>.
IN MEMORIAM
FLOR1AN CAJORI
&*-,•
THE ELEMENTS
GRAPHICAL STATICS
AND THEIR APPLICATION TO
FRAMED STRUCTURES,
WITH NUMEROUS PRACTICAL EXAMPLES OF
A
CRANES— BRIDGE, ROOF AND SUSPENSION TRUSSES— BRACED
AND STONE ARCHES— PIVOT AND DRAW SPANS-
CONTINUOUS GIRDERS, &c.,
TOGETHER WITH THE BEST
METHODS OF CALCULATION,
AND CONTAINING ALSO
NEW AND PRACTICAL FORMULAE
FOR THE
Pivot or Draw Span— Braced Arch— Continuous Girder, Etc,
A. JAY r>U BQI^r £,£.;, p Ph.D.
PROFESSOR OF CIVIL AKP. iI1ECHA.'SITrA<iJ ENGrV-SFHiWC., LETI5GH' UNIVERSITY, PENNA.
"Peu a peu les connaissances aleg6briques d6viendront moins indispensables, et la science, re-
duite a ce qu'elle doit etre, a ce qu'elle devrait etre deja, sera ainsi mise a la port6e de cette classe
d'hommes, qui n'a que des moments fort rares a y consacrer." [Poncelet-Traite des proprietea
projectives des figures, Paris, 1828.]
With an Atlas of 32 Plates.
NEW YOKK :
JOHN WILEY AND SON,
15 ASTOR PLACE.
1875.
COPYRIGHT, 1875, BY
JOHN WILEY & SON,
JOHN F. TROW & SON,
PRINTERS AND STEREOTYPERS,
205-213 Rast izth St.,
MEW YORK.
ELEMENTS OF GRAPHICAL STATICS.
NOTE TO STUDENTS AND TEACHERS.
TEACHERS examining the work with a view to introduction will at
once perceive that it was not intended to be read through in sequence
by any class, but that it is required to be used intelligently with
reference to the degree of preparation of their students, with a knowl-
edge of the relative importance and relation of the various subjects
treated, and with a view to the end desired. Though the logical
order of presentation requires a certain order in the development of
the subject as a whole, it by no means follows that this order should
be preserved by the student, or that he should even be acquainted
with the whole. In fact some of the subjects treated of are best
taken up by the student at a later period, when better prepared for
their comprehension, others are best omitted at least in the first
reading, and others again may even be omitted entirely as of minor
importance, and the student left to pursue them for himself, as taste
or the exigencies of practice may demand. In this latter respect, as
well as in the completeness with which the several topics are dis-
cussed, the work is intended to serve as a book of reference. As a
Text Book, it is designed to teach those students possessing the
knowledge of mathematics and mechanics usual to the senior classes
in our technical schools and colleges, how to find the conditions of
stability in every kind of structure of common occurrence • and this
not alone by graphical construction but also by calculation as well.
Structures of less common occurrence may or may not be then taken
up according to the ability of the class and the time at disposal.
With the above end in view, the teacher will find it not merely
desirable but even essential to depart from the plan of the work as
laid down, and to supplement it largely by various examples illustra-
tive of the general principles and bringing out as clearly and
repeatedly as may be necessary the various points noticed in the
text,
In the work four separate and distinct methods of solution are
given for such structures as bridge and roof trusses, and the student
XIV ELEMENTS OF GRAPHICAL STATICS.
should become familiar with all. Thus there are two methods by dia-
gram, viz., by resolution of forces (Maxwell) and by the equilibrium
polygon ( Culmann) ; and two corresponding methods by calculation,,
viz., by composition and resolution of forces, and by moments (Hitter).
To give these in such manner and with such emphasis that the student
shall be conversant with all, and able to use them with discrimina-
tion where in any case they best apply, we recommend the following
order of perusal :
FIRST. Chap. I, which gives the first method by diagram (Max-
welPs)y and such of the Appendix as relates to this chapter, is read.
The class may then go into the drawing room, and under the super-
vision of the teacher actually solve a variety of roof trusses from
simple to most complex, both for dead load and ivind force in each
case. In each and every case also the results should be checked by
calculation by the method of moments (.Ritter's), at first thoroughly
and in detail, and afterwards only a few test pieces to check the
accuracy of the diagram. From Roof Trusses we then pass on ta
Bridges, and here also a series of selected examples of every class
used in practice is solved, and the method of tabulation of apex weights-
referred to in Art. 12 and Appendix to Chap I, brought out repeat-
edly until the student has thoroughly mastered it, and appreciates
fully the fact that for each form of truss the strains due to only two-
weights are really necessary to be found, and that the others may
then readily be found directly from these. Here also each example
should be checked by calculation by the method of moments. In
the case of curved flanges the various lever arms may first be
measured directly to scale from the frame, and then trigonometrically
computed. At this point the student is then already in possession 01*
two independent methods of solution for any kind of framed roof or
bridge which occurs in general practice.
There are in fact only two framed structures, the continuous
girder and the braced arch remaining, which he is not able to solve.
Should it be deemed undesirable to consider these, he can at once
pass to Chap. II and the stone arch. If, however, a knowledge of
these is desired, he is now ready to extend his principles and methods
to them also. Thus he has already recognized that provided only all
the outer forces are known he can both diagram and calculate any
framed structure. In such structures as he has hitherto had, these
outer forces are either given or are easily found. In the cases now
considered they are not all given, and must, therefore, first be found.
Once known, however, his way is clear. Kecognizing clearly, then,
what is aimed at, the supplements to Chaps. VII and XIII are first
ELEMENTS OF GRAPHICAL STATICS. XV
taken and he is now able to find for the continuous girder, the outer
forces required. Chap. XIII will then give exercises in finding these
forces, and handling the formula he has just deduced. Finally, Chap.
XII resumes again in the light of his present knowledge, the same
old two methods with which he is already so familiar, of diagram
and calculation, and a few examples actually worked out by both
methods complete his mastery of the continuous girder and draw
span. He can now pass on to the braced arch, and in Chap. XIV
will find all that he needs. Here he must take at first the formulae
and constructions for finding the outer forces, on trust. Afterwards,
if deemed desirable, he can follow out the development of these
formulae as given in supplement to Chap. XI Y.
In the case of the parabolic arch, at least, the constructions are so
simple that it is well adapted to .class instruction. The draw span is
of such importance as to render some atttention to it, at any rate,
desirable in any full course. Thus the student is now able to solve
any case whatever of framed structure in two ways, by diagram and
calculation. The same simple principles have been applied throughout,
and formulae have been called in only in a subsidiary way to deter-
mine certain forces which are necessary to be first known before
these principles can be applied.
Thus, Chap. I, appendix to Chap. I, supplements to Chaps.
VII and XIII, then Chaps. XIII, XII, with appendix, and
XI Y with appendix, form by themselves and in the order a
complete and systematic course.
If, however, it is deemed undesirable to consider the continuous
girder and braced arch, the course indicated in the preceding para-
graph may be omitted, and then Chap. I, with appendix, constitutes
a course as thorough as can be desired, the student taking first those
examples given in the book, and then such others as the teacher
may select, and always solving in two ways, by diagram and calcula-
tion. He may then take Chaps. II- Y, and now possesses a second
method of diagram (Culmann's) by which he may check any or all of
the previous cases. In general, a few examples of application to
bridge girders, drawing the parabolas for total load (moments) and
moving load (shear) will be sufficient. He is now ready to pass on
to the Stone Arch, where again suitable problems should be proposed
by the teacher and solved under his supervision. The remainder of
the work, including moment of inertia and continuous girder treated
by the second method of diagram, will in general be found unneces-
sary and rather advanced, except in a very full course.
Xvi ELEMENTS OF GRAPHICAL STATICS.
The course recommended is then as follows, in the order given,
chaps, in brackets being omitted or not at option of teacher :
Chap. I and appendix, supplements to Chaps. YII (and XIII,
Chaps. XIII, XII and appendix, XI V and appendix), Chaps.
II-Y, XY (XVI, VI, XI). It will be seen that the order in-
tended for class instruction is quite different from that of the
work itself.
We would ask teachers examining the book with a view to adop-
tion, to look it over in the order above given.
PREFACE.
IT is now ten years since the appearance of the Graphical
Statics of Culmann,* during which time the method has been
greatly extended in its applications, and has met with such
acceptance that there is now scarcely a Polytechnikum in Ger-
many where it is not a prominent feature in the regular course
of instruction.
This rapid spread of a new discipline is the more remarkable
when we consider the obstacles which it encountered. Cul-
mann, with a boldness which we might almost term rash, based
his development upon the modern geometry of Yon Staudt, and
assumed in his re iers a familiarity with this very terse^)resen-
tation of a subject then, as indeed now, but little known, and
which, therefore, but few possessed. To practical engineers,
therefore, to whom his methods specially recommended them-
selves, his presentation of those methods was almost unintelli-
gible.
At a time when the students of the Zurich Polytechnic were
already overburdened, the new discipline was introduced ; while,
owing to want of familiarity with the fundamental principles
premised, they were unable to understand his lectures or read
his work. Yet such was the intrinsic value of the new method
that, notwithstanding these obstacles, even in spite of them, it
made rapid headway ; found friends everywhere ; crept into
other departments of the Polytechnic ; and finally the aim of
Culmann was completely attained when the modern geometry
was itself introduced, and a special lecturer in that branch ap-
pointed. Thus, as a direct result of the Graphical Statics of
Culmann, appeared the first and, till now, only complete text-
book upon the modern geometry, viz., Reye's " Geometric der
Lage" Hannover, 1868. Since then, hand in hand and with
remarkable rapidity, these two studies have made their way,
* Die Graphische Statik. Culmann. Zurich, 1866. Second Edition, 1st
vol., 1875.
v PEEFACE.
until, as already remarked, they now form a notable feature in
the course of every technical institution in the land.
The acceptance which the method has found in France, and
the attention which it has there excited, is sufficiently indicated
by the work of Levy (La Statique Graphique et ses Applica-
tions, Paris, 1874), which contains a very clear and elegant
presentation of the principles, though the applications are of
the simplest character, while, as was perhaps not unnatural in
the author, the German origin of the system is very imper-
fectly indicated, and the special methods of Culmann but little
more than hinted at.
In Italy also the method has found an ardent expounder in
the distinguished mathematician Cremona (Le figure recip-
roche nelle statica grafica, Milan, 1872), and to his efforts and
labors its introduction and acceptance is due.
In England, Prof. Clerk Maxwell, in the Trans, of the Royal
Society of Edinburgh, 1869-70, has contributed a paper upon
i% Reciprocal Figures, Frames and Diagrams of Forces/' and,
among others, Jenkin, JRanken, Bow, and Unwin have contrib-
uted to the popularity and spread of " Maxwell's Method."
Maxwell and his followers give, however, only the very simplest
applications, based upon the resolution and composition of
forces, such as will be found in our first chapter. The entire
system developed by Culmann, the properties of the " equilib-
rium polygon," upon which the fruitfulness and value of the
graphical statics wholly depend, are unnoticed both by our
English and French authors.
The author feels, therefore, that no apologies are needed for
the present work. Whatever its shortcomings and defects, he
claims at least the honor of making the first attempt to intro-
duce among American Colleges and American Engineers a
knowledge of a subject of approved interest and practical value
to both, whether regarded as a geometrical discipline or as a
most efficient aid in investigations of stability. Nor is he with-
out hope that the next ten years may find the method as uni-
versally accepted at home as now abroad.
The same difficulties certainly have not here to be encoun-
tered. The subject as here presented requires only a knowl-
edge of the elements of geometry as universally taught, and
can thus be readily introduced into our schools as well as read
by those practical engineers for whose benefit the method
PREFACE. V
seems so especially designed. A subject of such importance,
which has already endured successfully so severe a test, and
made headway against such obstacles, we cannot certainly af-
ford any longer to ignore, and it is hoped that the present
work may serve to excite a more general interest in the method.
For the practical engineer, the importance of graphical
methods needs, indeed, to-day no demonstration. Such methods
are everywhere in use. But a simple and general system
which shall include all special solutions — the development of
the few principles upon which all such solutions are based, and
from which they all flow — is at least in this country unknown.
Even in English literature there is to be found little more than
the very elementary deductions of our first chapter, so that it
may justly be said that the entire method owes its- existence
and development to the labors of German scholars and the en-
lightened appreciation of German engineers. How thorough
have been these labors, how widespread this appreciation, and
how various are the applications of the method itself, the reader
may gather from the Introduction to this work, and from the
appended list of literature upon the subject. A glance at this
list will also show that the selection of what was of most value,
and the omission of those applications of minor importance,
necessary to bring the present work within reasonable limits,
and at the same time preserve the logical unity and complete-
ness of the whole, was not the least difficult portion of our
task. It would, indeed, have been easy to have given the work
twice its present dimensions, though without a corresponding
increase in value sufficient to justify the additional cost. As
it is, no application of real and practical value to the engineer
strictly deducible from the graphical statics has been over-
looked, and discrimination has been chiefly exercised in those
departments where graphical and analytical processes are still
of necessity combined. Here we have selected only those cases
where such union shows itself most advantageous, and the
graphical constructions most simplify, illustrate, or interpret
the purely analytical process, and where such cases, moreover,
presented a useful, practical, and not merely theoretical value.
As to the plan of the work, a word of explanation is neces-
sary. We have endeavored to keep always in view the re-
quirements, of both students and practitioners, of technical
schools and practical engineers, and thus to combine a text-
VI PREFACE.
book for school instruction, and a book of reference and manual
for practice as well. The attempt is a difficult, if not a dan-
gerous one, and one which, in other departments, has met with
more failure than success. If we venture to indulge a hope that
in this case at least partial success has been attained, and that
the attempt to occupy the two stools at once has not been dis-
astrous, our belief is due to the nature of the subject itself,
and not to any overweening estimate of our own abilities to
succeed where so many have failed. The subject seems, indeed,
especially suited to such a method of treatment. In fact, no
other would appear at this period to properly meet the necessi-
ties of the case. Its geometrical principles are simple, its ap-
plications eminently practical. To present the principles alone
would be to deprive the- study of its chief interest and attrac-
tion. To rest content with a few practical applications would
be to sacrifice, in a great measure, system and clearness of pre-
sentation. In the accomplishment of our double task we are
fortunate to have had at our disposal such works as those of
Bauschinger in the one, and Culmann in the other direction.
Our obligations to both authors are great, and are fully indi-
cated in the text. The same acknowledgment is due, in greater
or less degree, to Mohr and Winkler, Hitter and Reuleaux. In
every case where such assistance has been received, due ac-
knowledgment has been made.
For the historical and critical Introduction, we are indebted,
with few alterations, to the pen of Weyrauch* It will, we are
sure, prove of value to the student, and serve to awaken an in-
terest in those highly important developments which geometry
has within the last decade undergone.
Thus collecting in a connected form the scattered results and
researches of various authors, it has been a pleasurable duty to
recognize the labors of those men who have chiefly contributed
to' this new branch of geometrical statics, and to whom our own
obligations are so great. While thus crediting fully that which
others have done, we have felt the more justified in calling at-
tention to any deviations of our own. We have especially
sought to extend the application of the method by resolution of
'forces (known best, perhaps, as MaxweWs Method ) — a method
* Veber die gra/phische Statik — zur Orientirung. Yon Dr. phil. Jacob J.
Weyrauch, Privat docent an der polytechnischen schule zu Stuttgart. Leipzig,
1874.
PREFACE. Vll
which bids fair to obtain widespread recognition, in direc-
tions in which it has hitherto been supposed of little service.
This often, indeed, by the aid of analytical results, we have
been enabled to do, and not, as we conceive, without a degree
of success. The formulae used are always simple and of ready
application, and this union of analytical results and graphical
processes the practical engineer will, we think, find of value.
Thus, in the braced arch (Chap. XIY.) and continuous girder
(Chap. XII.) new constructions will be found, and both these
important and difficult cases may thus be solved with an ease,
completeness and accuracy far : superior to that of the pure
graphical method itself. Those acquainted with the analytical
investigations of the " braced arch" as contained in Oapt. Eads>
Report to the III. and St. Louis Bridge Co., May, 1868 (App.),
will not, we feel sure, be slow to recognize the advantages of
the present method. The subject in its present state is thus
fairly brought within the reach of the practical Engineer and
Constructor.
To simple girders, contrary to usually received opinions, by
the means of apex loads, the above method applies directly, and
without the aid of analytical results — a fact which has been too
generally passed over without sufficient notice by. writers upon
the subject.
We have devoted considerable space to the subject of the
continuous girder, but not, we feel sure, more than its impor-
tance demands. The subject deserves more attention at the
hands of the practical engineer and constructor than it has
hitherto received. That the present indifference upon the sub-
ject is due chiefly to lack of information can hardly be doubted,
when the opinion is current, and is even endorsed by those who
are considered as authorities, that the complete solution of the
/problem is " probably impossible by reason of its complexity,"
and " too complex for mathematical investigation." * Opin-
ions like these are best met by the complete solutions of par-
ticular examples, and in Chapter XII. will be found the com-
plete calculation and tabulation of the strains in every piece
due to every apex load, for the central span of seven continu-
ous successive spans, and, as far as any inherent difficulties are
concerned, we might as well have taken 50 or 100 spans.
* OrapJiiccd Method for the Analysis of Bridge Trusses. Greene.
Vlll PREFACE,
When engineers shall have become convinced of the fact that
there is in the continuous girder a saving of material amount-
ing usually to from 25 to 30 per cent, per truss, and in the ex-
treme case even reaching as high as 50 per cent., as compared
with the si^rle girder ; and that the only objection which can
be urged — viz., the influence of small variations in level of the
supports — has, when properly considered, no force whatever,
we shall probably hear less often of designs contemplating many
successive and independent spans of considerable length — such
as, for instance, for a bridge over the Hudson at Poughkeep-
sie, consisting of five separate spans of 525 ft. each. Such a
design would find little favor in France or Germany, where
continuous girders are more favorably considered, possibly be-
cause the ability to calculate them is less rare, and reflects in
this respect little credit upon the American profession. About
the facts in the case there can now be no dispute ; the subject
has been too thoroughly investigated to admit of it, and we
refer the reader to the Appendix for the results. The mathe-
matician and theoretical engineer have done their part ; it
remains for the practical engineer and constructor to do theirs.
The present work contains the only complete graphical and
analytical presentation of this subject in English professional
literature, and should it succeed in causing a change of view
in the above respect alone, will not have been in vain. In this
connection the list of literature upon the continuous girder
appended to Chap. XIII. may also be of service.
We notice with pleasure in this direction the admirable little
treatise of Clemens Herschel, C.E., upon draw spans.* This
subject is at least of admitted practical value, and we have
treated it with a fullness which, in our opinion, leaves little to
be desired. We have borrowed from the above work the con-
ception of the " Tipper" or draw with secondary span, which
is both new and, as it would seem, most adequately represents
the true state of the case, and alluded to the idea, also original
with Mr. Herschel, of weighing off the reactions at the supports
of a continuous girder, instead of measuring the differences of
level. In this case, as in that of the continuous girder gene-
rally, we have clearly brought out the method of calculation ~by
* Continuous, Revolving Drawbridges. Little, Brown and Company, Bos-
ton, 1875.
PREFACE. IX
apex weights, and here, indeed, lies the whole secret of thorough
practical solution. In fact, from this point of view* the com-
plete solution of a continuous girder for any number of spans,
equal or unequal, offers no more essential difficulty than the
calculation of so many separate simple girders. That this is
not exaggeration, but accurate statement of fact, a perusal of
Chaps. XII. and XIII. will suffice to prove.
We cannot leave this part of the subject without acknowledg-
ing our indebtedness to Mansfield Merriman, C.E., Assistant in
Engineering in the Sheffield Scientific School of Yale College,
for the formulae of the latter chapter. Mr. Merriman has
done for the practical solution of the continuous girder what
Weyrauch has for its theoretical discussion. We refer the
student to the Supplement to Chap. XIII. for a specimen of
his method of discussion.
By the proper use of " indeterminate multipliers," the whole
analytical discussion is most remarkably simplified. The only
one of the many writers upon the subject known to us who
seems to have hit upon this treatment is WinMer (Die Lehre
von der Elasticitdt und festigJceit). In Art. 144 of the above
work he gives formulae similar to Mr. Merriman's for the
moments at the supports of a continuous girder for all spans
equal only. He seems, however, to have failed to realize the
true significance of the method, or the important part played
by the Clapeyronian numbers. Independently of Winkler,
Mr. Merriman has reproduced these formulae in their true
light, and applied the method to any lengths and number of
spans, with any differences of level and any method of loading.
His formulae are simple, entirely free, even in general form,
from integrals, and are given in just the shape required in
practice. This compactness renders it possible for the engineer
to enter upon a couple of pages of his note-book all the for-
mulae required for the thorough calculation of a continuous
girder of any number of spans, equal or unequal ; and this cal-
culation in any particular case proceeds iu a manner precisely
similar to that of the simple girder, directly and without refer-
ence to authorities, tables, points of inflection, elastic line,
methods of loading, or any of the " other paraphernalia with
which the subject is usually encumbered."
It will be observed that here and throughout we have no-
where left out of sight analytical processes or methods. The
X PREFACE.
reader who considers the present work as an attempt to super-
cede^ or even subordinate analytical investigation, misjudges
entirely our aim. So far from this, we indulge the hope that
its perusal cannot fail to render familiar the use of both
methods, to bring out their points of difference and relative
advantages, to illustrate the one by the other, to enable the
reader to check the results of the one by the other, and in any
case apply one or both, or a judicious combination of both, as
may in such case be most advantageous or desirable. This will
be especially noticed in the discussion of the simple and con-
tinuous girder and of the braced arch. (Chaps. XII., XIII.,
XIY. and XVL, and Appendix.)
As to the use of the work, the practical engineer will find in
Chap. I., and that portion of the Appendix relating to this
chapter, an easy and simple method of solution applicable to
any framed structure having simple reactions, and including
thus all varieties of bridge and roof trusses of single span. In
the Appendix he will find detailed examples calculated to illus-
trate every practical point of importance, and also a full expo-
sition of Hitters "method of moments" The principles of
this chapter alone will enable him to solve readily, both by cal-
culation and diagram, every case usually arising in practice.
In problems involving the moment of inertia of areas, in the
case of the continuous girder, the braced arch and stone arch,
as also the suspension system, he will find Chaps. VI., XII.,
XIII., XIV. and XVI. of value; and in the perusal of any or
all of these he will, it is hoped, find no trouble by reason of
logical connection with preceding principles. They are in this
respect, as far as possible, complete in themselves. We may
also call his attention to Chap. XV._, upon the stone arch,
though it is to be regretted that the practical importance of the
subject, in the present age of iron, renders the ease with which
it is graphically treated of less importance than formerly. For
his benefit also frequent practical examples are given in detail,
so' that in all important applications he can easily select a
parallel case, and follow it out, step by step, in the case in
hand, without studying up the whole process of development
in order to place himself in a condition to make use of the
methods employed. We would also refer him to THE NEW
METHOD or GRAPHICAL STATICS (Van Nostrand, 1875) — a reprint
of a series of articles contributed by the author to Van Nos-
PREFACE. XI
trand's Engineering Magazine during the present year, where
lie will find such a condensed presentation of the more essen-
tial principles of the subject as will enable him to follow the
practical examples in the present work without the perusal of
the more lengthy preparatory portion here given.
For the student much of the practical applications may well
l>e at first omitted. Notably Chaps. VII.-XIL, inclusive.
Chaps. I.-IY. and XIII.-XVL will put him in complete pos-
session of the method, and, moreover, enable him to solve with
ease any structure, including the continuous girder, braced
arch, suspension system, and stone arch, as well as all the more
ordinary forms of bridge and roof trusses, cranes, etc. Indeed,
if the first-named structures, which are of comparatively rare
occurrence, are at first omitted, Chaps. I.-IY. alone will con-
stitute a complete course upon framed structures so far as
usually taught in our schools at the present day. Afterwards,
in practice, and in the solution of the particular problems
treated of, he will, in common with the practical engineer, find
in the other portions of the work and in the Appendix just
such assistance as he needs. We would also call the attention
of the mathematician more especially to the investigation in
Chap. V., Arts. 47-51, of the effects of a given recurring system
of moving loads., the analytical treatment of which would be
almost impracticable by reason of the complexity of the for-
mulas obtained, and in this respect certainly worthless, even if
possible, but the geometrical treatment of which gives rise to
some of the most elegant constructions of the graphical statics;
also to the Supplements to Chaps. /£IIL and XIV., in which
the analytical treatment of the continuous girder and braced
arch is given.
Finally, if our purpose in writing these pages is accom-
plished, the principles and methods here set forth will be found
easily acquired, accurate in their results, and amply sufficient
for the ready determination of the strains in the various pieces
of any framed structure which the civil engineer can legiti-
mately be called upon to design.
With this much of introduction and explanation, we present
our work to the engineering profession in America and to
American technical colleges, in the hope that the spirit which
has led to its production, if not the method of its execution,
may win for it a favorable reception.
Xll PREFACE.
In this spirit and in this hope we may, we trust, be allowed
to appropriate the closing lines of Culmanrfs preface — " Und
nunftlhre hin — gem hdtte icJi dich zum Fundament einer auf
wissenschaftlicherer Basis gegrundeten IngenieurJciinde ge-
mac/itj allein Jcaum darf ich die Hcffnung hegen, so viel Kraft
in mir zu finden, um das Game dieses umfangreichen Fachen
umzuarbeiten : das ist ein Werk, das mir vor Augen sc/iwebt,
wie einer jener alien mittelaUerlicJien Dome sick vor dem
Kunstler erhob^ der ihn entwarf und der der Hoffnung sick
nicht hingeben konnte, ihnje in seiner Vollendung zu scJiauen.
" Dock es mogen dich A.ndere benutzen und weiter bauen"
nnd was ich nicht kann, werden nieine Nachganger voll-
bringen.
NEW HAVEN, April nth, 1875.
GElsTEEAL CONTENTS.
PAOB
INTRODUCTION xxv
PART I.— GENERAL PRINCIPLES.
CHAPTER I.
FORCES IN SAME PLANE — COMMON POINT OF APPLICATION. ILLUS-
TRATED BY PRACTICAL EXAMPLES 1
CHAPTER II.
FORCES IN SAME PLANE — DIFFERENT POINTS OF APPLICATION. PRO-
PERTIES OF FORCE AND EQUILIBRIUM POLYGON 16
CHAPTER III.
. CENTRE OF GRAVITY 29
CHAPTER IV.
"MOMENT OF ROTATION OF FORCES IN SAME PLANE IN GENERAL 33
CHAPTER V.
MOMENT OF ROTATION OF PARALLEL FORCES— PRACTICAL APPLICA-
TIONS 36
CHAPTER VI.
MOMENT OF INERTIA OF PARALLEL FORCES — PRACTICAL PROBLEMS. . 61
PART II.— APPLICATION TO BRIDGES.
A. THE SIMPLE GIRDER.
CHAPTER VII.
THE SIMPLE GIRDER OR TRUSS SUPPORTED ONLY AT ENDS 84
SUPPLEMENT TO CHAPTER VII.
THE THEORY OF FLEXURE.
CHAP. I. Methods of calculation 98
CHAP. II. Principles of the calculus 102
CHAP. III. Theory of flexure , . . 110
XIV GENERAL CONTENTS.
B. THE CONTINUOUS GIRDER OF CONSTANT
CROS&SECTION.
CHAPTER VIII.
PAGE
GENERAL PBINCIPLES 125
CHAPTER IX.
LOADED AND UNLOADED SPANS. 141
CHAPTER X.
SPECIAL CASES OF LOADING 147
CHAPTER XI.
METHODS OP LOADING CAUSING MAXIMUM STRAINS 155
CHAPTER XII.
COMBINATION OP GRAPHICAL AND ANALYTICAL METHODS 168
CHAPTER XIII.
ANALYTICAL FORMULAE 198
SUPPLEMENT TO CHAPTER ,XIIL
DEMONSTRATION OF ANALYTICAL FORMULAE. . . 239
PAET III.— APPLICATION TO THE AKCH.
CHAPTER XIV.
THE BRACED ARCH. COMBINATION OF GRAPHICAL AND ANALYTICAL
METHODS 251
SUPPLEMENT TO CHAPTER XIV.
DEMONSTRATION OF ANALYTICAL FORMULAE 271
CHAP. I. General considerations and formulae 271
CHAP. II. Hinged arch in general 277
CHAP. III. Arch hinged at abutments only 283
CHAP. IV. Arch fixed at ends— no hinges 287
CHAP. V. Influence of temperature 299
CHAP. VI. Partial uniform loading 304
CHAPTER XV.
THE STONE ARCH 31 1
CHAPTER XVI.
THE INVERTED ARCH— SUSPENSION SYSTEM. . 327
APPENDIX.. 339
TABLE OF CONTENTS.
INTRODUCTION.
ABT. PAGE
I. Upon mathematical investigations generally xxvi
II. Analytical and geometrical mechanics xxvii
III. Geometrical statics xxix
IV. The graphical calculus xxxi
V. Graphical representation xxxiv
VI. Graphical statics xxxv
VII. The methods and limits of the graphical statics xxxvii
VIII. The modern geometry xxxix
IX. The modern geometry in engineering practice xlii
X. Practical significance of the graphical statics xliii
XI. Literature upon the graphical statics xlv
XII. Graphical dynamics
PAET I.— GENERAL PRINCIPLES.
CHAPTER I.
FORCES IN SAME PLANE— COMMON POINT OF APPLICATION.
1. Notation — Representation of forces by lines 1
2. Resultant of two forces 2
3. Resultant of any number of forces 2
4. Conditions of equilibrium 3
5. Properties of force polygon 3
6. Order of forces in force polygon a matter of indifference 4
7. Forces acting in same straight line 5
8. PRACTICAL APPLICATIONS 5
9. Braced semi-arch 6
10. Roof truss 8
11 . Diagram for wind force 8
12. Application of method to bridges 11
13. Braced arch 12
14. Ritter's u Method of Sections " 14
CHAPTER II.
FORCES IN SAME PLANE— DIFFERENT POINTS OF APPLICATION.
16. Resultant of two forces 16
17. Case of forces parallel 18
XVI TABLE OF CONTENTS/
AKT. PAGE
18. Perpendiculars let fall upon components from intersection of outer
polygon sides, are inversely as the components 18
19. Equilibrium polygon 19
20. Case of a couple — Conditions of equilibrium 20
21. Properties of a couple 21
22. Force and equilibrium polygons for any number of forces 22
23. Influence of a couple 24
24. Order of forces in force polygon, a matter of indifference 25
25. Pole upon the closing line — Failing case for equilibrium.. 25
26. Relation between two equilibrium polygons with different poles .... 26
27. Mean polygon of equilibrium 27
28. Line of pressures in an arch 28
CHAPTER III.
CENTRE OF GRAVITY.
30. General method for determination of centre of gravity 29
31. Reduction of areas 30
32. Reduction of triangle to equivalent rectangle of given base 31
33. Reduction of trapezoid to equivalent rectangle 31
34. Reduction of quadrilateral generally 31
CHAPTER IT.
MOMENT OF ROTATION OF FORCES IN THE SAME PLANE
IN GENERAL.
35. Definition of moment of rotation 33
36. Culmann's principle 33
37. Application to equilibrium polygon 34
CHAPTER Y.
MOMENT OF RUPTURE— PARALLEL FORCES.
38. Equilibrium polygon — Ordinates to give the moments of rupture. . . 36
39. Beam with two equal and opposite forces beyond the supports 38
40. Beam with two equal and opposite forces between the supports 39
41. Special cases of importance 41
1st. Beam — load inclined to axis. ... 41
2d. Force parallel to axis 42
3d. Forces, in different planes ' 42
4th. Combined twisting and bending moments 43
5th. Application to crank and axle 44
42. Loading continuous — Load area 45
43. Beam uniformly loaded 46
44. Moment curve a parabola 46
45. Beam continuously loaded and also subjected to action of concen-
trated loads 48
46. Influence of moving load 50
47. Load systems 53
TABLE OF CONTENTS. XV11
ABT. PAGE
48. Properties of the parabola included by the closing line 54
49. Application of the above principles 54
50. Most unfavorable position of load system upon a span of given
length 56
51. Greatest moment at a given cross-section 59
CHAPTER VI.
MOMENT OF INERTIA OF PARALLEL FORCES.
52. Application of moment of inertia in proportioning any cross-section. 61
53. Graphical determination of moment of inertia 62
54. Signification of the area of the equilibrium polygon 63
55. Radius of gyration 63
56. Inertia curves— Ellipse and hyperbola of inertia 66
57. Construction of curve of inertia 69
58. Example 69
59. Central curve— Central ellipse 71
60. Centre of action of the statical moments considered as forces 73
61. Cases where the direction of the conjugate axes of the inertia curve
can be at once determined 75
62. Practical applications 76
1st. The parallelogram 76
2d. The triangle 77
3d. The trapezoid 78
4th. The parabolic segment 80
C3. Compound or irregular areas 81
64. Cases where there is no axis of symmetry 83
PAET II.— APPLICATION TO BRIDGES.
A. THE SIMPLE GIRDER.
CHAPTER VII.
THE SIMPLE GIRDER.
66. Forces which act upon a bridge 84
67. Bridge loading 85
68. Shearing force — Moment of rupture, etc 85
69. Concentrated loads — Invariable in position 87
70. Concentrated loads— Variable in position 88
71. Position of a given system of loads causing maximum shear. 88
72. Construction of the maximum shear 89
73. Maximum moments 90
74. Construction of maximum moments 91
75. Absolute maximum of moments 92
76. Continuous loading 93
77. Total uniform load 93
78. Method of loading causing maximum shear 94
79. Live and dead loads. . 95
XV111 TABLE OF CONTENTS.
SUPPLEMENT TO CHAPTER VII.
CHAPTER I.
METHODS OP CALCULATION.
ABT. PAGE
2. Hitter's method 99
3. Method by resolution of forces 100
CHAPTER II.
PRINCIPLES OP THE CALCULUS.
4. Differentiation and integration 102
5. Powers of a single variable 105
6. Other principles 106
7. Illustrations 107
a First differential coefficient 109
CHAPTER III.
THEORY OF FLEXURE.
9. Coefficient of elasticity 110
10. Moment of inertia 110
11. Change of shape of axis 112
12. Beam fixed at one end, load at other 113
13. Beam as above — Uniform load 117
14. Beam supported at both ends — Concentrated load 118
15. Beam as above — Uniform load 119
16. Beam fixed at one end, supported at other — Concentrated load 120
17. Beam as above — Uniform load 121
18. Beam fixed at both ends— Concentrated load 122
19. Inflection points, etc. 124
B. THE CONTINUOUS GIRDER OP CONSTANT
CROSS-SECTION.
CHAPTER VIII.
GENERAL PRINCIPLES.
80. Mohr's principle 125
81. Determination of tangents to the elastic curve 127
82. Effect of the moments at the supports 128
83. Division of the moment area 129
84. Properties of the equilibrium polygon 130
85. Polygon for the positive moment areas 131
86. Construction of the fixed points, and of the equilibrium polygon ... 132
87. Construction of the moments at the supports 135
88. The second equilibrium polygon 136
89. Determination of moments at the supports.. 138
90. Comparison with girder fixed horizontally at both ends 139
TABLE OF CONTENTS. XIX
CHAPTER IX.
LOADED AND UNLOADED SPANS.
ABT. PAGE
91. Unloaded span 141
92. Two successive unloaded spans , 141
93. The fixed points 142
94. Shearing- force — Reactions at the supports, and moments in the un-
loaded spans 143
95. Loaded spans 143
96. Two successive loaded spans 144
97. Arbitrary loading 145
CHAPTER X.
SPECIAL CASES OF LOADING.
98. Total uniform load 147
99. Practical examples — Girder of four spans 148
100. Partial uniform load 149
101. Concentrated load 153
CHAPTER XI.
METHODS OF LOADING CAUSING MAXIMUM STRAINS.
102. Maximum shearing force 155
103. Maximum moments 157
104. Determination of the maximum shearing forces 159
105. Determination of the maximum moments 161
106. Practical simplifications of the method 162
107. Approximate practical constructions 163
1st. Beam of two spans— Moments 163
Beam of two spans — Shearing forces 164
2d. Beam of three or more spans — Moments 164
Beam of three or more spans — Shearing forces 165
108. Method by resolution of forces — Draw spans 166
CHAPTER XII.
CONTINUOUS GIRDER CONTINUED— COMBINATION OF
GRAPHICAL AND ANALYTICAL METHODS.
111. Method of finding shearing forces, when inflection points are
known 168
1st. Loaded span 168
2d. Unloaded span 169
112. Determination of inflection points— Inflection verticals 170
113. Beam fixed horizontally at ends 171
114. Example 172
1 15. Counterbracing 174
XX TABLE OF CONTENTS.
ART. PAGR
116. Beam fixed at one end, supported at the other 175
117. Practical construction 178
118. Beam continuous over three level supports 178
119. Practical construction 170
120. The pivot draw with secondary central span 181
121. Supports not on a level — Reactions 182
122. Beam over four level supports 184
123. Practical construction 186
124. Pivot span— Example 187
125. Method of passing in construction, from one span to another 190
126. Method of procedure for any number of spans 192
127. Example — Central span of seven spans 193
128. Method of calculation by moments 196
CHAPTER XIII.
ANALYTICAL FORMULA FOR THE SOLUTION OF CON-
TINUOUS GIRDERS.
129. Introduction 198
130. Notation 200
131. Theorem of three moments 201
132. Example— Total uniform load— Moments 202
133. Triangle of moments 203
134. Total uniform load — All spans equal — Reactions 205
135. Triangle for reactions 205
136. Clapeyron's numbers 206
137. Uniform live load over any single span — Moments at supports 207
138. Triangle for moments — Loaded span 208
139. Unloaded spans — Moments 209
140. Practical rule and table for finding the above 210
141. Reactions at supports — Loaded span 212
142. Triangle for reactions 213
143. Reactions for unloaded spans — Tables for 214
144. Concentrated load in any span— Moments 216
145. Application of above formulae 216
1 46. Triangle and table for moments 218
1 47. Reactions at supports — Concentrated load , 219
148. Shear at supports 219
149. Recapitulation of above formulae 221
150. Continuous girder with variable end spans 224
151. Application of formulae for 225
152. Continuous girder with fastened ends 22t'>
153. Beam of single span — Fastened at both ends — Fastened at one end,
etc. — Examples 22f>
154. Tables for moments — End spans variable 229
155. Continuous girder — All spans different— General formulae — Exam-
ples 233
156. General method of calculation . . 237
TABLE OF CONTENTS. XXI
SUPPLEMENT TO CHAPTER XIII.
DEMONSTRATION OF ANALYTICAL FORMULA FOR THE
CONTINUOUS GIRDER.
IRT. PAGfi
1. Conditions of equilibrium 239
2. Equation of elastic line 240
3. Theorem of three moments 241
4. Determination of moments — Supports all on level £42
5. Uniform load w 244
0. Formulae for the " Tipper " [Art. 120] 244
7. Example of two span tipper 245
8. LITERATURE UPON THE CONTINUOUS GIRDER. . 247
PART III.— APPLICATION OF THE GRAPHICAL
METHOD TO THE ARCH.
CHAPTER XIV.
THE BRACED ARCH.
157. Different kinds of braced arch 251
158. Arch hinged at both crown and abutments 251
159. Hinged at abutments only— Continuous at crown 253
1st. Parabolic arch 254
2d. Circular arch, Tables for solution of 255
160. Arch fixed at abutments — Continuous at crown 257
1st. Parabolic arch 258
2d. Circular arch 260
162. General method of solution 262
163. Analytical formulae for horizontal thrust, and vertical reactions 264
164. Arches with solid web 266
165. Strains due to temperature, Formulae for 267
166. Effects of temperature 269
SUPPLEMENT TO CHAPTER XIV.
DEMONSTRATION OF ANALYTICAL FORMULA FOR THE
BRACED ARCH.
CHAPTER I.
GENERAL, CONSIDERATIONS AND FORMULAE.
1. Fundamental equations ; 271
2. Displacement of any point 275
CHAPTER II.
HINGED ARCH IN GENERAL.
3. Notation — The outer forces in general 277
4. Intersection line 278
5. Parabolic arch — Concentrated load 278
6. Circular arch — Concentrated load 280
7. Integrals used in above discussion 282
XX11 TABLE OF CONTENTS.
CHAPTER III.
ARCH HINGED AT ABUTMENTS ONLY.
A. Parabolic arc.
ART. PAGK
8. Horizontal thrust 283
9. Intersection curve 283
B. Circular arc.
10.- Horizontal thrust 283
11. Intersection curve 285
CHAPTER IV.
ARCH FIXED AT ENDS.
12. Introduction 287
13. Concentrated load — General formulae 287
A. Parabolic arc.
14. Determination of H, V and M0 289
15. Intersection curve 291
16. Direction curve 291
B. Circular arc.
17. Fundamental equations 291
18. Determination of H, V and M0 293
19. Intersection curve 297
20. Direction segments 297
20 (b). Transformation series 297
CHAPTER V.
INFLUENCE OF TEMPERATURE.
21. General considerations 299
22. Influence of temperature on the arch 299
23. Fundamental equations — General 300
24. Arch with three hinges 301
25. Arch hinged at ends 501
26. Arch without hinges 301
CHAPTER VI.
PARTIAL UNIFORM LOADING.
27. Notation 304
A. Arch hinged at crown and ends.
28. Vertical reaction 304
39. Horizontal thrust 305
B. Arch hinged at ends only.
30. Vertical reaction 305
31. Horizontal thrust— Parabolic arch 305
32. Horizontal thrust— Circular arch 306
O. Arch without hinges — continuous at crown.
33. Parabolic arch. Formulae for V, H and M 307
34. Circular arch. Formulae for V, H and M 308
TABLE OF CONTENTS. XX111
CHAPTER XY.
THE STONE ARCH.
ART. PAGB
167. Definitions, etc ;
168. Pressure line 311
169 . Sliding of the joints 311
170. Forces acting upon a cross-section — Neutral axis 312
171. Kernel of a cross-section 313
172. Position of kernel for different cross-sections 314
173. Proper position of resultant pressure 316
174. Pressure line — True pressure line 317
175. Support line 318
176. Deviation of support from pressure line 319
177. Dimensions of the arch 320
178. Construction of the pressure line 322
179. Practical example " 324
180. Proper depth of arch at crown 324
181. Increase of depth due to change of form 326
CHAPTER XYI.
THE INVERTED ARCH— SUSPENSION SYSTEM.
183. Methods of construction 327
184. Rear chains, and anchorages 327
185. Cable with auxiliary stiffening truss 329
186. Method of loading causing maximum strains 330
187. Example ,, 333
188. Analytical investigation of the forces acting upon the stiffening
ooo
truss 066
189. Concluding remarks 335
APPENDIX.
NOTE TO CHAP. VIII. OP THE INTRODUCTION— UPON THE MODERN
GEOMETRY.
NOTE TO CHAPTER 1 346
2. Bent crane 346
3. Character of strains in the pieces as indicated by the strain diagram, 347
4. Pieces in equilibrium — Points to be avoided in constructing the
strain diagram 347
5. Roof truss — Method of checking the accuracy of the diagram 348
6. The French roof truss — Solution apparently indeterminate — Method
of solution — Method of calculation by moments 348
7. Application to bridges — Bowstring girder — Method of tabulation
illustrated. 350
8. Strains in the flanges— Table 353
9. Method of calculation by moments 354
10. Girder with straight flanges — Howe or Murphy Whipple system of
. bracing 355
XXIV TABLE OF CONTENTS.
ART. ,PAGB
11. Lenticular girder, or system of Pauli 358
12. Remarks upon above system — The most economical system for long
spans 359
NOTE TO CHAPTER II.
14. Equilibrium polygon considered as a, frame 362
NOTE TO CHAPTER V., ART. 51.
15. Beam subjected to given force system — Maximum moment at any
cross-section 364
NOTE TO CHAPTER XII.
16. Pivot or draw span — Practical example — Solution by diagram — By
method of moments — By resolution and composition of forces .... 365
17. Relative economy of the continuous girder 374
18. Continuous girder — supports out of level 377
NOTE TO CHAPTER XIV. THE BRACED ARCH.
19. Practical example of the braced arch 382
20. Arch hinged at both abutments and crown 383
21. Arch hinged at abutments only 386
23. Temperature strains for above 391
26. Arch continuous at crown — Fixed at ends 394
27. Temperature strains for above 398
28. Advantage of arch with fixed ends for%long spans — Comparison with
the St. Louis arch. . 399
INTRODUCTION.
HISTORICAL AND CRITICAL.* I ^/t? I
X^ ' °
THE subject of Graphical Statics has, since the appearance of (/ulrnann's
work (Die grapliisclie Statik, Zurich, Meyer and Zeller, 18#t>), excited
considerable attention, but an accurate and just estimate of its methods
and practical value is still wanting. Thus there are some who oppose it;
others willingly accept it as an efficient and valuable aid in practical inves-
tigations of stability ; still others even profess to see in it a future rival of
Analytical Statics. This last somewhat remarkable claim seems apparently
justified by a passage in Culniann's preface, where it is asserted " that the
Graphical Statics will and must extend, as graphical methods find ever
wider acceptance — but in such case, however, its treatment will soon escape
the hands of the practitioner, and it will then be built up by the geometer
and mechanic to a symmetrical whole, which shall hold the same relation
to the new geometry that analytical mechanics does to the higher analysis."
These various and conflicting opinions find their supporters in technical
schools and among engineers throughout Germany.
In the consideration of the subject, we shall endeavor especially to give
an objective presentation, but shall also feel at liberty to present our own
opinions as well, and generally to venture such reflections as seem suited
to throw light upon the matter. For both reasons it will sometimes be
necessary to make apparent deviations, in order to point out the various
fields in which these new investigations take root, to define their limits, and
to decide in what directions and to what extent impulse and sustenance
for further development may exist. In such a manner only can we satis-
factorily ascertain how far the graphical statics may safely count upon
more than a passing recognition and brief existence.
We have therefore to ask of the reader who wishes to obtain a just and
accurate estimate of this new and, as we venture to think, highly important
subject, patience for the following general considerations.
* Ueber die graphische Statik— zur Orientirung. By J. I.Weyrauch. Leip-
zig, 1874.
XX Vl MATHEMATICAL INVESTIGATIONS. [iNTIiOD.
L
y
UPON MATHEMATICAL INVESTIGATIONS IN GENERAL.
Mathematical truths may be attained in two essentially different methods
— by synthesis or by analysis, by composition or by resolution. In synthe-
sis, we ascend from particular cases to general ones ; in analysis, we descend
from general cases to particulars. By synthesis we pass from the simplest
or admitted truths, by combination and comparison, to more complicated
phenomena. Analysis seeks to refer back such phenomena to their fun-
damental relations, or to deduce special properties from the general con-
ditions.
The analysis of a phenomena presupposes, then, an accurate comprehen-
sion of all its elements. So far as these last stand in relations of cause
and effect to the whole and its parts, or so far as such relations exist be-
tween the parts themselves, they may be expressed by equations. Thus
the operations which are necessary in analysis become independent of con-
crete phenomena, and are governed only by the laws of abstract quantities
as included by algebra in the widest sense of the word. Algebra, then, is
not analysis itself, but only its instrument, " instrument precieux et neces-
saire sans doute, parce qu'il assure et facilite notre marche,"mais qui ri>a par
lui meme aucune vertu propre ; qui ne dirige point I' * esprit, mais que V esprit
doit diriger comme tout autre instrument'1'' (Poinsot, Theorie nouvelle de la
rotation, pre"s a" T Acad., 1834). Ordinarily the higher branches of algebra,
with which numberless really analytical investigations are connected, are
designated as analysis. More properly, all investigations which rest upon
equations of condition may be termed analytical investigations.
Synthetic investigation rests mainly upon geometrical conceptions, and
attains to the knowledge of phenomena through concrete conditions, which
latter may be designated as space relations and processes. Hence the usual
division into analytical and geometrical methods, even in applied mathe-
matics. We have thus with equal appropriateness an analytical geometry
as also a geometrical analysis. When pure geometry (in distinction from
analytical) makes use of the symbols and operations of algebra, it is only
to express with corresponding generality and more concisely than in words
truths attained to by abstraction, and independent of the dimensions of the
auxiliary figure ; or so to formulate such truths that they may be applied
in analytical investigation. Accordingly, such use of algebraic formulae
has as little effect upon the synthetic process as from the above it would
seem essential to the analytic treatment. In either case, algebra is but the
instrument, the method lies back of and directs it.
If analytical formulae and operations are entirely excluded from the
more complicated geometrical investigations, we are at once restricted to
general laws of metrical relation. There remains only the faculty of
INTROD.] ANALYTICAL AND GEOMETRICAL MECHANICS. XXV 11
abstraction and graphical construction. The power of abstraction alone
suffices, indeed, to comprehend in full generality metrical relations in ele-
mentary geometry and its simplest applications, but fails when the relations
sought must be attained step by step by the application of a number of
principles, or in the auxiliary figure by a number of constructions. If,
indeed, we take the relation sought directly from the auxiliary figure
itself, and even if it were possible to take out the required distances with
absolute accuracy, still this result obtained would stand to the general law
desired only in the same relation that the result of a particular numerical
computation does to the more general algebraic formula.
Investigations by the aid of graphical figures can, however, make known
general relations of form and position, and have in this respect their special
advantage. So far also as by them metrical relations are sought, then, by
the exclusion of algebraic formulae, only the process of deduction — the
routine of construction — remains of general significance. Sciences, then,
which proceed in this manner, furnish indeed, with respect to metrical
relations, no general laws, but for the deduction of these relations do give
general methods. In this category we may place descriptive geometry and
the more recent graphical statics.
-it
ANALYTICAL AND GEOMETRICAL MECHANICS.
It is hardly necessary in these days to call attention to the advantages
of a geometrical treatment of mechanical problems. This, however, was
not always the case, and the most important developments of geometrical
mechanics belong to the present century. It is to Poinsot, Chasles, Afobius,
etc., that these developments are due.
By the Calculus of Newton and Leibnitz (1646-1714), and its subsequent
development, analysis became such a powerful instrument that the activity
of mathematicians was for a long time solely directed towards analytical
investigations. The power of analysis was in mechanics carried to its
highest point by Lagrange (1736-1813), in his Mechanique analytique. He
undertook the problem of reducing mechanics to a series of analytical
operations : " On ne trouvera point de figures dans cet outrage. Les
methodes que fy expose ne demandent ni constructions ni raisonnement geo-
metrique ou mecanique, mais seulement des operations algebriques assujeties
a une marche reguliere et uniforme" (Mechanique analytique. Paris, 1788.)
The principle of virtual velocities formed his point of departure. A
number of text-books upon theoretical mechanics still follow the method
of Lagrange.
The revival of pure geometrical investigations by Monge (1746-1818),
the creator of descriptive geometry, and his followers, could not well have
been without its influence upon mechanics. In the year 1804 appeared
the Elements de Statique, by Poinsot, in which, in contrast to Lagrange,
XXviii ANALYTICAL AND GEOMETRICAL MECHANICS. [iNTKOD.
we find : " que tons les theoremes de la Statique rationeUe ne sont plus au
fond que des theoremes de Geometric" This work was the beginning of a
series of treatises in which the advantages of the synthetic development
and geometrical treatment of mechanics were defended and, by most
important results, strikingly demonstrated.
At this time the views as to the best method of treating mathematical
problems were sharply opposed. Carnot (1753-1823), to whom, however,
the modern geometry itself owes no slight impulse, gives the preference
to analysis. For synthesis " est restreinte par la nature de ces procedes ;
elle ne peut jamais perdre de vue son objet, ilfaut que cet objet s^offre tou~
jours d V esprit, reel et net, ainsi que tons les rapprochements et combinaisons
qu'on en fait"1 (Oeometrie de position. Paris, 1803.) That which here Car-
not considers as a defect in the synthetic and geometrical method, Poinsot
claims as its special advantage : " On peut bien par ces calculs plus ou moins
longs et compliques parvenir d determiner le lieu ou se trouvera le corps au
bout d'un temps donne, mais ou le perd entierement de vue, tandis qu'on vou-
drait V observer et le suivre, pour ainsi dire, des yeux dans tout le cours de sa
rotation'1'1 (Theorie nouv. d. I. rot. d. corps}.
The example of Poinsot found numerous followers. In Germany, Mo-
bius followed with his " Lehrbuch der Statik." Mechanics as well as
geometry thus received enrichment. Mobius gives the preference always
to the synthetic method, and also endeavors to interpret geometrically,
analytically deduced formulae — " because in investigations concerning
bodies in space the geometrical method is a treatment of the subject itself,
and is therefore the most natural, while by the analytical method the sub-
ject is concealed and more or less lost sight of under extraneous signs "
(Lehrb. d. StatiTc. Leipzig, 1837.)
Even in analytical operations, geometrical considerations came more and
more in the foreground. On all sides the development of Kinematics, the
theory of motion without reference to its cause, was prosecuted. But.
neglecting the cause of motion, there remains only its path ; that is, geo-
metry proper (Kinematical geometry, or the geometry of motion}. The in-
vestigations of Chasles, Mobius, Rodrigues, Jouquiere, and others, may yet
be still further pursued ; and when by the aid of geometry a certain com-
pleteness has been given to the theory of the motion of invariable systems,
the geometrical theory of regular variable systems (to which the flexible
and elastic belong) will be possible. For the discussion of such branches
of mathematics, the synthetic geometry is necessary ; for their foundation
lies in a theory of the relationship of systems.
The advantage of the synthetic method in mechanics is denied by no
one. Wherever it is possible, we obtain more comprehensive conclusions
as to the nature of the phenomena, while all the properties of the same fol-
low directly from the simple and known truths premised. In analytical
investigations it is necessary, even when definite equations are obtained, to
deduce the actual laws singly and in a supplementary manner, although
they are indeed all contained in the equations themselves.
It is not, however, always possible to preserve the synthetic process
throughout. From the first truth the ways diverge in all directions, and
TNTROD.] GEOMETRICAL STATICS. Xxlx
a special ingenuity is often needed to reach the goal. Just here analysis
comes to our aid with its rich treasures of developed methods, and here it
is most certainly not for geometry to " undervalue the advantage afforded
by a well-established routine, that in a certain degree may even outrun the
thought itself " (F. Klclu : Vrrglaichemle JRetrachtungen aber neuere geome-
trische Forschung^en. _ Erlanggn-, 1 ft72T p. 41). Algebraic operations are
thus, however, not the chief thing, but only the instrument — a most excel-
lent instrument indeed, which can be almost universally applied, and
which, by reason of its connection with an extensive and independent
mechanism, often needs only to be set in action in order to work of itself.
Geometrical mechanics, moreover, can never entirely free itself from
analytical formulae and operations. For though it may be both interesting
and useful to follow, with Poinsot, the body during its entire rotation, yet
practically this is of minor interest, and the chief problem remains still,
" d determiner le lieu ou se trouvera le corps au T)out cTun temps donne"
In the present day all those familiar with both methods of treatment
hold fast the good in each ; they supplement each other. Often in the
course of the same investigation we must interrupt the general analytical
process with synthetic deductions, and inversely. Thus we may well close
these considerations with the sentence with which Schell begins his " Theo-
rie der Bewegung und der Krdfte " — both methods, the analytic and the
synthetic, can only, when united, give to mechanics that sharpness and
clearness which at the present day ought to characterize all the mathemati-
cal sciences.
III.
GEOMETRICAL STATICS.
Statics is a special case of dynamics, though earlier treated as indepen-
dent of the latter. The principle of d'Alembert furnishes the means of
passing from one .to the other. In technical mechanics the distinction is
still preserved, and indeed, in view of the distinct branches in which the
applications on either side, are found, not without propriety.
After the mechanics of the ancients, as comprised in the mathematical
collections of Pappus; the first great step towards our present geometrical
statics was made by Simon Stemnus (1548-1603), when he represented the
intensity and direction of forces by straight lines. Stevinus himself gave
a proof of the importance of his method, in the principle, deduced from it,
that three forces acting upon a point are in equilibrium when they are pro-
portional and parallel to the three sides of a right-angled triangle.
A main discovery was the parallelogram of forces by Newton (1642-
1727). The composition of two velocities in special cases was long famil-
iar. Galileo made use of it for two velocities at right angles, and exam-
ples also occur in Descartes, Roberval, Mersenne, and Wallis, but the funda-
mental principle was first established when Newton replaced the theories
XXX GEOMETRICAL STATICS. [iNTEOD.
of special by that of universal causation (Philosophic^ naturalis principia,
mathematica. London, 1687).
Varignon in his "Projet d'unenouvelle mecanique," in the same year (1687),
and independently of Newton, applied for the first time the general princi-
ple of the composition of motions. From this he passes, in the Nouvelle
mecanique ou Statique, dont le projet fat donne en 1687 (published after
his death, Paris, 1725), by means of the axiom that " les effets sont toujours
proportionnels d leurs causes ou, forces productrices " to the composition of
forces also.
The Statique of Varignon is purely geometrical. He postulates nothing
beyond books 1-6 and 11 of Euclid, and even explains the significance of
+ and — signs. In this work, the first founded upon the parallelogram of
motion and of forces, we find also the force and equilibrium polygons
(Funiculaire, Section II.), to the application and development of which
almost the whole of Graphical Statics is to be attributed. Varignon recog-
nized the value of the equilibrium polygon, and gave it as the seventh of
the simple machines.
After the great Interim of Geometry, Monge wrote a Traite elementaire
de Statique (Paris, 1786). The work claims to contain for the first time
everything in statics which can be synthetically deduced. In a later edi-
tion we learn that synthetical statics must be taken up as preliminary to
analytical, just as elementary geometry before analytical geometry. Thus
the work of Monge contains the necessary preparation for Poisson1 s "Traite
de mecanique" (Paris, 1811).
The greatest influence upon the development of geometrical statics was
exercised by Poinsot. By the introduction of force pairs, he solved in the
most elegant manner the fundamental problem of any number of forces
acting upon a body (Elements de Statique, Paris, 1804, and Memoire sur
la composition des moments et des aires dans la mecanique).
Chasles completed the solution by the proof that the contents of the
tetrahedron, which is determined by the resultant forces, is constant, how-
ever the forces may be composed.
In the hands of Mobius, geometry and geometrical statics were most com-
pletely developed.
Of the greatest importance, for later applications, was the introduction
of the rule of signs.
The germ of this had existed already in the preceding century.* Mobius
recognized its significance, extended it to the expression of the contents of
triangles, polygons, and three-sided pyramids, and applied it systemati-
cally (Barycentrischer Calcul. Leipzig, 1827).
A new impulse, extended field of action, and numerous additions were
given to geometrical statics by the Graphical Statics of Culmann.
* Mobius alludes to this, and we find, for example, in Kastner ( Geometrische
Abhandlungen, I. Saml., 1790, p. 464), the equation A B + B A = o.
INTROD.] THE GRAPHICAL CALCULUS. XXxi
IV.
THE GRAPHICAL CALCULUS.
The most extended applications of statics are in the Held of engineering.
Here, not only general properties of form and position are required, but in
a large number of cases numerical relations are also necessary. General
results of the latter character can, as we have seen, only be embraced by
algebraic formulae (I). The pure graphical theory of construction is there-
fore in this respect lacking in completeness, as it is unable to furnish gen-
eral metrical relations.
The practical engineer has almost always, however, to do with special
problems; dimensions and acting forces are numerically given. Geometry
in such cases could give no general relations, because the results desired
are the consequences of the special proportions of the figure. In any de-
terminate case, however, we may obtain a result holding good for that case,
and it only remains to show how generally to obtain such a result. The
graphical calculus treats of such methods, and so, although not exclusively,
does graphical statics. As soon now as practical use is made of the actual
proportions of the figure, everything depends upon the exactness of the
drawing. One condition for the application of the graphical method is,
therefore, skill in geometrical drawing — a requisition, indeed, which the
practical engineer can most readily meet.
The idea at bottom of the graphical calculus is simple. The modifica-
tions of numbers in numerical calculations correspond always to similar
modifications of the quantities represented by these numbers. The measure
of a quantity can be as well given by a line as a number, by putting in
place of the numerical the linear unit. In order for a graphical calculus,
then, the modifications of lines answering to corresponding numerical
operations are necessary, and these are furnished by geometry. They con-
sist of graphical constructions, and rest upon the known properties of
geometrical figures. The scale furnishes the means of converting directly
any numerical quantity into its corresponding linear representation, and
inversely any graphically obtained result can be at once transformed into
numbers.
The graphical determination of desired or computable numbers is natu-
rally nothing new. From the " Traite de Q-nomonique " of de la Hire
(1682) to the " Geometric descriptive'1'1 of Monge (1788), many examples
are to be found. The graphical calculus, however, goes further than this.
It aims to found a method, a routine, which shall not only apply to bodies
in space, but which shall also, like the arithmetical or algebraic calculus,
be independent of concrete relations and of general application. It seeks
further to obtain its results (products and powers) in the shape of lines
convertible by scale into numbers. (Hence the important part which area
XXXM THE GRAPHICAL CALCULUS. [iNTROIX
transformation plays in the graphical calculus.) Such was the problem
which Cousinery proposed, and whose solution he attempted in his " Calcul
par le trait" (Ses Elements et ses applications. Paris, 1839).
Cousinery applied the graphical calculus to powers, roots, proportion
and progression; to the measure of lines, surfaces, cubes, graphic inter-
polation, and the strength of retaining walls. The presentation is nat-
urally by no means complete, and labors also under a prolixity and
minuteness of detail to which the results obtained are by no means com-
mensurate. It sounds somewhat comic when Cousinery, in his " Calcul par
le trait," claims the then already-existing graphical solutions of Poncelet
(" Memoire sur la stdbilite des revetements, in Memorial de Toff du genie")
as an elegant example of the application of his graphical calculus.
While Cousinery thus sought to apply geometry in a direction where
until then analysis had held sway, he acted in entire accordance with the
spirit of his age, though without making use of those means for aid which
lay at his disposal. " Without effect upon him," says Culmann, " were
the researches of Steiner, already published in 1832, as well as those of
his predecessor; and instead of simply premising the elementary prin-
ciples of the modern geometry, he laboriously sought to deduce them in-
dependently by the aid of perspective." The works, at least, of the French
predecessors of Steiner were, at any rate, well known to Cousinery. In his
preface we read: " Peut-gtre mihne nos efforts eussent-ils 6t6 complete-
ment infructueux, sans les ressources que nous ont procurers et les annales
de M. Gergonne et les travaux de M. Brianchon, et ceux plus re"cents de
M. Poncelet. Nous avons envers M. Chasles une obligation encore plus
droite, car outre les pr6cieux documents que reufernie son ' Histoire des
methodes en geometric? nous avons a lui faire agre"er un tgmoignage par-
ticulier de reconnaissance pour la maniere dont il a bien voulu mentionner
nos premiers essais sur le systeme de projection polaire."
Why Cousinery made use of perspective and not of the modern geome-
try, is easily understood. The development of geometry at that time, as to-
day, proceeded in various almost independent directions, and Cousinery
himself had the pleasure of seeing his " Geometric perspective " (Paris,
1828) designated by the reporters for the Academy, Fremel and Matthien,
as new and ingenious, as well as favorably noticed by Chaxles.* He
sought, therefore, naturally to develop and render fruitful his own method,
so much the more as the true significance and value of the various growing
branches of geometry could not then, as now, be correctly estimated. Ac-
cordingly, the Inge"nieur-en-chef. B. E. Cousinery, wrote avowedly for his
colleagues, and did not feel justified in directly premising a knowledge of
the newest investigations, more especially of his own.
i We have noticed the above somewhat in detail, because it bears directly
* Its newness, at least, is not without doubt. According1 to Fiedler, the
principles are completely given in Lambert's celebrated work, " Die freie
Perspective" (Zurich, 1759). Poncelet also takes issue with the estimation of
the " Geometrie perspective" by Chasles (" Traite des propr. proj.^ II. , ed.
1865, p. 412).
INTKOD.] THE GRAPHICAL CALCULUS. XXXI 1 1
upon a point of our discussion ; for the introduction of the modern
geometry in the graphical method by Culmann, is still, thirty years after
Cousinery, a chief hindrance to its rapid spread.*
After Cousinery, no one occupied himself with the graphical calculus
till Culmann gave it a place in his GrapJiische Statih The presenta-
tion is here far better, and especially shorter. The rule of signs, which
was unknown to Cousinery, is at once brought out. Instead of such
long and tedious applications as the graphical interpolation, a few
examples from engineering practice are given, among which we may
especially notice earth-work calculations. In the extensive earth works of
roads, canals, and railways, the method shows not only most plainly the
extent and best arrangement of transport, but also allows, with the aid of
the planimetre, the cost of transport to be determined.
As to the rest, it would appear as if the graphical calculus should play
an important part in engineering practice. This circumstance, as well as
the interesting problems which present themselves in connection, has
gained for the Arithmograpliy many friends. Several publications
have since sought to win for it a wider recognition without furnishing
anything essentially new. [IT. Eggers : " Grundziige einer graphischen
Arithmetic," Schaffhausen, 1865. J. ScUesinger : " Ueber Potenzcurven,"
Zeitschr. d. osterr. Arch. u. Ing. Ver., 1866. E. Jdger : " Das graphischen
Rechnen," Speier, 1867. K. von Ott : " Grundziige des graphischen Rech-
nens und der graphischen Statik," Prag, 1871.]
Recently the method of the graphical calculus has been applied to Dif-
ferentiation and Integration. A treatise by Solin show§ the first exact, so
far as possible in a construction, the last approximate only (" Ueber graph.
Integr. ein Beitrag z. Arithmographie, Abhand. d. konigl. bohm. Gesellsch.
d. Wissenbach." VI. Folge, 5 Bd. Separate reprint by Rivnac, Prag,
1871). It is to be remarked, also that examples of double integration and
differentiation were given by Mohr in 1868. The graphical construction
of the elastic line, and the determination of the moments at the supports
of a continuous girder, are essentially examples in point (Mohr: "Bei-
trag zur Theorie der Holz und Eisenconstructionen,1' Zeitschr. d. Hannov.
Ing. und Arch. Ver., 1869 ; or W. Hitter : " Die elastische Linie," Zurich,
1871.)
As to the importance of the graphical calculus as an independent study
or discipline, it is, as we believe, often exaggerated. The theoretical value is
but little, and for graphical constructions, as given by the graphical calculus,
offer in no respect anything new. That which pertains to practical applica-
tions may be easily based directly upon geometry, and is nowhere found
as a consequence of the method itself. If it is considered advisable to call
special attention to a few general points before making such applications,
all that can be desired can be easily presented in ten or a dozen pages
octavo.
* See Preface ; also Chaps. VII. and VIII. of this Introduction.
c
GRAPHICAL REPRESENTATION. [iNTROD.
V.
GRAPHICAL REPRESENTATION.
Graphical representation, in the widest sense of the word, includes every
visible result of writing or drawing. The written sentence is the graphi-
cal representation of a thought — the drawn line the graphical indication of
an idea. In such generality we naturally do not here regard graphical
representation. In a narrower sense we understand the graphical represen-
tation of the diversity or dependence of numerical quantities. In this
sense we cannot speak of the graphical represention of pure geometry.
This last was introduced into analysis by Vieta (1540-1603). Here 'the
figure merely aids the conception, while the equation embraces the charac-
teristics of the phenomena (I.), and ensures the independent character of
the drawn lines. Thus the clearness of geometry is combined with the
fruitf ulness of analysis.
If the graphical representation is constructed frcm a number of suitably
chosen and calculated values, the intermediate values can be directly meas-
ured and. by means of the scale, reconverted into numbers. The graphical
representation, then, replaces numerical tables. Illustrative examples often
occur in practice. We instance, for example, the graphical representation
of maximum moments and shearing forces in the continuous girder. If
the several values are calculated from a formula, their graphical union gives
a simultaneous view — a picture — of the law which the formula represents.
If these values are merely known — observed, for example their graphical
combination may enable us to deduce the law which connects them. Thus
the graphical representation is of assistance in the deduction of empirical
formulae, and indirectly in the discovery of exact relations. Illustrations
of such application occur frequently in applied mathematics, especially in
astronomy and meteorology.
In this connection we may also remark that graphical representation
plays also an important part in statistics. By its aid a comprehensive view
is obtained of a series of separate results. Or it may be applied to still
higher problems — for example, from comparison of simultaneous but differ-
ent series of observations to determine an inner connection.
In engineering practice, graphical representations have in recent times
notably multiplied. All graphical constructions, so far as they do not de-
pend upon analytical formulae, and therefore are not directly given by
geometrical laws, are nothing more than consequences of graphical repre-
sentation.
INTROD.] GRAPHICAL STATICS. XXXV
VI.
GRAPHICAL STATICS.
The few text-books upon graphical statics and the more numerous works
upon its applications, afford us no definition, and can afford none, because
neither the method nor scope of this new study are anywhere sufficiently
indicated.
If, following Culmann, we speak of it in contradistinction to the appli-
cations of a pure graphical statics, we may define it somewhat as follows :
Graphical Statics comprises the theory of those geometrical constructions which
occur in the graphical solution of statical engineering problems ; it treats
further of the general relations deducible from such constructions. This
limitation, so far as it does not follow from the preceding, we shall seek
in the course of these remarks still further to establish.
Graphical representations of analytically obtained results have, as has
been already noticed, long been used in engineering practice. They served
also the purposes noticed in the preceding chapter. Often also certain
values, whose analytical determination is somewhat complicated, have
been sought by graphical constructions. Examples of this may be found
in many text-books upon the theory of structures, and we notice only, as
one of the most notable of recent date, the construction of lever arms and
limits of loading in A. Ritter's "Theorie und Berechnung eiserner Dach
und Briickenconstructionen " (Hannover, 1862). Poncdet applied analy-
sis in general to practical investigations, but sought in several complicated
cases to elucidate the deductions of formulae by geometrical constructions,
and to deduce graphical solutions from analytical relations. This pro-
cedure found considerable acceptance, and the investigations of Poncelet
were afterwards resumed upon more general assumptions by Saint Guil-
liem (Memoir e sur la poussee des terres avec on, sans surcharge, aim. des
ponts et chauss., 1858, sem. 1, p. 319).
The first, however, to give pure geometrical determinations of stability
in structures was Cousinery. He gave a number of examples as applica-
tions of his graphical calculus, but his ideas appear to have found in
France little acceptance. On the other hand, the graphical construction of
the curve of pressure in the arch by Mery (Memoir e sur V equilibre des voutes
en bwceau — ann. d. ponts et chauss., 1840, sem. 1, p. 50) was extensively
used, and has since been extended by Durand-Claye to iron arches also
(Ann. d. ponts et chauss., 1867, sem. 1, p. 63, and 1868, sem. 1, p. 109).
Special prominence was given to graphical investigations of stability by
Culrnann's " Graphische Statik " (first part, Zurich, 1864, entire work,
1866 ; second edition, 1st part, 1875.)
This work of Culmann must be considered as original in all those parts
relating to structures. Poncelet and Cousinery, beyond the general idea,
furnished only unessential contributions. Culmann recognized the fruit-
XXXvi GRAPHICAL STATICS. [iNTROD.
fulness of the relations between the force and equilibrium polygon, upon
which most of the practical solutions depend. He developed these rela-
tions, applied them in the theory of moments by the introduction of the
closing line (Schluss Linie), and, accepting the rule of signs, obtained gen-
eral points of view for the discussion of the most diverse figures which
could arise in the same problem. In this and in many other respects even
geometrical statics can profit from Culmann' s work, as, for instance, in the
investigation of the projective relations between the force and equilibrium
polygon.
The fundamental importance of the force and equilibrium polygon was
also recognized by those who, after Culmann, occupied themselves with
the graphical method. Here we may notice two works of special influence
upon the development of the graphical statics— those of MoTir and Cre-
mona. The idea of Molir, that the elastic line is an equilibrium polygon or
curve (" Beitrag zur Theorie der Holz und Eisenconstructionen." Zeitschr.
d. Hannov. Ing. und Arch. Ver., 1868) is of special significance forgraphi.
cal statics.
That from it MoJir obtained the graphical determination of the moments
at the supports of a continuous girder, is an example both useful as well
as interesting. Already it has been endeavored to utilize the same idea in
other cases (Frankel : " zur Theorie der Elastischen Bogentrager," Zeitschr.
d. Hannov. Ing. u. Arch. Ver., 1869, p. 115), and by it an impulse has been
given to similar investigations.
Cremona, has kept more especially in view the geometrical side of graphi •
cal statics. Starting from the theory of reciprocal polyhedrons, he gave
the reciprocal relations between the force and equilibrium polygon with a
generality and elegance to be expected from this distinguished Italian
mathematician (Le figure reciproche nelle statica graficu,. Milan, Langer,
1872). By this investigation the theoretical development of the graphical
statics is essentially anticipated.
It was under the most unfavorable circumstances that Culmann intro-
duced his graphical statics in the engineering department of the Ziirich
Polytechnic in the year 1860. It was finally, indeed, admitted as a regular
study, but not the geometry of position which he premised. It was not
till 1864 that this last was given in a series of lectures by Reye, and then
the time at disposition for both courses was insufficient. Meanwhile the
method spread, crept into the construction department of the engineering
school, and wherever it came, even in the other departments of the Poly-
technic, gained friends. Finally, at the present time, it is to be found, to-
gether with the modern geometry of position, upon which it was based, in
every Polytechnic throughout Germany.
According to the above given definition of graphical statics, the methods
of the graphical calculus, as far as applied in statical investigations, may
also be regarded as belonging to graphical statics, and justly so ; for
these methods follow directly from geometrical principles, and can be ap-
plied by any one acquainted with geometry, without being collected under
the special name of the " graphical calculus." Thus, for instance, Bausch-
inger, in his " Elemente der graphischen Statik" (Miinchen, 1871), disre-
INTROD.] METHODS AND LIMITS OF GRAPHICAL STATICS. XXXvil
gards entirely the graphical calculus, and also cuts loose from the modern
geometry ; he develops the elementary principles of the subject in a logi-
cal and easily comprehended, if not purely geometrical manner, and thus
brings the subject within the reach of those persons for whom it seems so
especially designed. The work is remarkable for clear presentation, but
expressly avoids all special investigations and practical applications, for
which it is merely intended to prepare the way. In the present work, also,
a similar plan is pursued, but all such applications as are of most value to
the engineer or mechanic find likewise a place. Thus, combining the
method of presentation of Bauschinger and the practical applications of
(Julmann, it has been endeavored to make it a practical manual, as well as
a text-book of elementary principles — to serve the wants of the practical
engineer, and also meet the requirements of the engineering student. How
far this twofold design has been realized, the judgment of the reader
must decide.
VII.
THE METHODS AND LIMITS OF THE GRAPHICAL STATICS.
The most perfect method of the graphical statics is the synthetic or geo-
metric; since in geometrical statics the solution must always, when possi-
ble, rest upon pure mechanical or geometrical reasoning. Culmann pre-
sents his graphical statics to practitioners "as an attempt to solve by the
aid of the modern geometry such problems pertaining to engineering prac-
tice as are susceptible of geometrical treatment."
The graphical statics, however, is not in and of itself the product of
endeavors to make the modern geometry of service in applied mechanics ;
graphical solutions merely were required. How to obtain these, was
another question. Thus it is that Poncelet's solutions consist almost en-
tirely of graphical representations of analytical relations ; that Cousinery
avoided all use of formulae ; that Culmann made use of the new geometry
wherever it was possible ; that Bauschinger and others make use only of
the ancient geometry ; and that the latest graphical solutions — in a certain
degree, those of Mohr also — entirely in the spirit of Fencelet's, rest again
upon analysis. The pure geometric solution is, indeed, desirable, but is not
always attainable.
If now we review all the cases in which direct and exclusively geomet-
rical solutions are not possible, we see at once that this occurs when it is
required to make use of the physical properties of bodies, as elasticity, co-
hesion, etc. Why? The actual condition of a body after equilibrium 18
attained, is a consequence of the motion of a variable system of points.
The theory of the motion of variable systems has, however, by no means, as
yet, been brought to practical efficiency (II.). We are therefore obliged to
start from an hypothetical condition or state of the body (in the theory of
flexure, for instance, we rest upon the assumption that all plane cross-sec-
tions made before the action of the outer forces remain plane after their
XXXVlii METHODS AND LIMITS OF GRAPHICAL STATICS. [iNTEOD.
action). To deduce now from this general condition the special relations
necessary for solution, demands an essentially analytical process (T)
Hence the dependence of the graphical solutions in such cases upon ana-
lytical relations — relations which, when the body is assumed to be rigid,
as in the arch, in frame work, or the simple girder, no longer exist.
The sphere of action of an independent graphical statics is, then, con-
fined to those problems which, under the assumption of inflexibility, are
determined by a sufficient number of conditions. Beyond this point we
have chiefly graphical interpretations only.
It has been already noticed that graphical statics, without the application
of algebraic operations, can furnish no general laws (IV.). From relatively
simple figures, indeed, here and there, general formula of metrical relations
have been derived, as is, in fact, not theoretically impossible (I.), but such
formulae were always previously known. Such a result holds, in general,
immediately good only for that form of figure which has been discussed,
Or, according to the terminology of Carnot, only for the existing " primi-
tive figure," and must be proved or transformed for all " correlative
figures " which can occur in accordance with the conditions of the prob-
lem. When the graphical investigation is guided by analytical opera-
tions, it is these last which render possible the deduction of general metri-
cal relations.
Thus, in the theory of structures, there remains subject to pure graphical
treatment only the general relations of form and position. Here we have
the elegant deductions upon unfavorable loading, and here the graphical
method often attains its end in a more elegant manner than the analytical.
A complete exploration and development of such form and place relations,
without a geometry of position, would evidently be impossible (IX.). The
scientific future of the graphical statics, therefore, rests essentially upon
the influence of the modern geometry. To endeavor to separate the higher
geometry from the graphical method would be as unwise and fruitless as
the attempt to exclude the higher analysis from analytical investigations.
As, however, for certain purposes an elementary presentation of analytical
theories relating to engineering practice will ever be acceptable, so also an
elementary development of graphical methods is not without justification,
the more so as long as the modern geometry itself is not sufficiently well
known.
Culmann says of the graphical statics : " It includes, thus far, only the
general part which we need in the investigation of problems in construc-
tion, but it must and will extend, as graphical methods find ever wider
acceptance. Then, however, it will escape the hands of the practitioner,
and must be built up by the geometer and mechanic to a symmetrical
whole, which shall bear the same relation to the new geometry that analyti-
cal mechanics does to the higher analysis." Such an estimation does not
appear to be entirely correct. It is geometrical statics (or mechanics) for
which the above relation may subsist, and to this, indeed, Culmann's valu-
able work has itself greatly contributed. It was, moreover, developed
quite independently of and much earlier than graphical statics (III.). In
this respect, therefore, the spread of graphical methods is of less inipor-
INTROD.] THE MODERN GEOMETRY.
tance than that of geometrical views and knowledge ; for when practical
calculations are disregarded, and the deduction of general truths alone
occupies us, then, first of all, we must exclude from the drawn figure all
special relations — that is, strike out of graphical statics the essentially
graphical part. A truth comprehended only in the abstract holds good
for all figures which can be drawn in accordance with the given condi-
tions.
We place, then, in one line geometry and geometrical statics (mechanics).
From geometry we obtain a method of construction, or descriptive geome-
try, which finds its practical applications in architecture and machine
drawing. From geometrical statics we obtain also a construction method
or routina — viz., graphical statics — which finds its practical applications in
the graphical calculation of structures and machines. Both descriptive
geometry and graphical statics have still, with reference to these practical
ends, to develop and make use of the general relations which subsist be-
tween the geometrical constructions to which they give rise, and thus each,
according to its means, contribute to the discovery and spread of geo-
metrical and mechanical truths.
From this co-ordination of descriptive geometry and graphical statics
we must not, however, infer an equal importance ; for, while in geometri-
cal drawing we have always to represent an ideal image, and the graphical
method is therefore directly suggested, we have for statical calculations
the analytical process also at our disposal, and everything depends then
upon the relative advantages and disadvantages of the graphical and ana-
lytical methods. We have thus noticed all the most important points
which occur in a theoretical consideration, and there only, remains to make
a comparison from a practical standpoint (X.).
VIII.
THE MODERN GEOMETRY.
Geometry treats of figures or constructions in space. These figures and
their properties are not always regarded and treated in equal extent and
generality.
Geometrical knowledge found its origin in practical needs, and the
ancients confined themselves almost exclusively to special investigations
of individual figures and bodies of definite form, such as presented them-
selves to the eye. In the pTiorisms of Euclid (-285), according to Pappus
(end of the fourth century), the mutual relations of the circle and straight
lines were, indeed, given with a certain degree of completeness, but these
have not come down to us.
Properties thus determined had naturally only a limited significance,
and could neither count upon permanence nor give satisfactory conclu-
sions. Investigators sought, therefore, assistance where it was best afforded,
in analysis. This was, in the sixteenth century, by the algebra of Vieta
(1540-1603), notably enriched.
Xl THE MODERN GEOMETRY. [iNTROD.
From this period geometry, for a long time, served merely as an aid to
analysis, interpreting graphically its results (V.). From this union the
greatest advantages were derived, as analysis led to the infinitesimal cal-
culus of Newton and Leibnitz, and geometry to the analytical geometry of
Descartes (1596-1650).
But the extension and generality which geometrical truths received by
this great creation of Descartes was essentially due to analysis. Desaryucts
(1593-1662) and Pascal (1623-1662) extended pure geometrical considera-
tions, and made the first step towards the modern geometry when they
regarded the conic sections as projections of the circle, and deduced the
properties of the first from those of the last. Then De la Hire (1640-1718),
Le Poivre (1704) and Hnygens (1629-1695) occupied themselves with geo-
metrical investigations. While the two first developed the methods of
Desargues and Pascal, Huygens and, later, Newton (1642-1727) applied
pure geometry in optics and mechanics. Soon, however, the Calculus of
Newton and Leibnitz (1684 and 1687) showed itself so wonderfully fertile
in analytical geometry, that geometry proper was put in the background.
Only a few, as Lambert (1728-1777), still regarded it with favor.
Then appeared Monge (1728-1777), and gave the impulse to a complete
revolution in geometrical views, and to the reconstruction of the science
upon a new basis. In his Lemons de Geometric descriptive (Paris, 1788), all
those problems previously treated in a special and uncertain manner in
stereotomy, perspective, gnomonics, etc., were referred back to a few gen-
eral principles, and, without the aid of analysis, the most important prop-
erties of lines and surfaces were deduced. While descriptive geometry
taught the relations between bodies in space and drawn figures, it strength-
ened the power of abstraction ; introducing into geometry the transforma-
tion of figures, it gave to its deductions an advantage till then possessed
only by analysis; and while, finally, it owed its comprehensive results to
the application of projections, it pointed the way for the further develop-
ment of geometry itself.
Meanwhile, in the field of analytical geometry, the conclusion had been
reached that the desired truths admitted of a still more general compre-
hension. All properties had been obtained only with respect to and by
means of a determinate co-ordinate system. But already Godin (1704-
1760) had announced " que I" art de decouvrir les proprietes dss courses est
a proprement parler, I1 art de changer le systeme de co-ordonnees" (Traite den
proprietes communes a toutes les courses). This idea Car not seized upon
(1753-1823), and in the sixth chapter of his Geometric de position (Paris,
1803) he sought to obtain a more general comprehension of figures by
analysis, and to avoid the indeterminancy of this last by the introduction
of the idea of position, and by many solutions after the method already
pointed out by Liebnitz and d'Alembert.
Now began a veritable race in the condensation and promulgation of
geometrical truths, in which the pure geometrical method obtained the
palm. The scholars of Monge — Brianchon, Servois, Chasles, Poncelet, Ger-
gonne — working with him and in his spirit, filled the Annales des mathe-
matiyues and the Correspondance sur Vecole polytechnique with new re-
INTBOD.] THE MODERN GEOMETKY. xli
suits — the two last named discovering the general law of reciprocity or du-
ality.
The foundation proper of the modern geometry was
his Traite des proprietes projectiles des figures (Paris, l828fi"Aggrandir les
resources de la simple Geometrie, en generaliser les conceptions et le langage or-
dinairement assez restreints, lesrapproclier deceuxde la Geometrie analytique,
et surtout offrir des moyens generaux, propres a demontrer et a fair e decouvrir,
d'une maniere facile, cette classe de proprietes dont jouissent les figures quand
on les considere d'une maniere purem'ent abstraite et independamment d'au-
cune grandeur absolue et determinee, tel est Vobjet qu'on s'est speciolemejit
propose dans cet ouvrage."1"1
The new ideas found in Germany especially fruitful soil. Mobius,
Pliicker, Steiner, Grassman, and many others, proceeding in part from
entirely different points of view, opened out an abundance of new direc-
tions which have not yet been thoroughly explored, and which, in union
with other investigations, have caused a thorough change in our concep-
tions of space relations, whose latest phases are indicated by the names of
Riemann, Helmlioltz and Lie-klein.
In this development period, also, still existed the two parties in analyti-
cal and synthetic, or pure geometry. Plucker held the analytical relations
as the most general, and which were with advantage to be illustrated and
interpreted geometrically ; while Steiner recognized in the space figure
itself the true object and most efficient aid of investigation. Both direc-
tions— the modern analytic and synthetic — lead naturally to the same results.
"With reference to the methods, however, they diverge the nearer the ideas
and transformations of geometry approach the generality and ease of the
alg3braic method, thus rendering possible an abandonment of this last.
Thus, while analytical geometry, through the theory of determinants of
Hesse, came into ever closer connection with analysis — a direction in which
English and Italian investigators — as Salmon, Cayley, Cremona — brilliantly
assisted, the Erlangen Professor von Staudt cut loose from algebraic formu-
la and metrical relations, and gave us the geometry of position (Nilrnberg,
1847, Beitr. z. Geom. d. Lage).
After von Staudt, the strict geometry of position remained a long time
disregarded, while the synthetic geometry of Steiner has enjoyed, without
intermission till the present day, a special preference on the part of mathe-
maticians. One reason may indeed be that mathematicians take little in-
terest in an independence of geometry to which analysis can lay no claim ;
but another, still more potent, is the extremely condensed, almost schematic
presentation of von Staudt, which has not exactly an encouraging effect
upon every one.
Culmann gave the impulse to a change in this respect. In his graphical
statics he rests directly upon the work of von Staudt, and, with something
more than boldness, assumes a knowledge of the geometry of position
among all practical men. Such a course was not indispensable for the
foundation of his method, and impeded the spread of the graphical stat-
ics ; but by it the geometry of position gained. This last had next, of
necessity, to be introduced into the Zurich Polytechnic, and thus' arose the
MODERN GEOMETRY IN ENGINEERING PRACTICE. [iNTROD.
first, until now, only complete text-book upon the subject, the " Geometric
der Lage," by Reye (Hannover, 1868), as the direct result of the graphical
statics of Culmaun.
Since then, the modern geometry has been introduced into all technical
institutions throughout Germany, and thus placed at the disposal of the
arts and sciences.
As, according to its founder, Poncelet, it reaches the highest range of
speculation, so also in the most practical relations it acts to simplify and
condense : " Pen d peu les connaissances algebriques deviendront mains in-
dispensdbles, et la science, reduite d ce qu'elle doit etre, d ce quSelle devrait etre
dejd, sera aimi mise d la portee de cette classe d' homines, qui n'a que des mo-
ments fort rares d y consacrer."
[For illustrations of the method of the modern geometry, the reader may
consult the Appendix to this chapter.]
IX.
THE MODERN GEOMETRY IN ENGINEERING PRACTICE.
One who should infer that a science created thus from its very inception
with reference to the needs of practice* must have found access, above all,
in technical circles, would be much mistaken. As Culmann sent out his
graphical statics, deep silence prevailed, and if the modern geometry ap-
peared here and there in the lecture plan of one and another polytechnic, it
was, without doubt, due to the zeal of some enthusiastic privat docent who
had undertaken the thankless task of holding forth to empty benches.
Whence came this indifference to a discipline proceeding from the Ecole
polytechnique ? It is hard, indeed, to find a sufficient reason. We often
hear it said that by reason of the colossal extension which engineering
sciences have experienced, students are already overburdened. Most true !
and it is just here that the modern geometry comes to our assistance. It
is precisely to this that the learned critic of Monge, Dupin, alludes : " II
(terrible que dans Vetat actuel des sciences mathematiques le seul moyen d'ern-
pecher que leur domaine ne devienne trop vaste pour notre intelligence, c'est
de generaliser de plus en plus les theories que ces sciences embrassent, afin
qu'un petit nombre des verites generates et fecondes soit dans la tete des
Tiommes I1 expression abregee de la plus grande variete des f aits particuliers"
The modern geometry in its present form starts with a small number of
elementary constructions whose properties are first set forth, and then, pro-
ceeding from these by combination and comparison, it covers the entire
department of space. The engineer, during and after his preparation, has
to do with space problems, with geometrical principles and constructions ;
* Poncelet himself set upon the title-page of his work : " Ouvrage utile d
ceux qui s^occupent des applications de la Geometric descriptive et d' operations
geometriques sur le terrain."
INTBOD.] PRACTICAL SIGNIFICANCE OF GEAPHICAL STATICS. xliii
"how many superfluous definitions and demonstrations could not be
spared, if they were already completely comprehended and recognized by
the scholar as parts of a higher whole" (Gulmann — "Die Graphische
Statik "). At no very distant day it will no* longer be possible to read a
scientific work upon applied mathematics without familiarity with- the
principles of the modern geometry.* Permitting pure graphical applica-
tions, without the aid of analytic symbols, it forms the common point of
view for descriptive geometry, practical geometry, and graphical statics.
Descriptive geometry existed before the modern, and this last has sprung
from it. Now, reversely, the geometry of position comes to the aid of
descriptive geometry, and offers in return its most fruitful principles and
efficient aid. Thus in descriptive geometry we may refer to the works of
Pohlke, tichlesinger, and Fiedler. The effect of the geometry of position in
this direction to simplify and condense may be seen from the work of
Staudigl ("Ueber die Identitat von Constructionen in perspective, schiefer
und orthogonaler Projection'1), where it is proved that "all problems of
the descriptive geometry, in which neither linear nor angular measure are
considered — therefore all problems which belong to the geometry of posi-
tion— can in similar manner and by precisely similar constructions be solved
as well in perspective as in oblique and orthagonal projection." In shades
and shadows and in geometrical drawing, Burmeister and Paulus owe to
the modern geometry the simplicity of their constructions.
In the department of practical geometry also, in geodesy, perspective,
surveying, we mark the influence of the modern geometry in the works of
Mutter and Spangeriberg, of Franke and Baur.
In mechanics and physics, we see it again in the works of Lindemann,
Burmeiater and Zech.
PEACTICAL SIGNIFICANCE OF THE GBAPHICAL STATICS.
We have already remarked (VII.) that the importance of graphical
statics is in great part dependent upon its advantages as compared with
the analytical method, and have reserved for this place a comparison from
a practical point of view.
Here, first of all, we have to notice the independence of the graphical
construction of the regularity or irregularity of the given relations.
Whether the forces are equal or not, whether they act at equal or varying
distances, even their relative position, are matters of indifference. Centre
of gravity, central ellipse, kernel — for all, even the most irregular figures,
are found in similar manner, with equal ease, even when exact analytical
solutions are hardly conceivable. Thus a process, a routine almost
mechanical is rendered possible in many investigations of stability, with-
out losing sight of interior relations ; for in the repeated and independent
compositions of the forces we always perceive the origin, connection and
* Well illustrated in Gillespie's Land Surveying. New York, 1870.
PRACTICAL SIGNIFICANCE OF GRAPHICAL STATICS. [iNTKCD.
reason of the result obtained, which, in the substitution of numbers in a
formula, is not always the case.
With this advantage goes hand in hand a disadvantage. This very
regularity of the process is a consequence of its special, we might almost
say numerical, character (I.). In a numerical analytical example greater or
less regularity has also but little effect. This numerical character has also
for consequence that we can never attain to general laws and relations
(IV., VII).
The practical engineer becomes with time ever more familiar with the
dividers and rule, while facility in analytical operations gradually disap-
pears. A graphical construction once completed is not easily forgotten, or
a single glance at a similar figure suffices to recall the whole process. It is
indeed easy in clearly given formulae to substitute special numerical values ;
but formulas unfortunately are not always clearly given, in some cases can-
not be so given, without presuming upon the thorough familiarity of the
reader with the processes involved ; these and the very many and various
systems of notation in use leave to the constant, easily acquired and
remembered graphical solution many advantages.
But here we may remark that graphical solutions can only be easily
acquired, retained or quickly recovered when the constructions are based
upon methods purely geometric, and not when they are simply the interpre-
tation of previously obtained analytical results. In the latter case we
must recall the process of development of the formula as well as the
graphical construction, and the method is thus too often confusing instead
of simple.
Often it is desired to make visible the results of an investigation, as in
the case of the arch, where the graphical method is especially advan-
tageous, and has in France been long used (VII.).
Errors relating to the mutual relation of strains are more easily discov-
ered in graphical solutions than in analytical, as a certain law of regularity
is always visible, which breaks abruptly for an error in construction. By
calculation, on the other hand, we can more easily select any one place in
the structure, and determine the strain there independently of the others.
As to which of the two methods demands the least time is a matter of
minor importance. In a construction costing from thousands to millions,
it matters little whether the calculations require one or several days, more
or less, if only the results are clear and correct. It is a question also
which can hardly be decided in favor of one or the other, dependent as it
is upon elements other than those pertaining to the methods themselves —
such as varying individual skill and capacity in either direction. The
declaration which is already sometimes encountered, that the numerical
calculation of a continuous girder requires about three times as much time
as the graphical solution, sounds questionable. Why not at once furnish
the statement with decimal places f In general, for ordinary cases, the ana-
lytical solution requires less time ; for irregular and more complicated cases,
the graphical.
The exactness of the graphical solution is sufficient, but it, too, depends
upon the care and skill of the draughtsman. The greater the forces and
IXTEOD.] LITERATURE UPON GRAPHICAL STATICS. xlv
dimensions with which one works, the better the results obtained. The
scales should not, then, be taken too small.
It is hoped that these considerations, now drawing to a close, will suffice
to give the reader clear ideas upon the nature and origin, advantages and
disadvantages, of the graphical statics. The determination whether he will
enter more fully into the subject — it may be, even take part in its develop-
ment (there is abundance of room for workers), and in this case the choice
of direction may thus be facilitated.
The graphical statics is certainly suited, especially in extended applica-
tions of the geometry of position, to furnish many new points of view, and
in a practical respect it can often greatly simplify. Whoever has really
studied the new methods must admit this.
On the other hand, the importance of the graphical statics is sometimes
exaggerated. It appears out of place when in works designed for practice
graphical solutions are given of problems which any reasoning being can
almost solve in his head.
Such solutions may find a place in special text-books upon the subject,
where they may, indeed, be desirable for completeness.
If it is desired to make two independent investigations of stability, as
for large and important constructions is always desirable, it will be found
of advantage, if a suitable graphical solution exists, to make the first deter-
mination graphically. Nothing more ensures a conviction of the correct-
ness of an investigation than a correspondence of the graphical and cal-
culated results.
XI.
LITERATURE UPON GRAPHICAL STATICS.
We have already referred in VI. to the most important contributions in
the branch of graphical statics, and now annex a list of the literature upon
the subject so far as known to us.
Where several works treat of the same subject, we have allowed ourselves
a brief critical notice. Opportunity is thus given to those who would take
part in the development of graphical statics to make themselves acquainted
with all existing works, and at the same time the practical man is enabled
in any case that may come up to inform himself as to where assistance
may best be sought. A short remark to specify the contents may in this
respect often help in the right direction. The succession is in each division
chronologically arranged
Although the literature of the subject would seem from the following
tolerably extensive, still the number of pure geometrical solutions in
which no analytical formulae appear is much less. Publications upon the
subject would, moreover, beyond doubt, be still more numerous were it
not for the difficulty and cost of production of lithograph plates.
LITEKATUliE UPON GRAPHICAL STATICS.
I. TEXT-BOOKS UPON GRAPHICAL STATICS.
Culmann, K.—" Die graphische Statik." With Atlas of 36 Plates. Zurich,
Meyer and Zeller, 1866. [I. Part, 1864: Elements and Graphical
Investigations of Structures. Also, second edition, first volume,
1875, with 17 Plates. General Principles, second volume, to follow
shortly.]
Bauschinger — "Elemente der graphischen Statik." With Atlas of 20
Plates. Miinchen, 1871. [Without the aid of modern geometry, and
without practical applications. Admirable exposition of the Princi-
ples.]
lleuleaux. — An outline of the graphical statics is to be found in " Der Con-
structeur," by Reuleaux, third ed. Braunschweig, 1872.
Levy — " La Statique Graphique et ses Applications." Paris, 1874. With
Atlas of 24 Plates. [Principles and several applications; clear and
elegant exposition of the subject.]
n. PAPERS UPON THE GRAPHICAL STATICS.
Most — " Ueber eine allgemeine Methode, geometrisch den Schwerpunkt
beliebiger Polygone und Polyeder zu bestimmen." Archiv d. Math,
und Phys., IL. (1869), p. 355. [Also applicable to curve areas, with-
out equilibrium polygon.]
Culmann, K. — "Ueber das Parallelogram und iiber die Znsammensetzung
der Krafte." Vierteljahrsschr. d. Naturforsch. Ges. zu Zurich, 1870.
[Correspondence of the graphical statics with the Statics of Pliicker.]
Mohr — " Bsitrag zur Theorie der Holz- und Eisenconstructionen." Zeitschr.
d. Hannov. Arch. u. Ing. Ver., 1870, p. 41. [Relation between the
neutral axis and centre of strains.]
Grunert, J..A. — "Ueber eine Graphische Methode zur Bestimmung des
Schwerpunktes eines beliebigen Vierecks." Arch. d. Math. u. Phys.,
LII. (1871), p. 494. [Simple and brief. Compare also L., p. 212.]
Cremona, B. — " Le figure reciproche nelle statica grafica." With 5 Plates.
Milan, 1872. German translation in Zeitschr. d. Ost. Arch. u. Ing.
Ver., 1873, p. 230. [Force and equilibrium polygon as reciprocal
figures.]
Du Bois, A. J.— " The New Method of Graphical Statics." Van Nostrand's
"Engineering Magazine," Vol. XII., Nos. 74, 75, 76, 77, 78. [General
properties of force and equilibrium polygons, with practical applica-
tions to bending moments, and several important mechanical problems.
Also, Maxwell's Method applied to bridges, roof trusses, etc.] Sepa-
rate reprint, 1875. Van Nostrand, New York.
LITERATURE UPON GRAPHICAL STATICS.
III. APPLICATION TO THE SIMPLE GIRDER.
Culmann, K.— "Der Balken." Third chap, of d. graph. Statik, 1866.
[Contains also the construction of the inner forces.] I
Vojdcek — " Graphische Bestimmung der Biegungsmomente an kurzen
Tragern." Zeitschr. d. Vereins Deutsch. Ing., 1868, p. 503. [Graphi-
cal interpretation of analytical relations.]
Cotterill, J. H. — "On the Graphic Construction of Bending Moments."
"Engineering," 1869 (VII.), p. 32. [Equilibrium polygon for the sim-
ple truss, with references to Rouleaux and Culmann.]
Winkler, E. — "Einfache Trager," "Theorie der Brticken," "Aeussere
Krafte gerade Trager." Wien, 1872. [Simultaneous presentation of
analytical and graphical methods.]
Ott, K. von — "Wirkung paralleler Krafte auf einfache Trager mit Gerade
Langenachse." In die Grundziige d. graph. Reclmens u. d. graph.
Statik. Prag, 1872. p. 28. [The most elementary principles pertain-
ing to composition of forces in a plane are prefaced.]
IV. APPLICATION TO THE CONTINUOUS GIRDER,
Culmann, K. — "Der continuirliche Balken." Fourth chap, of the Graph.
Statik, 1866. [With examples — the moments at the supports are
analytically determined.]
MoJir — " Beitrag zur Theorie der Holz- und Eisehconstructionen." Zeitschr.
d. hannov. Arch. u. Ing. Ver., 1868, p. 19. [Completion of Culmann' s
method — the moments at the supports are graphically determined.]
Lippich — "Theorie des continuirlichen Tragers Constanten Querschnitts."
Wien, 1871. Separate reprint from Forster's Bauzeit., 1871, p. 103.
[Graphical method, together with elementary analytical.]
Hitter, W. — " Die elastische Linie und ihre Anwendung auf den continuir-
lichen Balken." Zurich, 1871. [Mohr's method— given as a supple-
ment to the Graph. Statik of Culmann.]
Winkler, E. — " Continuirliche Trager. Theorie der Bracken — aeussere
Krafte gerade Trager." Wien, 1872. [The Mohr-Culmann method,
together with analytical.]
Solin, J. — " Geometrische Theorie der continuirlichen Trager." Mitth. d.
Arch. u. Ing. Ver. in Bohmen, 1873.
Greene, Chas. E. — " Graphical method for the analysis of Bridge Trusses ;
extended to Continuous Girders and Draw Spans." New York, 1875.
[Moments at supports found by successive approximation, or balancing
of moment areas.]
V. APPLICATION TO FRAME WORK.
Culmann, K.—" Das Fachwerk." Fifth chap. Graph. Statik, 1866. [Most
general form of parallel truss, suspension truss, Pauli's truss, roof
trusses.]
LITERATURE UPON GRAPHICAL STATICS.
Keck, W.— " Ueber die Erinittclung der Spannungen in Fachwerk tragern
mit Hulfe der graphischen Statik." Zeitsclir. d. hannov. Arch. u. Ing.
Ver., 1870, p. 153. Separate reprint, Hannover, 1872. [Presentation
of the method with reference to practice.]
Jenkin — " On the Practical Application of Reciprocal Figures to the Cal-
culation of Strains in Frame-work. Transact, of the R. Soc. of Edin-
burgh, 1870, (XXV.) p. 441.
Maxwell, Prof. Cleric — " Reciprocal Figures, Frames, and Diagrams of
Forces." Trans, of R. Soc. of Edinburgh, 1869-70.
Uhwin — " Iron Bridges and Roofs.'1 London, 1869. [Application to roof
trusses, wind force, etc.]
Banfon,F.A.—"The Strains in Trusses." New York, Appleton, 1872.
[Examples of simple trusses drawn to scale.]
Bow, Robert H. — "Economics of Construction in Relation to Framed Struc-
tures." London, 1873. [Application of Maxwell's Method only to
roof trusses, etc.]
Ott, K. von — " Das Fachwerk." In Grundzuge d. graph. Rechnens u. d.
graph. Statik. Prag, 1872. [Roof trusses, truss fixed at one end and
free at the other, bridge trusses.]
Reuleaux — " Hilfslehren aus der Grapho statik." Second chap, of the Con-
structeur, third ed., 1872. [Compound truss, roof trusses, etc.]
Scliaffer — "Graphische Ermittelung der Ordinaten des Schwedler'schen
Tragers." Zeitschr. fur Bauwesen, 1873, p. 237. [Proceeding from
the equation for the same.]
Heuser — "Graphische Ermittelung der Ordinaten des Schwedler'schen
Tragers." Zeitschr. f. Bauwesen, 1873, p. 523. [Preceding method
simplified — another by means of equilibrium polygon.]
VI. APPLICATION TO THE IRON ARCH.
Culmann, K.—" DerBogen." Sixth chap, der graph. Statik, 1866. [Con-
tains also the inverted or suspended arch. The arch as a rigid body.]
Durand-Claye, A. — " Sur la verification de la stabilite des arcs m&alliqucs
et sur 1'emploi des courbes de pression." Ann. d. ponts et chauss.,
1868, sem. 1, p. 109. [Mery-Durand pressure curves, but with refer-
ence to the absolute resistance of the material.]
Frankel, W. — " Zur Theorie der elastischen Bogentrager." Zeitschr. d. han-
nov. Arch. u. Ing. Ver., 1869, p. 115. [Following out Mohr's idea of
the equilibrium polygon as elastic line. ]
MoJir — " Beitrag zur Theorie der elastischen Bogentrager." Zeitschr. d.
hannov. Arch. u. Ing. Ver., 1870, p. 389. [Criticism of the preceding
method, and giving another.]
Beitrage zur graphischen Berechming elastischer Bogentrager mit
Kampfergelenken." Mitth. d. Arch. u. Ing. Ver., in Bohmen, 1873.
LITERATURE UPON GRAPHICAL STATICS.
VII. APPLICATION TO THE ARCH.
Cousinery, E. B. — " Application des proce'de's du calcul graphique a" divers
problemes de stability." Fourth chap, of Calcul par le Trait. Paris,
1839. [With especial reference to the strength of abutments — pure
graphical treatment.]
M&ry — " MSmoire sur I'Squilibre des voutes en berceau." Ann. d. ponts
et chauss., 1840, sem. 1, p. 50. [Geometrical determination of every
possible pressure curve.]
Culmann, K. — " Der Bogen.". Sixth chap, of Graph. Statik, 1866. [Con-
taining also arch centerings ; exact discrimination of support and
pressure line.]
Durand-Claye, A. — "Sur la verification de la stability des voutes en
maconnerie et sur 1'einploi des courbes de pression." Ann. d. ponts
et chauss., 1867, sem. 1, p. 63. [Reference to relative resistance of
material. ]
Harlacher, A. R.— "Die Stiitzlinie im Gewolbe." Tech. Blatter, 1870, p
49. [Practical method by inscription of support line, according to
Culmann.]
Heuser — "Zur Stabilitatsuntersuchung der Gewolbe." Deutsche Bauzeit.
1872, p. 365. [Also methods for unsymmetrical form and load.]
VHI. APPLICATION TO RETAINING WALLS.
Poncelet. J. V. — " Me'moire sur la stabilite des reve'tements et leur fonda-
tion." Mem. de 1'off. du Ggnie, 1838 (XIII.) ; separate reprint,
Paris, 1840. [First analytical graphical theorie.]
Cousinery, E. B. — " Application des precedes du calcul graphique & divers
problemes de stability." Hourth chap, of " Calcul par le Trait," 1839.
[Pure graphical, without formulae.]
^aint-G-uilhem — " MSmoire sur la poussSe des terres avec ou sans sur-
charge." Ann. d. ponts et chauss., 1858, sem. 1, p. 319. [Further
development and generalization of Poncelet's Theory.]
Rarikine — "Manual of Civil Engineering." London, fourth ed., 1865.
[Containing graphical construction of pressure parallel to earth sur-
face upon vertical wall.] i
Culmann, K. — " Theorie der Stutz- und Futter-Mauern." Eighth chap, of
Graph. Statik, 1866. [With use of equilibrium polygon, pressure
upon tunnel arches.]
JMzhey, E. — " Beitrage zur Theorie des Erddrucks und graphische Bestim-
mung der Sta'rke von Futter-Mauern." Mitth. iiber Gegenst. d. Artill.
und Geniewesens; separate reprint, with two plates, Wien, 1871.
[Point of application of earth pressure for complicated contour.]
MoJir — "Beitrage zur Theorie des Erddrucks." Zeitschr. des hannov.
Arch. u. Ing. Ver., 1871, p. 344. [Point of application of earth pres-
sure and new analytical theory.]
1 GRAPHICAL DYNAMICS.
WinTder, K-" Neue Theorie des Erddrucks." Wien, 1872. [Containing
graphical methods according to the old theory.]
Haseler, 0. — " Beitrage zur Theorie der Futter- tmd Stiitz-Mauern." Zeit-
schr. d. hannov. Arch. u. Ing. Ver., 1873, p. 36. [Graphical deter-
mination of earth pressure according to Culmann.]
MISCELLANEOUS APPLICATIONS.
Reuleaux—" Die graphische Statik der Axen und Wellen." Published by
polytech. Ver. in Zurich, 1863. [Autograph copy of lectures.]
Culmann, K. — " Der Werth der Constructionen." Seventh chap, of Graph.
Statik, 1866. [Best and cheapest systems under given conditions,
especially for bridges.]
Reuleaux — " Graphostatische Berechnung verschiedener Axen, Kranpfosten,
Kurbeln," in the Constructeur, third ed., 1872.
Scattering graphostatical constructions are to be met with in many text-
books upon construction, especially since the appearance of Culmann's
work, a second edition of which is in course of preparation, and expected
soon to appear.
XII.
GRAPHICAL DYNAMICS.
The scientific or practical value of graphical solutions once recognized,
there remains no reason for limiting them to statical problems only, and
endeavors in the above direction are already forthcoming. We limit our-
selves to a passing notice.
First, we have an attempt to employ graphical constructions in the
theory of the overshot and breast- wheel (Qeeberger, " Arbeitung der Theo-
rie der oberschlachtigen Wasserrader auf graphischen Wege." Civil Ing.
1869, p. 398, and 1870, p. 339). We cannot here notice the value of the
solutions given, but the very sparing applications of geometry hardly jus-
tify the title of the work.
A short article, which gives the graphical determination of the force at
every position of a moving point, may also be noticed. (Rapp, " Zur
graphischen Phoronomie," in Zeitsch. f. Math. u. Phys., 1872, p. 19.)
The genuine foundation of a graphical dynamics has been first attempted
by Proll (" Begriindung graphischer Methoden zur Losung dynamische
Probleme," in Civil Ingenieur, 1873). From the fact that the effects of
forces in dynamics are measured by the changes of velocity of any point or
points of a machine system, Proll concluded that it must be possible to
represent these force effects by geometrical relations, such as kinematic
geometry teaches.
His investigations, since published in independent form ("Versuch
einer graphischen Dynamic," with 10 plates, 1874), fall into thr.ee parts.
The first part treats of the action of the " outer forces " in machines whose
GRAPHICAL DYNAMICS. 11
motion is in a plane, the outer forces being also in this plane. In the sec-
ond part he subjects to graphical treatment the action of outer forces upon
a free movable material point. The third part, finally, considers the motion^
of rigid- invariable systems acted upon by given forces.
In the course of the development extended use is made of analytical
formulae. The work is but the beginning of the future structure, but this
beginning will be thankfully received by all those with whom graphical
methods have found acceptance.
PART I.
GENERAL PRINCIPLES.
CHAPTEE I.
FORCES IN THE SAME PLANE COMMON POINT OF APPLICATION.
1. Notation, etc. — In order that a force may be " given " or
completely determined in its relations to other forces, we must
know not only its intensity, but also its direction, and the posi-
tion of its, point of application. These three being known, the
geometrical expression of our knowledge is very simple. We
have only to assume a certain length as the unit of force, and
then any force is at once given by the length, direction, and
position of a straight line. This method of force representa-
tion is so obvious, that it is in fact used in mechanics, even
where the treatment itself is essentially analytical.
Unless expressly stated, all the forces with which we have to
do, will be considered as lying and acting in the same plane.
Graphically then, any force is completely determined by a
straight line, the beginning of which represents the point of
application, and the length and direction of which give the in-
tensity and direction of the force.
We shall indicate a force in general by the letter P, its point
of application by A. When we have several forces we repre-
sent the points of application by A1? A2, A3, etc., and the ends
of the corresponding lines by P1? P2, P3, etc. The direction in
which a force is supposed to act is thus unmistakably indi-
cated.
When, however, lines representing several forces are laid off
one after another, the beginning of each at the end of the pre-
ceding, it will be sufficient to put 0 at the beginning of the
first, and 1, 2, 3, etc., at the end of each. No confusion can
arise, as each force acts and reaches from the point indicated
by the figure which is one less than its index, to the point indi-
cated by that index.
When, finally, we designate a force by the two letters or fig-
ures which stand at the beginning and end, we shall always
indicate by the order in which the letters or figures are written,
2 FORCES IN THE SAME PLANE. [CHAP. I.
the direction of action of the force, first naming the point of
application, and then the end.
A force due to the composition of several forces, as Pl5 P2, P3,
we denote by Pi.3 or R^. Thus R^ denotes the resultant of
the forces Pl5 P2, and P8.
2. Parallelogram of Force§. — If two forces, P^ and P2,
given in direction and intensify by the lines OPX OP2 [Fig. 1,
PL 1], have a common point of application O, the resultant
Ri.2 is found by the well known principle of the " parallelo-
gram of forces," by completing the parallelogram as indicated
by the dotted lines, and drawing the diagonal. OR then gives
the resultant of the forces Pl and P2. If this resultant acts in
the direction from O to R, as indicated by the arrow, it replaces
P! and P2 ; that is, it produces the same effect as both forces
acting together. If it were taken as acting in the opposite
direction — i.e., from O outwards, away from R — it would hold
the forces pt and P2 in equilibrium.
Now, we see at once that it is unnecessary to complete the
parallelogram. It is sufficient to draw from the end of the
force P2 the line P2 R in the same direction that Pl acts in, and
make it equal and parallel to Plf The point R thus found is
the end of the resultant R, or is a point upon the direction of
the resultant prolonged through O.
As to the direction of action of the resultant — if we follow
round the triangle from O to P2 and from P2 to R and R to O
— i.e., if we follow round in the direction of the forces — the
direction for the resultant from R to O thus obtained is, as we
have already seen, the direction necessary for equilibrium.
3. If, instead of two forces, we have three or more, as P1? P2,
P3, P4 [Fig. 2] we still have the same construction. Thus com-
pleting the parallelogram for Px and P2 we find R^. Complet-
ing the parallelogram for R^ and P8, we find RM, and again,
with this and P4 we obtain Rw. Again, we see it is unneces-
sary to complete all the parallelograms. We have only to draw
lines P! Rj.2, Ri_2 R^, Ri_3 RM, parallel to the forces P2 P3 and
P4 respectively, and equal in length to the intensities of these
forces, and then, no matter what may be the number of forces,
the line drawn from the point of beginning to the end of the
last line laid off will give the intensity and position of the
resultant. As to direction, the same holds good as before.
If the end of the last line laid off as above, should coincide
CHAP. I.] COMMON POINT OF APPLICATION. 3
with the point of beginning, there is, of course, no resultant,
and the forces themselves are in equilibrium.
4. The polygon formed by the successive laying off of the
lines parallel and equal to the forces, we call the "force poly-
gon" Hence we have the following principles established :
If any number of forces having a common point of appli-
cation and lying in the same plane, a,re in equilibrium, the
" force polygon" is closed.
If the "force polygon " is not closed, the forces themselves
are not in equilibrium, and the line necessary to close it gives
the resultant in intensity and direction.
This resultant, if considered as acting in the direction ob-
tained ~by following round the " force polygon " with the forces,
will produce equilibrium — acting in the opposite direction, it
replaces the forces.
The resultant thus found in intensity arid direction can be
inserted in the force diagram at the common point of applica-
tion.
5. Thus, required the position, intensity, and direction of the
resultant of the forces P1? P2, P3, P4, P5.
These forces are given in position, direction, and intensity
by the force diagram, Fig. 3 (a). The resultant of all these
forces must have of course the same point of application A as
the forces themselves — it remains to find then its relative posi-
tion and the direction of its action, so that we may properly
insert it in the force diagram.
We have simply to draw the force polygon, Fig. 3, (b) by lay-
ing off successively O P1? Pt P2, etc., equal, parallel, and in the
same direction as the forces P1? P2, etc., as given by Fig. 3 (a).
Then the line P5 O necessary to close the force polygon gives
the intensity of the resultant, and in order to replace P^ it
must act in the direction from O to P5 ; i.e., contrary to the
order of the forces. If then in Fig. 3 (a) we draw A Rw equal
and parallel to O P5, we have the resultant applied at the com-
mon point of application A, and given in position, intensity
and direction.
Moreover, it is evident that any diagonal of the force poly-
gon as R^ [Fig. 3 (b)] is the resultant of P8^, and acting in the
direction from P4 to P2, it holds P^ in equilibrium. But it is
also the resultant of Pj, P2, P5, and R^, and acting in the same
direction as before, it replaces these forces. The force polygon
FORCES IN THE SAME PLANE. [CHAP. I.
thus shows that the force which replaces P1? P2, P5, and Rw, at
the same time holds P3 and P4 in equilibrium, just as it should
do.
If, on the other hand, we had originally only P1? P2, R^, P5,
and RLS forming a system of forces in equilibrium, we could
decompose Rg^ into two components by simply assuming any
point as P3 [Fig. 3 (£)] and drawing P3 P4, P3 P2. Then follow-
ing round this new polygon in. the direction of the forces, or,
what amounts to the same thing, taking the direction of the
components P3 P4, opposed to the direction of R34 for equilibri-
um, we obtain the direction of action of P3 and P4 as shown by
the arrows in Fig. 3 (b). These forces inserted in Fig. 3 (a), in
the place of R3.4 and in these directions, will not disturb the
equilibrium.
Hence, any diagonal in the force polygon, is the resultant
of the forces on either side, holding in equilibrium those on
one side and replacing those on the other, according to the
direction in which it is conceived to act.
Also, any force or number of forces may be decomposed into
two others in any desired direction, ~by choosing a suitable
point in the plane of the force polygon and drawing lines
from this point to the beginning and end of the force or force
polygon.
6. It matter§ not in what Order we lay off the Forees in
the Contraction of the force Polygon. — Thus, in Fig. 1,
whether we draw from the end of P2 the line P2 R^ equal and
parallel to P! or from the end of Px the line P! R^ equal and
parallel to P2, in either case we obtain the same resultant and
the same direction for the resultant. But by a similar change
of two and two, we can obtain any order we please. For exam-
ple, we lay off in Fig. 3 (c) the same forces in the order P3 P2
PI, P5 P4, and obtain precisely the same resultant, in the same
direction as before. For, the resultant of P3 and P2 must be
the same as that of P2 and P8 in the first case. The resultant
of R3.2 and Pt must then be the same in both polygons, and so
on.
Generally, then, no matter what the order in which the
forces are laid off, the line necessary to close the force polygon
is the resultant of the forces, and the diagonals of the force
polygon give us the resultants of the forces on either side.
By assuming a point at pleasure, and drawing lines from this
CHAP. I.] COMMON POINT OF APPLICATION. 5
point to the beginning and end of any side of the force poly-
gon, and taking the direction of these lines opposed to the
direction of that side, we can decompose any force in the force
polygon into its components. Thus the force polygon gives ns
complete information as to the action of the forces.
7. If the Force§ act in the same straight I-ine, the force
polygon of course becomes a straight line also, and the result-
ant is the sum or difference (algebraic sum) of the forces.
Thus, if we have Pl5 P2, P3, all acting at the point A, as
shown by the force diagram Fig. 4 (a), we form the force poly-
gon by laying off from 0, Fig. 4 (&), the intensity of P1? from
the end of this line Pl P2 equal to A P2 and from P2, P2 P3
equal to A P3. Then the line necessary to close the polygon is
evidently 0 P3 — P^ + P2 — P8. A single force acting then at A
in the direction of and having the intensity represented by the
line 0 P3 would replace P1? P2, and P3. If acting from P3 to 0,
it will produce equilibrium.
If we again choose an arbitrary point as C [we shall hereaf-
ter call this point the "pole" of the force polygon], and draw
lines S0 S3 from this pole to the beginning and end of the force
polygon, we can decompose the resultant into two forces in any
required direction. If the resultant is supposed to act down,
then the arrows show the direction in which these components
must act in order to replace the resultant. If then at A we
draw lines parallel arid equal, we have these components in posi-
tion, direction, and applied at the common point of application.
§. Practical Applications. — Simple and even self-evident
as all the preceding may seem, we have already acquired all
the principles requisite for a rapid, accurate, and very elegant
method of finding by diagram the strains in the various mem-
bers of all kinds of framed structures, such as roof trusses,
bridge girders, cranes, etc., no matter how complicated the
structure, or what special assumptions are made as to the load-
ing, provided only, that all the exterior forces are known. A
complicated or unsymmetrical arrangement of parts increases
greatly the labor of calculation, but has no effect upon the ease
or accuracy of the graphical method. The method moreover
checks its own accuracy, does not accumulate errors, and shows
in one view the relation of the strains to each other, and the
variations which would be caused by a change in the manner
of load distribution, or in the form of construction.
6 FORCES IN THE SAME P^ANE. [CHAP. I.
As this method is not as well known as it deserves to be, it
will perhaps be of advantage to pause for a moment in the
development of our subject, and make this direct application of
the principles already established.
BEACED SEMI-AKCH.
9. Stoney, in his " Theory of Strains," Yol. I., page 123,
gives the following example of a " braced semi-arch," repre-
sented by Fig. 5, PL 1. The dimensions are as follows: pro-
jecting portion, 40 ft. long, 10 ft. deep at wall. Lower flange,
circular, with a horizontal tangent 2 ft. below the extremity of
girder. Radius of lower flange, 104 ft. Load uniform and
equal to one ton per running foot supposed to be collected into
weights of 10 tons at each upper apex, except the end one,
which has only 5 tons.
Fig. 5 shows the arch drawn to a scale of 10 ft. to an inch.
This scale is too small in this case to ensure good results ; in
general the larger the scale to which the frame can be drawn,
the better; but for the purpose of illustration it will answer
well enough. With a large scale for the frame diagram, a
scale of 10 tons to an inch will in general be found to answer
well. Fig. 5 (a) gives the strains in the various members to a
scale of 10 tons to an inch and Fig. 5 (b) 20 tons to an inch ;
the first for the load at the extremity alone, the second for a
nniform load.
Fig. 5 (a) is thus obtained. We first lay off the weight, 5
tons, to scale, in the direction in which it acts ; i.e., down-
wards. Now this weight and the strains in diagonal 1, and
flange A, are in equilibrium ; therefore by article (4) the force
polygon must close. Drawing lines therefore from the ends of
the line representing the weight of 5 tons, parallel to these
pieces and prolonging them to their intersection, we obtain
the strains in A and 1. Commencing with the beginning of
the weight line and following down around the triangle thus
formed, we find that A acts from right to left, as shown by the
arrow. A acts then aioay from the apex ; it is therefore in
tension. Diagonal 1 acts towards the apex and is hence com-
pressed.
We pass now to apex a, of the frame. Here we have the
strains in E and diagonals 1 and 2, and these three strains hold
each other in equilibrium. The strain in 1 we have already,
CHAP. I.] COMMON POINT OF APPLICATION. 7
and know it to be compressive. We have then simply to draw
lines from 0 and b parallel to E and 2, and follow round the
triangle, to obtain the intensity and quality of the strains in E
and 2. We must remember that as 1 is in compression, and
we" are now considering apex #, we must follow round from o
* to b in Fig. 5 (#), and so round. We thus find 2 acting away
from apex a and therefore in tension, and E acting towards
this apex, and hence com/pressed.
Pass now to apex c. We have the strains in A and 2 in
equilibrium with B and 3. [No weights are supposed to act
except the one at the end.] But A and 2 we already have.
We draw 3 and B. Diagonal 2 has been found to be in ten-
sion. With reference to apex c it must therefore act away
fromc; i.e., from d to b in the force polygon. This is suffi-
cient to give us the hint how to follow round. We pass from
d to b for 2, from b to e for A, then from e to B and from B to
d for B and 3. B is therefore tension and 3 compression.
And so we proceed. For the next apex g, we have E and 3 in
equilibrium with F and 4. AVe draw parallels to F and 4 so
as to close the polygon of which we have already two sides, E
and 3, given, and remembering that as 3 is in compression, it
must therefore act towards g, we follow round the completed
polygon with this to guide us, and find 4 tension and F com-
pression. Thus we go through the figure, and when all is
ready we can scale off the strains. The strains in the lower
flanges it will be observed all radiate from o. The upper
flanges are all measured off on the horizontal e C, and the dia-
gonals are the traverses between. We see at once that however
irregular the structure, we can always easily and readily deter-
mine the strains at any apex, provided no more than two un-
known strain^ are to be found. If more than two pieces, the
strains in which are unknown, meet at an apex, we can evi-
dently form an indefinite number of closed polygons. The
problem is indeterminate, and the structure has unnecessary
or superfluous pieces.
Fig. 5 (b) gives the strains for a uniform load, taken, for con-
venience of size, to a scale of 20 tons to an inch. Here until
we arrived at apex c of the frame the strains are evidently the
same as before. Observe the influence of the weight at c.
Here we have the strains in A and 2 given in the diagram, in
equilibrium with B, 3 and the known weight acting at c\ viz.,
8 FORCES IN THE SAME PLANE. [CHAP. I.
10 tons. We lay off therefore 10 tons downward from e, Fig.
5 (£), and follow down from e around the polygon. We thus
find B tension and 3 compression. Then 4 and F are found as
before for apex <7, 4 tension and F compression ; and then we
come to the next apex and the next weight. This is laid off
downwards from the end of the preceding, and then we follow
round, finding C tension and 5 compression ; and so on.
1O. As another example, let us take the
given in Fig. 6, PI. 2. This truss is given by Stoney, Yol. I.,
page 128. Dimensions : span, 80 ft. : rise of top and bottom
flanges, 16 and 10 ft. respectively. Kadii, 58 and 85 ft. The
figure shows two different kinds of bracing. In the left-hand
part the extreme bay of the lower flange is half as long again
as the others. The upper flange is divided into 4 equal bays.
In the right-hand section, both flanges are divided into 4 equal
bays, and every alternate brace is therefore nearly radial. Each
upper apex in both cases is supposed to sustain a weight of
one ton.
The strains in the various pieces are given in Fig. 6 (a).
We form the force polygon by laying off the weights from 0
to Y and then laying off the reactions 3.5 apiece, upwards, we
come back to 0, and the force polygon is closed as it should be,
since the sum of the reactions must be equal and opposite to
the sum of the weights. Starting then with the reaction at the
left support A, we go through from apex to apex in a manner
precisely similar to the previous case. The operation is so
simple that it is hardly necessary to detail it again, but we
recommend the reader to go over it with the aid of Fig. 6 (#),
lettering the figure as he proceeds. The dotted part gives the
strains for the right-hand half.
DIAGRAM FOR WIND FORCE.
11. It is of considerable importance to investigate the influ-
ence of a partial load, such as that caused by the wind blowing
on one side of the roof, and this by the aid of our method we
can easily do.
From the experimental formulae of Hutton,*
* Iron Bridges and Roofs. Unwin. p. 120.
CHAP. I.] COMMON POESTT OF APPLICATION.
Pn=P sin. i
Ph=:P sin. i 1-84cos-i
Pv=Pcot. isin, i1-84008-1
where P is the intensity of the wind pressure in Ibs. per sq. ft.
upon a surface perpendicular to its direction, i is the inclination
of any plane surface to this direction ; Pn is the normal pres-
sure, Ph the horizontal component of this normal pressure, and
Pv its vertical component.
That is, if the wind blows horizontally, Ph is the horizontal
and Pv the vertical component of the pressure on the roof. If
we take P=40 Ibs., which probably allows sufficient margin
for the heaviest gales, we have the following values of the nor-
mal pressure and its components for various inclinations of
roof surface :
Angle of
Koof
Lbs.
per square foot
of surface.
Pa
Pv
Ph
5°
5.0. ...
4.9.
0.4
10°
9.7....
9.6.
1.7
20°
18.1
17.0.
6.2
30°
26.4....
22.8.
13.2
40°
33.3....
25.5.
21.4
50°.......
38.1
24.5.
29.2
60°
40.0....
20.0.
34.0
70°
41.0....
14.0.
38.5
80°
40.4....
7.0.
39.8
90"
40.0..
0.0.
40.0
The load at each joint may be taken as equal to the pressure
of the wind striking a surface whose area is equal to that por-
tion of the roof supported by one bay of the rafter, and inclined
at the same angle as the tangent to the rib at the joint. Thus
we can calculate P1? P2, P3, P4, (Fig. 6), resolve these forces into
their horizontal and vertical components, and find the reactions
at the supports as well as the horizontal force at the left abut-
ment, which in our construction is supposed to be fixed. Should
the wind be supposed to blow from the right side, the strains
would be entirely different, and it would be necessary to form
a second diagram. Each piece must be proportioned to resist
the strains arising in either case. The forces P^ and their
horizontal and vertical components, as also the reactions, being
known, we can now form the force polygon.
Thus in Fig. 6 (&), we lay off the forces P4.l5 make a c equal
10 FOKCES IN THE SAME PLANE. [CHAP. I.
to the vertical reaction at A, a b = the sum of the horizontal
components, or the horizontal force at A, and o 1) the vertical
reaction at the right snppprt. This last line should close the
force polygon and bring us back to o.
Now starting at the left support, we have the vertical reac-
tion a c, the horizontal force a b, and the wind force Pl5 in
equilibrium with A and E. Closing the polygon by lines par-
allel to A and E, we obtain the strains in these pieces, E ten-
sion and A compression. At the next apex we have A and P2
in equilibrium with 1 and B. Completing the parallelogram,
we find 1 compression and B compression. At the next apex
1 and E are in equilibrium with* 2 and F, and we find F and 2
tension and so on. The upper flanges are in compression and
start from the ends of the forces Pl5 P2, etc. The lower flanges
radiate from b. If we were to carry out the construction for
the rest of the frame, the upper flanges after D would radiate
from o.
A comparison of Fig. 6 (a) and (b) shows that whereas under
uniform load the strain in 1 is tension, for wind force the same
brace is in compression. In fact in the first case all the braces
are in tension, while in the second 1, 3, and 5 are compressed,
and 3 and 5 quite severely. The strains in the bracing gener-
ally are much greater in the second case.
Were we to consider the wind as blowing from the other
side, or what is the same thing, suppose the right end fixed and
the left supported on rollers, then the horizontal reaction a b
will be applied at the right abutment. In this case the lower
flanges will radiate from a instead of &, and the first upper
flange will start from o. Supposing the first two lines of this
new diagram drawn, as indicated by the dotted lines, and fol-
lowing round from b to o, and so round to a and back to 5, it
may easily happen that the last upper flange is in tension and
the last lower flange in compression; that is, a complete reversal
of the ordinary condition of strain.
For an excellent presentation of the above method, we refer
the reader to Iron Bridges and Roofs, by W. C. Unwin, pp.
128-140. The above method is there referred to as "Prof.
Clerk MaxweWs Method" and as such is known and used in
England.*
* Phil. Mag., April, 1864, and a Paper read before the British Association for
the Advancement of Science, by Prof. Maxwell, in 1874.
CHAP. I.] COMMON POINT OF APPLICATION. 11
BRIDGES.
12. For bridges the strains due to a uniform load are of
course easily found. In most cases a rolling load can be man-
aged also, without making a separate diagram for each position
of the load. Thus, if we diagram the strains for the load at
the first and last apex, the strains due to intermediate loads
will be multiples or submultiples of these. A calculation for a
simple Warren girder of small span, and a consideration of the
reaction for each position of the load, will at once illustrate
what is meant. [Compare Stoney, Theory of Strains. Pp.
99-111, Yol. I.]
Thus Stoney, in his Theory of Strains, Yol. I., p. 99, gives
the girder represented in Fig. 7, PL 2, span 80 ft., depth of
truss, 5 ft., 8 equal panels in upper flange, 7 in lower.
For the first weight of 10 tons, P1? the strains are given by
Fig. 7 (a] to a scale of 10 tons to an inch. We form first the
force polygon by laying off from o, 10 tons, to P^ From the
end of this line we lay off upwards the reaction at right abut-
ment = •$• of 10 tons, or 1.25 tons ; and then the reaction at the
left abutment = -J of 10 tons, back to 6>, thus closing the force
polygon. [Note. — In any structure which holds in equilibrium
outer forces, the force polygon must close. If it does not, there
is no equilibrium, and motion ensues (see Art. 20).] Com-
mence now with the reaction at a in the frame diagram, Fig. 7,
because here we have a known reaction, a o (force polygon),
and only two unknown strains to be determined. Drawing
lines parallel to A and 1, we obtain the strains in A and 1.
Then pass on to apex b. With the now known strain in 1, we
can determine 2 and E.
Passing now to the next apex, we have A and 2 known, and
also the weight Pt. Join therefore Pl and E [Fig. 7 (a)~\ by
lines parallel to B and 3. B and 3 are both in compression.
We find diagonal 2 also in compression, and 1 in tension. That
is, ~both the diagonals under the weight are compressed, as evi-
dently should be the case. From 4 on we have tension and
compression alternately.
Fig. 7 (b) gives the strains due to the last position of the load
P7. The strains in the diagonals are evidently all equal, and
alternately tension and compression.
Now it is not necessary to construct more than these two dia-
grams. From these two alone we can determine the strains fur
12
FORCES IN THE SAME PLANE.
[CHAP. I.
any intermediate weight. Thus scaling off the strains in Fig.
7 (a) and (&), we can tabulate them under Px and P7, as shown
by the table.
DIAGONALS.
Pi
P2
P3
P4
P5
P6
PT
C
+
T
1
—12 4
—10 6
— 89
— 71
— 53
— 35
1 8
49 6
2
+ 12 4
+ 10 6
+•89
'
+ 71
+ 53
+ 35
+ 18
+ 49 6
3
+ 1.8
—10.6
— 8.9
— 71
— 53
— 3.5
— 18
+ 18
—37 2
4
— 18
+ 10.6
+ 89
+ 7.1
+ 53
+ 35
+ 18
+ 37 2
— 18
5 ....
+ 18
+ 35
— 89
— 7 1
— 5.3
— 35
— 1.8
+ 53
26 6
6
— 18
— 35
+ 89
+ 7.1
+ 53
+ 35
+ 18
+ 26 6
53
7
+ 18
+ 35
+ 53
— 71
— 53
— 35
— 18
+ 10 6
—17 7
8
- 1.8
— 35
— 53
+ 7.1
+ 53
+ 35
+ 1.8
+ 17.7
—10 6
Now the reaction at the left abutment due to P6 is twice
as great as that due to P7. Hence the values in the column
for P6 will be twice as great ; in the column for P5 three times
as great, and so on. For similar reasons the strain in 5 for
P2 will be twice that for P^ In column P2, then, from 5
down we multiply the strains in P1 by 2. In P8 from 7 down
by 3. Thus we fill out the table of strains completely, and find
the maximum tension and compression. A similar procedure
will give the flanges.*
APPLICATION TO AN AECH.
13. For a " braced arch " (Stoney, p. 136) as represented in
Fig. 5 (c) PL 2, the strains in every piece due to any load are
in similar manner easily found by first finding the components
of the load acting at the abutments, and then proceeding as
above. Thus for a load P2, the left half of the arch is in equi-
librium with the forces acting upon it ; viz., a horizontal and a
downward force at a, and a horizontal and an upward force at
A. The resultant of the forces at a must then pass through
* The reader not familiar with the above method of tabulation will find it
further illustrated in Art. 7 of the Appendix. He cannot do better than to
refer to it here and now.
CHAP. I.] COMMON POINT OF APPLICATION. 13
a and A, and be equal and opposite to the resultant at A. The
resultant at the right abutment must pass through that abutment,
and also through the intersection of P2 with A a. So for any
other force, as P6, we have simply to draw B a to intersection
with P6, and then P6 A. We can now decompose P6 or P2 along
the resultants through the abutments thus found. Thus resolv-
ing P2 along A a and P2 B, Fig. 5 (e\ we find the force acting at
apex a. This force resolved into A and 1 gives the strains on
these pieces both compressive. Passing then to the next apex,
we obtain the strains in 2 and E. Then to the next, and we
get 3 and B, compression and tension respectively, and so on,
as shown by diagram, Fig. 5 (<?), which, it will be seen at once,
is similar to Fig. 5 (a), already obtained for the " semi-arch,"
except that the strain in A is less than for the semi-arch and
compressive, while B C and D are in tension. The reason is
obvious. At a [Fig. 5 (c)] the resultant lies between A and 1,
and therefore causes compression in both, while it passes out-
side of the arch entirely, to the right of the apex for diagonals
3 and 4, and hence causes tension in B C and D. Fig. 5 (d)
gives the strains due to P6. Here the resultant or reaction at
A is first found and resolved into 9 and H, and then we go
through the frame as before. We see that 4 and 5 under the
load are both compressed, that E and F are in tension and G-
and H, as also the entire upper chord, in compression. The
work checks from the fact that the line closing the polygon
formed by E and 2 should be exactly parallel to and give the
strain in diagonal 1, or A and 1 should be in equilibrium with
the resultant through a [see Fig. 5 (d)~\.
In every case of the kind we first, then, have to draw the
frame diagram. Then lay off \\\Q force polygon which should
close. Finally we construct the strain diagram. The frame
diagram should be taken to as large a scale as possible consist-
ent with reasonable size, and the scale for the force and strain
diagrams as small as possible, consistent with scaling off the
strains to the requisite degree of accuracy. A small frame
diagram does not give with the proper accuracy the relative
positions and inclinations of the various pieces, so as to ensure
the proper direction for the lines of the strain diagram. A
slight deviation from parallelism causes sometimes considerable
variation. Nevertheless with practice, care, and proper instru-
ments the accuracy of the method is surprising ; even in com-
14 FORCES IN THE SAME PLANE. [CHAP. I.
plicated structures, the variation resulting from performing
the operation twice being inappreciable. Every symmetrical
frame gives also a symmetrical strain diagram, and the accu-
racy of the work is tested at every point by this double sym-
metry, and finally by the end or last point of the second half,
exactly coinciding with the last point of the first half. Thus
in Fig. 6 (a), if we had but one system of triangulation carried
through the frame, the strain diagram for the right half would
be precisely similar and symmetrical to that already found for
the first, and the end of the last line would fall, or should fall,
precisely upon the point b of the first. If it does not, and the
error is too great to be disregarded, then by checking corre-
sponding points in each half, we can find the point where the
error was committed. In any case errors do not accumulate.
Thus, armed with straight edge, scale, triangle, and dividers,
we can attack and solve the most intricate problems, without
calculation or tables, with ease, accuracy, and great saving of
time.
METHOD OF SECTIONS.
14. The results obtained by the above method are best
checked in general by Hitter's u method of sections," or the
use of moments.* This consists in supposing the structure
divided by a section cutting only three pieces. We can then
take the intersection of two of these pieces as a centre of mo-
ments, and the sum (algebraic) of the moments of all the
exterior forces, such as reaction, loads, etc., upon one of the
portions into which the structure is divided by the section, with
reference to this centre of moments, must be balanced by the
moment of the strain in the third piece, with reference to this
same point. Thus in Fig. 6, PI. 2, required the strain in D.
Take a section through D, 7 and H (right half of Fig.), and let
a be the centre of moments. The moments of the strains in 7
and H are then, of course, zero, since these pieces pass through a.
The moment of the strain in D with reference to a must then
be balanced by the sum of the moments of all the outer forces
acting upon the portion to the left (or right) of the section.
Thus, strain in D multiplied by its lever arm with respect to
#, is equal to moment of reaction at A, minus sum of the mo-
ments of loads between A and #, all with reference to a. If
* Dock- und Brucken- Constructional. Ritter. Hannover, 1873.
CHAP. I.] COMMON POINT OF APPLICATION. 15
we take the direction of rotation of the forces on the left of the
section when in the direction of the hands of a watch as posi-
tive^ and find the moment of strain in D negative, it shows i
negative rotation about a, and the strain in D to resist this rota-
tion must act away from &, or be tensile. If the resultant
rotation of the outer forces is on the other hand positive, the
strain in D must act toward b, and D is therefore compressed.
This method of calculation, it will be observed, is both sim-
ple and general. It can be applied to any structure, when the
outer forces are completely known, and only three pieces are
cut by the ideal section.
15. It is unnecessary to give here further applications of our
graphical method. The reader can easily apply it for himself
to the " bowstring girder," bent crane, etc., and satisfy himself
as to its accuracy, and the ease with which the desired results
are obtained.
Enough has been said to indicate the many important appli-
cations which even at the very commencement of our develop-
ment of the graphical method we are enabled to make, and
here we shall close our discussion of forces lying in the same
plane and having a common point of application. As we pass
on to forces having different points of application, we shall
have occasion to develop new principles and relations not less
fruitful and useful in their practical results.*
* We refer the reader here to the Appendix to this chapter for further
illustrations of the application of the above principles, as well as for informa-
tion upon several points of considerable practical importance. We would also
remind him here once for all, that the Appendix to this work was NOT in-
tended to be disregarded, but has been thought desirable in order to avoid
encumbermg the general principles with too much of detail in the text. We
earnestly request him to neglect no reference to it which may be made in the
text.
He will do well in the present case, after first making himself familiar with
the above points, to solve for himself with scale and dividers a number of
similar problems, checking his results always by the method of moments.
He will thus in a very short time master the method, and be able to solve
readily and accurately every problem of usual occurrence in practice.
Though the method is very simple, actual practice with the drawing board is
here indispensable.
16 FORCES IN THE SAME PLANE. [CHAP. II.
CHAPTER II.
FORCES IN THE SAME PLANE DIFFERENT POINTS OF APPLICATION.
16. Re§ultant of Two Forces in a Plane— Different
Points of Application. — Heretofore we have considered
forces having a common point of application, and have seen
that in any case the direction and intensity of the resultant is
easily found by closing the force polygon.
But suppose we have two forces Pl P2 having different
points of application At A2 ; required the position and direc-
tion of the resultant [PI. 3, Fig. 8].
Any force acting in a plane may be considered as acting at
any point in its line of direction.
P! and P2 may then be supposed to act at their common
point of intersection #, and through this point the resultant
should pass. The case reduces therefore to a common point
of application. The resultant is given in intensity and direc-
tion as before by the force polygon (&), and its position is deter-
mined by the point of intersection a. At this point, or at any
point in the line through a, parallel to 0 2, the resultant may
be supposed to act.
But the direction of the forces may not intersect within
reasonable limits, or the forces may be supposed parallel to
each other, so that they may not intersect at all. In any case
the force polygon will still give the intensity and direction of
action of the resultant, but its position in the plane of the
forces remains yet to be determined. Now we have se.en [Art.
5] that we can decompose a force into two components in any
desired directions, by choosing a "pole " and drawing lines to
the beginning and end of the force in the force polygon. Let
us choose then a pole C [Fig. 8 (5)] and decompose the result-
ant thus into two forces given in intensity by the lines 0 C
and 2 C. The forces Px P2 being supposed to act at the
points A! A2 in the common plane, at what point in the plane
and in what direction must the resultant 0 2 be applied to keep
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 17
this plane and hold the forces in equilibrium? The direction
of action of the resultant is given at once from the force poly-
gon [Art. 5 (&)]. It must act in a direction from 2 to 0, and
must be equal to 2 0, taken to the scale of force. Now at any
point in the line of direction of P1? as for instance 1, let us
suppose the component given by C 0 to act. What is then the
resultant of P! and CO? A glance at the force polygon gives
us 1C, because this line closes the polygon made by C 0, 0 1
and 1C. At I then, the three forces S0 (parallel and equal to
C 0) St (parallel and equal to 1 C) and P! are in equilibrium,
and there is no tendency of the point 1 to move. But 1 C or
51 may be considered as acting in the plane at any point in its
line of direction ; therefore at 2 its intersection with P2 pro-
longed. Suppose at 2, S2 or 2 C to act. "We see at once from
the force polygon that 2 C, C 1 and P2 are in equilibrium.
There is therefore no tendency of the point 2 to move, and the
two forces Pl P2 are then in equilibrium with C 0, 1 C, C 1
and 2 C. But since the resultant of C 0 and 2 C or of S0 and
52 is also the resultant of the forces, and since it must there-
fore act through the point of intersection of S0 and S2 : we
have only to prolo ng these lines to intersection b. Through
this point the resultants R^2 must pass and acting downwards
(from 0 to 2) as indicated in the Fig., it replaces Pl P2. Act-
ing upwards it would hold them in equilibrium. We thus
easily find the point 2 in the plane at which 2 C or S2 must
be applied, when C 0 or S0 acts at 1, and S0 ?2 are thus found
in proper relative position. The position, intensity, and direc-
tion of the resultant are thus completely determined.
Had we taken ai*y other point than 1, as the point of applica-
tion of C 0, we should have found a different corresponding
point for application of 2 C, but in any case the prolongations
of 2 C and C 0 would intersect upon the line a b, prolonged if
necessary. The same holds true for any position of the "pole "
C. This construction is evidently general whatever the posi-
tion or whatever the number of the forces. We may thus
obtain any number of points along the line a b ; that is, the
resultant also, may act at any point in its line of direction.
[NOTE. — That b is a point in the resultant of Pt andP2 can
be proved in a method purely geometrical. In the two " com-
plete quadrilaterals " 0 1 2 C and 1 b 2 #, the Jive pairs of
corresponding sides 0 1 and a 1, 1 2 and a 2, 2 C and b 2, C 0
18 FORCES IN THE SAME PLANE. [CHAP. II.
and b 1, C 1 and 1 2, are parallel each to each, therefore the
sixth pair 0 2 and a b must also be parallel / b is therefore a
point of the resultant passing through a, parallel to 0 2.]
17. The above Construction holds good equally well for
Parallel Forces. — By means of it we find in PL 3, Fig. 9 (a)
and (b) and Fig. 10 (a) and (&), the resultant of a pair of paral-
lel forces, in the first case, both acting in the same direction ;
in the second, in opposite directions.
In both cases we have simply to choose a pole C, and draw
S0 Sj and S2. Then taking any point c in the line of direc-
tion of Pl5 as a point of application for, S0, draw through this
point Sl5 thus finding d, the point of application for S2. S0
and S2 prolonged, intersect upon the resultant, whose intensity,
direction, and position thus become fully known.
18. Property of the Point b. — It is plain that thus a point
of intersection J, through which the resultant must pass, can
always be found, provided S0 and S2 do not fall together in
the force polygon, or intersect without the limits of the draw-
ing. By properly choosing the position of the pole C, this can
always be avoided if the points 2 and 0 in the force polygon do
not themselves coincide, i.e., if the force polygon does not close.
The point &, Figs. 8, 9, and 10, which by reason of the arbi-
trary position of the pole may lie anywhere upon the resultant,
has a remarkable property. If we draw a line m n through
this point parallel to S^ and let fall from it perpendiculars pt
and p2 upon Pl and P2, then in all three cases, and therefore
generally, the triangle c m b is similar to 0 C 1, and d b n is
similar to 1 C 2. Hence we have the proportions —
0 1 : 1 C I ". c m : m b, and
1 C : 12 ; : n b : n d.
From these proportions we find
0 ~L : I 2 \\ c m x nb : mb x n d.
Now the triangles c m b and d n b have the same height
above the base m n ; the bases m b and b n are therefore pro-
portional to their areas. But their areas are equal to half their
sides cm and n d multiplied byj?i and^>2 respectively. Hence
we have from the above proportion, since c m = n d,
0 1 : 1 2 ; : n d x pz : n d x pt or
01:12 ::pz:p,.
That is, the perpendiculars let fall from any point of th/>,
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 19
resultant upon the components, are to each other inversely as
the components. Regarding any point of the resultant as a
centre of moments, the moments of the forces then are equal,
and of course the forces themselves are inversely as their lever
arms.
19. Equilibrium Polygon. — If we consider the forces Pt
P2, Figs. 8, 9, and 10, held in equilibrium by their components
C 0, 1 C, and 2 C, C 1, which act parallel to the lines S0 Sx
and S2 ; then regarding the line St or c d as part of the mate-
rial plane in which the forces act, C 1 and 1 C balance one
another, and cause either tension or compression in c d. Sup-
pose the resultant R is to act so as to cause equilibrium, or
prevent the motion of the plane due to Pt and P2. Then R
must act upwards in Figs. 8 and 9, and downwards from 2 to
0 in Fig. 10. In Figs. 8 and 9 then, S0 and S2 act away from
c and d (Art. 4), and in Fig. 10 towards c and d. Following
round the force polygon, we find in the first two cases c d in
tension, in the last c d in compression.
In the first two cases, the points of application c and d of S0
P! and S2 P2 if* connected by a string stretched between c
and d will be perfectly fixed and motionless ; while in the lat-
ter case, the string must be replaced by a strut. In case of
three or more forces the polygon or broken line which we thus
obtain, by choosing a pole, drawing lines to the beginning and
end of the forces in the force polygon, and then parallels to
these lines intersecting the lines of direction of the forces in the
force diagram, we call the " string " or "funicular polygon"
or the " strut polygon" according as the forces act to cause
tension or compression along these lines. We can apply to
both cases the general designation of polygon of equilibrium or
" equilibrium polygon" The perpendicular let fall from the
pole C upon the direction of the resultant in the force polygon,
we call the "pole distance "and shall always designate it by
H. The straight line joining the points c and d, or the begin-
ning and end of the equilibrium polygon, we call the "strut"
or " tie line " or generally the " closing line " and designate it
by L. The convenience and ' application of these terms and
conceptions will soon appear. In the present case of only two
forces, the equilibrium polygon becomes a straight line and
coincides with L, or c d.
[XoTE. — We repeat that in order to determine the quality of
20 FORCES IN THE SAME PLANE. [CHAP. II.
the strain in c d, we have only to follow round the force poly-
gon in the direction of the forces, and then refer to the force
diagram. Thus Fig. 9, at c, Px S0 and Sx act, and are in equi-
librium. The corresponding closed figure is given in the force
polygon (a). S0 acts away from c, Px acts downwards from 0 1.
Continuing this direction we find Sx acting from 1 towards C.
Reversing this direction (Art. 4), we find that the resultant
which replaces S0 and Pt acts from C to 1. Referring now to
the force diagram (&), and transferring this direction to the
point c, we find this resultant acts to pull c away from d or
contrary to the direction of the force 1 C which replaces S2 and
P2. The strain in c d is therefore tension.
A much better way of arriving at the same result is to con-
sider the triangle c b d as a jointed frame which holds in equi-
librium the forces P! P2 and R^. Then the strains in any two
pieces c d, c ft, meeting at a point, are in equilibrium with the
force or forces acting at that point.
We have then the force Pj acting at apex c, decomposed into
strains along c b and c d (Art. 5). represented by C 0 and 1 C in
the force polygon. All three are in equilibrium. Pl acts
down. Follow down then from 0 to 1 from 1 to C and C to 0.
Refer back now to apex c of the frame and transfer these
directions. The strain in c d acts away from the apex c and is
therefore in tension, while the piece c b would be in compres-
sion, since the direction of C 0 is towards apex c.
See also " practical applications " of the preceding chapter
for illustrations of this. In the same way follow round 0 1 C
Fig. 10 (a) and refer to (b) and S0 is in compression^
2O. Ca§e of a Couple. — In Article 18 we remarked that the
pole can always be chosen in such a position as to give S0 and
S2 intersecting within desired limits, provided that S0 and 83 or
the point 0 and 2 do not coincide. This case however actually
happens, with a pair of equal and opposite forces — that is, with
a couple.
Thus in Fig. 11, PI. 3, we have two equal and opposite force's
PI, P2-
The force polygon closes : therefore the resultant is zero.
S0 and 83 are parallel, hence their point of intersection in the
equilibrium polygon is infinitely distant. By changing the
position of the pole, we see that S0 and S2 may take any posi-
tions in the plane.
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 21
Two forces therefore which form a couple cannot be replaced
by a single force. Their resultant is an indefinitely small force
situated in any position in the plane of the forces, at an infinite
distance.
Conditions of Equilibrium. — If then, similarly to Art. 4,
any number of forces lying in the same plane and having differ-
ent points of application, are in equilibrium, the force polygon
always closes.
For this reason, as already repeatedly seen in the practical
applications of our last chapter, the force polygon form'ed by
the exterior forces must always close.
But inversely, if the force polygon closes, it does not follow
that the forces are in equilibrium — a couple may result.
To determine whether this is the case inspect the " equilibri-
um polygon." If this also closes [i.e., if S0 and Sn intersect]
the forces are in equilibrium. If this does not close [i.e., if S0
and Sn are parallel] there is no single resultant, but the
forces can be replaced by a couple, and this couple, as we have
seen, may have any position in the plane.
21. Thus if we suppose in Fig. 11, PI. 3, Pt and P2 decom-
posed into their components S0, S1? and S1? S2, the compressive
strains in Sl at c and d are equal and opposite [see (a)]. We
have then S0 and S2 remaining, which again form a couple
which must have the same action as the first.
Hence we see that one couple can be replaced by another with-
out changing the action of the forces.
It is easy to determine a simple relation between any two
couples.
If from c we lay off c a equal to o 1, and c o equal to Co, we
have o a parallel to C 1 or S1? and therefore to c d. Join a d and
o d. The triangles c d a and c d o having a common base c d
and their vertices o and a in a line parallel to c d, are equal in
area. The side c a of one is known, and the opposite apex lies
in the line of the force P2. Its area is then ca.= Pl multiplied
by half of the perpendicular distance of Pt from P2, and is
therefore completely determined. So also for the other trian-
gle, one side of which o c is one force of the new couple, and
the opposite apex of which lies in the other force S2.
Hence — a couple can be turned at will in its plane of action,
and the intensity and direction of its forces can be changed at
will if the area of the triangle the base of which is one of the
22 FOSCES IN THE SAME PLANE. [CHAP. II.
new forces, and whose opposite apex lies in the other force, is
constant; or when the product of the intensity of the forces
into their perpendicular distance remains the same. The di-
rection *of rotation, of course, must also remain the same.
We shall see further on the significance of this area, or of
this product — so much is clear, that a couple (or infinitely small,
infinitely distant force) is completely determined in its plane
when the direction of rotation is given, and the area of the tri-
angle or value of the product to which it is proportional, is
known. The couple itself can be replaced by any two parallel
equal and opposite forces whatever, if only the triangle having
one force as base, and the opposite apex in the other, has a given
constant area.*
22. Force and Equilibrium Polygons for any Number
of Force§ in a Plane.
In PL 3, Fig. 12 (b) we have the forces Pw acting in various
directions and at different points of application. P2 and P3
form a couple ; that is, are equal, parallel, and opposite in di-
rection. Required the position, intensity and direction of action
of the resultant.
First, form the force polygon, Fig. 12 (a\ by laying off the
forces to scale one after the other in proper direction. Thus
we have 0 1, 1 2, 2 3, 3 4, 4 5 in Fig. 12 (a) parallel respec-
tively to P! P2 P3, etc., in Fig. 12 (b). The line necessary to
close the polygon, 0 5, is the resultant in intensity and direc-
tion. In intensity because the length of 0 5 taken to the scale
of force, gives the intensity of the resultant ; in direction
because acting from 5 to 0 it produces equilibrium, while act-
ing in the opposite direction, from 0 to 5, it replaces the forces.
"We have, therefore, only to find the position of the resultant
in the plane of the given forces in Fig. 12 (b). Hence :
Second, choose anywhere a "pole " as C, and draw the lines
or rays, or " strings " S0 St S2 S3 S4, etc. S0 and S5 are evi-
dently components of the resultant, since they form with it a
closed figure in the force polygon.
Third, form the equilibrium polygon a b cde o', Fig. 12 (J),
as follows :
Draw a line parallel to S0 intersecting T?± (produced if neces-
sary) at any point as a. From this point draw a line parallel
* Eltmente der Graphischen Statik. Bauschinger. Munchen. 1871. Pp.
11, 12.
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 23
to Si to intersection with P2 (also produced if necessary) at b.
From b parallel to S2 to c, then parallel to S3 to d, and finally
parallel to S4, to intersection e with P5. Through this last point
draw a line parallel to the last ray S5. Now S0 and S5 are com-
ponents of the resultant 0 5 [Fig. 12 (a)~\ and are found in
proper relative position. Produce them, therefore, to intersec-
tion o '. Through this point the resultant must pass. Drawing
then through o', a line parallel to 0 5, we have the resultant in
proper position, and acting in the direction indicated in the fig-
ure, it produces equilibrium.
Any other point than a, upon the direction of P1} assumed as
a starting point, would have given a different point o' ; so also
for any other assumed position of the pole C. But in every
case we shall obtain a point upon the line of direction of R^
already found. The reader may easily convince himself of this
by making the construction for different poles, and points of
beginning.
ISTow the polygon or broken line, a b c d e, we call the equi-
librium polygon — that is, it is the position which a system of
strings or struts, S0 Si S2, etc., would assume under the action
of the given forces at the assumed points of application.
Thus Pt acting at a, is held in equilibrium by the forces along
S0 and S1? P2 acting at £, by Si and S2 and so on. If we join
any two points in the line of direction of S0, and S5, as m n by
a line, we have then a jointed frame, which acted upon at the
apices a. . .e by the forces P!. . .P5, and at m and n by S0 and
S5 is in equilibrium.
For S0 acting at m, we see from the force polygon may be
replaced by a force a 0 parallel and opposed to the resultant R
and a force C a acting along the line L. In like manner S5 may
be replaced by a C and 5 a parallel and opposed to the result-
ant. The two forces a C and C a being equal and opposed
balance each other through m n, while the sum of 0 a and 5 a
is equal and opposed to the resultant 0 5. There is, therefore,
equilibrium, and m and n may be considered as the points of
support of the frame acted upon by the forces Pt. . .P5 at the
apices a. . .e, a Q and 5 a being the upward reactions at the
points of support.
As to the quality of the strains in the different pieces ; as
before the reaction at m, viz., a 0, is in equilibrium with the
strain in m n and m a. Following round, then, in the force
24: FOKCES IN THE SAME PLANE. [CHAP. II.
polygon from a to 0, 0 to C and C to a, and referring back to
the frame, we find strain in in n acting towards apex ra, there-
fore compressive ; strain in m a acting away from m, theref ore
tensile. In like manner Sx S2 S3 are in tension, while S4 or d e
and S5 or e n are compressed.
Hence we m&jfix any two points of the equilibrium polygon
by joining them by a line. The forces acting at these points
are at once found by drawing from C in the force polygon a
parallel to this line to intersection with resultant. Thus a C
(since we have taken m n parallel to Sx) is the force in m n and
a 0, 5 a, are the forces opposed to the resultant at m and n.
23. Influence of a Couple. — Among the forces in Fig. 12
there are two, P2 and P3 which are equal, parallel and opposite,
the direction of rotation being as indicated by the arrow. Ex-
amining the equilibrium polygon, we see that the influence of
the couple is to shift St through a certain distance parallel to
itself, to S3. Now suppose the forces composing the couple
were not given, but the value of the couple known, from the
direction of rotation and the area of the triangle A2 P2 P3,
which has its base equal to one of the forces and a height equal
to their perpendicular distance. In this case the lines 1 2, and
S2 in the force polygon, would disappear, but we can none the
less find the point d, and from this point continue the polygon
by drawing S4 and S5, and thus find the same points e and o' as
before. To do this we have simply to apply the principle
deduced in Art. 21, that one couple can be replaced by another
provided the area of the triangle is constant.
In the present case we must replace the given couple by
another whose forces are Si and S3, having the same direction
of rotation.
Lay off then from a, a i equal by scale to S1 as given in the
force polygon. Describe upon St the triangle a g h equal to
the given area A2 P2 P3. Draw g i, and then through h, h 7c
parallel to g i. The point Jc is upon the line of direction of
S8, or in other words the area of the triangle i J& a is equal to
a g h. The proof is easy. The two triangles ig hvxAigk
are equal, since they have the same base i g, and height. But
if from the triangle a i g we subtract i g h, we obtain a g h.
If from the same triangle a i g we subtract i g k, which is equal
to i g A, we obtain i k a. Equals subtracted from equals leave
equals. Hence ik a is equal to a g h.
CHAP. II.] DIFFERENT POINTS OF APPLICATION. 25
If then through 7c we draw a line parallel to S3 and produce
it to d. we have the same point as before, and thus from d, can
continue the polygon.
\_Note that the direction of rotation shows the side of Sj upon
which the point 7c must fall. St acts away from a [from 1 to
C in (a)] hence for rotation as shown by the arrow, g must fall
above S1? and $! is shifted upwards.
24. Order of Forcc§ Immaterial. — As in the case of a com-
mon point of application, so also here, the order in which the
forces are laid off is immaterial. To prove this for two forces
is sufficient, as by continued interchange of two and two, we
can obtain any desired order.
Let the two forces be P4 and P5 (Fig. 13, PL 3) existing either
alone, or in combination with others preceding and following.
Taking the forces first in the order P4 P5, we have the equi-
librium polygon S3 S4 S5, (b) giving the point a in the result-
ant. Taking them now in reverse order, P5 P4, we have the
polygon S3 S'5 S'4 giving the same point a in the resultant. The
resultant in the force polygon (a), viz., 0 5, is of course un-
changed in intensity and direction in either case. It is required
to prove that in the second case the last string S'4 is not only
parallel to S5 in the first, but coincides with it.
This is easy. The resultant of P4 P5 goes through a, the in-
tersection of S3 and S5. The same resultant in the second case
must also pass through the intersection of S3 and S'4. But S3 is
the same in position and direction in both cases. If the second
point of intersection does not coincide with a, still it must lie
somewhere upon S3. Hence as the resultant must pass through
both points, it must coincide with this last line ; viz., S3. But
this is not possible, as the resultant must also pass through d,
the point of intersection of the forces, or when these do not
intersect must be parallel to them. As therefore S'4 must be
parallel to S5 (shown by the force polygon), the intersections in
each case must coincide, as also the lines S'4, S5 themselves, and
the polygon from e on has the same course in either case.
25. Pole taken upon closing line. — We have seen (Art.
20) that when any number of forces are in equilibrium both
the force and equilibrium polygon must close. There is one
exception to this statement. Since the pole may be taken any-
where, suppose it taken somewhere upon the line closing the
force polygon. This line, as we know, is the resultant, and
26 FORGES IN THE SAME PLANE. [CHAP. II.
holds the other forces in equilibrium. But now the equilibrium
polygon evidently will not dose. On the contrary the first and
last strings will be parallel. This position of the pole should
then be avoided. For any other position of the pole our rule
holds good ; viz.,
If the force polygon closes as also the equilibrium polygon,
the forces are in equilibrium. If the equilibrium polygon
however does not close, the forces cannot be replaced by a single
force but only by a couple. The forces of this couple act in
the parallel end lines of the equilibrium polygon, and are given
in intensity and direction of action by the line from the pole
to the beginning of the force polygon [beginning and end coin-
ciding] .
26. Relation between two equilibrium polygons with
different poles. — We may deduce an interesting relation be-
tween the two equilibrium polygons formed by choosing differ-
ent poles, with the same forces and force polygon.
Thus with the forces P! P2 P3 P4, we construct the force
polygon Fig. 14 (a), PL 4. Then choose a pole C and draw SM,
and thus obtain the corresponding equilibrium polygon S0 a b c d
S4 Fig. 14 (b). Choose now a second pole C'. Draw S^ and
construct the corresponding polygon S'0 a! b' c' d' S'4. [In our
figure c and c' fall accidentally nearly together.]
Join the two poles by a line CO'. Then — any two corre-
sponding strings of these two polygons intersect upon the same
straight line M N parallel to C C'. Thus S0 and S'0 intersect
at g, S'i and Sj at k, S'2 and S2 at I, S'8 and S3 at n, S'4 and S4 at
m — and all these points g, k, I, n and in, lie in the same
straight line M N parallel to the line C C' connecting the
poles.
The proof is as follows.* If we decompose Pt into the com-
ponents S0 Sj and S'0 S'^ these components are given in inten-
sity and direction by the corresponding lines in the force poly-
gon. If we take the two first as acting in opposite directions
from the two last, they hold these last in equilibrium. The
resultant therefore of any two as S0 and S'0 must be equal and
opposed to that of the remaining two, St and S'^ and both re-
sultants must lie in the same straight line. This straight line
must evidently be the line gk joining the intersections of S0 S'0
* Ekmente der Graphischen Statik. Bauschinger. Miinchen, 1871. Pp.
18-19.
CHAP, n.] DIFFERENT POINTS OF APPLICATION. 27
and Si S\. But from* the force polygon we see at once that the
resultant of S0 and S'0 is given in direction and intensity by
C C', and this is also the resultant of Si and S\. The line join-
ing g and k must therefore be parallel to C C'. For the second
force P2 we can show similarly that the line joining k and I is
parallel to C C'. But k is a common point of both lines — hence
g k and I lie in the same straight line parallel to C C'.
[NOTE. — The pure geometric proof is as follows : The two
complete quadrilaterals 0 1 C' C and g k a' a have five pairs of
corresponding sides parallel, viz., 0 1 and a a', a a' 1 C' and
a! k, C 0 and a g, o C' and a' g, 1 C' and ak' hence the sixth
pair are also parallel, viz., C C' and g k. In like manner for
1 2 C C' and lkbr b and so on.~\
We can make use of this principle in order from one given
equilibrium polygon S0 a b c d S4 and pole, to construct another,
the direction of C C' being known. For this purpose, having
assumed the position of the first string S'0 we draw through its
intersection g with S0 a line M N parallel to C C'. The next
string must therefore pass through the intersection a' of S'0 and
P! and through the point k, of intersection of the second string
of the first polygon and the line M N. It is therefore deter-
mined. The next side must pass through V and Z, and
so on.
[Note. Observe that the intersections r and r' of the first and
last lines of both polygons must lie in a straight line parallel to
0 4, the direction of the resultant.]
27. Mean polygon of equilibrium.— Since the pole may
have any position, let us suppose it situated in one of the angles
of the force polygon. It is evident that the first line of the
corresponding equilibrium polygon, then coincides with the first
force. If now the pole be taken at the beginning of the first
force in the force polygon, then the first side of the correspond-
ing equilibrium polygon will coincide with the first force, and
the last line will be the resultant itself in proper position.
Take for instance, the pole at o in the force polygon, Fig. 15
(a\ PL 4. The first side S0 reduces to zero. The next St coin-
cides with 0 1. In (b) 'therefore Pt is the first side of the equi-
librium polygon. The next side S2 corresponds with S2 in (a).
Thus we obtain the polygon a b c d e, the last side of which S7,
is the resultant itself. That is, S2 is the resultant of Pt and P2,
S:>, of P^, S4 of P14 and so on. Every line in the polygon then
28 FORCES IN THE SAME PLANE. [CHAP. II.
is the resultant of the forces preceding, and we call such a
polygon the mean polygon of equilibrium.
If we wish to find the mean polygon for P^ we have only to
take the new pole C' at 2 in the force polygon (a). According
to the preceding Art., each side of the new polygon must pass
through the intersection of the corresponding side of the first
with the line S2 which passes through a and is parallel to C C'.
Thus S'4 must pass through ~b' and o. S'5 through d and n, and
so on. S'7 is the resultant of PS7, and since S2 is the resultant
of P^; S7, the resultant of P1Jr, must pass through the intersec-
tion m of S'7 and S2.
We observe here again the influence of the couple P5 and P6.
S4 and S 4 are simply shifted through certain distances, without
change of direction, to S6 and S'6 ; and as we have seen above,
knowing the direction of rotation, and the moment of the couple,
we might have omitted it in the force polygon and still obtained
S7 and S'7 as before.
2§. Line of pressures in an arch. — The practical applica-
tion of the above will be at once seen in the consideration of
an arch. Thus with the given horizontal thrust applied at a
given point of the arch, and the forces Pw, we construct the
force polygon C o 5, and then the line of pressures abed.
[Fig. 16, PI. 4.]
Required with another thrust H' = o C' acting at another
point, and the same forces Pw, to construct the corresponding
line of pressures. To do this we have only to lay off o C' equal
to the new horizontal thrust, then choose a point of the force
line, as 3, as a pole and draw the corresponding polygon,
Jc op Jc • the point of intersection, &, is a point upon the line
m n parallel to o C, and upon this line will be found the inter-
section of corresponding sides of the two polygons. Thus from
the intersection of the side ap of the first polygon with m n,
draw a line to o and we have a'. From the intersection ~b of
the second line of the first polygon draw a line to a', and we
have b' #', and so on.
29. — The preceding articles comprise all the most important
principles of the Graphical Method which can be deduced in-
dependently of its practical applications. Future principles
will be best demonstrated, and at the same time illustrated, by
considering the various special applications of the method, and
to these applications we shall therefore now proceed.
CHAP. III.] CENTRE OF GRAVITY. 29
CHAPTER III.
CENTRE OF GRAVITY.
30. General Method. — One of the most obvious applica-
tions of the new method as thus far developed, is to the deter-
mination of the centre of gravity of areas and solids. We
shall confine ourselves to areas only, merely observing that all
the principles hitherto developed apply equally well to forces
in space. The forces being given by their orthographic pro-
jections upon two planes after the manner of descriptive geo-
metry, the projections upon each plane may be dealt with as
forces lying in that plane, and thus the projections of the force
and equilibrium polygons, the resultant, etc., determined.
A body under the action of gravity may be considered as a
body acted upon by parallel forces. The resultant of these
forces being found for one position of the body [or the body
being considered as fixed, for one common direction of the
forces] may have its point of application anywhere in its line
of direction.
For a new position of the body [or another direction of the
forces] there is another position for the resultant. Among all
the points which may be considered as points of application of
these two resultants there is one which remains unchanged in
position, whatever the change in direction of the parallel forces.
This point must evidently lie upon all the resultants, and is
therefore given by the intersection of any two.
It is hardly necessary to give illustrations of the method of
procedure.
Generally, we divide up the given area into triangles, trapez-
oids, rectangles, etc., and reduce the area of each of these fig-
ures to a rectangle of assumed base. The heights of these
reduced rectangles will then be proportional to the areas, and
hence to the force of gravity acting upon them ; i.e., to their
weights. Consider then these heights as forces acting at the
centres of gravity of the partial areas. Construct the force
30 CENTRE OF GRAVITY. [CHAP. III.
polygon by laying them off one after the other. Choose a pole
and draw lines from it to the beginning and end of each force.
These lines will give the sides of the funicular or equilibrium
polygon. Anywhere in the plane of the figure, draw a line
parallel to the first of these pole lines (S0). Produce it to inter-
section with the first force (P^, prolonged if necessary. From
this intersection draw a parallel to the second pole line (S^, and
produce to intersection with second force (P2). So on to last
pole line, which produce to intersection with first pole line.
Through this point the resultant must pass, and of course it
must be parallel to the forces.
Now suppose the parallel forces all revolved say 90°, the
points of application remaining the same. Evidently the new
force polygon will be at right angles to the first, as also the
new pole lines, each to each. It is unnecessary then to form
the new force polygon. The directions of the new pole lines
are given by the old, and this is all that is needed.
Anywhere then in the plane of the figure, draw a line (S'0)
perpendicular to the first pole line (S0) previously drawn, and
prolong to intersection with new direction of first force (P/).
Through this point draw a perpendicular (S/) to second pole
line, to intersection with new direction of second force (P2')
and so on. We thus find a point for new resultant, parallel to
new force direction. Prolong this resultant to intersection
with first and the centre of gravity is determined.
[NOTE. — If the area given has an axis of symmetry, that can
of course be taken as one resultant, and it is then only necessary
to make one construction in order to find the other.]
The given area of irregular outline must, as remarked above,
be divided by parallel sections into areas so small that the out-
lines of these areas may be considered as practically straight
lines. The forces are then taken as acting at the centres of:
gravity of these areas. This division will give us generally a
number of triangles and trapezoids.
It is therefore desirable to reduce graphically to a common
base the area of these triangles and trapezoids, and for this pur-
pose the following principles will prove of service :
32. Reduction of Triangle to equivalent Rectangle of
given Base. — Let b be the base and h the height. Then area
= — . Take a as the given reduction base, and let x represent
CHAP. III.] CENTRE OF GRAVITY. 31
the height of the equivalent rectangle. Then
bfi k x
ax = — or — = — -.
2, a %b
Now a, £, and h being given, it is required to find x graphi-
cally.
Let A B C be the triangle, and D the middle of the base.
[Fig. 17, PL 5.] Lay off A E = h and A F = a. Draw F D,
and parallel to F D draw E x. Then A a? is the required
height.
-r, A a? A E x h
For:AD = AF°rp = a
As to the centre of gravity of the triangle, it is evidently at
the intersection of the lines from each apex to the centre of
the opposite side ; since these lines are axes of symmetry.
33. Reduction of Trapezoid to equivalent Rectangle.
— In the trapezoid A B C D, Fig. 18, PL 5, draw through the
middle points of A D and B C perpendiculars to D C, and pro-
duce to intersections E and F with A B produced.
Then lay off F g — a — the given reduction base, and draw
g E intersecting D C in a?. Then H x is the required height.
T, EF H x EF x
H Yyp (\Y •
F^~HE a ~HE'
hence ##=:EFxHEi= area.
To find the centre of gravity , draw a line through the mid-
dle points of the parallel sides A B and D C. This line is an
axis of symmetry. Prolong A B and C D and make C a —
A B and A b = C "D and join a and ~b. Then the intersection
of a ~b with the axis of symmetry gives the centre of gravity.
The construction for the reduction of a parallelogram is pre-
cisely similar. [Fig. 18 (5).]
The points F and E here coincide with A and B, and we
have
A x AB
T- = = — , or # # — h x A B =: area.
h, B g
The same construction also holds good, of course, for a rect-
angle or square. The centre of gravity in each case is at the
intersection of two diameters, since these are axes of symmetry.
31. Reduction off Quadrilaterals Generally. — In general
32 CENTRE OF GRAVITY. [CHAP. III.
any quadrilateral may be divided into two triangles which may
be reduced separately, or into a triangle and trapezoid.
It is also easy to reduce any quadrilateral to an equivalent
triangle, which may then be reduced by Art. 32 to an equiva-
lent rectangle of given base.
Thus we reduce the quadrilateral A B C D [Fig. 18 (c)]
to an equivalent triangle by drawing C Cx parallel to D B to
intersection Ct with A B, and joining Ct and D. The triangle
D B Ci is then equal to D B C, and hence the area A D Cx is
equal to A B C D. The triangle A D C^ can now be reduced
to an equivalent rectangle of given base by Art. 32.
The centre of gravity of the quadrilateral may be found as
follows :
Draw the diagonals A C and B D and mark the intersection
E. Make A Ex = C E and B E2 = D E, also find the centres
Q! and O2 of the diagonals A C and B D. Join O2 Et and Ol
E2 ; the intersection S of these two lines is the centre of gravi-
ty required.
The above is sufficient to enable us to find the centre of gravity
of any given area of regular or irregular outline. The method
may be applied to finding the centre of gravity of a loaded
water-wheel (as given in Der Constructeur, Reuleaux, Art. 47),
and many similar problems. The reader will have no difficul-
ty, following the general method indicated in Art. 30, in mak-
ing such applications for himself. The method itself is so sim-
ple that it is unnecessary to give here any practical examples
in illustration. We shall, moreover, have occasion to return to
the subject in the consideration of moment of inertia of areas.
"We pass on therefore to the moment of "rotation of forces in
a plane.
CHAP. IV.] MOMENT OF ROTATION OF FORCES.
CHAPTEE IY.
MOMENT OF ROTATION OF FORCES IN THE SAME PLANE.
35. The "Moment" of a Force about any Point is the
product of the force into the perpendicular distance from that
point to the line of direction of the force. The importance
and application of the " moment " in the determination of the
strains in the various pieces of any structure will be evident by
referring to Art. 14, where Ritter's " method of sections " is
alluded to. In general, when the moments of all the exterior
forces acting upon a framed structure are known, the interior
forces, or the strains in the various pieces, can be easily ascer-
tained.
As we shall immediately see, these moments are given
directly in any case by the " equilibrium polygon"
36. Ciilmann's Principle. — If a force P be resolved into
two components in any directions as b C, b C^ (Fig. 19, PL 5),
and these components be prolonged, it is evident that the
moment of P with reference to any point as a situated any-
where in the line c d parallel to P, is P x b a. But if from C
we draw the perpendicular H to P, then by similar triangles,
P : H ; : c d : I a ;
Pxb a = Hxc d.
That is, the moment ofP with respect to any point a is equal
to a certain constant H multiplied by the ordinate c d, paral-
lel to P and limited by the components prolonged. The con-
stant H we call the "pole distanced
This holds good for any point whatever, and we have only to
remember that if we assume the ordinates to the right of P as
positive, those to the left are negative.
"We can choose the pole C where we please, and thus obtain
various values for H, but for any one value the corresponding
ordinates are proportional to tlie moments.
34 MOMENT OF KOTATION OF FOKOES. [CHAP. IV.
The above principle is due to Culmann, and will be referred
to hereafter as Culmann' s principle.
37. Application of the above to Equilibrium Polygon.
—Let P14 be a number of forces given in position as repre-
sented in Fig. 19 (a) PL 5. By forming foa force polygon Fig.
19 (b), choosing a pole C, and drawing S0 S1? S2, etc., we form
the equilibrium polygon abed ef, Fig. 19 (a).
The resultant of the forces P14 acts in the position and direc-
tion given in the Fig. How, as we have seen in Art. 22,
regarding the broken line a b c d e as a system of strings, we
may produce equilibrium by joining any two points as a and/"
by a line, and applying at a and f the forces S0 and S4. Let us
suppose this line a f perpendicular to the direction of the
resultant.- Since we can suppose the broken line or polygon
fastened at any two points we please, this is allowable, and
does not affect the generality of our conclusion.
Then the compression in the line a f is given by H, the
"pole distance" or the distance of the pole C from the result-
ant in the force polygon. We have therefore at a the force
H and Vi = H 0 acting as indicated by the arrows. At a then
Y! acting up, H and S0 acting away from a, are in equilibrium,
or Vx is decomposed into H and S0, as shown by the force
polygon.
According to Culmanrts principle then, the moment of Vt
with reference to any point, as m or o, is equal to H x o m.
Therefore H being known, the ordinates between a f and S0
are proportional to the moment of Vt at any point. Vx acting
upwards gives positive rotation (left to right) with respect
to m.
At the point b, Pt may be replaced by a force 0 K parallel to
R and a force K 1 along St [see force polygon]. This we see
at once from the force polygon where 0 K and K 1 make a
closed polygon with P1? and taken as acting from 0 to K and
K to 1, replace Plt But these two forces are in equilibrium
with Si and S0, or 1 C and C 0 [see force polygon], and since
K 1 and 1 K balance each other, all the forces acting at b may
be replaced by S0, 0 K and K C. We have then at b the force
0 K resolved into components in the directions S0 and S^
By Culmanrfs principle, therefore, the moment of O K
about any point as m, is proportional to the ordinate n m, and
since 0 K acts downward this moment is negative. Hence the
CHAP. IV.] MOMENT OF ROTATION OF FOKCES. 35
resultant moment at m or o of the components at a and b par-
allel to R, is proportional to the ordinate o n.
So for any point, the ordinate included by the polygon ab c
d ef, and the dosing line af, to the scale of length multiplied
by the "pole distance " H to the scale of 'force ;, gives the mo-
ment at that point of the components parallel to the resultant.
The practical importance and application of this principle
will appear more clearly in the consideration of parallel forces
in the next Chapter.
36 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V.
CHAPTEE Y.
MOMENT OF RUPTURE OF PARALLEL FORCES,
3§. Equilibrium Polygon. — Since the forces acting upon
structures are generally due to the action of gravity, these
forces may be considered as parallel and vertical, and in all
practical cases therefore, we have to do with a system of paral-
lel forces.
Given any number of parallel forces P^, PL 6, Fig. 20 ;
required to find the direction, intensity and position of the
resultant, and the moment of rotation at any point.
1st. Draw the force polygon, in this case it is, of course, a
straight lino.
2d. Choose a pole C, and draw the lines S0. S1? S2, etc.
3d. Draw the string or equilibrium polygon a b c d e f.
Considering this polygon as a system of strings, the forces will
be held in equilibrium if wre join any two points, as a and g^
by a strut or compression piece, and apply at a and g the up-
ward forces Vj and V2.
4th. Prolong a 1} and f g to their intersection o. Through
this point the resultant must pass. It is of course parallel and
equal to the sum of the forces.
Now, if a g is assumed horizontal, the perpendicular H to
the force line, or the "pole distance" divides the resultant 0 5
into the two reactions Y! and V2 (Art. 22).
All the forces in the equilibrium polygon have the same
horizontal projection H, in the force polygon.
Let a g represent a beam resting upon supports at a and g.
We have then at once the vertical reactions V1 and V2 or ~k 0
and 5 &, which, in order to cause equilibrium, must act up-
For the moment at any point, as 0, due to V1? we have, by
Culmann's principle, m o multiplied by H. The triangle formed
by a 5, a g, and Pl5 gives then the moment of rupture at any
point of the beam as far as Pj. For a point o, beyond Pj, the
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 37
moment due to V1? must be diminished by that due to P1? since
these forces act in opposite directions, and rotation from left
to right upon the left of any point is considered positive. We
see at once from the force polygon that Pt is resolved into S0
and Sj or into a b and b c. Hence the moment at o due to P!
is m n multiplied by H. The total moment at o is then mo —
m n = n o, multiplied by H.
Hence we see that the ordinates to the equilibrium polygon
from the closing line a g, are proportional to the total mo-
ments ; while the ordinate at any point between any two adja-
cent sides of this polygon, prolonged, represents the moment at
that point of a force acting in the vertical through the inter-
section of these two sides.
[The reader should make the construction, changing the order in which
the weights are taken, and thus satisfy himself that the order is a matter '
of indifference. As to the direction of the reactions Vi, Va, it must be
remembered that a & is to be replaced by Vi and H, hence Vi must be op-
posed to C 0, the direction obtained by following round in the force poly-
gon the triangle 0 1 O. Force and distance scales should also be assumed.
Thus the ordiuates to the equilibrium polygon scaled off say in inches, and
multiplied by the number of tons to one inch, and then by the "pole dis-
tance " taken to the assumed scale of distance, will give the moments of
any point.]
The resultant of any two or more forces must pass through
the intersection of the outer sides of the equilibrium polygon
for those forces (Art. 16). Thus, the resultant of Pt arid P3
must pass through the intersection of a b and c d. Of Vt and
Pb through the intersection of a g and b c ; of Pt P2 and P3,
through intersection of a b and d e, and so on. In every case
the intensity and direction of action of the resultant is given
directly by simple inspection of the force polygon.
Thus from the force polygon we see that the resultants Tc 2
and Jc 3 of Vj. P! P2 and Vx P! P2P3, act in different directions.
Their points of application are at the intersection of c d and d e
respectively with a g, or upon either side of d in the equilibrium
polygon. At d the ordinate and hence the moment is greatest,
and at this point the tangent to the polygon is parallel to a g.
If we had a continuous succession of forces ; if a g, for in-
stance, were continuously or uniformly loaded ; the equilibri-
um polygon would become a curve, and the tangent at d would
then coincide with the very short polygon side at that point.
38 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V.
The points of application of the resultants of all the forces
right and left of d are then at the intersection of this tangent
with a g, or at an infinite distance.
At d then we have a couple, the resultant of which is as we
have seen (Art. 20), an indefinitely small force acting at an
indefinitely great distance. That is, with reference to d, the
forces acting right and left cannot be replaced by a single
force.
Hence generally : at the point of maximum moment (" cross
section of rupture"), the resultant of the outer forces on either
side reduces to an indefinitely small and distant force, the
direction of which is reversed at this point, and the point of
application of which changes from one side to the other of the
equilibrium polygon.*
The " cross section of rupture " then, is that point where the
weight of that portion of the girder between it and the end is
equal to the reaction at that end, or where the resultant changes
sign.
The value of the moment at this point, is therefore equal to
the product of the reaction at one end into its distance from
the point of application of the equal resultant of all the loads
between that end and the point.
Thus for a beam uniformly loaded with w per unit of length,
the reaction at each end is -^-* From the above, the cross sec-
tion of rupture is then at the middle. The point of application
of the resultant of the forces acting between one end and the
A I , . w I I w Z2
middle is at -r, hence the maximum moment is-^- x~: = -5-.
4: A 4: O
39. Beam with Two Equal and Opposite Forces beyond
the Supports. — The ordinates to the equilibrium polygon thus
give, as it were, a picture or simultaneous view of the change
and relative amount of the moments at any point. The point
where the moment is greatest, i.e., where the beam is most
strained, is at once determined by simple inspection.
Let us take as an example a beam with two equal and oppo-
site forces leyond the supports. Thus, Fig. 21, PL 6, suppose
the beam has supports at A and B, the forces being taken in
the order as represented by P! P2. We first construct the force
* Die GrapMscJie Statik.— Culmann, p. 127.
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 39
polygon from 0 to 1, and 1 to 2 or 0. Next choose a pole C,
and draw S0 Si and S2. Draw then a parallel to S0 till intersec-
tion with first force, Pt, then parallel with Sj to second force,
P2, then parallel to S2 or S0 to intersection with vertical through
support B, and finally draw the closing line L. A line through
C, parallel to L, gives as before the vertical reactions. Follow-
ing round the force polygon, we find at A the reaction down-
wards, since S0 acts from C to 0 and is to be replaced (Art. 4)
by L and Vx ; at B reaction upwards, since P2 acts up, and fol-
lowing round, S2 acts from 0 to C. .Both reactions are equal to
a 0. At A then the support must be above, and at B below the
beam. The shaded area gives the moments to pole distance H.
Had we taken the pole in the perpendicular through o, S0 would
have been parallel with the beam itself. This is, however, a
matter of indifference. The moment area may lie at any in-
clination to the beam. We also see here again the effect of a
couple (Art. 23). S0 is simply shifted through a certain distance
to S2, parallel to S0, and therefore the moment at any point be-
tween P2 and B is constant. This is generally true of any
couple, as we have already seen, Article 21, and may be proved
analytically as follows :
Let the distance between the forces be a = A B, Fig. 22.
Then for any point o, we have P X (& + B 6>)— PxBo — P [a + ^B o
— B o] = P a. For o' between A and B, P x A o'+P x o' B=
P [A </ + <?' B] =Pa.
So also for any point to the left, the same holds true.
Graphically the proof is as follows :
Decompose both forces into parallel components, Fig. 23.
Then for any point, as o, we have the moment M = H x m n—
Hxmp or M = — H x n p. But n p is the constant ordinate
between the parallel components A n and A p.
We see, therefore, by simple inspection, that the distance of
P! and P2 from the support B, Fig. 21, has no influence what-
ever upon the moment or strain in A B, provided the distance
between the points of application remains the same, and that
the moment at all points between P2 and the support B is con-
stant and a maximum. From B and P2 the moments decrease
left and right, and become zero at A and Pj.
1O. Beam with Two Equal and Opposite Forces be-
tween the Two Supports. — Let the beam A B, Fig. 24, PL 6,
be acted upon by the two equal and opposite forces P1 P2.
4:0 MOMENT OF KUPTURE OF PARALLEL FORCES. [CHAP. V.
Construct the force polygon 012. Choose a pole C and draw
C 0, C 1, C 2. Parallel to C 0, draw the first side of the equi-
librium polygon to intersection with first force Pt; then paral-
lel to C 1 to second force P2 ; then parallel to C 2 to d. Join d
and 0. Parallel to this draw C a in force polygon. Then 0 a
is the vertical reaction at A, which acts upwards, since it niust
with C a replace C 0 ; and C 0, when we follow round from o to
1 and 1 to C, acts from C to 0.
We have the same vertical reaction at B, but here, since we
must follow from 1 to 2 and 2 to C, C 2 acts from 2 to C, hence
following round, the reaction at B is downward. The shaded
area gives the moments to pole distance H, as before.
We see at once that at a certain point e the moment is zero.
Left and right of this point the moment is positive and nega-
tive. At the point itself we have a point of inflection, and
here, since the moment is zero, there is no longitudinal strain.
At jb and c the moments are greatest ; here the beam is most
strained, and at these points, therefore, are the " cross sections
of rupture." Here again, if we had taken the pole C in the
perpendicular through a, the closing line of the polygon o d
would have been horizontal. It is, however, indifferent at
what inclination a* d may lie, but we may if we wish make it
horizontal now, and then lay off from its new intersections with
P! and P2 along the directions of these forces, the ordinates
already found at £ and c, and join the points thus obtained with
the ends of o d (i.e., with its intersections with the verticals
through the supports). The ordinates of the new polygon
thus found will be for any point the same as before, and will
also be perpendicular to the beam.
[NOTE. — Had we taken the forces precisely as above but in reverse order,
the force line would be reversed, and we should have 0 and 2 in place of 1,
and 1 in place of 0 and 2 ; that is, in place of O 1 we should have O 0 and
C 2. Constructing then the equilibrium polygon by drawing a line paral-
lel to new O 0 to intersection with new Pi, then parallel to new C 1 to in-
tersection with new P2, then parallel with new O 2 to intersection with
vertical through B, and finally joining this last point with intersection of
the first line drawn (C 0) with vertical through A, we have at first sight a
very different equilibrium polygon. This new polygon will consist of two
parts. If the ordinates in one of these parts are considered positive, those
iii the other must be negative. The difference of the ordinates in these two
portions for any point, will give the same result as above. This, by mak-
ing the above construction, the reader can easily prove.]
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 41
41. Many other problems will readily occur, which may in a
similar manner be solved. The weights may have any position,
number and intensities desired ; in any and every case we have
only to construct with assumed pole distance the corresponding
equilibrium polygon, and we obtain at once the moments at
every point. By the use of convenient scales, numerical results
may be obtained which may be checked by calculation, and
the practical value and accuracy of the method thus demon-
strated.
The above principles will be sufficient for the solution of any
such problem wrhich may arise, and we shall therefore content
ourselves with the above general indication of the method of
procedure, and pass on to the consideration of a few cases
where the above needs slight modification, and which, from
their practical importance, and the ease with which they may
be treated graphically, seem worthy of special notice.
1ST. BEAM OR AXLE — LOAD INCLINED TO AXIS.* [Fig. 25, PL 6.]
We have here simply to draw the " closing line "AC paral-
lel to the beam or axle. From d draw d B parallel to the force
P, then draw A B in any direction at. pleasure, and join B C.
We have thus the equilibrium polygon ABC, the ordinates to
which, as d B, parallel to the force P, will give the moments,
provided we know the corresponding pole distance.
But this can easily be found. As we have already seen, the
force polygon being given, the equilibrium polygon may be
easily constructed. Inversely, the equilibrium polygon being
given, the force polygon may be constructed. Thus from A
draw A c equal and parallel to P, and then draw c C^ parallel
to B C. A a and J c are the vertical reactions P! and P2 ; a b
is the horizontal component of the force which must be resisted
at one or both of the ends ; and the moments at any point are
given by the ordinates parallel to P multiplied by the perpen-
dicular distance from CL to A c. If we suppose the force P, as
in the Fig., as causing two opposite vertical forces, instead of
acting directly upon the axis, we have only to prolong A B to
B! and join B! 62, and then the ordinates of A B! B2 C parallel
to P or A c, multiplied by H (perpendicular distance from Ct
to A c) will give the moments.
* See Der Constructeur, Reuleaux.
4-2 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V.
2D. FORCE PARALLEL TO AXIS. [Fig. 26, PL 6.]
We have an example of this case in the " bayonet slide " of
the locomotive engine.
We have here two pairs of forces, the reactions Vt and V2
and the forces over B! and B2. The points of application of
these last change of course periodically, but for any assumed
position the moments are easily found. Thus draw A B! at
pleasure, and C B2 parallel to it, and join Bt B2 arid A C, and
we have at once the equilibrium polygon. To find the corre-
sponding force polygon^ suppose Px applied at 5, and join ~b with
the other support. Make b c equal to P then c d = V2. Lay
off then A a = c d — V2 and draw a Cl5 which is the pole dis-
tance. Draw GI e parallel to B! B2. Then A e and e A are the
forces acting over Bg and B1? and A a is the reaction V^ The
case is, indeed, precisely similar to that in Art. 40.
[NOTE. — The moment area should properly be turned over upon A C as
an axis, so that A a should be laid off and e fall lelow A. This can, how-
ever, cause no confusion.]
The application of the method to car axles,* crane standards,
and a large number of similar practical cases in Mechanics is
obvious. The formulae for many of these cases are too com-
plex for practical use ; in some, no attempt at investigation of
strain is ever made, the proportions being regulated simply by
" Engineering precedent " or rules of thumb. Those familiar
with the analytical discussion of such cases will readily recog-
nize the great practical advantages of the Graphical Method.
3D. BEAM OR AXLE ACTED UPON BY FORCES LYING IN DIFFERENT
PLANES.
The analytical calculation in such a case for instance is of
considerable intricacy, but by the graphical method, on the
contrary, the difficulty of investigation is scarcely greater than
before.
Thus, let Fig. 27, PL 7, represent a beam acted upon by two
forces P! and P2 not in the same plane.
First, we draw the force polygons A C^ M and D O2 2 for the
forces P! and P2, having both the same pole distance G Ol =
O2 H, the pole O2 being so taken that the closing lines of the
* Der Constructeur, Reuleaux, pp. 215-222.
CHAP. V.J MOMENT OF RUPTURE OF PARALLEL FORCES. 43
corresponding polygons A V D and A c" D coincide. This is
easily done, as if the closing line of the second polygon for any
assumed position of O2 (O2 H being equal to G OJ does not co-
incide with A D, the ordinate at c' can be laid off from C and
A G" D thus found in proper position, and then the pole O2 can
be located. It will evidently be at the intersection of the ver-
tical O2 O'2 with G" D.
The two force polygons being thus formed, we construct the
polygon A C" D by drawing lines B B", E E", C C", etc., so
that their angles with the vertical shall be equal to the angle
between the planes of the forces, and making them equal to
the ordinates B 5", E e'1 ', C c", etc., respectively. Join V B",
e' E", /' F", G' C", etc., "and lay off the ordinates B b, E e, F/,
C c, etc., respectively equal. The ordinates to the polygon
thus obtained, viz. : A 5 efc D multiplied by the pole distance
Oi G or O2 H, give the moments at any point. A 1) and G D
are straight lines, ~b efc\& a curve (hyperbola). If we drop
verticals through C^ and O2, and draw the perpendiculars O/ M,
O'2K ; A M is the reaction R1; and D K the reaction R2, both
measured to the scale of the force polygon. Their directions
are found by the composition of A G and H 2 and D H and
G M respectively, under the angle of the forces.
4TH. COMBINED TWISTING AND BENDING MOMENTS.
In many constructions pieces occur which are subjected at
the same time to both bending and twisting moments. Both
can be represented and given by moment areas. Thus., Fig.
28, PL 7, represents an axle turning upon supports at A and B
and having at C a wheel upon which the force P acts tangenti-
ally. We have then a moment of torsion Mt = P R and reac-
* o ft
tions Pt — P - - and P2 = P — - ; s being the distance of P
a i S a -\~ S
from B, and a of P from A.
Let the bending moments be represented by the ordinates to
the polygon a C ~b ; then laying off a o equal to P and drawing
o O parallel to ~b c, we find the corresponding pole distance
O &, and the reactions P! and P2 equal to Jc a and o Jc respec-
tively.
Now, in the force polygon O a o thus found, at a distance
from O equal to R, draw a line m n parallel to P. This line
m n evidently gives for the same pole distance the moment of
44 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V.
torsion P x R. Laying off C'C^ = b &', — in n, we have the
torsion rectangle Cx b' b C'.
. Now the combined moment of torsion Mt and bending Mb is
| Mb-f f v/Mg + Mp. We make then C' C0 equal to f C' Ct =
£ m n and C C2 equal to f C C' = Mb, and draw C2 b. Then
any segment of any ordinate, as/yj, is -f offf. Revolve now
C' C0 with C' as a centre, round to C'0 and join C'0 C2. Then
C2 C'0 is equal to f V^b H-M?> and therefore with C2 as centre
revolving C2 C'0 to C3, we find the point C8, C C3 being equal to
f Mb + f \/M^ 4- MjL In the same way we find any other
point as/3, by laying off//',, equal to//0, joining/3 and/'0 and
making f2 fs equal to f% fQ. The line C3 f$ b0 thus found is a
hyperbola, and the ordinates between' it and b C give the com-
bined moments [for pole distance O Jc] at any point.
[NOTE. — We suppose the axle to turn freely at A, and the working point
or resistance beyond B ; hence the moments left of the wheel are given by
the ordinates to a O.]
/
5TH. APPLICATION TO CRANK AND AXLE.
The above finds special and important application in the case
of the crank and axle.
Thus in PL 8, Fig. 29, let E D C B be the centre line of crank
and shaft. Lay off a P equal to the force P acting at A, choose
a pole o and draw o a o P and the parallels o a and a E. Join
E and d and draw o Pj parallel to E d. Then P P! is the
downward force at E and P1 a the upward reaction at D. The
ordinates to E d a to pole distance o P, give the bending mo-
ments for the shaft. Make a F equal to the lever arm R, then
F G is the moment P R, and we unite this as above with the
bending moments and thus find the curves c' d' e' the ordinates
to which give the combined moments at every point of the
shaft [see 4th].
For the arm B C, make the angle aQ B C equal to D a d,
and then the horizontal ordinates to aQ B give the bending mo-
ments for the arm. Make C c0 equal to C G and we have the
torsion rectangle C CQ bQ B, and as in the previous case we unite
the two and thus find the curve b0 h F, the horizontal ordinates
to which from B C give the required combined moments, to
* Der Constructeur, Reuleaux, p. 52, Art. 18.
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 45
polo distance o P. Thus h'h^ = f H A0j H i — f B 50, and
H h = h, h' + ti i = f Mb + | /MS + M?.
The application of the method when the crank is not at right
angles to the shaft, as also when the»crank is double, and gener-
ally in the most complicated cases, is equally simple and satis-
factory. Our space forbids any more extended notice of these
applications, and we must refer the reader to Der Constructeiir,
by F. Reuleaux, Braunschweig, 1872, for further illustrations
and applications of the method to the solution of various practi-
cal mechanical problems.
42. Continuous Loading— Load Area. — Thus far we have
considered only concentrated loads. But whatever may be the
law of load distribution, if this law is known, we can represent
it graphically by laying off ordinates at every point, equal by
scale to the load at that point. We thus obtain an area bounded
by a broken line, or for continuous loading, by a curve, the
ordinates to which give the load at any point. This load area
we can divide into portions so small that the entire area may
be considered as composed of the small trapezoids thus formed.
If, for instance, we divide the load area into a number of trape-
zoids of equal width, as one foot one yard, etc., as the case may
be, then the load upon each foot or yard will be given by the
area of each of these trapezoids. If the trapezoids are suffi-
ciently numerous, we may consider each as a rectangle whose
base is one foot or one yard, etc., as the case may be, and whose
height is the mean or centre height. The weight therefore for
each trapezoid acts along its centre line. We thus obtain a
system of parallel forces, each force being proportional to the
area of its corresponding trapezoid, and equal by scale to the
mean height or some convenient aliquot part of this height.
We can then form the force polygon / choose a pole ; draw
lines from the pole to the forces ; and then parallels to these
lines, thus forming the string or equilibrium polygon; and so
obtain the graphical representation of the moments at every
point.
Since, however, the polygon in this case approximates to a
curve, that is, is composed of a great number of short lines, the
above method is subject to considerable* inaccuracy, as errors
multiply in going along the polygon.
This difficulty can, however, be easily overcome.
Thus we may divide the load area into two portions only, and
46 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V.
then draw the force and equilibrium polygon, considering each
portion to act at its centre of gravity, and so obtain an equili-
brium polygon composed of three lines only. These lines will
be tangents to the equilibrium curve. (Art. 76.) We thus have
three points of the curve, and its direction at these points.
In this manner we may determine as many points as may be
necessary, without having the sides of the polygon so short or
so numerous as to give rise to inaccuracy.
43. — The above will appear more plainly by consideration
of a
BEAM UNIFORMLY LOADED.
The curve of load distribution becomes in this case a straight
line. The load area is then a rectangle, and hence the load per
unit of length is constant. Let us now divide this load area
[Fig. 30, PI. 8], into four equal parts, and considering each por-
tion as acting at its centre of gravity, assume a scale of force,
and draw the force polygon. Since in this case the reactions at
the supports must be equal, we take the pole C, in a perpendi-
cular to the force polygon at the middle point. This causes the
closing line of the equilibrium polygon to be parallel to the
beam itself, which is often convenient. We now draw C 0, C 1,
etc., and then form the polygon 0 a c e g h. The lines 0 a, a c, •
c e, etc., of this polygon, are tangent to the moment curve at
the points 5, d,f, 0 and A, where the lines of division prolonged
meet the sides. The curve can now be easily constructed, as
will appear from the next Art.
44. moment Curve a Parabola. — Suppose we had divided
the load area into only two parts, of the length x and I— x [Fig.
30, PL 8]. Then the moment polygon would be o a k h, and
the horizontal projection of the tangent a Jc would be % x 4- i
(t±xj-ll.
That is, the horizontal projection of any tangent to the mo-
ment curve is constant. But this is a property of the parabola.
The moment curve for a uniform load is therefore a parabola,
symmetrical with respect to the vertical through the centre of
the l>eam.
If, then, we divide o C and C h into equal parts, and join cor-
responding divisions above and below, we can construct any
number of tangents in any position.
["NOTE. — We may prove analytically that the moment curve is a parabola,
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 47
and hence that the line a Tc must le a tangent. Thus the moment at any
point is
p being the load per unit of length, I the length, and the reaction at sup-
f) 1 f)
port therefore ~-. Hence y = — — (Ix — x1} for origin 0.
a a H
When the origin is at d, representing horizontal distances by y' and ver-
tical by x1, we have x = - — y', and y = h — #', k being the ordinate at
9
^ 72
middle =
Hence by substitution
or reducing
2H
f = T x
which is the equation of a parabola having its vertex at <#.]
We may of course take the pole anywhere, and hence H may
have any value. It is in general advantageous in such cases
(i.e., for uniform load) to take H = *—• We have then
f = I so,
and for y — •%, or for the middle ordinate. we have x = j*
To draw the moment curve we have then simply to lay off
the middle ordinate equal to Jth the span. The curve can then
be constructed in the customary way for a parabola. Any
pi
ordkiate to this curve multiplied by H = ^— will then give the
moment at that point.
Enough has probably now been said to illustrate the applica-
tion of our method to the determination of the moment of rota-
tion, bending moment, or moment of rupture. The reader
will have no difficulty in applying the above principles to any
practical case that may occur.
It will be observed that the customary curve of moments in
the graphic methods at present in general use, comes out as
a particular case of the equilibrium polygon for uniform
load.
This polygon has other interesting properties, which we shall
notice hereafter. For instance, just as its ordinates [Fig. 30]
are proportional to the bending moments or moment of rotation,
4:8 MOMENT OF RUPTURE OF PARALLEL FORCES. [dlAP. V.
so also its area is proportional to the moments of the moments,
or the moment of inertia of the load area.
As to the shearing force at any point of a beam submitted
to the action of parallel forces, the reactions at the ends being
easily found as above by a line parallel to the closing line in
the force polygon, we have only to remember that the shear at
any point is equal to the reaction at one end, minus all the
weights between that end and the point in question.
Thus for a uniformly distributed load we have simply to lay
off the reactions which are equal to one-half the load, above
and below the ends, and draw a straight line, which thus passes
through the centre of the span. The ordinates to this line are
evidently then the shearing forces. If we have a series of con-
centrated loads, we have a broken line similar to A\ 1' 1" 2',
etc., Fig. 32, PI 7, where each successive weight as we arrive at
it, is subtracted from the preceding shear.
44. Beam continuously Loaded and also Subjected to
the Action of Concentrated Loads. — In practice we have
to consider not only a continuously distributed load, such as the
weight of the truss or beam itself, but also concentrated forces,
such as the weight of cars, locomotives, etc., standing upon or
passing over the truss.
In PL 8, Fig. 31, we have a continuous loading represented
by the load area A a b B, and in addition four forces P'w.
Now, since the total moment about any point is equal to the
sum of the several moments, we can treat each method of load-
ing separately and then combine the results. Thus with the
force polygon (b) we obtain the equilibrium polygon A' 1 2 3
B' for the continuous loading, and with the force polygon
(a) the equilibrium polygon A' I" 2" 3" B" for the con-
centrated loads. If now in (b) we draw C L parallel to the
closing line A' B', and in (a) C" L' parallel to the closing line
A' B", we obtain at once the reactions at the supports for each
case.
Thus for continuous loading we have L 0 for reaction at A,
and 10 L for reaction at B ; for the concentrated loads, L' 0'
at A and 4' L' at B. These reactions hold the beam in equi-
librium.
For any cross-section ^, the shear to the right is composed of
the two components L 7 and I/ 3' (i.e., is equal to the reactions
minus the forces between cross-section and support). The mo-
CHAP. V.] MOMENT OF KUPTUKE OF PARALLEL FORCES. 49
ment of L 7 is given by the ordinate o y to the corresponding
polygon, and we may consider L 7 as acting at the point of
intersection a of the side 7 8 with A! B' (Art. 38). In the same
way L' 3' acts at b. We may unite both these reactions and
find the point of application of their resultant £, by laying off
in force polygon (b) 7 b equal to I/ 3', and then constructing
the corresponding equilibrium polygon e a d c. The resultant
R passes through c. This construction remains the same evi-
dently, even when the points a and b fall at different ends of
the beam, as may indeed happen. The components will then
have opposite directions, and must be subtracted in order to
obtain the resultant.
The total moment of rotation at y is proportional to the sum
of m n and o y. The greatest strain is where this sum is a
maximum. In order to perform this summation and ascertain
this point of maximum moment it is advantageous to construct
another polygon instead of A' 1" 2", etc., whose closing line
shall coincide with A' B'. This is easy to do, by drawing in
force polygon (a), I/ C' parallel to A' B', and taking a new pole
C' the same distance out as before, that is, keeping H constant,
and then constructing the corresponding polygon A.' V 2' 3', etc.
Thus the ordinate p y gives the total moment at y. We can
make use here also of the principle that the corresponding sides
of the two polygons must intersect upon the vertical through
A' (Art. 26). We have thus the total moment at any point, and
can easily determine the point of maximum moment or cross-
section of rupture. This point must necessarily lie between
the points of maximum moments for the two cases, or coincide
with one of them. In the Fig. this point coincides with the
point of application of P'2.
45. Case of Uniform Load. — If the continuous load is uni-
formly distributed we can obtain the above result without
being obliged to draw the curve. As in this case we have a
very short construction for the determination of the point of
greatest moment, it may be well here briefly to notice it.
If we erect ordinates along the length of the beam as an axis
of abscissas, equal to the sum of the forces acting beyond any
cross-section, the line joining the end points of these ordinates
has a greater or less inclination to the axis according as the
uniform load is greater or smaller. At the points of applica-
tion of the concentrated loads this line is evidently shifted
4
50 MOMENT OF RUPTTJKE OF PARALLEL FOKCES. [CHAP. V.
parallel to itself. Since at the point of maximum strain the
sum of the forces either side is zero, this point is given by the
intersection of the broken line thus found with the axis.
Thus in PL 7, Fig. 32, let A B be the beam sustaining a uni-
form load, and also the concentrated loads Pt P2 P3 P4. The
reaction of the uniform load at the supports is equal to half
that load. To find the reactions for the concentrated loads we
draw the force polygon 01234, choose a pole C, then con-
struct the equilibrium polygon A' 1 2 3 4 B', and parallel to
A' B' draw C L. L 0 and L 4 are the reactions at A and B.
ISTow through L draw A0 L horizontal, make it equal to the
length of the beam, and take it as axis of abscissas. [It is of
course advantageous here to lay off the forces along the verti-
cal through B, as done in the Fig. Then A0 falls in the vertical
through A and 10 20 30 40 are directly under the forces them-
selves.]
The ordinate to be laid off at A0 is equal to L 0 + half the
uniform load. Between A0 and 10 the line A'x 1' is inclined to
the axis at an. angle depending upon the uniform load. Lay off
L U equal to this load and draw A0 U. A\ 1' must be parallel
to this line. At 1' the line A.\ I' is shifted to 1", so that 1 1 "
is the load Pt. Then 1" 2' is parallel as before to A0 U, and
2' 2" is the load P2, and so on. The intersection 20 with A0 1*
gives the point of maximum moment or cross-section of rup-
ture. The force P2 at this point in our Fig. is divided, as shown
by L in the force polygon, into two portions, one of which is to
be added to the forces left, the other to the forces right. The
ordinate y0 y' at any point gives the shear or sum of the forces
acting at that point. This force acts up or down according as
the ordinate is above or below the axis.
Moreover, the area between the broken line and axis A0 L,
limited by this ordinate, gives the moment of rotation of the
forces beyond the section y, areas below the axis being nega-
tive. For a section at 2, therefore, we have area A0 A\ 1' 1"
2' 20, minus 20 2" 3' 3" z' 20, or what is the same thing, the area
ZQ z' 4' 4" B\ L, since the sum of the moments of all the forces
is zero.
46. Influence of a Concentrated Load, pa§§iiig over the
Beam. — If in addition to the already existing uniform and
concentrated loads, a new force operates, we have by (44) simply
to construct for this new force its force and equilibrium poly-
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 51
gon, and unite the forces and moments thus found with those
already existing.
In PL 7, Fig. 32, we have assumed a new force P\ near the
left support. The force polygon is 0' 1' C', the pole distance
being taken the same as before. For any one position of this
force we have then the equilibrium polygon A' 1' B;/, and
drawing a parallel C' L' to A' B" we obtain the reactions 0' I/
and L' 1', which must be added to the reactions already ob-
tained.
If now we take a section y between P\ and the point of max-
imum moment 20 before found, the sum of the forces either
side of this section undergoes the following changes : Upon
the side where P^ lies, and the point 20 does not lie, where
therefore the sum was originally an upward force, we have the
downward force I/ V (equal to algebraic sum L' 0' 4- 0' 1').
The sum of the forces at the section, or the shearing force, is
therefore diminished.
The total rotation moment is, however, increased by the
amount indicated by m n. Both changes, that of the sum of
the forces and the moment 'of rotation, increase as P\ ap-
proaches y, and are therefore greatest when P\ reaches y.
If P\ passes ?/, this point is in the same condition as z with
reference to the former position of P\ ; that is, the force and
point 20 are now both on the same side of the section. For z,
then, the original downward force to the left is increased by the
force I/ V. To the right the upward force is increased by 1' L'.
In like manner the moment of the forces beyond z is increased
by the amount indicated by op. This change is greatest when
P\ reaches z.
Therefore when a load passes over the beam the sum of the
shearing forces is diminished in all sections between it and the
original point of greatest moment, and increased m sections be-
yond this point, while the moment of rotation, or bending
moment, for all cross-sections is increased. These changes
moreover increase for any section as the load approaches that
section. The shear at any point is therefore least, and the mo-
ment greatest, when the load reaches that point. As soon, how-
ever, as the load passes this point, the shear passes suddenly
from its smallest to its greatest opposite value, and then dimin-
ishes as the load recedes, together with t'he moment of rotation.
On the other side of the point 20 of original greatest moment,
52 MOMENT OF RUPTURE OF PARALLEL FORCES. [CHAP. V.
the shear and moment increase as the load approaches, and
become greatest for any point when the load reaches that point.
At the moment of passing, these greatest values pass to their
smallest values, and increase afterwards, as the load recedes.
Since by the introduction of the load the shear for points
upon one side of 20 is diminished (between 20 and the load), and
on the other side increased, and the greatest moment is at the
point where the shear is zero, it follows that the point of greatest
moment moves in general towards the load. At a certain point,
then, both meet. As the load then advances this point accom-
panies it, passes with it the original position, and follows it up
to the point where it would have met the same load coming on
from the other side. From this point, as the load continues to
recede, it returns, and finally reaches its original position as the
load arrives at the further end.
It is evidently of interest to learn the position of these two
points, where the load meets and leaves the point of greatest
moment, or cross-section of rupture, and this in Fig. 32 we can
easily do.
When P\ arrives at 1', we have evidently the reactions by
laying off L E equal to P'1? drawing A0 E, and through its
intersection with the vertical through the weight drawing the
horizontal A'0 B'0. L B'0 is then the increase of reaction at B due
to P\. The entire reaction is B'0 B'1? and the broken line A\
1' 1", etc., holds good still, if we merely change the axis from
AO L to A'0 B'0. The point of greatest moment, which is still
the intersection of the broken line with the new axis, in the
present case is not changed by reason of the overpowering in-
fluence of P2. It does not move to meet the load, but awaits it
until it reaches P2, and until, therefore, the new axis takes the
position A"0 B"0.
If, however, the force P\ comes on from the right, we have
the reactions for any position as z, by laying off A0 E' equal to
•P'u drawing L E', and then the horizontal A'"0 B'"0 through the
intersection of L E', with the vertical through z. Then A0 A'"0
is the reaction at A, due to this position of the load. The in-
tersection #', corresponding to a?, shows the point to which the
point of greatest moment 20 moves to meet the load. As the
load passes towards the left, this point moves towards the right,
and both come together evidently at the point V^ correspond-
ing to the new axis A0iv B0iv. The point of greatest moments
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 53
passes then from 20 to V0, and beyond these two limits it can
never pass.
Our construction, then, is simply to lay off the load in oppo-
site directions perpendicularly from each end of the axis A0 L,
and join the end points A0 E and L E'. The intersections of
these lines with the diagram of shear give the points 20 and
V0 required.
47. Load Systems.* — Concentrated loads occur in general
in practice in a certain succession, as for instance the forces
acting at the points of contact of the wheels of a train of cars
passing over the beam, and it is necessary then to investigate
the influence of different positions of the train. It evidently
amounts to the same thing whether we suppose the weights to
move over the beam, or suppose the weights stationary and the
beam to move. In either case we obtain every possible posi-
tion of every weight relatively to the ends of the beam.
The severest load to which we can subject a railway truss, for
example, is when the span is filled with locomotives. If we
suppose, for illustration, in round numbers, the distance between
the three axles of the locomotive 3 ft. 6 in., between the
axles of the tender 5 ft. 6 in., between the foremost tender
and the back locomotive axle 4 ft., and the entire length of
locomotive and tender 34 ft. 6 in., and then suppose the weight
upon each locomotive axle 13 tons, and upon each tender axle
8 tons, we have a system of weights in fixed order and at fixed
distances, and the truss should be investigated for a series of
these systems, as many as can be placed upon the span, passing
over it from one end to the other.
In PL 9, Fig. 32 (a), we assume two such locomotives as
shown by P!_IO, and construct the force and equilibrium poly-
gons. The forces are symmetrically arranged with respect to
a central point, and the pole in the force polygon is therefore
taken perpendicular from the middle of the force line.
Now the system of forces being as represented, suppose the
span to shift. Thus suppose the span of a given length repre-
sented by Sj Si in the Fig. Then 0 6 is the line closing the
polygon for this position of the span, and a parallel to 0 6 in
the force polygon, viz., C L gives the reactions at the ends.
Let now the span move from Si Sx to s5 s5 ; we have a new po-
* Elemente der Gra/phitchen StatiJc, Bauschinger.
54: MOMENT OF KUPTTJKE OF PAEALLEL FOKCES. [CHAP. V.
sition for the line closing the polygon and new reactions. As
the span continues to shift to the right, the lines closing the
polygon revolve, and as their projections are always constant,
viz., equal to the span, they are all tangent to & parabola, which
they therefore envelop.
48. Properties of tills Parabola. — This parabola has sev-
eral important properties which will aid us in the investigation
of the case above proposed.* In PL 9, Fig. 32 (d), let XX be
the line along which the span is shifted ; a M and a N the
outer sides of the polygon, intersecting at a, along which the
closing lines slide as they revolve. For a given position s s of
the span, or cr is the corresponding line. SQ s0 is the position of
the span, for which the centre, C0, lies in the vertical through a.
In this position cr0 cr0 is tangent to the parabola at o>0, its middle
point, and upon this line lie the centres of all the other lines
(taken of course as reaching from a N to a M). Now the
point of tangency, /3, of any other line, as cr cr, with the parabola,
is as far from the centre of that line, 7, as the centre of that
line is itself from CQ. We have then only to make c b equal to
c CQ, and drop a perpendicular through 1} to find /3. Thus for
the position ^ st and the line ^ ar^ to find the point of tangency
c\, make c± dl equal to c± CQ, and draw d± c^ perpendicular to
intersection with cq ar^
Inversely we may find that position for the span s s, for which
the vertical through a given point, b, shall pass through the
point of tangency.
We have only to move the span so that its middle point c
shall be as far from CQ as it is already from the given point, or
make c c0 equal to c b. (See Art. 75.)
If we shift now the span s s, and at the same time the point
ft through an equal distance, the intersections of the vertical
through b, with the corresponding closing lines of the polygon,
will all lie upon the same line a cr.
If therefore b\ is such an intersection, b has been moved from
b to b\, and hence the span from s s to sl s^.
49. Different Cases to be Investigated. — We are now
ready to investigate the effect of a live load such as represented
in PI. 9, Fig. 32 (a). For the determination of the proportions
of the truss the following points are specially important :
* See Memente der Graphischen Statik, Bauschinger, pp. 108-114. Also,
Die GrapliiscJie Statik, Culmarm, pp. 136-141.
CHAP. V.] MOMENT OF RUPTURE OF PARALLEL FORCES. 55
1. When a certain number of wheels pass over the truss, but
without any passing off, or new ones coming on ; what position
of the system gives the maximum moment at any given cross-
section not covered by the system, and how great is this
moment?
2. Under the same supposition as above, what position of the
load gives the greatest moment for a given point covered by one
of the load systems ?
3. Among all the various points of the span, at which is found
the greatest maximum moment, for what position of the load
does it occur, and how great is it ?
4. If the number of wheels is indeterminate, how many must
pass on, and what position must they have to give at any point
the greatest maximum moment; where is the corresponding
cross-section, what position must the load have, and how great
is this maximum moment ?
The three first questions are easily solved by the aid of the
above properties of the parabola, enveloped by the closing lines
of the equilibrium polygon, corresponding to different positions
of the span.
Thus, as regards the first question, let the given cross-section
be £, PL 9, Fig. 32 (df), and suppose the span s s in the position
where the vertical through ~b intersects a a at the point of tan-
gency 0. When now the span shifts, the intersection of the
ordinate through 5, with the corresponding tie line, will always
lie upon a cr. But this ordinate gives the reduced moments for
1) (reduced to pole distance H.) The greatest of these moments
will then be simply the greatest of the ordinates between a cr
and the polygon, and will always be found at an angle of the
same. When found, we have at once the position of t>, and of
course of the span with reference to the given loads. This is
always such that a wheel stands over the given section.
Thus in Fig. 32 (a\ supposing the * four wheels P6 to P9 to
pass over the span t± ^, we seek the position of the load to give
the greatest moment at a point £ of the span from the left,
therefore -g-th from the middle.
We lay off the span in such a position, ^ tly that its centre is
distant from the intersection a of the outer lines of the poly-
gon by Jth of the span.
The ordinate through the given point now passes through the
point of tangency of the tie line and parabola. We draw this
56 MOMENT OF KUPTURE OF PARALLEL FORCES. [CHAP. Y.
tie line ^ 9, and seek the greatest ordinate between it and the
polygon. This we find at 7, and directly above 7 the given
point must lie, and hence we have the position of the span, viz.,
1 1. If the scale of tons is ten tons to an inch, of distance 5 ft.
to an inch, and the pole distance H is assumed 12^ ft. = 2-J
inches, the scale of moments will be 10 x 2.5 x 5, — 125 ft, tons
to an inch.
As to the second question; the position of the span required,
is that where the vertical through the given point of the system
S Fig. 32 (a), intersects the corresponding tie line at its point
of tangency with the parabola ; all other tie lines intersect this
vertical in a point between the tangent point and the polygon.
The middle of the span must then lie midway between the in-
tersection a of the outer polygon sides and the point s, where
the vertical through S meets the line X X. Thus the span has
the position t2 t2.
The third question, finally, is easily solved if the parabola en-
veloped by the tie lines is drawn. The greatest ordinate be-
tween this parabola and the polygon gives the greatest moment,
and the point and the position of span required, since the
middle of the span must be half-wray between the point given
by this ordinate and a.
The greatest moment is always found upon an ordinate
through an angle of the polygon.
If, however, the parabola is not drawn, we find by trial at
several angles, drawing the tie lines and comparing the corre-
sponding ordinates, the ordinate required. Here the following
considerations may aid :
When the load is uniformly distributed, the maximum mo-
ment is in the middle of the span, and at the same time in the
vertical through the intersection a of the outer polygon sides.
The polygon itself becomes a parabola. The less uniform the
load is. the more this point approaches the heaviest loaded side,
as also the intersection a, though not in the same degree. For
loads not exceedingly unsymmetrical the point may be sought
for, then, in the neighborhood of a, i.e., near the resultant of the
forces acting upon the truss. Thus in our example we are jus-
tified in selecting the corner 7 of the polygon, nearest the point
of intersection a.
5O. Mo§t unfavorable Position of Load upon a Beam of
given Spun. — The fourth question above requires a somewhat
CHAP. V.] PARALLEL FOECES. 57
more extended consideration. The most unfavorable position
of a system of given concentrated forces is when it causes the
greatest moment at the cross-section of rupture. This position
is from the preceding, given by taking the centre of the beam
midway between the vertical through the point of intersection
of the outer sides of the equilibrium polygon and the nearest
angle of the same. If with this centre we increase the span,
the maximum moment increases until the span has the greatest
length possible without more wheels coming on.
Thus for the two wheels P4 and P3, PL 9, Fig. 32 (a), a is
the intersection of the outer polygon sides, and 4 the nearest
polygon angle. The almost equally near angle 5 gives at any
rate no greater moment. In order then that these two weights
may cause the greatest maximum moment, the middle of the
beam must lie half-way between a^ and 4 ; and as the span
increases in length this moment increases, and is then greatest
when the span reaches to $i or P3.
If now the span still increases so as to also include P3, the
point of intersection of the outer polygon sides recedes to a2,
where in our Fig. it coincides almost exactly with the polygon
angle 4. Here then, approximately at 4, we must locate the
centre of the beam. If we take the same length of span a£
before, that is, make the half span 0% s2 equal to the distance
from Si to the point midway between a^ and 4, we see by draw-
ing the closing lines for these two positions of the span, that
the maximum moments measured upon the vertical through 4
are almost exactly equal in each case. For a smaller length of
span including the three weights, the maximum moment de-
creases, and is less therefore than the maximum moment already
caused by the t\vo wheels. The span ^ ^ may then be regarded
as the greatest for which the two wheels P4 P5 give the greatest
possible maximum moment. As the space s2 &>, upon which we
have now three wheels, increases, the moment increases, and is
greatest when the span, its centre always remaining now at «2
reaches to s'2 or to P2.
If now it still increases so as to also include P2, the intersec-
tion of the outer polygon sides retreats to a%. The nearest
polygon angle is still 4, and midway then between a$ and 4 we
must now locate the middle of the beam. If from this centre
we lay off the half span equal to a% s'%, to s3, and draw the clos-
ing line for this position of the span, we see as before that the
58 MOMENT OF KUPTFRE. [CHAP. V.
moment given by the ordinate at 4 is for either case almost
exactly the same. Any less span including the four weights
would give a less moment ; less, therefore, than the moment
already caused by the three weights. The span s'2 s'2 then
precisely as before, is the extreme limit upon which the
three wheels P3 to P5 cause the greatest possible maximum
moment.
In a precisely similar manner we find that the span s'3 s's
with a centre midway between a3 and 4 is the limiting span for
the four wheels P2 to P5.
If now the span still increases so that P! comes on, the inter-
section of the outer polygon sides falls in our Fig. nearly at s1?
and since this point also happens to correspond almost exactly
with the angle 3, we take the centre of the beam at s^. The
greater the span now becomes, the greater the maximum
moment. The greatest length, however, which the span can
have without including P6, is twice ^ 6, or twice the distance
between ^ and P6. If P0 also comes on, the intersection of the
polygon sides is found at «5, and the nearest polygon angle is 4.
Midway then between a5 and 4 is the new centre of the beam,
while before P6 came on, it was nearly at s±. But for centre
st the half span was Si 6, while now it is somewhat less than
4 6 ; therefore considerably smaller. Since, however, we wish
to follow the span as it continues increasing, we must compare
those two spans which are equal before and after the coming
on of P6. The right-hand ends of these spans, viz., s'4 and s5
must evidently be distant each side of 6, by the half distance
of their centres sl and 4, or 0% (more accurately the point half-
way between a5 and 4, but a5 and 4 lie in our Fig. so nearly to-
gether that the centre cannot be indicated more exactly). We
make then ^ «'4 = o^ = Mx 6, provided that ML is taken half-
way between the centre ^ and 0%.
An exact construction shows that the maximum moments for
these two spans, the one given by the ordinate through 3, the
other by the ordinate through 4, are almost exactly equal, and
moreover, that the maximum moment for the span Si Sl of equal
length whose centre is at Mt is also almost exactly equal, when
measured upon the vertical through M!. We can therefore
take Si St as the limit of those spans for which the five wheels
Pt to P5 cause the greatest maximum moment.
Taking on now the seventh wheel, the intersection of the
CHAP. V.] PARALLEL FORCES. 59
outer polygon sides is at a6 and the nearest polygon angle is 5.
Half-way between a6 and 5 we must then take the centre, while
before it lay at a^ (nearly). If we take then M2 half-way be-
tween 0-2 and this new centre, we find precisely as before the
span S2 S2 with centre M2, and right end at P7, as the limiting
span for the six wheels Pl to P6. The same holds good for the
span S3 S3 with centre M3, for the seven wheels Pi to P7, and
so on. If, according to supposition, P8 P4 P5 are 3 ft. 6 in.
apart, P2 and P3 4 ft., and P! and P2 5 ft. 6 in. apart; then for
spans up to ^ SL = say 8 ft., the two wheels P4 P5 will give the
greatest maximum moment, and their place upon the beam is
given by the position of the centre (half-way between a^ and 4).
From about 8 ft. to 15 ft. span, or s2 s2 the three wheels P3 to P5
give the greatest maximum moment, and the centre of the span
is located at a%. For spans from 1 5 ft. to 19 ft. span, or s'z 8\,
the four wheels P2 to P5 give the maximum moment, and the
centre is at ^ ; and so on. Thus for a span of any given length
we have at once the weights and their position, in order to
cause the greatest maximum moment, as also the -place of this
moment, viz., the point vertically over that angle of the equi-
librium polygon nearest the centre of the span. The ordinate
through this point included by the equilibrium polygon, and
the closing line for the given span, taken to the moment scale
gives this moment at once ; or this ordinate taken to the scale
of force must be multiplied by the previously assumed jpale
distance.
51. Oreate§t Moment of Rupture caused by a Sy§tem of
Moving ILoads at a given Cross-Section of a Beam of given
Span. — For beams or trusses of long span, which are as a rule
caused to vary in cross-section, it is not sufficient merely to find
the greatest maximum moment which a given system of con-
centrated forces can cause ; we must also know for a number
of individual cross-sections, the maximum moments which can
ever occur.
For this purp6se the force and equilibrium polygons being
first constructed, we shift as above the given span along a
horizontal line, and draw for each successive position of the
span the corresponding closing line in the equilibrium polygon,
marking the point where each closing line is intersected by a
vertical through the given cross-section, which of course moves
with the span, keeping always the same position with reference
60 MOMENT OF RUPTURE. [dlAP. V.
to the ends. The points thus obtained form a curve, and the
greatest ordinate between this curve and the polygon gives the
greatest moment which can act at the given cross-section.
This greatest ordinate will always be found at an angle of the
polygon, and hence a weight must always rest upon the cross-
section. Since the cross-section itself must lie upon this ordi-
riate, we have directly the position of the span with reference
to the given forces. The closing line for this position being
then drawn, a parallel to it in the force polygon gives the reac-
tions for this position.
The reader will do well to make the construction indicated
for an assumed span and system of weights, to convenient scales,
checking the results by computation.*
The above method applies more particularly to solid or
"plate " girders, beams, or trusses. It may of course be applied
to framed structures also, such as those illustrated in chapter
first. Thus the moment at any point, divided by the depth of
truss at that point, gives the strain in flanges. The more pre-
ferable, as perhaps also the simplest method of determining the
strains in such cases, however, is to find the reactions due to
each individual weight. Each reaction can then be followed
through the structure, as explained in that chapter, and the
strains in every member for every weight in every position can
thus be obtained and tabulated. An inspection of the table
\^11 then give at once the strains due to the united action of
any desired number of these weights.
We have thus two methods for the solution of such cases ;
first, by the composition and resolution of forces, and, second,
by the equilibrium polygon and moments of' rupture, and may,
if we choose, check the results obtained by one method by the
other. In most practical cases involving framed structures,
however, the first method is preferable as being simpler, quicker
of application, and of superior accuracy.
For solid-built beams or "plate girders," etc., the second
method comes more especially into play. The determination
of the strains in a structure of this kind from the known mo-
ment of rupture at any point, requires a knowledge of the
moment of inertia of the cross-section at that point, and this
may also be found by the Graphical method.
* This construction is given in Art. 15, Fig. VIII., of the Appendix.
CHAP. VI.~| MOMENT OF INERTIA. 61
CHAPTER VI.
MOMENT OF INERTIA OF PARALLEL FORCES.
52. THUS far we have seen that by the graphic method we
can in any practical case determine the moment of the exterior
forces acting upon a piece at any cross-section of that piece.
But the exterior forces give rise to and are resisted by molecu-
lar or interior forces. Now the moment of the exterior forces
being found, the cross-section of the piece at any point being
known, and one of the dimensions of this cross-section being
assumed, it is required to find the other dimension, so that the
strain per unit of area of cross-section shall be less than the
recognized safe strain of the material as found by experiment.
The moment of the exterior forces at any cross-section we
call the moment of rupture ; and designate it by M. Let d =
the depth of cross-section.*
y = the variable distance of any fibre above or below the
neutral axis.
/3 =; 'the breadth of the section at the distance y from the
neutral axis, and consequently a variable, except in the case of
rectangular sections.
s = the horizontal unit strain exerted by fibres in the cross-
section at a given distance c from the neutral axis.
o
Then since the fibres exert forces which are proportional to
their distance from the neutral axis or to their change of length,
the unit strain in any fibre at a distance y from the neutral
o n i
axis will be — — . Let the depth of this fibre be d y, then, since
c
the breadth of section is /9, the total horizontal force exerted
o
by the fibres in the breadth /?, will be — f3y dy. The moment
C
Q
of this force about the neutral axis will be— /3 y* d y, and the
c
Thewy of Strains, Stoney, p. 43, Art. 07.
62 MOMENT OF INERTIA. [CHAP. VI.
integral of this quantity will be the sum of the moments of all
the horizontal elastic forces in the cross-section round the neu-
tral axis, that is, equal to the moment of rupture of the section
in question. We have therefore
For a rectangular cross-section, for instance, (B is constant
and equal to the breadth b. Representing the depth by d we
have M = y^ — , or if we make c the distance of the extreme
d
fibres — -
from which M being known, as also s, if we assume b we can
find d or the reverse.
The integral //? y* dy is the moment of inertia of the cross-
section, and may be defined as the sum of the products obtained
~by multiplying the mass of each elementary particle by the
square of its distance from the axis. [See Supplement to Chap-
terVIL.Art. 10.]
From the above, we see its importance in determining the
strain at any distance from the neutral axis, or in proportioning
the cross-section, so that the resulting strain shall be less than
a given quantity at any point. We see also that for a rectan-
~b dz
gular cross -section the moment of inertia is -^-, where b is the
breadth and d the depth.
53. Graphical Determination. — "We have already seen
that the moment of a force, as Pt (PI. 6, Fig. 20) with reference
to any point, as 0, is given by the ordinate n m multiplied by
the constant H (Art. 38). The ordinate n m then represents
the product of Pt multiplied by the horizontal distance of b
from n. But the area of the trianle ~b n m\$>mnK b n =
PI x -^ b n*, that is, the area of the triangle b n m represents
one-half the moment of inertia ofl?i with respect to o. Just
as the exterior ordinates of the equilibrium polygon have been
shown to have a certain significance, and to represent the mo-
CHAP. VI.] MOMENT OF INEETIA. 63
ments of the forces, so the exterior areas of the equilibrium
polygon represent the moments of the moments, or the moments
of inertia. Thus in PI. 8, Fig. 30, the exterior parabolic area
oC/i should be one-half the moment of inertia of the rectangle
or load area o p r h.
Let us see if this is so. The area of the triangle o h C is —
o h x the ordinate S C. This ordinate S C gives, as we have
seen, the moment, with respect to S, of the reaction. "We can
therefore find its value. Thus if p is the load per unit of length,
and I is the length, ^~ is the reaction, and ~— this moment.
I pi2 p 1B
The area of the triangle o c h is therefore -~ X —r— = --— .
The parabolic area odh is f of the circumscribing rectan-
gle. This rectangle is I x S d. The ordinate S d is equal to
S C — dC. We have already found SC and dC is the sum of the
7} I I* 'D 1? T) ft
moments of P! and P2, or ~* x -. — ^r~. Hence S d — -—- —
A 4: O 4:
p p p p
^— = Q~~~- The area °f *ne circumscribing rectangle is then
7} 1? 2 V 1? T) 1?
£p Two-thirds of this is -~9 which subtracted from &&-
gives for half the moment of inertia ~-j p Z8. Hence the
2i^c.
moment of inertia is — p P, as should be.
54. We see therefore the significance of the area of the equi-
librium polygon.
If, when a number of forces are given, we form the force
polygon, and then the equilibrium polygon, the ordinates to
this last give the moments to the assumed pole distance. If
now we take these moments themselves as forces applied at the
same points, form a new force polygon with new pole distance,
and new equilibrium polygon, the ordinates to this new polygon
to the new pole distance will give the moments of the moments
or the moments of inertia of the forces. The same method is
applicable to moments of a higher order, but in practice we
have only to do with those of the second order alone.
55. Radius of Gyration. — The moment of inertia of a
system of parallel forces P! P2 etc., in a plane, with reference
64 MOMENT OF INERTIA. [dlAP. VI.
to an axis from which the points of application are distant ql $>,
etc., is then 2 P <f. This is the product of three .quantities,
one of which is measured by the scale of force, and the other
two by the scale of length. We can therefore regard it as the
product of the square of a certain length by the sum of the
given forces, or 2 P <f = P 2 P. We call k the radius of
gyration.
In order to find the moment of inertia of a system of parallel
forces then, we must by the preceding Art. construct two force
and equilibrium polygons. If the pole distances are H and H',
and the segments into which the axis is divided by the produced
sides of the polygons are P\ P'2 and P'^ P"2 etc., respectively,
then
2 P (f = H H' 2 P"
and the radius of gyration is given by
HH' 2P" = L* 2P
,
or Js-.
This expression is easy to construct. Thus for example in
PI. 11, Fig 33, let o n C be the first force polygon, o n the force
line, containing the forces P ; C the pole, and H the pole dis-
tance. Make o b equal to the second pole distance H/ and draw
b c parallel to n c and c t parallel to H. Then
HH'
whence k =
If, therefore, in Fig. 33 (&), m"0 m"n is the segment of the
axis cut off by the outer sides of the second equilibrium poly-
gon, that is, if m"0 m"n — -2* P", we have only to prolong m"0
m"n to L, making m" L = A, and describe a semicircle upon
m"0 L, and erect the perpendicular m" 7t, which will be equal
to %. In general, the pole distance H and H' can be taken ar-
bitrarily, but it is often advantageous to take H (sometimes H'
also) equal to 2 P. Then
We should then have in Fig. 33 simply to increase m"''0 m"^
by the second pole distance H', and then proceed as above to
find L
CHAP. VI.] MOMENT OF INEKTIA. 65
It is to be remembered that q± q^ etc., the distances of the
points of application of the forces from the axis, may be meas-
ured in any direction, and H is parallel to this direction, and
is not therefore necessarily perpendicular to o n.
The above will be rendered plain by reference to Fig. 34-,
PI. 10. We suppose four forces applied at the points At A2 A3
A4 respectively, and acting parallel to XX. Required the mo-
ment of inertia of these forces and the radius of gyration, the
distances gl q^ etc., being measured parallel to Y Y. First we
form the force polygon by laying off along X X, 0 1, 1 2, 2 3,
3 4, parallel, and in the direction of action of the forces, cligos-
ing a pole C, and drawing C 0, C 1, C 2, etc. We now construct
the corresponding equilibrium polygon, CI, III, II III, in IV,
etc. The segments 01', 1'2', 2' 3', etc., represent the statical
moments of the forces with reference to X X. That is, these
segments to the scale of force multiplied by the pole distance C y
parallel to Y Y to the scale of distance, give the statical moments
of the forces. Now we take these segments themselves as forces,
and suppose them acting at the former points of application.
With the same pole as before we draw CO, C 1', C 2', etc., and
form the corresponding equilibrium polygon CI, III', II III'', etc.
The sum of the segments of XX cut off by the outer lines of this
polygon, or o y, to the scale of force multiplied by H H' or C if
gives the moment of inertia of the forces with respect to XX.
This moment then is M = 0 y x Cy*
where Oy "= 2P" and"Cp = HH'.
The radius of gyration Jc is, as we have seen, given by
CT rrr ^ T)fr
, 5P being equal to 0 4 in the Fig. Hence
04
If, then, we lay off 0 d = 0 4, and make 0 c = C y, and make
the angle dee a right angle, we shall find a point e to the right
/"I ~ 2 TT TT/
and Oe will be equal to -- = -y- Upon ey now describe
a semi-circle, the point of intersection y with the perpendicular
through 0 will give (Art. 55)
/Oy x Cy3 /HH'SP"
= V " Q4 c = V " ^£P — = ^ — radius of gyration.
The square of this line, then, multiplied by 2 P or 0 4, will give
5
66 MOMENT OF INERTIA. [CILAP. VI.
at once the moment of inertia of the given four forces with
reference to X X and Y Y as axes. If we were to suppose the
same forces with the same points of application to act parallel
to Y Y instead of X X, the distances qt q2 being measured par-
allel to X X instead of Y Y, we should have the force polygon
G! O li %i 3t 4X instead of 001234, and a precisely similar
construction would give us o x multiplied by pole distance for
the moment of inertia, and 0 a' for the radius of gyration. We
recommend the reader to follow through the construction as
shown by Fig. 34.
56. Curve of Inertia— Ellipse and Hyperbola of Inertia.
— II having found the radius of gyration as above, we lay it off
from the axis on either side, in a direction parallel to the direc-
tions in which <2i $2? etc., are supposed measured, and through
the points thus determined draw two parallels to the axis M'
and M" on either side, and then suppose the axis to revolve in
the plane of the forces about any point as O situated in the
axis ; the lines M' and M" also revolve and enclose a curve of
the second degree, whose centre coincides with O. Thus, if in
PL 10, Fig. 34, we lay off O I along Y Y on both sides of X X
equal to o V = k already found, and then let X X revolve
about O, K J and J K will also revolve, and enclose either an
ellipse or hyperbola.
In order to prove this, take O as an origin of co-ordinates.
Let the co-ordinates of the points of application of the forces Ax
A2, etc., be xt y}, a?2 y» etc. From each of these points A draw
parallels to the axis of y, intersecting the axis of x in the points
C. Then O C = a?, A C = y. Now pass through the point O
an axis of moments M in any direction, and project for each
point OCA parallel to this axis upon the line q, which meas-
ures the distance of each point from the axis of moment (not
necessarily perpendicular distance). This projection is evi-
dently equal to q. Denote by a and /3 the ratios by which dis-
tances along X and Y must be multiplied, in order to obtain
their projections upon £, by lines parallel to M. Then
q=a%+ @y
for each point of application, and hence
or since for one and the same axis M, and direction £, a and
are constant,
CHAP. VI.] MOMENT OF INERTIA. 67
In this expression a and /9 will vary with the position of M
and the direction of ^, but 2 P a?2, 2 P y3 remain unchanged.
These last expressions are, however, nothing more than the mo-
ments of the second order (moments of inertia) of the given
force system with reference to the co-ordinate axis, the distances
of the points of application being measured in the direction of
the axis. They are known if the force system is given and the
co-ordinate system assumed.
If we put 2 P ar1 = a2 2 P, 2 P f = tf 2 P, 2 P x y = f2 2 P,
£ and #, etc., are the radii of gyration of the moments of inertia
with reference to x and y, and the above eqiiation becomes
^P (f = 2 P [a2 a? + /32 &2 + 2 a /9/2]
If we conceive for the assumed position of M, the radius of
gyration "k to be found, and M' and M" drawn on either side
at a distance ± &, measured parallel to £, and indicate the dis-
tances cut oft' by these lines from the co-ordinate axes by ± xe
± 2/e, and then project these distances parallel to M upon the
direction of q> or &, we have Jc = a xe = ft y^ whence
K Q K
a — — p = —
®e y*
and these values substituted in the above equation give
where ^ is essentially positive in the second term.
Hence,
If we suppose the axis M to change its position revolving about
O, the segments xe y& cut off from the axes of x and y by
M7 and M" alone will change in this equation. It is therefore
the equation of the curve enclosed by M7 M". If this curve is
known for a given force system, then the moment of inertia for
any axis passing through its centre is easily found. We have
only to draw parallel to the axis two tangents to this curve, one
on either side, and measure their distance from M, in the direc-
tion in which the distances q of the points of application from
the axis are taken. This distance is the radius of gyration,
and the moment of inertia is simply the product of its square
by the algebraic sum of the forces.
68 MOMENT OF INERTIA. [CHAP. VI.
We call the curve represented by the above equation there-
fore, the curve of inertia. If we refer the curve to co-ordinate,
axes which coincide with the conjugate diameters, the equation
becomes
^+I2 = ±i
o ' o - —1— *•
a? y2
where x and y are the new ordinates, and A, B, the conjugate
semi-axes of the curve. A and B are therefore the radii of
gyration of the force system, measured in the direction of the
co-ordinate axes, and hence
where x and y are the co-ordinates of the points of application
of the given forces.
. Since 2 P <f = Z? 2 P if the sign of -£P <f is the same as
2 P, J£ is positive. When, on the other hand, these signs are
different, #* is negative. That is, when all the forces act in
the same direction W is positive, and we have
A2 "R2
— + — 1
X* + y*-
which is the equation of an ellipse.
If, however, the parallel forces act in different directions, kt
may be positive or negative. For cases where ^ is negative,
either A2 or B2 will be negative, and we shall have
'
or
Both cases coincide. The double curve consists of two hyper-
bolas with common assymptotes, common centre, and equal
semi-axes. For every axis M passing through the common centre
O, we have a pair of parallel tangents either to one or the
other hyperbola. The corresponding ^ is positive for the one,
negative for the other.
If, then, in the method of construction to which we shall
presently refer, the square of the semi-axis B, which lies in the
axis of Y, is negative, that hyperbola whose imaginary axis lies
in Y gives ^ positive, the other gives ^ negative, and reversely
for the other case. If the axis of moments M coincides with
CHAP. VI.] MOMENT OF INERTIA. 69
one of the common assymptotes, the radius of gyration and
moment of inertia with respect to it of the given force system
is zero.
57. Construction of the Curve of Inertia. — The curve of
inertia for a given system of parallel forces and given centre
O, is determined by the direction of any two conjugate diame-
ters, since as we have seen in Art. 55, PL 10, Fig. 34, these
directions being assumed we can find the radii of gyration with
respect to XX and Y Y, and can thus determine O a and O 5,
the semi-diameters. We have then to develop a principle by
means of which these directions may be determined.
If we denote the distances of the points of application of the
forces from the axis of M measured in any direction by y, then
the statical moments of the forces, P y, are indeed dependent
upon the direction in which y is measured, but their relative
values remain the same. If then being found for any direction
of y, these statical forces are considered as being themselves
parallel forces acting at the points of application, and their
centre of action is found (for gravity — centre of gravity) for
some other value of y, this centre of action remains unchanged.
For any axis passing through this centre of action the sum of
the moments of the forces is zero. If therefore we take a point
O in the axis M as origin of a system of co-ordinates, whose
axis O X may lie at will in the plane of the forces, while O Y
passes through the centre of action ; the sum of the moments
of the statical moments P y, considered as forces acting at the
points of application, with reference to O Y, will be zero.
These moments however, provided that the distances of the
points of application are measured along the co-ordinate axes,
are the moments of 'inertia, viz., 2 P y x. If these are zero we
see that the general equation of the curve of inertia (1) Art. 56,
becomes that of a hyperbola referred to its conjugate diameters
as axes. With the centre O therefore, the line joining O with
the centre of action, gives the direction of the conjugate di-
ameter of the curve.
This is the principle required. By means of it we can find
the conjugate diameters of the inertia curve, for a given centre
O, and thus construct it.
58. Con§truetion of the Curve of Inertia for four paral-
lel force§ in a Plane. Example. — As an example let us
take the four parallel forces, in PL 10, Fig. 34, supposed
70 MOMENT OF INERTIA. [CHAP. VI.
to act in different directions, parallel to X X at the points
A! Ag, etc.
As before we have the force polygon C 0 1 2 3 4 for an arbi-
trary axis as X X, and from the corresponding equilibrium
polygon, we determine the statical moments with reference to
X X, 01' 1/2', etc., to the basis C 0. These moments we again
consider as parallel forces acting at At A2, etc., for which we
have CO V 2' 3' 4' and corresponding equilibrium polygon
C I IT ffl', etc. We then determine the centre of action S, by
a second polygon 0" I II" III", etc., the sides of which are
respectively perpendicular to the first, according to the process
for finding the centre of gravity, Art. 30. The line joining O
with S gives th& direction ofYY, the diameter of the curve
conjugate to X X. To find the length of the semi-diameters
0 1} and O #, we must find the moments of inertia of the forces
with reference to X X and Y Y, taking the distances of the
points of application as measured parallel to these lines.
Therefore instead of C 0, we must take C y as basis or pole
distance, and then find the radii of gyration as already indi-
cated in Art. 55, viz., O bf and O a' '. These distances laid off
along Y Y and X X give the semi-conjugate diameters of the
curve of inertia.
From the Fig. we see that the force Pl whose direction from
left to right we shall always consider positive, and 5* P — 0 4
have the same sign. On the other hand the total moment of
inertia 0 y and the moment of inertia of Pl5 viz., 0 l"_have dif-
ferent signs. The square of radius of gyration #* = ~ y p —
is therefore negative, the radius itself or the semi-diameter O b
is imaginary.
In similar manner, we see that O a the radius of gyration for
Y Y is real, since the total moment of inertia O x and S P — 0 41?
have the same signs. The curve is then a double hyperbola
with the conjugate semi-diameters O a and O b.
It is then easy to find the assymptotes K K and J J, and by
bisecting the angle which they make, the principal axes A A
and B B. In order to find the length of these axes, we have
the well-known principle that for any point as &, the product
of a k and k O (a ~k being parallel to the assymptote J J) is equal
to -r the sum of the squares of the semi-axes ( — j — ). If then
CHAP. VI.] MOMENT OF INEKTIA. 71
we find Jc I, the mean proportional of O k and Jc a, and lay it off
twice from O to D along the assymptote O K, O D is the diag-
onal of a rectangle whose sides are the principal axes. We
thus find the vertices A, A, B, B.
We can thus construct the curves. Then for any position
of the axis X X as it revolves about O, we can find the cor-
responding radius of gyration and consequently the moment of
inertia, by simply drawing tangents to the curve above and
below the new position of X X w&& parallel to it. The radius
of gyration thus obtained measured to the scale of length and
multiplied by the algebraic sum of the forces, or 0 4 to the
scale of force, will give the moment of inertia required for the
assumed position of the axis.
59. Central Curve. Central Ellipse. — If the point O about
which the axis turns coincides with the centre of action (or
gravity) of the forces, we call the curve enclosed by the paral-
lels M' M" at the distance k on either side, the central curve.
When the parallel forces all act in the same direction this curve
is always an ellipse.
For the central curve the principle proved in Art. 57 and
the method of construction given in Art. 58, are no longer
applicable, for the algebraic sum of the statical moments of
the given forces is zero for every axis through the centre of
gravity. We cannot therefore find the centre of gravity of the
moments of the forces, wrhen considered as forces themselves
and applied at the given points of application.
If we divide, however, these moments considered as forces
into two portions or groups, and find the centre of gravity of
each group, the line joining these two points has an important
property, viz., that for every moment axis parallel to it, the
algebraic sum of the moments of the statical moments consid-
ered as forces, that is, the algebraic sum of the moments of
inertia of the forces, is zero. In other words, 2 P e e' is zero,
e being the distances of the points of application from the first
axis, which passes through the centre of gravity of the forces,
and e' the distances from the axis parallel to the line joining
the two centres of gravity of the two groups of statical moments
considered as forces. If we draw then through the centre of
gravity of the forces themselves the moment axis X X, and
take it as the axis of abscissas of a co-ordinate system whose Y
axis passes also through the centre of gravity of the forces and
72 MOMENT OF INERTIA. [CHAP. VI.
is parallel to the line joining the two centres of gravity of the
statical moments considered as forces, then the moments of
inertia 2 P y x are zero, and hence as in the preceding Art.
this axis ofYis conjugate to X X.
This holds good not only for the central curve, but also for
every inertia curve, whose centre O instead of coinciding with
the centre of gravity of the forces, lies in the axis passing
through that centre. In this case also the axis through the
centre O parallel to the line of union above, is a conjugate to
X X. Still more, the half length of this conjugate diameter is
in both cases the radius of gyration of the force system for the
axis X X and the direction of Y.
Hence in every inertia curve of a system of parallel forces,
whose centre lies in an axis passing through the centre of grav-
ity of the forces, the diameters conjugate to this axis are paral-
lel and equal. All these inertia curves are therefore touched
by two lines parallel to this axis and equally distant on. either
side. This distance is the radius of gyration for this axis.
For any such inertia curve, whose centre O is distant i from
the centre of gravity S of the forces, we call E and (£ the par-
allel conjugate axes to S O for this curve, and the central curve
respectively; q and q the distances from them of any point of
application, these distances measured parallel to S O, and con-
sidered positive when the point of application lies on the same
side of E or (S respectively as the centre of gravity S from E.
Then *, the distance apart of E and (£ is essentially positive,
and if we indicate by a and a the lengths of the semi-conj ngate
diameters for the inertia and central curve respectively, we
have
where q and q stand in the simple relation
q = q+i.
Hence
Since (§: passes through the centre of gravity $P q = o, and
therefore
2 P q2 = 5 P/ +«* S P = a* 2 P.
Hence
# = aa+£,
an equation which gives the relation between the lengths of
the semi-conjugate diameters of the central and. any inertia
CHAP. VI.] MOMENT OF INERTIA. 73
curve, whose centre lies upon an axis through the centre of
gravity of the forces, at a distance \' from this centre.
Any two curves at equal distances either side of the centre of
gravity are therefore equal. If the semi-diameter of the central
curve a is real, and therefore a2 positive, a2 is also positive and
greater than a2. All the inertia curves are therefore of the
same kind as the central curve, and enclose the centre of grav-
ity. If, however, a2 is negative, and the central curve there-
fore an hyperbola; all those inertia curves whose centres are
distant from the centre of gravity by a distance i less than a
are hyperbolas also. For a distance i equal to a, the curves re-
duce to straight lines equal and parallel to the conjugate diame-
ter of the central curve. For i greater than a, the curves be-
come ellipses.
6O. Centre of Action of tlie Statical Moments of the
Force§.* — We again suppose, through the centre of gravity of
the forces S [Fig. 35, PL 11] a line N N drawn which cuts the
central curve at A and A'. Two such points we have in every
case, except when the curve is an hyperbola, andN N coincides
with an assymptote.
Let (5 be the conjugate axis to N N in the central curve, E a
parallel to it through any point o distant i from S, and also
conjugate to N N in the inertia curve whose centre is o. Then
since the statical moments' of the forces with reference to N N
is zero, the centre of gravity of the statical moments with re-
spect to E? considered as forces acting at the points of applica-
tion, will be somewhere upon N N. It is required to find
where.
We call q the distance of any point of application from E,
measured parallel to N N, and positive when upon the same
side of E as S, then i is essentially positive.
As before, q is the distance of the points of application from
(S, also measured parallel to N N, and positive in the same
direction as q.
Then we have always
q = q+i.
and for the moments of inertia of the forces with respect to E
and(£ 2P<? = ZP (q+i)2 = 2Pf+#2P
or when a is the semi-diameter of the central curve, S A = S A'
and ^Pq2= a*+$ 2P
* See Supplement to Chap. VIL, Art. 10, latter part.
74 MOMENT OF INERTIA. [CHAP. VI.
Let now m be the distance of the centre of gravity or action,
of the moments of the force's with respect to E, from E, and in
its distance from (5, positive the same as q and q. Then
m = tn + i
and since the sum of the moments is equal to the moment of
the resultant :
^•Pq2 — m^Pq.
But the sum of the moments P q of the forces with reference
to E, is equal to the product of the sum of the forces into the
distance i of the centre of gravity of the forces from E. Hence
and therefore
m i 2 P = 2P q2 = (a
or, m. i — a2 + r.
Introducing the value for m
(m + i) i = a3 -f ^2
or tit i — a2.
If now a2 is positive, which is always the case for an ellipse
as central curve, m is also positive, and is therefore to be laid
off from S along N N on the opposite side of Qj from o. If
then we conceive an axis E' drawn parallel to E, and symmet-
rical with reference to S, which axis we shall call for conven-
ience the symmetrical axis to E, we see from the above relation
that M is the pole of this axis in the central curve.
If, however, a2 is negative, therefore a imaginary, nt is nega-
tive, and must be laid off from S towards 0, and the point M
thus found is therefore the pole of the axis E itself, or in the
case of an hyperbola is the pole of E' in that hyperbola which
is not cut by N N, and for which therefore A A! is imaginary.
Hence we have the principle —
If we consider the statical moments of the forces with refer-
ence to any axis as E as themselves forces acting at the given
points of application, the centre of gravity of these moment
forces does not coincide with the centre of gravity of the origi-
nal forces, but is the pole * in the central curve of an axis E'
parallel and symmetrical to E.
In those cases where the central curve becomes an hyper-
* POLAR LINE OF A POINT, in the plane of a conic section, is a line such,
that if from any point of it two straight lines be drawn tangent to the conic
section, the straight line joining the points of contact will pass through the
given point, which is called a pole.
CHAP. VI.] MOMENT OF INERTIA. 75
bola, we must observe whether the diameter conjugate to the
moment axis is real or imaginary. In either case the centre of
gravity is the pole of the line symmetrical to the moment axis
in that hyperbola for which that diameter is real or imaginary.
The construction is given in PL 11, Fig. 35.
Upon S o' = S o we describe a semi-circle. With S as cen-
tre, and S A' = a = semi-diameter of the central curve, describe
an arc, and from the intersection with the semi-circle drop a
perpendicular upon So'. The point M thus found is the centre
of gravity of the moments. For : a2 = a M* + m2 and a M2
= ra (i—m) hence a2 = m2+m i— m2 — m i. The central curve
being known as also the distance i, the point M can be readily
found.
61. Cases where the Direction of the Conjugate Axis of
the Inertia Curve can be at once Determined. — There are
certain special and practical cases in which the conjugate direc-
tions or axis of the inertia curve can be at sight determined, so
that only the length of the semi-diameters remains to be found.
The most important of such cases are as follows :
(1.) When in a system of parallel forces, these forces can be
so grouped in pairs, that the lines joining the points of appli-
cation of each pair are all parallel, and the centres of gravity
of each pair all lie in the same straight line. Then for the
central curve and all inertia curves whose centres lie upon this
straight line, the direction of the axis conjugate to this line is
the same as that of the lines joining the points of application
of each pair.
This is easy to prove. For, for each pair, the sum of the
moments with respect to the line joining their centres of gravity,
is zero. These moments regarded as forces and applied at the
points of application, give therefore for each pair two parallel
opposite and equal forces, the sum of the moments of which
for any line parallel to the line joining the points of applica-
tion, is zero. This is the case for all the pairs, and therefore
the direction of the lines joining the points of application is
that of the axis conjugate to the line joining the centres of
gravity, for the central curve as also all inertia curves whose
centres lie upon this last line.
(2.) When the forces can be so grouped that the points of ap-
plication of each group lie in parallel lines, and the centres of
gravity of the groups lie in the same straight line. Then this
76 MOMENT OF INEBTIA. [CHAP. VI.
straight line gives the direction for the central curve and every
inertia curve whose centre lies upon it, of the diameter conju-
gate to an axis passing through the centre and parallel to the
lines joining the points of application.
For if we take any such axis, the points of application of the
forces in each group are equally distant. The statical moments
for each group are then proportional to these distances. If,
therefore, they are considered as forces, their centre of gravity
coincides with that of the forces themselves, and lies therefore
in the line joining the centres of gravity of the groups. The
centre of gravity of the whole force system lies then in this
line, which is therefore the direction of the axis conjugate to
the line parallel to the lines joining the points of application,
in the central curve, and also all curves whose centres lie upon
this line.
(3.) When the forces can be so grouped that the centres of the
central curves of each group lie in the same straight line, and
the diameters in each curve conjugate to this line, are parallel.
Then in the central curve of the entire system, the diameter
conjugate to this line is also parallel to these diameters. For,
for any axis parallel to these diameters, the centres of gravity
of the moments of the forces in each group lie upon the line
joining the centres of the curves. The centre of gravity of the
moments for the entire system lies then also upon this line,
which is therefore the direction of the axis conjugate to an axis
parallel to the diameters of the curves, for any inertia curve
whose centre lies upon this line.
In all these cases, if the directions thus found are perpendicu-
lar, we have to do with the principal axes.
62. Practical Applications. — We can now apply the above
principles to practical cases, and as in the determination of the
moment of inertia of irregular figures, we have to deal with
triangles, parallelograms and trapezoids, we have first to con-
sider these three cases.
1st. The Parallelogram. PL 11, Fig. 36.
The moment of inertia of a parallelogram is, as is well known,
^ == 12 a ^** a ke*ng tne breadth and 5 the depth.
ax2 dx = — ab*
CHAP. VI.] MOMENT OF INERTIA. 77
Hence T& = — b2 = radius of gyration, or k= *J -bx — b.
That is, the radius of gyration is a mean proportional between
1 , . 1 _
-b and -b.
The centre of gravity of the parallelogram is at O the inter-
section of the diagonals, and this is therefore the centre of the
central curve.
If we suppose the parallelogram divided into laminae parallel
to D C, and suppose each lamina divided by G H parallel to
B C, the centres of gravity of each will lie upon G H. Right
and left of G H we then have a group of forces whose points of
application lie in lines parallel to G H, and the lines joining
any pair, one on each side of G H, are parallel. By (1) of the
preceding Art., therefore, GH and EF are conjugate axes of
the central curve. For the lengths of the half diameters, we
find the mean proportional between -^ b and -^ 5, -^ a and -^ #,
respectively, by the half circles B F and BH. We thus find k
and &', and can then construct the central ellipse directly, or
find the principal axes, and then construct it. The centre of
gravity of the moments of the parallelogram, with reference to
any axis parallel to A B, is as we have seen, Art. 60, the pole of
a line parallel and equally distant from O on the other side. If
we draw this line then, as D C, then from G draw two tangents
to the central ellipse, and unite the points of tangency by a
line ; the intersection of this line with O G is the centre of
gravity of the moments of the forces themselves considered as
forces, or area of the parallelogram, with reference to A B.
Zd. Triangle. PL 11, Fig. 37.
The moment of inertia of a triangle for the axis B C is
.H5 a A3 * whence # = -~ A2, and for an axis E F distant i =
L£ O
•o A, which passes through £he centre of gravity,
a* = tf-# = tf- (Art. 59.)
I
a '±=f tfdx = -^-ah3,h being the line A D, a = B O.
78 MOMENT OF IKERTIA. [CHAP. VI.
The conjugate axes of the central curve are by principle 1 or
2 of the preceding Art. E F and A D.
The above value of a is then the length of the semi- diameter
along A D, or a — V -^ h x -= h. That is, a is a mean propor-
tional between -^ h and -~ h. This is found by the semi-circle
ODFig. 37.
The moment of inertia of the triangle with respect to A D is
•g h (~o^)3- The radius of gyration then is y -^ {-^af =
' "2 \%a)x ^ \va) or a mean proportional between —, and
|offorDC.
This is given by the semi-circle on D G = ^ D C, and we
thus have the four points 1 2 3 4 of the central ellipse, and the
semi-diameters 0 1 and 0 3, and can therefore construct it.
From the central ellipse as before, we can find the centre of
gravity of the moments considered as forces for any axis par-
allel to B C or A D, as also in either case, the radius of gyration
and therefore moment of inertia, for any axis passing through O.
3d. Trapezoid. PL 11, Fig. 38.
Here the lines E F joining the centres of the parallel sides,
and G H parallel to these sides, aird passing through the centre
of gravity 0, are the conjugate axes of the central ellipse.
For the axis A B and direction E F, the moment of inertia is
a and 5 being A B and C D, and h = E F. The square of
radius of gyration is then
1 . ,. , ~6 a+b
2
-
CHAP. VI.] MOMENT OF INEKTIA. 79
For the radius of gyration for G- H, at a distance i = -r- h
- we nave
+ H
9
la + Sb I ,2^ + 2^\2
0? = %? — $ =-£• - — T- A2 — rr A2 I - --T- 1 =
6 a + b 9 V a + b /
1 /a
J *'•
This radius a is half the diameter along E F.
To construct it, put (3 a)2 = -^ A2 + -/ — T~T\2 ^2-
Describe a semi-circle upon E F, and at the centre 01? arid
at the intersection of the diagonals &, erect perpendiculars
Ol J and K L. Then F"!2 = A2 and iTL2 = 7-^ A2, since
^
E K = -- j h and K F = --- 7 A
a + b a + b
K L equal to J M from J, we have
E K = -- j h and K F = --- 7 A. If therefore we lay off
a + b a + b
and hence the half diameter sought is one-third F M. "We
thus find 0 1 and 0 2.
To find the other semi-diameter we have the moment of in-
ertia for E F and direction G H, jg- (a* -f- a* I + a
hence the square of the radius of gyration is
-^
This last expression is easily constructed^ In the right-angled
triangle F B N, the hypothenuse F N = A/ (- a\2 + (^ b\*>
B N being made equal to C E. If we describe then a semi-
J / ' 7iy*dy+ I *h a-^-y^y
80 MOMENT OF INERTIA. [CHAP. VI.
circle upon F U = -~ F N, and make F W — ^ F N, F V is the
semi-diameter sought. We thus find 0 3 and 0 4, and can now
construct the central ellipse. This being constructed we can
find the centre of gravity of the moments with reference to any
axis parallel to A B or E F, according to Art. 60, or the moment
of inertia for any axis through 0, by drawing a parallel tangent
to the ellipse. The distance from 0 to the point o£ tangency
gives then the radius of gyration for that axis.
Uh. Segment of Parabola. PL 11, Fig. 39.
Let the segment be limited by B C = 2 A, and A D = I.
Then it is evident that these two axes are conjugate (Art. 61),
and the centre of the central curve is 0, the ratio of A 0 to
OD being as 3 to 2. Hence AD and E'F', parallel to CD
through 0, are conjugate axes of the central curve. To find the
length of the semi-diameters along these axes we find first the
moment of inertia of the segment with reference to an axis Y Y
parallel to E' F' and tangent to the parabola at A. We have
then for this moment of inertia
/'
Jo
where p is the parameter of the parabola, and I — A D. Since
4
3
4
the area of the segment is -~ h I, we have for the square of the
radius of gyration
The square of the radius of gyration then for E' F' whose
distance from A is i = — I is
o .
a being the semi-diameter along A D. It is easier here to com-
pute a, viz., a = 0.26186 I, and lay it off from O, thus finding
3 and 4.
For the other, semi-diameter we find the moment of inertia
for A D and the direction E' F'. Thus
CHAP. VI.] MOMENT OF INERTIA. 81
The radius of gyration squared is, therefore, •
and hence the radius of gyration is & = 0.44721 A. Laying
this off from 0, we obtain 1 and 2, and can therefore now draw
the central ellipse.
63. Compound or Irregular Cro§§-Section§. — Every cross-
section may be divided up into trapezoids, triangles, parallelo-
grams and parabolic segments, and the above cases will aid us,
therefore, in the application of the graphic method to compound
or irregular cross-sections. The engineer is often called upon
to determine the moment of inertia of such sections as the T,
double T, or different combinations of these in proportioning
the different pieces of bridges, such as chords, struts, floor-beams,
etc., as also in many other constructions. The calculation for
such cross-sections is sometimes very laborious. As an example
' of the application of the graphical method best illustrating the
above principles, we take the cross-section shown in Fig. 40,
PL 12.
First we divide the cross-section into a series of trapezoids.
The first segment, bounded by a curve, we may consider a para-
bolic area. These trapezoids we reduce to equivalent rectangles
of common base a [Art. 32], and take the corresponding heights
as forces. These forces we lay off in the force polygon and
choose a pole C at distance H from force line, drawing CO, 01,
C 2, etc. Parallel to these lines we have the first equilibrium
polygon I II III VIII, the intersection of the two outer
sides of which gives the point of application of the resultant.
The intersection S of the resultant with the axis of symmetry
gives the centre of gravity of the cross-section [Art. 30]. The
segments o 1', 1'2', 2'3', etc., cut off from o S, give the statical
moments of the forces with reference to o S to the basis H.
We now choose another pole C' at distance H', and form another
force polygon, considering these moments as forces, and applied
at the centres of action of the moments of the separate areas
into which the whole cross- section has been divided. These
centres of action can be determined by forming the central
curve for each area according to Art. 62, and then applying the
6
82 MOMENT OF INERTIA. [CHAP. VI.
principle of Art. 60. A little consideration will show that these
centres of gravity will coincide approximately with the centres
of gravity of the areas themselves, except for areas (3) (4) (5)
and (6). Finding then for these areas the centres of action of
the moments considered as forces, we construct the equilibrium
polygon O' I' IT. . . .VIII'. The distance 0" 8" cut off by the
first and last sides of this polygon gives the moment of inertia
to the pole distances H and H' and the reduction base a. Thus
0" 8" measured to scale of force and multiplied by a H H' is
the moment of inertia of the cross-section with reference to o S.
In TI H' (V'8"
The radius of gyration is then k=\J = .
a 0 8
The division will be performed if we take H' = 0 8 = 2 P.
This we can easily do now without drawing a new polygon,
since what is required is the intersection of the outer sides only.
Thus take a new pole C/ distant from o S, H' = 0 8. Now we
know that each side of the new polygon for this pole distance
will intersect the corresponding side of the first in a line paral-
lel to o GI [Art. 27]. Since the new polygon may start from
any point, we may take the first side to coincide with O VIII'.
Then the line of intersection of any two sides is O VnT 8".
Produce any side as IV' V7 to intersection e with this line ; from
e draw e a/ parallel to C/ 4'.
Through a! the intersection of o' I' and V IV, the resultant
of (1) (2) (3) and (4), must pass. The change of pole cannot
affect this resultant, which must therefore pass through a/, the
intersection of <? a/ with the vertical through a' parallel to o S.
Hence 0/ at' is the direction of the last side of the new poly-
gon, and Sf/0i" is the moment of inertia for the new pole dis-
tance o C/ = 0 8. The radius of gyration then is k = V H O/'S"!
In other words, Jc is a mean proportional between H and O/'S".
The construction of k is given by the semi-circle described upon
O/'S" + H. The ordinate to this semi-circle through O/7 per-
pendicular to <?S gives Jc. We thus find the semi-diameter
S a = S a' of the central ellipse.
In order to find the other semi-diameter S b = S £', we might
divide the cross-section into areas by lines parallel to S X, and
then proceed as above. This is, however, unnecessary. "With
the same areas as before, we can find the central curve for that
area on each side of X X, and then the centre of application of
CHAP. VI.] MOMENT OF INERTIA. 83
the moment of each of these areas with respect to X X itself,
considered as a force. The method of procedure is then pre-
cisely as before. We draw a polygon the sides of which are
respectively perpendicular to those of the first polygon, and
thus find the statical moments 0"' I'" V" 2'", etc., to basis H.
Choosing then a pole C'" at distance H'" and drawing the
corresponding polygon, we have 0 8 1V for the moment of in-
ertia. The radius of gyration is then Jc = \ aHH"' Q8IV
a. 0 8
We have taken H'" = g 08, hence k = 1 H QQ- iv Hence k
1 _
is a mean proportional between g 0 8 1V and H. The construc-
1 _
tion is given in the Fig. by a semi-circle upon H+ ^ 0 8 IV. We
thus find the semi-axis S V — S £, and can now construct the
central ellipse. We have thus found graphically not only the
moments of inertia of the cross-section with respect to X X and
Y Y, but, by means of the central ellipse, for any other axis in
the plane of the Fig. passing through S.
64. — The above method of procedure holds good generally
for any cross-section, except that, when there is no axis of sym-
metry, the centre of gravity must be found by a second equili-
brium polygon whose sides are respectively perpendicular to
those of the first. When the moment of inertia with reference
to a single axis only is required, the above method becomes
quite short and simple, as well as accurate. In our Fig. the
scale used as also the number of divisions taken make the pro-
cess appear more complicated than it really is.
With this we shall close our discussion of moment of inertia,
merely observing, that all the principles deduced in this chap-
ter for forces acting in a plane hold equally good for forces in
space. The central curve then becomes an area, we have a mo-
ment plane instead of moment axis M, and the ellipse and hyper-
bola of inertia become ellipsoid and hyperboloid respectively*.
For a much fuller discussion of the subject than is possible
here, we refer the reader to Culmanrfs Graphische Statik, pp.
160-206 ; also Bauschinger's Elemente der Graphischen Statik,
pp. 116-168. To the latter we are largely indebted in the
preparation of the present chapter ; Plates 10 and 12 are, with
slight alteration, reproduced from that work.
84- APPLICATION TO BRIDGES. PART -II.
PART II.
APPLICATION TO BRIDGES.
65. — Under the head of Parallel Forces we have already
given the general application of the graphical method to the
determination of the moments and shearing forces in beams
resting upon two supports only. We shall now take the sub-
ject up more in detail, and show the methods of determining
the maximum strains for all the possible conditions of loading
which may occur in Bridge Girders. In the following we shall
adhere closely to the development of the subject as given by
Winkler. {DerBruckenbau, Wien, 1872.]
66. Force§ wliicli act upon a Bridge. — The forces which
act upon a bridge may be enumerated as follows :
1st. The weight of the bridge itself. — This, previous to the
calculation of the strains, is unknown, since it depends upon the
intensity of the strains themselves. It is customary to assume
the weight to begin with, by comparison with existing struc-
tures of similar character, and then to find the resulting strains.
The weight answering to these strains can then be easily ascer-
tained ; the strength of the materials used being known, and
compared with the assumed weight. According as it is less or
greater, the weight was then assumed too great or the reverse,
A second approximation to the true weight may then be made,
and the strains proportionally diminished or increased. As
rules for estimating the weight of bridge girders under 200 feet
span, we have, for weight of girder G-,
where W = the assumed approximate total distributed load in
tons, including the weight of girder ;
I = length in feet ;
d = depth in feet ;
PAKT II.] APPLICATION TO BRIDGES. 85
y= the working strain in tons per sq. foot of cross-section.
(See Stoney, Theory of Strains, vol. ii., p. 441.)
We have also the rule : " Multiply the distributed load in
tons by 4 ; the product is the weight of the main girders, end-
pillars and cross-bracing in pounds per running foot." Iron is
taken at 5 tons per sq. inch tension, and 4 tons per sq. inch
compression.
2d. The moving or live load; which is determined by the
purpose of the bridge. This load can take various positions
upon the bridge, and may even be divided into several por-
tions. It is therefore an important problem to determine that
distribution which shall cause the maximum strains.
The live load is, as the term implies, in motion, so that, in
combination with the deflection, there is a centrifugal force,
or increase of pressure. This is, however, in practice disre-
garded, while such a coefficient of safety is chosen in propor-
tioning the parts, that the increase of strain due to this cause is
fully covered.
3d. Horizontal forces, caused by the wind and the passage
of loads.
^th. Pressures at the supports. The known forces cause re-
actions at the supports, which evidently must also be considered
as forces acting upon the bridge girder. For straight girders,
these reactions are vertical, while in suspension and arch sys-
tems they are inclined.
67. Bridge Loading. — The heaviest load to which a railway
bridge can be subjected is when it is covered from end to end
with locomotives. " The standard locomotive is assumed to be
24 feet long, and to have six wheels with a 12-foot base ; to
have half its weight resting on the middle wheels, and one-
fourth on the leading and trailing pairs respectively, which are
supposed to be at equal distances on either side of the middle
wheels." (See Stoney, vol. ii., p. 405.) The standard engine
is assumed to weigh 24 tons. 30 tons and 32 tons, according to
the construction. This makes the standard load 1 ton, 1J ton, or
1^ ton per foot of single line. Short bridges of less than 40
feet span must be considered as subject to concentrated loads
from single engines.
The maximum load for public bridges is recommended by
Stoney at 100 Ibs. per sq. ft.
68. In the Straight Truss all the Outer Forces act in a
86 APPLICATION TO BRIDGES. [PAET If.
Vertical Direction. — The strain in any cross-section depends
upon, first, the resultant of all the outer forces acting either
side of the cross-section ; and second, the statical moment of
these forces with reference to the cross-section. The first, or
the algebraic sum of all the forces acting between the cross-
section and either end, we call the shearing force for this cross-
section, and indicate it by S. It is also designated as vertical
force, or transverse force. The moment of the resultant, or
the algebraic sum of the moments of all the exterior forces,
with reference to any cross-section, we call the moment for this
cross-section, and indicate it by M. It is also called bending
moment, or moment of rupture. For example, in a lattice
girder with horizontal flanges the strains in the web are pro-
portional to the shearing forces, those in the flanges to the
bending moments.
The shearing force is considered positive when it acts on the
left side upwards, or on the right side downwards. The mo-
ment M is positive, when on the left side the tendency of rota-
tion is to the left, on the right side to the right, or when it tends
to make the girder convex upwards, that is, causes compression
in the lower fibre or flange.
CHAP. VH.] SIMPLE GIKDEKS. 87
CHAFTEE VII.
SIMPLE GIKDEKS.
69. Action of Concentrated Loads — Invariable in Posi-
tion.— By "simple girder" we understand a girder resting
upon two supports only, in opposition to a continuous girder
which rests upon more than two.
Suppose a number of forces Pl . . . P5 acting at various points.
[Fig. 41, PI. 13.] We form the force polygon by laying off the
forces to scale one after another ; then choose a pole O, and
draw O 0, 0 1, 0 2, etc., to the points of division. Parallel to these
lines we draw the lines of the equilibrium polygon between the
corresponding force lines prolonged. If now we dose the poly-
gon thus formed by the line A B, and draw through O the
parallel O L to A B, the segments 0 L and L 5 of the force
line give the reactions Vj_ and V2. Further, the shearing force
between A and Pt is St *= V^ = L 0 ; between Pl and P2, S2 =
V\— P! ; at P3, S3 = Vt— P,,— P2, etc. That is, the shearing forces
are the' distances of the points of the force polygon fro in~L>. It
is easy, then, to construct them, as shown in the lower shaded
area of the Fig. (See also Art. 46.)
If in the equilibrium polygon we let fall at any point a ver-
tical as I K, and from K draw K L perpendicular to A B, and
indicate by H the horizontal pull, by L the strain in A B, and
by M the sum of the moments of all forces left of I K, then,
for equilibrium about K, we have M^LxKL^LxIK cos
I K L, or, since the angle I K L = L O H in force polygon,
L x cos I K L = H, and hence M = H x I K, or representing
the variable ordinate I K by y :
M = Hy.
But H is the distance of the pole O from the force line ;
the moment at any point is therefore proportional to the verti-
cal height of the equilibrium polygon. (See also Art. 38.) If
we take H equal to the unit of force, we have
88 SIMPLE GIRDERS. [CHAP. VII.
so that in this case the moment at any point is directly given
by the ordinate of the polygon at that point. It is this impor-
tant property of the equilibrium polygon which renders it espe-
cially serviceable in the graphical solution of this and similar
problems.
70. Concentrated Load — Variable Position— Shearing
Force. — If the load lies to the right of any given cross-section,
then the shearing force at this cross-section will be S^V^ or,
since we regard a force to the left acting up as positive, S is
positive. As the load P moves towards the left, Vt or S in-
creases. When the load is to the left of the cross-section, the
shearing force at the cross-section is S = V^ — P, and since P
is always greater than Vl5 S is negative. The nearer P ap-
proaches the cross-section, the smaller is S.
Hence : a concentrated load causes a positive or negative
shear, according as it is to the right or left of the cross-section
considered, and the shearing force is greater the nearer the load
is to the cross- section.
Moments. — If the load lies to the right of the cross-section,
the moment is M = — V\ x, x being the distance of the cross-
section from the left support. M is therefore negative and in-
creases with Y! ; that is, as the load .approaches the cross-sec-
tion. If the load is on the left of the cross-section, M — — V2
(I — #), V2 being the reaction at the right support. Here also
M is negative and increases with V2 ; that is, as the load ap-
proaches the cross-section.
Hence : a concentrated load wherever it lies causes in every
cross-section a negative moment, which for any cross-section is
a maximum, when the load is applied at that cross-section.
71. Position of a given System of Concentrated Loads
causing Maximum Shearing Force. — If P! is the sum of all
the loads to the left of any cross-section, the shear at that cross-
section is S = Vj — Px. As the system moves to the left with-
out any load passing off the girder or any load passing the cross-
section, Y! and therefore S increases as long as S is positive, or
as long as Vx > Pl8 If a load passes off the girder, then for the
remaining loads S increases anew as the system moves to the
left, until a load of the system passes the cross-section in ques-
tion. The same holds good for a system moving to the right,
where S is negative.
Hence : the shearing force is a maximum for any point,
CHAP. VII.] SIMPLE GIRDERS. 89
when there is a load of the system at that point, and the maxi-
mum is positive or negative, according as the load lies just to
the right or left of the point.
Since for a single load (Art. 70) S is positive or negative, ac-
cording as the load is to the right or left, S will be in general
a positive or negative maximum when all the loads lie to the
right or left, and the heaviest nearest the cross-section. Only
in cases where a small load precedes, can S be greatest when
the second load lies upon the point in question.
If P is the resultant of all the loads and j3 its distance from
the right support,
Vt = P £ and therefore S = P £ — ple
L L
Now the position of the loads of the system or @ remaining
unchanged, Px will vary as the first power of x, the distance of
the cross-section from the left support. Therefore, between any
two cross-sections for which the load on the girder remains the
same, the shear S is represented by the ordinates to a straight
line.
72. Construction of the Maximum Shearing Forces.—
Construct the force polygon with the given loads ; choose a
pole O [PL 13, Fig. 42 (a)~\ and draw the corresponding equi-
librium polygon. It is required to determine the shear S at a
crossrsection distant x from the left support, under the suppo-
sition that the first load PL of the system, moving towards the
left, acts at this cross-section.
Determine upon the outer side P! A of the polygon passing
through the point Pl5 a point A distant from Plt by the distance
x, and then find the point B upon the polygon distant from A
by I, the length of span, and draw A B. Parallel to A B draw
O L in the force polygon, then A L = Vt = S, the shear at Pt.
Drop a vertical through B intersecting P! A produced, in M ;
then the triangles O A L and Pl M B are similar, and there-
fore S = A L = B M — , when a is the pole distance. If we
if
choose a = I, then S = B M.
Hence : the maximum shearing forces are proportional to
the vertical segments between the equilibrium polygon and the
prolongation of the outer side taken at the end of the system,
or are equal to these segments if the pole distance is taken equal
to the span ; provided that the last load is at the cross-section.
90 SIMPLE GIKDEKS. [CHAP. VII.
We have, therefore, the simple construction given in PL 13,
Fig. 42 (b). The broken lines are parallel to the various posi-
tions of A B for corresponding positions of Pt. The positive
and negative values of S equally distant from the right and left
supports are equal, so that it is only necessary to construct S
for one value.
If the second load is to be at the cross-section, and if e is the
distance between the first and second, we draw first a line
& _-_.-L- np
whose equation is 2/ — PI — -, — , and construct, as above, a'
I/
polygon, for which the second load lies on the right support B,
and whose second side (between second and third loads) coin-
cides with the above line. The ordinates to this line above the
axis of abscissas will give maximum of + S.
73. Maximum Moments. — Since, according to Art. 70, a
concentrated load causes a negative moment at any point,
wherever it may lie, we must have evidently loads upon both
sides of any point, in order that the moment may be a maxi-
mum. Since a single load causes a greater moment at any
point the nearer it lies to that point, the greatest load must lie
nearest the cross-section in question. The method of loading,
causing maximum moments, can be best determined for a dis-
tributed load (not necessarily uniform). In this case the equi-
librium polygon becomes a curve [PL 13, Fig. 43]. If in this
curve we draw A B, and take C so that A C : C B \;x : I — #,
then C D = M f or x. Suppose A B to take the position A! B',
the horizontal protection of C C' being indefinitely small,
then C' D' = M + d M. In order now that M may be a
maximum, C' D' must be equal to C D or C C' parallel to D D'.
If in the force polygon O Al is parallel to A A', O Bx to B B'.
and O Dt to D D', then A^ ^ and E>! B1 are the loads upon
A C and B C.
Draw through C a vertical, and through A, A', B, B', paral-
lels to C C7 or D D' intersecting this vertical in E, E', F, F'.
Then CE:CF::AC:BC;:a;:Z-tf,
C E' : C F' ; ; A' C' : B7 C' : ; x : I - x;
therefore
C E : C F ; ; C E' : C F', or C E' : C E ; : C F' : C F ;
also CE'— CE:CF'-CF;:CE:CF,
that is, E E7 : F F' \\x\l- x.
If now we draw through A' and B' parallels to C C', or D D' to
CHAP. VII.] SIMPLE GIRDERS. 91
intersections H and I, we have A H = E E', I B' — F F'. Since
the triangle A A' H is similar to O A! ^ and B B' 1 to O B1 T>ly
and since A7 H = B I, we have
A! Dt : B! D! ; ; A H : B' I ; ; E E' : F F7 ; ; x : l—x.
Since A1 Dx equals the load Pt on A C, and B! "D^ the load P2
on B C, we have Pl : P2 ; ; x : l—x.
The same will hold true approximately for concentrated
loads. Hence, in order that the moment at any point may be
a maximum, the system of loads must have such a position
that the loads either side of this point are to each other as the
portions into which the span is divided.
In PI. 13, Fig. 44, let C D give the moment at C. If the line
A B moves so that the horizontal projections of A C and B C
remain equal to x and I— x, then as long as the ends A and B
move on the same straight lines, the point C will also move in a
straight line. The point C describes, therefore, a broken line.
The verticals between this line and the polygon correspond to
the moments for various positions of the load and a given value
of x. Evidently the greatest ordinate will be over an angle of
the equilibrium polygon which is not under an angle of the
line described by C— that is, for M maximum a load must lie
upon the cross-section.
For any cross-section, then, the moment is a maximum when
a load is applied at this cross-section. Which of the loads
must be so applied is determined by the preceding rule.
74. Contraction of Maximum Moments. — After the
equilibrium polygon has been constructed, in order to find M
for a point C (PI. 13, Fig. 45), we determine two points P and
G upon the polygon which are distant horizontally from the
load on the given. cross-section corresponding to the angle E by
distances A C, B C. Then draw F G-, and the vertical K E is
equal to M when the pole distance is unity. We make C I = E K.
In this way we can construct the moments for different loads
of the load system at the given cross-section, and thus determine
that position of the load which gives the maximum moment at
the cross-section.
Generally when K E = y, and the pole distance is a, we have
JML = ay. The pole distance a is measured to the scale of
force, and then y is given by the scale of length. The unit for
M, in order that M may be equal to y, is evidently -th part of
92 SIMPLE GIRDERS. [CHAP- VIT-
the unit of length (when the pole distance is a force units), or,
what is the same thing, one unit of length is equal to a moment
units. The same equilibrium polygon can be used for any
number of girders of various spans, hence the method is of very
rapid application.
75. Ab§olute maximum of Moments. — Since for any cross-
section M is a maximum when a load lies at that section, a load
must also lie upon the cross-section for which M is an absolute
maximum. ,
If the line A B slides upon the equilibrium polygon, altering
its length so that its horizontal projection is constant and equal
to I, it will envelop a portion of a parabola so long as its ends
move in the same sides of the polygon. [PL 13, Fig. 46.] The
curve thus produced is therefore composed of portions of a
parabola. Let the ordinate D C correspond to the moment at
the point of application of the load P. DC will be evidently
greatest when A B is tangent to .the curve at C, so that the
maximum of the moments occurring at D is given by the dis-
tance C D between the polygon and curve enveloped by A B.
Let the prolongation of the sides upon which A B slides meet
in E, and F G be the tangent to the parabola at the point H in
the vertical through E, so that F H = H G-, and let I be the in-
tersection of A B and F G. Draw through A a parallel to E B,
intersecting F G in K. Then the horizontal projections of A F
and A K are equal, since those of E F and E G are equal.
Since, however, the projections of F G and A B as also of
A F and G B are equal, A K must be equal to G B. Hence
A I = B I. In a parabola the distances of the three diameters
passing through two points and the point of intersection of the
corresponding tangents are equal, hence the projections of H I
and C I are equal.
The middle point I of the tangent A B lies, then, half way
between the angle D vertically below the point of tangency and
the intersection E of the sides upon which it slides.
Since the projection of A B is I, its construction is easy.
The construction must, of course, be repeated for each angle, in
order to determine that for which M is an absolute maximum.
The above principle may, then, be thus expressed : The mo-
ment at any load is a maximum, when this load and the result-
ant of all the loads are equally distant from the centre of the
girder. (See also Art. 48.)
CHAP. Vn.] SIMPLE GIRDERS. 93
76. — In Arts. 46 to 50 the above principles have been already
deduced so far as relates to the moments alone, and a reference
to Art. 49 will show their application to the investigation of the
effect of a system of loads moving over the girder. We pass
on, therefore, to
CONTINUOUSLY DISTRIBUTED LOADING.
Suppose the load p per nnit of length laid off as ordinate.
The area thus obtained we call the load area. PI. 13, Fig.
46 {£).
The equilibrium polygon becomes here a curve, for which
the same law holds good. If we draw tangents to the curve at
the points D/ and E' corresponding to D and E, intersecting in
C', then the resultant of the load upon D E passes vertically
through C', or C' is vertically under the centre of gravity of the
area D D" E" E.
If we consider the load area divided into a number of parts,
the resultant for each will pass through the intersection of the
tangents at the points vertically under the lines of division.
Since these tangents are parallel to the lines in the force poly-
gon corresponding to these lines of division, they form the
equilibrium polygon for the concentrated loads, or resultants of
the portions into which the load area is divided.
Hence : if we divide the load area into portions, and replace
each ~by a single force, the sides of the corresponding polygon
are tangent to the equilibrium curve at the points correspond-
ing to the lines of division. (Art. 42.)
77. Total Uniform Load. — In this case the reactions at the
supports are Vt = V2 = — p I. Hence, for any cross-section dis-
tant x from the left support, the shearing force is
For a? = -~ Z ; S = 0. Sis greatest for x = 0 and for x = I ;
that is, maximum S = -f -p Z, and S = — ^ p I.
The moment at any cross-section is
M = — V x + -px* = — -^px (l—x).
M will be greatest for x = -= l> and
94 SIMPLE GIRDERS. [CHAP. VII.
Max.
s^
The shearing forces are, then, given by a straight line inter-
secting the span in the middle, the ordinate at either end being
Ip I [PI. 14, Fig. 47.]
The moments, as we have already seen [Art. 44, Fig. 30], are
given by a parabola whose vertex is in the centre of the span
and whose middle ordinate is -p 1?. Since we have seen [Art.
o
70] that a load at any point causes at every point a negative
moment, the maximum moment at any point will be when the
whole span is loaded.
78. Method of Loading causing Maximum Shearing
Force. — We have seen [Art. 70] that a single load causes at
any point a positive or negative shear, according as it lies upon
the right or left side of the cross-section at that point. Hence,
for a uniform load,
The shearing force will be a positive or negative maximum
according as the load reaches from the right or left support to
the cross-section in question. For the positive maximum we
have Y! —p (l—%) ^r-y —~^P - — T~^- Therefore, max. + S =
a L 2i o
1 (l-xf
For the graphical determination we can apply the method
given in Art. 72, Fig. 42, by which we have for max. + S and
max.— S two parabolas whose vertices are at the ends of the
span, and whose ordinates at these points are +~r and — •*-—•
2i 2i
Since, however, each point is found thus from the preceding,
the construction is not very exact. We may deduce a better
construction as follows. [PI. 14, Fig. 48.] Through any point
F of the curve drop a vertical intersecting A B in C and the
line B K parallel to the tangent at F in G. Let the tangent
at F intersect A B in H. Then C H = B H ; hence, C F = - C G-.
2
We have, then, A E = i A D =-1^ Z. Since C F = - C G-,
a A 2
we have also AI =- . AK ; therefore, A I : A E ; ; AK : A D.
a
CHAP. VII.] SIMPLE GIRDERS. 95
Hence the following construction :
Make A E = \.p I.
\
Divide A E and A B into an equal number of equal parts, and
draw lines from B to the points of division of A E, and verticals
through the points of division of A B. The curve passes
through the points of intersection of corresponding lines.
79. Live and Dead Loads. — Let p be the load per unit of
length for dead, and m for live load. The maximum moment
for any point will be as before.
M— — -(j?-f ra) x (£—#); that is, will be
a
given by a parabola whose middle ordinate is — - (p+m) P.
o
For the shearing force, we have
Max.+S = Ip (Z-2
2
Indicate A C [Fig. 49, PI. 14] by x^ for which max.— S = 0.
then
0 =p I (I— 2 Xi)— m a?!2,
or
m
hence
I m m2 m
For the point D for which max. +S = 0, B D = a^. The
shearing force within A C ia positive, within B D negative,
while within C D it is both positive and negative.
For Z = 5, 10, 20, 50, 75, 100, 150.
5=0.12 0.19 0.31 0.64 1.05 1.55 3.12
m
:0.24 0.29 0.33 0.38 0.42 0.44 0.46Z
CD =-.0.52 0.42 0.34 0.24 0.16 0.12 0.08Z;
that is, C D diminishes with increasing span.
§O. Recapitulation. — For girders of a length of about 100
feet or more, then, we may consider the live load as distributed
per unit of length. The maximum shearing force can then be
96 SIMPLE GIRDERS. [CHAP. VII.
easily found according to the preceding Art., while the maxi-
mum moments will be given by the ordinates to the parabola
for full live and dead load [Fig. 30, Art. 44]. For a framed
structure, we have simply to multiply the shear at any point
by the secant of the angle which the brace at that point
makes with the vertical, in order to find the strain in that
brace. The moment, divided by the depth of truss at the point
in question, gives the strain in the flanges. For opiate girder,
the moment being found as above, and one dimension as the
depth given, we can, from Art. 52, so proportion the other di-
mension as that the strain in the outer fibre shall not exceed
the amount allowable in practice. The preceding Art. as also
Arts. 78 and 44 and 52 are all that we need to refer to for all
practical cases of parallel flange girders of large span.
The preceding will complete our discussion of the simple
girder. We have only to remark here that the strains due to
rolling load will, in general, be most satisfactorily found by the
method of resolution of forces, as illustrated in Art. 12. By
this method we first find the reactions at the supports for a sin-
gle apex load, either graphically or by a simple calculation
, and then follow this reaction through the
girder, arid find the resulting strains. We can thus find and
tabulate the strains in every piece due to a weight at each and
every apex. The maximum strains can, then, be easily taken
from the table thus formed. When the live load is supposed
thus concentrated at each apex, it is, as we have seen in Art. 12,
unnecessary to follow through every reaction. The reactions
due to the first and last weights are sufficient to fill out the
table. For solid-built beams or plate girders, the principles of
the present Chap., therefore, come more especially into play.
(See also remarks at close of Chap. Y.)
The preceding principles will, it is hoped, be found sufficient
to enable the reader to find the maximum moments and shear
at each and every cross-section of a beam of given span rest-
ing simply upon two supports, and acted upon by any given
forces or system of forces in any given position. The reader
will do well to take examples of simple trusses, and check the
results obtained by the method given in Chap. I. by the above
principles. The method of tabulation of single apex loads
CHAP. VII.] SIMPLE GIRDERS. 97
upon which we lay so much stress is fully given by Stoney
[" Theory of Strains," vol. i.], and the examples there given will
be found of service.
Finally, then, the strains in upper and lower chords are great-
est for full load over whole span. We have, therefore, only to
erect upon the given span a parabola whose centre ordinate is
— — -~- — , where p is the load per unit of length for dead, and
8
m for live load [Art. 44]. ,The ordinates to this parabola at
any point give at once the maximum moment at that point.
The depth of truss at this point, if a framed structure, or the
moment of inertia of the cross-section at this point, if it is a
solid beam [Art. 52], being known, the strain in the flanges or
outer fibres may be easily determined. The strain in the web is
given by the maximum shear. For dead load alone this is
given by the ordinates to a straight line passing through the
centre of span, whose extreme ordinates are^— - [Art. 77]. The
2i
maximum shear due to live load alone- (m I) will be given by
the ordinates to two semi-parabolas, convex to the span, having
their vertices at each end, and the extreme ordinates - - [Art.
A
78]. At any point, the greatest of the two ordinates to these para-
bolas is to be taken. For live and dead loads together, Art. 79
may also be useful. The shear being known, the strain in any
diagonal is equal to the shear multiplied by the secant of the
angle made by the diagonal with the vertical [Art. 10 of Ap-
pendix] i.m parallel flanges. For flanges not parallel, we must
find the resultant shear as given in Art. 16 (4) of Appendix,
or, better still, the flanges once known, the diagonals can be
diagrammed according to the principles of Chap. I.
For the investigation of load systems, the principles of
Arts. 70-75 will be found sufficient, and the application of
these principles we have already sufficiently illustrated in Arts.
49-51.
7
$8 SUPPLEMENT TO CHAP. VII. [CHAP. \.
SUPPLEMENT TO CHAPTEK VII.
CHAPTER l'
METHODS OF CALCULATION.
1. — In Chapter I. of the text we hare already obtained a method of dia-
gram which will be found both simple and general, and by which we can
readily determine the strains for any given loading in any framed struc-
ture, no matter how irregular in its shape or dimensions, provided only that
all the outer forces are Tcnown.
In Chap. VII. we have also been put in possession of another method of
diagram, by which we may for any structure of the above class, framed or
not, determine the moment at any point, and can then properly proportion
the cross-section.
Thus far, indeed, we are unable to apply these methods to the continuous
girder or braced arch, as in these cases there are not only upward reactions
but also end moments, and in the latter case a thrust also, which must first
be determined. The determination of these requires that the elasticity of
the material and cross-section of the structure be taken into account. But
with these exceptions, and they are of rare occurrence in practice, we can
already solve any case which may present itself.
In the Appendix, if he has attended to our numerous references to it,
the reader will have already become familiar with two corresponding meth-
ods of calculation, viz., that by resolution of forces and that by moments.
It is, however, in many cases desirable to know not only the strains in
every piece of a structure, but also the deflection of the structure, and this
also requires a knowledge of the theory of flexure or of elasticity. For the
sake of completeness, therefore, aiming as we do to put the reader in pos-
session of methods of calculation as well as of graphic determination, we
shall devote a few pages here to a brief notice of these two above-men-
tioned methods of calculation, and then pass on to the theory of elasticity
itself. This latter has been too generally considered by those unacquainted
with the methods of the calculus as difficult and abstruse. It is true that
the calculus must be called into requisition ; but so simple are the processes
for beams of single span — and it is with these only we have at present to
do — that we indulge the hope that by going back to first principles we may
enable even those at present unacquainted with the calculus to follow our
CHAP, l'] SUPPLEMENT TO CHAP. VII. 99
I
demonstrations intelligently, and to comprehend perfectly and even apply
readily the method for themselves.
We cannot, indeed, make the reader familiar with all the principles of
the calculus, but all these principles are by no means needed. Its funda-
mental idea, a few of its terms and applications, are all that he need be
familiar with in order to perform the simple integrations we shall encoun-
cer, as readily as the most skilled mathematician. This portion of the
present Supplement may, perhaps, be considered by many as unnecessary and
superfluous. We are, indeed, justified in assuming such knowledge. But
as we believe our plan practicable, we cannot resist the desire of making
our development intelligible to all, and thus rendering our treatment of the
simple girder at least complete.
The practical man as well as the mathematician may thus have at his
disposal the powerful aid of the calculus, so far at least as his purposes
require it, and be able to deduce for himself the formulae which hitherto
he has accepted " upon faith." It may also not be improbable that here
. and there one may be found who, pleased with the simplicity of the prin-
ciples and the f ruitf ulness of their application, may be led to further prose-
cute the study for his own satisfaction.
We shall first, then, notice briefly the two methods of calculation above
referred to ; then devote a few pages to the development of those prin-
ciples and rules of the calculus of which we shall make use, and finally
apply these principles to the discussion of the curve of deflection of loaded
beams.
2. Hitter's Method. — This method is referred to in Art, 14. It
rests simply upon the principle of the lever, or the law of statical moments ;
requires no previous knowledge, and converts the most difficult cases of
strain determination into the most elementary problems of mechanics.
Ritter, 'in his " Theorie eiserner Dach- und Briicken-Constructionen," has
applied this simple principle in such detail and fullness, and so clearly set
forth its elegance and simplicity, that it very generally, and justly, goes by
his name.
" Its results are clear and sharp as the results of Geometry, and of direct
practical application. There is hardly another branch of engineering
mechanics which, for such a small amount of previous study, offers such
satisfactory results, and which is so suited to engage the interest of the
beginner."
We have given in the Appendix to Chap. I. (Arts. 6, 9, 10) detailed ex-
amples of its application. Throughout this work similar illustrations of
its use will be met with, so that it is only necessary here to state more fully
than in the text its general principle.
If any structure holds in equilibrium outer forces, it does so by virtue of
the strains or inner forces which these outer forces produce. Now the
outer forces being always given, we wish to find the interior forces or
strains. If, then, the structure is framed, and we conceive it cut entirely
through, the strains in the pieces thus cut must hold in equilibrium all the
outer forces acting between the section and either end. Thus, in Fig. 6,
PL 2, a section cutting D, 7 and H completely severs the truss. Then the
100 SUPPLEMENT TO CHAP. VII. [CHAP. 1.'
strains in these three pieces must hold in equilibrium the reaction at A and
all the forces between A and the section.
Now the principle of statical moments is simply that, when any number
of forces in a plane are in equilibrium, the algebraic sum of their moments
with respect to any point in that plane must be zero.
The application of this principle is simply so to choose this point of
moments as to get rid of all the unknown strains in the pieces cut, except one
only ; and then the other forces being known in intensity, position, and
direction of action, we can easily find this one ; since, when multiplied by
its known lever arm, it must be equal and opposite to the sum of the
moments of the known forces.
In a properly constructed frame it will, in general, always be possible to
pass a section cutting only three pieces. Then, by taking as a centre of
moments the intersection of any two, we can easily find the strain in the
third.
Even if any number of pieces are thus cut, if all but one meet at a com-
mon point, the strain in this one can be determined.
Thus, in Fig. IV., PL 1 of the Appendix, a section may be made cutting
2 3, d h, h c and c Y. But all these pieces, except the last, meet in 2, and
the strain in this last piece may, therefore, be easily determined.
The above is all that is necessary to be said as to this method. The ex-
amples already referred to will make all points of application and detail
plain as we proceed. We see no reason why the reader who has mastered
Chapter I. and diligently followed out the examples as given in the Appen-
dix, should not now be able to both calculate and diagram the strains in
any framed structure all of whose outer forces are known.
3. Method by Re§olutioii of Forces. — We have also yet another
method of calculation, based upon the principle that, if any number of
forces in a plane are in equilibrium, the sum of their vertical and hori-
zontal components are respectively zero. In structures all the forces acting
upon which are vertical, and such are all bridge and roof trusses, etc., of
single span, we have only to regard the vertical components.
In this connection we have to call attention to the following terms and
considerations. The shear or shearing force at any point is the algebraic
sum of all the outer forces acting between that point and one end. These
outer forces are the weights and reactions at the ends. At any apex of a
framed structure, where several pieces meet, the horizontal components of
the strains in these pieces must balance, or the structure would move ; and
for the same reason, the algebraic sum of the vertical components must be
equal and opposite to the shear. The shear being known, if the strains in
all the pieces but one are also known, that one can be easily found. Thus
the algebraic sum of all the vertical components of the strains in the other
pieces being found, and added or subtracted from the shear, as the case
may be, the resultant shear, multiplied by the secant of the angle made by
the piece in question with the vertical, gives at once its strain.
This method is also fully explained in the Appendix, Art. 16 (4), and a
practical rule is there given for properly adding the vertical components
and determining whether the result is to be added to or subtracted from
J.
CHAP. I.] SUPPLEMENT TO CHAP. VII. 101
the shear. This rule we owe to Hiuriber* We have thus two methods of
calculation, which, for the sake of convenience, we may speak of as Bitter's
and Humberts. Corresponding to Humberts method we have also a graphic
solution, based upon the same principles precisely. This we have set forth
in Chapter I., and may call Prof. Maxwell's method. In Chapter II. and
the following we have also become acquainted with the graphic solution
corresponding to Hitter's method, or the method of moments, which we
may speak of as Culmanrit*. It is to this method, based upon the proper-
ties of the equilibrium polygon, that the graphical statics properly owes its
value and fruitfulness, and to it is due whatever pretensions it can claim
as a system. It will be seen hereafter that it alone can furnish a general
method applicable to all structures, whether framed or not ; whether all
the outer forces are known or not. By the same general method we are
enabled to find the centre of gravity and moment of inertia of areas, and
to solve thus a great variety of practical problems —through which, how-
ever different, runs one universal method, one simple routine of construc-
tion.
* Strains in Girders, calculated ~by Formulas and Diagrams.
102 SUPPLEMENT TO CHAP. VII . [CHAP. II.
CHAPTER II.
PRINCIPLES OF THE CALCULUS NEEDED IN OUR DISCUSSION.
4. Differentiation and Integration.— We need but a very few
simple ideas and conclusions in order to have at our disposal the whole
theory of flexure for beams of single span. Those to whom these ideas are
not familiar already may find them indeed new, but will not find them
difficult or even abstruse, and with attention to the following will, we
venture to think, make a valuable acquisition.
The sign / is called the " sign of integration,1' and integration
means
simply summation. It arises merely from the lengthening of the original
letter S, first used by Leibnitz for the purpose. The letter d is called the
"sign of differentiation-;'' in combination with a letter, as d x, it reads
" differential of #," and signifies simply the incitement which has been
given to the variable x. So much for terms.
Now suppose we have the equation
y = 5x\ . . ....... (1)
in which x and y, although varying in value, must always vary in such a
way that the above equation holds always true. This being the case, let
us give to y an increment — that is, supposing it to have some definite value
for which, of course, x is also definite in value, increase this value by d y.
Then x will be increased by its corresponding amount d x, and as the
above relation must always hold true, we have
y + dy = 5 (x + dxy . . . ... . (2)
or y +dy = 5 (x*+Zxdx + d a;2).
Inserting in this the value of y from (1), we have
dy = 5 (2xdx + dx*), ...... (3)
which is the value of the increment of y or d y, in terms of x and the in-
crement of x or d x. That is, the increments are not connected by the same
law as the variables. The variable y is always 5 times the square of the
variable x, but the increment of y is greater than 5 times the square of the
increment of x by an amount indicated by 5 x 2 x d x. From (3) we
have
(4)
which gives the value of the ratio of the two increments. Now, if we
assume a certain value for x, we find easily from (1) the corresponding
value of y. If we increase this value of x by a certain assumed increment,
d x, we find easily from (3) the corresponding increment of y, or d y. Then
(4) would give us the ratio of these two increments.
CHAP. II.] SUPPLEMENT TO CHAP. VII. 103
Now we see at once from (4) that the smaller we consider d x to be, the
nearer this ratio approaches the limiting value 5 x 2 x. We may suppose
d x as small as we please, and then this ratio will differ as little as we
please from 5 x 2 a. This value, 5 x 2 x, forms, then, the limit towards which
the value of the ratio ~ approacfies as d x diminishes, but which limit evi-
Ou X
dently it can never actually reach or exactly equal. Because, in order that
this should bo the case, d x must be zero. But if d x is zero, that is, if # is
not increased, y also is not increased; d y is, therefore, zero, and there is no
ratio at all.
Now, just here comes in what we may regard as the central principle of
the calculus.
If two varying quantities are always equal and always approaching certain
limits, then those limits must themselves he equal.
The principle is too obvious to need demonstration. "Two quantities
always equal present but one value, and it seems useless to demonstrate
that one variable value cannot tend at the same time towards two constant
quantities different from one another. Let us suppose, indeed, that two
variables always equal have different limits, A and B ; A being, for ex-
ample, the greatest, and surpassing B by a determinate quantity A.
The first variable having A for a limit will end by remaining constantly
comprised between two values, one greater, the other less than A, and hav-
ing as little difference from A as you please; let us suppose this difference,
for instance, less than — A . Likewise the second variable will end by re-
6 f
maining at a distance from B less than — A. Now it is evident that, then,
the two values could no longer be equal, which they ought to be according
to the data of the question. These data are then incompatible with the
existence of any difference whatever between the limits of the variables.
Then these limits are equal." *
Now let us apply this principle to equation (4). In this equation -y— is
a variable always equal to 5 (2 x + d x). But 5 (2 x+d a>), as we diminish
d x, approaches constantly the limit 5 x 2 x; and as -= — is always equal to
5 (2 x + d x), it also constantly approaches the same limit. These limits,
dy
then, are equal, and the limit of -=— = 5 x 2 x.
d x
Now, if we conceive, and such a conception is certainly possible, d x to
lie the difference between x and its consecutive or very next value, such that
between these two values there is no intermediate value of d x ; then d y
will be the difference between two consecutive values of y ; and regarding,
then, d x and d y in this light, -=— will be the limit of the ratio of the in-
CL X
* The Philosophy of Mathematics. Bledsoe.
104 SUPPLEMENT TO CHAP. VII. [CHAP. II.
crements, since the increments are then limiting increments, and can be no
smaller without disappearing.
We have thus
^ = 5x2,,
d x
which is an exact relation between the increments upon this supposition.
From this we have d y = 5x2 x d x.
If now we sum up all the increments d y, then by virtue of the supposi-
tion we have made, / d y must equal y. We thus suppose y to flow, as it
were, unbrokenly along by the consecutive increments d y, just as the side
of a triangle moving always parallel to itself, and limited always by the
sides, describes the area of that triangle, while the change d y of its length
is the difference between two immediately contiguous positions. Upon
this supposition, we repeat, -= — is the limit of the ratio of the increments,
a os
which limit is, as we see from (4), equal exactly to 5 x 2 x. We do not re-
ject or throw away d x from the right of that equation " because of its
small size with reference to 2 a,'1 but simply pass to the limit, and then, ac-
cording to our fundamental principle above, equate those limits them-
selves. But if / d y = y, then the integral of 5x2xdx, or / 5x2 xdx=y
=5 &2. By " differentiating," as we say, equation (1) we get (5), and by
" integrating " (5) we obtain (1).
Hence we see the appropriateness of the term "fluent" given by New-
ton to the quantity d y or 2 x d x. So also we see the appropriateness of
the term "ultimate ratio " * for — - itself.
d x
* Liebnitz undoubtedly discovered the calculus independently of Newton,
but he considered d x as a quanity so " infinitely" small that in comparison
with a finite quantity it could be disregarded " as a grain of sand in compari-
son with the sea." We see, indeed, from eq. (4) that if d x upon one side be
zero, we get the same value for - — as before. But if d x is zero on one side,
it should be zero on the other side also. No matter how small we suppose d x
to be, we have no right to get rid of it by disregarding it. That Liebnitz rec-
ognized this cannot be doubted, and he was therefore inclined to consider his
method as approximate only. But to his surprise he found his results exact,
differing from the true by not even so much as a " grain of sand." There was
to him ever in his method this mystery, nor could he conceive what these
quantities could be which., though disregarded, gave true results. Bishop
Berkeley challenged the logic of the method, and adduced it as ac evidence of
k' how error may bring forth truth, though it cannot bring forth science."
Strange to say, even the disciples of Newton were unable to answer Berkeley
without taking refuge in the undoubted truth of their results. And yet New-
ton in his Principia lays it down as the corner-stone of his method, that
44 quantities which during any finite time constantly approach each other, and
CHAP. II.] SUPPLEMENT TO CHAP. VII. 105
The whole of the calculus is but the deduction of rules for finding from
given equations as (1) their " differential equations " as (5), or inversely
of finding from the differential equation by "integration," or summation,
the equation between the variables themselves.
Such of these rules as we need for our purpose we can now deduce.
5. Differentiation and integration of powers'of a single
variable. — We have already seen that the / d y = y and / 2 xd x = a?,
hence d (x2) = 2 x d x.
If we should take y = x?>, we should have, in like manner, as before,
y + d y — (x + d x)3 = x* + 3 a8 d x + 3 x d xz 4- d z3,
or dy= 3x*dx + 3xdx2 + dx*,
or ~ = 3z2 + 3xdx + dx'2,
dx
and passing to the limits, as before,
- — = 3 x2, or d y=3 x2 d x. Hence the differential of x3 or d (a?3) =3 x2 dx,
it 3}
and reversely, the integral of 3 xz d x or / 3 xz d x = x3. In similar man-
ner, we might find
d (x5) = 5 x* d x and A x*dx = x5.
Comparing these expressions, we may easily deduce general rules which
will enable us at once upon sight to " differentiate," that is, find the rela-
tion connecting the increments ; and " integrate " or sum up the successive
consecutive values of the variable; for any expression containing the
power of a single variable.
These rules are as follows :
To differentiate:
" Diminish the exponent of the power of the variable by unity, and then
multiply by the primitive exponent and ~by the increment of the variable."
Thus, d(xz) = 2xdx, d (x*) = 3 x* dx, d (a?7) = 7*6 dx, d(x$) = ^X2dx,
d (#") — n a;0"1 d x, etc.
To integrate :
"Multiply the variable with its primitive exponent increased "by unity, by
the constant factor, if there is any, and divide the result by the new exponent."
before the end of that time approach nearer than any given difference, an eguaV
There can be little doubt that Newton saw clearly that although the quantities
might never be able to actually reach their limits, yet that those limits them-
selves were equal, and hence the increment could be left out in the equation,
but not because by any means it was of insignificant size. His terms ' ' ultimate
ratio" and '•''fluent'1'1 are alone sufficient to indicate that he understood the
true logic of the method he discovered ; while Liebnitz seems to have stood
gazing with wonder at the workings of the machine he had found, but whose
mechanism he did not understand. [See Philosophy of Mathematics. Bledsoe.
Lippincott & Co., 1868.]
1U() SUPPLEMENT TO CHAP. VII. [CHAP. TI.
Thus A xdx=2-j-=x* f$afdx = -|" = *' /V dx = X~
/i as* 2 f /• n x°
x dx=- = 5-s / nx*-idx = = a», etc.
i a / »
It is of this latter rule that we shall make especial use in what follows,
6. Other Principles — Integration between limits, etc. —
We may observe from (1) and (4) that a constant factor may be put out-
side the sign of integration. Thus / 5x2 x d x = 5 I 2 x d x = 5 x\
It is also evident without demonstration that the integral of the sum of
any number of differential expressions is equal to the sum of the several
integrals.
Thus / he d a + 2s d z + y*dy + x*
is the same as / x d x + I z3 d z + I y* d y, etc.
If in (1) we had
y = 5 x*+a,
where a is a constant, we should have
y + d y = 5 (x + d x)* + a = 5 (x* + 2 x d x+d a?) + a,
or d y = 5 (2 x d x + d afy or ^| = 5 (2 x + d x) ;
whence
— -^ = 5x2 a?, ordy = 5x2xdx, or
d x
just the same as before.
The integral of this will then be y = 5 x1 as before, whereas it should be
y = 5 x* + a.
If two differential equations, then, are equal, it does not necessarily follow
that the quantities from which they were derived are equal.
We should, then, never forget when we integrate to annex a constant. The
value of this constant will in any given case be determined by the limits
between which the integration is to be performed.
We indicate these limits by placing them above and below the integral
sign. Thus the integral ofx*dx between the limits of x = + h and x= — h is
r+h r x*
I x3 dx. If we integrate a;2 d x, we have, then, / a2 d x = — + C,
J-h J
where O is a constant whose value must be determined by the conditions
of the special case considered. If we introduce the value of x = h for one
limit, we have — + O. For x = 2 h for another limit, we have -— + O.
We have, then, two equations, viz. :
when x — h,
c
Jr~h x ~ 3~
CHAP. H.] SUPPLEMENT TO CHAP. VII. 107
rx = 2h
JZ.J*
n the other, w
rh 7
x = h, I x'dx=-
*s h
and when x = 2h I x*dx= — - + C ;
and by subtracting one from the other, we have for the integral between
r2h 7
the limits x = 2 h and x — li, I x'2 dx — — 7*s, and O thus disappears.
We have, then, only to substitute in succession the values of the variable
which indicate the limits, and subtract the results.
If also there is but one limit, we could determine O if there were also a
condition, such, for instance, as that / x1 d x should equal h when x = 2 h.
The ratio — ^ is called the "first differential coefficient ; " if it were to
d x
be differentiated again, the next ratio, viz., that of the differential of the
differential of y to differential of or2, or — -, is the " second differential co-
d x
efficient," and so on.
dy
Thus, y = x5 ; dy = d (x5) = 5 x*d x, or.-=— = 5 x* ; differentiating again,
— -i = 20 x3 d x, or — - — 20 x3, and so on to third differential coefficient, etc.
dx dx*
7. Example. — As an example of the application of our principles, let
it be required to determine the area of a triangle. Let the base be b and
the height h. Take the base as an axis, and at a distance of x above the
base draw a line parallel to &, and at a very small distance d x above this
line draw another, thus cutting out a very small strip. (Let the reader draw
the Fig.) Now for the base y of this strip we have the proportion h — x : y
7) T ~h 'JC (L i/1
:: h : 5, or y = b , hence the area of the strip is b dx — . But
n n
the area of this rectangular slip is not equal to the area of that portion of
it comprised within the triangle. It projects over at each end, and is,
therefore, somewhat greater. Thus for the small trapezoid actually within
the triangle we have for the upper side y', h—(x+d x) : y'::h : b, or y' =
b— - (x + d x). Hence y—y' — , and the area of the projecting portion
h, h
of the rectangle, that is, its excess over the trapezoid, is then (y—y') d x, or
bdx"1 __ , ,, bxdx b d x* , da , bx bdx
Therefore, bdx — — — = d «, or - — = b — — — , where
h h Ti d x h h
d a is the area of the small trapezoid itself. Now these latter two quanti-
ties are always equal for any value of d x. But as d x decreases, one side
of the equation approaches the limit &— — — , and — — , therefore, approaches
h d x
this same limit. The rectangle itself is, then, the limit of the ratio of the
area of the small trapezoid to its height, and we can then equate the limits
themselves, remembering that in this case d a is the area passed over by the
108 SUPPLEMENT TO CHAP. VII. [CHAP. II.
side y in passing from one position to the consecutive or very next. We
have, then, da = bdx — , and if we integrate this expression, that is,
sum up all the d a's, we have the area of the triangle. Therefore,
x d x 5 x1
=ji -IT- )m.-ji+*
where O is the constant of integration, which we must never forget to annex.
Now, in the present case we wish to sum up all the areas d a, or " integrate,"
between the limits x — o and x = Ji. But for x — o, A must be zero, and
hence we have O = o for the condition that x starts from the base. If in
addition to this condition we make x = h, we have the sum of all the areas
between x = o and x = h.
, , & h & h
A = & h . = — , as should be.
The above reasoning is somewhat prolix.
If we thoroughly appreciate that d x is the difference between two con-
secutive values of x, we see at once that we obtain the limiting value of the
rectangle directly by multiplying its base by d x. The sum of all these
must be the area. This conception ofdx enables us to curtail much of our
reasoning.
Let us take the same problem again, but this time take the axis through
the centre of gravity of the triangle ; that is, at $h above the base. Then
for the base y at any distance x above this axis, we have
-A-av "h'l or = -l>-*J!L
3 3 h
Multiply this by d x upon the above conception of d x, and we have at
once not for the rectangle upon y, but for its limiting value, that is, for the
area of that portion of the rectangle included within the triangle,
, -. 2, b x d x
~y =3 h '
Integrating this, then, we have
•2 , & x d x 2 , bx*
where O is a constant to be determined by the limits as before. For one
limit, x = — 7i, and hence we have
3
A'=-l
2
For the other limit, x = + —h, and hence we have
3
If we subtract the first from the second, O disappears, and we have A =
9 1
A"— A' = — & h = — & A, as before.
18 2
o
We might also have integrated first between the limits x = 0 and x — —h.
3
CHAP. H.] SUPPLEMENT TO CHAP. VII. 101)
4
For x = 0, C = 0, and the area above the axis is then — b Ji. For x = 0 and
lo
x = h, we have for the area below the axis — — b h. This area has a dif-
3 18
ferent sign because below. If we give it the same sign as the other, and
then add it, we have the total area. If it also had been above, the total
area would have been the difference. Generally, then, we subtract accord-
ing to our rule.
8. Significance of the first differential coefficient.— Any
equation between two variables of the first degree is the equation of a
straight line. If of the second degree, it represents one of the conic sec-
tions, an ellipse, circle, parabola, or hyperbola, Of a higher degree, a
aurve generally. If, then, we take the axis of x horizontal and y vertical,
and if d y and d x are the consecutive increments of y and x, that is, the dif-
ference between any value and the very next, the ratio ~ is evidently the
d x
tangent of the angle which a tangent to the curve at any point makes with the
horizontal.
If, then, we make — = 0, and find the value of the variabb x corrc-
d x
spending to this condition, we find evidently the value of x for which the
tangent to the curve is "horizontal. If now the curve is concave towards the
axis, this value of x, substituted in the original equation, will give the maxi-
mum or greatest value of the ordinate y ; because for the point just one
side of this the tangent slopes one way, and for the point just the other
side it slopes the other. The point where the tangent is horizontal must
then be the highest.
If the curve is, on the other hand, convex to the axis, the value of x, which
makes — - = 0, substituted in the original equation, will give y a minimum
d x
value for similar reasons. By setting the first differential coefficient, then,
equal to zero, we may find that value of x which corresponds to the maxi-
mum or minimum value of the ordinate, as the case may be. In the case
of the deflection of simple beams upon two supports, the curve is always
concave to the axis, and hence we obtain by this process always the maxi-
mum deflection.
The above comprises all the principles of which we shall make use in the
discussion of the theory of flexure. With a little study, we believe that
any one familiar with analytical operations, even although he may never
have studied the differential or integral calculus, can follow us intelligently
in what follows. Whatever points may still be a little obscure will clear
up as he sees more plainly than now their application.
110 SUPPLEMENT TO CHAP. VH. [CHAP. III.
CHAPTER III.
THEORY OF FLEXURE.
9. Coefficient of EIa§ticity. — Let us now take up the theory of
flexure, and see if it is not possible so to present the subject that, in the
light of the preceding principles, we may be able to solve all such prob-
lems as present themselves.
If a weight P acts upon a piece of area of cross-section A, and elongates
or compresses it by a small amount I, we know from experiment that,
within certain limits, twice, three times, or four times that weight will
produce a displacement of 21, 31, 4 I, etc. These limits are the limits of
elasticity. Within them practically, then, the displacement is directly as
the force. If we assume this law as strictly true for all values of the dis-
placement, and if we denote the original length by L, then, since the
force per unit of area is -r, and since this unit force causes a displacement
I, in order to cause a displacement L equal to the original length, this
unit force must be T time* as great, or equal to — T. This force we call
I At
the modulus or coefficient of elasticity. It is always denoted by E. Hence
The coefficient of elasticity, then, is the unit force which would elongate a
perfectly elastic body BY ITS OWN LENGTH. It is a theoretical force then ;
but as the law upon which its value is based is true practically within cer-
tain limits, by experiments made within those limits, knowing P, A, and
L, and measuring I, we can find what the force would have to be if the law
were always true. Such experiments have been made, and the values of E
for different materials are to be found in any text-book upon the strength
of materials.
From (6) we have for the unit force of displacement
These expressions will be found useful as enabling us to replace often
expressions containing an unknown displacement by a definite or experi-
mentally known value.
1O. Moment of Inertia. — This is also a convenient abbreviation,
and enables us to replace unknown expressions by a, in any given case,
perfectly determinate value.
The moment of inertia, with respect to any axis, is the algebraic sum of the
CHAP. III.] SUPPLEMENT TO CHAP. VII. Ill
products obtained ~by multiplying the mass of every element of a given cross-
section T)y the square of its distance from that axis.
If a parallelogram stand on end, and then its support be suddenly pulled
away from under it, it will fall over backwards. But to knock it over
thus requires force. The force which, in this case overturns it is that of
inertia. At every point of the surface there is, then, a force acting, depend-
ing upon the mass of this point. But not alone upon the mass. A force
at the top acts evidently with more effect to turn the body over than one at
the bottom, which merely tends to make it slide. The moment of each ele-
ment of the area is. then, a measure of the force which at each point causes
rotation, and the sum of these moments is, then, the measure of the over-
turning action of the whole force of inertia upon the surface. The moment
of this latter force, or the sum of the moments of the moments, is, then, the
moment of inertia of the cross- section. Each element of the surface must
then be multiplied by the square of its lever arm, and the sum of all the
results thus obtained taken. In other words, the moment of each element
is itself considered as a force, and then its moment again taken. The sum
is denoted by I. ' For any given dimensions and axis it is a perfectly defi-
nite quantity, and may thus often replace expressions containing unknown
quantities.
The principles of the calculus just developed will enable us to deter-
mine it in some cases, at least, very readily. Its value for various forms of
cross-section, in terms of the given dimensions, is given in every text- book
upon the strength of materials.
Let us suppose a rectangular cross-section of breadth 5 and height h, and
take the bottom as axis. The area of any elementary strip is, then, b d x.
If its distance from the bottom is x, we have for its moment b x d x, and for
its moment of inertia, then, & #2 d x. Integrating this expression, we have
_ &>
- 3 +0.
This integral is to be taken between the limits x = 0 and x — h. For x = 0,
1) x' d x = 0, and hence C — 0. For x = h, then, we have -. If the axis
3
had been taken through the centre of gravity, we should have the above
integral between the limits + -and— — . For+— we have h O. For
2 2 2 24
— — , + O. Subtracting one from the other (Art. 6), we have
for the moment of inertia. For a triangle of height h and base Z>, we have
for axis through centre of gravity, from Art. 7, for the area of the very
small strip at distance a?, -Idx — - x d x. Multiplying this by x*, we have
o fl
for its moment of inertia -lxldx — -x*dx. The integral of this is
o h
2 T) x4
O A
For x = — h, this becomes I
o 243
112 SUPPLEMENT TO CHAP. VII. [CHAP. ITT.
For x =——71, we have — — — & 7i3 + C.
27 1
Subtracting one from the other (Art. 0), we have -^ & A3, or _— & A3 for
the moment of inertia. The moment of inertia of the rectangle I = —
may be written — = — — - x— Ax — , or the moinent of inertia of the half
parallelogram is equal to its area, into the distance of its centre of gravity
multiplied by f ds. of its height. We see at once that when we consider,
then, the statical moments as themselves forces, the centre of action, of these
moment forces does not coincide with the centre of gravity of the area. This
principle we have already noticed in Chap. VI., Art. 60.
I A2 /I \2 1
We can also put --=_ = (_ - A ) . This value A is called the
A 12 \2V3 ) 3y£
radius of gyration. It is evidently the distance from the axis to that point
at which, if the mass were concentrated or sum of all the forces were con-
sidered as acting, their moment of inertia would be that of the cross-sec-
tion itself. The value of — is, in general then, the square of the radius of
JA
gyration. We have already shown in Chap. VI. how to find it graphically
for various cross-sections.
We are now ready to take up the case of a deflected beam, and to find
the differential equation of its curve of deflection.
11. Change of SUiape of the Axis. — In the Fig. given in the
Supplement to Chap. XIV., we have represented a beam deflected from its
original straight line by outer forces. Let the two sections A C, B D be
consecutive sections, parallel before flexure, and remaining plane after. Let
the length of the axis m a be s, then n a — d s, and let d <p be the very
small angle between the sections after flexure.
If the deflection is small, s will be approximately equal to x, and d s to
d x. The elongation of any fibre at a distance « from the centre is. then,
v d (f). The unit force corresponding to this elongation is from (7) T =
E — - v. If d a is the cross-section of any fibre as d c, then the whole force
d x
of extension is
Ev da d $
dx
The moment of this force is, then, — — = — - — -. The integral of this be-
tween the limits + -- and will give the entire moment of rupture. But
this is equal and opposite to the moment M of ail the outer forces ; hence
CHAP. III.] SUPPLEMENT TO CHAP. VII. 113
But, as we have just seen, this integral is the moment of inertia I of the
cross-section with reference to the axis through the centre. Hence,
M = — - — ® . Since rf> is a very small angle, it may be taken equal to its
d x
tangent, or equal to ; hence - and M = B I .
dx d x d or d x*
But v d cf) : v :: d x : r, where r is the radius of curvature ;
v d & d x d <b 1
hence • = — or -r-^ = — .
v r dx r
Therefore, M = Hli = EI^|=— ;* v . * . (8)
r dx* v
and T = 5-B (9)
r
Equation (8) is our fundamental equation.
In any given case we have only to write down the expression M for the
dz y !
moment of the outer forces at any point, and equate it with E I — -^.
• •• Cu <Xs ,
Integrating once we shall then have for I constant, of course, E I — - and,
d x
integrating again, E I y in terms of a?, or the equation of the deflection
curve itself. Making E 1~ = 0, we can then find the point of maximum
deflection, and inserting in the value for Ely the value of * thus found,
can find the maximum deflection itself. The discussion of any case reduces
thus to a simple routine, and every case is in many respects but a repetition
of the same processes.
12. Beam fixed at one end and loaded at the other—
Constant cro§s-section. — We shall always consider a moment positive
when it causes compression in the lower fibre ; negative when it causes ten-
sion in that fibre. Distances to the right of the origin are always positive,
to the left negative. Hence on the left of any section an upward force is
negative, a downward force positive ; while on the right of the section the
upward force is positive and the downward one negative. The reader
should always draw the Fig. for each case discussed, and in the beginning,
at least, review these conventions each time.
Now let a beam of length I have the weight P at the free end, and let it
be fixed horizontally or " walled in " at the right end. Then the moment
at any point distant x from the left or free end is M — + P x.
(a) Change of shape.
From (8) we have now
114 SUPPLEMENT TO CHAP. VII. [CHAP. III.
Integrating once (Art. 5) we have
BI^L^l
dx 2
where C is the constant of integration to be determined (Art. 6) by the
given conditions. Now by the condition in this case, when x = I, — -
a x
must be zero, because the end is fixed, and the tangent there must therefore
P 1?
be horizontal (Art. 8). Hence O = — - , and
E 2
_
dx ~ 2 2'
We have thus introduced the condition that x cannot be greater than I.
Integrating again (Art. 5)
P a? PVx
Here again we have a constant to be determined, and here again we have
the condition that for x — l,y must be zero, since at the fixed end there can
P 1?
be no deflection. Therefore, O = — - and
o
P / \ P
E I y == — [2 lz — 3 Z2 a; + a;8 i = — (21 + x) (I — a)2.
6 y j 6
The deflection will evidently be greatest at the free end, and here, therefore,
for x = 0, we have
_ _ p*3 .
If the cross-section is rectangular, I = — 5 A3 (Art. 10), and the maximum,
deflection A = — — — -•
(5) Breaking weight.
T I
We have also from equation (8) M = , where T is the tensile strain
t>
in any fibre distant « from the centre. For v — — , T is the tensile strain in
2
O *p T Jj
the outer fibre, and M = . For v = — — we have the compressive
7i 2
2 C T
strain in the outer fibre upon the other side, or M = — - — . Theoretically
the two should be equal. Practically they are not. In fact, if we put for
.Pa =
Tbh2
2 T I
M its value, we have P x = — - — , or for a rectangular cross-section P x =
h
i— T & h2. This is greatest for x = I, hence the breaking weight P = • .
ft TTJ 7
From this we have T = -r-rr' Now experimenting with beams of various
o fi
CHAP. III.] SUPPLEMENT TO CHAP. VII. 115
materials, known dimensions and given weights, we may find experimen-
tally T. It would seem that this value thus found should equal either the
tenacity or crushing strength of the material, but the results of experiment
show that it never equals either, but is always intermediate between T and
C. Calling this intermediate value R, we have
The formula is based upon the condition of perfect elasticity, while R is
determined by experiments made at the breaking point when the condition
of perfect elasticity is no longer fulfilled. In the following table the tabu-
lated values of R are correct for solid rectangular beams, and sufficiently
exact for those which do not depart largely from that form. If instead of
we use the values of T or C, whichever is the smaller, we shall always
T)e on the safe side, since R is invariably intermediate between these.
In general we shall refer to the equation
when we have occasion to find the breaking strength. But it must be
always remembered that in any practical example we should replace T by
R for rectangular beams, or by T or O, whichever is the smaller, for others.
We give also the values of the coefficient of elasticity E. (Wood's Resist.
of Materials.)
TOR E
Cast-iron ....................... 16.000 96,000 36,000 17,000,000
Wrought-iron .................. 58^200 30,000 33.000 25,000,000
English Oak .................... 17,000 9,500 10,000 1,451,200
Ash ............................ 17,000 9,000 10,000 1,645,000
Pine ........................... 7,800 5,400 9,000 1,700,000
All in pounds per square inch.
2. Beam of uniform strength.
Suppose the cross-section or I is not constant, but varies so that at every
point the strain T is constant. From (11) we have
2 T I
M = P x — — ,' - for the outer fibre, whence
h
T = — -. For a rectangular cross-section T = , ' . Now suppose the
21 o h
'
2
breadth and height at the fixed end are &i and hi. Then at this end T =
n ~p 7
But this must be equal to T at any other point ; hence
6Pa QPl IV x
i A,2
If we suppose the height constant, we have for the varying breadth at any
point b = li X. That is, the breadth must vary as the ordinates to a straight
I
line, and the plan of the beam is a triangle with the weight P at the apex.
116 SUPPLEMENT TO CHAP. VII. [CHAP. III.
If the breadth is constant, h = hi./ £., or the elevation of the beam is a
parabola with the weight at apex. If the cross-section is always similar,
that is, if — = -, we have 5 = -\—, and substituting in the equation above
hi h hi
h = hi ? / £. , which is a paraboloid of revolution.
(a) Change of shape
From (8) we have
dz y _ P x _ P x
where & and h are variable. If we suppose the height h constant and
always equal to hi, then, as we have seen, & = &i yj hence for rectangular
cross-section
d2 y _ 12 PI
d x2 E hi* 5i
/I nj
Integrating, since for x = I, — — = 0, we have
(L (K
dy _ 12 P Ix _ 12 F Z2
d x E hi* &i E hi3 &i*
Integrating again, since f or x = Z, y = 0. we have
_ 6P?a2 _12PZ2a 6P
E hi3 &i E hiz bi E h^
For the maximum deflection x — 0, and
. ^ ^^
The above value of y can be written
but is — , the deflection of a beam of constant cross-section &i Alt
as already found. Calling this deflection A0, we have
Q
for the deflection at any point, or A = — A0 for the maximum deflec-
tion.
In a similar manner, for constant breadth, we have
CHAP. HI-.] SUPPLEMENT TO CHAP. VII. 117
For similar cross-sections, we have
9 r 5 x 3 „ I \5 ~ 9 36 P I3
If we call the volume of the beam of constant cross-section V, then in
1 2
the first case the volume Vj = — V ; in the second, V2 = — V ; in the third,
6 O
V3=?-V; or
V : V2 : V3 : V: = 30 : 20 : 18 : 15.
The maximum deflections, as we see above, are as
2 Ao, - Ao, - Ao, or as 20, 18, and 15.
o /*
That is, the deflections at the ends for a beam of uniform strength in the
three cases are as the volumes.
1.3. Beam a§ before fixed at one end— Uniform load-
Constant cro§s-sectioii.— If p is the load per unit of length, we have
for the moment at any point distant x from the free end,
x p xz p xz dz y
M = *»x Sf = -a-.and hence — — =^r.
;.:.,. . p p
This moment is greatest for x = I, and hence Max. M = — — .
For the breaking weight, then, from (11)
pi2 2TI 4 T I
T = — or pl=-hT>
or twice as great as for an equal weight at the end.
For the change of shape, we integrate twice, precisely as before, the ex-
» d2 y p xz
pression -= — j — and obtain thus
* 24 B I
The maximum deflection, then, is
_ 4 p x
or
A = 8EI'
or only fths as great as for an equal load at the end.
2. €on§tant strength.
We have, as before, from (11) M = *£ = ^, whence T = ^f-
for rectangular cross-section, I = i I & and T = *-~. If h A, are the
breadth and heighth of the fixed end section, then, since T must be always
constant,
3 px2 _ 3p I2 I h* __ x2 •
5 A2 far*? or&7^7= Is""
For lieight constant, 5 = &j | X- \
118 SUPPLEMENT TO CHAP. VH. [CHAP. HI.
<y*
For breadth constant, h = hi -=• .
, V
For similar cross -sections, h = fa
The first is in plan a parabola ; the second, in elevation a triangle ; the
third, a paraboloid of revolution.
For the change of shape, -we have, by proceeding in the same manner as in
Art. 12, A = 2 Ao, A = 4 Ao, and A = 3 A0 in the three cases, where A0 is the
deflection of a similar beam of constant cross-section &i hi.
14. Beam supported at both the ends — Constant cross-
section — Concentrated load. — Let the weight P be distant from the
left end by a distance Zi and from the right end by Za. Let the distance of
any point from the left end be x. For the upward reaction at the left end,
— P — k
T
The moment, then, at any point between the left end and F, for x less
than Zi, isM — — ""?""• ^or any point to the right of P, or # greater
than llt M' = . — + P i x — Zi I. Instead of this, however, we may
take the reaction at the other end, Va = P~; and then for x greater than Zi,
The moment is evidently greatest at the point of application of the load,
or for x = li Hence the maximum moment is — — T— -.
(a) Breaking weight.
2 T I P I I
From (11) M = - — = -- '— , or, for the breaking weight, P =
2 T I Z 1 T 5 A? Z
-, 7 7 « For rectangular cross-section, I = 77: 5 h* and P — a -, 7 -
ti ti It t 12 o Li L-i
For a load in the middle, Zi = Z2 = — I and Max. M = — -r P I, and P =
& 4:
r 4 times as great as for a beam of same length fixed at one end
and free at the other.
(5) Change of shape.
We have, then, from (8), for x less than Zi,
& y P Z2 x , & y' P |j (7— x) £ ., ,
d* = -HIT and 5/ "irr2 for z greater than '••
Integrating, we have
• d+y_ PZ,** dtf _ PI, r *fl,0.
-~ = " "
For x = lit these two values of — ^-are equal, and hence, since Z, = I — llt we
a a;
CHAP. III.] SUPPLEMENT TO CHAP. VII. 119
P I 2
have C' = C H — -^-. We have then the two equations
d y Pl*x2 , d y' P I,
^=-^TI+C and jf • -«i
containing both the same constant C.
Integrating these, we have
P Z2 y? P Zi rZ a;2 a*n P Z,2 0
In the first of these, f or x — 0, y — 0 ; hence d = 0.
P Zi3
For x = Zi, y = y' ; and hence O2 = — .
D £! I
For x — Z, y' — 0 ; and hence, finally, C = * ' (2 Z — Zi]
We have, therefore, by substitution of these constants,
P Zi2 Z22
For x — Zi, we have the deflection at the load y = — —7.
O E I 6
>7 rti
Inserting the value of C in the value for — — above, and placing the
value of — equal to 0, we have for the value of #, which makes y a maxi-
d x
mum, x =* /— (2 Z— Zj) Zi, an expression holding good only for x less than
Zj. Inserting this in the value for y, we have for the maximum deflection
itself
If the load is in the middle, we have for the curve of deflection
P Z3
and for the deflection itself A =
48 El
The greatest deflection is not, then, at the weight, except when the load is
in the middle. When this is the case, the deflection is only ^-th of the
deflection for the same length of beam fixed at one end and loaded at the
other free end.
15. Beam as before supported at the end§ — Uniform
load.— For a load p per unit of length, the entire load is p Z. The reac-
tions at each end are — -, and the moment at any point is
is evidently greatest at the centre, and hence
Mar. M = -££
120 SUPPLEMENT TO CHAP. VH. [CHAP. IH.
For the breaking weight, then, from (8)
16 T I
or 4 times as much as for a beam of same length loaded uniformly and
fixed at one end.
For the change of shape, we have
dzy _^px (l—x}
dx*~ 2EI
The constants of integration are determined by the conditions that, for
X = W'dx=®' ^ — °' y~®'-> and x = l,y = 0. Integrating, then, twice
under these conditions, we have
This is greatest at" the centre, or for x = — ; hence the maximum deflection is
2
A = — - •—-, or only i&ths of a beam of the same length fixed at one
end and uniformly loaded.
16. Beam supported at one end and fixed at the other
— Constant cross-section— Concentrated load. — Let the left
end be fixed horizontally so that the tangent to the deflected curve at that
point is always horizontal, and therefore — = 0.
a x
Let the distance of the weight P from left be a, and the distance of any
point x.
Then, for x less than a, we have
M = — V (l-x) + P (a-x) •
for x greater than a,
M' = -V(Z-aO,
where V is the reaction at the free end, and is so far unknown.
If we put M = — -£ and M' = '{, and integrate as usual, and remem-
d x d x*
ber that f or x = 0, ^ = 0, and for x = a, -^ ' = ~, we have
d x dx dx*
Integrating again and determining the constants by the conditions that,
for x = 0, y = 0, and for x = a, y — y\ we have
V = ~~ [V (3 «-aO-P (3 «-*)]
y1 = -^ [V *« (3 Z_a)-P (3 x-a} a*].
CHAP. III.] SUPPLEMENT TO CHAP. VII. 121
Now, for x = I, y = 0 ; hence V = P — L_ — - -. If the load is in the
middle, V = — P.
, . V, or the reaction at the free end, is now known, and substituting it in
, the value of y' above, we have the equation of the deflection curve between
the weight and the free end.
y' — — — --\ — — (3 x — a) a2
Substituting it also in the value of — — above, and placing then — ^- equal
dx dx
to zero, we find for the value of x, which makes the deflection a maximum,
when x is greater than #, x = I — 1\/ — j .
' 3 & — o>
Substituting this value of x in the value of y' above, we have for the
maximum deflection itself
P Z3 1
When the weight is at the middle, this becomes A = - x — = , or
48 E I y/5
only — — = , as much as for a beam of same length fixed at end and with
tC y 5
load at other end, and only — - as much as for same beam simply sup-
y5
ported at ends.
Breaking weight.
Having now V, we know M and M'. Rupture will occur where the
moment is greatest, that is, either at the fixed end or at the weight. Now
the moment at P is — V (I— a) — — V 1 + V a. The moment at the fixed
end is — V l + P a. Now, as V is always less than P, we see at once that
for any value of a less than I, the moment at the weight is greatest.
We have for the moment at the weight from (8)
2(3Z~a)/7 , 2TI ...
i—j — - (I- a) = — — , and hence
4 T 1 13
for the breaking weight P = - — —
ha* (3 I— a) (l-o)*
64 T I
If the weight is in the middle, P — — -=-.-,
o fi l
or fths as much as for the same beam supported at the ends.
IT. Beam as before fixed at one end and supported at
the other — Uniform load. — In this case the moment at any point is
M— — V (l—®)-r-p (l-x)* = E I --Jj. Integrating twice and determin-
ing the constants by the conditions that, for x = 0, — - = 0 and y = 0, we
d x
easily obtain
122 SUPPLEMENT TO CHAP. VH. [CHAP. III.
(B l-**-P (6 Z2~
o
For x = 1, y = 0, and hence V = — p 1.
8
Substituting this value of V
•j r _ V'-iR
Putting this last equal to zero, we find x — — - - Z, or x = 0.5785 Z,
lo
for the value of a*, which makes the deflection a maximum, and this in-
serted in the value of y gives for this maximum deflection itself,
164 El El
For the "breaking weight, we have, since the greatest moment is at the fixed
1 72 __ 1 72 2 T I , . 16 T I
end and equal to — p I2, M = —p I2 — — - — ; hence p I = - - — .
8 O fl hi
The strength is, then, f times as great as for the same load in the middle,
but no greater than for a beam of same length and load supported at both
ends.
18. Beam fixed at botli euds— Coii§tant cros§-section—
Concentrated load. — Taking our notation as before (Art. 12), we
have in this case not only a reaction at the right end, but also a positive
'moment there as well, both of which must be found. If h be the distance
from left end to weight, and h from weight to right end, and if Vj and
V2 are reactions, Mi and M2 the moments' at left and right ends respec-
tively, then for equilibrium we must have — Vi + V2 = P, Mj + Vi ^ =
M2-V3Za.
For x less than ^ we have M = Mj +V1x = EI — |. Integrating
cL x
once, since the constant is zero, because, for 0 = 0, ~ — 0, we have
a x
d x 2 E
Integrating again, since, for x = 0, y = 0, and the constant is zero,
x\
For the distance from the right end to the weight we may obtain similar
expressions, if we take that end as the origin, only we should have — V9
and M2 in place of Vi and MI. At the weight itself — - and y must in
(L x
each case be equal, but *=-£ of opposite sign. Therefore we have the equa-
CL $
tions (2 MX + V, Zx) li = -(2 M2-V2 Z2) Z2,
(3 Mi + Vx Zx) Zx2 = (3 M2-V2 Z2) Za2.
CHAP. III.] SUPPLEMENT TO CHAP. VII. 123
From these two equations, and the two equations above, viz., ~ Vi + V2 = P
and Mi + Vi li = Mo— V2 Z2, we can determine Vi, V2, Mi and M2.
Thus from the last two we have
Vi Zi+V2 ?2 = M2-M: =V, k + PZa + Vi Z2=Vi Z + FZ2,
or Vi I = Ma-Mi-P Z2.
So also V2 Z = M2 -Mi +P Zi, and substituting these in the equations above,
we have
(Mi+Ma) ZrrFZj Z2,
M! Z (2 ZX-Z2)-M2 Z (2 Za-ZO = P Zi Z2 (k - Z2) ;
and from these we have, finally,
and then from the values of Vi Z and V2 Z above
Vl = -P"2 v"^/ v^ =
Z3 Z3
Change of shape.
Substituting these values, we can now find
077
Hence y is a maximum for x = ^-, and the maximum deflection
, 3 Zi + ly
itself is
2 P Zx3 Z22
A =
3 E I (3
This expression will be itself a maximum for Zi = Za or Zi = \ Z, that is,
the maximum deflection for a weight in the middle is at the weight and
equal to
PZ3
A =
192 E I
This deflection is greater than the maximum deflection for any other
position of the weight, which in general is not found at the weight itself,
but at some other point between the weight and farthest end.
We see above that the deflection in this case for load in middle is only
one-fourth as much as for same beam and load when supported at the ends.
Breaking weight.
For the greatest moment, which we easily find to be at the end, we have
_PZ1Z22_ PZ, (Z-ZO
~~ ~
This is a maximum for Zi = £ Z. That is, the greatest moment at the end
occurs wlien the load is distant one-third of the length from that end. The
124 SUPPLEMENT TO CHAP. VII. [CHAP. in.
value of this greatest moment is — - P 1. Hence we have from (11) P I
A t 27
2 T I 27 T I 27
= — 7 — or P = , or — as great as for the same beam supported at
/i d> fi I lo
p 7
the ends only. If the weight is in the middle, however, we have — —
8
2 T I 1 6 T I
— — - or P = — — — , or twice as much as the same beam supported at the
tl ii L
ends.
19. The above is sufficient to introduce the reader to the theory of
flexure. He can now discuss for himself the above case for uniform load,
and prove that the maximum deflection is at the centre and equal to
pi2 1
. That the greatest moment is at the end and equal to — p l\
24 T I
and that the breaking weight is p I = — . We may also observe that
ft v
both in the beam fixed at one end and supported at the other, and fixed at
both ends, the moment at the fixed end is positive. From this end it de-
creases towards the weight, and finally reaches a point where the moment
is zero. Past this point the moment becomes negative, and in the case of
the beam, free at the other end, increases gradually to a maximum and then
decreases to zero. In the beam fixed at both ends, it increases to a maxi-
mum, then decreases to zero, then changes sign and becomes positive and
increases to the other end. These points at which the moments are zero
are points of inflection, because here the curvature changes from convex to
concave, or the reverse.
They can be easily found from the equations for the moments by finding
the value of x necessary to make the moments zero.
Thus, for a beam fixed at one end and supported at the other, uniform
load, the inflection point is at a distance from the fixed end x — — . For
both ends fixed, we make M — — p [I* — 6 (I — x) x] = 0, and find x =
12
V (3 T VW) I = 0.21131 I and 0.7887 Z. The reader will also do well to
b
discuss the curves of moments. He will find the moments represented by
the ordinates to parabolas, and limited by straight lines similarly to Figs.
73 and 75, PI. 18.
We shall give in the Supplement to Chap. XIV. much more general
formulae, from which, for one or both ends fixed or free, the moments and
reactions at the supports may be found, when any number of spans of vary-
ing length intervene, for single load anywhere upon any span, or uniformly
distributed over any span.
CHAP. VIII.] CONTINUOUS GIRDERS. 125
CHAPTER VIII.
APPLICATION OF THE GRAPHICAL METHOD TO CONTINUOUS
GIRDKRS GENERAL PRINCIPLES.
8O. Mohr's Principle. — Thus far, in addition to the general
principles of the Graphical method, we have noticed more or
less in detail its application to the composition and resolu-
tion of forces, and the corresponding determination of the
strains in the various pieces of such framed structures as Bridge
Girders, Roof Trusses, etc. We have also illustrated the
graphical determination of the centre of gravity and moment
of inertia of areas, as also of the bending moments and shear-
ing forces for simple girders, including several important cases
in practical mechanics. (See Art. 41.) Lastly, we have taken up
the subject of Bridge girders more in detail, and developed in
order the principles to be applied 'in the solution of any par-
ticular case. Although brief, it is hoped that this portion will
be found sufficient to illustrate fully the method of procedure
to be followed in practice.
As regards simple girders, the principles referred to are so
easy of application that the reader will find no difficulty in
diagraming the strains in any structure of the kind, as explained
in the " practical applications " of Arts. 8 to 13 ; or he can find
the maximum moment at any cross-section for given load-
ing according to the last chapter. In the case of beams or
girders continuoii,s over three or more supports, however, we
meet with difficulties which for some time were considered in-
superable.
Thus Cidinann, in the work which we have so often quoted,
says : * " The determination of the reactions at the supports for
a continuous beam, which depend upon the deflection, the law
of which is given by the theory of the elastic line, is impossi-
ble by the graphical method, at least so far as at present de-
veloped. The theory rests upon the principle that the radius of
*Culmann?8 GrapMsche8ta.Uk, p. 278.
126 CONTINUOUS GIKDERS. [CHAP. VIII.
curvature of the deflected beam, for any cross-section, is in-
versely proportional to the moment of the exterior forces.
'Now the deflection at any point is so small, and the radius of
curvature so great, that its construction is impracticable, and
will so remain until Geometry furnishes us with simple rela-
tions between the corresponding radii of curvature of pro-
jected figures whose projection centre lies in the vertical to the
horizontal axis of the beam. If such relations were known, we
could by projection exaggerate the deflection of the beam
until the radius of curvature became measurable. Since we
are not yet able to do this, we must have recourse to calcula-
tion" He then enters into a somewhat abstruse analytic dis-
cussion of the continuous girder, and deduces formulae for the
reactions at the supports. These being thus known, the graph-
ical method is then applied.
Concerning this difficulty, Mohr* remarks that it has but
little weight, and may be easily overcome if the same simplifi-
cation of the graphical method is made which is considered
allowable in the analytical investigation, viz., when we take in-
stead of the exact value of the radius of curvative
as given by the calculus, the approximate value
Thus, let PL 14, Fig. 50 represent a perfectly flexible cord
A B D loaded by arbitrary successive forces. The variation of
these forces per unit of horizontal projection dvwe represent by
p. Take the origin of co-ordinates at the lowest point B. If
the cord is supposed cut at B and D, we have at B a horizontal
force H, and at D a strain S, which may be resolved into a
horizontal force Hj and a vertical force V. Since these forces
are in equilibrium with the external forces, the conditions of
equilibrium are
(1) ....... H = H!
and px
(2) ..... V= / pdx.
Jo
* Zeitsoh. fas kannov. Arch,-u. Ing. Vereim. — Band xiv., Heft 1.
CHAP. TUT.] CONTINUOUS GIRDERS. 127
Moreover, Cx
I p dx
(3) . . *La = — = —
dx H! H
Differentiating :
d2 y p d x
«1 • ± ryp
~7
6? X
Now, had we formed a force polygon by laying off the forces,
then taken a pole at distance H and drawn lines from pole to
ends of forces, the corresponding equilibrium polygon would,
as we have seen, Art. 43, be tangent to the curve A B D at the
points midway between the forces. The greater the number
of forces taken, the shorter, therefore, the sides of the polygon ;
the nearer it will approach the curve A B D. This curve is
therefore the equilibrium curve, found according to the graph-
ical method. Its equation is given above by (4).
But the equation of the elastic line is, as is well known,
(y) »•>'**• 7 'o ~r~ ?
a (E> J-
where E is the modulus of elasticity of the material, M the
moment of the exterior forces, and I the moment of inertia of
the cross -section.
Comparing now this equation with equation (4) above, we
see that the elastic line is an equilibrium curve whose horizon-
tal force H is E, and whose vertical load per unit of length p
is represented by the variable quantity -«-•
This simple relation, first given by Mohr, renders possible
the graphical representation of the elastic line, and not only
solves graphically almost all problems connected with it, but in
many cases simplifies considerably the analytical discussion
also.
81. Elastic Curve. — If we choose the pole distance H at -th
n
E instead of E, the ordinates of the elastic line will be n
times too great. If the scale of the figure is, however, - th the
* Stoney— Theory of Strains, p. 146. Wood— Resistance of Materials, p. 98.
Also Supplement to Chap. VII , Art. 11.
128 CONTINUOUS GIRDERS. [CHAP. VIII.
real size, then in the diagram the ordinates of the elastic line
will l)e given in true size.
Equation (5) may also be written
that is, the elastic curve is an equilibrium curve, or catenary,
whose horizontal force H is E I or F, and whose corresponding
variable load per imit of length is M or — respectively.
If we divide, then, the moment area by verticals into a number
of smaller areas, and consider these areas as forces acting at
their centres of gravity, these forces determine, as we have seen
(Art. 43), an equilibrium polygon which is tangent to the elas-
tic curve at the verticals which separate the areas. Thus we
can construct any number of tangents to the elastic curve ; areas,
which are positive or negative, must, of course, be laid off in the
force polygon in opposite directions.
If we divide the moment area by lines which are not vertical
[PL 14, Fig. 51], the directions of the outer polygon sides are
the same as for vertical divisions, because the vertical height
between the corresponding outer sides in the force polygon is
in any case always equal to the total load.
The two outer polygon sides for any method of division are,
therefore, tangents to the elastic curve at the- ends of the same.
Here also we can, of course, have negative areas.
82. Effect of End Moments.— A beam or girder continuous
over three or more supports differs from a beam simply resting
upon its supports, in that, in addition to the outer forces, we
have acting at each intermediate support a moment or couple.
But, as we have seen, Art. 23, the effect of these moments or
couples will be simply to shift the closing line of the equilibrium
polygon through a certain distance. Thus [PL 14, Fig. 52 («)],
if the span ^ were uniformly loaded and simply supported at
the extremities A and B, the equilibrium curve, or curve of mo-
ments, would, as we know (Art. 44), be a parabola A D B. If,
however, the beam is continuous, we have at A and B moments
or couples acting, and the closing line A B is shifted to some
position as A' B'. If now wre consider the moment area, we
see that by the shifting of the closing line the former moment
area, which we shall call \\\Q positive area, is diminished, while
to the right and left we have negative areas A A' C and B B' C.'
CHAP. VIII.] CONTINUOUS GIRDERS. 129
It is evident that these areas have also a corresponding action
upon the elastic line. For a positive moment area this last is
concave upwards, while for negative areas it is convex upwards.
At the points of transition C and C' we have the inflection
points. This follows easily if we only hold fast the manner in
which the elastic line is constructed, viz., by dividing the mo-
ment area into laminae and regarding the area of each as a
force. The forces thus obtained must plainly act, some upwards
and some downwards, and the corresponding equilibrium poly-
gon or elastic line must be in part convex upwards and in part
convex downwards, and hence at the points of transition we
must hav re points of inflection where the moment is zero.
83. Division of tne Moment Area. — We shall assume the
cross-section of beam constant. Regarding the elastic line
simply as an equilibrium polygon, we can apply the principle
that the order in which the forces are taken is indifferent (Art.
6) when the resultant only is desired. Since in the considera-
tion of a single span only the first and last sides are of impor-
tance, we can, so long as we consider a single span only, take
then the laminae or divisions of the moment area in any order
we please. More than this, we can, as we have seen in Art. 81,
divide the moment area into laminae not vertical ; for example,
we may in any span distinguish three parts, one positive and
two negative, and consider each as a force acting at the centre
of gravity of the corresponding area. [This holds good only
for constant cross-section. For variable cross-section the hori-
zontal force E I is variable.] Still further, we can divide the
moment area for a single span into a positive area, which is pre-
cisely the same as for a non-continuous beam, and into a nega-
tive area, which will be evidently a trapezoid.
This is of great importance. To understand it fully we refer
to PL 14, Fig. 52. Here, in the second span, we see that the
real moment area consists of a positive part, viz., the parabola
C D C', and two negative parts A A' C and B B' C'. Instead of
these we may take the entire parabolic area A D B and the tra-
pezoid A A' B' B, or, finally, instead of this trapezoid, we may
take the two triangles A A' B' and B B' A'. The parabolic
area is positive, the triangular areas are negative.
If we assume the load as uniformly distributed, the first area
will be always parabolic, and we may, therefore, call it the
parabolic area.
130 CONTINUOUS GIRDERS. [ciIAr. Vm.
By this division of the moment area we have obtained a great
advantage. While the three areas C D C', A A' C and B B' C'
are all three dependent upon the moments at the supports A A!
and B B', we have by £his new division to do with three areas,
of which the first is entirely independent of the moments at the
supports, the second depends only upon that to the left, and the
third only upon that to the right.
84. Properties of the Equilibrium Polygon. — Let us con-
sider now the case of a beam over four supports, that is, of
three spans — Z0, 4 and 4 — the first and last being, as is usually the
case, equal, and the two first loaded with both live and dead
load, the last with dead load only. The parabolas for the ver-
tical loads [PL 14, Fig. 52] may be constructed by means of a
force polygon, or the ordinates at the centre calculated, and the
parabolas then drawn. The moments at the supports are A A'
and B B'. Although these are unknown, it is not necessary to
assume them at first. They may be directly constructed.
Thus, if we conceive the moment areas in each span divided
into positive parabolic areas and negative triangles, we have in
the first and last span one, in the middle two triangles. If we
consider these areas as forces acting at the corresponding centres
of gravity, we shall obtain an equilibrium polygon of the form
given in Fig. 52 (b). That is, this polygon must have eight
sides, and its angles must be somewhere on the verticals through
the centres of gravity of the parabolic and triangular areas.
The parabolic areas act downwards, the triangular areas up-
wards. The problem is, to make these last so great that this
polygon shall pass through all the points of support.
One of the properties of the polygon we have, therefore, just
noticed, viz. : its angles must lie in the verticals through the
middle points of the spans and through the points distant from
A and B one-third of the spans on each side (i.e., the centres of
gravity of the triangles).
If we prolong the second and fourth sides of the polygon,
they intersect in a point M, the point of application of the
resultant of the two contiguous triangular area forces (Art. 44).
The areas of these two triangles are - A A' Z0 and - A A' ^, that
is, the areas are as the spans 4 and ^.
Then by the principle of Art. 18 the resultant divides the
distance between the forces into two portions, which are to each
CHAP. VHI.] CONTINUOUS GIRDERS. 131
other as ^ to 4> or inversely as the forces. Since the entire dis-
tance is - £0+ 7: ^i> the distance of the resultant or of the point
o o
of intersection M from L is - ^ ; from N it is - 1Q. The point
O t>
M, therefore, must lie somewhere in the vertical at - ^ from L,
o
the point of application of the triangular area force for the
span 4- The verticals through the centres of the parabolic
areas we call the parabolic or middle verticals / those through
the centres of gravity of the triangular areas, the third verti-
cals / those through a point as M, the point of application of
the resultant of two contiguous triangular area forces, the lim-
ited third verticals. Upon these verticals two sides must always
intersect.
85. Polygon for tlie Positive Moment Areas. — It will be
found best to take as the reduction base for areas — ^, i.e., half
2
the second span, and for pole distance — l±. Reducing the
3
areas of the parabolas to this basis, and considering the heights
thus obtained as forces, we can form a force polygon with pole
distance - \. It is not necessary to draw this polygon ; our ob-
3
ject is to find the corresponding equilibrium polygon. This
last, since we consider the entire parabolic area as a'force act-
ing at its centre of gravity, consists of two lines which inter-
sect in the vertical through the middle of the span. We pro-
long each of these lines and obtain two lines as shown in PL 14,
Fig. 52 (c). The segments cut off by these lines from the verti-
cals through the supports are the moments of the parabolic
areas with respect to the supports. These moments we can
easily find.
Thus, let the deflection of the parabola in the second span at
o
the centre be f, then its area is — fl^. This reduced to the
3
1 4:
basis - 4 gives -/ as the force. The moment of this force
A o
4: 1 1
with reference to the supports is — f x —li = 2fx—li. This
32 3
moment is equal to the segment sought multiplied by the pole
132 CONTINUOUS GIRDERS. [CHAP. VIII.
distance. This last is - ^. The segment, therefore, is 2/! "We
o
do not need, therefore, to draw the force polygon, but have sim-
ply to take off with the dividers the middle or din ate of the para-
/D $
bola f = ~rr , and lay it off twice on the verticals right and left
8
through the supports, and join the four points thus obtained by
lines crossing each other under the centre of the span. The
equilibrium polygon for the positive parabolic area is then
ready for the middle span. ["We advise the reader to construct
it for himself.]
For the two side spans the construction is different. Here
2 4 I
the area of the parabola is -f 10) the reduced area -f j, and
3 3 L\
the moment x = 2' .
Dividing by - ^, we have for the segment required
3
Therefore, in the end spans, or generally in any span not equal
to the standard span, or that which furnishes the constants - l±
2
and - Zj, we must multiply the middle ordinate of the parabola
o
by the square of the ratio of the " standard n span to the span
in question, and then lay the product off twice upon the verti-
cals through the supports. This multiplication is easily per-
formed graphically. If from the middle of span 4 we lay off
If horizontally and join the end with the end of y, then lay off
42 in same direction and draw a parallel to the first line, the
lz
segment on/"7 will bejf'~& For we shall have
h
79 /» / • • V O /*/ ''0
42:/ :: tf:^ or os=f 'A.
Since these cross-lines depend upon given quantities, they
can be constructed for every span, and thus we have Fig. 52, c.
They give us not only the moments over the supports, but also
the moments for any point of the parabolic area forces. We
shall hereafter make use of them.
$6. Construction of the Fixed Points, and of the Equi-
librium Polygon. — We have thus all the given and known
CHAP. Vni.] CONTINUOUS GIRDERS. 133
quantities, have deduced the general properties of the equili-
brium polygon, and will now endeavor by their aid to draw the
polygon itself. We shall then be able to find the actual mo-
ments A A' and B B' at the supports. It is impossible at first
to draw any single side of the polygon in true position, and we
must, therefore, endeavor to find certain /points of the same suf-
ficient to determine it.
Lay off first the three spans, PL 14, Fig. 52 (d). Suppose
the second side of the polygon prolonged till it intersects the
vertical through the end support a, in a point K, Fig. 52 (£).
This point is known. It is given by the moment of the para-
bolic area in the first span with respect to this end support.
This moment we have already by the cross-lines in Fig. 52 (c).
We have then simply to take it off in the dividers from (c) and
lay it off from d to K7 in Fig. 52 (d). We have now in Fig. 52
(J), two points of the polygon known, namely, the end support
and K, which last must be in the second side prolonged.
The triangle L M N is now of special importance. What-
ever may be the position of KM and M N, we have already
seen that the intersection M must always lie somewhere in the
limited third vertical. The first side KM must, however ',
always pass through K, a known point. The second must pass
through the support, also a known point. The points L and N
must, moreover, always lie in the third verticals, distant from
A, 5- \ and ^ Z0 respectively.
If the line K M takes up various positions under these con-
ditions, the line M N will revolve about a fixed point which is
given ~by the intersection of a line through K and the support
A with M N.
If, then [Fig. 52 (d)], we draw a line in any arbitrary direction
through K7, and note the intersections L' and M7 with the first
third vertical and the limited thtrd, then through L' and the
support draw a line to intersection N7 with second third verti-
cal, and join M7 N7, and finally through K' and the support
draw a line intersecting this last in I, the point I thus deter-
mined is & fixed point, and remains the same for any position
of K' M7. It is therefore a point on the fourth side M N of
the polygon. For the triangle L7 M' N' may have any posi-
tion, yet so long as its angles lie in three parallel fixed lines,
ui id two of the sides- pass through two fixed points, the other
134 CONTINUOUS GIRDERS. [CHAP. VIII.
side must also pass through a fixed point.* Out of all possible
positions of the triangle, one of these positions must coincide
\vith the polygon sides, and hence this fixed point is a third
known point, since we have already K' and the end support.
Although, then, we are as yet unable to draw any of the sides
of the polygon in true position and direction, still from the
hitherto known properties we have deduced a new one. We
know now a point through which the fourth side must pass.
But this is not all. We proceed still further. The fourth and
fifth sides must intersect upon the vertical through the centre
of the second span. These sides, moreover, cut off upon any
vertical the moment of the parabolic area with respect to any
point in that vertical. We know this moment thus for the
point I just found. It is found by taking the segment cut off,
from the vertical through I, by the cross-lines for the parabolic
areas found above in Fig. 52 (c).
Laying this segment off from I, we thus find I', a point in the
fifth side prolonged. From this point we proceed as before to
find the next fixed point I". We then lay off from 1" the mo-
ment of the parabolic area for this point and find F", a point
upon the eighth side. We can now draw the polygon itself.
Thus the eighth side passes, of course, through the last sup-
port and also I"'. It is therefore determined. Through the
intersection of this line with the vertical through the middle of
the span and the point I" the seventh side passes. The sev-
enth side is therefore determined. Through the intersection of
this with the third vertical and the support the sixth side
passes and continues till it intersects the third vertical on the
other side. Then from this point towards I' to intersection
with vertical through centre of middle span. From this last
point towards I to intersection with third vertical. From this
last point again through support to intersection with third verti-
cal on other side ; then towards K' to intersection with vertical
through centre of end span ; and lastly, from this last point of
intersection through the end support, and the polygon is com-
plete as given in Fig. 52 (b).
In this manner, however, inaccuracies may occur. To avoid
these we may start from the rig/it end support and also find
four fixed points as above. It is unnecessary to make the con-
* This proposition the reader can easily prove geometrically or analytically.
See Art. 112.
CHAP. VIII.] CONTINUOUS GIRDEKS. 135
striiction. We see at once that we shall thus obtain in each end
span two points, in the middle span four points, which last, be-
ing joined by lines crossing each other, give in the middle span
two sides in proper position. It is also evident how the poly-
gon may then be completed.
87. Construction of the Moments at the Support.*. — Thus
we are able to construct the equilibrium polygon, or rather the
extreme tangents to the elastic line for each span. "We have
now to determine the two moments over the supports. This is
very simple. The first moment to the left is cut off by the
fourth side, the second by the fifth side of the polygon, from
the verticals through the supports. We have therefore only to
prolong these two sides, take off the segments in the dividers,
and lay them off in Fig. 52 (a) in A A! and B B'. We have,
then, in PL 14, Fig. 52 (a) the moments for the given case and
loading at any point, as shown by the shaded area.
The proof is simple. The two lines N M and N L [Fig. 52,
Z*] evidently cut off upon the vertical at the support the moment
of the force acting at N. This force is the area of the triangle
A A.' B' equal to - A. A! ^. This reduced to the basis - l± gives
A A!. If. we multiply this by the lever arm of the force, we
have its moment. This moment is, however, equal to the seg-
ment A A' multiplied by the pole distance, and since this pole
distance is itself ~ h the segment itself to the assumed pole
o
distance gives us the moment.
We see that it is not necessary to draw the line N L as it
passes through the support. We have simply to prolong the
side M N to intersection with the vertical through the support.
It is to be observed that the moments at the supports are cut off
at the supports only by those lines which pertain to the " stand-
ard " span, or that span from which we take our reduction basis
and pole distance. For lines in the other spans the above does
not hold good without modification. It is, however, always
possible, at least for from two to five symmetrical spans, to
observe the above conditions. In those cases where this is not
possible, an easy graphical multiplication of the segments by
the square of the ratio of the spans will give the moments.
We see also the reason why, for four symmetrical spans, the
and not the first must be taken as the standard span.
136 CONTINUOUS GIBBERS. [CHAP. VIII.
If the construction of the moments over the supports is our
sole purpose, as is in practice the case, the polygon need not be
drawn. We have only to find our fixed points, and note the
intersection of the sides with the verticals through the supports,
without drawing the sides themselves. In the preceding Arts-
we have purposely considered only the particular case of uni-
form loading, and have taken only three spans, in order to
familiarize the reader with the nature of the problem and the
method of its solution. In oi'der to attain a clear understand-
ing of the subject as thus far developed, he would do well to
take some particular case, as, for instance, that of a girder of
two or three or four spans of given length, the end spans being
equal, and intermediate spans equal and say one-fourth longer
than the ends, and work out by diagram the moments at the
supports for a uniform load over the whole length of girder.
For two spans the moment at the centre support should be
g p P, I being the length of span,^> the load per unit of length.
For three spans the moment at the two inner supports is
1 -f- n3
4 (3 i o — \ P ft, where n I = the length of end spans. Thus, if
3 91
n = -, we have TT^Q & ^' "^or ^our sPans ^e moment at the
1-4-2 w3
second and fourth supports is . Q , — rp P, and at the middle
4: \O ~r 4: 7l>\
1 _l_ 2 n nz
support 4/014 — r P P- By these formulae the graphical
results may be checked.
When the reader has thus become thoroughly familiar with
the principles of the preceding Arts, and their practical appli-
cation, he will be ready to resume at this point the more gen-
eral development which follows.
88. The Second Equilibrium Polygon. — We see, there-
fore, that the actual form of the elastic line is not required to
be known. Only the outer forces and their moments are sought,
and to determine these it is sufficient to know the position of
the tangents to the elastic line at the supports. Thus the first
line of the equilibrium polygon [Fig. 52, &] being given in
position, by the aid of the middle, third, and limited third
verticals and the known point K, all the other sides may be
CHAP. VIII.] CONTINUOUS GIKDEKS. 137
drawn, and the moments at the supports found. We conceive,
therefore, the moment area as the difference of the trapezoid
A' A" B" B' [PI. 15, Fig. 53] and the parabolic area A" C" B",
or equal to A" C" B" minus triang. A7 A" B' 'minus triang.
B/ B// A//_ rp}ie area A// c// B// we cau the wrwpiQ or parabolic
moment area.
If we indicate the moments at the supports A7 A." and B' B"
by M' and M", then, for a given span Z,
A' A"- B7 = J M' Z and
A" B' B" = J M" I.
If we indicate further the height of a rectangle of base Z and
area A7/ C" B", that is, the mean value of the 'moments of the
corresponding simple girder by 9)2, we have
area A" C" B" = 3tf Z.
The verticals through the centres of gravity of the triangles
divide the span into three equal parts. We call these the third
verticals. The load 9ft I acts at the centre of gravity of the
parabolic area A" B" C".
The four-sided equilibrium polygon A U S V B corresponding
to these forces we call the second equilibrium polygon. The
pole distance must be (Art. 81) -El. Instead of this, we take,
itf
which amounts to the same thing, the forces G F = - M' - ,
EH—- M"- and F E = 9ft -, and the pole distance 1} —
2i A» A
E I
— -, where \ is any assumed length. For \ we may take the
n A/
arithmetical mean of all the spans, or, as we have seen, one of
the actual spans. If the outer spans are both equal to 10 and the
other spans equal to 11? we should naturally choose \ = ?1? since
then the forces would be - M', - M" and 9ft.
2t 2t
If the position of the tangents at the supports were known
or found, the equilibrium polygon could be easily drawn as
follows : Upon the two verticals distant each side of the centre
E I
S [Fig. 53] by the pole distance ~b — — - lay off the distance
n A»
138 CONTINUOUS GIBBERS. [CHAP. VIII.
A — %)l - and join the points thus obtained by two lines cross-
ing each other. These cross-lines are the lines OP O E of the
force polygon. If now we make U IT and V V equal to the
ordinates of the cross-lines vertically under U and V, then the
sides of the equilibrium polygon U S and V Q prolonged, pass
through U' and V'. This will at once appear from an inspec-
tion of Fig. 53.
In this form the equilibrium polygon was first repre-
sented by Mohr. (Zeitschrift des Arch, und Ing. Ver. zu
Hannover, 1868.)
89. Determination of the Moments over tlie §upports.
— If we draw in the force polygon, lines parallel to the four
sides of the second equilibrium polygon, then the segments of
the force line between the lines parallel to A U, B V [PI. 15,
Fig. 53] and those parallel to S U, S V, are respectively F G =
- M' - and E H = - M" -.. If we prolong S U and S V to
A A/ 2t A
intersection^ M and N with verticals through the supports, and
represent A M and B N by y' and y", we have from the simi-
larity of the triangles U A M and V B N with O G F and O H E
hence
' - ?^L£ " - M"^
y ~ 6 I A y ~ 6 I A*
The segments A M and B N are, therefore, proportional to
the moments at the supports Mx and M".
These moments themselves can now be determined in vari-
ous ways.
\st. It is in general best to choose the second pole distance
b = - \. We have then
o
M' = y' ( M" = y" (Art. 70).
If, then, at a distance from U and V either way equal to
2bl-\ = --\ we draw verticals, the segments A! M! and
\ LI 06
Bj Nj cut off from these verticals will evidently be equal to
the moments required, viz., 1VT and M".
CHAP. VIII.]. CONTINUOUS GIRDERS. 139
%d. If we take X = I, we have at once M/ = y' and M" = y" .
In the inner spans, therefore, we have directly, as we have al-
ready seen, for the special case (Art. 72), when \—l, the mo-
ments at the supports.
3d. For a span adjoining a span whose length is X, we have
the moment for the intervening support directly from this last
span. If the inner spans have the same length I, and the two
outer the same length £1? we can accordingly, by making \ = I,
obtain directly the moments at the supports.
90. Comparison with Girder fixed horizontally at both
ends. — If the ends are fixed horizontally, the lines in the force
polygon parallel to A U and B V coincide, i.e., H and G fall
together. Accordingly, if we designate the end moments now
by W W" we have (Fig. 53)
") = W.
Therefore, in the first equilibrium polygon, the moment areas
on each side of the dosing line are equal.
Indicating the points for this case by the index 0 [Fig. 54,
PL 15], we have the triangle A U0 M0 equal to U0 V0 V'o,
and therefore A M0 = V0 V0 — V V, as also, in like manner,
B N0 = UolTo = U U', or, taking b = 1 X,
U U' = 3JT /-\2 V V =
Therefore, the ordinates between the cross-lines at the verti-
cals passing through U and V are, for girders fixed horizon-
tally, proportional to the end moments Wl' and 9ft".
For X = I, fi = - I, and these ordinates give the moments
6
directly.
If we draw through U7 and V a straight line intersecting
the end verticals in Q and R, and prolong N U' and N V to
intersections S and T, then Q M — 2 U U7, Q S = V V', and
hence M S = 2 U IF+V V' = (2 2U'+3B") (^Y; and in the
same way N T = (2 2tt" +2ft') {-j\
Therefore, the segments cut off upon the end verticals ly the
cross-lines are proportional to2W+ W and 2 2ft" + W.
140 CONTINUOUS GIEDEES. [CHAP. VIII.
The quantities SDc' and $31" being known, we can easily con-
struct the cross- lines.
If we draw a line through U and V to intersections O and P,
we have O M = V V, P N == U U'. Therefore, A O and B P
are equal to (>Itt'-M') ^ and (2JT-M") f -)*, where M' and
M" are the moments at the supports for a continuous girder,
and 9ft' W those for a girder horizontally fixed at its ends.
CHAP. IX.] LOADED AND UNLOADED SPANS. 141
CHAPTER IX.
CONTINUOUS GIRDER — LOADED AND UNLOADED SPANS.
91. Unloaded Span. — If the span is unloaded, we have to
construct the second equilibrium polygon, only the two forces
- M7 I and - M" I. If the position of the end tangents is known,
the polygon is completely determined. If we prolong the
middle side U V to intersections. M and N with the end ver-
ticals [PL 15, Fig. 55], then, by the preceding Art, A M ==
MM , B N = M" - ) ; therefore, A M : B N ; : M/ : M".
\A// \A//
If now w^e draw A B intersecting U V in I, the moment at this
point is zero. That is, the intersection I of the line joining the
supports with the middle side of the polygon is the point of in-
flection of the elastic line.
92. Two §ucce§§ive Unloaded Spans. — Prolong the two.
middle sides U V and Ut Vt [PI. 15, Fig. 56] of the equilibrium
polygon for the two spans Z0 and ^. The point of intersection
W is a point in the resultant of the forces at V and U^ Since
these forces are - JV^ 1G and - M: ^, we have W0 V0 : W0 U0 : :
li : 4- But the horizontal projection of V0 U0 is - (70 + k),
3
therefore that of V0 W0 is - ^ and of U0 W0, - 4 ; while that of
o 3
B W0 is - (li—lo). The vertical through W we have called the
o
limited third vertical. Its position is, as we see, easily found,
and depends simply upon the length of the spans.
Let us now consider more closely the intersections I and L of
the middle sides with the straight line joining the supports.
We have
14:2 CONTINUOUS GIEDEK. [CHAP. IX.
But we have also
UoU^VoV-BUoiBVo:: l,:l^
hence
L U0 : L W0 ; ; I V0 X \ : I W0 x 10.
The ratio of the parts into which W0 U0 is divided by the
point L depends, therefore, for given spans only upon the ratio
of I V0 to I W0, or upon the position of I alone.
If, therefore, we were to draw through the point I another
polygon, the point L would be unchanged, or still more gener-
ally, if the point I for different heights of the supports and
different polygons moves in the same vertical^ the point L will
also move in a vertical.
If the supports are in the same straight line, the points I and
L are the points of inflection of the elastic line. We have
therefore the principle, that if for different polygons the in-
flection point I remains the same, the inflection point L re-
mains also the same.
The point I being given, we can easily construct the point L.
We have only to draw through I at will any line intersecting
the third vertical through V0 and the limited third at, say, V
and W. Through V and the support B draw a line to inter-
section Ui with third vertical through U0. Join now U^ with
W. The line U^ W cuts the line through the supports A and
B in the point L. (See also Art. 86, Fig. 52, d.)
93. The "Fixed Points."— Suppose that, starting from the
left support A [PI. 15, Fig. 57], we have a number of unloaded
spans. The end A then is an inflection point, since the mo-
ment there is zero. Starting from this point, therefore, we can
construct, according to the preceding Art., the inflection point I2
for the next span. Then starting from this we may construct
the point I3 for the third span, and so on. Since these points,
under the assumption that the supports all lie in the same
straight line, do not change their position, whatever may be the
loading of the loaded spans, and whatever spans be loaded, we
call tlwm fixed points.
A second series of fixed points may be in similar manner
constructed, when a number of spans from the right are un-
loaded, so that there are two series of fixed points. In the
end spans the end supports are fixed points.
It follows directly from the construction that the fixed points
are always within the outer third of the span.
CHAP. IX.] LOADED AND UNLOADED SPANS. 143
The construction of the fixed points is the first operation in
the graphical treatment of the continuous girder.
The above construction was first given by Mohr.
91. Shearing Force, Reactions at the Supports, and UIo-
meitts in the Unloaded Spans. — The moments in the unloaded
spans are given, then, by the ordinates to a broken line whose
angles lie in the support verticals, and which, for the case of
supports on a level, passes through the corresponding fixed
points.
It follows directly that the moments at the supports are
alternately positive and negative, and increase from the
end, so that any one is more than three times the preceding.
(See Art. 111.)
Since now this polygon has alternately angles down and up,
the reactions at the supports must be alternately positive and
negative. From the corresponding force polygon it follows
that they must increase from the end.
The shearing forces are, therefore, also alternately positive
and negative, and increase from the end on.
95. Loaded Span. — Let now the span A B [PL 15, Fig. 58]
be arbitrarily loaded. It can be proved here also, as in Art.
92, that the prolongation of the sides XT' V and S U, as also of
V" U" and S V, intersect in the limited third verticals.
When the supports are in a straight line, then, by the con-
struction of Art. 92, the fixed points I and K are the intersec-
tions of S V and S U with A B. We can, therefore, at once
assert, that the sides S U and S V of the second equilibrium
polygon pass through the fixed points I and- K, when the sup-
ports are on a level. ^
For known position of the fixed points and for given load, it
is, therefore, easy to draw the second polygon by drawing ver-
ticals H! and KKt equal to the corresponding ordinates of the
cross-lines. S U and S V pass, then, through I K^ and K It re-
spectively. Then, by Art. 89, the moments at the supports may
be determined.
Since A I < -Z, U must lie to the right of I, and the angle
3
AUS is concave downwards. Accordingly, the force at U,
viz., - M' I, acts upwards. The same holds good for V. Hence,
144: CONTINUOUS GIRDER. [CHAP. IX
the moments M' M" for the loaded span are always posi-
tive* If we draw the lines M N and U V cutting the middle
vertical inP and Q, then, for b = \l, P O = 1 (M' + M") and
o 2
P Q = 2 (W + 2ft")- (See Arts. 89, 90.) Since now the points
U and V must lie under A B,
P O < P Q' or M' -f M'' < 3R'+ 3B".
If the supports are not upon a level, it follows from this Art,
and Art. 92, that the intersections of S U and S V prolonged,
with the prolongations of the lines A A' B B', joining the sup-
ports of two adjacent .spans, lie in the VERTICALS THROUGH THE
FIXED POINTS.
96. Two successive Loaded Spans.— PL 15, Fig. 59.
1. Here also, as in Art. 92, we can prove that the prolongar
tions ofSV and St Uj. intersect in the limited third vertical.
2. Draw through B a line which intersects S V and Sj U^ in
I' and ri? and the verticals through V, W and Ut in V0, W0
and U0.
Then
U0 Ut : V0 V ; ; U0 B : V0 B ; : Z0 : Zx
v0 v : w0 w : : r v0 : r w0.
Hence by composition
U U : W W
: r v0 x 4 : i' w0 x i± ;
or since U0 1^ : W0 W : U0 1\ : W0 1\
UT • W T ' • T' V v 7 • T "W v 7
01* 01 • ~Q A Cfl • * " o "*> *!•
If, then, the point I' moves in a vertical, the ratio I' V0 to
I' W0 does not change, therefore the ratio of U0 1/ to W0 I/ also
remains unchanged, and accordingly I\ must also move in a
vertical. If I' coincides with I, it follows from the construction
of Art. 93 that the point I\ becomes the fixed point 1^ Hence,
the intersections T and I\ of verticals through the fixed points
I and Ii with the sides S V and St U\, or with the middle
sides of the two polygons adjacent to the support, lie always
in a straight line through that support, for any heights of
supports.
* A positive moment always indicates compression in lower flange.
CHAP. IX.] LOADED AND UNLOADED SPANS. 14-5
This property of the second equilibrium polygon was first
made known by Culmann.
97. Arbitrary leading. — According to the above properties
of the second equilibrium polygon, the general course of pro-
cedure for any given case of loading is then as follows [Fig. 60,
PI. 16] :
1. Construct all the fixed points A I2 13 --- I^Ka, etc. (Art.
93), and draw verticals through them.
2. Construct the cross-lines for every span, Art. 88.
3. Make A C equal to Ot Qi as given by the cross-lines, and
draw a line through C and Al to intersection D2, with vertical
through I2. Then make D2 C2 equal to O2 Q2, and draw a line
through C2 and A2 to D3, and so on. Precisely the same con-
struction holds for the other way from the right end. Thus
A4 E4 is equal to R4 P4, etc.
4. In this way we obtain for each of the middle sides of the
second equilibrium polygon two points, C and Fl5 C2 and F2)
etc. ; A and Et, D2 and E2, and so on ; so that now we can ac-
tually draw these middle sides.
The intersections of these lines with the support verticals
give, according to Art. 88, the moments at the supports. For
spans whose length 13 X these moments are given directly; for
other spans the ^ construction of Art. 74 must be applied. The
following simple construction may also be applied. Let
I K be the intersections of the verticals through the fixed
points, with the line A B joining the supports [Fig. 61, PL 16].
Make I D' = I D *, K F' = KF Y, C' D' = O Q -
E'F' = RP (^Y, and draw C' F' and E' D'. These lines cut
\v /
the support verticals in M' and N', so that A M' and B N' are
the moments.
By the construction errors accumulate from one span to the
next, so that the diagram must be made with care. We have
also several checks, viz. : 1. The intersection of the middle sides
must lie in the vertical through the intersection of the cross-
lines. 2. The prolongation of the middle sides must intersect
in the limited third vertical. 3. The corresponding intersec-
tions of the middle sides with the third verticals must lie in a
straight line through the support.
10
146 CONTINUOUS GIRDEK. [CHAP. IX.
If any span is unloaded, the cross-lines coincide. The above
method of construction holds good when the supports are not
upon a level. If the difference of height of the supports is
represented — th of the real amount, the unit for the moment
m
scale is , as is easily seen by reference to Arts. 81
6 m E I
and 88.
The above method of construction of the moments at sup-
ports was first given by Culmann.
CHAP. X.] SPECIAL CASES OF LOADING. 147
CHAPTEE X.
CONTINUOUS GIRDEK - SPECIAL CASES OF LOADING.
98. Total uniform Load. — If a span is loaded with a uni-
formly distributed load of p pounds per unit of length, the
simple moment area is a parabolic segment whose vertical axis
passes through the centre of the span. [PI. 16, Fig. 62.]
1 -.-,.?••,
The ordinate D C" is p P, and hence the area
It will be advantageous here to take p X2 as the unit of the
moment scale ; and therefore
The vertical height of the cross-lines at the pole distance o
from the middle is 8ft -. If, then, we take b = - X, we have
X 6
and therefore
2. Moments.
If the moments A' A" and B' B" are known, we can find
the end tangents of the first equilibrium polygon by drawing
A" B", dropping a vertical through the middle D, and laying
off D E = 2 x \p P = \p P = \p X2 (|y. The lines A" E
and B" E are, then, these end tangents. With the help of
these we may easily construct the parabola.
3. Shearing force.
If we draw in the first force polygon, lines parallel to the
148 CONTINUOUS GIEDEE. [dlAP. X.
end tangents and the closing line A' B', the distances on the
force line are the reactions at the ends of the span. Instead
of this we may lay off from A and B, A G and B H equal to
the first pole distance #, and through G and H draw parallels
to the end tangents, intersecting the verticals through A and B
in A' and B^ We thus obtain the reactions A A1? B Bt. The
ordinates to Ax B1 then give the shearing force at any point.
If the line p X2 representing the moment units is equal to m,
and that representing the force units p\ — n^ then the first
pole distance must be a = r-ra X = — X.
2 m
Accordingly, it is now easy, from the general construction
given in Art. 97, to construct the shearing force and moments
for uniform or dead load of girder in any case. Let us pass
on to an example illustrating more fully the above principles.
99. Example. — As an example of the application of the
above principles, we take a girder of four spans, as given in
PL 17, Fig. 63. The two interior spans are each 96 ft., the
exterior spans 80 ft. each ; that is, l± : I : : 5 : 6. Choose any
scale of length convenient, as, for instance, 50 ft. to an inch,
lay off the spans and construct first the fixed points. For this
purpose we draw the third and limited third verticals. These
last are easily found from the principle already deduced, that
they must divide the distance between the third verticals into
segments inversely as the corresponding spans [Art. 92]. Lay-
ing off. then, from the third vertical in the first span, -J I to the
right, or from the third vertical in the second span -J l± to the
left, we have the first limited third vertical. The same at the
other end gives .the other. For the centre support, of course,
the limited third, since the adjacent spans are equal, passes
through the support itself. We can, therefore, now construct
i\\e fixed points according to Art. 93.
Let the load per unit of length p be -J ton per ft. Then
taking X [Art. 8§] equal to Z, we have n =p\=pl = 48 tons
and m=j>\2=pP= 4608 ft. tons [Art. 98 (3)]. It remains
to assume a scale of force. Let this be 20 tons per inch, then
our moment scale is 20 x 50 = 1000 ft. tons per inch. The
values of which we shall need to make use are, then, to scale
^ = 1.6 inches, X = I — 1.92 inches,
|)2 = 0.6944 inches, (^ = 1.44 inches, (|)* = 0.4823 inches.
CHAP. X.] SPECIAL CASES OF LOADING. 149
These values are repeated upon the PL for convenience of
reference. Also, p I = 48 tons — 2.4 in. = n, p fi = 4608 ft.
tons = 4.608 in. = m. For the first pole distance [Art. 98 (3)]
n 2.4 25
we have a = — X = . -A:> I — -^ I = 1 in. Second pole dis-
m 4.608 48 r
tance [Art. 98] b = ^ X - 0.32 in.
According to Art. 98, we have now, for the cross lines,
:?X»/ and O' P' = Q' P' =
Laying off these distances under the supports, we have thus the
cross-lines.
We have next to construct the second equilibrium polygon.
This, by the aid of the cross-lines and fixed points already con-
structed, we can easily do, as detailed in Art. 97 (3). Then
the moments at the supports are given directly to moment
scale in the interior spans, or we can find them from the end
spans by laying off £ j I [Art. 89].
h
Finally, the moments thus found and laid off at the sup-
ports, we can construct the moment curve by making D' E' —
/ \2 fl\2
r and DE = i^X2- [Art. 98 (2)], and thus draw-
ing the end tangents and corresponding parabolas.
According to Art. 98 (3), we can then find the shearing
A-I 4
forces by laying off a = — A and drawing parallels to the end
tangents to intersection with verticals through supports, as
shown in Fig.
We thus have both moments and shearing forces for uniform
load. By careful attention to the above, the reader will have
no difficulty in solving any case. We recommend him earn-
estly to perform the entire construction for himself, referring
to the proper Arts, at every step. [For convenience of size,
we have not observed our scales strictly in the Figs. TJie
reader should therefore ?wt attempt to check results with the
dividers.'}
1OO. Partial uniform Load. — 1. When the girder is only
partially loaded, as, for instance, a certain portion of the span
id I from B, the simple moment area consists of a triangle ABC
150 CONTINUOUS GIEDEK. [CHAP X.
and a parabolic segment C E B. [PI. 16, Fig. 64.] If in the
first force polygon B' D' is the total load upon the span, the
line O A' parallel to the end tangent A G divides B' D' in the
same ratio as the end of the load divides the span, or denoting
the length B0 C0 by /3 1.
B' A' :B'D ' :;/3l:l.
The intersection G of the end tangents lies in the vertical
G H, which halves B F. Since the triangle B C T is similar to
O A! B', we have
BT:B' A' :;££:#; •
a
or since B' D' =p I, B' A' =p j3 = <pl.
It is therefore easy to construct B T as in Fig. 64, where
/Q
B! D! —p I, A! Bt = B' A' = p I 7, and pole distance = i £ ;
then B2 D2 = B T.
If B T is thus found, we can easily, when A and B are given,
construct the end tangents, and then construct the first equili-
brium curve itself.
2. Make G K — £ GI, then the triangle CKB is equal to
the parabolic area C E B. If, then, through K we draw a
parallel to C B, intersecting C G in L, and through L the ver-
tical LM, the triangle ALB is equal to the entire simple
moment area. This last is therefore proportional to L M, or
2ft I = $ I x L. M ; hence %tt = J L M. It is easily proved
also that F M = i F B. Thus, as GK is £ of GI, GL is
J- of G C ; hence M H is $ of F H, and therefore F M is f of
F H, or | of J- F B = ^ F B. L M can therefore be easily
drawn.
3. Let N be the middle of A B. Make N O equal to $ F N.
Then the centre of gravity of the triangle A C B is in the ver-
tical through O, while the centre of gravity of the parabola is
in H G. If through L we draw a parallel to AB, intersecting
the vertical through C in P, the two areas ^are to each other
as FC to C^P. If in the vertical through 6 we make OQ =
J CP, then, since IH = J CF, we have
IH:OQ::FC:CP.
CHAP. X.] SPECIAL CASES OF" LOADING. 151
The intersection R of the line Q I with A B lies, then, in the
vertical through the centre of gravity of the simple moment
area.
Thus the construction of the cross-lines is now easy.
4. It is most convenient in the application of the above to
construct or calculate the distances of the cross-lines under the
supports once for all, for load over various parts of the span.
The necessary formulae can be directly deduced. Thus the
Triangle B AT = BT x l = ~ X
\$ 1=^ x
Triangle BC T = BT x $ 1= x 01.
Triangle BLT = BTX/3Z =xjS I
6 A (t, 6
The entire area is equal to the triangle ALB.
But A L B = B A T-B L T = 2& (I Z-l ft
2t d \^ O
The triangle A C B = B A T— B C T ; hence
The parabolic area is equal to the entire area minus A C B,
7} BP 1
or parabolic area = TT— *•# P I-
2i Cb O
F is distant from B T by a distance ft I, N by a distance =
\l, NO is \ of N F ; hence O is distant from B T
2i o
Therefore, the moment of the triangular area, with reference
to the right support B, is
The moment of the parabolic area is
The total moment, with reference to the right support, is,
therefore,
152 CONTINUOUS GIRDER. [CHAP. X.
In a similar manner we find for the left support
When ft = 1 the span is completely covered, and we have,
then, right and left
24 a
If we compare this value with those for partial loading, we
see that they differ only by certain coefficients, and that these
coefficients depend only upon the length of the loaded portion.
If, then, we have the distance between the cross-lines for total
load, we have only to multiply by certain factors to obtain the
distances for partial loading. For uniform or total load over
1 / A4
the whole span, this distance is given by -p\2 I - ) (Art. 98).
4: \ A//
If we divide this distance in certain proportions we have at
once the distances for partial loading. These proportions are
given by (2— 0*) $ for the right support, and (2-/3)2 /3 for the
left, under the supposition that .the load comes on from the
right. The reverse is the case for load coming on from left.
~| ~| Q
If we take /3 = -,-,- of the span, we can calculate these pro-
4 2i 4
portions once for all. We thus have the following table :
Support under load Support for wwloaded end
(2-02) /5». (2-p)» p.
- span loaded — — = 0.1211 . . .
i span loaded — |— = 0.4375 .... — = 0.5625.
3 207 225
- span loaded — = 0.8086. ..:-&&** 0.8789.
The division of the distance between the cross-lines for uni-
form load over the whole span into these proportions is easily
accomplished graphically. Thus, from the end of the line to
be divided, draw a line in any direction, and lay off upon it the
six numbers above, to any convenient scale. Join the end of
the last division with the end of the line to be divided, and
then draw parallels through the other points.
It is important to observe some definite system of numera-
tion, otherwise, especially in the first attempt at construction,
confusion is apt to arise.
CHAP. X.] SPECIAL CASES OF LOADING. 153
We can thus find the cross-lines for any position of the load,
and for each position can, if we wish, draw the equilibrium
polygon and determine the moments according to the general
method of Art. 97.
1O1. Concentrated Load. — The simple moment area is in
this case a triangle [PL 16, Fig. 65] whose area is - I h, h be-
2
ing the height C D. Therefore
If from the centre of the span E we lay off E F = - E D, D
o
being the point of application, a vertical through F passes
through the centre of gravity.
As the height C D is proportional to 9Cft, we may take C D as
second force polygon. Since h =. 2 £0?, the distance of the pole
N must be 2 x \ I = i Z, when X = I (Art. 89). Draw N P
6 3
parallel to A B, then is N P = - A B.
3
Parallel to N C and N D we may draw the cross-lines. A
simple construction may be given for them when they are made
to pass through A and B. Let A M and B L be the cross-lines.
Then the triangle S B M is similar to N C D, and
B M : C D ; ; B F : N P. But
and N P = £ A B; therefore,
B M : C D ; ; (2 A B-A D) : A B.
*Make D G = A B, then B G = 2 A B-A D.
The point M is accordingly found by drawing a straight line
through G and C, D G being equal to A B. In the same way
make D H = A B, and draw a line through H and C. We
thus obtain L.
The prolongations ofM.C and I* C, therefore, intersect A B
prolonged in the points G and H, distant from D ~by A B.
If, then, we have to investigate a concentrated load in various
positions, we draw the first equilibrium polygon C X and C Y
[PI. 16, Fig. 66], and lay off in the same the closing lines (for
154 CONTINUOUS GIRDER. [CHAP. X.
the simple moment area) for the different positions of the load.
The distances cut off on the vertical through C by these lines
give, then, the various values of h or 2 %R. If we draw from.
these intersections lines to the pole N, at the distance - I from
3
D C, the cross-lines are parallel to these lines. It will be best
to keep for each pair the common line P Q parallel to N C.
"When the point of application of the load divides the span into
two equal parts, the point of intersection of the cross-lines di-
vides the middle third of P Q into equal parts.
The centre of gravity of the simple moment area cannot pass
beyond the middle third of the span. Since any load can be
considered as made up of a number of concentrated loads, it
follows generally that far any method of loading the centre of
gravity of the simple moment area lies between the third verti-
CHAP. XI.] MAXIMUM STRAINS. 155
CHAPTEE XI.
METHODS OF LOADIN0 CAUSING MAXIMUM STRAINS.
1O2. Maximum Shearing Force— -Uniformly distributed
Jloviiii? Load. — Suppose, first, the span in question loaded
with a concentrated weight. The simple moment area is
A' C' B'. [PL 18, Fig. 67.]
In the force polygon let O A1? O Bx and O Ot be respectively
parallel to C' A', C'B' and A7 B7. Then ^ Aj and C^ are
the reactions at A and B. Since, according to Art. 80, the mo-
ments A A7 and B B' are always positive, and the middle sides
A' S, B7 S pass through the fixed points 1 and K, it follows
from the construction of the preceding Art. that the intersec-
tions O and P of the sides A' C' and B' V of the first equili-
brium polygon with the closing line A B must always lie within
A I and B K. That is, the points of inflection O and K are al-
ways between the fixed points and the ends. Therefore A', B'
and the point C' must lie on opposite sides of the closing line
A B, and consequently Cj. in the force polygon must lie between
A! and B^
Accordingly the shear At C^ at A is positive^ and the shear
G! B1 at B is negative.
Let the distance of the load from the left support be /3, from
the right support &. The load itself is P, and the moment
A A' at the left support M', B B' at the right M". Kequired,
the shearing force S at a point distant x from the left support.
The partial reaction at the left is R'. Then
R' Z = M' - M" +P&,
M'^M"
~~ ~
M' and M" are always positive. If, therefore, M' > M", R'
is positive ; if M'< M", then M" - M'< M" + M7, or, since
by Arts. 75 and 80, M' + M"< 2R' + 3JT, M" - M^aft'+aR".
Now it can be easily proved analytically that for a girder hori-
zontally fixed,* mf = ~ and 2R"= 3 hence 3B' + 3R"
Supplement to Chap. VII., Art. 18.
156 METHODS OF LOADING [CHAP. XI.
since £ + & = J. Therefore M" - M'
Since, however, /8 < Z, we have also M"— M' < P $_. Hence, if
M' is < M", we have also R' positive. R' is therefore always
positive whatever may be the position of the load. In the same
way it may be shown that R" is always negative.
If now the load is to the right of the point distant x from
the left support, then for this point the shearing force S' — R',
and is therefore positive. If the load is to the left of this
point, the shearing force S = R' — P — R", and is therefore
negative. S for any point is therefore positive or negative,
according as the load lies right or left of this point. Hence
for a uniform load we deduce directly —
The shearing force at any point is a positive or negative
maximum when the load extends from this point to the right
or left support respectively.
The same principle holds good for the simple girder.
2. Thus far we have considered the load in the span itself.
Suppose now the load is in some other span, and the span in
question is unloaded, then
M' - M" „ M"-M'"
R =- -IT R = r •
As we pass away from the loaded span the moments at the
supports are alternately positive and negative, and each is
greater than the one following (Art. 94). Since the moments
M' and M" are alternately positive and negative, R' will have
the same sign as M', and R" as M", and generally R^ as Mm.
Adopting, then, the. notation shown in PL 18, Fig. 68, we
have for the span lm_±
.t - Mm
where Rra has the same sign as Mm.
In the same way
and therefore
1 _
CIIA1'. XI.] CAUSING MAXIMUM STRAINS. 157
M
But rc+ is negative and greater than 2 (Art. 94-). There-
fore in the preceding expression the numerator is positive and
ivi 1
> 3. Further, -«^— L is negative and less than -, hence the de-
1V1m
• 3
nominator of the above expression is negative and < -. There-
R I
fore R+1 is negative and > 2 -y1^, that is, the shearing forces
at the supports are alternately positive and negative, and in-
crease (when 2 lm_± is not less than Zm) towards the loaded span.
We have then
For any span, then, the shear at the left support R' will be
positive when the left adjacent span is loaded, the right adja-
cent span unloaded, and all the other spans each way alternately
loaded. The shear R' will be negative when the remaining
spans are loaded. Hence :
The shearing force is a maximum (positive) at any point
when the load extends from this point to the right support, and
the other spans are alternately loaded, the adjacent span to the
right ~being unloaded, that to the left, loaded. The negative
maximum, on the contrary, occurs when the load extends from,
the point to the left support, when the right adjacent span is
loaded and the left unloaded ; the other spans alternately
loaded.
PL 18, Fig. 69, gives these two cases.
In practice such a loading can never occur. If we suppose
the rolling load divided into two portions only, the above rule
reads as follows :
The shearing force at any point will be a positive maximum
when the load reaches from the right support to this point, and
when the left adjacent span is covered. The negative maximum
occurs when the load reaches from left support to the point,
and the right adjacent span is covered,.
1O3. Maximum Moments. — 1st. Loaded Span. Let a weight
act at the point D, PL 18, Fig. 67. Then A U S V B is the
second. A' C' B' the first equilibrium polygon, and A A', B B'
158 METHODS OF LOADING- [CHAP. XI.
are the moments at the supports. We have already seen that
the inflection points O and P (for which the moment is zero)
lie outside of the fixed points. We can therefore assert that
WITHIN the fixed points the 'moments are negative wherever the
weight may Replaced. From the Fig. we see at once that the
inflection points O and P move to the right or left as the weight
moves to the right or left. Accordingly when for the weight
at D the moment at O is zero, the moment at this point will be
positive when the load moves to the right of D, negative when
it moves to the left of D.
Hence for the maximum moment we have at once the follow-
ing principle :
For any point O outside the fixed points the moment will ~be
a positive or negative maximum when the load reaches from
the point D, where a load must Replaced to cause the moment
at O to ~be zero, to the right or left support respectively. For
the negative maximum, therefore, the load reaches from A to D;
for the positive, from D to B.
If the point O is given, it is indeed possible to determine by
construction the point D to which the load must reach. It is,
however, simpler to assume D and then construct O.
If we choose for the different positions of D an arbitrary
length for C' D' (Fig. 67), so that the point C' falls in a parallel
Q R to A' B' (Fig. 70), and, moreover, take D at equal intervals,
then the points L and M will be at equal distances (Fig. 67),
and hence the points I and K (Fig. 70), in which the verticals
through the fixed points are intersected by the lines A' M and
B' L (Fig. 67), will be at equal distances. We have, then, the
following simple construction [Fig. 70, PL 18] :
Between the verticals through the supports draw two parallel
lines A B and Q R at any convenient distance apart, and divide
Q R into a number of equal parts ; four or five are sufficient.
Draw lines from A to R and the middle S intersecting the ver-
tical through the fixed point I in Ix and I2. In the same way find
K! and K2. Divide It I2 and Kt K2 into the same number of
equal parts as Q R has been divided into, and join these points
in reverse order by lines. The intersection of these lines with
the lines drawn from A and B to the points upon Q R give the
points O, for which the moment is a maximum when the load
is limited by the corresponding point upon Q R.
This construction was first given by Mohr.
CHAP. XI.] CAUSING MAXIMUM STRAINS. 159
IO 1. Determination of the maximum Shearing Forces.
— According to the general method of construction given in
Art. 97, we can now determine by reference to Arts. 98 and 99,
which treat of total and partial distributed loading, the shear-
ing forces corresponding to the methods of loading which cause
maximum strains.
As a review of the preceding principles, we take the same
example as before, as given in PL 19, Fig. 71. Here again the
reader should construct the Figs, for himself. The scales are
as before, Art. 99.
Fig. a shows the method of loading for positive maximum
shear in first span ; and the second Fig. below, the same for
the second span. [Art. 102.] ,
We first find, precisely as in Fig. 63, the shearing forces in
the third and fourth spans for the total loads over those spans,
and lay off the shear thus obtained in the first and second spans,
as indicated by the broken lines in Fig. b in those spans. Thus
having first found the fixed points, which we may here take
directly from Fig. 63, we construct as in that Fig. also the cross-
lines for total load in third and fourth spans. Thus laying off
54 equal to Oj. P1? and drawing a line through 4 and support to
intersection with vertical through fixed point in second span
[Fig. 60], we determine D', and then from the cross-lines find
at once D". In like, manner, supposing for the moment the
load on the other two spans, we have e a, a D, D a and a F',
and then at once F''. F" D' cuts off then the moment at the
right support, and joining 1, 2 and 5, we find, according to Art.
98 (2), precisely as in Fig. 63, the end tangents ; and then from
these, with the first pole distance a, find the shear. This is
given by the broken lines in third and fourth spans. Lay off
these lines in first and second spans, remembering, since the
shears at the supports alternate, that the positive shear at left
of fourth span must be laid off as negative (down) at right end
of first, etc. [See Fig. 63.]
Now for the positive shears in first and second spans for the
different methods of loading, we have only to determine the
direction of the tangents through end of load [Fig. 64, Art.
100]. The lengths of the segments cut off upon the verticals
through the supports by these tangents are given for first and
second spans by Figs, /and e, for each position of load. [Art.
100 (1).]
160 METHODS OF LOADING [dlAP. XI.
We have first, then, to construct the cross-lines [Art. 100 (4)]
as shown in Fig. d.
. Take first the second span. For this span, as shown in Fig.
dj the first span is fully loaded, as also the last. Make, there-
fore, e a = Oi P1? draw a D, and we thus find the point D, com-
mon to all the middle sides of the second equilibrium polygon
for different positions of load. So also make on right 54 —
Ot P! and draw 4 D', and then, since third span is empty, D' F,
and we thus have P. Now from D and F lay off the cross-
line distances, as "D a, D 5, D <?, equal to e a, e b, e <?, etc. Draw
lines through these points and D and F respectively, and note
their points of intersection abed with the verticals through
the supports. [NOTE. Be careful to preserve an orderly nota-
tion.] These points give the moments for each position of
load in second span. Take the length a a from Fig. e and lay
it off from a on the right support vertical, and join the end
with a on left. This is the tangent for full load in second
span. A parallel to it at distance a from left gives the shear
in Fig. b. Then lay off b b taken from Fig. e on right, and join
with b on left. This is tangent for load over three-fourths
second span from right. A parallel to it at distance a from
foot of perpendicular one-fourth of span from left cuts off
shear for this position of load. So for tangents c c, d d. We
thus obtain the curve for positive shears in second span. The
negative shears are obtained by subtracting these from the
shear already found for full load. We thus have the lower
curve, and the shear diagram for second span is complete.
For first span, only the third is loaded. We lay off, then,
78 equal to the distance between cross-lines corresponding,
draw 8 F1? and thus find Ft. Lay off now at left end ed, ec,
e by draw lines from these points through second support, and
note intersections with vertical through D. Through each of
these intersections draw a line through F1? and produce to
intersections abed with vertical through second support. It
is from these last points that the distances a a, bb, etc., taken
from Fig. f must be laid off respectively in order to find the
tangents e a, e b, e c, e d, e e.
Parallels to these tangents above in Fig. b give, as before,
the positive shear for each position of load. The negative
shear is, as before, found by subtracting the positive from total
load shear. Thus shear diagram for first span is complete.
CHAP. XI.] CAUSING MAXIMUM STRAINS. 161
Of course, the same circumstances can hold good for third
and fourth spans as for first and second, except that positive
shears on one side of centre support are the corresponding
negative shears upon the other, and vice versa.
Following carefully the above with the aid of the Fig., the
reader cannot fail to grasp the method. An independent con-
struction for a similar case will make both principles and de-
tails familiar. Once thoroughly understood, the method is
rapid, accurate, simple, and of general application.
1O5. ]>etcr 111 illation of the Maximum Moments. — In
like manner, it is easy, according to the general construction
given in Art. 97, and referring to Arts. 98 and 99, to determine
the maximum moments. In Fig. 72, PL 20, we have the same
example as before, concerning which we have but little addi-
tional to remark. Fig. 64, Art. 100, shows that the end tan-
gents give the moments within the unloaded portion of the
girder. These tangents are constructed precisely as before in
the several spans, except it will be noticed that in the first
span we have made use of the construction given in Art. 97,
Fig. 61. Thus the point F' is determined so that K F' =
(A2
j-] , and thus the moments are measured directly at the
end vertical. Also upon the left support vertical we have laid
off the distances between the cross-lines in the second span
(7 \2
7/ '
The only thing new in the PL is Fig. c, which, as we have
seen in Art. 103, Fig. 70, gives the points at which the positive
moment is a maximum for each position of load.. The positive
moments can then be taken directty off upon the verticals
through these points, and are limited by the horizontal through
the supports and the tangents, as above.
Thus, at second support the vertical distance to a, for first
span, gives the moment at the support. La}r it off in Fig. a
from the support line. For a load over f span, wre see at once
the point for which the positive moment is a maximum from
Fig. c. Follow up the vertical through this point. The dis-
tance on this vertical in Fig. J, between the support line and
tangent b b^ gives the moment to be laid off in Fig. a upon
this vertical. So, for load over £ span, we have next vertical
and tangent cc^ and so on. We thus obtain the curve for
11
162 METHODS OF LOADING [CHAP. XI.
•
positive moments at the right end of first span, aticde. From
e draw a line to left support. At the other supports, in like
manner, we determine the 'positive moments, and join the
points ee in second and third spans, and e and right support
in fourth.
We have already seen (Art. 103) that within the fixed points
the moments are negative wherever a load may be placed.
The maximum, therefore, occurs for full load. We have, there-
fore, found for 3d and 4th spans the parabola for full load,
precisely as in Fig. 63. These parabolas are given in broken
lines in the Fig. a. By subtraction of the positive moments
outside the fixed points from the positive moments at the same
points for total load given by these parabolas, we obtain directly
the lower curves as far as the points e in each span.
The second parabola (partly full, partly broken) is all that is
needed to complete our Fig. To obtain this we have simply,
in 2d and 3d spans, to make the vertical through centre of
line e e equal to its length already laid off for total load, viz. :
Z\2
-J, Art. 98 (2), produce e e to intersections with sup-
port verticals, and join these intersections with the extremities
of the first verticals above. We can then construct the para-
bola, which completes our diagram, and gives us Fig. a.
1O6. Practical simplifications. — In practice, the construc-
tions given in Figs. 63, 71 and 72, admit of many simplifica-
tions, which, in order to avoid confusion at first, have been di&
regarded. The whole solution, given for the sake of clearness
in three separate Plates, can be performed upon a single sheet,
since the Fig. for the second equilibrium polygon in Figs. 63,
71 and 72 may be combined in one. Indeed, the lines neces-
sary for the construction of the maximum shearing forces can
be applied directly to the determination of the maximum
moments. It is therefore unnecessary to divide the construc-
tion into separate sheets.
2. The cross-lines in the end spans can be omitted, since
all that is required are the distances to be laid off upon the
end verticals, and these when found can be laid off at once.
3. We can apply the second equilibrium polygon directly
in order to find the moments for the dead and moving load.
Thus the transferring of ordinates from one Fig. to another is
avoided.
CHAP. XI.] CAUSING MAXIMUM STRAINS. 163
4. It is evidently unnecessary to actually draw all the vari-
ous lines. We need only to mark the different points of inter-
section.
5. The construction for dead and live loads can be per-
formed at once, thus avoiding the necessity of a subsequent
addition.
1O7. Approximate Practical Constructions — If the suc-
cessive steps of the preceding development are carefully fol-
lowed, the method will be found simple and easy of appli-
cation. Indeed, the complete and accurate solution of the
difficult problem of the continuous girder by a method purely
graphical, is the most important extension of the system since
the date of Culmann's treatise, and well illustrates the power
and practical value of the Graphical Method.
Humber gives the following constructions, " which may be
relied upon for safety without extravagance." * As rapid
means of obtaining approximate results, they may not be with-
out value to the practical engineer, and we therefore append
them here. It must be remembered that the constructions
hold good ONLY for end spans three-fourths the length of the
others.
I. Beam of Uniform Strength, continuous over one
Pier, forming two equal Spans, subject to a fixed Load
uniformly distributed, and also to a moving 'Load.
Maximum Moments.* PI. 18, Fig. 73.
The greatest moment at the pier (positive) will be when both
spans are fully loaded.
The greatest negative moment will obtain in the loaded span
when the other span bears only the fixed load.
(A moment is positive when the upper fibres or flanges are
extended, negative when the upper flange is compressed.)
Construction. — Let A B C be the beam. On A B draw the
parabola whose centre ordinate D E is (p + m) ^-, and on
p 1?
B C the parabola whose centre ordinate G F is "—-.
i • *
At the pier B erect the perpendicular B H = — — ^ — —-.,
and make B L = ^ 0 - — . Join AH, A L, and L C.
\A
* " Strains in Girders, calculated by Formulas and Diagrams." Humber.
New York : D. Van Nostrand, publisher.
164: METHODS OF LOADING [CHAP. XI.
Then the vertical distances between the parabolic arc A E B
and the lines A H and A L, the greatest being taken, will give
the maximum moments,— positive in the first case and negative
in the last. The points of inflection approach as near the pier
as K and recede as far as M.
If 4r~ is less than — — , the beam must be latched
2i \.2i
down at the abutments. The load comes on from the left;
p and m are the loads per unit of length of the permanent or
fixed and the moving or live loads.
Shearing Forces (PI. 18, Fig. 74). — The maximum shearing
force at either abutment will obtain when its span only sus-
tains the moving load. The maximum shear at the centre pier
will obtain when both spans are fully loaded.
Construction. — Lay off AC = (4:p + 5m) and A D =
o (p + m). At B lay off B P = twice A D. Take a point
M distant -J I from A, and join D and F to M. Draw C N
parallel to D M. Sketch in a curve similar to that dotted in
the figure, giving an additional depth to the ordinates at the
point of minimum shear of m -. Then the vertical ordinates
between AB and COPF may be considered to give the
maximum shearing force for either span.
II. Beam as above — continuous over three or more
Pier§. li = end spans. I = the other spans.
Moments.
The maximum moment (positive) will obtain when only the
two adjacent spans, and every alternate span from them, are
simultaneously loaded with the total load — the remaining spans
sustaining only the fixed load.
The maximum moment at the centre of any span will obtain
when it and the alternate spans from it are fully loaded — the
remaining spans sustaining only the fixed load.
Construction (PL 18, Fig. 75). — Let A B C be part of the
beam. On B C draw the parabola whose centre ordinate E F
= — ~ , and on A B the parabola whose centre ordi-
nate C D = (P + ™Wt At B and c make B H _ c L =
8
CHAP. XI.] CAUSING MAXIMUM STRAINS. 165
Join A Hand I, Make B G = C S = *l.
o
Join A G and S. The maximum vertical ordinates between
the two parabolas and the lines A H, A G, H L, and G S, as
shown in the figure, give the maximum moments.
The points I, K and O, P or M, N show the limits of devia-
tion of the points of inflection.
If -^— is less than - — - , the beam will require to be
2i oA
held down at the abutments.
If the beam be continuous for three spans only, I in the
expression for B H = — ( —=- + — ) must have a value given to
Shearing Forces.
The maximum shear at any pier (B or C) will obtain simul-
taneously with the maximum moment over that pier.
Construction.— PL 18, Fig. 76.
Let A B C be part of the beam. First, for any inner span as
Z__. At B and C erect B G = C H = i&^J&A. Make B D
and C E each = - (p + m). Join D and E to midspan F, and
2
draw G K and H K parallel to D F and F E respectively.
Second, for either end span as \ — . At B erect a perpen-
2 7 3
dicular = — - (p+m), which, if ^ = - I, will coincide with B D.
3 4:
At A make AL = - B D. Join D and L to M distant - \ from A.
2 o
Make A O =| (p + m)- ^-2, and draw ON parallel to L M.
2i ' 32 1-± •
Sketch in curves as shown by the dotted curves in the figure,
giving additional depth to the ordinates there of —^ and --ir-
respectively. Then the vertical distances between O a b D and
A B give the maximum shearing forces for either end span, and
those between G c d H and B C, the shearing forces for the re-
maining spans.
166 METHODS OF LOADING [CHAP. XI.
If the beam be continuous for three spans only, B G and C H
must be made equal to (18 * + 16 "*>*+" (2-# + ?), where
oz o I \ 7 2i '
L = —~. Further, the value given to B D for the inner span
3
1O8. JUethod by Resolution of Forces— Draw Spans.—
The most usual cases of continuous girders which occur in prac-
tice are draw or pivot spans, which when shut must be consid-
ered as continuous girders of two spans. The graphical method
becomes for such cases short and easy of application. In the
case of framed structures of this character, it may, however, be
more satisfactory to first find the maximum shearing forces
(Art. 104), and then follow the reactions thus obtained through
the structure from end to end by the method of Arts. 8-13.
As a check upon the accuracy of the work, we may apply the
" method of sections " referred to in Art. 14. In either case
we must, of course, start from an end support where only two
pieces intersect and the moment is zero.
Still again, we may find the reactions by calculation, and
then apply the method of Arts. 8-13. In the case of two spans
only, the formulae for the reactions are sufficiently simple, and
the ready and accurate determination of the strains offers, there-
fore, no difficulty.
We shall give here, therefore, the analytical formulae requi-
site for our purpose, referring the reader to treatises upon the
subject for their demonstration.*
Formulae for Reactions. — Continuous Girder of two un-
equal /Spans, I and n I. 1st. Concentrated weight P, in first
span Z, distant /3 from left end support. Reaction at left end
support :
P
Reaction at middle support :
* Bresse — La Flexion et la Resistance, and Cours de Mecanique Appliquee.
Weyrauch — Theorie der Trdger. CoUignon — Theorie Elementaire des Poutres
Droites, etc. Also Supplement to Chap. XIII.
CHAP. XI.] CAUSING MAXIMUM STRAINS. 167
Reaction at right support :
0= _P f-^4-^1
2tt(l + 7l) L J "*>_!•
2d. Uniformly distributed load extending to a distance /3
from left support. Load per unit of length =p.
o=
For two equal spans we have only to make n = 1 in the
above equations. For a uniform load over whole span /3 = L
From the above formulae we can find the reactions for any
case, and then proceed as indicated above.
109. By means of the graphical method, as we have now
seen, we are enabled to solve completely the problem of the
continuous girder, and that too without the aid of analytical
formulae, tables, or tedious computation. The method can also
be applied to continuous girders of variable cross-section, or of
uniform strength. We shall not, however, proceed further
with the development of the method in this direction. The
preceding will, we think, be found to contain all that is practi-
cally serviceable. For the application of the method to girders
of variable cross-section, we refer the reader to Winlder — " Der
Bruckenbau" Wien, 1873 — where will be found a thorough
presentation of the subject, both analytically and graphically,
to which we are greatly indebted in the preparation of the
preceding pages. Plates 17, 19 and 20, are, with but few alter-
ations, reproduced from that work.
* These formal EG are demonstrated in Van Nostranffs Eng. Mag., July,
1875.
168 GRAPHIC AND ANALYTIC [CHAP. XH.
CHAPTEE XII.
CONTINUOUS GIRDER (CONTINUED) COMBINATION OF GRAPHICAL
AND ANALYTICAL METHODS.
1 1O. In the present chapter we shall develop a method for
the solution of continuous girders not purely graphical, but
based upon the method of resolution of forces illustrated in
Arts. 8-13, together with well-known analytical results, which
method for accuracy, simplicity, and ease of application will,
we think, be found superior to any hitherto proposed. The
method is, of course, applicable only to framed structures, but
for such cases is the most satisfactory of any with which we are
acquainted.
111. The Inflection Points being known, the Shearing
Forces and Moments at the Support* can, by a simple
construction, be easily determined. — ~Lst. Loaded Span —
Fig. 77, PL 21.— Thus in the span B C = I, let the distance of
the weight P from the left support be a, and let i and i' be the
distances of the inflection points from B and C respectively.
Then if through any point P of the weight we draw lines, as
P D, P E, through i and i', intersecting the verticals at B and C
in the points D and E, the vertical ordinates between these lines
and B C will be proportional to the moments. For, as we see
from the force polygon, the equilibrium polygon must consist
of two lines as D P, P E, parallel to O 0 and O 1, and because
of the moments at the ends, the closing line D E is shifted to
B C (Art. 23). Since the moments at the points of inflection
are zero^ the ordinates to P D and P E to pole distance H will
give the moments. Now the points of inflection being known,
and P J3 and P E drawn, we can easily find the pole distance
H and the shearing forces L 0 and 1 L by laying off P verti-
cally, and drawing from its extremities lines parallel to P D
and P E intersecting in O. A perpendicular through O upon
0 1 gives H and the reactions L 0 and 1 L. In other words,
we have simply to decompose P along P D and P E.
CHAP. XII.] METHODS COMBINED. 169
The construction, then, is simply as follows : Take any point
on the direction of P, and draw P D, P E through the points
of inflection. Lay oft P to the scale of force as A P, and draw
A O parallel to P E or P D. We have thus the pole distance
H, and the shearing forces P H and H A at B and C.
B D or C E to the scale of distance, multiplied by H to the
scale of force, give the moments at B and C. That is, D and
E may be regarded as the points of application for H. The
forces along P D and P E considered as acting at these points
are held in equilibrium by the reactions P H and H A = L 0
and 1 L and H. Since H, acting as indicated in the figure
with the lever arm B D or C E, causes tension in the upper
fibres, the moments at B and C are positive.
2d. Unloaded Span—Fig. 78, PL 21.
As we have already seen in Art. 93, the inflection points in
the unloaded spans are independent of the load, and are found
by the simple construction there given for the "fixed points"
Since, each fixed point lies within the outer third of the span,
we have in Fig. 78 the broken line a b c, referred to in Art. 94,
where the moments are alternately positive and negative, and
increase from the end, so that any one is more than twice the
preceding. Lines drawn parallel to these lines in the force
polygon, cut off from the force line the reactions at the sup-
ports. Thus, in • Fig. 78, c b in the force polygon gives the
reaction at D, a b the reaction at C, and if B were an end sup-
port— that is, if b a went through B — a H would be the re
action at B. For the resultant shear at D, we should then have
aH — ab+-cb = Hc. So for any number of spans ; the in-
flection points in the loaded span being known, we can easily
find the fixed or inflection points in the other spans, which are
independent of the load, and depend only on the length of
these spans. Then draw the broken line a b c P d. Then find
the pole distance H by laying off c P = P to scale, and draw-
ing c O parallel to c P, and through the point O thus deter-
mined drawing H O. Then find the reactions at the other
supports, or the shear at any support, by lines in the force
polygon parallel to a b, b c, etc. Thus the shear at B is the
distance H a cut off by H and O a parallel to a b. Since the
shear at D is plus and alternates from D, we have at B the
shear + H a. The shear at C is — H b ; at D, H- H c, etc. O c
being parallel to PC; O a, to b a ; O b, to c b, etc.
170 GRAPHIC AND ANALYTIC [CHAP. XII. *
112. Inflection Verticals, — Draw a line from P through
the support D, and through its intersection with cb draw a
vertical 3 (Fig. 78). This vertical we call the inflection ver-
tical.
The equation of the line c ft is
where ml = D C, n I = C D, i = Ci'. The origin being at D.
For the line C,
where ^ = D i±.
If in this last equation we make x = a, we have for the or-
dinate at P,
1ft! fa — a)
h
and hence for the line P D,
mt (*, — a)
y = — *; - -x.
For the intersection of P D with b c then
m± m± (^ — a)
— ^ - . x + m± = — ^ - - x.
nl — v i^a
Hence
- /. \ / 7 -\ • «...
(^ — a) (n I — i) — \ a
We see at once that the value of x is independent of ml or
B C, hence the intersection of P D and c b lies always in the
same vertical, whatever be the position ofPC. In other words,
if the three sides of a triangle pass always through three fixed
points (a1, D, ^), and two of the angles (P and c) be always in
the same verticals, the third angle must also always lie in the
same vertical.
For the distance of the inflection vertical on the other side
of the loaded span (beyond E), we have similarly
(I — a) («B + 4)
where I is the loaded span and iB the distance of the inflection
point to the right of E.
Equations (1) and (2) give the distances of the inflection ver-
ticals from the supports D and E.
CHAP. XII.] METHODS COMBINED. 171
113. Beam fixed horizontally at both ends— Supports
011 level. — Consider the span D E (Fig. 78) as fixed at the sup-
ports so that the tangent to the deflection curves at D and E is
always horizontal. Conceive the span prolonged right and left
beyond the supports a distance equal to the span I. It is re-
quired to find the position of the inflection verticals.
From equation (1) of the preceding Art. we have, since
n = 1, i = 0,
ii a \
t/U •
""• 7 -w • •
tj — a) l — ^la/
and from equation (2), since is — I,
i% I (I — a)
x =
Now for a beam fixed at the ends the distances of the points
of inflection are
a I I (I — a)
Substituting these values in the equations above, we have
x = — - and x = 4- o- That is, the position of the inflection
6 o
verticals is in this case independent of the load, and always
equal to ~ from the supports*
o
This remarkable property of the beam fixed at both ends
enables us to find the inflection points by a construction similar
to that for the fixed points in the unloaded spans, as given in
Art. 86.
Thus we have simply to draw from C distant I from A (PI.
21, Fig. 79) a line in any convenient direction, as C ~b intersect-
ing the inflection vertical I, which is distant from A, - Z, at a.
Through a and the fixed end support A draw a line to inter-
section with the weight P. Then draw P b. The intersection
it of this last line with A B is the inflection point. A similar
construction gives i%.
We can now find the reactions and moments. Thus H O to
* This important result, which renders possible a complete graphical solution
of this case, has, so far as we are aware, never before been published.
172 GRAPHIC AND ANALYTIC [CHAP. XII.
the scale of force, multiplied by A b to the scale of distance,
gives the moment at A, while H G is the reaction at A (Fig. 79).
114. Beam fixed at both ends— Example. — Since when
the points of inflection are once determined, we may draw P b
or P c at any inclination (Fig. 79), provided we afterwards find
the corresponding pole distance HO; if A b or B c be made
equal to the height of the truss, H O will ~be the strain in the
upper or lower flange at the wall (the flange in question being
always that for which there is no diagonal at its union with the
wall). Thus in PL 21, Fig. 80, we lay off D E = I, draw the
vertical I at ~ I from D, and for the given position of the load
o
P find the inflection point 4 by the preceding Art. A similar
construction on the other side gives %. Now laying off P M
equal by scale to the weight P, and decomposing it along P D
and P C, we find O H the pole distance which to the scale of
force will give directly the strain in the lower flange B m at
the wall, provided P D is made to pass through the intersection
of the upper flange with the wall. If the triangulation were
reversed, O H would be the strain in the upper flange at the
wall. In any case it is the strain in that flange at whose junc-
tion with the wall there is no diagonal.
The reaction at D is also H M, at C it is P H. Lay off then
B B' in Fig. 80 (a) equal to P M, and make B A = H M and
A B' = P H. Now draw m B parallel to O H and m A paral-
lel to O M, and produce both lines to intersection at m. Then
m B to scale of force is evidently the strain in the lower end
flange at the wall. We assume the following notation.*
Let A represent all the space above the girder, B all the space
below, and a h c d7 etc., the spaces within the girder included
by the flanges and diagonals. Then, for instance, A b is the first
upper flange at the left, B a the first below ; a b the first diag-
onal at the left, and so on.
Now draw in Fig. 80 (#), m I and A I parallel to the corre-
sponding lines in the frame, and we have at once the strains in
these pieces to scale. Following round the triangle according
to our rule (Arts. 8-13) from m to A, A to I and I «to m, we
* See an excellent little treatise on " Economics of Construction in Relation
to Framed Structures," by R. H. Bow, to whom this method of notation is
due.
CHAP. XII.] METHODS COMBINED. 173
find A I tension and I m tension. [The strains in the upper
flanges must always be tension, since the moments at the sup-
ports for loaded span are always positive.] Moreover, the Fig.
thus far shows that A I, I m and m B are in equilibrium with
the shear B A = H M, as evidently should be the case ; hence
the strain in B m is compressive.
We have thus the strains in the three pieces at the right, and
can proceed from these to find all the others. Thus the strains
in B k and k I are in equilibrium with m I and B m. Lines
parallel to B It. and k I, therefore, which close the polygon com-
menced by B m and m I, give us the strains in B k and k I.
Observe that the line I k crosses A B, thereby making B k op-
posite in direction, consequently in strain from B m. This may
also be seen by following round the triangle m 1 k B, remem-
bering that, as m I is always found to be in tension, it must act
away from the new apex, that is, from m to I. We thus find
k I in compression, and k B acting away from this apex, or in
tension, therefore of opposite strain from the preceding flange
B m, which, as we have seen, is in compression. The reason is
obvious. The inflection point i% falls in the flange B k. If the
beam wrere solid, the strain at i2 would be zero ; to the right of
4 we should have compression, to the left, tension. In the
framed structure the strains can only change at the vertices.
The crossing of A B by I k indicates such change, and B k gives
its amount by scale.
Now taking the upper apex, we have here A I and I k in
equilibrium with k h and A h. As we already know, k I is in
compression. We must, therefore, now take it acting from k to
I, and following round the triangle we find A h compression,
and h k tension. From h on, the traverses between A h and
B k produced towards the right [Fig. 80 (&)] will give the di-
agonals, while the upper and lower flanges will be given by the
distances to them from A and B respectively, until we arrive
at the weight P. Observe the influence of the weight. We
have k h and B k in equilibrium with h g and g B, and also the
weight P = B' B. We must take, therefore, A B' = P H, and
then draw h g and B' g. Distances to the right of B' along
B'<7 are compressive lower flanges, to the left, tensile; while to
the right of A we have compressive upper, and to the left of A
tensile upper flanges. The two diagonals at the weight k h and
h g are in tension. From h g on, the diagonals are alternately
174 GRAPHIC AND ANALYTIC [CHAP XII.
tension and compression. Moreover, the diagonal e d passes
through A, that is A d is zero. The weight P causes no strain
in A d, and for this one position of P, A d might be omitted
from the structure. The reason is again obvious. The point
of inflection i± coincides with the apex under A d. Since at
^ the moment of rupture is zero, if the flange A.~d were cut
there would be no tendency to motion. We have at ?'t the
shearing force only, A B' giving the strains in the diagonal
e d and d c. The upper flange A £', we see again, is in tension,
which is also shown by its lying to the left of A.
Thus we have the strains in every piece by a very simple
construction for any position of P, without any calculation what-
ever. The method in this case is purely graphical. We have
only to find the points of inflection and then proceed as above.
Fig. 80 (b) gives the strains for the same girder and position
of weight P, merely supported at the ends. For this case P D
in Fig. 80 not only passes through D, but P O also passes through
the upper left-hand corner at C. Hence A B will be less than
H M, and A B' greater than P H. Moreover, the end lower
flanges B a and B m no longer act, and must be removed.
Starting now with the reaction B' A [Fig. 80 (£)], we go along
to the weight, from which point at h k we go ~back towards the
force line, and the reactions are such that the last diagonal
must pass exactly through B, just as in Fig. (a) e d passed
through A, because the points of inflection or zero moments are
now at the ends C and D. A careful comparison and study of
the two cases and their points of difference will be advanta-
geous to the reader.
115. Counterbracing. — The objection may arise that the
above method applies only to a system of bracing such as rep-
resented in the Fig., where the diagonals take both compres-
si ve and tensile strains. In case, as in the Howe or Pratt Truss,
for instance, we had vertical pieces as also two diagonals in
each panel, then the strain in any diagonal as m I and flange as
B m, even if found, are apparently in equilibrium with three
pieces, viz., Jcl,~Bk and a vertical strut or tie at the intersec-
tion of these pieces. Hence, having only two known strains
and three to be determined, the method would seem to fail, as
any number of polygons may be constructed with sides par-
allel to the forces, and hence the problem is indeterminate
(Art. 9).
CHAP. XII.] METHODS COMBINED. 175
jSTow in any framed structure of the above kind, the counter
ties are inserted to prevent the deforming action of the Tolling
load only. For the dead load but one system of triangulation
is required, and the strains in every piece due to this dead load
can therefore easily be determined.
We have then only to determine the strains in the same pieces
due to the rolling load also. If now in any diagonal the strain
due to this rolling load exceeds the constant strain due to the
O
dead load, and is of opposite character, and if the diagonal is
to be so constructed as to take biit one kind of strain, then a
counter diagonal must be inserted in that panel, and propor-
tioned to this excess of strain only. For instance, if a diagonal
takes only the compressive strain (a condition which is easily
secured in practice) due to the dead load, and the live load
would cause in that diagonal a tensile strain, then the excess of
this tensile strain over the constant compressive strain due to
the dead load must be resisted by a counter diagonal, which
also takes compressive strain only. The method is precisely
the same as by calculation (see Stoney and other authors 011
the subject), and we only notice the point here, as in all our
examples we have taken a single system of triangulation only —
a system which, we may here remark in passing, has many ad-
vantages, and is worthy of more general attention * than it has
hitherto obtained.
[See also on this point Art. 10 of Appendix.]
116. Beam fixed horizontally at one end, supported at
the other— Supports on Level. — In this case, equation (1),
Art. 112, becomes for left end fixed, since n = 1, i% = 0, i — 0,
i-i a I
(^ — a) l — ^\ a
But for this case the distance of the point of inflection from
the fixed end is
. _ (2 I— a) la ,
* See " A Treatise on Bracing." By R. H. Bow. D. Van Nostrand, pub-
lisher.
f The values of the distance of the inflection points which we assume above
as known, may easily be deduced by the theory of elasticity. See Supplement
to Chap. VTL, Arts. 16 and 19. See Wood, Strength of Materials ; Bresse,
Mecanique Appliquce ; or other treatises on the subject.
176 GRAPHIC AND ANALYTIC [CHAP. XII.
Inserting this value of ^ in the value for x above, we have
x - _ l (2 t-a)
5 I— a
for the distance of the inflection vertical to the left of the left-
hand support, which is supposed fixed.
Now this is the equation of an hyperbola, as shown in PL 21,
Fig. 81, whose vertex is at/J the distance Ay being 2 I, whose
assymptotes are respectively parallel and perpendicular to the
span, the perpendicular distance of E above the span A B be-
5 9
ing 1 I, and which intersects A B at ^ I from A. The ordinate
2i 5
d e, A d being equal to Z, is - 1. The diameter passes through
3
E and f, and E f is, therefore, the semi-transverse axis. The
hyperbola can, therefore, be easily constructed. We need only
to construct that portion between A B and the point e.
The construction for the point of inflection i± is, therefore,
simply as follows :
Lay off A k vertically upwards and equal to the distance of
the weight P from A, and draw the horizontal h b to intersection
b with the curve. Now make A a = I and draw a b to inter-
section c. Draw b A to intersection P with weight, and then
P c intersects A B at the point of inflection ilm Decomposing
P along P B and P c, as in Art. 114, we have at once the re-
actions at A and B. Here also we see that, by a construction
purely graphic and abundantly exact, we can find the inflection
point and the reactions.
The method detailed in Art. 114 can then be applied to de-
termine the various strains in the different pieces. It is un-
necessary to give an example, as the process is precisely similar.
We have simply in this case to start with the reaction at the
free end B and follow it through. Observe only that, as this
reaction must be less than for a girder with free ends for the
same position of P, the point h will lie nearer the force line
B' AB (Fig. SO, &), hence Im will not pass exactly through B,
but will lie to the right of it, giving thus a reversal of strain in
the flanges, as by reason of the inflection point should be the
case.
Instead of constructing tho hyperbola, we may calculate its
ordinates from the equation for x above, for different values
of a.
CHAP. XII.] METHODS COMBINED. 177
Thus, for
x=—OAl %=-0.3SSl x=-0.375l o?=— 0.35YZ aj=— 0.333 J
This will be sufficient to construct the curve in any given
case. The inflection vertical moves, therefore, between the
narrow limits of x = f I and % = % I, or within J^th of the span,
as the load passes from A to B.
Inasmuch as all that is needed for the determination of the
strains in the various pieces are the reactions at the supports,
and (for girder fixed at both ends) the moments at the supports
also, and as the formulae for the two cases above are very sim-
ple, we may determine these quantities at once by interpolation
of the given distance of the weight P in the formulae, and then
apply the graphical method for the strains, as illustrated in
Art. 114.
Thus, for a horizontal beam fixed at both ends, we have for
the moment at the left support A,
- — V
At the right support B, ..
MB = ^ (I - a).
For the reaction at the left,
For the reaction at the right,
RB = ? (3 a2l - 2 a3).
In the case of a horizontal beam fixed at left end and merely
Besting upon the right support, we have
MA = (3 a^l - 2 a 1? - a*\ MB = 0,
a being always the distance of the weight P from the left.
These formulae are simple, and easily applied to any case.
We may also observe that in Figs. 79 and 81 the ordinates to
the lines P 5, P c, and P c, P B, from A B, are proportional to
the moments (Art. 110). These ordinates to the scale of dis-
178 GRAPHIC AND ANALYTIC [cilAP. XTT.
tance, multiplied by the pole distance to scale of force, give the
moments at any point. ' Onr construction, therefore, gives the
moments also at every point, and we may thus check the re-
sults obtained by Art. 114 by the results obtained by the
method of moments.
117. Approximate Construction. — It will be readily seen
that the portion of the hyperbola in Fig. 81, PI. 21, needed for
our construction, is nearly straight. In most cases it will be
practically exact enough to lay off f I to the left of A, and -J I
also to the left of A at a vertical distance equal to Z, and join
the two points thus obtained by a straight line. This line can
be taken instead of the curve, and the construction is then the
same as above. The error due to thus considering the curve
as a straight line is greatest for a weight in the middle of the
span, where it does not exceed To^th of the span for the posi-
tion of the inflection vertical, and diminishes from the centre
both ways.
118. Girder continuous over three Level Supports-
Draw Spans. — This case is perhaps of the most frequent
practical occurrence, and an accurate and simple method of
solution is therefore very desirable.
In the first place, the formulae for the reactions are very
simple and easy of application. Thus, for left end support A,
the load being in the second span, or to the right of the middle
support B,
RA = j?(3«^-2«?-«3);
•*• ^
for the reaction at middle support,
for reaction at right end C,
where a is always the distance of the weight P from the mid-
dle support.* We are therefore already in a position to solve
completely the case under consideration. We have only to
* As already remarked, the development of the formulas assumed in this
chapter must be sought for in special treatises on the subject. We assume
them as known, and then apply them graphically as above.
See also Supplement to Chap. XIII.
CHAP. Xn.] METHODS COMBINED. 179
find the reactions and follow them through by the method of
Art. 114.
From the above reactions we can, however, easily determine
the distance of the inflection point. This will, of course, be
found only in the loaded span, at a distance from the middle
support.
VU - ~^ TQ \ ^T 7 O ' -" v«
4&&+ 2 al— a2
We can find the values of x corresponding to different values
of a, and thus plot the curve for the inflection points. Thus,
for
a — :0 a = ^l a = $l a = %l a = I
x = 0 cc — -fa I x = -f^l x = ^l x = \l.
This curve being drawn for any particular case, we can
easily find the position of the inflection point for any given
value of #, and hence the reactions, and then find the strains in
the various pieces.
Thus, in PL 21, Fig. 82, the curve B e d being drawn, we
can at once find the inflection point i for any position a of
the weight P. We have simply to make B J — a and draw 5 e.
b e is the distance of the point of inflection from B. We can
now, as explained above, draw any line as P i, and then P C
and A h. The ordinates to the broken line A h P C from A C,
to the scale of distance, multiplied by the pole distance H to
scale of force, will give the moments at any point. Moreover,
H E is the shear at B. E a is the reaction at B, H a the re-
action at A, and H P the reaction at C. The reactions at B
and C are, of course, positive or upwards, that at A negative
or downwards. Hence E « — H # -f- H P = P, as should be.
The value of x for the inflection vertical is by Art. 112
i a I
X = -T-. - r-j - ; — j
(i — a) I — ^ a
or, substituting the value of i above,
a I (2 I - a)
Since, therefore, in this case the value of x is no simpler than
that for i given above, it will be preferable to plot the first
curve directly as represented in Fig. 82.
119. Approximate Construction.— In practice it will be
180 GRAPHIC AND ANALYTIC [CHAP. XII.
found abundantly accurate to assume the curve for i between
the required limits, as a parabola whose equation is i = x =
a ^ — -. The greatest error for a = - will then be about
5 t 4:
— of the span, and decreases both ways to a = o and a = -.
100 * 2
From a = — to a = I the parabola coincides closely with the
2
q -i
true curve. The difference for a = — I is only -^^-l, and we
4 500
have, therefore, a very simple practical construction for both
reactions and moments. We have only (Fig. 82) to erect a
vertical at the centre support B and make it equal to I, and
then construct a parabola passing through B whose ordinate
c d = - I. The horizontal ordinates to this parabola for any
5
vertical value of &, give the distance out from B of the inflection
points. For the load in first span A B, of course this parabola
lies on the other side of B c, and b e is laid off to the left. The
remainder of the construction is as in Art. 118 for the reactions
and moments. When great accuracy is required, we can find
the reactions from the equations of Art. 118. In any case, the
reactions being given, we can follow them through the structure
by the method of Art. 114, and thus determine the strains in
every piece due to every position of each apex load. A tabu-
lation of these strains will then give by inspection the maximum
strain in any piece due to the live load. All the weights taken
as acting simultaneously will then give the strains due to uni-
form total live load. The strains due to dead load will be
multiples or sub-multiples of these. Thus if total live load
causes, say, 100 tons compression in a certain piece, and if the
Q O
dead load is - of the live load, then we shall have - of 100 ==
2 2
150 tons compression in the same piece due to the dead load
alone. If now the live load causes a maximum tension in the
same piece of 200 tons, then the piece must be made to resist
both tensile strain of 200 — 150 — 50 tons and compressive
strain of 150 + 100 — 250 tons. If a diagonal, the counter tic
is strained 50 tons, while the maximum strain on the diagonal
CHAP. XII.] METHODS COMBINED. 181
is 250 tons compression. It is only necessary, therefore, to find
the strains due to each weight of the live load. From the
tabulation we can, then, by means of the ratio of the dead to
live load, find the strains due to dead load alone, and then by
a comparison of the two find the maximum compressive and
tensile strains. If the maximum strains due to live load are of
opposite kind, but less than the constant strains clue to dead
load, we shall need no counterbracing. The resultant strains
will then always be of the character given by the dead load.
If greater, we must counterbrace accordingly. The process is
the same as by the methods of calculation, and the reader may
refer to Stoney — Theory of Strains — for illustrations.
12O. The " Tipper," or Pivot Draw, with secondary cen-
tral Span. — We have said that a pivot draw may be considered
as a beam continuous over three supports. In practical con-
struction this statement needs some modifications which deserve
special notice. Thus practically that portion of the beam over
the central support forms a short secondary span D D [Fig. 83,
PL 22] the reactions at the supports D and D being always
equal and of the same character. If a weight acts, say, on the
first span A B, and the beam itself is considered without weight,
the end C must be held down, that is, the reaction there is neg-
ative. ISTow as the weight P deflects the span A B (Fig. 83), it
causes one secondary support D to sink, and the other to rise
an equal amount. In practice D and D may be the extremities
of the turn-table, and the reactions are then evidently different
from those given by the formulas of Art. 118.
If in this case we take a as the distance of the weight P from
the left support A, the reaction for load in A B will be given
by the following formulae :
Where the ratio -^ = &, a being the distance of the weight P
from the left support A (for load in the span A D), I = span,
A D = D C, and n 1= span D D, and where the constant
(4 + 8 n + 3 n2) is put for convenience = H, then
\ 2 H - (10 + 15 n + 3 n*) & + (2 + n) & I
|_ J
RA = — 2 H - (10 + 15 n + 3 n*
2 H
* See Supplement to Chap. XIII., Art. 6.
182 GRAPHIC AND ANALYTIC [CHAP. XH.
R° = 2^ r (2 + n\ ® ~~ (2 + 3 n + 3 ^ ^ \
These reactions, it will be observed, when added together
RA 4- 2 RD + Rc are equal to P, as should be the case.
By the application of these formulae, which are for any par-
ticular case by no means intricate, we can find the reactions at
A and C as also at D or D ; and then starting, say, from A, can
follow the reaction there through the frame by the method of
Art. 114. A negative reaction indicates that the support tends
to rise, and unless more than counterbalanced by the positive
reaction due to uniform load, the end where this negative reac-
tion occurs must be latched down.
121. Supports in Pivot Span are not on a level— Reac-
tions for live load, however, are the same as for level sup-
ports.— The three supports of a pivot span should not be on a
level. It is evident that if this were the case, the first time the
draw is opened the two cantilevers deflect and it would be diffi-
cult to shut it again. The centre support should therefore be
raised until the reactions at the end supports are zero, that is,
until they just bear. The centre support is then raised by an
amount equal to the deflection of the learn when open, due to
the dead load. Even when shut, then, there are no reactions at
the end supports except when the moving load comes on. Now
this being the condition of things, it may seem, strange to assert
that these reactions %XQ precisely the same as for three level sup-
ports, and yet such is the fact. If the beam, originally straight
were held down at the lower ends by negative reactions, then
the reactions would have to be investigated for supports out of
level, and a load would diminish these negative reactions, or
might even cause them to become positive. But such is not
the state of things. The end reactions are in the beginning
zero, and any load gives, therefore, at once positive reaction at
its end support. This positive reaction is just what it ivoulo,
be for the same beam over three level supports.
An analytical discussion of the case would be out of place
here, but assuming the expression to which such a discussion
would lead us, we may show that this is so.
Thus, for a beam over three supports A, B, and C, not on a
level. GI being the distance of A below B, and c% the distance
of C below B, the modulus of elasticity being E and the mo-
CHAP. XII.] METHODS COMBINED. 183
inent of inertia I, we have for the moment at the centre sup-
port B due to any number of weights in both spans,
a being always measured from the left support.
Now in this expression the last two terms are precisely the
same as for supports on a level ; the influence of the different
levels is contained in the first term on the right only. Now by
the supposition, ct and c2 must be taken equal to the deflection
due to the dead load, and the value of this term will therefore
be entirely independent of the live load, which enters only in
the last two terms.
A particular case may perhaps render this plainer. If a
girder of two equal spans over three level supports is uni-
formly loaded, the reaction at an end support is, as is well
known, |ths of the load on one span.
Now let us take the girder over three supports not on a level,
and from our formula above find the reaction at one end due
to uniform load when ^ and c2 have the proper values given
to them. First the dead load p I over each span causes a de-
flection at each end of the two cantilevers — ii&s-i This,
then, is the value for ^ and <?2 m the formula. Now let us
take an additional moving load of ml over the whole beam,
and with this value of c1? C2 find the reaction. We have from
our formula
4 MB I = - lp 1? - \ (p + m) P,
or MB = -£>Z2-
Now we have by moments,
I2
RAX Z-(^ + m)-=
hence, inserting the value of MB above,
* Theorie der Trdger: Weyranch. Also Supplement to Chap. XIII., Art. 3.
f Supplement to Chap. VII., Art. 13.
184 GEAPHIC AND ANALYTIC [CHAP. XII.
That is, the reaction at A is due to the moving load alone,
as evidently should be the case, and is, moreover, just what it
should ~be for a girder with level supports; viz., \ml. (See
also Appendix, Art. 18, Ex. 5.)
The raising of the centre support, then, will not affect our
construction for the reactions as given in Figs. 81 and 82, pro-
vided there are only three supports.
We have deemed it well thus to call special attention to the
considerations of the last two articles, both on account of their
practical importance and because they are not brought out
clearly, nor indeed, so far as we are aware, ever alluded to in
any treatise upon the subject.*
122. Bctiiii continuous over four Level Supports. — We
thus see" that a draw or pivot span is more properly considered
as a beam of three spans instead of two, of which the centre
span is very small compared to the end spans ; it may be only
two or three panels long. Moreover, we must often in practice
consider the beam as a " tipper," and therefore apply the formulae
for reactions of Art. 120. If, however, by reason of the method
of construction, as often happens, for instance, by the under
portion of the beam coming in contact with the frame below,
this tipping of D D (Fig. 83) is confined between certain limits,
beyond which the supports must be considered fixed, it will be
necessary to find the reactions as for a beam over four fixed
supports, and determine the corresponding strains in this case-
also.
Comparing, then, the strains obtained each way, we take only
the maximum strains from each.
The formulae for the reactions at \h.Q fixed supports A B C D
are as follows (PI. 22, Fig. 84) :
1st. Load P in left end span A B at a distance a from left:
support A, the end spans being n I and the centre span B C=L
We put Jc = —~ and H = 3 + 8 n + 4 n\ Then
n L
RA=g| H-(H + 2^-h2^)&-f(2rc+2rc,2)F
* Clemens Herschel, in his treatise upon " Continuous, Revolving Draw-
bridges" (Little, Brown & Co., Boston, 1875), notices this fact for the first
time.
CHAP. XII.] METHODS COMBINED. 185
RB=j|| (3 + 10^+9™2+2^3)&-(27i + 5^2+27i3)&3
Rcr=-| -(n+3n* + 2n*)fc+(n + Sn*+%n*)&
RD=H| njc~n^
These reactions add up, as they should, equal to P.
In practical cases of pivot spans, we have only to consider the
the outer spans ; as a load in the middle span B C = I rests
directly upon the turn-table. The above formulae are then all
we need. For a load in the right end' span the same formulae
hold good, only remembering to put now RD in place of R^
Rc in place of RB, RB in place of Rc, and RA in place of RD.
If, however, neglecting the particular case of pivot spans, we
suppose the middle span B C = I loaded, we have — a being
now the distance of P from B, and & being now -j instead
C
of — ^, as above, H remaining the same.
2d. Load in B C.
1
LH 3 4n-M-4n* *- 6 15 62 H 1
[si
J
r ^ ' _ ^ ^ i
L J
These reactions should also add up to P, as is the case. The
number n may be taken at pleasure, so that the end spans may
be as much larger or less than the centre spans as is desired.
H, P and the quantities in the parentheses, it will be observed,
are for any given case, constants which may be determined and
inserted once for all.
"We have, then, only to insert the values of 7c for different
positions of the load P. Thus the equations for any particular
'•ase'are very simple and easy of application.
186 GRAPHIC AND ANALYTIC [CHAP. XII.
123. Construction. — We may, if desired, apply our method
of construction to the determination of the reactions. Thus
from the above reactions we may easily determine general ex-
pressions for the inflection points. For the case of a load in
C D = n I (PI. 22, Fig. 85), we have, when i is the distance of
the inflection point from C,
-RD x (nl-i) +P(a-i) = 0;
P a - RD n I
whence ^ = —= - = — .
P — KJJ
For the inflection point distant i from B in the unloaded
span,
hence
For the second case of load in B C = I, we have for the in-
flection point between B and P
— RA (n l+i) + Rji i = 0, or
i=K±nl_
For the point between P and C
— RD (n I + i) + Rc i = 0, or
The insertion of the proper values of the reactions for each
case, as given above, will easily give general expressions for the
inflection points, which the reader may, if desired, deduce for
himself.
Our construction is, then, as follows [PL 22, Fig. 84] :
\st Case. Load in C D. -.
Having found %, draw a line at any inclination, as ^ d through
&1? intersecting P at d, and the vertical through C at c^ Then
lay off B i and draw d D, c± b and ~b A.
Make d c = P by scale, and c D drawn parallel to ^ d then
gives the pole distance H. The ordinates, then, to the broken
line A b c: d D taken to scale of distance, multiplied by H to
scale of force, give the moments at every point. Moreover, H d
is the reaction at D. Draw D b parallel to c± 5, then c b is the
reaction at C. In like manner a b is the reaction at B, and H a
the reaction at A. The moment at C, and reactions at C and
CHAP. XII.] METHODS COMBINED. 187
D, are positive. Reaction and moment at B negative / reaction
at A positive ; as a little consideration of what the curve of the
deflected beam must be, will show. The shear at C is, there-
fore, + H a — a b + b c — + H c. The shear at B is — H b or
4- H a— a b, and so on. The shear being always given by the
segment between H and lines parallel to A b, b cl} c{ d and d D.
2d Case.— PL 22, Fig. 85.— Load in B C.
Having found the distance B i± from the equation for this dis-
tance of the point of inflection above, we lay off B c\ = B -&t
and thus draw c± E at an angle of 45°. Finding then the value
of C i2 from its equation above, we can draw E c% and then
c2 D and Ci A. The construction is then the same as before.
Thus H is the pole distance, H c the negative reaction at D,
H b the negative reaction at A, and cl b, c E the positive re-
actions at B and C. The shear at B is H cly etc. Thus the outer
forces are completely known for a weight at any point. It will,
however, in general, in practice, be found more satisfactory to
use the formulae for the reactions which we have given than to
find these reactions by the above construction.
We shall now illustrate the preceding principles by an exam-
ple taken from actual practice.
124. »raw Span— Example.— In PL 22, Fig. 86, we have
given to a scale of 20 ft. to an inch the elevation of one of the
trusses of the pivot draw over the Quinnipiac Eiver at Fair
Haven, Conn.*
Length of span A B = 89.88 ft. B C = 21.666 ft, divided
into seven panels of 12.84 ft. and two of 10.833 ft. respectively.
Height at B and C, 16 ft. ; at A and D, 12.1 ft. Diagonal
bracing as shown in Fig. Line load 9 tons per panel.
21 (
In this case n = — '— - or n = 0.24106, hence the equations
of Art. 120 become
A = P (l-1.1298 ?+0.1836 ^
B
= C = P /0.6S36 2-0.1836
* Designed and erected by Clemens Herschel, C.E., and probably the only
structure of the kind in this country for which the strains have been accu-
rately and thoroughly determined. For the above data I am indebted to M.
Merriman, assistant engineer in charge.
188 GRAPHIC AND ANALYTIC * [CHAP. XII.
D = -P fo.2374 —0.1836 -V
\ \ I 1? I
Now -' is -, -, -, -ths, etc., according to the position of the
L 7 i 7 *
weight at lst% 2d, 3d apex from end. So also — is — r, — , — —
/ Otto 343 343,
etc. The above equations for the reactions, then, may be
written A = P (1-0.1614 £ + 0.000535 £3),
B = C = P (0.09766 £-0.000535 £3),
D = -P (0.03391 £-0.000535 £3),
where £ has the values 1, 2, 3, 4, etc., for P!, P2, P8, P4.
Thus, if we wish the reactions due to a weight P4 of 9 tons at
the fourth apex, as shown in Fig., we have only to make P — 9
and £ — 4, and we find at once A = 3.498 tons, B = C =
3.207 tons, D = —0.912 tons. The sum of all these reactions
exactly equals P, as should be.
The middle supports are supposed raised by an amount equal
to the end deflections of the open draw, therefore the strains
due to dead load are easily found, as in the " braced semi-arch,"
Art. 9.
The reactions due to live load, according to Art. 121, will not
be affected by this raising of the supports.
To find the strains due to P4, we draw the force line F2 F
[Fig. 86 (a)] by laying off P4 — 9 tons down from F to Fl3 then
F! E2 downwards equal to the negative reaction at L , viz.,
—0.912 tons. Then from E2 lay off upwards E2 Et = to positive
reaction at C — +3.207 tons. Then ^ E = reaction at B =
+ 3.207 tons, and finally E F equal to reaction at A = +3.498
tons, which should bring us back exactly to point of beginning
F, since the reactions and the weight P must be in equilibrium.
\JN~ote. — When we wish to begin at the left end of the frame,
it is best, as in this case, to lay off the reactions in order, com-
mencing at the right.'] We have taken the scale of force 4
tons per inch.
The weight P4 acts upon the triangulation drawn full in the
figure. Using now the notation of Art. 114, and representing
all the space above the truss by E, all lelow by F, we have at
A the reaction E F [Fig. 86 (a)] in equilibrium with E 1 and
F 1, and drawing parallels to these lines from E and F, we find
the strain in F 1 = 3. 54 tons tension, and E 1 = 5.1 compression.
CHAP. XII.] METHODS COMBINED. ISO
So we go through the truss and find the strains in every
piece. Heavy lines in the strain diagram denote compression.
We see at once that for this position of the weight, all the
upper flanges in span A B are compressed, the last lower flange
P 7 is also compressed, and all the other lower flanges are in
tension. At the point of application of the weight P4, the two
diagonals 3 4 and 4 5 are in tension, and either side they alter-
nate in strain as far as C or diagonal 8 9. Diagonals 8 9 and 9 10
are both tension, and then the strains alternate to support D.
All the upper flanges of the right half are tension and increase
towards the middle. All the lower are compression and like-
wise increase towards the middle.
If we go through the whole truss from A to D, the last diago-
nal 15, 16 should evidently pass exactly through E2, thus check-
ing the accuracy of the construction. The diagonal 6 7 crosses
the force line, thus causing the strain in the lower flange to
change from tension in Ft 5 to compression in F! 7. The point
of inflection, therefore, falls to the right of diagonal 5 6.
The reaction at B diminishes greatly the strain which would
otherwise take effect in 7 8 and E 8 ; while the reaction at O
reverses the strain which would otherwise take effect in 9 10
and diminishes E 10. We recommend the reader to follow
through carefully the strain diagram, Fig. 86 (a).
A series of figures similar to Fig. 86 (a) (in the present case
seven separate figures) will give completely the strains due to
the rolling load. A table may then be drawn up containing
the strains due to dead load, and the maximum strains due to
live load in every piece, and the total maximum tension and
compression in every piece may then be found. [Compare
Art. 12, Fig. 7.]
For the supports fixed, instead of B and C tipping, the pro-
cess is precisely similar, except that we have to make use of
the formulae of Art. 12i4. The reaction at A will then be
somewhat less than in the present case ; the inflection point is
therefore found further from the right support B ; it may be
even to the left of diagonal 5 6, in which case (see Fig. 86, a)
we should have tension in upper flange E 6. The reaction at
B would then be still positive, but greater than E E1? while C
would be negative and no longer equal to B, and D would be
positive. We should thus have 7 8 tension and E 8 tension ;
F 7, as before, compression, 8 9 compression, and 9 10 com-
190 GRAPHIC AXD ANALYTIC [CHAP. XII.
pression, and E 10 compression ; while F 9 would be tension.
From 9 10 to the right the diagonals would alternate in strain,
the compressed upper flanges, as also the tensile lower flanges,
would diminish towards D, and the last diagonal should pass
exactly through new position of F2, thus closing the strain
diagram and checking the work. The reader will do well to
construct the diagram.
The strains should be found for both cases, and the maximum
strains taken from each, which, compared with the permanent
strains due to the dead load, will give the total maximum
strains.
"We have taken for convenience of size too small a scale for
the frame to ensure good results. With a large and accurately
constructed ymw e diagram, dealing as we do with only single
weights, and consequently small strains, the above force scale
of 4 tons per inch would give very accurate results.
If the strains due to uniform load (no end reactions) are
found by addition of the strains for each apex load diagramed
separately, the same scale may be employed ; but if all the
loads are taken as acting together (Fig. 5, b\ a smaller scale
for strains will have to be adopted, as the force line will other-
wise be too long. [See Art. 16 of Appendix for the method of
calculation.]
125. Method of passing direct from one Spun to next.
— By inspection of Fig. 86 we see that we might find the strains
in the intermediate span B C without first going through the
whole of A B or C D. Thus, if we knew the moment at B,
this moment^ divided by depth of truss at B, would give the
strain in flange F 7 for the system of triangulation indicated
by the full lines. If then we knew also the shear at B =
P — A — B = E! F! (Fig. 86, a), we could at once lay off Fl 7
and E! Ft (Fig. 86, a\ and then proceed to find E 8 and 7 8,
just as before. In the same way the moment at C, divided by
height of truss at C, would give us strain in F 9, and with
shear at C = P - A - B - C = E2 Fl = D, we could find E 10
and 9 10, as before. As we know already, a load anywhere
upon a beam causes positive moments at a fixed end — i.e.,
makes upper flange over support tension and lower flange com-
pression. But as we see from the last case, owing to the tri-
angulation, the last upper flange may also be compression (see
E 6 in Fig. 86) if the inflection point lies between diagonal 5 6
CHAP. XH.J METHODS COMBINED. 191
and the support. The known moment gives, then, the charac-
ter of the 'strain only for that flange which does not meet a
diagonal at the support. The moment at B, therefore, being
positive, gives us compression here in lower flange, because, for
the system of triangulation corresponding to the weight, that
flange does not meet a diagonal at B. For a weight upon the
other system of triangulation (dotted in Fig.), the same moment
would give us the tension in E 7. The construction assumes
equilibrium between F 7, 7 8, and E 8, and the shear at B ;
that is, between the pieces cut by an ideal section to the right
of B through the truss and the shear at that section. That this
is so is shown by the strain diagram, since there we see that the
strains in these pieces form a closed polygon with the shear at
B = E! Fx. This must evidently be so if these are the only
pieces cut by such a section, since then the horizontal com-
ponents of the strains in these pieces must balance, and the
resultant vertical component must be equal and opposite to the
shear.
It is important to know which side of Et Pj_ to lay off F! 7,
since, if we had laid it off in this case to the right, we would
have obtained a very different value for Et 8. For this pur-
pose we have only to suppose the strain in the flange (either
upper or lower, as the case may be) to be applied at the point
of junction or apex of the other two pieces, and then lay it off
in the direction with reference to that apex corresponding to
the known character of its strain. The direction of the shear
is always known from the reactions.
Thus in our Fig. the shear between B and C acts down from
E! to Fl5 because P4, which also acts down, is greater than the
sum of the upward reactions at A and B. The strain in F 7
is also known to be compressive, and therefore, in following
round the strain polygon commencing from Ex to F1? it must
act towards apex at 7. We must, therefore, lay it off to the
left of E! Fj. In similar manner, for the other triangulation,
the strain in flange E 7 is, in span B C, in equilibrium with 7 8
(dotted diagonal) and P 8 and shear Et F1? and is, moreover,
known to be tension. Consider it acting then at B ; and then,
since it is tension, we go round the polygon from Et to Pt, and
then to the right of Ex F1? or away from B, the point at which
it is supposed to act.
Now for the case of the " tipper : " the reaction at D, and
192 GKA.PHIC AND ANALYTIC [CHAP. XII.
therefore the moment at C, is also positive. The lower flange
F 9 is therefore compression, or for the dotted system of trian-
gulation E 9 is tension. The shear to the left of C, Et F! acts
down, since — P + A + B = — ^ Flt Consider F 9 acting at
apex 9, and then, since it is compression, it must act towards 9
(from right to left), and passing down then from Ex to F1? we
must lay off F 9 to the left of Ex F^ For similar reasons, for
the other system, E 9 must be laid off to the right.
For fixed supports B and C, the moments alternate from B,
and the moment at C is therefore negative — that is, gives com-
pression above and tension below. Flange F 9, for the system
of triangulation of P, would then be tension instead of com-
pression, as above ; P will, however, still be greater than A + B,
and hence the shear is to be laid off down, and F 9 must be
laid off to the right.
If, then, it were required to find the strains in the span B C,
preceded and followed, it may be, by many others, it is suffi-
cient to know the moment and shear at one support. We can
then commence and continue the strain diagram, without being
obliged to go off to a distant free end and trace all the strains
through till we arrive at the span in question.
126. Method of procedure for any number of Spans.
Let us take, then, any number of spans, say seven [PL 22, Fig.
87], and let it be required to find the maximum strains in the
span D E. It is not, as we have just seen, necessary to com-
mence at the extreme end A or H, and follow the reaction there
through, from span to span, till we arrive at D. As we have
seen from the preceding Art., we may start directly from D,
provided we know the moment and shear there. Now, since a
load in any span causes positive moments and reactions at the
two ends of that span, and since either way from these ends the
moments and shear at the other supports alternate in character
[Art. 102], any and all loads in A B cause positive moments
and reactions at D. So also for loads in C D and in F G.
Loads in B C, E F and G H, on the other hand, cause negative
moments and reactions at D. [See Fig. 87.]
To find the maximum positive moment and shear at D due
to the other spans, we must then suppose the method of
loading shown in Fig. 87 (a). For the maximum negative
moment and shear at D, we have the system of loading shown
in Fig. 87 (b).
CHAP. XII.] METHODS COMBINED. 193
these two moments and shears being once known, we
can find by diagram and tabulate the respective strains in every
piece of the span D E. Thus dividing the moment at D for
either case by the height of truss, we have at once the strain in
either upper or lower flange at D depending upon the system
of triangulation as explained in Art. 125. With this strain and
the shear at D properly laid off to scale, we can commence the.
strain diagram precisely as though we had traced all the loads
through from the extreme end A or H to D or E.
We must next find and tabulate the strains in D E due
to each apex load in the span itself, and for this we must
know to begin with the moments and shears for each separate
load.
[Note. — Distinguish carefully between shear and reaction at
a support. The shear at D, or at a point just to right of D, is
the algebraic sum of all the reactions and weights between that
point and A. See also Fig. 84 (Art. 123), where the reaction
at B is — l> a, but the shear atB is— b a + H. a = — H b.
So also the reaction at C is -f ~b c, but the shear at C is
+ I c — ba + Ha — HC, etc.]
Conceiving n-ow that we have found and tabulated the strains
due to the first and second systems of loading as shown in Fig.
87, and also the strains for each load P in D E, the sum of these
strains will give the strains due to live load o-ver the whole
length of girder, and taking the proper proportion of these, we
shall have the strains due to the dead load. Combining then
these strains with those first found, we can easily find the total
maximum strains which can ever occur in D E.
127. Example. — Let us take, as an illustration of the pre-
ceding, the girder shown in Fig. 87, of seven equal spans, and
seek the maximum strains which can ever occur in the middle
span D E. Let Fig. 88, PI. 23, represent the span D E — length
80 feet, divided into 4 panels ; and let the live load at each
apex be 40 tons,* the uniform load being half as much, or 20
tons per apex. Height of truss = 10 feet.
Now the quantities which for the present we must suppose
known or already found are as follows :
* A very great load : half the resulting strains would give more nearly the
strains in a single truss.
13
194 GRAPHIC AND ANALYTIC [CHAP. XII.
Positive moment at D (1st system of loading),
as shown by Fig. 87 (a) ....— + 788.2 feet tons.
Corresponding shear at D = -1- 14.63 tons.
Negative moment at D
(2d system of loading) = - 382.54.
Corresponding shear ~ — 14.63.
Also for the loads in D E :
For the first load P1? moment — -f 158.92, shear = + 36.17.
P, « = + 271.96 « = + 25.88.
P3 " = + 203.36 « = + 14.16.
P4 " = + 62.88 " = + 3.82.
In Fig. 88 we have found by diagram the strains due to P3.
[For notation, see Art. 114.]
We l&y off to scale the shear 14.16 upwards, since it is posi-
tive, and then, since the moment 203.36 at D is positive, and
hence the strain in A a must be tension, we lay off A a =
?H^- = 20.3 tons to the right of B A (Art. 125). With B A
and A a thus given, we can rapidly and accurately find all the
other strains. Thus from our diagram we have, representing
tension by minus and compression by plus,
A a = — 20.3 A c — + 8.0 A e = + 36.4 A g = + 24.4
A ]c = - 27.2
B6= + 6.0 B<Z=:-22 B/=-50.8 B A = + 1.2 tons ;
and for the diagonals :
a &=+19.6 be =-19.6 cd = + 19.6 de=-l9.6
ef = + 19.6 fg =4- 36.4 g h =-36.4 h Jc = + 36.4 tons.
Heavy lines in the diagram represent compression.
In a manner precisely similar we can find the strains due to
the other weights, as also to the two systems of loading shown
in Fig. 87. Suppose all these strains thus found. Then the
method of tabulation is as follows :
CHAP. XII.]
123
METHODS
56
10
11
195
12
DIAGONALS. FLANGES.
'l
Live Load in D E -
Interior Loading
Maximum
Strains.
Exterior Load-
ing.
Dead
Load
=»#
Live.
Total
Maximum
Strains.
P:
* 2
p3
?4
Tens.
Comp.
+
1st
Case.
2d
Case.
Tens.
Comp.
+
Aa
- 15.6
- 27.2
_ 20.4
- 6.4
- 69.6
-78.8
+ 38.2
-55.1
- 203.5
+ 63.6
Ac
+ 16.4
+ 24.4
+ 8.0
+ 1.2
+ 50.0
- 49.5
+ 8.9
+ 4.7
-44.8
Ae
4- 8.8
+ 36.4
+ 36.4
+ 8.8
....
+ 90.4
-20.3
- 20.3
+ 24.9
- 15.7
+ 115.3
Afir
A*
+ 1.2
+ 8.0
+ 24.4
+ 16.4
+ 50.0
+ 8.9
- 49.5
+ 4.7
-44.8
+ 63.6
- 6.4
- 20.4
-27.2
- 15.6
- 69.6
+ 38.2
- 78.8
- 55.1
_ 203.5
B6
-20.4
+ 1.2
+ 6.0
+ 2.8
-20.4
+ 10.0
+ 64.2
- 23.6
+ 15.1
-28.9
+ 89.3
B<1
-12.8
-50.8
-22.0
- 5.2
-90.8
+ 34.9
+ 5.6
-25.1
-115.9
+ 15.4
B/
- 5.2
- 22.0
- 50.8
-12.8
-90.8
+ 5.6
+ 34.9
- 25.1
- 115.9
+ 15.4
B/,
+ 2.8
+ 6.0
+ 1.2
- 20.4
-20.4
+ 10.0
- 23.6
+ 64". 2
+ 15.1
-28.9
+ 89.3
a 6
+ 51.2
+ 36.4
+ 19.6
+ 5.2
+ 112.4
+ 20.7
- 20.7
+ 56.2
+ 189.3
fo c
+ 5.2
- 36.4
- 19.6
- 5.2
-61.2
+ 5.2
-20.7
+ 20.7
- 28.0
_ 109.9
<: d
- 5.2
+ 36.4
+ 19.6
+ 5.2
- 5.2
+ 61.2
+ 20.7
- 20.7
+ 28.0
+ 109.9
d e
+ 5.2
+ 19.6
- 19.6
- 5.2
-24.8
+ 24.8
-20.7
+ 20.7
0.0
- 45.5
+ 45.5
ef
— 5.2
-19.6
+ 19.6
+ 5.2
-24.8
+ 24.8
+ 20.7
- 20.7
0.0
- 45.5
+ 45.5
f(J
+ 5.2
+ 19.6
+ 86.4
- 5.2
- 5.2
+ 61.2
-20.7
+ 20.7
+ 28.0
....
+ 109.9
gh
- 5.2
-19.6
-36.4
+ 5.2
- 61.2
+ 5.2
+ 20.7
- 20.7
-28.0
- 109.9
hk
- 5.2
+ 19.6
+ 36.4
+ 51.2
+ 112.4
-20.7
+ 20.7
+ 56.2
+ 189.3
Having found and tabulated the strains clue to each weight
as shown by the first five colums of the table, add all the ten-
sions and compressions for each piece and place the results in
columns 6 and 7.* ^Ve thus have the maximum strains of each
kind which can be caused by the weights in span D E alone.
In the next two columns, 8 and 9, place the strains due to the
two cases of loading of Fig. 87. Now if the uniform or dead
load is taken at one half the live, we have simply to take the
algebraic sum of the strains in columns 6, 7, 8, and 9 horizon-
tally, and divide by 2. We thus find column 10. Finally,
from columns 6, 7, 8, 9 and 10, we can find the total maximum
* A more convenient form of table, perhaps, is obtained by putting the
weights Pj_4 in the vertical left-hand column, and the pieces A a, Ac,
etc., in the top horizontal line. The numbers are thus more conveniently
placed for add tion.
196 GKAPHIC AND ANALYTIC [CHAP. XII.
strains as given in the last two columns. Thus take the piece
A c. In this piece there is a constant compression due to dead
load of 4.7 tons. The second system of loading adds to this
for live load 8.9 tons, and the loading of D E itself causes 50
tons compression. Since all three cases may exist together, we
have 4.7 + S. 9 + 50 — 63.6 tons compression. Again the first
system of loading, which may act also alone, causes 49.5 tons
tension in A c. Diminishing this by the 4.7 tons compression
dne to dead load already existing, we have —49.5+4.7 =—44.8
tons tension. These two strains are the greatest which can ever
occur in A c.
For A a we see that the greatest compression is due to the
second system of loading acting alone, but this compression =
'38.2 tons, is less than the tension always existing in A a from
the dead load, which is 55.1 tons. Compression, then, can never
occur in A a. So also for A Jc. In like manner diagonals a £,
b c, c d, y <7, g A, and h k do not need counterbracing, as the
constant dead load strain overbalances any strain of opposite
character which can ever come upon them.
As we have in the present example taken a middle span,
observe that the strains of Pt and P4, P3 and P2 are similar. Thus,
strain in A a due to P4 is the same as in A k due to Pl5 and so on .
1£§. Method of Moments. — We can very easily check our
results by the method of moments of Art. 14.
Thus, for the first system of loading, we have the moment at
D = +788.2 ft. tons and the shear =14.63. For any upper
flange, then, as A g, since the positive moment causes tension
in upper fiange, and positive shear causes compression, we have,
taking apex g as centre of moments, — 788.2 + 14.63 x 60 =
- 788.2 + 877.8 = + 89.6. Dividing this by depth of truss
= 10 ft., we have 8.96 tons compression in A g.
For the strain in By due to the weight P2, since the moment
at D for this case is +271.96 and shear =+25.88, and the
positive moment and shear cause compression and tension re-
spectively in lower flanges, while the weight P2=40 causes
compression in By we have, taking f as centre of moments,
+ 271.96-25.88x50+40x20 = -222.0:1:, rind dividing this by
10, we obtain 22.2 tension in By, and so on.
For the diagonals, the shear at any apex, multiplied by the
secant of the angle with the vertical, gives at once the strain.
In this case the angle is 45°; therefore the secant is 1.414.
CHAP. XII.] METHODS COMBINED. 197
Hence, for strain in c d due to P2, we have
25.88x1.414:= +36.6 tons.
The calculation, then, of the strains in any span of a continu-
ous girder, as also the diagraming of these strains, is simple and
easy, and offers no more difficulty than in the case of a simple
truss, provided we know or can find the moments and shearing
forces at the supports for the various cases of loading. The
method of finding these necessary quantities will form the sub-
ject of the next chapter.
The reader will do well to compare the strains in the above
table with those for the same simple girder similarly loaded.*
It will be found that there is a saving of material in the
flanges of about eleven per cent, over the corresponding simple
girder. Some of the flanges are, to be sure, subjected to both
tensile and compressive strains, instead of being always of the
same character ; but in wrought-iron girders this is of little con-
sequence.
It is worthy of remark that a slight relative difference of level
of the supports of a continuous girder may cause very great
changes in the strains, and hence, in structures of the kind, the
foundations must be secure from settling, and the pier sup-
ports accurately on level. Under these circumstances, .where
long spans are desirable, the continuous girder is to be preferred
to a succession of single spans.
If the greatest negative reaction at any support, as E (Fig. 87),
is greater tnan the constant positive reaction at that point due
to the dead load, the girder will require to be latched or held
down at that support.
* See Appendix, Art. 16, where this comparison is made.
198 CONTINUOUS GIEDEE. [CHAP. XIII.
CHAPTER XIII.
ANALYTICAL FORMULAE FOE THE SOLUTION OF CONTINOUS GIEDEES.
129. Introduction. — As we have seen in the preceding
chapter, the complete and accurate determination of the strains
in the continuous girder, both for uniform and moving loads, is
easy, provided we can find the moments and shearing forces at
the supports for the various states of loading,, and for each apex
load. Now this we are able to do with mathematical accuracy,
and without much labor. The formulae necessary for the pur-
pose, when put into proper shape for use, are neither difficult of
application nor more complicated than many which the practi-
cal engineer is often called upon to manipulate. Since the
publication of Clapeyron's paper * in 185 7, in which, for the
first time, his well-known method was developed, and his cele-
brated "theorem of three momenta" made known, the subject
has engaged the attention of many mathematicians. In 1862
WinJder f first developed a general theory, and gave general
rules fo? the determination of the methods of loading causing
greatest strains, together with tables for the maximum values
of the moments, shearing forces, etc., for various numbers of
spans of varying length. In the same year Bresse \ followed
with a similar work. In 1867 WinHer § gave a general ana-
lytical theory, and, finally, in 1873 Weyrauch \ has treated the
subject with a degree of completeness and thoroughness which
leaves but little to be desired. He discusses the subject in its
most general form, for any number of spans of varying length,
* Clwpeyron — Calcul d'une poutre elastique reposant librement sur des ap-
puis inegalement especes. — Compte rendus, 1857.
f Beitrage zur Theorie der continuirlichen Briickentrager — Civil Ingenieur,
1862.
\ Bresse — Cours mechanique appliquee. Paris, 1862.
§ Winkler— Die Lehre von der Elasticitat und Festigkeit. Prag. 1867.
H Weyrauch — Allgemeine Theorie und Berechnung der continuirlichen und
einfachen Trager. Leipzig, 1873.
CHAP. XT!!.] ANALYTICAL FORMULA. 199
and for all kinds of regularly and irregularly distributed and
concentrated loads — both for constant and varying cross-section
of girder. His formulae are mathematically exact, and for
given loading are free from integrals.
The above is but a very imperfect sketch, and we have named
but a few of the many writers who have been occupied with
the subject. Clapeyrorfs Theorem above alluded to, as origi-
nally given by him, applied only to uniform load over whole
length of girder, or over an entire span. But as early as JBresse's
Treatise, it had been extended to include concentrated and
local loads as well, and Winkler has also given a very complete
and practical discussion of the subject.
Notwithstanding the labors of these and many other mathe-
maticians, there seems to be a wide-spread idea, even among
those who are supposed to have considerable familiarity with
mathematical literature, that the results deduced are unpracti-
cal. It is not uncommon to meet with even recent publica-
tions* in which it is stated that the authorities pass over such
problems with "judicious silence;" that the mathematical in-
vestigations are intricate, and the formulae deduced trouble-
some in application ; that even a a partial solution of the prob-
lem by mathematical calculation is attended with considerable
difficulty, and that a complete solution for the bending moment
and shearing force at every section, under moving partial and
irregular loads, taxes the powers of the best mathematicians,
and is well-nigh impossible, so far as any practical application
of them by the engineer is concerned." How far such ideas
are justified may be seen from the following pages. That the
authors and works above referred to can only be read by good
mathematicians is not to be denied. It may also be admitted
that the subject is an intricate one, and when treated mathe-
matically in its most general form the results are naturally in an
unpractical shape. But that these results are, therefore, worth-
less, or that the formulae, when applied to any particular case,
are " too intricate for practical use," by no means follows.
The desirability of formulae for the application of our graph-
ical method as developed in the preceding chapter ; the erro-
neous ideas prevalent on the subject which we have just noticed ;
* Graphical Method for the Analysis of Bridge Trusses : Greene. D. Van
Nostrand, publisher, New York. 1875.
200 CONTINUOUS GIRDER [CHAP. XIII.
and the deplorable fact that the " authorities " do but too often
treat the subject with "judicious silence," and that, therefore,
there exists in our engineering literature no collection of prac-
tical and useful formulae for this important class of bridges,
though such formulae are, and have been for years, free to all
for the asking, — all these facts may serve as apology for the in-
troduction of the present chapter, in a work which professedly
treats only of Graphical methods. The apologies of those who
professedly treat the subject analytically, and have yet omitted
such formulae, are not so numerous.
"We propose to give the analytical results necessary for com-
plete solution of a girder of uniform cross-section over any
number of level supports, with end spans any desired ratio of
intermediate ; for uniform load over whole length from end to
end of girder, for uniform load over any single span, and for
concentrated load in any single span at any point of that span.
These three cases, as we have seen in the preceding chapter,
are sufficient for the complete solution of framed Bridge
Trusses.*
Many of these results are here given for the first time, at least
in their present shape, in any published treatise, though, as re-
marked, some of them in more or less practical form have
long been common property for all who may have desired to
make use of them.
The formulae only will be given, in such shape and with
such illustrations of their application that, we trust, they will
be found free from complexity, and of considerable practical
importance. In the Supplement to this chapter a demonstra-
tion of the formulae is presented.
13O. Notation. — The notation which we shall adopt is as fol-
lows [see Fig. 89, PL 23] :
* The formulae for concentrated loads are alone all that is really necessary.
Their addition gives, as we have seen in our tabulation, Art. 127, the strains
for uniform load also. In fact, for strict accuracy, only single isolated loads
should be considered, as the results given by the formulae for uniform load are
not perfectly accurate. This may be seen from the well-known fact that, for
a girder fixed at one end, and supported at the other, the reaction at the free
5
end for a load in the middle is -TO. of the load, while if the same load were uni-
3 6
formly distributed, the reaction is ^ths, or ^ of the load. The difference,
however, for any practical case, where there are a number of panels, is very
Blight.
CHAP. XIII.] ANALYTICAL FOKMULJE. 201
Whole number of spans is indicated by s ;
Hence, whole number of supports is s + 1 — numbered from
left to right.
Number of any support in general, always from left is m.
The supports adjacent to a loaded span left and right are
indicated by r and y + 1.
When extreme end spans vary in length from the interme-
diate, they are always denoted by n I, where n is a given fraction
or ratio for any particular case. Thus, if intermediate spans
are all 70 feet and end spans 50 feet, I = 70, n I = 50, and n —
When spans next to ends also vary, they are similarly denoted
All other spans are of equal length and denoted by I.
The length of the span in which the load is supposed to be
is in general 4> where the value of r for any particular case in-
dicates the number of the loaded span from left.
A concentrated load is indicated by P.
Its distance from nearest left-hand support, by a.
The ratio of a to length of loaded span Zr, is It = — .
4
Moment at any support in general is MmJ where m may be
1, 2, 3, r, r 4- 1, s, etc., indicating in every case the moment at
corresponding support from left.
In same way reaction at any support is Rjn, shear Sm.
At supports adjacent to loaded span, then, we have Mr, Mr+1,
Rr) R^, Sr, Sr+1, for the moments, reactions, and shears at those
supports.
A dead uniform load is u per unit of length.
A uniform live load, w per unit of length.
w 4x, then, indicates a uniform live load over any span.
These comprise all the symbols we shall have occasion to use*
By reference to Fig. 89, the reader can familiarize himself with
their signification, and will then find no difficulty in under-
standing and using the following formulae. Certain symbols
which we shall use for expressions of frequent occurrence, will
be best explained as we have occasion to introduce them.
131. <4 Theorem of Three Moments.'' — This remarkable
Theorem, due to Clapeyron, expresses a relation between the
moments at any three consecutive supports, both for uniform
202 CONTINUOUS GIRDER. [CHAP. XIII.
load over whole length of girder from end to end, and for uni-
form load over the whole of any single span. It may be writ-
ten as follows :
Mm lm+ 2 Mm+1 (lm+ lm+1) + Mm+2 Zm+1 = [ + +J.
If we suppose only one of the two adjacent spans as lm to
contain the full live load -10, while all the spans are of course
covered with the dead load u, the above equation becomes
+1
m ' •*-"Am -t- 1 i_-iu • -1U + J.J • in +4 wia + i
If both spans bear the same uniform load u alone,
]\Im 1+2 ]VTm i\l + I i"] + Hff I = — \f-\-f "I
If the spans are equal, the above two equations become re-
spectively
n.. l O » .
MmZ + 4Mm+1 1 +
and
Mm I + 4 Mm+1 I + Mm+2 I — r&?~*
Now in every continuous beam, whose extreme ends are not
fixed, two moments are always known, viz., those at the ex-
treme supports, which are always zero. Hence, by the applica-
tion of this theorem, we can" form in any given case as many
equations as there are unknown moments, and then, by solving
these equations, can determine the moments themselves.
132. Elx&mple— Total uniform Load — all Spans equal. —
Thus let it be required to find the moments at the supports for
a beam of seven equal spans, uniformly loaded over its whole
length. The moments at the end supports Mt and My are zero.
We have then, by the application of the last equation above, the
following equations :
Fdr the first three supports 1, 2 and 3, m — 1, and
4 Mg I 4- Ms I — — ^— , or 4 3YT2 -f- M3 = — .
For supports 2, 3 and 4, ny, = 2, and
M2 + 4 M3 + M4 = "
CHAP. XIII.] ANALYTICAL FORMULAE. 203
For supports 3, 4 and 5, m — 3, and
or since in this case the moments equally distant each way
from the middle are equal, this last equation becomes
M3 + 4 M4 + M4 •= ?tf.
a
We have therefore three equations between three unknown
moments, M2, 3YT3 and M4, and by elimination and substitution
can easily find
1 9
If, as in our example of Art. 127 in the preceding chapter,
we take u = 1 ton per ft., I = 8'0 ft., then u 1? — 6400, and the
moment at the fourth support becomes 540.8. If the height of
truss is ten feet, this gives [Fig. 88] 54.1 tons strain in the
upper flange A a. By reference to our tabulation, Art. 12T,
we see that this agrees closely with strain in A 1 due to uni-
form load, found in a manner entirely different, viz., by sum-
mation of the strains due to first case of loading, and the several
loads in -the span itself, and serves therefore as a check upon
our results.
133. Triangle of Momciit§. — For the benefit of the practi-
cal engineer, who may object to the algebraic work involved in
elimination of the unknown moments from the equations above,
when the number of spans is great, we offer the following tabu-
lation, from which he may easily and directly determine the
moments at the supports for any "desired number of spans
without formulce or calculation.
Thus, if we were in the above manner to find the moments
for a number of spans, and tabulate our results as given in the
annexed table, an inspection of the table will show us that we
can produce it to any extent desired without further calcula-
tion.
204
CONTINUOUS GIRDER.
[CHAP. XIIL
MOMENTS AT SUPPORTS TOTAL UNIFORM LOAD ALL SPANS EQUAL.
Coefficients of u I2 given in triangle.
The Roman numerals along the sides of the triangle indicate
tfye number of spans, and the horizontal line to which they be-
long give the moments. Thus, for our example of seven spans
just worked out, we have the extreme moments M! and M8 = 0,
15
M2 and M7 = — u Z2, etc.
Now, a simple inspection of this table will show us that for
any even number of spans, as VIII., for example, the numbers in
the horizontal line are obtained by multiplying the fraction
above in any diagonal column, both numerator and denomina-
tor, by 2, and adding the numerator and denominator of the
fraction preceding that.
Thus,
15 x2 + 11
41
142 x 2 + 104 388 ' 142 x 2 + 104
30 _ 15 x 2
388 "" 142x2
12 x 2 + 9 33
m the other diagonal column ; j^~2 + 104 == 388 °r
11 x 2 + 11 .
= :r77r~ T777 m the other diagonal column ; and so on.
142 x 2 + 104
For any odd number of spans, as IX., we have simply to add
numerator to numerator and denominator to denominator, the
two preceding fractions in the same diagonal column.
CHAP. XIII.] ANALYTICAL FORMULAE. 205
41 H- 15 56 33 + 12 30 + 15 45
I n n Q — — __ Ol* _ _ — _
' 388 + 142 7 530' 388 + 142' " 388 + 142 530'
and so on.
We can, therefore, independently of the theorem and analyt-
ical method by which the above results were deduced, produce
the table to any required number of spans.*
134. Total uniform L,oad— all Span§ equal— Reactions.
— The moments being known, the reactions at the supports can
be very easily found.
Thus, the reaction at the first or last support is
_ul Ms
':T~ T;
at any other support
Thus, in our example in Art. 127, we find
56 161 _ 137 143
Hence, the shear at the fourth support is
56 161 137 143 . . 71
71
or when u I = 80 tons, -— ul = 4:0 tons.
Multiplying this shear by 1.414 (the secant of the angle with
vertical), we find for the strain in diagonal a b (Fig. 88) due to
uniform load + 56.5 tons, the same nearly as already found in
our tabulation.
135. Triangle for Reactions. — The reactions for a number
of spans being found, and tabulated, as above, in the case of the
moments, we sjiall have a triangular table precisely similar to
the one above, in which the same rule holds good for odd and
even numbers of spans.
* The above relations between the moments can be shown analytically to
be a result of the properties of the well-known " Clapeyronian numbers."
For the table above, as also the others which follow, we are indebted to the
kindness of Mr. Mansfield Merriman, Instructor in Civil Engineering in the
Sheffield Sci. School of Yale College. They are given, so far as we are aware,
in no treatise upon the subject yet published.
206
CONTINUOUS GIRDER.
[CHAP. xm.
REACTIONS AT SUPPORTS TOTAL UNIFORM LOAD ALL SPANS
EQUAL.
Coefficients of u I given in triangle.
vni.
VII.
VII.
VIII.
"We are thus able to find both moments and reactions at the
supports for any number of spans, so far as uniform loading is
considered, and may then either diagram the strains in the
various pieces or calculate them as explained in Arts. 127 and
128. No formulae are required. Any one who understands
the method of moments as applied to simple girders can, by
the aid of the two tables above, find accurately the strains in
every piece of a girder, continuous over as many equal spans as
is desired, and uniformly loaded over its entire length, all sup-
ports being on the same straight line.
As we have seen, Art. 127, this is one of the cases which
must be considered in order to find the maximum strains in
any span,* and the results above given for its solution will, we
trust, be found by the practical engineer to be neither " com-
plex " nor " difficult of application."
136. Clapeyroiiian Numbers. — In the analytical discussion
of continuous girders, certain numbers having many remarka-
ble properties play a very important role.
We have seen that the theorem of three moments furnishes
us with as many equations between the moments as there are
moments to be determined. For a small number of supports,
* See note to Art. 129.
CHAP. XIII.] ANALYTICAL FORMULAE. 207
these equations can, be solved by the ordinary rules of algebra ;
but for a great number, or in the general analytic discussion of
any number, we must have recourse to a special artifice. Tims
we multiply our equations, beginning with the last, by numbers
indicated by <?1? <?2, c3, . . . . c^_i, and then choose these num-
bers such that, by the addition of all the equations, all the JVC's,
with the exception of Ml5 disappear. We thus easily determine
M! without the tedious process of substituting from one equa-
tion to the other, through the entire list.
The following relations must then evidently hold between
these numbers, as is evident from the theorem of three mo-
ments of Art. 131 :«
2<
\ (4-i + 4) + <?2 i
4-i + 2 C2 (4_2 +
Li = o.
4-2 = o.
tfs-3 4 + 2 cs^ (4 + 4) + cs_, 4 = o.
If the first number is chosen at will, say ± 13 the other num-
bers can be found from these equations.
Now in the present case of all spans equal, we have between
any three of these numbers the relation :
Cm-l + 4 Cm + Gm + l = 0.
If we take the first, ct — 0, and the next, c% = 1, we have
for the others the following values :
Gl = 0 c, = + 15 c, = - 780 CM = + 40545
c2 = -f 1 c5 = - 56 <% = + 2911 cu = - 151316
c3 zn — 4 c6 = + 200 c9 = - 10864 c12 = + 564719
These are the so-called Clapeyronian numbers. They alter
nate, as we see, in sign, and each is numerically 4 times the
preceding minus the one preceding that. We shall always indi-
cate these numbers by the letter <?, the index denoting the par-
ticular one. Thus, c7 is the seventh number, counting 0 and
1 as the two first.
No table of these numbers is needed. The index being
given, any one can write down the series for himself, till he
arrives at the desired number.
137. Uniform Live Load over any single Span — Moments
at Supports of Loaded Span. — These numbers being pre-
mised, we can now give the following formulse for the moments
at the supports r and r + 1 of the uniformly loaded span :
208 CONTINUOUS GIRDER. [CHAP. XII'.
For tl.e left support,
M --- WP |gr gS-r+2 + 4. 6%_r + 1 |
L c«+i J
For the right support,
Mr+1 = - -W P pr gs-r + l
These formulae, it will be seen, are very simple an<} easy of
application.
Thus, for seven spans, load over the fourth from left, we
have s = 7, r = 4, and hence
1 72 F^ <"5 + <°4 '4!
--wP ~
Both moments are equal, as should be the case for a middle
span. Inserting now the proper values for the Clapeyronian
numbers from the preceding Art., we have
x— 56 4-152"! 615
So for any desired number of spans, the values of r and s
being known, the corresponding Clapeyronian numbers can be
easily found, and, inserted in our formulae, give us at once the
moments at the supports.
Turning again to our example, Art. 127, and making w = 2
tons, and I = 80 ft., we have w Z2 = 12800, and therefore M4 =
676, and dividing by depth of truss — 10 ft., we find the strain
in A a (Fig. 88) 67.6 tons, nearly what we have found by the
summation of the strains due to the loads Pt_4 in our tabulation.
138. Triangle of Moments— Uniform Live Load over any
single Span. — If from the above formulae we find the moments
at supports for a number of spans, and tabulate as before, we
shall have a triangle of moments similar to those already given,
which may be produced to include any desired number of
spans. We have only to observe that the numerator or de-
nominator of any fraction in the table follows the law of the
Clapeyronian numbers — that is, is four times the preceding in
the same diagonal column minus the one preceding that.
CHAP. XIII.] ANALYTICAL FORMULAE. 209
MOMENTS AT SUPPORTS OF LOADED SPAN. — UNIFORM LIVE LOAD
OVER ANY SINGLE SPAN.
Coefficients of w I'2 given in triangle.
m, - 153 x 4- 41 571 ,
Thus for seven spans, - _ — = —, and so on.
The triangle above gives the moments for uniform load over any
span, both right and left. For left supports we have simply
to count the span from right to left. Thus for seven spans for
load in the sixth span from left, we have moment at left-hand
627
support = - w Z2, counting the spans from left to right in
triangle. For the moment at right support of same loaded
span, we count six the other way from right to lejct, and find
571
13d. Moments at Supports of Unloaded Spans. — The
triangle and formulae above give the moments at the supports
of the loaded span only, both positive — that is, always tending
to cause tension in upper flange and compression in lower.
If m represents the number of any support counting from
the left, the moments at any support generally may be found
by the following formulae :
When m<r + l, Mm == \w V pm Gs~r+* + ^m Cg-r+11
£g_l + 4: CB
When m>r M^l^Z2 r cs-m+2 + <Vn 0M+i1
<?s_i + 4 <?a J
If we make in these formulas m = r in the first, and m =
14
210 CONTINUOUS GIKDEK. [CHAP. XIII.
r 4- 1 in the second, we obtain the formulse of Art. 137. For
any other support left of r, or right of r + 1, we have only to
give the proper values to m, s and r for any given case, and
find the corresponding Clapeyronian numbers.
14O. Practical Rule, and Table. — The moments at the
supports of the loaded span having been found by the formulse
of Art. 137, or the triangle of moments of Art. 138, instead of
using the above formulse, we may find the moments at the
other supports as follows :
For all supports left of the loaded span: Commencing at
the left end support, place over each support the Clapeyronian
numbers
1 4 15 56 209 780, etc.
Take the last number thus obtained, before reaching the left
support of the loaded span, as a common denominator. Then
the moment at the left end is of course zero. At the second
support 1, at the third 4, at the fourth 15, at the fifth 56, etc.,
all divided by this common denominator, will express the frac-
tional part of the moment at the left support of the loaded
span, which the moment at the support in question is. For the
moments at the supports right of the loaded span, proceed simi-
larly, only count from the right end.
Thus, for a girder of ten spans, sixth span from left loaded :
The moments M6 and M7 due to load being found, suppose we
wish the moments left of M6. Commencing at left end, num-
ber the supports 1, 4, 15, 56, 209. (Let the reader draw a figure
representing the case.) The number 209 is the last before
reaching the sixth support. We take this, therefore, for a^com-
mon denominator. Then we have M^ = 0 M2 = — — M6
209
So for supports to the right of support 7, we have
Remembering that M6 and M7 are both positive, and that
the moments alternate either way from these supports, we find
easily the proper signs for the moments right and left.
We can now, therefore, find the moments at D and E due to
the first and second cases of loading of Fig. 87 (Art. 126).
CHAP. XIII.]
ANALYTICAL FORMULAE.
Let us take the first case. For load on A B, we have from
/-rorv
our triangle or formulae the moment at B = w P. At D,
, 56 780
then, we have — *
56
wl?.
For load on C D, we have at once from triangle the moment
Finally, for load on F G, we have for moment at F, from
* • 1 627
triangle = ^r^-rr w
11644
therefore at D, ^ x ^ «» P =
45
11644
All these moments at D are positive ; we have therefore, for
the first case of loading, the total moment at D — 4-
717
w
If we make I = 80 ft. and w = 2 tons, we find the moment
at D = 788, and dividing by 10 we obtain 78.8 tons as the
strain in A a, Fig. 88, corresponding with our tabulation, Art.
127.
Table for all the Moments.
All the moments may be found from a simple table similar
to the following, which will be found perhaps preferable to the
triangle of Art. 138.
TABLE FOE MOMENTS. — UNIFORM LOAD IN SINGLE SPAN.
Support counted from Left. Denominator A
1
2
3
4
5
6
1
0
1
4
15
56
209
I.
0
3
12
45
168
627
II.
III.
0
11
44
165
616
0
41
164
615
IV.
0
153
612
V.
0
571
VI.
I
Spans.
iA.
1
1
2
4
3
15
4 -
56
5
209
6
780
7
2911
212 CONTINUOUS GIEDEK. [CHAP. XIII.
This table, it will be observed, can be produced to include
any number of supports desired. The law of the Clapeyroniaii
numbers runs both horizontally and vertically. The smaller
table gives the denominator, the larger the numerator of the
coefficient of w ft for any case. Thus, for seven spans we have
four times 2911 = 11644 for the common denominator. For
load on second span from right, moment at sixth support from
627
left, we have then directly w P ; for fourth support from
45
left, w P, the same as above.
For load in fifth span from right, the table gives us at once
612
and + ^— -r-r w P, for supports 1, 2 and 3.
For the other supports, since if now we were to continue count-
ing from left we should have to pass a loaded support, we
must count the loaded span from left, and count the supports
in reverse order. For fifth support from right, then, the num-
ber required is at intersection of III. (instead of Y.) and 5, or
(\~\ A
5-j-j w P, as found above. Thus the tables above cover all
11644
cases, giving supports at loaded span itself, as also right and
left of this span. We have only to remember to count sup-
ports from left, and loaded span from right, for all supports
left of load, and inversely for all supports right of load. [The
reader should always, when using Table, make a sketch of the
given number of spans, indicate the loaded span, and number
the supportsl\
141. Reaction* at Supports — Live Load over single Span.
— For the reactions at ends of loaded span, we have
For reactions at extreme ends, when end spans are loaded, or
w
When any other spans are loaded, or
when r > 1 and < s, R, = - ??? Rs + 1==_^?!.
CHAP. XIII.] ANALYTICAL FORMULA. 213
For all other reactions,
Thus for load covering the first span of seven spans, we find
from the known moment, given in preceding Art., for the fourth
support,
6 x 56 336
R^n^wl = u^wL
For load over third span from left,
6 x 616 1 , 6607
For load on sixth span,
Hence, total reaction at fourth support for first case of loading
7213
is jpj-rr; w &• In the same way we can find the reactions at the
first, second, and third supports, for the second case of loading,
as shown in Fig. 87, and then can easily find the shear at any
support, as D, by taking the algebraic sum of all the reactions
and loads between that support and the end.
We can now, therefore, find the shear and moment at D, and
thus determine the strains in the span D E for both cases of
loading, as given in our tabulation, Art. 127.
142. Triangle for Reactions — Single Span loaded. — If
we calculate from our formulae the reactions at supports of
loaded span, for a number of spans, we can tabulate the results,
as on next page, in a triangle, where each number is four
times the preceding minus the one preceding that, all in the
same diagonal column.
214 CONTINUOUS GIRDER. [CHAP. XIII.
REACTIONS AT SUPPORTS — LIVE LOAD OVER SINGLE SPAN.
Coefficients of w I given in triangle.
This triangle, similar to the preceding one for moments,
gives the moments at the left support of the loaded span, when
we count from left to right. Counting the other way, we have
the reactions at the right support of the loaded span.
Thus for six spans, fourth span from left loaded, we count
1770
four from left in horizontal line for VI., and find w I for
3120
reaction at left support. For reaction at right, we count four
also from right end, and find w I.
143. Reaction§ in unloaded Spans— Load over one Span
only— Table. — The formulae of Art. 141 for the reaction at
any unloaded support are sufficiently simple and easy to apply ;
still we may, if thought preferable, also draw up tables for
these, to be used in connection with the triangle of the pre-
ceding Art. The following tables give the coefficients of w I
for the reactions not adjacent to the loaded span. The denomi-
nator of the fraction is to be taken from the triangle above ; the
tables referred to give only the numerators.
CHAP. XIH.]
ANALYTICAL FORMULAE.
215
REACTIONS AT UNLOADED SPANS.
Supports counted from left. Supports counted from right.
1
2
3
4
5
6
1
6
24
90
336
1254
r.
3
11
18
72
264
270
990
1008
ii'.
66
in1.
41
153
571
246
918
984
IV.
V.
VI'.
2131|
VII'.
B -g
6'
5'
4'
a-
2'
1'
I.
II.
III.
1254
336
1008
90
270
24
72
6
18
1
~3~
11
990
264
66
IV.
V.
VI.
VII.
984
246
918
41
153
571
2131
Tables give the numerators of the coefficients of w L
triangle on page opposite.
Denominators from
These tables may be carried out to any desired extent by the
law of the Clapeyronian numbers in the vertical columns.
As an example of their use, take seven spans load in fifth
from left, that is, in third from right. (Make sketch.) From
the triangle we take the common denominator 11644. Then
from first table in the horizontal column of III/ we have for
left end
Ri-
wl, Ra = — .
66
264_
11644'
990
11644 w 11644 ~ 11644 ~ 11644'
For supports right of loaded span, we must take the second
table, and look in horizontal column for Y. We thus obtain
153 918
^-11644^ *»- ~ 11644**
We can now, therefore, either by our tables or formulae, or
both, find the moments, reactions, and shearing force at any
support for both cases of loading given in Fig. 87. The reader
will do well to take the example of Art. 127, and find the mo-
ment and shear at D for both cases, and thus check our results
as giren in Art. 127, viz., + 788.2 ft. tons and — 382.5 ft. tons
for the moments, and ± 14.63 tons for shear.
216 CONTINUOUS GLBDEE. [CHAP. XIII.
144. Concentrated Load in any Spun— Moments at Sup-
ports. — It only remains to consider a concentrated load at any
point. If the formulae for this case do not prove to be too com-
plex or intricate for practice, we may consider the case, so far
as equal spans are concerned, as fully solved.
We have seen that the " theorem of three moments," so far
as uniform loads are concerned, enables us to solve the case
thoroughly. It is more especially as regards concentrated or
partial loads that the opinion widely prevails as to the impossi-
bility of obtaining practically useful formulae ; and this, not-
withstanding that it has been shown by Bresse, WinJder, Wey-
rauch, and many others, that the theorem of three moments
can be extended to include concentrated loads also.
The Theorem as thus extended is as follows:
^-1 + 2 Mm /^ + Zm + Mm+1 lm =
Pm-l a'
where, by our notation (Fig. 89, Art. 130), a! and a are the dis-
tances of Pm_i, Pm, from the nearest left supports.*
By the aid of this theorem, we are able to deduce the follow-
ing formulas :
For moments left of r, and including support r, that is
when m <*• + !, Mm = - cm A Cs~r+2 + A' CS~T+I .
C&+1
For moments right of r + 1, including support r + 1, or
•u TUT A Cp + A' cI+l
when m > r, Mm = — ca_m+2 - - - *&*
CB + 1
In these formulae, c represents, as above, the Clapeyronian
number, and A A' stand for the following expressions :
A = PZ(2£-3#J + P) A' = PZ (&-#),
k being the fraction -, or the ratio of the distance of the weight
L
P from the left support, to the length of span.
145. Illustration of Application of above Formulae. —
These formulae are by no means difficult of application. Let
* For demonstration of this Theorem, see Supplement to this chapter.
CHAP. XIII.] ANALYTICAL FORMULA. 217
us take the example of Art. 127 (Fig. 88), where P = 40 tons,
1=80 ft., and a becomes 10, 30, 50 and 70 ft. respectively.
First, as regards the expressions A A' :
These become in the present case 3200 (2 & — 3 ^ + ^) and
3200 (k — %?) respectively, where Jc has the values -J-, f , -|, and -J
successively. Now as the denominator is in each term always
the same, in the first 8, in the second 64, in the third 512, and
only the numerators of the values of Jc vary for the different
positions of P, we may put these values of A and A' in the
forms
or
A = 800 h — 150 A2 4- 6.2305 A3,
A' = 400 h - 6.2305 A3,
where A has successively the values 1, 3, 5 and 7, for P15 P2,
P3 and P4 respectively. These are then the practical formulae
for substitution in the present case.
We can now apply the formulae for M above. Thus, sup-
pose for seven spans we have P3 in the fourth, as shown by
Fig. 88, and wish the moment due to P3 at the fourth support
D. Then s = 7, r = 4, and m = 4, and we have
M4=-o4
or, referring to Art. 136 for the Clapeyronian numbers,
-56 A + 15 A' 840 A - 225 A'
M4 = - 15
2911 2911
Now for P3 we have k = -, or A = 5, and therefore
o
A = 1028.81 A' = 1221.2.
Hence
= 202.4 ft. tons.
This divided by 10 = height of truss gives tension in A a =
20.2 tons, nearly what we have already found in our tabulation,
Art. 127.
In like manner we may easily find the moment at D due to
218 CONTINUOUS GERDEE. [CHAP. XIII.
every weight, or by giving the proper value to m in our for-
mulae, we may find the moment at any support we please.
The moments at the supports of the loaded span being found,
the moments at the other supports may be obtained according
to the rule given in Art. 140 for uniform live load over single
span.
146. Triangle of Moments. — The reader may also by the
aid of the formulae above form a triangle similar to those al-
ready given, containing the coefficients of P I for the moments
at the supports of the loaded span.
Thus for two spans, for moment at left support, we should
obtain 0 and J [2 Jc - 3 Z? + F] P I, and this last value will
run down the right diagonal column without change, except in
its coefficient J, which will become successively T4^, -J|-, •££$,
for three, four and five spans respectively. For three spans
we shall have, 0, -^ [7 k — 12 I2 + 5 %?] P Z, and, as above,
TS- t2 ^ — 3 ^ + ^] P I. The second of these will run down
the second diagonal column from the right without change,
except in its coefficient, which will be -fa, ^V5?? etc., for five
and six spans.
So, for four spans we have
0, .fa [26 Jc - 45 & -f 19 %?] P /, ^ [7& - 12 W + 5 ^] PI,
^ [2 Jc - 3 J<? + #] P I,
for moments at left, for load in 1st, 2d, 3d and 4th span re.
spectively. The second of these runs down the third diagonal
column from the right, changing coefficient as above.
If the triangle be now drawn, and these expressions properly
inserted, we shall observe that along the diagonal columns
sloping down and to the left, the values of Jc in the parenthesis,
as also the denominators of the outside fractions, follow the law
of the Clapeyronian numbers. The numerators of these out-
side fractions in these columns remain unchanged. The outer
left column is of course always zero.
Another triangle must be found for moments to the right of
load, and then the moments at the unloaded supports -may be
found by the rule of Art. 140.
All the moments may also be found from a couple of tables
formed similarly to those of that Art. It is unnecessary to
give such tables here. From the above the reader can form
them for himself, if desired. The formulae for moments given
CHAP. Xin.] ANALYTICAL FORMULA. 219
above are so simple, and with a little practice so readily worked,
that tables are scarcely needed.
147. Reactions at Support§ for concentrated Load in
§ingle Span. — For the reactions we have the following formulae :
1st. Abutment reactions.
When the end span contains the load, that is, when r = 1 or
When the load is not in the end spans, i.e., when r > 1 and
r <s,
R . M2 Ms
Rl- -J-, Rs+l = —.
2d. Reactions at supports of the loaded span itself (not end
span),
3d. For all other reactions.
•
The above formulae, in view of what has been said in Art.
145, are sufficiently simple to need no illustration.
For load in fourth span of seven spans we find easily for the
reaction at left support,
R4 = g [2911 - 3 Jc - 6387 & + 3479 #].
This can be put in working order as explained in Art. 145,
and the reader can check the results which we have given in
the example of Art. 127 for himself.
A triangle and two subsidiary tables for the reactions at the
supports of loaded spans may be formed similarly to the tri.
angle and tables of Arts. 142 and 143. We leave this for the
reader to accomplish for himself, if thought desirable.
148. Shear at Supports of loaded Span. — We are now in
possession of all the formulae necessary for the complete solution
220 CONTINUOUS GIRDER. [CHAP. XIII.
of a girder over any number of supports, all spans equal. For
any desired span, we can find the maximum positive and nega-
tive moments by the cases of Fig. 80, as also the moments due
to various positions of the weight P. We can also find the re-
actions at all the supports due to these cases. From the reac-
tions and known forces, we can then easily find the algebraic
sum, or shear , at any support. The moment and shear at any
support due to any case of loading are, as we have seen, the
quantities required for calculation.
Now it is not necessary to find all the reactions in order to
obtain the shear. The moments at the supports being known?
we can find the shear directly.
Thus, for concentrated load in a span 4 .(Fig. 89) we have for
any point x
Mr - Sr x + P (4 - a) - m = 0,
where Sr is the shear at the left of the loaded span, and m is
the moment at any point. We see at once that, to determine
this moment, it is the shear that we wish, and not the reaction.
For a uniform load we have similarly,
Mr - Sr x + ~ - m = 0.
2i
If in both these equations we make x = 1T) m becomes M^,
and we have
where q = P (1 — Jc) for a concentrated load, and q = -g£ for
uniform load.
In an unloaded span at the left support, or when m < r,
<] disappears, and we have
For the shear at the right support of the loaded span we have
simply Sr — P or Sr — w I, and hence
CHAP. XIII.] ANALYTICAL FORMULA. 221
where qr — P k for concentrated load, and q' = ^-^- for uni-
2i
form load. For any other span at the right support
S' — Mm - Mm-l
^m-1
Thus, Sm and S'm are the shears at any support just to right and
left of that support respectively. The reaction at any support
is then Rm = Sm + S'm.
The moments, then, at two successive supports being known,
we can readily find the shear at any support, and these two,
moment and shear, we repeat, are the quantities required for
calculation. The reactions, and the tables for the reactions
above, are only useful as enabling us to find the shear. It is
this last, together with the moment at the support, which gives
us the moment m at any point of the span in question, as is
evident from the above equations. It is only in the case of the
simple girder that the reactions at the ends are the same as the
shears. In the continuous girder only the latter should ~be
used, except for ends of end spans, where the two are identical.
We have only to remember, then, that the shear at any support
is the algebraic sum of all the reactions and loads from that
support to the nearest extreme end, and then, knowing these
reactions and loads, the determination of the shear is easy.
We might give tables for shears directly, as above, for reac-
tions; but this is unnecessary. Having taken the reactions
from our tables already given, and found the moments either by
our formulae or tables, we can then find the shears both by means
of the reactions and also directly from the moments them-
selves, and thus check at once the accuracy of our determina-
tion of both. From what has already been given, the reader
can easily construct tables of shears similar to those alreadj^
given for reactions for himself, if desired.
149. Recapitulation of Formulae— Continuous Girder
over any Number of Level Supports, all Spans equal.*—
For notation, see Art. 130, Mg. 89.
* These formulae are given in similar form in Winkler's "Der Lehre von
der Elasticitaet und Festigkeit," Art. 144, p. 122. They were also indepen-
dently deduced by Mr. Merriman, to whose kindness we are indebted for much
of this chapter. The above methods of tabulation were communicated by
him, and are given in no treatise upon the subject.
222 CONTINUOUS GIRDER. [CHAP. XIII.
1st. Moments at supports.
When m<r + l, M = - C A gg~r+2 + A/ GS-*+I
when m > r, Mm = -
in which & = * and A = P J (2& - 3/P-f /fc3), A' = P J (^- #)
for concentrated loads, and A = A' = J w Z2 for a uniform
load over any one span.
2d. Shear at the supports.
In the loaded span, to the right of the left support,
To the left of the right support,
S'r+1 = »!
« r
In the wMloaded spans, to the right of the left support,
8 _
To the left of the right support,
' Mm ~ Mm-1
For the reaction at any support, Rm = S'm + Sm.
3d. Reactions.
(a) Abutment reactions :
when r = 1, R1=-?^ + ^; when r = s, Rs+1 = - — s+/;
when y > 1 and < s, Rt =: — ^ Rs+1 = — ???.
^ ^
(5) Eeactions at supports adjacent to loaded span (when this
span is not an end span) :
CHAP. XIII.] ANALYTICAL FORMULAE. 223
(<?) All other reactions :
Where for concentrated load,
A = P I (2 Jc- 3 & + IF) A' = Pl(%-X*)
q = P (1 — Jc) qf = P Jc.
For uniform load in single span,
A = A' = J w P, q = c[ = \ w I, Jc being always y, and
Cl = 0 c, = 15 c7 = - 780 cw = 40545
c2 — 1 c5 = - 56 c8 == 2911 cu = - 151316
c3 — - 4 e6 = 209 c9 = - 10864 c12 — 564719, etc.
We give also, for sake of completeness, although not needed
for calculation, the formulae
FOR UNIFORM LOAD OVER ENTIRE LENGTH OF GIRDER.*
Moment at any support,
Reactions at abutments,
Reactions at other supports,
where, as before, A = ^ wP, q — \ w I,
and the numbers indicated by 5 are as follows :
&! = 0 &4 = - 3 58 = - 41 &a = - 571
5a = 1 £5 = - 4 J9 = - 56 518 = - 780
J3 = 1 56 = + 11 J10 = 4- 153 etc. ,
J7 = + 15 ln = + 209
* The above equations were first given by Mr. Merriman, in the Jour.
Franklin Institute, April, 1875.
224 CONTINUOUS GIRDER. [CHAP. XIII.
These numbers change signs by pairs alternately, and every
other one follows the law of the Clapeyronian numbers. In
fact those with the odd indices are those numbers, and the even
ones, commencing with 0, 1, 3, follow the same law. '
The above comprises all the formulae thus far given, in a
shape very convenient for reference. The reader who has fol-
lowed attentively our explanation of their use, needs nothing
more to solve the case of equal spans completely. The expres-
sions, however, for the reactions are unnecessary. As we have
seen from Art. 148, we need only the moments and shears at
any support in practical calculations. The practical formulae
necessary and sufficient for any case will be found in the next
Art.
15O. Girder continuous over any Number of Level Sup-
ports ; Symmetrical with respect to the Centre, and with
two variable end Spans n I and p I on each side. — [Fig.
89.]*
Moments at Supports :
Mm =
when m> r, IV^ = %*» - A <* + .
I p CB_! + 2 (n + p) ca
Shear at Supports — loaded span,
4
sWi = - ~±V~ ~"~ "*" ^' '
wwloaded spans,
• B^=M,-M^t
Mm - m^^
For uniform load,
A = A' = J w fr ; q — q' = \ w lr
* Jour. Franklin Institute, March and April, 1875.
CHAP. XIII.] ANALYTICAL FORMULA. 225
For concentrated load,
A = P g (2 k - 3 & + #), A7 = P £ (& - If),
and ^ = P (1 — &), £' = P k, k being always •=-.
^r
The quantities denoted by c are also as follows :
_
+ 12 j9) - n (14 +
j> (52 + 45 j>) + ^(52 + 60 p)
T
_ -p (194 + 168 p) - n (194 + 224 p)
-J~
p (724 ^ 617^) + w (724 + 836^)
c8 — - — — — — , etc.,
following the law of the Clapeyronian numbers.
151. Application of the above Formulae. — The for-
mulse of the preceding Art. comprise in a most compact form all
the formulae hitherto given, and are all that is necessary for the
complete solution of any practical case.
Thus, by making p — unity and retaining only 71, we have
the case of a girder with variable end spans n I, of different
length from the others, which latter are all equal and repre-
sented by I. The reader will find no difficulty in using the
above. For any particular case, when w or P and Z, k, n and
p are given, A, A7, q and q' can be easily found, and the prob-
lem is solved. If n and^) be both unity, we have the formulae
for all spans equal. The expressions for Mm will then reduce
to those already given in Art. 144. Thus, in Art. 145 we have
already found for seven equal spans, I = 80 ft., load P = 40
distant 50 ft. from left ; the moment M4 = 202.4. Now, from
our formulae above, we find for MS making m — 5, 5 = 7,
r = 4; M5 = 272.8.
Then by our formulae for shear, S4 = + 14.12, or nearly
15
226 CONTINUOUS GIKDER. [CHAP. XIII.
what we have assumed in Art. 145. We may also find the
same shear by finding the algebraic sum of the reactions at
ABC and D from the formulae of Art. 147. This is more
tedious, and, as we see, unnecessary. The moments can be
easily found, and then the shear obtained directly from these.
We must bear in mind that lr always denotes the span the
load is upon, whether nl,pl, or I, while lm is any span in gen-
eral, according to the value of m.
152. Continuous Girder with fixed ends.— It is worthy of
remark that if n be made zero in the formulas of Art. 149, we
have a girder with fastened ends and variable end spans p I.
If in addition^? is unity, then all the spans become equal. We
must, however, remember that when we thus make n = 0, the
number of spans is s — 2 instead of s, as before/ and the end
spans are p I ; the end supports are also 2 and s instead of 1
and s + 1.
153. Examples. — As illustrations of the use of the formulae
of Art. 150, we give a few examples.
Ex. 1. A beam of one span is fixed horizontally at the ends.
What are the end moments and reactions for a concentrated
weight distant k \from the left end?
Here the two outer spans of three spans are supposed zero.
Therefore, 5 — 2 = 1, and s = 3. The left end is 2 instead of
1, and the right end 3. Hence, r — 2,^> = 0, and n = 1 in the
formulae of Art. 150. We have, then,
d = 0, c2 = 1, CQ = — 2, £4 = 4> and hence,
or, inserting the values of c above,
M2 = 1(2A-A'), M3=-1(A-2A').
For a concentrated load, A = P Z2 (2 k - 3 # + #),
and A' = P P (& - ;&»). Hence, M2 = P I (k - 2Z? + .#),
and M3 = P I (1$ - Vs).
For the reaction at the left end, which is in this case the
same as the shear, we have
CHAP. Xm.] ANALYTICAL FORMULAE. 227
52 = M2-Ms + P (1 _ k) or S2 = P (1 - 3 # + 2 #),
f
53 = M3-Ms + P k or S3 - P (3 # - 2 #).
^
For a load anywhere, we have simply to give the proper value
to k, and we have at once the reactions and moments. Thus,
for a load at \ the span from the left, Tc, = £, and
s2 = fjp, SS = AP; M2 = &Pi,mi = &Pi
For a load in centre, k — J, and
[Compare Supplement to Chap. VII., Arts. 16 and IT.]
Ex. 2. J^r # uniform load over the same beam, what are the
end moments and reactions f
We have simply to introduce the proper values of A and A'
for this case, and we have at once
Mg — T^- w Z2 = M3 and S2 = S3 = £ w I
Ex. 3. A girder of three equal span's is " walled in " at the
ends, and has a concentrated load in the first span. What are
the moments, shears, and reactions at the ends and intermediate
In this case, s — 2 = 3, and hence s = 5, r = 2, n — 0,
j9 = 1, and therefore
C2 A i5 + A' c4
= T *4 + 2* '
frA^ + A'fr ^AC^ + A^
2 c4 4- 2 c?5 ^ c4 + 2 c5
also, ct = , 0 c2 = 1, <k — — 2, c4 = 7, c5 = — 26, etc.
Inserting these values and the values of A and A' for con-
centrated load, we have
P 7 91
= - (3 & - 3 #), M5 = (3 ^ - 3
4bO 4:0
228 CONTINUOUS GIEDEE. [CHAP. XIII.
Observe that the moments are positive at each end of the
loaded span and alternate in sign from that span, varying as
the numbers 1, 2, 7, or as the numbers c [Art. 140]. A positive
moment always denotes compression in lower fibre. For the
shears we have, then, from Art. 150,
S2 = -?- (45 - 99 # + 54 #% S'8 = -^ (99 & - 54 Jf).
4o 45
Ss = ^ (27 ^ - 27 #), S'4 = ^ (- 27 /fc2 + 27 #),
4:0 4:5
S4 = ^ (- 9 # + 9 #), S'5 = ^ (9 #> - 9 #).
4-0 4:0
For the reactions, then,
Ra = S2, R3 = S'8 + S8 = ^ (126 # -81.&3),
4-O
R4 = S'4 + S4 = ?- (- 36 & + 36 #), R5 = S'5.
4:0
Observe that the reactions as also the shears are positive at
the supports of the loaded span, and alternate in sign from
those supports. A positive shear or reaction acts always up-
wards. Disregarding, then, for the present, the weight of the
beam itself, it would have to be held down at first pier from
right end.
If the weight is in the centre of first span from left, Jc — J, and
36 _ 9
360 ' - 360 ' 360 ' ^ ~ 860
..« _p B R _
^ ~ ' *• - ' 4 ' ^ ~ *
The reactions add up to P, as they should.
If P = 100 tons, and 1 = 16 ft., we have
M2 = 237.5 ft. tons, M3 = 87.5, M4 = - 50, M5 = 25 ft. tons.
R2 = 60 tons RS = 47.5, R4 = - 10, R5 = 2.5 tons ;
S2 = 60 tons, S'8 = 40, S8 = 7.5, S'4 = - 7.5,
S4 = -2.5, S'5 = 2.5.
Ex. 4. A beam of five spans, free cut ends ; centre and adja-
cent spans 100 ft., end spans each 75 ft., has a uniform load
extending over the whole of the second span from left. What
are the moments at the ends and supports f
CHAP. Xin.] ANALYTICAL FORMULAE. 229
Here s = 5, n = f , p — 1, r = -2 ; therefore, from Art. 150,
.2_A.5 + A'*4 ^A.2+ A'.3
* ^4 + 1 4, Z C4 + J C5
and <?! — 0, £2 = 1, <?3 = — f, c4 — 13, £5 = — 48.5.
w Z3
Since, then, A = A' = — — for uniform load, we have
! = 0, M2 = -g-VV ™ P> M3 - Tf I* w ?, M4 = - y 35^ w
M ^ 0.
If the load is two tons per ft., w 1? = 20,000, and
Mt = 0, M2 = 414.7, M3 = 1036.6, M4 = - 279.5, M5 = 79.7.
Find the shears and reactions at each support.
Ex. 5. A beam of four equal spans, has the second span from
left covered with full load. What is the moment and shear at
left of load f
Ans. M2 = Tg^j- w I*, S2 = JJJ- w I.
What at right of load ?
Ans. M3 = 3^- w I, S'3 = % Jf w I.
What are the formulae for concentrated load ?
5o
TV/r — r'r Z> 1 0 Z»2 i K L3~| ~p 7
J-TXo f /*» I * tv ~~" -L^ 'v ~i *-^ *v I ^ v*
OD
S2 = ^ [56 - 58 ^ + 3 t? - ^3],
S'8= [58^-3^ + ^] ^>-
Examples might be multiplied indefinitely.
The above is sufficient to show the comprehensiveness of our
formulae, and the ease with which results may be obtained,
which, by the usual methods, would require long and intricate
mathematical discussions. The points of inflection and the
deflection may also in any case be easily determined, and gen-
eral equations similar to the above deduced, but, as we have
seen, the above are sufficient for full and complete calculation.
154. Table* for Moment*. — From the formulae of Art. 150
we can easily find the moments for both uniform and concen-
trated load in a single span for various numbers of spans. If
these results are tabulated we, shall obtain tables from which
230
CONTINUOUS GIEDEE.
[CHAP,, xm.
the moments may be at once taken. The formulae for the
shears are so easy when for any case the moments are known,
that it is unnecessary to give tables for these.
The reader will do well to make himself perfectly familiar
with the formulae by calculating the moments for various cases,
and comparing with the following tables. We give the prac-
tical case of variable end spans n I and equal intermediate
spans I.
TABLE FOB MOMENTS — UNIFORM LOAD OVER ANY SINGLE SPAN.
Coefficients 'of w I2 from talk. End spcuns n 1.
Supports counted from left.
1 2
3
4
5
6
0 n*
(2 + 3n)n3
(7 8«)»»
(26 + 3071)^3
(97 + 11271)7*3
I.
0 1+271
(l + 2ra) (2 + 27i)
(1+270(7+871)
(l+27i)(26 + 80n)
etc.
II.
0 5 + 0 n
(5 + 6ra)(2 + 3/i)
(5 + 6 n) (7 + 8 n)
etc.
III.
0 19 + 2271
(19 + 22»)(2 + 2»)
(19 + 227i)(7 + 87i)
IV.
0 71 + 8271
(71+83n)(2 + 8n)
etc.
V.
0 365+ 304 n
etc.
VI.
Number of
spans.
One-fourth of Denominator.
1
2
3
3 + 8 Ti + 4 7ia
4
12 + 2871 + 16 n»
5
45 + 104 71 + 60 7i2
6
168 + 3837* + 224 r
I*
The above table can be easily extended to include any num-
ber of spans. It is precisely the same as the table of Art. 140,
and, in fact, includes that table. We have only to make n = 1
and we have at once the table for equal spans. Suppose we
take five spans, load in second from right. From the smaller
table we have at once for the denominator of the coefficient
of w Z2, 4 (45 + 104 n + 60 n2). Then from the other table
we have at support 1 from left Mx = 0,
(l+2n)wP
at support 2,
4 (45 + 104 ?i
ANALYTICAL FORMULA.
CHAP, xm.]
, at support 3, M3 •— (1 4- 2 n) (2 + 2 n) w
231
, etc.
4 (45 4- 104 n + 60
If the load were in second span from left, and supports to
right of load were required, we have simply to count the sup-
ports the other way in the table.
Thus, 3YE6 = 0, 3YC3 = — - — - — — - ., or same as 3VT2
in first case, etc.
TABLE FOR MOMENTS CONCENTRATED LOAD IN ANY SPAN, k = — .
End spans n 1. Coefficients of P \Tfrom table.
a = a'
(P
y
0=0'
<i
V'
11.
3 + 271
1
2n
1
2 + 271
3 + 4»
II'.
III.
9 + 10/1
2 + 2 n
3 + 471
2 + 271
7 + 871
12 + 1471
III'.
IV.
33 + 38%
7 + 871
12 + 14w
7 + 87i
26 + 307*
45 + 5271
IV.
v.
123 + 142 n
26 + 3071
45 + 52n
26+3071
97 + 11271
168 + 19471
V.
VI.
459 + 53071
97 + 11271
168 + 194 n
97+11271
362 + 41871
627 + 72471
VI'.
VII.
etc.
etc.
etc.
etc.
etc.
etc.
VII'.
Number of
spans.
Denominator A.
2
3
3 + 87i + 4n2
4
12 + 2S?t + 167i2
6
45 + 10471+60712
6
168 + 388 »+ 224712
7
etc.
The above tables give only the moments at the supports of
the loaded span. The Roman numerals L, II., III., etc., denote
the number of this span from left, and I'., II'., III'., the num-
ber of the loaded span from right. The expression for the
moment at left support is
M
= P lr^ [7' & - 3 ff
CONTINUOUS GIRDER. [CHAP. XIII.
For the right support,
where the expressions for 0, 0', A, 7, 7'? £ £', a, a', are to be
taken from the tables and inserted.
Thus, for five spans load in second span from right, or third
from left, we have at once A = 46 + 104 n + 60 n\ For the
moment at the left support, to find 0, we must take horizontal
line for III., and thus find (2 + 2 *). For 7', /3' and a' we
must take II'., and find, therefore, 3 + 4 n, 2 + 2 n and 3 + 2 n.
Hence, moment at left support is
For moment at right support, we must take line II'. for 6' and
line III. for 7, ft and a, and hence
Since the span in question is not an end span, 1^ = 1 and
For a load in an end span, use the formulae of Art. 150.
For a load in middle span of five spans, i.e., third span from
each end, we have
+- n n) *~3 (7+8 n) *' +
M '=«•=
When Jc = -=, both these moments become, as they should, equal
2i
for any assumed value of n, as the reader may readily prove by
insertion.
For the other moments not adjacent to the loaded span, the
rule of Art. 140 holds good.
Thus, Mt = 0, M, = I M,, M8 = J M,, M4 = -J M,, etc.,
P " "
CHAP. Xin.] ANALYTICAL FOKMUL^EJ. 233
and similarly on the other side,
M8 = 0, M^ = * Mr+1, Ms_2 = Jz? M;^, etc.
C/r+1 UI+l
We must remember always to give the proper signs to the
moments, viz., positive for extremities of loaded span, and
alternating each way from these for the others. From the
formulae of Art. 140 we can then easily find the shear at any
support.
155. Continuous Girder— I^evel Supports — Spans all dif-
ferent— Oeneral Formulae.* — The preceding formulae com-
prise the case of one or, at most, two variable end spans. We
give below the general formulas for all spans different. These
formulas include all the others as special cases. Thus, if we
make all spans equal, we have the formulae of Art. 149. If end
spans \ and ls are made zero, and we take the number of spans
equal to s— 2, and first support 2, we have the continuous gir-
der with fixed ends, in which the intermediate spans may or
may not be equal, as we choose. If we make ^ — 0 or Z8 = 0
alone, and s — 1 = No. of spans, we have a continuous girder
fixed at one end only. In short, the formulae comprise the
entire case of level supports. They are as follows :
Let s = number of spans, 4=length of loaded span, 7c = — ,
Ly
a being distance of load from left support ; l^ 4> 4 • • • • 4-i> fc
the lengths of the various spans counting from left.
Then, when m < r + 1,
TVT — /« A
m4<-i
when m > r,
M ±A A er + B cr+l
A™-m — "'s— m+27 — — ; — o (1 i — 7 - \ — *
V_i tfs-i + 2 (4 + ^.j ca
For the shear at supports of loaded span,
gr = Mr-Mr+1 + g SWi^M -Mr +
6r lt
For -zmloaded spans,
* These formulas were first given by Mr. Merriman, and may be found In
the London Phil. Magazine, Sept., 1875.
234 CONTINUOUS GIEDEE. [CHAP. XIII.
For the reaction at any support, Rm = S'm + Sm. In which
we have always Jc = - , q = P (1 — #), q' = P Jc.
i>r
A = P lr2 (2 k - 3 & -f If) and B = P I* (k - F) for concen-
trated load ; and q = q' = ^ wliy and A = B = % wl* for uni-
form load entirely covering any one span.
Also for G and d we have the following values :
= 1,
2 ft +
"
_ 4 ft 4- 4) (4 + 4) - V
— ""'
or, generally,
_ 4: ft + d) (4-1 + 4-2) ~
74-7 7
"s— m+3 i ^s— m+2 J ^s— m+3
d5 = — 2 d±
or, generally,
dm= —2d
As an illustration of the use of the above formulae, let us
take three unequal spans, load in the first. Then s = 3,r = ~L,
Tc, = y. For moment at second support, m — 2, or m > ^ ; hence
^i
M -- /7 A ^i + B g2
^ *
CHAP. Xm.] ANALYTICAL FOKMUL^E. 235
But Ci = 0, C2 = 1,
hence, since B = P If (k — If),
_ 2 P y (* - Jf) (4 + 4)
* (4 + 4) (k + 4) -V
If in this we make \ = 4? we have the extreme spans equal,
and then
M _ a P y (* - #) ft + 4)
* (4 + 4)2 - V
If we make in this, again, \ = 4, we have for all spans equal
just what we should have from Art. 149.
For the reaction at the end support, we have
or, since Mt = 0,
For all spans equal, or 1^=1^=1^ = 1^ this reduces to
S! = Z (15 - 19 Jc + 4 #»),
15
as we should have found from Art. 149.
Ex. 1. — A beam of one span is fixed horizontally at the
right end ; what are the reactions and the moments for concen-
trated load ?
Here 5 — 1=1 or s = 2, r = 1, 4 = 0, and from the for-
mulae of Art. 155, cl = 0, c2 = 1, and <h = 0, d% = 1, d$ = — 2,
, A GI + B c2 B c%
.= -2 + P (!-&) = (2 -
'a = ? (3 *-£>).
236 CONTINUOUS GIRDER. [CHAP. XIII.
Ex. 2. — A 'beam of three spans of 25, 50 and 40 feet respec-
tively is fixed horizontally at the right end, and has a concen-
trated load of 10 tons at 12 feet from the third support from
left. What are the moments at the supports f
Here I, = 25, ^ = 50, 4 = 40, 4 = 0, P = 10, k 4 = 12,
Jc = 0.3, 5-1 = 3. s = 4, 4 = 0 and r = 3. Also, cl = 0,
c2 = 1, GS = — 3, c4 = 12.25 and ^ = 0, ^ = 1, 4 = — 2,
<Z4 = 6.4, d5 = - 32.4.
When, then, m < 4,
Inserting & = 0.3 and the values of <?,
for m = 1, M! = 0 ; m = 2, M2 =- 8.20 ; m = 3, M3 = 24.62 ;
for n = 4,
M4 = (_ 3 A + 12.25 B) = (6.25£ + 9#- 15.25 #),
or M4 = 42.29 ft. tons.
Find the shears. Also moments and shears for uniform load
over third span.
Ex. 3. — A beam of four spans \ = 80, 13 = 100, 13 = 50,
14 = 40 ft., free at the ends, has a load of 10 tons in the sec-
ond span at 40 ft. from left. What are the moments?
Here s = 4, ^ = 0, c* = 1, <?3 = — 3.6, c4 = 19.6, c5 = -83.7,
dj. = 0, 4 = 1, <& = - 3.6, d, = 10.3, 4 = - 41.85, r = 2.
For m < 3, Mm = - ^- (A ^4 + B 4),
m>3,
Hence, Mj = 0,
Us <17 * - 30-9 * + 13-9 *> =
M5 = 0.
Find the shears. Also find the moments and shear for uni
form load over second span.
CHAP. XIII.] ANALYTICAL FORMULA. 237
156. Thus we see that, as in Art. 150, a few short and sim-
ple formulae, which may be written on a piece of paper the size
of one's hand, are all that we need for the complete solution of
any case of level supports — whether the spans be all equal or
the end ones only different, or all different ; whether the girder
merely rest on the end supports or be fastened horizontally at
one or both ends. We have only to remember that a positive
moment causes tension in upper flange at support, and there-
fore compression in lower; inversely for negative moment.
Also, that a positive shear acts upwards, and a negative shear
downwards. Also, that both moment and shear are positive at
supports of loaded span, and alternate in sign both ways. This
is all that we need to form properly the equation of moments
at any apex, and determine the quality of the strains in flanges
and diagonals. We can thus solve any practical case of framed
continuous girder which can ever occur with little more diffi-
culty than in the case of a simple girder.
Thus, for the span DE (Fig. 87) we have only to find the
moments at D and E due to every position of P in the span
D E, and the corresponding shears at D. These once known,
and, as we have seen, they can be easily obtained from our
formulae, we can find and tabulate the strains in every piece
due to each weight, as shown in Art. 127. An addition of these
strains gives, then, the maxima of each kind due to interior
loading.
We have, then, to find, in like manner, the strains due to the
two cases of exterior loading as represented in Fig. 87. From
the four columns thus obtained, we can deduce the dead load
strains, and then finally the total maximum strains of each kind
for every piece. [See, for illustration of the above, Art. 127.]
Thus, the whole subject is solved with the aid of but four
simple formulas, and for a problem generally considered impos-
sible by reason of its " complexity," our results will, we trust,
be found sufficiently simple and practical.
In view of the fact that the necessary formulae for practical
computations have been often given in the later works of
French and German authors, although perhaps never before
in so compact and available a shape as above, it is indeed sur-
prising that they should have been so completely ignored by
English and American writers.
The tables and formulae which we have given will, we trust,
238 CONTINUOUS GIRDER. [CHAP. XIII.
bring the subject fairly within the reach of the practical engi-
neer, and should they be the means of calling more general at-
tention to this important class of structures, will not, we hope,
be considered as out of place in the present treatise.
For the influence of difference of level of the supports, as well
as for variable cross-section and the relative economy of the
continuous girder, see Arts. 17 and 18 of the Appendix.
ART. 1.] SUPPLEMENT TO CHAP. XIH. 239
SUPPLEMENT TO CHAPTER XIII.
DEMONSTRATION OF ANALYTICAL FORMULAE GIVEN IN TEXT.
IN the following we shall give the complete development of the general
formulae of Art. 155. As these formulae include, as we have seen, all the
others as special cases, it is sufficient to show how they are obtained in
order to enable the reader to deduce all the others.
1. Conditions of Equilibrium. — In the rth span of a continuous
girder, whose length is IT (see Fig.), take a point o vertically above the rih
support as the origin of co-ordinates, and the horizontal line o I as the axis
of abscissas. At a distance x from the left support pass a vertical section,
and between the support and this section let there be a single load Pr
whose distance from the support is a.
Now all the exterior forces which act on the girder to the left of the
support r we consider as replaced, without disturbing the equilibrium, by
a resultant moment Mr and a resultant vertical shearing force Sr. This
moment is equal and opposite to the moment of the internal forces at the
section through the support r ; while the vertical force is equal and oppo-
site to the shear.
Not only over the support, but also at every section, the interior forces
must hold the exterior ones in equilibrium, and therefore we have the con-
ditions :
1st. The sum (algebraic) of all the horizontal forces must be zero.
2d. The sum (algebraic) of all the vertical forces must be zero.
3d. The sum (algebraic) of the moments of all the forces must be zero.
Thus, for the section #, we have from the third condition
2 M = Mr - Sr x + Pr (x - a) - m = 0 . . . . (1)
where m is the moment at the section. From this we have
m = Mr — Sr x + Pr (x — a) . (2)
If in this we make x = 1& m becomes Mr+i, and we thus have for the shear
just to the right of the left support of the loaded span
24:0 SUPPLEMENT TO CHAP. XIII. [ART. 2.
For an unloaded span the weight P disappears, and
For the shear just to the left of the right support of loaded span,
For unloaded span, the weight P disappears, and
S'm is then the shear to the left of any support m, and Sm that to the
right. The reaction at any support is therefore
Rm = S'm + Sm.
These are the formulae already given in Art. 148.
2. Equation of tlie Elastic Line.— We can now easily make
out the equation of the elastic line for the continuous girder of constant
cross-section, or constant moment of inertia.
The differential equation of the elastic line is,*
where E is the coefficient of elasticity, and I the moment of inertia.
If now we insert in (3) the value of m, as given in (2), we have
d? y _ Mr — Sr x + PT (x — a)
dH?~ El
Integrating f this between the limits x = 0 and .?, and upon the condition
that x cannot be less than a, the constant of integration ~~ = tr = the tan-
gent of the angle, which the tangent to the deflected curve makes with the
horizontal at r ; and we have, since we must take the / Pr (x — a) simul-
taneously between the limits x — a and z for x — 0 and x ;
dy_ 2Mrg-Sta? + Pr(g-a)«
cTx-tr+ 2EI ' (3'^
If we take the origin at a distance hT (see Fig.) above the support r,
then integrating again, the constant is hT, and we have
which is the general equation of the elastic curve. If in this we make
* See Supplement to Chapter VII., Art. 12.
f Notice that when x = 0, a — 0, and hence (x — a) — 0 also.
ART. 3.]
SUPPLEMENT TO CHAP. XIII.
x — lr, y becomes hr+i. If also we put - = &, or a = k Zr, and insert also
for Sr its value as given in (2 a), we find for
_ _ 2 Mr Zr -f Mr+1 Zr - Pr Zr2 (2 * - 3
..(5)
We see, then, that the equation of the curve is completely determined,
when we know Mr and Mr+1, the moments at the supports. These, as we
shall see in the next Art., are readily found by the remarkable "theorem
of three moments," already alluded to in Art. 144.
3. Theorem of Tliree moments.
In the Fig. we have represented a portion of a continuous girder, the
spans being li Z2 . . . /r? etc., and Ihe supports 1, 2 ... r, etc. Upon the
spans IT-I and lr are the loads Pr_i and Pr, whose distances from the near-
est left-hand supports are k /r_i and k lr ; k being any fraction expressing
the ratio of the distance to the length of span.
The equation of the elastic line between Pr and the r + 1th support is
given by (4), and the tangent of the angle which the curve makes with the
axis of abscissas is given by (3 a). If in (3 «) we substitute for ST its
value from (2 a), and for tr its value from (5), and make at the same time
x — lr, then -£ becomes
a x
1, the tangent at r + 1, and we have
2 M
r+1
- pr
Remove now the origin from o to ft, and we may derive an expression
for tr by simply diminishing each of the indices above by unity ; therefore
7, T, ^ I- -I
--
^ Zr_i + 2 Mr ^-Pr,! Zr2_! (*-
r— 1 -
Now, comparing these two equations, we may eliminate the tangents,
and thus obtain
Mr_i ?r_! + 2Mr (7r_! + Zr) + Mr+l t =
_ 6 E
(2 yfc-
which is the most general form of the theorem of three moments for a
girder of constant cross-section.
When the ends of the girder are merely supported, the end moments are,
of course, zero. Then, for each of the piers, we may write an equation of
the above form, and thus have as many equations as there are unknown
moments.
16
242
SUPPLEMENT TO CHAP. XIII.
[ART. 4.
4. Determination of the Moments— Support§ all on
level.— When all the supports are in the same horizontal, the ordinates
hij hi, #r, etc., are equal; and hence the term involving E I disappears,
and we have simply
Mr_! Zr_! + 2 M? (Zr_i +lr) + Mr_! Zr =
Pr_i IJ-i (Tc - F) + Pr V (2 k - 3 #» + &3),
as already given in Art. 144.
Now let s = number of spans, and let a single load P be placed on the
rth span. [PL 23, Fig. 89.]
From the above theorem, since M! and Ms+1 are zero, we may write the
following equations :
2M2 (I, + Z2) + M3Z2 = 0;
M2 Z2 + 2 M3 (Z3 + 13) + M4 Z3 = 0.
Zr)
Mr+1 ^ =
= A
Mr_! Zr-l + 2 Mr (Zr_
PZr2(2&- 3
Mr Z,. + 2 Mr+1 (Z,. + Zr+i) + Mr+2 Zr+i =
P Z,-2 (& - P) = B.
Ms_2 Zs_
Ms ^ = 0;
=, 0.
(6)
2 Ms_! (7S_2
Ms_i Zs_! + 2 Ms (Zs_! +
The solution of these equations can be best effected by the method of
indeterminate coefficients, as referred to in Art. 136.
Thus we multiply the first equation by a number c2, whose value we
shall hereafter determine, so as to satisfy desired conditions. The second
we multiply by c3j the third by c4, the rth by Cj.+i, etc., the index of c cor-
responding always to that of M in the middle term. Having performed
these multiplications, add the equations, and arrange according to the co-
efficients of M2, M3, etc. We thus have the equation
[2 C2 (li + Z2) + C3 Za] M2 + [ca Z2 + 2 Ca (la + Z3) + C4 Z3] M3 + . . .
+ [Cr_i Zr_i + 2 Cr (Zr_i + ZT) + CT+I Zr] Mr + . . . .
+ [cs_2 Zg_2 + 2 cs_i (Zs_2 + Zs_i) + CB Zs_i] Ms_!
+ [Cg_i Zs_i + 2 C8 (Zs_! -f ?„)] Ms = A Cj. + B Cr+1.
Now suppose we wish to determine Ms. We have only to require that
such relations shall exist among the multipliers c that all the terms in the
first member of the above equation, except the last, shall disappear. We
have then evidently, for the conditions which these multipliers must
satisfy,
21 c2 (Zi + Z2) + Z3 Z2 = 0 ;
c2 Z2 + 2 c3 (Z2 +Z3) + c4 Z8 = 0 ;
Cr_l Zr_! + 2 C,. (?r_! + Zr) + CT+I Z,. = 0 J
Cs_2 Z8_2 + 2 C8_! (Zs_2 + Z8_!) + CB Z?_i = Oj
ART. 4.] SUPPLEMENT TO CHAP. XIH. 24:3
while for Ms we have at once,
__Acr
- _
Cs_i Zs_i + 2 Cs (la_-L + Zg ) CB+i I*
If, in like manner, we should multiply the last of equations (6) by the
number d2, the last but one by d3, the rth by <4_r+1, etc. ; then add, and
make all terms, except that containing M2, equal to zero ; we should have
the conditions :
2ds(k + Zs_i) + <ZsZB-i=0;
d* 4-1 + 2 d3 (4_i + 4-2) + dt 4_2 = 0 ;
2 <4-r+2 (4 + 4-l) + ^s-r+3 4-1 = 0 ;
4_i Z3 + 2 ds_i (lt + Za) + <4 Z2 = 0 ;
while for the moment we have
The values of M2 and Ms are thus given in terms of the quantities
A and B and c and d.
A and B depend simply upon the load and its position in the rth span.
Thus A=PZr2(2£-3&2 + P), B = PZr2(fc-F).
As for the multipliers c and d, they depend only upon the lengths of the
spans, and need only satisfy the conditions above. Hence, assuming
d = 0, ca = 1, and dl = 0, d* — 1, we can deduce the proper values for
all the others. Thus,
d = 0, d, = 0,
ci --1, d* = 1,
0 Za + ?3 Za ^ 0 T Zg_i -f Zs_2 , Zs_
C4 = — 2 C3 — - -- C2 -, »4 = — 2 rfs - 5 - — rfa ^ —
Zs Z3 Z8-2 4-
0 Zs + Z4 Zs , n J 4-2 + 4-3 * 4-
C6 = — 2 C4 — = - — C3 — , a*, = — 4 Ci4 - - - — «3 j
O ^ ^4 + ^5 ^ ^4
Ce = — 2 Cs — = - — C4 — ,
65 ZB
4-
etc., etc. etc., etc.
Now from equations (6) we see at once that M3 = c3 M2, M4 = c4 M2,
etc., or, universally, when n < r + 1,
. . (7)
Also taking the same equations in reverse order, M^ = d* M8, Ms_2 =
d* MB, etc., or, universally, when n > r,
... (8)
SUPPLEMENT TO CHAP. XIII. [ARTS. 5, 6.
Equations (7) and (8) are the general equations given in Art. 155, which,
as we have seen, include the whole case of level supports.
5. Uniform Load. — For uniform load the same equations hold
good. We have only to give a different value to A and B.
Thus, for several concentrated loads we should have
A = S P Zr2 (2 lc - 3 F + &3).
For a uniform load over the whole span I r, let w be the load per unit
of length, then
ri ri
2P = I wd a; or since a = lc Zr, 2 P = / wlTdlc.
Jo Jo
Inserting this in place of 2 P above, and integrating, we have
A = B = i w Zr3.
Thus the equations of Art. 155 hold good for concentrated and uniform
load in any span, for any number and any lengths of spans.
The above formulae were first published in an article on the Flexure of
Continuous Girders, by Mansfield Merriman, C.E., in the London Phil,
Magazine, Sept., 1875.
6. Formulae for the Tipper. — The expressions for the reactions
in this case, already given in Art. 120, may be easily deduced. The solu-
tion is tedious by reason of lengthy reductions, but the process of deduc-
tion is simple. -
The construction in this- case is indicated in Fig. 83, PL 22. We sup-
pose, as shown there, a weight upon the first span only. Under the action
of this weight the beam deflects, and one centre support falls and the
other rises an equal amount. Thus, if we take the level line as reference,
A2 = — ha. Moreover, the reactions at these two supports must always be
equal.
We have, then, as representing this state' of things, A2 = — A3, and calling
the supports 1, 2, 3 and 4, we have from Art. 1, since Mj — M4 = 0, and
Zi = Za,
R, = Si = S, = - ^ + P (1 - jfe),
•B — a' ms
**-St = --!7'
These reactions will evidently be known, if we can determine the mo-
ments.
Let Yr = 6 E I F^r - ftr-l + ftr-ftr + ri Then the gen_
L 4r-l TT J
eral equation of three moments of Art. 3 becomes, when we neglect Pr, that
is, suppose only the first span loaded :
Mr_i Zr_i 4- 2 Mr (7r_! + Zr) + Mr + 1 Zr = - Tr 4 Pr-l V-l (^ ~ *")•
This expresses a relation between the moments at three consecutive
AET. 7.] SUPPLEMENT TO CHAP. XIII. 245
supports for load between the first two. Let r — 1 = 1, or r = 2. Then,
since Mj = M4 = 0, we have
2Ma(fa + fa)+M3fa = -Ya + Pfaa(;&-A;3)=R . . (10)
where R stands for convenience equal to the expression on right.
Let r — 1 = 2, or r = 3. Then the weight disappears, and since fa = fa,
M3fa + 2M3(fa + fa) = -Y3 ...... (11)
From (11) we have
But since R2 must always equal R3, we have from (9)
M.-M. »_».-- 8M. = _Pit
fri ta
Substituting (12) in (10), we have
_ R k „ a Y, (fa + fa)
3fa'+8fafa + 4fa'
Substituting (12) in (13),
_fafa«Pft-Ya(fa + 2fa)
8fa'+8fafa + 4fa'
From (14) and (15), we have then
- Y3 - R = I, fa P k.
Insert in this the value of R from (10), and
Ya - Y3 = P I? (k - If) + P I, fa T& = P (IS k - fa2 ¥ + fa fa A).
Now in the present case hi = 0, 7it = 0, and h2 — — A3, and since also
fa = fa,
and Y2 = 6Eir3--+-l- That is, Y2 = - Y3.
Hence, from our equation above,
Substituting these values of y2 and y3 in (11) and (13), we can obtain at
once M2 and M3, which finally substituted in eq. (9), will give us the re-
actions as already given in Art. 120, when we put n I in place of fa.
7. In similar manner we can solve other problems. Thus — what are the
reactions for a girder continuous over three supports, the two right-hand ones
resting upon an inflexible body which is pivoted at the centre ?
This is the case of the tipper when raised at the centre so that the ends
just touch, and then subjected to a load at any point of first span — the
other end not being latched down, so that it rises freely, as though without
weight of its own.
246 SUPPLEMENT TO CHAP. XTTT. [ABT. 7.
In this case we have from (9), since now M3 = 0,
By the conditions Ra must equal R3, hence
2 I, M2 + Z2 M3 = - ll I, P TQ . . . *• . . (16)
From the equation of three moments above, we have, making r — 1 = 2,
or r = 3, since then P disappears,
M2Z2 = - Y3 . . ,•> , .... . . (17)
ivr Ts
or M'=-T
Substituting this value of M2 in (16), we find
Y, = ^*; hence M. = - ^
2 ii + ta <« 61 + ta
and therefore, at once,
Putting w Z in place of Z2, we have
R! + 2 R2, it will be observed, equals P, as should be.
The conception of a beam tipping, as in the last two Arts., is due to Clem-
ens Herschel (Continuous, Revolving Drawbridges, Boston, 1875), and the
above formulae were first deduced by him in the above work.
LITERATURE UPON THE CONTINUOUS GIEDEE. 247
LITERATURE UPON THE CONTINUOUS GIRDER.
"We give below, for the benefit of students and those interested in the
subject, a list, chronologically arranged, of works upon the continuous
girder. A glance at this list will convince the reader as to the thorough-
ness with which the problem has been treated.
1. REBHANN. — "Theorie der Holz-und Eisenconstructionen." Wien,
1856. — [Treats the continuous girder of constant cross-section and equal
spans according to the old method ; first determining the reactions at
the supports. A load in any single span only is considered, either total
uniformly distributed, or concentrated and acting at the centre.]
2. KOPKE. — " Ueber die Dimensionen von Balkenlagen, besonders in
Lagerhausern." Zeitschr. des Hannov. Arch. u. Ing. Ver., 1856. [The
simple and continuous girder. Attention is here first called to the advan-
tage gained from sinking the supports.]
3. SCHEFFLER. — " Theorie der Gewolbe, Futtermauern und Eisernen
Bracken." Braunschweig, 1857. [Continuous girder with total uniformly
distributed load, and invariable concentrated loading. Advantage of
sinking the supports.]
4. CLAPEYRON. — Calcul d'une poutre elastique reposant librement gui-
des appuis inegalement especes." Comptes rend us, 1857. [Here, for the
first time, the well-known Clapeyronian method is developed, by which a
series of equations between the moments at the supports is first obtained.
Application to total distributed loads, but varying in different spans.]
5. MOLLINOS ET PRONIER. — " Traite theoretique et practique de la con-
struction des ponts metalliques." Paris, 1857. [Treatment of the con-
tinuous girder of constant cross-section, according to Clapeyron.]
6. GRASHOF. — "Ueber die relative Festigkeit mit Riicksicht auf deren
moglichste Vergrosserung durch angemessene Unterstiitzung und Einmau-
erung der Trager bei constantern Querschnitte." Zeitschr. des Deutsch.
Ing. Ver., 1857, 1858, 1859.
7. MOHR. — " Beitrag zur Theorie der Holz-und Eisenconstructionen."
Zeitschr. des Hannov. Arch. u. Ing. Ver., 1860. [Theory of continuous
girder, with reference to relative height of supports. Application to gird-
ers of two and three spans. Best sinking of supports for constant cross-
section. Disadvantage of accidental changes of height of supports. In-
fluence of breadth of piers.]
8. II. — " Continuirliche Briickentrager^" Bornemann's Civil-Ingenieur,
1860. [Continuous girder of constant cross-section of three spans. Best
ratio of spans, and sinking of supports.]
9. WINKLER. — "Beitrage zur Theorie der continuirlichen Briicken-
triiger." Civil-Ingenieur, 1862. [General Theory. Determination of
methods of loading causing maximum strains; and, for the first time,
general rules for the same given. Best ratio of end spans.]
248 LITERATURE UPON THE CONTINUOUS GIRDER.
10. BUESSE. — "Cours mecanique applique*e professe a" Fecole imperiale
des ponts et chaussees." Seconde Partie. Paris, 1862. [Analytical treat-
ment of the continuous girder of constant cross-section. The transverse
forces are not considered. The exact determination of the most dangerous
methods of loading, with reference to the moments in the neighborhood of
the supports, is also wanting.]
11. ALBARET. — "Etude des ponts metalliques a poutres droits reposant
sur plus de deux appuis." Ann. des ponts et chaussees, 1866. [Continu-
ous girder of constant cross-section, treated after Clapeyron.]
12. RENAUDOT. — "Memoire sur le calcul et le control e de la resistance
des poutres droites a" plusiers travees." Ann. des ponts et chaussees, 1866.
[Continuous girder, treated according to Clapeyron.]
13. CULMANN. — "Die Graphische Statik." Zurich, 1866. [Graphical
treatment of simple and continuous girder of constant and variable cross-
section. Moments at the supports are determined analytically.]
14. H. SCHMIDT.—" Ueber die Bestimmung der ausseren auf em Briick-
ensystem wirkenden Krafte." Forster's Bauz., 1866. [Data for the amount
of live load for Railroad and Way Bridges. Determination of the equiva-
lent uniformly distributed load. Data for dead weight and wind pres-
sure. ]
15. GRASHOF. — "Die Festigkeits Lehre." 1866. [General analytical
treatment of the girder without special reference to bridges. Continuous
girder of uniform strength.]
16. WINKLER. — " Die Lehre von der Elasticitaet mid Festigkeit." Prag.,
1867. — [General analytical theory of the continuous girder of constant and
variable cross-section. Application to total uniformly distributed loading.
Influence of difference of height of supports.]
17. FRANKEL. — "Ueber die ungiinstigste Stellung eines Systems von
Einzellasten auf Tragern iiber eine und iiber zwei Oeffnungen, speciell auf
Tragern von Drehscheiben." Bornemann's Civil-Ingenieur, 1868.
18. MOHR. — "Beitrag zur Theorie der Holz- und Eisenconstructionen."
Zeitschr. des Hannov. Arch. u. Ing. Vcr., 1868. [Here, for the first time,
the elastic line is regarded as an equilibrium curve, and the graphical
treatment of the continuous girder founded.]
19. H. SCHMIDT. — " Betrachtungen iiber Briickentrager, welche auf zAvei
und mehr Stiitzpunkte frei aufliegen, sowie iiber den Einfluss der unglei-
chen Hohenlage der Stiitzpunkte." Forster's Bauz., 1868.
20. COLLIGNON. — " Cours de mgcanique applique*e aux constructions."
Paris, 1869. [Continuous girder of constant cross-section and uniform
load.]
21. LAISSLE and SCHUBLER. — "Der Bau der Briickentrager init beson-
derer Riicksicht auf Eisenconstructionen." III. Ami., I. Theil. Stutt-
gart, 1869. [Treatment of the continuous girder, according to Clapeyron.]
22. LEYGUE. — " Etude sur les surcharges a conside"rer dans les calculs
des tabliers metallique d'appr§s les conditions generales d'exploitation
des chemins de fer." Paris, 1871.
23. LIPPICH. — " Theorie des continuirlichen Triigers constanten Quer-
sehnittes. Elementare Darstellung der von Clapeyron und Mohr begriin-
LITEKATUKE UI>ON THE CONTINUOUS GIKDER. 249
deten Analytischen und Graphischen Methoden und ihres Zusammen-
hanges." Forster's Bauz., 1871, also separate reprint. [The geometrical
constructions are deduced from the analytical formulae.]
24. SEEFEHLNER, G. — " A tobbnyngpontu vasr&cstartokrdl — A magyar
m6rrok es Spitesz— egylet kozlonye," 1871 [Hungarian].
25. HITTER, W. — " Die elastische Linie und ihre Anwendung auf den
continuirlichen Balken. Ein Beitrag zur graphischen Statik." Zurich.
1871. [This and the preceding work treat the continuous girder after the
Culmann-Mohr method.]
26. OTT.— " Vortrage uber Baumechanik," II. Theil. Prag, 1872.
[Analytical determination of the shearing forces and moments for the sim-
ple and continuous girder of constant cross-section and level supports.]
27. WEYRAUCH, J. I. — " Allgemeine Theorie und Berechnung der con-
tinuirlichen und einfachen Trager." Leipzig, 1873. [A work well deserv-
ing to close the list. Gives the general theory for constant and variable
cross-section for any number of spans from 1 to oo, and for all kinds of
regular or irregularly distributed and concentrated loads. The formulae
are general, and for given loading free from integrals. Difference of level
of supports ; most unfavorable position of load ; exact theory of the fixed
and movable inflexion and influence points, etc. Examples illustrating
use of formulae, and complete calculations of girders.]
This last work leaves but little to be desired in thoroughness and com-
prehensiveness.
It will be observed that England and America have contributed but lit-
tle to the literature of the subject. Indeed the standard works of both
countries show scarcely a trace of the influence of the labors of French and
German mathematicians in this field. The only works which, so far as we
are aware, can be mentioned in this connection are as follows :
RANKINE, W. J. M. — " Civil Engineering." 1870. [Very brief and in-
complete.]
HUMBER. — " Strains in Girders." American Ed. New York : Van
Nostrand. 1870. [Graphical constructions, holding good only under the
supposition that the end spans are so proportioned that the girder may be
considered as fixed at the intermediate supports, for full load.]
STONEY, B.— " Theory of Strains." London, 1873. [Very brief notice
of the subject. Points of inflection are found for full load, and the flanges
then cut at these points.]
HEPPEL, J. M.— Phil. M^. (London), Vol. 40, p. 446.
Also Minutes of the Proceed, of the Inst. Civ. Eng. [Excellent papers,
which might well have been followed up.]
In the latter publication also :
BELL, W.— Vol. 32, p. 171.
STONEY, E. W.— Vol. 29, p. 382.
BARTON, JAMES.— Vol. 14, p. 443.
In American literature :
FRIZELL. — "Theory of Continuous Beams." Jour. Frank. Inst., 1872.
[Davelopment of the subject according to Scheffler. See 3.]
250 LITEEATUEE UPON THE CONTINUOUS GIEDEE.
GREENE, CHAS. E. — "Graphical Method for the Analysis of Bridge
Trusses." Van Nostrand. 1875. [Force and equilibrium polygons are
used, but the moments at the supports are found by an original method of
approximation, or balancing of moment areas. ,]
HERSCHEL, CLEMENS. — " Continuous, Revolving Drawbridges." Boston,
1875. [The formulae of Wey ranch are made use of. The case of " second-
ary central span " is for the first time investigated, and the appropriate
formulas given. The fact that the live load reactions for supports out of
level are unchanged, provided the dead load reactions are zero, is also for
the first time clearly stated. The draw span is thoroughly treated, and the
idea of weighing off the reactions at the piers of a continuous girder sug-
gested.]
MERRIMAN, MANSFIELD. — " Upon the Moments and Reactions of the
Continuous Girder" — Journal of the Franklin Inst..for March and April,
1875; Van Nostrand's Eng. Mag., July, 1875; also the London Phil.
Mag., Sept., 1875, as well as the formulae contained in Chapter XII. of this
work. [By the aid of the properties of the Clapeyronian numbers, Mr.
Merriman has deduced new and general formulae eminently suited for
practical use. Also relations are deduced from which tables for moments
and reactions may be drawn up to any desired extent by simple additions
and subtractions, independently of the general formulae. (See Chap. XII.)
The simple girder appears as a special case of the continuous girder. The
formulae are, in respect to simplicity and ease of application, superior to
any heretofore given.]
CHAP. XIV.] THE BEACED AECH. 251
PART III.
APPLICATION OF THE GRAPHICAL METHOD TO THE ARCH.
CHAPTEK XIV.
THE BKACED AECH.
157. Different kinds of Braced Arches. — Just as in gird-
ers, we may distinguish between the solid beam, or " plate gird-
er," and the open work, or framed girder ; so, regarding the arch
as a bent beam, we may distinguish the braced arch and the
solid arch, or arch proper. The strains in the various pieces
composing the braced arch may be easily found by the method
of Arts. 8-15, or by calculation by the method of moments of
Art. 14 for any loading, if only all the outer forces acting
upon the arch are known : that is, so soon as, in addition to the
load, we know also the reactions at the abutments, or the hori-
zontal thrust and vertical reactions at the points of support, and
the moments, if any, which exist at these points.
We may distinguish three classes of braced arches : viz., 1st.
Arch hinged at both crown and springing ; 2d. Arch hinged
at spring line only — continuous at crown ; 3d. Arch continu-
ous at crown &&& fixed at abutments.
158. Arch hinged at both Crown and Abutments. — This
form of construction [PL 23, Fig. 90], owing to the hinges at
crown and abutments, affords for live load but little of the
advantage of a true arch. It is, in fact, an arch only in form,
but in principle is more nearly analogous to a simple triangular
truss of two rafters, these rafters being curved and braced ; the
thrust being taken by the abutments, instead of resisted by a
tie line A B.
The case presents no especial difficulty, and may be easily
calculated or diagramed, provided that not more than two
pieces, the strains in which are unknown, meet at any apex.
Thus, in PI. 23, Fig. 90, the resultant at the abutment due to
any weight P being known, it may be directly resolved into tnc
252 THE BRACED AKCH. [CHAP. XIV.
two pieces which meet there. The strains in these two pieces
being thus found, those in two others in equilibrium with each
of them may be obtained. In Art. 13 we have already illus-
trated the method of procedure for such a case, as also the
method of finding graphically the resultant at crown and abut-
ments due to any position of the weight.
Thus the resultant at the crown for the unloaded half must,
for equilibrium, pass through the hinge at B also. Its direc-
tion is thus constant for all positions of P upon the other half.
The resultant for the other half must then pass through a and
the hinge at A (Fig. 90).
We have then simply to draw a B, prolong P to intersection
a, and draw a A. A a and B a are the directions of the resul-
tant at A and B, and by resolving P along these lines, we may
find the vertical reaction V = d l> and the horizontal thrust
H = cb.
We can thus easily find the reactions at the abutments in
intensity and direction, and following these reactions through
the structure, as illustrated in Arts. 8-13, Chap. I., can deter-
mine the strains upon all the pieces for any position of the
weight. A tabulation of the strains for each weight will then
give us the strains for uniform load as well as live load, as al-
ready explained in the preceding chapter, Art. 156.
There must be only two pieces meeting at the abutments.
Thus the pieces in Fig. 90, represented by broken lines, can
serve only to support a superstructure, or transmit load to the
arch, and have no influence upon the strains in the other pieces.
If the span A B = 2 «, the rise of the arch is A, and the dis-
tance of the weight P from the crown is x, positive to the left ;
then taking moments about the end B, we have
2 V0 = P (a + a), or V = P (» + ^-
A Cb
Similarly, taking moments about the crown,
„, n V a— Px P (a — x)
— v # + H A = — P #, or H= =— - — — ^— ^ — '-.
fi 2t fi
The same formulae apply for a weight upon the other half, for
V and* H at the other end.
The values of V and H can easily be found from these for-
mulae, and the strains then calculated by moments, thus check-
ing the diagrams. If these reactions are found for the given
CIIAr. XIV.] THE BRACED ARCH. 253
dimensions of the centre line, we may, if we choose, suppose
the depth of the arch to vary above and below the centre line
equally, from the crown to ends. The lever arms of the pieces,
and hence their strains, will be different, but V and H are the
same as before. Thus, whatever the shape of the arch, we can
easily find the strains both by diagram and calculation. If we
draw a line through A and the hinge at crown, we may easily
prove that the greatest vertical ordinate between this line and
the arch is
y = —
O /v
where r is the radius.
Now if the depth d of the arch is made greater than this or-
dinate, it may be shown that both flanges will always be in
compression. This condition serves, then, to determine the
proper depth of circular arch, which should not be less than
It is unnecessary to give here an example.* The method is
so simple that the reader will find no difficulty in applying the
principles above to any case. He will do well to calculate or
diagram the strains in an arch similar to that shown in the Fig.
for comparison with the two cases which follow.
159. Arcli hinged at Abutments — continuous at Crown.
— If we suppose the hinge at the crown removed — those at the
abutments being, however, retained — then, for any position of
the weight, the resultant at each end must for equilibrium
pass, as before, through the end hinges. In the preceding
case, a, for load on left half, was always to be found at intersec-
tion of weight with the line through B and hinge at crown, and
was therefore fully determined. Now, however, &, the com-
mon intersection of weight and resultant abutment pressures,
has a different position, and hence the resultants and horizon-
tal and vertical reactions are different.
If we can find or know the locus or curve in which this point
a must always lie, we can easily find, as before, the resultants
or reactions by simply prolonging the line of direction of the
weight till it meets this locus, and then drawing from the point
* See Note to this Chap, in Appendix.
254:
THE BRACED ARCH.
[CHAP. xiv.
of intersection lines to A and B, and resolving P in these direc-
tions.
The equation of this locus can be found analytically without
much difficulty.
1st. PARABOLIC ARC. — Thus, for & parabolic arc, we have*
32 a? h
y ~ 5 (5 ^ - ar8)'
Where [PL 23, Fig. 91] a is the half span, and h the rise of
the arc ; x the distance of the weight from the crown, and y
the ordinate N dot the locus cd eiJc.
For a given arc, then — that is, a and h given — we have only to
substitute different values for a?, as x — 0, 0.1, 0.2, etc., of the
span, and we can easily find the corresponding ordinates y, and
thus construct the locus cdeiJc. It is then easy to find the
reactions at A and B for any position of P, as above indicated.
The vertical reaction at the abutment may also be easily
found by moments — thus,
Vt x 2 a = P (a + a?), or Vt = — (a + x).
2i a
The horizontal thrust is
5
(5 a9 - a?) (a2 - a?) »
a3 A
These values, though not needed for the construction above,
may be of use, and are therefore given. In the following
tables we give the values of H and y for different values of x :
X
H
y
x
H
y
0
0.3906
1.280
0.5
0.2783
1.347
0.1
0.3859
1.283
0.6
0.2320
1.379
0.2
0.3706
1.290
0.7
0.1797
1.415
0.3
0.3490
1.304
0.8
0.1226
1.468
0.4
0.3176
1.322
0.9
0.0622
1.527
0.5
0.2783
1.347
1.0
0
1.620
.a
£
.h
.«
ft
.A
* For the demonstration of the analytical results made use of in this chap-
ter, we refer the reader to Die Lehre von der Elasticitat und Festigkeit, by E.
Winkler. Prag, 1867. See also the Supplement to this chapter.
CHAP. XIV.] THE BRACED ARCH. 255
From the table, a and h and P being known, H and y can be
found for the successive positions of P at 0.1, 0.2, etc., of #, or
the half span, by multiplying P — by the tabular number for
H, and h by the tabular number for y.
2d. CIRCULAR ARC. — For a circular arc we have for the equa-
tion of the locus cdeik [Fig. 91],
where K = -r— 3, I being the moment of inertia of the constant
cross-section, A its area, and r the radius of the circle : also
where
(sin2 a— sin2 j3) (a— 3 sin a cos a + 2 a cos2 a)
sin a [sin2 a— sin'2 /3 + 2 cos a (cos /3— cos a) — 2 cos a (a sin a— £ sin 0)]'
a being the angle subtended at centre by the half span, and
. __ 2 cos a (a sin a — /3 sin /3)
B =
sin2 a - sin2 /3 + 2 cos a (cos @ — cos a — a sin a H- /3 sin
2 a cos2 a
2 (a — 3 sin a cos a + 2 a cos2 a)'
or, approximately,
24 6 aa
5 A2
= IT^4 = 64 P
where $ is the angle from crown to weight.
— — the square of the radius of gyration, or, approximately,
the square of the half depth, hence K = -_ approximately.
256 THE BRACED ARCH. [CHAP. XIV.
For the exact values of A and B, we have the following table :
fi
a= 0
a=10°
a=20°
a-30-
a-=40°
a=50°
a=W
a=90°
0
1.20
1.19
1.17
1.14
1.08
1.00
0.88
0
0.2
1.21
1.20
1.18
1.15
1.10
1.01
0.90
0
A
0.4
0.6
1.24
1.29
1.24
1.29
1.21
1.27
1.18
1.24
1.13
1.20
1.05
1.13
0.94
1.02
0
0
-
0.8
1.88
1.38
1.36
1.34
1.30
1.24
1.18
0
1.0
1.50
1.50
1.49
1.47
1.45
1.41
1.36
0
B
a
0.234
0.233
0.221
0.203
0.178 0.146
1
0.107
0
a4
For the values of yQ we have the following table
£
a=0
•=10-
a=20°
a=30°
«=«•
a=50°
.=60-
a=W
0
1.280
1.282
1.288
1.300
1.316
1.340
1.375
1.571
0.2
1.290
1.292
1.298
1.309
1.327
1.348
1.380
1.571
0.4
1.322
1.324
1.320
1.340
1.354
1.374
1.403
1.571
0.6
1.379
1.380
1.385
1.393
1.405
1.421
1.443
1.571
0.8
1.468
1.4G9
1.471
1.476
1.483
1.490
1.504
1.571
1.0
1.600
1.600
1.599
1.597
1.594
1.591
1.588
1.571
.a
.h
It will be seen that for the semi-circle the locus is a straight
line, for which y = ^irr = 1.5708r. Thus, for any given case
—that is, I, A and r given — we can easily calculate K. Then
from our tables, for given value of a, we can find A, B and yQ for
values of ft of 0, -j^ths, T\ths, etc., of a. These values inserted
in the equation for y above, will enable us to plot the curve or
locus cdeik) which being once known, the rest is easy. We
have thus by a union of analytical results with our graphical
method a very easy and practical solution of this important
case. We may, jf we choose, only use our method to determine
the horizontal thrust and vertical reaction as shown by the Fig.
91, and then calculate the strains by the method of moments.
The availability and ease of the method here given, as com-
pared with calculation, will be seen from a consideration of the
CHA.P, XIV.] THE BRACED ARCH. 257
analytical formulae for the horizontal thrust and vertical reac-
tion at A. Thus, for the vertical reaction, we have, as before,
simply
For the horizontal thrust, however, we have the following
very clumsy formula :
sin* a — sin2 p + 2 cos a (cos {I— cos a) — 2 (l-f-ic)cosa (a sin a— £sin/9)
2 [a — 3 sin a cos a + 2 (1 + *) a cos2 a]
For the semi-circle, this reduces to
K being, as before, = -r— ^ ; where A is the area and I the
moment of inertia of the cross -section, r the radius of the arch,
and the angles a and /3, as represented in Fi^. 91, viz., the
angle of the half span, and the angle to the load, subtended by
x. The first of the above formulae is sufficiently simple, and
by it we may check the accuracy of our construction. Thus
having plotted the curve cdeiJc by the aid of our expression
for y and the tables above for any position of P required, we
draw d A d B, and resolve P along these lines, thus finding V
and H [Fig. 91]. We can then calculate V from the formulae
p
above, viz., V = — (a + x). If this calculated value agrees
2 Ob
with that found by diagram, we may have confidence that the
curve is properly plotted, and hence that the value of H is also
correct. Thus, with very little calculation and great ease,
rapidity and accuracy, we can find the reactions at the end A
for any given position of P in any given case. These reactions
once known, we can easily find the strains either by diagram,
as illustrated in Chap. 1., or by calculation by the method of
moments of Art. 14.
16O. Arch fixed at Abutments— continuous at Crown.
This is by far the most important case of braced arch, as by the
continuity of the crown and fixity of ends we obtain all the
advantage possible due to the combined strength and elasticity
of the arch. It is also the most difficult case of solution, as the
formulae obtained by a mathematical investigation are complex,
17
258 THE BE AGED ARCH. [CHAP. XJV.
and give rise to tedious and laborious computations in practice.
A method combining simple analytical results with graphical
construction similar to the preceding, will, however, obviate
these difficulties, and bring the subject fairly within the reach
of the practical engineer.
In the present case, as before, the common intersection of the
weight and the reactions lies in a curve, the equation of which
may be found, and the curve itself thus plotted for any given
case.
But this curve, or locus, ILK [PI. 24, Fig. 92] being con-
structed, in order to find the directions of the reactions which
now no longer pass through the ends of the arc A and B, it is
necessary to find and construct also the curve enveloped ~by these
reactions for every position of P ; that is, the curve to which
these reactions are tangent. If, then, these two curves are con-
structed, we have only to draw through L [Fig. 92] lines tan-
gent to this enveloped curve, and we have at once the reactions
in proper direction, and by resolving P along these lines, can
easily find their intensities, and therefore V and H, as before.
\st. PARABOLIC ARC.
For a parabolic arc we have for the locus ILK, y = ^ h ;
that is, the locus is a straight line at \th the rise of the arch
above the crown • since we now take y as the ordinate to the
locus measured above the horizontal tangent at the crown. The
origin is, therefore, at the crown instead of at the centre of the
half span, as in the previous case.
For the second curve, or curve enveloped by the reactions, we
have,* taking v as the abscissa and w as the ordiuate of any
• 4. n?- oon 2 <# (23 a2 + 20 ax + 5 a?) k
point [Fig. 92], v = , w = ^— , . ,- f- ,
3 a + x ' 15 (a + x) (3 a + x)
where, as before, a is the half span, h the rise, and x the dis-
tance of the weight from crown. For x = 0, v = f a, and
w = ff h. For x = a, v = £ a, and w = f h. For x = —a,
v = a, and w = — oo. Eliminating x from both equations, we
5 a2 — 5 av + 2v* t
have ^K - h.
15 a (a — v)
Hence the curve enveloped by the reactions is on each side an
* For the proof of all the expressions assumed, see the Supplement to this
chapter.
CHAP. XIV.]
THE BEACED ARCH.
259
hyperbola, which has for asymptotes the vertical through the
abutment and a straight line which cuts the axis of symmetry
of the arch at the point b [PL 24, Fig. 93], ^ h under the crown,
the tangent at the crown at f a from the crown, and the chord
of the arc at — 6 a from the centre. The centre of the hyper-
bola is at e, yV h below the horizontal through the crown. The
two hyperbolas osculate at the point -5- a vertically below the
crown. [See Fig. 93.]
As an aid to the construction of these hyperbolas, we give the
following ^table :
oc
V
w
X
V
w
1
1.0000
00
0
0.6667
0.5111
0.9
0.9524
2.7721
0.1
0.6452
0.4897
0.8
0.9091
1.5455
0.2
O.B249
0.4722
0.7
0.8695
1.1065
0.3
0.6061
0.4577
0.6
0.8334
0.9999
0.4
0.5882 .
0.4463
0.5
0.8000
0.7600
0.5
0.5714
0.4349
0.4
0.7693
0.6756
0.6
0.5555
0.4258
0.3
0.7407
0.6155
0.7
0.5405
0.4160
0.2
0.7144
0.5714
0.8
0.5263
0.4102
0.1
0.6897
0.5377
0.9
0.5128
0.4053
0
0.6667
0.5111
1.0
0.5000
0.4000
.a
.a
.h
.a
.a
.h
From the table it is easy to construct the hyperbola for any
given case. We have,, of course, a perfectly similar hyperbola
for the other half, its centre e being similarly situated with
respect to the crown, to the right of c. "We have then simply
to draw a line through the intersection m of the weight P
[Fig. 93] with the line i k, at -g- h above <?, tangent to the hyper-
bola, and we have at once the direction of the resultant. This
tangent may be drawn by eye, or geometrically constructed if
THE BRACED ARCH. [CHAP. XIV.
desired.* A similar tangent to the hyperbola on the other side
determines the direction of the other reaction. We can then
resolve P in these two directions, and find at once V and H.
The problem, then, so far as a parabolic arc is concerned, is
sufficiently simple and easy of solution. We have only to draw
a straight line and two easily constructed curves. The formulae
for V and H and moment at crown M0 are for this case also
simple, and may be used for checking our results. They are :
TVr - IP (a ~ XY (3 ^ ~ 10 a x ~ 5 ^
°~ *» ~>
H- !*p- V -
"J -
(a ~
where, as before, a is the half span, h the rise, and x the dis-
tance of weight P from crown. A negative moment always
indicates tension in lower or inner flange.
2d. CIRCULAR ARC.
In this case we 'have for the locus ILK [Fig. 92], for small
central angles a, the equation :
#, A, and x being as above, and <f — —r — the square of radius
of gyration ; A being the area and I the moment of inertia of
cross- section.
[Note. — In all the cases hitherto considered, or which we
shall consider hereafter, the cross-section is assumed constant.']
According to the exact formula, which is too complicated to
make it desirable to be given here, we have the following
* For the construction of a tangent to a conic section, see Appendix, Fig. 4.
CHAP. XIV.]
THE BRACED ARCH.
TABLE, FOR VALUE OF y.
261
0
a = 0°
a = 30°
a = 60°
a =90°
0
0.200
0.211
0.252
0.329
0.2
0.200
0.210
0.246
0.312
0.4
0.200
0.307
0.236
0.280
0.6
0.200
0.202
0.217
0.228
0.8
0.200
0.197
0.200
0.158
1.0
0.200
0.188
0.151
0.082
a
h
Instead of determining the curves enveloped by the reac-
tions, the expressions for which are in this case somewhat com-
plicated, it will be found preferable to find the distances c± c2
of the intersections <p and ty of the reactions [see Fig. 92] with
the verticals through the centres of gravity of the end cross-
sections. For small central angles a, we have
C, —
15 (a + x) I A A2
___2A_r
15 (a + x) I
where 0, A, A and a?, have the same signification as above.
Since I divided by A equals the square of the radius of gyra-
tion = ^, we have
2A
4-
15 (a + a?)
[
For braced arches when the material is nearly all in the
flanges, the material in the bracing being very small, we may
call the radius of gyration half the depth of the arch measured
upon the radius from centre to centre of flanges ; or represent-
ing this depth by d,
262 THE BRACED AECH. [CHAP. XIV.
2A .
— 5a? —
[A negative result indicates that the distance is to be laid off
below the centre of cross- section.] These formulae are easy of
application, and sufficiently exact for arches whose rise is small
compared to the span ; when — - is, say, not greater than -^.
2 a>
All the above formulae are for constant cross-sections. Exact
formulae for variable cross-section give results but little less,
and are much more complicated. The effect of using the above
formulae is therefore, merely, to increase slightly the coefficient
of safety.
161. We are now able to determine readily and accurately
the strains in the various pieces of braced arches hinged at
crown and abutments, and hinged at abutments only. We
have only to construct in each case the reactions at the abut-
ments, as explained in Arts. 158 and 159, Figs. 90 and 91, and
then, by the method already detailed in Arts. 8-13, we can fol-
low these reactions through the structure, and thus find the
strains in each piece due to every position of the load. We
may also, having found the reactions for given position of
weight, calculate the strain in each piece by moments.
For the case of the arch continuous at the crown w&& fixed at
the abutments, we must remember that we have also a moment
at each end tending to cause either tension or compression in
the inner flanges according as it is negative or positive. The
case is precisely analogous to the continuous girder, or girder
fixed at ends. As in that case [see Fig. 77, Art. Ill] the
moment at one end, as B, was the product of H into the vertical
distance B D, so here the moment at A (Figs. 92 and 93) is the
product of H into ct, found by the formulae above. This
moment can, then, be easily found when cl and H are known.
We can then lay it off, according to the directions of Art. 125,
for " passing from one span to another of a continuous girder,"
and thus commence our diagram of strains; or we can cal-
culate the strains by the method of moments.
162. Illustration of method of Solution. — As an illustra-
CHAP. XIV.] THE BRACED AECH. 263
tion, take a portion of a braced arch, as represented in PI. 24-,
Fig. 94. We have first to plot the upper curve or locus of m
for the given dimensions of the centre line of the arch. This
curve once plotted, then, for any position of the weight, we have
only to prolong P to m, and draw a line from m to the end of
centre line if the arch is hinged at ends, or to <£ at a distance
<?!, above or below the end of centre line if the arch is fixed at
ends ; cl being easily found from our formulae above. In sim-
ilar manner, we draw a line from m to the other end, or C2.
Now these two lines are the resultants of the outer forces P,
and by simply resolving P in these directions, we have at once
V and H, while the moment at the end Mt = — H c1? positive
if it tends to cause compression in lower flange, or since c± is
negative down, if it acts below the end.
We can now easily find the strain in any flange, as D,
whether the arch vary in depth or not, provided only it is sym-
metrical with respect to its centre line. Thus for D, take the
opposite apex a as the centre of moments. The moment of H
with reference to #, as shown in the Fig., tends to cause tension
in D, while that of V causes compression. We have then,
representing tension by minus,
moment of V — moment of H
strain in D = = „ — ,
lever arm or D
all with reference to a. If the result is minus, it indicates thus
tension, if plus, compression ; if it is zero, the two moments are
equal, and at «, therefore, no moment exists ; hence a must be
a point of inflection. Note that H and V must be taken as act-
ing at $, Fig. 94 We can also evidently take them as acting
at the centre of the end cross-section, if we take into account
the moment H c^
In similar manner, for C we take 5 as centre of moments, and
then, since H now causes compression in C and V tension, we
have for V and H, acting at <£,
. . ^ moment of H — moment of V
strain in C = - — — .
lever arm or C
For V and H, considered as acting at the end of centre line, we
„ moment of H + H^ — moment of V
have C = - — =— , — .
lever arm or
taking <\ without regard to its sign, but simply to the kind of
264: THE BRACED ARCH. [CHAP. XIV.
strain it tends to cause in the piece in question. Properly,
since when H is belo\v ^ is negative, we should have — H ^
for moment causing compression in C.
Thus we may proceed till we pass P, and then the moment
of P, with its proper sign, as producing tension or compression
in the piece in question, must also be taken into account, or we
may instead take the moments of V and H at the other end,
that is, the same side of the weight as the piece itself.
The diagonals may be similarly found by moments. It will,
however, be best to determine them by diagram, one of the
flanges being first calculated (in this case the first upper flange),
as explained in Art. 125. They may also be calculated from
the resultant shear at any apex. Thus, for diagonal 3 find the
vertical components of the previously determined strains in D
and C. These vertical components, together with the vertical
component of the strain in 3, must for equilibrium be equal and
opposite to the total shear at ft.
Calling this shear F, and a, fi and y the inclinations of D
and 3, we have for the strain in 3,
S8 = (F — Sl sin a — S2 sin ft) cos 7.
If either of the vertical components of the strains in D or C
acts opposite to the shear F, it must, of course, be subtracted ; if
in the same direction, added to F. For the ready determina-
tion of the proper signs, see Appendix, Art. 16 (4).
The moment H ^ is the moment at the fixed end, and is con-
stant throughout the arch for any one position of the load. It
causes tension in outer and compression in inner flanges, pro-
vided, as in the Fig., <f> fall below the centre of the end section.
This moment is increased (or diminished if <j> is above) by the
varying moment of H for each apex.
The above method of determining the strains in the braced
arch, though not strictly graphical, but rather a combination of
analytical and graphical methods, offers such a ready solution
of this important and difficult case, that we have not thought it
out of place to notice it somewhat in detail. We consider it by
far the simplest and easiest method which has yet appeared.
163. Analytical Formulae for V and H. — A comparison
of our method with the long and involved analytical expres-
sions to which the theory of flexure conducts us, will render its
advantages still more apparent.
CHAP. XIV.] THE BRACED AKCH. 265
Thus, for a load of w per unit of horizontal length, reaching
from left end to a point whose angle from vertical through
crown is ft (Fig. 92), a being the angle subtended by the half
span, we have *
H = & sin (B-Jci +.k + sin
where R is radius of arch, and
7 . 2 sin a , sin a cos a . sin2 a
fc = a -r sm a cos a — . , \ = -f — s— ,
a a 2
7 sin a 7 sin a r n -,
#2 = — — , #3 = — — a — sin a cos a , and
/z, = /3 + 2 yS sin2/3 + 3 sin ^Q cos j3.
For V we have
v_ RF cos a sin2 8 sin/3 K 3 cos a— cos3 a . "I
L.2 (a— sin a cos a) 2 a— sin a cos a 6 (a —sin a cos a) J'
„, cosyS 8 sin 8 coss/3
Where -^ + '__J yZt
For a concentrated load P for any point [Fig. 92], we havef
or, more correctly,
__ _ p a — /3 — sin a cos a — sin /3 cos /3 + 2 cos a sin ft
2 (a — sin a cos a)
For the semi-circle, this becomes
v _ p TT — 2 /3 — 2 sin /3 cog ft
2 TT
For H we have
H
— p 2 sip « [C03 3 — cos a + (1 + «) /3 sin 3] — (1 + «) a (sin2 a + sin2 ff)
2 [(1 + K) (t (a + sin a cos a) — 2 sin2 a]
* Taken from Capt . Bads' Report to the lUinois and St. Louis Bridge Co. ,
May, 1868.
f Die Lehre von der Eiasticitdt und Festigkeit. Winkler. Prag. 1867.
266 THE BKACED AKCH. [CHAP. XIV.
where K = — — g ; I being the moment of inertia, and A area of
the cross-section, and r the radius of circle.
These formulae, it will be observed, involve much labor in
any particular case. Where the number of weights is large,
the computation is tedious in the extreme. A method which
shall give accurate results and avoid such formulae as the above
is certainly very desirable, and such we believe to be the
method which we have given.
For the analytical investigation of arches, and the demon-
stration of the formulae for the curves of which we have made
use, the reader may consult Die Lehre von der Elasticitaet und
Festigkeit, by Dr. E. WinJder, to which we have already re-
ferred, and which contains a thorough discussion of the whole
subject. The tables which we have given, as well as the for-
mulae for £/, G! and <?2, will, it is hoped, give the method here
presented a practical value, and render the solution of any par-
ticular case easy and rapid.
164. For a solid or plate girder arch of given cross-section,
we may also determine the proper proportions by finding, as
above, the moment M of the exterior forces at any point.
TI
Then M = _,
t
where T is the strain per unit of area in any fibre distant t
from. the axis, and I the moment of inertia of the cross-section.
Thus, for a rectangular cross-section I = ^ b d3, where b is the
breadth and d the depth.
Hence M = % T I d\
if we take t = -.
The strain, then, per unit of outer fibre will be
T_6M
" bd2'
The safe working strain should not exceed for iron 5 tons
per sq. inch for tension and 4 tons for compression, and there-
fore d being assumed, we can easily proportion b so as to sat-
isfy this condition.
As examples of braced arches, such as we have considered,
viz., continuous at crown and fixed at abutments, we may men-
CHAP. XIV.] THE BRACED AKCH. 267
tion the Bridge over the Mississippi River at St. Louis, ~by
Copt. Eads / one over the Elbe near Hamburg on the Paris-
Hamburg R. R., in which, however, the outward thrust of the
arch is balanced by a precisely similar inverted braced arch,
or suspension system. Thus the piers have to support a verti-
cal reaction only, and the necessity of large and expensive abut-
ments of masonry for resisting the horizontal thrust is obviated.
The strains in the inverted arch of this character are found
in a precisely similar manner. The only difference is that the
reactions, and therefore the vertical and horizontal components,
act now in a direction opposite to the direction for the upright
arch, and the strains, though the same in amount, are of re-
verse character in each piece.
The bridge over the Rhine at Coblenz is an illustration of
the braced arch pivoted at the abutments only.
Examples of the solid or cast-iron arch of all kinds are
common.
165. Strains due to Temperature. — In the first class of
braced arches, viz., pivoted at both abutments and crown, there
are evidently no strains due to changes of temperature. The
arch can accommodate itself to any change of length by rising
at the crown and turning at the abutments, and no strains are
induced.
We represent by e the coefficient of linear expansion for one
degree [about 0.000012 for iron, for every degree centigrade],
and by t, the temperature above or below the mean tempera-
ture £0, for which no strain exists.
Then for arch pivoted at abutments only, we have for the in-
crease of thrust,*
__ 2 El et sin a
~ 7* (a — 3 sin a cos a + 2 a cos2 a) + 2 K r* a cos3 a,
where E is modulus of elasticity, I moment of inertia of cross-
section, and K = — — — ; A being area of cross-section, r radius of
arch, a the angle of half span, or, approximately,
8 A A2 + 15 I'
where h is the rise of arch.
*Lehre von der Elasticitdt. Winkler. Also Supplement to this chapter,
Art. 26.
268 THE BKACED ARCH. [CHAP. XIV.
„ For the moment at any point, then, due to change of temper-
ature, we have
M = H r (cos /3 — cos a),
ft being the angle from vertical to that point.
This moment, if positive, causes tension in outer and com-
pression in inner flanges, and we can, as before, easily find the
corresponding strains either by diagram or calculation.
For an arch fixed at ends and continuous at crown, we have
TT
2 Ele t (1 + K) a sin a
[(1 + K) (a2 -4- a sin a cos a) — 2 sin 2a] '
or, approximately,
45 E I e t 45 E I A e t
~~ r2 (a4 + 45 K) ~ 4= A A2 + 45 I'
But this thrust does not act at the abutment, since, if it did,
there would be no moment there. It must be considered as
acting at a distance for rise of temperature, below the crown of
(1 + K) a — sin a
e'--= (l + «)a -*•'•
or, at a distance above the end abutment of h — eQ.
Approximately, we have
_ a2 + 6 K _ ( A a2 + 6 I) h
6 3Aa? '
a being the half span.
This thrust and its point of application being known, we can
easily find the moment, and hence the strains at any point.
We see that the horizontal thrust is about six times as great as
for the case of an arch pivoted at both ends.
The constant moment acting at the abutment, which may be
considered as acting at every point, is
_ /sin a \ „
MA = — 1 -- cos a t H r ;
it acts to cause compression in outer and tension in inner
flanges at abutment. If we find this moment, we can then con-
sider H as acting at the end, and then we have for the moment
at any point
* Capt. Eads' Report to the Illinois and St. Louis Bridge Co., May, 1868.
CHAP. XIV.] THE BRACED ARCH. 269
a positive result, giving tension, a negative, compression in the
outer flanges.
166. Effects of Temperature. — We are now able to solve
accurately and thoroughly any class of braced arch, both for
variable loading and changes of temperature, and here the fol-
lowing remarks upon the latter subject may not be without
interest. We quote from Culmann — Die Graphische Statik,
p. 487 :
" The question arises whether the fears which the additional
strains, due to variations of temperature, have given rise to, are
well founded. Before the construction of the Arcole Bridge
in Paris the Engineer Oudry made various experiments with
a rib of about the same span as the bridge itself, of which the
following seems decisive as regards the present question. By
driving in the wedges upon which the rib rested above and be-
low, he could raise and lower the crown much more than the
distance due to variation of temperature without diminishing
its supporting capacity Oudry, having thus assured
himself of the harmlessness of temperature variations, decided
upon broad and firm bearing surfaces.
" Interesting observations have also been made upon the
changes of form of the cast-iron arch of 60 metres span over
the Rhone at Tarascon, published in the Annales des Ponts et
Chemins, 1854, from which, however, it only appeared that the
changes of form followed slowly the temperature; that they
were less than the received coefficients would have led us to
expect, and were nowhere found to be prejudicial.
" Since, then, this question appears to have been settled more
than ten years ago, may we not fear that those who still wish to
pivot iron may some day seize upon the idea of pivoting stone
arches also !
" Stone, as is well known, expands not much less than iron
for equal changes of temperature, and, moreover, its modulus of
elasticity is much less. The expanded stone arch cannot accom-
modate itself to the given span, therefore, as easily as the iron
arch, and it would then be clearly more advantageous to pivot
the stone arch ! As, however, such a clumsy contrivance would
give no great impression of stability, we feel justified in recom-
mending a broad and solid bearing surface for all arches."
As the opinion of an eminent engineer, the above may not be
without interest. We would only add that, according to the
270 THE BRACED AKCH. [CHAP. XIV.
accepted formulae for temperature strains already given, the
results are of more importance than the above remarks would
indicate. As will be seen in the Appendix, the temperature
strains in the braced arch, fixed at ends and continuous at
crown, are very considerable, and, if the formulae are accepted
as correct, can by no means be disregarded. By comparison of
our numerical results for the three cases of braced arch there
given, it appears that the one hinged at crown, and springing,
is by far the best form of construction, but it must be remem-
bered that a different proportion of span to height and depth
may considerably affect this conclusion. Upon this point we
refer the reader to Art. 28 of the Appendix.
With the above, we conclude our discussion of braced arches,
or arches whose weight is not so great that the effect of the live
load can be disregarded, and pass on to the stone arch, or arch
proper.*
* See Appendix, Art. 17, for a practical application of the principles of this
chapter.
CHAP. I.] SUPPLEMENT TO CHAP. XIV. 271
SUPPLEMENT TO CHAPTEE XIY.
DEMONSTRATION OF ANALYTICAL FORMULA GIVEN IN TEXT.
In order to complete our discussion of the braced arch, we shall now
give the analytical development of the formulae of which we have made
use in the preceding chapter. We do this the more readily, as in no book
of easy access to the student are these formulae made out. In the work of
WinUer, already referred to in the text, will be found a very thorough
discussion of the subject. We shall confine ourselves at present to the
case of a single concentrated load.
CHAPTER I.
GENERAL CONSIDERATIONS AND FORMULAE FOR FLEXURE.
1. Fundamental Equations — The resultant of all the forces act-
ing upon a curved piece in a common plane may be decomposed into a
force normal to the piece N, and into a compressive or tensile force in the
direction of the axis or of the tangent to the axis G ; and this latter force,
if taking effect above or below the axis, acts to bend the piece, and gives
rise to a moment as well as to a compressive or tensile force G. These
forces cause corresponding strains. Thus, if P is the tangential strain per
unit of area d a, then
(1)
while, if v is the distance of any fibre from the axis,
M (2)
(a) Coefficient of elasticity.
Let the length of a piece be s, its area of cross-section A, and, as above,
G
the force acting upon this area G. Then ^ will be the force per unit of
area. Let the displacement [elongation or compression] produced by this
G
A
force be As; the sign A indicating and reading "elongation.1" Now
272
SUPPLEMENT TO CHAP. XIV.
[CHAP. i.
experiment shows that within narrow limits, i.e., within the elastic limits,
the elongation or compression is directly as the force of extension or com-
pression. Supposing that this held true always for all values of A «, then,
since a force — produces a displacement A s, the force necessary to produce
a displacement *, will be —
A s
E =
as great. Calling this force E, we have
Gs
The force E, then, is the force which would le necessary to produce a dis-
placement s equal to the original length, if the law of proportionality of the
displacement to the force always held good for all values of A s. This
value
E =
_
AA
(3)
we call the coefficient of elasticity.
From (3) we easily obtain for the force in the direction of the tangent
to. the axis
= EA
A s
and for the relative displacement
A a
G
E A
(4)
(5)
(&) Fibre strain P, and moment M.
As seen in eq. (1), the longitudinal strain upon an element of cross-sec-
tion da is called P. In a curved piece conceive two cross-sections, as
CHAP. I.] SUPPLEMENT TO CHAP. XIV, 273
shown in the Fig., as A O B D perpendicular to the axis of the piece m n.
Let these sections be infinitely near; then, the distance J>a upon the axis is
d s. Let d *v be the length of any fibre, as d c, before the change of form.
Then, after deformation, its length is = d sv + A d sv. But if d (f> is the
small angle between the normals, dsv = ds + vd(f), where v is the distance
a c of any fibre from the centre of gravity of the cross-section. After
deformation, d s becomes ds + Ads, and d <j> becomes d $ + Ad<f>, and
d sv becomes d sv + A d sv. Hence the length of any fibre after deforma-
tion is d sv + A d sy = d s + A d s + v (d <£ + A d <£).
Subtracting this from the eq. for d sv above, we have
Therefore the ratio of the change of length to the original length of fibre
Ads
is
dsv
If r is the radius of curvature, then rdd> = ds, — — = — ; hence
d s r
A d sv r A d s A d 0-i r ...
-^=h^+Bir^i— (0)
G
From eq. (1) we have the strain on a fibre P = -r .
From eq. (4), G = E A — . Hence P = B — . In the present
case
— is given by (6) ; therefore
ds d s
Since now from (1) G = / Pda, we have from (7)
G__ Ads r da Ad<f) f*v d a
E = r ~d7J 7^o + ' ~~d~»J r~TV
Since from (2) M = / P v d a, we have again from (7)
M_ Ads rvda Ad<f> /V d a
B = "~d~s~J 7T^+ "TT'J 7+~V
But / - — is, when 9 is very small compared to r, equal to
/ 02 d a, which is the moment of inertia of the cross-section I. Also,
da
7^ =
r
18
274: SUPPLEMENT TO CHAP. XIV. [CHAP. I.
or T I — — = I da— - I vda + - I = A -f —3, because since
1 J r + v J rj rjr + v r*
v is measured from the centre of gravity, / v d a = 0.
rvda f* fvzda I
Again, r I — — = I v a a — I — • — = — — -.
/ T + 1) I I T + V T
Therefore the insertion of these values of the integrals in the equations
for _ and _ above gives
E E
Gr Ads r~Ad(f> AflJsnl
E ~" ds ~ [_ d s ~ r ds\ r
M rA d ^ A d «~j I
E r~ L d s rdsj r'
M
G- A a s
Inserting the second in the first of the above equations, ^- = — v- A — =— ,
and hence
_- — - + — —- . (8)
ds EA + EAr
Inserting this in the second equation above,
M M G
E I + E~A~^ + E~A~7
(c) Change of length and position of axis.
From (8) we have at once for the elongation of axis,
or, when v is very small compared to r,
1
From (9) we have for the change of direction of the tangent to the axis
A^ = E/>(T + A~7* + A^r)d8' or from (10) for r
constant, that is, for a circle,
When D is very small compared to r, we have
ds ....... (13)
CHAP. I.] SUPPLEMENT TO CHAP. XIV. 275
If the piece had been originally straight, d A (j£> would be equal to d <£, and
dA(b dd> 1 El
—3-^ = — — = -; hence from (13) we have M = - .
as rd r r
From the calculus we have the radius of curvature
r -- * ' or> aPProximatelv> r =
hence M = B I ~ . . . . - . . . (13 &)
This is the equation assumed in the Supplement to Chap. XIII. , Art. 1.
2. Displacement of any point. — We indicate the horizontal dis-
placement of any point of the axis along the axis of a by A x, along y by
A y. The corresponding changes of d x, d y, and d s are A d x, Ady, Ads.
The total horizontal displacement is then dx + Adx = (ds + Ads) cos
A <£). The total vertical displacement is d y + A d y = (d s + A d s) sin
+ A $). Hence, since Adx = d A z,
d Ax — (d s + Ads) cos (<£ + A$) — dx,
n (0 + A </>) — d y.
By Trigonometry, cos (0 + A $) = cos </> cos A <£ — sin <£ sin A 0,
sin (0 -f A 0) = sin 0 cos A <£ + cos 0 sin A <£, or if cos A <£ = 1,
<?« ^ <w
sin A 0 = A <£, cos </> = -r- , sin <£ = -=— .
Substituting these in the equations above,
. / Ads\
dAx = (dx — A(pdy) II +—5 — I — d x,
\ I
(Ads\
1 + — j — I — d y,
as /
or, removing the parentheses, and neglecting quantities of the second order
with respect to A (f> and — = — ,
dAx= — A<f>dtM — ^ — dx
-di~dy
276 SUPPLEMENT TO CHAP. XIV. [CHAP. T.
When the radius of curvature is very great with reference to the thick-
ness of the beam, and the relative change of length — , — is disregarded,
CL 8
we have simply
dAy = A<}>dx.
But from (12) A <£ is equal to / _ _
Mds Ads
H
/WLds
E j , hence, for » very small with
reference to r,
We shall have frequent occasion to refer to formulae (8), (12), (13), (14)
and (15) in the following discussion.
CHAP. H.] SUPPLEMENT TO CHAP. XTV. 277
CHAPTER II.
HINGED ARCH IN GENERAL.
3. Notation — Tlie outer fbrce§ in general. — We suppose the
ends of the arch to be hinged at the abutments at the centre of gravity of
the end cross-sections. Then the end reactions must pass through these
points. These end reactions and the loads constitute, then, the outer forces.
For equilibrium, then, the horizontal components of these reactions must
be equal. Each of these components we call the horizontal thrust.
We use the following notation [PL 23, Fig. 91] :
R and R', the reactions at ends A and B.
V and V, their vertical components.
H, their horizontal component, or the thrust.
<z, the half span.
A, the rise of arch.
a, the half central angle, if arch is circular.
The origin of co-ordinates we take at crown, x horizontal, y vertical. The
angle of radius of curvature at any point with y, or of tangent to curve at
any point with #, we call 0.
THE OUTER FORCES IN GENERAL.
Suppose a single load P to act at any point E. Let its horizontal dis-
tance from crown be 2, the corresponding central angle E O C be /3.
Then the conditions of equilibrium are :
V + V ' = P,
Va — V a — Pz = Q.
From these last, we have
(16)
For a circular arc, since a = r sin a, z = r sin 0,
sma + sinff sma-sin/3
2sina ' I ~~
We distinguish three segments in the arch [Fig. 91], viz., A E, or from end
to the load ; E O, or from load to crown ; C B, or from crown to right
end. Quantities referring to the second we indicate by primes, to the
third by double primes. Thus for the tangential component of the result-
ant at any point within A E, we put G ; for the normal component, N.
In E O, then, we have G' and N' ; in O B, G" and N*. We have then
278 SUPPLEMENT TO CHAP. XIV. [CHAP. II.
G = — H cos (f> — V sin <£, G' = — H cos 0 + V sin $ ) ^
N = — H sin 0 + V cos $, N' = — H sin 0 — V cos $ )
M = H(fc-y)-V(0-aO, M' = H(fc-y)- V (« + *) • • (19)
In the case of a circular arc, a = r sin a, h = r (1 — cos a), # = /• sin <£,
y = r (1 — cos (/>), and hence
M =
JVT
= H r (cos 0 — cos a) — V r (sin a — sin <£) )
= H r (cos 0 — cos a) — W (sin a + sin <£) j
4. Intersection Line.— We call the locus of d [PL 23, Fig. 91], or
the curve cdeik, the intersection line.
If now there are three hinges, one at crown and one at each abutment,
then the resultant for each half must pass through the crown O. If, there-
fore, for the crown (*=0, y=0), we make in (19) M'=0, we have H=V -,
or inserting the value of V from (16),
H = ^|^ ......... (21)
If the load lies to the left of O, then only the resultant If acts upon the
right half, and must pass, as above, through the crown O. The point d
lies then always upon B C or A O prolonged. Hence, the intersection lines
are two straight lines, which pass through the crown and ends.
5. Parabolic Arc— concentrated Load. — For a parabola we
have y — — xz ; hence, d y — — x d x, and, approximately, ds =d x.
a' a
(a) Change of direction of tangents.
Inserting this value of y in equations (18) for M and M', we have from
equation (13), Art. 1, since r d 0 = d s = d x for the change of direction of
the tangent at any point before and after flexure,
Integrating this, we have for the three segments A E, E C and O B,
. . (22)
where A, A', A" are constants of integration, to be determined by the as-
signment of the proper limits. Thus, if we make x = z, the two first of
equations (22) are equal; hence,
CHAP. II.] SUPPLEMENT TO CHAP. XIV. 279
and accordingly A - A' = (V - V) a z - \ (V + V) z. Since V + V = P,
and from (16) V — V = P — , we have,
a
A-A'=lPz* ....... (I.)
(&) Horizontal displacement.
Inserting the value of dy for the parabola, viz., ay = — xdx, and the
value of M from (19), and inserting in this last the value of y, viz.,
y = — x*, we have from equation (15), Art. 2,
2
Integrating this, we have for the three divisions of the arch, as before,
For the point B, or x = z, we have from the two first of these equations,
B - B' = 1 (V - V) a z* - i (V + V) z4 - £ (A - A') z*, that is,
B-B=-2\-Ps* ....... (II.)
For the crown, x = 0, and the second and third equations are equal,
hence
B' = B* ........ (III.)
For the left end, that is, for x = a, since the end of the arch must not slip,
we must have x — x = 0. So also for the right end, for x = — a. There-
fore, from the first and third equations, putting B' for B", we have
&Ua3h--*sVa4' + iAa2 + B = 0,
- A- H a3 h + -fr V a4 + i A" a2 + B' = 0.
By the addition and subtraction of these equations, we have, since
V + V = P, and (V - V') a = Pz,
A")»2-(B + B') = 0 .... (IV.)
^ + 24) + i(A-A")a2=0 . . (V.)
We might, in a precisely similar manner, form three equations similar to
(23) for the vertical displacement A y. This would introduce three more
constants and four more equations of conditioh between them. By the
nine equations I. to IX. thus obtained, these constants may be then deter-
mined in terms of the known quantities H, A, P, a and z, and thus the
change of shape at any point may be fully determined.
The complete discussion, as indicated, is unnecessary for the purpose we
2SO SUPPLEMENT TO CHAP. XIV. [OHA.P. II.
have in view, and we shall not, therefore, pursue it further. We have
already all the general formulae of which we shall need to make use in the
discussion of the parabolic arch.
6. Circular Arc — concentrated Load. — In a perfectly similar
manner we may make out analogous formulae for the circular arch. Thus,
referring to equation (8), Art. 1, and inserting for G and M their values as
given in (18) and (19), Art. 3, we have for the force in the direction of
the axis (see eq. 4),
.
E A - = — H cos a — V sin a
<E A - — = — H cos a — V sin a
ds
Putting for Hr Vand V their values from (21), Art. 4, and (17), Art. 3, we
have,
A d s _ _ sin a + sin /3 — 2 cos a sin $
=
. . (25)
A d s' _ — sin a — sin /3
~dT' 2(1 -COS a)
(a) Change of direction of tangents.
Referring to equation (12), Art. 1, we have, since rd<f) = ds and r </> = «,
/M Ads
EIr** + -dT*-
The value of M is given in (20), of ^U-J in (24). Inserting these values,
we have
A <£ = ^j-j / |~H (cos $ — cos a) —V (sin a — sin
— ==-r- (H cos a -H V sin a) $.
Performing the integration,* and putting, for brevity, * = — — -, we have for
the three segments of the arc, as before,
ElA£ = r2 pK (sin <£ - <£ cos a) - V Casino + cos <f>)~^ - K T* (H cos a + V sin a) $ + A "|
El A<£' =ra [~H (sin</>— £cosa) — V(<£sina— cos </»)"! — K r2 (H cos a + Vsina)0 + A' t 26
E I A $" = r* [H (sin <£- 0 cos a) - V (<£ sin a - cos<£)]- K r2(Hcos a + V sin a)<f» + A" J
For the point E, <£ = 3, and the first and second equations become simul-
taneous. Hence, after reduction,
A-A' = (V- V) r2/3sina + (V + V) r* cos 0 + K (V-V) rz /3sin«.
But V + V = P, and from (6) V - V = P , hence
sin a
. . . (I.)
. * See Art. 7, f allowing.
CHAP. II.] SUPPLEMENT TO CHAP. XIV.
(b) Horizontal displacement.
According to eq. (14), Art. 2, since x = r sin 0, y — r (1 — cos 0),
dx = r cos 0 d 0, d y = r sin 0 d 0, we have
E I A x = r I [H r'2 (sin 0 — 0 cos a) — W (0 sin a + cos 0),
— K r* (H cos a + V sin a) 0 + A J sin 0 <Z 0,
— K r2 / (H cos a + V sin a) cos 0 d 0.
The integration gives us the three equations,
E I A x = — r* H ( £ 0 — \ sin 0 cos $ — cos a sin 0 + 0 cos a cos 0)
— V (sin a sin 0 — 0 sin a cos 0 + •£ sin2 0)1
— K r3 (H cos a + V sin a) 0 cos 0 — A r (1 — cos 0) + B.
£ I A #' — — r3 1 H (•£ 0 — -J sin 0 cos 0 — cos a sin 04-0 cos a cos 0)
— V (sin a sin <^> — 0 sin a cos (f> — i sin2 0) I
— K r3 (H oosa + V sin a) 0 cos 0 — A' r (1 — cos 0) + B'.
E I A x" = — r* I H (f 0 — | sin 0 cos 0 — cos a sin 0 + <j> cos a cos 0)
—V (sin a sin 0 — 0 sin a cos 0 — \ sin2 <^>)
— K r3 (H cos a + V sin a) 0 cos 0 — A" r (1 — cos 0) + B".
For 0 = £, that is at the load, A a? must equal A x'.
Hence, after reduction,
B - B' = -$Pr3 (2 + sina/3- 2003/3-2)8 sin /3) + * P r3 /3 sin /3 . . (II.)
For the crown 0 = 0, and A x' — A x" ; hence
B' = B" (III.)
For the left end, 0=a and A» = 0; for the right end, 0 = — a and
A a" = 0; that is,
— iHr3 (a — 3 sin a COS a -f 2aCOS2a) 4- i V r3 (3 sin2 a — 2 a sin a COS a)
— K r3 (H cos a + V sin a) a cos a — A r (1 — cos a) + B = 0, and
4- i H r3 (a — 3 sin a cos a + 2 a cos2 a) — | V r3 (3 sin2 a — 2 a sin a cos a)
•f K y3 (H cos a + V sin a) a cos a — A" r (1 — cos a) + B" = 0.
The subtraction and addition of these equation gives, after reduction,
H r2 (a — 3 sin a cos a + 2 a cos2 a)
— iPr2 (3 sin2 a — 2 a sin a cos a — 2 — sin2 3 + 2cos/3+ 2/3sin/3)
+ K rz 12 H a cos2 a + P (a sin a cos a — /3 sin /3) I
+ (A-A")(l-cosa) = 0 (IV.)
and -J Pr3 sin 3 (3 sin a — 2 a cos a)
- (A + A") r (1 - cos a) - K Pr3a cosa sin/3 + B + B' = 0 . . (V.)
282 SUPPLEMENT TO CHAP. XIV. [CHAP. II.
Here, as, before, we shall leave the discussion, as we have already all the
equations of which we shall make use.
7. Integrals used in the above Discussion.— For convenience
of reference, we here group together the known integrals employed in the
preceding discussion.
/ sin x d x = — cos x, / cos x d x — sin x.
I sin2 ado: = $x — isinxcosx, I cos2 xdx = ix + •£ sin x cos x.
I sin x cos x dx — i sin2 x.
I sin3 x dx — — \ cos x (2 + sin2 x), I cos3 x d x = $ sin x (2 -+- cos2 x).
I sin2 x cos x dx = $ sin3 x, I sin x cos2 x d x = — $ cos3 x.
I xsmxdx = smx — x cos x, I x cos x d x = cos x + x sin x.
I x sin2 xdx = ±x*+i sin2 x — | x sin x cos x.
I x cos2 xdx = ±x* — i sin2 x + %x sin # cos «.
I xsinx cos # tZ a? = J (2 « sin2 x — # + sin x cos #).
CHAP. III.] SUPPLEMENT TO CHAP. XIV. 283
CHAPTER III.
AEOH HINGED AT ABUTMENTS ONLY - CONTINUOUS AT CROWN.
A. PARABOLIC ARC — CONSTANT CROSS-SECTION — CONCENTRATED LOAD.
8. Horizontal Thru§t.— We can apply here directly the results of
Art 5. Thus, in equations (22) for x = 0, A $ = 0, and A 0" = 0, hence
A' = A". If then in eq. (V.) of that Art. we put A' for A", and then for
A — A' its value from (I.), we have at once
H_ 5 p5a*-6aV + g*_ (5 a* - z*) (a* - z>)
H=r*P~ a* h ~64P ~^h~ "(27)
This is the formula which we have given in Art. 159 of the text, without
demonstration. The thrust is greatest when the load is at the crown. We
have then z — 0 and H = |f P % The value of V is given in Art. 3.
n
9. Inter§ection Curve. — Denote the ordinate of the curve cdei/k
(PL 23, Fig. 91), taken above the line A B by y. Then we see from the
•y
Fig. that y = A N tang, d A N — (a — z) =. The value of H is given
above, that of V has already been found in Art. 3, eq. (16). viz.,
V = P - . Hence we have, after reduction,
™
(28)
which is the equation already given in Art. 159, and from (18) and (19)
the values of the table in that Art. have been calculated.
The above values of H and V are simple and of easy application, not in-
volving much calculation in any special case. Hence we can readily com-
pute H and V, and thus check the accuracy of our method of construction
given in Chap. XIV.
B. CIRCULAR ARC — CONSTANT CROSS-SECTION — CONCENTRATED LOAD.
1O. Horizontal Thru§t. — Here we can apply directly the results of
Art. 6. Thus, inserting in eq. (IV.) of that Art. A — A' for A — A", and
taking the value of A — A' from (I.), we have an equation for the deter-
mination of H. This, after reduction, becomes
_ sln2 a— sin- ft + 2 cos a (cos /3 — cos a) — 2 (1 + K) cos a (a sin a — )3 sin g)
2 [a — 3 sin a cos a + 2 (1 + *) a cos2 a]
which is the equation given in Art. 159 (2) of the text.
For the semi-circle, a = 90°, sin a — 1, and cos a = 0, and this becomes
284: SUPPLEMENT TO CHAP. XIV. [CHAP. III.
If we put
A! = sin2 a — sin2 £ + 2 cos a (cos /3 — cos a — a sin a + /3 sin j3),
A2 = 2 cos a (a sin a — /3 sin $),
B! — 2 (a — 3 sin a COS a + 2 a COS2 a),
B2 = 2 a COS2 a,
we have
A1-A2^_ *~A/ /A,
Bx+B, • *V B, \B,
H*
A TJ A
or, if we put A = -^, B= g2, and H0 = P ~, we have
, 1 — A K
But Ho = P .g-1 is the value of H from the formula above, when the terms
containing Ic are disregarded.
We have also, by series (see Art. 20, following),
A! = -,V («2 — £2) [ (5 a2 - /32) - ^o (49 a4 + 34 .a2 /32 - 11 £*) + . . .]
A2 = 2 (a2 - /32) [1 - i (4 a2 + ^2) + . . .]
B: = ^ a6 (1 - -/r a2 + . . .) B2 = 2 a (1 - a2 + . . .)
Approximately, therefore, since for h small with respect to a, the tan-
gent may be taken for the arc, and hence — = — , or a = — , we have
rah 2 h
24 6 a2 15 _ 15 a*
~~-
Hence, when rise is small compared with span, we have the approximate
expression
. 24 -
5 A2 1-- -
15 15
By means of a table calculated for H0, for various values of a and /3 = 0,
0.2, 0.4, etc., of a, the thrust can be readily found in any case from the
above formula.
We give in the following Tables the values of H, , A and B, calculated
from the exact formulae. The formula for H above is thus made of easy
practical application, without tedious calculation, and the results given by
the method of Chap. XIV. may easily be checked.
The value of V is given in Art. 3.
CHAP. HI.]
SUPPLEMENT TO CHAP. XIV.
TABLE FOB H0.
285
a
a = 0
a = 10°
a = 20°
o = 30°
a = 40°
a = 50°
a = 60°
a = 90°
0
0.391
0.391
0.388
0.385
0.380
0.373
0.364
0.318
0.2
0.372
0.372
0.369
0.365 0.359
0.352
0.342
0.288
0.4
0.318
0.317
0.315
0.309
0.301
0.292
0.278
0.208
0.6
0.232
0:231
0.228
0.222
0.213
0.202
0.187
0.110
0.8
0.123
0.122
0.119
0.115 0.108
0.099
0.086
0.030
1
0
0
0
0
0
0
0
0
a
Coefficients of P^.
h
Thus, for /3 = 0,
, etc., of a, the numbers in the table give the co-
efficients of P — for a = 0, 10°, 20°, etc.
h
For the values of A and B, we have the following
TABLE FOR A AND B.
a
gj
a = 10°
a =20"
a =30°
a = 40°
a =50°
a = 60°
a =90°
i
0
1.20
1.19
1.17
1.14
1.08
1.00
0.88
0
Coeffi-
0.2
1.21
1.20
1.18
1.15
1.10
1.01
0.90
0
of
A
0.4
1.24
1.24
1.21
1.18
1.13
1.05
0.94
0
a*
0.6
1.29
1.29
1.27
1.24
1.20
4.13
1.02
0
0.8
1.38
1.38
1.36
1.34
1.30
1.24
1.18
0
1'
1.50
1.50
1.49
1.47
1.45
1.41
1.36
0
B
0.234
0.233
0.221
0.203
0.178
0.146
0.107
0
a4
,0 2
Thus, for various values of — , we have the coefficients of ^, which give
A for a = 10°, 20°, etc., and for the values of a have the coefficients of
a4
— , which give B.
11. Intersection Curve.— Indicating, as before, by y the ordinate
286 SUPPLEMENT TO CHAP. XIV. [CHAP. III.
Nd of the curve cdeilc [Fig. 91], we have, as before, y=ANtan^.
d A N = (a - z) ^, or since a = r sin a, z = r sin fl. V = P sm « + sin ff
1=1 2 sin a
from eq. (17),
2 sin a H
Inserting the value of H above, we have
sin2 a — sin2 /3 ~1 + B » B 4
2 sin a l - A AC A
/sin2 a — sin'/SXBi
or if yo = r I -- g-jjT^ - 1^- ; that is, if y0 is the value of y when & i
neglected,
which is the value of y given in Art. 159 (2) of the text. In that Art. we
have already tabulated the values of A and B, as also of y0 for various
values of a and /3.
For /3 = a, that is, for the end ordinate, our expression for y reduces to
— In this case, by differentiating numerator and denominator, we have
a — 3 sin a cos a + 2 (1 + K) a cos" a
sin a — a COS a — K (sin a + a COS a)'
For the semi-circle, a = 90° = — , sin a = 1, cos a = 0, and hence
2
y = \ nr = 1.5708 r. Hence, for the semi-circle the intersection curve be-
comes a horizontal straight line at 0.5708 r above the crown. In all cases
for small central angle c, K may be disregarded.
The above results are sufficient to enable us to either diagram or calcu-
late the strains in every piece for any given position of load.
CHAJP. IV.] SUPPLEMENT TO CHAP. XIY. 287
CHAPTER IV.
ARCH FIXED AT ENDS.
12. Introduction. — In the previous case, the end reactions pass
always through the ends. If, however, the ends are " walled in," so that
the end cross-sections remain unchanged in position, and cannot turn, these
reactions pass then no longer through the centres of the end cross-sections.
In the first case, the moments at the ends are zero ; now, however, we have
end moments to be determined, viz., Mj and M2, left and right. For their
determination we have the condition that the tangents to the curve at the
ends must always remain invariable in direction, or for the ends, A ^> = 0.
In the arch above with hinges at ends, we have always considered a por-
tion lying between the end and any point. In the present case, however,
we shall consider the portion between the crown and any point. Both
methods lead, of course, to the same results, but the latter, in the present
case, is somewhat simpler.
Accordingly, we conceive the arch cut through at the crown [PI. 24,
Fig. 93]. The total resultant force exerted upon the one-half by the other,
we decompose into a vertical force V at the crown, and a horizontal force
H. The distance c & of this last from the centre of gravity of the section
at crown is e0, and hence the moment at crown is M0 — — H e0.
13. Concentrated Load— General Formulae. — Let a weight
P act at any point; then representing, as before, by primes, quantities
relating to the portion between the load and right end, we have, as in (18),
G = — H cos <j> — (P — V) sin 0, G' = — H cos 0 + V sin <£ |
N ; = - H sin 0 + (P - V) cos 0, N' = - H sin $ - V sin $ \" ^
Also, M = - H (e0 + y) - Vx + P (x - z), M' = - H (e0 + y} - V x,
or, since — H e0 — M0 = moment at crown,
M = Mo - H y - V x + P(x - «), M' = M0 - H y - V x . . (32)
(a) Intersection curve.
The two reactions, R and R', intersect, as before, in a point L (Fig. 92),
which must lie upon P prolonged, as otherwise R, R' and P could not be
in equilibrium. The locus of the point L we call, as before, the intersection
curve. The equation of this curve can be easily found when V, H and M0
are known.
The force acting upon the portion B E (Fig. 92) is the resultant of V
and H. The component H acts at the point of intersection o of this
resultant L ^ with the vertical through C. The vertical distance of this
point o from c is, as above, e0 ; its horizontal distance from P is z. Then
z cot L ^ Tc is the vertical distance of this point from L, and e0 4- z cot
288
SUPPLEMENT TO CHAP. XIV.
[CHAP. iv.
= e0 + z ~ = y, where, as in the Fig., e0 is negative. In any case, e0
IB
jyr
is given by e0 = — ^F 5 hence, for the intersection curve,
H
(33)
(&) Direction curves and segments.
The direction of the resultants R and R' can be determined in two ways
First, by the points of intersection $ and ^ with the verticals through the
centres of the end cross-sections ; second, by means of the curves enveloped
by these resultants for every position of P. We call the first distances
A (f) = Ci, B \|r = c2, the direction segments, and the enveloped curves the
direction curves.
Taking Ci and c2 as positive when laid off upwards above the ends, we
have MI = — H c\ M2 = — H c2 ; therefore
(34)
We may also easily determine the equation of the direction curves. Let
the co-ordinates with reference to the crown of any point be v and w (Fig.
92). If the load P is now moved through an indefinitely small distance,
the new resultant cuts the former in a point of the curve required. These
two resultants intersect the vertical through C in two points. Let the dis-
tances of these points from O be c and c + d c, and let y and y + d y be
the angles of the resultants with the vertical. Then v — (w + c) tan y,
v = (w + c + d c) tan (y + d y).
Eliminating v,
(c +dc) tan (y + d y) — c tan y _ d (c tan y)
tan (y + d y) — • tan y d tan y
From the first of the equations above we have then
dc
v = — -= tan2 y.
d tany
But tan y = -, c = - -, c tan y = — -, hence
. . (35)
w =
d
Thus we see that in any case we have only to determine H, V and M0, and
we can then from (33) and (34) or (35) determine at once the intersection
CHAP. IV.] SUPPLEMENT TO CHAP. XIV. 289
curve, and the direction segments or curves. These are all we need for our
method of construction as given in Chap. XIV. ; once given, we can then
easily construct H, V and M9 or Mi for any position of weight.
A. PARABOLIC ARC — CONSTANT CROSS-SECTION — CONCENTRATED LOAD.
#2
14. Determination of H, V and Mo.— We put y = h — ,
dy = 2 -jdx, ds = dx, as before. Then from the values of M given in
(32) we have, according to equation (13), Art. 1, after inserting the values
of y and d s above, and integrating,
and
r- T3T 7. -i
+ A'.
For x = z, A $ = A $', hence | P (z — 2 z) z + A = A', or
A-A' = iPs2 ....... (L)
For x = a, A $ — 0, and for x = — a, A 0' = 0, hence
0 = a
0 = -ajMo -iHA + *Vo + A',
and by addition and subtraction,
A + A' = Va't-lP(a-2z)a .... (II.)
2M0a-fHaA + iP(a-2)2 = 0 . . . (IE.)
From I. and II. we have
A = |V«2-iP(a2 - 2as-22) )
A'^-LVa2~iF(a2-2as + s2) j
For the horizontal and vertical displacement of any point, we have from
( 5), Art. 2, after integration,
E I A x =- i Mo z3- rf-i Vz<+iP Q^- | sz3) + i A «2 + B
],
J
El Art =- ^-[i Mo z3-^ V-* V «4+i AV+B'l
and
E I A y = iM0 «2- ?^2 ^4-| (V-P) z'-
i« a
B L A y»= i Mo x9- ?s a;4-i V aj«+ A'
19
290
SUPPLEMENT TO CHAP. XIY.
[CHAP. rv.
For x = z, A x = A a?', and A y = A y', hence
B-B' = /4-Ps4-i(A-A>2, and C - C' = $P zs - (A - A')z, or
(IV.)
(V.)
For a = a, A a? = 0, A y = 0, and f or 35 = — a, A *' = 0, A y' = 0.
hence,
0 = + I Mo a2 - nfeH Ji a* - i (V - F)a3 - $ Pzo? + Aa + O,
0 = + i Mo a2 — ff H h a2 + % V a3 — A' a + O'.
The addition and subtraction of the two first and two last of these equa-
tions gives, when we put for A +A', A — A', B — B', O — C', their
values above :
2 M0 a3 — | H Ji a3 + •£ P (3 a4 — 8 a3 z 4- 6 a2 s2 — 24) = 0 . . (VII.)
O + C' = — Mo a? + | H h a? — £ P a (a8 — 3 a z + 3 s2) . . (VIII.)
4 f V a9 — i P (2 a9 — 3 a8 s + z*) = 0 (IX)
Equations HE. and VII. contain only H and M0 unknown. Their solu-
tion gives
(36)
a3 h
M0 = -
From IX. we find directly
. . . (37)
(38)
These are the equations given in Art. 160. From them we have the fol-
lowing table :
z
H
V
Mo
z
H
V
Mo
0
0.4688
0.5000
- 0.09375
0.5
0.2637
0.1563
+ 0.02539
0.1
0.4594
0.4252
- 0.04936
0.6
0.1920
0.1040
+ 0.02400
0.2
0.4320
0.3520
- 0.01606
0.7
0.1219
0.0607
+ 0.01814
0,3
0.3882
0.2818
+ 0.00689
0.8
0.0607
0.0280
+ 0.01025
0.4
0.3308
0.2160
+ 0.02025
0.9
0.0169
0.0073
+ 0.00314
0.5
0.2637
0.1562
+ 0.02539
1.0
0.
0.
0.
.a
•*f
.P
.Pa
.a
•-:
.P
.Pa
CHAP. IV.] SUPPLEMENT TO CHAP. XIV. 291
From (34) and (35) we can now find the intersection and direction curves.
The preceding table gives us sufficient data for complete calculation by
moments according to Art. 162. The intersection and direction curves
will, as already explained, enable us to find the above quantities graphi-
cally.
15. Intersection Curve. — From (33) we have y = — — , or m-
H
serting the values of H, V and M0 above, and reducing,
Hy = &P (a ~ ^ (a + z? = iHft, hence y = *fc
a
For the parabolic arch with fixed ends, then, the intersection curve becomes
a straight horizontal line, i h above the crown.
16. Direction Curve.— From (36), (37) and (38) we have
-22g dV SP'-s8
dz ~ 8a*h ' dz 4 a3
P (a - z) (4aa - 5az - 5g2)
dz Sa*
Inserting these in (35), as also the values of H, V and M0 themselves, and
reducing, we have
(39)
15 (a + z) (3 a + 2)
For z = 0, 0 = | a, w = £f h. For z = a, v = %a, w — £ h.
For z — — a, v = a, w = — oo . Eliminating s, we have
x
15 a (a — v)
This is the equation of an hyperbola. Hence, for the parabolic arc with
fixed ends, the direction curve is upon each side of the crown an hyperbola.
This hyperbola is described in Art. 160 of the text, (Fig. 93), and a table
to facilitate its construction is there given.
B. CIRCULAR ARC — CONSTANT CROSS-SECTION — CONCENTRATED LOAD.
17. Fundamental Equations. — From eq. (32) we have, since
x = r sin $, y = r (1 — cos <£), z = r sin 0,
M = Mo — Hr(l — cos$) + (P — V)rsin$ — Ffsin/3
' = M0 - Hr(l - '
The expressions for G, Art. 13, eq. (12), apply here directly.
Therefore, from eq. (8), Art. 1, we have
292 SUPPLEMENT TO CHAP. XIV. [CHAP. IV.
Hence from (12), Art. 1, since ds = rdfa
^rM0--Hr(l-cos0) + (P-V)rsin(/)-P7'sm/3]^0-f — d<p,
-^rMo-Hr(l-cos<£)-Vrsind>l dd) + Sds' 0.
E I L J d s
Substituting the values of above, integrating, and putting, as be-
fore, for brevity, K = —^ we have
E I A <£ = r [MO 0 — H r (<f> — sin <£) — (P — V) r cos $ — P r $> sin /3 J
+ K r (M0 - H> - Gr sin /3) $ + A.
E I A <£' ' = r [MO <j>- Hr ($ — sin <£) + Vr cos $J + K r (M0 — H>) $+A'.
For 0 = 0, A ^ = A 0', and we obtain
A - A' - Pr2 [cos ft + (1 + K) j3 sin £ J . . . (I.)
For <p = a, A <£ = 0, and for <J> = — a, A #' = 0. Adding and subtracting
the equations thus obtained, and eliminating A — A', we have
A + A' = P r2 1 cos a + (1 + K) a sin /3~| - 2 V r8 cos a . . (II.)
2 Mo a - 2Hr(a- sina)-Pr [cos a - cos 0 4- (a - 0) sin /3 J
+ K [2 Mo a — 2H r a - P r (a — j3) sin /3j = 0 . . (HI.)
From eq. (14), Art. 2, we have, as before, after integrating, for the hori-
zontal displacement,
E I A a?— —Mo ?*2 (sin <j>— <p cos 4>)+£ H>3 (2 sin <p — 2 $ cos </> — <j> + sin <j> cos 0)
— i V r3 sina <p + i P r3 (sin2 <j> + 2 sin /3 sin ^> — 2 ^ sin /3 cos <J>)
+ K T*2 (M0 — Hr — P r sin /3) $ cos 0 + A r cos 0 + B.
E I A x' = — Mo r2 (sin <J>— <j> cos 0) 4- £ H r3 (2 sin <f>— 2 <£ cos 0 — <£+ sin 4> cos ^>)
— i V r3 sin2 0 + AC r2 (M0 — H r) <p cos ^ 4- A' r cos 0 + B'.
For * = /3, A a; = A #', hence
B-B' = -iPr(2 + sin2/3) .... (IV.)
Further, for $ = a, A* = 0, and for </> = — a, A x' = 0. Hence, by add-
ing and subtracting,
B + B' = Vr (2 - sin2 a) - £Pr (2 - sin2 a + 2 sin a sin /3) . . (V.)
2 Mo (sin a — a COS a) — HT* (2 sin a — 2 a COS a — a + sin a COS a)
+ iPr [2 - sin2 a + sin2 3 - 2 cos (a - /3) + 2 (a - 0) cos a sin /3j
- K f"2(M0 - Hr a cos a - Pr(a - 0) cos a sin /3J = 0. .(VI)
CHAP. IV.]
SUPPLEMENT TO CHAP. XIV.
293
Multiplying HI. by cos a, and adding to VI., we haye
2 Mo sin a — H r (2 sin a — sin a cos a — a) + | P r (sin a — sin /3)2 =0.
In similar manner, we have from eq. (14), Art. 2, for the vertical dis-
placement
E I A y = MO r2 (cos 4> + </> sin #) — i H r3 (2 cos <f> + 2 $ sin <f> — sin2 0)
^P?>3(4> + sin4>cos4>4-2sin/3cos4> + 2 <£ sin /3 sin <?>)
(Mo — H r — Pr sin /3) <£ sin 0 4- Ar sin <J> + O.
2 (cos # + 4> sin 0) — |Hrs (2 cos<|> -+- 2 <£ sin <£ — sin8 <j>)
+ £ V r3 (<£ + sin <£ cos <p) + K r2 (M0 — H r) <£ sin ^> + A' r sin # + C'.
For $ = 0, A y = A y', hence
C-C'^Pr'O -f sin /3 cos 0) . . . . (VIH.)
Finally, f or ^» = a, Ay = Q. For $ = — a, Ay' = 0, and hence
C + O' = — SMor2 (cos a -I- a sina) + Hr3 (2 cos a + 2a sin a — sin2 a)
+ £ P r3 1 a + sin a cos a — 2 sin (a - - j8) + 2 (a — 3) sin a sin 3|
- 2Hr2(M0-Hr)asina-|-KPrJ(a-/3)sinasin/3 . . .(IX.)
Vr (a — sinaCOSa) = £Pr(a— 13— sina cos a— sin /3 cos /3 4- 2 cos a sin /3). .(X.)
18. Determination of H, V and M0.
(a) Vertical ^Reaction.
The vertical force V is given directly by eq. X. Thus
_a — /3 — sina cos a — sin /3 cos /3 + 2 cos a sin /3
2 (a — sin a cos a)
an expression independent of AC.
Transforming by means of series, we have, approximately,
For the semi-circle
TT — 2 3 — 2 sin (3 cos 3
27T
(43)
(44)
(45)
From the exact formula (43) we have the following table :
3
a = 0
a = 10°
a = 20°
a = 30°
a = 40°
a = 50°
o = 60°
a = 90°
0
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.2
0.3520
0.3515
0.3500
0.3475
0.3439
0.3392
0.3332
0.3065
0.4
0.2160
0.2152
0.2130
0.2092
0.20B7
0.1966
0.1876
0.1486
0.6
0.1040
0.1033
0.1014
0.0981
0.0934
0.0874
0.0799
0.0486
0.8
0.0280
0.0277
0.0269
0.0255
0.0238
0.0211
0.0182
0.0065
1
0
0
0
0
0
0
0
0
.a
.P
. — .
294
SUPPLEMENT TO CHAP. XIV.
[CHAP. iv.
(5) Horizontal thrust.
Eliminating M0 from HI. and VI. we obtain, after reduction,
2 sin a Fcos ft - cos a + (1 + *) /3 sin /3~| - (1 + *) a (sin2 a + sin2 /3)
H = P L= — J . . (46)
2 I (1 + K) a (a + sin a cos a) — 2 sin2 al
If we put Ai = 2 sin a (cos /3 — cos a + /3 sin /3) — a (sin2 a + sin2 /3),
A2 = a (sin2 a + sin2 /3) — 2 /3 sin a sin /3, ,
Bi = 2 a (a + sin a cos a) — 4 sin2 a,
B2 = 2 a (a + sin a COS a),
A ip A
and let A = -^-2-, B = =*-, and H0 = =•* P, we have
AI J3l JD!
. 1- AK
° 1 + B *'
where H0 is the value of H from the above formula, when terms containing
K are disregarded.
Transforming by series, we have
H =
From the exact formula (46) above, we have the following tables :
TABLE FOB H0.
/B
a=0
a = 10°
a =20°
a = 30°
n = 40°
a =50°
a = 60°
a =90°
0
0.4688
0.4687
0.4683
0.4678
0.4671
0.4661
0.4610
0.4592
0.2
0.4320
0.4317
0.4309
0.4291
0.4272
0.4243
0.4173
0.4017
0.4
0.3308
0.3301
0.3281
0.3244
0.3196
0.3128
0.3012
0.2601
0.6
0.1920
0.1912
0.1887
0.1845
0.1784
0.1703
0.1578
0.1087
0.8
0.0608
0.0603
0.0590
0.0566
0.0534
0.0490
0.0421
0.0181
1
0
0
0
0
0
0
0
0
.a
|
CHAP. IV.]
SUPPLEMENT TO CHAP. XIV.
295
For the values of the quantities A and B we have the f ollowing table :
VALUES OF A AND B.
*
. = 0
a =10°
a =20°
«=30°
a =40°
a=50°
a=60°
o=90°
0
3.00
3.01
3.02
3.06
3.10
3.16
3.25
3.66
0.2
3.12
3.15
3.16
3.19
3.26
3.34
3.45
4.07
0.4
3.57
3.58
3.63
3.70
3.81
3.79
4.20
5.65
a*
A
0.6
4.69
4.70
4.81
4.96
5.20
5.57
6.14
10.57
h*
0.8
8.33
8.42
8.67
9.10
9.78
10.87
12.93
35.62
1
CO
CO
CO
CO
CO
CO
CO
CO
B
.a
2.813
2.825
2.861
2.931
3.039
3.198
3.417
5.279
a*
From the above tables it is easy to find the thrust for any given position of
load, and any given span and rise. The preceding table gives the reac-
tion ; it only remains to determine the moment M0 at crown.
(c) Moment at crown.
From VII. we have
(sin a — sin B)2
v
/~ a \
Mo = JHr [2 — cos a- - - )-
\ sin a/
sin a
Substituting the value of H, already given, eq. (46), we obtain
2 Mo f"(l + K) a (a + sin a cos a) — 2 sin2 a~|
= Pr I sina— sina cos (a— £)+2 sina (cos #— cosa)— sina (sina— sin /3) sin ,8
— a (cos /3 — cos a) — (1 + K) { a (sin2 a -t- sin2 [3) — 2 ^ sin a sin /3 J
+ (1 + K) (a — /3) (a + sin a cos a) sin /3 | .
In similar manner, as before, for H we have
0*
MO=MO
l+B/c'
where M00 is the value of M0 when terms containing Tc are disregarded,
and B has the same value as above. By series we have
[~3a2-
L
10a/3-3/32
and
^4 (3a4 + 6a* 0+ 45 a2 /32 + 308a/33 + 154/34)!,
0 =
360
a2 (3a2 -10a/3-5/32)
296 SUPPLEMENT TO CHAP. XIV. [CHAP. IV.
From the exact formula above we have the following tables.
VALUE OF M00.
&
a = 0
0=10°
a = 20°
o = 30°
a =40°
a =50°
a = 60°
a = 90°
0
0.09375
0.09426
0.09582
0.09849
0.10241
0.10777
0.11613
0.15108
0.2
0.4
0.01606
0.02025
0.01615
0.02021
0.01658
0.02008
0.01742
0.01973
0.01846
0.01942
0.02063
0.01888
0.02280
+
0.01719
0.03083
+
0.01441
0.6
0.02400
0.02393
0.02369
0.02326
0.02267
0.02181
0.02001
0.01454
0.8
0.01025
0.01019
0.00999
0.00962
0.00917
0.00850
0.00716
0.00332
1
0
0
0
0
0
0
0
0
.a
.Pa
VALUES OF B AND C.
•
a=0
a = 10°
a = 20°
a = 30°
a =40°
a = 50°
a =60°
o = 90°
0
7.50
7.52
7.57
7.78
7.94
8.52
8.61
11.12
0.2
28.13
28.09
27.99
28.14
28.43
27.99
27.76
31.81
0
0.4
12.50
12.61
13.05
14*08
15.40
17.35
21.11
37.47
a4
0.6
4.69
4.77
5.00
5.51
6.24
7.31
9.10
20.27
0.8
2.74
2.85
3.20
3.89
4.94
6.61
9.77
42.53
1
1.88
00
00
00
00
00
00
00
B
.a
+2.81
+2.83
+2.86
+2.93
+3.04
+3.20
+3.42
+5.28
a4
Here, as always, a negative moment denotes tension in lower or inner
flange. We see at once from the table that the maximum compression in
this flange at crown does not occur for full load, but for load extending
from both ends towards the crown as far as about gths of the span or fths
of the half span. "Within the middle half of the arch, then, a load any-
where causes tension in lower flange at crown— outside of this middle half
a load anywhere causes compression in the lower flange at crown. For
large central angles, K may be disregarded, and we have simply M = M00.
CHAP. IV.] SUPPLEMENT TO CHAP. XIV. 297
19. Intersection Curve.— From Art. 13 we have
V r sin 3 - M0
y= — ^ — — •
Hence, by substitution of the values of V, H and M0,
- rj£ l~l - 12°(a* -2afl-/32) 1
y "" 10 L " a4 (a + ft)2 *J
which is the equation given in the text, for which a table is there given.
2O. Direction Segments.— It will in the present case be found
most convenient to determine the directions of the resultants by d and e2
equation (34).
M! M2 -
Thus, Ci = - -|p c2 = — -jj-.
But M! = Mo - H h + (P — V) a — P z, M2 = M0 — H h + V a.
We have, by series, then the approximate formulae,
45Ififl 2h T 451
15 («+«) L A7i2 J 15 («-+-
where positive values of Ci and c2 are laid off upward above, negative
values downward below, the centres of gravity of the end cross- sections.
From the preceding tables we can calculate easily in any case H and V
an^. Mo, and thus check the results obtained by the method of Chap. XIV.
The formulae above for d and c2 do not admit of tables, nor, in fact, are
such needed. They are sufficiently simple for ready insertion.
Thus, by the aid of our tables, having computed V and H, and, if neces-
sary, Mo and 0o, we can by the method of moments, as explained in Chap.
XIV., Art. 162, readily calculate the strains in the braced arch, whether
continuous at crown and fixed or hinged at the ends, or hinged at both ends
and crown.
2O (ft). Transformation Series.— We have in the preceding repeat-
edly made use of series in the transformation of angular functions, such as
sin, cos, etc., into functions of the arc itself. We group here, for conve-
nience of reference, the series thus used :
sin x = x (I - i»2 + 7^0 aj4 - <roL4-u «6 + Wainr & - uinrunn) o *
COS X = 1 - i X2 + & X* - Tib «? + Toku «" ~ TSjtv 0 0 ^ °
sin 2x = x2 (I - -U2
cos2 0=1- x
sin x cos x = x (I — I x2 -t- -ft x* — 3-73- x° + dW «9 — ITT&TS a?1 ° + ...)
298 SUPPLEMENT TO CHAP. XIV. [CHAP. IV.
COS3 X = 1 - t X9 + U4 - #jj Xs + Tfffcj X8 - TTWoo Z1 ° + ...
tan a; = x (1 + 1 x2 + -fr a* + & x« + 7f|F zs + rHfle «10 4- ...)
1 #
cot * = - - (1 + -iV x2 + ^ tf + -nky a;6 + ^if «? «8 + ...)
sin x siny = xy [l
+ 7 32
cos a; cos y=l-
+15
sin « cos y=a [l-
+35
CHAP. V.] SUPPLEMENT TO CHAP. XIV. 299
CHAPTER V.
INFLUENCE OF TEMPERATURE.
21. Oeneral considerations.— When the temperature of a per-
fectly free body, which possesses in every direction the same coefficient of
elasticity and expansion, changes equally at all points, there can be no
strains in the body. For were there such strains, then, as there are no
outer forces, there could be no equilibrium.
If, however, the change of temperature is not the same at all points ; or
if the body is not free, so that it is possible for outer forces to act, there
are strains.
In the following we assume the change of temperature to be everywhere
the same, but that the body is not free.
We assume that at a certain temperature t0 no strain exists in the body,
and call this the mean temperature. The deviation above or below the
mean temperature we call + t or — t, and denote the coefficient of expan-
sion for one degree by e.
The determination of the strain in a straight beam held at both ends, is
very simple. If the length is I, its relative change of length is e t. Since,
however, it cannot expand, the strain S per unit of area is precisely as great
as the force which would be required to produce this relative elongation,
or from eq. (4) s = + E e t ........ (48)
If the area of cross-section is A, then the strain at each end is
In equation (48) it is assumed that a compressive strain, due to + t, is
positive, a tensile strain, due to — t, is then negative.
22. Influence of Temperature on the Arch.— Since by a
change of temperature the length of the arch varies, while the span remains
always the same, the shape or curvature must change, which naturally must
give rise to strains and outer forces. In the following we have only to
determine these outer forces, since, as shown in Chap. XIV., these are all we
need to determine the strains themselves.
*f / TVT\
The relative change of length is from eq. (8), Art. I., ^-^ JG + — J.
This change is caused by the outer forces. The relative change of length
due to temperature alone is e t. Hence the total relative change of length is
Gr + M
300 SUPPLEMENT TO CHAP. XIV. [CHAP. V.
Hence the change of length of the axis is
— / d s — f t s (50)
The change of the angle between two infinitely near cross-sections, and the
actual turning of a cross-section is from (9) and (12), Art. 1, given by
M G r + M
d 0 - El EA?>
Gr
1 /»M 1 /^
l-(j> = Ej Tds+<B~rJ
A
Finally, from (14) we have
d s
ds
/(*&d s
&<!>dy + I -j^
/
^tX
Substitute in these last two equations for A <j> and - their values from
(.1 8
(49) and (52). The double integral thus arising can be resolved by par-
tial integration.
Thus /* //(*)<&
Applying this, we obtain
r r&
Ax— — yA0+ / yd&+ I — — dx
J J *' .... (5S)
/r&ds
zdA<f>+ I -j—dy
We shall assume in the following the axis always circular.
23. Fundamental Equations — General. — Upon this assump-
tion of a circular axis we have generally
G = H cos <£, N = H sin <f> "J
M = M0 -f Hr(l-cos^) I ..... (54)
Gr + M = M0 + Hr j
Hence, from the preceding Art,
H r + Mo ' /**
A s = — — — -Id + —Ttt$ ..... (55)
E A Jo
AEy. °
A a? = — r A 0 (1 — cos <j>) H / M (1 — cos $>) d $
r + Mo r _
EA J * *««?*+
CHAP. V.] SUPPLEMENT TO CHAP. XIV. 301
24. Arch with three Hinges.— If there are three hinges, the mo
ment M0 at the crown must be zero, and therefore M = H?*(l — cos <£).
But for 0 = a, M must also be zero, hence H r (1 — cos a) = 0, and therefore
H is zero. Then for any point G = 0, and M = 0, and N = 0. That is,
for the arch with three hinges there are for a change of temperature no outer
forces, and hence no strains.
25. Arch hinged at Ends.— Here, since f or 0 = a, M = 0, we
have from. (54)
Mo = — H r (1 — cos a) \
M = - H r (cos 0 - cos a) I (58)
G r ~\- M = + H r cos a J
. From (56), since for 9 = 0, A 0 = 0,
A ^ = — / (cos 0 — cos a) d 0 + cos a / d $ . . (59)
J o Jo
From (57), since for 0 = 0, A * = 0,
Hr'f f* C* 1
A 02= I (1 — COS 0) / (COS 0 — COSa)^0— / (1— COS 0) (COS 0 — COS a) d 0 I
L Jo Jo J
H r cos a cos 61,
•I ^r~a I d ^ — r e t sin (j>.
E A /
Jo
For ^ = a, this becomes zero, and we have for the horizontal thrust
_, _ E c t sin o
r2 Ca >cos« /"" cos2 a /*" '
— I cos <4 (cos 0— cos a) d <h — I (cos0— cos a)d6-\ i — I a$
I / I / ^ A /
«/0 i/0 t/0
.... (60)
Performing the integrations indicated (Art. 7), and putting, for brevity,
K = , we have
Ar2
„_ 2EIf£sina
r2 (a — 3 sin a cos a + 2 a cos'2 a) + 2 x?*2 a cos2 a
By series (Art. 20), we obtain the approximate formula
r2 (2«4 + 15/c) 8AA2 + 151
The above are the expressions given in Art. 165 without proof. The less
h, the greater for equal dimensions is H. For h = 0, we have
H = E A e t, as we should have for a straight beam.
26. Arch without Hinges.— In this case we have the general equa-
tions (54) and (55), which apply directly without change.
From (56), since for 0 = 0, A 0 = 0, we have
302 SUPPLEMENT TO CHAP. XIV. [CHAP. V.
From (57), since for $ = 0, A x — 0, we have, inserting the value of A <£,
above,
«."r A r , i
A x = 1 cos & I dfy — I cos 0 a 0 I
H :JF A A
~^F I C1 ~ cos <?) / (1 - cos 0) <Z0 - / (1 -
EIL Jo Jo
r*
» / d $ —
Jo
-+- — =-. - — - cos 0 / d $ — r 1 1 sin
E A
For 0 = a, A ^ = 0, hence from the first of these expressions
r2 C" • 1 Ca
=
r* c* i r
— I dQ + j Ic
ljo AJo
If the distance at which the horizontal thrust H acts from the crown is
«o, we have M0 = — H e0, whence we see at once that e0 is the fraction in
(63) multiplied by r. For </> = 0, A a must also be zero, and we thus
obtain another relation between Mo and H which does not contain A. If
we multiply the expression thus obtained by r cos a, and then subtract the
result from that previously obtained for 0 = 0, A $ = 0, we have
^ /"a /*a
| / (l-cos0)d0- / (l-
.... (64)
Performing the integrations indicated in (63), we have (Art. 7)
where, as before, K = - .
Ar2
From (64) we obtain
M0 r sin a — iHr2 (a — 2 sin a + sin a cos a) = — E I f t sin a.
Inserting the value of M0 above, we have
H = — 8BI,<(l + ,).rin.
2 [(1 + K) (a2 + a sin a COS a) — 2 sin2 a]
r
and hence
M — — -
""»•[(! + f) (a2 + a sin a cos a) — 2 sin2 a]
From these two we obtain for the point of application of H
« = -igt = + (1+')''-rin"r .... (88)
(1 + K) a
CHAP. V.] SUPPLEMENT TO CHAP. XIV. 303
By series, we hare (Art. 20) ,
without reference to K, e0 = $ h.
For small central angles, then, for which < may be disregarded, the
thrust given above by (66) acts at } Ji below the crown for a rise of tem-
perature of t degrees above the mean. For a decrease of temperature be-
low the mean it acts above, M0 is negative, and the strain in the lower
flange tensile.
Further, we have, by series, the approximate formulae
r2(a4 + 45*) 4 A A2 +451
These are the expressions given in Art. 165 without proof.
304 SUPPLEMENT TO CHAP. XIV. [CHAP. VI.
CHAPTER VI.
PAETIAL UNIFORM LOADING.
27. Notation. — In the preceding discussion of the -arch we have con-
sidered the influence of a> single concentrated load only, and this, as we
have repeatedly seen in the case of the simple and continuous girder, etc.,
is sufficient for full and accurate solution. When once we are able to find
and tabulate the strains in every piece due to a single load in any position,
the thorough solution becomes simply a question of time.
It may often happen, however, that we may wish to determine the strains
for a full load only, or for a uniformly distributed load extending from
one end to some given point. In such case it would be unnecessarily tedi-
ous to obtain our result by the successive determination and addition of all
the intermediate apex loads. We may easily deduce from the preceding
the general formulae for partial loading also.
As before, we shall let a = the half span, h = the rise, I the moment of
inertia, and A the area of the cross-section. But we shall represent by p
the load per unit of length of horizontal projection, and by z the distance
of the end of the load extending from the left, from the crown. This dis-
tance z, from the crown to the end of load, is then positive towards the
left. In the circular arch the angle subtended by this distance -z we call /3.
The angle /3 is then positive to the left of the vertical. The angle sub-
tended by the half span is, as before, a. For /3 = a, then, or for z — a,
there is no load upon the span. For /3 = 0, or z = 0, the load extends
from the left to the centre. For /3.= — a, or z = — a, the load covers the
whole span. Pis. 23 and 24, Figs. 91 and 92, still hold good, therefore, for
our notation. We have only to conceive, instead of the concentrated load
P, a uniformly distributed load, per horizontal unit, extending from left end
as far as the position of P. This much being premised as to notation, we
shall treat, as before, the three cases of arch hinged at crown and ends,
hinged at ends only, and without hinges.
A. ARCH HINGED AT CROWN AND ENDS.
28. Reaction. — This case is too simple to demand any extended
notice, in view of what has already been said. We have from eq. (16), Art.
3, for the reaction at the left or loaded end, for concentrated load,
2a
If now we put P = p d z, and integrate, we have
r ., a + z 2az + z
V = / p d z — — = p — -. -- h C,
J * 2a ±a
where O is the constant of integration. By taking the proper limits, we
can eliminate this constant, and thus obtain the reaction for load covering
CHAP. VI.] SUPPLEMENT TO CHAP. XIV. 305
any desired portion of the span. As we shall in every case suppose the
load to extend from the left end up to any point, we shall take the limits
of z = a and 2, and therefore obtain
For z — a, this is zero, as it should be, since then the load has not come
on. For z = — a, the load extends over the whole span, and V = p a, or
half the whole load, as it should. We might have obtained this result at
once by moments. Thus,
V x 3a = a -
but have preferred the above method as showing how uniform loading is
deduced directly from concentrated by inserting pdz for P and inte-
grating.
29. Horizontal Thru§t. — In precisely similar manner we have
from (21), Art. 4, for the thrust due to concentrated load P, H = P ^ ~ e\
2 h
Put P — p d 3 and inte-grate between the limits z — a, and 2, and we have
For z = a, this is zero, as should be. For z = — a, or for full load over
'D QJ^"
whole span, H = ^T. We may also deduce the above equation directly
tii fi
by moments.
The above formulse (71) and (72) are all that we need either for calcula-
tion or diagram. They apply evidently equally well, whether the arch be
circular or parabolic, or, in general, whatever its shape may be. The form
has no influence upon either the thrust or the reaction.
For the moment at any point whatever, whose distance horizontally
from crown is x and vertically below crown y, we have at once
M = H (h - y) - V (a - x) + 2- (a - a)2.
2
If this point is an apex, then the moment divided by depth of arch at
this point is the strain in flange opposite that apex. A positive moment
throughout this work always indicates compression in the inner or lower
B. AKCH HINGED AT ENDS ONLY.
30. Reaction.— The vertical reaction at the end is precisely the same
as before for three hinges, and is given by equation (71). This reaction is
evidently independent of the shape of the arch, and the above formulae
holds good generally.
31. Horizontal Thr«§t— Parabolic Arch.— We must here
distinguish the shape of the arch, and treat first the parabola. We have
already from eq. (27), Chapter III., Art. 8, for a single load,
5
~
20
306 SUPPLEMENT TO CHAP. XIV. [CHAP. VI.
We have, as before, simply to make P = p d z, and then integrate between
the limits z = a and z indeterminate.
We thus find at once
n r -i
. . (73)
For z = a, this reduces to zero, as it should. For z = — a, the load covers
the whole span, and we have H —^-. For z = o, the load reaches from
2 h
2
the left as far as the crown, and H —-r-^-. The formulae is simple, and re-
4ft
quires no table. Numerical values may be easily inserted.
32. Horizontal Thrust— Circular Arch.— As already noticed,
the vertical end reaction for this case has been given in eq. (71). It re-
mains to determine the thrust. We have, as before, simply to insert pdx =
p r cos /3 d 3 in place of P in the expression for the thrust for concentrated
load of Art. 10, and then integrate between the limits /3 = a and /3 inde-
terminate.
We have thus similarly to that Art.
where H0 is the value of H when terms containing K are neglected, or
H.=?;|-;andA = ^B = B1
12 BI AI BI
The quantities Ai, Bi, A2 and B2 are as follows :
A, = 7 sin3 a + 3 a cos a — 3 sin a — 6 a cos a sin2 a — 6 sin2 a sin /3
+ 2 sin3 /3 — 3 /3 cos a — 9 cos a sin £ cos /3 + 12 cos2 a sin /3
+ 12 a cos a sin a sin £ — 6 £ cos a sin2 /3.
A2 = 3 [2 a cos a sin2 a + a cos a — sin a cos2 a — 4 a cos a sin a sin j3
+ 2 |3 cos a sin2 0 — /3 cos a + cos a sm £ cos 3].
Bj = a — 3 sin a cos a + 2 a cos2 a. B2 = 2 a cos2 a.
These expressions can be tabulated as in Art. 10, or developed into
series as in that Art., and the formula thus made practically available.
For £ = a, we have H zero, as should be. For ]3 = — a, we have the
load covering the entire span.
For this case we have
H_l f sin3 q - 3 (1—2 sin2 a) (sin a -acos a) - 3 K cos a (a + 2 a sin» a — sin a coa a)
' ~ 6 "a •+• 2 a cos'-i a — 3 sm a cos a + 2 K a cos1^ a
For the SEMI-CIRCLE, this reduces simply to
H = — pr = 0.424 pr.
3 7T
In any case where exact results are desired, eq. (74) must be used, and a
table calculated for the central angle a. We have approximately by
series also, more especially for small central angles, or for a large in respect
to hj for total load over whole span :
j>a2 8 Aft* _pa* 8ft2 „
2 ft 15 I + 8 A ft2 ~ 2 ft 15 g* + 8 ft2
CHAP. VI.] SUPPLEMENT TO CHAP. XIV. 307
where A is the area and I the moment of inertia of cross-section, and g
the radius of gyration. In framed arches this may be taken as approxi-
mately equal to the half depth from centre to centre of flanges.
C. ARCH WITHOUT HINGES — FIXED AT ENDS, CONTINUOUS AT GROWTH.
33. Parabolic Arch — Formulae for V, H and M. — In this
case the reactions no longer follow the law of the lever, and eq. (71), there-
fore, no longer holds good.
(a) Vertical reaction at unloaded end.
We have from eq. (28), Art. 14, for the reaction at the right end for a
single load,
1 (a - zY (2 a + e)
~4P ~~^~~
Making P =pdz, and integrating between the limits z = a and a, we
find the reaction for a load coming on from left,
V =
L. 3 a4 - 8 a:J z + 6 a2 z2 - z*\ . . . (76)
for a full load z = — a and V=pa, as should be.
(5) Horizontal thrust.
In like manner we have for the horizontal thrust at end from (36), Art.
14,
15 p g* - 2 a2 z* + g*
H = 32 a*h
Replacing P by p d z, and integrating as before, we obtain directly
H= -— -15a«2 + Wa*z*-3z* . . (77)
in $2
for a full load z = — a, and H = £— — .
to h
(c) Moment at unloaded end.
In precisely similar manner we have from (37), Art. 14,
_ 1 (a — zY (3 a2 - 10 a z - 5 g2)
~32 ~~^~~
Putting P = p d z, and integrating, we harve for the moment, always at
the right end, or for load not extending past the centre, at crown
M
= -£- [3a4z-8a3z* +6a*z* - g8] . . (78)
32 a L J
For z = a, this is zero, as should be. For z — 0, or for load extending as
far as crown, it is also zero. For z = — a, the moment at the end is — ^— .
2
A negative moment, as always, denotes tension in lower flange.
Just as for concentrated load, as shown in Art. 14, as the load comes on,
the moment at crown is positive, and increases with increasing load up to
a certain point, beyond which any load causes a negative moment, and be-
yond which the moment at crown, therefore, decreases, until, when the load
303
SUPPLEMENT TO CHAP. XIV.
[CHAP. vi.
reaches the crown, it becomes zero. This point, which gives M0, the mo-
ment at crown, a positive maximum, is at a distance z = — a + aVf —
0.264911 a, or nearly i a from the crown.
The values of V2, H and Ma (M2 and V? being always the moment and
reaction at unloaded end), for various values of z, are given in the following
table:
z
V2
H
M2
z
V2
H
M2
1
0
0
+
—0.1
0.2349
0.29656
-0.01206
0.9
0.0002437
0.000579
0.0001096
-0.2
0.3024
0.34128
-0.03024
0.8
0.0014
0.00421
0.00076
-0.3
0.3707
0.38242
-0.055611
0.7
0.00624
0.01643
0.002185
-0.4
0.4459
0.41846
-0.08918
0.6
0.0144
0.02889
0.00432
-0.5
0.5273
0.44824
'-0.131836
0.5
0.0273
0.051757
0.006836 i
-0.6
0.6144
0.47104
-0.18432
1
0.4
0.0459
0.08346
0.00918
-0.7
0.7062
0.48667
-0.24718
0.3
0.07074
0.11777
0.010611
-0.8
0.8014
0.49884
-0.32076
0.2
0.1024
0.15872
0.01024
-0.9
0.90024
0.49942
-0.405109
0.1
0.1412
0.20315
0.007062
-1
1.0000
0.5000
-0.5000
0
0.1875
0.25
0
a
pa
pa*
p a2
a
pa
p az
p a2
h
k
It will be seen that the moment at the unloaded end, which, as long as
the load is left of crown, is the moment at crown also ; increases as the
load passes on, is positive and increases up to about z = .25 a. Then it
diminishes, becomes zero when the load reaches the crown, changes to
negative as the load passes the crown, and this negative value increases up
to full load when it is — i p a2. For full load, then, the lower end flanges
are in tension. At the crown the moment is zero, and the compression
there in both flanges is due to H only.
34. Circular Arch— Formulae for V, H and H.
(a) Vertical Reaction.
Here we have r sin /3 = x, r cos j3 d /3 = d $, and P=pdx = prcosfidp.
Inserting this in place of P in eq. (43), Art. 18, and integrating between
the limits /3 = a and /3 indeterminate, we have for the reaction at w?aloaded
end, or for reaction at crown when load does not extend past the crown,
V = — ^L |COS a — cos a — a sin /3 + cos ft + /3 sin |8
2 (a — sin a cos a) L 3
+ sin a cos a sin 0 - 22jl5 _ cos a sin2 01. . . .(79)
J
CHAP. VI.] SUPPLEMENT TO CHAP. XIV. 309
For 0 = a, this is zero, as should be, since then the load is not upon the
span. For 0 = — a, V = p r sin a, as should be, for full load over whole
span. For the semi-circle, a = 90° = — , sin a — 1, cos a = 0, and
2
TT nr»H3 R
— - sin 0 + cos ft + 0 sin 0 —
V — p r — — .
7T
If the semi-circle is uniformly loaded over whole span, 0 = — a = — 90°
= — — , sin 0 = — 1, cos 0 — 0, and V — pr, as should be. The for-
2
mula (79) above is precisely the same as that given by Capt. Eads in his
Report to the Illinois and St. Louis Bridge Co., May, 1868.
(5) Horizontal Thrust.
In similar manner, from eq. (46), Art. 18, by inserting p d x = p r cos 0 dj3
in place of P, and integrating between 0 = a and 8, we have similarly to
Art. 32
H = Ho^A_* (80)
and
, ,
12 -t»i AI Uj
A! =3 a— 3 sin a cos a— 2 a sin2 a— 3 0— 9 sin 0 cos 0+12 COS a sin 0
—6 0 sin2 0+6 a sin a sin 0+2 a ^?— -,
sin a
A2=3a— 3 sinacosa+2a sin2 a+60sin2 0— 3 0+3 sin/3 cos0
— 6 a sin a sin 0—2 a — - ,
sin a
Bi=a (a+sin a COS a) —2 sin2 a, B2=a (a+sin a COS a).
Formula (80) agrees exactly with that given by Capt. Eads in the Report
above quoted, if terms containing K are neglected. Since K = - -, where I
is the moment of inertia and A is the area of cross-section ; r being the
radius ; for small central angle r is very large in proportion to — , or the
A
square of the half depth. In such case, then, K. may be neglected. For
0 == a, we have H = 0, as should be. For 0 = — a, we have load over
entire span, and
H_i. • 3 a— 2 n sin- «— 3 sin a cos a— K (3 a+2 a sing a— 3 sin a cos a)
(1+K.) a (a+sin a cos a) —2 sin2 a
Approximately we have, by series, for full load, from (80) :
_
2h 2 22
310 SUPPLEMENT TO CHAP. XIV. [CHAP. VI.
For o = 90°, or for semi-circle, we have from (80)
t [37r-6/3+18sin/3cos/3-12/3sin2£+67rsin/3
•hK (-3 7T-12 ft sin2 /3+6 £-6 sin ft cos ft+Q ir sin /3+2 TT sin3 /3j.
For /3 = — 90°, or for full load upon semi-circle,
PL- V«,-9.
(c) Moment at unloaded end.
From Art. 18 (c) we have, for concentrated load,
(sin a — sin /3)2
/2-
\
cosa- --
sin a/ Sill a
The value of H we have already given in (80).
Inserting in the second term p d x = p r cos ft d ft for P, and then inte-
grating, we have
M = AH-B ....... (82)
where A = | r 1 2 — cos a -- ^— ) = JT-? — (2 — cos a — -^— \
y sin a/ 2 sin a \ sin a/
and B = P f (sin a — sin ft } .
12 sin3 a \ /
For a uniform load over whole span, ft = — a, and
^ . (83)
sm «/ 3
"We have from (83), by series, the approximate formula for moment at
crown
175y« a* .
where A is the area and I the moment of inertia of cross-section, g the
radius of gyration, or, approximately, the half depth for framed arch — as
always a negative moment indicates tension in lower or inner flange.
Equations (79), (80) and (82) may, if desired, be tabulated as in Art. 18.
For small central angles, or for h small with respect to a, K may be disre-
garded, and the results already given for parabola (Art 33) may be taken
as sufficiently exact.
CHAP. XV.] THE STONE AECH. 311
CHAPTEE XY.
THE STONE ARCH.
167. Definitions, etc. — In the stone arch we have a system
of bodies in contact with each other, and so supported between
certain fixed points, that they are not only in equilibrium
among themselves, but also with the exterior forces. The sur-
faces of contact we call the bed-joints ; the fixed points are
the abutments ; the central or highest arch stone is called the
key -stone, and those resting upon the abutments, the imposts.
The inner and outer limiting surfaces of the arch, generally
curved, are designated as the intrados and extrados, and the
arch stones generally are called voussoirs.
16§. L.IIIC of Pre§sure§ in Arch. — We have already indi-
cated (Art. 28, Fig. 16) the manner in which a number of suc-
cessive forces are resisted by an arch. We see from the force
polygon in that Fig. that the horizontal pressure is the same at
every point, and that the vertical pressure is equal to the sum
of the weights between the crown and any point. The pres-
sure line is then an equilibrium polygon formed by laying off
the weights of the arch stones, choosing a pole, and drawing
lines from this pole, etc., as described in our second chapter.
If the weights are very small, and their number very great,
the equilibrium polygon becomes a curve. This curve for
equilibrium should never pass outside the limits of the arch.
169. Sliding of the Arch Johns. — The arch is properly,
then, nothing but a curved wall. Upon a vertical wall, which
may also support loads, but which has no horizontal thrust, only
vertical forces act, and the resultant is known in position and
direction. We may, then, investigate the stability of an ordi-
nary wall, and apply the results directly to the arch.
We assume the wall divided by plane bed-joints extending
through its entire breadth, whose distances apart depend upon
the dimensions of the stones. These joints are the weak places
of the wall, since separation here is not resisted by the greatest
strength of the stone. Neglecting the influence of the mortar,
312 THE STONE ARCH. [CHAP. XV.
we assume that any section along a bed-joint resists only a per-
pendicular pressure due to the parts above, and a force paral-
lel to the joint which must not exceed the resistance to sliding
due to friction. If this parallel force is greater than the resist-
ance of friction, the upper part will slide upon the joint.
If we represent the greatest angle of repose by <£, then the
resultant of the vertical forces, acting upon the joint in ques-
tion, must make an angle with the normal to the joint less than
the angle <£. Thus at the joint A (Ph 24, Fig. 95), this angle is
greater than </>, and the upper part will slide along this joint.
At B this angle is less than </>, and no sliding can take place.
The ratio of the force of friction due to the component of P
normal to the joint, to the component of P parallel to the joint,
we call the coefficient of safety against sliding. It is evidently
equal to — , or to the distance G-N divided by PN.
tan P N
Since we can dispose the bed-joints at pleasure, we may,
always make them perpendicular to the direction of the pres-
sure, for instance in Fig. 95 horizontal ; or at least so place
them that their normals vary from the direction of the resultant
of the outer forces, at most by an allowable angle P N.
The sliding of the joints can then always be prevented by the
position of the bed-joints.
17O. Force§ acting upon a Cros§-§ection— Neutral Axis.
— Let us consider what happens when the resultant of the outer
forces acting upon a joint, instead of acting at the centre of
gravity, approaches the edge of a joint, under the assumption
that sliding cannot take place, or that the direction of this re-
sultant is perpendicular to the joint. There is no reason for
assuming the distribution of pressure upon the joint surface
any different from the case of a beam. The stone, as well as
the mortar, is elastic, though in a less degree than wood or iron,
and accordingly the pressure at any portion of the joint is pro-
portional to the approach of the limiting surfaces of the upper
and lower portions of the wall. If, then, we assume that these
surfaces are plane before and after loading, if the resultant
pressure does not act at the centre of gravity, but near to one
edge, the pressure at different points will vary, and there will
be a neutral axis, or line of no pressure, either within or wholly
t without the joint surface.
Every cross-section is therefore acted upon by a system of
CHAP. XV.] THE STONE ARCH. 313
parallel forces whose intensities are directly as their distances
from a certain axis.
Now, neglecting the influence of the mortar, the wall can
resist compression only. No tension can exist at any point of
the joint surface.
Clearly, then, the neutral axis should lie wholly without the
cross-section, or at most only touch it. It should never be
found within the cross-section, as in that case all the material
on the other side is useless, and might be removed entirely
without affecting the pressure upon the actual bearing surface.
The neutral axis, then, should always lie without the cross-
section of the joint.
IT I. System of Parallel Forces wlu»se Intensities are
proportional to their Distances from a certain Axis— The
Kernel of a Cross-section. — If in a system of equal and paral-
lel forces we find the moment of each of these forces with
reference to a certain axis, and then consider these moments as
themselves forces, we shall have a system of the kind referred
to, since each moment force will be directly proportional to its
distance from a given axis.
Now, as we have seen in Art. 60, Chapter VI., the centre of
action of such a system of moment forces does not coincide
with the centre of gravity of the original simple forces, but for
any given axis is found from the central curve of the cross-sec-
tion. In PL 11, Fig. 35, we have already given the construc-
tion for finding; this centre of action, the semi-diameter of the
O '
central curve being known, for any given axis.
Suppose now this axis to envelop in all its different posi-
tions the outline of the given cross-section, and find the corre-
sponding positions of the centre of action of the moment forces.
These different points lie in a closed figure which we may call
the kernel of the cross-section. Then, in order that we may
always have compression in every part of the joint surface of
our wall, the resultant of the forces acting upon it should
always act within the kernel.
In Plates 11 and 12, Figs. 36, 37, 38 and 40, we have con-
structed the kernels of the various cross-sections represented.
Thus in Fig. 36, according to the construction of Art. 62, for
an axis at A, we describe upon O C a semi-circle. Then with O
as a centre and radius equal to semi-diameter of the central
ellipse on A C, describe an arc intersecting the semi-circle in a.
314 THE STONE AKCH. [dlAP. XV.
From a drop a perpendicular upon A C, and we obtain the
centre of action for axis at A. A similar construction for other
axes, as A B, B C, etc., give us other points, and we thus find
the small central parallelogram, which is the kernel or locus
of the centres of action of the moment forces for all positions
of the axis enveloping the parallelogram A B, C D. A similar
construction gives us the kernel for the other figures.
We have from Art. 60
a'
m = — >
^
where m = the distance of the resultant P of the forces acting
upon the cross-section from the centre of gravity, and a = the
semi-diameter of the central curve, and i = the distance of the
neutral axis from the parallel diameter of the central curve.
If we call c the distance of an outer fibre from this diameter
measured on the side of P, its distance from the neutral axis
is i + c. If the strain in this fibre is S, we have
P
s -i + c ; ; -r- : i,
-A.
where A is the area of the cross-section. Hence
If P acts at the centre of gravity of the cross-section, i= oo
p
(Art. 60), the neutral axis is infinitely distant, and S = — . If
A.
P moves away from the centre of gravity, the neutral axis
approaches, and is always parallel to the conjugate diameter
in the central ellipse. When P reaches the perimeter of the
kernel, the neutral axis touches the perimeter of the cross-
section, and at least, then, in one point of this perimeter, the
pressure is zero. If P passes beyond the kernel, the neutral
axis enters the cross-section, and tensile strains enter on one
side to balance the compressive strains on the other. The ker-
nel then forms a limit beyond which the resultant P must 'not
act.
172. Position of Kernel for different Cross-sections.—
If the cross-section is symmetrical with reference to the cen-
c P
tre of gravity, we have — . = 1, and therefore S = 2 — ; that is,
1 A.
CHAP. XV.] THE STONE AKCH. 315
when the neutral axis touches the cross-section, or P acts in
the kernel, the strain S is twice as great as when P passes
through the centre of gravity of the joint surface and is uni-
formly distributed.
As P passes beyond the kernel, the neutral axis, as we have
seen, enters the joint area, and on the side away from P occa-
sions, or would occasion in a beam, tensile strains. But as the
assumption is that the joint (neglecting mortar) cannot resist
tensile strains, we may remove all that portion on the opposite
side of the neutral axis without increasing the pressure on the
other side.
In this case, then, the central ellipse is not that for the whole
joint area, but only for that portion 011 the same side as P, and
P is upon the kernel for that portion.
This portion can be determined directly for a certain posi-
tion of P only in a few individual cases ; generally, it must be
found by trial. We must first find 'for the central ellipse of
the entire joint area the neutral axis corresponding to given
position of P, and then draw a parallel cutting off somewhat
more of the area. Then determine the central ellipse of the
cut-off portion, and see if the pole lies symmetrically to the
pole of the cutting line.
The parallelogram is one of the areas in which we can de-
termine directly the amount cut off when P acts at a point
upon the line joining the centres of two opposite sides. For if
we cut off by a parallel to these sides a portion so that P is at
£d of the line joining the centres of the opposite sides of the
new parallelogram, then P lies upon the kernel for this new
area. The proof is easy. The moment of inertia of the
parallelogram is -£% b A3, with reference to the diameter b. The
square of the radius of gyration aa is then y1^ A2. The distance
of the point of application of P from one of the sides is
a"
^ -h m = ^ + —. Hence
The half height of the kernel is, then, -J-th the height of the
parallelogram, or the kernel occupies the inner third. (See Fig.
36 ; also Woodbury : Theory of the Arch, p. 328, Art. 3.)
For any given position of P, then, three times its distance
316 THE STONE ARCH. [CHAP. XV*.
from the nearest side on a line parallel to the other two, gives
the position of the fourth side of the parallelogram for which
P is upon the kernel.
173. The resultant pressure should therefore act
within the middle third of the joint area. — As this prin-
ciple is most important, and the demonstrations of Chapter VI.,
upon which the above result is based, may appear to some too
purely mathematical, we give here the demonstration of the
same principle as given by Woodbury, in the work above
cited.
" Suppose the pressure to be nothing at the in trad os #, and
to increase uniformly from that point to the extrados b (PL
24, Fig. 96). It is plain that the pressure at any point along a ~b
will be represented by the brdinate of a certain triangle. The
whole pressure will be represented by the surface of that tri-
angle ; and the point of application of the resultant of all the
pressures will be at c opposite the centre of gravity of that
triangle. We have then c b = ^ a ft. Vice versa, if the point
of application be at c, c I — % a 5, we know that the pressure
is nothing at a.
" If the point of application be at c, c 7> being less than -J a J,
c being still opposite the centre of gravity of the triangle
whose ordinates represent the pressure, we know that the ver-
tex of that triangle and point of no pressure are at e}be = 3
xbc.
" In this case, the joint a b will open at a as far as e ; the
adjacent joints will also open until we come to one where the
curve of pressure passes within the prescribed limit.
" This reasoning is, of course, applicable to all the joints ;
and we readily conclude that the curves of pressure should lie
entirely between two other curves which divide the joint into
three equal parts."
Thus, in PL 24, Fig. 97, suppose .the resultant P of the
upper part of the wall to have the position as represented, so
that it intersects the joint B D in C outside of the middle third
of the cross-section. The entire pressure is distributed over
3 C B = A B, and the area D A does not act at all. Moreover,
the pressure at B is twice as great as when P passes through
the centre of gravity and is uniformly distributed over A B, or
is f ds of the uniformly distributed pressure of 'P upon C B.
Beyond A the pressure is zero, and the conditions of load
CHAP. XV.] THE STONE AKCH. 317
and equilibrium would not be changed if the stone beyond A
were removed.
If C approaches still nearer B, so that the pressure is distrib-
uted upon an ever-decreasing area, the resistance of the mortar
will be finally overcome ; it will be forced out, and stone will
come in contact with stone, and there will be rotation about the
edge at B. This rotation can never occur if the pressure P is
distributed over the whole joint area. If, then, we consider
rotation to commence at the moment when P is no longer dis-
tributed over the entire area — when, therefore, the neutral axis
just enters the joint — then, in order that no rotation may occur, y
P must pierce the joint area inside the kernel.
174. LJne of Pressures in tlie Arch. — When the dimen-
sions and form of a wall are given, we can determine directly
the resultant P of the outer forces acting upon a joint, and then
by the two preceding Arts, can determine the condition of sta-
bility of the wall. In the arch, however, we cannot determine
P directly for a given cross-section, but must first make certain
assumptions.
In the first place, it is clear that an arch is stable when it is
possible in two joints to take two reactions P1 and P2 (PI. 24,
Fig. 98) such that, with the weight of the intervening portion
of the arch and its load, the resulting line of pressure shall lie
so far in the interior of the arch that rotation about a joint edge
cannot take place. If the arch is so feeble and the resistance
of the material so slight that only one such assumption of P^ and
P2 can be made, and only one such pressure line drawn, this is
plainly the true pressure line for stability, and by it Pt and P2,
as also the pressure at every joint, are determined.
If, however, the arch is so deep and the resistance of the
material so great that by variation of P1 and P2 several such
pressure lines may be drawn, none of which causes rotation
about a joint edge, which of all these possible pressure lines is
the true pressure line of the arch ?
We assert : That is the true pressure line which approaches
nearest the axis, so that the pressure in the most compressed
joint edge is a minimum.
If we assume the material so s}ft that the pressure line ap-
proaches the axis so near that only one assumption of P, and
P, is possible, then this would evidently be the true pressure
line. If now the material hardens without altering any of its
318 THE STONE ARCH. [CHAP. XV.
other properties, such as its specific weight or modulus of elas-
ticity, then the position of the pressure line is not altered. As
there is no reason for supposing the pressure line different in
an arch built of hard material from that in one originally soft
which has afterwards gradually hardened, it follows that the
pressure line in all arches of same form and loading has the
same position which it would have had if the arch had been
originally of the softest material ; that is, that position which
makes the pressure in the most compressed joint edge a mini-
mum.
In order to draw the pressure line in an arch, we may then
seek by means of the formula
o •*
this pressure in the joint, where the pressure line approaches
nearest the edge, and ascertain whether it can be still further
diminished by change of position of the pressure line. This is,
however, not necessary. We have only to ascertain whether it
is possible to draw a pressure line whose sides cut the corre-
sponding joint a.rea, within the kernel, for then, since we know
that there can be a still more favorable position, there is no
danger of rotation.
175. The Line of Support. — The curve formed by joining
the intersections of the sides of the pressure line with the joint
areas we call the support line, or line of support.
If the joints of an arch answer to the condition of Art. 169,
so that sliding of the joints cannot occur, we see at once from
the position of the support line on what side and where rota-
tion will take place. If at any point this line passes beyond
the kernel, we have theoretical beginning of rotation ; if it
passes outside of the arch, there is actual rotation, and if it lies
within the kernel, there is no rotation.
The manner of determining from the position of the support
line all the possible motions of an arch is illustrated in the fol-
lowing Figs.
In PI. 24:, Fig. 99, we have a possible support line touching
the extrados at crown and springing, and the intrados between
these points. We have accordingly rotation at crown, and at
the points between crown and springing, so that the joints at
these points open on the sides of the arcli opposite the support
line. The crown will sink, and as at the crown and flanks the
CHAP. XV.] THE STONE AECH. 319
support line is approximately parallel to the extrados and intra-
dos, there will be several joints in the same condition, and
several will open, as indicated in the Fig.
In PI. 25, Fig. 100, we have the condition of stability of a
pointed arch, not loaded at the crown. The support line ia
horizontal at crown, and there is no angle there, as in the arch
itself. The rotation at various points is indicated in the Fig.
We shall soon see that the support line deviates but very little
from the pressure line. From the direction of the tangent to
the support line at any point, therefore, we may conclude as to
the conditions of sliding.
From Fig. 101 we may conclude that the arch will slide out-
wards upon the right abutment. The rotation at various points
is given by the Fig. It is sufficient, as we see, to make the
abutment surface more nearly perpendicular to the support
line, as shown in the left abutment, to prevent this sliding, and
at the same time a more favorable support line can be drawn.
Since, as we have seen in Art. 100, sliding can and must in
this manner be always prevented, we shall give no more exam-
ples of arches unstable in this particular.
The arches of Figs. 99 and 100 can be made stable by suffi-
ciently increasing their thickness, or conforming their shape
more nearly to that of the support line.
176. Deviation of the Support from the Pre§sure Line.
This deviation is not great. In order to make it apparent, we
must draw a pressure line for slight pressure in the lower part
of an arch with very long and inclined voussoirs [PL 25, Fig.
102]. Thus, if we combine the weights of the voussoirs 1, 2, 3,
4, etc., acting at their centres of gravity, with the pressure^Q in
the first joint, we have the pressure line shown by the broken
line 1, 2, 3, 4, 5, 6, 7, 8, whose sides 1 2, 2 3, 3 4, etc., give the
direction of the pressure in the corresponding joints between
the voussoirs 1 and 2, 2 and 3, etc. Thus 5 6 is the direction
of the pressure upon the joint between voussoirs 5 and 6. This
direction cuts the joint at 5', which is therefore the point of
application of the pressure, or a point upon the line of support.
Thus we find 3', 4', 5', 6', and the line joining these points is
the support Iww. In general, then, the support and pressure
lines coincide when the vertical through the centre of gravity
of any very small element coincides with the joint, and they
deviate when this vertical does not coincide with the joint.
320 THE STONE ARCH. [CTIAP. XV.
In the ordinary form of joint, as shown in Fig. 103, the sup-
port line varies from the pressure line, since the vertical through
the centre of gravity S does not coincide with the joint under
S. If, however, we should conceive the arch divided into ver-
tical laminae, then the support and pressure lines fall together.
This is precisely the assumption always made in the analytical
discussion of the theory of the arch.
Thus we take the area A = / y dx, and this expression sup-
poses the arch divided into vertical laminse.
The first to make clearly this distinction between the lines of
pressure and support, was Mosely (Civil Eng.). Other authors
have after him adopted this distinction, and then proved that
the two lines always coincide, without remarking that this coin-
cidence is only because of the adoption of the above integral.
The same assumption simplifies greatly the graphical construc-
tion also (the analytical treatment is without it well-nigh im-
possible). We shall therefore assume vertical lam i 1133 where
it is at all permissible. This is always permissible at the crown
of arches with horizontal tangent, because there the joints are
vertical, and over all, when the pressure line lies below the
axis of the arch ; for the support line lies always above the
pressure line, and therefore, in this case, the conditions of sta-
bility are more favorable for it than for the pressure line itself,
when considered as the line of support.
Moreover, it is easy at any point of the pressure line con-
structed with vertical laminae to pass to that line for another
form of joint, and to the corresponding support line. Thus, if
for the point A (PL 25, Fig. 104) we have found the pressure
Q, and if now we wish to pass to the joint ABC, we prolong
Q till it meets P, the weight of the voussoir A B C D, and re-
solve P and Q at this point into Q'. Then Q' is a side of the
new pressure line, and it cuts A B in a point of the support
line.
In this way we can easily determine whether the error com-
mitted when we substitute the pressure line for vertical laminae
for that for the actual joints, which is given by the segment of
the joint A B between QJ. and the pressure line, can be disre-
garded.
177. Dimension* of the Arcli. — The object of the con-
struction of the pressure or support line in the arch is to deter-
CHAP. XV.] THE STONE ARCH. 321
mine the stability and the joints of the abutments. When, the
live load of the arch can be neglected with respect to its own
weight, and when the material of the arch possesses the usual
strength, and the pressure line lies within the inner third, then
the lower point of rupture lies so low that the back masonry
reaching from this point beyond the pressure line completely
encloses it.
There is, therefore, nothing arbitrary, when the form of the
arch is given, except the depth. Since in an arch of less depth
than is allowable in practice a support line can still be in-
scribed, the graphical method is unable to determine the
proper depth. We must then leave to theory the development
of formulae by which this can be determined, and assume that
not only the form of the arch is given, but also its proper
depth and the lower joint of rupture. It is required to deter-
mine the stability of the abutments.
The stability of the abutments can be regarded from two
points of view. We may consider it as a continu|tion of the
arch, as in many English and French bridges, in which the
arch is continued as such, clear to the foundation ; or we may
regard it as a wall whose moment about the joint of rupture
resists the rotation about this joint due to the thrust. Both
views are identical, as the entire theory of the support line rests
upon the investigation of the rotation. They differ only in the
method of expressing the safety of the abutment.
If the arch is continued to the foundation, and the space be-
tween it and the road line filled up by spandrels ; or if the
thickness of the abutment increases from above as the support
line requires ; or, as is often the case in England, the abutment
consists of walls parallel to the crown, separated by hollow
spaces; still, in every case the abutment is not to be distin-
guished from the arch proper — it is stable when the support
line lies in the interior. If the prolonged arch is separated
entirely from the adjacent masonry, there is no reason for not
giving the axis of the prolongation the form of the support
line itself.
If, on the other hand, there is no separation of the arch and
abutment, as in the English hollow abutments, it is sufficient
that the support line lie in the inner third, and the abutment
will be certainly stable.
The supposition that the resistance of the rnortar is suffi-
21
322 THE STONE ARCH. [CTIAP. XV.
cientlj great to unite the whole abutment as a single block
which turns about its under edge, gives too small dimensions.
To ensure safety it is assumed that equilibrium exists with
reference to rotation about the lower edge, when the thrust of
the arch is 1.5 greater than the actual. Investigations of
French engineers have shown that this coefficient of safety for
very light arches is not less than 1.4. The old tables of Petit
give 1.9. We assume it, therefore, = 2.
If, therefore, the double thrust of the arch at the lower point
of rupture is united with the weight of the abutment, the re-
sultant should still fall within the base. Since it is indiiferent
in what order the elements of the abutment are resolved, it is
best to divide it into vertical laminae, and unite these with the
double thrust. The equilibrium polygon thus obtained should
cut the foundation base within the edge of the abutment.
When the thickness of the abutment is thus determined, we
must construct the actual pressure line for the simple thrust
in order to Determine the joints. In drawing this second pres-
sure line, we should properly take the divisions of the arch by
the joints themselves. If, however, we take the division in
vertical laminae, the deviation, as we have seen, is insignificant.
The normals to the actual joints must, then, not deviate from
the sides of this pressure line by more than the angle of
repose.
178. Construction of the Pressure Line. — In PL 25, Fig.
105, we give the method of construction of the proper width of
abutment for an arch. We first divide the arch into vertical
laminse, and determine their weight. If the surcharge has
vacant spaces, or is generally of different specific weight from
the material of the arch itself, it must first be reduced. Thus,
if the surcharge (spandrel filling, etc.) weighs, for instance, only
f ds as much as an equal area of masonry in the arch, we have
simply to diminish the vertical height above the arch by -Jd.
We thus obtain the dotted line given in the Fig., which forms
the limit of the reduced laminae, and we can treat the areas
bounded by this line — the vertical lines of division and the
intrados — as homogeneous. We have then only to determine
the centres of gravity of the various laminae according to the
construction for finding the centre of gravity of a trapezoid
(Art. 33), and suppose at these points the weights, which are
proportional to the reduced areas of the trapezoids to act.
CHAP. XV.] THE STONE ARCH. . 323
Laying off these weights in their order, we have the force
line (Fig. to left). The weights of the abutment laminae 9, 10
and 11 are laid off to same scale one-half 'of their proper in-
tensities. The reason will soon appear.
1st. To determine the thrust H, and also the joint of rup-
ture.
We first inscribe a pressure line by eye, and assume the
point of the intrados to which this line most nearly approaches
as the edge of the joint of rupture. Draw next from the cor-
responding point of the force line a line parallel to the assumed
pressure line at this point. This line will cut off from the
horizontal through the beginning of the force line our first
approximate value of H.
Thus, suppose we have inscribed by eye the pressure line 1,
2, 3, 4, etc., which gives us the point a for the position of the
edge of the joint of rupture. Then a line drawn from 5 on
the force line, parallel to the side 4 5 of the pressure line, gives
us our first value for H.
Now assuming this value of H, we erase the first assumed
pressure line, and proceed to construct the pressure line cor-
responding to this value of H, and the force line divisions 1, 2,
3, 4, etc. If this pressure line lies always within the middle
third of the arch, it may be taken as the proper pressure line,
and H as the true thrust. In general, however, this will not
be the case. The pressure line thus determined may even pass
without the arch entirely. We then determine the new point
of rupture, as given by the point of exit of this pressure line,
and produce the side at this point back to intersection with H
prolonged. From this point of intersection draw a line which
does lie within the middle third of the arch at the lamina of
rupture, and then in the force polygon from the corresponding
point of the force line draw a parallel to this line, thus cutting
off a new value for H. Erasing now the preceding pressure
line, we construct a third with this new value of H, which will
in general give us a pressure line lying everywhere within the
middle third of the arch. If not, another approximation may
be made. We thus find by successive approximation the true
joint of rupture and the corresponding thrust.
2d. Width of abutment. — Since we have laid off the arch
weights to scale in their true value, the pressure line thus ob-
tained is the true pressure line for the arch. But we have laid
324 THE STONE ARCH. [CHAP. XV.
off the abutment laminse 9, 10 and 11, one-half \\\G\Y true value,
and the pressure line thus obtained with the same thrust and
pole O is the same as if we had taken their true value and twice
H. Its intersection with the foundation gives us, then, the
proper width of the abutment for stability, according to our
assumption of 2 for the coefficient of stability (Art. 177).
179. Thus we can easily determine for any given case of
arch and surcharge the horizontal thrust and the proper width
of abutment, and then from the pressure line can easily so dis-
pose the joints as to prevent sliding. If the dimensions of the
arch as given are not such as to be stable, it will be found im-
possible to inscribe, as above, a pressure line which shall lie
within the middle third, and the curve of 'extrados or intrados
will have to be altered so that this shall be the case. The pres-
sure line thus obtained, it is true, does not exactly correspond
with the true one, as it is still possible to inscribe another which
shall deviate still less from the true line. We have also taken
the double thrust for the abutment laminse alone, instead of
for all laminse from the joint of rupture of the arch. Both
deviations are made on account of the far greater ease and
rapidity of construction. It would be found very tedious to
take first the force polygon up to somewhere about the section
of rupture, then by long trial find the innermost support line,
and finally, after the section of rupture is by this line deter-
mined, to lay off the remainder of the force polygon, and pro-
long the pressure line through the abutments.
It is far simpler to 'proceed, as above, by assuming the point
of application of the horizontal thrust, as also temporarily the
section of rupture. We obtain thus a somewhat smaller value
for the width of abutment, but, on the other hand, we have
taken the coefficient of stability at 2 instead of 1.9, as assumed
in Petitfs tables.
Moreover, the widths of abutment thus obtained are greater
than those obtained by these tables, as it is assumed in them
that the point of application of the horizontal thrust is at the
upper edge of the abutment. Thus in every respect the con-
struction gives results reliable and even more accurate than the
tables, as we take the arch as it really is in any given case,
while in the tables suppositions are made with reference to
spandrel filling, etc., which do not hold good for every case.
1§O. Proper Thickness of Arch at Crown. — The proper
CHAP. XV.] THE STONE AKCH. 325
depth of the arch at the key depends not only upon the rise
and span, but also upon the load. The pressure at the extrados
at the key, which, is in general, the most exposed part of joint,
should not, according to the best authorities, exceed -^th the
ultimate resisting power of the material. If P is the pressure
per unit of surface, H the thrust, and d the depth of key-stone
2H
joint, then P — -y,
since, on the assumption that the curve of pressure does not pass
outside the kernel, the maximum pressure is twice the mean
TT
pressure -j. This mean pressure, then, should not exceed ^th
CL
the ultimate resistance of the material. In the best works of
Rennie and Stevenson the thickness at key varies from -^-th to
•gJg-d the span, and from ^-th to ^th tne radius of the intrados.
The augmentation of thickness at the springing line is made by
the Stevenson's from 20 to 40 per cent., by the Rennie's at
about 100 per cent.
Perronet gives for the depth at crown the empirical formula
d= 0.0694 /* + 0.325 meters,
in which r is the greatest length in meters of the radius of cur-
vature of the intrados.
For arches with radius exceeding 15 meters, this gives too
great a thickness. According to Rankine,
d = 0.346 Vr
for circular arches, and
d = 0.412 Vr,
where r is the radius of curvature of the intrados at the crown
in feet.
" The London Bridge is in its plan and workmanship per-
haps the most perfect work of its kind. The intrados is an
ellipse, the span 152 ft., the rise £th as much, the depth of key
-gig-th the span. The crown settled only two inches upon remov-
al of centres." — [ Woodbury : Theory of the ArchJ]
In general, we must first assume the depth at key in view of
the strength of the material, the character of the workmanship,
the load, etc. Then the thrust being found, we find the mean
pressure per unit of area as above. If this mean pressure
exceeds ^th the ultimate resisting power of the material, make
326 THE STONE ARCH. [CHAP. XV.
a new supposition, increase the thickness, find the thrust and
pressure anew, and so on, till the results are satisfactory.
The ultimate resisting power of granite may be taken at
6,000 Ibs., brick 1,200, sandstone 4,000, limestone 5,000 Ibs.
per square foot. These values are, of course, very general,
and subject to considerable variations, according to the kind
and quality of the stone. The strength of the material to be
used must, for any particular case, be determined by actual
experiment.
The weight of a cubic foot of stone may, in general, be as-
sumed at 160 Ibs., brick masonry at 125 Ibs.
IN!. Increase of thickness due to change of form.—
Having obtained a thickness which satisfies all the conditions,
we must, if the arch be very light, make some further provi-
sion for the change of form which is sure to take place af ter
the removal of the centres. By this change of form the pres-
sure line is altered, and the thickness may need to be increased.
In general, we need only to increase the depth from the key to
the springing. This increase need not exceed fifty per cent, at
the joint of rupture and weakest intermediate joint. [ Wood-
l>ury : Theory of the Arch.]
182. Thus, by a simple and rapid construction, we can de-
termine, for any particular case, the thrust, joint of rupture,
and proper thickness of the abutments, without the use of
tables or the intricate formulae usually employed. There is no
difficulty in laying down on paper and verifying all the ele-
ments of the most complex case. The method is entirely in-
dependent of all particular assumptions, and is therefore
especially valuable when irregularities of outline or construc-
tion place the arch almost beyond the reach of calculation. It
is general, and may be applied with equal ease to loaded and
unloaded, full circle, segmental, or elliptical arches with any
form of surcharge.
CHAP. XVI.] THE INVESTED AKCH. 327
CHAPTEK XYI.
THE INVERTED AKCH SUSPENSION SYSTEM.
183. The inverted arch forms the supporting member of
chain or cable suspension bridges. Whether the cable be com-
posed of chains, links, or wires, we suppose them so flexible
that they can perfectly assume the curve of equilibrium. As,
therefore, disregarding the dead weight, any partial load would
cause a change of shape, the cables must be stiffened in order
to prevent the motion which would otherwise take place.
We may stiffen the chains, as shown in PL 26, Fig. 106, by trian-
gular bracing, thus making a rigid system ; or we may have two
parallel chains and brace them to each other, as shown by Fig.
90 inverted ; or we may introduce an auxiliary truss, the office
of which is not to add in any degree to the supporting power
of the combination, but simply to distribute a partial load over
the whole span, so as to cause it to take effect as a distributed
load, and thus prevent change of shape.
As in the first and last cases the structure is commonly
hinged at the centre in order to eliminate the effects of tem-
perature, the method of resolution of forces explained in Arts.
8-13 will, in general, be applicable for the determination of
the strains.
In the second case, we can apply the principles of Arts. 158-
161. !
The rear chains, anchorages, and stiffening truss deserve,
however, special notice.
1§4. Rear Chain* and Anchorages. — The greatest ten-
sion in the main chains occurs, of course, for full load. To
find the tension^at top of tower, as also the horizontal pull, we
have simply to lay off half the whole load vertically from o to
d [PI. 26, Fig. 106], and then draw O o horizontal and O d
parallel to the last side at tower. Then O d is the tension in
that side, and o O the horizontal pull. This pull is neutralized
by the opposite and equal pull of the rear chain leading to the
anchorage ; provided, as should always be the case, it makes an
328 THE INVERTED ARCH. [CHAP. XVI.
equal angle with the vertical. We have thus acting upon the
tower simply the half load ; and the tension in the rear chain
is equal to that in the last link, O d.
If from O we draw parallels to the other links, we have at
once the strains in these links, O <?, O 5, O a, etc.
Now, if the anchorage is a solid block of masonry, its con-
dition of stability is, of course, very easily determined. The
moment of the tension in the rear cable, with reference to the
edge of rotation, must be more than balanced by the moment
of the weight of the block acting at its centre of gravity, with
reference to this edge. The case is too simple to need further
notice.
It is, however, more economical to make the anchorage hol-
low— that is, in the form of an arch. The preceding method
for determining the stability of the arch has then here direct
application.
Thus, laying off along the vertical through the centre of the
tower the weights of segments of the arch, we form with these
segment weights and the double tension in the chain an equi-
librium polygon. For this we have the pole On A O1 being
double the tension Od already found. We then draw Otl, (^2,
Oj3, etc., and then from A parallels to these to the segment
verticals 1, 2, 3, etc. We thus have the polygon A 1, 2, 3, 4, 5.
\_N~ote. — We take the double tension, as before, for the arch, we
took 2 H instead of H, in order to ensure stability.]
The last line of this polygon 4 5 prolonged must, for sta-
bility, pass within the pier abutment, and its resultant, when it
is combined with the weight of the pier and pier abutment,
must pass within the abutment foundation. Through its inter-
section with the vertical line through the axis of the tower the
curve of pressure for the arch must pass.
Drawing now Oa 4 parallel to the rear chain, and making it
also equal to the double tension, or twice O d, we find the pole
O2, and from it draw O2 1, O2 2, O2 3, etc., and then construct
the pressure line for the arch. It must, for sfeaiulity, lie within
the middle third.
To ensure stability when the tension in the rear chain dimin-
ishes, or when the bridge is unloaded, the arch must be stable
lyy itself. We must, therefore, construct the curve of pressure
f < >r the arch alone, neglecting the tension of the rear chains,
as explained in the preceding chapter.
CHAP. XYI.] THE INVERTED AKCH. 329
If this also passes within the middle third of the arch, as
represented by the dotted line, the arch is, under all circum-
stances, stable, and can fully resist the tension of the rear
chains.
We can now, finally, so dispose the joints as to prevent
sliding.
185. Stiffened Suspension System. — We have already re-
ferred to the methods of stiffening the cable or chain so as to
prevent the changes of shape due to partial loading. Of these
methods, it only remains to notice particularly the last, viz., by
means of an auxiliary truss. The office of this truss is to dis-
tribute a partial load over the whole length. We have now to
investigate the forces which act upon the truss.
In PL 27, Fig. 107, let the chain be acted upon by the truss
represented by A B, which is called into action only by a par-
tial load, and not at all by a total uniform load. We can neg-
lect then the weight of the truss itself, as this is borne by the
cable. At the apices 3, 4, 5, 6, 7 let us suppose partial loads
indicated by the small arrows pointing down. Then, at every
point of connection with the chain, we have the reactions 1', 2',
3', etc., acting upwards.- Now the truss must prevent deforma-
tion, and hence these forces are dependent upon the form of
the cable itself. Indeed, if we take any point, as O, as a pole,
and draw lines parallel to the respective sides of the cable,
these lines will cut off upon a vertical P' these forces. -The
absolute value of these forces will, it is true, vary according to
the position of the pole assumed, but their relative proportions
remain always the same. The resultant P7 of all these forces
passes then through the intersection of the two outer sides of
the catenary.
Since the truss distributes its load P upon the cable, the reac-
tion B at the right support is here zero. The reaction, however,
at A cannot be zero unless P and P' coincide, as is the case for
total uniform load. These, then, are all the forces which are
kept in equilibrium by the truss. If P is given, P' and the re-
action at A can be easily found, and if we then divide P', ac-
cording to the form of the chain, into the portions I/, 2', 3',
etc., we have the forces at each apex.
Thus we lay off to scale the given forces 3, 4, 5, 6, 7 — P, and
with a pole distance any convenient multiple of the height of
truss draw lines to these points of division, and then construct
330 THE INVEKTED ARCH. [CHAP. XVI.
the corresponding equilibrium polygon A 3, 4, 5, 6, 7, B. Pro-
long then the outer side B 7 to intersection with P', and draw
the closing line A P'. A parallel through O to this line cuts
off from the force line P the reaction at A and the cable reac-
tion P'.
Now P' being thus found and the form of cable given, we
have only to lay off P' vertically, draw from its extremities
lines parallel to the two outer sides of the given cable arc, and
from the pole thus determined, lines parallel to the other sides
will give us the forces 1', 2', 3', etc. These when thus found
we lay off on our force line for the pole O, as shown in the
Fig., and then construct the corresponding equilibrium poly-
gon A 1', 2' 9', 10', B.
Thus the vertical ordinates between A P', P' B and this
polygon give us the moment at any point for a truss acted
upon by the forces I/, 2', 3', etc., alone. The ordinates between
A P', P' B and the polygon A 3, 4, 5, 6, 7, give, in like manner,
the moments for a truss acted upon by the forces 3, 4, 5, 6, 7,
whose reactions are A and P'. The ordinates, then, included
between both polygons give us the moment at any point of the
stiffening truss. Thus the ordinate y, multiplied by the pole
distance, gives us the moment in the truss at the point o. If
we had taken the pole distance O equal to the height of the
truss, then these ordinates would give us at once the strain in
the flanges. We can thus easily find the strains in the stiffen-
ing truss for any weight or system of weights in any position.
186. Most unfavorable method of Loading. — Let us in-
vestigate the action of a single weight P at any point. In PI.
27, Fig. 109, we have a single weight P acting between A and
P'.
The Fig. is nothing more than a repetition of Fig. 108, only
we have a single load P instead of a system of four loads, and
therefore the polygon for P consists only of two straight lines
instead of having as many angles 3, 4, 5, etc., as there are apex
loads in the first case. All lines have the same position as in
Fig. 108, and hence the construction needs no further explana-
tion.
We see at once from the Fig. that any load between A and
P' increases the moment at every point of the span A B, and
therefore at the point of rupture or of maximum moment also.
So also for the shearing force. When, therefore, the moment
CHAP. XVI.] THE INVERTED ARCH. 331
at any point, and the sum of the forces between that point and
A, is a maximum, at least the entire distance from A to P'
must be covered with the load.
In Fig. 110 we have the weight P on the other side of the but
centre P'. The construction is identical with Figs. 109 and 108,
the position and the direction of action of the forces is now
different. Since the resultants A and P' now lie on the same
side of P, A and P' act in opposite directions, and since P'
must still act upwards, A must act downwards. In the neigh-
borhood of 5' the moment is zero. Between this point and B
the moments have the same signification as in Fig. 109 ; on
the other side the moments have then a different siom. In
O
order, then, that the moment at 5 ' shall be a maximum, the load
must cover the length from A to P, this last point being the
point at which a load causes no moment in 5'; for if any
point between A and P were not loaded, as we have seen, a
load at that point would increase the moment at 5'. A load
beyond P, however, would diminish the moment at 5'.
The above holds good for every point between A and P', and
therefore for the point of rupture or of maximum moment it-
self. In order that this maximum moment can be no more
increased, the load must extend from A beyond the centre to
that point at which a load being placed causes no moment at
the cross-section of rupture.
As for the shearing force, at the end A it will evidently be
greatest for load from A to P', or over the half span, since
every load on the other side of P' diminishes this reaction.
Hence we have the following principles established :
The moment cut any Gross-section of the stiffening truss is a
maximum, when the load reaches from the nearest end beyond
the centre to a point for which the moment at this cross-section
is zero.
The above condition holds good, therefore, for the maximum
of all the maximum moments, or for the cross-section of rup-
ture itself.
The maximum shearing force is at one end of the truss
when the adjacent half span is loaded.
If the arc is unsymmetrical, we must understand by " half
span " the distance between the end and vertical through the
intersection of the outer arc ends produced.
1ST. Example. — As an illustration of the above principles,
332 THE INVERTED ARCH. [CHAP. XVI.
let us take the structure represented in PL 28, Fig. 111. Span,
60 ft. ; depth of truss and panel length, 5 ft. Scale, 10 ft. to an
inch. We suppose the live load to be 2 tons per ft., giving
thus 10 tons for each lower apex, and take the scale of force 50
tons per inch.
On the left we have laid off the force lines for the loads 2, 3,
4, 5, 6 and 7 to 11, and have taken the poles 'for each, so that
the first lines are all parallel to each other and to the first link
of the cable ; the common pole distance being 2 j- times the
height of trass, or 1.5 inches. The moment scale is then
1.5 x 10 x 50 = 750 ft. tons per inch. Since the full load is
entirely supported by the cable, we have only to investigate the
effect of the live load upon the truss.
Precisely as in Fig. 108, we construct the polygons for forces
2-11, 3-11, 4-11, etc., and draw the closing lines as indicated
by the broken lines radiating from the centre O. Parallels to
these from the poles cut off from the force lines the end and
chain reactions. The upper portions are the chain reactions,
the lower the reactions at the right end for the loads 2-11,
3-11, etc.
Now we have to divide these chain reactions into as many
parts as there are load apices by lines parallel to the sides of
the chain. This we have done by drawing two lines parallel to
the two chain ends, inserting the chain reactions between these
lines, and then drawing parallels to the chain sides. If, as in
this case, the curve of the chain is a parabola, these reactions
are divided into 11 equal parts. If the chain has any other
form, the parallels to the chain sides determine the relative
lengths of these portions.
It will only be found necessary to construct the moment
polygons for 4, 5 and 6-11 ; the other polygons already drawn
are necessary for the determination of the shearing forces only.
Thus, on the force line for loads 4 to 11 we can now lay off
the 11 equal parts just found, into which the chain reaction is
divided. So for 5-11 and 6-11. These portions we have indi-
cated by Roman numerals. We can now draw the correspond-
ing polygons precisely as in Fig. 108, which are indicated also
by Roman numerals.
It is then easy with the dividers to pick out the maximum
moment at any apex. These moments, laid off as below, give
the curve of moments for the truss, which being scaled off and
CHAP. XVI.] THE INVERTED ARCH. 333
divided by the depth of truss, give at once the strains in the
flanges. Since the moment scale is 750 tons per inch and the
depth of truss 5 ft., the moment ordinates scaled off at 150 tons
per inch will give at once the strains in the flanges, without
division.
For the shearing forces, we know from the preceding that
the maximum, reaction at right end is for loads 6-11. This
reaction we have already found in the corresponding force line
by means of the closing line already drawn. We lay it off then
right and left, half-way between the ends and first apex, that
being the effective length of load, the two half-end panels rest-
ing directly upon the abutments.
The maximum shear at any point is evidently when the load
reaches from right support to that point, and is equal to the
sum of the chain reactions at the unloaded apices. Thus, max-
imum shear at 3 is equal to the interval I II for the line
3-11; at 4, I III for line 4-11; at 5, I IV for line 5-11;
and at 6, I V for line 6-11. Laying off the shear at 6, we
can draw the line 6-11, as indicated in the diagram, and thus
determine the shear at 2. This we cannot find, as above, for 3,
4, etc., as for the load 2-11 ; owing to the shape of the chain
as represented, there is no upward reaction at 1, as there is no
angle of the chain at 6.
The shear diagram is, of course, symmetrical on each side of
the centre. We can therefore construct it as represented, and
then the determination of the strain in the diagonals is easy.
We have only to multiply the shear at any apex by the secant
of the angle which the diagonals make with the vertical. This
we may do by properly changing the scale at once, and thus
scale off the strains directly.
188. Analytical Determination of the Forces acting
upon the stiffening Tru§s. — Assuming that the truss distrib-
utes the partial loading uniformly over the whole arc, we may
deduce very simple formulae for the forces acting upon the
truss. As we have already seen, for a maximum moment at
any point, the load must always extend out from one end.
Let us represent, then, the ratio of the loaded part from left
to the whole span by k.
Let the entire span be 2 Z, then the loaded portion is 2 k I.
Let m be the load per unit of length ; then the whole load
[Fig. 108].
334 THE INVERTED ARCH. [CHAP. XVI.
The distance of P from the left is then half the loaded por-
tion, or k I. Its distance from P', which acts at the centre of
the span, is I (1 — &).
Hence we have for the left reaction A
Axl = Pl(l-k) or A = -p(l-fy
Also P'l = P x Tel or P' = 2k*l
The chain reaction per unit of length is then
Now let x be the distance to the point of maximum moment.
Now since at this point the shear must be zero, the weight
of the portion x must be equal to A (Art. 38).
We have then
Ax - ~ = P (kl - x) = 2Jclm(Jcl - x),
Zi
whence, by substituting the value of A,
x = ^.
1 + k
But the maximum moment is Ax — - = -— , and there-
fore, substituting the value of x above,
_ _ fC \ X "^ fC ) /-\ 70
M max. = — j =—£ . 2 1 m.
1 + K
This becomes a maximum for 1 — k — %? = 0, or for
Jc = i 1/5 - i = 0.618034.
Therefore, the greatest moment occurs when 0.62 of the span
is covered with the load.
We have then the
Length of the loaded portion, = 2 k I = 0.61803 x 2 .1
Reaction, A = 2 & Z ra (1 - &) = ( 4/5 — 2) 2Zm = 0.23607.
Chain reaction, P' = *2&?lm = (J- — ^ 1/5)2 lm = 0.38196.
Load per unit in loaded portion, or the difference between
the load m and the chain reaction m 1£ per unit of length
= m (1 - &') = i ( 1/5 - 1) m ?= 0.61803 m.
The distance of the point of maximum moment is
A
x = - = (t - 4 ^5) 2 ^ = 0.38196 . 2 Z =
CHAP. XVI.] THE INVESTED AECH. 335
The maximum moment itself is
- TO\ -4 7 u- . * v \ v ' " --- *. F v * * v
2 m (1 — F) 1 + &
For a simple girder uniformly loaded, the maximum moment!
is \p 1?. The maximum moment is then reduced from \ to
0.18, or to about £d, or is -f^ths the maximum moment for a
simple girder of same span and load*
If we represent the dead load by p, then, since the stiffening
truss sustains only the moving load, we have
0.18 ml? _ 36 m \ m (0.6 If
That is, the maximum moment in the stiffening truss is the
same as for a simple girder of -f^ths the span, loaded only with
the moving load.
189. Summary, — The reaction at the end abutment and the
chain reaction at each apex having been found, as above, for
any given load, we might have found the strains in every
piece by the method of Arts. 8-13. This would, however, in
this case have proved long and tedious. The construction of
the curve of maximum moments and shear is preferable.
We can therefore readily determine the strains in such a
combination as that represented in Fig. 111. We have already,
Arts. 90-94, given practical and simple methods for the deter-
mination of the strains in braced arches of the usual forms of
construction.
It will be observed that it is by no means necessary that the
arrangement of bracing and flanges should be the same as that
shown in Figs. 90 and 94.
Thus we may treat the arch represented in Fig. 5 (c) accord-
ing to Art. 158. as hinged at both abutments and crown, or,
making the lower flange continuous at the crown, we may find
the resultant pressures at the abutments by Art. 159, and then
follow these pressures through precisely as shown in the Fig.
The combination of Fig. Ill being of considerable impor-
tance, as the more usual form of construction of suspension
bridges, and not falling under our classification of " braced
arches," we have considered it desirable to discuss it somewhat
* Rankine gives -/7-ths for a girder whose ends are fixed, the greatest mo-
ment occurring for a load over Jds the span.
336 THE INVERTED ARCH [CHAP. XIV.
fully. A better form of construction is that shown in Fig.
106, which is perfectly rigid, and the strains in which are
easily found by Art. 158 or 159, according as we hinge it in
the centre or not.
Reviewing now the preceding, we see that the graphical
method, as here developed, furnishes us with a simple, accurate
and practical solution of nearly every class of structure occur-
ring in the practice of the engineer or builder. In our first
chapter we have a method by the resolution of forces applica-
ble to any framed structure, however irregular or unsymmetri-
cal, provided only there are no moments at the ends to be
determined.
In Art. 125 we have explained fully the application of the
method for this case also, when these moments are known, and
in Chaps. VIII. to XIV. inclusive we have given practical con-
structions for the determination of these moments for all the
important classes of structures in which this condition occurs,
such as the continuous girder, braced arches, etc.
When the structure is not framed, or composed of pieces the
strains in which can be definitely determined, we have the
method of moments of Chap. V., which, as we have seen, may
be extended so as to completely solve the difficult case of the
continuous girder, and which may, of course, be applied to
framed structures also, as illustrated in Fig. Ill (Art. 187) in
the case just discussed. Thus we have two distinct graphical
methods by which our results may be checked. The first-
method includes a great variety of the most important and
usual structures, such as bridge girders, roof trusses, cranes,
etc., and in view of its ease and accuracy will undoubtedly be
found of great service by the engineer and architect. The
second method has important mechanical applications, as no-
ticed in Art. 41 ; and aside from these, and its application to
structures having end moments, such as the continuous girder,
etc., furnishes us with ready determinations of the centre of
gravity of areas (Chap. III.), the moment of inertia of areas
(Chap. VI.), and also gives us a very complete solution of the
stone arch (Chap. XV.).
We have also the analogous methods of calculation, viz.,
CHAP. XVI.] THE INVEKTED ARCH. 337
both by resolution of forces and by moments (Arts. 9 and 16
of Appendix). The latter being so general and simple in its
application, we have not felt justified in leaving it entirely out
of sight, and in those cases where it seemed of especial service,
or assisted the graphical solution, we have illustrated it more or
less fully (Chap. XII.). In this latter chap, we have also given
constructions as well as formulae, and developed principles
which, it is believed, render possible, for the first time, the com-
plete and accurate solution of the important .case of the " draw
xpan." (Arts. 118-121.)
The formulae of Chap. XIII. in connection with the method
of calculation by moments, render the calculation of the con-
tinuous girder generally as simple, and but little more tedious
than for the simple girder itself. Whatever may be thought
of the advantages or disadvantages of this class of structures
by engineers generally, it is at least time that such structures
as draws or pivot spans should be calculated under suppositions
which approach somewhat more nearly the actual case than is
at present the practice. As to the relative economy of con-
tinuous girders, we have endeavored to enforce the fact that
the saving over the simple girder is from 15 to 20 and even 50
per cent. We give in the Appendix a tabular comparison of a
few cases sufficient to show the point beyond dispute, and any
one may easily add to the list, or verify the calculations.
The " graphical arithmetic," as it might be called, such as
graphical addition, subtraction, multiplication, division, extrac-
tion of roots, determination and transformation of areas, etc.,
we have entirely omitted in the present work, judging it of but
little practical value, except in rare cases, when we have ex-
plained the necessary constructions as they occur, and unneces-
sary for the development of the graphical method proper. [See
Chap. IV. of Introduction.]
22
APPENDIX.
GRAPHICAL STATICS.
A. JAY DUBOIS.
APPENDIX.
NOTE TO CHAPTER VIII. OF THE INTRODUCTION - UPON THE
MODERN GEOMETRY.*
IT is to be regretted that, notwithstanding its beauty of form,
simplicity, and many happy applications in the technical and
natural sciences, the Modern Geometry is yet hardly known,
scarcely by name even, in our schools and colleges.
The work of Gillespie upon Land Surveying, already cited in
the Introduction, and a treatise on Elementary Geometry by
William Chauvenet (Phil., 1871), are the only ones which
occur to us in this connection.
It has already been stated that the modern orj)ure geometry
of space differs essentially from the ancient, and from analytical
geometry, in that it makes no use of the idea of measure ny r>f
We find in it no mention of the bisection
of lines, of right angles and perpendiculars, of areas, etc., any
more than of trigonometrical quantities, or of the analytical
equations of lines. We have nothing to do with right-angled,
equilateral, or equiangular triangles, with the rectangle, regular
polygon, or circle, pxcept in a supplementary manner. So also
for the centre, axes, and foci of the so-called curves of the second
order, or the conic sections.
On the contrary, we obtain much more general and compre-
hensive properties of these curves than those to which most
text-books upon analytical geometry are limited.
A new path is thus opened to the conic sections, without the
aid of the circular cone, after the manner of the ancients, or of
the equations of analytical geometry.
As a direct consequence, the principles and problems of the
modern geometry are of great generality and comprehensive-
* The following- remarks and illustrations are taken from the Geometrie der
Lage, by Reye. Hannover, 1866.
342 NOTE TO CHAP. VIH. OF THE INTKODUCTION. [APPENDIX.
ness. Thus the most important of those properties of the conic
sections which are proved in text-books of analytical geometry
are but special cases of its principles. A few particular ex-
amples taken from the Geometrie der Lage, by Reye, which
could not well have been inserted in the Introduction to this
work, will best explain and illustrate our general remarks — the
more so as these examples are of special interest and value to
the engineer.
It is a problem of frequent occurrence in surveying to pass a
line through the inaccessible and invisible point of intersection
of two given lines. The Geometry of Measure, or ancient
geometry, gives us any required number of points upon this line
by the aid of the principle, that the distances cut off from par-
allel lines by any three lines meeting in a common point are
proportional. The__Geometry of Position furnishes us with a
simpler solution.
FIG. 1.
Thus the two lines «, 5 being given [Fig. 1.], we have sim-
ply to choose any point we please, as P. From this point draw
any number ;of lines desired, in any direction intersecting the
given lines. Now, in any quadrilateral which any two of these
lines form with the two given lines a and &, we have simply to
draw the diagonals. The intersections of all these diagonals
lie in the same straight line passing through the intersection A
of the two given lines, and therefore determine the line re-
quired. Observe that the construction is entirely independent
of all metrical relations, and depends solely upon the relative
position of the two given lines.
Again : If we take upon any straight line three points, A, B
and C [Fig. 2.], and construct any quadrilateral, two opposite
sides of which pass through A, one diagonal through B, and the
APPENDIX.]
THE MODERN GEOMETRY.
343
other two opposite sides through C, then will the other diagonal
intersect the line in a point D, which for the same three points,
FIG. 2.
A B C D
A, B and C, is always the same for every possible construction.
Moreover, these four points, A, B, C and D, are always harmonic
points, so that D is harmonically separated from B by the
points A and C. Thus, A B : B C ; ; A D : C D. This construc-
tion may also be applied in surveying, as in passing around an
obstruction, as a wood, etc., into the same line again.
Again : We may notice the following principle concerning
the triangle [Fig. 3] ;
FIG. 3.
u.
If two triangles, ABC and A! Bt C1? are so situated that the
lines joining corresponding angles, as A A1? B B1? C C1; meet in
a common point S, then will the intersections of corresponding
sides, as A C and Ax Gl9 A B and At B1? B C and B! C1? meet in a
common line, as u u. The inverse also, of course, holds good :
that if the sides intersect on a line, the lines through the angles
intersect in a point.
Another series of principles are connected with the curves of
the second order, or conic sections. From analytical geometry,
344 NOTE TO CHAP. VIII. OF THE INTRODUCTION. [APPENDIX.
as is well known, a curve of the second order is completely de-
termined by five points or five tangents. But the length of the
calculation or construction of a curve thus determined is also
well known. The geometry of position, however, proves two
very important principles, which render it easy to construct to
the five given points or tangents any number of new points or
tangents, and thus quickly draw the curve itself. The reader
already acquainted with these principles will also probably re-
member how much auxiliary demonstration their proof in the
analytical geometry requires. The first of these, due to Pascal,
is, that the three pairs of opposite sides of a hexagon inscribed
within a conic section intersect upon a straight line. The
second, due to Brianchon^ is, that the three principal diagonals
of the circumscribing hexagon, which unite every pair of oppo-
site angles, intersect in one and the same point. Both prin-
ciples are easily deduced from the circle. It will be observed
that they are independent of the relative dimensions, centre,
axes, and foci of the curves. For this very reason they are of
the greatest generality and significance, so that an entire theory
of the conic sections can be based upon them. Thus Pascal's
principle solves the important problem of tangent construction
from a given point, even when the curve is given by five points
only, without completely constructing it.
This problem of tangent construction to curves of the second
order can in many cases be solved by the aid of a principle
which expresses one of the most important properties of the
conic sections, but which, nevertheless, is seldom found in text-
books upon analytical geometry, because its analytical proof is
somewhat complicated, and little suited to set forth the property
in its proper light.
For example : If through a point A [Fig. 4] in the plane of
but not lying upon a curve of the second order, we draw se-
cants, every two secants determine four points, as K, L, M, N,
upon the curve. Any two lines joining these four points, as
L M and K N or K M and Li l\r, intersect in a point of a
straight line a a, which is the polar of the given point A ; that
is, which intersects the curve in the two points of tangency
G G. Thus the lines through A and the intersections of a a
with the curve are the tangents to the curve through A. If
the point A were within the curve, this line a a would not in-
tersect it. This construction can be used in order to draw
APPENDIX.] THE MODEEN GEOMETRY. 345
through a given point tangents to a conic section by the sim-
ple application of straight lines. Upon every secant through
FIG. 4.
\L
A, moreover, there are four remarkable points, viz. : the point
A itself, the first intersection B with the curve, the intersection
with the polar, and, finally, the second intersection D with
the curve. These four points are harmonic points, and the
polar a a contains, then, every point which is harmonically
separated from A by the two curve points. The important,
principles relating to the centre and conjugate diameter of
conic sections are merely special cases of the above important
principles. These last can be easily extended to surfaces of
the second order, as the intersection of these by a plane is, in
general, a curve of the second order.
From these few examples, which might be indefinitely multi-
plied, it may easily be seen how very different, but not less im-
portant than those of analytical geometry, are the theorems of the
geometry of position. Thus the latter are generally proved by
aid of the angle which the tangents make with the line through
the focus, or by the distances cut off from the axes — that is, by
metrical relations. We refer, of course, to the elements of
analytical geometry as contained in most text-books, and not to
those most fruitful and later methods .whose existence are
chiefly due to the sagacity of Plucker (Introduction, VIII.).
346 NOTE TO CHAP. I. [APPENDIX.
NOTE TO CHAPTEE I.
1. The method by the resolution of forces developed in
Chapter I. is so simple and easy of application, and its principles
are so few and self-evident, that we have not considered it ad-
visable to tax the patience of the reader by any great variety
of practical applications. A large number of such applica-
tions are to be found in a most excellent little treatise by
Robert H. Bow, entitled The Economics of Construction in
Relation to Framed Structures. There are, however, a few
important practical points- of detail, and a few general consid-
erations, which we think it well to notice here, and to which,
in illustration of the remarks in Chap. I., the reader will do
well to attend.
2. In PL 1, Fig. I. (Appendix), we have represented the
"Bent Crane " given by Stoney in his Theory of Strains, p.
121, Art. 200.
We assume the following method of notation. Let all that
space above the Fig. be indicated by X, and all that space
below by Y, and the triangular spaces enclosed by the flanges
and diagonals by the numbers 1, 2, 3, 4, etc. The first upper
flange is then denoted by X 2, the second by X 4, and so on.
So also the first lower flange is Y 1, the next Y 3, etc. The
first diagonal is then X 1, the next 1 2, the next 2 3, etc.*
The flanges are equidistant, forming quadrants of two cir-
cles whose radii are respectively 20 and 24 feet. The inner
flange is divided into four equal bays, on which stand isosceles
triangles, and a weight of 10 tons is suspended from the peak.
The scale for this and all the Figs, of PI. I. is 20 tons to an incji
and 10 feet to an inch. Laying off, then, the weight X Y = 10
tons, we form, according to the method of Chapter I., the strain
diagram. It will be seen at once that all the lower flanges, Y 1,
Y 3, etc., radiate from Y, all the upper flanges, X 2, X 4, etc.,
from X, and everywhere the letters in the one diagram indi-
cate the corresponding pieces in the other.
* For this very elegant niethod of notation, we are indebted to the work of
B. H. Bow, above alluded to.
APPENDIX.] NOTE TO CHAP. I. 347
"We can now at once take off the strains to scale in the vari-
ous pieces.
A comparison of our method with that given by Stoney for
the same case will be instructive, as illustrating the compara-
tive merits of the two.
3. Character of the Strain* in the Pieces. — One of the
most important points of our method is the ease and certainty
with which the character of the strains in the pieces may be
determined. We have only, as detailed at length in Chapter 1.,
to follow round any closed polygon in the direction of the
forces, and then refer back to that apex of the frame where for
the moment we may happen to be.
Thus for the peak, since we know that the weight acts down,
we follow down from X to Y, and then from Y to 1, and 1
back to X. Referring back now to the frame, and remember-
ing that a force acting away from the apex means tension, and
towards, compression, we have at once Y 1 compression and
X 1 tension.
Now for apex a, since X 1 is tension, with respect to this new
apex, it must act away. We go round then from X to 1, 1 to
2, and 2 back to X, and then, referring these directions to the
corresponding pieces meeting at &, we have 1 2 compression
and X 2 tension.
We find thus all the outer flanges in tension, as evidently
should by simple inspection be the case. Also all the inner
flanges compression. As for the diagonals, they alternate, the
first being tension, the next compression, until we arrive* at 4 5,
which we find to be also compression.
A glance at the strain diagram shows how this comes about.
The line X 4 crosses Y 5, and thus gives us a reverse direction
for 4 5.
In such a simple structure as the present, the character of the
strains would present no especial difficulty in any case ; but in
more complicated ones, the aid of such a simple and sure crite-
rion as the above is indispensable, and we have been thus even
prolix upon this point, the more so as it is not so much as
alluded to, as far as we are aware, in those few works which
notice the above method at all.
4. There are other points which we may here illustrate by
our Fig.
According to our first principle (Art. 3, Chapter I.), when
348 NOTE TO CHAP. I. [APPENDIX.
any number of forces are in equilibrium, the force polygon is
closed. Inversely, then, a closed force polygon indicates forces
which, if applied at a common point, would hold each other in
equilibrium.
Thus Y 3, 34, X 4, and the weight are, or would be, if all
applied at a common point, in equilibrium. This we see
directly from the Fig. Thus we know that when any num-
ber of forces are in equilibrium, the algebraic sums of their
vertical and horizontal components must be zero, otherwise
there must, of course, be motion. Now the vertical component
of Y 3 plus that of 3 4 minus that of X 4 is exactly equal and
opposed to the weight, while the horizontal component of Y 3
plus that of 3 4 is equal and opposed to that of X 4, and there
is then equilibrium.
Again, according to the principle of Art. 5, Chap. I., any
line, as the one joining 2 and 6 (broken line in Fig.), is the
resultant of X 2 and X 6, as also of 2 3, 3 4, 4 5 and 5 6.
The Fig. also well illustrates the points to be avoided in
making a strain diagram, already alluded to in Art. 13, Chap.
I. The scale to which the frame is taken is here altogether
out of proportion to the scale of force. The first should be
increased or the second diminished, or both. The present
length of the diagonals and flanges is inadequate to give
with sufficient accuracy the directions of strain lines of such
length.
Nevertheless we have experienced no difficulty in checking
to tenths of a ton the results given by Stoney for this structure.
5. In PI. 1, Fig. II., we have represented a roof truss, span
30 ft., rise 8 ft., camber 1 ft. ; and the strain diagram illustrates
in its two symmetrical halves (one full, the other dotted) the
remarks of Art. 13, Chap. I., upon the check which in such cases
our method furnishes of its accuracy.
We lay off the weights 1, 2, 3, 4, 5, and then the reactions at
A and B, which should bring us back to the point of beginning,
and thus complete the force polygon. The strains are then easily
found, and the two halves should be perfectly symmetrical, and
give the same results.
In Fig. III. we have given another form of truss with strain
diagram, the other half of which the reader can complete and
letter for himself.
6. In Fig. IY. we have a form called the French roof truss
APPENDIX.] NOTE TO OH A P. I. 349
and two strain diagrams — the larger for vertical reactions, the
smaller for inclined reactions.
This last brings out the force polygon in perhaps a clearer
shape than before. The weights 1 to Y being laid off down-
wards, the two reactions must always bring us back to the
starting-point, and thus close the polygon — in this case a trian-
gle, in the preceding case a straight line, and in the case of Fig.
6, Art. 10., Chap. I., a true polygon. Both strain diagrams
illustrate the check we have upon the accuracy of the work.
The second half should be perfectly symmetrical with the first,
and the lines Y k and points Jc in each should coincide.
We have here also to notice a point which in roof trusses is
of frequent occurrence, and may, if not noticed, cause diffi-
culty.
We have already observed in Art. 9, Chap. I., that we can
always find the strains in the pieces which meet at an apex,
provided only two are unknown. Now in the strain diagram
to Fig. IY., we readily determine the strains in X «, Y a, X b,
a b, Y c and b c successively, and arrive finally at apex 2, where
we have the two known strains in X 5 and b £, and wish to find
the strains in three pieces, viz., X cZ, d Ji and c h. At first sight
this seems impossible. If, however, we assume that the pieces
of the frame can take only strains of a certain kind, as, for
instance, hd only tension, and not compression, the problem is
perfectly determinate. This assumption is easily realized in
practice. Thus if h d is a rod of small diameter, it cannot
act as a compression member at all. Moreover, the strain
of tension in h d must evidently be precisely equal to that in 1 1
b c, already found. We have then to form a closed polygon u
with the weight at 2 and the known strains in X b and b c, whose *
other three sides shall be parallel to X d, h d and c h respec-
tively, and in which, moreover, the strain in h d shall be equal
to that in b c, and where both these strains must be, when the
polygon is followed round according to rule, tensile. We have
evidently, then, in accordance with these conditions, only tlio
polygon 2 X d h c b X, thus finding the point d, from which
we can now proceed to find e, etc. The points a, b, d and e are
evidently in the same straight line parallel to c h. This point
is one of importance, and the reader should carefully follow
the above remarks with the aid of the Fig.
The strain diagram thus constructed shows us many facts
350 NOTE TO CHAP. I. [APPENDIX.
about the system not otherwise apparent. Tims ~b c, c h and h d
are in equilibrium with the load at 2. Again, a b and b c are
in equilibrium with Y a minus Y <?, as also are hd and de with
Jc e minus kh. Also Teh, h c, Y c and Y Jc are in equilibrium,
and Yc, cb and X b are in equilibrium with the reaction minus
the weight at 1, or with the shear to the right of 1. This last
principle is general. When a section can ~be made entirely
through a structure, the strains in the pieces cut are in equi-
librium with the shear at the section. If only three pieces are
cut, then, by taking as a centre of moments the point of inter-
section of any two, we can easily find, knowing the moment of
the shear, the strain in the third.
Thus we have the general and easy method of calculation
given in Art. 14, Chap. I. The moment of the shear is, of
course, the sum of the moments of all the exterior forces be-
tween the section and one end.
We have then two methods, one graphic and one by calcula-
tion, by which we can find the strains in every kind of simple
truss which can ever occur in practice. By li simple " we mean
merely resting at the supports, or not acted upon at the ends by
a couple or moment, as is the case, for instance, in the continuous
girder.
When the structure is unsymmetrical, or complex, the deter-
mination of the different lever arms is often very tedious, involv-
ing a good deal of trigonometrical computation. On the other
hand, the frame can always from its known proportions be
easily and accurately drawn to scale, and then the exterior
forces, whatever their relative intensity or directions, can be
laid off, and the strains at once determined. Here we see, then,
one of the great advantages of our graphical method. An
unsymmetrical frame and different directions of the forces
requires no more time or labor than a more simple case.
7. Application to Bridges— Bow-string Girder. — In Art.
12, Chap. I., we have alluded to this application, and shown
how by two strain diagrams only we can completely calculate a
bridge of any length. As this application is so important, and
as the method is stated by several authors to be inapplicable to
bridges,* or, at best, to be unsatisfactory, we will here call more
* Iron Bridges and T&w/s— Unwin — p. 143. Economics of Construction —
Bow- -p. 61.
APPENDIX.]
NOTE TO CHAP. I.
351
special attention to the points to be observed in the tabulation
of the strains. There is, indeed, no more satisfactory, complete
and rapid method for the solution of bridge girders generally
than that afforded by the graphic method.
As an example, let us take the Bow-string Girder given by
Stoney, p. 131. Span, 80 ft., divided into 8 panels ; rise of
bow, 10 ft. Load, 10 tons at each lower apex.
We construct the two strain diagrams* given in Fig. V.,
PL 2, viz., one for the load P7 at the first apex, and one for the
load at the last apex, P,. Referring, if necessary, to Art. 12,
Chap. I., the reader can easily follow out these diagrams. We
then scale off the strains, and obtain, for the strains in the
diagonals —
ab
be
cd
de
ef
fff
ffh
FT
-2.7
-11.4
+ 4.8
- 4.3
+ 2.4
-2.3
+ 1.4
Pi
-0.4
+ 0.23
-0.56
+ 0.51
-0.9
+ 0.88
- 1.4
Now from the strains thus obtained for these two weights
we can easily obtain all the others.
Thus, as the end reactions are inversely as the distances of
the weight from the ends, the reaction at the left end due to
P2 will be twice that due to Pa. For Ps three times that due
to P,. The strains will therefore be twice and three times
those due to P1? until we arrive at the weights P1 and Pa re-
spectively. So also for P6 the reaction at the right is twice
that due to P7, and the strains are therefore double up to the
weight P6. To the right, then, of P6 the strains are twice
those due to P7, and to the left of P6 they are six times those
due to P!. Take, for instance, P5. The right reaction is f ths
of the apex load, and the right reaction of P7 is -|th of that
load. For P5, then, the strains in all pieces to the right of that
weight are 3 times those due to P7. Again, the left reaction
is for P5 |ths the apex load. But the left reaction for Pl
is -Jth the same load. The strains then in all the pieces to the
left of P5 are 5 times those due to P!. So for any other load.
We can therefore form at once the following table :
* Strain diagrams in Fig. V., and also in Fig. VL, are, for obvious reasona,
drawn to different scales.
352
NOTE TO CHAP. I.
[APPENDIX.
«
PI
P2
P3
P4
£5
P6
P7
Uniform
Load.
Max.
Comp.
+
Max.
Tens.
Total
Strains.
ab
-0.4
- 0.8
- 1.2 - 1.6
-2.0
-2.3
-2.7
- 8.25
1-n.o
- 19.25
be
cd
+0.23
^.56
+ C.5
— 1.1
+ 0.7
— 1.7
+ 0.9
- 2.2
+ 2.0
+ 1.1
+ 1.4
-11.4
+4.8
-4.9
+ 4.8 - 11.4
+ 4.bj - 11.8
- 16.3
- 2.8
-3.4
-5.2
- 17.0
de
ef
+0.51
^0^
+ 1.0
- 1.8
+ 1.5
+ 2.6-8.6
-4.3
+ 2.4
-3.97
-4.8
+ 7.6
+ 7.1
- 12.9
- 16.9
- 18.3
- 2.7 - 3.6
-4.5
+ 4.7
- 13.5
fa\
d
+0.88
+ 1.8
+ 2.6 + 3.5
-6.9
-4.6
-2.3
— 0.55
- 4.2
+ 8.8
- 13.8
-20.5
-1.4
-2.8
-4.2
-5.6
+ ,2
+ ,8
+1.4
+ 8.4
-14.0
- 18.2
In the columns for Pt and P7 we put the strains already
found by diagram. The strains for P2 on the entire left half
will be double those for Px ; for P3 three times, and for P4 four
times those for Px. We have therefore at once the columns
for P1? P2, P3 and P4. Now for P5 we see from the Fig. that
the strains in diagonals ef and fg must both be tension.
From the left end, then, as far as ef, the strains are 4 times
those due to Pl5 and from the right, as far asfg, 3 times those
due to P7. We thus obtain the column for P5. In the same
way for P6, all above or to left of cd are 6 times Pl5 all below
or to right of de twice P7. Thus we fill out the whole table.
Adding now all the tensions and compressions in each piece, we
obtain the maximum strains of each kind due to the live load,
as given in the last two columns but one. Suppose now the
dead load or weight of the girder itself to be fths of the roll-
ing or live load. We have only to take, then, f ths the sum of
these last two columns and we have the strains due to uniform
or dead load, as given in the fourth column from the right.
We can now easily obtain the total strains. Thus the ten-
sion in a 1) due to the live load only is 11 tons. The tension
due to the dead load is 8.25 tons. Total greatest strain which
can ever come upon a 5, then, is 19.25 tons tension. No com-
pression can ever come on this |)iece ; it does not need, there-
fore, to be counterbraced. On the other hand, all the other
diagonals, except perhaps cd, must be counterbraced, as the
maximum compression due to the live load overbalances the
constant tension of the dead. Had the dead load been taken
much greater than the live, the diagonals might always have
been in tension. Hence the appropriateness of this class of
girder for long spans.
APPENDIX.]
NOTE TO CHAP. I.
353
We see also from the table just what weights, and where
placed, give the greatest strain of eacli kind- in any piece.
§. Strains in tlie Flanges. — The method is precisely simi-
lar for the flanges. Thus we scale off from our diagrams —
Xa
Xd
JLd
*/
Th
P!
+ 2.82
+ 3.08
+ 3.47
+ 4.11
+ 5.11
PT
+ 19.7
+ 21.6
+ 10.4
+6.8
+ 5.1
Ya
Yc
Ye
?9
Pi
- 2.52
-3.01
-3.62
- 4.46
P7
- 17.6
- 13.1
-7.9
-5.7
Tabulating these, we obtain the following table :
Flanges.
PI
!
P2
P3
P4
P5
P6
P7
Uniform
live
load.
Uniform
dead
load.
Total
Strains.
Xa
+2.82
+ 5.6
+ 8.5
+11.3
+ 14.1
+ 16.9
+19.7
+ 78.9
+ 59.1
+ 138
X6
+3,08
+ 6.2
+ 9.2
+12 3
+ 15.4
+18.5
+21.6
+ 86.3
+ 64.7
+ 151
Xrf
+3.47
+ 6.1)
+ 10.4
+13.9
+ 17.3
+20.8
+10.4
1^.8
+5.1
+ 83.2
+ 62.4 + 145.6
x/
~XA
+4.11
+ 8.2
+ 12.3
+16.4J+20.5
+ 13.7
+ 10.2
+ 82.0
"+81.6
+ 61.5
+ 61.2
+ 143.5
+5.11
+ 10.2
+ 15.3
+20.4
+15.3
+ 142.8
Ya
-2.52
-3.01
-5.0,
-7.G
-9.0
-10.1
-12.0
-12.6
-15.0
-15.1
^-IsT
-17.6
-70.5
-76.2
-52.8
-128.3
Yc
-6.0,
-13.1
-57.1
- 133.3
Ye
Y7
-3.62
^A6
-7. 2, -10.9
-14.5
-18.1
-15.9
-7.9
-78.1
-58.5
- 136.6
-8.9 -13.4
-17.8
-17.1
-11.4
-5.7
- 78.8
-59.1
— 137.9
This table is obtained precisely as before. Thus for P6 the
strains in Xa and Xb are multiples of P1? while those in the
other flanges are multiples of P7. So also for Y c and Y e. We
f see at once that the greatest strains are for full load, since for
all loads the upper flanges are always compressed and the
lower extended.* The above is sufficient to illustrate fully the
application of our method to bridges. It is evidently appli-
* A more convenient form of tabulation is to put the weights in the left
vertical column and the pieces in the top horizontal line. The numbers can
then be more easily added.
23
354 NOTE TO CHAP. I. [APPENDIX.
cable to any structure where the reactions are inversely as the
distances from the end. The strains due to the first and last
weights are all that we need in order to thoroughly solve any
case of the kind. It is advisable, however, to construct a
third diagram for an intermediate weight, in order to serve as
a check upon the others.*
9. Method of Calculation by Moments. — AVe may illus-
trate here the method of calculation by moments referred to in
Art. 14, Chap. L, a little more fully. Thus in the example
above, Fig. V., suppose we wish the strains due to P{. Reac-
tion at left end is evidently £th of 10 tons = 1.25 tons. Con-
ceive the lower flange Y a cut. Rotation would evidently take
place about apex a, and we have, therefore, strain in Y a x its
lever arm from apex a = 1.25 x 5. The depth of truss, or
lever arm of Y$, from apex a, is 2.58 feet. Hence we have
' , K0 — = strain in Y a — 2.42 tons.
a.Do
This strain is evidently, by reason of the direction in which
the two portions of the truss would rotate about a, tension. In
like manner, for upper flange X &, if we know the lever arm of
this flange from the opposite apex, we can easily find the strain ;
for the diagram shows that X 5, b c and Y c are in equilibrium
with the reaction, and hence, if we take the point of moments
at the intersection of the two pieces b c and Y c, the moments
of these pieces are zero, and we have remaining only the mo-
ment of the strain in X b balanced by the moment of the reac-
tion.
Again, if X b and Y<? are thus found, and if these two strains,
together with the reaction, are in equilibrium with the diago-
nal b c, we can find the strain in this diagonal by taking the
apex d as a centre of moments. The moment of Y c then is
* It may also be well to notice here that the practice of deducing in the
tabulation the dead load from the live load strains is not strictly accurate, as
the live load acts at the lower apices only [or at the upper apices only, if the
bridge is under grade],, while the dead load is distributed along both flanges,
and acts at both upper and lower apices.
In every case, however, the greater portion of the dead load, say, for in-
stance, fds of the whole, owing to the track, platform, cross-girders, etc.,
acts at the same apices as the live load itself ; and the error is in any case
very slight, and practically of no account.
APPENDIX.] NOTE TO ClfrAP. I. 355
zero, and we have the moment of the strain in ~b c balanced
by the moment of the reaction and the moment of X 5 ; the
first causing compression, the second tension, and the difference
then giving the resultant moment strain, which, divided by the
lever arm of & <?, gives the strain itself.
The method is easy of application, but, as we have already
remarked, the determination of the lever arms for each case is
frequently tedious. " These may, however, be scaled off from
the frame diagram with sufficient accuracy in practice.
As before, we need only the strains due to the first and last
weight, and can then form our tabulation as above. This
tabulation we can also check by finding the strains due to uni-
form load independently, and seeing whether it agrees with
the sum of the separate apex weight strains.
Thus, for all the weights acting, suppose we wish the strain
in Y g. The lever arm of Y g is 9.85 feet. We have then re-
action = 35 tons, multiplied by 35 feet = 1225. This must be
diminished by P7 X 25 = 250, P6 x 15 = 150, and P5 x 5 =
50. We have then 1225 - 450 = 775, which, divided by 9.85,
gives 78.7 tons tension in Y </, agreeing with our tabulation
above.
For the method of calculation by resolution of forces, see
Art. 16 of this Appendix.
10. Girder with Straight Flanges. — In such a case, as the
lever arms are at once known and are constant, the above
method is of very easy application. In this case the strains in
the diagonals are best found by multiplying the shear by the
secant of the inclination of the diagonal with the vertical.*
Thus, if this angle is 45°, we have simply to multiply the shear
at any point by 1.4142, and we have at once the strain in the
* This is but a particular result of the general method of moments. Thus,
for any diagonal, as ab (Fig. VII.), according to our rule, we take the centre of
moments at the intersection of the two other sides cut by a section through
the truss, viz. , the flanges. But these two sides are here parallel, hence their
intersection is at an infinite distance. The lever arm of a b is then GO x cos </-,
6 being the angle with the vertical. If the weight PI acts, we have then,
calling the reaction R, R x GO — P GO = S cos </> x GO, where S is the strain in
a b. This can be put (R — P) GO = S cos 0 x co, hence S = - . But
GO 1
R — P is the shear at 5, C08 > x 00= T = sec 0 ; hence we have only to
multiply the shear by the secant.
356 NOTE -TO ciiAr. i. [APPENDIX.
diagonal at this point. The shear is always in such cases the
reaction at the end minus the weights between that end and
the apex in question.
The flanges are easily obtained by moments, as above.
The following points need attention, however. First, if
there are two or more systems of diagonals, as represented in
PL 2, Fig. YIL, by the full and dotted diagonals (omitting the
upright lines), we must find the strains for each system sepa-
rately, and then add them together. Thus, if the strains found
in a c and c e* etc., for one system, are 50 and 60 tons, and
those in df and fg, for the other, are 40 and TO tons, we have,
when the two systems are combined, d c = a c + df = 50 + 40
= 90, cf= df+ce = ±0 + 60 = 100, fe = ce +.fg =
60 + 70 = 130, and so on. This holds true, of course, whether
the strains are obtained by calculation or diagram. Thus, for
a lattice girder, we calculate or diagram each system ~by itself,
and then the strain in any flange, when the two are combined,
is equal to the sum of the strains on that flange due to each
system of triaiigulation which includes it.
There is another point to be observed in connection with
the system known as the Howe or Pratt Truss. Inserting the
dotted verticals into our Fig., we have this system of square pan-
elling. Let us suppose that the diagonals take tension only,
and the verticals compression only.
Now for a weight at apex 9 of 10 tons, we have a right re-
action of 1 ton, which, running through the system, causes strain
in the diagonal of f P4. For the flange D, then, our point of mo-
ments is at/, and if the height of truss is equal to panel length,
1 x 50
viz., 10 feet, we have the strain in D = — — — = 5 tons, for
P9. In the same way for P8, we have for D 10 tons ; for P7, 15
tons ; for P6, 20 tons ; for P5, 25 tons. For P4, on the other
hand, we have a left reaction of 4 tons, which causes strain in
diagonal e &, and for this weight and all succeeding weights
our point of moments for D is then at e. We have then P4
4 x — = 24 tons ; for P3, 18 tons ; for P2, 12 tons ; and for
P!, 6 tons.
For all these weights, then, acting together, we have 135 tons
strain in D.
APPENDIX.] NOTE TO CHAP. I. .357
But for all tlie weights acting together, it is evident that
only all the braces sloping each way from the centre are
strained. Hence e Tc is not strained, and our point of moments
is for D always at/! Thus for total load we have strain in
45 x 50 - 10 x 40 - 10 x 30 - 10 x 20 - 10 x 10 '
D = - 1Q - = 125 tons,
whereas we found by addition of the several weights 135 tons.
There is thus an ambiguity in this class of bracing as to the
way in which the strains may go. Two symmetrical weights,
as 9 and 1, may either go left and right directly to the abut-
ments or a portion of each go towards the centre. The inter-
mediate diagonals may be either all strained or not strained at
all. The strains may go partly in one way or partly in the
other. We should then not rely on our summation of the sepa-
rate weights, but always check them by calculation or diagram
for the total load also, and take the greatest strain*. Practi-
cally, for long spans, it is very rare that the difference is of any
importance.
In diagraming by our method such a system of bracing as
the above, we should consider but one series of braces, viz.,
those strained by the uniform load alone. Thus, for our Fig.
and loads on the lower apices, we should take only the diago-
nals parallel to fh on the left of centre, and yP4 on the right.
If, on the other hand, the verticals are ties and the diagonals
struts, we should retain only those parallel to c & on the left,
and those parallel to k e on the right of centre. The others
are to be omitted. Then, the tabulation being formed, if in
any diagonal a strain may occur of reverse character to that
which it is intended to resist, a counterbrace must be inserted
in this panel to take this reverse strain.
As in our examples we have taken always a triangular system
of bracing, it is important that the reader clearly understand
the method to be pursued in other forms. For the rectangular
system of bracing generally, the point where for uniform load
the shear is zero is the point from which the braces must slope
both ways. The other diagonals, or the count erbr aces, are then
omitted in both calculation and diagram, and replaced from the
tabulation when necessary to replace a strain of the reverse
character to that which the braces are intended to sustain.
Attention to the above points will enable us to both calculate
353
NOTE TO CHAP. I.
[APPENDIX.
and diagram with ease and accuracy any form of truss which
occurs in engineering practice.
11. In the bow-string girder represented ir Fig, Y. it is
evident that the bottom flange serves merely to resist the thrust
of the bow and keep it from spreading. It adds nothing to the
supporting power of the combination. We might remove it
entirely and replace it by abutments which would equally well
sustain this thrust, and if we then introduced a horizontal flange
at crown, and inserted diagonals between for stiffness, we should
have the form of braced arch given in Chap. I., Fig. 5 (c). If,
however, we should resist the thrust of the bow by an inverted
arc, it would answer the same purpose as the bottom flange,
and we should, in addition, double the supporting power.
We have illustrated this in Fig. YI.
The span is the same as before. The lower apices only are
supposed to be loaded, for comparison. [Properly, we should
have distributed the load over both upper and lower apices.]
The rise of each arc is one-half as great as before, or 5 f t% only,
thus making the total depth the same as in the preceding
case.
By means of two strain diagrams, we find the strains due to
PiandP7. Thus:
xa
Xb
Xd
x/
Xh
Ya
Y c
Y e
Y<7
Pi
+2.3
+2.6
+3.2
+3.7
+5.0
-2.25
-2.87
-3.48
-4.32
P7
+16.1
+17.8
+9.1
+5.7
+4.5
-15.75
-12.35
-7.3
-4.8
Then, precisely as in the preceding Art., we can fill out our
table of strains. This the reader can now easily do for himself.
We thus find, for a uniform dead load f ths the live load, the
total maximum strains below.
Xa
Xb
Xd
x/
Xh
Y a
Yc
Y<5
Y£
+112.7
+126.7
+133.5
+127.
+134.7
-110.1
-126.7
-129.3
-125.6
Comparing these with the corresponding strains for the bow-
APPENDIX.] NOTE TO CHAP. I. 359
string, we find that they are very much less in every piece. In
fact, there is a total gain of over 10 per cent., and that, too, not-
withstanding that the rise of each arc is only half that in the
first case. Had we taken a double depth, the saving would
have been very great, and as in this case also, for a long span
and relatively large dead load, the diagonals would always be in
tension, the increased length of these last would be no dis-
advantage.
12. The above construction is worthy of the careful consid-
eration of the bridge builder. It peculiarly recommends itself
for long spans, and has several important advantages possessed
by no other form of truss. For long spans the strains in the
flanges are nearly uniform. The diagonals are less strained
than in any other system, and are always in tension. Every
member acts to support, as well as to strengthen. The height*
is everywhere proportional to the maximum moment of the ex-
terior forces. The load is distributed along the neutral axis,
thus securing the maximum of rigidity ; while the neutral axis
itself passes through the points of support.
This construction is known in Germany, from the name of its
inventor, as Paulas Truss. Upon this system are the double
track bridge over the Isar at Grossheselohe, 2 spans of 170.6
ft. ; a large number of smaller bridges, such as one over the
Rodach, 109 ft. span ; over the Main in Schweinfurt, 116.4ft.
span ; and especially one over the Rhine at Mayence, of 32
spans, 4 of 345 ft., 6 of 116 ft., 20 of 50 ft., and 2 of 82 ft. ; all
upon the same system.
In England, we might notice the famous bridge over the
Tamar at Saltash, near Plymouth, whose two principal spans
are 455 ft., which is also constructed upon this system.
Finally, we may mention the bridge over the Elbe, near
Hamburg, the three principal spans of which are 325 ft. each.
In this latter structure both the upper and lower members
are braced or ribbed arches, of 'a constant depth of about 10 ft.,
a combination which, for long spans, seems most excellent. A
single arch alone, similar, for example, to the steel arch over the
Mississippi, by Capt. Eads, would have required heavy abut-
ments.
The same arch inverted would have required equally heavy
anchorages. The combination does away with both. The
360 NOTE TO CHAP. I. [APPENDIX.
thrust of the upright arch is opposed by the pull of the inverted
one, all the advantages of Pauli's system are obtained, and
there are no temperature strains such as occur in the single
arch, while the bracing is reduced to a minimum. At the same
time all the rigidity due to the arch is obtained.*
13. In the construction of the diagrams, care should be exer-
cised in the selection of the scales, that the frame diagram may
be large enough to secure the desired accuracy. Lines should
be drawn very fine with a hard, sharp-pointed pencil, so as to be
scarcely discernable, and their intersections accurately marked
by needle point.
With an accurate scale and good instruments, strains can be
taken off in nearly every practical case to hundredths of a ton
* Compare Long and Short Span Railway Bridges, by John A. Roebling, C.E.
— In this work, Mr. Roebling proposes a system in principle essentially the
same as the above, to which he gives the name of "Parabolic Truss." He,
however, constructs the arch of channel irons bolted to the sides of a straight
truss, the sole office of which is to give rigidity to the system. Also, claiming
that iron in the shape of wire will safely sustain three times as much as in the
shape . of bars or rods, he introduces a wire cable in place of the inverted
braced arch.
It will thus be seen that for rigidity the system is wholly dependent upon
extraneous members, such as the auxiliary truss and the tower stays, which are
liberally introduced. By dividing the material composing the upright arch
into two portions, bracing between them, and thus forming a braced arch sim-
ilar to Capt. Eads, the stays and stiffening truss might be entirely dispensed
with, the construction greatly simplified in the number of its members, and
the bracing reduced to a minimum. If, also, as claimed by Capt. Eads, the
conditions for cast steel are just the reverse of iron, and it is most advantageous
to use it in compression, then it seems that such a modification of Mr. Roeb-
ling's design with wire cable and a cast-steel braced arch would better sustain
the thesis with which his work, above quoted, opens, viz. : that "the greatest
economy in bridging is only to be obtained by a judicious application of the Para-
bolic Truss."
Such a combination of the suspension and upright arch would seem to avoid
the principal objections urged against each separately. ' The anchorages and
abutments are dispensed with, the greatest rigidity is secured with the mini-
mum of bracing, and the material is used in the most advantageous way. In
addition to the advantages of Paulas system-being secured, we have the ease
of erection of the suspension system combined with the rigidity of the arch.
The system is self -balancing, and practically unaffected by changes of temper-
ature.
For the practical details of construction of such a system, the reader can
with profit consult Mr. Roebling' s work, above quoted. They will be found
to be neither expensive nor difficult of execution.
APPENDIX.] NOTE TO CHAP. I. 361
with perfect accuracy. The use of parallel rulers is not to be
recommended. The T square, triangle and drawing-board are
far preferable. It should be remembered, finally, that careful
habits of manipulation, while they give constantly increased
skill and more accurate results, affect in no degree the rapidity
and ease with which those results are obtained.
362 NOTE TO CHAP. H. [APPENDIX.
NOTE TO CHAPTER II.
14. The reader will observe that in Chapter I. we had given
forces acting at certain points of a given frame, and we found
by simple resolution of forces the strains in the pieces of that
frame. In Chapter II. we have given forces acting in certain
directions, and having assumed the strains, we find the equi-
librium, polygon or frame, which, having its angles on these
force directions, and having these strains, will hold the given
forces in equilibrium. Thus in Figs. 12 (b) and (c), PL III., of
the text, by choosing a pole and drawing lines to the forces in
the force polygon (a), we virtually assume the strains which
are to act upon our frame. Then lines parallel to these strains
in (b), forming a polygon whose angles are upon the forces,
must give us the frame which holds these forces in equilibrium,
provided we close the polygon by a line and apply at the ends
forces which balance each other horizontally, and whose com-
ponents parallel to the resultant of the forces balance the
forces.
Thus the polygon maficdenm is a frame along whose
sides the forces S0 Sl5 etc., act, and whose reactions at the sup-
ports m and n must then be a o and 5 a, as given in (a).
This frame — keeping the same pole, that is, the same strains
— we may put anywhere in the plane, its angles being always on
the forces, and its sides always respectively parallel, though
varying in length according to the position assumed.
We might also have assumed different strains, that is, taken
a different pole, and constructed a different frame; but evi-
dently the end reactions will not be altered, and will be always
equal to a 0 and 5 a, as given in (a).
The peculiarities of the frame thus obtained are, as we see
further on, that its end sides always intersect upon the result-
ant of the forces ; its depth is always proportional (for paral-
lel forces) to the moment at any point ; its area to the moment
APPENDIX.] NOTE TO CHAP. II. 3G3
of inertia of the forces ; while, finally, in a loaded beam the de-
flection curve itself is but a polygon or frame of this character,
when the curve of loading follows the law of the moments in
the beam.
It is upon this polygon and its properties that the entire
system of Graphical Statics is based.
364 NOTE TO CHAP. V. [APPENDIX.
NOTE TO CHAPTEE V., AET. 51.
15. In Fig. YIIL (Appendix) we have given the construction
referred to in Art. 51 of the text for a system of loads of given
intensities. The span s0 s0 is supposed to shift to sx s1? s% s2, etc.,
and a certain cross-section &0 to shift with it to Tc^ &%, etc. The
intersections of the respective closing lines with verticals
through &0, &u &2> etc-? gives us a curve between which and the
polygon the greatest ordinate gives the maximum moment for
the assumed cross-section. The place of this ordinate is the
position of the cross-section from which we determine the ends
of the span, and thus have its position with reference to the
loading when the moment in k is the greatest possible.
Thus if this greatest ordinate is- at the angle YIIL in the Fig.,
the weight P8 must rest upon the cross-section. The distance
then from P8 to the left end of span s, is the distance from SQ
to &o on left, and to right end of span s, is the distance from
#o to SQ on right.
The ends s and s being thus found, perpendiculars through
them determine the closing line L, and the parallel to this in
the force polygon gives the end reactions L 0 and 20 L for the
position of span which makes moment at Jc a maximum.
APPENDIX.] PIVOT SPAN. 365
NOTE TO CHAPTEE XII., ART. 124.
16« In Arts. 120 and 121 we have given the formulae and
principles necessary for the complete solution of the pivot span.
We propose here to illustrate more fully their application by a
simple example.
Fig. IX. represents such a structure. The two outer spans
A B — C D = 40 f t. The central or turn-table span, B C =
20 ft. Centre height at B and C = 10 ft. End height = 6 ft.
Panel length, 10 ft. ; each apex live load, 10 tons, or 1 ton per
foot. Dead load, half 'as much. Two systems of triangulation,
as shown in the Fiff.
O
Our proportions are taken for the sake of illustration merely,
and not as an example of actual practice. All the points to be
observed are, however, illustrated as well as by a much longer
span, and more usual proportions.
It is to be observed that the end verticals are compression
members only, and cannot take tension. This is necessary -to
prevent ambiguity as to the way in which the strains go. A
negative reaction might otherwise cause tension in 1 2, and
compression in F, or tension in 1 5, compression in 5 6, and
tension in A. If 1 5 cannot take tension, we have but one
course for the strains, and the problem is determinate.
We also, for similar reasons, construct the centre span so
that the diagonals take tension only, and the verticals compres-
sion only. These points as to construction being settled, let us
proceed, first, to determine the reactions.
1st. REACTIONS.
We shall consider the case of the " Tipper" or secondary
central span only [Art. 120], as this case most nearly ap-
proaches the true state of things. The method of procedure
for fovrjixed supports is precisely similar, only taking the for-
mulse for that case from Art. 122.
The less the span B C, the nearer the case approaches to three
fixed supports ; and when the distance B C is zero, n is zero, and
our formulas are the same as for beam over three supports.
For a load in the left span distant a from A, these formulae
are as follows [Art. 120] :
366 NOTE TO ART. 124. [APPENDIX.
RA = j 2H - (10 + 15 w, + 3 n>) Jc + (2 + n)ji\,
C ^g |(6 + 9
"
RD=2H
1
J
I (2 + ri)k* — (2 -f 3 n + 3 ft2) k I,
in which Jc=^ I =A B=C D, n I = B C and H = 4 -f 8 7* + 3 ft2.
£
We have first to put these formulse into the most convenient
shape for use in the particular case under consideration. Thus
"1 Q K
in this case I = 40, n I = 20 ; hence n = - and H = -— , and
2 . 4
Jc = — , where a has the successive values of 10, 20, 30, 40 for P15
Pa, P3, P4. J& is therefore successively -, -, - and -.
Our equations for reactions are then, after reducing,
J 45 £-10 ¥ I
[1
1A7i8 17^" I
107, 17^J
Now, as we may notice, the denominator of & is always 4, of
¥ always 64 ; the numerator only changing according to the
position of the weight. These equations can then be written
RA - 224
^J2240-584a+5a3 I
A |~72 «-</],
L -I
RD =
where a has the values 1, 2, 3 for P1? P2, P3, etc.
These, then, are the practical formulas for this case, and from
APPENDIX.] PIVOT SPAN. 367
them we can easily find the reactions for the apex loads of 10
tons each.
Thus, for P! make a = 1, and we have
RA = 7.415, RB = RC = 1.58, RD = — 0.584
For Pa make a — 2, and
RA = 4.964, RB = Rc = 3.035, RD = — 1.035.
For P3 make a = 3, and
RA = 2.78, RB = Rc - 4.22, RD = - 1.22.
For P4 make a = 4, and
RA = lj -^B — RC == 5, RD — — !•
Loads upon the centre of the span B C acting, that is, at
apex 10, give no reactions, but are supported directly by the
turn-table. Hence, for P5 ; RA, RB, Rc and RD are zero. For
the first load, P7 to the right of C, the reactions at A and B are
the same as for P3 at D and C, already found. For the next
load, P8, the reactions at A and B are the same as for P2 at D
and C, already found. For P9, the same as for Pt. For P6, as
for P4, etc.
We thus have the reactions at A and B due to every indi-
vidual apex load, and can now proceed to find the strains.
Our formulae, it will be observed, thus become very simple
and easy of application for any particular case.
2c7. FLANGES — BRIDGE SHUT.
Let us first find the strains in the flanges. We have only to
apply the method of moments, and the work is so simple that
an example or two will suffice.
We repeat again the rule. Conceive a section cutting only
three strained pieces. Take the intersection of two of these as
the centre of moments for finding the strain in the third. The
moment of the strain in this last about this point must be equal
to the algebraic sum of the moments of all the forces acting
between the section and one end. Take Px for example. Its
upward reaction at A is 7.415. [A negative reaction acts
down. Thus, for P7 above, the reaction at A is, from our for-
mulae, — 1.22. The minus sign indicates that the reaction is
down, and that, neglecting the dead load, the girder must be
held down to the support A. If the reader will draw roughly
the curve of deflection, he will see that this is so.]
368 NOTE TO AKT. 124:. [APPENDIX.
Conceive a section through the girder at, say, the centre of
flange A. It cuts 4 pieces, but, since the weight Pt acts only
through its own system of diagonals, only three are strained.
The point of moment for A is then at 6, the intersection of the
other two strained pieces. The strain, then, in A x by its lever
arm = 7.415 x 10. The lever arm of A is 6.965 ; hence
A x 6.965 = 7.415 x 10,
or A = + 10.64 tons compression,
because the upward reaction acting with 6 as a centre of rota-
tion tends to compress A.
This strain evidently acts through both A and B, since both
these flanges are included by the two diagonals of the system
for P! ; hence also, B = + 10.64 tons.
For flanges C and D, since 78 is the strained diagonal, 8 is
the centre of moments. The same reaction acts now with the
lever arm 30 to cause compression, and Px acts with the lever
arm 20 to cause tension. We have then
C x 8.955 = + 7.415 x 30 - 10 x 20,
or . C = D = 2.5 tons compression.
Now we come to the centre span, and must carefully observe
the following points. Since D has been found to be compres-
sion for Pt, we see at once that the whole upper flange for the
span A B is for this weight in compression. Diagonal 8 9 is
therefore in tension. Were there no vertical strut at B, this
would cause compression in 910. But brace 910 cannot by
construction take compression. The strained pieces cut by a
section through E are then E, B 11 and K, which give us the
centre of moments at B for strain in E. Observe, that were
it not for the vertical, we should have had 10 for the centre
of moments; or, with the vertical, had D been found tension,
8 9 would have been compression ; there would then have been
no strain in the vertical, that being incapable of tension, and
diagonal 9 10 would have been strained, thus giving us also 10
for the centre of moments. Attention to the above is necessary
in order to properly pass from the span A B into the middle
span.
We have then for strain in E
E x 10 = 7.415 x 40 - 10 x 30, or E = - 0.34,
APPENDIX.]
PIVOT SPAN.
369
or in tension, as indicated by the sign, since the moment of P,
overbalances that of the reaction.
\Note. — The different lever arms are easily obtained from
the known dimensions of the truss. We have considered it
unnecessary to detail how they are to be found. They may
either be measured to scale from the frame or computed trigo-
nometrically.]
The lower flanges are found in similar manner.
Thus, strain in F is zero, since it passes through the point of
moments.
For G and H, we have
G x 8 = - 7.415 x 20 + 10 x 10, or G — - 6.04 tension.
In like manner, for I,
I x 10 = - 7.415 x 40 + 10 x 30 = + 0.34.
For K, for similar reasons as above for E, we have centre at
11, and therefore the reaction at B also enters into the equa-
tion of moments, and
K x 10 = - 7.415 x 50 + 10 x 40 — 1.58 x 10, or K = + 1.34.
We have then, finally, for the strains in the flanges due to Pt
A
B
C
D
E
F
G
H
I
K
Pi
+10.64
+10.64
+2.5
+2.5
-0.34
0
-6.04
-6.04
+0.34
+1.34
In a precisely similar manner we find the strains due to P2,
P3 and P4.
We have only to observe that for P7, the first weight to the
right of C in the other span, the reaction at A is negative and
equal to the reaction of P3 at D, already found, or — 1.22.
Now as we suppose the end A bolted down, this reaction acts
as a weight of 1.22 tons suspended from the end. So for the
reactions of P8 and P9, viz., — 1.035 and — 0.584. These reac-
tions, moreover, must all take effect through diagonal 1 2 and
flange F, as the end vertical cannot take tension.
Finding then the strains due to each of the other weights,
we can, finally, tabulate our results as on next page :
24
370
NOTE TO ART. 124.
[APPENDIX.
STRAINS IN FLANGES LIVE LOAD BRIDGE SHUT.
A
B
0
D
E
F
G
H
I
K
PI
+10.64
+ 10.64
+ 2.5
+ 2.5
— 0.34
0
— ?.r
— 6.04
— 6.04
+ 0.34
-1.34
pa
0
+ 12.47
+ 12.47
— 0.14
— 0.14
— 7.1 j— 5.32
— 5.82 +2.14!
P3
+ 3.9
+ 3.9
+ 2.5~
+ 9.31
+ 9.31
+ 1.12
0
— 6.95 — 6.95
— 1.12
+1.88
P4
P6
0
+ 2.5
+ 4.0
+ 4.0
—1.42
— 1.42 — 3.;£i
— 3.33
0
0
0
0
0
0
0
0
0
0
0
P6
0
— 2.5
— 2.5
— 4.0
— 4.88
— 4.0
+ 1.42
+ 1.42 + 3.33
+ 3.33
0
PT
0
~~0
— 3.1 | —3.1
— 4.88
+ 1.74
+ 1.74
+ 4.U6
+ 4.U6
+l.bb
+2.14
P8
-2.6
-2.6
— 4.4
— 4.14
+ 1.46
+ 1.46
+ 3.45
+ 3.45
P»
0
— 1.4
— 1.4
— 2.34
— 2.34
+ o.as
+ 0.83
+ 1.95
+ 1.95
+1.34
Total
Strains
+14.54
+ 29.51
+ 26.78
+ 15.81
+ 5.12
+ 5.45
+ 5.45
+ 12.79
+ 13.13
+9.38
;
-9.6
- 9.6
— 15.76
— 15.84
- 8.52
— 21.61 1 — 21.64
— 9.77
-1.34
In the two horizontal lines at bottom, we have the total
strains of each kind caused by the live load.
3d. FLANGES — BRIDGE OPEN — DEAD LOAD.
We have next to find the strains due to the dead load when
the span is open.
We have then 5 tons at each apex, except the ends, where
we have P0 — 2.5 tons.
These strains are easily found by moments as above, and we
have then the following table :
A
B
C
D
E
F
G
H
I
K
PO
0
0
— 6.2
— 6.2
— 10.1
— 10.1
+ 3.57
+ 3.57
+ 8.3
+ 8.3
+ 10
P>
0
— 11.1
— 11.1
— 15.0
0
+ 6.1
+ 6.1
+ 15.0
+ 15.0
P2
0
0
0
— 10.1
— 10.0
0
0
+ 5.5
+ 5.5
+ 10.0
P3
0
0
0
0
— 5.0
0
0
0
+ 5.0
+ 5.0
Total
Strains
0
— 6.2
— 17.4
— 31.2
— 40.0
+ 3.6
+ 9.7
+ 20.1
+ 33.8.
+ 40.0
If now, as should be the case, we suppose the centre sup-
ports raised above the level of the ends, so that the ends just
bear, then these strains above act even when the bridge is shut.
As we have already seen in Art. 131, our formulae for the
reactions are not affected by this state of things. The strains
due to live load will then be increased by those above, and we
thus have for the total maximum strains which can ever occur,
APPENDIX.]
PIVOT SPAN.
371
A
B
0
D
* E
F
a
H
I
K
+ 14.5-1
+ 29.51
+ 26.78
+ 15.81
+ 5.12
+ 9.05
+ 15.15
+ 32.89
+ 46.93
+ 49. S8
— 15.8
— 27.0
— 46.96
— 55.84
— 8.52
— 21.51
— 21.64
-9.77
— 1.34
Of course, for this condition of things the ends must always
be bolted down.
It is sometimes customary to raise the ends by an apparatus
for that purpose, after closing the draw, until the proper pro-
portion of the dead load takes effect also as a positive reaction.
We can easily find the strains in this case also by adding
the numbers in the last horizontal line of our table for bridge
shut, with their proper signs, and taking half the results for a
new line for dead load strains. The resulting strains can
then be found precisely as in the table of Art. 8 (Appendix).
We must also find the strains for bridge open as above, and then
take the greatest strains of each kind from these two tables.
In this case 'the strains would be differently distributed.
Flange E will be always in tension, A and K always in com-
pression ; the compression in B C and D will be somewhat
greater than above, and the tension in the same flanges less.
The reader can easily deduce the strains for this case from the
two preceding tables.
If the truss may act as a girder over four fixed supports,
we should, in order to be certain of the maximum strains,
make the calculation for this case also, using the formulae of
Art. 122. This is unnecessary, however, if the supports B and
C can never sink far enough to strike the turn-table, or be im-
peded in their motion.
4cth. STRAINS IN THE DIAGONALS.
We may find the strains in the diagonals also for each
weight separately, both for bridge open and shut ; and a pre-
cisely similar method of tabulation will give the strains.
It will here be found preferable to make a series of dia-
grams, as illustrated in Fig. 86, Art. 124, for each weight and
its own system of triangulation. We obtain thus the diagonal
strains, and at the same time check the results obtained for the
flanges above.
If we wish to calculate the diagonals, it will be better to find
the resultant shear acting upon the diagonal, and multiply it by
the secant of the angle the diagonal makes with the vertical.
372 NOTE TO ART. 124. [APPENDIX.
We can also, if we wish, apply, the method of moments.
Thus, if we determine the point of intersection in the present
case of the inclined upper flange with the horizontal lower
flange, this point will be a common centre of moments for the
diagonals. The lever arms of the diagonals with reference to
this point must next be determined, and then we are ready.
This point above for centre of moments is easily found ; thus
4 : 40 : ; 10 : 100.
It is therefore 60 ft. to the left of A, or 100 ft. left of B.
Take now any diagonal, as 3 4. Its angle with the horizontal
is very nearly 42°, and with the vertical 48°. Its lever arm is
then 80 sin 42° = 53.5, and sec. of angle with .vertical is 1,49.
Now take the weight P3. Its upward reaction at A is 4.964,
P2 being 10.
We have then
[str. in 3 4] x 53.5 = 10 x 80 — 4.964 x 60 = + 502.16.
The resultant rotation is then positive, or from left to right.
The point P2 then sinks and 4 rises, and 34 is in tension and
502.16
— "KO~K~ = ~~ 9-38 tons.
oo.o
This is sufficient to illustrate the method.
For the first method referred to above, viz., that by resultant
shear, the following points are to be observed :
When a piece slopes towards the nearest support, we say it is
sloped as a strut, whatever the real strain in it may be.
When it slopes away from the nearest support, it is sloped as
a tie.
The simple shear is the reaction at the support minus the
weights between any point and that support.
If any three strained pieces are cut by a section through the
structure, the strains in these pieces are in equilibrium with the
simple shear at this section. Hence the algebraic sum of the
vertical components of these pieces must be equal and opposite
to the shear itself.
In order to add these vertical components with proper signs,
we must remember that if a flange is in tension and sloped as a
strut, or in compression and sloped as a tie, we add the vertical
component of the strain in it to the simple shear already
obtained. If in compression and sloped as a strut, or tension
and sloped as a tie, we subtract.
APPENDIX.] PIVOT SPAN. 373
The resultant shear thus obtained then, multiplied by the
secant of the angle with vertical, gives the strain in diagonal.
If the sign of the result is negative (— ), it shows that the
strain on the diagonal is contrary to that indicated by its slope.
To illustrate, let us again take the weight P2 and consider
diagonal 34.
The simple shear at apex 4 is 4.964 — 10 = — 5.036. The
strain in C for Pa we have found to be compression, and equal
to -h 12.47. It is sloped as a strut, and its vertical component ia
therefore to be subtracted from the shear above. Since its
angle is nearly 5° 43' with the horizontal, this vertical com-
ponent is
12.47 x sin 5° 43' = 1.24.
Since H is in this case horizontal, it has no vertical com-
ponent.
The resultant shear is then
- 5.036 - 1.24 := - 6.276.
As the secant of the angle of 3 4 with the vertical is 1.49,
we have for the strain in 34, •- 6.276 x 1.49 = — 9.35.
This result being minus, and 3 4 being sloped as a strut,
the strain is 9.35 tons tension, agreeing closely with the value
found above by moments.
The above method is preferable to the method by moments
for the diagonals, as we have only to determine the secants for
the verticals and the sines for the flanges, which is in most
cases easier than to find the lever arms for the diagonals and
the points of intersection of the upper and lower flanges in each
panel. It is, like the method of moments, of general applica-
tion to any framed structure whose outer forces are known.
The method of diagram in Art. 124 will be found preferable
to both.
It is unnecessary to pursue our example further. With the
mutual checks of the two methods of calculation explained
above, as well as the diagrams, correct results cannot fail to be
obtained. The diagrams should always be made first, as they
settle by mere inspection many points which may at first cause
trouble — such as whether the shear in a piece is subtract! ve or
not according to our rule, the character of the strains in dif-
O '
ferent pieces, etc. It is well to indicate on the diagrams com-
pressive strains by double or heavy lines.
374
NOTE TO ART. 128.
[APPENDIX.
NOTE TO AKT. 128, CHAPTER XII.
17. We wish here to call more particular attention to the
relative economy of the continuous as compared with the
simple girder. This, we think, is greater than is generally sup-
posed. It may reach from 18 to 25, and even as high as 50
per cent.
Take the example worked out in Art. 128, Fig. 88. We
have obtained the maximum strains in that Art. upon every
piece.
We give them below, compared with the strains in the same
pieces for a simple girder of same dimensions anb load :
Aa Ac Ae Ag A* B6 Ed Bf BA
Continuous..— 203.5 +63.6 +115.3 +63.6 —203.5 +89.3 —115.9 —115.9 +89.3
Simple 0 +180 +240 +180 0 —90 —210 —210 —90
a& be cd de ef fg gh hk
Continuous.. +189.3 —109.9 +109.9 +45.5 +45.5 +109.9 +109.9 +189.3
Simple +127.3 —127.3 +56.5 +56.5 —56.5 +56.5 —127.3 +127.3
It will be seen at once that there is a saving in the flanges —
about 11 per cent, in all — but the bracing is heavier, giving lit-
tle or no saving. The span is too short to properly represent
the relative economy of the two systems.
If we take a truss such as represented in PI. 2, Fig. VII.,
Appendix, by the full lines only, omitting the dotted verticals and
diagonals — height 6 ft., span 50 ft., panel length 10 ft,, dead
load 5 tons per panel, live load 7 tons per panel — and calculate
the strains in the pieces for a simple girder, and then as a con-
tinuous girder of two spans and three spans, we have the fol-
lowing results :
1 span
strain in tons.
2 spans
3 spans
end.
3 spans
middle.
205 2
231.6 .
229.8
239.4
Lower Chord .
200.
168.7
152.9
122.6
Upper Chord ....
200.
^158.7
159.9
135.2
Total
605 2
559
542 6
497 2
Per cent, saving. .
8 per cent.
10 per cent.
18 per cent.
APPENDIX.] THE CONTINUOUS GIRDEK. 375
We have then, in the first case, a saving of 8 per cent., in the
second, 10, and in the third, 18 per cent, over a simple girder.
Quite a notable saving, although the spans are very short, and
although, in the first two cases (the spans being end spans), we do
not obtain the full advantages of continuity. If, then, instead
of three simple girders of the above dimensions, we should con-
struct the girder continuous over the piers, we should save in
strain, and hence in material, 10 per cent, in each end span, and
18 per cent., or nearly twice as much, in the centre span.
The advantage of continuity is rendered still more apparent
by taking a longer span. Thus for a girder of 200 ft., height
20 ft. — 10 panels, and double system of triangulation, similar to
Fig. YII. — for a live load of 20 tons per panel, and dead load
of 10 tons — we have the following results :
One span. Two spans. Five spans.
r centre.
Bracing 1398.6 1428.2 1596.2
Lower Chord., 2400. 1793.2 1395.7
Upper Chord . . 2550. 1981.6 1622.6
Total 6348.6 5203.0 4614.5
Per cent, saving 18 per cent. 27 per cent.
That is, we have a saving of 18 per cent, instead of only 8
per cent, as before, for two spans, and of 27 per cent, for the
centre span of five spans. For three spans, then, of this length
we should save 18 per cent, on the end, and at least 30 per cent,
on the centre span.
If we suppose the same girder as above fastened or fixed
horizontally at the ends, we shall have the case of a middle span
in a very great number, and may expect to find the greatest
saving possible for this length.
The formulae of Chapter XIII., as also the simple graphical
method for this case, given in Chapter XII., Art. 114 [Fig. 80],
enable us to solve this case easily. The reader will find, on mak-
ing the calculation, the following strains :
Strain in tons.
Diagonals 1279.2
Upper Chord 940.2
Lower Chord 965.4
Total 3184.8
Per cent, saving 49.8
376 NOTE TO ART. 128. [APPENDIX.
The above will serve to illustrate the point in question quite
as well, perhaps, as an extended theoretical discussion. We see
that the saving increases rapidly with the length of the span,
and may easily rise as high as 30 or 40 per cent., while in some
cases even 50 per cent, may be realized.
THE DISADVANTAGES OF THE CONTINUOUS GIRDER ABE I
1st. The fact that the various pieces, especially the chords,
undergo strains of opposite character.
This, in wrought- iron structures, we venture to think of little
importance. The extra work and cost of chords and chord con-
nections necessary to secure the flanges against both compressive
and tensile strain, can hardly amount to 10, 18, 30, or even 50
per cent, of the cost of girder !
%d. Difficulty of calculation.
We have, we trust, in what precedes, and in Chapter XIII.,
succeeded in removing this objection.
The opinion is widespread among engineers that the deter-
mination of strains in the continuous girder is impracticable
and involved in mystery. No opinion could well be more un-
founded. The accurate and complete calculation for all pos-
sible loading, live or dead, is precisely similar to and offers no
more difficulty than the simple girder itself.
The formulae for moments and shears are, as we have seen,
simple and easy of application.
The graphic method here developed offers also a thorough
solution. In view of both, and of the extensive literature upon
the subject (which seems, by the way, to have been so generally
ignored), we can finally pronounce the problem to be fully
solved.
3d. The changes of strain, unforeseen and often considerable,
which a small settling of the piers or change of level of the
supports may occasion.
This, be it observed, is only of importance when the piers
settle after the erection of the superstructure. If piers are to
be considered as settling indefinitely, or continuously during a
succession of seasons, continuous girders are not to be thought
of. If, however, as is generally the fact, the piers take their
permanent set during the first season,, and afterwards are im-
movable, the above objection has no weight. It is not necessary
APPENDIX.] SUPPOETS OUT OF LEVEL. 377
that the piers should be exactly on level or even on line, or even
that the differences of level be known.
As shown in Art. 121, these differences produce no effect,
provided the girder be built to the profile of the supporting
points.
If in any case these differences are required, and it is con-
sidered difficult to determine them over water with sufficient
accuracy, then the proper reactions at the several piers may be
weighed off* and the girder thus left in position under pre-
cisely the circumstances for which it has been calculated.
THE PRINCIPAL ADVANTAGES OF THE CONTINUOUS GIRDER ARE :
1st. Ease of erection, where false works are difficult or ex-
pensive. The girder may be built on shore, and then pushed
out over the piers.
2d. Saving in width of piers, as compared with width re-
quired for separate successive spans. The girder may be placed
upon 'knife edges at the piers. In fact, such a construction is
preferable, as better ensuring the calculated strains. Width of
piers is undesirable.
3d. Saving in jnaterial — usually from 25 to 30 per cent.
18. Continuous Crirder— Supports not on a level. — In
Chapter XIII. we have all the formulae required for the solution
of the continuous girder for supports on a level, or all on line,
when the deviation from level is small, whatever may be the
number or relative length of the spans. If for a continuous girder
of constant cross-section, the supports are properly lowered after
the girder is placed upon them, we may obtain a saving of 23
per cent., or more in material over the same girder with sup-
ports all on level. If, however, the cross-section varies accord-
ing to the strain — in other words, if the girder is of constant
strength — no advantage is gained from thus lowering interme-
diate supports. Such disposition of the supports may even act
injuriously.
The formulae for shear and moments which we have given
are, indeed, based upon the hypothesis of constant cross-section,
but the strains in every piece of the girder being found for the
shears and moments thus obtained, each piece is proportioned
* An idea first suggested by Clemens Herschel, C.E.: Continuous, Revolving
Draw Spans. Little, Brown & Co. , Boston, 1875.
378 NOTE TO ART. 128. [APPENDIX.
to its strain, and the actual girder erected is not of constant
cross-section, but more nearly one of uniform strength. Formu-
lae for the case of supports out of level, as well as determina-
tions of the best differences of level, are hence of but little prac-
tical importance, and have not been given. If, however, it be
required to find the effect due to the sinking of any one pier,
the following may be found of service.
Let the nih support be depressed below the level of the others
by the distance A. Then the moments at all the supports are
changed. The moments at n and at each alternate support
from n are diminished, and at the others increased.
Let H = ?=i
» Then, when all the spans are equal, the following formulae
give the moment at any support :
1st. All spans equal, n — number of lowered support, from
left,
when m<n, Mm =
when m=n, Mm = - 5 - °n~l C*-»+2 + °n G*-«+1 H,
when m > n, Mm = HLtl H,
CB+l
where ^ = 0, cz = + 1, c3 — — 4:, c4 = + 15, c5 = — 56,
c6 = + 209, etc.
From the moments at the supports the shears can be readily
determined from the formula of Art. 148, viz. :
Mm+l a, _Mm-Mm_1
-
where q = P (1—^) for concentrated load and — r; for uniform,
2d. Spans att unequal.
when m < n, cm\ ^-°+i ~ ^-n+a + »s_J+8 — «»-n+2 I
L 4 ^-' J
APPENDIX.] SUPPORTS OUT OF LEVEL. 379
when m =
+ *-«
4 4-1 ^ (4—1 + 4)
[Cn-1~gn + ^+1~Cn1 6AnEI,
4-i 4
s— m+2
when m > n* M = 7 n /7
i_i <v-i + 2 (
in which expressions
— — - ^ - , 4 - — 7 7
=_2o4^-c^,etc.
64 64
_ A /7—1 /7 - 9 ^s + 4-1
= U, <^2 = Ij ^3 — — ^ - 7 -- >
^a-1
_ 4: ft + Cl) (Cl + ^-^s-l
— -- 7 - 7— — J
V-l ^&-2
= _ 2 <Z4 ^ + ^ - d. etc.
The reader who has learned the use of the formulae of Chap-
ter XIII. will have no difficulty in applying the above to any
particular case. In the same way as there explained, by mak-
ing 4 and 4 zero, we may fix the girder at the ends, etc. The
formulae for shear at any support are, of course, the same as
before (Art. 150).
Ex. 1. Let a beam df two equal spans be uniformly loaded
throughout its whole length, and let the centre support be low-
ered by an amount h% = * . What are the moments and
reactions f
The moments due to the full load alone before the support is
lowered are M! = 0, M2 = — , M3 — 0 (Art. 150). For
8
the moment due to the lowering of the support alone, we have
from the above formulae, since
J " J — — — — • y
Hence the total moment is
.__ w Z2 w P wfi
IyL0 — _ — _ — _
8 16 " " 16 '
380 NOTE TO ART. 128. [APPENDIX.
For the shears, then, we have
M2 wl 7 , 9
7 18 7
Hence, R1= ~ w I, R2 = — w I, RB = --w I.
ID lo lo
Ex. 2. Zfow mw?A m^stf w<2 lower the second support in the
above example, in order that the reaction at the centre support
may he just zero f
In this case we have
_wP_ H__ <z0P 3
~~ ~Z:~
g/ _ 2 M2 , _ wl __ 6 A2 E I 5 7 6 AS E I
T~ " T Z3 ~ 4 ^ ~~p "
If this is to be zero, we have
_ 5wl* and M = _ 1 ^ ?
and R! = w Z, Rg = 0, R8 = w I,
or precisely as for a beam of single span ^and length 2 I.
Ex. 3. J. beam of four equal spans is unloaded, and the
third support is lowered by an amount h% = _ .
the reactions f
Ans. Ri =
Ex. 4. ^i J^^m of five equal spans rests as a continuous gir-
der over six supports. Having given the dimensions of the
beam, length of spans, and coefficient of elasticity ; to deter-
mine the reactions due to a sinking of the third support one-
eighth of an inch.
Let the beam be of wood, 1 foot wide, 1.5 deep,
I = 20 feet, 5 = 5, r = 3, E = 288,000,000 Ibs. per sq. ft.,
A, = £ in. = 0.010417 ft.
APPLNDIX.] SUPPORTS OUT OF LEVEL. 381
Then, c, = 1, c3= - 4, c, = 15, £5 = - 56, c6 = 209,
540EI&, __ 906EI^ _576EU3
and M2 = 209 P ' M3= 209 /* ' M4-~~~~
1 3 375
or, inserting the constants above, and I = — b dB = ' ,
M! = M6 1= 0, M2 = 5448, M3 = - 9142, M4 = 5812,
M5 = - 1453 ft. Ibs.
If all the spans are unloaded. For the reactions necessary
to bend it and keep it down to the supports, Rt = — 272 Ibs.,
112 = 1002, R3=-1477, R4 = llll, R5 = - 436, R, = 73.
If the beam weigh 75 Ibs. per foot^ what deflection of third
support will raise the left end from the abutment f
Ans. Rt = w I = ? or 7^ = 0.0226 f t.=0.2712 in.
It will be observed that a small difference of level has then
considerable effect.
Ex. 5. Two equal spans are uniformly loaded. How high
must the centre be raised in order that the ends may just
touch f
This is the case of the pivot span with centre support raised.
(See Art. 121.)
The reactions at the ends are zero. At pier, then, Rg = 2 w I,
hence moment at pier JYT2 = \ w P. But the moment when the
supports are on level is M2 = -J- w~P, hence f wff must be due
to the elevation of the support. Then from our formulae,
3 E I ^ , w P
-— , or A,= -
which is precisely the same as the deflection of an horizontal
beam, fastened at one end, and free at the other (Supplement
to Chap. VII., Art. 13).*
* The calculations and formulae given in this Note (pp. 374-381) are taken,
by permission, from a manuscript work on the Theory and Calculation of Con-
tinuous Bridges, by Mansfield Merriman, C.B., which, we hope, will be soon
given to the public.
382 NOTE TO CHAP. XIV. [APPENDIX.
NOTE TO CHAPTEE XIV.
THE BRACED AKCH.
19. The subject of braced arches is an important one, and
is treated in no work with the fulness and completeness it
deserves. The methods and formulae of Chapter XIY. will,
we believe, render the determination of the strains in this class
of structure easy, and we propose in the following to illustrate
their use, so far as may be necessary to render their application
clear.
In PI. 4, Fig. X., we have represented a braced circular
arch with parallel flanges. Span of centre line = 175 ft. ;
radius, 201.4 ft. ; rise, 20 ft. In practice, the panels would be
taken of equal length ; for convenience of calculation, however,
we suppose the panel length to vary so that the horizontal pro-
jection is constant, and equal to 25 ft. Depth of arch, 10 ft.
Hence, span of lower flange = 170.6 ft. ; rise, 19.5 ft. ; radius,
196.4 ft. Span of upper flange, 179.34 ft. ; rise, 20.5 ft. ; radius,
206.4 ft.
Since the flanges are, in practice, broken lines, and not true
curves, the depth or lever arm for upper flanges is 9.43 ft., for
lower flanges, 10.4 ft.
The determination of the other dimensions required is then
easy, and a simple question of trigonometry.
Thus we have for the half central angle a = 25° 45', and for
the distances of the apices from the chord of the centre line :
For 1...— 4.5 3.... 4.7 5. ...11.3 7 14.6ft.
" 2.... 10.8 4.... 18.5 6.... 23.4 8.... 25 "
We suppose the load at each apex 10 tons, and shall consider
~Lst. Arch hinged at crown or ap6x 8, and at the ends of
the lower flange — the flanges H and A being removed.
2d. Arch hinged at apex 8, and at the ends of the centre
line — the flanges A and E butting against a skew ~back or pivoted
plate, and the flange H only being removed.
3d. Arch continuous at crown — the flange H being retained,
and hinged at ends of lower flanges.
APPENDIX.] THE BRACED ARCH. 383
kth. Arch, as in 3d, but pivoted at ends of centre line.
5th. Arch without hinges, or continuous at crown and fixed
at abutments.
These cases will illustrate all the principles of Chapter XIY.,
and a comparison of the results obtained in each case may
prove instructive.
2O. Arch hinged at apex 8, and at the extremities of the
lower flange— flanges H and A being removed.
From Art. 158 we can easily find the reaction and horizon-
tal thrust at left end either by construction or formula for
every weight. Thus
For the first weight P,, then,
For the weight P1
In similar manner, we find
P! = 0.603, P2 = 1.33, P3 = 2.06,
H! = 2.74, H2 = 3.84, H3 = 5.9,
P4 = 2.8, P5 = 3.53, P6 = 4.2,
H4 = 8.1, H5 = 10.2, H6 = 12.1,
P7 = 5.0, P8 = 5.7, P9 = 6.47, P10=7.2,
H7 = 14.4, He = 12.1, H9 = 10.2, H10= 8.1,
Pu=7.94r, Pu=S.6, PIS- 9.4,
Hu= 5.9, H12= 3.84, H13= 1.74.
It will be at once seen that the reaction of P8 at A is the
same as of P6 at B, or equal to 10 — P6 ; while the horizontal
thrust is the same for both. We need them only to find P and
H for weights 1 to 7, and can then at once write down the
others. We are now ready either to calculate or diagram the
strains.
384: NOTE TO CHAP. XIV. [APPENDIX.
Thus, for instance, for P10 (see Fig. X.), we lay off the reaction
at A upwards to scale from C to D, then the horizontal thrust
at A from D to 1, then the equal thrust at B from 1 back to D,
then the reaction at B from D to 8, and finally the weight
down from 8 back to C, thus closing the polygon for the exterior
forces. Lines parallel to the pieces then give the strains.
Thus the thrust and reaction at A are in equilibrium with E
and 1 2. Then 1 2 is in equilibrium with B and 2 3, and so on.
Observe that the diagram checks itself. Thus the last diagonal
7 8 must be in equilibrium with 6 7 and G (flange H being re-
moved), and that this is so is shown by the strain in 7 8 passing
exactly through 8, thus making the strain in H zero. We can
also check the work by calculating the strain in the last flange
D by moments. Thus for P10
D x 9.43 = 7.2 x 72.8 - 8.1 x 19.1 - 10 x 25 = 119.45,
or D = + 12.6.
If this agrees with D as found by diagram, and if the diagram
also checks, we may have confidence in the accuracy of the
work, and at once scale off the strains. Observe that diagonals
45 and 56 are both tension ; also that F and G are tension.
We have given also the diagram for Pn, which the reader
can easily follow through for himself. F and G are both ten-
sion, 3 4 and 4 5 both compression. The horizontal thrust is
8 &, and the reaction at A = b 1.
We thus make a diagram for each separate weight, and then
taking the dead load at -J the live, we can form the following
table of strains. Since we wish only the maximum strains on
one-half the arch, those on the other half being precisely simi-
lar, we can diagram the strains due to all the weights upon the
right half at once by taking the sum of their reactions and
thrust at A. We have then the following table :
APPENDIX.]
THE BRACED ARCH.
385
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NOTE TO CHAP. XIV. [APPENDIX.
21 . Arch hinged at Apex § and at the Extremities of the
centre Une ; FIange§ A being retained, and only H re-
moved.— The method of solution is precisely the same as
before, the only difference being that the span is now 175 ft.
instead of 170.6, and the rise 25 ft. instead of 29.5. The reac-
tions and thrust will then be somewhat different. Thus, for
the left abutment A,
P1=0.71, Pa =1.42, P8 =2.14, P4=2.85, P5 = 3.57,
H!=2.48, H2 = 4.97, H8 = 7.5, H4 = 9.97, H5 = 12.5,
P6= 4.28, P7= 5.00, P8= 5.72, P9 = 6.43, Plo = 7.15, etc.
H6 = 14.98, H7 = 17.5, H8 = 14.98, H9 = 12.5, H10 = 9.97, etc.
We have therefore the following table :
APPENDIX.]
THE BRACED ARCH.
387
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388 NOTE TO CHAP. XIV. [APPEKDIX.
A comparison of this table with the preceding shows that we
have gained nothing by introducing two end flanges at A at
each end, and pivoting the arch at the extremities of the centre
line. We have indeed slightly diminished the strains in the
lower flanges E and F, as also in the bracing, but the other
strains are much greater than before — a result which might
have been anticipated, since the effect of hinging at the centre,
instead of at the extremities of the lower flange, is simply to
reduce the effective height or rise from 29.5 ft. to 25 ft. In our
example, since the depth of arch is half the whole rise of the
centre line, this reduction is considerable.
For a much longer span and smaller proportional depth the
difference would not be so marked, but it would seem that the
strains in the second case must always be greater than in the
first. The best construction, then, seems to require the hinge
in the upper flange at the crown, and at the extremities of the
lower flange at the abutments. By this means, the greatest ef-
fective rise is obtained, and both ribs aid in supporting the
load. Were the hinges all three in the same rib, then, for
uniform load, that rib alone is the sole supporting member, and
is unassisted by the other. This should then be avoided.
22. Arcli continuous at Crown, and hinged at End§ of
Lower Rib. — For this case, referring to Art. 159, we have sim-
ply to interpolate from our table there given the values of A,
B and y0 in the equation,
1 +B/c
y = r=ST*y»
and thus plot the curve cdeik. Fig. 91. The construction of
the reactions and horizontal thrust for each weight is then easy.
These once known, we proceed as above, in order to find the
strains.
Now, in the formulae above K = -5, and since we can put
A T"
for — the square of the radius of gyration, and this radius is
A.
approximately the half depth of the arch, we have
25 25 1
1 (170.6)2 - 2910436 ~~ 1164'
Now B and A are, as we see from the table, small, and he^nce
in our present example the terms containing K can be disre-
APPENDIX.] THE BEACED ABCH. 389
garded, and the value of y can be taken directly from the
table for y0, given in Art. 159. For a — 25° 45', then we have
at once, since h = 19.5,
for 0 = 0, y = 1.295 h = 25.25 ft.,
0 = 0.2 a, y = 1.304 A = 25.42 ft.,
ft = 0.3 a, y = 1-335 A = 26.03 ft., etc.
The corresponding value of x is R cos ft.
Having thus plotted the curve, and constructed the reaction,
and thrust for each weight, the diagram for strains proceeds as
before. We thus form the following table :
390
NOTE TO CHAP. XIV.
[APPENDIX.
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APPENDIX.] THE BRACED ARCH. 391
Comparing these resiflts with those in our table above, Art.
19, for the same case not continuous at the crown, we see that
the strains in the upper flanges are much less, and are, more-
over, of opposite character ; while the strains in the lower
flanges are greatly increased, and nothing is gained. This re-
sult might also have been anticipated, since the effect of insert-
ing the flange H is to reduce the effective height from 29.5 to
19.5 ft., and, moreover, for total dead and live load, nearly the
whole weight comes directly upon the continuous lower rib, and
the upper aids but very little.
23. Strain* due to Temperature. — We have, in addition,
strains due to change of temperature to be taken into account
in determining the total maximum strains.
For the present case we have, from Art. 165, for the thrust
due to change of temperature,
15 El A et
~8AA2 + 15T
or, substituting in the place of — the square of the radius of
A.
gyration = <f, we have
15 in A ^2 */
H = —
Now g is approximately the half depth of arch ; hence cf1 —
25 sq. ft. = 3600 sq. inches. Taking 5 tons to the square inch
as our unit strain, we may take the area of our flanges, as de-
termined from the above table of strains at about 25 square
inches. Hence A = 50. Taking E = 14,000 tons per sq. inch,
A2 — 19. 52 = 54,756 sq. inches, and supposing the temperature
to vary 25° (Centigrade) on each side of the mean, we have,
assuming e at 0.000012, the thrust H — about 25 tons.
It is easy to find either by moments or diagram, or both, the
strains due to this thrust. Since the temperature varies between
25° on both sides of the mean, this thrust can be both positive
and negative, and the corresponding strains have, therefore,
double sign. We find, therefore,
B = =F 24.5 C = T= 40.4 D = =f 49.1 E = ± 37.5
F = i 55.9 G- = ± 66.9 H = j_ 69.6
12==Fl?.l 23 = ± 15.0 34=^8.6 45 =±12.5
56 = ^2.2 67 = i 9.7 7 8 = =p 5.7
392 NOTE TO CHAP. XIV. [APPKNDIX.
Hence we have, for the total maximum strains for the case of
the preceding article,
B = + 57.4-63.4 C= + 81.1-77.1 D= + 64.2-91.4
E=+ 279.4 F = + 286.3 G= + 286.5
H=+ 285.3 12 = + 53.4-39.723=+ 43.7-28.8
34= + 37.3—22.445=+ 40.4-28.456=+ 23.4-12.5
67=+ 36.4-22.778=+ 42.5-23.8
24. Arch continuous at the Crown, pivoted at the
extremities of the Centre Line. — The method of solution is
precisely similar to the preceding. We have only to take the
rise of the centre line, or h — 20, instead of h =19.5, as before,
and the radius and span of centre line, instead of the radius and
span of the lower rib.
One point only needs to be noticed. Having found the reac-
tion and thrust for any weight, these forces now act at the ex-
tremity of the centre line. "We can therefore form the strain
diagram as follows :
First calculate by moments the strain in A and E. Then, in
diagram (c), Fig. X., having laid off the thrust o C and the reac-
tion C 5, draw from o and b lines parallel to A and E, and lay
off o A equal to the strain in A, and b d to the strain in E.
Then, if these strains have been correctly found, the line A d
must be parallel to and give the strain in diagonal 1 2. The
diagram thus commenced, can then be continued as shown, and
the strains in all the pieces determined.
We may also form the strain diagram without calculating A
and E. Thus o b is the resultant acting at the end of the
centre line. Since it acts then half way between A and E,
bisect it in 0, and draw a A perpendicular to flange A. Then
o A is the strain in A, and drawing A d parallel to diagonal 1 2,
we have at once the strain in E and 1 2.
Performing the operations indicated, we obtain the following
table of strains :
APPENDIX.]
THE BRACED ARCH.
393
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394: NOTE TO CHAP. XIV. [APPENDIX.
A comparison of the above with the same case hinged at
the ends of the lower rib, shows a decided gain. The effective
height is increased, being now 20 ft., in place of 19.5 ; in
addition to which both ribs under total load bear their proper
proportion. If, then, we wish the arch continuous at crown,
both ribs should butt against an end plate, pivoted in the
centre. This is preferable to hinging the lower rib at its ex-
tremities, and removing the end flanges A.
25. Temperature Strains. — For the strains due to tem-
perature, we may take, as before, the thrust H = 25 tons, and
rind thus
A=±11.2 B=q=13.2 C=q=29,4 D = qp 37.6
E = ± 27.0 P = ± 45.5 G=± 56.4 H == ± 59.2
12 =T 17.8 2 3 =±13.9 3 4 = =F 9.8 4 5 = ± 11.3
5 6 = =F 2.8 6 7 = ± 8.1 7 8 = ± 4.3
— all somewhat less than in the previous case, as they should be,
since the point of application is at the centre between the
flanges.
We have then from the preceding table the total maximum
strains :
A =
+ 135.7
B = + 135.2 C = + 144.3
D = + 155.5
E = + 158.2
„ +176.7 r +173.9
- 7.2 - - 19.2
H + 172.3
ti~ - 25.5
12 =
+ 49.0
-44.1
o o + 33.4 o A + 28.5
- _ 33.0 - _ 25.4
45 _ +24.8
~ -26.1
56 =
+ 27.8
— 28.5
Rf7 + 29.4 ^ft + 33.0
67-_30.7 78:"-34.7
26. Arch continuou§ at Crown and fixed at Ilie Ends
—From our table, Art. 160, we have directly for a = 25° 45',
and Ti — 20, y being now measured above the horizontal tangent
at crown of centre line,
for ft = 0, y = 0.209 h = 4.18 ft.
/3 = 0.2a, y = 0.208 h = 4.16
ft = 0.4a, y = 0.206 h = 4.12
£ = 0.6a, y = 0.201 h = 4.02
ft = O.So, y = 0.198 h = 3.96
ft = l.Oa, y = 0.189 h = 3.78
APPENDIX.] THE BRACED ARCH. 395
Also from our formulae of Art. 160, we have
40
Cl ~
15 (87.5 + x)
40
I 87.5 — 5 x — 246.1 I
I 87.5 + 5 x - 246.1 I
15(87.5+0)1
Hence for
P! x = 75 d = + 8.7 c2 = - 3.5
P2 x = 62.5 ot = + 8.3 c2 •= - 2.7
P3 a = 50 ^ = + 7.9 es, = - 1.7
P4 cc = 37.5 d = + 7.3 <fe = - 0.61
P5 z = 25 ^ = + 6.7 c2 = + 0.8
P6 ,7? = 12.5 d = + 5.8 e2 = + 2.5
P7 x = 0 cl = 4- 4.8 a* = + 4.8
P8 #=— 12.5 G! = + 2.5 <?2 = + 5.8
P9 «rrz_25 ^=+0.8 c2 = + 6.7, etc.,
negative values of c being laid off above the ends of centre
line.
We can therefore easily construct the left reactions, as
explained in Art. 160, Fig. 92. We thus obtain
Pt = 0.5, P2 = 0.9, P3 = 1.2,
U, = 4.4, H2 = 8.6, H3 = 13.1,
M! = + 38.3, M2 = + 71.4, M3 = + 103.5,
P4 = 2.3, P5 == 3.1, P6 = 4.4,
H4 = 15.9. H5 = 18.9, H6 = 21.1.
M4 = + 116.1, M5 = + 126.6, M6 = + 122.4,
P7 = 5.0, P8 = 5.6, P9 = 6.9,
H7 = 22.4, H8 = 21.1, H9 = 18.9,
M7 = + 107.5, M8 = + 52.7, M9 = + 15.1,
P_77 p=88 P=91 P=95
H10=15.9, Hn = 13.1, HM = 8.6, H1S = 4.4,
Mj0=-9.7, Mu=-22.3, M^ = — 23.2, M13 = - 15.3.
A positive moment indicates tension in the lower flange at
abutment, and compression in upper.
396 NOTE TO CHAP. XIV. [APPENDIX.
We can therefore easily calculate the strains in. flanges A
and E for each weight.
Thus, for u
A x 9.7 = 8.8 x 2.2 + 13.1 x 4.5 - 22.3,
or A = + 5.7.
E x 10.4 = - 8.8 x 12.5 + 13.1 x 10.8 + 22.3,
or E — + 5.1.
The strain diagram can then be commenced, as shown in dia-
gram (<?), Fig. X., and explained above. We can find by cal-
culation the flange H, and thus check our diagram.
Proceeding thus, we obtain the following table of strains :
APPENDIX.]
THE BRACED AECH.
1
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1
598 NOTE TO CHAP. XIV. [APPENDIX.
The strains in the present case are, we see, much greater
than for any of the others. Unless the maximum of stiffness
is essential, it would appear, then, undesirable to fix the arch at
the ends for an arch of the above dimensions.
27. Temperature Strains. — The strains due to tempera-
ture are also very great. Thus, from Art. 165, we have
45 EIA et
H =
4 A A2 + 45 I
and for the distance of the point of action of this thrust, below
the crown of the centre line,
_ (A a2 + 6 I) h
~ 3 A a* >
or since - = g2 = 25 ft.,
A
_ _
~
_ h(a?
For A = 60 square in., a = 87.5 ft., h = 20 ft. = 240 in.,
g = 60 in. = 5 ft., e = 0.000012, E = 14,000 tons, t = 30°,
we have
H = 125 tons and e0 = 6.7 ft.
Hence we have the strains
A = ± 228, C = d= 29.0, E = =F 30.0, G = ± 118.5,
B = ± 112.5, D = =p 13.0, F = =F 65.0, H = ± 137.5,
12=^96.0, 34=^67.0, 5 6 ='T 29.5,
2 3 = d= 51.0, 4 5 = ± 39.5, 6 7 = ± 25.0, 7 8
Therefore the total strains are
A + 48o.4 -D + 318.3* /-, , -ic\-i K T\ , i K-I c\
XI.
— 142.9
~ - 43.9
" '
E =
+ 108.2
- 5.7
•p + 185.4
~ - 28.1
r + 242.1
~ - 77.4
w + 284.7
~- 88.4
12 =
+ 103.8
- 167.4
2 o + 86.3
- - 63.4
34-+ 72'3
- 116.0
A K + 67.9
-45.3
5 6 =
+ 46.2
-46.3
ft » + 51.6
6 7 = _ 35.3
7 8 - + 22'1
7 8 - _ 31.7
With the above we close our discussion of the braced arch.
Our design has been to illustrate the application of the for-
mulae and methods of Chapter XIY., and to show that by their
aid such a structure can be calculated with ease and certainty.
APPENDIX.] THE BRACED AKCH. 399
In short, the difficulty is but little if any greater than for a
simple girder, only for a long span and many panels the work
becomes tedious and wearisome.
In such a case, perhaps the method of moments will be found
preferable to diagrams. Thus, for any condition of loading, we
can easily find the strains at certain given intervals or portions
of the span, as TVth, T20-ths, etc- These strains being plotted to
scale along the span, we have a curve, from which we can
readily determine the strain at other points.
The strains in the flanges being thus known, we can readily
determine the transverse force, or force at right angles to the
rib, at any point. This force causes strain in the diagonals, and
has simply to be multiplied by the secant of the angle made
with it by any diagonal.
As to the effects of temperature, the remarks of Art. 166 do
not seem to be substantiated by our results. It would seem
that, according to the received formulae, the strains due to tem-
perature are very great, and that by far the best form of con-
struction for short spans is that in which the arch is hinged at
both abutments and crown.
28. Advantage of Arch with fixed Ends for long Spans.
We cannot conclude from our results above anything as to the
comparative advantages or disadvantages of the arch with fixed
ends. Different proportions will give altogether different re-
sults. We can only say that for small spans the arch with
three hinges is undoubtedly the best construction. The ad-
vantages of continuity will be apparent only for long spans
where the point of inflexion is distant from the ends by a
greater proportion of the span. We have already seen the
same to be true of the continuous girder. If we were to judge
from comparisons of short spans only, we should be inclined
to discredit any great advantage for continuity. If, however,
we take longer spans, so as to bring the points of inflection well
out, we find a marked saving.*
We had intended to give here a comparison of the strains in
a hinged arch with those in the central span of the St. Louis
bridge, as given in the Report of Capt. Eads to the Illinois and
St. Louis Bridge Co. for May, 1868.
As this goes to press, however, our attention has been called
* Art. 17 of this Appendix.
400 NOTE TO CHAP. XIV. [APPENDIX.
to an article in the Trans, of the Am. Soc. of Civil Eng. for
May, 1875, by Mr. S. H. Shreve, which, although written with
precisely the opposite intention, seems to prove so clearly the
superiority for long spans of the arch without hinges, that it is
unnecessary to give a comparison here. We have only to take
Mr. Shreve's results and properly interpret^ them.
Thus, while ostensibly investigating the strains in the centre
arch of the St. Louis bridge — an arch which is continuous at
crown and fixed at the end — Mr. Shreve uses the formula given
Q
in Art. 27 of the Supplement to Chap. XIV., viz., H = TTT'
That is, he considers the arch as having hinges at both crown
and ends.
Then, supposing the arch to be affected by temperature, he
applies the above formula to an arch hinged at crown in lower
chord and at ends in upper chord of the same dimensions as
the St. Louis bridge. It is hardly necessary to point out here
that if the arch is really thus hinged, or can be supposed thus
hinged, there can be no temperature strains. If, however, it is
not hinged, then the above formula does not apply. The one
assumption contradicts the other. The formula H = ~. can
2i fi
be applied to no arch which is strained by temperature. Such
a treatment would seem justified on Mr. Shreve's part in view
of the statement of Capt. Eads, that for the greatest rise of
temperature above the mean, the lower arch does all the duty
at crown, and the upper at the ends. If this were accurately
so, then Mr. Shreve's results would give the true strains. All
that Capt. Eads evidently intended to imply was, that a rise of
temperature relieved the upper chord at crown of a great part
of its compression and increased that of the lower. It does not
by any means follow that the upper chord is entirely relieved,
under which supposition only can the lower chord be supposed
hinged. On the contrary, for an equal fall of temperature
below the mean, the lower chord is relieved and extra strain
brought in the upper chord at crown. If the adjustment were
just such that the previous compression in the lower chord
should be exactly neutralized, then the arch might be consid-
ered as hinged at the upper flange and lower ends, and thus
Mr. Shreve should increase the rise of his arch by the depth,
APPENDIX.] THE BRACED ARCH. 401
which would decrease greatly his strains. The one supposition
is as much justified by the remarks of Capt. Eads, which he
quotes, as the other — and neither are correct. Apart, however,
from the merits of the controversy, with which we have nothing
to do, Mr. Shreve's results are undoubtedly correct for an arch
of the same dimensions as the St. Louis — uniformly loaded and
hinged at the ends in upper flange and at the crown in lower.
If, then, a comparison of these results with those given by Capt.
Eads shows them all too large, then, since Capt. leads' formulae
are, as we have seen, undoubtedly correct, it clearly shows the
superiority of the arch without hinges. This is the only legiti-
mate deduction which can be made.
Mr. Shreve's formulae are undoubtedly as " true as the prin-
ciples of the lever," and apply, beyond question, to an arch
hinged as he supposes. Our formulae in Art. 27 of the Sup-
plement to Chap. XIY. are also as true as these principles ; but
to apply correctly even so simple a principle as that of the
lever, demands a knowledge of all the forces and their points
of application. From our formulae, as we have shown in Art.
34 of the above Supplement, we may easily deduce Capt. Eads',
thus proving the accuracy of both. Though the " calculus will
not determine the strains affecting a truss, whether arched or
horizontal," it may nevertheless be exceedingly serviceable in
determining the forces which act upon the truss — without an
accurate knowledge of which the '"'principle of the lever" can
only mislead. This principle, upon which Mr. Shreve lays such
stress, is precisely that which we have employed so often in
this work, and s'hown to be of universal application. In Art.
36 of this Appendix we have made use of it, just as Mr. Shreve
does, in the calculation of an arch similar to the St. Louis.
Our results differ from those he would obtain, simply because
we take into account a force and lever arm whose existence he
ignores. Mr. Shreve assumes that V and H and the load are
all the forces which act, and these are all of which his formula
takes account. In common with Capt. Eads, we take in addition
a moment due to the continuity of the ends, while V and H them-
selves, by reason of this continuity, have very different values.
Thus, for full load, we have from eq. (81), Art. 34 of Sup-
plement to Chap. XIY.,
u__pa? 4 A2
~ 2A 45
26
402 NOTE TO CHAP. XIV. [APPENDIX.
and from eq. (84)
Mfl- -
70
Thus, instead of H = ^-— . , as given by Mr. Shreve, we have
this into a certain coefficient which is less than unity.
Taking pa = 936,000 Ibs., g = 6.025 ft., a = 257.88 ft,
and h — 46.65 ft. for centre line, we have H = 2,178,317 for
thrust at crown, instead of 2,586,184.9 Ibs., as given by Mr.
Shreve. This thrust alone would cause, then, 1,089,158 Ibs.
compression in each flange. But due to continuity of ends and
crown, we have also a moment at crown M0 — — 6,587,335,
which being negative causes tension in lower flange at crown.
Dividing by 12.05, the depth of arch, we have 546,666 Ibs.
tension, and therefore only 1,089,158—546,666 or 542,492 Ibs.
resulting compression. This at 27,500 Ibs. per square inch,
requires 19.72 square inches area, while Mr. Shreve requires
in his arch 126.42 square inches area. It is, however, but just
to notice, that while this loading (uniform) causes the maxi-
mum compression in lower flange at crown for Mr. Shreve's
arch, it does not for the arch fixed at ends and continuous at
crown.
In this latter case, as we may see at once from the table for
M0 of Art. 18, Supplement to Chap. XIV., a load within the
centre half anywhere causes tension in the lower flange, and
the maximum compression is when the flanks are loaded and
this portion is empty. It is with the maxima that the com-
parison must be made, and as Capt. Eads has, very properly,
taken the rolling load into account, it is with these maxima
that the comparison has been made. From such comparison
Mr. Shreve h'nds that " every member of the two tubes is
deficient in area, many containing much less than half the
material that is necessary." As his results are correctly calcu-
lated for a hinged arch, and Capt. Eadsr results are also correct
for an arch without hinges, we can only conclude — not that
" the great importance of immediately strengthening the ribs
of the St. Louis bridge can no longer be ignored," but rather
that, for long spans of small relative rise, the arch without
hinges is much preferable and more economical. The case
APPENDIX.] THE BRACED ARCH. 403
is, indeed, perfectly analogous to that of the continuous girder.
Here also we have end moments, and here also for long spans
the advantage over the simple girder is marked.
In Mr. Shreve's arch it is, indeed, perfectly true that, " when
one segment is loaded, any weight whatever in any other
position on the other segment will lessen the tension on the
lower arc of the loaded segment." In the arch without hinges
the case is altogether different, owing to the influence of the
end moments, which Mr. Shreve so persistently ignores.
The two cases have, indeed, nothing whatever in common,
and from the strains in one no conclusion whatever can be
drawn as to what should be the strains in the other. ' With the
same propriety might one comparing the strains in the same
girder fixed at ends and free at ends, as given in Art. 17 of
this Appendix, infer that the strains in the first were unduly
small. The only legitimate conclusion from such comparison
is the one there drawn, viz., that the one in which the strains
are least is the one most economical of material. In this
respect, and in this only, Mr. Shreve's results are valuable, and
we can only thank him for having saved us the labor of
making the comparison for ourselves.
As a case in point bearing out our conclusion above, we
may instance the ' Coblentz bridge, which, as originally con-
structed, was continuous at the crown, but pivoted at the ends
of the centre line, as in our example, Art. 20. But unlike
that example, owing to the length of span being much greater,
and the rise and depth much less in proportion, it was found
advantageous to block up the ends after erection, and thus fix
it at the ends.
If Mr. Shreve's deductions are to be believed, this was a
very dangerous thing to do; but, as experience has proved,
greater rigidity has thereby been secured, and no evil effects
have as yet been perceptible. It is, however, quite possible
that before thus blocking the ends, the effect of the end
moments thus brought into play was duly considered; and in
view of the result, it would appear as if they really had some
influence upon the character and distribution of the strains.
It would seem, therefore, that, for the present at least, the
" strengthening " of the arches of the St. Louis bridge by hinging
them (!) at crown and ends may be safely postponed until it
can be satisfactorily shown in what manner, for rise of tern-
4:04 NOTE TO, CHAP. XIV. [APPENDIX.
perature, the end moments mysteriously disappear, and the
previously existing compression, due to load in upper flange at
crown and lower at ends, is exactly and entirely neutralized.
Meanwhile it would seem that the St. Louis arch, as con-
structed, is far superior to the same arch hinged, more eco-
nomical of material and more rigid, and sanctioned alike by
theory and precedent.
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