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This is to certify that the

thesis entitled

DYNAMIC BEHAVIOR OF ELASTO-
INELASTIC BEAMS SUBJECTED TO
MOVING LOADS

presented by

l, Teoktistos Toridis

has been accepted towards fulfillment
of the requirements for

JCKLL fry ([LI‘IL

Major professor

Datezé/f 47,47} /(/A[¢

0-169

  
     

  

LIBRARY

Michigan State
University

ABSTRJXCT

DYNAMIC BEHAVIOR OF ELASTO-INELASTIC
BEAMS SUBJECTED TO MOVING LOADS

by Teoktistos Toridis

An analytical study is made of the elasto-inelastic behavior of
beams subjected to moving loads. The structure analyzed consists of
a slender beam resting on rigid supports. The types of moving loads
used in this study are a single unsprung mass, a sprung mass, and a
combination of a sprung and unsprung mass connected by an elastic
spring and a viscous damper .

Although the method of analysis can be applied conveniently to
continuous beams, only simply supported beams have been considered
in obtaining the numerical results for this investigation. The analysis
is based on‘a discrete beam model consisting of massless rigid panels
connected by flexible joints with point masses . The flexibility and mass
of the joints are obtained by lumping the corresponding continuous
properties of the actual beam. This model facilitates considerations
of the nonlinear deformation characteristics of the beam and possible
non-uniform distribution of the beam mass.

The solution of the problem is carried out by numerical tech-

niques. The results were obtained by the use of a high speed digital

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Teoktistos Toridis

computer with the objective of gaining an insight into the physical
problem, illustrating certain features in the workings of the mathe-
matical model, and assessing the influences of certain parameters
on the beam response.

For an elastic perfectly plastic beam, it is found that the beam
could suffer permanent damage due to the passage of a load, whose
weight is less than the static yield load. On the other hand, loads
much heavier than the static yield load can cross the beam, if large
deformations are allowed. However, in general, the inelastic
deformations increase at a rapid rate when the load is increased
beyond the static yield load. For unsprung loads, an increase in the
load speed generally results in an increase in the permanent deflec-
tion. The opposite is true for sprung loads.

In the case of beams having an "inelastic stiffness, " it is found
that even a small positive inelastic stiffness reduces the permanent
displacement significantly. For such beams, an increase in load

speed decreases the amount of permanent displacement.

DYNAMIC BEHAVIOR OF ELASTO-INELASTIC

BEAMS SUBJECTED TO MOVING LOADS

By ’

3" ’

Teoktistos Toridis

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Civil and Sanitary Engineering

1964

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ACKNOWLEDGEMENTS

The study reported herein was made as part of a research
project on the elasto ~inelastic behavior of beams subjected to moving
loads conducted in the Department of Civil Engineering of Michigan
State University under the direction of Dr. R. K. Wen. The work is
supported by the National Science Foundation under Grant No. (3.12.143
administered by the Division of Engineering Research.

This report also constitutes the author's doctoral dissertation.
It has been written under the direction of Dr. R. K. Wen, the author's
advisor, to whom gratitude is extended for his valuable guidance dur-
ing the course of this study.

The author also wishes to express his thanks and appreciation
to the members of his guidance committee, to Dr. C. E. Cutts, for
his encouragement during the course of the author's graduate. studies,
to Dr. L. E. Malvern, Dr. W. A. Bradley, and Dr. C. S. Duris, for
their valuable advice.

Thanks are also due the staff of the Computer Laboratory of

the University for their help and cooperation.

ii

 

TABLE OF CONTENTS

ACKNOWLEDGEMENTS

LIST OF FIGURES

II.

III .

IV.

INTRODUCTION
1. 1. General
1. Z . Notation

ME THOD OF ANAL YSIS

Z.
2.

l.
20

System Considered
Analysis of System

SIMPLY SUPPORTED BEAMS SUBJECTED TO

WWWUJ

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IprI—o

O‘U‘l

A SINGLE MOVING LOAD

General

Dimensionless Form of Equations

Bilinear Moment—Curvature Relation

Properties of Actual Bridges and Parameters
of Problem

Procedure of Numerical Solution

Use of Computer

NUMERICAL RESULTS

I-hhllI-hvhrliI-PnI-F‘I-h
mNOWIbWNI-a

General

History Curves for Unsprung Mass
History Curves for Sprung Mass

History Curve for a. Single -axle Load Unit
Final Shape of Deformed Beams

Effect of Speed Parameter

Effect of Stiffness Ratio

Effect of Weight of Load

iii

Page

ii

0‘.—

10

10
ll

23

23
23
26

29
31
33

35

3 5
36
42
43
44
46
47
47

Page

V. DISCUSSION, SUMMARY, AND CONCLUSION 48
5. 1. Discussion 48
5.2. Summary 51
5. 3. Conclusion 53
LIST OF REFERENCES 54
FIGURES (1-29) 56

APPENDIX. COMPUTER PROGRAM 85

iv

Figure

10

ll

12

13

14

LIST OF FIGURES

System Considered

Nonlinear Bending Moment-Curvature Relation
Discrete Beam System

Free Body Diagrams

Bilinear Bending Moment-Curvature and Bending
Moment—Rotation Diagrams

Properties of a Highway Bridge

History Curve of Deflection--Unsprung Mass
(R = 0, a = 0.10, B = 0.90, y = 2.676)

History Curve of Interaction Force--Unsprung
Mass (R = 0, a = 0.10, [5 = 0.90, y = 2.676)

History Curve of Acceleration--Unsprung Mags
(R = o, a = o. 10, p = 0.90, y = 2.676)

History Curve of Deflection--Unsprung Mass
(R = 0. a = 0.30, (3 = 0.90, y = 2.676)

History Curve of Interaction Force--Unsprung
Mass (R = 0, a = 0.30, B = 0.90, y : 2.676)

History Curve of Deflection--Unsprung Mass
(R = 0, a : 0.20, [3 = 1.20, y = 3.568)

History Curve of Interaction Force--Unsprung
Mass (R = 0, a = 0.20, [3 =1.20, y = 3.568)

History Curve of Deflection--Unsprung Mass
(R = 0.1, a = 0.20, [3 =1.20, y = 3.568)

Page
56
57
58

59

60

61

62

63

64

65

66

67

68

69

Figure

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

History Curve of Interaction Force--Unsprung
Mass (R = 0.1, a = 0.20, 8 =1.20, y = 3.568)

History Curve of Deflection--Unsprung Mass
(R = 0.5, a = 0.20, [3 = 1.20, y = 3.568)

History Curve of Deflection--Sprung Mass
(R = 0, a = 0.20, [3 = 1.00, y = 2.973, k = 33.57)

History Curve of Interaction Force--Sp1_‘ung Mass
(R = 0, a = 0.20, [3 21.00, y = 2.973, k = 33.57)

History Curve of Deflection--Sprung Mass
(R = 0, a = 0.20, [3 = 1.20, y = 3.568, k = 33.57)

History Curve of Interaction Force--Sp_rung Mass
(R = 0, a = 0.20, (3:1.20, y = 3.568, k = 33.57)

History Curve of Deflection--Single -ax1e Load
Unit (R = 0:1, a = 0.20, B 21.20, y = 3.568,
X = 0.132, k = 33.57)

Deformed Beam Shapes--Unsprung Mass
Deformed Beam Shapes-~Unsprung (Mass
Deformed Beam Shapes--Sprung Mass

Deformed Beam Shapes--Sing1e -ax1e Load Unit

Effect of Speed on Maximum Deflections-~Unsprung
Mass

Effect of Speed on Maximum Deflections-~Sprung
Mass

Effect of R on Maximum Deflections --Unsprung
Mass

Effect of Weight of Vehicle on Maximum Deflections--
Unsprung Mass

vi

Page

70

71

72

73

74

75

76

77
78

79

80

81

82

83

84

I. INTRODUCTION

1.1. General

In recent years the elasto-inelastic analysis of structures
subjected to dynamic loads has received much attention. Due to
the nonlinear nature of the deformation characteristics of the
structure, rigorous mathematical solutions of problems of this type
are usually hard to obtain. In order to arrive at workable methods
of solution, researchers have been forced to make major or even
drastic simplifying assumptions regarding the physical character-
istics of the structure and/or the loading, thus often significantly
limiting the applicability of such analyses .

Past work in this field has mainly been concentrated on elastic
perfectly plastic structures. The only work known to the author
dealing with a more general moment-curvature relation is that of
Bohnenblust1 who developed a method for calculating the response of
a semi-infinite elasto -inelastic beam to an impact at one end of the
beam. However, it does not seem feasible to extend this method to
beams either with a finite length or with a lateral loading. Bleich
and Salvadori2 presented a variation of the normal mode approach
for the analysis of elasto -plastic structures. This method, although

theoretically sound, is unwieldy to apply to practical problems,

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except the simplest ones.

A relatively simple solution for large plastic deformations by
the "rigid-plastic" theory for the dynamic analysis of beams was
presented by Lee and Symonds . 3 This solution can be applied only to
the particular case of very large plastic deformations as compared
to the elastic ones. The work was followed by a number of applica-
tions and extensions of the theory. For example, the influence of
strain rate was considered by Ting and Symonds.

In all of the above studies the continuum nature of the structure
was maintained; namely, a continuous distribution of both flexibility
and mass was considered. A different direction during the more
recent research efforts in this area has been the use of a discrete
model representing the actual structure. Thus, Berg and DaDeppo5
developed a method for the analysis of multi-story elastic perfectly
plastic structures responding to lateral dynamic forces by lumping
the mass of the actual structure at the floor levels. This method was
based on a "predictor-corrector" approach which had previously been
proposed by Bleich. 6 The "predictor" is, in essence, an elastic
solution that satisfies dynamics but generally violates the material
property at some parts of the structure by requiring resisting moments
larger than the yield moment. The “corrector” consists of also an
elastic solution with hinges inserted at certain appropriate points; at

these hinges, equal and opposite moments having the same magnitude

.u.
.4;

.i

8v

‘lh

as the yield moment of the section are applied. If the "predictor"
and "corrector" solutions are superposed, the result is a system
which satisfies both the laws of dynamics and the material proper-
ties (including the effects of past history of deformation such as the
direction of plastic yielding) .

A finite difference approach based on the "predictor-
corrector" procedure was presented by Baron, Bleich, and Weid-
linger. 7 The method is easy to use. However, again, it deals with
elastic perfectly plastic structures only. All of the above mentioned
methods of analysis, dealing with systems of finite dimensions, are
limited to structures with specialized moment-curvature relations:
either the elastic perfectly plastic, or the rigid plastic type. A
method applicable to structures having a more general type of non-
linear moment-curvature relation was proposed by Wen and Toridis .8
Several discrete models, based on the lumping of the mass and/or
the flexibility of the structure have been considered. Extensive
numerical results were presented to demonstrate the credibility of
the approach.

The methods of analysis mentioned above, in general, have
been developed to handle dynamic problems involving stationary
loading in the sense that the position of the load does not change with
time. The problem of a beam subjected to a moving load has attracted

a great deal of attention through the years, mainly because of its

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practical importance in bridge engineering. In fact, the first funda-
mental paper by Stokes15 was published as early as 1849. A great
amount of work has since been done on this general problem. A
bibliography on this subject can be found in Reference 9. However,
all the papers listed in that bibliography have dealt with the elastic
behavior of beams subjected to moving loads.

Since the end of the Second World War, the concept of ultimate
strength, as applied to the design of structures, has gained increas-
ing attention. For an application of this concept to the analysis and
design of bridges, the study of the inelastic behavior of beams sub-
jected to moving loads seems to be of basic importance. It is of
interest to note that although the earliest work on the elastic behavior
of beams under moving loads was reported in 1849, the first paper on
the inelastic behavior under a similar loading did not appear until
1958.10 Then, Parkes analyzed the inelastic behavior of a beam
traversed by a single unsprung mass. The beam is assumed to be
massless, and to possess a rigid-plastic moment-curvature relation.
One of his major findings was that there is an upper limit to the load
that can cross the beam, irrespective of the magnitude of the speed.
The limit given is 1.228 times the static yield load.

Symonds and Neal, 11 in 1960, reported a similar study in
which the effect of the mass is taken into consideration, but the beam'

is again assumed to be rigid-plastic. The analysis showed that

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there is no upper limit to the magnitude of the load that can cross
the beam provided the load moves at a sufficiently high speed.

In passing, it may be mentioned that more recently,
Heidebrecht, Fleming and Lee12 have described a procedure that is
applicable to the analysis of elastic perfectly plastic beams subjected
to a moving force of constant magnitude. However, it seems that an
extension of their approach to problems involving moving masses
would give rise to insurmountable difficulties.

The work presented herein may be regarded as an extension
of that reported in Reference 8. However, whereas the work in
Reference 8 was confined to problems of prescribed dynamic loading,
the moving load problem treated here involves certain major diffi-
culties in the analysis because of the interactions between the
mechanical system of the load and that of the beam. In. relation to
the work of Parkes, and Symonds and Neal mentioned earlier, the
present analysis considers more realistic moment-curvature rela-
tions and load characteristics, such as a load unit consisting of a
combination of a sprung and an unsprung mass. As in Reference 8,
in this study the continuous beam is also replaced by a discrete
modelfi; and solutions are obtained by using numerical techniques.

In Chapter II, the physical system considered is first speci-
fied. Also, in the same chapter, the general method of analysis

including the derivation of the equations of motion is presented. In

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Chapter III, the method of analysis is specialized to the case of a
simply supported beam subjected to a single axle load unit; the
variables are made dimensionless, and the parameters of the prob-
lems defined. The detailed steps of the numerical solution are dis-
cussed therein also. In Chapter IV are presented typical numerical
solutions obtained by using a computer program developed for this
investigation; the results are examined and interpreted. In the same
chapter, a discussion of the results in the light of previous works is
given. The final chapter comprises a summary and conclusion of

the pre sent study.

1.2. Notation

The symbols used in this report are defined in the text where
they first appear. For convenience, they are listed here in alpha-
betical order, with English letters preceding Greek letters.

Dots appearing over certain variables represent differentia-
tions with respect to time. Similarly a bar above a letter is used to

refer to the dimensionless version of the variable in question.

c, = damping constant of spring connecting the j-th sprung
J mass to the j-th unsprung mass;

E = modulus of elasticity;

g = gravitational acceleration;

h = L/N;

h = i-th panel length;

r.
uIIn

. ..rh

.\

urn

i,j

ll

p(X.t) =

P.(t)
i

P.(t)
J

subscripts;
moment of inertia;
curvature of beam, or spring constant;

spring constant of spring connecting the j-th sprung mass
to the j-th unsprung mass;

elastic and inelastic slopes of a bilinear moment-
curvature relation, respectively; see Figure 5a;

elastic and inelastic slopes of a bilinear moment-rotation
relation, respectively; see Figure 5b;

flexibility distribution of beam;

span length of beam;

mass per unit length of a uniform beam;

lumped mass at joint (i);

mass distribution of beam;

bending moment;

elastic limit moment; see Figure 5a;

moment at joint (i);

ratio of modulus of elasticity of steel to that of concrete;
number of rigid panels into which a beam is divided;

positive and negative transition moments, respectively;
see Figure 5b;

distributed external load on beam;
lumped external load at joint (i);

interacting force between the j-th unsprung mass and the
beam;

II

the magnitude of the concentrated static force causing a
yield moment at mid-span when applied at mid-span;

total mass of beam;

the sprung mass of a moving load;

the unsprung mass of a moving load;

the j-th unsprung mass of a moving load;
total mass of moving load;

stiffness ratio = k2 /k1;
time coordinate;

fundamental elastic period of vibration;

constant horizontal velocity of moving load;

resultant shear force to the left and to the right of mi,
respectively;

absolute vertical displacement of 08;
total weight of moving load;

length coordinate of beam;

x/L;

length coordinate of joint (i);
vertical deflection of beam;
deflection of joint (i);

deflection of unsprung mass, or beam point in contact with
moving load;

distance of moving load from left end of rigid panel between
joints (i-l) and (i);

(h, - z.) 2‘ complement of z,;
1 1 1

zi/h

or.

W /P ;

V Y
Qv/Qb;
PyL3/48EI;

shortening in the spring connected to a sprung mass at its
static equilibrium position;

prefix indicating increment;
angle of rotation of joint (i);
Q /Q ;
u v
position of moving load from near support of beam; and

tv/L.

11. METHOD OF ANALYSIS

In this chapter, the characteristics of the physical problem
considered are described first, followed by a presentation of the

method of analysis .

2.1 . System Considered

 

The structure considered is shown in Figure 1. It consists of
a slender beam, resting on rigid supports. Both the mass and
flexibility of the beam are continuously distributed along the beam.

For beams considered herein, the relation between the bend-
ing moment, M, and curvature k is, in general, nonlinear and de-
pends on the history of deformation. Such a relation is a rather
complex one, particularly when loading and unloading are involved.
A general shape of a moment-curvature diagram is illustrated in
Figure 2. Starting from the origin, for some of the most common
structural materials, the correspondence between moment and
curvature is linear between the points (ke, Me) and (-ke, -Me),
where ke and Me refer to the elastic limit curvature and bending
moment, respectively. The relation beyond these points, however,
depends on the history of the deformation and usually exhibits

hysteretic behavior. For all the numerical results obtained in this

10

11

study, the moment-curvature relation has been taken to be of the
bilinear type discussed in detail in Chapter III.

The beam shown in Figure l is subjected to load units that
traverse at a constant horizontal speed, v. The types of moving
loads considered in this study are: a "single -axle load unit, " a
"sprung mass, " and an "unsprung mass. " A single -axle load unit
consists of a sprung mass connected to a single unsprung mass
through a linear spring and a viscous damper. Shown in Figure l
are also special cases of a single -axle load such as a single sprung
mass and a single unsprung mass. In addition to the moving loads,
of course, the beam may be subjected to other prescribed time

dependentloads.

2.2. Analysis of System

 

2. 2.1. Discretization of Beam. In order to effect an analysis

 

of the system described above, the continuous properties of the beam
are lumped or discretized. The manner of discretization corresponds
to "Model B" discussed in Reference 8, but for the sake of complete-
ness, is given below.

This model, as illustrated in Figure 3a, replaces a beam of
continuum by a series of rigid and massless panels connected by
joints. At these joints both the flexibility and the mass of the beam

are lumped. It is not necessary for the panels to be of equal length;

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12

the symbol h1+ is used to denote the length of the panel between the

1
joints (i) and (i+l), as shown in the figure.

The relation between the flexibility of the continuum and that of
the discrete model may be illustrated by considering joint (i) at xi,
in Figure 3a. At this joint the rotation represents all the curvature
changes in the actual beam over a length tributary to this joint; i.e.,
hi h1+1

from x! = x. - — to x‘.‘ = x, +
1 1 2 1 1 2

 

, as shown in the figure. Let Oi
denote the rotation at joint (1) . Assuming k(x) = k(xi) within the

tributary interval, one obtains:

XII
i

9, = k(x)dx = (l/2)(h.+h. )k (x,) (l)
1 1 1+1 1

x!
1

The above shows that the moment-rotation diagram for joint (i)
may be obtained directly from the moment-curvature diagram for the

cross-section at Xi' Thus, the abscissas of the moment-rotation

diagram are equal to the abscissas of the moment-curvature diagram

multiplied by (1/2) (hi+ hi+ ).

1

Assuming that the angle of rotation is small, the relation

between the vertical deflections y.1 1, yi, y and Bi may be derived

1+1

from Figure 3b as:

 

 

9‘ iv +(-1-+ 1)Y 1
1 hi 1-1 hi hi+l 1 hi+l

y1+1 (Z)

The deflections are taken as positive downward, while the rotations

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13

are considered positive if the joint in question is concave upward as

shown in the figure.
The lumped mass mi at joint (1) for the model is related to the

distributed mass m(x) of the continuum by the equation

m = m(x) dx (3)

2.2.2. Assurrlptions. The assumptions made in the derivation

 

of the differential equations governing the motion of the system in
consideration are outlined below.

(1) The deformation of the beam is assumed to be due to bend-
ing only; the effects of shear, strain rate and geometry
changes on the beam deformation are not considered.

(2) The unsprung mass of a moving load is assumed to be in
contact with the beam at all times.

(3) The horizontal component of the load velocity is assumed
to remain constant during the passage of the load over the
beam, even at relatively large deformations of the beam.
Also, the interaction force between the moving load and the

beam is supposed to act vertically.

2.2. 3. Derivation of Equations of Motion. Consider the general

 

case of a number of single -axle load units moving across a beam. The

14

position of the j-th unsprung mass in contact with the beam, at any
time, is denoted by the variable Qj (see Figure 3a), where I; is a
function of time, t. Assume that at a certain time the unsprung mass
in question is on the panel between the joints (i-l) and (i) . Consider-
ing the fact that each beam panel between two flexible joints (or a
support and a flexible joint) is rigid, the vertical deflection (y§)j,
of the j-th unsprung mass in contact with the beam can be expressed
as:
zi <

(y§)j=yi_1+1-1-i(yi-yi_l). x.1_1 3(éj) -X. (4)

in which zi denotes the distance of the unsprung mass from the left

end of the rigid panel between the joints (i-l) and (i) .

The velocity and acceleration of the unsprung mass are:

a, z,
' ° 1 1 . .
221 zi

where each dot represents a differentiation with respect to time.
Let Wj denote the absolute vertical displacement of the j-th
sprung mass, (Qs)j' from its original static equilibrium position.

The equation of motion of (QS)j in the vertical direction is:

d
on = -k - _ — -
(Qsljwj J.(Wj (y§)j) Cj dt (wj (y§)j) (6)

where kj and cj represent the spring constant and damping constant,

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15

re spectively, of the system connecting the j-th sprung mass to the
j-th unsprung mass.

To facilitate the use of the equations for other cases of load-
ing (e.g. a number of unsprung masses, time -dependent distributed
pressure, etc. ), the reaction between the j-th unsprung mass and
the beam will be denoted by Pj(t) , a time dependent force (see
Figure 3a).

Considering the free body of a joint (1), as shown in Figure 4a,
one obtains the equation:

_ .. + ,_ :0
miyi Vi Vi (7)

in which ‘yi represents the acceleration of the mass at joint (1), and
Vi and V; the resultant shear force to the left and to the right of
joint (1), respectively.
Additional equations may be obtained by a consideration of the
equilibrium of moment. Referring to the free body of the panel
( i —l) - (i) in Figure 4b, the requirement for equilibrium of moments
ar ound the left end of the panel yields:

M, -M.+V.h.+P.(t) z.=0 (8)
1-1 1 1 1 j 1

Where Mi and Mi are the bending moments at the joints (i-l) and

-l

( i ) , respectively. In deriving the preceding equation the rotational

ine rtia of the panel is neglected.

Similarly, considering the free body of Panel (1) - (1+1) shown

16

in Figure 4c, and summing moments around the right end of the
panel one obtains:

14. - 14. -+‘V:IL - P’ (t) z' = 0 (9)
1 1+ 1

1 1+1 j+l 1+1

in which z'1+1 is the complement of the distance z1+1 as shown in the
figure.

Assuming that hi = hi+l = h, or equal panel length throughout
the beam, and subtracting Equation (9) from Equation (8) the follow-

ing equation is obtained:

- _ I a :
Mi_1 2M1+Mi+1+ (V,1 Vi)h + ij zi + Pj+1(t) 2 1+1 0 (10)

Finally, substitution of Equation (7) into Equation (10) and

solution of the resulting equation for yi yields:

 

1-1 1+1] (11)

- 2Mi + Mi+l) + Pj(t) zi+ Pj+1(t)z'
in which,

Pju) = (Qv)jg - (Qu)j (ypquJIwj - (ypj) +cJ. ft— [wj - (3%] (12)
where (Qv)j denotes the total mass of the j-th single -axle load unit,
(Qu)j represents the j-th unsprung mass, and g the gravitational
acceleration. The expression for Pj+l(t) may be obtained from
Equation (12) by replacing j by j+l.

An equation of the form of Equation (11) can be written for

each joint, resulting in a set of equations which are coupled in the

yi's. The fact that the equations are coupled may be seen by an

..-w“"‘-
clsnuu‘

r wr‘ 3‘
,.;... I
~

‘ '9
the I'llI

ETCES

l7

examination of Equations (5b), (11), and (12), and the expression

for Pj+ (t).

1

If no load unit is located on the panel either to the left or to the

right of joint (1), Equation (11) reduces to:

.. _ 1
Y1 ' hmi (Mi-l

) (13)

 

- 2M. + M,
1 1+1

The differential equation for a beam having equal panels and
subjected to a time -dependent distributed pressure can be derived
easily by use of Equation (13) . Let p(x, t) denote the distributed
external load on the beam to be lumped in the form of concentrated
forces, P.1(t), acting at each joint (1) . Thus:

XII
i
P.1(t) =[ p(x,t)dx (14)
XI
1
It is obvious that Pi(t) enters into Equation (7) as an addi-

tional term and this in turn changes Equation (13) to

 

.. _ 1
yi — hmi [(M - 2Mi + Mi+1) + Pi(t)] (15)

1-1

Returning to the case of a beam traversed by a load, Equation
(11) may be used to derive certain differential equations correspond-
ing to particular conditions. Due to practical considerations, only
one unsprung mass at most is assumed to be located at any time
either on the rigid panel immediately to the left or to the right of any

joint (i). (The case of an unsprung mass exactly over a joint is

O.

r»

.—

:1)

Oc-
6-

u,

I.

18

treated in Art. 2.2. 4.) This assumption may be justified by the
fact that for the case of a multiple -ax1e load unit the above condition
can usually be met by choosing a panel length that is small enough.
Thus, if the j-th unsprung mass is on the panel immediately to the

left of joint (1), Equation (11) becomes:
1

m.
1

 

71: h [(M1-

1 - 2Mi + Mi+l) + Pj(t)zi] (16)

Similarly, if the j-th unsprung mass is on the panel immediately

to the right of joint (1), Equation (11) reduces to:

.. _ 1
yi-hm.
1

 

[(M1-

_ I
1 2Mi + Mi+l) + Pj(t)zi+l] (17)

where Pj(t) is given by Equation (12) . Of course, (Y§)j and its time
derivatives appearing in Equation (12) are to be found from Equations
(4), (5a) and (5b), in each case with due consideration given to the
location of the unsprung mass. For example, if the j-th unsprung

mass is to the right of joint (1), Equations (4), (5a) and (5b) take

the form of:

 

Z
1+1
= ' 18
”83' yi+ h. ”1+1 Y1) ( )
1+1
9 = . . _ h o - o 1
(y§)j Yi+(zit+1/h:1+1)(yi+1 Vi)+ (zi+1/ i+1Hyi+1 Vi) ( 9)
.. :.. . , _, .. _ 20
(Y§)j yi+(.7.zi+1/hi+1)(yi+1 yi)+(zi+1/hi+1)(yi+1 Vi) ( )
where hi+l can be replaced by h.

The differential equation for each joint is obtained from one

f I

I“.

Ira:
v.
.

..n

u-

(“t
(j

vow!

.4

v‘f
0..

(t:
n )

0*-

C)

'1;
I ~’

*.

19

of the three Equations (l3), (16) or (17) depending on the location
of each unsprung mass and the joint in consideration.

It should be noted that, using Equation (16) and (17) for the
joints immediately to the left and to the right of the point where an
unsprung mass is located causes the resulting equations to be coupled
in the accelerations of the two joints in question. However, this
coupling is of local nature, and if a single-axle load unit alone is
considered to cross the beam, only two of the equations are coupled,
while the remaining ones are uncoupled and can be solved
independently.

Equations (16) and (17) apply also the case of a beam traversed
by a single sprung mass. For this purpose (Qu)j in Equation (12)
should be set equal to zero. The solution for this case is substan-
tially easier to obtain since the accelerations are not coupled.

In passing, it may be noted that the above analysis may be
easily adapted to the case of a multiple -axle load unit. A major
change involved would be the consideration of the rotatory properties
of the sprung mass.

2 . 2 . 4. Discontinuities in the Velocity and Acceleration

 

Functions . If a mass moves on a beam with a continuous distribution
of flexibility, so long as it is in contact with the surface of the beam,
its vertical velocity and acceleration are continuous functions of

time. Such is not the case for the discrete beam model used in the

20

present analysis . On account of the discontinuity in the slope at,

say the i-th joint, jump discontinuities in the vertical velocities and
accelerations of the lumped mass at joint (i) and the moving mass
occur when the latter passes over the joint. The magnitudes of these

jumps may be computed by using previously derived equations for

Y1;

joint (1) . Denote by ti the time corresponding to the j-th unsprung

and y; when an unsprung mass is to the left or to the right of

. . . - + .
mass being exactly over jomt (1) , and by ti and ti the times an
infinitesimal amount before and after ti' respectively. For ti-'

application of Equations (5a) and (5b) with zi = h. = h gives:
1

 

Z.
. _ ,- _1_
[(Yt)j]t=tl' - y.l + h (Vi-Yi_1) (21a)
_ Zii _
[(v§)j]t=ti':vi + h (vi ”3-1) (21b)

+
Similarly, for ti use of Equations (5a) and (5b) with

 

 

z1+1 = 0 and hi+l = h yields:
[(‘ )] -'++zi+1( 4“) (22)
’2 jt=ti+ ‘ Yi h Y1+1 Vi a
+ Zéi+l + +
[(Y§)j]t=ti+ 2" Vi + h ”1+1 ‘ Y1) (22b)

Note that when t = ti (or g = xi), 57.1: z = v, where v is the

1+1
Velocity of the moving load. Therefore, the magnitudes of the jumps

in the (graj and ('yg)j functions in the neighborhood of joint (i) are

IV

SN

V0

21

found to be:

(Av ). = [(yg)j]t:t: - [(yé)j]

1’; J tzti

.+ .- v
(Yi - Yi) + h ”1+1" 233+ 3'14)

. V
Ayi + if (Vi+'1‘ 2Y1+ yi-l) (233”

A- _ .. _ ..
( 31)]. [(Y§)j]:1 [(Y§)j]t:t{

..+ ..- 2v . .- . .
“'1 ”’1 ) + h ”1+1" 233+ yi-l "Al?

.. 2v . .- . .
Ayi + h ”1+1 ‘ Zyi + yi-l ' Ayi) (23b)

in which (Ayaj and (A3: ). represent the jumps in the (yg’)j and

C. J
(’yg)j functions, respectively; A91 and Ayi represent the jumps in
the yi and 'yi functions, respectively.

The conservation of the linear momentum in the vertical direc-

tion furnishes an additional expression for the determination of (Aflr ).

 

 

L J
and A91:
(Qu)j (Ayglj + miAy.1 = 0 (24)
Solving Equations (23a) and (24) simultaneously one obtains:
(Air) =3- l—l—‘—](Y -2Y +Y ) (25a)
g3 h (Q). 1+1 1 14
1+ __u_.l
m.
1
(A')- XI 1 ]( 2 + ) (25b)
Vi ‘ “h mi y1+1 ’ Vi V1-1
1+
(£2qu

22

- +
Similarly, use of Equations (16) and (17) at ti and ti’ re-

. . : , z . .
spectively, w1th zi zi+1 h, yields.

(Qu)j(A'yr )J. + mils-vi + CJ(A)’L)j = 0 (26)

15.

After a substitution of Equation (25) into Equation (26) the resulting
equation may be solved simultaneously with Equation (23b) to de-

termine (Ay ), and A",. Considerin the simple case of c, = 0, one
I. J v1 g J

 

 

obtains:
(A") =2—V [——]L 1(' -2° +' -A') (27a)
yéj h (D). YI+1 Va. 3’14 VI
1+ u
m,
1
(A")=-Z—"[ 1 ](° -2'+' -A‘) (27b)
Y1 h mi y1+1 Y1 yi—l Vi
l
+ (Q )

When using Equations (6) and (16) or (17) the above jumps
should be applied to the values of 9i, 'yi, 9;, and 'yg, each time an
unsprung mass moves over a flexible joint.

2.2. 5. Boundary Conditions . The boundary conditions for the

 

model used in this analysis can be handled in a very simple manner.
Thus, at a simple support both the deflection, y, and the bending
moment, M, are equated to zero. At an interior support of a con-
tiIluous beam the deflection is zero; however, in general, at such a

Sllpport neither the moment nor the rotation is zero.

3...

.Pl.

III. SIMPLY SUPPORTED BEAMS SUBJECTED TO
A SINGLE MOVING LOAD

3.1. General

The equations of motion derived in the previous chapter are
applied here to the special case of a simply supported beam subjected
to a single moving load unit. Thus, Equations (6) , (12), and one of
the Equations (13) , (16), or (17) can be used to determine the
dynamic response of the system. To consider simpler types of
moving loads, the above equations need only be simplified accordingly.
For example, when the moving load consists of a single sprung mass,
Qu appearing in Equation (12) is simply set equal to zero. On the
other hand, if the moving load is a single unsprung mass, the spring
constant k and damping constant c appearing in Equation (12) are

equated to zero. Of course, in this case Equation (6) does not apply.

3. 2. Dimensionless Form of Equations

 

In what follows, the differential equations for a simply sup-
Ported beam of uniform cross-section are rendered dimensionless.
A list of the physical quantities used for this purpose and not defined
So far, is given below:

L = span length of beam;

23

24

P = concentrated load which if applied statically at mid-
y span causes a yield moment at mid-span;

m = mass per unit length of a uniform beam;

E1 = elastic stiffness of cross-section of a uniform beam;

Wv’Qv: weight and mass of moving load, respectively; and

Qb = total mass of beam (= mL) .

In addition, to render the equations dimensionless, the following

notation is introduced:

6 :

P L3/48EI;
Y

P L/4
Y

2
fundamental period of elastic vibration = Efi—fi;

 

 

tv/L

d' d2“ 2 2
iii _L311. “1-3.233
6 d7 v6 dt d7 26 dtz
v_v 915-: <1”. <1va 1.2 dzw
6 dr v6 dt d'rz v26 dtz
x/L
M. M.
...1. = 1
M P L/4
e

Us

25

 

W Q g
_ V _ V
l3 - $— - —p_ 9 Y : QV/Qb
Y Y
Qu
h = b—
V
k(T )2
12 = 1
Qb
C(Tl)
5 = (Z8)
Qb

Substituting Equation (12) into Equations (16) and (17), and
using the above defined quantities, Equations (6), (l3), (l6), and

(17) are reduced to the dimensionless form shown below:

2 - - d;
212:. 1 1_- — _ - _ 9!,_ _JL
dTZ - - a(1-)\)Y [a k (w (Y;)10r l+l)+c(d1- ( d1. )1 or i+l)] (29)

(The subscript "i" or "1+1" should be used according as the load is

to the left or to the right of joint (1).)

 

 

 

 

2-
d y. 2
1 48N - - -
2 ‘ 2 2 (Mi-l ‘ 2M1+ Mm) (30)
(IT Tl’ a
2-
d y.
1 48N - - _ _
2 ' 2 2 '[N(Mi—1' 2Mi+ Mi+1)+4pzi]+
d‘r 1r a
d2; d?
g, 1 — - - 1 _. d“? g -
- — - — —- 31
N[ \IM 2 )1+ 2 km (ydih‘addt (C1T (1)21 ( )

(it a

26

 

 

 

 

 

 

 

 

 

 

d2"
Y
). 48N - - - _l
2 ‘ 2 2 [N(Mi-l ’ 21\’11H"11+1)Jr 4‘32 +1] +
(IT TT (1
2- ..
d y — dy
4’ .1— - - ' .1. - 93- _§' '1
NH’M 2 (1+1Jr 2 k("V‘Wg’iH’ + a “<17 (<17 )i+l)zi+1 (32)
d'r a
where
(yg)i = V14 + zi (Yi -yi_1) (33a)
db; ). dY dy. (137.
CI _ 1-1 - - __l 1-1
2 - 2- - Z- 2-
d (Y ). d y. dy, dy d y. d y
I; 1 1-1 1 1 l - 1 1-1
2 = 2 +ZN(d - d )+z( 2 - 2) (33C)
d-r d7 T T 1 dT dr
and (3.74.)1+1 and its derivatives are obtained from Equations (33a) -(33c)

by replacing (i) by (1+1) . Again, in the case of a single sprung mass,
x appearing in Equations (29), (31), and (32) is set equal to zero.
Similarly, for a single unsprung mass, k and E in Equations (29), (31),
and (32) are equated to zero; Equation (29) is not applicable in this

case.

3.3. Bilinear Moment-Curvature Relation

 

Although the mathematical model used in this analysis is
adaptable for use with a large class of nonlinear, history-dependent
moment-curvature relations, for simplicity, a bilinear bending

moment-curvature relation has been used for the numerical solutions

27

to be presented. This relation is illustrated in Figure 5a, in Which
k1 and k2 represent the elastic and inelastic stiffnesses, respectively.
The appropriateness of such a bilinear relation obviously depends on
the actual shape of the bending moment-curvature diagram of the
beam cross section for which this relation is used. In Reference (8)
it was shown that for two particular problems the numerical solu-
tions obtained by using both an actual curvilinear moment-curvature
relation and a bilinear moment-curvature relation are in good A
agreement.

As was mentioned in the preceding chapter, the moment-
rotation diagram of joint (i) at xi may be obtained directly from the
corresponding moment-curvature diagram which is known. In the
numerical solution of a problem, the moment-rotation diagram is
used to find the change in one of the variables due to a change in the
other. Specifically, given a change, AGi, in the rotation Bi of joint
(1) , the corresponding change AMi in the bending moment, Mi’ is
found as follows.

Referring to Figure 5b, to facilitate the determination of the
incremental moment, AMi' it is convenient to introduce the term
"transition moment." Thus, associated with each joint (1) of the
beam, in each phase of external loading there is a positive transition

+ . . . -
moment, Ni' and a negative tran31tion moment, Ni . They play the

following role in determining the moment-rotation relation of joint (1):

28

For A01? 0:
AM, = k' A9,, if NI 2k' A9. (34a)
1 l 1 1 l 1
+ I"; + +
AM, = N, + ‘— (k' A9, -N-. ). if N, Sk'Ae, (34b)
1 1 k' l 1 1 1 l 1
1
For A913 0:
AM, = k'A9,, if N." <_ k'AG. (35a)
1 l 1 1 1 1
kl
AM, = N,-+ -—2 (k'A9,-N.-), if N.- 3 k' A0, (35b)
1 1 k'1 l 1 1 1 l 1
where
k'zl—(h+h )k' k':l(h+h )k (36)
l 2 i i+l l’ 2 2 i 1+1 2

It may be noted that the transition moments are essentially the
elastic limit moments at a particular stage of deformation history.
For example, for a section at its virgin state (no prior history of
bending), represented by the origin in Figure 5b, the positive and
negative transition moments are equal to Me and -Me, respectively.
However, during the various stages of loading the magnitudes of the
positive and negative transition moments are not, in general, equal
to each other. Nevertheless, the transition moments for the j-th phase
of loading can always be determined from the transition moments and
the change in bending moment of the preceding (j-l) -th phase. It is

seen from Figure 5b that:

29

(N ),=(NTL—AM,), , 1103(NI-AM) SZM
j 1 1 j-l 1 1 j-l e

(NT), = o, if (NI - AM). 5 0 (37a)

1 j 1 1j-1
(NI).=2M, if(Nl-AM.). 2 2M

1 j e 1 1 j-l e

and,

(N ).=(N,'—AM.). , if0>.'(N,--AM.). 3-2M

1 j 1 1 j-l 1 1j-l e

_ ___ . -_ 7
(Ni)j 0, 1f (NiL AMi)j_1 2 o (3 b)
(N.").=-2M, if (N.- -AM.). 5-2M

1 j e 1 1 j-l e

Obviously, at any phase of the loading, the sum of the absolute
values of N: and Ni is equal to 2Me' Therefore, if either one is
known, the other can be determined easily as the complement to 2Me'
These relations are, of course, defined for j >1. For a given problem
(Nil)1 and (Ni-)Ishould be specified.

In conclusion, the above procedure allows one to determine for
each joint (i) the change in bending moment, when the corresponding
change in the rotation of the joint is known. The history of loading and
deformation is taken into account by using the transition moments

described above .

3.4. Properties of Actual Bridges and Parameters of Problem

 

To determine a relatively realistic range of parameters, a
study of seven simply-supported bridges considered in Reference 13

was made. A typical cross-section and its properties are shown in

30

Figure 6. In this particular case, the ratio, 11, of the modulus of
elasticity of steel to that of concrete was taken as 10.

The speed parameter, a (= T v/L), incorporates in it the effect

1
of the stiffness of the beam and its span length. However, for a
given beam, the only variable in the expression of a is the horizontal
component of the velocity, v, of the moving load. In the case of the
typical bridges under consideration, a speed range of 10 to 80 mph
corresponds approximately to a range of values of a from 0. 042 to
0. 333. However, if the properties of standard bridges given in
Table l of Reference 9 are used, the above range of a corresponds
to approximately 7 to 55 mph.

The load parameter, (3 (= Pl)’ is a measure of the weight of
tihe moving load with respect to th: static mid-span yield load, Py.
A value of f) = l. 00 corresponds to a moving load the weight of which
is equal to that of the static mid-span yield load for the beam.

The weight parameter, y, represents the ratio of the weight of
t1'Tle moving load to that of the beam. In the case of a load consisting
of both an unsprung and a sprung mass, the parameter X denotes the
I'a.t:i.o of the unsprung mass to the total mass of the weight of the load,
and k and d are the dimensionless versions of the spring and damp-
ing constants, respectively.

The stiffness ratio, R, is the ratio of the inelastic stiffness to

the elastic stiffness of the beam. (See Figure 5a.) If R = 0, the

31

beam is elastic perfectly plastic; R = l. 0 corresponds to the per-

fe ctly elastic case .

3 . 5. Procedure of Numerical Solution

For simplicity, in what follows, the method of numerical
solution for a beam traversed by a single unsprung mass is outlined.

Let the time when the moving unsprung mass is exactly over
the entry support be denoted by to. For all numerical solutions
obtained, at t = to both the moving load and the beam have been as-
sumed to be in static equilibrium. With given initial conditions, it
is possible to compute the dynamic response of the beam when the
unsprung mass has moved a short distance on the beam, provided the
time elapsed is small enough. Proceeding in a similar manner one
Can determine the response of the beam at any time t, where t) to.

For the numerical solution, the problem may be stated as

follows: Given at time t = t the values of the variables yi(t1),

l
Yi(t1), yi(t1) , 01(t1) and Mi(t1) it is required to determme the values

of these variables at t = t2 = t1 + At, where At is a small increment

of time.
It may be said that, the numerical solution hinges on the
numerical integration of the differential equations. In this study,

th-e numerical integration has been carried out by using the following

expres sions:

32

Wt?) = y1(ti)+(At)§'(t1) +1;— (At)2')"(t1) (38a)
fitz) = W1) + %- At [vi(t1)+vi(tz)] (38b)

The procedure of solution is outlined below:

1) Obtain the value of yi(t2) for each joint (1) by using Equation

(38a), and form:
Avi = Yi(t2) -y.1(t1)

2) From Equation (2) obtain A0i by substituting therein Ayi for
y, and A0, for 0,.

1 1 1
3) Compute AM.1 using the value of A01 according to the bending
moment-rotation relation of the individual joints (see Art. 3. 3) .
Having AMi' form the quantity:

M,(t ) =M,(t ) +AM,
1 2 11 1

4) Write out the expressions for y using Equations (4) -

I." 51’ y'r.
(5b) . Note that these expressions cannot be written out explicitly,
since some of the variables involved are still unknown.

5) With the quantities found in steps 1, 3 and 4, obtain a set of
equations in the 3) (t2), by using one of the Equations (l3), (16), or
(17) depending on the joint in consideration and the location of the
moving load. Solve the above set of equations simultaneously for the

unknown 3) (t2)'s.

6) Obtain 91(12) from Equation (38b) .

f’

‘-

f.-

I
9.4

.-.

(I)

U]

be

33

7) Repeat the previous steps to extend the solution of the problem
from tZ to t2+At, etc.

For the numerical integration method used in the present study,
the time increment, At, should be small enough to satisfy the cri-
teria of stability and convergence. In Reference 15, the upper bound
of the time increment was given as At = 0. 389 'I; where T is the
shortest natural elastic period of vibration of the system. A smaller
increment will in general give a more accurate answer, if round off
errors do not tend to become critical. An increment of At = 0.200 T
was used in obtaining the numerical results in this study. Using a

smaller value of At for a few cases did not change the results

significantly.

3.6. Use of Computer

 

The numerical results presented in this study have been ob-
tained by means of a computer program developed for use on the
"CDC 3600" digital computer system of Michigan State University.
The program, which is based on the equations and procedures dis-
cussed in this chapter, computes the response of a simply supported
beam subjected to a single moving load. The moving load may con-
sist of either a single unsprung mass, a single sprung mass, or a
Single -axle load unit.

Upon being supplied the parameters of the problem to be solved,

34

the program prints out the following dimensionless quantities at
regular intervals of 1- (each interval is a multiple of AT, the interval
of numerical integration):

1) the value of T, which also specifies the position of the load;

2) both the dynamic and static deflections of all the joints of the
discrete beam system;

3) the interaction force; and

4) the permanent deformation angles at every joint of the beam.

In addition, at the end of the solution of each problem, the
maximum values of the displacements incurred during the passage
of the load, and the ordinates of the final shape of the deformed beam
are printed out.

The computer time necessary for the solution of a typical
problem, including the free vibration response, is about 7 1/2 min-
utes for a = 0.10, and about 41/2 minutes for a = 0.30.

The format of the parameters to be supplied, and some other

details, as well as a copy of the Fortran Program have been com-

piled in the Appendix.

IV. NUMERICAL RESULTS

4. 1. General

The numerical results presented in this chapter comprise the
elasto -inelastic dynamic response of simply-supported beams sub-
je cted to heavy moving loads. The types of moving loads considered
are limited to a single unsprung mass, a single sprung mass, or a
single axle load unit consisting of a sprung mass and an unsprung
mass . For nearly all the numerical results obtained, the beam is
divided into 20 equal panels .

First, typical time history curves of deflections, acceleration,
and interaction forces are presented to depict certain typical pat-
terns of the behavior of the beam-load system under critical com-
binations of the parameters. The objectives are (i) to gain an in-
sight into the physical problem, and (ii) to illustrate certain features
in the workings of the mathematical model.

Following the discussion of the history curves, typical shapes
0f the final deformed beam are shown to illustrate the permanent
daJinage that the beam undergoes due to the pas sage of the load.
After that, the influences of such parameters as the speed and weight
of 1ihe load, and the inelastic stiffness of the beam on the maximum

beam deflection are briefly considered. Finally, the general trends

35

36

of the results are compared with certain numerical data available in

the existing literature .

4. 2. History Curves for Unsprung Mass

The term "history curve" designates a plot of the dimensionless
response of the beam as a function of the dimensionless time, T,
which is also a measure of the location of the moving load. As all
the variables mentioned in the following discussions are dimension-
less, this adjective will be omitted for the sake of simplicity.

In order to form a clear picture of the connection between the
beam response and the various parameters, the history curves for
an unsprung mass have been grouped together, depending on the com-
bination of the speed parameter a , and the load parameter (3. Terms
Such as "low" and "medium" used in describing the relative values of
these parameters within their ranges are, of course, qualitative.

4.2.1. "Low" Speed and "Medium" Load. Figure 7 shows a
time history curve of the mid-span (:2 = 0.5) deflection, (7, for:
R = 0, a = 0.10, (3 = 0.90, and y = 2.68. The values of the parameters
cOrrespond to the case of an elasto -perfectly plastic beam subjected
to a load slightly less than the yield load, moving at a "low" speed
( about 25 mph). It is seen that for this set of parameters no inelastic
action has taken place. In fact, the dynamic deflection curve is very

close to the static deflection curve, which is also shown in the figure.

37

This curve corresponds to the (static) influence line for mid-span
deflection due to the load; it has been computed on the basis that the
beam remains perfectly elastic. The free vibration after the load has
left the span is also seen to have small amplitudes.
In Figure 8 is plotted the history curve of the interaction force
15, for the same problem as considered in Figure 7. The curve is not
sniooth; it exhibits many oscillations around the static value. This is
5 a characteristic feature of history curves of P for an unsprung mass.
The reason is that P depends directly on the acceleration of the un-
sprung mass (see Equation (12)), which, in turn, depends on the
accelerations, as well as the velocities (see Equation 5b), of the
flexible joints adjacent to the instantaneous position of the load.
Although in this case, so far as magnitudes are concerned, the
dominating term in the expression for P is the constant static or
gravitational component, among the time dependent terms those
related to accelerations, in general, predominate over those related
to velocities. These accelerations generally oscillate a great deal,
as illustrated in Figure 9, and bring about the osCillations in P.

4. 2.2. "High" Speed and "Medium" Load. A representative
history curve of deflection for this case is shown in Figure 10. The
Values of the parameters are the same as in the preceding case ex-
cept that a has been increased to 0. 30 (corresponding to about 70

I7":1ph) . Furthermore, the point under consideration is )2 = 0. 55, at

38

which the maximum displacement is larger than that at Si = 0. 50. In
contrast to the history curve in Figure 7, one may observe that in
this case inelastic action has taken place. The equilibrium position
of free vibration is below the initial horizontal configuration of the
beam, indicating a "permanent set" or permanent deformation. The
permanent set is 0. 788, and the maximum dynamic deflection is 1.70.

It is also noted from the figure that, although the displacement
being considered is for .2 = 0.55, its maximum value occurs when
T = 0.817 , i.e., when the load is at x = 0.817. In subsequent dis-
cussions, the symbol Tmax is used to designate the time at which the
maximum dynamic response occurs. Similarly, ymax denotes the
maximum dynamic deflection (at 'r = Tmax) .

Figure 11 shows the history curve of P for the same problem.

In general, this curve also exhibits an oscillatory pattern. However,
it is to be noted that initially the dynamic P curve remains consist-
ently below the static one, for a relatively large interval of T.
I“filter on, a trend for a gradual increase develops, resulting in
VEllues of P appreciably larger than the static value. However, as
the load is approaching the end of the span, after 1' = 0.85, P drops
Su(:ldenly, and even becomes negative. It is to be noted that for
physical realization of a negative interaction force, a tensile force

Would be required to maintain the contact between the moving load

and the beam (that is, to keep the load from leaving the surface of

39

the beam). Thus, the present solution involves a hypothetical tensile
force which does not exist in such practical cases as a bridge-
vehicle system. However, the effect of such a negative force has
been found to be insignificant, since it occurs near the departure end
of the beam; and its numerical value is not exceedingly large.

4. 2. 3. ”Medium" Speed and "Heavy" Load. The history curve
in Figure 12 corresponds to the displacement at point )2 = 0. 8 with the
following values of parameters: R = 0, a = 0.20, (3 = 1.20, and y =
3 . 568. The distinguishing feature in this case is, of course, the
heavier load, which is 20% greater than the yield load. Consequently,
as expected, the beam exhibits a behavior quite different than the
previous cases. It is worth noting that, while the moving load is on
the beam there is no rebound or "snap back" of the beam, but rather
a tendency for increasing velocity of the downward movement, and
ul‘lczontained deformations .

The corresponding P curve is shown in Figure 13. For the
most part this P curve closely resembles the preceding one (Figure
1 1). However, shortly after the load crosses the third quarter point
of the beam, P increases very rapidly without a reversal that took
Place in the preceding case. Near the support, P attains an exceed-
ingly large value of 49. 38. This behavior is explained as follows.

For an unsprung mass, the value of P varies with its vertical

acceleration which, as given by Equation 5b, depends on the difference

40

of 9i and yi+1, and the difference of yi and 'y Generally, these

i+1°
differences are relatively small, due to the fact that the correspond-
ing quantities have the same sign. When P is on the last (N-th)
panel, 9N“ and yN+1 are zero, since the (N+l) -th joint corresponds
to the end support. Consequently, the value of P depends only on

9N and 9N.

Under certain combinations of the parameters, such as for the
problem being considered, the beam yields a great deal, and the
yielding (plastic hinges) moves with the load. When the load is at
the last panel, the contribution of 'yN to P is relatively small, but the
yielding of joint N results in large values of 9N which apparently is
responsible for the large value of P.

It may be thus conjectured that the large value of P in this case
could be a consequence of the discrete nature of the beam model.
Hence, the numerical results after the load moves on the last panel
should be viewed with some caution. However, it is worth noting
again that, in general, the major damage to the beam is done before
the load passes over that panel.

When the parameters are such that the deformations are con-
tained, i.e. , they do not become exceedingly large, the results for
P for an unsprung mass do not exhibit the kind of behavior described

above. One of the parameters that may change the response from an

uncontained deformation to a contained one is the stiffness ratio R.

41

This is considered in the next section.

4.2. 4. History Curves for R > 0. Retaining the values of all

 

the other parameters in Figure 12 but changing the value of R from 0
to 0.1 yields the history curve of y shown in Figure 14. In this case,
the beam rebounds before the load has left the span, and ymax is much
less than the one in Figure 12. Thus, at .2 = 0.8, when R = 0, yma =

X

12.58 at Tmax = 1.000, and when R = 0.1, {rmax = 2.43 at Tmax =
0.833. This shows that even a small positive value of R causes the
deformations to be contained. In passing, it may be pointed out that
for the parameters in Figure 14, the absolute maximum ymax occurs
at the point i = 0.6 (Figure 14 is for :2 = 0.8), and has the value of
3. 32.

The corresponding history curve of P is shown in Figure 15.
This curve is similar to the one shown in Figure 11. As the moving
load approaches the departure end, P attains a relatively large posi-
tive value with a subsequent drop to a negative value; and P does not
build up to any exceedingly large value that appeared in the case of
R = 0 as shown in Figure 13.

The effect of having an inelastic stiffness is very pronounced

for the history curve of (7, shown in Figure 16. The parameters are

the same as before, except that R has been increased to 0. 5. In this

case, the absolute maximum y occurs at x = 0. 5 when -r
max max

0. 641, and has the value of 1.63. The permanent set at this point of

42

the beam is 0. 309. The value of ymax at the point 32 = 0.8 is 1.02 as

compared to the corresponding ymax = 12. 58 for R = 0.

4.3. History Curves for Sprung Mass

 

Figure 17 shows the history curve of point 32 = 0. 5 when the
moving load is a sprung mass. The parameters have the following
values: R = 0, a = 0.20, 8 =1.00, y = 2.973, andk = 33.57 (this
corresponds approximately to the stiffness of the springs of a heavy
duty truck) . Even though the static value of the load is just equal to
Py' there is a considerable amount of inelastic deformation resulting
from its passage. The maximum response ymax is 2.33 and 1-
= 0.813; the permanent set is 1.38.

The history curve of P for the above set of parameters is shown
in Figure 18. Since in this case the interaction force depends on the
vertical deflections rather than the accelerations, the graph for P is
much smoother in comparison to similar curves for an unsprung mass.
However, in this case also, as the load approaches the departure end,
P attains a relatively large value and drops quickly, but it remains
positive as the load leaves the span. This is also true for other data
involving sprung loads.

In Figure 19 is plotted the deflection history curve for a case in
which considerably large deformations occur. The point considered

is atxz 0.7, andR= 0, a = 0.20, 8:1.20, y: 3.568 andk= 33.57.

43

The reason for choosing the point >2 = 0. 7 is that the value of ymax is
largest at that point. It is seen that ymax = 10.18, Tmax = 0. 972, and
the permanent set is 9.14, while at :2 = 0.8 the results are ymax=8.87,
Tmax = 0. 964, and the permanent set is 8. 07. It is interesting to
note that it is only when the load is about to leave the beam that the
structure begins to rebound.

Figure 20 shows the history curve of P for the same problem
dealt with in Figure 19. Again, the curve is seen to be relatively
smooth. As the load approaches the far support, P grows con-
siderably, and attains a maximum value of 6. 15. This value is still

considerably less than the maximum of 49. 38 reached by 15 in the

unsprung load case shown previously in Figure 13.

4. 4. History Curve for a Single -axle Load Unit

A typical history curve of y for a beam traversed by a single-
axle load unit is shown in Figure 21. The response is for the point
x = 0. 6. The unsprung and sprung part of the moving load are,
respectively, 13% and 87% of the total load. All the other parameters
are the same as for the proceding problem. It may be observed that
the inelastic action which has taken place during the passage of the
load over the beam is of considerable magnitude. In this case,

y = 3.68, -r = 0.825, and the permanent set is equal to 2.472.
max max

44

4. 5 . Final Shape of Deformed Beam

 

The final shape of the deformed beam is of obvious interest as
it represents the apparent damage incurred during the passage of the
load. In the present study, this shape is obtained by plotting the final
equilibrium position of each flexible joint. This equilibrium position
is estimated by taking the average of the maximum and minimum
deflections of each joint during free vibrations. This is done some-
time after the moving load has left the span, at which time the beam
can be assumed to be vibrating elastically.

In the following, shapes of deformed beams are presented for
typical sets of parameters and different types of moving loads .

4. 5.1. Unsprung Loads. In Figure 22 are shown, for an elastic

 

perfectly plastic beam (R = 0), the shapes of three deformed beams
traversed by a load 8 = 1.10 for three values of a: 0.15, 0,25, and
0. 30. The permanent damage is seen to increase with an increase in
speed. It is of interest to note that, as the load speed increases, the
region of the greatest permanent distortion moves closer to the
departure end.

Figure 23a shows the deformed shape of the beam for 8 = l. 30.
Although the load in this problem is only 18% heavier than the pre-
ceding one, the maximum permanent set is 4. 9 times larger than
that in the preceding case.

The deformed shapes of a beam with an inelastic stiffness

pJ

...

(I!

(.0

45

R = 0. 5 are shown in Figure 23b for 8 = 1.20 and two values of a:
0.10 and 0.20. In this case, in contrast to the elastic perfectly
plastic beams, an increase in speed is seen to reduce the permanent
damage.

4. 5.2. Sprung Load. In Figure 24 are shown typical deformed

 

shapes of an elastic perfectly plastic beam due to a sprung load for
the different load speeds as indicated in the figure.

The shapes of the three deformed beams in Figure 24a corre-
spond to a very heavy load 8 = l. 50. It is of interest to note that, in
contrast to the case of an unsprung load, an increase in speed in this
case causes a decrease in the permanent deformation. For this set
of parameters, the maximum set occurs, for all three load speeds,
at 32 = 0. 7.

A similar pattern may be observed in the shapes of the

deformed beams shown in Figure 24b, which is for a lighter load,

8:1.20.

4. 5. 3. Single~axle Load Unit. Some typical shapes of the

 

deformed beam due to a load consisting of a sprung and unsprung
mass are shown in Figure 25, for the parameters indicated in the
figure.

The two deformed beam shapes in Figure 25a correspond to
R = 0 and R = 0.1. It may be observed that a value of R = 0.1

not only cuts down the amount of deformation, but it also causes

46

the position of the maximum set to shift towards the entry end of the

beam.

4.6. Effect of Speed Parameter

 

For a beam of given geometrical and physical properties, the
effect of the speed parameter on the dynamic response is considerable.
Its effects on permanent sets have been mentioned in the preceding
section. In Figures 26 and 27 are plotted the maximum dynamic
deflections of some particular point on the beam as a function of the
speed parameter a . The values of the other parameters in each case
are also noted in the figures.

Figure 26 shows the effect of a on ymax for the case of un-
sprung loads. Figure 26a is for R = 0, and [3 = 0. 90, and Figure 26b
for R = 0.1, and [3 = l. 20. Within the range of values of a for which
results have been obtained, both plots indicate that the maximum
dynamic deflections increase as (1 increases.

Similar plots for the case of sprung loads are shown in Figures
273. and 27b, for {3 = 1.20 and [3 = 1. 50, respectively. An elastic
perfectly plastic beam is considered for both figures. Contrary to
the results for unsprung loads, in this case the maximum dynamic
deflection decreases as (1 increases. A similar trend was, of course,

nOted previously in connection with permanent sets.

47

4. 7. Effect of Stiffness Ratio

The importance of the stiffness ratio, R, has already been men-
tioned (Figure 25). In Figure 28 is plotted the maximum dynamic de-
flection at :2 = 0. 5 as a function of R for the case of an unsprung load.
It may be observed that a small value of R, say R = 0.1, reduces
drastically the value of the deflection. However, the rate of the
reduction of the value of the maximum deflection decreases rapidly
with an increase in R. For R > 0. 5, further increases in its value

does not result in any significant reduction in the response.

4.8. Effect of Weight of Load

For a beam of given geometrical and physical properties and a
fixed value of the speed parameter a , the effect of the weight of the
moving load on the response of the beam may be studied by varying B,
and simultaneously varying proportionately the parameter y. Numer-
ical results obtained in this manner are plotted in Figure 29. This
figure corresponds to a graph of the quantity ymax/fi as a function of £3.

The vertical scale in Figure 29 represents the ratio of the
maximum dynamic deflection to the maximum static deflection which
is found under the assumption of perfect elasticity. As expected, this
quantity increases as {3 increases. However, it is to be noted that

the relation is not linear--the rate of increase of y H3 increases
m x

3.

with an increase in [3.

V. DISCUSSION, SUMMARY, AND CONCLUSION

5. 1. Discussion

 

5.1.1. UnspruniLoad. As mentioned in the Introduction, there

 

have been two analytical studies reported in the literature dealing

with the inelastic behavior of beams subjected to moving loads that
possess mass. The analysis of Symonds and Nealll has included the
influence of the mass of the beam, which has been neglected in Parkes'
work. 10 However, in both of these studies a rigid-plastic moment-
curvature relation has been assumed. It is of interest to compare

the results presented in the preceding chapter with those given in
these references. Of course, such a comparison can be made only
with the data related to elastic perfectly plastic beams subjected to
unsprung loads.

A parameter that is present in all existing analyses is the load
speed. While the past two works indicated a decrease in the damage
to the structure with an increase in speed, results of the present
analysis indicate the contrary (see Figure 22) .

A result of the present study that is in agreement with an obser-
vation made by Symonds and Neal is that a load greater than Parkes'
"limit load" of l. 228 Py can cross a beam. However, it should be

pointed out that the ensuing deformations may be very large .

48

49

In the numerical computations by Symonds and Neal, the assump-
tion was made that a single plastic hinge existed in the beam at its
mid-span. The results of the present study indicate that the point of
the greatest plastic deformation has a tendency to move toward the
departure end of the beam (see Figure 22) .

In order to explain the differences mentioned above, it is to be
noted that the present analysis has sacrificed the continuum nature of
the structure for a more realistic representation of the deformation
characteristics, including the elastic part. Although the previous
works are confined to rigid-plastic systems, they have maintained the
continuum character of the system. The advantage of either approach
depends on the nature of the problem being investigated. For prob-
lems in which the loading is prescribed, i.e. , unaffected by.the
deformation of the structure, the elastic part of the deformations may
be neglected, if the deformations are large. In the‘present case of a
moving mass, however, the magnitude of the loading is significantly
affected by the beam deformation, regardless of it being elastic or
inelastic.

It is of interest to note further that under the assumption of a
rigid-plastic behavior, there can be no displacements, much less
permanent deformations, if 8 is less than 1. 0. On the other hand, it
is found in this study that an appreciable amount of permanent damage

can be done to a beam, even when 8 is less than 1. 0, e.g. 8 = 0.90

50

(see Figure 10). In this respect, it seems that the approach used in
the present study is a significantly more realistic one.

The discrete system, however, has its shortcomings, mainly
loss of details and that some of its operations can be justified only on
an intuitive physical basis. As a consequence of the rigid-panel
assumption, there is on the beam no centrifugal force due to the
moving mass, which must be present in a model with continuous
flexibility and consequent curvature. This, however, is compensated
by the corrections at the joints as discussed in Art. 2.2. Another
seemingly questionable phenomenon seen in the discrete model, is
the large values of 15 near the departure end of the beam, as discussed
in Art. 4. 2. Fortunately, as already mentioned, generally this is not
too significant so far as the beam deformations are concerned, be-
cause it happens near the end of the beam.

5.1.2. Sprung Load. Probably the real advantage of the dis-

 

crete model used herein is the fact that it can be employed to deal
more realistically with more complex problems. In this connection,
the sprung load representation of a vehicle, and the analysis of
structures with a more general elasto-inelastic moment-curvature
relation may be mentioned. It seems simply not practicable to treat
such problems from the continuum point of view.

The numerical results obtained in this study seem to indicate

that the present approach yields better results for sprung loads than

51

for unsprung ones. This is on account of the fact that, for sprung
loads 15, depends on the deflections, while for unsprung loads it de-
pends on the accelerations, which are not smooth functions (see Art.
4. Z) . It is also of significance to note that numerical solutions for the
sprung mass case are also easier to obtain than those for the un-

sprung mass case .

5. Z . Summary

An analytical study of the elasto-inelastic behavior of beams
subjected to moving loads has been presented. The structure analyzed
consists of a slender beam resting on rigid supports. The types of
moving loads used in this study are a single unsprung mass, a sprung
mass, and a combination of a sprung and unsprung mass, referred to
as a single -ax1e load unit.

In obtaining the numerical results, only the case of a simply
supported beam traversed by a single load has been considered in this
study. The analysis is based on a discrete beam model, consisting
of massless rigid panels connected by flexible joints with point mass.
The flexibility and mass of the joints are obtained by lumping the
corresponding continuous properties of the actual beam. This model
facilitates considerations of the nonlinear deformation characteristics
of the beam as well as possible non-uniform distribution of the beam

mass. The nature of the discrete system, however, gives rise to

52

discontinuities in the vertical velocities and accelerations which, of
course, have been taken into account in the analysis.

The solution of the problem has been carried out by numerical
techniques. In nearly all cases, the beam has been divided into 20
equal panels. The numerical results have been obtained by the use
of a high speed digital computer. They consist mainly of (i) typical
time -history curves for deflections and interaction forces, (ii) final
shapes of deformed beams, and (iii) the influence of certain
parameters on the beam response. The major findings of this study
are summarized as follows:

1) A beam may suffer permanent damage due to the passage of a
load whose weight is less than the static yield load.

2) The effect of a positive inelastic stiffness on deformation is of
considerable significance; even a small inelastic stiffness reduces
the deformations substantially.

3) The "inelastic deformations increase at a rapid rate when the
load is increased beyond the static yield load. However, loads much
heavier than the static yield load can cross the beam, if large
deformations are allowed.

4) If permanent deformations occur, with an increase either in
the weight or in the speed of the load, the location of the maximum
permanent deformation shifts toward the departure end of the beam.

5) The model used in this analysis yields smoother variations in

53

the interaction force for a sprung load than for an unsprung load.

5. 3 . Conclusion

 

In this thesis a procedure for the dynamic analysis of the
elasto-inelastic behavior of beams subjected to moving loads is
presented. It has been demonstrated that the method is practical,
and yields solutions of the physical problem that seemingly cannot be
obtained by any other existing method. It may be pointed out also
that, although the method has been applied herein to simple beams
and single -axle loads only, extension of it to continuous spans and
more complex moving load systems should present no difficulty.

Although the numerical results presented in this thesis seem
generally reasonable from the physical viewpoint, their validity,
strictly speaking, should be verified by experimental data. This
verification represents a most important direction for future work
for the general problem under consideration. The application of the
method of analysis to problems that are more closely related to the
present practice in bridge engineering--for example, in the specifi-

cations for allowable overloads--should be worthwhile and rewarding.

10.

LIST OF REFERENCES

"The Behavior of Long Beams Under Impact Loading, " by P. E.
Duwez, D. S. Clark, and H. F. Bohnenblust, Journal of Applied
Mechanics, Vol. 72, 1950, p. 27.

"Impulsive Motion of Elasto-Plastic Beams, " by H. H. Bleich
and M. G. Salvadori, Trans. ASCE, Vol. 120, 1955, p. 499.

"Large Plastic Deformations of Beams Under Transverse
Impact, " by E. H. Lee and P. S. Symonds, Journal of Applied
Mechanics, Vol. 20, No. l, 1953, p. 151.

"Impact of a Cantilever Beam with Strain Rate Sensitivity, " by
T. C. Ting and P. S. Symonds, Proceedings. the Fourth U.S.
National Congress of Applied Mechanics, 1962, p. 1153.

"Dynamic Analysis of Elasto-Plastic Structures, " by G. V.
Berg and D. A. DaDeppo, Journal of Engineering Mechanics
Division, ASCE, Vol. 86, No. EMZ, April 1960, p. 35.

"Response of Elasto -P1astic Structures to Transient Loads, "
by H. H. Bleich, Transactions, New York Academy of Sci-
ences, Ser. II, Vol. 18, No. 2, December 1955, p. 135.

"Dynamic Elasto -Plastic Analysis of Structures, " by M. L.
Baron, H. H. Bleich, and P. Weidlinger, Journal of Engin-
eering Mechanics Division, ASCE, Vol. 87, No. EMl,
February 1961, p. 23.

"Discrete Dynamic Models for Elasto -Inelastic Beams, " by
R. K. Wen and T. Toridis, to be published in the October 1964
issue of the Journal of Engineering Mechanics, ASCE.

"Dynamic Behavior of Simple -Span Highway Bridges, " by R. K.
Wen and A. S. Veletsos, Bulletin 315, 1962, Highway Research
Board, Washington, D.C.

How To Cross an Unsafe Bridge, " by E. W. Parkes, Engin-
eering. Vol. 186, November 7, 1958, pp. 606-608.

54

11.

12.

13.

14.

15.

55

"Travelling Loads on Rigid-Plastic Beams, " by P. S. Symonds
and B. G. Neal, Journal of Engineering Mechanics Division,
ASCE, Vol. 86, No. EMl, January 1960, p. 79.

"Dynamic Analysis of Inelastic Multi-degree Systems, " by

A. C. Heidebrecht, J. F. Fleming and S. L. Lee, Journal of
the Engineering Mechanics Division, ASCE, Vol. 89, No. EM6.
December 1963, p. 193.

"Vibration Susceptibilities of Various Highway Bridge Types. "
by L. T. Oehler, Proc. Paper No. 1318, ASCE, Vol. 83, No.
ST4, July 1957.

"A Method of Computation for Structural Dynamics, " by N. M.
Newmark, Trans. ASCE, Vol. 127, 1962, p. 1406.

"Discussion of a Differential Equation Related to the Breaking
of Railway Bridges, " by G. G. Stokes, Mathematical and
Physical Papers, Cambridge, Vol. 2, p. 179.

56

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59

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é -

 

Simply -Suppo rte (1 Beam

Span Length, L = 64.25' c.c.
36 WF 170 1b. at 5' 0” c.c.
Slab Thickness:

Steel Section:

7”

Composite De sign

I
transformed
transformed

a
yield

m(x)

18,270 in.4

30 x 106 lb./in.2

33, 000 1b. /in.2

o. 127 lb. -sec.Z/in.z
2.169 x 107 lb./in.
1.125 x105 1b.
0.157 sec.

1.96 in.

Fig. 6--Properties of a Highway Bridge

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R = o, 13:1.10, y = 3.271

 

 

Fig. 22--Deformed Beam Shapes--Unsprung Mass

 

.0

 

.20.0

17

De fle ction/ 6

>7

Y

De fle ction [6

De fle ction /6

 

 

120 ..

 

(a) Permanent Sets for R = 0, a = 0.10, 8 = 1.30, y = 3.865

 

x = x/L
0 0'2 0:4 06 018 1.0
O v' I U a
0'2 v ‘ ' a =0.20
(2 =0.10
0.4 a.

 

(b) Permanent Sets for R = 0.5, 8 = 1.20, y = 3.568

Fig. 23--Deformed Beam Shapes--Unsprung Mass

Y

>7

Deflection /6

Deflection I6

 

 

20.50
/——a

4.01-

’ =o.4o
6.01- , / a ‘

 

(a) Permanent Sets for R = o, [3 =1.50, y = 4.460, 12 = 33.57

 

4..

 

 

(b) Permanent Sets for R = o, p=1.20, y = 3.568, 12 = 33.57

Fig. 24--Deformed Beam Shapes--Sprung Mass

Deflection /6

37:

De fle ction /6

,7:

 

 

 

(a) Permanent Sets for
a = 0.15, 8 21.20, y = 0.472, x = 0.132, k = 33.57

 

 

x = x/L
O 0 0.2 0.4 0.6 0.8 . 1.0
ufiw ' r r ' .9,»
4.0....
8.0..
12. 0...

(b) Permanent Sets for R = 0, a = 0.20,
8 = 1.20, y = 0.472, X = 0.132, k .= 33.57

Fig. 25--Deformed Beam Shapes --Single-Axle Load Unit

81

 

 

 

 

 

 

 

 

2.007?
R = O, 8 = 0.90, y = 2.676
)2 = 0.5
g ,0 1.50..
to .9
2 ‘23
II a
x o #4»—
ng 1.00.1 - .0 L
|>~
0.50 § ; : % fi;
0.05 0.10 0.15 0.20 0.25 0.30
Speed Parameter a
(a) Effect of a on Maximum
Deflections (R = 0, 8 = 0.90)
6.00"
8
so R=0.l, 8:1.20,y=3.568
§\ -
3; "g x = 0.6
2 1'; 4.00..
n 5%."
(>515 / -
E
1>~
2-00 .L ; TL F fi‘
0.05 0.10 0.15 0.20 0.25 0.30

Speed Parameter a

(b) Effect of a on Maximum Deflections
(R = 0.1, 8 = 1.20)

Fig. 26--Effect of Speed on Maximum Deflections--Unsprung Mass

82

  

 

 

  

 

 

 

 

12.0 ‘F’ _
x = 0 6
R=o, 13:1.20, 1, = 3.568, 12:33.57
5 to 8.0 .
.§ .1
3; .2
H
2 8
n 1;:
)4 Q)
«1 D
4.0 ..
l>.E
0 1‘ l : 1
0.10 0.20 0.30 0.40 0.50
Speed Parameter a
(a) Effect of a on Maximum Deflections
(R = 0, 8 21.20)
40.01.
E -
E 3 x = 0.7
"-1 G ..
f; ,2 R=0, 8:1.50, y = 4.460, k=33.57
H
2 3’, 20.0..
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356’
E
|>~
+
—o
o 1 1 : A.
0.20 0.30 0.40 0.50 0.60

Speed Parameter a
(b) Effect of a on Maximum Deflections
(R = 0, 8 = 1.50)

Fig. 27--Effect of Speed on Maximum Deflections --Sprung
Mass

Maximum De fle ction /6

ymax

83

 

 

 

 

10.0
a = 0.20, (3 = 1.20, y = 3.568
8.04» x = 0.5
6.0--
4.0 +-
2,0...
A

0 'r Jr % 4. i

0 0 2 0 4 0.6 0 8 1 0

R = kZ/kl

Fig. 28--Effect of R on Maximum Deflections--Unsprung Mass

84

12.0}?

10.04

max/‘3
0‘
O
1

 

 

I.
--

l
I T

0.80 0.90 1.00 1.10 1.20

p = Wv/Py

Fig. 29--Effect of Weight of Vehicle on Maximum
Deflections --Unsprung Mass

APPENDIX

COMPUTER PROGRAM

Presented in this Appendix are the computer programs used in

this stud}r and certain information useful to a potential user.

1. Parameters of the Problem

 

The programs can be used to solve a series of numerical prob-

lems involving a range of values of the parameter a . The parameters

are:
R = R(=k2/k1)
TALPHA = initial value of a
TBETA = B
TGAMMA = y
TLAMBDA = X
TKBAR = 12
TCER = increment in a
TMED = 0. 98 times the final value of a
NP = an integer such that NP times AT is the interval
between print-outs.
2. Computer Time

 

The computer time required for the solution of a given problem

85

86

depends primarily on the value of a . The approximate amounts of

computer time necessary for some typical values of a are listed

 

below:
1) for a = 0.10, 71/2 minutes
2) for a = 0.20, 5 minutes
3) for a = 0.30, 41/2 minutes
4) for a = 0.40, 4 minutes
3. Identification of Important Variables in the Program
TTAU = 'r
ZBAR = 2
REMD = 72'
DPA(I) = [Ema/1.25] {ri

VEA(I) = [61r2/1.25]dyi/d-r

A Z 2 -
ACA(I) = [6 1r /l.25]d2yi/d-rz

SYA = [61r2/1.25] v5

VSA = [5172/1.25]dv7/d-r
ASA = [62w2/1.25] dzv'ir/d-rz
ANGLAu): 61

BMA(I) = Mi

AMM = n2/1.25 x c1237; / d-rz

PBAR :5

87

4. Quantities Printed Out

 

At the start of the program N, NP, and AT are printed out.
During the process of solution, the following quantities are printed

out at regular intervals as determined by NP:

1) ‘r

2) (Yi)dynam1c

3) P3

4) (yi)static

5) (6.) (=6. -1\7I./k')
1 permanent 1 1 1

In addition, in the case of a single -axle load unit, the accelera-
tion of the unsprung mass and the spring force are printed out simul-
taneously with P.

At the end of the problem, the following quantities are printed
out:

1) absolute maximum 371

2) “Ti corresponding to absolute maximum yi

3) numerical values of parameters

4) maximum 371 during free vibration

5) minimum yi during free vibration

6) yi corresponding to the final shape of the

deformed beam

7) Bi corresponding to the final angles of the

8)
9)
10)

11)

88

deformed beam

maximum value of P

'r corresponding to the maximum value of P
minimum value of P

T corresponding to the minimum value of P

 

 

H

H

1-! h! g—a

(1()O f)0(1()0

(3(30

89

***** COMPUTER PROGRAM *§****

PROGRAM MOVLINK
***** UNSPRUNG MASS **‘**

DIMENSION ACA¢355oACB(35)oACD(35)OVEA(35)0VEB(35)oDPA(35)ODPB(35)v
IDDP(35)OBMA(35)OBMB(35)OTMPA(35)OTMpB(35)OTMNA(35)OTnNB(35,0DPAMX
2(35)9TMAX(35)oANGLA(35)QANGLB(35)cDA(35)0P(35)oDP5(35)ODIFC35)0
BSTDE(35)oSOF(35)oEF‘35)oTHPL(35)cRAC(35)

N820

M820

TOLER‘OoOOOOOOOS

BETA=OO

ZNAN

ii§ii PARAMETERS *****
R300
TALPHA'OQ3O
TBETA3009O
TGAHMA8206761

TCER30010
TMED=Oo39
NP8N**2/4

RK=R
VBAR=TALPHA*ZN
TMED=TMED*ZN
CERSOO
FBAR3005*TBETA
RSTGAMMA*ZN

***** CLEAR 0R SET STORAGE *****
1001 VBARSVBAR+CER

RANGE=ZNIVBAR

NN8N+2

DO 2 ISIoNN

ANGLA(I)=OoO

ACA(!)IO¢O

1000

278

90

ACD(I).OOQ

VEA(I)'000

VEB(II.OOO

DPACTI8000

DPB(I)3000

BMA(I)8000

BMB(II8000
DPAMX(II8000
TMPA(I)'I¢0
TMPB(I)‘I-0
TMNACII3IOO
TMNB(II‘IOO

KT=1

TKN‘IOO

TNLE‘O.

TAUSOOO

VNxN-I

DELTA‘O.2/VN**Z
STFA=40.*ZN**2/(384o*30141592654**2I
STEPSRANGE/(DELTA*ZNI
ISTEP'STEP

STEPBISTEP
DELTA-RANGE/(STEP*ZN)
PANGE‘RANGE+500
TAPN3100/VBAR

PRINT IOOOoNcNPQDELTA
FORMAT(IHOQI205X0ISQSXQFIZOIOI
TEPCSVBAR/ZN
AGRSZoiFBAR

PHYSSP/ZN

PRINT 2780PK9TERC9AGPOPHYS
F0RMAT(IH004FIO¢5)
GBAR=8.*3B.4*ZN*FBAP/R
PBMA=(R*GBAR/ZN)/I53o6
PBMI=(R*GBAR/ZN)/I5306
TPMA=00

TPMI=0¢

TLENG=100IZN
TMEASsTLENG

IOTSI

INSNP

TAUSTAU+DELTA
ZBAR=VBAR*TAU
IF(PANGE-TAU) 109100300

C “an;

10

279

20
21

270

271

1002

300

299

302

298
297

296

301

PRINT
PRINT
PRINT
PRINT
PRINT
PRINT
PRINT
PRINT
PRINT
PRINT

PRINT MAXIMUM VALUES
21
2710(DPAMX(IIOI-2vll)
939(DPAMXCIIOI=120NI
2710(TMAX(II¢I32vIII
989‘TMAX(I)OI=120NI
2780RK9TERC9AGR0PHYS
2710‘50F(I)!I.2011I
969(SOFIII¢I=120NI
2719(EF(II9I=20111
989(EF‘IIoI3120N)

DO 279 I=KoJ

SOF(I
PRINT
PRINT
FORMA
FORHA
DO 27

THPL(I)=ANGLA(I)-BMA(I)/STFA

1'0.5*(SOF(I)+EF(I)I
2719(50F(I)9I820111
980(SOF(I1918120N)

T(1HOo9F10¢5)

TIIHO0IOX03IH END. MAX.

0 I320N

91

AND DEFORMED SH‘PE *****

DISPLACEMT. AND TIME)

THPL(I)=THPL(I)/3.l§1592654**2

PRINT
PRINT
FORMA
PRINT
PRINT
PRINT
PRINT

2719(THPL‘IIOI3ZOII)

209(THPL(I)OI8129NI
T‘IHOOIOFIOOSI

979PBMA

970TPMA

970PBMI

970TPMI

NP=7*NP/12

CERaT
IF(TM
STOP

LOCAT
IF(TK

CER*ZN
ED-VBAR)1002.IOOI.1001

ION OF MOVING MASS
N-ZBAR1296v2980299

GAM=ZBAR+O-01*DELTA*VBAR

IF(TK
ZBAR=
ASSIG
GO TO
ZBAR=
ASSIG
GO TO
ZETSZ
IF(TK

N-GAM) 298.298.302
ZBAR-TNLE

N 290 TO LIL

5

TKN-TNLE

N 291 TO LIL

5
BAR-0.01*DELTA*VBAR
N-ZET) 30102980298

TKNBTKN+IOO

()0

(TO

102

18

SO
51

52
53

55

56

65
7O

71
73

75

76

92

TNLEaTNLE+IoO
KT=KT+1
GO TO 302

***** INTEGRATE BY BETA METHOD *****
DO 102 I320"

ACB(I)3ACD(1I

X8005*ACB(I)+005*ACA(I)

VEDCII'VEA‘II+X

X8005*ACA(II'BETA*ACA(II
DDP‘II8VEACI)+X+BETA*ACB(II
DPB‘I18DPACII+DDP(II

DPB(N+21=DPB(NI

***** COMPUTE ANGLE OF ROTATION *****
DO 18 I329"
ANGLB‘I)8-DPB(I’I)+2¢0*DPB(I)-DPB(I+II
DA(II‘ANGLB(1)-ANGLA(II

***** CONPUTE BENDING MOMENT *****
DO 79 1329M

IF(DA(III 50050065
IF(TNNA(II+STFA*DA(III 51052955
C=RK

GO TO 53

c=OoO

TMNB(I)=0.0
XSSTFA*DA(I)+TMNA(II
BMB(IIBBMAIII‘TMNA(II+C*X

GO TO 56
TMNBCII'TMNA(I)+STFA*DA(II
BMB(I)=BMA(I)+STFA*DA(II
TMPB(II‘200-TMNB(II

GO TO 79

IF(TMPA(II‘STFA*DA(III 70071975
C3RK

GO TO 73

C3000

TMPB(1)=OOO
XBSTFA*DA(I)’TMPA(II
BMB(I133MA(I)+TMPA(II+C*X

GO TO 76
TMPB‘II‘TMPA(II“5TFA*DA(II
BNB(II8BMA(1)+STFA*DACII
TMNB(II‘200-TMPB(II

93

79 CONTINUE

***** AUXILIARY ROUTINE *****
DO 23 I329N
XBZN**2*(BMB(I-l)-2.0*BMB(I)+BMB(I+1))
23 ACD(I)SDELTA*3894*DELTA*X
IF(N-KT) 31093229322
322 IF(KT-1) 30493049305
304 X8190/(190+R*ZBAR**2)
ACD(KT+1)=X*DELTA*(3894*ZN*(ZN*(8MB(KT)‘290*BMB(KT+1)+BMB(KT+2))+
1890*F8AR*ZBAR)*DELTA-290*R*VBAR*ZBAR*VEB(KT+1)1
GO TO 310
305 IF(KT-N) 30693079308
308 PRINT 309
309 FORMAT(IH29IOX912HKT=TOO LARGE)
307 REM08190-ZBAR
X=190/(190+R*REMD**2)
Y-BMB(KT-1)-2.0*BMB(KT)+BM8(KT+II
ACD(KT1=X*DELTA*(3894*ZN*DELTA*(ZN*Y+890*FBAR*REMDI-290*R*VBAR*
1REMD*(VEB(KT+1)-VEB(KT)))
GO TO 310
306 REMD=190-ZBAR
S'IoO/R+REMD-ZBAR*REMD
T8R*(ZBAR*REMD)**2
U=R*ZBAR**2
V81.0/(190~T/$+U)
UsBMB¢KT-1)-2.0*BMB(KT)+BMB(KT+1)
X=ZN**2*ZBAR*REMD*U
YSZN*ZBAR*REMD**2*FBAR
Z=BMB(KT)-290*BMB(KT+I)+BM8(KT+2)
YE=2.0*R*VBAR*ZBAR*(REMD**2/S-1oO)*(VEB(KT+1)-VEB(KT))
ACD(KT+1)=V*DELTA*(3894*ZN**2*Z+890*3894*ZN*F8AR*ZBAR-3894*X/S-890
1*38.4*Y/S)*DELTA+V*DELTA*YE
TIM=190/(190+R*REMD-R*ZBAR*REMD)
ACD(KT)3TIM*(DELTA!(3894*ZN**2*U+89O*3894*ZN*FBAR*REMD)*DELTA
1-R*ZBAR*REMD*ACD(KT+1)~290*DELTA*R*VBAR*REMD*(VEBCKT+1)-VE8(KT)))
310 00 35 1:29N
X8095*(ACD(I)+ACA(I))
35 VE8(I)=VEA(I)+X
GO TO LIL
290 A0AC=09
AUTSO o
YOJ‘O.
GO TO 289
291 AUT:-DELTA*R*VBAR*(DPBCKT+2)-29*DPB(KT+1)+DP8(KT))/(lo+R)

94

ADAC=-29*R*VBAR*(VEB(KT+2I‘29*VEB(KT+:)+VEB(KT)-AUT1*DELTA/(19+R)
YOJ=(29*VBAR/(19+R))*(VEB(KT+2)‘29*VEB(KT+1)+VE8(KT)’AUTI*DELTA
IFCN-KTI42594269426

425 GO TO 289

426 218809

289 LOT=KT+1
IF(N—LOT) 28892879287

287 ACD(KT+1)=ACD(KT+1)+ADAC
VEB(KT+I)=VEB(KT+1)+AUT

288 IF(N*KT) 4092549254

254 AMM:ACD(KT)/DELTA**2+29*VBAR*(VEBIKT+1)‘AUT-VEBIKTII/DELTA
I+(ZBAR/DELTAI*(ACD(KT+I)-ADAC-ACD(KTII/DELTA
TYOJ=YOJ/DELTA**2
AMMaANM+TYOJ
PEAR:(R*(GBAR‘AMM)/ZN)/15396

***** PREPARE FOR NEXT STEP UP INTEGRATION *****
40 DO 45 1:29M

BMAKI)=BMB(I)

TMPA(I)=TMPB(I)

TMNA(II‘TMNB(I)

ACA¢I18ACDKII

VEACII=VEB(I)

DPAtI)=DPB(I)

ANGLA(I)=ANGL8(I)
45 CONTINUE

***** COMPUTE MAXIMUM RESPONSE *****
JsN
K82
SIK=ZN/VBAR+3.
IF(TAU-SIK) 110192859285
1101 DO 109 1=K9J

DPS(I)81.25*DPA(11/39141592654**2
SOF(1)=DPS(I)
EF(I)=DPS(1)
IF!DPS(I)-DPAMX(1)) 10991099112

112 DPAMX(1)=DPS(I)
TMAX(I)=TAU*VBAR/ZN

109 CONTINUE
IFCPBMA-PBAR126192629262

261 PBMAsPBAR
TPMA=TAU*VBAR/ZN

0f)

()0

1262
263

1264
285

283
282
281
280

48

I6

115

97
98

266

260
267

323
398
397

399
400

401

802

95

IF‘PBMI-PBAR126492649263
PBMISPBAR

TPMI=TAU*VBAR/ZN

GO TO 48

DO 280 I=K9J
DPSCIISI.25*DPA(11/39141592654*§2
IF($0F(II‘DP$(III 28392829282
SOF(II=DPS(II
IF(EF(II-DPS(I)) 28092809281
EF(I)=DPS(II

CONTINUE

***** TEST TIME TO PRINT *****
IFCIN‘II 1159115916
IN=IN-I

GO TO 4

TTAU=TAU*VBAR/ZN

PRINT 979TTAU

PRINT 2719(DPS(1)9I=2911)
PRINT 989(DP51119I=129N)
FORMATC1H09F1297I
FORMAT11H099F1095I
IFITAU-ZN/VBAR126692669267
PRINT 2609PBAR
FORMAT(IHO95HPBAR=F1096)
218300

***** STATIC DEFLECTION *****
IF(N-KTI 32493239323
ABAR8‘ZBAR+TNLEI/ZN

IF(IoO-ABAR) 39893999399

PRINT 397
FORMATK1H2920HABAR=LARGER THAN ONE)
IF‘TMEAS-ABARI 40094019401
TMEAS=TMEAS+TLENG

IOT=IOT+I

GO TO 401

BBAR=190-ABAR

DIST'OQ

RENDBOO

DO 802 1329IOT

DIST=DIST+TLENG
DIFCI131298*BBAR*DIST*(ABAR*(BBAR+I90)-DI$T**2)
TIT8IOT

DIST'TIT*TLENG-TLENG

96

JJ=IOT+1
DO 803 I=JJ9N
DISTBDIST+TLENG
REMD=IoO-DIST
803 DIF(I1812.8*A8AR*REMD*(BBAR*(A8AR+IoOI-REMD**21
DO 804 1:29N
804 STDE(II‘I¢25*FBAR*DIF(II
JEN
K82
PRINT 2719(STDE(II9I=29III
PRINT 989(STDE(II9I=I29N)
DO 570 I=Z9N
THPL‘IIgANGLACII'BMA(I)/$TFA
570 THPL1II=THPL(11/39141592654**2
PRINT 2719(THPL(I)91=29III
PRINT 209(THPL(II9I3129N)
324 GO TO 7
END
END

C
C
PROGRAM MOVLINK
C ***** SPRUNG MASS *****

DIMENSION ACA(35)9ACB(35)9ACD(35)9VEA(35)9VEB(3519DPA(35)9DP8(35)9
IDDP(35)9BMA(35)9BMB£3519TMPA(35)9THPB(35)9TMNA(35)9TMNB(35)9DPAMX
2(35)9TMAX(35)9ANGLA(35)9ANGLB(35)9DA(35)9P(35)9DPS(35)9DIF(35)9
38T0E(35)9SOF(35)9EF(35)9THPL(35)9RAC(35)

N820

M=20

TOLER=0900000005

BETA=Oo

ZN=N

0(10

{*iii PARAMETERS *****
R300
TALPHA‘OQZO
TBETA=1920
TGAMMA8395681
TKBAR=330573

TCER=0910
TMED=O929
NP=N**2/4

(10

1001

97

RK=R
VBAR=TALPHA*ZN
TMED‘TMED*ZN
CER'O.
FBAR=O.5*TBETA
QS‘TGAMMA*ZN
$K=TKBAR*ZN

***** CLEAR OR SET STORAGE *****
VBAR=VBAR+CER
RANGE‘ZN/VBAR
NN3N+2

DO 2 1819NN
ANGLAKII=OOO
ACAIII‘OOO
ACD‘I)=OOO
VEA‘II‘OOO
VEB1II=OOD
DPA(II=OOO
DPB(II=OQO
BMA1II’OOO
BMB(II3OOO
DPAMX(1)=OOO
TMPACII=IOO
TMPBCII‘IOO
TMNA‘I)=IOO
THNBCII’IQO
5YA=Oo

V5A=Oo

A5130.

KT'I

TKNSIOO

TNLE=Oo

TAU'OOO

AMMSOO

TIM‘O.

VNaN-l
DELTA‘OOZ/VN**2
STFA=400*ZN**2/(384o*39I41592654**2)
RANGE‘RANGE+5OO
TAPN=IoO/VBAR
PRINT IOOO9N9NP9DELTA

1000 FORMAT(1HO91295X9I395X9F129IO)

C
C
C

278

98

TERC‘VBAR/ZN

PRINT 2789RK9TERC9TBETA9TGAMMA9TK8AR
F0RMAT(1HO95F10.5)
GBAR=8.*38.4*ZN*FBAR/OS
PBMA8(OS*GBAR/ZN)/153.6
PBMISCQSiGBAR/ZN1/153.6
TPMA=O.

TPM1=O.

TLENG=1.0/ZN
TMEAS=TLENG

IOT=1

IN-NP

TAU=TAU+DELTA
ZBAR=VBAR*TAU
TTAUITAU*VBAR/ZN
IF(RANGE-TAU) 109109300

*§§*§ PRINT MAXIMUM VALUES AND DEFORMED SHAPE

10

279

20
21

270

271

PRINT 21

PRINT 2719(DPAMX(I)9I=2911)
PRINT 989(DPAMXIII913129N)
PRINT 2719(TMAX(II9I=29III
PRINT 989(TMAX(I)9I=129NI
PRINT 2789RK9TERC9TBETA9TGAMMA.TKBAR
PRINT 2719(SOF(I)9I329111
PRINT 989(SOF(II9It129N)
PRINT 2719(EF(II9I=29111
PRINT 989(EF(I)913129N)

DO 279 I=K9J
SOFIII=0.5*($OF(II+EF(III
PRINT 2719(SOFIII9II29III
PRINT 989(SOF(I)9I=129N)
FORMATI1H099FIO.5)

F0RMATIIHO9IOX.31H END. MAX. DISPLACEMT. AND TIME)

DO 270 I=29N
THPLII)=ANGLA(II-BMAIII/STFA
THPL(II8THPL(I)/3.I41592654**2
PRINT 2719(THPL(I)9I=29111
PRINT 209(THPL‘II9I=129NI
FORMATI1H09IOFIO.5)

PRINT 979P8MA

PRINT 979TPMA

PRINT 979P8MI

PRINT 979TPMI

*“I-‘I'

1

002

300

302

301

321

102

18

50
51

52
53

99

NP=7*NP/12

CER:TCER*ZN
IFITMED-VBAR110029100191001
STOP

LOCATION OF MOVING MASS
IF(TKN-ZBAR)3OI93029302
ZBAR=ZBAR-TNLE

GO TO 5

TKN=TKN+1.

TNLEITNLE+1.

KT=KT+1
IFIZN-TKN132193029302

GO TO 302

***** INTEGRATE BY BETA METHOD *****
DO 102 I=29M

AC8III=ACD(I)
X80.5*ACB(I)+O.5*ACA(I)
VEB(II=VEA(I)+X
X80.5*ACA(I)‘BETA*ACA(I)
DDPII)8VEAII)+X+BETA*ACBII)
DPB(I)BDPA(I)+DDP(I)
DP8(N+2)=-DPB(N)

VEBIN+2)=‘VEB(N)

ASBSASI

VSB=VSA+0.5*(ASB+ASI)
SYB=SYA+VSA+O.5*ASI+BETA*(ASB~ASII

*§*** COMPuTE ANGLE OF ROTATION *iiii
DO 18 I=Z9M
ANGLB(1)=-DPB(I-l)+2.0*DPB(1)-DPB(1+1)
DA(I)=ANGLB(1)-ANGLA(I)

***** COMPUTE SENDING MOMENT *****
DO 79 I=2.M

IF(DA(II) 50950965
IF(TMNA(I)+STFA*DA(III 51952955
C=RK

GO TO 53

C3000

TMNBIII=OOO

XSSTFA*DA( I )+TMNAI I I
BMBIII'BMA(II“TMNA(I)+C*X

GO TO 56

100

55 TMNB(I)8TMNA(I)+STFA*DA(I)
BMB(I)8BMA(I)+STFA*DA(I)

56 TMP8(I)=2.0-TMN8(I)
GO TO 79

65 IF(TMPA(II-STFA*DA(I)) 70971975

70 C=RK
GO TO 73

71 C3000

73 TMPBII)=0.0
X=STFA*DA(I)‘TMPA(I)
BMBII)=BMA(II+TMPA(I)+C*X
GO TO 76

75 TMPBII)=TMPA‘II-STFA*DA(I)
BMBII)=BMA(I)+STFA*DA(II

76 TMNB(I)=2.O-TMPB(II

79 CONTINUE

***** AUXILIARY ROUTINE *****

DO 23 1329N

XBZN**2*(8MB(1-11-2.O*BM8(II+8MBII+III
23 ACD(II=DELTA*38.4*DELTA*X

IFIN-KTI 31093229322
322 REMD=1.-ZBAR

YMI=DPBIKTI+ZBAR*(DPB(KT+II-DP8(KT))

IF(KT-1)50095009501
500 W8-2.*BMB(KT)+BMB(KT+II

GO TO 502
501 W88MBIKT~I)-2.*BMB(KT)+BM8(KT+11
502 Z=8MBIKT)-2.*BMB(KT+I)+8M8IKT+21

ACD(KT)=38.4*ZN*DELTA*(ZN*W+8.*F8AR*REMDI*DELTA

I+SK*DELTA*(SYB-YM1)*REMD*DELTA

ACDIKT+1I=3894*ZN*DELTA*(ZN*Z+8.*FBAR*28AR)*DELTA

1+SK*DELTA*(SYB-YMI)*ZBAR*DELTA

ACDI I 1300

ACDIN+II=0.

ASD=~IDELTA*(SK*(SYB-YMI)1/OS)*DELTA
310 DO 35 1329N

X=O.5*(ACD(II+ACA(III

35 VEBIIISVEAIII+X
VSBSVSA+O.5*(ASD+ASI)
PBAR=I(05*GBAR+SK*(SYB-YM1)I/ZNI/153.6

***§* PREPARE FOR NEXT STEP OF INTEGRATION *****
40 DO 45 1829M
BMAIIIIBMBII)

C
C

45

101

TMPA(I)=TMPB(I)
TMNACI)'TMNB(1)
ACA(1)=ACD(I)
VEAII)=VEB(I)
DPA(1)=DPB(1)
ANGLA(I)=ANGL8(I)
CONTINUE

ASIIASD

VSA=V$B

SYA=SYB

***** COMPUTE MAXIMUM RESPONSE ii!!!

1101

112

109

510

261

262
263

264
285

283
282
281
280

as
16

J8N

K82

SIK=ZNIVBAR+3.

IFITAU~SIKI 110192859285

DO 109 I=K9J
DPSII)=1.25*DPA(I)/3.141592654**2
SOFIIIBDPSII)

EFCII=DPSIII
IFIDPSII)-DPAMX(I)) 10991099112
DPAMXIII=DPS(I)
TMAXIII=TAU*VBAR/ZN

CONTINUE
IF(TAU-ZN/VBAR151095109264
21880.

IF(P8MA-PBAR)26192629262
PBMA=PBAR

TPMA=TAU*VBAR/ZN
IF(PBMI-P8AR)26492649263
PBMISPBAR

TPMIBTAU*VBAR/ZN

GO TO 48

DO 280 I=K9J
DPSII181.25iDPA(I)/3.141592654**2
IFISOFIII-DPSIIII 28392829282
SOFIIISDPS(I)

IF(EF(I)-DPS(I)) 28092809281
EF(II=DPS(I)

CONTINUE

***** TEST TIME TO PRINT *****
IF(IN‘1) 1159115916

IN81N“1

GO T04

(10

115

98
97

266

260
267

323

398
397
399
400

401

802

803

804

570

102

TTAU=TAU*VBAR/ZN

PRINT 979TTAU

PRINT 2719(DPS(I)9I=2911)
PRINT 989(DPS(I)9I8129N1
FORMATI1H099F10.5)
FORMATIIHO9F12.7)
IF(TAU-ZN/VBAR)26692669267
PRINT 2609PBAR
FORMAT(1HO95HPBAR=FIO.6)
218800

9999* STATIC DEFLECTION .9...
IF(N-KT) 324.323.323
ABAR=(ZBAR+TNLE)/ZN

IF(1.0—ABAR) 393.399.399

PRINT 397
FORMATI1H2920HABAR=LARGER THAN ONE)
IFtTMEAS—ABAR) 400.401.401
TMEAS=TMEAS+TLENG

10T=IOT+1

60 TO .01

BBAR=1.0—ABAR

DIST=Oo

REMD=0.

00 802 1=29IOT

DIST=DIST+TLENG
DIFII)312.5*BBAR*DIST*(ABAR*(BBAR+1.01-DIST**2)
TITnIOT

DISTsTIT*TLENG—TLENG

JJ=IOT+1

no 803 I=JJ9N

DIST=DIST+TLENG

REMDaloo-DIST
DIFTI1:12.3*ABAR*REM0*(BBAR.(A5AR+1.0)~REMo**2)
00 804 1:2.N
STDE<IJ=1.25*FBAR*DIF(I)

J=N

K82

PRINT 2719(STDE(I)9I=2911)

PRINT 989(STDE(I)9I=129N)

DO 570 teaoN
THPL(I).ANGLA(I)‘BMA(I)/STFA
THPL(I)=THPL(1)/3.141592654**2
PRINT 2719(THPLII191-2011)

PRINT 209(THPLIII913129N)

OIWC) 0

0(3()

103

324 GO TO 7
END
END

PROGRAM MOVLINK

***** SINGLE AXLE LOAD UNIT *****
DIMENSION ACA(35)9ACBI3519ACDI35)9VEAI3519VEB(35)9DPA(3519DP8(35)9
IDDP(35)9BMA(35I9BMB(35)9TMPAI3519TMP8(35)9TMNA‘3519TMN8(35)9DPAMX

2(3519TMAXI35)9ANGLA(35)9ANGLB(3519DA(3519P(35)9DPS(35)9DIF(35)9
3STDEI35)9SOF(35)9EF(35)9THPL(35)

N310

M310
TOLER=0900000005
BETA=Oo

ZN8N

R3091
TALPHA'OOZO
TBETA'IOZO
TGAMHA8305681
TLAMDA'OOI323
TKBAR=330573

TCER=0910
TMED=Oo29
NP8N**2/4

RK=R
VBARSTALPHA*ZN
TMED=TMED*ZN
CER‘O.
FBAR‘O.5*TBETA
RSTLAMDA*TGAMMA*ZN
QSSTGAMMA*ZN-R
SKSTKBAR*ZN

DC=Oo

***** CLEAR OR SET STORAGE §****
1001 VBARSVBAR+CER
RANGE'ZN/VBAR

1000

278

104

NNIN+2

DO 2 1319NN

ANGLA(II'OOO

ACAIII‘0.0

ACDIII‘OOO

VEAIII3OOO

VEB(I)'OOO

DPAII13OOO

DPBII)3O.O

BMA(II‘0.0

BMB(II=OOO

DPAMX(II:0.0

TMPAIII=IOO

TMP8(II3IOO

TMNAIII‘IOO

TMNBIII‘IOO

SYA=OO

VSABO.

A5130.

KT=I

TKN=IoO

TNLE=Oo

TAU=OOO

VN‘N“I

DELTA=O.2/VN**2
STFA=40.*ZN**2/(384.*3.141592654**21
STEP=RANGE/(DELTA*ZNI
ISTEP'STEP

STEP=ISTEP
DELTAzRANGE/(STEP*ZNI
RANGE=ZN/VBAR+50

TAPN= 1 oO/VBAR

PRINT IOOO9N9NP9DELTA
FORMATIIHO9IZ95X9I395X9FIZQIO)
TERC‘VBAR/ZN

PRINT 2789RK9TERC9TBETA9TGAMMA9TLAMDA9TKBAR
FORMAT(1HO96FIO.51
GBAR=8.*38.4*ZN*FBAR/(OS+RI
PBMA=((GS+RI*GBAR/ZN)/15306
PBMI=((OS+RI*GBAR/ZNI/153.6
TPMA‘O.

TPMI'O.

TLENGtI90/2N

TMEAS‘TLENG

IOTSI

105

7 INcNP

4 TAUsTAU+DELTA
ZBAR=VBAR*TAU
IFIRANGE-TAU) 109109300

C
C iiiii PRINT MAXIMUM VALUES AND DEFORMED SHAPE *****
10 PRINT 21
PRINT 209(DPAMXIII9I8K9J)
PRINT 209(TMAXIII9IBK9JI
PRINT 2789RK9TERC9TBETA9TGAMMAoTLAMDAvTKBAR
PRINT 209(SOFIII9IBK9J)
PRINT 209(EFIII9I=K9JI
DO 279 I=K9J
279 SOF(1130.5*(SOF(I)+EF(III
PRINT 209(SOF(II918K9JI
2O FORMATIIHO99F10.5)
21 FORMAT(IHO9IOX93IH END. MAX. DISPLACEMT. AND TIME)
00 270 I=29N
THPLII)=ANGLA(I)-BMA(I)/STFA
270 THPL(I13THPL(I)/3.141592654**2
PRINT 2719(THPL(I)9I=2911)
PRINT 209(THPLIII913129N)
271 FORMAT‘IHO9IOFIO.5)
PRINT 979PBMA
PRINT 979TPMA
PRINT 979P8MI
PRINT 979TPMI

NP=7*NP/12
CERsTCER*ZN
IF(TMED-VBAR)10029100191001
1002 STOP
C
C LOCATION OF MOVING MASS

300 IF(TKN~ZBAR)29692989299
299 GAM=Z8AR+0.01*DELTA*V8AR
IF(TKN-GAM) 29892989302
302 ZBAR=ZBAR-TNLE
ASSIGN 290 TO LIL
GO TO 5
298 ZBAR=TKN-TNLE
297 ASSIGN 291 TO LIL
GO TO 5
296 ZET=ZBAR-0.01*DELTA*VBAR
IFITKN—ZET) 30192989298
301 TKNSTKN+1.O

OI?

102

18

50
51

52
53

55

56

65
7O

71
73

106

TNLE=TNLE+1.0
KT=KT+1
GO TO 302

***** INTEGRATE BY BETA METHOD *****
DO 102 1329M

ACBIIISACDIII
X=O.5*AC8(II*O.5*ACA(II
VEB(I)=VEA(II+X
XBOO5*ACA(II‘BETA*ACA(II
DDPIII'VEAII)+X+BETA*ACB(II
DPSII)=DPA(I)+DDP(I)
DPBIN+2I=DPBINI

ASBBASI

VSB‘VSA+O¢5*(ASB+ASII
SYBBSYA+VSA+O.5*ASI+8ETA*(A58“ASII

***** COMPUTE ANGLE 0F ROTATION *****
DO 18 1829M
ANGL8(I)=-DPB(I-1)+2.0*DPB(I)-DP8(I+II
DA(I)=ANGLB(I)-ANGLA(I)

***** COMPUTE SENDING MOMENT *****
DO 79 1:29M

IFIDAIIII 50950965
IFITMNA‘I)+STFA*DA(III 51952955
CaRK

GO TO 53

CgOoO

TMNB‘II‘OOO
X=5TFA*DA(II+TMNA(II
BMBII188MAIII’TMNAIII+C*X

GO TO 56
TMNB‘IIzTMNAIII+STFA*DA(II
BMBIIIRBMAIII+STFA*DAIII
TMPB(I)=2.o-TMNB(I)

GO TO 79

IFITMPAIII“STFA*DA(III 70971975
C=RK

GO TO 73

C3000

TMPBIII‘OOO
XBSTFA*DA(I)‘TMPAIII
BMBIII'BMAIII+TMPA(II+C*X

GO TO 76

0(3

107

75 TMP8(II'TMPA(I)“STFA*DAIII
BMBIII'BMA(I)+STFA*DA(II

76 TMNB(1)'2.O-TMPB(II

79 CONTINUE

iiiii AUXILIARY ROUTINE *****

DO 23 I=29N

X82N**2*(BMB(I-I)-2.O*8MB(II+8MB‘I+1)I
23 ACDII)=DELTA*38.4*DELTA*X

IF(N~KT)31093229322
322 IF(KT‘1) 30493049305
304 X81.O/(1.0+R*ZBAR**2)

YM1=ZBAR*DPB(KT+II

VMBVBAR*DPB(KT+1)*DELTA+ZBAR*VEBIKT+1I

CH=X*DELTA*(SK*(SYB-YM1I*ZBAR*DELTA+DC*(VSB-VMI*ZBARI

ACD(KT+1)8X*DELTA*(38.4*ZN*(ZN*(BM8(KT)‘2.0*8M8(KT+1)+BMBIKT+2)I+

18.0*FBAR*ZBAR)*DELTA-Z.0*R*VBAR*ZBAR*VEB(KT+111+CH

GO TO 310
305 IF(KT~N) 30693079308
308 PRINT 309
309 FORMAT(1H29IOX912HKT8TOO LARGE)
307 REMD=1.0-ZBAR

X81.0/(1.0+R*REMD**2)

Y=BMBIKT-1)-2.0*BMB(KT)+8MB(KT+1I

YM1=REMD*DPB(KTI

VM=REMD*VEB(KT)‘VBAR*DP8(KT)*DELTA

CH2X*DELTA*(SK*(SYB~YM11*REMO*DELTA+DC*(VSB-VMI{REMDI

ACD(KT)aX*DELTA*(38.4*ZN*DELTA*(ZN*Y+8.0*FBAR*REMDI-2.0*R*VBAR*

1REMD*(VEB(KT+1)“VEB(KT)))+CH

GO TO 310
306 REMD=I.O-ZBAR

S‘I.0/R+REMD“ZBAR*REMD

T8R*(ZBAR*REMD)**2

U8R*ZBAR**2

V81.O/(1.O-T/S+U)

U=8MB¢KT-1)-2.0*BMB(KT)+BMB(KT+1)

XIZN**2*ZBAR*REMD*W

YBZN*ZBAR*REMD**2*FBAR

ZIBMB(KT)-2.O*BMB(KT+II+8MBIKT+21

YE=2.0*R*VBAR*ZBAR*(REMD**2/S-1.0)*(VEB(KT+1)“VEBIKTII

YM1=DPBIKT)+ZBAR*(DP8(KT+1I-DPBIKT)’

VM2VEBIKT)+VBAR*DELTA*(DP8(KT+1)-DP8(KT))+ZBAR*(VE8(KT+1I-VEBCKT)1

XIX=1.-(REMD**2)/S

CH8DELTA*(SK*(SYB-YM1)*ZBAR*DELTA+DC*(VSB-VM1*ZBAR)

ACDIKT+1)=V*DELTA*I38.4*ZN**2*Z+8.0*38.4*ZN*F8AR*ZBAR~3894*X/S-890

108

1*38.4*Y/SI*DELTA+V*DELTA*YE+V*XIX*CH
TIM:1.0/(1.0+R*REMD-R*ZBAR*REMD)
CH=DELTA*(SK*(SYB-YMI1*REMD5DELTA+DC*(VSB’VM)*REMDI
ACD(KTI=TIM*(DELTA*(38.4*ZN**2*U+8.O*3894*ZN*F8AR*REMDI*DELTA
l-R*ZBAR*REMD*ACD(KT+1I‘2.0*DELTA*R*VBAR*REMD*(VEBIKT+I)‘VEBIKTIII
2+TIM*CH
310 DO 35 1:29N
X30.5*(ACD(II+ACA(I)I
35 VE8(I)8VEA(II+X
VSB=VSA+O.5*(ASD+ASII
GO TO LIL
290 ADAC=0.
AUTSOQ
YOJ=OO
GO TO 289
291 AUTz-DELTA*R*V8AR*(DPBIKT+21'2.*DP8(KT+1I+DP81KTIIIIIQ+RI
ADAC=‘2.*R*V8AR*(VE8(KT+21‘20*VE8(KT+1,+VEB(KT)-AUTI*DELTA/II.*R)
YOJ=(2.*V8AR)/(1.+R)*(VEB(KT+2)“29*VE8(KT+II+VEBIKTI’AUTI*DELTA
IF(N-KT)42594269426
425 GO TO 289
426 21830.
289 LOT=KT+I
IFIN’LOT) 28892879287
287 ACD(KT+1)=ACD(KT+II+ADAC
VEBIKT+II=VEB(KT+I)+AUT

288 IFIN-KT14O92549254
254 AMM=ACDIKT)/DELTA**2+2.*V8AR*(VEBIKT+1I‘AUT-VEBIKTII/DELTA
1+(ZBAR/DELTA)*(ACD(KT+1)-ADAC-ACD(KT)I/DELTA
TYOJ=YOJ/DELTA**2
AMMaAMM+TYOJ
PBAR=(R*(I(QS+R)/R)*G8AR-AMM+SK*(SYB-YM1)/R)/ZN)/153.6
40 ASD=-DELTA*(SK*(SYB-YM1)*DELTA+DC*(VSB-VMII/OS

***** PREPARE FOR NEXT STEP OF INTEGRATION *****
DO 45 1:29N
BMAIII=BMB(II
TMPA(II=TMP8(I)
TMNAIIISTMNB(I)
ACA(II=ACD(I)
VEAIIISVEBII)
DPAIII=DPBIII
ANGLA(II=ANGL8(I)
45 CONTINUE

C

109

ASISASD
VSASVSB
SYA=SYB

C ***** COMPUTE MAXIMUM RESPONSE {iii}

1101

109

261

262
263

264
285

283
282
281
280

48
16

97

98

266

J=N

K82

SIK=ZN/VBAR+3.

IFITAU‘SIK) 110192859285

DO 109 I=K9J
DPS(I131.25*DPA(II/3.14I592654**2
SOFII)‘DPS(II

EF(I)=DPS(II
IF‘DPSII)-DPAMX(II) 10991099112
DPAMXCII=DPS(II
TMAXIII'TAU*V8AR/ZN

CONTINUE
IFIPBMA-PBARI26192629262
PBMA=PBAR

TPMA=TAU*V8AR/ZN
IFIPBMI‘PBAR126492649263
P8MI=PBAR

TPMI=TAU*V8AR/ZN

GO TO 48

DO 280 I=K9J
DPSII)=1.25*DPA(II/3.I41592654**2
IFISOFIII‘DPSIII) 28392829282
SOFIII'DPSII)

IFIEFCII‘DPS(III 28092809281
EF(I)=DPS(II

CONTINUE

***** TEST TIME TO PRINT *****
IF(IN-1) 1159115916
IN=IN-1

GO TO 4

TTAU=TAU*VBAR/ZN

PRINT 979TTAU
F0RMAT€1HO9F12.7)

PRINT 989(DPS(I)9I=K9JI
FORMATI1H099FIO.5)
IF(TAU-ZN/VBAR)26692669267
SHO=1.25*AMM/3.141592654**2
COV=(SK*(SYB-YM1I/ZN1/153.6

260
267

323
398
397

399
400

401

802

803

804

570

324

110

PRINT 2609P8AR9SH09COV
FORMAT(IHO95HPBAR=F10.692X9F109692X9F10.6)
21.830.

***** STATIC DEFLECTION *****
IF(N-KT) 32493239323
ABAR=IZBAR+TNLE)/ZN

IF(1.0-A8AR) 39893999399

PRINT 397
FORMATI1H2920HABAR=LARGER THAN ONE)
IFITMEAS-ABAR) 40094019401
TMEAS=TMEAS+TLENG

IOT=IOT+1

GO TO 401

BBAR=1.0-ABAR

DIST=O.

REMD= O 0

DO 802 I=2910T

DIST=DIST+TLENG
DIF(I)312.8*88AR*DISTiIABAR’IBBAR+I.OI’DIST**2I
TIT=IOT

DIST=TIT*TLENG-TLENG

JJ=IOT+1

DO 803 I=JJ9N

DIST=DIST+TLENG

REMD=190~DIST
DIF(I1812.8*ABAR*REMD*(BBAR*(ABAR+1.OI-REMD**2)
DO 804 I=Z9N
STDE(I)=1.25*FBAR*DIF(I)

PRINT 989(STDEIII9ItK9J)

DO 570 1829N
THPLII)=ANGLA(I)-8MA(I)/STFA
THPL(I)=THPL(I)/3.141592654**2
PRINT 989(THPL(1)91829NI

GO TO 7

END

END

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