Embed Size (px)
Citation preview
) .. )
,_DEVELOPMENT OF INTERACTIVE COMPTJI'ER PROGRAl'iS FOR
MECHANICAL ENGINEERING DESIGNyFATIGUE ANALYSIS,
SECTION PROPERi'IES, AND BEAM ANALYSIS
by
Yiu. Wah,,Lukt'l
Thesis submitted to the Graduate Faculty of the
Virginia Polytechnic Institute and State University
in partial fulfillment of the requirements for the degree of
MASTER OF SCIENCE
IN
MECHANICAL ENGINEERING
APPROVED:
--~------------i;::::.::-------
L. D. Mitchell, Chairman
H. H. Mabie
________________ [;" ______ _ N. s. Eiss
May 1978
Blacksburg, Virginia
DEDICATION
To my beloved parents, and , and my brothers
and sister who have constantly encouraged me to pursue a higher
education.
11
ACKNOWLEtGEMENT
The author wishes to thank his major ad.visor
for introducing him to the field of computer-aided design for all his
advice and for his help in carrying out this research. This work could
not been completed without the assistance and careful guidance of his
major advisor,
The author also wishes to thank the other members of his committee,
for their valuable
suggestions.
111
TABLE OF CONTENTS
lJedica tion • •.•••••••••.•••• , .•••••• , ••••••••••• , ••••••••.•••••••• , •• ii
A ckn.owledgement • , •••••• , , •••• , ••••••••••• ', •••••••••••• , ••••••••• , •• iii
Table of Contents,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, iv
List of Tables ••••••••••••••••••••••••••••••••••••••••••••••••••••• vii
List of Figures •••••••••••••••••••••••••••••••••••••••••••••••••••• viii
Nomenclature ••••••••••••••••••••••••••••••••••••••••• ,,,,,,,,,,,,,, ix
Chapter One
1.0 Introduction................................................. 1
Chapter Two
2.0 Literature Review•••••••••••••••••••••••••••••••••••••••••••• 5
Chapter Three
3.1 Theory --- Fatigue Analysis.................................. 13 J .1 .1 Fatigue Failure Line. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 1.J
3.1.2 Parametric Method•••••••••••••••••••••••••••••••••••••••• 14
3.1.3 Significant Endurance Limit•••••••••••••••••••••••••••••• 17
3.1.3.1 Surface Factor••••••••••••••••••••••••••••••••••••••• 17
3.1.3.2 Size and Shape Factor•••••••••••••••••••••••••••••••• 18
3.1.3.3 Reliability Factor••••••••••••••••••••••••••••••••••• 20
3.1.3.4 Temperatu.-""'9 Factor••••••••••••••••••••••••••••••••••• 21
3.1.3.5 Fatigue Strength Reduction Factor •••••••••••••••••••• 21
3.1.3.6 Miscellaneous Factor•••••••••••••••••••••••••••••••••
iv
v
3.1.3.7 En.qurance Limit for a R. R. Moore Rotating beam
specimen.~ ......... ~.................................. 23
3.1.4 Significant Endurance Limit for finite life.............. 24
3.2 Theory --- Section Properties•••••••••••••••••••••••••••••••• 25 3.2.1 Polygonal Cross Section•••••••••••••••••••••••••••••••••• 25
3.2.z Circular Cross Section••••••••••••••••••••••••••••••••••• 35
3.2.3 Radius of Gyration••••••••••••••••••••••••••••••••••••••• 36
3.3 Theory --- Beam Analysis••••••••••••••••••••••••••••••••••••• 38 3.3.1 Derivation of Transfer Matrix•••••••••••••••••••••••••••• 40
Chapter Four
4.1 Discussion of Results
4.2 Discussion of Results
Fatigue Analysis •••••••••••••••••••
Section Properties •••••••••••••••••
51 51
4.J Discussion of Results --- Beam Analysis•••••••••••••••••••••• 53
Chapter Five
5.0 Conclusion................................................... 55
C~..apter Six
6.1 Recommendation
6,2 Recommendation
6.J Recommendation
Fatigue Analysis ••••••••••••••••••••••••••
Section Properties••••••••••••••••••••••••
Beam Analysis •••••••••••••••••••••••••••••
References......................................................... 60
Appendix A
A.1 User's Guide
A.2 User's Guide
A.) User's Guide
Appendix B
B.1 Program Listing
B.2 Program Listing
B.3.1 Program Listing
B.3.2 Program Listing
vi
Fatigue Analysis••••••••••••••••••••••••••• 65
Section Properties••••••••••••••••••••••••• 96
Beam Analysis•••••••••••••••••••••••••••••• 111
Fatigue Analysis••••••••••••••••••••••• 146
Section Properties••••••••••••••••••••• 164
Beam Analysis (Part!) ••••••••••••••• 184
Beam Analysis (Pa.rt II) •••••••••••••• 212
Vita. •••••••••••••••••••••••••• I ••••••••••••••••••••••••••••••• I •• 218
Abstract
LIST OF TABLES
A-1. Parameters for the selection of the desired fatigue failure
line. • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 67
A-2. Kececioglu factor••••••••••••••••••••••••••••••••••••••••••••• 68 A-J. Results of Example A-1 obtained by six fatigue failure lines.. 87
vii
1.
2.
4.
5. 6.
LIST OF FIGURES
Area of Triangle ABC ••••• • .•••••••••••••••••••••••••••••••••••
Coordinate Definitions --- Section Properties Program ••••••••
Trapezoid ABCD•••••••••••••••••••••••••••••••••••••••••••••••
Massless straight beam•••••••••••••••••••••••••••••••••••••••
Diagrams for Field and Point Transfer Matrices •••••••••••••••
A C&ntilever beam W1 th end. · lQ&d ••••••••••••••••••••• • ••••••••
26
28
29
41
44
48
A-1 to A-12. Output for Example A-1••••••••••••••••••••••••••••••••75-86
A-13. Flat steel spring under fatigue loading•••••••••••••••••••••• 90 A-14 to A-15. Output for Example A-2 ••• ••• •••••• •••••• ••• · •• ••••• ••• 94-95
B-1. A hollow hexagonal cross section•••••••••••••••••••••••••••••• 99
B-2 to B-6. Output for Example B-1·••••••••••••••••••••••••••••••101-105
B-7. L-shaped cross section with a circular hole ••••••••••••••••••• 107
B-8 to B-10. Output for Example B-2·•••••••••••••••••••••••••••••108-110
C-1. Cantilever beam with end load••••••••••••••••••••••••••••••••• 117
C-2 to C-9. Output for Example C-1·••••••••••••••••••••••••••••••118-125
.C-10. Statically loaded, statically 1ndeterm1nant beam. • • • • • • • • • • • • 12?
C-11 to C-13. Output for Example C-2•••••••••••••••••••••••••••••128-130
C-14. Complex beam loading and supports•••••••••••••••••••••••••••• 135
C-15 to C-19. Output for Example C-3·••••••••••••••••••••••••••••1)6-140
c-20. Dynamically loaded shaft ••••••••••••••••••••••••••••••••••••• 142
C-21 to C-23. Output for Example C-4•••••••••••••••••••••••••••••14-3-145
viii
NOMENCLATURE --- FATIGUE ANALYSIS
b = Kececioglu factor.
D = Dimension, in or m.
Ka = Surface finish factor.
Kb = Size and shape factor.
Kc = Reliability factor.
Kd = Temperature factor.
Ke = Fatigue Strength Reduction factor.
Kf = Miscellaneous factor.
Kt = Theoretical stress concentration factor.
L = Load line.
M = Moment, lb-in or N-J11.·
N =Number of cycles. I
n = Safety factor, dimensionless.
p = Pro 'ta bili ty' %.
p = Exponent on dimensionless alternating stress term, see F.q. 1 and
Table A-1.
Q = Notch sensitivity factor.
q =Exponent on dimensionless mean stress term, see Eq. 1 and Table A-1.
R1 =A variable, see Eq. 1 and Table A-1.
~ = A variable, see Eq. 1 and Table A-1.
r = Notch radius, in or m~
S = Strength of a material, psi or Pa.
se•=-Endurance limit for a R. R. Moore rotating beam specimen.
ix
x
Se'' =Significant endurance limit for infinite life, psi or Pa.
Se'''= Significant endurance limit for finite life, psi or Pa.
T =Torque, lb-in or N-m. 0 ~
t = Temperature, F or C.
Z =Section modulus, 1n3 or mJ.
ZR = Standard deviation of the endurance limit.
't = Shear stress, psi or Pa.
Subscripts 1 a = al terna. ting.
e = equivalent except 1n Se', Se'', or Se''' where it
indicates endurance.
m = mean.
u =ultimate.
y =yield.
Superscriptsa ' = indicates a load including a measure of overloading,
or a proportionality factor, usually n, safety factor,
NOMENCLATURE --- SECTION PROPERTIES
A = Area, in2 or mm2 •
b = Base, in or mm.
h = Height, in or mm. 4 4 I = Area moment of inertia, in or DUil •
i = Index number of a point. 4 4 J = Polar area moment of inertia, in or mm •
M = Sta tic moment of an area, in3 or mm 4 •
n = Number of points.
R = Radius of gyration, in or mm.
r = Radius of a circle , in or mm.
x = Coordinate of x-axis.
y = Coordinate of y-axis.
9 = Angle between the original axis and arbitrary axis.
cp = Angle between the original axis and principal axis.
xi
NOMENCI~TURE --- BEAM ANALYSIS
E = Modulus of Elasticity, psi or Pa.
G = Structural damping constant, dimensionless.
H = A constant for rotating shaft or rotor in vibration, dimensionless. 4 4 I = Area moment of inertia, in. or m •
i = Number of sections.
j =Mass moment of inertia, lb-sec or Kg-m2 •
L = Length, in or m.
M = Moment, lb-in or N-m.
m = Mass, lb-sec/in2 or Kg.
P = Concent:rated load, lb or N.
q = Magnitude of uniformly distributed load, lb/in or N/m • •
q1 = Magnitude of linearly varied distributed load, at the left end,
lb/in or N/m.
q2 = Magnitude of linearly varied distributed load, at the right end,
lb/in or N/m.
S = Slope , :radian.
V =Shear, lb or N.
W = Deflection, in or m.
w =Forced circular frequency, :rad/sec.
x = Variable length, in or m.
[P]= Point matrix.
[FJ= Field matrix.
[u]= Product of matrices.
[z]= State vector.
xii
xiii
Superscripts: L = Ieft.
R = Right.
CHAPTER I
1,0 INTRODUCTION
The computer is an essential tool that engineers have used to
increase the effectiveness and efficiency of their work. New
technologies have become feasible because of it. In order to get a
quick, accurate, and optimized solution to a complicated design
problem, there is little alternative but to use the computer.
Therefore, the computer and the designer should work as a. teami the
computer does the analysis and the engineer supplies the creativi~y
and decision-making components. This interactive process is called
computer-aided design. But the engineer is not using the computer as
f-requently nor as fully as he should. Currently, major coq_>uter-a.ided
design applications --- in order of use and market aize --- are:
electronics, drafting, cartography, architecture, and last and least,
engineering.1* If the computer were more readily available to the
engineer, and if the 19canned" programs were tailored to the more
general uses in engineering, the engineers would be more willingly
to use the computer.
*All numbers subscripted refer to references listed at the 'tack of
this thesis.
1
2
A large number of "canned" computar programs have been developed
for engineering as listed L'"l the Co~uter Program Abstracts2 in batch
form; i.e., using computer cards for input. In order to use these
programs, one has to have some knowledge of programming and also has
to know exactly what data have to be input card by card. Therefore,
one has to read pages upon pages of instruction manual before one: even
attempts to use such programs. One small error in the input, such as
wrong sequence of cards, data being put in the wrong location, or data
in the wrong format will invalidate the entire run. The interactive
computer system will eliminate this problem. Since all the instructions
are given to the users step by step via the screen or printout of the
terminal in a conversational form, all the user need to do is to
follow the simple instructions. The user enters the data by answering
the questions that are asked by the computer. Even one who has no
previous programming knowledge will be in a position to produce
quality work if the interactive program is written properly.
This thesis deals with three programs in mechamcal engineering
desigm Fatigue Analysis, Section Properties, and Beam Analysis. The
Fatigue Analysis program will size a mechanical component, circular,
rectangular or any shape, to prevent fatigue failure. User can select
one of the six most generally accepted fatigue failure linesa modified
Goodman fracture line, modified Goodman yield line, Soderberg line,
Gerber line, Quadratic line, and Kececiogl.u line. There is also a
routine built in to find the significant endurance limit provided that
the user can supply the value of the theoretical stress concentration
.'.3
factor along with other physical a.."ld envll'cnmental parameters. If the
theoretical stress concentration factor ia not known, the program will
supply a list of references where the user can find the.theoretical
stress concentmtion factor. The user also has to supply a subroutine
for the stress equations applicable to his problem where ariy equivalent
stress theory can be used.
The cross sectional properties program is designed to find twenty
different section properties, such as area, area moment of inertia,
and radius of gyration about different axis of any shape plane cross
section. Several computer and calculator programs exist for finding
these properties, but they are complex. Data entry erxori.s highly
probable and cross seetion :verliicat1on is· not generally: available for a
specified shape. But the program developed here can fini section
properties of any shape that can possibly be thought of as well as
graphically verifying its shape on an interactive basis.
The Beam Analysis program uses the transfer matrix method to find
and also plot the graphs. of defiection, slope, moment, and shear of
ariy continuous beam with ariy kind of loadings, such as uniformly
distributed load, concentrated load, concentrated moment, and so on.
This can be done provided that the beam is deflected within the linear,
elastic range. It will do static as well a.s forced, undamped, dynamic
response analysis.
All th.-ee programs provide the option of using English or SI uni ts.
The micro-processor used was a Teketrcnix, model 4051, with 32 K
memor/, a CRT output and a hard copy facility. The prog:ra.mming
4
language used was BASIC, an acronym for ~eginners All-purpose §ymbolic
Instruction Qod.e, and was developed at Dartmouth College, New Hampshire
by Professors J. G. Kemeny and T. E. Kurt.z under the terms of a grant
from the National Science Foundation in 1965'3. It is a high level
computer language that is easy to use and is applicable to scientific
and mathematical work.
The symbols used for the variables are different for each program.
They are defined in the nomenclature of the individual program.
CHAPTER II
2.0 LITERATURE REVIEW
A large number of .. canned .. computer programs have been developed.
Thei-r number is increasing enormously. They are mainly designed for
large computers, wr.1 tten in batch mode FORTRAN. Recently, some
progmms have been developed for the pocket size programmable
calculators. These applications are getting increasing attention
because of their advantages such as small size, low cost, and
accessibility to almost anybody. Owing to their size limitation, they
are used only for small program with small data storage. Sanderson3
in 1973 recommended the .. use of BASIC in engineering design, but very
seldom are engineering programs written in BASIC la.ngua.ge.
Mischke4' 5 used IOWA CAm:I' (Qomputer Augmented 12esign !ngineering 6 .
Iechnique) program as an illustration of computer application in
mechanical design. This program had a number of fatigue analysis
routines such as Marin's deterministic fatigue modification factors,
sigrMicant strength with generalized Gerber failu..'""'e diagram, and
fatigue damage. All the routines were written in FORTRAN.
For fatigue analysis, Hewlett-Packard Company? in 1976 had developed
one program that calculated aey one of the seven variables (yield
strength, significant endurance limit, cross sectional area, stress
concentration factor, maximum load, minimum load, and safety factor)
in the Soderberg's equation, provided that the nUlllerloa1 values of the
5
6
other six variables were known. This program used only Soderberg's
equation to size a mechanical component under fatigue, loading. This
program was designed for HP-67 or HP-97 series programmable calculators,
Shigley8 in 1976 provided the equations and £low-charts for six
short programs for programmable calculators. Only those that are
related to this thesis are described here, The program included
modified Goodman fracture line for prediction of a fatigue failure,
the log S - log N diagram to determine the significant endurance
limit for finite life when the life in number of loading cycles was
given or vice versa, and the theoretical stress concentration factor
for a rounded, shouldered shaft in bending. The program that computed
the theoretical stress concentration factor used the curve-fitting
. technique which converted the immense amount of experimental data to
a regression line form for the programmable calculator.
Pilkey and Ja;/] in 1974 had compiled a list of the existent
section properties programs. Programs included SASA, developed by
Structui'a.l Dynamics Research Corporation, Cincinnati, Ohio, which
calculated both cross sectional properties and stresses using the
finite element method proposed by Herrmann.10•11 Such properties as
moment of inertia, torsional constant, and warping constant could be
found for a cross section of any shape,
TRW Systems Group, STRU-PAK, Redondo Beach, Calli'ornia.9, had
developed three section properties programs: GENSECT1 , GENSEC'l'2, and
STANSECT. GENSECT1 used an integration technique to compute the section
properties of arbitrary plane cross section. The user had to input the
x and y coordinates for points which defined the outline of the section.
7
Voids in a section were handled similarily. GENSECT2 only computed the
section properties of a cross section ihat could readily be subd.ivded
into recta.ngl.es, right triangles, and circles. Voids were handled in
the same way and had to be composed of these same geometric figures.
The user had to divide the cross section into rectangles, right
triangles, and circles, specifying their dimensions and eentroids.
STANSECT calculated the section properties of eighteen types of
standard cross section.
AREA$$, developed by General Electric, Bethesda, Ma.ryland,9 computed
the section properties for any- aµ:ea bounded by straight lines and
circular arc segments. This program used a problem-oriented language
designed especially for geometric property calculations.
SECTION1, developed by United Computing Systems, Kansas City,
Missouri,9 calculated the area, centroid, moment of inertia, section
modulus, extreme fiber distance, and radius of gyration of any section
provided that it could be resolved into rectangles.
SECTION PROPERTIES, developed by North American Aviation, Canoga
Park, california,9 determined the section properties of any area
defined by peripheral coordinates. Areas were restricted to those
bounded by convex curves.
In 19'75, Pilkey12 developed BEAMSTRESS, a program which computed the
section properties and stresses for an arbitrar,r cross section of a 'tar
using the finite element method proposed by Hermann.10•11 The cross
section had to be modelled as an assemblage of quadrilateral finite
elements of any size. The accuracy of the results would depend upon
8
the fineness of these elements used. Properties computed included
area, centroid, moment of inertia, radius of gyration, shear center,
shear deformation coefficients, torsional constant, and warping
constant. If the internal shear forces, bending moments, axial force,
and axial torque were given, this program would compute the normal
and shear stress distribution.
Wojc1echowski13in 1976 developed a technique that replaced
integration by summation of finite elements to find the section
properties of a plane cross section. This technique applied only to
areas bounded by straight lines, but because curves could be
approximated by straight line segments, the method could be used on
any shape cross section. The equations for computing area, static
moment, and moment of inertia were derived.· This method could be
used by a programmable calculator. 14 Hewlett-Packard Company in 1976 developed a section properties
program using the same method developed by Wojciechowski13 but had
expanded it to include more properties, such as centroid, product of
inertia, and.moment of inertia about the centroid, the principal
axis and also an arbitrary axis. The x and y coordinates of the
vertices of the section were input sequentially for a complete,
clockwise path. Holes in the section were input in the same manner but
in a counter-clockwise path.
Using the transfer matrix method, P1lkey15 in 1969 developed a
generalized structural analysis program which could an&lyze static,
stability, free dynamic, and forced dynamic motion. The user had to
9
supply four subroutines: transfer matrix, initial parameter, time
dependent loading, and time dependent displacements subroutines for
the beam to be analyzed. Therefore, one had to be very familiar with
the transfer matrix method in order to be able to write these
subroutines. The output gave the deflection, slope, moment, and shear
along the beam.
Paz and Cassaro16 in 1974 presented the theory and described a
computer program in Fortran IV for the analysis of continuous beallltl
on discrete elastic supports using the five moment equations. This
analysis permitted direct determination of the redundant moments at
the supports of a continuous beam. This program accepted uniformly
distributed loads and any number of concentrated loads in the
cantilever or at any span of the beam. Output gave both bending moment
and reactions of the supports only. Two sample problems were
provided for demonstrating the capabilities of the program.
Pilkey a~ Jay17 in 1974 compiled a list of beam analysis programs.
SPIN, a program developed by Structural Dynamics Research Corporation,
Cincinnati, Ohio,17 calculated the critical SPeeds of rotating shafts
and the natural frequencies in bending of multispan beams of arbit:rary
cross section. It also calculated the response due to sinusoidally
applied forces. The deflections, bending moments, shear forces, and
stresses created by static forces could also be found. by forcing the
shaft to zero speed. SPIN used a distributed mass method for dynamic
analysis.
TRW Systems Group, STRU-PAK, Redondo Beach, California17, developed
10
two programs: MULTISPAN and STANBEAM. MULTISPAN used a static,
Euler-Bernoulli analysis of mult:tple span beams with no dynamic
motion. It allowed up to ten spans having uniform or piecewise
variable cross sections. In~span.supports were pinned; end supports
could be fixed, pinned, or free. STANBEAM allowed only static bending
analysis of single span beams with either unif'orm or piecewise
variable cross sections. Internal forces and displacements were
found using an integration procedure, Maximum shear and bending
stresses were calculated by the usual VQ/IE and MC/I formulas.
COM/CODE Corporation, Alexandria, Virginia17, developed LINKI, a.
beam analysis program which used a special purpose language called
LINKI. It allowed static, stability, and free dynamic analysis of
Euler-Bernoulli, Rayleigh, or Timoshenko beams and accepted inspan
supports, varlabl.e cross sections, and foundations.
Dow Engineering Company, Houston, Texas17 had a general two-
dimensional beam analysis program called GENERAL ANALYSIS which
could deal with static beams, beams on elastic foundations, and
simple frames. The method used was finite differences and outputs
were deflection, shear, and moment.
Southwest Research Iiistitute17 had two programs& DANAXXO and
DANAXX4. DANAXXO computed the frequencies and eigenvectors of a
beam with lumped mass using a stiffness matrix method of analysis.
Response due to static loads could also be analyzed. DANAXX4 computed
the time history of the response of a beam to applied force pulses
and.applied torque pulse which was represented by a lumped parameter
11
system. The program allowed any combination of hinged, clamped, free,
or guided flexural boundary conditions, but no damping was included.
The response was determined by a step-by-step integration of the
equations of motion using the linear acceleration method.
Pilkey12 in 197 5 developed another program called BF.AMRESPONSE
using transfer matrix method. Static, stability, and dynamic analysis
could be performed for beams of uniform or variable cross section
with any kind of loadings. Any number of in~span supports we.re
acceptable, including extension springs, rotary springs, rigid
supports, guides, shear releases, and moment releases. The program
calculated the deflection, slope moment, and shear for static and
steady ;state conditions, the critical load and mode shape for
stability, and the natural f.requencies and mode shapes for transverse
vibrations.
Hewlett-Packard Comparry18 in 1976 also developed your programs
for the beam. analysis used in their programmable calculators. All
four programs were designed only for static response with the mass
of the beam neglected. Each program dealt with only one type of beam.
The four types of beam that could be applied were cantilever beam,
simply supported beam, beam fixed at both ends and propped cantilever
beam. Each program calculated the deflection, slope, moment, and
shear at any specified point along the beam of uniform cross section.
Distributed loads, point loads, applied moments, or combinations of
all three might be applied. The equations used were based on elementary
beam theory and applied only to simple beams.
12
A report19 in 197.5 described an interactive computer program,
KINSYN III, for designing complex linkage. This program had both
synthesis and analysis capabilities. It relied on continuous
communication between the designer and the computer to arrive at a
final solution. This program was developed by the kinematicians at
Massachusetts Institute of Technology under the direction of Roger
Kaufman, who is now at George Washington University. TlE designer
created and altered a linkage on a "data. tablet", a graphic device
continually sensed by the computer. In turn the computer responded
via a display screen. In this way, the designer experimented with the
linkage design, while the computer guided and instructed him by
performing the required calculations and by flashing pertinent
design information on the display screen. Therefore, the designer
would know immediately if the design was possible.
Shore, Wilson and Semsarzadeh20 in 197.5 developed ST.ACRB, an
interactive computer program with graphical output which analyzed
horizontally curved and straight aligned bridge structures. This
program used the finite element modelling method and allowed the user
to: define the parameters that characterized the structures mj!)dify
an already discretized structure; and obtain graphical or tabular
results on various structural components such as structure geometry,
cross section geometry, and various results of analysis such as
deflections, reactions, and stress resultants.
CHAPTER III
J.1 THEORY --- FATIGUE ANALYSIS
This program computes the dimension of a mechanical component
whose cross section is either circular, rectangular, or expressable
as a function of one variable by proportions. The dimension computed
is the diameter, 1:f the cross section is circular; otherwise, it is
the dimension of the cross section which has all its other dimer.sions
described in terms of this basis dimension.
J.1.1 FATIGUE FAILURE LINE
When one analyzes a fatigue failure, there are a number of fatigue
failure lines available. This program provides the six most generally
accepted fatigue· failure linesa modified Goodman fracture line,
modified Goodman yield line, Soderberg line, Gerber line, Quadra~ic
line, and the Kececioglu line. Marin21 has developed the genera,l form
of the fatigue failure lines. It is given in modified form a~ follows:
= 1
Speci:fically, the exponents p and q control the fatigue line being
used. Table A-122 gives tm controlling parameters that select the
13
(1)
14
fatigue failure line desired in the design.
For the Kececioglu factor, b, it depends on the type of material
being used. From experimental data obtained by the application of 23 24 the prol:abilistic ''Design for Reliability" method, Kececioglu '
found the values of b for different materials. These are given in
Table A-2.
J.1.2 PARAMETRIC METHOD
When using this Fatigue Analysis program, the user has to supply
a subroutine program where any equivalent stress theory may be used.
The stress equations in this subroutine have to be derived using the
Pa:rametric method developed by Mitchell and Zinskie2S. In developing a fatigue design equation, it is necessary to solve
the simultaneous equations of two lines: the general load line and
the material fatigue safety failure line. This parametric method can
apply to both straight and curved load lines. The parametric pair of
equations which describes the general load line are of the forma
S' = f (L , n, D) a a a (2)
S' = f (L , n, D) m m m
as a function of the parameter, n. Alternately, n is the safety factor
as defined by Juvina.1126• It is redefined here ass
n = Maximum allowed externa.1 load Design external load
This safety factor, n, can also be considered as a proportionality
constant that signifies overload. It will scale the externally
applied loads if they are changed from the design load during the
loading process. Therefore, any load that is changed directly
proportional to the overload is a function of n and should be
multiplied by the proportional factor, n. These are the para.metric
equations of Eq. 2. Substituting F.q. 2 into Eq. 1, one gets Eq. 4
(J)
which is the generalized fatigue equation for a mechanical component.
a a m m _ ( f (L , n, D))P (R1 f (L , n, D))q \ ~ Se''' + Su - 1 (4)
To illustrate the concept of the parametric method, an example of
a rotating shaft subjected to a transmitted torque, T, and an applied
moment, M, is used. The shaft is assumed to be loaded such that, if
more torque is demanded, the bending moments applied to the shaft will
increase proportionally by the same factor, n. The design equation for
this case has been found.22 • 27, as
16
D 3£..n(M = T( Se'" + 0,8~ T) i/J (5)
Using the maximum distortion energy theory, the equivalent stresses
are:
.!.
= ( a2 + JC2i >2 ) i s =(s! +Jt;)2 = o.866 ~ em
s = (s! +Jt! Ji =(<~>2 + a2 )i :11 ea z
But upon overload, both the torque, T, and the moment, M, go up
proportionally by the same factor, n, that is&
T--_..,n T
thus S' 0,866 n T. em= Z '
and
and
M---4~,..n M
s' -· .n.11 . ea - Z
Equation 8 is the stress equation using the para.metric method, a.s
required for the subroutine.
(6)
(7)
(8)
Substituting Eq. 8 into F.q. 1 w-lth the para.meters R1, R2 , p, and q
set at 1, 1, 1, and 1 (see Table A-1), respectively for the modified
Goodman fracture line, one gets&
n M Se"' z + 0,866 n T = 1 Su Z
Substituting the section modulus for a circular cross section,
(9)
17 z = rrDJ
32 and solving for shaft diameter, one gets:
D = 3Trn{ M Se" I
which is identical to F.q. 5.
3.1.3 SIGNIFICANT ENDURANCE LIMIT
This routine for computing the sign1£icant endurance limit uses a
combination of data presented in: Section 6-13 to Section 6-22 of
Shigley28, Section 3-24 to Section J-29 · of Deutschman29, and C.hapter
three of SorsJ0•
The significant endurance limit is found by the eqiation,
Se" = Se' x Ka x Kb x Kc x Kd x Ke x Kf
J.1.J.1 SURFACE FACTOR, Ka
The data for the surface factor which is a function of the
ultimate tensile strength were taken from Fig. 6-2? of Shigely28 •
Five types of surface fir.ish are provided. For each type, an
equation in terms of the ultimate tensile strength is obtained by
polynomial regression. The equations used for each type of surface
finish are given below.
(10)
; ~.,
For polished finish, Ka = 1.0
For ground finish, Ka = 0.89
For machined or cold drawn,
18
Ka = -2.91 X 10-l7 SJ + 2 X. 10-1! ~ - 4.95 X 10-6 S + 1.064 u u . u
For hot rolled,
Ka = -5.77 x: 10-11 sJ + J.41 x. 10-11 if- - ax: 10-6 s + 1.066 u u u For as forged,
Ka = -6.45 x. 10-11 sJ + J.6J x. 10-11 if- - 7.8? x. 10-6 s + o.a9 u u u
{11)
where the ultimate tensile strength, Su' must be in psi. If SI units
are used, the program automatically converts the tensile strength to
English unit.
J.1.J.2 SIZE AND SHAPE FACTOR, Kb
The data for the size and shape factor, Kb, were taken from Fig. 42,
Part II of SorsJO.
The factor, Kb, is determined by size, shape, and material. Since
the size of the mechanical component is not known until the end of the . computation, this factor, Kb, has to be treat~ separately while all
the other factorsa Ka, Kc, Kd, Ke, and Kf are calculated first. The
dimension of the component is found by solving the simultaneous
equations of the general load line and the selected fatigue failure
line. An iteration process called the half interval search is used.
For each iteration, a new site factor and significant
19
endurance limit are computed. Using this value of the signi£icant
endurance limit, the simultaneous equations of the general load line
and the fatigue failure line can be solved by iteration. If the root
of these equations is found, then this set of size factor and
significant endurance limit are C.QrreCt for this design•
The cross section of the mechanical component must be either
circular or rectangular. Moreover, it must be made of steel or light
alloy, because only the experimental .data of these two are available.
Using a curve fitting technique, ~he data were converted to equations
by regression.31 • These equations are given below where the dimension,
D, must be in mm. If English units are used, the program automatically
corrects the dimension to SI units.
For steel,
Circular cross section,
If D<23, Kb = 1
- D If D>2J but<1JO, Kb - _18•75 +!,802 D
If D>t.30, ~ ~ 0.59 ·
Rectangular cross section,
If D<19,
If D>19 but< 150,
u n>150,
For light alloy,
Circular cross section,
Kb = o.aa Kb = 0.5061 + 7 .214
D
Kb = 0.55
(12)
(13)
20
If D<7, Kb = 1.0
If D>7 but <41, Kb = 0.515 + 3jj24
If D>41, ·-- Kb = 0.59
Rectangular cross section,
If D<7,
If D>7 but <47, .
If D> .47,:
Kb = o.aa Kb = 0.5061 + 2-n25
Kb = 0.55
3.1.3.3 RELIABILITY FACTOR, Kc
(14)
(15)
The reliability factor was computed using the method descr1,bed'in
Shigley28• This uses the equation
Kc = 1 - o.08 ZR (16)
where ZR is the standard deviation of the endurance limit. The data
of Table 6-2 in Shigley were fitted to a curve by regression. The
relation between the reliability and the proba.bility is given by
P = 100 - reliability
A regression on data in terms of l:ase ten logar.ltnm:. was used to
obtain the standard deviation, zR31 ,
(17)
21
ZR = 2.37 - o.885(log P) - o.193(1og P)2 - o.0502(1og P)J
- 0.00489(log P)4 (18)
The reliability factor, Kc, is then computed using Eq. 16.
J.1.J.4 TEMPERATURE FACTOR, Kd
The method used for.the temperature factor, Kd, is described by
Shigley28• From Shigley,
620 Kd = 460 + t
0
where t = operating temperature in F. 0 0
The above equation is used only when t> 160 F (71 C). '!be user
(19)
0 0 should be wamed that at operating temperatures above S?O F (JOO C),
creep and reduced yield strength may cause problems which are not
accounted for by Eq. 19.JO
3.1.3,5 FATIGUE STRENGTH REDUCTION FACTOR, Ke
The user must supply the theoretical stress concentration factor,
Kt. If the user doesnot know the theoretical stress concentration
factor, the references32' )J, 34 , 35 where tables or charts of Kt
are given, will be helpful.
The user may supply the notch sensitivity factor, Q, or he can
instruct the micro-processor to calculate it. The data for the notch
22
sensitivity factor were taken from Fig. B-2 on page 892 in Deutschma.n29.
Again, using regression, F.q. 20 is obtained. The notch radius, r,
must be in inches. The program automatically corrects the radius data
to English units if SI units are used.
Under bending or axial loadings
If S <50,000 psi (345 MPa) u Q =-8828 r 4 + 3345.3 .r'J - 440.94 r 2 + 24.62 r + 0.18
If s·<::6o,ooo psi (414 MPa) but~50,ooo psi (345 MPa) u
Q = -7031.25 r 4 + 2671.9 .r'J - 353.13 r 2 + 20.2 r + 0.28
Ii' s <80,000 psi (5.52 MPa) but:::=: 60,000 psi (414 MPa) u
Q = -10,156.25 r 4 + 3,825 r3 - 497.5 r 2 + 27.05 r + 0.23
Ii' S <::100,000 psi (696 MPa) but3 80,000 psi (5.52 MPa) u
Q = -15,057.38 r 4 + 5,165.4. r3 - 606 r 2 + 29.23 .r + 0.3
If S <140,000 psi (965 MPa) but~l00,000 psi (690 MPa) u Q = 5,431,250 r5 - 1,236,125 r 4 + 104,242.5 r3 - 4,010.7 r 2
+ 71.06 r + 0.33
If S~200,000 psi (1379 MPa)
Q = -271,319 r 4 + 37,276.5 r3 - 1,771 r 2 + 35.03 r + 0.67
For torsional loading,
If S <:60,000 psi (414 MPa) u
Q = -10,156.25 r 4 + 3,825 r3 - 497.5 r 2 + 27.05 r + 0.23
If s <80,000 psi (5.52 MPa) but~6o,ooo psi (414 MPa) u
Q = -15,057.38 r 4 + 5,165.4 r3 - 606 r 2 + 29.23 r + 0.3
If su<12o,ooo psi (827 MPa) but~8o,ooo :psi (5.52 MPa)
1 (20)
I I
2J
Q = 5,4)1,250 r:5 - 1,2J6,125 r4 + 104,242,5 r3 - 4,010.7 r2
+ 71 • 06 r + O. 33
If S <180,000 psi (1241 MPa) but ~120,000 psi (827 MPa) u Q = -271,319 r 4 + 37,276.5 r3 - 1,771 r 2 + 35.03 r + 0.67
For aluminum alloy (based on 2024-T6 data) 4 1 2
Q = -8,815.2 r + 3,411.J r - 462.64 r + 27.85 r + 0.013
After the theoretical stress concentration factor and the notch
sensitivity factor are known, the fatigue strength reduction faet.dr. :is
dbtained by
Ke = 1 1 + Q(Kt - 1) (21)
If the stress concentration factor is accounted for on the stress side
of the equation, Q should be set to zero to make Ke = 1.
3.1.3.6 MISCELLANEOUS FACTOR, Kf ·
This factor accounts for any miscellaneous effect that is not
considered above. If there is no miscellaneous effect, Kf will be·set
equa.1 to unity by the program.
3.1.J. 7 ENDURANCE LIMIT FX>R A R~. R'"~· MOORE ROTATING BEAM SPECIMEN, Se'
24
The calculation of' the· endurance limit for a R. R~. Moore
rotating beam specimen, Se', is described in Shigley28•
If S <:200,000 psi (1379 MPa), Se' = 0.5 Su u If s ~200,000 psi (1379 MPa) t se I = 100,000 psi u
J.1.4 SIGNIFICANT ENDURANCE LIMIT FOR FINITE LIFE, Se'"
(22)
For finite life, there are two methods to compute the significant
endurance limit, Se'''• One is the log S - log N approach and the
other is log S - linear N approach.
The equation for log S - log N approach is
Se"' =!OB (23) where B = (10~ N - 1) [1og Se" - log(0.9 Kd Se•)]+ log{0.9 Kd Se')
And the equation for log S - linear N approach is
Se''' = 0.9 Kd Se' + (10~ N - !)(Se'' - 0.9 Kd Se') (24)
When a life of greater than one million cycles is desired, the
in:f'inite life significant endurance lil'.:it is used.
J.2 THEX:lRY --- SECTION PROPERl'IE3
This method uses a technique that replaces integration by summation
of finite elements to find the section properties of a plane cross
section. Basically, the method divides a cross section into a series of
trapezoids or rectangles, then adds or substracts the properties of
the elemental areas to find the composite properties of the total
area. It applies only to area bounded by straight lines, but because
curves can be approximated by straight line segments, it can be used
on any shape plane cross section.
To speed up the computation process, the program uses different
methods· for ·.the ·circUJ.ar· cross section· and· holes. · It· xeqliires · the
radius and the x and y coordinates of the center o:f' the ci.-rclll.ar
se~tion or hole. It will give an exact result. The previous method
requires a large number of da.ta points in order to get a good
approximation of the curve, which in turn requires many computations.
J.2.1 POLYGONAL CROSS SECTION
The method for computing the section properties o:f' polygonal
cross section uses difference equations which replace integration by
summation of finite elements. To illustrate how this summation
method works, an example of an arbitrary triangular cross section ABC
shown in Fig. 1 is used. First the area of the trapezoid DFBC is
found. Then the trapezoid FBAE and EACD are substracted. And the net
25
-26
F _____ _.._._.._ - -
E
D -·-·----------.---------·----...-...---- C
Figure 1. Area of Triangle ABC.
27
result will be the area of triangle ABC.
Since there are a few axes involved, they are clarified with the
help of an irregular shape as shown in Fig. 2, where all the axes
are labelled.
All plane cross sections are made up of connected straight lines
(curved bol,llldaries may be approximated by straight line segments).
For each straight line, a trapezoid, a rectangle, or a triangle can
always be formed with the axis, like the shaped area ABCD, as shown
in Fig. J. The area of such a trapezoid can be calculated from the
bi.sic rules of plain geometry as
/lA = -(yi~ - yi)(xi:f1. + xi)/2
which is essentially the area of a trapezoid.
(2.5)
The general formula of the static moment of an area is defined as
My = J x dA (26)
where x is the distance between the centroid of the area and y-axis.
The static moment of /lMY of the trapezoid ABCD about the y-axis
can be considered to consist of contributions from the rectangle AEGD
and triangle EBF minus the contribution from triangle CFG. Using F.q.
26, the static moment /}.My can be found to be
x:i + xi:f1. 2 ~ - xi 11. /lMY = ( 4 )(yi:f1. - yi)(xi + xi:f1.)/2 + t(xi - 3( 2 ))
(Yi+J.2
- Yi)(xi; xiii)_ j(yi11.2
- Yi)(Xi ~ xi-+1)
(27)
I
where l I J I
x-y Crigina.1 axis.
28
y' I
I
Centroid (x·', y I) I -------- --- - - - - --- x·
XI I I
~x"
x'-y' Translated axis (axis translated to the centroid)
x ' '-y' ' Rotated, principal axis.
x:' ' '-y' ' ' Rotated, arbi tra.:ry axis (can be located at any point with any
angle of rotation) •
cp Angle between the translated axis and the principal axis.
9 Angle between the arbi tra:ry axis and the original axis.
Figure 2. Coordinate Definitions --- Section Properties Program.
29
id ABCD -:i Trapeze Figure ..1•
JO
which, after simplification and rearrangement, becomes
The centroid of an area with respect to x-axis is defined as
Dirlding F.q. 28 by the area gives
The formula for calculating the area moment of inertia with
respect to the y-axis is
Similiar to the static moment, the area moment of inertia, Iy' of
the trapezoid ABCD about y-axis consists of contributions from
(28)
(29)
(JO)
(31)
rectangle AEGD and triangle EBF minus contribution of triangle CFG.
To transfer all the area moment of inertia to the y-axis, the parallel
axis theorem is used.
I = I , + Ax2 y y (.32)
31
And the area moment of inertia of a rectangUlar and a triangular
sections with respect to the a."tis at their bases are given by
I of. rectangle = (b h3)/3
I of triangle = (b h3)/36
Using F.qs. 31, 32 and .33, the area moment of inertia, /i I , can be y
found as
x +x y -y x -x flry = - !(yi+! - yi)( i+12 i)3 + ~6( 1+12 1)( i+12 1)3
+.!.(Yi+! - Y1)(xi+! - xi)( _ g_(Xi+! - x-l))2 2 2 2 xi+! 3 2
_ 1-(yi+! - Y1)(i+1 - xi)3 _ .!.(Yi+! - Y1)(i+! - xi) .'.36 2 2 2 2 2
which, after simplification and xearrangemen~, becomes
(33)
(.'.34)
(J.5)
To find the total area, A, centroid, x', and area moment of inertia,
I , of a plane cross section, the F.qs. 25,30, and 3.5 are first solv~d y
for each line segment and then the individual results are summed, that
is
or n
A = -i~(yi-fi - yi)(xi-fi + xi)/2 (36)
x' = Lllx' or
x' = - tit((yiii - Y1)/8)({xii1 + x1)2 +(xiii - x1)2/J) (J?)
Iy =l:~ry or
Iy = -k({yiii - yi)(xiii + xi)/24)((xii1 + x1)2 + (xl."'1 - x1)2)
(JS)
Also, the area moment of inertia about the axis through the centroid
(x'-y') can be found by
I =I - Ax'2 y' y
(39)
which is essentially the parallel axis theorem that transfers the
area moment of inertia from the original axis (x-y) to the axis through
centroid (x'-y').
The equations for static moment, Mx' centroid y', area moment of
inertia with respect to x-axis, I , and area moment of inertia of the x axis through centroid, I , are found in the same manner. Their x formulas are given as Eqs. 40 through 42.
Static moment,.liM is given by x
Mx = J y clA
LlMX = -((xi-fi - x1)/8){{yi-f1 + Y1)2 + (yi+!- yi)2/J) (40)
33
y' = L: fly' n
y' = ! ~((xi-11 - xi)/B)((yi-11 + yi)2 + (yi-11 - yi)2/3) (41)
Area moment of inertia with respect to x-axis, I , is given by x
I =" flr x L x
The area moment of inertia with respect to the axis through centroid,
Ix'' is given by the pa:rallel axis theorem.
I , =I - Ay 12 x x (4J)
The product area moment of inertia with respect to the axis x-y
is given by
!icy =it (1/(xi-11 - xi))((yi-11 - y1)2(xi-11 + xi)(x~-11 + xi)/8
+ (y i -11 - Yi )(xi -11 Yi - xi Yi -11 )(x~ -11 + xi -11 xi + xi) /3 . 2
+ (xi-11.yi - xiy i-t1) (xi+1 + xi)/4) (44)
The product area moment of inertia about the axis translated to the
centroid (x'-y') is given by
I = I - Ax'y' x'y' xy (45)
The ailgle,4>, between the tra~ted axis
axis (x''-y'') is given by
(x'-y') and the principal
<P . 1 2I,, = t tan- (- x y )
I I - I ' x y (46)
The area moments of inertia about the translated, rotated, principal
axis (x' '-y' ') are given by
(47)
(48)
The arbi trar,y axis (x' ' '-y' ' ') is the axis that can be fixed
anywhere by specifying the x and y coordinates of the origin of the
arbitrary axis. The angle between the original axis (x-y) and the
arbitrary axis (x' ''-y' '') is e. The area moments of inertia about the
arbitrary axis (x'''-y''') are given by
Ix"'a =Ix' + A{y' - y"')2 (49)
I = I , + A (x' - x' • ')2 y' •'a Y (50)
I , , , , , , = I , , + A (x • -x'' • )(y • - y' '') x y a x y (51)
where x' ' ' a:r.d y' ' ' are distances from the arbitrary axis to the
original axis (x-y) •
35
Now, rotating to the angle e, the area momients of inertia and the
product area moment of inertia of the arbitrary axis become
I = I , , , cos29 2 - I , , , , , , sin 29 x'.' +I"' sin 9 x a Y a x y a (.52)
I - 2 2 + I , , , , , , sin 29 y' •• - I , , , cos 9 + I , , , sin 9 Y a x a x y a (53)
Ix'y' = t(I ,,, - I ,,, )sin 29 +I ,,, ,,, cos 29 x a ya x ya (.54)
The polar moment, J, a bout the arbitrary axis (x ' ' '-y ' ' ·• ) is
J=I ,., +I,., x y (55)
3.2.2 CIRCULAR CROSS SECTION
For circular cross section, the formulas are commonly known. They
are list~d below.
Area of circle is
A = 7trf- (.56)
The centroid is always at the center.
The area moment of inertia and product area moment of inertia with
respect to the axis translated to the centroid (x'-y') are
(57)
(.58)
I = 0 x'y' (59}
36
Then the area moment of inertia and product area moment of inertia
about any other axis are given by the parallel axis theorem. For the
original axis (x-y), they are given by F.qs. 60, 61, and 62.
I = I , + Ay'2 x x
I =I , + Ax'2 y y
I = I + Ax'y' xy x'y'
(60)
(61)
(62)
Now, all the other properties, such as angle <f:>, the area moment of
inertia about the translated, rotated principal axis (x"-y"), the
area moment of inertia about the arbitrary axis (x" '-y" '), and the
polar moment of inertia about the arbitrary axis can be calculated by •
the same equations as for the polygonal section, F.qs. 46-55.
J.2.J RADIUS OF GYRATION
Tm radius of gyration· of an area is the square root of the area
moment of inertia of that area divided by the area itself. TIE ref ore,
radius of gyration with respect to the original axis is given as
(63)
And the radius of gyration about the axis through the centroid are
given as
J.. r , = {I /A )2 x x'
.1. r , = {I /A)2 y y'
{64)
3.3 THEORY --- BF.AM ANALYSIS
The.b!t.sic philosophy 0£ the transfer_matrix method is l::ased on the
idea that a continuous and complicated system can be broken up into
component parts with simple elastic and dynamic properties that can be
expressed in matrix form. These component matrices are considered
building blocks that, when fitted together and evaluated with the
proper boundary conditions, will give the static and dynamic responses
of the entire system. A continuous beam can be considered to have a
number of elements linked together end to end in the fo:rm of a chaln.
F.a.ch element, represented by a transfer matrix, can be fitted together
as a system by successive matrix multiplications. This method is so
generalized that it can deal with any kind of continuous beam with al"'..y
combinations of loadings. This method has few restrictions• Howeve:;-,
the .system must be loaded. Within the elastic range. ·This
method is well documented in Chapter 3 of Pestel and Leckie36•
This program, using transfer matrix method, computes and also plots
th.e curves of deflection, slope, moment, and shear along the beam.
Static and forced, undamped dynamic analysis can be performed for beams
of uniform or variable cross section. Uniformly or linearly varied
distribu+.ed loads, concentrated loads, concentrated applied moments, or
combinations of all three may be applied. This program allows any
combination of pinned, fixed, free, or guided flexural boundary
conditions. A normally kinematically unstable condition can be
handled if sufficient internal elastic supports are provided. rn~~pe.n
39
support can be elastic springs and/or elastic moment springs •. Modelling
for dynamic responses uses lumped mass. Rigid in-span indeterminants are
not treated by this program.
For a given continuous beam with several sections, say i, each
element or section is represented by the appropriate field and point
transfer ma.trices. The state vectors from one em,(z] 0, to the other
end, [ Z Ji, .are related by the equati~n,
[z]i = [P]i [F]i [P]i-1 [Fli-1 ....... (P]j [F]j ....... [P]1 [F]1 [z]o
= (u] [z]0 (65)
where (F]j =a field transfer matrix that describes the jth section of
distributed stiffness with or without distributed loads.
[P] j =a point transfer matrix that describes. thejth element at
a point with no finite length.
The state vector, (z] , has five components defined ass
w [z] s = M
v 1
The boundary conditions are as follows:
For pinned end, W=O, M=O
For fixed end, W=O, S=O
For free end, M=O, V=O
For guided end, S=O, V=O
(66)
(67)
The field and point transfer matrices in Eq. 65 are known and the
40
boundar/ conditions of both· ends should be applied to &!,. 68.
(68)
Solution of F.q. 68 yields all the variables in the state vectors [z]0•
Once. (z]0 is known, the matrix multiplication process is repeated to
yield the states at each desired point along the beam. In fact, matrix
multiplications for many more stations can be determined the second
time through in order to get a better :resolution of the states within
the beam.
3.3.1 DERI.VATION OF TRANSFER MATRIX
Since the transfer matrices are the essential building blocks for the
systems, some of them will be derived.
Field Matrix for a Massless Bea.ma
Figure 4 shows a massless straight beam with two displacements; the
deflection W and the slope S, and the two corresponding forces, the
shear force V and the bending moment M. Using the equilibrium condition
requires that the sum of forces in vertical direction be zero ar..d also
the sum of the moments about point i-1 be zero, that is
or
LF = O, gives y
LM1_1 =O, gives
~ -v:-1 = 0
~ - r{_1 - ~ Li = 0
~ = ~-1 (69)
(70)
y
41
c~~::Js~-1 ~-1
Figure 4. Massless straight beam.
From elementary beam theory, for a cantilever beam with concentrated
moment at.its-·free.end, the eni deflection.and slope are. given as
ML2 w =--2EI
ML s = --EI
(71)
(72)
And for a cantilever beam with a .concentrated load at its free end,
the end.deflection and slope are
(73)
(74)
Using superposition and noting that the point 1-1 has an initial
deflection w1_1 and an initial slope Si_1 , the following equations can
be obtained.
(75)
(76)
Substituting Eq. 70 into Eq. 75, gives 2 3
~ _ R R Li R Li R i - w;._1-tisi-1 - 2(EI) Mi-1 - 6(EI) vi-1
i i (77)
Substituting F.q. 69 into F.q. 76, gives
_L R l -_R L; R S! = 5i-1 + (EI)i M""i-1 + 2(EI)1 v i-1 (78)
Equations 69,70,77 and 78 can be put in a ma.tri.i: form. Adding an
extension column, yields
4.3
w 1 -L -L - L.3 0 WR 2EI 6EI
s 0 1 ....L. _:£__ 0 s EI 2EI
M. = 0 0 1 L 0 M (79)
v 0 0 0 1 0 v
1 i 0 0 0 0 1 1 i-1
F.quation 79 is the field matrix for massless beam, see Fig • .sa.
The other field matrices are derived in a similiar way making use of
elementary beam theory, other elastic a.nd dynamic properties. Other
ma.trices are listed below.
Field matrix for a massless beam with uniformly distributed load, q,
see Fig. ,5b.
2 =2_ 4 1 -x ---3,_ gx
2EI 6EI 24 EI 2 -~ 0 1 .....L x
EI 2EI 6 EI 2
0 0 1 x - 92' (80) 2
0 0 0 1 - qx
0 0 0 0 1
Field matrix for a massless beam with linearly varied distributed
load, q1 , ~·see Fig. 5c, .5d.
44
q
I ••
1--L (a) {b)
(d) (e)
f:::...-,.. .. J S sin wt
T
r (g) (h)
I I
-{c)
M
" --0--.
(f)
==O== i V sin wt
' (1)
Figure 5, Diagram for Field and Point transfer ma.tries.
45
2 -2- 4 q XS q XS 1 -_L_ ~- 1 + 2 -x 2EI 6EI 24EI 120LEI 120LEI
x2 3 4 4
0 1 ..A.. ~ q1x q2x EI 2EI 6EI + 24LEI 24LEI
2 q1x3 q XJ -q x 0 0 1 1 2 (81) x 2 + 6L - 6L
2 2 0 0 0 1
qlx q2x -q1x + 2L 21.
0 0 0 0 1
The field ma. trix of Eq. 81 , when qt = q2 , will become the same as
the field matrix for a massless beam with uniformly distributed load.
Point matrix for a concentrated load, P, see Fig. Se.
1 0 0 0 0
() 1 0 .o 0
0 0 1 0 0
0 0 0 1 -P
0 0 0 0 1
Point matrix for a concentrated moment, M, see Fig. Sf. 1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
-M
0
1
(82)
(83)
46
Point matrix for an elastic spring support with stiffness K, see Fig • .5g.
1
0
0
K
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
·o
0
0
0
0
1
(84)
Point matrix for an elastic moment spring support with moment stiffness
T, see Fig. 511.
1
0
0
0
0
0
1
T
0
0
0 0 0
0 0 0
1 0 0 (85)
0 1 0
0 0 1
Lumped mass is used for modelling of dynamic response. Point matrix for
a lumped mass, m, with frequency w, see Fig • .51.
1
0
0
2 -mw
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
(86)
An overall point transfer matrix can be obtained by combinµtg .all. the
point matrices listed above and also adding in the moment term, (HJw2).
47
This term.'.36 describes the gyroscopic or rotatory L~ert.ia effects of a
beam or rotor in the dynamic case. ·The combined_ point_ transfer
matrix becomes
1
0
0
2 K - mw
0
0
0
0
0
0
1
0
0
where H= -1 for bending vibration.
0
0
0
1
0
0
0
-M
-P
1
(87)
H= i1. for rotating shaft (equal angular direction of" whirl and
rotation).
H= -3 for rotating shaft (opposite angular direction of whirl and
rotation).
If any element of the combined point transfer matrix is not needed, that
particular element should be set to zero to give the required point
transfer matrix.
An example of a simple cantilever beam with a concentrated load at
its free end, shown in Fig. 6, is used to illustrate the transfer
matrix method. 'This same example is done numerically by this program.
It is shown together with more complex examples in the user's guide in
Appendix A • J.
There is one field and one point transfer matrices L'"l this proble111.
The field matrix is for the massless beam and the point matrix is for
the concentrated load. The eql,l&tion that describes this beam is
48
p
,,. , r
- u 1 , , L ,
"-"
[ ~ 1 [P 11
Figure 6. Cantilever beam with end load.
49
(88)
where [P] 1 is the point matrix shown in F.q. 82, and [FJ1 is the one
shown in F.q. 79. Substitution yields
w 1 0 0 0 0 1
s 0 1 0 0 0 0
M=O 0 1 0 0 0
v 0 0 0 1 -P 0
1 1 0 0 0 0 1 0
or
w -L2 -L''.3 2EI 6EI 1 -L
_k_ L2 EI 2EI 0 1 s
M = 0 0 1 L
v 0 0 0 1
1 1 0 0 0 0
-L2 -L 2EI
1
0
0
0
0
0
0
-P
1
l EI
1
0
0
w
s
M
v
1 0
The boundary conditions at the ends are
0 w
0 s
L 0 M
1 0 v
0 1 1 0
(89)
(90)
Applying the boundary conditions to Eq. 89 gives the 1110ment and shear
at point 0 as
MO =-PL
v = p 0
(91)
"·
Substituting the values of state vectors, w0, s0, M0, and v0 at point O
into F.q. 89 gives the state vectors at point 1.
w1 PL3 PL3
= 2EI - 6EI PL3
= 3EI
s1 -PL2 PL2 PL2
=~+2EI :.--EI (92)
M1 =O
v1 = 0
The intermediate results of deflection, slope, moment, and shear along
the beam can now be computed using F.q. 89, where L is replaced by the
variable length, x, Along._ with. F.qs. ·90. and 91, the results are
p 2 W = 6Ex (3L - x) x I
Sx = & (x - 2L)
M = P(x - L) x
V· = p x where 0 <x <L.
(93)
Using F.q. 93, the curves of deflection, slope, moment, and shear can
be plotted as a filnction of the length of the beam.
For the forced, undamped, dynamic case, a forcing circular frequency,
w, has to be given. The same analysis procedure can be followed. It is
similiar to the static case because static case is just a. special
dynamic case with the forcing circular frequency equal to zero.
CHAPI'ER IV
4.1 DISCUSSION or RESULTS --- FATIGUE ANALYSIS
The application of this program is unlimited. It can be applied to
shaft, bolt assembly, spring, and other problems that need a
fatigue analysis. The component can be either circular, rectangular,
or any shape, provided that the dimensions are all given 1n terms of
the height of the cross section. Any equivalent stress theory can be
used in the analysis. Any one of the six fatigue failure lines,
which are the most generally accepted 1n the present, can be chosen.
One can even use all six fatigue failure lines one by one and compare
the results obtained from different theories.
Tre significant endurance limit is computed by using the known
experimental data presented 1n the form of graphs. Since all the
experimental data have been converted into equations by regression
analysis, the computation of the significant endurance limit is
easily accomplished.
4.2 DISCUSSION OF RESULTS --- SECTION PROPERI'IES
Finding the sectional properties of an irregularly shaped cross
section is a tedious task. This program provides a means of
calculating twenty sectional properties of any shape cross section
describable by straight lines. It is especially useful for
extruded shapes which are usually not simple, but a:re frequently
51
.52
encountered in; the design. where· volume and low· cost.is desired. An
example of a hexagonal extruded bar is shown in the user's guide of
this program in the Appendix A in Section A .2.
This program treats the polygonal section and circular section
separately. For a polygonal section, the method of replacing
integration by summation of finite elements is used to find the
sectional properties. Although this method can apply to circular
section by approximating the curves with straight line segments, it
is time consuming and difficult to get an accurate result. Therefore,
to speed up the computation, to make the input easier, and to obtain
a more accurate result, the properties of circular cross section and
circular holes are computed using the ordinary formulas of area and
area moment of inertia about the centroid of a circle. Tm area
moment of inertia with respect to other axis is computed using the
parallel axis theorem. In this way, a circular section or a circular
hole inside a cross section ma.y be input by giving only the radius
of the circle, and x and y coordinate of the center. This is done
instead of approximating the curves by straight line segments and
inputting the x and y coordinates of each line.
Another special featUX"e of this program is that it provides a
graphical verification of the input cross section in addition to the
data list. This is especially useful in checking to see whether the
holes are in the correct position and completely within the perimeters
of the section. One good example of this error can be found in the
Hewlett-Packard14 page 02-10, example 4 in the Section Properties
program. The original drawing of the L-shaped cross section with a
5'.3
circular hole is shown in Fig. B-7. If one examines the location of
the hole carefully, one will find that the hole, with the given
dimensions, is actually located partly outside the perimeter of the
cross section instead of inside as shown in the figure. The error can
be elim.1nated if the cross section is drawn to scale first. This
obvious mistake can be observed and rectified. The corrected drawing
together with the results are given in the user's guide in the
Appendix A .2 I
This program also computes the area moment of inertia about an
arbitrary axis, that is, this axis can be located at any specified
point with any angle of rotation. This is very useful in design.
4.J DISCUSSION OF RESULTS --- BEAM ANALYSIS
This is a generalized beam analysis program which can analyze
any kind of beam with any loadings. Even static indeterminate beam
or beams that are normally kimatically unstable but hAve sufficient
internal supports can be analyzed. The different types of beam that
can be analyzed are unlimited. A few exa.aples are given in the user's
guide in Appendix A in Section A.). This program has a list of sixteen
different boundary conditions that cover all of the boundary cases.
The output includes a drawing of the beam and its loadings together
with a list of input data as a check. If any input error is present,
corrections can be ma.de before going into computation. The deflection,
slope, moment, and shear along the beam are given in numerical valur,s.
These are also given in the form of graphs. A general feeling of how
the beam behaves can easily be obtained for this output. Completion
of such problem by hand can be very time consuming and difficult
because of so many. loadings...involved.
This program can also analyze a beam with different cross section
and flexural stiffness in each section by inputting the different
area moment of inertia and modulii of elasticity for each section.
CHAPI'ER V
5.0 CONCLUSION
The interactive system when properly programmed provides the
engineer with in£ormat1on regarding a proposed design. This is done
by providing results of an analysis in a graphic, comprehensible form
and at a much faster rate than previously achieved through l:atch
operation. Interactive computing is also easier to use especially by
those who have no previous programming knowledge. Also, the use of
graphical verification of the input data provides a easy check before
the computation takes place, in this way, a correct result can be
assured •.
Another advantage is that the designer is able to exercise greater
control of his computer-bl.sed system, and to execute a design or an
analysis much faster. Apart from obvious savings in time, he is also
able to consider alternative designs in greater depth and to achieve
a greater degree of optimization. Thus, economies in all respects of
the design should result.
55
CHAPTER VI
6.1 RECOMMENDATION --- FATIGUE ANALYSIS
The following points are recommended to be added to improve this
program in the futures
1. The program now will only find the dimension of a mechanical
component which will prevent fatigue failure, but, in some
occasions, a safety factor of the component may be desired. Thus
the future program should provide the option to find either the
dimension or the safety factor of the component to prevent
fatigue failure.
2. A tutorial section designed to help the user to write the
subroutine of stress equations in Parametric for.a should be
included. Step-by-step simple instructions and examples for
illustration would be helpful. This way, the user Who. is not
familiar with the Parametric method needs not to refer to the
user's ~;uide and its appendix. Thus, this program could stand alone.
Thereby, the full assets of the interactive system could be
utilized.
3. In the present program, the finite life region of the S-N curves
is fixed between 0.9 Su@ 1 x 1oJ cycles to Se'' @ 1 x 106 cycles.
The future program should be able to have this region be varied, to
suit some purposes.
4. Moreover, it is sometimes desired to determine the life of a given
component at a specified safety factor. This analysis could be
completed by fixing the safety factor and dimension, and sclving
for the significant endurance strength 0£ fhe .component •. This.
significant endurance strength can be used in the S-N equation to
predict life.
5. A separate program should be considered to deal with the design
under cumulative damage situations.
6. This program should be modified or a new program should be
generated to replace the safety factor with a proeability of failure.
Statistical interference theory should be utilized.
7. A modification to the method should be considered so that different
stress concentration factors may be used on different type stresses.
8. The theoretical stress concentration factor has to be supplied by
the user in the present program, but with the help of graphics and
more computing efforts, this factor can be computed with the user
providing the necessary geometric information. This will increase
the usefulness of the program.
6.2 RECOMMENDATION --- SECTION PROPERTIES
The twenty sectional properties obtained by this program are quite
sufficient for general design, but another useful property --- tne
location of shear center should be added to the future program. In the
case of some non-symmetrical cross sections, like a channel section
which has only one axis of symmetr;y, the plane of the bending moment
must pass through the shear center if twisting of the section is to be
avoided. This problem usually arises when a thin open section is used
and under these conditions, local failure or buckling may happen.
Another program should be considered which will compute the mass
moments of inertia of three dimensional configu.""'8.tions. A generalized
three dimensional version of this should be considered for such
computations.
6.J RECOMMENDATION --- BEAM ANALYSIS
For this program, the following points are recommended for
improvement:
1. For the dynamic case, plots of amplitude and the phase angle of
deflection, slope, moment, and shear as a function of exciting
frequency should be included. In this way, the critical frequency
can be recognized and avoided in the design.
2. In-span .. indeterminates in the form of fixed support shoW.d be
allowed. These are not provided in the present prognilll. This
feature, although such in-spa.n.indeterminarits exiSt only··theoreti-
cally, will enlarge the application of this program. It can be
approximated by assigning a large value for the stiffness of the
elastic spring support; e.g., 1 x 1020 lb/in (1.13 x. 1019 N/m).
This will give a result that approaches the fixed support case.
J. The present progzam provides for the analysis of forced, undamped
59
dynamic response, but a damped dynamic case is desirable aome.~
times. The addition of this feature will require new complex
transfer matrices for t.he beam and its loadings, which can be
derived by replacing EI by EI( 1 + jG), where j is the imaginary
number and G is the structural damping constant. A new scheme of
multiplying the complex matrices and also new boundary conditions
are needed,
REFERENCES
1. "CAD/CAMi The Computer in Action"• Mechanical Engineering, Vol. 100,
No. 2, Feb., 1978, PP• 105-106.
2. Computer Program Abstract, Washington, U. s. National Aeronautics
and Space Adminstration, Office of Technology Utilization, prepared
by the Technical Information Services Co., July 15, 1969 - present.
J. Sanderson, P. c., Interactive Computing with BASIC: An Introduction
to Interactive Comouting and a Practical Course in the BASIC Lar..guage •
1st Ed., Petrocelli Book, New York, 1973.
4. Mischke, c. R., An Introduction to Comouter-aided Design, Prentice-
Hall Inc. , Englewood Cliffs, N. J. , 1968.
5. Mischke, C, R., "Organizing the Computer for Mechanical Design",
Design Technology· Transfer, ASME., 1974, pp. 51-64.
6. Mischke, c. R., IOWA CADET Documentation Volume, Engineering Research
Institute Project 628 1 Computer-aided Design, Iowa State University,
Ames., 1973.
7. User's Manual of Mechanical Engineering, Pac I, for HP-6?/HP-97
Programmable C3.lcu1ator, Hewlett-Packard Company, Corvallis, Oregon,
197 6 ,' pp~ 04-01 to . 04-04.
8. Shigley, J. E., "Mechanical Design and the Programmable Calculator",
ASME Paper 76-DET-91 for meeting Sept., 26-29 1 1976
9, Pilkey, W. D. and Jay, A • , "Structa.""8.1 Members and Mechanical
Elements", Structural Mechanics Computer Prom.ms. University Press
of Virginia, Charlottesville, Vmµiia, 19741 pp. 556-562.
60
/
61
10. Herrmann, L. R., "Elastic Shear Analysis of C~neral Prismatic Beams",
Journal of the Engineering Mechanics Division: ASCE, Vol. 94, No.
EM4, Aug., 1968, PP• 965-983.
11. Herrmann, L. R., "Elastic Torsional Analysis of Irregular Shapes",
Journal of the Engineering Mechanics Divisions ASCE, Vol. 91, No.
EM6, Dec., 1965, PP• 11-19.
12. Pilkey, w. D., Formulas and Comouter Routines for Stresses. Stability
and Vibrations --- Bars· Beams. Plates. and Shells, The Structural
Members User Group, C!harlottesville, Virginia, 1975.
13. Wojciechowski, F., "Properties of Plane Cross Sections", Machine
Design, Vol. _48,. No• ·1, Jan.,. 22·,. 1976, PP• 105-107.
14, Reference 7, pp. 02-01 to 02-10.
15. Pilkey, W. D., Manual for the Resoonse of Structura1 Members, Vol.
.L ITT Research Institute, Chicago, Illinois, Aug. 1969.
16. Paz, M. and Cassaro, M,, "Analysis of Continuous Beams on Discrete
Elastic Supports by the Five Moment Equation", Computer and
Structure, Vol. 4, Oct,, 1974, pp. 1005-1023,
17. Reference 9, pp. .541-.547.
18. Reference 7, PP• 05-01 to 08-05.
19. "Mechanical Design: Let the Computer do it", Ma.chine Design,
Vol. 4?, No. 24, Oct;., 20, 1975, PP• 174-176.
20. Shore, s., Wilson, J. L. and Semsarza.deh, G. A., "Interactive
Techniques with graphical output for Bridge Analysis", Computer
Methods in Applied Mechanics and Engineering, Vol. 5, No. 2, North-
Holland. Publishing Ca., Amsterdam, 1975.
62
21. Marin, J., "Design for Fatigue Loading, Part 3", Machine Design,
Vol. 29, No. 4, Feb., 21, 1957, PP• 128-1)1, and series of same
title Parts 1-5,.Ma.chine Design, Vol. 29, No. 2-6, Jan., - April,
1957.
22. Mitchell, L. D., and Vaughan, D. T., "A General Method for the
Fatigue-Resistant Design of Mechanical Components, Pa.rt 2,
Analytical," Journa.l of Engineering for Industry, .Vol. 97.', .. _Series B,
No. J, Aug., 1975, PP• 970-975.
23. Kececioglu, D. B., and Chester, L. B., ''Fatigue Reliability Under
Combined Mean a.nd. Alternating Axial Stresses for AISI 1018 and
10)8 Steels", ASME Paper No. 75-DET-128.
24. Kececioglu, D. B. and Chester, L. B., "Combined Axial Stress Fatigue
Reliability for AISI 4130 and 4340 Steels", ~ Paper No. 75-WA/DE
-'!?. 25. Mitchell, L. D. and Zinskie, J. H., "A Man-Machine Interactive
Method for the Development of Fatigue Design Equations", Unpublished
paper, available from Prof. L. D. Mitchell, Department of Mechanical
Engineering, Virginia Polytechnic Institute and State University,
Blacksburg, Virginia.
26. Juvinall, R. c., Engineering Consideration of Stress. Strain, and
Strength, McGraw-IU.11,_ New .York,: N; Y., 196;, Se_c. 14.13,
pP ·- 29~297.
27. Lehnhoff, T. F., "Sha.ft Design Using the Distortion Energy Theory",
Mechanical Engineering News (MEN), Vol. 10, No. 1, Feb., 1973,
PP• 41-4J.
63
28. Shigley, J. E., Mechanical Er.gineering Design, 2nd Ed., McGraw-Hill,
New York, N. Y., 1972.
29. Deutschman, A. D., Michels, W. J., Wilson, c. E., Machine Des!gn,
Macm1J Jan Publishing Co., Inc., New York, N. Y., 1975.
30. Sors, L., Fatigue Design of Machine Components, Pergamon Press,
Oxf o:rd, England, 1971.
31. •Jones,. D. H., "Significant Endurance Limit --- Se'•', Computer
Program", Term Paper of ME 4080, available from Prof. L. D.
Mitchell, Department of Mechanical Engineering, Virginia Polytechnic
Institute and State University, Blacksburg, Virginia.
,32. Peterson, R. E., Stress Concentration Factor, John Wiley & Sons,
New York, N. Y., 1974.
33. Shigley, J. E. , Mechanical Engineering Design, 3:rd -Ed.., McGraw-Hill,
New York, N. Y., ~ pp. 663-670.
J4. Reference 29, pp. 894-901.
JS. Reference 30, pp. 42-84 of Part II.
36. Pestel, Edua:rd c., and Leckie, F. A., Matrix Method.s 1n
Elastomechanics, McGraw-Hill, New York, N. Y., 1963. •
APPENDIX A
USER'S GUIDES
64
A .1 USER'S GUIDE --- FATIGUE ANALYSIS
INTRODUcrION
Machine members a.re often found to have failed under the action
of repeated or nuc.tuating stresses, and yet an analysis reveals that
the actual maximum stresses were below the ultimate strength of the
material and quite frequently even below the yield strength. These
failures usually are under stresses repeated for large number of
times. This failure is called a fatigue failure.
There are many theories that predict the fatigue failure. The six
most generally accepted ares modified Goodman fracture line, mod.ifiod
Goodman yield line, Soderberg line, Gerber line, Quadratic line, and
Kececioglu line. These six fatigue failure lines are available in the
program to size a mechanical component, circular, rectangular, or
any shape, to prevent fatigue failure. Any equivalent stress theories
are allowed. It can al.so compute the significant endurance limit with
the theoretical stress concentration factor and other physical and
environmental parameters supplied by the user.
The program is written in BASIC and runs on a Teketronix micro-
processor model 4051 with J2K memory.
THEORY
When one analyzes a fatigue failure, there are a number of fatigue
6.5
66
failure lines. The general form of the fatigue failure line is
where Sa = al term. ting stress·, psi or Pa.
S = mean stress, psi or Pa. m S = ultimate tensile strength, psi or Pa. u
Se' ' ' = significant endurance limit, psi or Pa.
R1 ·~ =variables depend on the fatigue failure line being
selected, see Table A-1.
p, q = exponents control the fatigue failure line being
selected, see Table A-1.
(A-1)
Specifically, the exponents p and q control the fatigue failure line •
being used. Table A-1 lists the controlling para.meters that select
the desired fatigue failure line.
For the Kececioglu line, a Kececioglu factor, b, has to be supplied.
This factor depends on the type of material being used. From
experimental data obtained by the application of the prol:a.bilistic
"Design for Reliability" method, Kececioglu found the values of b
between o.8914 and 1.0176, see Table A-2.
A subroutine program of stress equations where any equivalent
stress theory may be used has to be supplied by the user. The stress
equations must be written using the Para.metric method described in the
paper, ·~ Man-Ma.chine Interactive Method for the Development of
67
Table A-1. Paxa.meters for selection of the desired fatigue failure line.
Type Fatigue Failure Line p q R1 R2
Modified Goodman fracture line 1 1 1 1
s s Modified Goodman yield line 1 1 __lL -1L s Se I It
y s
Soderberg line 1 1 u 1 s y
Gerber line 1 2 1 1
Q.uadi'a.tic line 2 2 1 1
Kececioglu line b* 2 1 1
*See Table A-2
68
Table A-2. Kececioglu factor.
Type of Diameter Brinell Tensile Signif'icant Kececioglu Steel of Hardness Strength Endurance factor
Specimen Number Limit
AISI inches BHN psi psi b
1018 0.375 130 60,400 26,200 0.8914
1038 0.375 164 68,600 31,000 0.9266
4130 0.375 207 106,100 39,700 1.0176
4340 0.375 233 116,400 48,700 0.9685
AISI mm BHN MP a MP a b
1018 9.53 130 416.44 180.64 o.8914
1038 9.53 164 472.98 213.74 0.9266
4130 9.53 207 731.53 273.72 1.0176
4340 9.53 233 802.55 335.77 0.9685
li'a:tigue Design Equa.tions",.'by L. D. Mitchell and J. H. Zinskie.25
The ta.sic concept of the Parametric method is to introduce a
proportionality factor, n, to any loads that will change during the
process of overloading. This factor n can be considered as a
proportionality constant that signifies overload, so any load that is
affected directly and in the same proportion to the overload is a
function of n and should be J11ultiplied by n.
The fatigue failure equation and the stress equations are solved by
Half Interval Search. This method converges slower but it avoids
difficulties which other methods, like Newton's method or false
position method, may encounter. It searches for sign changes of the
equation. The interval between sign changes will be continuously
halved until the desired accuracy or the tolerance of the root is
reached.
The significant endurance limit is found by the equation,
Se'' =Se' x Ka x Kb x Kc x Kd x Ke X. Kf (A-2)
where Se•' = significant endurance limit for infinite life, psi or Pa. -Se' = eridu:ta.nce·limft· of a, R ... R~ Moore specimen,
psi or Pa.
Ka = surface factor.
Kb = size and shape factor.
Kc = reliability factor.
70
Kd = temperature factor.
Ke = fatigue strength reduction factor.
Kf = miscellaneous effects factor.
The correction factors, Ka, Kb, Kc, Kd, and Ke are obtained from
material presented in Section 6-13 to Section 6-22 of Mechanical
Engineering Design, 2nd Ed., McGraw-Hill, N. Y., 19?2, by J. E. Shigley,
Chapter three of Fatigue Design of Ma.chine Components, Pergamon Press,
19?1, by L. Sors, and Section J-23 to S;ection J-29 of Machine Design,
Ma.caillan Publishing Co. Inc., N. Y., 19?5, by A. D. Deutschman, w. J.
Michels, and c. E. Wilson. The experimental data of the correction
factors are converted into regression lines, from which t~se factors
are computed. After the correction factors are obtained, the
significant endurance limit for infinite life, Se'', is computed
using F.q. A-2. 'Then it is corrected for finite life to obtain the
significant endurance limit for finite life, Se''', if required.
INPUT DATA ~UIRED
Units can be either in English or SI system. This depends on the
option chosen by the user. The following data are needed.
1. Ultimate tensile strength of the material, psi or Pa.
2. Yield strength of the material, psi or Pa.
J. Significant endurance limit, psi or Pa.
4. Moment causing alternating stress, lb-in or N-m.
71
5. Moment causing a steady stress, lb-in or N-m.
6. Alternating ax±al force, lb or N.
7. Steady axial force, lb or N.
8. Alternating torque, lb-in or N-m.
9. Steady torque, lb-in or N-m.
10.Safety factor, dimensionless.
11.Lower limit of the dimension, in or m.
12.Upper limit of the dimension, in or m.
If the significant endurance limit is not known, the following
physical and environmental para.meters are required for computing the
significant endurance limit.
1J.Type of surface finish.
14.Reliability, %. 15.0perating temperature, • 0 F or c. 16.Theoretical stress concentration factor.
17.Notch sensitivity factor (optional).
18.Notch radius (optional), in or m.
19.Type of material.
20.Type of loading.
21.Miscellaneous effect factor (optional).
22.Shape of cross section.
2J.Endw:ance limit for R. R. Moore· rotating beam
specimen (optional), psi or Pa.
24.Number of cycles.
72
EXAMPLE A-1
A rotating shaft is subjected to a transmitted, steady torque of'
1,200 lb-in (135.58 N-m) and an applied stationary moment of' 2,400
lb-in (271.16 N-m). Overload is caused by an increase of' torque and
moment. The saf'ety factor is 1.8. The shaft is circular, ma.chined, and
made of steel with tensile strength of 200,000 psi (1.397 X 109 Pa)
and yield strength of' 150,000 psi (1.048 X 109 Pa). The shaft has a
notch radius of 0.02 in (5.08 X 10-4 m) and a theoretical stress
concentration factor of 2.5, and is operating at a temperature of
190 °F ( 87. ~ °C). The reliability used is 99%. Determine the
diameter of the shaft.
First, the stress equations using the Parametric lll:ethod are derived.
Using the ma.ximUJ'll distortion energy theory, the equivalent mean and
alternating stresses ares
+ -l(-T )2 )t T ..1 2z = 0.866 z
but upon overload, both the torque, T, and moment, M, go up proportion-
ally by the same factor n. That is
T _ __.,._ nT, and M --- nM
thus, S' _ 0.866 n T
em - Z and S' = n M ea Z
73
_rrr)J where the section moaulus, Z - J2 • Substituting in the section
modulus, the stress equations become
s• em _ o.866 n T - TT-DJ and s• ea
nM = --------ID!.
32 J2
These two equations are the :required stress equations for the
subroutine program.
The procedure of inputting the data is as followsa
English units are used.
Tensile strength
Yield strength
= 200000 psi
= 150000 psi
Instruct the micro-processor to find the significant endurance limit.
Type of surface finish, enter 3 for machined or cold dJ:a.wn.
Reliability = 99%
Opera.ting temperature = 190 •F
Theoretical stress concentration factor = 2.5
Notch radius = 0.02 in
Material is steel.
The shaft is under bending.
No miscellaneous effect factor is needed.
Method to be used for. computing the .fini t.e. life .significant.
endurance limit, Se' ' ', is log - log method. The code is 1.
The cross section of the shaft is circular.
Ne:> el'.ldurance limit :t"or a R. R. Moore rQtat.ing. :beam specimen
74
is known.
Finite life, number of cycle = 500000
The following is the subroutine program for stress equations, using
the Para.metric method, and written in BASIC. Using the variables
defined by the program, the numerical values of these variables are
entered before the stress equations. The variables should be numbered
starting with line 6000 incrementing by 10'a. The stress equations
should start with line 6100 incrementing by 10's. After the stress
equations are entered, a numbered return statement is typed. Then type
RUN 600.
6000 T2 = 1200
6010 M1 = 2400
6020 N = 1.8
6100 A1 = N*M1/(PI*D1'J/J2)
6110 A2 = 0.866*N*T2/(PI*Dt3/J2)
6120 RE:l'URN
RUN 600
Modified Goodman fracture line is selected for the design. Enter MGF,
the code for this fatigue failure line.
Establish the limits on the half interval search for the solution.
Enter lower limit = 0.01 in
E.."1.ter upper limit = 10.00 in
The above procedure of entering the data is also shown in Figs.
A-1 to A-6. The result together with the input data, the correction
factors for computing the significant endurance limit, and the
************************************************************** * * * FATIGUE ANALYSIS PROGRAM, ANALYTICAL * * * t By Yiu Wah Luk, UPI & SU, Spring 1978 * * * ************************************************************** You have entered a FATIGUE RESISTANT, INTERACTIVE DESIGN routine. CoRponents will be sized to prevent fatigue failure. Do you want to use English Units? <V or H>V We will use English Units throughout this routine. Please enter the strength of the Material to be used. Tensile Strength, in psi, Su Yield Strength, in psi, Sy
= 200009 • 150000
Do you know the Significant Endurance Li"it of the Material? <Y or H>H
Figure A-1. Output for Example A-1.
....., IJ\
Thi
s se
ctio
n of
the
pro
gra"
wil
l ca
lcul
ate
the
Sig
nifi
cant
En
dura
nce
Li"
it.
Ent
er t
he I
fo
r th
e ty
pes
of s
urfa
ce f
inis
h us
ed.
11
for
poli
shed
fin
ish.
12
for
gro
und
fini
sh.
13 f
or M
achi
ned
or c
old
draw
n.
14 f
or h
ot r
olle
d.
15 f
or a
s fo
rged
. 3 W
hat
is t
he r
elia
bil
ity
in
%?9
9 W
hat
is t
he o
pera
ting
te"
pera
ture
in
Deg
ree
F 11
90
Do y
ou k
now
the
The
oret
ical
Str
ess
Con
cent
rati
on F
acto
r (K
t>?Y
T
heor
etic
al S
tres
s C
once
ntra
tion
Fac
tor
• 2.
5 Do
you
kno
w N
otch
Sen
siti
vity
(Q
)?
<V o
r N>
N W
hat
is t
he n
otch
rad
ius
in i
nche
s?0.
02
Is
the
"ate
rial
ste
el?
CY o
r N>
Y Is
it
und
er b
endi
ng o
r ax
ial
load
ing?
<V
or N
>Y
Is
ther
e an
y M
isce
llan
eous
-eff
ect
fact
or?
<Y o
r H>
N Th
e S-
N c
urve
is
used
for
th
is d
eter
"ina
tion
. A
Log
-Log
or
a L
og-L
inea
r S-
N c
urve
wil
l be
use
d.
The
fin
ite
life
reg
ion
is 0
.9*S
u t1
E3 c
ycle
s to
Se'
' @
1E6
cycl
es.
Wha
t "e
thod
do
you
wan
t to
use
for
coM
putin
g th
e S
igni
fica
nt E
ndur
ance
LiH
it <
Se'
'')?
Ent
er 1
1 fo
r Lo
g-Lo
g M
etho
d.
Ent
er 1
2 fo
r L
og-L
inea
r M
etho
d.
1 Fi
gure
A-2
. O
utpu
t fo
r Ex
ampl
e A
-1.
--.J °'
Is t
he c
ross
sec
tion
cir
cula
r? <
V or
H>V
Do
you
kno
w th
e E
ndur
ance
Li"
it <
Se')
for
rota
ting
-bea
M
spec
iMen
? <V
or
N>H
Is t
he d
esig
n li
fe i
nfi
nit
e? <
V or
H>H
H
uRbe
r of
cyc
les
• 59
9000
Fig
ure
A-)
. O
utpu
t fo
r Ex
ampl
e A
-1.
:j
You
wil
l no
w be
req
uest
ed t
o su
pply
the
PAR
AMET
RIC
DESI
GH S
TRES
S EQ
UATI
ONS
for
your
des
ign
prob
le".
Yo
u "u
st w
rite
the
se e
quat
ions
in
BASI
C.
Use
th
e fo
llow
ing
inst
ruct
ion
s.
If
you
«re
unfQ
Mili
Qr
wit
h th
e de
velo
pMen
t of
suc
h eq
uati
ons,
se
e u
ser'
s gu
ide
and
its
appe
ndix
fo
r gu
idQ
nce.
E
nter
coM
pone
nt
load
s st
arti
ng
wit
h li
ne 6
000
incr
eMen
ting
by
10
's.
Use
Ml=
MoM
ent t~using
Qlt
ernQ
ting
str
ess
<lb
-in)
. U
se M
2=M
oMen
t '~using
a st
eady
str
ess
(lb-
in>
. U
se F
1=A
n Q
lter
nati
ng a
xial
fo
rce
<lb
f).
Use
F2=
A st
eady
axi
al
forc
e <
lbf)
. U
se T
1=An
alt
ern
atin
g t
orqu
e <
lb-i
n).
Use
T2=
A st
eady
tor
que
(lb
-in
).
Use
H
=Sof
ety
fact
or.
Ent
er t
he n
uMer
ic
valu
e fo
r ea
ch o
f th
e va
riab
les
used
in
your
str
ess
equa
tion
s.
Do
this
bef
ore
you
ente
r yo
ur s
tres
s eq
uati
ons.
E
nter
you
r st
ress
equ
atio
ns s
tart
ing
wit
h li
ne
6100
in
cre"
enti
ng b
y 1
0's
. U
se A
t=A
lter
nati
ng s
tres
s.
Use
A2=
Hea
n st
ress
. U
se P
I=Pi
. U
se D
=Bas
ic d
iMen
sion
. N
.B.,
All
di
Men
sion
s sh
ould
be
give
n in
ter
Ms
of
D.
In c
ase
of r
ecta
ngul
ar c
o"D
onen
t, us
e pr
opor
tion
s.
Aft
er y
our
stre
ss e
quat
ions
are
ent
ered
, ty
pe a
nuM
bere
d re
turn
st
ate"
ent.
Th
en
type
run
600
. ~>
:ftt
1f'l
-E •••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
• Th
e fo
llow
ing
equa
tion
s ar
e in
Eng
lish
Uni
t.
6000
T2=
1200
. 60
10 H
1=24
00.
6020
H=l
.8
6100
A1=
N*M
1/(P
I*D
tJ/J
2)
6110
A2•
0.86
6*H
*T2/
(PI*
DtJ
/J2)
61
20 R
ETUR
N RU
H 60
0 F1
.gur
e A
-4.
01.i:
tput
fo
r Ex
ampl
e A
-1.
'-.J
())
6000
T2=
1200
60
10 M
1=24
00
6020
H=t
.8
6100
At=
H*M
1/(P
l*D
t3/3
2)
6110
A2=
0.86
6*H
*T2/
(PI*
Df3
/32>
61
20 R
ETUR
H RU
N 60
0
Figu
re A
-5.
Out
put
for
Exam
ple
A-1
.
....., '°
Sel
ect
the
fGti
9ue
fnil
ure
line
to
be u
sed
in t
he d
esig
n.
If M
odif
ied
Goo
d"an
Fra
ctur
e L
inc,
en
ter
MGF
If M
odif
ied
Goo
d"an
Yie
ld L
ine,
en
ter
HGV
If S
oder
berg
Lin
e,
ente
r S
If G
erbe
r L
ine,
ent
er G
If
Qu«
drG
tic L
ine,
en
ter
Q
If K
ecec
iogl
u L
ine,
en
ter
K
MGF
The
Fai
lure
Lin
e se
lect
ed i
s M
odif
ied
Good
Man
Fra
ctur
e L
ine.
Th
e fQ
ilur
e eq
uati
on i
s <
Sa/(
R2*
Se'''
)tM
+ <
Rl*
SM/S
u)tp
= 1.
w
here
M=1
w
here
P=1
w
here
R1=
1 an
d w
here
R2=
1 Do
you
wis
h to
cha
nge
any
para
Met
ers?
<V
or H
>H
The
foll
owin
g en
trie
s w
ill
esta
blis
h th
e li
ffit
s on
a H
alf
Inte
rval
Sea
rch
for
the
solu
tion
to
your
pro
bleM
. W
hat
is t
he s
"all
est
basi
c di
Men
sion
tha
t yo
u w
ish
to t
ry?
Giv
e yo
ur a
nsw
er i
n in
ches
. Do
not
ans
wer
0.0
. 0.
01
Wha
t is
the
lar
gest
diM
ensi
on,
in i
nche
s? 1
0
Figu
re A
-6.
Out
put
for
Exam
ple
A-1
.
~
~
******
******
******
******
******
******
******
******
******
******
******
******
T
ensi
le S
tren
gth,
Su
•••
••••
••••
••••
••••
••••
••• •
20
0,00
0 ps
i Y
ield
Str
engt
h, S
y •••
••••
••••
••••
••••
••••
••••
• •
159,
000
psi
Sig
nifi
cant
End
uran
ce L
iMit
, S
e'''i
5000
09 e
ye.=
21,
395
psi
SMal
lest
diM
ensi
on t
ried
••••
••••
••••
••••
••••
•• =
0.0
1 in
ches
L
drge
st d
iMen
sion
tri
ed ••
••••
••••
••••
••••
••••
• = 10
.00
inch
es
Safe
ty F
acto
r,
H ••
••••
••••
••••
••••
••••
••••
••••
= 1.
80
HoM
ent
caus
ing
alte
rnat
ing
stre
ss,
Ml •
••••
••••
= 2,
400
lb-i
n St
eady
tor
que,
T2
•••
••••
••••
••••
••••
••••
••••
•• =
1,20
0 lb
-in
The
foll
owin
g fa
ctor
s ar
e us
ed f
or·c
oMpu
ting
Se'''
: Su
rfac
e fa
ctor
<Ka
> = 0
.64
Size
and
sha
pe f
acto
r <K
b>
= 0.
81
Rel
iabi
lity
fac
tor
<Kc)
= 0
.81
TeH
pera
ture
fac
tor
<Kd>
=
0.95
Fa
tigu
e St
reng
th R
educ
tion
fac
tor
<Ke>
= 0.
42
Mis
cell
aneo
us f
acto
r <K
f>
= 1.
90
Endu
ranc
e L
i"it
for
Rot
atin
g-be
aM S
peci
"en
<Se'
) =
109,
000
psi
Sig
nifi
cant
End
uran
ce L
iMit
for
infi
nit
e li
fe <
Se'
')a
16,9
60 p
si
The
Fai
lure
Lin
e se
lect
ed i
s M
odif
ied
Good
Man
Fra
ctur
e L
ine.
Th
e fa
ilur
e eq
uati
on i
s (S
a/(R
2tS
e''')
tM +
<R
ltSM
/Su>
tp=
1.
whe
re M
=1
whe
re P
=1
whe
re R
1=1
and
whe
re R
2•1
The
desi
gn d
iMen
sion
is
1.29
11
inch
es
******
******
******
******
******
******
******
******
******
******
******
******
Do
you
wis
h to
con
vert
the
des
ign
diM
ensi
on t
o SI
uni
t? (
V or
H>Y
Th
e de
sign
di"
ensi
on i
s 0.
9328
"
Figu
re A
-?.
Out
put
for
Exam
ple
A-1
.
f!l
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
T
ensi
le S
tren
gth,
Su •
••••
••••
••••
••••
••••
••••
• •
200,
000
psi
Yie
ld S
tren
gth,
Sy •
••••
••••
••••
••••
••••
••••
••• =
150
,000
psi
S
igni
fica
nt E
ndur
ance
Li"
it,
Se'''
@50
0000
eye
.• 2
5,77
9 ps
i SM
alle
st d
iMen
sion
tri
ed ••
••••
••••
••••
••••
••••
= 0.
01
inch
es
Lar
gest
di"
ensi
on t
ried
••••
••••
••••
••••
••••
••• =
10.0
0 in
ches
Sa
fety
Fac
tor,
H ••
••••
••••
••••
••••
••••
••••
••••
= 1.
80
HoM
ent
caus
ing
alte
rnat
ing
stre
ss,
Hl •
••••
••••
= 2,
400
lb-i
n St
eady
tor
Que
, 12
••••
••••
••••
••••
••••
••••
••••
• =
1,20
0 lb
-in
The
foll
owin
g fa
ctor
s ar
e us
ed f
or c
oMpu
ting
Se'''
: Su
rfac
e fa
ctor
<Ka
) =
0.64
Si
ze a
nd s
hape
fac
tor
<Kb)
=
1.00
R
elia
bili
ty f
acto
r <K
c>
= 0.
81
TeM
pera
ture
fac
tor
<Kd>
=
0.95
Fa
tigu
e St
reng
th R
educ
tion
fac
tor
<Ke>
= 9.
42
Mis
cell
aneo
us f
acto
r <K
f) =
1.09
En
dura
nce
Lif
tit
for
Rot
atin
g-be
aM S
peci
"en
<Se'
) =
100,
000
psi
Sig
nifi
cant
End
uran
ce L
i"it
for
in
fin
ite
life
<S
e'')
= 2
9,86
5 ps
i Th
e F
ailu
re L
ine
sele
cted
is
Mod
ifie
d Go
odM
an Y
ield
Lin
e.
The
fail
ure
equa
tion
is
<S
a/(R
2tS
e''')
t" +
<R
1*SH
/Su)
tp=
1.
Yhe
re H
=l
whe
re P
=l
whe
re R
l=l.3
3333
3333
33
and
whe
re R
2=5.
8187
2093
284
The
desi
gn d
iMen
sion
is
9.74
91
inch
es
• **
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
**
Do y
ou w
ish
to c
onve
rt t
he d
esig
n di
Men
sion
to
SI
unit
? <V
or
H>Y
The
desi
gn d
iMen
sion
is
0.91
90 "
Fi
gure
A-8
. O
utpu
t fo
r Ex
ampl
e A
-1.
~
******
******
******
******
******
******
******
******
******
******
******
******
T
ensi
le S
tren
gth,
Su •
••••
••••
••••
••••
••••
••••
• •
200,
000
psi
Yie
ld S
tren
gth,
Sy
•••
••••
••••
••••
••••
••••
••••
• •
150,
000
psi
Sig
nifi
cant
End
ur4n
ce L
i"it
, Se
'''@
5000
00 e
ye.=
21,
345
psi
SM41
1est
di"
ensi
on t
ried
••••
••••
••••
••••
••••
•• =
0.01
in
ches
L
arge
st d
iMen
sion
tri
ed ••
••••
••••
••••
••••
••••
• = 1
0.00
inc
hes
Saf
ety
Fac
tor,
N ••
••••
••••
••••
••••
••••
••••
••••
= 1.
80
Mo"
ent
caus
ing
41te
rnat
ing
stre
ss,
H1 •
••••
••••
= 2
,400
lb-
in
Stea
dy t
orqu
e, 1
2 •••
••••
••••
••••
••••
••••
••••
•• =
1,20
0 lb
-in
The
foll
owin
g f4
ctor
s ar
e us
ed f
or c
o"pu
ting
Se''':
Su
rfac
e fa
ctor
<Ka
> =
0.64
Si
ze a
nd s
hape
fac
tor
(Kb>
=
0,81
R
elia
bili
ty f
acto
r <K
c>
= 0.
81
TeM
pera
ture
fac
tor
(Kd>
=
0.95
Fa
tigu
e S
tren
gth
Red
ucti
on f
acto
r <K
e>=
0.42
M
isce
llan
eous
fac
tor
CKf)
= 1.
00
Endu
ranc
e Li
Mit
for
Rot
4tin
g-be
a" S
peci
"en
<Se'
) =
100,
000
psi
Sig
nifi
cant
End
uran
ce L
iMit
for
in
fin
ite
life
<Se'~)=
16,9
16 p
si
The
Fai
lure
Lin
e se
lect
ed i
s So
derb
erg
Lin
e.
The
fail
ure
equa
tion
is
<Sa/
CR
2*Se
''')t
M +
<R
l*SM
/Su)
tp•
1.
whe
re H
=1
whe
re P
=1
whe
re R
1=1.
33JJ
J333
J33
and
whe
re R
2=1
The
desi
gn d
i"en
sion
is
1.29
83 i
nche
s
******
******
******
******
******
******
******
******
******
******
******
******
Do
you
wis
h to
con
vert
the
des
ign
di"c
nsio
n to
SI
unit
? CV
or
H>V
The
desi
gn d
i"en
sion
is
9.93
39 "
Fi
gure
A-9
. O
utpu
t fo
r Ex
ampl
e A
-1.
-· ~
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
T
ensi
le S
tren
gth,
Su •
••••
••••
••••
••••
••••
••••
• •
200,
000
psi
Yie
ld S
tren
gth,
Sy •
••••
••••
••••
••••
••••
••••
••• •
15
9,00
0 ps
i S
igni
fica
nt E
ndur
ance
LiH
it,
Se'''
@59
0000
eye
.• 2
1,54
7 ps
i S
Hal
lest
di"
ensi
on t
ried
••••
••••
••••
••••
••••
•• =
0.0
1 in
ches
L
arge
st d
iHen
sion
tri
ed ••
••••
••••
••••
••••
••••
• =
10.0
0 in
ches
Sa
fety
Fac
tor,
H ••
••••
••••
••••
••••
••••
••••
••••
= 1.
80
MoH
ent
caus
ing
alte
rnat
ing
stre
ss,
Ht •
••••
••••
= 2,
400
lb-i
n St
eady
tor
que,
T2
•••
••••
••••
••••
••••
••••
••••
•• =
1,2
00 l
b-in
Th
e fo
llow
ing
fact
ors
are
used
for
co"
puti
ng S
e'''
: Su
rfac
e fa
ctor
<Ka
) = 0
.64
Siz
e an
d sh
upe
f4ct
or
<Kb)
= 0
.82
Rel
iabi
lity
fac
tor
<Kc)
=
0.81
Te
Mpe
ratu
re f
acto
r <K
d)
= 0.
95
F4ti
gue
Stre
ngth
Red
ucti
on f
acto
r <K
e>=
0.42
M
isce
llan
eous
fac
tor
<Kf>
=
1.00
En
dura
nce
Lif
tit
for
Rot
atin
g-be
aH S
peci
Men
<S
e')
= 10
0,00
0 ps
i S
igni
ficu
nt E
ndur
4nce
LiH
it f
or i
nfi
nit
e li
fe <
Se''>
= 17
,094
psi
Th
e F
ailu
re L
ine
sele
cted
is
Ger
ber
Lin
e.
The
fail
ure
equu
tion
is
<Sa
/(R
2*Se
''')f
H +
<R
1*SH
/Su)
tp=
1.
whe
re H
=1
whe
re P
=2
whe
re R
1=1
and
whe
re R
2=1
The
desi
gn d
iHen
sion
is
1.26
96 i
nche
s
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
****
Do
you
wis
h to
con
vert
the
des
ign
diH
ensi
on t
o SI
uni
t? <
V or
H>V
Th
e de
sign
diM
ensi
on i
s 0.
0322
M
Figu
re A
-to.
Out
put
for
Exam
ple
A-1
.
~
******
******
******
******
******
******
******
******
******
******
******
******
T
ensi
le S
tren
gth,
Su
•••
••••
••••
••••
••••
••••
••• = 2
00,0
00 p
si
Yie
ld S
tren
gth,
Sy •
••••
••••
••••
••••
••••
••••
••• =
150,
000
psi
Sig
nifi
cont
End
uron
ce L
iMit
, Se'''~500000
eye.
= 2
1,55
1 ps
i SM
alle
st d
i"en
sion
tri
ed ••
••••
••••
••••
••••
••••
= 0
.01
inch
es
Lar
gest
di"
ensi
on t
ried
••••
••••
••••
••••
••••
•••
• 10
.00
inch
es
Saf
ety
Fac
tor,
H
••••
••••
••••
••••
••••
••••
••••
•• =
1.90
M
oHen
t C
Qus
ing
41te
rn4t
ing
stre
ss,
Ht •
••••
••••
= 2,
400
lb-i
n
Stea
dy t
orqu
e,
T2 •
••••
••••
••••
••••
••••
••••
••••
= 1,
200
lb-i
n
The
foll
owin
g fQ
ctor
s Q
re u
sed
for
co"p
utin
g S
e''':
Su
rfQ
ce f
4cto
r (K
Q)
~
0.64
S
ize
ond
shap
e fQ
ctor
<K
b)
= 0.
82
Rel
iab
ilit
y f
Qct
or
<Kc>
=
0.81
T
e"pe
ratu
re f
acto
r <K
d)
= 0.
95
Fat
igue
Str
engt
h R
educ
tion
fG
ctor
<Ke
>= 0
.42
His
cell
aneo
us f
acto
r <K
f) =
1.00
E
ndur
ance
LiM
it fo
r R
otat
ing-
beaM
Spe
ciM
en <
Se')
=
100,
000
psi
Sig
nifi
cant
End
urQ
nce
LiA
it f
or
infi
nit
e li
fe <
Se'
')•
17,0
98 p
si
The
Fai
lure
Lin
e se
lect
ed i
s Q
uadr
atic
Lin
e.
The
fail
ure
equ
atio
n is
<S
a/(R
2*S
e''')
tM +
<R
l*SM
/Su)
tp=
1.
whe
re H
=2
whe
re P
=2
whe
re R
1=1
and
whe
re R
2=1
The
desi
gn d
iHen
sion
is
1.26
91
inch
es
******
******
******
******
******
******
******
******
******
******
******
******
Do
you
wis
h to
con
vert
the
des
ign
diM
ensi
on t
o SI
uni
t? <
V or
H>V
Th
e de
sign
diH
ensi
on i
s 9.
0322
M
Figu
re A
-11.
Out
put
for
Exam
ple
A-1
.
CX>
\11
******
******
******
******
******
******
******
******
******
******
******
******
T
ensi
le S
tren
gth,
Su
•••
••••
••••
••••
••••
••••
••• = 2
00,0
00 p
si
Yie
ld S
tren
gth,
Sy •
••••
••••
••••
••••
••••
••••
•••
• 15
0,00
0 ps
i S
igni
fica
nt E
ndur
ance
Li"
it,
Se'''
@50
0000
eye
.= 2
1,54
9 ps
i SH
G11
est
di"e
nsio
n tr
ied
••••
••••
••••
••••
••••
•• =
0.01
in
ches
L
arge
st d
iMen
sion
tri
ed ••
••••
••••
••••
••••
••••
• •
10.0
0 in
ches
S
afet
y F
acto
r,
H ••
••••
••••
••••
••••
••••
••••
••••
= 1.
80
MoM
ent
caus
ing
alte
rnat
ing
str
ess,
Mt
•••
••••
•• =
2,40
0 lb
-in
Stea
dy t
orqu
e,
T2 •
••••
••••
••••
••••
••••
••••
••••
= 1,
200
lb-i
n Th
e fo
llow
ing
fact
ors
are
used
for
coM
putin
g S
e''':
Su
rfac
e fa
ctor
<K
a)
= 0.
64
Siz
e an
d sh
ape
fact
or <
Kb)
= 0.
82
Rel
iab
ilit
y f
acto
r <K
c>
= 0.
81
TeM
pera
ture
fac
tor
<Kd)
=
0.95
F
atig
ue S
tren
gth
Reducti~n
fact
or
<Ke>
= 0.
42
Mis
cell
aneo
us f
acto
r <K
f) =
1.09
En
dura
nce
Li"
it f
or R
otat
ing-
beaM
Spe
ciM
en
<Se'
) =
190,
009
psi
Sig
nifi
cant
End
uran
ce L
iMit
for
in
fin
ite
life
<Se
''>=
17,0
96 p
si
The
Fai
lure
Lin
e se
lect
ed i
s K~cecioglu
Lin
e.
The
fail
ure
equa
tion
is
(Sa/
(R2
*S
e''')
t" +
<R
1*SM
/Su)
tp=
1.
whe
re H
=l.5
w
here
P=2
w
here
R1=
1 un
d w
here
R2=
1 Th
e de
sign
diM
ensi
on i
s 1.
2693
inc
hes
******
******
******
******
******
******
******
******
******
******
******
******
Do
you
wis
h to
con
vert
the
des
ign
diM
ensi
on
to S
I un
it?
<Y o
r H>
Y Th
e de
sign
diM
ensi
on i
s 0.
0322
"
Figu
re A
-12.
Out
put
for
Exam
ple
A-1
.
~
87
Table A-3. Results of Example A-1 obtained by six fatigue failure lines.
Design Dia.meter Type of Failure Line used
in m
Modified Goodman fracture line * 1.2911 0.0328
Modified Goodman yield line * 0.7491 0.0190
Soderberg line 1.2983 0.0330
Gerber line 1.2696 0.0332
Quadratic line 1.2691 0.0322
Kececioglu line (with b=1.5) + 1.269.3 0.0322
* If the user subscribes to the Modified Goodman theory, the largest
dimension of the two Goodman solutions must be selected..
+This is an arbitrary choice of exponent used for this example only.
88
parameters of the fatigue failure line selected for this design a.re
output and shown in Fig. A-7. The design diameter of the shaft is
found to be 1.2911 in. (0.0'.328 m).
Using the same input data., the component is redesigned by the other
five fatigue failure lines. The results using the other five fatigue
failure lines a.re shown in Figs. A-8 to A-12. All six results are
listed in Table A-3 .. for comparsion.
Normally-, all these analyses are not carried out. Only the analysis
that corresponds to the user's selected theory is computed. But the
program provides the capability to redesign the component using other
fatigue failure theories in order that the user can choose among
alternatives.
From the results listed in Table A-3, the smallest dimension is
obtained by modified Goodman yield line. This result is not correct
for a fatigue design because the modified Good.man theory demands that
one considers the yield line and the fmcture line. Thus, the largest
dimension of the two Goodlllan solutions must be selected. The most
conservative solution is obtained using Soderberg line. This is not
always true. It depends upon the location of the load line. In this
example, the solutions from modified Goodman line, Soderberg line,
Gerber line, Quadratic line, and Kececioglu line a.re quite close.
This is true only for this particular example because of the load
line. Other examples may give more significant differences between
these theories.
89
EXAMPIE A-2. (Taken from problem 5-38a, Mechanical Engineering Design
3:rd Fd., McGraw-Hill, New York, N. Y., 1977, by J. E. Shigley.)
Figure A-13 shows two views of a flat steel spring loaded 1n bending
by the force F. The spring is assembled so as to produce a preload
Fmin = 900 N. The :force then varies from this minimum value to a
maximum of 3000 N. The spring is forged of AISI 1095 steel having
the following properties:
s = 1,400 MPa u
Sy = 950 Mfa
1i3 = 398
12 percent elongation 1n 50 mm.
Find the thickness t 1f Kt = 2.50 and a margin of safety of 90 percent
is to be used.
let D be the thickness of the :flat spring.
Since no reliability factor is mentioned, 50% is assumed.
Room temperature is assumed when operating,. 24
Notch sensitivity factor is asswned to be 0.97.
Let the spring be designed for 100,000 cycles.
SI units are used.
0 c.
The stress equations using the Parametric method are desired.
90
(0.30.5 m) I. --12"~ .. ,
F
0.3" DIA (0.0076 m DIA)
-Si
'° : S' (""'\ .
0 -
'
_________ _,_.......,. ________ ....._J_ ' I I I I I
T
Figure A-1J. Flat Steel Spring under Fatigue Loading.
Alternating force, F a
91
- 3000 - 900 - 2
Mean force, F _ 3000 + 900 m - 2
Using the maximum distortion energy theory, the equivalent mean and
alternating stresses are obtained.
= cs2 + 3t2)t M o.1s (JOOO - 200) s = _J!L = em m m z 2Z
= cs2 + Jt2)t M o.1s (JOOO + 200} s = A = ea a a z 2Z
Bu:t upon overload, maximum load will increase by factor n, so
Fmax.. n F max.
Substituting in the section modulus, Z = 0 ·~S if , yields
s• em _ 6(JOOO n - 200} - rl-
S' ea - 6(JOOO n + 200} - rl-
These two equations are the required stress equations for the
subroutine program.
The procedure of inputting the data is as follows1
SI units are used,
Tensile strength = 1400000000 Pa
Yield strength = 950000000 Pa
Instruct the micro-processor to find the significant endurance limit.
Type of surface finish, enter 5 for as forged.
Reliability = 50% Operating temperature = 24 °C
Theoretical stress concentration factor = 2,5·
Notch sensitivity factor = 0.97
Material is steel.
The spring is under bending.
No miscellaneous effect factor is needed.
Method used for computing the finite life significant endurance limit,
Se''', is log - log method, the code is 1.
The cross section of the spring is rectangular.
No endurance limit for a R. R. Moore rotating beam.
specimen is known.
Finite life, number of cycles = 100000
The following is the subroutine program for stress equations, using
the Parametric method and written in BASIC. Following ~he instructions
provided by the program, the subroutine is
6000 N = 1.9
6100 Al = 6*(JOOO*N - 900)/Dt2
6110 A2 = 6*(JOOO*N + 900)/Df2
6120 RETURN
RUN 600
Soderberg line is selected for this design. Enter S, the code for
Soderberg line.
93
Establish the limits on the half interval search for the solution.
Enter lower limit = O. 001 m
Enter upper limit = 0.10 m
The input data and result are shown in Figs. A-14 to A-15. The
thickness of the fiat spring is found to be 0.0146 m (0.5747 in).
******
******
******
******
******
******
******
******
******
******
******
******
T
ensi
le S
tren
gth,
Su
•••
••••
••••
••••
••••
••••
•••
• 1,
400,
000,
000
Pa
Yie
ld S
tren
gth,
Sy •
••••
••••
••••
••••
••••
••••
••• =
950,
000,
000
Pa
Sig
nifi
cant
End
uran
ce L
iMit
, Se
'''@
1000
00 e
ye.•
167
,994
,384
Pa
SMal
lest
di"
ensi
on t
ried
••••
••••
••••
••••
••••
•• =
0.00
M
Lar
gest
diM
ensi
on t
ried
••••
••••
••••
••••
••••
•••
• 0.
10 M
S
afet
y Fa
ctor
, H
••••
••••
••••
••••
••••
••••
••••
•• =
1.90
Th
e fo
llow
ing
fact
ors
are
used
for
coM
putin
g S
e'''
: Su
rfac
e fa
ctor
<Ka
) =
0.25
S
ize
and
shap
e fa
ctor
<Kb
) = 0
.88
Rel
iabi
lity
fac
tor
<Kc>
=
1.00
T
eHpe
ratu
re f
acto
r <K
d)
= 1.
00
Fati
gue
Stre
ngth
Red
ucti
on f
acto
r <K
e)=
0.41
M
isce
llan
eous
fac
tor
<Kf>
=
1.00
En
dura
nce
LiM
it fo
r R
otat
ing-
beaM
Spe
ciM
en <
Se')
= 68
9,47
5,70
0 Pa
S
igni
fica
nt E
ndur
ance
LiM
it fo
r in
fin
ite
life
<S
e'')
= 6
1,34
1,85
1 Pa
'8-
The
Fai
lure
Lin
e se
lect
ed i
s So
derb
erg
Lin
e.
The
fail
ure
equa
tion
is
<Sa/
(R2*
Se'''
)tM
+
<R1*
SM/S
u)tp
= 1.
w
here
M=l
whe
re P
=1
whe
re R
l=t.4
7368
4210
53
and
whe
re R
2=1
The
desi
gn d
iMen
sion
is
8.81
46 "
******
******
******
******
******
******
******
******
******
******
******
******
Do
you
wis
h to
con
vert
the
des
ign
di"e
nsio
n to
Eng
lish
uni
t? <
Y or
N>V
Th
e de
sign
di"
ensi
on i
s 8.
5747
inc
hes
Do y
ou w
ish
to r
edes
ign
the
coM
pone
nt u
sing
ano
ther
Fi
gure
A-1
4. O
utpu
t fo
r Ex
ampl
e A
-2.
fail
ure
lin
e? W
arni
ng!
A M
odif
ied
Goo
d"an
app
roac
h re
quir
es
both
a
frac
ture
and
a y
ield
ana
lysi
s.
<V o
r H>
H Do
you
wan
t to
des
ign
a ne
w co
Mpo
nent
? <V
or
H>H
EHD
Figu
re A
-1.5
. O
utpu
t fo
r Ex
ampl
e A
-2.
~
A.2 USER'S GUIDE --- SECTION PROPERTIES
INTRODUCTION
The cross sectional properties of a machine member are usually needed
in the design. To find these properties for an irregular shape cross
section is time consuming and difficult. This program provides a means
of computing twenty sectional properties of any shape cross section.
The sectional properties computed are: area, location of centroid,
area moment of inertia about different axis, and radius of gyration.
This program also provides a graphical verification of the input
cross section together with the list of data. In this way, the cross
section can be checked easily before the computation takes place.
The program is written in BASIC and used on a Teketronix micro-
processor model 4051 with 32 K memory.
THEORY
Part of the theory is documented in the paper ''Properties of plane
cross sections", Machine Design, Jan., 22, 1976, pp. 105-107, written
by F. Wojciechowski. The technique used is to replace integration by
summation of finite elements to find the section properties of a plane
cross section. This technique applies only to areas bounded by straight
lines, but because curves could be approximated by straight line
segments, the method can be used on any shape.
96
-
9?
·Basically, the method divides a crosa section into series of
trapezoids or rectangles, then the properties of each elemental area
are added or substracted to give the composite properties of the desired
cross section.
In order to speed up the input and computation, and also to obtain
a more accurate result, the properties of circular cross section and
circular holes are not computed using the above mentioned method. It
uses the ordinary formulas of area and area moment of inertia about the
centroid of a circle. Then the area moment of inertia with respect to
other axis is computed using the parallel axis theorem. In this way, to
input circular section or a circular hole inside a cross section, only
the radius of the circle, and x and y coordinates of the center are
needed, instead of approximating the curves by straight line segments
and inputting the x and y coordinates of each line.
INPUT Il4 T.A NEEDED
Units can be either in English or SI system. This depends on the
option chosen by the user. The following data are needed.
For circular section or circular holes
Radius of circle, in or mm.
X and y coordinates of the center, in or mm.
For polygonal section or polygonal hole:
'
98
X and y coordinates of each vertices, in. or mm.
The following points should be noted when inputting a polygonal section
or polygonal hole.
1. The polygonal section must be located entirely within the first
quadrant, that is, both x and y coordinates must be positive.
2. The x and y coordinates of the vertices of the polygonal section or
hole must be entered sequentially for a complete, clockwise path
around the section or hole.
3. Be sure to end the path with the first vertice.
EXAMPLE B-1. (Taken from example 1, pp. 106, of the paper "Properties
of ·plane cross sections", Machine Design, Jan., 22, 19?6, written by
F. Wojciechowski.)
A hollow hexagonal cross section is shown in Fig. B-1. Determine the
area, centroid, area moment of inertia about the x-y axis and principal
axis, and radius of gyration.
English units are used.
The outside perimeter is not a circular section.
Enter the x and y coordinates of the vertices of the hexagonal cross
section sequentially for a complete, clockwise pa th around the :section..
#1 x, y = 1 t 0
#2 x, y = o, 1.732
#J x, y = 1 t J.464
#4 x, y = ), J.464
#5 x, y = 4, 1.732
99
y
.3.
0.866
0
1.0 1.5 2.5 J.O 4.0 (inches)
Figure B-1. A hollow hexagonal cross section.
#6 x, y = 3, 0
#7 x, y = 1, 0
100
For the hexagonal hole, enter x and y coordinates of the vertices
sequentially for a complete clockwise path around the hole.
#1 x, y = 1.5, o.866
#2 x' y = 1, 1. 732
#J x, y = 1.5, 2.598
#4 x, y = 2.5, 2.598
#5 x, y = J, 1.732
#6 x, y = 2.5, o.866
#7 x, y = 1.5, o.866
Since the properties with respect to an arbitrary axis are not required,
answer "N" when ask if the area moment of inertia about an arbitrary
axis is desired.
The input data, the drawing of the cross action, and the results are
shown in Figs. B-2 to B-6.
EXAMPLE B-2. (Taken from pp. 02-10, example 4, in the Section Properties
program in User's Guide of Mechanical Engineering Programs, Pac I. for
HP-67 or HP-97 programmable calculator, Hewlett-Packard Company, Oregon,
1976.)
For the part shown in Fig. B-7, compute the polar moment of inertia
about point A. Point A denotes the center of a hole a.bout which the
part rotates. The area of the hole must be deleted from the cross
section. (No unit was given, so English unit is assumed.)
******
******
******
******
******
******
******
******
******
******
** *
* *
SECT
ION
PROP
ERTI
ES O
F PO
LVOH
AL S
ECTI
ON
* *
* *
By Y
iu W
ah L
uk,
UPI
~
SU,
Spri
ng 1
978
* *
* ***
******
******
******
******
******
******
******
******
******
*****
Do y
ou w
ant
to u
se S
I un
its?
<V
or H
>H
We
wil
l us
e E
ngli
sh U
nits
thr
ough
out
this
pro
gra"
. Is
the
out
er p
eri"
eter
a c
ircu
lar
sect
ion?
CV
or H
>H
Ple
ase
ente
r th
e X
and
V c
oord
inat
es o
f th
e ve
rtic
es o
f th
e po
lygo
n <w
hich
Hus
t be
lo
cate
d en
tire
ly w
ithi
n th
e fi
rst
quad
rant
) se
Que
ntia
lly
for
a co
Mpl
ete,
cl
ockw
ise
path
aro
und
the
poly
gon.
U
nits
sho
uld
be i
nche
s.
Be s
ure
to e
nd w
ith
the
firs
t po
int.
X
<t>,
V<
l> =
1,0
X
<2>,
V
(2)
= 0,1
.732
X
(3),
V
(J)
= 1,
J.46
4 X
(4),
V<
4> = 3
,J.4
64
X<S>
, VC
S> =
4,1
.732
X<
6>,
V(6
) =
J,0
X<7>
, V<
7> =
1,0
A
re
ther
e an
y ho
les
in t
he s
ecti
on?
<V o
r H>
V A
re
ther
e an
y ci
rcul
ar h
oles
? <V
or
H>H
How
Man
y po
lygo
n ho
les
are
ther
e in
the
sec
tion
?!
For
poly
gon
hole
s:
Figu
re B
-2.
Out
put
for
Exam
ple
B-1
.
.... a
Ent
er t
he X
and
V c
oord
inat
es o
f ea
ch v
erti
ces
in a
coM
plet
e,
cloc
kwis
e pa
th.
Uni
ts s
houl
d be
inc
hes.
Be
sur
e to
end
wit
h th
e fi
rst
poin
t of
eac
h ho
le.
For
poly
gona
l ho
le 1
1:
X<1
), V
<l>
= 1.
5,9.
866
X<2
>,
V<2>
=
1,1.
732
X<3
>,
V(3
) =
1.5,
2.59
8 X
<4>,
V
(4)
= 2.
5,2.
598
X<5>
, V
(5)
= J,
1.73
2 X
(6),
V
(6)
= 2.
5,0.
866
X(7
),
V<7>
=
1.5,
0.86
6
Figu
xe B
-3.
Out
put
for
Exam
ple
B-1
.
.... 2
All
data
are
in
inch
es.
v GR
APH
OF I
NPUT
SEC
TION
Po
ly.
D4t
4 <X
,V>:
11
= 1,
8
12=
9,
1. 7
32
4.ee
13
= 1,
3.
464
14=
J, 3
.464
15
= 4,
1.
732
3.50
16
= 3,
0
Poly
.Hol
e'A
'Dat
4 CX
,V>:
3.
00
11=
1.5,
0.8
66
12=
1,
1. 73
2 #3
= 1.
5,
2.59
8 2.
58
#4=
2.5,
2.
598
15=
J,
1.73
2 16
= 2
.s,
0.86
6 2.
ee
Tl
, 1.
59
1. 0
0
0.50
0.00
e.ee
1.00
2.
00
Do y
ou w
ant
to c
hang
e an
y da
t4?
<V o
r H>
H Fi
gure
B-4
. O
utpu
t fo
r Ex
ampl
e B
-1.
\. I 3.09
\.
s ...
I 0 \,.)
x 4.
99
*** SE
CTIO
N PR
OPER
TIES
OF
THE
REQU
IRED
SEC
TIOH
***
AreG
•••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
a X
coo
rdin
Gte
of
the
cent
roid
••••
••••
••••
••••
••••
••• •
,, co
ordi
no.te
of
the
cent
roid
••••
••••
••••
••••
••••
••• =
Ar
eG "
or1e
nt o
f in
erti
G a
bout
X-a
xis •
••••
••••
••••
••• =
Ar
ea.
Hor1
ent
of i
nert
ia a
bout
Y-a
xis •
••••
••••
••••
••• =
A
rea
prod
uct
of i
ner
tia •
••••
••••
••••
••••
••••
••••
••• =
A
rea
'10M
ent
of i
nert
ia a
bout
X'-o
.xis
tra
nsla
ted
to
the
cent
roid
••••
••••
••••
••••
••••
••••
••••
••••
••••
••• =
Are
a. Ho
r1en
t of
ine
rtia
abo
ut V
'-o.x
is t
rans
late
d to
th
e ce
ntro
id ••
••••
••••
••••
••••
••••
••••
••••
••••
••••
• ~
Are
a pr
oduc
t of
ine
rtia
abo
ut t
rans
la.t
ed a
xis •
••••
• =
Ang
le b
etw
een
tran
slat
ed a
xis
a.nd
prin
cipa
l ax
is
(in
degr
ee>,
pos
itiv
e is
cou
nter
-clo
ckw
ise •
••••
••••
= 7.
79 i
nt2
2.ee
inc
hes
1. 7
3 in
ches
31
.59
int4
39
.29
int4
27
.00
int4
e.12
int
4
e.12
int
4 0.
00 i
nt4
0.00
A
rea
HOHe
nt of
ine
rtia
abo
ut t
he t
rans
late
d,
rota
ted,
pr
inci
pal
X''-
axis
••••
••••
••••
••••
••••
••••
••••
••••
• •
8.12
int
4 A
rea.
Mo"
ent
of i
nert
ia a
bout
the
tra
nsla
ted,
ro
tate
d,
prin
cipa
l V
''-o.
xis •
••••
••••
••••
••••
••••
••• =
8.12
int
4
Do y
ou w
ant
to g
et a
rea.
'10
Hen
t of
ine
rtia
. ab
out
o.n
arbi
trar
y a.
xis?
<V
or N
>H Fi
gure
B-5
. O
utpu
t fo
r Ex
ampl
e B
-1.
... i
Rud
ius
of g
yrat
ion
abou
t X
-axi
s •••
••••
••••
••••
••••
••••
• R
udiu
s of
gyr
atio
n ab
out
Y-a
xis •
••••
••••
••••
••••
••••
•• •
Rci
dius
of
gyrc
itio
n ab
out
X'-
axis
tra
nsla
ted
to
the
cent
roid
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
•• =
Rad
ius
of g
yrat
ion
abou
t '/
'-ax
is t
rans
late
d to
th
e ce
ntro
id ••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
=
Do y
ou w
ant
to f
ind
the
prop
erti
es o
f an
othe
r se
ctio
n?
<V o
r N)
N
END
------~---~--------~
Figu
re B
-6.
Out
put
for
Exaa
ple
B-1
.
2.01
2.25
1. 0
2
1. 0
2
inch
es
inch
es
inch
es
inch
es
... 0 \n
106
English units are used.
The outside perimeter is not a circular scetion.
For the L-shaped cross section, the x and y coordinates of the vertices
are entered in a complete clockwise path.
#1 x, y = o, 0
#2 x, y = o, 2
#3 x, y = 5, 2
#4 x, y = 5. 1.4
#5 x, y = 0.8, 1.4
#6 x, y = o.a, 0
#? x, y = o, 0
For the circular hole at A:
Radius = 0.25
x, y = 0.2, o.6
To find the properties about the arbitrary axis at A with zero degree
of rotation, enter:
x, y coordinates of the origin of the arbitrary axis = 0.2, 0.6
Angle of rotation of the arbitrary axis = 0 degree
The drawing of the cross section and the results are shown in Figs. B-8
to B-10. There is a difference between the original drawing, Fig. B-7,
and the one from the output, Fig. B-8, because the hole was actually
located partly outside the L-shaped cross section. The error can be
eliminated if the cross section is first drawn to scale. This obvious
mistake can be observed and rectified. However, it is not rectified for
this example so that the solution obtained may be compared to the
Hewlett Packard results.
107
(0, 2) (5, 2)
0 • .5D 0 .8' 1. (5.0, 1.4)
Figure B-7. L-shaped cross section with a circular hole.
A11
data
are
in
inch
es.
Poly
. OQ
tQ
<X,Y
>:
11=
e, e
12=
e, 2
v
13=
s, 2
14=
s, 1.
4 5.
09
15=
0.8,
1.
4 16
= 0.
8,
0
Cir
. H
ole
'A'
DQ
ta:
4.00
R
adiu
s =
0.25
x,
v =
0.2,
0.
6
3.0
0
2.00
1. ee
GRAP
H OF
INP
UT S
ECTI
ON
..... _
__
__
__
__
__
__
__
__
__
__
__
3
__
__
__
__
__
__
4
0. ee
. I
• •
" I
I I
I I
e.ee
1.ee
2.
ee
J.00
4.00
s.00
Do y
ou w
ant
to c
han1
e an
y da
ta?
<V o
r H>
N Fi
gure
B-8
. O
utpu
t fo
r Ex
ampl
e B
-2.
x
..... ~
*** SE
CTIO
N PR
OPER
TIES
OF
THE
REQU
IRED
SEC
TION
***
Are
a. ••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
• •
x co
ordi
na.te
of
the
cent
roid
••••
••••
••••
••••
••••
•••
• y
coor
dina
te o
f th
e ce
ntro
id ••
••••
••••
••••
••••
••••
• =
. Ar
· ea
"o"e
nt o
f in
erti
a <l
bout
X-a
xis •
••••
••••
••••
••• =
Are
a. HO
Men
t of
ine
rtia
. 4b
out
V-4
xis •
• ~ •
••••
••••
••••
=
Are
a. pr
oduc
t of
ine
rtia
. •••
••••
••••
••••
••••
••••
••••
• =
Are
a. M
OP1e
nt of
ine
rtia
. 4b
out
X'-
4xis
tra
.nsl
4ted
to
the
cent
roid
••••
••••
••••
••••
••••
••••
••••
••••
••••
••• =
A
rea.
MoN
ent
of i
nert
ia.
abou
t V
'-a.x
is t
r4ns
late
d to
th
e ce
ntro
id ••
••••
••••
••••
••••
••••
••••
••••
••••
••••
• =
Are
a. pr
oduc
t of
ine
rtia
. a.
bout
tr
ansl
ated
a.x
is ••
••••
=
Ang
le b
etw
een
tran
slat
ed a
xis
and
prin
cipa
l ax
is
<in
degr
ee>,
po
siti
ve i
s co
unte
r-cl
ockw
ise •
••••
••••
• A
rea
HOHe
nt of
in
erti
a a.
bout
th
e tr
ansl
ated
, ro
t4te
d, J.
92 i
nt2
2.02
inc
hes
1. 47
in
ches
9.
42 i
nt4
25.2
3 in
t4
13.0
4 in
t4
0.94
int
4
9.29
int
4 1.
42 i
nt4
9.38
prin
cipa
l X
''-a.
xis •
••••
••••
••••
••••
••••
••••
••••
••••
• 0.
72 i
nt4
Are
a. HO
Hent
of
iner
ti4
a.bo
ut t
he t
rans
late
d,
rota
ted,
pr
inci
pal
V''-
a.xi
s •••
••••
••••
••••
••••
••••
• •
9.52
int
4
Do y
ou w
ant
to g
et a
rea.
"o"
ent
of i
nert
ia.
a.bo
ut a
.n a.
rbit
ra.r
y ax
is?
<V o
r H>
V Fi
gure
B-9
. O
utpu
t fo
r Ex
ampl
e B
-2.
.... 0 '°
Ple
ase
ente
r th
e an
gle
betw
een
the
orig
inal
ax
is <
X-V>
an
d ar
bit
rary
axi
s <
X'''
-V'''
),
in d
egre
e,
posi
tive
is
coun
ter-
cloc
kwis
e •••
••••
••••
••••
••••
••••
••••
••••
••••
•• •
0 E
nter
X-a
xis
coor
dina
te o
f th
e ar
bit
rary
axi
s <i
nche
s>=
.2
Ent
er Y
-axi
s co
ordi
nate
of
the
arb
itra
ry a
xis
<in
ches
)• .
6
Are
a ffO
Men
t of
in
erti
a ab
out
the
arb
itra
ry a
xis
,X'''
•• =
J.9
1 in
t4
Are
a HO
Men
t of
in
erti
a ab
out
the
arb
itra
ry a
xis
,V'''
•• ~
22.2
2 in
t4
Pol
ar a
rea
"oM
ent
of i
ner
tia
abou
t th
e ar
bit
rary
Q
xl·-
(V'''-
'i''')
-
:::. r.
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• •
• • -
Are
a pr
oduc
t of
in
erti
a ab
out
the
arb
itra
ry
Qx
l.S (V
'''-
V''')
-
" '
......
......
......
......
......
......
.. -R
adiu
s of
gyr
atio
n ab
out
X-a
xis •
••••
••••
••••
••••
••••
•• =
R
adiu
s of
gyr
atio
n ab
out
V-a
xis •
••••
••••
••••
••••
••••
•• =
Rad
ius
of g
yrat
ion
abou
t X
'-ax
is t
rans
late
d to
th
e ce
ntro
id ••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
= R
adiu
s of
gyr
atio
n ab
out
V'-
axis
tra
nsla
ted
to
the
cent
roid
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
•• =
Do y
ou w
ant
to f
ind
the
prop
erti
es o
f an
othe
r se
ctio
n?
<Y o
r H)
H
END
-~~~-----~--~~--~---
Figu
re B
-10.
Out
put
for
Elca
mpl
e B
-2.
26.1
3 in
t4
7. 61
i n
t4
1. 55
in
ches
2.
54 i
nche
s
0.49
inc
hes
1.54
inc
hes
.... .... 0
A .3 USER'S GUIDE ---BEAM ANALYSIS
INTRODUCTION
One of the most frequently encountered engineering designs is beam
because it can be used for modelling many structures. This program,
using transfer matrix method, computes and also plots the curves of
deflection, slope, moment, and shear along the beam. Static and forced,
undamped dynamic analysis can be performed for beams of uniform.. or
variable cross section. Unifomly or linearly varied distributed loads,
concentrated point loads, applied moments, or combinations of all three .
may be applied. This program allows arry combination of pinneQ., fixed,
free, or guided flexural boundary conditions., even normally ·
kinematically unstable conditions can be handled i:f sufficient internal
supports are provided. In-span--suppcrt :can be ela-Stic ·springs-anti/or
elastic moment spring. Mi:idelling for dynamic response uses lumped mass.
The programming language used is BASIC and the micro-processor used is
a Teketronix model 4051 with 32 K memory.
THBDRY
The theory, ilsing transfer. matrix method, . is well documented .in
chapter J of the Matrix Methods in Elastromechanics, by Eduard c. Pestel and Frederick A. Leckie, McGraw-Hill, 196J. The method is based
on the idea that a continuous beam can be broken up into component
111
112
parts with simple elastic and dynamic properties that can be expressed
in matrix fo:rm. These component matrices, when fitted together by
successive matrix multiplication and evaluated with the proper boundary
conditions will give the response of the entire beam.
For a given continuous beam with several sections, say i, each
element or section is represented by the appropriate field and point
transfer matrices. The state vectors from one end, [z]0 , to the other
end, [z]1 , are related by the equation:
where [FJj =a field transfer matrix that describes the lh section
of distributed stiffness with or without distributed loads
(P]j =a point transfer matrix that describes the jth element at
a point with no finite length.
i = nwnber of sections.
[zJ 0 =state vector at beginning, usually the left end.
[ zJi = state vector at the temination end, usually the right end.
[uJ =product of all field and point transfer ma.trices.
The state vector, [z], has five components, defined as:
w s
[zJ = M (C-2)
v
1
11.'.3
where W = deflection
S =slope
M =moment
V =shear
The boundary conditions are as follows:
For pinned end, W=O, M=O
For fixed end, W=O, S=O
For free end, M=O, V=O
For guided end, S=O, V=O
where the symbols are defined same as Eq.C-2.
(c-3)
The field and point transfer matrices in Eq. C-1 are known and the
boundary conditions of both ends should be applied to Eq. C-4.
(zJ~ = [uJ[zJ0 (C-4)
All the variables in the state vectoxs[z] 0 and (z]i can now be found.
Once [z] 0 is known, this matrix multiplication process is repeated to
yield the states at each desired point along the beam.
INPtn' DATA ~UIRED
The beam should first be divided into several sections, such that
each section has the same distributed stiffness and a point element at
the end if a.ny.
Units can be either in English or the s. I. system •. This depends on
114
the option chosen by the user. For each section, the following data
are needed:
1. Length of the section, inches or metres.
2. Modulus of elasticity, psi or Pa, 4 4 J, Area moment of inertia, in or m ,
4. Magnitude of unifomly distributed load, lb/in or N/m. (N.B.;
Load is positive downward.)
5. M!ignitude of linearly varied distributed load, on the left and on the
right, lb/in or N/m. (N.B.; Load is positive downward.)
6. &gnitude of concent:rated load, lb or N, (N.B.; Load is positive
downward.)
7. ~agnitude of moment, lb-in or N-m.(N.B.; Moment is positive in
counter-clockwise direction.)
8. Stiffness.of the suppo~, lb/in or N/m.
9. Support moment stiffness, lb-in/:rad or N-m/:rad.
10. &gnitude of concent:rated weight, lb or N. 2 2 11. Weight moment of inertia, lb-in or N-m •
12. Forced circular fx-equency, :rad/sec.
13, Enter the type of vibration of the beam or rotor,
H, a factor for determining gyroscopic inertial effects in the
dynamic case, given as:
-1 for bending vibration
-+t for rotating shaft (equal angular direction of whirl and
H = rotation)
-3 for rotating shaft (opposite angular d~ction of whirl
and rotation)
115
Enter 1 for H = -1
2 H=-+1
J H = -J
EXAMPLE C-1.
A cantilever beam, 8 inches (0.2032 m) long, has a concentrated
load of 100 lb (444.82 N) at its free end, see Fig. C-1. The moment of 4 ( -6 4) inertia, 4.7 in 1.956 X 10 m is constant. The material used is
6 . 11 steel and the modulus of elasticity is JO X 10 psi (2.068 X 10 Pa).
Neglect the weight of the beam. Determine the def1.ection, slope, moment,
and shear along the beam.
English units are used.
This is a static response.
Section #1
For field matrix, enter 1 for massless beam.
length of this section = 8 inches. 6 Modulus of elasticity = JO X 10 psi.
Area moment of inertia = 4.7 in4•
For point matrix,
Concentrated load = 100 lb.
Moment = 0 lb-in.
Stiffness of support = 0 lb/in.
Support moment stiffness = 0 lb-in/rad.
Section #2
For field matrix, enter 0 for the last section.
116
Choose the boundary condition of fixed-free, enter 7. Enter 10 increments for the section.
The micro-processor outputs the input data, the numerical values, and
curves of deflection , slope, moment, and shear, see Figs. C-2 to C-9.
EXAMPLE 2 (Taken from page 87, Example 3-9, of Matrix Methods in
Electromechanics, by Eduard c. Pestel and Frederick A. lecke, McGra.w-
Hill, 1963, numerical values have been assumed for this example.)
A beam of uniform flexural stiffness EI is simply supported at one
end, built-in at the other, and supported by a spring one third of the
distance along its length, see Fig. C-10. It is subjected to a uniform
distributed static load of 100 lb/in (1.7.5 X 104 N/m) between points O
and 1 and a concentrated moment of -10000 lb-in (-1130 N-m) at point 1.
Determine the deflection, slope, moment, and shear diagrams. Steel is
the material used,
English units are used.
This is a static :response.
Section #1
For field matrix, enter 2 for uniformly distributed load on massless beam.
Length of this section
. Modulus of elasticity
Uniformly distributed load
For point matrix,
Concentrated load
Moment
Stiffness of support
= 10 inches 6 = 30 X 10 psi
= 100 lb/in
= 0 lb
= -10000 lb-in
= 282000 lb/in
117
P=100 lb
, , ,
~a· I . (0.2032 :;---1
. (444.82 N)
Figure c-1. Cantilever beam with end load.
•
******
******
******
******
******
******
******
******
******
******
** *
* *
THIS
IS
A B
EAM
ANAL
YSIS
PRO
GRAM
WHI
CH S
OLVE
S ST
ATIC
OR
* *
DVHA
MIC
<FOR
CED,
UN
DAMP
ED U
IBRA
TIOH
> RE
SPON
SE.
* *
* *
by V
iu W
ah L
uk,
advi
sed
by P
rofe
ssor
Lar
ry D
. M
itch
ell
* *
Vir
gini
a Po
lyte
chni
c In
stit
ute
and
Sta
te U
nive
rsit
y *
* D
epar
tMen
t of
Mec
hani
cal
Eng
inee
ring
*
* Sp
ring
197
8 *
* *
******
******
******
******
******
******
******
******
******
******
** Do
you
wan
t to
use
Eng
lish
uni
ts?
<V o
r H>
V
Eng
lish
uni
ts w
ill
be u
sed
thro
ugho
ut t
his
pro
graH
. Is
thi
s a
stat
ic a
naly
sis?
<V
or H
>V
Plea
se d
ivid
e yo
ur b
eaM
in
to s
ever
al
fiel
d se
ctio
ns,
and
ente
r th
e in
forH
atio
n re
quir
ed f
or e
ach
sect
ion.
H
.B.
Alw
ays
star
t fro
A
the
left
end
.
Pre
ss R
ETUR
N to
pro
ceed
. Fi
gure
C-2
. O
utpu
t fo
r Ex
ampl
e C
-1.
... ... O>
For
sect
ion
11:
The
foll
owin
g F
ield
Mat
rice
s ar
e av
aila
ble.
0
the
last
sec
tion
. 1
Mas
sles
s be
aM.
2 un
ifor
Mly
dis
trib
ute
d l
oad
on M
assl
ess
beaM
. 3
line
arly
var
ied
dis
trib
ute
d l
oad
on M
assl
ess
beaM
. E
nter
I
for
the
requ
ired
Fie
ld M
atri
x= 1
E
nter
len
gth
of t
his
sec
tion
<in
) <H
.B.
Do n
ot e
nter
zer
o.)
=8
Ent
er H
odul
us o
f E
last
icit
y (
psi>
=J9E
6 E
nter
Are
a M
oMen
t of
In
erti
a C
int4
>=4.
7
9t~u~u1Ai£d
Tra
nsfe
r M
atri
x in
clud
ing
sect
ion
11
is a
s fo
llow
s:
w
1.00
E+00
0 -8
.00E
+900
-2
.27E
-007
-6
.05E
-007
0.
00E+
000
~
.;J
0.00
E+00
0 1.
09E+
000
5.67
E-0
08
2.27
E-0
07
0.00
E+00
0 t1
0.00
E+00
0 0.
00E+
000
1.00
E+00
0 e.
00E
+eee
0.
00E+
000
u 0.
00E+
000
0.00
E+00
0 0.
00E+
000
1.00
E+00
0 0.
00E+
000
1 0.
00E+
000
0.00
E+00
0 0.
00E+
000
0.00
E+00
0 1.
00E+
000
Pre
ss R
ETUR
H to
pro
ceed
. Figu
re C
-.).
Out
put
for
Exam
ple
C-1
.
... ... '°
For
sect
ion
11:
ls t
here
any
con
cent
rate
d lo
ad,
MOM
ent,
luM
ped
Has
s <f
or d
ynaM
ic
anal
ysis
onl
y),
or e
last
ic s
uppo
rt i
n th
is s
ecti
on?
<V o
r H)
V E
nter
Mag
nitu
de o
f th
e co
ncen
trat
ed l
oad,
<l
b>
<H.B
. Lo
ad
is p
osit
ive
dow
nwar
d)
•100
E
nter
Mag
nitu
de o
f th
e M
oMen
t (l
b-i
n)
<H.B
. M
oMen
t is
pos
itiv
e in
cou
nter
-clo
ckw
ise
dire
ctio
n.
Ple
ase
put
in t
he r
igh
t si
gn.>
=0
Ent
er s
tiff
nes
s of
the
sup
port
<lb
/in)
=0
Ent
er s
uppo
rt M
OMen
t st
iffn
ess
<1b-
in/r
Qd)
=0
0''U
8UlA
i£d
Tra
nsfe
r H
dtri
x in
clud
ing
sect
ion
11
is Q
S fo
llow
s:
w
1.00
E+00
0 -8
.00E
+000
-2
.27E
-007
-6
.05E
-007
0.
00E+
000
s 0.
00E+
000
1.00
E+00
0 5.
67E
-008
2.
27£-
007
0.00
E+00
0 H
0.
00E+
000
0.00
E+00
0 1.
00E+
000
8.00
E+00
0 0.
00E+
000
u 0.
00E+
090
0.00
E+00
0 0.
00E+
000
1.00
E+00
0 -1
.00E
+002
1
0.00
E+00
0 0.
00E+
000
0.00
E+00
0 0.
00E+
000
1.00
E+00
0
Pre
ss R
ETUR
N to
pro
ceed
.
Figu
re C
-4.
Out
put
for
Exam
ple
C-1
.
... N
0
For
sect
ion
12:
The
foll
owin
g F
ield
Mat
rice
s ar
e av
aila
ble.
0
the
last
sec
tion
. 1
Mas
sles
s be
a".
2 un
ifor
MlY
dis
trib
ute
d l
oad
on M
assl
ess
beaM
. 3
line
arly
var
ied
dis
trib
ute
d l
oad
on M
assl
ess
bea"
. E
nter
I
for
the
requ
ired
Fie
ld H
atri
xa 0
Figu
re C
-S. O
utpu
t fo
r Ex
ampl
e C-
1 •
•
.. N ..
*** BO
UNDA
RY C
ONDI
TION
S ***
R
I G
H T
E
N D
PINN
ED
FIXE
D FR
EE
GUID
ED
PINN
ED
1 2
3 <K
.U.)
4 L E
FIXE
D 5
6 7
8 F T
FREE
9
<K.U
.) 10
11
<K
.U.>
12
<K
.U.)
E N
D
GUID
ED
13
14
15 C
K.U
.) 16
<K
.U.)
whe
re K
. U.
=
Kin
e"at
ical
ly U
nsta
ble,
unl
ess
inte
rnal
su
ppor
ts e
xis
t.
Use
th
ese
boun
dary
con
diti
ons
at y
our
own
risk
. Do
no
t an
swer
K.
U.
to t
he q
uest
ion
belo
w.
Ent
er I
fo
r th
e re
quir
ed B
ound
ary
Con
diti
on •
7
Figu
re C
-6.
Out
put
for
Exam
ple
C-1
.
.... ~
Dra
win
g of
the
Bea
M
~ ,. , ,. DATA
: Fo
r se
ctio
n 11
: Le
ngth
of
this
sec
tion
••••
••••
••••
••••
••••
• =
Mod
ulus
of
Ela
stic
ity
••••
••••
••••
••••
••••
•• •
Are
a M
oHen
t of
In
erti
a ••
••••
••••
••••
••••
••• =
Mag
nitu
de o
f C
once
ntra
ted
Load
•••
••••
••••
•• =
Do
you
wan
t to
cha
nge
the
data
? <Y
or
H>N
8.00
E+00
0 in
3.
00E+
007
psi
4.70
E+00
0 in
t4
1.00
E+00
2 lb
How
Man
y in
cre"
ents
wou
ld y
ou
like
to
have
for
eac
h fi
eld
sect
ion?
10
Figu
re c
-7.
Out
put
for
Exam
ple
c-1.
p
'
... N
\,)
LENG
TH
DEFL
ECT
I OH
SLOP
E MO
MEHT
SH
EAR
<in>
<i
n>
<Rad
ian>
(l
b-in
) (l
b)
0.00
E+00
0 0.
00E+
000
0.00
E+0
00
-8.0
0E+0
02
t.00E
+002
~
8.00
E-0
01
t.76
E-0
06
-4.3
1E-0
06
-7.2
0E+0
02
1.00
E+00
2 1.
60E+
000
6.78
E-0
06
-8.1
7E-0
06
-6.4
0E+0
02
1.00
E+00
2 2.
40E+
000
1.47
E-0
05
-1.1
6E-0
05
-5.6
0E+0
02
1.00
E+00
2 3.
20E+
000
2.52
E-0
05
-1.4
5E-0
05
-4.8
0E+0
02
1.00
E+00
2 4.
00E+
000
3.78
E-0
05
-1.7
0E-0
05
-4.0
0E+0
02
1.00
E+00
2 4.
80E+
000
5.23
E-0
05
-1.9
1E-0
05
-3.2
0E+0
02
1.00
E+00
2 5.
60E+
000
6.82
E-0
05
-2.0
7E-0
05
-2.4
0E+0
02
1.00
E+00
2 6.
40E+
000
8.52
E-0
05
-2.1
8E-0
05
-1.6
0E+0
02
1.00
E+00
2 7.
20E+
000
1.03
E-0
04
-2.2
5E-0
05
-8.0
0E+0
01
1.00
E+00
2 8.
00E+
000
1.21
E-0
04
-2.2
7E-0
05
-t.0
9E-0
11
1.00
E+00
2 8.
00E+
000
1.21
E-0
04
-2.2
7E-0
05
-1.0
9E-0
11
0.00
E+00
0 .... N
.i:-
Do y
ou w
ish
to s
ee t
he g
raph
s fo
r de
flec
tion
, sl
ope,
M
OHen
t an
d sh
ear?
<Y
or H
>V
Figu
re C
-8.
Out
put
for
Exam
ple
C-1
.
Support moment stiffness
Section #2
126
= 0 lb-in/rad
For field matrix, enter 1 for massless beam.
Length of this section
Modulus of elasticity
Area moment of inertia
= 20 inches
= 30 X 106 psi
= 4.7 1n4
No pcint matrix for this section.
Section #3
For field matrix, enter 0 for the last section.
Choose the boundary condition of pinned-fixed, enter 2.
Enter 10 increments for each section.
The micro-processor outputs the data and deflection, slope, moment, and
shear diagrams, see Figs. C-11 to c-13.
EXAMPLE C-3.
The beam with 6 elastic supports of stiffness 1200 lb/in (2.10 X 105
N/m) each is shown in Fig. C-14. The moment of inertia of the beam is
constant, 144 1n4 (6 X 10-5 m4), and the modulus of elasticity is
20 X 106 psi (1.38 X 1011 Pa). Determine the deflection, slope, moment,
and shear diagrams. Neglect the weight of the beam.
English units are used.
This is a static response.
Section #1
For field matrix, enter 1 for massless beam.
q=100 lb/in (1.75 X 104 N/m)
0 1
127
M=-10,000 lb-in (-1,130 N-m)
K~82,000 lb/in~
(4.94 X 107 N/m)
10" __ ..,..,.... _____ 20'' (0.2.54 m) (0.508 m)
2
~----""------~'-,,.-''--------...------------~
Figure C-10. Statically loaded, statically indeterminant beam.
Dra
win
g of
the
Bea
"
DATA
: Fo
r se
ctio
n 11
: L
engt
h o
f th
is s
ecti
on ••
••••
••••
••••
••••
••• =
1.00
E+00
1 in
M
odul
us o
f E
last
icit
y ••
••••
••••
••••
••••
••••
= 3
.00E
+007
psi
A
rea
HoH
ent
of I
ner
tia
••••
••••
••••
••••
••••
• =
4.70
E+00
0 in
t4
Mag
nitu
de o
f un
ifor
"ly
dis
trib
ute
d l
oad •
••• =
1.00
E+00
2 lb
/in
M
agni
tude
of
HoH
ent •
••••
••••
••••
••••
••••
•••
=-1.
00E+
004
lb-i
n S
tiff
ness
of
Supp
ort •
••••
••••
••••
••••
••••
•• =
2.82
E+00
5 lb
/in
Fo
r se
ctio
n 12
: L
engt
h of
thi
s se
ctio
n ••
••••
••••
••••
••••
••• = 2
.00E
+001
in
M
odul
us o
f E
last
icit
y ••
••••
••••
••••
••••
••••
= 3.
00E+
007
psi
Are
a H
oHen
t of
In
erti
a ••
••••
••••
••••
••••
••• = 4
.70E
+000
int
4
Do y
ou w
ant
to c
hang
e th
e da
ta?
<Y o
r H>
H
How
"any
inc
reM
ents
wou
ld y
ou
like
to
have
for
eac
h fi
eld
sect
ion7
10
Figl
ll':9
C-1
1. O
utpu
t fo
r Ex
ampl
e C
-2.
.... N
CX>
LE HG
TH
DEFL
ECTI
ON
SLOP
E HO
r1EHT
(i
n)
<in>
<R
adia
n)
<lb
-in)
0.
00E+
000
0.00
E+00
0 -9
.32E
-005
0.
00E
+000
1.
00E+
000
9.30
E-0
05
-9.2
7E-0
05
1.08
E+00
2 2.
00E+
000
1.85
E-0
04
-9.1
9E-:
005
1.16
E+00
2 3.
00E+
000
2.77
E-0
04
-9.1
3E-0
05
2.42
E+00
1 4.
00E+
000
3.68
E-0
04
-9.1
8E-0
05
-1.6
8E+0
02
S.00
E+00
0 4.
61E
-004
-9
.39E
-005
-4
.60E
+002
6.
00E+
000
S.57
E-0
04
-9.S
SE-0
05
-8.5
2E+0
02
7.00
E+00
0 6.
59E
-004
-1
.06E
-004
-1
.34E
+003
8.
00E+
000
7.71
E-0
04
-1.1
8E-0
04
-1.9
4E+0
03
9.00
E+00
0 8.
96E
-004
-1
.34E
-004
-2
.63E
+003
1.
00E+
001
1.04
E-0
03
-1.5
5E-0
04
-3.4
2E+0
03
1.00
E+00
1 1.
04E
-003
-1
.SSE
-004
6.
58E
+003
1.
20E+
001
1.26
E-0
03
-6.9
8E-0
05
5.48
E+00
3 1.
40E+
001
1.JJ
E-0
03
2.34
E-0
07
4.39
E+00
3 1.
60E+
001
1.27
E-0
03
5.47
E-0
05
3.29
E+00
3 1.
80E+
001
1.12
E-0
03
9. 3
5E-0
05 -
2.19
E+0
03
2.00
E+00
1 9.
09E
-004
1.
17E
-004
1.
10E+
003
2.20
E+00
1 6.
64E
-004
1.
25E
-004
-2
.07E
+000
2.
40E+
001
4.20
E-0
04
1.17
E-0
04
-1.1
0E+0
03
2.60
E+00
1 2.
08E
-004
9.
34E
-005
-2
.20E
+003
2.
80E+
001
5.71
E-0
05
5.45
E-0
05
-3.2
9E+0
03
3.00
E+00
1 0.
00E+
000
-3.4
7£-0
18
-4.3
9E+0
03
Do y
ou w
ish
to s
ee t
he g
raph
s fo
r de
flec
tion
, sl
ope,
H
offe
nt a
nd s
hear
? <V
or
N>V
Fig
ure
C-1
2 • O
utpu
t fo
r E
xaap
le C
-2.
SHEA
R (l
b)
1.58
E+00
2 5.
81E+
001
-4.1
9E+0
01
-1.4
2E+0
02
-2.4
2E+0
02
-3.4
2E+0
02
-4.4
2E+0
02
-5.4
2E+0
02
-6.4
2E+0
02
-7.4
2E+0
02
-8.4
2E+0
02
-5.4
9E+0
02
-5.4
9E+0
02
-5.4
9E+0
02
-5. 4
9E +
002
· .... N
-5
.49E
+002
'°
-5.4
9E+0
02
-5.4
9E+0
02
-5.4
9E+0
02
-5.4
9E+0
02
-S.4
9E+0
02
-S.4
9E+0
02
E-3
DEF
L.,
W 1
.00
<in
)
e.ee
E-4
SLOP
E,
S 0.
00
(rQ
d)
-2.0
0 ....
E+4
'5
1'10r1
EHT,
M 0
.09
(lb
-in
)
-1.0
9
E+2
9.89
SH
EAR,
V
( 1
b)
-5.0
0
-10.
00
e.ee
9.58
1.
00
1.50
2.
ee
2.59
3.
00
E+l
LENG
TH <
in>
Figu
re C
-1).
Out
put
for
Exam
ple
C-2
.
131
Length of this section = 1 X 10-20 inches (Do not enter 0)
Modulus of elasticity = 20 X 106 psi
Area· moment of inertia = 144 in 4
For point matrix,
Concentrated load = 0 lb
Moment = O lb-in
Stiffness of support = 1200 lb/in
Support moment stiffness = 0 lb-in/rad
The use of an almost zero length beam allows the immediate installation
of a spring support.
Section #2
For field matrix, enter 3 for linearly varied distributed load on
massless beam.
Length of this section
Modulus of elasticity
Area moment of inertia
I,inearly varied distributed load, on the left
= 480 inches
= 20 X 106 psi
= 144 in4
= 75 lb/in
Linearly varied distributed load, on the right = 100 lb/in
For the point matrix,
Concentrated load = 0 lb
Moment = 0 lb-in
Stiffness of' support = 1200 1 b/in
Support moment stiffness= 0 lb-in/rad
132
Section #3
For field matrix, enter 1 for massless beam.
Length of this section
Modulus of elasticity
Area moment of inertia
For point matrix,
Concentrated load
Moment
Stiffness of support
Support moment stiffness
Section #4
= 480 inches
= 20 X 106 psi
= 144 in4
= 0 lb
= 0 lb-in
= 1200 lb/in
= O lb-in/rad
For field matrix, enter 2 for uniformly distributed load on massless
beam.
Length of this section
Modulus of elasticity
Area moment of inertia
Uniformly distributed load
For point matrix,
Concentrated load
Moment
Stiffness of support
Support moment stiffness
Section #5
• = 480 inches
= 20 X 106 psi
= 144 1n4
= 100 lb/in
= 0 lb
= O lb-in
= 1200 lb/in
= O lb-in/rad
For field matrix, enter 1 for massless beam.
Length of this section = 480 inches
Modulus of elasticity
Area moment of inertia
For point matrix,
Concentrated load
Moment
Stiffness of support
Support moment stiffness
Section #6
133
6 = 20 X 10 psi /to = 144 in
= 0 lb
= O lb-in
= 1200 lb/in
= O lb-in/rad
For field matrix, enter 3 for linearly varied distributed load on
massless beam.
length of this section = 480 inches
Modulus of elasticity = 20 X 106 psi 4
Area moment of inertia = 144 in
Linearly varied distributed load, on the left. = 100 lb/in
Linearly varied distributed load, on the right = 75 lb/in
For point matrix,
Concentrated load
Moment
Stiffness of support
Support moment stiffness
Section #7 For field matrix, enter 0 for the last section.
= 0 lb
= 0 lb-in
= 1200 lb/in
= O lb-in/rad
Choose the boundary condition of free-free, enter 11.
Enter 10 increments for each section.
1J4
The micro-processor output of the data and deflection, slope, moment,
and shear diagrams are shown in Figs. C-15 to c-19.
EXAMPLE C-4.
A shaft with two lumped mass, 800 lb (.'.3559 N) and 1200 lb (5.'.3.'.38 N)
in weight, vibrates at a frequency of 10 rad/sec, see Fig. C-20. The
area moment of inertia is constant, I=l.2 in 4 ( 5 X 1 o-6 m 4 ). It has a
fixed end and the other end is an elastic support with stiffness,
K=! X 105 lb/in (1.75 X 107 N/m), and support moment stiffness,
T=1. X 104 lb-in/rad (11.'.30 N-m/rad). Find the def1.ection, slope, moment,
and shear diagrams at this frequency. Steel is the material used. Weight
moment of inertia is 100 lb-in2 (0.287 N-m2).
SI units·a:re used.
This is a dynamic response problem forced by the concentrated load
P sin wt where w=1.0 n.d/sec. P is shown on Fig. C-20.
Section #1
For field matrix, enter 1 for massless beam.
length of this section
Modulus of elasticity
Area moment of inertia
For point matrix,
Concentrated load
Moment
Stiffness of support
s~'Pport moment stiffness
= 0.254 m
= 2.068 X 1011 Pa
= 5 X 10-6 m4
=ON
= 0 N-m
= 0 N/m
= 0 N-m/rad
135
q=100 lb/in (17,513 N/m) q
q
K K
480 .. _..._480'' __........_ 480"----.--480" ___. ...... _ 480" (12.19 m) (12.19 m) (12.19 m) (12.19 m) (12.19 m)
~-~'---v---'-Y~~' ...- ,,,,,.I' '"' ''"""'" .., 'i...,J
~11 L~ [FJ2 ~ 1 [Fh [PJJ r~ ~ L [~5 [P Js J~6 [P 16 4 where q1 = 75 lb/in (1.313 X 10 N/m)
q2 =100 lb/in (17,513 N/m)
Figure C-14. Complex beam loading and supports.
=1200 lb/in
(2.10 x 105N/m)
Dra
win
g of
the
Bea
"
A:
For
sect
ion
11:
Leng
th o
f th
is s
ecti
on ••
••••
••••
••••
••••
••• =
1.00
E-0
20 i
n M
odul
us o
f E
last
icit
y ••
••••
••••
••••
••••
••••
= 2.
00E+
007
psi
Are
a M
oMen
t of
Ine
rtia
••••
••••
••••
••••
••••
• = 1.4
4E+0
02 i
nt4
Sti
ffne
ss o
f Su
ppor
t •••
• ~ •
••••
••••
••••
••••
• = 1.2
0E+0
03 l
b/i
n
For
sect
ion
12:
Leng
th o
f th
is s
ecti
on ••
••••
••••
••••
••••
••• •
4.
80E+
002
in
Mod
ulus
of
Ela
stic
ity
••••
••••
••••
••••
••••
•• •
2.00
E+00
7 ps
i A
rea
MoM
ent
of I
nert
ia ••
••••
••••
••••
••••
••• =
1.44E
+002
int
4 L
eft
end
of l
inea
rly-
vari
ed d
ist.
lo
ad ••
••• •
7.
50E+
001
lb/i
n
Rig
ht e
nd o
f li
near
ly-v
arie
d d
ist.
lo
ad ••
•• •
1.00
E+00
2 lb
/in
S
tiff
ness
of
Supp
ort •
••••
••••
••••
••••
••••
•• =
1.20
E+00
3 lb
/in
Fo
r se
ctio
n 13
: Le
ngth
of
this
sec
tion
••••
••••
••••
••••
••••
• •
4.80
E+00
2 in
M
odul
us o
f E
last
icit
y ••
••••
••••
••••
••••
••••
• 2.
00E+
007
psi
Are
a H
onen
t of
Ine
rtia
••••
••••
••••
••••
••••
• •
1.44
E+00
2 in
t4
Sti
ffne
ss o
f Su
ppor
t •••
••••
••••
••••
••••
••••
• 1.
20E+
003
lb/i
n
For
sec ti
on 1
4:
Leng
th o
f th
is s
ecti
on ••
••••
••••
••••
••••
••• •
4.
80E+
002
in
Mod
ulus
of Elasticit~··•••••••••••••••••••••
2.00
E+00
7 ps
i A
rea
Ho"
ent
of I
nert
ia ••
••••
••••
••••
••••
••• •
1.
44E+
002
int4
M
agni
tude
of
unif
or"l
y di
stri
bute
d lo
ad ••
•• •
1.00
E+00
2 lb
/in
Fi
.gur
e C
-15.
Out
put
for
Exa
aple
C-)
.
.... \..> °'
Sti
ffn
ess
of S
uppo
rt ••
••••
••••
••••
••••
••••
• •
For
sect
ion
15:
Len
gth
of
this
sec
tion
••••
••••
••••
••••
••••
• •
Mod
ulus
of
Ela
stic
ity
••••
••••
••••
••••
••••
•• •
Are
a M
oMen
t of
In
erti
a ••
••••
••••
••••
••••
•••
• S
tiff
nes
s of
Sup
port
••••
••••
••••
••••
••••
•••
• Fo
r se
ctio
n 16
: L
engt
h of
th
is s
ecti
on ••
••••
••••
••••
••••
•••
= M
odul
us o
f E
last
icit
y ••
••••
••••
••••
••••
••••
=
Are
a M
oMen
t of
In
erti
a ••
••••
••••
••••
••••
••• =
Lef
t en
d of
lin
earl
y-va
ried
dis
t.
load
••••
• =
Rig
ht e
nd o
f li
nea
rly
-var
ied
dis
t.
load
••••
=
Sti
ffn
ess
of S
uppo
rt ••
••••
••••
••••
••••
••••
• =
Do y
ou w
ant
to c
hang
e th
e da
ta?
<Y o
r N>
H
t.20E
+003
lb
/in
4.80
E+0
02 i
n 2.
00E
+007
psi
1.
44E
+002
i r
1t4
1.20
E+0
0J l
b/i
n
4.80
E+0
02 i
n 2.
00E
+007
psi
1.
44E+
002
int4
1.
00E+
002
lb/i
n
7.50
E+00
1 lb
/in
1.
20E+
003
lb/i
n
How
Han
y in
creH
ents
wou
ld y
ou
lik
e to
hav
e fo
r ea
ch f
ield
sec
tion
?10
Figu
re C
-16.
Out
put
for
Exam
ple
C-)
,
.... ~
LEHG
TH
DEFL
ECT
I OH
SLOP
E MO
MEHT
SH
EAR
(in
) (i
n)
<Ra.
dian
> <
lb-i
n)
( 1 b
)
0.00
E+00
0 t.5
0E+0
01
-1.2
3E-0
01
0.00
E+0
00
0.00
E+0
00
1.00
E-0
21
1.50
E+00
1 -1
.23E
-001
0.
00E
+000
0.
00E
+000
2.
00E
-021
1.
50E+
001
-1.2
3E-0
01
0.00
E+0
00
0.00
E+0
00
3.00
E-0
21
1.50
E+00
1 -1
.2JE
-001
0.
00E
+000
0.
00E
+000
4.
00E
-021
1.
50E+
001
-1.2
3E-0
01
0.00
E+0
00
0.00
E+0
00
5.00
E-0
21
1.50
E+00
1 -1
.23E
-001
0.
00E+
000
0.00
E+0
00
6.00
E-0
21
1.50
E+00
1 -1
.23E
-001
0.
00E
+000
0.
00E
+000
7.
00E
-021
1.
50E+
001
-1.2
3E-0
01
0.00
E+00
0 0.
00£+
000
8.00
E-0
21
1.50
E+00
1 -1
.23E
-001
0.
00E
+000
0.
00£+
000
9.00
E-0
21
1.50
E+00
1 -1
.23£
-001
0.
00E+
000
0.00
E+0
00
t.00
E-0
20
1.50
E+00
1 -1
.23£
-001
0.
00E+
000
0.00
E+00
0 1.
00E
-020
1.
50E+
001
-1.2
3£-0
01
0.00
E+00
0 1.
80E+
004
4.80
E+00
1 2.
08E+
001
-1.1
6£-0
01
7.78
E+0
05
1.44
E+00
4 ....
9.60
E+00
1 2.
60E+
001
-9.7
8£-0
02
1.38
E+00
6 1.
06E+
004
\,,/,)
1.44
E+00
2 3.
01£+
001
-7.1
1£-0
02
1.79
E+00
6 6.
70E
+003
O>
1.92
E+00
2 3.
27E+
001
-3.9
0E-0
02
2.02
E+00
6 2.
68E
+003
2.
40E+
092
J.38
E+00
1 -4
.84E
-003
2.
05£+
006
-1.4
6E+0
03
2.88
£+00
2 3.
32E+
001
2.92
E-0
02.
1.88
E+00
6 -5
.72E
+003
3.
36£+
002
3.11
E+00
1 5.
66E
-002
1.
50E+
006
-1.0
1E+0
04
3.84
E+00
2 2.
79E+
001
7.69
E-0
02
9.06
E+00
5 -1
.46E
+004
4.
32E+
002
2.39
E+00
1 8.
56E
-002
9.
44E+
004
-1.9
2£+0
04
4.80
E+00
2 1.
99E+
001
7.98
E-0
02
-9.4
2E+0
05
-2.4
0E+0
04
4.80
E+00
2 1.
99E+
001
7.88
E-0
02
-9.4
2E+0
05
-4.2
9E+0
01
5.28
E+00
2 1.
65E+
001
6.JI
E-0
02
-9.4
4E+0
05
-4.2
9E+0
01
5.76
E+0
02
1.39
E+00
1 4.
74E
-002
-9
.46E
+005
-4
.29E
+001
6.
24E
+002
1.
20E+
001
3.16
£-00
2 -9
.48E
+005
-4
.29E
+001
6 .
. 72E
+002
1.
98E+
001
1.SS
E-0
02
-9.5
9E+0
05
-4.2
9E+0
01
7.20
E+00
2 1.
05E+
091
-7.2
8£-9
05
-9.5
2E+0
05
-4.2
9E+0
01
7.68
E+00
2 1.
09E+
001
-1. 6
0£-.0
02
-9.5
4E+0
05
--4.
29E
+001
8.
16E+
002
1.20
E+00
1 -3
.19E
-002
-9
.56E
+005
-4
.29E
+001
8.
64E+
002
1.39
£+00
1 -4
.78E
-002
-9
.58E
+005
-4
.29E
+001
9 .
. 12E
+092
1.
66E+
091
-6.J
SE-0
02
-9.6
0E+0
95
-4.2
9£+0
91
Figu
re C
-1?.
Out
put
for
Exam
ple
C-J
.
9.60
E+0
02
2.00
E+00
1 -7
.98E
-002
-9
.62E
+005
-4
.29E
+001
9.
60E+
002
2.00
E+00
1 -7
.98E
-002
-9
.62E
+005
2.
40E
+004
1.
01E+
003
2.41
E+00
1 -8
.69E
-002
7.
47E
+004
1.
92E+
004
1.06
E+00
3 2.
81E+
001
-7.8
6E-0
02
8.81
E+0
05
1.44
E+00
4 1.
10E+
003
3.15
E+00
1 -5
.SB
E-0
02
1.46
E+0
06
9.60
E+0
03
1.15
E+00
3 3.
J7E
+001
-3
.13E
-002
1.
80E
+006
4.
80E
+003
1.
20E+
003
3.44
E+00
1 -1
.07E
-014
1.
92E
+006
2.
JJE
-008
1.
25E+
003
3.37
E+00
1 3.
13E
-002
1.
80E
+006
·-
4.80
E+0
03
1.30
E+00
3 3.
15E+
001
5.88
£-00
2 1.
46E
+006
-9
.60E
+003
1.
34E+
003
2.81
E+00
1 7.
86E
-002
8.
81E
+005
-1
.44E
+004
1.
39E+
003
2.41
E+00
1 8.
69E
-002
7.
47E
+004
-1
.92E
+004
1.
44E+
003
2.00
E+00
1 7.
98E
-002
-9
.62E
+005
-2
.40E
+004
1.
44E+
003
2.00
E+00
1 7.
98E
-002
-9
.62E
+005
4.
29E+
001
1.49
E+00
3 1.
66E+
001
6.38
E-0
02
-9.6
0E+0
05
4.29
E+00
1 1.
54E+
003
1.39
E+00
1 4.
78E
-002
-9
.58E
+005
4.
29E+
001
1.58
E+00
3 1.
20E+
001
3.19
E-0
02
-9.5
6E+0
05
4.29
E+00
1 1.
63E+
003
1.09
E+00
1 1.
60E
-002
-9
.54E
+005
4.
29E+
001
.... 1.
68E+
003
1.05
E+00
1 7.
28E
-005
-9
.52E
+005
4.
29E+
001
~
1.73
E+00
3 1.
0SE+
001
-1.S
SE-0
02
-9.5
0E+0
05
4.29
E+00
1 t.7
SE+0
03
1.20
E+00
1 -3
.16E
-002
-9
.48E
+005
4.
29E+
001
1.82
E+00
3 1.
39E+
001
-4.7
4E-0
02
-9.4
6E+0
05
4.29
E+00
1 1.
87E+
003
1.65
E+00
1 -6
.31E
-002
-9
.44E
+005
4.
29E+
001
1.92
E+00
3 1.
99E+
001
-7.8
8E-0
02
-9.4
2E+0
05
4.29
E+00
1 1.
92E+
003
1.99
E+00
1 -7
.88E
-002
-9
.42E
+005
2.
40E+
004
1.97
E+00
3 2.
J9E+
001
-8.5
6E-0
02
9.44
E+0
04
1.92
E+00
4 2.
02E
+003
2.
79E+
001
-7.6
9E-0
02
9.06
E+0
05
1.46
E+00
4 2.
06E
+003
J.
11E+
001
-5.6
6E-0
02
1.50
E+0
06
1.01
E+00
4 2.
11E
+003
J.
32E
+001
-2
.82E
-002
1.
88E+
006
5.72
E+0
03
2.16
E+0
03
J.38
E+0
01
4.84
E-0
03
2.05
E+0
06
1.46
E+00
3 2.
21E
+003
J.
27E
+001
3.
90E
-002
2.
02E
+006
-2
.68E
+003
2.
26E
+003
3.
01E+
001
7.11
E-0
02
1.79
E+00
6 -6
.70E
+003
2.
J0E
+003
2.
60E+
001
9.78
E-0
02
1.38
E+00
6 -1
.06E
+004
2.
J5E
+003
2.
08E+
001
1.16
£-00
1 7.
78E+
005
-1.4
4E+0
04
2.40
E+0
03
t.50E
+001
1.
23£-
001
-4.0
6£-0
04
-1.8
0E+0
04
2.40
E+0
03
1.50
E+00
1 1.
23E
-001
-4
.06E
-004
2.
80E
-006
Fi
gure
C-1
8, O
utpu
t fo
r Ex
ampl
e c~J.
... g
1.se
2.
00
E+J
j
141
Weight of concentrated load = 3.5.59 N
Section #2
For field matrix, enter 1 for massless beam.
Length of this section
Modulus of elasticity
.Area moment of inertia
For point matrix,
Concentrated load
Moment
Stiffness of support
Support moment stiffness
Weight of concentrated load
Weight moment of inertia
= 0 • .508 m
= 2.068 X 1011 Pa
= 5 X 10-6 m4
= 7500 N
= 0 N-m
= 1.75 X 107 N/m
= 1130 N-m/rad
= 5338 N 2 = 0.287 N-m
For type of vibration, enter 1 for bending vibration.
Section #3
For field matrix, enter 0 for the last section.
Choose the boundary condition of fixed-free, enter 7.
Enter 10 increments for each section.
The micro-processor outputs the data and the deflection, slope, moment,
and shear diagrams, see Fig. C-21 to c-23.
142
Wt.=800 lb (J,559 N)
K=1 X 105 lb/in
(1.75 X 107 N/m)
P=1,686 lb (7,500 N)
Wt.=1200 lb (5JJ8 N)
T=1 X 104 lb-in/rad (1,1JO N-m/rad)
10'-' _.,....f--___ 20" -------~ (0.254 m) (0.508 m)
~---.... ~--""' "'"¥-' ...._...., ____ ~-----" '\.yJ
Figure C-20. Dynamically loaded shaft,
Dra
win
g of
the
Bea
M
DATA
: Fr
eque
ncy •
••••
••••
••••
••••
••••
••••
••••
••••
• =
1.00
E+00
1 ra
d/se
c Fo
r se
ctio
n 11
: L
engt
h of
th
is s
ecti
on ••
••••
••••
••••
••••
••• =
2.54
E-0
01 "
M
odul
us o
f E
last
icit
y ••
••••
••••
••••
••••
••••
= 2.
07E+
011
Pa
Are
a M
oHen
t of
In
erti
a ••
••••
••••
••••
••••
•••
= 5.
00E
-006
Mt4
M
agni
tude
of
Con
cent
rate
d W
eigh
t •••
••••
••••
= 3.
56E
+00J
H
For
sect
ion
12:
Len
gth
of t
his
sec
tion
••••
••••
••••
••••
••••
• =
5.08
E-0
01 "
M
odul
us o
f E
last
icit
y ••
••••
••••
••••
••••
••••
= 2.
97E+
011
Pa
Are
a M
oMen
t of
In
erti
a ••
••••
••••
••••
••••
••• =
5.99
E-0
06 M
t4
Mag
nitu
de o
f C
once
ntra
ted
Load
••••
••••
••••
• = 7
.50E
+003
H
Sti
ffne
ss o
f Su
ppor
t •••
••••
••••
••••
••••
••••
= 1.
7SE+
007
H/M
Su
ppor
t M
oMen
t S
tiff
ness
••••
••••
••••
••••
••• =
1.13
E+00
3 H
-M/ra
d M
agni
tude
of
Con
cent
rate
d W
eigh
t •••
••••
••••
m 5.
34E+
003
H
Wei
ght
MoR
ent
of I
ner
tia
<WRt
2)
abou
t a
diaH
eter
of
disc
or
eleM
ent •
••••
••••
••••
••• =
2.87
E-0
01 H
-Mt2
Do y
ou w
ant
to c
hang
e th
e da
ta?
<V o
r H>
N
How
Man
y in
cre"
ents
wou
ld y
ou
like
to
have
for
eac
h fi
eld
sec
tion
?10
Figu
re C
-21,
Out
put
for
Exam
ple
C-4
.
p
.... ~
LE HG
TH
DEFL
ECT
I OH
SLOP
E M
Ot1E
t .. T
SHEA
R (M
) (M
) (R
Gdi
Gn)
(H
-M)
(H)
0.00
E+00
0 0.
00E+
000
0.00
E+0
00
-1.6
4E+0
03
2.54
E-0
02
5.05
E-0
07
-3.9
6E-0
05
-1.5
9E+0
03
5.0B
E-0
02
2.00
E-0
06
-7.7
8E-0
05
-1.5
3E+0
03
7.62
E-0
02
4.45
E-0
06
-1.1
5E-0
04
-1.4
8E+0
03
1.02
E-0
01
7.81
E-0
06
-1.5
0E-0
04
-1.4
2E+0
03
1.27
E-0
01
1.21
E-0
05
-1.8
4E-0
04
-1.3
7E+0
03
1.52
E-0
01
1.72
E-0
05
-2.1
7E-0
04
-1.3
1E+0
03
1.78
E-0
01
2.31
E-0
05
-2.4
9E-0
04
-1.2
6E+0
03
2.03
E-0
01
2.98
E-0
05
-2.7
9E-0
04
-1.2
0E+0
03
2.29
E-0
01
3.73
E-0
05
-3.0
8E-0
04
-1.1
5E+0
03
2.54
E-0
01
4.54
E-0
05
-3.3
5E-0
04
-1.0
9E+e
03
2.54
E-0
01
4.54
E-0
05
-3.J
SE
-004
-1
.09E
+e03
3.
05E
-001
6.
JSE
-005
-3
.86E
-004
-9
.83E
+002
3.
56E
-001
8.
46E
-005
-4
.32E
-004
-8
.74E
+e02
4.
06E
-001
1.
08E
-004
-4
.72E
-004
-7
.6SE
+e02
4.
57E
-001
1.
32E
-004
-5
.07E
-004
-6
.55E
+e02
5.
08E
-001
1.
59E
-004
-5
.36E
-004
-5
.46E
+f02
5.
59E
-001
1.
87E
-004
-5
.60E
-004
-4
.37E
+002
6.
10E
-091
2.
16E
-004
-5
.79E
-004
-J
.27E
+002
6.
60E
-001
2.
46E
-004
-S
.93E
-004
-2
.18E
+002
7.
11E
-001
2.
76E
-004
-6
.01E
-004
-1
.09E
+002
7.
62E
-001
3.
07E
-004
-6
.03E
-004
6.
64E
-001
7.
62E
-001
3.
07E
-004
-6
.03E
-004
-1
.92E
-011
Do ~ou
wis
h to
see
the
gra
phs
for
def
lect
ion
, sl
ope,
M
OMen
t an
d sh
ear?
<Y
or H
>Y
Figu
re C
-22.
Out
put
for
Exam
ple
C-4
.
2.15
E+00
3 2.
15E+
003
2.15
E+00
3 2.
15E+
003
2.15
E+00
3 2.
15E+
003
2.15
E+00
3 2.
15E+
003
2.15
E+00
3 2.
15E+
003
2.15
E+0
0J
2.15
E+00
3 2.
15E+
003
2.15
E+00
3 2.
15E+
003
2.15
E+00
3 2.
15E+
003
2.1S
E+00
3 2.
15E+
003
2.1S
E+00
3 2.
1SE+
003
2.15
E+00
3 -1
.46E
-010
.... ~
Freq
uenc
y=10
rQ
d/se
c E-
4 2.
00
DEF
L.,
W
( ,., )
0.00
-2.0
0
E-4
SLOP
E,
S -5
.00
<rad
>
-10.
00
E+J
0.00
HO
HEHT
, M
<H
-M)
-1.0
0
-2.0
0
E+J
2.00
SH
EAR,
lJ
<H>
9.00
-2.ee
- -0.
00
2.00
LE
NGTH
(")
Fi
gure
C-2
3. O
utpu
t fo
r Ex
ampl
e C
-4. '·
·
.... ~
4.88
6.
00
E-1
APPENDIX B
APPENDIX.. B .1
PROGRAM LISTING --- FATIGUE ANALYSIS
146
150
IHIT
16
0 PA
GE
161
PRI
•*****
******
******
******
******
******
******
******
******
******
***"
162
PRI
•* *"
163
PRI
"* FA
TIGU
E AN
ALYS
IS P
ROGR
AM,
ANAL
YTIC
AL
*" 16
4 PR
I "*
tu
165
PRI
•* By
Yiu
Wah
Luk
, UP
I &
SU,
Spri
ng 1
978
*" 16
6 PR
I "*
*" 16
7 PR
I "**
******
******
******
******
******
******
******
******
******
******
" 17
0 1=
0 18
0 t1
1=0
190
M2=
0 20
0 F1
=0
210
F2=0
22
0 T1
=0
230
T2=0
....
240
J=0
-:S 25
0 V=
0 26
0 J5
=0
265
N1=0
27
0 DI
M A
$(1)
,8$(
6),C
$(J)
,0$(
5),E
$(J)
,F$(
31),
G$(
8)
280
PRIN
T "J
JVou
hav
e en
tere
d a
FATI
GUE
RESI
STAN
T,
INTE
RACT
IUE
DESI
GH
";
290
PRI
•rou
tine.
_CoM
pone
nts
wil
l be
siz
ed t
o pr
even
t fa
tigu
e fa
ilur
e."
300
PRIH
T "J
Do
you
wan
t to
use
Eng
lish
Uni
ts?
CY
or N
>";
310
INPU
T A$
32
0 IF
Af=
"V"
THEN
420
33
0 IF
Af<
>"H
• TH
EN 3
00
340
PRIH
T "J
Me
wil
l us
e S
I un
its
thro
ugho
ut t
his
rout
ine.
• 35
0 J=
J 36
0 8$
=•1"
111
370
C$=
11Pa
11
380
0$="
H-r1
" 39
0 E$
="H
11
400
G$=
"0eg
ree
C"
"'to:
:.t:.1
o
+-
incn
e::
:>--
430
C$=
11p
si11
440
D$=
11lb
-in
11
450
E$=
11lb
f11
460
G$=
11D
e9re
e F"
47
0 PR
INT
"JW
e w
ill
use
Eng
lish
Uni
ts
thro
ugho
ut
this
ro
uti
ne.
" 48
0 PR
INT
''JP
leas
e en
ter
the
stre
ngth
of
the
Mat
eria
l to
be
used
.J"
4 9
0 PR
It~ T
11 T
e n s
i 1 e
S
tr e n
g th
, i r
1 11
; C
$ ;
" ,
Su
= " ;
50
0 IN
PUT
S1
510
PRIN
T "Y
ield
Str
engt
h,
520
INPU
T S2
in
11
; C
$; "
, Sy
-
II
• -
' 53
0 PR
INT
"JD
o yo
u kn
ow
the
Sig
nif
ican
t E
ndur
ance
LiM
it o
f th
e ";
54
0 PR
INT
"Mnt
eria
l? C
Y or
H)"
; 55
0 IN
PUT
A$
560
IF A
$="Y
" TH
EN
590
570
GOSU
B 27
30
580
GO
TO
596
590
PRIN
T "S
igni
fica
nt E
ndur
ance
LiM
it,
in
";C
f;",
S
e'''
592
INPU
T SJ
59
4 J4
=1
596
GOSU
B 24
60
600
PAGE
64
0 PR
IHT
650
PRIN
T 66
0 PR
INT
670
PRIN
T 68
0 PR
INT
690
PRIN
T 70
0 PR
INT
?10
INPU
T 72
0 M
=1
· 730
p·
=t
740
R1=1
75
0 R2
=1
"Sel
ect
the
fati
gue
fail
ure
li
ne
to b
e us
ed
in
ulf
Mod
ifie
d G
oodM
an
Fra
ctur
e L
ine,
en
ter
MGF
" "I
f M
odif
ied
Goo
dMan
Y
ield
Lin
e,
ente
r M
GY"
"If
Sod
erbe
rg L
ine,
en
ter
S"
"If
Ger
ber
Lin
e,
ente
r G"
•J
f Q
uadr
atic
Lin
e,
ente
r Q"
"I
f K
ecec
iogl
u L
ine,
en
ter
K"
Rf
-".
-
'
the
desi
9n.J
11
...... &
769
IF R
t••H
GF•
TH
EN 9
40
770
IF R
S••M
GV"
THEN
870
78
0 IF
R$=
"S"
THEH
940
79
0 IF
R$=
"G"
THEH
980
80
0 IF
Rt=
"Q"
THEH
10
20
810
IF R
$="K
" TH
EH
1070
82
0 PR
INT
"JV
our
resp
onse
May
hav
e be
en i
n er
ror;
TR
Y A
GA
IN.•
830
GO T
O 64
0 84
0 F$
="M
odif
ied
Good
Man
Fra
ctur
e L
ine.
" 85
0 Y=
0 86
0 GO
TO
1140
87
0 F$
="M
odif
ied
Good
Man
Yie
ld L
ine.
" 88
0 IF
J4=
1 TH
EH 9
10
890
V=1
900
GO
TO 9
30
910
R1=
S1/S
2 92
0 R
2=S2
/SJ
930
GO T
O 11
40
940
F$="
Sode
rber
g L
ine.
" 95
0 R1
=S1.J
S2
960
V=0
970
GO T
O 11
40
980
F$="
Ger
ber
Lin
e.•
990
P=2
10
00 '
(=0
1010
GO
TO
1140
10
20 F
t="Q
uadr
atic
Lin
e.•
1030
H=2
10
40 P
=2
1050
Y=0
10
60 G
O TO
11
40
1070
Ft=
"Kec
ecio
glu
Lin
e.•
1080
PRI
NT
•for
Kec
ecio
glu
Lin
e,
you
have
to
supp
ly K
ecec
iogl
u's
• 10
90 P
RIHT
"C
onst
ant
whi
ch i
s de
pend
ed o
n th
e M
ater
ial
prop
erty
.•
1100
PRI
NT
"Kec
ecio
glu'
s C
onst
ant,
H
• "J
... $
1110
INP
UT H
11
20 Y
=0
1130
P=2
11
40 P
RINT
•J
The
FQi1
ure
Lin
e se
lect
ed i
s •1
Ft
1150
IF
Y=1
TH
EN 1
189
1160
PRI
"T
he f
ailu
re e
quQ
tion
is (
SQ/(
R2*
Se'''
)tM
+
<Rl*
SM/S
u>tp
•
1.•
11
70 P
RI
"whe
re H
=•JH
;u_w
here
P="
1P;"
_whe
re R
1="1
R11
•_an
d w
here
R2=
";R2
1180
PRI
NT
•oo
you
wis
h to
chQ
nge
any
para
Met
ers?
CV
or H
>";
1190
INP
UT A
$ 12
00 I
F A
$="V
• TH
EN 6
40
1201
PR
IHT
•JTh
e fo
llow
ing
entr
ies
wil
l es
tabl
ish
the
liM
its
on a
Hal
f "
1202
PRI
NT "
Inte
rval
Sea
rch
for
the
solu
tion
to
your
pro
bleH
."
1210
PRI
HT
"Wha
t is
the
sM
alle
st b
asic
diM
ensi
on t
hat
you
wis
h to
try
?n
1220
PRI
NT •
Giv
e yo
ur a
nsw
er i
n ";
8$;•
. Do
not
ans
wer
0.0
. •;
12
30 J
5=0
1240
IH
PUT
01
1250
PRI
NT
•wha
t is
the
1Q
rges
t di
Hen
sion
, in
"18
$;"?
•;
1260
IHP
UT 0
2 12
70 D
4=Dl
12
80 D
3=D2
12
90 D
=DI
1300
GOS
UB 2
440
1310
IF
J4=
1 TH
EN 1
360
1320
GOS
UB 4
890
1330
IF
V=0
THE
N 13
60
1340
Rl=
S1/S
2 13
50 R
2=S2
/S3
1360
X1=
<A1/
(R2*
SJ))
tH+<
A2*
R1/
Sl)t
P-1
1370
FOR
T=l
TO
10
0 13
80 D
=<D
4+D
J)/2
13
90 G
OSUB
244
0 14
00
IF J
4=1
THEH
145
9 14
10 G
OSUB
489
0 14
20
IF V
=9 T
HEH
1450
14
30 R
1=S1
/S2
... 'cS
1440
R2•
S2.tS
3 14
50 X
2•<A
1.t<
R2*S
3>>t
H+<
A2*
R1.tS
l>tP
-1
1460
IF
ABS
<X2>
<•1.
0E-5
THE
N 15
40
1470
IF
X1*X
2<=0
THE
N 15
10
1480
04=
D 14
90 X
1=X2
15
00 G
O TO
15
20
1510
D3=
D 15
20 H
EXT
T 15
30 J
5=1
1540
PAG
E 15
50 P
RINT
USI
NG
•72•
"*••
":
1560
IMA
GE C
FD,1
x,s
1570
IMA
GE C
FD.2
0,1x
,s 15
80 P
RIHT
•J
Ten
sile
Str
engt
h,
Su •
••••
••••
••••
••••
••••
••••
• =
";
1590
PRI
NT U
SIHG
15
60:S
1 16
00 P
RINT
C$
1610
PRI
HT
"Yie
ld S
tren
gth,
SY
••••
••••
••••
••••
••••
••••
••••
=";
~
1620
PRI
NT U
SING
15
60:S
2 ~
1630
PRI
NT C
$ 16
31
IF N
1<=0
THE
N 16
40
1632
IF
H1=
>109
0000
THE
N 16
40
1633
PRI
NT
"Sig
nifi
cant
End
uran
ce L
iMit
, S
e'''@
";N
t;"
eye.
= ";
16
34 G
O TO
16
50
1640
PRI
HT
"Sig
nifi
cant
End
uran
ce L
iMit
, S
e''
for
infi
nit
e li
fe=
";
1650
PRI
NT U
SIHG
15
60:S
3 16
60 P
RIHT
Cf
1670
PRI
NT
"SM
alle
st d
iMen
sion
tri
ed ••
••••
••••
••••
••••
••••
=";
16
80 P
RIHT
USI
NG
1570
:01
1690
PRI
NT 8
$ 17
00 P
RINT
"L
arge
st d
iMen
sion
tri
ed ••
••••
••••
••••
••••
••••
• =•;
17
10 P
RINT
USI
NG
1570
:02
1720
PRI
NT 8
$ 17
30 P
RINT
"S
afet
y F
acto
r,
H••
••••
••••
••••
••••
••••
••••
••••
= "
I 17
40 P
RINT
USI
NG 1
570:
N
1759
PRI
NT
1760
IF
H1•
9 TH
EN 1
809
1770
PRI
NT
•Ho"
ent
caus
ing
alte
rnat
ing
stre
ss,
Mt •
••••
••••
=
"I
1780
PRI
NT U
SING
156
0:M
1 17
90 P
RIHT
D$
1800
IF
H2=
0 TH
EH
1840
18
10 P
RIHT
"M
o"en
t ca
usin
g a
stea
dy s
tres
s, H
2 •••
••••
••••
• = "
; 18
20 P
RIHT
USI
NG
1560
:H2
1830
PRI
HT D
S 18
40 I
F F1
=0 T
HEN
1880
18
50 P
RIHT
"A
lter
nati
ng a
xial
fo
rce,
·Ft
•••
••••
••••
••••
••••
= ";
1860
PRI
HT U
SIHG
156
0:F1
18
70 P
RINT
E$
1880
IF
F2=
0 TH
EH 1
920
1890
PRI
HT
•ste
ady
axia
l fo
rce,
F2
•••
••••
••••
••••
••••
••••
• =
•; 19
00 P
RIHT
USI
HG 1
560:
F2
1910
PRI
HT E
$ 19
20 I
F T1
=0 T
HEN
1960
19
30 P
RIHT
"A
lter
nati
ng t
orqu
e,
Tl •
••••
••••
••••
••••
••••
••• =
"; 19
40 P
RIHT
USI
HG
1560
:T1
1950
PRI
NT D
$ 19
60 I
F T2
=0
THEH
200
0 19
70 P
RINT
•s
tead
y to
rque
, 12
••••
••••
••••
••••
••••
••••
••••
• = •1
1980
PRI
NT U
SING
15
60:T
2 19
90 P
RINT
0$
2000
IF
J4=
1 TH
EH 2
210
2010
PRI
HT
"JTh
e fo
llow
ing
fact
ors
are
used
for
co"
puti
ng S
e'''
:"
2020
IM
AGE
CFD
.2D
20
30 P
RINT
"S
urfa
ce f
acto
r <K
a>
= "J
20
40 P
RINT
USI
NG 2
020:
K1
2050
PR
INT
usiz
e an
d sh
ape
fact
or <
Kb)
• "J
20
60 P
RINT
USI
HG 2
020:
K2
2070
PRI~T
"Rel
iabi
lity
fac
tor
<Kc>
•
"J
2080
PR
INT
USIN
G 20
20:K
3 20
90 P
RINT
•t
eMpe
ratu
re f
acto
r <K
d>
= "I
.... ~
2190
PRI
NT U
SIHG
202
9:K
4 21
10 P
RINT
"F
atig
ue S
tren
gth
Red
uctio
n f4
ctor
<Ke
>•
•;
2120
PRI
NT U
SING
292
9:K
5 21
30 P
RINT
•M
isce
114n
eous
f4c
tor
<Kf>
=
•; 21
40 P
RINT
USI
NG 2
020:
K6
2150
PRI
NT
•End
ur4n
ce L
i"it
for
Rot
4tin
g-be
4M S
peci
Men
<Se
'>
= ";
2160
PRI
NT U
SIHG
15
60:S
5 21
70 P
RINT
C$
2180
PRI
NT
"Sig
nifi
c4nt
End
ur4n
ce L
iHit
for
in
fin
ite
life
<Se
''>=
•;
2190
PRI
NT U
SIHG
15
60:S
6 22
00 P
RINT
C$
2210
PRI
HT
"JTh
e F
4ilu
re L
ine
sele
cted
is
•1Ft
22
20 P
RINT
"T
he f
4ilu
re e
quat
ion
is <
Sa/(
R2*
Se'''
)tM
+
CR1*
SM/S
u)tp
= 1
."
2230
PRI
•w
here
M=•
;M;"
_whe
re P
=";P
;•_w
here
R1=
";R
t;•_
and
whe
re R
2=";R
2 22
40
IF J
5=1
THEH
226
0 22
50 G
O TO
229
0 22
60 P
RINT
•JA
sol
utio
n do
es n
ot e
xist
in
your
spe
cifi
ed r
ange
of
" ~
2270
PRI
NT
•diH
ensi
ons.
Sor
ry a
bout
tha
t,
CHIE
F! •
•••
Plea
se t
ry a
gain
."
~
2280
GO
TO
1210
22
90 P
RINT
"JT
he d
esig
n di
Men
sion
is
•;
2300
PRI
HT U
SIHG
"c
fd.4
d,1x
,s•:
o 23
10 P
RINT
8$
2320
PRI
HT
2330
PRI
NT U
SING
•7
2u•*
••n:
23
31
IF J
=3 T
HEN
2340
23
32 P
RINT
"J
Do
you
wis
h to
con
vert
the
des
ign
di"e
nsio
n to
SI
";
2333
PRI
NT
•uni
t? C
V or
H>•
• 23
34 D
=D*B
.025
4 23
35 H
f="M
" 23
36 G
O TO
234
5 23
40 P
RINT
•JD
o yo
u w
ish
to c
onve
rt t
he d
esig
n di
Mcn
sion
to
Eng
lish
•;
2341
PR
INT
•uni
t? <
V or
H>•
• 23
42 D
=D*3
9.37
0978
74
2343
Htm
•inc
hes•
23
45
INPU
T Af
2346
IF
At•
•N"
THEH
236
4 23
47 I
F AS
<>"V
• TH
EN 2
331
2350
PRI
NT
•JTh
e de
sign
di"
ensi
on i
s "I
23
51
PRIH
T US
ING
"cfd
.4d,
lx,s
•:o
2352
PRI
NT H
$ 23
64 P
RINT
"J
Do
you
wis
h to
red
esig
n th
e co
"pon
ent
usin
g an
othe
r "
2366
PRI
HT
0fa
ilur
e li
ne?
War
ning
! A
Mod
ifie
d G
ood"
an a
ppro
ach
reQ
uire
s"
2368
PRI
NT
"bot
h a
frac
ture
and
a y
ield
ana
lysi
s.
<V o
r N
)•;
2370
INP
UT A
$ 23
80
IF A
$="Y
" TH
EN 6
40
2390
PRI
NT
"JD
o yo
u w
ant
to d
esig
n a
new
coff
pone
nt?
<Y o
r N>
"; 24
00 I
NPUT
A$
2410
IF
Af=
"Y"
THEN
150
24
20 P
RIHT
"J
JJ
---
END
---"
24
30 G
O TO
272
0 24
40 I
F I=
0 TH
EN 2
460
2450
GO
TO 6
000
2460
PAG
E 24
61
PRIH
T "Y
ou w
ill
now
be r
eque
sted
to
supp
ly t
he P
ARAM
ETRI
C DE
SIGN
•;
24
62 P
RIHT
"S
TRES
S_EQ
UATI
OHS
for
your
des
ign
prob
leM
."
2463
PRI
"Y
ou M
ust
wri
te t
hese
equ
atio
ns i
n BA
SIC.
U
se
the
foll
owin
g ";
24
64 P
RINT
"i
nstr
ucti
ons.
_If
you
are
unfa
Mil
iar
wit
h th
e de
velo
pMen
t •;
24
65 P
RIHT
"o
f su
ch e
quat
ions
,_se
e u
ser'
s gu
ide
and
its
appe
ndix
•;
2466
PRI
HT
"for
gui
danc
e."
2470
PRI
"E
nter
coM
pone
nt
load
s st
arti
ng
wit
h li
ne 6
000
incr
eMen
ting
";
2480
PRI
HT
"by
10
's."
24
90 P
RIN
T" U
se M
l=M
oMen
t ca
usin
gHHH
HHHH
____
___
alte
rnat
ing
stre
ss•;
24
95 P
RIHT
•c
";D
t;•>
."
2500
PRI
HT
" Us
e M
2=M
oMen
t ca
usin
gHHH
HHHH
____
___
a st
eady
str
ess
";
2505
PRI
HT
"C";
Dt;
•>."
25
19 P
RINT
•
Use
Ft=A
n al
tern
atin
g ax
ial
forc
e <
";E
t;•>
."
2520
PRI
NT
• Us
e F2
•A s
tead
y ax
ial
forc
e <
";E
t;">
."
2530
PR
INT•
Use
Tl=
An
alte
rnat
ing
torq
ue <
"JD
t1•>
."
2540
PR
INT"
Use
T2=
A st
eady
tor
Que
<
"ID
tJ•)
."
2550
PRI
HT
•Use
N
=Saf
ety
fact
or."
~
'$.
2560
PRI
NT
•Ent
er t
he n
uRar
ic v
alue
for
eac
h of
the
var
iabl
es u
sed
in •
25
70 P
RINT
•y
our
stre
ss e
quat
ions
._D
o th
is b
efor
e yo
u en
ter
your
"I
2580
PRI
HT
"str
ess
equa
tion
s."
2590
PRI
HT
"JE
nter
you
r st
ress
equ
9tio
ns s
tart
ing
wit
h li
ne 6
100
" 26
00 P
RINT
"i
ncre
Men
ting
by
10
's.•
26
10 P
RI
" U
se A
1=A
lter
nati
ng s
tres
s.
_Use
A2•
Mea
n st
ress
. _U
se P
I=";
26
20 P
RINT
"P
i.
_Use
D=B
asic
diM
ensi
on.
H.B
., A
ll di
"ens
ions
sho
uld
";
2630
PRI
NT
"be
give
n in
te
r"s
of D
. In
cas
e of
rec
tang
ular
coM
pone
nt,
";
2640
PRI
"u
se p
ropo
rtio
ns.
_Aft
er y
our
stre
ss e
quat
ions
are
ent
ered
, ";
26
50 P
RINT
•ty
pe a
nuR
bere
d re
turn
_sta
te"e
nt.
Then
ty
pe r
un 6
00."
26
60 P
R I H
T "E
XAM
PLE •
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
•••
II
2670
PRI
HT
•The
fo
llow
ing
equa
tion
s ar
e in
Eng
lish
Uni
t.•
2680
PRI
HT
"600
0 T2
=120
0.
_601
0 M
1=24
00.
_602
0 N
=1.8
_61
00 A
l=H
tMl/
11;
2690
PRI
HT
"<PI
*DfJ
/32>
_61
10 A
2=0.
866*
NtT
2/(P
l*D
tJ/J
2) _
6120
RET
URN
" 27
00 P
RINT
"R
UH 6
00J"
27
10
I=I+
l 27
20 E
HD
2730
REM
***
SUBR
OUTI
NE F
OR F
INDI
NG S
E'~'.
2740
REM
***
THI
S SE
CTIO
N CA
LCUL
ATES
Ka
<Kt>
. 27
50 J
4=2
2760
PRI
11LT
his
sect
ion
of t
he p
rogr~H
wil
l ca
lcul
ate
the
Sig
nifi
cant
"
2770
PRI
HT
"End
uran
ce L
iMit
.•
2780
PRI
HT
"JE
nter
the
I
for
the
type
s of
sur
face
fin
ish
used
."
2790
PRI
HT
"11
for
poli
shed
fin
ish.
u 28
00 P
RINT
"1
2 fo
r gr
ound
fin
ish.
" 28
10
PRIH
T "1
3 fo
r M
achi
ned
or c
old
draw
n."
2820
PRI
HT
"14
for
hot
roll
ed."
28
30 P
RINT
"I
S f
or a
s fo
rged
."
2840
IH
PUT
11
2850
IF
J=3
THE
N 28
80
2860
S8=
S1
2870
GO
TO 2
890
2880
S8=
S1*1
.450
3773
97E
-4
2890
GO
TO
11
OF 2
930,
2950
,297
0,29
98130
10
2909
PRI
NT
"JE
rror
' Y
our
inpu
t sh
ould
be
fro"
1 t
o 5,
bu
t yo
u ha
ve "
..... ~
2910
PRI
NT
"ent
ered
•11
11•.
Pl
ease
try
aga
in."
29
20 G
O TO
278
0 29
30 K
1=1
2940
GO
TO 3
020
2950
K1=
0.89
29
60
GO T
O 30
20
2970
K1=
-2.9
1E-1
7*S8
*SS*
S8+2
.9E-
11*S
8*S8
-4.9
5E-6
*S8+
1.06
4 29
80 G
O TO
302
0 29
90 K
1=-5
.77E
-17*
S8*S
8*S8
+3.4
1E-1
1*S8
*S8-
8.0E
-6*S
8+1.
066
3000
GO
TO 3
028
3010
K1=
-6.4
5E-1
7*SB
*S8*
S8+3
.6JE
-11*
S8tS
8-7.
87E-
6*S8
+0.8
9 30
20
IF K
1<=1
TH
EN 3
030
3022
K1=
1 30
26 R
EM **
* THI
S SE
CTIO
N CO
MPU
TES
Kc
CKJ>
. 30
30 P
RIHT
"J
Wha
t is
the
rel
iab
ilit
y i
n %
?"1
3040
IN
PUT
12
3050
P1=
100-
12
3060
IF
P1>
0 TH
EN 3
100
3070
PRI
NT
"Err
orl
The
reli
abil
ity
sho
uld
be
in%
, an
d ca
nnot
be
" 30
80 P
RINT
"g
reat
er t
han
100.
Pl
ease
try
aga
in.•
30
90 G
O TO
303
0 31
00 P
2=LG
T<P1
> 31
10 2
1=2.
37-0
.885
tP2-
0.19
3*P2
*P2-
0.05
02*P
2t3-
0.00
489*
P2t4
31
20 K
3=1-
0.08
*21
3130
REM
*** T
HIS
SECT
ION
COM
PUTE
S Kd
<K
4>.
3140
PRI
NT
0JW
hat
is
the
oper
atin
g te
Mpe
ratu
re i
n ";
Gt;
" ?"
; 31
50
IHPU
T IJ
31
60
IF J
<>J
THEN
318
0 31
70 1
3=9*
13/5
+32
3180
IF
IJ<
=160
THE
N 32
10
3190
K
4=62
0/(4
60+1
3)
3200
GO
TO 3
220
3210
K4=
1 32
20 R
EH **
* THI
S SE
CTIO
N CO
MPU
TES
Ke
<KS>
. 32
30 P
RINT
"J
Do
you
know
th
e T
heor
etic
al
Str
ess
Con
cent
rati
on "
J
.... ~
3249
PRI
NT
"Fttc
tor
<Kt>
?"I
3250
INP
UT A
$ 32
60 I
F A$
="Y"
TH
EN 3
398
3270
IF
At<
>"N
" TH
EN 3
230
3280
PRI
NT
"JTh
e fo
llow
ing
refe
renc
es w
ill
help
you
to
fin
d th
e "
3290
PRI
NT "
The
oret
ical
Str
ess
Con
cent
rati
on F
acto
r:"
3300
PRI
NT
"JR
.E.
Pete
rson
, S
tres
s C
once
ntra
tion
Fac
torHH
HHHH
HHHH
HHHH
H";
3310
PRI
HT
"HH
HH
HH
HH
HH
HH
----
----
----
----
----
----
---'
John
Wiley~
Sons
,"
3315
PRI
HT
"N.
Y.,
1974
."
3320
PRI
NT
"JJ.
E.
Shig
ley,
Hec
httn
ical
E
ngin
eeri
ng D
esign
HHHH
HHHH
HHH"
; 33
25 P
RIHT
"H
HH
HH
HH
HH
HH
HH
HH
HH
H--
----
----
----
----
----
----
---'
3rd
Ed.
,"
3330
PRI
HT
"McG
raw
-Hill
, 19
77,
p.
663-
670.
" 33
40 P
RI
"JA
.D.
Deu
tschM
an,
W.J.
M
iche
ls,
C.E
. W
ilson
, M
achi
ne D
esig
n";
3345
PRI
NT
"HHH
HHHH
HHHH
HHH _
____
____
____
_ ,•
3350
PRI
NT
"Mac
Mill
an P
ubli
shin
g C
o.,
Inc.
N
.Y.,
1975
, p
. 89
4-90
1."
3360
PRI
NT
"JL.
So
rs,
Fati
gue
Des
ign
of M
achi
ne C
oHpo
nents
HHHH
HHHH
"; 33
65 P
RINT
"H
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
H---
------
------
------
-----"
; 33
66 P
RIHT
"--
----
----
•"
· 33
70 P
RIHT
"Pe
rga"
on P
ress
, 19
71,
p. 4
2-84
of
part
II.
•
3372
PRI
HT
"JJD
o yo
u w
ish
to l
eave
.the
pro
gra"
to
look
up
the
Str
ess
" 33
73 P
RIHT
"C
once
ntra
tion
Fac
tors
? <V
or
H>";
3374
IN
PUT
A$
3375
IF
A$="
Y" T
HEN
2420
33
76
IF A
f<>"
H"
THEN
337
2 33
90 P
RINT
"J
The
oret
ical
Str
ess
Con
cent
rati
on F
acto
r =
"; 34
00
INPU
T 14
34
10 P
RIHT
"J
Do
you
know
Not
ch S
ensi
tivi
ty (
q)?
<Y o
r H
>";
3420
IHP
UT A
$ 34
30
IF A
$="H
" TH
EN 3
490
3440
IF
A$<>
"Y"
THEN
341
0 34
50 P
RIHT
"N
otch
Sen
siti
vity
(q)
•
"J
3460
J2=
1 34
70 I
NPUT
Q
3480
GO
TO 4
230
3490
PRI
NT
•wha
t is
the
not
ch r
adiu
s in
";B
fJ-?
"J
.... ~
3508
INP
UT R
35
10 P
RINT
•J
is t
he "
ater
ial
stee
l? <
V or
H>"
J 35
20 I
NPUT
A$
35
30 I
F At~"H"
THEH
409
0 35
40 I
F A
f<>"
V"
THEH
349
0 35
50 J
2=1
3560
IF
J<>J
THE
N 35
80
3570
R=R
*39.
3700
7874
35
80 P
RIHT
"I
s it
und
er b
endi
ng o
r ax
ial
load
ing?
<V
or H
>"J
3590
IHP
UT A
$ 36
00 I
F A$
="N"
TH
EN 4
020
3610
IF
A$(
)uyu
TH
EH 3
580
3620
IF
J=3
THEH
365
0 36
30 S
4=S1
36
40 G
O TO
366
0 36
50 S
4=S1
*1.4
5037
7397
E-4
3660
IF
S4>5
0000
THE
H 37
20
3670
IF
R<=
0.16
THE
N 37
00
3680
Q=0
.8
3690
GO
TO 4
230
3700
Q=-
882B
*Rt4
+334
5.3*
RtJ
-440
.94*
R*R
+24.
62*R
+0.1
8 37
10 G
O TO
423
0 37
20 I
F S4
>600
00 T
HEN
3780
37
30 I
F R
<=0.
16 T
HEN
3760
37
40 Q
=0.8
37
50 G
O TO
423
0 37
60 Q
=-70
31.2
S*R
t4+2
671.
9*R
tJ-3
53.1
3*R
t2+2
0.2*
R+0
.28
3770
GO
TO 4
230
3780
IF
S4>8
0000
THE
N 38
40
3790
IF
R>0
.16
THEN
382
0 38
00 Q
=-10
156.
25*R
t4+3
825*
RtJ
-497
.5*R
*R+2
7.05
*R+0
.23
3810
GO
TO 4
230
3820
Q=0
.9
3839
GO
TO
4230
38
40 I
F S4
>100
000
THEN
390
0
.... '<£
3859
IF
R>0
.15
THEN
389
9 38
60 Q
a-15
957.
39*R
t4+5
16S.
4iR
t3-6
96*R
*R+2
9.23
*R+0
.3
3870
GO
TO 4
230
3880
Q=0
.92
3890
GO
TO 4
230
3900
IF
S4>1
4000
0 TH
EN 3
970
3910
IF
R>0
.08
THEH
394
0 39
20 Q
=S43
1250
*Rt5
-123
6125
*Rt4
+194
242.
5*R
t3-4
010.
7*R
*R+7
1.06
*R+0
.33
3940
GO
TO 4
230
3950
Q=0
.92
3960
GO
TO 4
230
3970
IF
R>0
.06
THEN
409
0 39
80 Q
=-27
1319
*Rt4
+372
76.5
*Rt3
-177
1*R
*R+3
5.03
*R+0
.67
3990
GO
TO 4
230
4000
Q=0
.965
40
10 G
O TO
423
0 40
20 I
F S4
>600
00 T
HEN
4040
40
30 G
O TO
379
0 40
40 I
F S4
>800
00 T
HEN
4060
40
50 G
O TO
385
0 40
60 I
F S4
>120
000
THEN
408
0 40
70 G
O TO
392
0 40
80 G
O TO
397
0 40
90 P
RINT
u
Jis
the
"ate
rial
A
lu"i
nuM
all
oy?
CV o
r H
>";
4100
IN
PUT
At
4110
IF
At=
"N"
THEN
419
0 41
20 I
F A
t<>"
V"
THEN
409
0 41
30 J
2=2
4149
IF
R>0
.16
THEN
417
0 41
50 Q
=-88
15.2
*Rt4
+341
1.3*
Rt3
-462
.64*
R*R
+27.
85*R
+0.0
13
4160
GO
TO
4230
41
70 Q
•0.8
5 41
80 G
O TO
423
0 41
90 P
RIN
T ~Jsorry!
You
have
to
supp
y th
e N
otch
Sen
siti
vity
Fac
tor.
" 42
00 P
RIN
T "N
otch
Sen
siti
vity
(q)
•
"I
.... ~
4210
J2=
0 42
20 I
HPUT
Q
4230
K5=
1/(1
+Q*(
14-1
>>
4240
PRI
HT
•Jis
the
re a
ny "
isce
llan
eous
-eff
ect
fact
or?
<Y o
r H
>"I
4250
IH
PUT
A$
4260
IF
A
$•"Y
• TH
EH 4
300
4270
IF
A$<
>"H
" TH
EN 4
240
4280
K6=
1 42
90 G
O TO
432
0 43
00 P
RIHT
"M
isce
llan
eous
-eff
ect
Fac
tor=
•1
4310
IN
PUT
K6
4320
PRI
HT
"JTh
e S-
N c
urve
is
used
for
th
is d
eter
"ina
tion
. •
4321
PR
IHT
•A L
og-L
og o
r a
Log
-Lin
ear
S-N
cur
ve w
ill
be u
sed.
"
4322
PRI
NT
•The
fin
ite
life
reg
ion
is-0
.9*S
u @
1EJ
cycl
es t
o S
e''
";
4323
PRI
NT
"@1E
6 cy
cles
."
4329
PRI
NT
"Wha
t "e
thod
do
you
wan
t to
use
for
co"
puti
ng t
he "
43
30 P
RINT
"S
igni
fica
nt E
ndur
ance
LiM
it <
Se'
'')?"
43
40 P
RINT
"E
nter
11
for
Log-
Log
Met
hod.
" 43
50 P
RIHT
"E
nter
12
for
Log
-Lin
ear
Met
hod.
" 43
60 I
NPUT
J1
4370
GO
TO J
I OF
442
0,44
20
4380
PRI
NT
"JE
rror
' Y
our
inpu
t sh
ould
be
1 or
2,
but
you
hove
ent
ered
";
4390
PRI
NT J
1 44
00 P
RINT
"P
leas
e tr
y ag
ain.
"
4410
GO
TO 4
320
4420
REH
*** I
HFOR
MAT
IOH
FOR
COM
PUTI
NG K
b <S
IZE
FACT
OR>.
4430
IF
J2=
1 TH
EN 4
560
4440
PRI
HT
"Jis
the
Mat
eria
l li
ght
allo
y? <
V or
H>"
; 44
59
IHPU
T A
t 44
60
IF A
$="Y
" TH
EN 4
550
4470
IF
Af<
>"H
" TH
EH 4
440
4490
PRI
NT
"JSo
rry,
I
can
only
coM
pute
th
ose
for
stee
l an
d li
ght
allo
y."
4550
J2•
2 45
60 J
3=1
4570
PRI
NT
"Jls
the
cro
ss s
ecti
on·c
ircu
lar7
<V
or N
>"I
... g-
4580
INP
UT A
$ 45
90 I
F A
t=•v
• TH
EN 4
629
4600
IF
Af<
>•H
• TH
EN 4
579
4610
J3=
2 46
20 P
RINT
•J
oo y
ou k
now
th
e E
ndur
ance
LiM
it <
Se')
for
rota
ting
-bea
M "
46
30 P
RINT
"s
peci
Men
? CV
or
N)M
; 46
40 I
HPUT
A$
4650
IF
At=
"H"
THEN
470
0 46
60 I
F A
f<>"
V"
THEH
462
0 46
70 P
RIHT
"S
e'
in "
;C$;
" =
•; 46
80 I
NPUT
SS
4690
GO
TO 4
780
4700
IF
J=J
THEN
473
0 47
10 S
5=0.
StS1
47
20 G
O TO
474
9 47
30 S
5=0.
5*1.
4503
8E-4
*S1
4740
IF
55<
1000
00 T
HEN
4760
47
50 S
5=10
9000
~
4760
IF
J<>J
THE
H 47
80
~
4770
55=
S5i6
994.
757
4780
PRI
NT
•Jis
the
des
ign
life
inf
init
e? <
V or
N>•
; 47
90
INPU
T A
$ 49
00
IF A
$=•H
• TH
EN 4
940
4910
IF
A$<
>"V
" TH
EH 4
780
4820
H1=
0 48
30 G
O TO
499
0 49
40 P
RINT
•H
u"be
r of
cyc
les
= "J
48
50
INPU
T H
l 48
60
IF H
1<10
0009
0 TH
EN 4
880
4870
H1=
9 49
90 R
ETUR
N 48
90 R
EH **
* SUB
ROUT
INE
TO C
OMPU
TE S
e'''
WHI
CH D
EPEN
DS O
N D.
49
90 R
EH **
* THI
S PA
RT C
OMPU
TES
Kb
<K2>
. 49
10
IF J
•J T
HEN
4949
49
29 H
2•D
i25.
4
4930
GO
TO 4
950
4940
N2=
D*1
000
4950
GO
TO J
2 OF
496
0,49
79
4960
GO
TO J
J OF
498
0,59
40
4970
GO
TO J
J OF
510
0,50
60
4980
K2=
H2/
(-18
.75+
1.80
2*H
2)
4990
IF
H2>
2J T
HEN
5010
50
00 K
2=1
5010
IF
H2<
130
THEH
522
0 50
20 K
2=0.
59
5030
GO
TO 5
220
5040
K2=
0.50
61+7
.214
/H2
5050
IF
H2>
19 T
HEH
5070
50
60 K
2=0.
88
5070
IF
H2<
150
THEH
522
0 50
80
K2=
0.55
50
90 G
O TO
522
0 51
00 K
2=0.
515+
3.24
/H2
5110
IF
N2>7
THE
N 51
30
5120
K2=
1 51
30 I
F N2
<41
THEN
522
0 51
40 K
2=0.
59
5150
GO
TO 5
220
5160
K2=
9.59
61+2
.25/
H2
5170
IF
H2>
7 TH
EN 5
229
5180
K2=
9.88
51
90 I
F H
2<47
THE
N 52
20
5200
1<2
=0.5
5 52
10 R
EM**
* TH
IS S
ECTI
ON C
OMPU
TES
Se'''
• 52
20 S
6=A
BS<K
1*K
2*K
l*K
4*K
S*K
6*S5
) 52
30
S7=0
.9*S
1*K
4 52
40 I
F H
1=0
THEN
530
9 52
50 G
O TO
Jl
OF 5
260,
5289
52
60 S
3=10
t<<L
GT<
Hl)/
J-1>
*<LG
T<S6
>-LG
T<S7
>>+L
GT<
S7>>
52
70 G
O TO
531
0
... Rl'
\
5280
SJ•
S7+C
LGTC
H1>
/J-1>
*<S6
-S7>
52
90 G
O TO
531
9 53
00 S
J=S6
53
10 R
ETUR
N 60
00 T
2=12
00
6010
H1=
2400
60
20 H
=l.8
61
00 A
t=H
*M1/
CPl
*D*D
*D/3
2)
6110
A2=
0.86
6*H
*T2/
CPI
*D*D
*D/J
2)
6120
RET
URH
6500
EHD
'
~ °' \,
,.)
~
APPENDIX :B. 2
PROGRAM LISTING --- SECTION PROPERTIES
164
109
IHIT
11
0 PA
GE
120
PRI
"*****
******
******
******
******
******
******
******
******
******
***"
130
PRI
•* t"
14
0 PR
I •*
SECT
ION
PROP
ERTI
ES O
F PO
LVOH
AL S
ECTI
ON
*" 15
0 PR
I "*
*" 16
0 PR
I "*
By
Viu
Wmh
Luk
, VP
I &
SU,
Spri
ng 1
978
*" 17
0 PR
I "*
t"
180
PRI
"*****
******
******
******
******
******
******
******
******
******
***"
190
SET
DEGR
EES
200
DIH
XC
30>,
V<3
0>,X
1C11
,30)
,V1C
1J,3
0),D
1C21
,3>,
T<5>
,I<9>
,A$(
1),B
$<1>
21
0 DH
1 J1
<11)
22
0 T=
1 23
0 T5
=0
240
I=0
250
PRIH
T "J
JDo
you
won
t to
use
SI
unit
s? C
V or
H>•
s 26
0 IH
PUT
A$
270
IF A
$=•v
• TH
EH 3
50
280
IF A
S<>"
H• T
HEH
.250
29
0 PR
INT
"JW
e w
ill
use
Eng
lish
Uni
ts t
hrou
ghou
t th
is p
rogr
oM."
30
0 T5
=1
310
C$=
"inc
hes•
32
0 D
S="i
nt2"
33
0 F$
=•in
t411
340
GO T
O 39
0 35
0 C$
="M
r1"
360
DS=11
rtf1t
2•
370
F$:::
1111
Mt4•
38
0 PR
INT
"JW
e w
ill
use
SI u
nits
thr
ough
out
this
pro
grof
1.•
390
REH
*** EN
TER
DATA
FOR
BOU
NDAR
Y 40
0 PR
IHT
"Jis
the
out
er p
erif
1ete
r 4
circ
u14r
sec
tion
? <V
or
H>"
; 41
0 IN
PUT
A$
420
IF A
$=11H11
TH
EH 4
88
430
IF A
f<>•
v• T
HEN
390
.... ~
440
T<l>
-=2
450
PRIN
T •J
Rad
ius
of t
he c
ircl
e <"
ICSI
")
• •1
46
0 IH
PUT
Dt<
l,1)
47
0 GO
TO
590
480
PRI
"JJP
leas
e en
ter
the
X a
nd V
coo
rdin
ates
of
the
vert
ices
of
the
• 49
0 PR
IHT
"pol
ygon
<w
hich
Mus
t be
lo
cate
d en
tire
ly w
ithi
n th
e fi
rst
" 50
0 PR
INT
"qua
dran
t> s
eque
ntia
lly
for
a co
Mpl
ete,
cl
ockw
ise
path
aro
und"
51
0 PR
IHT
"the
pol
ygon
. U
nits
sho
uld
be "
JC$
;"."
52
0 PR
IHT
"Be
sure
to
end
wit
h th
e fi
rst
poin
t."
530
GOSU
B 43
80
540
J1(1
)=J
550
FOR
H=l
TO
J1(1
) 56
0 Xl
< 1,
tD=X
01)
570
V1<1
,t1>=
V<t1>
58
0 HE
XT H
59
0 PR
INT
"JA
re t
here
any
hol
es i
n th
e se
ctio
n? <
V or
H>"
; 60
0 IN
PUT
A$
610
IF A
$=•H
• TH
EH 1
040
620
IF A
$<>"
Yu T
HEH
590
630
REH
*** EH
TER
DATA
FOR
CIR
CULA
R HO
LES
640
PRIH
T •A
re t
here
any
cir
cula
r ho
les?
<V
or H
>";
650
INPU
T A$
66
0 IF
Af=
"H"
THEN
840
67
0 IF
AS<
>•V"
TH
EN 6
40
680
PRIH
T "H
ow "
any
circ
ular
hol
es a
re t
here
ins
ide
the
sect
ion?
•;
690
INPU
T 02
70
0 T<
2>=2
71
0 PR
INT
•Jfo
r ci
rcul
ar h
oles
giv
e:•
720
FOR
T1=1
TO
D2
730
PRIN
T •J
Rad
ius
of t
he c
ircu
lar
hole
1•1
111•
("
JC$
J")
• "I
74
0 IN
PUT
D1<
T1+1
,1)
750
PRIN
T •x
and
V c
oord
inat
es o
f th
e ce
nter
of
the
hole
1•1
Tll
" <"
JC$;
76
0 PR
INT
•> •
•1
770
INPU
T D
1<T
1+1,
2),0
1<T
1+1,
J)
780
NEXT
Tl
... ~
790
PRIN
T "J
Are
the
re a
ny p
olyg
on h
oles
? CV
or
N)"
J 80
0 IN
PUT
AS
810
IF A
S="N
• TH
EN 1
949
820
IF A
S<>"
V"
THEN
790
83
0 RE
M ***
ENTE
R DA
TA F
OR P
OLYG
ONAL
HOL
ES
840
PRIN
T •JH
ow M
any
poly
gon
hole
s ar
e th
ere
in t
he s
ecti
on?"
J 85
0 IN
PUT
DJ
869
IF T
<2>=
2 TH
EN 8
99
970
T<2>
=3
889
GO T
O 90
0 89
0 T
(2)=
4 ,
900
PRIH
T "J
For
poly
gon
hole
s:•
910
PRIN
T "E
nter
the
X a
nd V
coo
rdin
ates
of
each
ver
tice
s in
a
";
920
PRIN
T "c
oMpl
ete,
cl
ockw
ise
path
. U
nits
sho
uld
be •
;cs;
"."
930
PRIN
T "B
e su
re t
o en
d w
ith
the
firs
t po
int
of e
ach
hole
.J"
940
FOR
M=2
TO D
J+l
950
PRIN
T "F
or p
olyg
onal
ho
le l
";M
-1;"
:"
~
960
GOSU
B 43
80
~
970
J1(t
1)=J
98
0 FO
R t11
=1
TO J
1<H
> 99
0 X1
<M,H
l>=X
<H1>
10
00 V
1CM
,M1>
=V<M
1) 10
10 N
EXT
Ml
1020
PRI
NT
1030
NEX
T M
10
40 T
C4)=
1 10
50 T
<5>=
1 10
60 R
EH **
* SET
UP
FOR
CIR.
OR
POL
V.
BOUN
DARY
AND
DRA
W OU
TER
BOUN
DARY
10
70 G
O TO
T(1
) OF
110
0,10
90
1080
GOS
UB 4
580
1090
GO
TO 1
110
1100
GOS
UB 5
590
1110
GO
TO T
<2>
OF 1
630,
1130
,137
0,11
30
1120
REM
*** D
RAW
CIRC
ULAR
HOL
ES
1130
WIN
DOW
R<l
),R(2
),RC
5),R
C6)
1140
UIE
WPO
RT v
1,v2
,uJ,
V4
1150
FOR
Tl•
2 TO
D2+
1 11
60 C
=Dl<
Tl,l
>iPI
/18
1170
C1=
D1<
T1,2
>-D
l<T1
,1>
1180
C2z
Dl<
Tl,3
> 11
90 H
OVE
c1
,c2
1200
FOR
T2=
360
TO 0
STE
P -1
9 12
10 R
OTAT
E T2
12
20 R
DRAW
0,C
12
30 H
EXT
T2
1240
MOU
E D
1<T1
,2>+
<R<2
>-R
C1)
)/50,
D1C
T1,J
>+<R
<2>-
R<1
>>/S
0 12
50 T
<4>=
T<4>
+1
1260
8$=
CHR<
63+T
(4))
• 12
70 P
RINT
8$
1280
MOV
E D
1<T1
,2>,
D1<
T113)
12
90 S
CALE
1,
1 13
00 R
t10UE
-2
, 0
1310
RDR
AW 4
,9
1320
Rt10
UE -
2,2
1330
RDR
AW 0
,-4
1340
WIH
DOW
R<1
>,R
<2>,
RC
5),R
(6)
1350
HEX
T Tl
13
60 G
O TO
T<2
> OF
16
30,1
630,
1370
,137
0 13
70 W
INDO
M R
<1>,
R<2
>,R
C5>
,RC
6)
1380
UIE
WPO
RT u
1,v2
,uJ,
V4
1390
REH
*** D
RAW
POLY
GONA
L HO
LES
1400
FOR
T2=
2 TO
D3+
1 14
10 T
<4>=
T<4>
+1
1420
HOV
E X
1CT2
,1>+
<R<2
>-R
C1)
)/50,
V1<
T2,1
>+<R
<6>-
R<5
))/5
0 14
30 B
t=C
HR
<63+
T(4)
) 14
40 P
RINT
8$
1450
HOV
E X
l<T
2,1>
,v1c
r2,1
> 14
60 F
OR T
3•2
TO J
1<T2
> 14
70 D
RAW
XIC
T2,T
3>,V
1<T2
,T3>
14
80 H
EXT
T3
.... ~
1499
REM
*** D
RAM
DIAM
OND
OH E
ACH
POIN
T OF
POL
VCOH
AL H
OLE
1500
MOU
E X
t<T2
,1>,
Vl<
T2,1
> 15
10 F
OR T
J=t
TO J
l<T2
>-1
1520
MOV
E X
l<T2
,TJ>
,V1<
T2,T
3)
1530
SCA
LE 1
,1
1540
RHO
VE 1
90
1550
RDR
AW -
1,-1
15
60 R
DRAW
-1,
1 15
70 R
DRAW
1,1
15
80 R
DRAW
1,-
1 15
90 W
INDO
W R<
1>,R
<2>,
R<5>
,R<6
> 16
00 N
EXT
TJ
1610
HEX
T T2
16
20 R
EM **
* PRI
NT D
ATA
OH T
HE G
RAPH
16
30 W
INDO
W 0,
130,
0,10
0 16
40 V
IEW
PORT
0,4
0,0,
100
1650
MOU
E 0,
109
1660
PRI
HT "
All
data
are
in
•;ct
;".J
" 16
70 G
O TO
T<1
> OF
171
0,16
89
1680
PRI
NT
"Cir
. Se
ctio
n D
ata:
• 16
90 P
RIHT
uR
adiu
s =
•;D
1<1,
1>
1700
GO
TO 1
759
1710
PRI
NT
"Pol
y.
Dat
a (X
,Y>:
" 17
20 F
OR T
2=1
TO J
1<1>
-1
1730
PRI
HT "
l";T
2;•=
•1x
1<1,
T2>
;•,
•;vt
<1,T
2>
1740
NEX
T T2
17
50 G
O TO
T<2
> OF
195
0,17
60,1
850,
1760
17
60 P
RINT
17
70 F
OR T
4=2
TO 0
2+1
1780
T<S
>=T<
5>+1
17
90 8
$=CH
R<63
+T<S
>>
1800
PRI
HT "
Cir
. H
ole
1•1
et1"
' D
ata:
" 18
10 P
RIHT
"R
adiu
s• •
1D1<
T4,1
> 18
20 P
RIHT
"X
, Y
•
•1D
l<T4
,2>1
•,
•;D
l<T4
,3>
1830
NEX
T T4
... $
1849
GO
TO T
C2>
OF
1958
,195
0,18
59,1
850
1850
PRI
NT
1860
FOR
T2
•2 T
O D3
+1
1870
T<S
>=T<
S>+l
18
80 8
$=CH
R<6J
+T<S
>>
1890
PRI
NT •
Pol
y.H
ole'
•;B
tl"'
Dat
a <X
,Y>:
• 19
00 F
OR T
3=1
TO J
l<T2
>-1
1910
PRI
NT •
1";T
J;•=
"IX
1<T2
,T3>
1•,
"JY
1<T2
,TJ>
19
20 H
EXT
TJ
1930
HEX
T T2
19
40 R
EM **
* CHA
NGE
DATA
19
50 W
INDO
W 0
,130
,0,1
00
1960
UIE
WPO
RT 0
,130
,0,1
08
1970
MOV
E 12
7,13
19
80 P
RINT
ux
• 19
90 M
OUE
0,4
2000
PRI
NT
"Do
you
wan
t to
cha
nge
any
data
? <V
or
N>"I
2010
IH
PUT
A$
2020
IF
At=
•H•
THEH
252
0 20
30 I
F A$
<>"Y
• TH
EN 2
000
2040
T<3
>=1
2050
PRI
NT
"LPl
ease
ent
er 0
whe
n yo
u ha
ve f
inis
hed
chan
ging
dat
a."
2060
PRI
HT
"JD
o yo
u w
ant
to c
hang
e th
e da
ta f
or
the
oute
r bo
unda
ry?
" 20
70 P
RIHT
"<
V or
H>0
;
2080
IN
PUT
A$
2090
IF
At=
"H"
THEN
220
0 21
00
IF A
t<>"
Y•
THEH
206
0 21
10 G
O TO
T<l
> OF
217
0,21
20
2120
PRI
HT
"JFo
r C
ircu
lar
Bou
ndar
y:•
2130
PRI
NT
"Pre
sent
Rad
ius
<";C
$J0
) •
"JD
1<1,
1>
2140
PRI
NT
•Ent
er n
ew R
adiu
s <"
JCt;•
> •
"I
2150
IH
PUT
D1<
1,1>
21
60 G
O TO
220
0 21
70 P
RINT
•J
For
Poly
gona
l B
ound
ary:
• 21
80 T
1=1
.... .....,
0
2190
GOS
UB 6
000
2200
IF
TC
2)•1
THE
H 25
10
2210
PRI
HT
"JO
o yo
u w
ant
to c
hang
e th
e da
ta f
or
the
hole
s?
<V o
r N
>";
2220
IHP
UT A
S 22
30 I
F A
t="H
" TH
EH 2
510
2240
IF
AS<
>"V"
TH
EN 2
210
2250
GO
TO T
<2>
OF 2
510,
2260
,244
0,22
60
2260
FOR
T2
=2 T
O 02
+1
2270
T<3
>=TC
3)+1
22
80 B
$=CH
RC63
+TC3
)) 22
90 P
RINT
"JF
or C
ircu
lar
Hol
e '"
;Bf;
n' :
• 23
00 P
RINT
"P
rese
nt R
adiu
s ( 11
; C
f; 11.)
=
"; 01
CT2
,1)
2310
PRI
HT
"Ent
er n
ew R
adiu
s <•
;C$;
•>
= ";
23
20 I
HPUT
Hl
2330
IF
H1=0
THE
N 23
50
2340
D1C
T2,1
>=H
1 23
50 P
RIHT
"P
rese
nt c
oord
inat
e of
cen
tre
<X,Y
>,
in "
;Cf;
" =
";
2360
PRI
HT D
1<T2
,2>;
",
•;D
1CT2
,3)
2370
PRI
NT "
Ent
er n
ew c
oord
inat
e of
cen
tre
<X,Y
>,
in "
;Ct;•
= ";
2380
INP
UT H
2,H
3 23
90 I
F H2
=0 O
R H3
=0 T
HEH
2420
24
00 D
1CT2
,2>=
H2
2410
D1<
T2,J>
=H3
2420
NEX
T T2
24
30 G
O TO
T(2
) OF
2s1
0,2s
1e,2
449,
2449
24
40 F
OR T
2=2
TO 0
3+1
2450
T<3
>=TC
3)+1
24
60 8
$=CH
R<63
+T<3
>>
2470
PRI
NT
•JFo
r Po
lygo
nal
Hol
e 1•1
et1"
' :•
24
80 T
1=T2
24
90 G
OSUB
60
09
2500
NEX
T T2
25
10 G
O TO
19
49
2520
REH
*** C
ALCU
LATE
SEC
TION
PRO
PERT
IES
2530
GO
TO T
<t>
OF 2
540,
2960
..... ~
2540
K1•
J1<1
> 25
59 T
1•1
2560
GOS
UB 6
180
2570
A1=
A2
2580
X4=
X5
2590
V4=
V5
2600
1(
1)=1
1 26
10 1
(2)=
12
26
20 1
(3)=
13
2630
GO
TO T
<2>
OF 2
750,
2750
,264
0,26
40
2640
FOR
T2=
2 TO
03+
1 26
50 K
1=J1
<T2>
26
60 T
1=T2
26
70 G
OSUB
618
0 26
80 A
1=A1
-A2
2690
X4=
X4-X
5 27
00 V
4=V
4-V
5 27
10 I
<t>=
I<1>
-11
2720
1(2
)=1(
2)-1
2 27
30 I
<J>=
I<J>
-IJ
2740
HEX
T T2
27
50 X
4=X4
/A1
2760
Y4=
V4/A
1 27
70 G
O TO
TC2
) OF
327
8,27
80,3
279,
2789
27
80 J
4=X
4*A
1 27
90 J
5=V
4*A
1 28
00 F
OR T
2=2
TO 0
2+1
2810
A3=
PI*D
l<T2
,1>t
2 28
20 J
3=PI
*D1<
T2,
l>t4
/4
2830
J6=
J3+A
3*D
1CT2
,J>t
2 28
40 J
7•J3
+A3*
Dl<
T2,2
>t2
2850
J8=
A3*
D1<
T2,2
>*D
l<T2
,3)
2869
J4=
J4-D
1<T2
,2>*
A3
2879
J5c
J5-D
1CT2
,3>*
A3
2889
A1•
A1-
AJ
.... i\l
2899
I<1
>=I<
1>-J
6 29
00
1(2)
•1(2
)-J7
29
10
1(3)
•1(3
)-J8
29
20 H
EXT
T2
2930
X4=
J4/A
1 29
40 '
f4=J
5/A
1 29
50 G
O TO
327
0 29
60 A
l=Pl
*D1<
1,1>
t2
2970
X4=
D1<
1,1)
29
80 V
4=D
1<1,
1)
2990
I<
1>=5
*PI*
D1<
1,1)
t4/4
30
00
1(2)
=1(1
) 30
10 1
(3)=
Al*
X4*
V4
3020
GO
TO T
<2>
OF 3
270 1
2780
,393
0,30
30
3030
A3=
0 30
40 J
3=0
3050
J4=
0 30
60 J
5=0
3070
J6=
0 30
80 J
7=0
3090
FOR
T2=
2 TO
03+
1 31
00 K
1=Jl<
T2>
3110
T1=
T2
3120
GOS
UB 6
180
3130
J6=
J6+X
5 31
40 J
7=J7
+V5
3150
A3=
A3+A
2 31
60 J
3=J3
+11
3170
J4=
J4+1
2 31
80 J
5=J5
+13
3190
NE
XT
T2
3200
I<
1>=I
<1>-
J3
3210
1(
2)al
(2)-
J4
3220
I<
J>=I
<3>-
J5
3230
X
4=<X
4*A
1-J6
)/(A
1-A
3>
... "'-l
\...>
3249
V4=
<V4*
A1-
J7)/C
A1-
A3)
32
59 A
l=A
1-A
J 32
69 G
O TO
T<2
> OF
327
9,27
90,3
270,
2790
32
79 I
<4>=
I<1>
-Al*
V4*
V4
3280
I<S
>=I<
2>-A
l*X
4*X
4 32
90 I
<6>=
I<J>
-Al*
X4*
V4
3300
R6=
SQR
(l(l
)/A
1)
3310
R7=
SQR
(l(2
)/A
l)
3320
R8=
SQR<
I<4)
/A1>
33
30 R
9=SQ
R(l(
5)/A
1)
3340
IF
I<4>
-I<5
>=0
THEH
337
0 33
50 G
1=0.
S*A
TH<-
2*IC
6)/(
I<4>
-I<5
)))
3360
GO
TO 3
390
3370
G1=
0 33
80 I
MAGE
CFD
.20,
1x,s
33
90 P
RIHT
"L
***
SEC
TION
PRO
PERT
IES
OF l
HE
REQU
IRED
SEC
TION
***
" 34
00 P
RINT
•JA
reo
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
•••=
";
3410
PRI
HT U
SING
339
0:A
1 34
20 P
RINT
D$
3430
PRI
HT
•Jx
coor
dina
te o
f th
e ce
ntro
id ••
••••
••••
••••
••••
••••
• = "
; 34
40 P
RIHT
USI
NG 3
380:
X4
3450
PRI
NT C
$ 34
60 P
RIHT
•J
v co
ordi
nate
of
the
cent
roid
••••
••••
••••
••••
••••
•••
= ";
34
70 P
RINT
USI
NG 3
380:
V4
3480
PRI
NT C
t 34
90 P
RIHT
"J
Are
a M
o"en
t of
in
erti
a ab
out
X-a
xis •
••••
••••
••••
••• =
";
3500
PRI
NT U
SING
33
80:1
(1)
3510
PRI
NT
Ft
3520
PRI
NT
"JA
reo
"o"e
nt o
f in
erti
a ab
out
Y-a
xis •
••••
••••
••••
••• =
•; 35
30 P
RIN
T US
ING
3380
:1<2
> 35
40 P
RIN
T F
t 35
50 P
RIN
T "J
Are
o pr
oduc
t of
in
erti
a ••
••••
••••
••••
••••
••••
••••
•• =
";
3560
PRI
NT
USIN
G 33
80:1
<3>
3570
PRI
NT F
t 35
80 P
RIN
T •J
Are
o M
OMen
t of
in
erti
a ab
out
X'-
axis
tra
nsla
ted
to •
.... ~
3590
PRI
HT
"the
cen
troi
d ••
•••
~ ••
••••
••••
••••
••••
••••
••••
••••
••• ~
"I
3600
PRI
HT U
SIHG
338
0:1<
4>
3610
PRI
HT F
f 36
20 P
RINT
"J
Are
a M
OMen
t of
in
erti
a ab
out
V'-
axis
tra
nsla
ted
to "
36
30 P
RINT
"t
he c
entr
oid
••••
••••
••••
••••
••••
••••
••••
••••
••••
•••
= ";
36
40 P
RINT
USI
NG 3
380:
1<5>
36
50 P
RINT
Ft
3660
PRI
HT
"JA
rea
prod
uct
of i
ner
tia
abou
t tr
ansl
ated
axi
s •••
••• =
";
3670
PRI
NT U
SIHG
338
0:I<
6>
3680
PRI
NT F
$
3690
PRI
HT
"JA
ngle
bet
wee
n tr
4nsl
ated
axi
s an
d pr
inci
pal
axis
"
3700
PRI
NT "
<in
degr
ee),
po
siti
ve· i
s co
unte
r-cl
ockw
ise •
••••
••••
= "
; 37
10 P
RINT
USI
NG
"FD
.2D
":G1
3720
K2=
I<4>
*CO
S<G
1)+1
(5)*
SIH
<G1>
t2-IC
6>*S
IH<2
*G1>
37
30 K
J=I<
5>*C
OS<
Gl>
t2+1
<4>*
SIH
CG1>
t2+I
<6>*
SIH
<2*G
1)
3740
PRI
NT
"JA
rea
"oH
ent
of i
ner
tia
abou
t th
e tr
ansl
ated
, ro
tate
d,
" 3'~0
PRIN
T •
. .
1 V
II
• H
• (J
pr1n
c1pa
n
-axi
s •••
••••
••••
••••
••••
••••
••••
••••
•• =
'
3760
PRI
NT U
SIHG
338
0:K
2 37
70 P
RIHT
F$
3780
PRI
NT
"JA
rea
"o"e
nt o
f in
erti
4 ab
out
the
tran
slat
ed,
• 37
90 P
RINT
"r
otat
ed,
prin
cipa
l Y
''-ax
is ••
••••
••••
••••
••••
••••
•• =
";
38
00 P
RIHT
USI
NG 3
380:
K3
3810
PRI
NT F
$ 38
20 P
RINT
"J
JDo
you
wan
t to
get
are
a M
o"en
t of
in
erti
a ab
out
an
" 38
30 P
RIHT
"a
rbit
rary
axi
s? C
V or
N>"
I 38
40
INPU
T A$
38
50 I
F A
f=•N
" TH
EN 4
160
3860
IF
A$<>~Y"
THEN
382
8 38
70 P
RIHT
•J
Plea
se e
nter
the
ang
le b
etw
een
the
orig
inal
axi
s <X
-Y>
and"
38
80 P
RIHT
•a
rbit
rary
axi
s C
X'''
-Y'''
),
in d
egre
e,
posi
tive
is
" 38
90 P
RINT
"c
ount
er-c
lock
wis
e •••
••••
••••
••••
••••
••••
••••
••••
••••
•• =
"; 39
00
INPU
T G2
39
10 P
RINT
"E
nter
X-a
xis
coor
dina
te o
f th
e arbitrar~
axis
<";
Ct;•
>= "
; 39
20
INPU
T X6
39
30 P
RIHT
.,E
nter
Y-a
xis
coor
dina
te o
f th
e ar
bitr
ary
axis
<"J
C$;">
= ";
... ...., \.n
I ..4
3940
IHP
UT V
6 39
50 K
7=I<
4>+A
1*<Y
4-V
6>t2
39
60 K
9=I<
5>+A
1*<X
4-X
6>t2
39
70 K
9=1<
6>+A
1*<X
4-X
6>*<
V4-
V6)
39
80 I
<7>=
K7*
COS<
G2>
t2+K
9*SI
HCG
2>t2
-K9*
SIH
<2*G
2)
3990
I<9
>=K
8*CO
S<G
2>t2
+K7*
SIH
<G2>
t2+K
9*SI
H<2
*G2>
40
00 K
1=1<
7>+1
(8)
4010
IC9
)=0.
5*<K
7-K
9>*S
IH<2
*G2>
+K9*
COS<
2*G
2>
4020
PRI
HT
•JA
rea
Mo"
ent
of i
ner
tia
abou
t th
e ar
bit
rary
ax
is,X
''' ••
= ";
4030
PRI
HT U
SING
338
0:I<
7>
4040
PRI
NT F
S 40
50 P
RINT
"J
Are
a "o
Hen
t of
in
erti
a ab
out
the
arb
itra
ry a
xis
,V'''
•• =
";
4060
PRI
NT U
SING
338
0:IC
8)
4070
PRI
NT F
$ 40
80 P
RINT
"J
Pol
ar a
rea
MOM
ent
of i
ner
tia
abou
t th
e ar
bit
rary
"
4090
PRI
NT
"axi
s <
X'''
-V'''
)•••
••••
••••
••••
••••
••••
••••
••••
••••
•··=
41
00 P
RINT
USI
NG 3
380:
K1
4110
PRI
NT F
$ 41
20 P
RINT
"JA
rea
prod
uct
of i
ner
tia
abou
t th
e ar
bit
rary
•
4130
PRI
NT "
axis
<X
'''-V
''')•
••••
••••
••••
••••
••••
••••
••••
••••
••••
•=
4140
PRI
NT U
SING
338
0:1<
9>
4150
PRI
NT F
$ .
4160
PRI
NT
"JR
adiu
s of
gyr
atio
n ab
out
X-a
xis •
••••
••••
••••
••••
••••
•• =
41
70 P
RINT
USI
HG 3
380:
R6
4180
PRI
NT C
$ 41
90 P
RINT
"J
Rad
ius
of g
yrat
ion
abou
t Y
-axi
s •••
••••
••••
••••
••••
•••
,= 42
00 P
RINT
USI
HG 3
380:
R7
4210
PRI
NT C
f 42
20 P
RINT
"J
Rad
ius
of g
yrat
ion
abou
t X
'-ax
is t
rans
late
d to
•
II ;
a• ' II
• '
II •
' 42
30 P
RINT
"t
he c
entr
oid
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
•• =
";
4240
PRI
NT U
SING
JJ8
8:R
8 42
50 P
RINT
Cf
4260
PRI
NT
"JR
adiu
s of
gyr
atio
n ab
out
V'-
axis
tra
nsla
ted
to •
42
70 P
RINT
"t
he c
entr
oid
••••
••••
••••
••••
••••
••••
••••
••••
••••
••••
•• =
•; 42
80 P
RINT
USI
NG J
380:
R9
.... ~ °'
4290
PRI
HT C
t 43
00 P
RINT
"J
JJD
o yo
u w
ant
to f
ind
the
prop
erti
es o
f Q
noth
er s
ecti
on?
• 43
10 P
RIHT
"<V
or
H>"S
43
20 I
HPUT
A$
4330
IF
A$=
"Y"
THEH
100
43
40 I
F A
$<>"
H"
THEN
430
0 43
50 P
R I H
T "J
J---
----
----
----
----
--EH
D --
----
----
----
----
--"
4360
EHD
43
70 R
EH **
* EHT
ER D
ATA
FOR
POLY
GONA
L SE
CTIO
N 43
80 P
RINT
"X
<l>,
VC
1)
= "I
43
90 I
NPUT
X<1
>,Y<
1>
4400
IF
X<1
>=>0
AHD
Y<1
>=>0
THE
H 44
50
4410
PRI
NT
"JE
rror
! Y
our
poly
gon
shou
ld b
e lo
c4te
d en
tire
ly w
ithi
n th
e "
4420
PRI
NT "
firs
t qu
4dr4
nt,
thQ
t is
, X
4nd
V b
oth
have
to
be p
osit
ive.
" 44
30 P
RIHT
"P
leas
e tr
y ag
Qin
."
4440
GO
TO 4
380
4450
J=2
44
60 P
RIHT
"X
<";J;
">,
vc•;
J;">
• "
; 44
70 I
NPUT
X<J
>,Y
<J>
4480
IF
X<J
>=>0
AH
D Y
<J>=
>0 T
HEN
4530
44
90 P
RIHT
•JE
rror
! Y
our
poly
gon
shou
ld b
e lo
cate
d en
tire
ly w
ithi
n th
e "
4500
PRI
HT "
firs
t qu
adr4
nt,
th4t
is,
X 4
nd Y
bot
h ha
ve t
o be
pos
itiv
e."
4510
PRI
NT
"Ple
ase
try
agQ
in."
45
20 G
O TO
43
80
4530
IF
X<J
>=X
<l>
AND
Y<J
>=Y
<l>
THEN
456
0 45
40 J
=J+1
45
50 G
O TO
446
0 45
60 R
ETUR
N 45
70 R
EM *
** S
ET U
P W
IHDO
W,
VIEW
PORT
AND
AXI
S FO
R CI
RCUL
AR S
ECTI
OH.
4580
W1=
0 45
90 M
2=2*
D1<
1,1)
46
00 M
3=0
4610
M4=
2*D
1<1,
1)
4620
V1=
40
4630
V2=
130
.... ~
4640
V3=
5 46
50 U
4•95
46
60 G
OSUB
476
0 46
70 W
5=<R
<2>-
R<l
))/2
46
80 W
6=(R
<6>-
R(5
))/2
46
90 R
EM **
* DRA
W CI
RCUL
AR B
OUND
ARY
4700
WIN
DOW
-W
5,W
5,-W
5,W
5 47
10 V
IEW
PORT
u1,
u2,u
3,u4
47
20 E
=D1<
1,1)
47
30 G
OSUB
554
0 47
40 R
ETUR
H 47
50 R
EM **
* LAB
EL
"HEA
T"
TIC
4760
PRI
NT
"L
Y
GRAP
H OF
IN
PUT
SECT
ION"
47
70 l
J1=l
J1+1
0.8
4780
UJ=
U3+
8.4
4790
UIE
WPO
RT u
1,v2
,u3,
u4
4800
DIM
R<
8)
4810
R<1
>=W
1 48
20 R
<2>=
W2
4830
R<J
>=IH
T<<l
J2-U
1)/(8
*1.S
>> M
AX
1 48
40 R
<S>=
W3
4850
R<6
>=W
4 48
60 R
<7>=
IHT<
<V4-
U3)
/(3*2
.8))
MAX
1 48
70 R
EM *
** CA
LCUL
ATE
HEW
LIM
ITS
AND
INTE
RVAL
S 48
80 R
S=J
4890
GOS
UB 5
040
4900
RS=
? 49
10 G
OSUB
504
0 49
20 W
INDO
W R
<1>,
R<2>
,R<S
>,RC
6>
4930
REM
AXI
S R<
J>,R
<7>,
R<l>
+R<J
>,RC
5)+R
(7)
4940
AX
IS
R<J>
,R<7
> 49
50 R
EM **
* LAB
EL T
HEM
4960
R5
=4
4970
E$=
"1:U
:lt:tJJ
" 49
80 G
OSUB
533
0
... ......
CJ)
4990
R5=
8 50
00 £
$a•tn
U:H:
U::I"
50
10 G
OSUB
533
0 50
20 H
OME
5030
RET
URN
5040
REH
*** R
<R5>
•
HIHI
HUM
HO.
OF
TIC
S 50
50 R
1=<R
<R5-
1>-R
<R5-
2))/R
(R5)
50
60 R
2=10
tlNT<
LGT<
R1>>
50
70 R
1=R1
/R2
5080
IF
R1>2
THE
N 51
20
5090
IF
R1=1
TH
EN 5
160
5100
R2=
2*R2
51
10 G
O TO
516
0 51
20 I
F R1
>5 T
HEN
5150
51
30 R
2=5*
R2
5140
GO
TO 5
160
5150
R2=
10*R
2 51
60 R
EM *
** A
DJUS
T DA
TA M
IN
5170
R1=
IHT<
R<R5
-2)/R
2>
5180
R3=
R2*<
R1+2
> 51
90 I
F RJ
<R<R
S-2)
TH
EN 5
220
5200
R3=
R3-R
2 52
10 G
O TO
519
0 52
20 R
<R5-
2>=R
3 52
30 R
EH *
** A
DJUS
T DA
TA H
AX
5240
R1=
IHT<
R<R5
-1)/R
2>
5250
R3=
R2t<
R1-2
) 52
60
IF R
<R5-
1><R
3 TH
EN 5
290
5270
R3=
R3+R
2 52
80 G
O TO
526
0 52
90 R
.( R5
-1 )
=R3
5300
REM
*** R
CR5>
•
ADJU
ST T
IC
INTE
RVAL
53
10 R
<R5>
-=R2
53
20 R
ETUR
N 53
30 R
EM L
ABEL
AXI
S
.... ~
5340
R4•
R<R5
-1)
5350
R(4
)•R
<l)
5360
R<8
)•R
C5)
53
70 R
3=AB
S<R<
R5-3
>+R4
) HA
X A
BS<R
<R5-
2>-R
4)
5380
R3=
IHT<
LGT<
R3)+
1.9E
-8)
5390
R2=
10t-
R3
5400
R1=
R<R
S-2>
-R4/
2 54
10 R
<RS>
=R<R
5)+R
4 54
20
IF R
<RS>
>Rt
THEN
547
0 54
30 M
OUE
R<4>
,R<B
> 54
40 P
RINT
E$;
54
50 P
RIHT
USI
HG
·-o.
20,s
•:R
<R5>
*R2
5460
GO
TO 5
410
5470
IF
R3=
0 TH
EH 5
520
5480
R<R
5>=R
1 54
90 M
OVE
R<4>
,R<B
> 55
00 P
RIHT
E$;
55
10 P
RIHT
USIH
G •J
A,+
Fo,s
•:•
E•;R
J 55
20 R
ETUR
N 55
30 R
EM **
* TO
DRAW
A C
IRCL
E 55
40 M
OVE
E,0
5550
FOR
T1=
10 T
O 36
0 ST
EP 1
0 55
60 D
RAW
E*CO
S<T1
>,E*
SIH<
T1>
5570
NEX
T T1
55
80 R
ETUR
N 55
90 R
EM **
* SET
UP
FOR
POLY
GONA
L SE
CTIO
N 56
00 G
OSUB
582
0 56
10 H
l=XB
HIN
VB
5620
M2=
X9 M
AX V
9 56
30 W
1=t11
56
40 W
2=H2
5l
i50
W3=
H1
5660
W
4=f12
56
70 V
1=40
56
80 l
)2=1
30
... ~
5690
t)3
•5
5700
lJ4
sa95
57
10 G
OSUB
476
9 57
20 M
OUE
X1<
1,1)
,V1<
1,1)
57
30 F
OR T
2=2
TO
J1(1
) 57
40 D
RAW
X1<
1,T2
>,V
1<1,
T2>
5750
HEX
T T2
57
60 F
OR T
2=1
TO J
t<l>
-1
5770
MOU
E X
1<1,
T2>,
V1<
1,T2
) 57
80 P
RINT
T2
5790
HEX
T T2
58
00 R
ETUR
N .
5810
REM
*** F
IND
THE
HIN
AND
MAX
OF
X1CJ
1>
AND
V1<
J1)
5820
REM
*** X
8,V8
ARE
MIN
I X
9,V9
ARE
MAX
5830
XB=
X1C1
,1>
5840
V8=
V1<1
,1>
5850
FOR
M=1
TO
J1<
1>-1
58
60
IF X
1(1,
H>=
>X9
THEN
588
0 58
70 X
8=X
1<1,
'1)
5880
IF
Vl<
l,H>=
>VS
THEN
590
0 58
90 V
8=V1
<1,M
> 59
00 N
EXT
M
5910
X9=
X1<1
,1>
5920
V9=
V1<1
,1>
5930
FOR
H=1
TO
J1<
1>-1
59
40
IF X
1<1,H
><=X
9 TH
EN 5
960
5950
X9=
X1<1
,H>
5960
IF
V1<
1,H
><•V
9 TH
EN 5
989
5970
Y9
:sV1<
1,r1>
59
80 N
EXT
M
5990
RET
URN
6000
REM
*** S
UBRO
UTIN
E FO
R CH
ANGI
NG D
ATA
OF P
OLVG
OHAL
SEC
TION
60
10 P
RINT
•E
nter
poi
nt I
•
•1
6020
IH
PUT
Ht
6030
IF
H1•
8 TH
EN 6
160
.... f!i
6040
IF
Ht<•Jt<T1)~1
THEN
609
0 60
59 P
RINT
•J
Err
or!
You
h4ve
onl
y •1
Jt<T
1>-1
1" d
4t4,
yo
ur i
nput
"JH
1;
6060
PRI
HT
•is
out
of
your
d4t
d r4
nge.
" 60
70 P
RIHT
•p
1e4s
e tr
y «9
4in.
" 60
80 G
O TO
691
0 60
90 P
RIHT
"P
rese
nt v
«lue
s <X
,V>,
in
"1C
s1•
•"JX
1<T1
,H1>
;•,
"JV
1<T1
,H1>
61
00 P
RINT
"E
nter
new
Vdl
ues
CX,V
>,
in "
IC$1
"•"1
61
10 I
HPUT
X1C
T1,H
l),V
1CT1
,H1>
61
20
IF H
1<>1
TH
EN 6
150
6130
X1<
Tt,J1
<T1)
)=X
1<T1
,1>
6140
V1C
T1,J1
<T1>
>=V
1CT1
,1)
6150
GO
TO 6
010
6160
RET
URH
6170
REM
***
SUBR
OUTI
NE T
O CA
LC.
A,x
,v,1
x,Iy
,lxy
FOR
POL
YGON
AL S
ECTI
OH
6180
A2=
0 61
90 X
5=0
6200
V5=
0 ~
6210
11=
0 N
62
20 1
2=0
6230
13
=0
6240
FOR
K=I
TO
Kl-1
62
50 A
2=A
2-<V
1<T1
,K+1
>-V
l<Tt
,K>>
*<X
1<T1
,K+1
)+X
1<Tt
,K>>
/2
6260
B1=
<V1<
T1,K
+1)-V
1<T1
,K>>
/8
6270
C1=
<X1<
Tt,K
+1)+
X1C
T1,K
>>t2
+<X
1<Tt
,K+1
)-X1<
T1,K
>>t2
/J 62
80 X
S=XS
-B1*
C1
6290
B2=
<X1C
T1,K
+1)-X
1<T1
,K))/
8 63
00 C
2=<V
l<T1
,K+1
)+V
1CT1
,K>>
t2+<
V1C
T1,K
+1)-V
1CT1
,K>>
t2/J
6310
V5=
V5+B
2*C2
63
20 B
J=<X
1<T1
,K+1
>-X
1<T1
,K>>
*<V
t<T1
,K+1
>+V
1<T1
,K>>
/24
6330
CJ
=CV
1CT1
,K+1
)+V
1<Tt
,K>>
t2+<
V1<
T1,K
+1)-V
l<T1
,K>>
t2
6340
11=
11+B
3*C3
63
50 B
4=<V
1CT1
,K+1
>-Y
1<Tl
,K>>
*<X
1CT1
,K+1
>+X
1CT1
,K)>
/24
6360
C4=
<X1<
Tl,K
+1>+
Xl<
T1,K
>>t2
+<X
1<T1
,K+1
)-X1C
T1,K
>>t2
63
70
I2=I
2-B
4*C
4 63
80
IF X
1<T1
,K+1
>-X
1CT1
,K>•
0 TH
EN 6
460
6390
B5•
<Vl<
T1,K
+1>-
Vl<
T1,K
>>t2
*<X
1(T1
,K+1
>+X
l<T1
,K>>
64
00 B
5=B
5*<X
l<T1
,K+l
)t2+X
1<T1
,K>t
2)/8
64
10 C
5=<V
l<Tl
,K+1
>-V
1<T1
,K>>
*<X
t<Tt
,K+l
>*V
l<T1
,K>-
X1<
T1,K
>*Y
1CT1
,K+l
)) 64
20 C
5=C5
*<X
1<T1
,K+l
>t2+
Xl<
T1,K
+1)*
Xl<
T1,K
>+X
1<Tt
,K>t
2>/3
64
30 C
6=<X
l<T1
,K+1
>*V
l<T1
,K>-
Xl<
T1,K
>*V
1<T1
,K+l
))t2
6440
C6=
C6*<
X1<
T1,K
+1>+
X1<
T1,K
>>/4
64
50 1
3=13
+CB5
+C5+
C6)/(
X1<
T1,K
+1>-
X1<
T1,K
)) 64
60 H
EXT
K
6470
RET
URH
... ~
• J
APPENDIX B.J.1
PROGRAM LISI'ING --- BF.AM ANALYSIS (PAR!' I)
184
100
HU
T 11
0 PA
GE
115
SET
DEGR
EES
120
DIM
T<S
,5>,
Tt<S
,5>,
T2<5
15>
,T3<
5>,S
<S>,
Af(
1),F
(30)
,L(3
0>,E
C30
> 12
2 DI
M
11<3
0>,P
<30>
,Q1<
30),
Q2(
30>,
K1C
30),
K2<
30),
p1(3
0),M
5<30
>,M
<30)
12
4 DI
M A
<2,3
>,I2
C30
) 13
0 GO
SUB
6000
13
1 T=
T1
132
P1=0
13
3 M
5=0
134
K1=0
13
5 K2
=0
136
Q1=0
13
7 Q2
=0
139
M=0
14
2 L=
0 15
0 S=
0 15
1 S<
5>=1
15
2 J$
=11ra
d.l's
ec•
200
PRI
"*****
******
******
******
******
******
******
******
******
******
***•
201
PRI
"* *"
210
PRI
"* TH
IS I
S A
BEA
M AN
ALYS
IS P
ROGR
AM W
HICH
SOL
VES
STAT
IC O
R *"
220
PRI
•* DY
NAMI
C <F
ORCE
D,
UNDA
MPED
VIB
RATI
ON>
RESP
ONSE
. *"
221
PRI
"* tu
22
2 PR
I "*
by Y
iu W
ah L
uk,
advi
sed
by P
rofe
ssor
Lar
ry D
. M
itch
ell
*" 22
3 PR
I •*
Vir
gini
a P
olyt
echn
ic
Inst
itu
te a
nd S
tate
Uni
vers
ity
*" 22
4 PR
I 11*
Dep
artR
ent
of M
echa
nica
l E
ngin
eeri
ng
*" 22
5 PR
I •*
Spri
ng 1
978
*" 22
6 PR
I "*
*" 22
7 PR
I •**
******
******
******
******
******
******
******
******
******
******
" 23
0 PR
INT
"JJD
o yo
u w
ant
to u
se E
ngli
sh u
nits
? <Y
or
N>"I
240
INPU
T A$
25
0 IF
Af=
"Y"
THEN
339
26
0 IF
Af<
>•N
• TH
EN 2
19
... &
270
PRIH
T "J
JSI
unit
s w
ill
be u
sed
thro
ugho
ut t
his
prog
raM
."
280
U=t
28
5 G
=9.8
06
290
8$=
11N,
,,.1•
300
C$="
H"
310
D$=u
t'111
311
E$="
Pa."
31
2 FS
="r1
t411
316
G$=
11H
-N11
318
K$=
11H
-11t
2•
319
L$=
11H-
f"1/rC
ld11
______
__ 32
0 GO
TO
380
330
U=2
335
G=3
86.0
88
340
8$=
11lb
/in•
35
0 C
$="1
b"
360
D$=
11in
11
362
E$=11
psi11
364
F$=
11in
t411
366
G$=
11lb
-in•
36
8 K$
=11lb
-int
2•
369
L$=
"1b-
in/r
ad"
------
.470
PRIN
T "J
JEng
lish
uni
ts w
ill
be u
sed
thro
ugho
ut t
his
prog
raH
."
380
PRIN
T •J
is t
his
a s
tati
c an
alys
is?
<Y o
r H
>";
390
INPU
T A$
39
5 U
1=1
396
W=0
40
0 IF
A$=
"V"
THEN
460
41
0 IF
At<
>"H
" TH
EN 3
90
420
U1=
2 43
0 PR
IHT
11JS
0 th
is i
s a
dyna
"ic
a.na
lysi
s.•
440
PRIN
T "E
nter
fre
Que
ncy
<"I
J$;"
>
a•J
------
-442
IN
PUT
W
460
PRIN
T "J
Plea
se d
ivid
e yo
ur b
eaM
in
to s
ever
al
fiel
d se
ctio
ns,"
47
0 PR
IHT
"a.n
d en
ter
the
info
rHa.
tion
reQ
uire
d fo
r ea
.ch
sect
ion.
"
... ~
475
PRIH
T •H
.B.
Alw
ays
std
rt f
ro"
the
left
end
."
476
PRIN
T "J
JPre
ss R
ETUR
N to
pro
ceed
."
477
INPU
T A$
47
8 IF
A$(
)"
" TH
EN 4
80
480
H=
l 49
0 PR
IHT
"LFo
r se
ctio
n l"
;H;"
:"
500
PRIH
T "T
he
foll
owin
g F
ield
Mat
rice
s ar
e Q
Vdi
labl
e.•
510
PRIH
T u9
th
e la
st s
ecti
on."
52
0 PR
IHT
"1
Mas
sles
s be
aM."
530
PRIH
T "2
un
ifor
MlY
dis
trib
uted
loa
d on
Mas
sles
s be
QM
." 54
0 PR
IHT
•3
line
arly
var
ied
dist
ribu
ted
load
on
Mas
sles
s be
aM."
55
0 PR
IHT
"JE
nter
I
for
the
requ
ired
Fie
ld M
atri
x= "
; 56
0 IH
PUT
F<N>
57
0 IF
F<N
>=0
THEN
201
0 58
0 GO
TO
F<H>
OF
620
,620
,620
59
0 PR
INT
"Err
or!
The
re a
re o
nly
J F
ield
Mat
rice
s,
but
you
have
ent
ered
" 60
0 PR
IHT
F<H
>;•.
Plea
se t
ry a
gain
.•
610
GO T
O 55
0 ~
620
PRIN
T •J
Ent
er l
engt
h of
thi
s se
ctio
n <"
tDt;
•>"
!3 62
5 PR
INT
"<H
.B.
Do n
ot e
nter
zer
o.>
="t
630
INPU
T L<
N)
640
PRIN
T "E
nter
Mod
ulus
of
Eld
stic
ity
<";E
t;">
=";
650
INPU
T E<
N>
660
PRIN
T "E
nter
Are
d M
o"en
t of
Ine
rtia
<";
Ft;"
>=";
67
0 IN
PUT
Il<N
> 67
5 X=
L<H>
67
7 E1
=E<H
>*I1
<N>
680
GO T
O F<
H>
OF 6
90,7
28,7
90
690
GOSU
B 61
10
710
GO T
O 86
0 72
0 PR
IHT
•Ent
er M
agni
tude
of
the
unif
orM
ly d
istr
ibut
ed l
odd,
<"
;Bt;
">"
722
PRIN
T •c
N.B.
Lod
d is
pos
itiv
e do
wnw
ard.
)•••
73
0 IN
PUT
Ql<H
> 75
0 Q2(t~j•Q1 <N
> 77
0 GO
SUB
6200
780
GO T
O 86
8 79
0 PR
I "W
hat
is
the
Mag
nitu
de o
f th
e li
nea
rly
var
ied
dis
trib
ute
d l
oad,
" 80
0 PR
INT
"<H
.B.
Load
is
pos
itiv
e do
wnw
ard.
) on
th
e le
ft <
";B
SJ">
="
; 81
0 IH
PUT
Qt<H
> 82
0 PR
INT
•on
the
rig
ht
<"IB
$J")
="I
83
0 IN
PUT
Q2<H
> 84
0_ G
_Q_$
_UB_
62_0
_0 ___
___ --
__ _
B6B
GOSU
B 66
00
-87
0 GO
SUB
6710
88
0 PR
IHT
•Lfo
r se
ctio
n l"
JH;"
:"
890
PRIN
T "I
s th
ere
any
conc
entr
ated
loa
d,
MOM
ent,
luM
ped
Mas
s <f
or •
; 90
0 PR
I "d
ynaM
ic_a
naly
sis
only
), o~
elas
tic
supp
ort
in
this
sec
tion
? ";
90
5 PR
IHT
"CY
or H
>";
910
INPU
T A$
91
5 P<
H>=0
91
7 H=
0 91
8 12
<H>=
0 92
0 IF
A$=
"N•
THEN
19
60
930
IF A
S<>•
v• T
HEN
890
935
P<H>
=1
1020
PRI
NT "
JEnt
er M
agni
tude
of
the
conc
entr
ated
loa
d,
<";C
t;">
" 10
22
PRIN
T "<
H.B
. Lo
ad
is p
osi
tiv
e do
wnw
ard)
=•
s 10
30
INPU
T Pl
<H>
1090
PRI
NT
•JE
nter
Mag
nitu
de o
f th
e H
oHen
t <"
;Gt;"
> "
1095
PRI
HT
"<H
.B.
MoM
ent
is p
osit
ive
in c
ount
er-c
lock
";
1096
PRI
NT
"wis
e di
rect
ion.
_Ple
ase
put
in t
he r
igh
t si
gn.>
=";
11
00
IHPU
T M
S<H>
11
50 P
RINT
"JE
nter
sti
ffn
ess
of t
he s
uppo
rt (
"JB
t;•>
=";
1160
IN
PUT
K1 <H
> 12
70
PRIN
T "J
Ent
er s
uppo
rt M
OHen
t st
iffn
ess
<"JL
$J">
=";
1280
IH
PUT
K2<H
> 12
90
IF U
1=1
THEN
148
0 12
92 P
RIHT
"E
nter
fta
gnit
ude
of c
once
ntra
ted
wei
ght
("JC
$J">
•"J
1293
IH
PUT
H<N>
12
95
IF A
BS<K
2<H
>><1
.0E-
20 T
HEN
1480
.... m
1298
PRI
NT
•Ent
er W
eigh
t M
o"en
t of
In
erti
a,
WRt
2 C
"IK
t;">=
" 12
99 I
NPUT
12
(N)
1300
PRI
"T
he
foll
owin
g ty
pes
of v
ibra
tion
of
the
supp
ort
are
avai
labl
e"
1310
PRI
NT
•1
for
bend
ing
vibr
atio
n.•
1320
PRI
"2
ro
tati
ng s
haft
(eq
ual
angu
lar
dire
ctio
n of
whi
rl
and
0;
1330
PRI
NT
"rot
atio
n."
1340
PRI
"3
ro
tati
ng s
haf
t (o
ppos
ite
angu
lar
dire
ctio
n of
whi
rl
&";
1350
PRI
NT
" ro
tati
on
.•
1360
PRI
NT
"Ent
er t
he I
fo
r th
e re
quir
ed v
ibra
tion
="J
13
70 I
HPUT
D2
1380
GO
TO D
2 OF
14
00,1
420,
1440
13
90 P
RI
"Err
or!
You
r in
put
shou
ld b
e 1,
2
or 3
. B
ut y
ou h
ave
ente
red
" 13
95 P
RINT
02
;•
• P
leas
e tr
y ag
ain.
" 13
98 G
O TO
13
00
1400
H=-
1 14
10 G
O TO
149
0 14
20 H
=l
1430
GO
TO
1480
--.
!AA
9 H
=-3
148E
f-GO
SUB
6340
19
00 G
OSUB
660
0 19
10 G
OSUB
671
0 19
60 H
=N+1
20
0f;
GO T
O...~--
~-_/
"-~f
0-·lf
f=t4
-1
, 201
3 DI
M
I1<N
1>,I2
CH
1>,P
<H1)
,Q1(
H1>
,Q2(
Hl>
,K1<
N1>
,K2<
Hl),
p1(H
1>,H
5<N
1>
2014
DIM
H<N
1>,F
<H1>
,L<H
1>,E
<Hl)
/201
5 DE
LETE
1,
2014
20
20 F
IND
7 20
25 A
PPEN
D 38
15
~035 P
RINT
•L
***
BOUN
DARY
CON
DITI
ONS
***"
2036
VIE
WPO
RT 0
,13e
,e,1
ee
2037
WIN
DOW
e,1
Je,e
,1ee
20
38 H
OVE
35,9
0 20
39 D
RAW
115,
99
... $
2040
HOV
E 35
,85
2050
DRA
W 11
5,85
20
60 M
OVE
10,7
5 20
70 D
RAW
115,
75
2080
MOU
E 15
,65
2090
DRA
W 11
5,65
21
00 M
OVE
15,5
5 21
10 D
RAW
115,
55
2120
MOU
E 15
,45
2130
DRA
W 11
5,45
21
40 H
OVE
10,3
5 21
50 D
RAW
115,
35
2160
MOU
E 11
5,90
21
65 D
RAW
115,
35
2170
MOU
E 95
,85
2180
DRA
W 95
,35
2190
MOU
E 75
,85
2200
DRA
W 75
,35
2210
MO~JE
SS, 8
5 22
20 D
RAW
55,3
5 22
30 M
OVE
35,9
0 22
40 D
RAW
35,3
5 22
50 t
10VE
15
, 75
2260
DRA
W 15
,35
2270
MOV
E 10
,75
2280
DRA
W 10
,35
2285
DRA
W 11
5,35
22
90 r
10lJE
58
, 86
2300
PRI
NT
"R
I G
H T
E
N o•
23
10 M
OlJE
40,
78
2320
PRI
NT
"PIN
NED
FIXE
D 23
30
HOVE
12
,63
2340
PRI
NT "LJ~EJHFJHTJJJHEJHNJHD•
2350
MOV
E 17
,68
2360
PRI
NT •
PI
NNED
1
•
.... '8
FREE
GU
IDED
•
2 3
<K.U
.>
4"
2365
MOV
E 17
,59
2370
PRI
NT
" FI
XED
5 6
7 8"
23
75 M
OUE
17,5
0 23
80 P
RIHT
"
FREE
9
<K.U
.>
10
11
<K.U
.>
12
CK.L
I.>"
2385
MOV
E 17
,40
2390
PRI
NT
" GU
IDED
13
14
15
CK
.LI.)
16
<K
.U.)"
24
00 P
RI
"JJJ
JJw
here
K.
U.
= K
ineM
otic
ally
Uns
t4bl
e,
unle
ss i
nter
nQl
";
2401
PR
INT
•sup
port
s ex
ist.
_U
se
thes
e bo
und4
ry c
ondi
tion
s Q
t yo
ur "
; 24
02 P
RIHT
•o
wn
risk
.•
2405
PRI
HT
"Do
not
4nsw
er K
. U.
to
the
que
stio
n be
low
.•
2410
PRI
NT
•JE
nter
I
for
the
requ
ired
Bou
ndar
y C
ondi
tion
= ";
2420
IH
PUT
D2
2430
IF
D2=
>9
THEN
243
8 24
35 G
O TO
D2
OF 2
460,
2490
,251
0,25
30,2
550,
2570
,259
0,26
10
2438
GO
TO D
2-8
OF 2
630,
2650
,267
0,26
90,2
710,
2730
,275
0,27
70
2440
PRI
NT
"Err
or!
The
I fo
r B
ound
ary
Con
diti
on i
s on
ly f
roM
1
to 1
6,"
2450
PRI
HT
"but
you
hav
e en
tere
d ";
D2;
".
Ple
ase
try
agai
n."
_ 24
55 G
Q___
I_Q __ 2
41_e
_
~4b8-r11=1
2462
112
=3
2464
M3=
2 24
66 H
4=4
· 24
70 G
OSUB
649
0 ...
2472
S<2
>=A
2 '
2474
S<4
>=A
3 24
76__
GD I
O 38
19 -
----
-'2
49
0 H
1=1
2492
r12
=2
2494
f-13
=2
2-24
96 1
14=4
24
98
GOSU
B 64
90
2500
S<2
>aA
2 25
02 S
<4>=
A3
2504
GO
TO 3
810
--25
10 '
11•3
... ~
2512
112
=4
2514
M3=
2 !
2516
M4=
4
4
~ (,, I
2518
GOS
UB 6
490
2520
S<2
>=A
2 25
22 S
(4)=
A3
2524
GO
_T0-
3810
2--
S30
M1=
2 25
32 1
'12=4
25
34 M
3=2
2536
M4=
4 25
38 G
OSUB
649
0 25
40 S
<2>=
A2
2542
S<4
>=A
3 25
44 G
O TO
381
0 25
50 '
11=1
25
52 t
12=3
25
54 "
13=3
25
56 M
4=4
2558
GOS
UB 6
490
2560
S<3
>=A
2 25
62 S
<4>=
A3
2564
GO
TO 3
810
2570
H1=
1 25
72 f
12=2
25
74 '
13=3
25
76 M
4=4
2578
GOS
UB 6
490
2580
S<3
>=A
2 25
82 S
<4>=
A3
2584
GO
TO 3
819
2590
M
1=3
2592
t12
=4
2594
113
=3
2596
t14
=4
... 'i6
2598
GOS
UB 6
490
2600
S<J
>=A
2 26
02 S
<4>=
A3
2604
GO
TO ~
$10
--ze;
-i 0
Mt=
2 .
2612
112
=4
2614
M3=
3 J
2616
t14
=4
2618
GOS
UB 6
490
2620
S<3
>=A
2 26
22 S
(4)=
A3
. 26-
24 G
O JO
381
0.
2630
M1=
1 26
32 M
2=3
2634
M3=
1 26
36 M
4=2
3 26
38 G
OSUB
649
0 26
40 S
<l>=
A2
.... 26
42 S
<2>=
A3
\()
\N
2644
GO
TO 3
810
2650
M1=
1 26
52 M
2=2
2654
M3=
1 I i>
26
56 '
14=2
26
58 G
OSUB
649
0 26
60 S
<1>=
A2
2662
SC2
>=A3
26
64 G
O TO
381
0 26
70.
M1=
3 26
72 f
12=4
26
74 t
13=1
II
2676
M4=
2 26
78 G
OSUB
649
0 26
89 S
<1>•
A2
2682
S(2
)=A
J
2684
GO
TO 3
819
2690
Mt=
2 -
2692
M2=
4 26
94 t
13=1
26
96 M
4=2
f L
2698
GOS
UB 6
490
2700
S<1
>=A
2 27
02 S
<2>=
A3
~~~~-~-:to. 3
81-E
L_
-·-27
12 t
12=3
27
14 "
13=1
-
2716
M
4=3
13
2718
G
OSU
B 64
90
2720
S<1
>=A
2 27
22 S
<J>=
AJ
-212
4 .. G
CL IO
_ 38
.10
2730
H1=
1 27
32 H
2=2
2734
t13
=1
. 27
36 t
14=3
1 '-1
2738
GO
SUB
6490
27
40 S
<1>=
A2
2742
S<J
>=A
J 27
44 G
O TO
38
10 .
. -
275-
0 H
1 =3
2752
H2=
4 27
54 M
3=1
1-27
56 M
4=3
) 27
58 G
OSUB
64
90
2760
S<1
>•A
2 27
62 S
CJ>=
AJ
~~~~
~? =
~p _
JiU9.
2772
H2•
4
.... 'g
2774
H3•
1 tG
2116
tt4
•3
2778
GOS
UB 6
499
2780
S
<t>
=A
2 27
82
S< J
)=A
J---
-38
10 G
OSUB
793
0 38
15 P
RIH
T 38
21
X=0
3822
X
4=0
3823
PRI
•J
JHow
Man
y in
creR
ents
wou
ld y
ou
lik
e to
hav
e fo
r ea
ch f
ield
";
38
24 P
RIHT
"s
ecti
on?•
; 38
30
INPU
T HJ
38
50
J2=0
38
52
IMAG
E 5<
2E,3X
> 38
53 P
RI
"L L
ENGT
H DE
FLEC
TION
SL
OPE
MOME
NT
SHEA
R"
3855
PRI
NT "
<"
;D$J
")
<";O
f;">
<R
adia
n)
<";G
$;">
";
3856
PRI
HT
" <"
;C$;
">"
~
3857
PRI
NT
~
3860
IF
P<N1
>=0
THEH
387
0 38
62 H
2=H3
tH1+
H1+1
38
64 G
O TO
387
5 38
70 H
2=H3
*N1+
H1
3875
DEL
ETE
2035
,382
2 38
77 D
ELET
E 79
30,1
0680
38
80 D
ELET
E s1
,s2,
s3,5
4,55
,x3
3882
DIM
S1<
H2>
,S2<
H2>
,S3<
H2)
,S4<
H2>
,S5(
H2)
,X3<
H2>
,R<8
> ----
3885
FO
R J1
=1
TO H
1 38
90
IF J
1=1
THEN
394
0 39
00 H
=Jl-
1 39
05
IF P
<H>=
0 TH
EN 3
940
3930
GOS
UB 6
340
3935
GOS
UB 6
800
3937
S=T
3 39
40 X
1=LC
J1)/H
3 39
42 H
=Jl
3944
E1•
E<J1
)*11
<J1)
~
3950
FOR
J4=
9 TO
L(J
l> S
TEP
Xl
3955
GO
TO
F<Jl
> OF
39
60,3
975,
3975
39
60 G
OSUB
611
0 39
65 G
OSUB
689
0 39
70 G
O TO
399
5 39
75 G
OSUB
620
0 39
80
GOSU
B 68
00
3985
J2=
J2+1
39
90 X
3<J2
>=X4
39
95 S
t<J2
>=T3
<1>
4000
S2<
J2)=
T3<2
> 40
05 S
J<J2
)=T
J(J)
40
10 S
4(J2
>=T
3(4)
40
12 P
RINT
USI
NG 3
852:
X3<
J2),
S1C
J2>,
S2C
J2),
SJ(J
2),S
4<J2
) 40
13 X
4=X4
+X1
4014
X=X
+Xl
~-4015
NEXT
J4
4016
X4=
X4-X
1 40
17 X
=0
4018
S=T
3 40
19 N
EXT
J1
4.02
0 G
OSU
B 78
00
4021
PR
INT
•JJJ
Do
you
wis
h to
see
the
gra
phs
for
defl
ecti
on,
slop
e,
" 40
22 P
RINT
"M
oHen
t an
d sh
ear?
<V
or H
>";
4023
IN
PUT
AS
4024
IF
A$=
"Y0
THEN
402
8 40
25
IF A
f<>"
H"
THEN
402
1 40
26 G
O TO
599
9 40
28 P
AGE
4030
MOV
E 60
,0
4032
PRI
HT "
LENG
TH <
"JD
tJ">
t"
4033
S5=
S4
4034
H6=
9 40
35 '
17=2
9
.... '8..
4036
H$=
"SH
EAR,
U"
40
37
l$=C
$ 40
45 G
OSUB
688
0 40
55 t
16=3
2 40
60 M
7=52
40
65 S
5=S3
40
66 H
$="M
OMEH
T,
M"
4067
l$
=G$
4070
GOS
UB
6880
40
80 H
$="S
LOPE
, S"
40
82 !
$="r
ad"
4085
'16
=55
4090
M7=
75
4095
S5=
S2
4100
GOS
UB 6
880
4110
H$=
"DEF
L.,
W"
4112
!$=
[)$
4115
M6=
78
4120
M7=
98
4125
SS=
Sl
4130
GOS
UB 6
880
4132
IF
U1=
1 TH
EH 4
140
4134
PRI
HT
"Fre
quen
cy="
;W;"
41
40 U
IEW
PORT
0,1
30,0
,100
41
50 W
IHDO
W 0
,130
,0,1
00
4160
MOV
E 0,
0 41
70 P
RIHT
"J
Do
you
wan
t to
41
80 I
t~PUT
A$
4190
IF
A$="
H"
THEN
424
0 42
00
IF A
$<>"
V"
THEH
417
0 42
10 F
IHD
6 42
20 O
LD
4230
RUH
42
40 P
RIHT
•J
JJJ
5990
END
..... ~
";J$
anal
yze
anot
her
beaM
? CV
or
H>"
;
******
******
EH
D ***
******
*** II
6900
REM
***
SUB
ROUT
INE
TO P
UT T
1 AS
AH
IDEN
TITY
MAT
RIX.
60
19 F
OR
1=1
TO 5
60
20 F
OR J
=l T
O 5
6039
IF
I<>J
THE
H 60
70
6040
Tt<
I,J>
=l
6050
GO
TO 6
080
6070
T1<
I,J>=
0 60
80 N
EXT
J 60
90 H
EXT
I 61
00 R
ETUR
N 61
10 R
EM *
** S
UBRO
UTIN
E FO
R M
ASSL
ESS
BEAM
TRA
NSFE
R MA
TRIX
61
20 G
OSUB
600
0 61
30 T
1<1,
2>=-
X
6140
T1<
1,J>
=-X
*X/<
2*E1
> 61
50 T
1<1,
4)=-
Xt3
/C6*
E1>
6160
T1<
2,J>
=X/E
1 61
70 T
1<2,
4>=X
*X/(2
*E1>
61
80 T
1<3,
4)=X
61
90 R
ETUR
H 62
00 R
EM *
** S
UBRO
UTIN
E FO
R UN
IFORM
LY O
R LI
HEAR
LV U
ARIE
D D
IST.
LO
AD.
6210
REM
*** T
RANS
FER
MAT
RIX.
62
20 G
OSUB
600
0 62
30 T
1<1,
2)=-
X
6240
T1<
1,3)
=-X
*X/(2
*El)
6250
T1(
1,4)
=-X
t3/(6
*E1>
62
60 T
1<1,
5)=Q
1<N
>*X
t4/(2
4*El
)-(Q
1(H
)-Q2<
H>>
*Xt5
/(120
*L<N
>*E1
> 62
70 T
1<2,
J>=X
/E1
6280
T1(
2,4)
=X*X
/(2*E
1>
6290
T1<
2,5)
•-Q
1<N
>*X
tJ/(6
*El)+
(Ql(N
)-Q
2CN
>>*X
f4/(2
4*L<
N>*
El>
6300
T1<
J,4>=
X
6310
T1C
J,5)=
-Q1C
N>*
X*X
/2+(
Ql<
H>-
Q2<
N>>
*Xt3
/C6*
L<N
>>
6320
T1<
4,5)a
-Q1C
N>*X
+CQ1
<N>-
Q2<N
>>*X
*X/(2
*L<H
>>
6330
RET
URH
6340
REM
***
SUB
ROUT
INE
FOR
CONC
. LO
AD,
MOME
NT,
TORQ
UE A
ND E
LAST
IC
6350
REM
***
SUP
PORT
, EL
ASTI
C SU
PPOR
T OR
LUM
PED
HASS
TRA
NSFE
R M
ATRI
X.
.... '£
6360
GOS
UB 6
909
6390
T1<
3,2)
•H*I
2<H
>*W
t2+K
2<H
> 64
00 T
l<J,
5>=-
"5<H
> 64
10 T
1(4,
1>=-
H<H
>*W
t2/G
+K1<
H>
6420
T1(
4,5>
=-P1
CH
> 64
30 R
ETUR
H 64
90 R
EM *
** S
UBRO
UTIN
E TO
CAL
C.
INIT
IAL
PARA
MET
ERS
USIN
G B.
C.
65
00 A
<1,1
>=T<
Mt,H
3>
6510
A<t
,2>=
T<M
t,H4)
65
20 A
<1,3
)=-T
<H1,
5)
6530
A(2
,1)=
T<M
2,H
3>
6540
A<2
,2>=T
<M2,H
4>
6550
A<2
,J)=
-T<H
2,5)
65
60 A
1=A
C1,
1>*A
<2,2
>-A
<2,1
)*A
<1,2
> 65
70 A
2=<A
<2,2
>*A
<1,3
>-A
<1,2
>*A
<2,3
))/A
l 65
00 A
3=<A
<1,1
>*A
<2,J
>-A
<2,1
)*A
<1,3
))/A
1 65
90 R
ETUR
N 66
00 R
EM**
* SU
BROU
TINE
TO
MUL
TIPL
Y 2
MAT
RICE
S, T
' T1
.<BO
TH 5
X5>
6610
FOR
I=
1 TO
5
6620
FOR
J=1
TO
5
6630
T2<
I,J>
=0
6640
FOR
K=1
TO
5
6650
T2<
I,J>=
T2<I
,J>+
T1<I
,K>*
T<K
,J>
6660
NEX
T K
66
70 N
EXT
J 66
80 H
EXT
I 66
90 T
=T2
6700
RET
URN
6710
REM
***
SUB
ROUT
INE
TO L
IST
OUT
ACCU
MULA
TED
TRAN
SFER
MAT
RIX.
67
20 P
RINT
"J
JAcc
u"ul
ated
HH
HH
HH
HH
HH
H--
----
----
-T
rans
fer
Mdt
rix
•;
6730
PRI
NT
"inc
ludi
ng s
ecti
on l
"IH
I" i
s ds
fol
low
s:J"
67
40 I
MAGE
5<2
X,2
E)
6750
FOR
Ja
l TO
5
6760
PRI
NT U
SING
674
0:T<
I,t>,
T<I,2
>,T<
I,J>,
TCI,4
>1T<
I,5>
6770
HEX
T I
.... ~
6780
PRI
NT
"KlK
HVKH
MKH
SKHW
JJJJJJ
" 67
82 P
RINT
uP
ress
RET
URN
to p
roce
ed."
• 67
84
IHPU
T A$
67
86
IF A
$<>"
"
THEH
679
0 67
9B
RETU
RN
6800
REH
*** S
UBRO
UTIN
E TO
MUL
TIPV
A 5
X5 A
ND A
1X5
MAT
RICE
S. 68
10 F
OR
1=1
TO 5
68
20 T
3<I)
=0
6830
FOR
K~1
TO 5
68
40 T
J<I>
=TJ<
I>+T
t<I,K
>*S<
K>
6850
NEX
T K
~l
--l1
2,;;
68
60 N
EXT
I QC
<l>:_
--68
70 R
ETUR
N 68
80 R
EM **
* DRA
W TH
E CU
RVES
OF
w,s,
H,V
68
90 R
EM **
* LAB
EL u
NEAT
• TI
C 69
00 B
1=1.
0E+3
00
6910
B2=
-1.0
E+30
0 69
20 F
OR 1
=1 T
O H2
69
30 B
t=SS
<I>
MIN
Bl
6940
B2=
S5<I
> MA
X 82
69
50 N
EXT
I 69
60 V
IEW
PORT
25,
130,
M6,
M7
6970
R<1
>=0
6980
R<2
)=XJ
<H2)
69
90 R
<J)=
IHT<
105/
(8*1
.8>>
MAX
1
7000
R<S
>=B1
70
10 R
(6)=
82
7020
R<7
>=IH
T<<M
7-M
6)/(3
*2.8
)) HA
X 1
7030
REM
*** C
ALCU
LATE
HEW
LIH
ITS
AND
INTE
RVAL
S 70
40 R
5=J
7050
GOS
UB 7
310
. ._;;
;T.t'~
+ 70
60 R
5=7
y ~o
"'
~
/ .....
7070
GOS
UB 7
310 ~-0 u
--~~
~"'-
v-'V";;.
\ 70
80 W
IHDO
W R<
1>,R
<2>,
R<5>
,R<6
> 70
90 R
EH A
XIS
R<J>
,R<7
>,R<
1>+R
<J>,
R<5>
+R<7
>
,.., 0 0
7190
AXI
S R
<J>,
R<7
> 71
10 R
EM **
* DRA
W TH
E CU
RVE
7120
HOV
E X
3<1>
,S5(
1)
7130
FOR
J=i
TO
H2
7140
DRA
W X
3<J>
,S5<
J>
7150
HEX
T J
7160
PRI
HT
7170
REM
*** L
ABEL
THE
M 71
80
IF M
7>29
THE
N 72
20
7190
R5=
4 72
00 E
$="i:U:U~JJ"
7210
GOS
UB
7600
72
20 R
5=8
7230
E$=
II iU
ll:tl:tl
:t ..
7240
GOS
UB
7600
72
50 V
IEW
PORT
0,1
30,M
6,M
7 72
60 W
IHDO
W 0,
10,0
,10
7270
MOl
JE 0
,5
7280
PRI
NT
Hf•
" ("
•!$
•")"
' -
' '
7290
HOM
E ~
7300
RET
URN
~-Rar-t** R
<RS>
= M
INIM
UM H
O.
OF T
ICS
7312
IF
ABS
<R<R
5-1>
-R<R
S-2>
>=>1
.0E-
12 T
HEN
7320
73
14 R
l=R<
RS-1
> 73
16 G
O TO
733
0 73
20 R
1=<R
<R5-
1>-R
<R5-
2))/R
(R5)
73
30 R
2=10
tIHT<
LGT<
R1)
) 73
40 R
l=R
1/R
2 73
50
IF R
1>2
THEN
739
0 73
60
IF R
l=l
THEN
743
0 73
70 R
2=2*
R2
7380
GO
TO
7430
73
90
IF R
1>5
THEN
742
0 74
00 R
2=S*
R2
7410
GO
TO 7
430
N ~
-'----
--..._
_
7420
R2•
19*R
2 74
30 R
EM **
* ADJ
UST
DATA
MIH
74
40 R
1=IHT<R<R5-2>~R2>
7450
R3=
R2*<
R1+2
) 74
60
IF R
J<R<
RS-2
> TH
EH 7
490
7470
R3=
R3-R
2 74
80 G
O TO
746
0 74
90 R
<RS-
2>=R
3 75
00 R
EH **
* ADJ
UST
DATA
MAX
75
10 R
1=1H
T<R<
R5-1
)/R2)
75
20 R
3=R2
*<R1
-2)
7530
IF
R<RS
-l><R
J TH
EN 7
560
7540
R3=
R3+R
2 75
50 G
O TO
75
30
7560
R<R
5-1>
=R3
7570
REH
*** R
<R5>
= A
DJUS
T TI
C IN
TERV
AL
7580
R<R
5)=R
2 75
90 R
E TU
RH
. --
. . --
..
7600
REM
LAB
EL A
XIS
7610
R4=
R<R5
-1>
7620
R(4
)=R<
1>
7630
R(8
)=R
(5)
7640
R3=
ABS
<R<R
5-3>
+R4)
HA
X A
BS<R
<R5-
2>-R
4>
7642
IF
R3>1
.0E-
7 TH
EN 7
650
7644
RJ=
ABS<
R<RS
-3>>
MA
X AB
S<R<
RS-2
>>
7650
R3=
IHT<
LGT<
R3)+
1.0E
-8>
7660
R2=
10t-R
3 76
70 R
1=R<
R5-2
>-R4
/2
7672
IF
RS=
4 TH
EN 7
680
7674
MO
UE R
<1>,
R<5>
76
75 P
RIHT
"H
HHHH
H"I
7676
PRI
NT U
SING
"-2D
.2D
,s•:
R<R
5>*R
2 76
80 R
<RS>
=R<R
S>+R
4 76
90 I
F R<
R5>>
R1
THEN
774
0 77
00 M
OUE
R<4>
,R<S
>
N 2
_,...,,--
7710
PRlt~T
ESS
7720
PRI
HT U
SIHG
•-
D.2
D,S
":R
CR
5>*R
2 77
30 G
O TO
768
0 77
40
IF R
3=0
THEH
779
0 77
50 R
<R5>
=R1
7760
MO
UE R
<4>,
R<8>
77
70 P
RIHT
E$J
77
80 P
RIHT
USI
NG
"3A
,+Fo
,s•:
n E"
;R3
7790
RET
URN
7800
IF
P<H
1>=0
TH
EH.79
20
7850
GOS
UB 6
340
7860
GOS
UB 6
800
7870
S1<
H2>=
T3<1
> 78
80 S
2<H
2)=T
3<2>
78
90 S
3CH2
>=T3
<3>
7900
S4<
H2>=
T3<4
> 79
10 X
JCH2
>=X3
<H2-
1)
7915
PRI
NT U
SING
38S
2:X
J<H
2),S
1<H
2>,S
2<N
2),S
J(H
2>,S
4<N
2)
7920
RET
URN
.. .
. 79
30 R
EH **
* SUB
ROUT
INE
TO D
RAW
THE
BEAM
. 79
40 P
AGE
7950
L2=
0 79
60 L
3=0
7970
FOR
J4=
1 TO
N1
7980
L2=
L2+L
(J4)
79
90 H
EXT
J4
8000
VIE
WPO
RT 3
,127
,75,
95
8010
WIN
DOW
0,L
2,e,
2e
8020
C1=
3 80
30 C
2=80
,
8040
GO
SUB
9460
: K
.c:n
1:>/
l;\v-
1 1~.c. ~
8050
UIE
WPO
RT J
,127
,75,
95
8060
WIH
DOW
e,L
2,e,
2e
8070
L1=
0 ~8080
FOR
K=1
TO H
t
"lo
")J.t
.OO
N
0 \,,.
)
8090
LJ•L
J+L<
K>
8100
L4=
<LJ-
Ll)/
50
8110
REM
***
TO
DRAW
MAS
SLES
S BE
AM.
8120
MOV
E L1
,9
8130
DR
AW L
3,9
8140
110
VE L
3 1S
8150
DR
AW L
1,5
8160
HO
VE
L3,5
_
8170
GO
TO F
<K>
OF 8
710 9
8190
,834
0 --
-818
0 RE
H **
* TO
DRA
W UH
IFOR
HLV
DIST
RIBU
TED
LOAD
. 81
90 M
OUE
Lt,
9 82
00 R
DRAW
L4,
2 82
10 R
MOUE
-L
4,-2
82
20 D
RAW
L1,1
8 82
30 D
RAW
LJ,
18
82
40 D
RAW
LJ,
9 82
50 R
DRAW
-L
4,2
8260
A2=
9 82
70 M
OUE
Ll+L
CK
)/4,
9 82
80 G
OSUB
10
630
8290
HOU
E Ll
+LC
K)/
2,9
8300
GOS
UB
1063
0 83
10 M
OUE
L1+3
*L<K
)/4,9
83
20 G
OSUB
10
630
8330
GO
TO 8
710
~---8340
IF Q
1<K>
>Q2C
K>
THEH
854
0 83
50 R
EM **
* TO
DRAW
LIH
.-UA
RIED
DIS
TRIB
UTED
LOA
D W
ITH
Q1(
Q2.
83
60
HOlJE
L
1 , 9
83
70 R
DRAW
L4,
2 83
80 R
HOVE
-L
4,-2
83
90 D
RAW
Lt,
14
8400
DRA
W L
J,18
84
10 D
RAW
LJ,
9 84
20 R
DRAW
-L
4,2
8430
MOU
E L
t+L
(K)/
4,9
I\) ~
8440
A2•
6 84
50 G
OSUB
10
630
8460
HOV
E L
l+L
(K)/
2,9
8470
A2=
7 94
80 G
OSUB
10
630
8490
MOV
E L
1+3*
L(K
)/4,
9 85
00 A
2=9
8510
GOS
UB
1063
0 85
20 G
O TO
871
0 85
30 R
EH **
* TO
DRAW
LIN
.-VA
RIED
DIS
TRIB
UTED
LOA
D W
ITH
Q1)
Q2.
85
40 M
OVE
L1,
9 85
50 R
DRAW
L4,
2 85
60 R
MOUE
-L
4,-2
85
70 D
RAW
Lt,
18
8580
DRA
M L
3,14
85
90 D
RAW
LJ,
9 86
00 R
DRAW
-L
4,2
8610
HOV
E Ll
+L<K
)/41
9 86
20 A
2=8
8630
GOS
UB
1063
0 86
40 M
OUE
Lt+
L(K
)/2,
9 86
50 A
2=7
8660
GOS
UB
1063
0 86
70 M
OUE
L1+L
<K>*
3/4,
9 86
80 A
2=6
a690
GOS
UB
1063
0 87
10 I
F P1
<K>=
0 TH
EH 8
820
8720
REM
*** T
O DR
AW A
POI
NT L
OAD.
87
30 M
OUE
LJ,
9 87
40 A
2=11
87
50 V
IEW
PORT
3,1
30,7
5,95
87
60 W
INDO
W 0
,L2+
L4,
0,20
87
70 G
OSUB
10
630
8780
UIE
WPO
RT 3
,127
,75,
95
8790
WIN
DOW
0,L
2,0,
20
N
0 \,n
.880
0 MO
VE L
J,18
98
10 P
RIHT
•
P"
8820
IF
M5<
K>=0
THE
N 90
19
8830
REH
***
TO
DRAW
A M
OMEH
T. 88
40 t
10VE
LJ,
9 88
50
VIEW
PORT
12
4*L
3/l2
,124
*L3/
l2+6
,85,
88
8860
WIH
DOW
-1,1
,0,1
88
70
A1=1
88
80 G
OSUB
10
580
8890
MOV
E -1
,0
8900
RDR
AW 0
.5,0
.5
8910
RMO
VE -
0.5
,-0
.5
8920
MOV
E 0,
1.3
8930
PRI
HT
11H11
8940
HOV
E -1
,0
8950
VIE
WPO
RT
124*
L3/
l2-0
.8,1
24*L
3/l2
+3,
86.5
,88
8960
MIH
DOW
-2,
1,0,
1 89
70 D
RAW
-1.5
,1
8980
VIE
WPO
RT 3
,127
,75,
95
8990
MIH
DOW
0,L
2,0,
20
9000
REM
***
DRA
W EL
ASTI
C AH
D TO
RQUE
SUP
PORT
. 90
10
IF K
1<K>
=0 A
ND K
2<K>
=0 T
HEN
9310
90
20 M
OUE
L3,
9 90
30
VIEW
PORT
12
4*L
3/L
2,12
4*L
3/L
2+6,
73,8
0 90
40 W
INDO
W 0,
5,0,
10
9050
MOU
E 2.
5,10
90
60
DRAW
2.5
,8
9070
RDR
AW -
1,-1
90
80 R
DRAW
1,
-1
9090
RDR
AW -
1,-1
91
00 R
DRAW
1,
-1
9110
RDR
AW 0
,-2
9120
RDR
AW
1.5,
0 91
30
RORA
W -3
,0
9140
RDR
AM
1,0
N &
9150
RDR
AW -
1,-1
91
60 R
HOlJE
1,
9 91
70 R
DRAW
1,
1 91
80 R
HOVE
1,
0 91
90
RDRA
W -1
,-1
9200
MO~JE
0, 4
92
10 P
Rit~T
•K11
9220
IF
K2(
K)=
0 TH
EH 9
310
9230
t10
VE
2.5,
9 92
40 R
DRAW
1,
0 92
50 R
DRAW
0,-
7 92
60 M
OlJE
4,4
92
70 P
RIHT
11T11
9280
VIE
WPO
RT 3
,127
,75,
95
9290
WIN
DOW
0,L
2,0,
20
9300
HO
VE L
3,7
9310
IF
H<K
)=0
THEH
937
0 93
20
MOVE
L3,
.7
9330
VIE
MPO
RT
124*
L3/
L2,
124*
L3/
l2+6
,79,
85
9340
WIN
DOW
-1
,1,-
1,1
93
50 A
1=1
9360
GOS
UB
1057
0 93
70
UIEW
PORT
3,1
27,7
5,95
93
80 W
IHDO
W 0
,L2,
0,20
93
90 L
1=L3
94
B0 H
EXT
K
9410
C1=
127
9420
VIE
WPO
RT 3
,127
,75,
95
9430
WIN
DOW
0,L
2,0,
20
9440
GOS
UB 9
460
_--9
450
_RET
U.RH
___
.. -
--
9460
REM
*** S
UBRO
UTIN
E FO
R DR
AWIN
G BO
UHDA
RV C
OHDI
TIOH
. 94
70
IF C
1>10
0 TH
EH 9
500
9480
IF
D2<
=4 T
HEH
9510
94
82
IF D
2<=8
TH
EN 9
660
I\)
~
9484
IF
D2<
=12
THEN
10
490
9486
IF
D2<
•16
THEH
996
9 94
90 G
O TO
951
0 95
00 G
O TO
D2
OF 9
510,
9660
,104
90,9
960,
9510
,966
0,10
490,
9960
95
02 G
O TO
02-
8 OF
951
0,96
60,1
0490
,996
0,95
10,9
660,
1049
0,99
60
9510
UI
EWPO
RT c
1-1.
s,c1
+1.
s,c2
-4,C
2+4
9520
WIN
DOW
0,3
,0,e
95
30 t
lOVE
0, 1
95
40 D
RAW
1.5,
4 95
50 D
RAW
3,1
9560
DRA
W 0,
1 95
70 M
OUE
0.5,
1 95
80 D
RAW
0,0.
5 95
90 H
Ol.JE
1.
5,1
9600
DRA
W 1,
0.5
9610
110
UE 2
.5,1
96
20 D
RAW
2,0.
5 96
30 1
10tJE
1.
5, 4
96
40 D
RAW
1.5,
8 96
50 G
O TO
105
59
9660
IF
C1>
100
THEN
981
0 96
70 U
IEW
PORT
c1-
1.5,
C1,
C2-
3,C
2+7
9680
WIH
DOW
0,1
,0,1
0 96
90 M
OlJE
1 ,
19
9700
DRA
W 1,
0 97
10 f
10l)E
1,
9
9720
RDR
AW -
1,-1
97
30 M
OlJE
1,
7 97
40 R
DRAW
-1,
-1
9750
MOV
E 1,
5 97
60
RDRA
W -1
,-1
9770
MOl
JE 1
, J
9780
RDR
AW
-1,-
1
9790
MOV
E 1,
3 98
00 G
O TO
10
550
N
0 O>
9810
VIE
WPO
RT c
1,c1
+1.
s,c2
-J,C
2+7
9820
WIH
DOW
0,1
,0,1
0 98
30 M
OUE
0,10
98
40 D
RAW
0,0
9850
MOU
E 0,
9 98
60 R
DRAW
1,
-1
9870
tlO
UE
0,7
98
80 R
DRAW
1,-
1 98
90 H
OVE
0,5
9900
RDR
AW
1,-1
99
10 M
OVE
0,3
9920
RDR
AW 1
,-1
9930
MOV
E 0,
3 99
40 G
O TO
10
550
9950
REH
*** T
O DR
AW G
UIDE
D EN
D 99
60
IF C
1>10
0 TH
EH 1
0230
99
70 V
IEW
PORT
c1-
J,c1
,c2-
3,c2
+7
9980
WIN
DOW
0,2,
0,10
99
90 M
OlJE
2, 7
10
000
DRAW
2,3
10
010
MOUE
1,1
0 10
020
DRAW
1,0
10
030
MOVE
1,9
10
040
RDRA
W -t
,-1
10
050
HOVE
1
,7
1006
0 RD
RAW
-1,-
1 10
070
t10VE
1,5
10
080
RDRA
W -1
,-1
10
090
MOVE
1,
3 10
100
RDRA
W -1
,-1
10
110
A1=1
10
120
MOUE
1.
5,5.
75
1013
0 VI
EWPO
RT c
1-1.
s,c1
,c2+
2,c2
+3.
5 10
140
WIN
DOW
-1
,1,-
1,1
10
150
GOSU
B 10
570
N ~
1016
0 UI
EWPO
RT c
1-J,
c1,c
2-J,
C2+
7 10
170
WIH
DOW
0,2
,e,1
0 10
180
MOVE
1.
5,4.
25
1019
0 VI
EWPO
RT c
1-1.
s,c1
,c2+
0.s,
c2+
2 10
200
WIH
OOW
-1
,1,-
1,1
10
210
GOSU
B 10
570
1022
0 GO
TO
1055
0 10
230
VIEW
PORT
C
1,C
1+3,
C2-
3,C
2+7
1024
0 W
IHDO
W 0
,2,0
,10
10
250
MOVE
0,7
10
260
DRAW
0,3
10
270
MOUE
1,
10
1028
0 DR
AW
1,0
1029
0 t10
VE
1,9
1030
0 RD
RAM
1,-1
10
310
MOVE
1,
7 10
320
RDRA
W 1,
-1
1033
0 MO~JE
1, 5
1034
0 RD
RAW
1,-1
10
350
MOVE
1,
3 10
360
RDRA
W 1,
-1
1037
0 MO
UE 0
.5,S
.75
1038
0 A1
=1
1039
0 UI
EWPO
RT c
1,c1
+1.
s,c2
+2,
c2+
J.5
1040
0 W
IHDO
W -
1,1,
-1,1
10
410
GOSU
B 10
570
1042
0 UI
EWPO
RT c
1,c1
+3,C
2-J,
C2+
7 10
430
WIN
DOW
0,2
,0,1
0 10
440
HOVE
0.5
,4.2
5 10
450
VIEW
PORT
c1,
c1+
1.s,
c2+
0.s,
c2+
2 10
460
WIHD
OW -
1,1
,-1
,1
1047
0 GO
SUB
1057
0 10
480
GO T
O 10
550
1049
0 IF
C1>
100
THEN
19
530
1050
0 MO
VE 0
,5
N .... 0
1951
0 DR
AW 0
,9
1052
0 GO
TO
1055
0 10
530
MOUE
L2,
5 10
540
DRAW
L2,
9 10
550
RETU
RN
1056
0 RE
M **
* SU
BROU
TINE
TO
DRAW
A C
IRCL
E 10
570
MOUE
A
1,0
1058
0 FO
R J4
=0 T
O 36
0 ST
EP 2
0 10
590
DRAW
A
1*CO
S<J4
),A1*
SIH
(J4)
10
600
NEXT
J4
1061
0 RE
TURN
_
. 10
620
REH
***
SUBR
OUTI
NE T
O DR
AW A
H AR
ROW
OF L
ENGT
H A2
10
630
RDRA
W -L
4,2
1064
0 RM
OVE
L4,
-2
1065
0 RD
RAW
L4,2
10
660
RMOV
E -L
4,-2
10
670
RDRA
W 0,
A2
1068
0 RE
TURH
N
.... ....
APPENDIX B.J.2
PROGRAM LISTING --- BF.AM ANALYSIS (PARl' II)
212
3815
REM
*** L
IST
DATA
38
25 U
IEW
PORT
0,1
30,0
,100
38
35 W
INDO
W 0
,130
,0,1
00
3845
HOV
E 0,
97
3855
PRI
HT
"Dra
win
g of
the
Bea
M•
3865
MOU
E 0,
72
3875
PR
IHT
"DA
TA:•
3885
IM
AGE
2E,1
X,F
A
3900
IF
U1=
1 TH
EN 3
905
3901
PR
IHT
"Fre
Que
ncy •
••••
••••
••••
••••
••••
••••
••••
••••
• ="
; 39
02 P
RIHT
USI
NG 3
885:
W,J
$ 39
05 F
OR K
=t
TO N
I 39
15 P
RINT
•F
or s
ecti
on l
"1K
;•:•
39
25 P
RINT
0L
engt
h of
th
is s
ecti
on ••
••••
••••
••••
••••
••• =
"; 39
35 P
RINT
USI
NG 3
885:
L<K
),D$
3945
PRI
NT
•Mod
ulus
of
Ela
stic
ity
••••
••••
••••
••••
••••
•• =
•; 39
55 P
RINT
USI
NG 3
885:
E<K
>,E$
39
65 P
RINT
•A
rea
HoH
ent
of I
ner
tia
••••
••••
••••
••••
••••
• =0
;
3975
PRI
NT U
SING
388
5:I1
<K),
F$
3985
GO
TO F
<K>
OF 4
155,
3995
,403
5 39
95 P
RINT
•M
agni
tude
of
unif
or"l
y di
stri
bute
d lo
ad ••
•• =•
; 40
05 P
RINT
USI
NG 3
885:
Q1(
K),
9$
4025
GO
TO 4
155
4035
PRI
NT
•Lef
t en
d of
lin
earl
y-va
ried
dis
t.
load
••••
• =";
4045
PRI
NT U
SING
388
S:Q
1(K
),8$
40
65 P
RINT
"R
ight
end
of
line
arly
-var
ied
dis
t.
load
••••
=•;
40
75 P
RINT
USI
NG 3
885:
Q2(
K),
8$
4155
IF
P1<
K>=0
THE
N 41
85
4165
PRI
NT
"Mag
nitu
de o
f C
once
ntra
ted
Load
••••
••••
••••
• a•;
41
75 P
RINT
USI
NG 3
885:
P1<K
>1
C$
4185
IF
H5<
K>=
0 TH
EN 4
215
4195
PR
IHT
"Mag
nitu
de o
f M
oMen
t •••
••••
••••
••••
••••
••••
• =•
; 42
05 P
RINT
USI
NG 3
885:
H5(
K),G
$ 42
15
IF K
1<K
>•0
THEN
424
5
I\) .... \,.
.)
4225
PRI
NT
•sti
ffne
ss o
f Su
ppor
t •••
••••
••••
••••
••••
••••
=";
42
35 P
RINT
USI
NG 3
885:
K1C
K),9
$ 42
45 I
F K2
<K>=
0 TH
EN 4
266
4255
PRI
NT
"Sup
port
Mo"
ent
Sti
ffne
ss ••
••••
••••
••••
••••
• =";
4260
PRI
NT U
SING
388
5:K
2(K
),L$
4266
IF
U1=
1 TH
EN 4
275
4267
PRI
HT
"Mag
nitu
de o
f C
once
ntra
ted
Wei
ght •
••••
••••
•• =
";
4268
PRI
NT U
SIHG
388
5:M
<K>,
C$
4270
IF
ABS
CK2<
K>>
<t.0
E-20
THE
N 42
75
4272
PRI
NT
"Wei
ght
HoH
ent
of I
nert
ia <
WRt
2>
abou
t a
• 42
73 P
RIHT
"d
io"e
ter
of d
isc
or e
leM
ent •
••••
••••
••••
••• =
";
4274
PRI
HT U
SING
398
5:I2
<K>,
Kt
4275
PRI
HT
4285
HEX
T K
42
95 R
EM **
* CHA
NGE
DATA
43
05 P
RIHT
"J
Do
you
won
t to
chong~ th
e do
to?
<Y o
r H>
•• 43
15 I
NPUT
A$
4325
IF
A$="
H"
THEN
510
0 43
35
IF A
f<>"
V"
THEH
430
5 43
45 P
AGE
4355
PRI
NT "
JWhi
ch s
ecti
on d
o yo
u w
ont
to c
hang
e?";
43
65
INPU
T N
43
75
IF N
=0 T
HEH
4378
43
77
IF H
<=H1
TH
EH 4
385
4379
PRI
NT "
Err
or!
You
only
hov
e •;
Ht;
• se
ctio
ns,
but
you
have
ent
ered
"
4379
PRI
NT H
;".
Plea
se t
ry a
gain
.•
4380
GO
TO 4
355
4385
IF
U1=
1 TH
EN 4
392
4386
PRI
NT
"Pre
sent
fre
quen
cy ••
••••
••••
••••
••••
••••
••••
••••
•• =
"; 43
87 P
RINT
USI
NG 3
883:
W,J
$ 43
98 P
RINT
"E
nter
new
fr
eque
ncy•
"•
4389
IN
PUT
W
4392
PRI
HT
•For
sec
tion
1•1
H;"
:"
4395
PRI
HT "
Pres
ent
Leng
th o
f th
is s
ecti
on •
••••
••••
••••
••••
•=";
44
05 P
RINT
USI
NG J
885:
L<H
>,D
$
N .... ~
4415
PRI
HT
•Ent
er n
ew L
engt
h an
; 44
25 I
HPUT
L<H
> 44
35 P
RIHT
•P
rese
nt M
odul
us o
f E
last
icit
y ••
••••
••••
••••
••••
•• =
";
4445
PRI
HT U
SIHG
388
5:E<
H>,
E$
4455
PRI
HT
"Ent
er n
ew M
odul
us o
f E
last
icit
y •
ul
4465
IN
PUT
E<H>
44
75 P
RIHT
"P
rese
nt A
rea
HoM
ent
of I
nert
ia ••
••••
••••
••••
••••
• ="
; 44
85 P
RIHT
USI
NG 3
885:
I1<N
>,Ff
44
95 P
RINT
"E
nter
new
Are
a M
oMen
t of
Ine
rtia
=";
4505
IHP
UT
I1<H
> ,
i ~
--45
15 G
O TO
F<N
> OF
479
5,45
25,4
585
-452
5 PR
IHT
"Pre
sent
Uni
forM
ly D
istr
ibut
ed L
oad •
••••
••••
••••
•• ="
; 45
35 P
RINT
USI
NG 3
885:
Q1<
H>,
B$
4545
PRI
NT
•Ent
er n
ew U
nifo
rMly
Dis
trib
uted
Loa
d =•
; 45
55 I
NPUT
Ql<
N>
4565
Q2<
H>=Q
1<N>
45
75 G
O TO
479
5 ~4585
PRIN
T •P
rese
nt l
eft
side
of
lin-
vari
ed d
ist.
lo
ad ••
••••
• =•;
45
95 P
RINT
USI
NG 3
885:
Q1<
N>,
C$
4605
PRI
NT
"Ent
er n
ew
left
sid
e of
lin
.-va
ried
dis
t.
load
=•;
46
15
IHPU
T Q
t(N)
4625
PRI
NT
"Pre
sent
rig
ht s
ide
of l
in.-
var
ied
dis
t.
load
••••
• =•;
4635
PRI
NT U
SING
388
5:Q
2<H
>,C$
46
45 P
RINT
"E
nter
new
rig
ht s
ide
of l
in.-
vari
ed d
ist.
lo
ad=
";
4655
INP
UT Q
2<N>
-..
4795
IF
P1<H
>=0
THEN
484
5 48
05 P
RINT
"P
rese
nt C
once
ntra
ted
Load
••••
••••
••••
••••
••••
••••
=";
48
15 P
RINT
USI
NG 3
885:
P1<N
>,C
$ 48
25 P
RINT
"E
nter
new
Con
cent
rate
d Lo
ad •
";
4835
IH
PUT
P1<H
> 48
45 I
F M
5CN>
=0 T
HEN
4895
48
55 P
RINT
"P
rese
nt M
oMen
t ••
••••
••••
••••
••••
••••
••••
••••
••··
•";
4865
PRI
HT U
SING
388
5:M
5<H
>,G
f 48
75 P
RINT
•E
nter
new
HoM
ent
••1
4885
IH
PUT
MS<
H>
N .... \.n
4895
IF
K1<
H>•0
THE
N 49
45
4905
PRI
NT
"Pre
sent
Sti
ffne
ss o
f Su
ppor
t •••
••••
••••
••••
••••
•• •"
; 49
15 P
RINT
USI
NG 3
885:
Kl<
H>,
Bf
4925
PRI
NT
"Ent
er n
ew S
tiff
ness
of
Supp
ort
=•1
4935
IH
PUT
K1<H
> 49
45
IF K
2<H>
=0
THEN
497
5 49
55 P
RIHT
"P
rese
nt S
uppo
rt M
oMen
t S
tiff
ness
••••
••••
••••
••••
• ="
; 49
60 P
RIHT
USI
HG 3
885:
K2<
H),L
$ 49
66 P
RIHT
"E
nter
new
Sup
port
MoM
ent
Sti
ffne
ss =
"I
4967
IH
PUT
K2<H
) 49
75
IF U
1=1
THEH
499
0 49
77 P
RIHT
"P
rese
nt M
agni
tude
of
Con
cent
rate
d W
eigh
t •••
••••
•• =
•; 49
78 P
RIHT
USI
HG 3
885:
M<H
>,C
f 49
79 P
RIHT
•E
nter
new
"ag
nitu
de o
f C
once
ntra
te W
eigh
t ="
; 49
80
INPU
T f1<
H>
4982
IF
ABS
<K2<
H>>
<1.0
E-20
THE
H 49
90
4984
PRI
HT
"Pre
sent
Wei
ght
Mo"
ent
of I
nert
ia <
WRt
2> •
••••
••••
• =";
4985
PRI
HT U
SIHG
388
5:12
CH
) 9K
$ 49
86 P
RINT
"E
nter
new
Wei
ght
MoM
ent
of I
ner
tia
<WRt
2)
=";
4987
IH
PUT
I2<H
> 49
90 P
RI
"JDo
you
wan
t to
cha
nge
the
data
of
anot
her
sect
ion?
<Y o
r H
>";
4992
IH
PUT
A$
4995
IF
A$=
"V"
THEN
434
5 50
05
IF A
f<>"
H"
THEN
499
9 50
10 G
OSUB
60~
5012
T=
T1
'fJ
5013
S=0
50
14 S
<S>=
1 50
15 F
OR K
4=1
TO N
I 50
25 E
t=E<
K4>
*I1<
K4>
50
30 X
=L<K
4)
5032
H=K
4 50
35 G
OSUB
F<K
4>
OF 6
189,
6270
,627
0 50
45 G
OSUB
660
0 50
55
IF P
<K4>
=0 T
HEH
5985
N .... °'
13) IS) If) .-tlS> M "It' IJ) IS) IJ) IJ)"lt'N
~ mm o :::> ::n- i-c.r".n x COLI.JO ..., '-' z..., If) If) II'> CD t.t: r .... oo •1"• IS) r.D a;,~ IO II'> II'> U"')
217
The vita has been removed from the scanned document
DEVELOPMENT OF INTERACTIVE COMPUTER PROGRAMS FOR
MECHANICAL ENGINEERING DESIGN: FATIGUE ANALYSIS,
SECTION PROPERTIES, AND BEAM ANALYSIS
by
Yiu Wah Luk
(ABSTRACJI')
This thesis presents the theory and describes three interactive
computer graphics programs for mechanical engineering designs Fatigue
Analysis, Section Properties, and Beam Analysis. The Fatigue Analysis
program sizes up a mechanical component, circular, rectangular, or any
shape, to prevent fatigue failure. Six most generally accepted
fatigue failure lines are available and any equivalent stress theories
are allowed. It can al.so calculate the significant endurance limit with
the theoretical stress concentration factor supplied by the user.
The Section Properties program finds twenty section propertie~,
such as area, area moment of inertia, and radius of gyration about
different axis, of any shape plane cross section.
Tm Beam Analysis program, using transfer matrix method, computes
and also plots the curves of deflection, slope, moment, and shear
along the beam. Static and forced, undamped dynamic analysis can be
performed for beams of uniform or variable cross section. Uniformly or
linearly varied distributed loads, concentrated point loads, applied
moments, or combinations of all three may be applied. This program
allows any combination of pinned, fixed, free, or guided flexural
boundary conditions, even normally kinematically unstable condition
can be handled if sufficient internal supports are provided. In-span
support can be elastic springs and/or elastic moment spring. Modelli~
for dynamic response uses lumped mass.
All three programs provide the option of using either English or
SI units. The programming language used is BASIC and the micro-processor
used is a Teketronix model 4051 with 32 K memory.
![My beloved Nabi (SAW) [Children series]](https://img.dokumen.tips/doc/110x75/55587fe8d8b42aad358b48cb/my-beloved-nabi-saw-children-series.jpg)
