR(61 ) * B M O M ( 6 1 ) r Q ( 6 1 ) r V O D ( 6 l ) ~ W D D ( 6 l ) r 4 5 S l C C 2 ( 6 l r 6 ) r C M T ( l 2 ) r T P I S 1 C ( 6 1 r 6 ) r R , T , D P , l F ~ T G
(3 1 ) r RD I SP ( 62 1 r WRD I S (62 1 * VRD I S (62 ) r WOP (62 1 * VOP ( 62 ) * O S I ( 6 1
V I T ( 62 ) r W I T (62 ) E P S T (61 9 6 1 r DEPS ( 6 1 e 6 1 r S 1 C T (61 r 6 1 SI GC 1 (61 9 6 1 9
D I M E N S I O N D A T A ( 8 0 0 ) 500 FORMAT (5E14.8) 501 FORMAT ( 1 2 A 6 ) 502 F O W A T (712 ) 510 FORMAT (49H1 D E F O R M A T I O N P R O F I L E OF A N I M P A C T L O A D E O R I N C r 4 2 X e
1 A 2 9 t H/ r A 2 r 1 H/r A 2 r 6 X S H P A G E 1 4 / 5 X 1 2 A 6 1 511 FORMAT ( l H O r B X * 4 H D T = r E 1 5 * 8 r 3 X * 3 H E ~ ~ E l S 0 8 r 3 X ~ 6 H S l G l s * E 1 5 0 8 r 3 X r
1 4 H T C = r E l 5 ~ 8 / 9 X 4 H T F = r E l 5 0 8 s J X , 3 H H = r E l S e 8 * 3 X r 6 H A L F A z tE l5.80 2 3X4HSM = r E l 5 * 8 / 9 X 4 H T = t E 1 5 e B r 3 X 1 3 H B = ~ E I ~ O B ~ S X I ~ H R I N ~ , € l S . 8 * 3 3 X 4 H E P = r E l S o B / 9 X 4 H T P = r E 1 5 0 8 * 3 X * 3 H R srE15.8)
512 F O R M A T ~ 1 H O r 8 X 4 H T l M E ~ l 4 X 6 H W l T ~ I ~ ~ l 4 X 6 H V I l ~ I ~ ~ l 4 X ~ 8 H € P S ~ I ~ 6 ~ ~ l 2 X ~ 1 B H E P S ( I r 1 ) r 7 X r l H I )
513 FORMAT ( 2 H O r 5 X E 1 2 . 6 ) 514 FORMAT ( 1 9 X r 4 E 2 0 r B r 2 X 1 2 ) 515 FORMAT ( B X l l H ( C O N T 1 N U E D ) )
1 14HT0 MASS P O I N T r I 2 ) 516 FORMAT ( 9 X r 4 5 H I N I T I A L D I S P L A C E M E N T E X T E N D S FROM M A S S - P O I N T 912rlX1
C A L L FPT (.TRUE. 9 0 1 C A L L SSWTCH ( 3 r J S S ) C A L L P L O T S ( D A T A r 8 0 0 ) C A L L P L O T (0 .0*500r-3) C A L L D A T E ( I D l r I D 2 9 1 0 3 ) NPGN= 1
1 R E A D ( 5 r 5 0 1 ) C M f 3 WRITE (6r510) I O 1 r 1 0 2 r f 0 3 * N P G N r C M T
R E A D (59500) D T ~ E ~ E P ~ S ~ C ~ ~ T C ~ B ~ H ~ A L F A ~ S M ~ T ~ R I N O ~ ~ T P ~ ~ ~ READ (5 r502) N P L r K l r K 2 r L l r t 2 r L 3 r L 4 T I r T C T G = T P WRITE (6r511) D T ~ E ~ S I G ~ ~ T C ~ T F ~ H ~ A L F A ~ S M ~ T ~ B ~ R I N ~ E P ~ T P O R WRITE (6.516) K19K2 W R I T E ( 6 0 5 1 2 ) NPGN=NPCN+ 1 L C r l O I F (NPL.EQ.0) GO TO 6 RCsO 0
DO 4 I t l r I O RC=RC+S IF (RoLEoRC) GO T O 5
4 C O N T I N U E
30
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
31
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
32
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
V1 ( I ) = V I T ( I 1 w O ( I ) = w l ( I )
47 W1 ( 1 ) = W I T ( I ) DO 51 I = l * b l D S I ( I ) r S O R T ( ( A B S ( V l T ( i + l ) - V I T ( 1 ) ) ) * * 2 + ( A B S ( W I T ~ l + l ~ ~ ~ I T ( I ) ) ) ~ * * 2 ) s ( I ) = ( W I T ( 1+1 ) - W I T ( I 1 ) /OS1 ( I 1 C ( I ) = ( V I T ( 1 + 1 1-V I T ( 1 1 1 /OS1 ( I 1 E l ( I ) = ( D S I ( I 1 -OS1 U ( I ) 1 I D S 1 U( I 1 D t I ( I ) = O o E2 ( 1 ) Z O O E2 (62 1 SO. DO 54 I- lrbO D T I ~ ! + l ~ ~ S ~ I + l ~ * C ~ I ~ ~ C ~ I + l ~ * S ~ l ~ E 2 ~ 1 + 1 ~ ~ ~ o * ~ D T I ~ I + ~ ~ ~ D T I O ~ I + I ) I / ( D S t U ~ I + l ~ + D S I U ~ I ~ ~ DO 53 J z l - 6
51
53 E P S f ~ l ~ J ) = E l ~ I ~ - P S I ~ J ~ + E Z ( r + l 1 54 C O N T I N U E
DO 541 J = l r 6
DO 56 1 = 1 , 6 l DO 58 J t l r 6 DEPS( I J )=EPST( 1 * J )-EPS( I * J 1 EPS ( I * J )=EPST( I * J 1 S1 GT ( I J ) = S 1 f( 1 9 J )+E*DEPS( I t J 1 S 1 G ( I 9 J ) IF ( D E P S ( I + J ) ) 6 1 * 6 0 * 5 9
58 C O N T I N U E 56 CONTINUE
541 EPST(61 r J ) = E P S T ( 6 0 r J )
=S1 GT ( I J 1
T = T + D f IF (NPLoEQ.0) GO T O 1 5 1 IF ( T G o C T o T ) GO T O 1 5 1 C A L L GRHR
1 5 1 IF ( T e G E o T C ) GO T O 72 70 IF ( T F o G T o T ) GO T O 68 69 GO TO 1 72 I P C E = I
LC=LC+4 IF ( L C o G E . 5 4 ) GO TO 87 LC=LC-2
73 W R I T E (6 ,513) T I P C E t 2 DO 74 I f l t 3 1 LC=LC+ 1 IF ( L C o G E . 5 4 ) GO TO 87
89 W R I T E (6,514) W I T ( I ) r V I T ( I ) * E P S ( I * 6 ) * E P S ( I ~ l ) * I 74 CONTINUE
f C = T C + T I
33
I
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
GO T O 70
WRITE (6r512) WRITE (6*513) T LC=6 NPGN=NPGN+ 1 GO TO (73*88)r!PGE
LC=LC+1 GO TO 89
IF(SlG(IrJ)-SlGC1 (I*J))33*33*34
GO T O 29
GO T O 29
GO TO 29 SI GC2 ( I r J )*-S1 G1 +EP*EPS( I r J 1 IF (SI G ( I r J 1-S 1 GC2 ( I * J ) )35+36* 36
GO TO 29
GO TO 29
87 WRITE ( 6 r 5 1 0 ) IDIrID2r103tNPCNrCMT
88 WRITE (6r515)
30 SlGC1(IrJ)=SIGl+EP+EPS(IrJ)
33 SlC(I*J)=SlG(I*J)
34 SlC(IrJ) =SlGCl(IrJ)
31 SlC(IrJ)=S1C(IrJl
32
35 SlG(IrJ)=S1GC2(lrJ)
36 S I G ( I I J I = S ~ G ( ~ ~ J )
59 SlGCl(IrJ)=SlGl+EP*EPS(IrJ) i F ( S l G ( I r J ) - S l G C l ( I * J ) ) 6 2 r 6 2 r 6 3
62 SlG( I r J)=SlG( I r J ) GO TO 58
63 SlG(IrJ)=SlGC1(IrJ) GO TO 58
60 SlG( I r J) =SlG( I r J 1 GO TO 58
61 SIGC2(IrJ)r-SlGl+EPwEPS(IrJ) IF(SlG(IrJ)-SIGC2(ItJ))64r65*65
64 SlC(IrJ)=SIGC2(IrJ) GO TO 58
65 S1 G ( I GO TO 58 END
SIBFTC GRHR L I S T SUBROUT I NE GRHR COMMON PSI (6 1 rVO(62 1 rWO(62 1 *V1(62 1 * W1 (62 ) *BETA (31 1 *CBETA (31 1 r SBETA
J) =SIC ( I J 1
- - - - - I . A . 1 (31 ) ~ W U I ~ F \ O ~ jrWR~:S!52)tVnD!S!32!.YOP(62)rVOP(62)*VoP(62)r~SI(6l)r 2 S(61 ) r C ~ 6 l ~ r O S I U ~ 6 l ~ r S O ~ 6 l ~ * C O ~ 6 l ~ r D T I O ~ 6 l ~ * E l ~ 6 l ~ r D T ~ ~ 6 l ) r 3 E 2 ( ~ 2 ~ r E P S ~ 6 l r 6 ~ r F N O R ~ 6 l ~ * ~ M O ~ ~ 6 l ~ ~ ~ ~ 6 l ~ r V D D ~ 6 l ~ r W D ~ ~ 6 l ~ r 4 5 SlGC2(61r6)rCMf(12)rTPrSlG(6lr6)rRrTrDPrTFrTG
V I T (62 ) r WIT (62 r EPST (61 * 6 1 r DEPS (61 r 6 1 r SlGf (61 9 6 ) +SlGCl(61 r 6 1
34
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
D I M E N S I O N X ( 6 2 ) * Y ( 6 2 ) 2 A D = R / D P XCO=RAD*2 C A L L S Y Y B O L ~ R A D ~ 0 ~ 0 ~ 0 0 1 4 * 3 * 0 ~ 0 * ~ 1 ~ C A L L S Y M B O L ( R A D * O o O * O o 1 2 * 4 * 0 0 0 * - 1 ) C A L L C I R C L E ( X C O ~ O ~ O ~ O ~ 0 ~ 3 6 0 ~ ~ R A D I R A D I - . 5 ) DO 2 I t l r 6 2 Y(I)=VIf(I)
2 X(I)=WlT(I 1 IF ( Q o L E o 5 . 0 ) GO TO 4 DO 3 1=1*62 X ( I ) = X ( I ) / D P
3 Y ( I ) = Y ( I ) / D P 4 DO 5 I s l e 6 2 5 C A L L S Y M B O L ~ X ~ I ) ~ Y ~ I ~ r O 0 1 0 * 3 * 0 ~ 0 * - 1 ~
C A L L S Y Y B O L ( 0 0 2 ~ - 2 0 5 4 r ~ 1 4 r 6 H T I M E = * 2 7 0 0 * 6 ) T I M E = T * l O o * + 6 C A L L NUMBER ( 002 - 3 . 2 6 9 0 14 T I ME e2700 e 1 1 C A L L PMUY ( 02 -3 098 0 1 4 1 C A L L SYMBOL ~ 0 0 2 r - 4 ~ 1 0 ~ 0 1 4 ~ 4 H S E C 0 ~ ~ 7 0 ~ ~ 4 ~ TG= TG+T P IF ( T G o G T o T F ) GO T O 8 R I P=20+2 o*R/DP
7 C A L L P L O T ( R I P * O o 0 * - 3 ) RETURN
8 R I P = l o 0 2 + 2 o + R / D P C A L L SYMBOL (RIP~4.44~021r44HDEFORMATION P R O F I L E O F A N I M P A C T L O A D
1 ED RING, 2700 e 44 1 R 1 P-0 I) 9 8 + R I P GO TO 7 END
35
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
TR65-08
u z 0 3
01 w - I z - 0 Y - O b 0 1 - l o c z 7 w 5 z - 3 c u 4 0 s o L O Y - -_ '2 - 0 %
m u 0 0 I I w w 00 a 3 0 0 O D O f - O d 00 - 7 4 . . . 0 5 0
I1 I1 II
u x n C V I W
V \ d N 0 0 0
I uuLu 0 0 0 0 0 0 00'0
~ ~ m m m m m m m m m m N N N N N N N N 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I I I I I I I I I I I I I I I I I I I I
Id w w w tu w w w w w w w w w w IU w w w .lJ I U - ~ N V \ ~ Q + O O ~ U e m m m u \ s m . r
- 9
e - c v r
- w 0 . . . . . . 4 . - z n
v) - d h v) . . . r . -TY a 0 3 c I1 I1 I1 c
cl O O ~ O - l d d d
-a. hl w UI w w w w Lu UI 3 u 7 hl I 4 . G o m + b - * L ?
m c c L - - h -: ;1: J + L? I? 7 - L 3 v c - ~ 3 . c - m - - . J a - - o G m h 3 c c r - c 6 m n + d m
m o 4 - n m L n Q b o m . 9 0 ' 0 0 0 0 > - . l d * t C d - l d r *
I v) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . u Y w ' - I v ) O O O Q O O O O C O O O O O O O O O O O O O O O O O O O O O O 00000000
I I I I 1 I I 0 0 0 0 4 O C O O l 0 0 C ) n 00002' 0 0 0 0 0 a L r 3 0 3 :
00000000 I l l
--I- m Q 3 t - z m t t c
o w o o u N V N ~ - r O O O 0 00000000
1u w Lu w 'U w w u f - 4 f - r - r c r y + - m
u s m 0 1 x - O Q Q h U N V \ h
4 H - m
0 0 0 0 ~ 7
. . . . v ) I I I I I I I
I1 I1 I@ I! u Q*+.aDIC6'hF c - m o 3 C o - l + r - *
w - c ' ) - m r 4 ( r - Z - r c ~ - d . ? . n ) r U - *
C r d - m w b m + a Q m n Z 7 * 0 4 ry 3 L? 3 0 o u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I 1 l r 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ~ 0 0 0 0 0 0 3 0 0 ~
w w - J U 00 o v 00 0 4 0 3 0-1 00 01 o m o m 0 0 0- m 3 0 0 0 0 m i . c ~ m I . . * * A UI o o o o c 0) - e I1 !I II II c w U
- E 0 c u a 7 - 0 3 c c - - c m
0 .
0
36
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
TR65-06
0 7
Y
c l i l Li
3
II
a
c u h z. a
U t
LL c3 Y
3 7 3 e!
z 3
m -
~ ~ ~ N N N N 0000000 I I I I I I I
37
I G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N I I I -
TR65-08
RING PROGRAM
Memory Required
Object Program (13523)*
Blank Common (10264)*
1/0 Buffers (112543
Time for Execution (sample problem)
9 min
Routines Required (those not included with IBM's IBSYS)
RING CIRCLE
GRHR FPTMOD
PMUY DATE
38
G M D E F E N S E R E S E A R C H L A B O R A T O R I E S @ G E N E R A L M O T O R S C O R P O R A T I O N
1
TR65-08 I
0 7
IT H
0 W
c3 _I
n a
i- o a
H
7 a
7 0
I-
E
(3 ii W 0
H
a a
+ + + + + +
m
0 0 m I I
w E
' + \
+ + + + + +
39