Engineering Library
This volume is damaged or brittle
and CANNOT be repaired
Handle with EXTREME CARE
iStiiy
j
illllllilllllli<i>H
2
ELEMEXTAKY
APPLIED MECHANICS
.Cs?
MACMILLAN AND CO., LiMiTEr.
LONDON • BOMBAY • CALCUTTA
MELBOURNE
THE MACMILLAN COMPANY
NEW YORK  BOSTON • CHICAGO
DALLAS • SAN FRANCISCO
THE MACMILLAN CO. OF CANADA, Ltd.
TORONTO
ELEMENTARY
APPLIED MECHANICS
T. ALEXANDER, C.E., M.Lvsr.C.E.L
M.A.I, [hoii. causa)
4th okdku meiji, j.\pan, 18S7, kogaku hakushi, 1915
PKOFESSOll OF EXGINKEKING, TUINITY COLLEGE, DUBLIN
AXD
A. W. THOMSON, D.Sc.
EMERITUS PKUFESSOK OK ESGI.NEERIXG. COLLEGE OF SCIENCE, POONA
TFJTH NUMEROUS DIAGRAMS, JNJJ A SERIES OF GRADUATED
EXAMPLES CAREFULLY WORKED OUT
MACMILLAN AND CO., LIMITED
ST. MARTIN'S STREET, LONDON
1916
COPYRIGHT.
FIRST EDITION.
Part I. By Thomas Alexander. Crown Svo, 1880.
Tart II. By Tlioinas Alexander and Arthur Watson Thomson.
Crotvn Sro, 1883.
SECOND EDITION. In one vol., Sro, 1902.
THIRD EDITION. In one vol., Sro, 1916.
f^ ^^ \,*'
William John Macqiorn Raxkine, ll.d.
From the Preface to La Statique Graphiqtie.
Bv ^I. Maurice Lew.
1st edition.
" Au moment meme oil nous ecrivons ces lignes, nous recevons de la famille
de M. Alacquorn Rankine une lettre de faire part de la mort de I'eminent
professeur de I'Universite de Glasgow. Qu'il nous soit permis de lui ofiFrir ici
notre tribut de regrets. Sa perte sera ressentie nonseulement par les hommes
de science, mais aussi par les ingenieurs et les constructeurs ; car ses recherches
ont prcsque toutes un caractere utilitaire. Son Manuel of applied Mechanics
notamment est le digne prolongement des Lecons sur la Mecanique industrielle
de Poncelet." (Mai 1875.)
TO THE MEMORY OF
DE. WILLIAM JOHX MACQUORN EANKINE
LATE PROFESSOR OF CIVIL ENGINEERING AND MECHANICS
IN THE IMVERSITT OF GLASGOW
THOMAS ALEXANDER
AND
ARTHUR WATSON THOMSON
BY THE SAME AUTHORS.
A SET OF GRAPHICAL EXERCISES, quarter Double Elephant
size, with Skeleton Data for the Student to practise upon. Many
are printed in two colours. An Essay forms a running com
iiientary on the Exercises, and deals with Reciprocal Figures.
Piice OS.
GRAPHIC STATICS. 8vo. 2?.
PREFACE.
This third edition has the honour to be induded in the Dublin
University Press Series.
Chapter XI is new work, dealing with a Standard Loco
motive on Short Girders. It was read to the Koyal Irish
Academy on June 24th, 1912, and includes a description of our
Kinematical Model by J. T. Jackson, m.a.i.
In the Analysis of the Warren Girder, Chapter XXI, we
have replaced Levy's rolling load by an uniform advancing
load.
The new Chapter XXIII shows diagrams of Eoof Stresses
reduced from the Exercises advertised opposite the titlepage,
with excerpts from the Essay.
The work forms an elementary consecutive treatise on the
subject of Internal Stress and Strain, based on the late
Professor Eankine's treatment of the subject in his Applied
Meclicmics and Civil Engineering. The end kept in view is the
Scientific and Practical Design of Earthworks, of Linkwork
Structures, and of Block work Structures. The whole is illus
trated by a systematic and graduated set of Examples. At every
point graphical methods are combined with the analytical, and
a feature of the work is, that the diagrams are to scale, and,
besides illustrating the text, each diagram suits some of the
numerical examples, having printed on its face both the data
and results. For the Student with little time to draw, the
fullpage diagrams should prove useful models, furnishing con
cisely the data and checking the results of his constructions to
a bold scale without delay.
X PREFACE.
In Chapter II a Moving Model of Eaukine's Ellipse of
Stress is shown at p. 49. It was exhibited to the Eoyal Irish
Academy. The Scientific Design of Masonry Retaining Walls
and of their foundations in Chapter IV is an extension of a
paper by the authors in Industries of 14th September, 1888.
The Rules given in Chapter X, p. 196, for the Maxima
Bending Moments caused by a locomotive crossing a bridge
were published in Engineering on January 10th and July 25th,
1879, in response to queries in Du Bois' early treatise on
Graphic Statics, and are quoted by him in his more recent work,
Strains in Framed Girders, 1883. Mention is made of them also
in the preface to Levy's La Gh^aphique Statique, second edition.'
Hele Shaw's Report on Graphic Methods to the Edinburgh
Meeting of the British Association for the Advancement of
Science refers to our use of the parabolic setsquare for the
rapid construction of bending moment diagrams, as introduced
in the first edition. We now reduce the solution of this impor
tant problem to the construction of a Diagram of Square Eoots
of Bending Moments, with arcs of circles only (see p. 188, and
for a cut of our Moving Model of a locomotive on a girder, see
pp. 175, 222. Two of those are figured in the correspondence
on Mr. Farr's paper on Moving Loads on Bridges published in the
Proc. of the Inst, of C. E., vol. cxli, 1900). The arcs of circles
replacing the parabolic arcs are also to be seen at page 146 for
the more general cases of fixed loads.
At the end of Chapter XV, on Stress at an Internal Point
of a Beam, we have printed an extract, by the kind permission of
the Council of the Phil. Soc. of Glasgow, from the late Professor
Peter Alexander's paper on The Uses of the Polariscope in the
Practical Determination of Internal Stress and Strain. The lines
of stress for a bentglass prism as drawn by his mechanical
pen in conjunction with the polariscope lantern are shown.
1 " Nous saisissons, avec plaisir, cette occasion tie mentionner les Ouvrages ou
M^moires suivants, parvenus a notre connaissance depuis la redaction de la
Prefiice qui precede : —
" Un tr^s beau travail de M. le professeur Thomas Alexander sur le problenie
du passage d'un convoi sur une poutre a deux appuis simples, public dans les
numeros des 10 Janvier et 25 juillet 1879 de V Engineering et dans son excellent
Ouvrage Elementanj applied Mechanics, publie en conimun avec M. le professeur
A. W^atson Thomson."
PREFACE. Xi
In Chapter XVII. the solutions given for the uniform
girder fixed horizontally at the ends, and subjected to the
transit of a concentrated rolling load in one case, and of a
movinti uniform load in the other case, will be found interestin2
(see pp. 321 and 327). The most recent graphical treatment
is followed, and the symmetrical form in which the results are
shown is new. The analytical result for the rolling load agrees
with the unsymmetrical results given by Levy and Du Bois,
but we have not seen anywhere an attempt at the solution of
the second and more practically important case of the transit of
the uniform load shown at p. 327.
In Chapter XIX we have greatly extended the part
on Long Steel Struts, bringing it up to the most recent
practice. We quote two of Fidler's tables for their design. On
p. 357 we give a formula (8) for the immediate design of the
economical doubletee section of required strength and required
local stiffness. Also on the table, p. 360, at iv., we give a close
approximate expression independent, like the others, of the
thickness of the metal, for the square of the radius of gyration
of the teesection, or angleiron constrained to bend like it, as
it usually is. This is an important addition to Rankine's list, as
these sections are of everyday occurrence.
The Steel Arched Girder in Chapter XX is treated by
Levy's graphical methods. Two numerical examples, one
hinged at the ends and another fixed at the ends, are worked
out in detail, and the scaled results written on the diagrams
(see pp. 376 and 380). In Chapter XXI we follow, on the
other hand, his beautiful analysis of the Triangular Trussing,
involving the three variables, the number of subdivisions in
the span, the form of the triangles, and the ratio of the depth
to span. The Table of Volumes of Trusses we show at one
opening of the book, and the minimum volumes are placed
among the others with a heavyfaced type in such a way that it
can be seen whether they occur at depths that can be adopted
consistent with the more important requirement of stiftness.
In Chapter XXII we have developed the method of con
jugate load areas, given by Eankine in mathematical form
difi&cult of practical application, to the equilibrium of arches.
Xll PREFACE.
We substitute a semigraphical method of constructing the load
areas, reducing the mathematics and extending the whole to
the complete design of segmental, semicircular, and semielliptic
arches, with their abutments, spandrils, and piers. Incidentally,
the design of sewers, inverts, shafts, and tunnels illustrates the
full scope of the method.
In a paper to the Eoyal Irish Academy, referred to at
bottom of p. 448, we demonstrated the true shape of the
equilibrium curve, dividing the family into two groups, the
more important of which we venture to call twonosed
catenaries. We then show that these two varieties ofifer a
philosophical explanation of the two distinct ways in which it
has been found by experiment that masonry arches break up
when the abutments are gradually removed.
Some of the leading writers in America upon Engineering
expressed approbation of our method, and Professor Howe, in
his treatise, referred to at bottom of p. 447, has done us the
honour of adopting it, and calls it the best method yet pub
lished, when the arch bears only vertical loads. In our present
treatment we have both vertical and horizontal loads duly
considered, but have insisted upon a central elastic portion of
the archring being wholly free from other than vertical loads,
except when the live load covers only half the arch when the
horizontal reaction of the light elastic spandrils of the opposite
side comes into play. Now, with the catenary tables, the
segmental arches were limited to this elastic part only, and
so the assumption of vertical loads only is justified, and the
more especially as the horizontal load indicated in this case
is necessarily outwards, and cannot be introduced in any
practical way.
CONTENTS.
CHAPTEK I.
LIMEAL STRESS AND 8TKAIX. paok
State of Simple Strain — Exs. 1 to 6— Elasticity, Young's Modulus —
Exs. 7 to 12— The production of Strain— Exs. 13 to 17— Resili
ence — Exs. 18 to 31, 1
CHAPTEK II.
INTERNAL STRESS AND STRAIN, SIMPLE AND COMPOUND.
RANKINE'S METHOD OF THE ELLIPSE OF STRESS.
Internal Stress at a point in a solid in a simple state of Strain —
Do., for a compound state of Strain — Uniplanar Stress — Direct
Problem, the Principal Stresses given — Equal Like and Unlike
Principal Stresses — Positions of third Plane, for maximum
Normal Stress, for maximum Shear, for maximum Obliquity —
Exs. 1 to 9 — Inverse Problem, to find the Principal Stresses —
Moving Model illustrating the Method— Exs. 10 to 13, . . 20
CHAPTEK III.
APPLICATION OF THE ELLIPSE OF STRESS TO THE STABILITY OF
EARTHWORK.
Friction among loose Earth Particles — Conditions of Equilibrium
and Angle of Repose — Earth spread in horizontal layers behind
a wall — Depth of Foundation — Earth spread in sloping layers
behind a wall — Construction of the Auxiliary Figure to the
Ellipse of Stress — Exs. 1 to 6, . . . . . .53
CHAPTEK IV.
THE SCIENTIFIC DESIGN OF .MASONRY RETAINING WALLS.
Practical considerations — Rectangular and Trapezoidal Walls — Do.
do. surcharged — Battering Wall with Stepped Back — Spread
and Depth of the Foundation of a Wall — Table of Thicknesses
required for Masonry Retaining Walls of moderate heights —
Profiles of Do. — Graphic Statics, composition of Plane Forces —
Graphical Methods of Designing Retaining Walls — Exs. 1 to 6, . 67
XIV CONTENTS.
CHAPTEK V.
TKANSVERSE STRESS.
Beams or Girders — Varieties of Loads — Conditions of their Equili
brium — Exs. 1 to 13 — Neutral Plane and Axis — Deflection,
Slope and Curvature — Elements of the Stress at an Internal
Point of a Beam — Resolution into Shearing Force and Bending
Moment — Resistance to Shearing and Bending — The Cantilever,
CHAPTEK VI.
THE PARABOLA.
Straight and Parabolic Slopes — Graphic Construction of Parabolic
Segment and its Tangent — Parabolic Template, its use like a set
square — Equations to arcs drawn with the Template in various
positions — Compounding Straight and Parabolic Slopes, . 104
CHAPTER YII.
BENDING MOMENTS AND SHEARING FORCES FOR FIXED LOADS.
Definition of Bending Moment and Shearing Force Diagrams — Beam
loaded with Unequal Weights at Irregular Intervals — Analytical
and Graphical Solutions— Cantilever with irregular Loads —
Beam with Load at centre or uniform Load — Cantilever do. do. —
Beam with equal Loads at equal intervals — Exs. 1 to 14, . . 115
CHAPTEK VIII.
BENDING MOMENTS AND SHEARING FORCES FOR COMBINED
FIXED LOADS.
Beam with Uniform Load and Load at Centre — The Continuous
Beam — Irregular Fixed Loads combined with an Uniform Load —
Diagram of Squareroots of Bending Moments for do. by means
of Arcs of Circles only —Partial Uniform Loads— Beam loaded
in any manner with Fixed Loads— Exs. 1 to 9, . . . 139
CONTENTS. XV
CHAPTER TX.
BENDING MOMENTS AND SHEARING FORCES FOR MOVING LOADS.
PACB
Classes of Moving Loads— Beam subjected to an uniform advancing
Load as long as the Span— Do., Load shorter than the Span
Beam subjected to a Rolling Load — Two wheeled Trolly confined
to a Girder like a Travelling Crane, Loads on the wheels equal,
Loads unequal — Moving Model illustrating the Bending Moments
on a Girder Bridge due to a Locomotive passing over it — Exs.
1 to 12, 150
CHAPTER X.
BENDING MOMENTS AND SHEARING FORCES DUE TO A TRAVELLING
LOAD SYSTEM.
Maximum Bending Moments for a beam under a Travelling Load
system of unequal weights fixed at irregular intervals apart,
the Load confined to the span — "Fields" commanded by the
different Loads — Nature of the Bending Moment Diagram and
its Graphical Construction with one Parabolic Template — How
the Scale is readily determined — Graphical Construction of a
Diagram of Squareroots of Bending Moments by means of
Circular Arcs only — Shearing Force Diagram — Particular Sys
tems of Equal Loads on wheels equally spaced — Exs. 1 to 5, . 182
CHAPTER XL
ON THE GRAPHICAL CONSTRUCTION OF MAXDIUM BENDING
MOMENTS ON SHORT GIRDERS DUE TO A LOCOMOTIVE.
Reference to Farr's paper — A typical Locomotive — First Method,
with one parabola only — Second Method, with circular arcs only
— Third Method, being Cul man's rendered precise — Loco, and
Dead Load combined — Kinematical Model — Example of the
Uniform Load combined with Travelling Load .system, . . 198
XVI CONTENTS.
CHAPTER XII.
COMBINED LIVE AND DEAD LOADS WITH APPROXIMATION BY MEANS
OF AN EQUIVALENT UNIFORM LIVE LOAD.
PAGE
Reduction of Live Loads to an equivalent uniform Dead Load —
Shearing Force Diagram for a beam with uniform Dead and
Live Loads — Its importance in the matter of counterbracing —
The Live Load reduced instead to a single equivalent Rolling
Load — Mr. Farr's paper on "Moving Loads on Railway under
Bridges" — His allowance for the Impulse of the Moving Loads,
with Tables— Exs. 1 to 9, 224
CHAPTER XIII.
RESISTANCE, IN GENERAL, TO BENDING AND SHEARING AT THE
VARIOUS CROSSSECTIONS OF FRAMED GIIiDERS AND OF
SOLID BEAMS.
Graphical application of Ritter's method of sections to the construc
tion of the Stresses on the booms and braces of Framed Girders
— Elevations of Flanged Girders of Uniform Strength — Solid
Beams — Their Moments of Resistance to Bending for the Cross
Section either rectangular or triangular — Rectangular Beams of
Uniform Strength and Uniform Depth— Do. do. of Uniform
Breadth — Approximate Plans and Elevations — Exs. 1 to 10 —
Area, Geometrical Moment, and Moment of Inertia — Theorems
regarding, and their representation by, the Volume and Statical
Moment of Isosceles Wedges, ....... 234
CHAPTER XIV.
CROSSSECTIONS : THEIR RESISTANCE TO BENDING AND SHEARING
AND DISTRIBUTION OF STRESS THEREON.
Rectangular and Hollow Rectangular Sections — Tabular Method for
Sections built of Rectangles— Graphical Construction of an
approximation to the Moment of Inertia of a Section built of
Rectangles — Absolute Correction in same by a further extension
— Triangular CrossSection and Sections partly built of Triangles,
Hexagon, Diamond, Trapezoid — Circular and Elliptic Cross
Sections — Resistance of CrossSections to Shearing and the Dis
tribution of the Shearing Stress thereon, including the Hollow
Rectangle, Doul)le TSection, Triangle. Circle— Exs. 1 to 20, . 252
CONTENTS. xvii
CHAPTER XV.
STRESS AT AX INTERNAL POINT OF A BEAM.
PAGB
Curves of Principiil Stress — Detenninjitioii of Internal Stress by the
Polariscope — Professor Peter Alexander's Method of drawing
the Lines of Stress with a mechanical pen on a revolving disc
upon which are thrown, by means of a lantern, the dark lines
due to a strained prism of glass rotating in the polariscope
attached to the lantern — Exercise, . .... 286
CHAPTEPt XVI.
CUrVATURE, SLOPE AND DEFLECTION.
Definition, depending on the skin coming to the proof strain at the
CrossSection where the bending moment is greatest — Beam
hinged at ends. Section uniform, Load at the centre and again,
Load uniformly spread — Two Loads symmetrical about the
centre, as a pair of locomotive wheels on a crossgirder — Canti
lever — Beams and Cantilevers of uniform strength — Proportion
of the depth of a beam to its span dominating the stiifness of the
beam — Slope and Deflection due to a given Load less than Proof
Load — Beam supported on three props — Uniform beam uniformly
loaded and fixed at one or both ends — Virtual and actual Hinges
— Exs. 1 to 12 293
CHAPTER XVII.
FIXED AND MOVING LOADS ON A UNIFORM GIRDER WITH ENDS
FIXED HORIZONTALLY.
A. Fixed Load placed unsymmetrically on the span — Theorems : (a)
The sum of the tips ^qj at the ends, had they been hinged, pro
portional to the Area of the BendingMoment Diagram and (6)
their ratio to each other inversely as the ratio of the segments
into which a perpendicular from the centre of gravity of that
Area divides the Span — Transit of a Rolling Load — Diagram of
the Maximum positive and negative Bending Moments at each
point of the Span, due to the transit — The corresponding posi
tions of the Rolling Load — Maximum of maxima at nearer
Abutment, when the Load trisects the Span — References to
Levy's La Statiqiie Graphiqxie — Shearing Force Diagram as
XVlll CONTENTS.
PAGE
modified due to the ends being fixed — Transit of an Uniform
Advancing Load — Diagram of Maximum positive and negative
Bending Moments and corresponding positions of the Load —
The Modified ShearingForce Diagram — Exs. 1 to 4, with a full
page diagram to scale, . . . . . . . . 31 6
CHAPTEK XVIII.
SKELETON SECTIONS FOE BEAMS AND STRUTS — COMBINED THRUST
WITH BENDING AND TWISTING WITH BENDING — LONG STRUTS.
Formulae for the Strength of thin hollow CrossSections of Beams —
CrossSections of equal strength, Rankine's approximate formula
for their design in castiron, the difference of the strengths being
great — Exs. 1 to 4 — Allowance for Weight of Beam— Resistance
to Twisting and Wrenching — Bending and Torsion combined —
Thrust or Tension combined with Torsion — Thrust and induced
Bending — Table of Loads on Pillars — Table of Breaking Loads
for castiron and wroughtiron Struts for ratios of 1 : h from 10
to 50— Exs. 5 to 19, . . 335
CHAPTEK XIX.
DESIGN OF LONG STEEL STRUTS.
The Economical Double T Section of Uniform Strength to resist
Bending — Formulas for its Direct Design with given fractional
thicknesses of Web and Flanges to secure a prescribed degree
of Local Stiffness — RankineGordon Formula — Table of Areas,
Moments of Inertia, and Radii square of Gyration, giving close
approximations for the Practical Application of the Rankine
formula to Struts of box fianged and T Section — Quotation of
two of Fidler's tabulated results of the formula — Exs. 1 to 8,
including a fullpage diagram. 354
CHAPTER XX.
THE STEEL ARCHED GIRDER.
As adapted to Roofs — Dimensions of, and test Loads prescribed for,
the St. Pancras Station Roof — Reference to Central Station
Roof, Manchester Machinery Halls, Paris and Columbian Exposi
tions, and other Roofs, illustrating different outlines as circled.
CONTENTS. xix
PIGB
parabolic. &c., and both with and without Hinges— The adapta
tion to Bridges, Niagara Falls Bridge, Oporto Bridge, and
others — Disposition of Loads^ Levy's Graphical Stress Diagram
— The curve of benders and the curve of ttatteners — Weight of
metal relatively to three types— Rib of Uniform Section, hinged
at the ends — Rib of Uniform Stiffness fixed at the ends — Two
pair of fullpage diagrams, to scale, with the Stresses scaled off
in detail, 361
CHAPTEK XXI.
ANALYSIS OF TRIANGULAR TRUSSING ON GIRDERS WITH
HORIZONTAL PARALLEL BOOMS.
Analyses following Levy in the main — His notation for the Triangu
lated Truss — The Dead Load — The Live Load — General expres
sions for the volume of steel required to resist these Loads —
Particular expressions for the Warren and Rectangular Trusses —
Discussion of the economical shapes of Triangle and of the
economical depth which is consistent with that demanded for
joint general and local stiffness — The Fink and Bollman Trusses
— Tables of the theoretical volumes of those Trusses, showing at
one opening of the book their relative advantages — Allowance
to stiffen the long Struts — Exs. 1 to 6, . . . . . 384^
CHAPTEE XXII.
THE SCIENTIFIC DESIGN OF MASONRY ARCHES.
The Aichring — Elastic portion from crown outwards to two joints,
practically the joints of Rupture — The segmental and semi
elliptic (false) as rival designs: their outstanding characters —
Line of Stress confined to a " Kernel " of the Archring — Its
equilibrium and stability — Strength for Granite, Sandstone, and
Brick rings — The balanced linear rib or chain — Conjugate Load
areas — Superposition of Loads — Fluid Load of equal or varying
potential — Thrust at the crown of a rib — Buried Arches, Shafts,
and Sewers — Exs. 1 to 10— The Stereostatic Rib — Horizontal
Loadarea for semicircular Rib, loaded uniformly along the
Rib — Do., loaded with the area between itself and a horizontal
straight line over it — Allowance for excess weight of uniform
Archring — Do., for ring thickening outwards — Rankine's point
and joint of rupture — Authors' location and modification of it —
p
XX CONTENTS.
PAGS
Semi circular Masonry Arch — Equilibrium Rib or Transformed
Catenary — The twonosed Catenaries — Table of do., modulus,
unity — Tables for the immediate design of Segmental Arches in
granite, sandstone, or brick, with a minimum real factor of
safety of 10 and with the Line of Stress confined to a " Kernel,"
a middle third, fifth, or ninth — Exs. 11 to 20 — Linear trans
formation of balanced rib — Geostatic Load uniform or varying
potential — Approximate Elliptic Rib — Hydrostatic and Geostatic
Ribs — Semielliptic Masomy Arch — Exs. 21 to 25 — Abutments
and Piers — The Tunnel Shell — Allowance for excessdensity of
Elliptic Archring — Exs. 26 to 31 — Quotations from Rankine's
CE.y and from Simms' Practical Tunndling, .... 413
CHAPTEE XXIII.
THE METHOD OF llECIPKOCAL FIGURES.
A rival of Ritter's Method of Sections — Suitable for roofframes and
irregular fixed loads — Already used in Levy's constructions,
Ch. XX, figs. 2 and 4 — The simplest pair of reciprocal figures —
One treated as a selfstrained frame and the other as its stress
diagram, the reciprocity being perfect — General application to
an iron roofframe — To be statically determinate it is assumed
that the storm end, only, is anchored — Reactions at the supports
constructed by a Link Polygon — The Stress Diagram constructed,
using Levy's scientific notation — Anchor shifted to the lee end,
and a new stress diagram inscribed in the first by inspection —
These two diagrams are maximum and minimum strains ; in the
* second it is the wind that has shifted and the frame viewed
from the reverse side — Such a stress diagram is only a part of a
reciprocal figure — is always a part if the frame be just rigid or
indeformable — rule for predetermining this — The frame and
link polygon made into one geometrical figure joined by inter
cepts on the lines of action of the force ; they mutually strain
each other — This figure^ being freed from all idea of extern
forces, has one redundant bar and is selfstrained — The one and
only reciprocal figure to it is the stress diagram augment by the
vectors from the pole — The reciprocity is imperfect in the
reverse order, as the selfstrained frame is one of its many
reciprocal figures— Ex. I, Queenpost roof double braced in the
centre and with an extra load between the supports — Ex. II,
Great curved iron Kingpost roof — do., double braced like
King's Cross Ry. station roof— Ex. III. Compound roofframe,
pull on main tie implicitly given — Note on Ch. X, . . . 498
ELEMENTARY APPLIED MECHANICS.
CHAPTER 1.
LINEAL STRESS AND STRAIN.
Elasticity is a projierty of matter. When dealing with the
equilibrium of a body under the action of external forces,
in order to find the relations among those external forces,
the matter of the body is considered to be perfectly rigid, or,
in other words, to have no such property as elasticity. When
external forces, the simplest of which are stresses acting really
on a part of the surface of a body, are considered to act at points
on the surface, it is taken for granted that the matter of the
body is infinitely strong at such points. But after considering
the equilibrium of the body as a whole,
we may consider the equilibrium of all
or any of its parts. If we take a part
on which an external stress directly acts,
equilibrium is maintained between that
external stress acting on the free surface
and the components parallel to it, of stresses
which the cut surface of the remaining
part exerts on its cut surface.
Let MONQP be a solid in equilibrium
under the action of the three external
uniform stresses acting on planes of its
surface at 0, P, and Q. Let MN be the trace
of the plane at under the uniform stress A.
The stresses aaa . . . hbb . . . ccc may be represented in amount
and direction by the single forces A, B, and C acting at the
points 0, P, and Q, rigidly connected. We know that, by the
triangle of forces, A, B, and C are proportional to the sides of a
triangle DFE drawn with its ^ides parallel to their directions.
Also that they are in one plane and meet at one point. Hence
B
Fig. 1.
APPLIED MECHANICS.
[chap I.
Fig. 2.
we infer that the stresses which they represent are all parallel
to the plane of the paper, and that the planes of action of h and c
are at right angles to the plane of the paper as well as that of o..
Thus we find the relation among the external forces.
Let a plane mn divide the solid into two parts (fig. 2).
Consider the equilibrium of the part
MNmn. Sx, s^, S3 . . . are the stresses
exerted at all points of the cut surface
of MNmn by the cut surface of the
other part. S is the sum of their
components parallel to the direction
of A, acting through P, the centre of
pressure. Because there is equili
brium, S is equal and opposite to A ;
and they act in one straight line.
Also the remaining rectangular com
ponents of Si, 52, S3 are themselves in
equilibrium. Thus we see there is a
stress on the plane mn, and know the
amount of it in one direction.
Had we been considering the equilibrium of the other part
of the solid, the stresses on mn (fig. 3) would have been acting on
the other surface as t^, U, U,... in opposite directions to Si, So, S3,...
and of equal intensities. Thus on the plane mn there are pairs
of actions, acting on all points of it, as
53, ^3, at q. These vary in intensity and
obliquity to mn at different points of the
plane. If another plane, as gh, dividing
the solid, pass through q, there will be,
similarly, pairs of actions at all points
of it, and a pair of definite intensity and
direction at the point q. If we know the
stress at the point q in intensity and
direction on all planes passing through q, we are said to know
the intenud stress at the point q of the solid. Similarly for all
points of the solid.
The pairs of actions as S3, ^3 act respectively on the cut
surfaces of the upper and under parts of the solid ; but mn may
be considered to be a thin layer of the solid with S3 and ^3 acting
on its under and upper surfaces. This layer of the solid must
l)e, however, infinitely thin ; otherwise its two surfaces would be
different sections of the solid, and S3 and ^3 not necessarily equal
and opposite. If gh be also considered a thin layer, and^ and K
be the pair of actions on it at the point q on the two sides of it,
CHAP. I.]
LINEAL STRESS AND STRAIN.
Fig. 4.
thou will the point q be a solid, in figure a parallelepiped, with
a pair of stresses acting upon its opposite pairs of faces. S3 and t^
being equal, *S is now put for each,
and H is put for both H and K ; and
since q, instead of being a point in
both planes, has small surfaces in
both, though so infinitely small that the
stresses over them do not vary from the
intensities at the point q, yet surfaces,
the stresses spread over which it is
more convenient to represent by sets
of equal arrows SSS . . . , HHH ....
There are two convenient ways of representing by a diagram
the iTiternal stress at q, a point
within a solid. One, as in figure 5,
in which the indefi
nitely smallparallel
epiped q is all of the
solid to be imme
diately considered ;
and the other, as
in figure 6, in which
sheaves of equal ar
rows stand on small
portions of the planes mn and gh in the neighbourhood of q.
State of Simple Stkain.
Thus we see that, in a solid acted upon by external forces,
every particle exerts stress upon all those surrounding it.
Such a body is said to be in a state of strain. In solids the
phenomenon is marked by an alteration of shape, but not
necessarily of bulk, or of both.
Let AB and CD (fig. 7) be acted upon by two equal and
opposite forces P and P in the direction of their length acting
in AB away from each other, and in CD towards each other.
If P be uniformly distributed over the area A, the section of AB
perpendicular to its length, the intensity of the stress on it is
P =
P
2"
Let the prism be of unit thickness normal to the paper ; then
b2
APPLIED MECHANICS.
[chap. I.
will the line MN be equal to the area of the section of the prism
perpendicular to its axis, and
P
P
A
At any internal layer
the prism, the intensity
is also 7?, for the equi
librium of the parts
requires it ; and not
only is the stress of this
same intensity at all
points of one such sec
tion, but also upon all
such sections. The
solids AB and CB are
said to be in a state of
simple strain, in the
case of AB of extension,
and in that of CD of
compression. It is usual
to consider the first as
MN
perpendicular
I.
to the axis of
H
pv
"^
^^
3f
11
771
LLU
J!r
PV PV
■r:
PP
Ap
Ph pP
'P
Fig. 8.
.Fig 7.
positive and the second as negative.
The change of dimensions due to a simple state of strain is
an alteration of the length of the solid in the direction of the
stress with or without an accompanying alteration of its other
dimensions. Thus a piece of cork in a state of simple compres
sion has become shorter in the direction of the thrust, yet with
scarcely any, certainly without a corresponding, increase of area,
normal to the thrust. Again, a piece of indiarubber grows
shorter in the direction of the thrust with an almost exactly
proportionate increase of area normal to it.
The increase of length in the case of an extension is the
aiigmentation, in that of a compression it is a negative augmen
tation, and in either case it is the amount of strain. The
measure of the strain is the ratio of the augmentation to the
original unstrained length.
^ ^ . ,. , . augmentation of length
Definition. — Longitudmal stram = — ^ , . ^—
° length
where both are in the same name, that is, both in inches or
feet, &c. The foot being the vmit of length, it is most convenient
to take both in feet; then 
., ,. , , . augmentation in feet
longitudmal strain = — —. r—. — z —
° length in feet
CHAP. I.] LINEAL STRESS AND STRAIN. 5
Suppose the denominator on the righthand side of the equation
to be unity, then
longitudinal strain = augmentation in feet of 1 foot of
the substance
= augmentation per foot of length,
expressed in feet.
Hence the total augmentation or amovnt of strain in feet equals
the length in feet multiplied by the strain.
If the augmentation equal the length, that is, if the piece be
stretched to double its original length or compressed to nothing,
then from the definition
strain = unity.
Examples.
In the following questions the weight of the material is neglected : —
1. A tierod in a roof, Mhose length is 142 feet, stretches 1 inch when bearing
its proper stress. What strain is it subjected to ?
augmentation = 1 in.
unstrained length = 1704 in.
augmentation 1 „„„
strain = — ^^ ; = — — or '0006.
length 1704
2. How much will a tierod 100 feet long stretcb when subjected to "001 of
strain r
augmentation
— ^ ; = strain ;
length
.•. augmentation = strain x length = OOl x 100 ft. = •! ft.
3. A castiron pillar 18 feet high shrinks to 17"99 feet when loaded. "What is
the strain ?
augmentation of length = — "01 ft.
augmentation  01 ft. 1
strain = ,  = — = or  0005.
length 18 ft. 1800
4. Two wire cables, whose lengths are 100 and 90 fathoms, respectively, while
mooring a ship, are stietched, ttie first 3 inches and the second 275 inches. What
strains do they sustain ? Which sustains the greater ? Give the ratio of the
strains.
For the 100fathom cable
augmentation 3 in.
strain = , = — = 000417.
length 7200 m.
APPLIED MECHANICS. [CHAP. I
For the 90fathom cable
augmentation 2*75 in. „„^ „
strain = ^. ,,  = ^.^^ ■ = 000424.
length 6480 in.
The 90fathom cable is the more strained.
Eatio of these strains is 417 to 424.
5. A 30feet suspension rod stretches ^ inch under its load. Find the strain
upon it.
strain = 00014.
6. How much does another of them, which is 23 feet long, stretch when equally
strained ?
augmentation = 039 in.
Elasticity.
The elasticity of a solid is the tendency it has when strained
10 reg'ain its original size and shape. If two equal and similar
prisms of ditterent matter be strained similarly and to an equal
degree, that which required the greater stress is the more
elastic — e.g. a copper wire 1000 inches long was stretched an
inch by a weight of 680 lbs. while an iron wire of the same
section and length required 1000 lbs. to stretch it an inch.
Hence iron is more elastic than copper. If they be strained by
equal stresses, that which is the more strained is the less elastic
— e.g. the same copper wire is stretched as before 1 inch by
a weight of 680 lbs., while the iron one is only stretched a g^th
part of an inch by 680 lbs.
Hence the elasticities of different substances are proportional
to the stresses applied, and inversely proportional to the accom
panying strains.
If similar rods of steel and indiarubber be subjected to the
same stress, the indiarubber experiences an immensely greater
strain, so that steel is very much more elastic than indiarubber.
If two similar rods of the same matter, or the one rod
successively, be strained by different stresses, the corresponding
strams are proportional to the stresses. Thus, if a 480 lbs. stress
stretch a copper wire one inch, then a 960 lbs. stress will stretch
it, or a similar rod, two inches.
Hooke's Law is " The strain is proportional to the stress." It
amounts to — " the effect is proportional to the cause." It is
only true for solids within certain limits — e.g., 2400 lbs. should
stretch the copper wire mentioned above five inches by Hooke's
law, but it would really tear it to pieces ; and although 1920 lbs.
CllAI'. 1.] LINEAL STRESS AND STRAIN. 7
applied very gradually will not tear it, yet it will stretch it more
than four inches ; and further, when that stress is removed the
wire will not contract to its original length again. Strain and
stress are mutually cause and effect. The efi'ect of stress upon
a solid is to produce strain ; and, conversely, a body in a state of
strain exerts stress. The expressions " Strain due to the stress,"
&c., and " Stress due to the strain," &c., are both correct.
If a solid be strained beyond a certain degree, called the
proof strain, it does not regain its original length when released
from the strain ; in such a case the permanent alteration of
length is called a set.
Def. — The Proof Load is the stress of greatest intensity
which will just producea strain having the same ratio to itself
which the strains bear constantly to the stresses producing them
for all stresses of less intensity.
If a stress be applied of very much greater intensity, the
piece will break at once ; if of moderately greater intensity,
the piece will take a set ; and although only of a little greater
intensity, yet if applied for a long time, the piece will ultimately
take a set ; and if it be applied and removed many times in
succession, the strain will increase each time and the piece
ultimately break. For all stresses of intensities less than the
proof load the elasticity is constant for the same substance, and
the
r, ., r T 1 n ^ .■ ■. iuteuslty of stress
Def. — Modulus of elasticity = r— =^^ ^
strain due to it
= stress per unit strain.
If the denominator on the righthand side of the equation
be unity, then the numerator is the stress which produces unit
strain, and
Mod. of elasticity = stress which would produce unit strain
on supposition of rod not experiencing a set and Hooke's law
holding.
For most substances the proof stress is a mere fraction of £,
the modulus of elasticity. For steel the proof stress is scarcely
loo^oo ^h part of F. Hence in the equation above, the word
would is employed, as it would be absurd to say that U equalled
the stress that will produce unit strain, that being an impossi
bility with most substances ; and even when possible, as in the
case of indiarubber, the strains at such a pitch will have ceased
to be proportional to the stresses producing them, and hence U
8 APPLIED MECHANICS. [CHAP. I.
will be no longer of a constant value. But the definition is
quite accurate and definite for all substances amounting to this,
that for any substance
^ = 10 times the stress that will produce a strain of fV^h,
if such a pitch of strain be possible and within the limit of
strain, that is, not greater than the proof strain.
But if not, then,
E = 100 times the stress that will produce a strain of xftt^^
if such a pitch of strain be possible and within the limit of
strain, that is, not greater than the proof strain.
Thus for steel E equals one million times the stress which
will produce a strain of onemillionth part. Pliability is a term
applied to the property which indiarubber possesses in a higher
degree than steel.
Examples.
7. A wroughtiron tierod has a stress of 18000 lbs. per square inch, of section
which produces a strain of 0006. Find the modulus of elasticity of the iron.
„ intensity of stress 18000 „ ^„„ ,, . ,
E = ■/—. = "— — = 30000000 lbs. per square inch.
strain 0006 ' ^
8. A tie rod 100 feet long has a sectional area of 2 square inches ; it bears a
tension of 32,000 lbs., by which it is stretched fths of an inch. Find the intensity
of the stress, the strain, and modulus of elasticity.
total stress 32000 lbs.
stress = = ; = 16000 lbs. per sq. m.
area 2 sq. m.
aug. of length 76 in.
strain = ^j —5— = — —  = 000625.
length 1200 in.
„ stress 16000
E = —r~ = ^^_,, = 25600000 lbs. per sq. in.
strain 000625 ^
9. A castiron pillar one square foot in sectional area bears a weight of 2000
tons ; what strain will this produce, E for cast iron being 17,000,000 lbs. ?
total stress = 2000 tons per square foot = 2000 x 2240 lbs. per sq. foot.
2000 X 2240
stress = — = 311111 lbs. per sq. in.
144
„ stress 31111
E = — . , or 17000000 = ;
strain strain
••• strain = ^^^^=0018 ft. per ft. of length.
CHAP. I.] LINEAL STRESS AND STRAIN. 9
irr 10. The modulus of elasticity of steel is 30,000,000. How much will a steel
rod 50 feet long and of Jth inch sectional area be stretched by a weight of one
ton?
total stress = 2240 lbs. ;
total stress in lbs.
stress = . : = 2240 i i = 17920 lbs. per sq. inch.
area in sq. in.
„ stress stress 17920
E = —: .. strain =———= — =000.'512:
strain £ 35000000
elongation
, , = strain ;
length
.. elongation = strain \ length = 000512 x ')0 = •02.')6 feet or ^ of an inch.
11. An iron wire 600 yards long and A'th of sq. inch in section, in moving
a signal, sustains a pull = 250 lbs. ; liow much will it stretch, assuming
£ = 25000000?
stress = 20000 lbs. per sq. inch ; strain = OOOS ; elongation = 1*44 feet.
12. Modulus of elasticity of copper is 17,000,000 : what weight ought to
stretch a copper thread, of 12 inches in length and 004 inches in sectional area,
i^T7th part of an inch. If after the removal of the weight the thread remains a
little stretched, what do you infer about the weight and about the strain to which
the thread was subjected ?
strain = TaVoth ; stress = 14167 lbs. per sq, inch ; weight = 56*668 lbs.
Since this weight causes a set, it is greater than the proof load.
The Production of Strain.
We have as yet only considered the statical condition of
strain, i. e. of a body kept in a state of strain by external forces,
these forces being balanced by the reactions of the solid at their
places of application due to the elasticity, and the forces exerted
on any portion of the solid being in equilibrium with the re
actions of the contingent parts. Thus when we found that
32,000 lbs. produced a strain of 00063 on a tierod 100 feet long
and 2 square inches in area, in all stretching it f ths of an inch,
we meant that the weight ke2}t it at that strain ; the rod is sup
posed to have arrived at that pitch of strain and to be at rest,
to be stretched the fth inch, and so (by its elasticity or tendency
to regain its original length) to balance the weight. We have
taken no notice of the process by which the rod came to the
strain, nor do we say it was the weight that brought it to that
state, the weight being only a convenient way of giving the
value of the stress on the rod when forcibly Jcejyt strained. In
fact, an actual weight of 32,000 lbs. is capable of producing
10 APPLIED MECHANICS. [CHAP. I.
greater strains on the rod in question, depending upon how it
IS applied to the rod as yet unstrained. The weight might be
attached by a chain to the end of a rod and let drop from a
height; when the chain checked its fall, it would produce a
strain on the rod at the instant greater the greater the height
through which it dropped. Still, if that strain were not greater
than the proof strain, the weight upon finally coming to rest
after oscillating a while could only keep the rod at the strain
•X)0063.
We come now to consider the kinetic relations between the
stress and the strain, that is, while the strain is being produced,
the matter of the body being then in motion, we are consequently
considering the relations among forces acting upon matter in
motion.
If a simple stress of a specific amount be applied to a body,
it produces a certain strain, and in doing so the stress does
work, for it acts through a space in the direction of its action
equal to the total strain. Bui if this stress is applied gradually,
so as not to produce a shock, its value increases gradually from
zero to its full value, and the work it does will be equal to its
mean value, multiplied by the space through which it has acted.
And since the stress increases in proportion to the elongation,
its average value will equal half of its full value. For example,
if a stress of 30,000 lbs. be applied to a rod and produce a strain
of I inch, it will do ^1^ x f = 11,250 inchlbs. of work, which
will be stored up as a potential energy in the stretched rod.
Suppose a scalepan attached to the top of a strut or bottom
of a tie and the other end fixed. Let a weight be put in contact
with the pan, but be otherwise supported so as to exert no stress
on the piece, and the next instant let it rest all its weight on the
piece, then will the weight do work against the resistance offered
by the straining of the piece till the weight ceases descending
and comes to rest, when the piece will be for an instant at the
greatest strain under the circumstances, at a strain greater than
the weight can keep it at; the unstraining of the piece will
therefore cause the weight to ascend again, doing work against
it to the amount that the weight did in descending, and so the
weight will return to its first position, then begin to descend
again, and so oscillate up and down through an amplitude equal
to the augmentation. Owing to other properties of the matter,
whereby some of the work is dissipated during each strain and
restitution, the amplitude diminishes every oscillation, and the
weight will finally settle at the middle of the amplitude.
A weight applied in this manner is called a live load. A
CHAP. I.] LINEAL STRESS A1«D STKAIN. 11
live load produces, the instant it is applieil, an augmentation of
length double of that which it can maintain, and therefore
causes an insfanta neons strain doul)le the strain due to a stress
of the same amount as the load.
Let now a weight W be applied in the following way : —
Divide W into n equal parts of weight ic each. If A be the
strain due to a stress of amount W, and a the strain due to a
stress u\ then W ^ mu,
and by Hooke's law. A = na.
Let the first weight iv be put into the scalepan. It will
produce a strain 2a at once, but the piece will settle at a
strain a. Now put on the second weight u\ It will produce at
once an additional strain 2^, but only of a additional after the
piece settles ; there being now a total strain 2a. Add the third
weight u\ It also will produce at first an additional strain 2a,
but only of ct after the piece settles, giving a total now of 3a ;
and so, adding them one by one, there will be a strain of (n  l)a
when the second last one has been added and the piece has
settled. Now, upon adding the nth. weight w, it will at first
produce an additional strain 2a, but only of a after the piece
settles, giving then a total strain na or A. Thus we have
brought the piece to a pitch of strain A by means of the weight
W, and only at the instant of adding the last part (iv) of it was
the piece strained to (n + l)a, or to a more than A. By making
the parts more numerous into which we divide W, and so each
part lighter and producing a lesser strain per part, we can make
the strain a the extent to which the piece is strained beyond A
at the instant of adding the last part, as small as we please.
By so applying the load W we can bring the piece to the
corresponding strain A without at all straining it beyond that.
A weight so applied is called a dead load.
A live load therefore produces, at the instant of its applica
tion, a strain equal to that due to a dead load of double the
amount. In designing, the greatest strain is that for which
provision must be made, so that live loads must be doubled in
amount, and the strain then reckoned as due to that amount of
dead load. The dead load, together with twice the live load, is
called the gross load.
The weights of a structure and of its pieces are generally
dead loads. Stress produced by a screw, as in tightening a
tierod, is a dead load. The pressure of earth or water gradually
filled in behind a retaining wall, and of steam got up slowly, of
water upon a floating body at rest in it, &c., are all dead loads.
The weight of a man, a cart, or a train coming suddenly upon
12 APPLIED MECHANICS. [CHAP. I.
a structure, is a live load ; so is the pressure of steam coming
suddenly into a vessel ; so is a portion of the pressure of water
upon a floating body which is rolling or plunging. The pressure
upon a plunger used to pump water is a live load, but that on
a piston when compressing gas is a dead load, the gas being so
elastic itself. A load on a chain ascending or descending a pit
is a dead load when moving at a constant speed or at rest, but a
live load at the starting, and while the speed is increasing, partly
a live and partly a dead load. The stress upon the coupling
between two railway carriages is a dead load while the speed is
uniform, and if the buffers keep the coupling chain tight, the
stress is a live load while starting; but if the buffers do not
keep it tight, but allow it to hang in a curve when at rest, then
the stress upon it at starting will be greater than a live load.
JIXAMPLES.
13. An iron rod in a suspension bridge supports of the roadway 2000 lbs., and
wben a load of 3 tons passes over it, bears onefourth part thereof. Find the
gross load. If the rod be!20 feet long, and f of a square inch in section, find the
elongation, E being 29,000,000.
dead load = 2000; live load = 1680 lbs., equivalent to a dead load of
33G0 lbs. ;
.. gross load = 5360 lbs. ;
gross load 5360 _,,_,,
stress = ° .  = —  = 7147 lbs. per so. in.
section '75
_ _ stress
strain '
stress 7147 „,^^ „ elongation
.. strain = = = 000246 = , ° , ;
E 29000000 length
.. elongation = length x strain = 20 x 000246 = 00492 ft. = 06 in.
14. A vertical wroughtiron rod 200 feet long has to lift a weight of 2 tons.
Find the area of section, first neglecting its own weight ; if the greatest strain to
which it is advisable to subject wroughtiron be 0005 and E = 30,000,000.
Let A be the sectional area in sq. in.
live load = 4480 lbs. is equivalent to a dead load of 8960 lbs.
8960 ^ stress „„ „„„ „„„ 8960
.. stress = — — ; E = — — , or 30,000,000 = — — .
A strain A x 0005
8960
•597 sq. in.
0005 X 30000000
15. Find now the necessary section at top of rod, taking the weight into
account, calculated from the section found in last.
200 ft. X 597 sq. in. gives 1433 cubic in. ; reckoned at 480 lbs. per cubic foot
gives 398 lbs.
CHAl'. 1.] LINEAL STHESS AND STRAIN. ]
Hence live load = 4480 lbs. ; dead load = 398 lbs.
9358 „ stress 9358
.. Rross load = 9358 ; stress = ; E= , or 30000000 =
areii strain area x •000.'
9358
 — = 62 sq. in.
•0005 X 30000000
The weight of the rod being greater when calculated at this section, a third
approximation to the sectional area might be made.
16. Taking now the sectional area at "G'i sq. in., find average strain and
elongation.
At lowest point
8960 . 8960
gross load = 8960 lbs. ; stress = — ; strain = — = 00048 ; while
strain at highest point is 0005 ;
.. average strain ^ 00049 ; elongation = 00049 x 200 = 098 ft. = M76 in.
17. A short hollow castiron pillar has a sectional area of 12 sq. in. It is
advisable only to strain castiron to the pitch 0015. If the pillar supports a dead
load of 50 tons, being weight of floor of a railway platform, and loaded waggons
pass over it, what amount should such load not exceed ? E = 20000000.
greatest stress = 30000 lbs. per sq. in. ; gross load = 360000 lbs.
deduct dead load = 112000 lbs. ; gives a dead load = 248000 lbs.
The live load must not exceed onehalf of this.
Note. — Other considerations limit the strength of the pillar if it be long.
Eesilience.
Def. — The Resilience of a body is the amount of work
required to produce the proof strain. A weight onehalf the
proof stress applied as a live load would produce the proof
strain ; therefore the work done is this weight multiplied by the
elongation at proof strain, the distance which the weight has
worked through ; or
the resilience of a body
= ^ amount of proof stress x elongation at proof strain.
For comparison among different substances the resilience is
measured by the resilience of one foot of the substance by one
square inch in sectional area.
.'. ifc = ^ proof stress x proof strain,
R being in footlbs. when the stress is in lbs. per square inch
and the strain in feet.
14 APPLIED MECHANICS. [CHAP. I.
And now comparing the amount of resilience of different
masses of the same substance : if two be of equal sectional area,
that which is twice the length of the other has twice the amount
of resilience (the elongation being double) ; also if two be of
equal length and one have twice the sectional area of the other,
then the amount of its resilience is double (the amount of stress
upon it being twice that upon the other). That is, the amounts
of the resilience of masses of the same substance are proportional
to their volumes. This is true not only for pieces in a state of
simple strain with which we are in the meantime occupied, but
can be proved to be universally true for those in any state of
strain, however complex.
For any substance R being the amount of resilience of a
prism of that substance one foot long by one square incli in
sectional area, it follows from the alcove that the amount of
resilience of a cubic inch of the substance will be jVK or that
of any volume will he ^R x volume in cubic inches.
The resilience of a piece, as defined, is the greatest amount
of work which can be done against the elasticity of the piece,
without injuring its material.
We can find the amounts of work done upon a piece in
bringing it to pitches of strain lower than the proof strain. For
brevity we will call this also resilience. Thus, for a piece 1 foot
long by 1 square inch in section
amount of resilience = ^ stress x strain is pro. to (stress)^
the strain being proportional to the stress ; hence
amount of resilience for any stress _ (stress)^
the resilience (proof stress)*'
.'. amount of resilience = R x i 1 •
Vp. stress/
For a piece of volume V cubic inches, at any stress we have
either —
V
amount of resilience = i stress x strain x — ,
or = I amt. of stress x amt. of strain.
The amount of resilience of a piece, at the instant a live load
is applied, will be the product of that load and the instantaneous
elongation. Let W be a load the elongation due to which is A.
CHAP. I.J LINEAL STRESS AND STHAIN. 15
If ]r be applied as a live load, the iiistaiitaneous elungatiou is
2A, and the
amount of resilience due to a live load W = W x 2 A.
If W be applied as a dead load, the amount of resilience is
steadily that of the piece elongated to an amount A, is the
same as what it would be for an instant upon the application of
W
a live load — , or
W
amount of resilience due to a dead load W = ^ y. A.
Therefore, a live load produces for an instant an amount of
resilience four times that produced by an equal dead load.
Examples.
18. A rod of steel 10 feet long and "5 of a square inch in section is kept at the
proof strain by a tension of 25,000 lbs., the modulus of elasticity for steel being
35,000,000. Find the resilience of steel, also the amount of resilience of the rod.
25000 „ . ,
proof stress = — ^ — = 50000 lbs. per sq. inch.
*o
„ proof stress
E =
proof strain =
proof strain *
proof stress 50000 1
E 35000000 700
= 00143 elongation in ft. per ft. of length.
resilience, R = h proof stress x proof strain = \ x 50000 lbs. x 00143 ft.
= 3575 ft. lbs. of work per vol. of 1 ft. in length by 1 sq. in.
in sectional area.
amt. of res. of rod = R x (vol. expressed in number of such prisms)
= — , i? X vol. in cub. in. = — x 3575 x 120 in. x 5 sq. in.
= 17875 footlbs. of work.
Otherwise, to find amount of resilience directly,
proof strain = —  ; total elongation = — ft.; amount of stress = 25000 lbs.
700 " 70
amount of resilience = t amount of stress x elongation,
25000 1
= — — X — , = 1786 ft. lbs. of work.
2 lO
16 APPLIED MECHANICS. [CHAP. I.
19. A series of experiments were made on bars of wrought iron, and it was
found that they took a set when strained to a degree greater than that produced by
a stress 20,000 lbs. per square inch, but not when strained to a less degree. At
that pitch the strain was 0006. Find the resilience of this quality of iron.
proof stress = 20000 lbs. per sq. in.
proof strain = 0006 ft. per ft. of length.
Jt = ^ X 20000 X 0006 = 6 ft. lbs.
20. Find how m\ich work it would take to bring a rod, of the above iron,
20 feet long and 2 square inches in sectional area, to the proof strain.
volume = 480 cub. in.
E ft. lbs. of work brings to proof strain a rod 1 foot long by 1 square inch
in area; that is, of volume 12 cubic inches, and amounts of resilience being
proportional to the volumes.
work required = ^R . vol. in cub. in. = ^^ x 6 x 480 = 240 ft.Ibs.
amount of res. 480 cub. in.
or = ; — : — = 40.
" It 12 cub. m.
.. amount of res. = iJ x 40 = 6 x 40 = 240 ft. lbs.
21. A wooden strut 18 square inches in section, and 12 feet long, sustains a
stress of 1000 lbs. per square inch. Find the amount of resilience of'the strut,
^heing 1,200,000 lbs.
half of total stress = 9000 lbs. ; elongation = 01 ft. ;
amount of resilience = 90 ft. lbs.
22. Steam at a tension of 600 lbs. on the square inch is admitted suddenly
upon a piston 18 inches in diameter. If the piston rod be 2 inches in diameter
and 7 feet long, what is the amount of its resilience at the instant? £ = 30000000.
for live load stress = 97200 lbs. per sq. in.
gives instant strain = 00324 ft. per ft. of length.
elongation = 02268 ft.
resilience of rod = live load x elongation = 486007r lbs. x 02268 ft.
= 11025r ft. lbs.
23. The chain of a crane is 30 feet long and has a sectional area equivalent to
A of a square inch : what is the amount of its resilience when a stone of 1 ton weight
resting on a wooden frame is lifted by the action of the crane ? £ = 30000000.
stress = 4480 lbs. per square inch, strain = 000149.
amount of stress = 2240 lbs., elongation = 00447 ft.
resilience of chain = ^ amount of stress x elongation = 5 ft.lhs.
24. If the chain be just tight, but supporting none of the weight of stone,
and if now the wooden frame suddenly gives way, what is the amount of resilience
of the chain at the instant?
Being now a live load, there is an instantaneous strain of double the former amount.
instantaneous strain = 000298. instantaneous elongation = 00894 ft.
resilience of chain = live load x elongation = 2240 x •00894 = 20 ft. lbs.
CHAF. I.] LINEAL STRESS AND STIiAIN. 17
25. The wire for moving a signal 600 yards distant bus, when the signal is
down, u tension upon it of 250 lbs., which is maintained by the back weight of the
hand lever; under the circumstances the wire is stretched an amount 144 feet,
and so the buck weight of the signul, which is 280 lbs., rests portion of its weight
upon its bed. The hand lever is suddenly pulled back and locks : the wire being
more intensely strained, the signal is raised by the elasticity of the wire partially
unstraining. 1 1 tlie point where the wire is attached to the signal moves through
■2 feet, find the range of the point where the wire is attached to the hand lever,
also the force which must be exerted there.
When the signal settles up, the amount of stress on the wire is 280 lbs.
elongation for 280 lbs. _ 280
elongation for 250 lbs. 250 '
280
elongation for 280 lbs. = —  x 1'44 = 1613 ; additional elongation = 173 ft.,
2?0
.•. range of point at lever = range of point at signal plus this additional elongation
= 2+ 173 = 373 feet.
Thus, when tiie lever is put back there is upon the wire for an instant before
the signal rises an additional elongation of •373 feet. Hence the tension on the
wire the instant the lever is put back will be that due to an elongation of
(1 44 + 373) ft. = ^"^'^ + '^'^ 250 lbs. = 3148 lbs.
144
This is the force which must be exerted at the point where the wire is attached
to the hand lever. That is, the instantaneous value of the force used to raise the
signal is 348 lbs. greater than its weight.
26. On a chain 30 feet long, f of a square inch in sectional area and having a
modulus of elasticity of 25,000,000 lbs. there is a dead load of 3900 lbs. and a live
load of 900 lbs. Find the amount of resilience of chain when dead load only is on,
also at instant live load comes on.
stress 5200
as strain = — — = — = 0002,
£ 25000000 '
elongation = 006 ft.,
amount of resilience for dead load = J amount of stress x elongation
= J X 3900 X 006 =117 ft.lbs.
Live load gives an additional elongation equal to that for a dead load of
1800 lbs.
elongation (due to 1800 lbs.) _ 1800
006 ft. ~ 3900 '
inst. elongation for live load = ~— — x 006 = 00277 ft.
jyuu
Now both the 3900 lbs. and the 900 lbs. worked through this 00277 ft.
.. additional resilience = 4800 lbs. x '00277 ft.' = 133 ft.lbs.
amount of resilience at instant live load comes on = 25 ft.lbs.
27. A rod 20 feet long and ^ inch in sectional area bears a dead load of 5000 lbs.
Find the live load which would produce an instantaneous elongation of another
i'oth inch. E = 30000000. Aus. 3125 lbs.
C
18 APPLIED MECHANICS. [r'HAP. I.
28. A rod of iron 1 square inch section and 24 feet long checks a weight of
36 lbs. which drops through 10 feet before beginning to strain it. If ^ = 25000000,
find greatest strain.
Let p = the stress at instant of greatest strain ; then
P 1 • 24»
strain = — , elongation = — ,
amount of resilience = \ amount of stress x elongation
= ?>^'^ = 12$ft.lbs.
2 E E
Work done by weight in falling = 36 lbs. x ^10 + ~^\ ft. = 360 + ^ ft. lbs.
Equating, ^ ^ ~E
p"  lip = 30£; p^  72jo + (36)2 = 750001296,
;>  36  27386; ;^ = 27422 lbs. per square inch,
strain = 001097 ft. per ft. of length.
29. If the weight in Ex. 28 had fallen through the 10 feet by the time it came
first to rest and E = 30000000 lbs., what is the greatest strain :
12;>*
amount of resilience = 360 ft. lbs., or —  = 360,
E
.. p = 30000 lbs. per square inch,
strain = "001 ft. per ft. length.
30. If the proof strain of iron be 001, what is the shortest length of the rod of
one square inch in sectional area which will not take a set when subjected to the
shock caused by checking a weight of 36 lbs. dropped through 10 feet?
IE = 30000000 lbs. per square inch.]
Let X = length in feet.
By hypothesis it comes to the proof strain ; hence
elongation = OOl x a; ft.
proof stress = E x proof strain = 30000 lbs. per sq. inch,
amount of stress = 30000 lbs.
(inst.) amount of res. = ^ amount of stress x elongation
= 15000 x ^ = 152: ft.lbs.
1000
Equating to work done by weight,
15t = 360 : x = 24 ft.
Note. — This is the shortest rod of iron one square inch in sectional area which
will bear the shock. The volume of this rod is 288 cubic inches, and a rod of iron
which has 288 cubic inches of volume will just bear the shock ; as 48 feet long by
I square inch in area or 12 feet long by 2 square inches sectional area.
The 10 feet fallen through by the weight includes the elongation of rod.
When the question is to find the shortest rod to sustain the shock in the case
where the weight falls through 10 feet before it hegim to strain the rod, the
volumes of the rods would not he exactly equal for difi"erent sectional areas ;
for a long thin rod will sustain a greater elongation than a short thick one,
and as the falling weight works through this elongation over and above the 10 feet,
the first rod wiU require a greater cubical volume than the second.
CHAr. I.] LINEAL STKESS AND STRAIN. 19
3L Find the shortest length of a rod of steel which will just bear without
injury the shock caused by checkinj; ii weight of 60 lbs. which falls through
12 feet before beginning to strain the rod. First for a rod of sectional area
2 square inches, and then for a rod of ^ square inch sectional area. Given that
for steel E = 30000000 lbs. and E = 16 ft. lbs.
Let A = sectional area in square inches and x = length in feet.
proof stress y proof strain = 2R def.
proof stress ^ , .
; r = £ def.
proof strain
Dividing (proof strain) = — •
I2S I , . X
proof strain = / = r— rr , elongation = ft.
^ V £■ 1000' ^ 1000
"Work done by the weight in falling
= 60 lbs. X (l2 + ^)ft. = 720+l.ft.lbs.
Amount of resilience of rod at proof strain
= iZ X I — vol. in cub. in. j
= i? X (length in ft. x sec. area in sq. ia.).
= I5xa;x2 = 30a; ft. Ibs.^/s^.
15
and lh>.xy.\ = —x ft. lbs, second.
4
Equating for first case,
3
30x = 720 + —X, 1497a; = 36000,
oO
X = 2405 ft. length of rod.
Equating for second case,
— a; = 720 + ^x, 738a: = 144000,
4 oO
X = 19512 ft. length of rod.
For the first case length is 2886 inches, and sectional area 2 square inches
gives
volume = 577*2 cub. inches.
While for second case length is 2341*44 inches, and sectional area \ square
inch, giving
volume = 5853 cub. inches,
which is a little greater. See Note to Ex. 30.
c2
( 20 )
CHAPTER 11.
INTERNAL STRESS AND STRAIN, SIMPLE AND COMPOUND.
<5i.
In this Chapter, except where specially stated, we premise
that —
(a) All forces and stresses are parallel to one plane.
(b) That plane is the plane of the paper in all diagrams.
Hence planes subjected to the stresses we are considering are
shown in diagrams by strong lines, their traces.
(c) The diagrams represent slices of solid, of unit thickness
normal to the paper ; hence, the lengths of the strong lines are
the areas of the planes.
(d) The stresses which are normal to the paper are supposed
constant both in direction and intensity,
or every point on a diagram is in the
same circumstances with respect to stress
normal to the paper. i>r,r,l.,r,^ "~vJr
(e) The relative position of two planes
is measured by the angle between their
normals.
(/) The obliquity of the stress to
the plane upon which it acts is the
angle its direction makes with the
normal to the plane.
Internal stress at a point in a solid zr^
in a simple state of strain.
Let the axis OX (fig. 1) be drawn
in the direction of the stress P. Let
A A be any section normal to this axis.
Since the stress is uniformly distributed
over AA, the intensity of the stress at
all points of the plane AA is the same
ppp
"A
Fig. 1.
Consider the point 0, the intensity of the stress at on the
plane normal to OX is
P =
total stress
area of plane
P
AA
Through draw any oblique plane, BB, whose normal, ON,
CHAP. II.]
INTERNAL STRESS AND STRAIN.
21
makes the angle 9 with OX. The stress on this plane is in the
direction OX, and the amount of stress upon it is P (for the
equilibrium of the parts). But the inten j
sityof the stress on 56 is less than;;, since f ^
P is spread over a larger area than A A.
Since
AOB = e,
OB.''''
and
OA
OB
 cos AOB,
or
BB
cos 6'
Intensity of stress on
total stress P
area of plane BB
AA
<
cos 6
BB =
P
AA
cos 9 = p
AA
^cos 9
cos 9.
Hence the internal stress at all points within a solid, in a
state of simple strain, is parallel ^
to the direction of that stress
— is greatest in intensity on the
plane normal to that direction —
on any other plane inclined at
an angle 9 to last, the intensity
is one (cosine 9)th part of that
intensity, and zero on any plane
parallel to the direction of the
stress.
The stress ^; cos 6 on BB
being oblique to BB, it is con
venient to resolve it into com
ponents normal and tangential
to BB respectively.
The arrow p cos (fig. 2)
represents the stress at the point
on the plane BB; from its
extremity perpendiculars are
dropped on ON and BB, which,
by parallelogram of forces, give
p„ and pt, the intensities of the
stresses upon B B, normal and tangential respectively.
22
APPLIED MECHANICS.
[chap. II.
Now
Vn
2) cos d
cos clef.
:Pn = p cos' e ; (fig. 3)
also
Pt
p cos
= sin 0.
Pt=p sin 6/ cos 0.
V
From the superposition of forces these two sets of forces
may be considered independently of
each other. For some cases in designing
it might only be necessary to consider
one set, if it were manifest that in pro
viding for it there would be more than
sufficient provision made for the other.
It is apparent from symmetry that
for the plane CC inclined at the angle on the
other side of the axis the stress is the same in
all particulars as that on £B.
On a pair of planes whose obliquities are
together equal to a right angle, the intensities of
the tangential stresses are equal, and the sum of
the intensities of the normal stresses equals the
intensity of the initial stress.
Let BB (fig. 4) be inclined at the angle 6, and
DD at the angle ^, where
e + <i>
2
On BB pn = p cos' 0, pt = p sin cos 0.
On DJD p'n = p cos^ (j), p't = p sin ^ cos (j>.
But sin = cos <p, and cos ^ = sin ;
therefore pt = p't, or the tangential component stresses
have the same intensity on both
planes.
Also Pn + p'n = P (cos d + COS' <p)
= p (cos 6 + sin^ ^) = P'i
or the sum of the intensities of the normal component stresses
equals the intensity of the primary stress.
CHAP. 11.]
INTEUNAL STRESS AND STRAIN.
23
There are therefore at one point (fig. 5) four planes
BB, DD, CC, and EE, two inclined on each side of OX, upon
which the tangential stress ^
has the same intensity. ,'■ "  .,
Grouping together the ,^ .^p'^'p'^^ r
pair of planes BB and EE, / JS^.. \\'\ ^ .">
the one inclined at on the  /'
one side of OX, and the^^ ,
other at upon the oppo ~~ ,
site side, and therefore at
^ I 0, or 90° to each other, we find
that at any point two planes being
chosen at right angles to each other, the
tangential or shearing stresses are of
equal intensity, and the sum of the in
tensities of the normal stresses is equal
to the intensity of the primary stress.
For all planes such as BB, DD,
&c., that which is inclined at 45°
sustains the tangential stress of great •
est intensity, for
TP
jO/ = j? sin cos = ^ sin 20.
Therefore pt is greatest when sin '2B is greatest,
when sin 20 = 1, when 2B = 90°, or = 45*.
The tangential stress on a plane such as BB is
called a sJiearing stress. Many substances fracture
under a shearing stress very readily. Notably cast
iron under a strain of compression fractures by shear
ing along an oblique plane, the one portion sliding
upon the other (fig. 6). The resistance then which
cast iron ofiers to shearing is that which must be
considered in designing short pillars to bear great
loads. The planes upon which the intensity of the
shearing stress is greatest, that is, planes inclined at
45° to the direct thrust, are those upon which it will
shear. As the texture of the material is never
homogeneous, it may shear along planes more or less w
inclined than 45°, also the toughness of the skin will j
cause great irregularity.
Brick stalks give way by the mortar shearing, and
the upper portion sliding down an oblique section like
24 applied mechanics. [chap. ii.
Internal Stress at a Point in a Solid in a
Compound State of Strain.
A solid is in a compowiid state of strain when subjected to two
or more simple stresses in different directions simultaneously.
We proceed to consider a solid in such a state of strain without
inquiring how it was brought into that state; all its parts
being supposed to be at rest, and all the parts into which it
may be divided in equilibrium under the stresses exerted
among each other, due to their elasticity, and those exerted at
the external surface : but at the outset we do not regard those
external stresses.
Upon any plane passing through a point within the solid,
the stress at that point is definite in intensity and direction ;
for if along that plane the solid were divided into two parts,
the mutual pressures between the cut surfaces at that point
(no matter how complicated) can be compounded into one
force, definite in amount and direction. Along this plane the
intensity and direction of the stress vary, and at the point
will only be constant over a very small part of the surface
round it. If the stress be stated in lbs. per square inch, the
total stress on this small surface which we are considering
would be a mere fraction of the intensity. It will be convenient
to consider the intensities of these stresses to be expressed,
say, in lbs. per millionth part of a square in., so that in the
diagrams tivo or three arrows {each representing the intensity)
may be drawn to represent the total amount of stress upon
such small planes, without leading us to the supposition thnt
they are of a few square inches in extent. And yet whatever
results we arrive at are equally true for intensities expressed
in the usual units, for the intensity at a point on a plane, upon
which the intensity varies, can be expressed to any degree of
accuracy in lbs. per square inch. Thus, at the point, the
intensity of the stress in lbs. per square inch equals roughly,
nearly, more nearly, &c.
Amount of stress on the square inch surrounding point,
roughly.
10 times amount of stress on the iVth of a square inch
surrounding point.
100 times amount of stress on the yoT^^ ^^ ^ square inch
surrounding point.
1,000,000 times amount of stress on the 10000 77^^ of a
square inch surrounding point, &c., &c.
CHAP. II.]
INTERNAL STRESS AND STRAIN.
25
Let OAO'B (fig. 7) be a small rectangular parallelepiped
at the point in a solid in a state of strain.
Let q = intensity of X*^ ,
stress on the faces OA and
CB at an obliquity a.
Let p = intensity of
stress on faces OB and O'A
at an obliquity j3. f
The normal components ^
are — •/
J)„ = p cos (3, q„ = q cos a.
The two sets of forces
})„ directly balance each
other, and may be removed,
and also the two sets q„,
leaving the parallelepiped
in equilibrium under the
action of the tangential
components.
Pt = p sin j3, qt = q sin a.
inUruvUej and. ohhquitLes iitO ff
stresj&t upon OA ami OB
^r
'•'
I I
'•
pCosfi
o'
_
f^ 1
</;t *
q=qCoo qe.

—
B
ifn.
NorrruxL componerus
A PtOS
qtOA
9tOA.
The amount of tangen
tial stress on each of the
faces OA and O'B is qt. OA.
Also the amount on each
of the faces OB and O'A is
Pt. OB.
The two forces qt . OA
form a couple with a lever
age OB tending to turn the
parallelepiped in the direc
tion in which the hands of
a watch turn, while the
two forces pt • OB form a
couple with a leverage OA
tending to turn it in the
opposite direction. Since
the parallelepiped is in equi
librium under these two
actions alone, the moments of these two couples must be equal.
o — Tfmr~B
Arrwujvcs of tanaaiUaJL. streM
A O'
Pt Pt Pt Pt^B*^V^'''a
/ruerhfUces of UmgcnuaL
ccmnonenCa
Fig. 7.
Force. Leverage. Force. Leverage.
qt • OA X OB = p, . OB ^ OA.
26
APPLIED MECHANICS.
[chap II.
Now the area OA multiplied by the length OB gives the
volume of the parallelepiped, and the area OB multiplied by
the length OA also gives the volume.
it
Ut=pu
Hence, at a point within a solid in a state of strain, the
tangential components of the stresses upon any two planes
through it at right angles to each other are of an equal intensity.
Cor. — If the stress upon any plane through a point be wholly
normal, then will the stress be wholly normal upon another
plane at right angles to that plane.
Such a pair of stresses are called principal stresses, and the
planes upon which they act are planes of principal stress.
When these two stresses are given, the uniplanar stress at a
point is completely given. There is also a third principal stress
acting on the plane at right angles to the other two.
DiKECT Pkoblem, the Pkikcipal Stresses given.
Eqiudlike principal stresses (fig. 8). — If the pair of principal
stresses at a point be like (both thrusts or both tensions), and
be of equal intensity, the stress
on any third plane through the
points is of that same intensity,
and is normal to the plane.
Let AA' and BB' be the
planes of principal stress at the
point 0, and let the intensities
of the principal stresses, p and
q, be equal and alike (both
thrusts).
CC is any third plane
through 0, inclined at B to
AA\
GAB is a small triangular prism at 0, having its faces in
those planes. This prism is in equilibrium under the three
forces — the total thrusts upon OA, OB, and AB.
The three total pressures F, Q, R on the faces, if represented
by straight lines, intersect at the middle point of AB. The
parallelogram of forces is, however, drawn from the point 0.
Lay off OD = total stress parallel to OX = p . OA,
and OH = total stress parallel to OY = q . OB.
Complete the parallelogram.
CHAP. II.]
RANKINE S ELLIPSE OF STRESS.
27
Then RO represents the total stress on AB in direction
and amount.
„^^ OE q.OB OB
tan ROD=^= j^ = ^ , since p = rj.
.: BOB = OAB = 6; .: OR is upon ON.
Hence BO is normal to AB.
And BO' = OB + OB" = f . OA' + q' . OB"
= p' {OA' + OB') = p . AB', as p = q.
.: .RO = p.AB.
amount of stress on AB RO
Now
area oi AB
AB
= p or q.
'^^JH
6'or.— Every plane through is a plane of principal stress.
Each point in a fluid is in this state of strain.
Equalunlike principal stresses (fig. 9). — If the pair of
principal stresses at a point be unlike (one a thrust and the
other a tension) and be
of equal intensity, the
stress on any thirdplane
through the point is of .< ^
thatsameintensity,and "'JT
is inclined at an angle to
the normal to the plane
of principal stress, equal
to that which the nor
mal to this third plane
makes therewith, but
upon the opposite side.
Let 44' and .5^ be Fig. 9.
the planes of principal
stress at the point ; p and q the intensities of the principal
stresses of equal value, p being a thrust and q a tension ;
and CC any third plane through inclined at 6 to AA'.
OAB, a small triangular prism at bounded by these
three planes, is in equilibrium under the three forces, viz.,
the amount of stress on its faces OA, OB, and AB.
Lay off OD = total stress parallel to OX = p . OA,
and OF = total stress parallel to F in the direction of q
q.OB.
28
APPLIED MECHANICS.
[chap* II.
Complete the parallelogram ODRE. Then RO represents
the total stress on AB in direction and amount.
tan ROD =
OE
OB
OA
OB
= Z—nA =02^ ^^^^ ^^^'
^ 9_
OD p
ROD = OAB = e.
That is, BO is inclined at the same angle to the axis OX as
OiV is, but on the opposite side. Hence the inclination of RO
to the normal ON is 20.
Again RO' = OD^ + OE' =^ f . OA" + q^ . OB'
=p'(OA' + OB')^f.AK;
amount of stress on AB RO
BO = p . AB,
area of AB
AB
= p or q.
PTPV
Y
Consider the triangle of forces OER, we have OE drawn
from in the direction of q, then ER drawn from E in the
direction of /; ; hence RO, taken in the same order, is the
direction of r.
If 6 be greater than 45°, r is like q.
If 6 equal 45°, r is entirely tangential to AB.
If B be less than 45°, r is like p.
Hence, if the principal stresses at
a point be equal and unlike, the stress
on a third plane is of that same in
tensity, is like the stress on the plane
it is least inclined to, and its direction
is inclined to the axis at the same
angle as the normal is, but upon the
opposite side. If the new plane be
inclined at 45°, the stress is entirely
tangential.
Unequal principal stresses (fig. 10).
— The principal stresses at a point
within a solid in a state of strain
being given, to find the intensity and
obliquity of the stress at that point
on a third plane through it.
A A' and BB' are the planes of
principal stress at ; p and q are
the principal stresses. Let p) be
the greater, and let them be both
positive, say both tensions. It is
pppp
Fig. 10.
required to find r, the intensity of the stress upon CO',
CHAP. II.]
H^V:s'KINES ELLIPSE OF STRESS.
29
and y, the angle it makes with 01^, the normal to C'C\
B is the inclination of C'C" to AA', the plane of greatest
principal stress.
Of two unequal quantities the greater is equal to the sum
of their half mm and their half difference, while the lesser
equals their dift'erence.
Therefore
p + q p  q .. ...
p = ^ + ^  , an identity,
and
? =
p+q pq
, an identity.
We may look upon the plane A A' (fig. 11) as bearing two
p + q 1 T ~ 9. ■
separate tensions of intensities — x and — ~ in lieu of the
tension of intensity p ; and on the plane BB' as bearing a
a. *
ea4i ^
inch Pt.Q
Z
'■A
Fis. 11.
<uuch.
\l
C$\
'It*;
^1^
rum '"lyterLfum.
tension of intensity ^^ and a thrust of intensity ^ — ^
in heu of the tension of intensity q. We may now group
these together in pairs, thus : the tension on AA' of intensity
'P + Q
~Y alo°R with the tension on BB of intensity ^^^, and
the tension on AA' of intensity ^^ along with the thrust
30
APPLIED MECHANICS.
[chap. II.
on BB' of intensity
jpq
Then find separately for each pair
the stress upon CC\ and finally compound these two stresses on
CC by means of the triangle of forces. The first pair is a pair
of equallike principal stresses ( both tensions of intensity —
So the consequent stress on CC will be a tension of intensity
^^, and normal to CC (fig. 12).
The second pair is a pair of equal
unlike principal stresses of intensity
^ (a tension on AA' and a thrust
on BB')^ so the consequent stress
on CC will be of intensity
A
2)q
Fig. 12.
and inclined at an angle vipon
the side of OX opposite from that upon which ON lies.
The diagram shows these partial resultant stresses on AB,
a very small part of CC , at 0. To find the total resultant
stress upon CC\ it remains to compound these by the triangle
of forces. From (fig. 13) lay off OM = ^^^ = the intensity of
X
^^
^^
Fig. 13.
the first partial stress and in the direction thereof, i.e., along OA^.
From M draw MB =
pq
= the intensity of the second partial
stress and in the direction thereof, i.e., parallel to OS, which
CHAP. II.]
KANKINES ELLIPSE OF STRESS.
31
direction is most conveuieutly found by describing from M as
centre with radius MO a semicircle QOP, and joining QMP.
Then will OR, the third side of the triangle OMR, taken in
the opposite order (see arrows) be the direction and intensity of
the resultant stress r on CC.
The preceding construction, as shown on last figure, is
geometrically all that is required, p and q being given to find
r\ the text and figures given before being the development
and proof.
From the construction note that
MP = MQ= OmJ^;
also
and
QR.MQ.MR = ^^^^^
PR = MP  MR =
p+q pq
= ?;
2 2
RMN =26; R03f =y, the obliquity of r.
Normal and tangential components of r, the stress on the
third plane CC.
Drop RT perpendicular to ON.
The tangential component of r is
rt = TR
= MR sin RMT
pq
sin 2B
= {p  q) sin 6 cos 6.
since sin 20 = 2 sin d cos 0.
Fig. 14.
Cor. — If DJ)' be the plane at right angles to CC, its
inclination to the axis OX being 0' = (0 + 90°), the sine of
which equals cos 0, and the cosine of which equals  sin ;
the value of Vt for DI/ will be the same as above, that is,
the tangential components of the stresses on any pair of
rectangular planes is the same.
32 APPLIED MECHANICS. [CHAP. II.
The normal component of r is
r„= OT=OMMT
= OM  MR cos EMT= 0M+ MB cos EMN.
(These two angles have the same cosine,
but of opposite sign.)
= OM ^ ME cos 26
= ^4^ (cos2 e + sin^ 6) + ^^ (cos^ Q  sin^ B)
= cos^ ( — ^ + ^—^ ) + sm^ ^(^} ~ 9"^ )
= p . COS" d + q . sin^ 6.
Cor. — If 5„ be the normal component of stress on DI/, the
plane at right angles to CC\ whose inclination to OJC is
0' = (0 + 90°), then
Sn = p cos 6' + q sin* 6'.
But cos 0' =  sin ©, and sin 0' = cos 6* ;
.•. Sn = p sin + g' cos'' d.
Now, r„ = JO cos^ + q sin^ 0,
and adding, we get
Sn + I'n = p (sin B + cos B) + q (sin + cos^ B) = /j + ry.
That is, the sum of the normal components of the stresses on
any pair of rectangular planes is equal to the sum of the
principal stresses.
As CC moves through all positions, M moves in a circle
round 0, and B moves in a circle round M, OM and MB
keeping equally inclined to the vertical on opposite sides of
it. The diagram shows their positions for eight positions of
CHAP. II.]
KANKINES ELLIPSE OF STKES8.
33
the plane CC. The locus of B is an ellipse, the major semi
axis being
ORi = OM, + MiR,
P + 9 . PfL
0, 9.
p;
and the minor semi ,'
axis is )
OR^ = OM,  M,R,
^p+q P1^
2 2^'
The moving model,
fig. 31 following, shows
these positions nicely.
This is called the
ellipse of stress for the
point within a solid
in a state of strain. Its
principal axes are the
normals to the planes
of principal stress, the
principal semiaxes be
ing equal to the inten
sities of the principal
stresses. The radius
vectors OR., OR,, &c.,
are the stresses in direc
tion and intensity upon
the planes at to which
OM:^, OM3, &c., are re
spectively the normals.
The ordinary tram
mel (fig. 16) for con
structing ellipses con
sists of a piece like FRQ,
whose extremities F and
Q slide in two grooves,
JCOJC' and YOY', at
right angles to each
other, while the point
R traces an ellipse whose semiaxes are PR and QR.
34 APPLIED MECHANICS [CHAP. II.
When Q arrives at 0, J? is at ^ and OA = QR = p ; when
P arrives at 0, ^ is at P and OB = PR = q.
Taking as origin, the coordinates of R are
A. = Ovi ; y = On; :. x = nR = QR .cos d = p. cos 0,
X V
and y = mR = PR sinB = q sin 0; .'.  = cos 0, and  = sin 6/ ;
p q
.. — +  = cos^fl + sin'0 = 1,
P ?
the ordinary equation to an ellipse in terms of the semiaxes
p and q.
If p and q are both thrusts, it is convenient to consider a
thrust positive, and the proof is exactly the same, all the sides
of OMR representing the opposite kind of stress to what they
did in the last case.
When p and q are unlike, the kind of stress of which the
greater p consists is to be considered positive.
Thus, it p > q, and p a tension while $ is a thrust, the
preceding proof will hold if q be considered to include its
negative sign ; but in this case if ( q) be substituted for q,
we have arithmetically
OM = P^. and Mll=l^
SO that now OM < MR (see fig. 18).
It is important to notice that although OM is always
positive, that is like p, the greater principal stress, yet MR is
now positive or negative according as d is less or greater
than 45°, and r = 031 is positive or negative according as b
is of a lesser or greater value than that shown on figure 18.
Hence, the proposition is proved generally.
An advantage of this geometrical method, the ellipse of
stress, is that we are now in a position to examine the value
and sign of r, the stress upon a third plane CC, and of its
normal and tangential components for special positions of that
plane. OM is always normal to CC, while MR generally is
resolvable into two components, one tangential to CC and the
other normal, which last has to be either added to, or subtracted
from, OM to give the total normal component according as
OMR is an obtuse or an acute angle.
CHAP. II.]
RANKINE S ELLIPSE OF STRESS.
35
(a) Portions of GC for which r, the stress upon it, will have
the greatest or least value.
Since OM and MB are constant, OR increases as the angle
0MB increases, is greatest when 0MB = 180°, and OM and
MB are in one straight line, and a continuation one of the
other, when
OB=OM+ MB;
,. P + 1 ^P Q
P
^=4.5°
and OB is least when z 0MB, is zero, and OM and MB are
again in one straight line, but MB lapping back on OM, when
OB=OMMB; r = ^ili  ^ = j.
Hence the planes of principal stress are themselves the planes
of greatest and least stress.
(b) Position of CC for which the intensity of the shearing
stress has the greatest value.
As OM is always normal
to CC, it does not give any
tangential component, whereas
MB assumes all positions as
CC changes, and will give a
component tangential to CC,
which will be the greatest
possible when MB is altogether
tangential to CC. Hence the
position of CC, which makes
MR parallel to CC, is that
for which the shearing stress
has the greatest possible in
tensity (fig. 17).
Fig. 17.
Hence intensity of shearing stress = MB =  —  •
And since MB is parallel to CC and OM normal to it,
.. 0MB = 90°,
and the triangle MOP being isosceles, we have
B = inclination of CC = MOP = 45°.
D 2
36
APPLIED MECHANICS.
[chap. II.
And we know that the tangential stress is the same on
the section perpendicular to CC ; that is, the planes of
greatest tangential stress are the two planes inclined at 45°
to the axes.
(c) Position of CC for which the total stress r wpon it will
he entirely tangential.
When ci is like 'p, it is impossible for the stress to be
entirely tangential to CC', because OM > MB, and, how^ever
acute 03IE may be, the normal component of MH, which has
to be subtracted from OM to give the total normal stress.
tenswn,
■posttbre
'drmst
w
Fis. 18.
cannot be greater than MB itself, and consequently is always
less than OM, and so there will always be a remainder ; that
is, for all positions of CC there is a normal component stress
and the total stress can never be entirely tangential.
But when ([ is unlike p, then OM < MB, and for the par
ticular position of CC (fig. 18), when the angle OMB is of
such an acuteness that the normal component of MB, which
has to be subtracted from OM to give the total normal stress,
is exactly of the same length as OM ; then the total normal
stress will be zero, and the total stress r entirely tangential.
This occurs when B is in one straight line with CC.
BOM is then a right angle, making MO, the normal com
ponent of MB, equal and opposite to OM, which it destroys,
CIIAl'. II.]
RANKINES ELLIPSE OF STRESS.
37
leaving the total stress OE tangential to CO' ; its magnitude
is found thus :
OE = 3IE'  OM'
or
r* =
fp_+g
\ 2
2
= p q.
i.e., the stress on CC is the geometrical mean of the principal
stresses. Also
26 = EMN; .'. cos 26 = cos E3m,
MO
cos 2e
cos EMO =  ;;r = _ {_
ME
p + g.
which determines 6, the position of CC for which the total stress
is tangential.
Here we must guard against supposing that the above is the
position of CC for which the tangential stress has the greatest
intensity ; for case (b) holds for all conditions of p and q ;
that is, the tangential stress on CC when inclined at 45°,
although only a component of the total stress, will be of greater
intensity than the total tangential stress in case (c).
(d) Position of CC for vjMch y, the oUiquity of the stress
tJiereon, is the greatest possible.
When q is unlike f), case (c) is the solution, for in it
7 = EON = 90°, the greatest possible.
When q is like jj, OM > ME ; and the obliquity of OE, the
stress on f^C, is greatest
when y = EOM is the
greatest possible of all
triangles constructed
with OM and ME for
two of their sides. This
occurs when OEM is a
right angle.
For suppose the tri
angle OME constructed
with OEM not a right
angle ; then drop ME'
at right angles to OE. It is evident that MR is less than ME.
Now sin EOM = ^iv^ is greatest when ME' is greatest ; that
38
APPLIED MECHANICS.
[chap. II.
is, when MB = MB ; that is, when 0E3£ is a right angle, and
BOM is greatest when its sine is greatest.
In this case the intensity of the stress is
OB' = OM'  MB' ;
fP + 2
p_q
2
= PI,
and
\ ^
a geometrical mean between the principal stresses.
Also 20 = BMN; cos 20 = cos BMN
cos 20 =  cos BMO = ~= ^^,
OM p + q'
which determines 0, the position of CC for which the stress
has the greatest obliquity possible.
Note that these values of r and cos 20 are the same as
those of (c), and that whether jj» and q are like or unlike. But
this is not the case with y, the obliquity, which is 90° when
p and q are unlike, and has
alike.
p + q
for its sine when p and q are
Examples.
1. At a point within a solid in a state of strain the principal 'stresses are
tensions of 255 lbs. and 171 lbs. per
square inch. Find the stress on a plane
inclined at 27° to the plane of greatest
principal stress (fig. 20).
Data.
p = 255,
hence
p + q
g = n\, and fl = 27° ;
= 213, and
= 42.
Construction. — OX and OY, the axes
of principal stresses ; draw ON the
normal to CC, making XON = 9 = 27°.
Lay off along it OM = ?^ = 213.
From M as centre with radius MO, de
scribe semicircle FOQ and join FMQ\
f> q
2
= 42. This construction makes MR to
be inclined to OX at an angle 9 = 27°,
l)ut upon the opposite side of it from
that of OM.
lay off from M towards P, MR =
Fig. 20.
Looking upon the principal stresses as a pair of like principal stresfes, tensions
CHAI'. II.]
RANKINES ELLIPSE OF STRESS.
.39
of intensities 213, together with a imir of unlike principal stresses, a tension and a
thrust of intensities 42. Then OM represents a tension 213 upon plane CC due to
first group, and Mli the tension 42 upon CC due to second group ; hence OJt, the
third side of the triangle, taken in the opposite direction, represents the total stress
upon CC in direction and intensity.
OH = OM + MR  20M . MB cos OMR,
cos OMR =  cos RMN =  cos 2fl :
.. 6»i?2 = OM + MR + lOM . MR cos 20 ;
r = 45369 + 1764 + 17892 cos 54° = 57649 ;
r = 240 lbs.
sin 7 sin ROM MR
sin~2e
but
Also
sill OMR
42
240
X sin 54°= 14158;
and figure shows that »• is upon the same side of the normal as OX.
tension, since OR is like OM.
2. In Ex. 1 find thf intensity of the tan
gential stress on that plane thioiigh the point
upon which the tangential stress is of greatest
intensity (fig. 21).
The plane is that which is inclined at 45° to
the axes of principal stress.
Since OMR is 90°, MR is the tangential
component of OR,
Also r is
= MR
= 42 lbs. per square inch.
3. In Ex. 1 find the obliquity to the plane
of greatest principal stress of that plane, through
the point, upon which the stress is more oblique
than upon any other ; also find the stress
(fig. 22).
OJ/and MR being constant, the angle MOR
has its greatest value when MRO is a right
angle.
Construction. — Upon OM describe a semi
circle ; from M as centre, with radius MR,
describe an arc cutting the semicircle in li ;
join OR.
cos 20 = cos RMN =  cos OMR
MR __P  q _ 42
~0M:~~ p + q ~~ 213
=  19718;
.. 26 = 101" 22' obtuse, cosine being nega
tive ;
.. e = 50° 41', obliquity of CC .
r2 = OR"" = OM  MR
Fig. 21.
H^yi'^)
p • q;
= V p . q = v 43605 = 2088 lbs. per square inch of tension like OM.
40
APPLIED MECHANICS.
[chap. II.
MR
And sin y = sm RON = —  = 19718 ;
MO
7 = 11° 22', obliquity of r.
4 . The principal stresses at a point being a tension of 300 lbs. and a thrust of
160 lbs. per square inch.
Find (a) the intensity, obliquity, and kind of stress on a plane through the
point, inclined at 30° to the
plane of greatest principal pp pp posituv
stress ; (*) find the intensity
of tangential stress on the
plane upon which that stress
is greatest ; and (c) find the "^
inclination to the plane of
greatest principal stress of
that plane upon which the
stress is entirely tangential
and the intensity thereof.
Data.
j? = 300; q =  160,
considering a ten
sion positive ;
.'. 7j— = 70 tension like p ;
pq
and — = 230 tension like p.
(a) Construction. — Draw Fig. 23.
ON at 30° to OX (fig. 23).
Lay off OM = 70. From ^as centre, with radius MO, describe semicircle TOQ.
Lay off MPR = 230. Then OR, the third side of the triangle OMR, taken in the
opposite order, is the stress on CO in direction
and intensity.
OR^ = Oif2 + MR^ lOM.MR cos OMR
= OM'' + MR" + 20M.MR cos 29,
r2 = 4900 + 52900 + 16100 = 73900,
r = 272 lbs. per square inch ;
sin 7 sin ROM MR
and
sin 2fl sin RMO OR
2.30
sin 7 = ;;;;7r sin 60° = 7323.
272
7 = 47° .5', being acute, OR is like OM,
a tension.
{b) Take
9 = 45° ; vt = MR
230 lbs.
(c) On MR (fig. 24) describe a semicircle, and from J/, with raiiin.s MO,
describe arc cutting it at 0.
CHAP. II.]
HANKINES ELLIPSE OF STRESS.
41
IiMX= 26, cos 20 = cos RMN ■■
cos RMO = ^=~ =  3044 ;
MR 230
28 = 107° 44' ; = 53° 52', obliquity of plane upon which the stress is
entirely tangential.
r2 = OR = MR  OM = 52900  4900 = 48000, r = 219,
or r = ^ p . q = v (300 x 160) = 219 lbs. per square inch.
Note that, though r is entirely tangential, it is less than r, was in (b).
5. At sam'fe point as in Ex. 4, find intensity, kind, and obliquity of a stress
on a plane inclined at 85° to the plane of greatest principal stress (fig. 25).
Since > 53° 52', > the obliquity
of plane upon which the stress was
wholly tangential, OR will make
with ON an angle greater than 90°,
and OR will be unlike OM, and
therefore a thrust.
Arts, r = 161*5 lbs. per sq. in. ;
y = 165° 41'.
6. The principal stresses on AA'
and BB' are thrusts of 60 lbs. per
square inc^h. Find direction and
intensity of the stress on a third
plane CC inclined at 65° to A A'.
Ans. A thrust of 60 lbs. per
square inch normal to CC.
7. The principal stresses on A A' and BB' are of the equal intensity of 34 lbs.
per square inch, being a thrust on AA' and a tension on BB'. Find the direction
and intensity of the stress on a third plane CC inclined at 65° to AA'.
Ans. A tension of 34 lbs. per square inch, its direction being inclined at 65°
upon the other side of OX from that to which ON is inclined.
8. The principal stresses on AA' and BB' at a point are a thrust of 94 lbs.
and a thrust of 26 lbs. Find kind, intensity, and obliquity of a stress on a third
plane CC inclined at 65° to AA'. Using results of Exs. 6 and 7,
r = 46*2 lbs. per square inch thrust, y = 34° 19'.
9. Two unlike principal stresses are: on A A' a thrust of 146 and on BB' a
tension of 96 lbs. per square inch. Find the stress on CC, a third plane inclined to
^^'at 50°.
p = 146, and g =  96.
Half sum is a thrust like p,
2 ^
p — g
half diflf. — r — is a tension like g, since > 45 .
Atis. r = 119'22 lbs. per square inch thrust, y = 88° 5'.
42
APPLIED MECHANICS.
[chap. II.
Inverse Pkoblem, to find the Peincipal Stresses.
Given the intensities, obliquities, and kinds of the stresses
upon any two planes at a point within a solid, find the principal
stresses and their planes.
In the general problem we know of the triangle OMR
(fig. 13), the parts OR and 7
for two separate positions of
the plane CC, and we also
know that OM and MR are
the same for both.
If the two given stresses
be alike and unequal (fig. 26),
let r and r' be their inten
sities, and 7 and 7' their
obliquities upon their re
spective planes CC and BB' .
Let r be greater than >■'.
Note that it is not necessary
to have given the inclina
tion to each other of CC
and BB'.
Choose any line ON and
draw OR = r, and making
the angle NOR = 7, also
draw OR! = ?■', and making
the angle NORil = 7'. Join
RR', and from S, the middle
point of RR', draw, at right
angles to it, SM meeting ON
at 31. Then will JIR = NR'.
Thus we have found OM
comparing
we have
Fig. 26.
and MR to suit both data, and
the construction of the direct problem (figs. 3, 4)
and therefore
0M.'^\^, and MB=PJ,
p = OM+ MR; q = OM MR.
Consider the triangle OMR' alone, and consider ON' the
normal to BB' : then R'JI'N = 29' ; hence OJC, drawn parallel
to M'T (the bisector of R'M'N), is the axis of greatest principal
CHAP. II.] kankine's ellipse of sthess. 43
stress. Thus we have found the principal stresses p and 7, and
the position of their axes OX and OFrelative to DD\ one of the
given planes.
Since
li'JIR = JifMN  JULY = 'IB'  26 ; .. liMS = 6'  B,
the inclination to each other of CO' and DU' ; hence if tlie other
triangle OMU be moved round through this angle, it and
consequently CC, to which ON is the normal, will also be in
their proper positions with respect to the axes l^X and OY.
This triangle might be further turned round till ON is
inclined at an angle XON = B on the other side of OX, when
CO' would again be in a position for which the stress would be
the same as given. This would increase the relative inclination
of BD' and CO' by twice JCON or by 2B. Adding this to B'  B
gives 6' + B. That is, the inclination of CO' and DD' to each
other is
(B'  B) = BMS on diagram,
or
(B' + B) = NlMS on diagram,
according as they lie on the
same or on opposite sides of
OJC, the axis of principal stress.
If the two given stresses ^'
be unlike and unequal (fig. 27), °'
considering r the greater as positive, /■' will be negative. Follow
the same construction, only OM' = r' must be laid off from in
the opposite direction. Complete the figure as before, and we
have from either figure —
TrigonometHcally
MR = OM' + OE"  20M. OR cos MOB
= OM^ + r^ 20M . r cos 7.
Similarly,
MR'^ = 03P + r" + 20M . r cos 7'
from figs. 26 and 27 respectively.
Subtracting, = r'  r"^  2031 (r cos 7 + / cos 7'),
and therefore ^^ = OM = — "^^ ;, (A)
2 2 (r cos 7  r cos 7 )
/ to include its sign ;
44
APPLIED MECHANICS.
[chap. II.
also
or,
P1
= MR = ^ (03/2 + 72  20M. r cos 7)
= ^(071/2 + /2 _ 2OM. r cos y)
(B)
a known quantity when the value of OM is substituted from
equation (Aj.
;p and q are now obtained by adding and subtracting
equations (A) and (B).
From R drop RL perpendicular to ON, then
ML = OL OM; MR . cos NMR = OR . cos ROM  OM,
^^cos20 = rcosy^:l;
cos 26/ =
1r cos y  p  q
(C)
This gives twice the obliquity of the axis of greatest principal
stress to the given plane CC\ and similarly for DD'
cos 2H' =
2r' cos y'  p  q
p q
These three equations (A), (B), and (C) are the general
solution of the inverse problem of the ellipse of stress. fA)
and (B) give the intensities of the principal stresses, which will
come out with signs showing whether they are like or unlike r,
the greater of the given stresses.
In some particular cases the construction gives a much
simpler figure from which the equations (A), (B), and (C) in
their modified form are readily calculated.
Particular case [a) (figure 28;. Given the intensities and
common obliquity of a pair of
conjugate stresses at a point ; find
the principal stresses and posi
tion of the axes of principal
stress. (Note — There are more
than sufficient data.)
In this case 7 = 7', and R,
S, and R' are in one straight
line with 0.
Draw any line ON; draw OR, making NOR = 7 = 7'; and
Fig. 28.
CHAP. II.] It^VNKINE'S ELLIPSE OF STRESS. 45
lay off OR = r and OE' = r in the same or opposite directions
according as it is like or unlike r ; and from S, the middle
point of BR, draw SM at right angles to it, meeting OiVat M.
Join M to R and K.
Then
— = cos MOS ; ai/ = ,,7^ = rpTH^ ,
X)M COSMOS COSMOS
or i^ = (A)
2 2 cos 7
Again MB' = MS' + BS" = {OM'^  OS') + BS"
= OM^  {OS'  BS') = OM^  {!l±l'J(^!:^'Jj
+ /V
(B)
= OM'  rr = I 1
\ 2 cos 77
9>.
1 o^ Ir COS J jpq
and cos 20 = — ~ (C)
pq
as in general case,
^OjV' =0' +9 = mis = MSO + MOS = ^ + 7.
Hence, the angle between the two normals to the sections
CC" and DD' (or the ohtuse angle between CC and 1)1/) exceeds
the obliquity by a right angle. This we know ought to be the
case from the definition of conjugate stresses.
It may be easier to calculate cos 26 =  cos B3I0 in terms
of the sides of the triangle 03IB when these have been already
found.
From M as centre, with radius MB, describe the semicircle
HBfBN; then (fig. 28)
OH = OM  MB =q; ON ^ OM + MB = p.
But
ON X OH = OB X OR (Euc. iii. 36), and pq = rr' (B,)
may be used instead of (B).
46
APPLIED MECHANICS.
[chap. II.
Particular case (b) (fig. 29). Given the intensities and
obliquities of the stresses on a pair of rectangular planes,
find the principal stresses
and the position of the axes
of principal stress. (Note
— There are more than suffi
cient data.)
If r and r be like
stresses, draw any line ON'.
Draw OR = r, making
NOR = y, also OR' = r
making NOR' = y.
Complete the figure as
before.
The given planes being at right angles are necessarily
inclined upon opposite sides of the axis of principal stress;
hence
NMS = inclination of given planes = 90°,
and RSR' is parallel to ON.
Therefore MS = RL = r sin y = R'K = r' sin y,
or r sin y = r' sin y'.
That, is the tangential components of r and r' are equal,
p + q
OM =^{OL + OK);
2
= ^{r cos 7 + / cos 7'). (A)
That is, the sum of the principal stresses is equal to the sum
of the normal components of r and r'. (Compare page 32.)
MR' = RS' + MS' = (^^Ll^Jy + MS^
(r cos y  r' cos 7')^^
+ r' sin'' 7 ;
pq
tan 20
(r cos y  r cos y)
4
RL
+ r sm7 I •
r sin 7
(B)
RL _
ML ~ ^{OL  OK) \ (r cos 7  r' cos 7')
2r sin 7
r cos 7  ?•' cos 7'
(C)
CHA.P. II.] RANKINE'S ellipse OF STRESS. 47
Putting Tt = MS = r sin y = r sin 7' = the common value
of the tangential components of r and r ; also
r„ = OL = r cos y = norm. comp. of r,
Tn = 0K=^ r cos 7' = norm. comp. of r ,
the equations become
and
2
P1
2
tan 20 =
>/
2 '
+ n^
2r,
?•„  ?
When T and ?'' are
unlike stresses, con
sider r, the greater,
as positive ; then must
OR be laid oft' in
the opposite direction
from (fig. 30).
^ovfR'K=RL = rt, 4,,^
the common tangen '.
tial component of r
and r ; hence LiR'
and KL bisect each fi
other at *S' or M, which
coincide.
Fig. 30.
0M=^ {OL  OK)
MR
p + q r cos 7  r cos 7
J/Z^ + RL = ( ^^ + RL'
or
^2^=J{
(r cos 7 + r' cos 7')
+ r' sin
^^ i^Z RL
tan 20 =  =
i^Z
J/Z UK liOL+OK)
2?' sin 7
r cos 7 + r' cos 7'
(Ax)
(Bx)
(Cx)
(A.)
(B,)
(C,)
These three equations (A2), (B2), and (Co) are identical with
(A), (B), and (C) with ( /) substituted for /.
48 APPLIED MECHAl^ICS. [CHAP. II.
The diagram (fig. 31) represents the kinematical model in the
Engineering Laboratory of Trinity College, Dublin, devised by
Professors Alexander and Thomson for illustrating Kankine's
Method of the Ellipse of Stress for Uniplanar Stress. On the
left hand of the diagram, ON is a T square, pivoted to the
blackboard at 0, and MR is a pointer pivoted to the blade of
the T square at M. On the pivot M a wheel is fixed at the
back of the blade, and round the wheel an endless chain is
wrapped, which also wraps round a wheel fixed to the board at 0.
The wheel at 6* is of a diameter double that of the wheel at M.
Hence, when the blade of the J square is turned to the right,
the pointer MR automatically turns to the left, so that the
angle RMN is always equal to 20 when PON equals Q ; or, in
other words, the bisector of RMN is always parallel to OP. A
cord is fastened to the pointer at ^ ; it passes through a swivel
ring at 0, and is kept tight by a plummet. It is easily seen
that the point R describes an ellipse.
CC, the headstock of the T square, represents the trace on
the blackboard of any plane through normal to the board,
and the vector of the ellipse, consisting of the segment RO of
the cord, represents the stress on the plane CC in intensity
and direction. The angle PON = 6 is the position of the plane
CC relative to the plane of greater principal stress AA, while
RON = 7 is the obliquity of the stress r upon CC.
The model* shows clearly the interesting positions of the
plane CC; thus CC may be turned to coincide with A A, when
the cord RO will be found to be normal to CC and to be of a
maximum length. On the other hand, if CC be turned to
coincide with BB, the cord RO is again normal to CC, but of a
minimum length. Again, when CC is so placed that the angle
RMO is a right angle, the component of RO parallel to CC is a
maximum ; and lastly, if it be turned till MRO is a right angle,
then ROM, the obliquity of the cord, is a maximum.
The auxiliary figure on the right of the diagram is for solving
the general problem of uniplanar stress at a point, viz. given
the stress in intensity and obliquity for two positions of CC, to
* Tlie model was exhibited to the Royal Irish Academy early in 1891, and
described for the first time in the Academy's Tratisactxons. Reference may be
made to Rankine's Civil Engineeriixj, or Applied Mechanics, and to Williamson's
Treatise on Stress. Numerical examples are liere worked by this method, and in
Howe's Retaining Walls. The model is made by Messrs Dixon and Hempenstal,
SuEFolk Street, Dublin.
In Williamson's Treatise on Elasticity, this model is shown in its proper place
relative to the complete systematic treatment of elasticity.
CHAP. II.]
RANKINES ELLIPSE OF STRESS.
49
find the stress for any third required position of CC. On the
auxiliary figure, the T squares for all positions of CC are super
imposed upon each other and are represented by one T square
fixed to the board, and only the pointer MR turns.
In solving questions on the stability of earthworks, some
linear dimension upon the auxiliary figure represents the
known weight of a column of earth, while another dimension is
the required stress on a retaining wall or on a foundation, &c.
Generally two angular quantities also are known, such as
KON = (p, the maximum value of 7, this being the steepest
possible slope of the loose earth, while HON = 7 may be the
known slope of surface of earth.
On the other side of the board, another T square is similarly
pivoted to the same pin 0, the only difference being that OM is
shorter than the pointer MR. It represents the case of Unlike
Principal Stresses, and is useful in demonstrating the composition
50
APPLIED MECHANICS.
[chap. II.
of stresses such as that of thrust and bending on a pillar, or of
bending and twisting on a crankpin.
By means of the slots on the easel the board can be placed
so that any desired vector of the ellipse may be vertical.
Examples.
10. If from external conditions it be known that the stresses on two planes
at a point in a solid are thrusts of 54 and 30 lbs. per square inch, and inclined at
10° and 26° respectively to the normals to these planes — find the piincipal stresses
at that point ; the position of the axis of greater principal stress relative to the
first plane ; and the inclination of the two planes to eacli other.
Make
and
Lay off
and
NOR =7 = 10°,
NOR = y = 26°.
OR = r = 54,
OS = r = 30.
Join RR\ bisect it in S, draw SM at
right angles to RR' , meeting ON at M:
complete figure.
Fig. 32.
Then
p + q
= OM, and
pq
= MR = MK,
also 2d = NUR.
2 ' 2
or p = (OM>r MR), and q = {OM  MR),
Trigonometricatly
MR' = OM + Oi?2 20M. OR cos MOR = OM^ + r^  20M . r cos 7.
SimHarly MR = OJP + r'^  lOM. r cos 7'.
Therefore subtracting
= r^  j'2 _ 20M{r cos 7  r' cos 7') ;
r^  r"^ p+q 2016
OM =  , „.
2 (»• cos 7  r' cos 7') 2 5243
MR^ = (3845)2 + (54)2 _ 2 x 3845 x 54 cos 10°
= 14784 + 291G  40888 = 3056 ;
^^ = \/305^ = 1748.
= 3845.
The principal stresses are
P
p + q p — q
+ ;r— = 55"93 lbs. per square inch thrust, like r,
2
2
pq
9 = —2~ ~ ~~^  '0'97 lbs. per square inch, thrust being .
(A)
(B)
CHAP. II. j
RANKINKS ELUPSE OF STRESS.
51
Drop RL perpendicular to ON,
ML = OL  0.\f, or MR cos LMR = OR cos LOR  DM;
p + q
pq
cos '10 = r cos 7 
(C)
e = 16° 17^'= XON,
,03 ,e = °^^^^f'^^ = 8426 ; .. 2. = 320 30';
1 1 '48
the inclinakion of OX, the axis of greatest principal stress, to ON, the normal to
the plane for which r was given.
Similarly,
cos 26' = — "6573 ;
.. 26' = 131° 6' (obtuse for  sign) ;
d' = 65° 33', inclination XON'.
And inclination of the two planes to each other
NON = RMS = {ff e) = 49° 15f ,
or = NMS = (6' + e) = 81°50r,
according as they are on the same or opposite sides of OX.
11. At a point within a solid a pair of conjugate stresses are thrusts of 40 and
30 lbs. per square inch, and their common
obliquity is 10°. Find the principal stresses
and tiie angle which normal to plane of
greater conjusiate stress makes with the axis
of greatest principal stress.
Draw OR, making NOR = 7 = y = 10° ;
lay off OR = >■ = 40, and OR' = r' = 30.
Bisect i?A' in S, draw SM perpendicular to
£R'; complete figure. Then
^i^ = OM, and ^^ = J/J?, and 20^RMN.
OS
OM
OM.
cos 7
!(>• + /) ^ p + q
cos 7 ' 2 9848
MH = MS + JiS = OM^  [OS^  RS^)
— [(^r(4^)i='
.. ^^ = /(12638  1200) = 8.
Adding and subtracting (A) and (B),
I? = 43*5 a thrust, and j = 27 5 a thrust,
788  71
cos 20 = ~ = 49.
16
(B)
(C)
E 2
52
APPUED MECHANICS.
[chap. II.
Therefore 28 = 60° 40', and = 30° 20'.
Or geometrically describe semicircle HL'RN (fig. 33),
ON = O.U + MR = p, and OH = OM  MR = 9.
ON .OK = OR . OR' (Eiic. iii. 36), or p . fj = rr' = 1200. (B')
Now p + ? = 711; (A)
.. p'^Jr2pq+ q^ = b0bb2\
bnt(B), 4j09 = 4800 ; ..jt?' 2;?^ + j^ = 2552; .. j»j=16:
adding to and subtracting from (A),
.. 2;? = 711 + 16, and 2^ = 71116: .. jo = 4355, and ? = 2755.
12. At a point witliin a solid, a pair of conjugate stresses are 182 (tension) and
116 (ihrust). common obliquity 30°. Find the principal stiesses and the pnsilion
of axes (fig. 34).
r' is negative
p + q
2
p  q
= C>ilf= 3814,
= MR = 1503 ;
and
p= 1884 (thrust),
.. q =  1122 (tension),
cos 20 = 7947 ;
.. e=lS°41'.
Fig. 34.
Fig. 35.
13. The stresses on two planes at right angles to eacli other being thrusts of
240 aTid 193 lbs. per squaie inch and at obliquities, respectively, 8° and 10°, find
tlie principal stresses and tlieir axes (fig. 35).
r,, = r cos 7, and »•'„ = r' cos 7' ; also r< = r sin 7 = r' sin 7' ;
= 2376 =190 =334
iLJ1 = OM =
2138,
P  <1
2"
= MR = l\ ( ''" ^''" \ +r,'j = v/(566+ 1115) = 41
j9 = 2548, and 9=1728; tan 26
26 = 54° 32'; .. d = 27° 16'.
l'V = 14034;
In — r„
CHAP. III.] STABILITY OF EARTHWORK. 53
CHAPTER III.
APPLICATION OF THE ELLIPSE OF STRESS TO THE STABILITY
OF EARTHWORK.
Loose earth, built up into a mass on a horizontal plane, will
only remain in equilibrium with its faces at slopes whose
inclinations to the horizontal plane are less than an angle 0.
If the earth be heaped up till the slope is greater, it will inin
till the slope is at greatest ^. Moist and compressed masses
of earth can be massed up into a heap with slopes greater
than ip. and will remain in equilibrium for some time, but
will ultimately crumble down till the slopes do not exceed <\,.
The surfacesoil, which is in a compressed state, may be cut
away, leaving banks with slopes much greater than ^. These
banks will only remain in equilibrium for a time. Slips will
occur till ultimately the slopes are not greater than <p.
This angle (p, which is the greatest inclination (of the slopes
to the horizontal plane) at which a mass of earth will remain
in equilibrium, is called the angle of repose. It has different
values for different kinds of earth, and also different values for
the same earth kept at different degrees of moistness. Average
values of (p for different kinds of earth have been ascertained by
experiment and observation, and are tabulated.
If two particles of earth are pressed together by a pair of
equal thrusts p and p normal to their surface of contact, it
requires a pair of equal thrusts q and q'
tangential to that surface to make them p
slide upon each other. For the same
material, when q is just sufficient to
make them slide, it is a constant frac .
tion of ;;. The fraction which § requires ^^ ^t ^ ^amtact
to be of ;; just to cause slipping is called
the coefficient of friction for that mate
rial. Hence the coefficient of friction
? Fig. 1.
P
Figure 2 is a section of two troughs enclosing earth, and
54
APPLIED MECHANICS.
[chap. III.
pressed together with a thrust of intensity^ normal to MN, the
plane where the troughs are just not in contact, and P is the
amount of this thrust. A thrust of intensity q tangential to
the plane MN tends to cause
the earth to slide in two parts
along MN, also Q is the amount
of this thrust. If Q be just
sufficient to cause slipping
along MN, then the coefficient
of friction of the earth is
u
f^ =
Q
P
If
U on AB and CD there be
a thrust of intensity p inclined
at an angle (j) to the normal,
we know that for equilibrium
of the prism ABCD there must be a stress q upon the faces AC
and BD, whose tangential component equals that oi p; but as
far as stability along the plane 3/iVis concerned, we may neglect
q, whose normal components destroy each other through the
material of the trough, and the
tangential ones are at right
angles to MN. Considering
the components of P, the
amount of jo, we have P cos ^
normal to MN. If slipping is
just about to take place, then
A* =
P sin ^
P cos ^
tan (p.
It is apparent that ^ is the
same angle we were before
considering; for, if P be due
to the weight of the material, the figure ought to be turned till
the direction of P is vertical, when MN, the plane of slipping,
will be inclined at ^ to the horizontal. The relation between
the coefficient of friction and the angle of repose is
f^i = tan <p.
Note. — If it were not upon the supposition that the two
troughs (being very rigid compared to the earth) transmitted
the equal and opposite forces tangential to Jl/^A'" without causing
CHAP. III.] STABILITY OF EARTHWORK. 55
lateral compression of the earth, we could not neglect 9. From
this result we learn that the tendency to slip along the plane
MN, due to p, depends entirely upon the obliquity of p, and not
at all upon its intensity. Thus, if p be inclined at an angle less
than (p, slipping will not occur though j) be ever so great ; but, if
p be inclined at an angle greater than ^.slipping will take place,
though p be ever so small.
Consider now the equilibrium of a small prism at a point
within a mass of earth in a compound state of strain. The
earth will have a tendency to slip along any plane through the
point (as there is no artificial envelope), except along the planes
of principal stress at the point ; and the tendency to slip will be
greater along the plane upon which the resultant stress is more
oblique, and greatest along the pair of planes upon which the
residtant stress is most oblique, it being of no consequence how
intense the stresses upon these various planes may be, but only
how oblique. If the stresses upon the pair of planes, for which
the resultant stress is more oblique than that upon any other
plane through the point, be themselves less oblique than ^, no
slipping will occur upon any plane through that point ; but if
more oblique than (f>, slipping will take place along one or both
of those planes.
The condition of equilibrium of a mass of earth in a com
pound state of strain is that, at every point, the obliquity of
the stress on the plane upon which, of all others through the
point, the resultant stress is most oblique, shall itself not be
greater than (f>.
Since earth can only sustain thrusts, the principal stresses
at a point will be both thrusts and so excludes case (c), and
if 7 be the obliquity of the resultant stress upon the plane
through the point upon which the stress is most oblique, then
by case (d) (fig. 19, Ch. ii),
p  Q p 1 + sin 7
sm 7  ' ' ; ..  = :, . — ' ■
p + g g 1  sin y
By increasing 7, the numerator of the term on righthand side
of equation increases, while the denominator decreases, and so
the ratio  increases. But ^ is the greatest value of 7 for which
equilibrium is just possible.
p 1 + sin
q 1  sin (f)
is the greatest ratio of ^ to ^r consistent with equilibrium ; hence
56
APPLIED MECHANICS.
[chap. III.
radam
The condition of equilibrium of a mass of earth is most
conveniently stated thus : that at every point the ratio of the
greatest to the least prhicipal stress shall not exceed that of
(1 + sin <p) to (1  sin <p).
Or geometrically,
let OM = ^4"^, niake MOB = <p.
Drop 3/R perpendicular to OR.
Describe the semicircle HEN.
Because MOR = obliquity of thrust on
plane which sustains
most oblique strain,
Fig. 4.
and
ORM = 90°.
and
MR=p;'^.
Therefore
ON = (OM + MR) =^ p,
[
OH = {OM  MR) = q ;
p ON
OM+MR OM+OM sin (f>
' q~ 0H~
OM  MR OM  OM sin ,p
1 + sin ^
1  sin ^
For earth whose upper surface is horizontal, the vertical
stress due to the weight and the horizontal stress are for all
points the principal stresses, and their intensities are the same
for all points on the same horizontal plane. Generally the
vertical is the greater principal stress in any ratio not exceeding
the above ; whenever it exceeds the horizontal thrust by a greater
ratio the earth spreads. But the horizontal thrust may be
artificially increased till it exceeds the vertical in any ratio not
exceeding the above. Whenever it exceeds the vertical by a
greater ratio, the earth heaves up.
The third axis of principal stress, which we are all along
neglecting, is also horizontal. When the earth is in horizontal
layers with a horizontal surface, all vertical planes are sym
metrical, and the three planes of principal stress are any two
vertical planes at right angles to each other and the hori
zontal plane. The stress on the two vertical planes being equal,
the ellipsoid of stress becomes a spheroid. When, howe\er, the
horizontal thrust on one vertical plane is artificially increased,
CHAP. III.]
STABILITY OF EARTHWORK.
57
surftLce. of earth
that plane becomes (me of the planes of principal stress, and the
stress may be different on all three. See the definition of a
geostatic load in a chapter followin<^, the numerical examples
there, and especially the quotations from Simras on Ihinnelling.
Earth in horizontal layer a loaded ivith its ovm weight, to find
the pressure against a retaining vmll with vertical hack.
Let
w = weight in lbs. of a
cub. ft. of earth,
^ = its angle of repose,
D = depth of cutting.
Consider a layer 1 foot
thick normal to paper,
and choose a small rect
angular prism at depth x
feet.
Let
If
p ^ intensity of vertical pressure at depth x feet in
lbs. per square foot
= weight of a volume of earth one square foot in
section, x feet high
= wx lbs. per square foot.
q = least horizontal stress which will give equi
librium, we have
? =
1 + sin0
1  sin
1  sin
? =
1  sin
P
1 + sin (p
w . X lbs. per square foot
1 + sin ,
intensity of horizontal pressure on wall at depth x.
On the right side of equation all is constant but x ; hence q is
proportional to x, is zero at the top, and uniformly increases to
? =
1  sin
1 + sin </»
w . D a.t the bottom,
and therefore
1  sin
1 + sin
wD
—^ = average intensity of pressure upon wall.
58 APPLIED MECHANICS. [CHAP. III.
And the area exposed to this pressure is D square feet. Hence
the total pressure on wall is
Q = average intensity of pressure x area.
1  sin d> roD^ ,,
= . —  . 7 lbs.
1 + sm 2
This tends to make the wall slide as a whole along MP : for
equiHbriurn the weight of the wall, multiplied by the coefficient
of friction at the bedjoint there, must be greater than Q.
If PM be laid off to represent the horizontal pressure at P,
and M be joined to A, then MA gives the liorizontal thrusts at
all points as shown by arrows ; Q, the resultant of all these, is
horizontal, and passes through the centre of gravity of the
triangle APM : it therefore acts at a point C called the centre
of pressure, and
•* 3
Q tends to overturn the wall with a moment
M = Qy. leverage about P.
^ D 1  sin A wD'^ P ^ IT.
= Q'o = ^TT'^^ • T footlbs.
6 1 + Sin o
Let K be the centre of the combined pressure (due to the
weight of the wall and horizontal pressure of earth) on the bed
joint at M\, also let the vertical line, drawn through G the
centre of gravity of the wall, cut the joint at S; then for
equilibrium the moment, weight of wall x leverage KS, must be
greater than M the overturning moment.
It is generally sufficient to ascertain if this lowest bedjoint
be stable : but for some forms of wall it is necessary to go
through all calculations for each bedjoint considered in turn as
bottom of wall.
In a wall of uniform thickness throughout its height, the
weight increases as D; whereas the force Q increases as i)', and
the lowest bedjoint is most severely taxed. Similarly, for
overturning, KS being constant, the product, KS x weight of
wall, increases as D while M increases as D\ K would be
the extreme outside of the wall if the material were perfectly
strong; for stone retaining walls SK the distance from the
middle of base of wall to the centre of pressure at that base
is Iths or ^ths of the thickness.
CHAP. HI.]
STAIJIUTY OF EARTH WOIIK.
59
Depth to which the foundation of a xvall mtist, at least, be
siink in earth laid in horizontal layers consistent with the equi
Uhnum of earth.
Consider one lineal foot
of wall, normal to paper.
V = vol. of wall in cub. ft.,
W = weight of wall per „
h = height of wall in feet,
b = breadth of wall in feet,
d = required depth of
foundation,
IV ^ weight per cubic feet
of earth,
<f> = its angle of repose.
P
"^
sartaceoffeirth
9'
^p' — Y — '
Fig. 6.
When the wall has just stopped subsiding, the earth on
each side is on the point of heaving up, so at the horizontal
layer at the depth of d, for points in contact with the bottom
of foundation, p exceeds q in the greatest possible limit, that
earth being on the point of spreading,
or
1 + sin </)
1  sin <^'
while, for points just clear of it, })' exceeds q in that limit ;
»' 1 + sin <6 , pp
—. = I : — .. and — ;
1  sin <^'
qq
1 + sin ^
1  sin ^
Now p' = q, being horizontal thrust on same horizontal layer,
cancel these and substitute the values
weight of wall WV
area exposed to y> b '
q = weight of column of earth = wd ; hence
bwd \1  sin 0/ ' ' ' wb \1  sm (f>/
60
APPLIED MECHANICS.
[chap. III.
Fig. 7.
Earth spread in layers at a uniform slope, and loaded with
its own weight, to find the pressure against a retaining wall with
a vertical hack.
The simplest (com
monest in practice) case
is when the vertical face
of wall is at right angles
to the section showing
greatest decHvity of free
surface. Let the paper
be that section ; then
AB is the trace of the
upper surface, and 7 is
its greatest inclination
to the horizon.
This inclination must be less than the angle of repose, or
the earth would run over the wall. In an extreme case they
may be equal.
Generally y < (p.
Take a slice one foot normal to paper ; suppose the earth
to be spread behind the wall in layers sloping at the angle y,
consider a small parallelepiped in the layer of depth D having
vertical faces. At this depth JD, the intensity in lbs. per square
foot of the vertical pressure due to the weight of earth above,
on a horizontal surface, would be the weight of a cubic foot of
earth multiplied by the depth I) in feet. Hence
w . D lbs. per square foot
= intensity of vertical pressure on parallelepiped had its surface
been horizontal ; but the sloping surface 3fN is greater than
the corresponding horizontal surface that supports the same
earth ; so the vertical stress thereon will be less than wB,
(fig. 2, Ch. II), and will be
r = wD cos y lbs. per square foot.
This is the intensity of the pressure upon the faces MN and
KL, and its direction is vertical and therefore parallel to any
pair of vertical faces of the parallepiped ; hence the pressure
on any pair of vertical faces is in its turn parallel to the
face llIN; that is, every vertical plane is conjugate to the free
surface.
CIIAI". III.]
STABILITY OF EARTHWOIiK.
61
Now, as we have selected the faces of MNLK, the pressure
on the faces parallel to the paper when drawn parallel to the
free surface will be horizontal, so that the stress normal to the
paper is a principal stress, and the plane of the paper is the
plane of the other two principal stresses. We can apply there
fore our preceding results.
Let r be the stress on the vertical faces ]\IK and NL : it
must be parallel to the free surface, and so its direction is that
of the sloping layer, so that every point in that layer is in the
same state of strain, and r' is transmitted along the layer to act
on the wall.
To find out the ratio of the pair of conjugate stresses r and r'
whose common obliquity is y. Consider
The Auxiliaky Figure to Ellipse of Stress.
Let (fig. 8)
OM = ^^ ; make MOK = 0,
Li
the angle of repose of earth. Drop Jf.S' perpendicular on 0K\
then
MR =
pq_
(Case (d), Ch. ii.)
i'iK. 8.
Draw semicircle
OH ^OM MR = ?,
ON = 0M+ MN^p.
Draw OR'R, making NOR = 7, the common obliquity of
the conjugate thrusts r and r', and
OR
ORf = r.
(Case (a), Ch. il)
62
APPLIED MECHANICS,
[chap. III.
The relations among those are easily expressed trigono
metrically by supposing OM proportional to unity, when
OM prop, to 1 ; radius p prop, to sin ^ ;
OS prop, to cos 7 ; MS prop, to sin y.
BS = ^{MR'  MS), or y{p^  MS')
prop, to v/(sin^  sin^y), or v^(cos^7  cos'^).
}) or ON = OM + p, prop, to (1 + sin<^),
q or OH = OM  p, prop, to (1  sin <^),
r or OR' = OS  BS, prop, to {cosy  v^(cosy  cos<^)!,
and r or OB = OS + BS, prop, to {cos y + ^/(cos^y  cos"^^)}.
r' cos y  \/ (cos"y  cos^^)
r cos y + v/(cOS'y  COS'^) '
p 1 + sin (fy
r cos y + v^(cos^y  cos*^) '
q 1  sin <f>
r cos y 4 v^(cos^y  cos^</>)'
The axis of ^? makes an angle = ^BMN, with OiV the
normal to the (sloping layer) plane upon which r acts and on
the same side.
Also
cos 26 =
2r cos y  p = q
Ijq
(Case (a), Ch. 11.)
Returning to the problem on page 60, and substituting
the value of r, we have the least intensity of the conjugate
thrust at the depth B,
r = wj) cosy
cos y  'v/(cOS^y  C0S(/))
cos y + v/'(cOS'y  COS^0) '
and its direction is parallel to the upper free surface.
On the righthand side of equation everything is constant
except D^ so that r' varies as the depth.
CHAP. 111.]
STABIUTY OF EARTHWORK.
63
Let D be depth of vertical face of wall. Lay off CT to
represent r'. Join AT, and this locus will represent the
thrust on the wall. The average intensity . r, dL
and the total thrust is
18
R' = average intensity x area exposed
r'
=  lbs. per sq. ft. x D sq. ft.
wD^ cos 7  v/(cos'7  cos</)) ,,
= 7rcos7 ' 77 r ttIds.
2 ' C0S7+ v/(cos7cos0)
and it is parallel to free surface, and passes through the centre
of gravity of the triangle A TC so that EC =  •
ResoMng B' into horizontal and vertical components,
H = R' cos 7, V = R' sin 7.
H tends to make the wall slide as a whole alon^ the bed
joint at C; and for equilibrium of the wall, weight of wall x
coefficient of friction at bedjoint must be greater than R.
H tends to overturn the wall with a moment
= h(^\ 
M = H
V3
wD^ J cos 7  v/(cos7  cos'^)
v/(cos'7  cos'^)
COS' 7
6 cos 7
ft.lbs.
For equilibrium of wall, its weight multiplied by KC feet
must exceed M. For position of K see fig. 5 and foot of
page 58.
Note. — V, the tangential component of the pressure of
earth on the back of wall multiplied by KC, tends to resist M
and to increase effective weight of wall, but the friction of the
earth there is liable to be destroyed by water lodging, and it is
not always safe to rely on it.
Geometrical Solution. — r = vjD cos 7, being the vertical con
jugate thrust, on a layer at depth I), due to the weight of the
earth, to find in terms of r,
r', the conjugate thrust parallel to layer.
p and q, the principal stresses in the plane of paper.
64
APPLIED MECHANICS.
[chap. m.
6  y, the inclination to the direction of r {i.e., the
vertical), of the axis of p.
And the tliird principal stress normal to plane of paper.
Since the earth is upon the point of spreading, the principal
stress normal to the paper will be the least possible, that is, it
will be equal to q.
r=OR
 r OR
Fig. 10.
Hence this is the horizontal stress on vertical face of a wall
running up the steepest declivity.
From a point on layer at depth D draw ON (^g. 10) the
normal to layer. Lay oft OM, &c.. complete construction as in
last, but now in its proper position, properly oriented and to
scale.*
Draw OF parallel to 3IT, the bisector of RMN\ this and
OQ are the axes of the ellipse of stress parallel to plane of
paper.
* In Professor Malvarn A. Howe's Treatise on Relainivg Walls, in which he
adopts tliis method of Rimkine's as developed by us, lie conii.ares B;iiiscliiiiger's
construction with that of tig. 10, supra, and siiows that they me ideniictil. Bau
Bchinger's construction is very arbiti iiry, and fails lo recommend itself as Uankine's
does by rational steps readily remembered.
CHAP. 111.]
STABILITY OF EARTHWORK.
65
Lay otf OP = ON and OQ = OH, and draw ellipse ; since
MR is always less than OM for like principal stresses, MliO > y,
.. NMR> 2y, .•. > y, and OF is always in the acute angle
EO W between the vertical and the line of greatest declivity,
and making (0  y) with vertical.
Fig. 11.
Since the third principal stress normal to paper is also OQ,
then if the ellipse revolves about POP it will sweep out a
spheroid. Its trace on a plane parallel to the free surface is the
ellipse R'QR' (fig. 11), some vector of which is the stress on any
vertical plane.
Examples.
1 . Th« weight of a certain earth is 120 lbs. per cubic foot, its angle of repose 2.5°.
It is spread in horizontal layers. Find the average intensity of the pressure against
a retaining wall with vertical back and 4 feet in depth. Also, find total pressure
against a slice of wall 1 foot in the direction of the length of the wall and the
overturning moment of the earth about the lowest point.
p = iw = 480 lbs. per square foot,
1 — sin <f)
, p = 194 '8 lbs. per square foot.
1 + 8in (^
average pressure = \q = 97 "4 lbs. per square foot,
total pressure Q = 974 lbs. per square foot x 4 square feet = 389'6 lbs. ^
overturning moment M = Q lbs. x f ft. = 519o ft. lbs.
F
66 APPLIED MECHANICS. [CHAP. III.
2. Gravel is heaped against a vertical wall to a height of 3 feet : weight of
gravel 94 lbs. per cubic foot : angle of repose 38°. Find horizontal thrust per
lineal foot of wall, also overturning moment.
Q = 1005 lbs. ; M = 1005 ft.lbs.
3. A ditch 6 feet deep is cut with vertical faces in clay. These are shored
up with boards, a strut being put across from board to board 2 feet from bottom
at intervals of 5 feet apart. The coefficient of friction of the moist clay is 287,
and it weighs 120 lbs. per cubic foot. Find the thrust on a strut ; al.«o find the
greatest thrust which might be put upon the struts before the adjoining earth
would heave up.
Since tan<^ = 287; .. sin (^ = 276;
therefore Q = 12255 lbs. per lineal foot.
Thrust per strut = 61275 lbs., just to prevent earth from falling in.
Greatest thrust which might be artificially put upon each strut before earth
would heave up = 19029 lbs.
4. A wall 10 ft. high and 2 ft. thick, and weighing 144 lbs. per cubic foot,
is founded in earth 112 lbs. per cubic foot, and whose angle of repose is 32°.
Find least depth of foundation.
p = intensity of vertical pressure below bottom of foundation
= 144 X 10 = 1440 lbs. per square foot,
q' = intensity of vertical pressure at same depth clear of foundation = 112 . «f,
d. 112
p \1 + sin<^/
, _ = 094; .. d = 121 ft.
1440
Note.— The height of wall above ground is 10  d = 879 ft.
5. The slope of a cutting being one in one and a half, weight of earth being
120 lbs. per cubic foot, and its angle of repose 36°, find average intensity,
amount of liorizontal component, and overturning moment of the thrust upon a
3 feet retaining wall at bottom of slope.
1
tan 7 =  — ^=6666; .. 7 = 33° 42', <?> = 36°, and m = 120 lbs.,
1 . o
i> = 3 feet; .. r = wL cos 7 = 299 lbs. per square foot.
»•' cos 7 — v'(cos^7 — COSrf))
 = ^7^ — ,r ^ = '62 ; .. r = 299 x 62 = 1854,
r COS 7 + v{cos''7 — cos^<p)
and average intensity of stress = 92*7 lbs. per square foot,
r'
2
M = H X  = 2316 ft.lbs.
o
6. A cutting having 3foot retaining walls is made on ground sloping at 20° to
the liorizon. Weight of earth is 120 lbs. per cubic foot, audits angle of repose
30°. Find the horizontal thrust and the overturning moment — 1st, when cutting
runs horizontal ; 2nd, when cutting runs up steepest declivity.
Data : D = 5 feet, 7 = 20°, w = 120 lbs., cp = 30°.
CHAP. IV.] KKTAlNlNti WALLS. 67
(^Itl) r = wD cos 7 = 338 lbs. per square foot
= stress on sloping layer at depth B, being vertical,
»■' _ cos 7  v/(cos7  C08^(^) _ 'oTG _
r cos 7 + V(tos''7 — cos<f)) 1304
.. >•' = 338 X '442 = 149'4 lbs. per square foot
= conjugate stress on vertical face of wall, being in sloping layer
inclined at 7,
»•' cos 7 = 140*4 11)S. per square foot
= horizontal thrust on wall, ut foot of wall.
Average do. = 702.
Total do. = average intensity x area = 702 x 3 = 210*6 lbs. per lineal
foot of wall.
Moment = 2106 x  = 2106 ft.lbs.
3
I. j\ 9 ^ ~ ^^^ 'P '^
r cos 7 + V(cos7  co.s2(f>) 1304
.. y = 338 x 383 = 1293 lbs. per square foot
= least principal stress in section on greatest declivity
= also third principal stress which is horizontal on face of vertical
wall.
Average do. = 65 lbs. per square foot.
Total do. = 65 x area = 65 x 3 = 195 lbs. per lineal foot of wall.
D
Moment = 195 lbs. x — = 195 ft.lbs.
CHAPTER IV.
THE SCIENTIFIC DESIGN OF MASONRY RETAINING WALLS.
In treating this subject analytically, we will consider retaining
walls as being built of blocks which touch each other at the
joints, and which can exert pressure and friction, but not
tension. Some cements are so strong that the whole structure
may be considered as one piece, in which case arise questions
of strength. In what follows we do not take account of this
action of the cement, but consider the joints as being able
to resist pressure only. The two conditions which must be
fulfilled for a joint of this kind are : (1) the resultant pressure
on the joint should fall well within that joint ; and (2) the
line of action of this pressure should not be inclined to the
normal to the joint at an angle exceeding the angle of repose
for masonry. When these two conditions are fulfilled, the
f2
68 APPLIED MECHANICS. [CHAP. IV.
joint is said to have stability of position and stability of
friction.
In order to find the direction and amount of earth pressure
on the wall, Eankine's method of the ellipse of stress is employed ;
and from the results obtained for earths whose natural slopes
are ^ = 30° and (/» = 45°, and whose free surfaces are horizontal
and inclined at the natural slope (p, the thicknesses of wails of
depth 20 feet are calculated.
The angle of repose for earth is its natural slope, and is the
greatest inclination to the horizon at which its free surface will
])ermanently remain ; and we assume for earth what is true for
a granular mass, that " It is necessary for stability that the
direction of the pressure between the portions into which a
mass of earth may be divided by any plane, should not at any
point make with the normal to that plane an angle exceeding
the angle of repose."
Rectangular Wall. — ACFK, in fig. la, represents the wall
in crosssection ; depth, c? = 20 feet ; length, 1 = 1 foot ;
it supports a bank of earth whose upper surface is horizontal,
and whose natural slope is 30°.
Let w = weight of masonry = 140 lbs. per cubic foot,
w' = weight of earth = 120 lbs. per cubic foot,
(p = angle of repose of earth = 30',
t = thickness of wall at base in feet.
In this case r and r' are principal stresses, and for distinction
may be replaced by p and q ;
^ ^ 1  sin ^ ^ 1
p 1 + sin 3 ■ ^ ^
That is, the horizontal pressure at any point of AC, the back of
the wall, is onethird of the vertical pressure due to that depth
of earth. At C, the base of the wall, the vertical and horizontal
pressures of earth are therefore
2J = 20 X 120 = 2400 lbs. per square foot,
q= \ = 800 lbs. per square foot.
o
This amount q is represented by OT. By drawing AT,d. triangle
is formed ; and the horizontal earth pressure at any point of
AG \^ given by the line drawn from the point to meet AT^
and parallel to CT. The area of ACT represents the total
overturning force axi AC due to the earth pressure, and it
CHAP. IV.J
RETAINING WALLS.
69
may be taken as acting through the centre of gravity of ACT;
thus
Q = «^i« . 8000 Ite.
acting at £',6^ feet above the base C.
The ellipse of stress for a point at the average depth of 10
feet is drawn ; OJC, OF(fig. lb) are the principal a.xes of stress ;
the semidiameters represent p and q, now of half the values
above ; OC represents a portion of the plane AC on which
the intensity of pressure is required. ON is drawn at right
angles to CC, and along it, OJf is taken equal in length
to ^'^ = 800 lbs. ; MB = ^^^ = 400 lbs. is drawn so that
BMN = 26 where the angle XON = 6 ; in this case 9 = 90°,
and JIR lies in MO ; the point B thus found lies on the ellipse
of stress, and the line BO represents the intensity and direction
of pressure on CC at the average depth of 10 feet.
To find the thickness of wall required for stability of position,
let G (fig. la) be the centre of gravity
of wall, and from C draw downwards a
vertical line ; produce the line of action
of Q through IJ, and let R be the point
of intersection ; from II draw HV = Q
= 8000 lbs., and ED = W = weight of
wall (not yet determined) ; complete the
parallelogram of forces and draw the
diagonal HZ, producing it, if necessary,
to cut the base in L; draw LU dX right ■i;^
angles to HV. The point L just found
must lie within the base FC\ and in
order that the bedjoints near C, the heel
of the wall, should not have any tendency
to open or to crush at F, the toe, it is
necessary to have the point L not further
from the centre of FC than about troths of
that base ; that is, the distance from centre
of base of wall to the centre of pressure v; *
should not exceed 'Zt. Taking moments
round L, we have * ^
Qx LU= TVx LI,
.,fl  sin d)\
\1 + sin (fij
t is nearly 8 feet.
(2)
^ X ' = c? X ^ X 140 X 'Zt.
Fig. 1*.
70
APPLIED MECHANICS.
[chap. IV.
For stability of friction, considering the masonry built in
horizontal courses, the angle LHI should not exceed the angle
of repose for masonry — 38°. Now,
tan LHI
Q ^ 8000
W~ 22400
= 357
< LHI is 20° nearly, a quantity well within the assigned limit
for friction.
Trapezoidal Wall (fig. 2a). — Front and back of wall inclined
at 80° to horizon; upper surface of earth horizontal. The
ellipse of stress (fig. 2b) for a point 10 feet deep is drawn in
position, and is similar to that shown in fig. \h ; CC is now
inclined at 80° to the horizon. When the triangle OM/i is
constructed as described for the previous case, OB = r = 445'8 lbs.
per square foot, and y = BOM = 17° 52'. The stress on AC
(fig. 2a) is represented by the triangle
ACT; it is zero at A and increases to
CT = 8912 lbs. per square foot at base of
wall ; AC = 203 feet, and the total pres
sure on ^6' is i^ = 9050 lbs.
Let LU and CJ be perpendiculars
on the line of action of B, and LS be
perpendicular to CJ; then taking moments
round L, as before, we have
WxLI,
(2)
Fi£
•3^,
Bx LU
observing that
LU=^CJ CS,
B X {GE cos 17° 52'  8^; sin 27° 52')
= [t 352) X 20 X 140 X
from which
t = 82 ft. = thickness at base of wall,
t  352 X 2 = 12 ft. = thickness at top
of wall.
For this case the angle LHI = 25°, a
quantity less than the angle of repose
for masonry.
Surcharged Bcctangular Wall. — The earth is surcharged
at its natural slope ^ = 30° (fig. 3rfc),and the conjugate pressures
are equal {OK, fig. 8, Ch. III). The ellipse of stress (fig. 36)
CHAP. IV. ]
RETAINING WALLS.
71
is drawn for a point 10 feet deep, and it may be noted^that
the major axis of the ellipse is midway between the directions
of r and r ; that is, the major diameter is inclined to the
vertical at the angle
45^  I = 30°.
The intensity of the earth pressure on a horizontal surface at a
depth d, due to the weight of the column above it, is wd ; on a
plane inclined to the horizon at the angle ^, the intensity is
diminished to ud cos ^ ; and thus the intensity of the two
equal conjugate pressures r and /, for a point 10 feet deep,
is 10392 lbs. per square foot.
In drawing the triangle OMR,
proceed as in first case :
<RMX= '29 = 120°;
<ROM=r^ = 30°;
OR = 10392 ;
OM = ^J = 1200 ;
2
600
as found by calculation or graphic
construction ; from which we
have;) = 1800, q = 600 lbs. per
square foot as the greatest and
least conjugate stresses at the
point (), that is, at a point 10
feet deep. The triangle ACT
(fig. 3a) represents the pressure
on^Cas before; (7r=207841bs.
per square foot at base, and the
total earth pressure on AC is
R = 207841bs. Taking moments
round X,
R X LU= W X LI, (2)
and, as before,
LU = CJ  CS = CBcosSO°
 St sin 30^
W^UOtd, Ll^'M
Fig. 3«.
Fig. 3*.
20784 (577  M) = 840^^ ^ = 8 ft.
72
APPLIED MECHANICS.
[chap. IV.
The angle LHI = 29°, a quantity only a few degrees within the
assigned limit for friction.
Surcharged Trapezoidal Wall.~ln fig. 4a the front and
back of wall are inclined at 80° to the horizon ; the earth is
surcharged at its natural slope rp = 30°. The ellipse of stress
(fig. 4:h) for a point 10 feet deep i.s drawn in position, and
IS similar to that shown in fig. 35 ; CC is now inclined at
80° to the horizon. When
the triangle OMR is con
structed as described for fig.
Ih, OM = 1200, MR = 600,
i?0 = 124.51bs., <XOiV=0=5O°,
RMN =29 = 100°, ROM = y
= 28° 20'.
In fig. 4a, the earth pres
sure on ^Cis represented as
before by the triangle ACT;
CT = 2490 lbs. per square
foot at base of wall, and the
total earth pressure R = 25286
lbs. inclined at 38° 20' to the
horizon.
Taking moments round L,
we have
RxLU=W xLI, (2)
 ^t sin 38° 20',
J^=(^352)x 20 xl40,
^ = 89ft. = thickness at base,
t  352 X 2 = 18 ft. = thick
ness at top of wall.
The angle LHI = 33^ a
quantity exactly equal to the
angle of repose for masonry ;
the courses of masonry should
therefore have their bedjoints
dipping from front to back of
wall at an angle of say 10° to
the horizon. '
Trapezoidal Wall. — In fia.
Fis. 4ff.
Fig. 4A.
5 the back of the wall
IS
CHAP. IV.]
RETAINING WALLS
73
vertical, and the face batters ; upper surface of earth is hori
zontal. The wall here represented is that shown in fig. la
with the wedge, whose crosssection is KK'F (fig. 5),
removed. The centre of gravity of the triangle KK'F is
vertically above L, and the moment of stability of the wall
is not altered if this wedge be removed. To find the position of
K in K A, where K is vertically above F, take K K= SFL = (it.
The thickness of the wall will there
fore be
/ = 8 ft. at base, t  Qt = 32 ft. at top.
The angle £HI = 27'', a few degrees
within the limit for friction.
Battering Wall of Uniform Thick
ness. — In fig. 6, the face and back of
wall incline backwards, making an angle
a = 10° with the vertical ; earth surface
is horizontal. If we suppose AC, the back
of the wall, to be made up of a number of
rectangular steps, vertical and horizontal,
the horizontal earth pressure on a vertical
face at a depth d will be, as before,
, ^ 1  sin (h
iv . d . : — ^ ;
1 + sin
this horizontal pressure will tend to cause
the earth to spread upwards, and the
vertical earth pressure on a horizontal
face looking downwards at depth d.
will be
«^£:
Fig. 6.
v/ . d .
lsin0 lsin'/< , , /lsin0
^ ^ — 7. q . — =w.d.( r^
l + sin0 1 + sm^ \l + sm0
The straight line AC may now be considered as the limit
of these steps, and the horizontal and vertical pressures on AC
will be represented by the triangles ACT and C'CT ; the
horizontal pressure is zero at A, and increases to C'T =120
X 20 X ^ = 800 lbs. per square foot at the base ; the vertical
pressure is zero at A, and increases to CT' = 120 x 20 x ^
= 267 lbs. per square foot at the base. The total horizontal
pressure on AC is Q = 8000 lbs. as for fig. la ; the total
vertical pressure on ^ C is
Q' = ^CC" xCT' = ^x 352 X 267 = 470 lbs.,
and the points of application are F and F'.
74 APPLIED MECHANICS. [CHAP. IV.
Taking moments round L,
W X leverage = ^ x leverage + $' x leverage
d
J J 04. ^ ^ \ , , 1  sin d> c?
w . a .t\'6t ^^  tan a\=wcl : —  z
\ 2 / 1 + sm ^ 2
, /I  sin aV c? . tan o . _^ ,,^ .
+ 10 . d.i ; — i (U + \d tan a),
\1 + sm <j)J 2 ^ ^ ^
taking w = 140, and ?// = 120 lbs. per cubic foot, <b = 30°, and
<^ = 20 feet,
t = 8 v/l + 4 tan^ a  154 tan a. (3)
The vertical through G, the centre of gravity of wall, must
fall within the base ; and if it be not allowed to deviate further
from the centre of the base than f^ths of the breadth of base,
we obtain
'3t
tan a = j^ = '03^, when d = 20 feet ;
putting this value in equation (3), we get
^ = 8 (1 + 4 X 0009^0^  154 X 03^,
t = 578 ft. = FC, the tlnckness of wall.
The angle a is 10° nearly ; that is to say, the back of the
wall should not be inclined to the vertical at an angle greater
than 10°.
In fig. 6, G' is the centre of vertical forces, viz., the down
ward weight of the wall, and the upward pressure of earth ;
HD is the vertical drawn through G', and is equal to W  Q'
or 15713 lbs.
The angle ZJII = 27°, a few degrees less than the limit for
friction.
Wall loith Vertical Face and Stepped Back. — In fig. 7,
the steps are taken at vertical intervals of 5 feet ; the upper
surface of earth is horizontal. The base of the wall FC sup
ports the masonry A'CFK, and the earth A' AC vertically
above that base; and the triangle ACT represents the hori
zontal earth pressure on the back of the wall ; G is the centre
of gravity, and W is the weight of masonry and earth vertically
above FC.
CHAP. IV.]
RETAINING WALLS.
75
FiK. 7.
Taking moments round L as before,
Q xLU = W X LI. (2)
To find the thickness of upper 5 feet of wall, proceed as for
tig. \a ; and obtain 1^ = 2 feet.
For /in, thickness at 10 feet deep, we
have ^10 = 2000 lbs. acting at LQ foot above
this assumed level; and taking moments
round the point corresponding to L
Q,, X Y = 10 X itg X 140 X (54  •2^,o)
+ (t,,  U) (140 + 120) (3^,0 + bh)
from which ^i,, = 4'1 ft.; similarly ^15 = 6"3 ft.;
and tw = 84 ft., the thickness at base.
The angle LHI = 20"", a quantity well
under the limit for friction.
Wall ivith Battering Face and Stepped Back. — In fig. 8
the steps are taken 2 feet wide, and at vertical intervals of
5 feet; the batter of face is 1 in 12, and the earth is surcharged
at its natural slope 9 = 30°. The ellipse of stress (fig. 3&)
applies to this case ; the
depth AC \Q inci eased to
23'5 feet by the earth slope ;
the pressure at base, CT
= 2440 lbs. per square foot ;
the total earth pressure on
AC\^R= 28600 lbs. acting
at E. On account of the
battering face, ihe point F
projects 1"7 feet beyond the
vertical through K; and on
account of the steps of the
back of wall, the base pro
jects 6 feet beyond the ver
tical through .j' ; the thick
ness of wall at base may be
represented thus
/ = 17 + ^' + 6 = ^' + 77 ft.
In order to find the value of t', take the moments round F,
thus : —
Weight of masonry = 2800^' + 10780.
76 APPLIED MECHANICS. [CHAP, IV.
Moment of masonry round F= UOOd'^ + 18160^' + 36500.
Weight of earth vertically over FC = 8400.
Moment of earth round F = 8400 + 44800.
Moment of earth and masonry round L = 840^' + 13436^'
+ 51950.
Equating this to
lixIU^ 28600 S677  "4 {f + 77) ) ,
we get t' = 2 feet, t = 97 feet, thickness of wall at base.
The angle LHI = 33°, a quantity exactly equal to the
angle of repose for masonry ; the courses of masoniy should
have their bedjoints dipping from front to back of wall at an
angle of, say, lO*^ to the horizon, or at right angles to FK the
battering face.
Tabulated Dimensions.
Since the stability of a wall is proportional to d . f, where
d = depth and t = thickness, and since the overturning pressure
of earth is proportional to d?, it follows that the thickness of a
wall should be proportional to its depth. Having calculated
the thickness for a given depth d = 20, it is easy to fix on the
thickness required for any other depth. If the depth of wall
be taken as 100, the accompanying table gives the other dimen
sions of walls, as deduced from the results given in this Chapter,
when the earth has for its angle of repose ^ = 30*^, or ^ = 45°, or
slopes of 1 vertical in 1"73 horizontal, and 1 in 1.
For comparison, corresponding values are given for water
whose heaviness is 62*5 lbs. per cubic foot.
It will be observed that the thickness of a wall for resisting
water pressure is much greater than for the varieties of earth
considered; this, of course, is caused by ^ becoming zero in
the case of water. It is, therefore, of the utmost impor
tance that the earth behind a retaining wall should be kept
dry; and for this purpose weeping holes through the walls
are formed near the level of the original surface of ground,
and a diy stone backing from 12 inches to 18 inches thick is
laid beliind the wall for conveying the water easily to the
weeping holes.
CHAP. IV.] RETAINING WALLS. 77
Spread and Depth of Foundation. .
Because of the obliquity of the downward thrust on the
foundation, the concrete or masonry forming the substantial
part of it must spread out in front of the wall in steps as it
goes deeper. The bottom of the trench may have to be con
solidated by driving packing 'piles, or hearing piles may have
to be driven to a firm stratum, and surmounted by a staging
generally dipping back as the friction is then also precarious.
In old consolidated earth two conditions serve to determine the
spread and depth of the trench (see first wall, fig. 12). The total
vertical tiirust of the bed of the trench on the base of the
concrete must equal the weight of the wall and the concrete.
Also the centre of this upward thrust must be vertically below
(i^'on fig. 12, but L on figs. 8, 7) the assumed centre of stress
at the lowest bed joint of the wall. But besides these two
conditions which concern the equilibrium of the wall we have
the limiting conditions concerning the equilibrium of the earth
surrounding the concrete, namely, the upward thrust of the unit
cube under the toe of the concrete is at most nine times a
column of earth the depth of the bed of trench below the
surface of the earth in front of wall, which is then on the point
of heavingup. And for the unitcube at heel of trench, it is as
a practical lower limit, onethird of the column of earth behind
the wall.
The two equations obtained can best be solved by trial and
error. Taking the height of the wall as sensibly 3^, where t is
its thickness, we will try the spread 45 per cent, of the thickness,
and the depth 20 per cent, of the depth, or 60 per cent, of the
thickness.
Weight of wall and concrete is wt^ (3 + 1'45 x 6) = 3"87 ivt^.
Upward stress at toe of trench is 9 x Qtvt = 432 wt ; while at
the heel it is ^ (3 + 6) w't = 96^^. Multiplying their average
value by 1"45^, we get the total upward stress = SSdwt^ which
satisfies the first condition. If the centre of this upward stress
be distant ya and yj, from the heel and toe of bed of trench
respectively, then for the centre of gravity of the trapezium of
stress drawn below first wall, fig. 12, we have,
ya.yb: (432 +  x 96) : (i x 432 + 96), and ya + yb = 145^,
so that ya = 'S8t or ^t nearly, which satisfies the second con
dition, as F on fig. 12 is at most ^t from back of wall.
H
P
O
O
CO
lJ
<1 .
P g
S"
o
in
<
P^
o
Q
W
P5
t— I
P
w
P5
C5
t— I
w
w
H
O
CO
H
w
CO W
o
tH
W
H
O
w
pq
H
« II
« "5
«N •<»• O C^
t« CO
H
*
c^
^ S
ja e
W
w
<M 00 lO CI
ri CO
*
o
(M
O
o
Tf
1—1
CO
CO O Q UJ
O 3t3
?5 e tr.
tc
O Tl
in
o
c/:
<
O
CO
O
^^^O.^
CHAP. IV.] RETAINING WALLS. 81
Graphical Solution.
Graphic statics is a study by itself, and in this book we only
intend to use the very first elements.
In graphic statics a force is represented by a pair of lines ;
one a short thick finite line giving the magnitude of the force
upon a scale of forces ; the ends of this thick line are marked
by the centres of two little rings ; in a handmade drawing the
centres are pricked on the paper. The other is a long thin line
giving the actual position of the force on the plane of the paper
relative to some structure or solid body upon which the force
acts, and relative to other forces also acting on it. It will be
seen then that a second scale of feet is necessary to measure the
distances along the solid body and among the forces. These
force lines are furnished with a barb or arrowhead to indicate
the sense of the force in that line ; near the barb stands a
numeral which is really the suffix of a letter such as P^ ;
the force is called 3 when simply being referred to, but when
speaking about it as a quantity it is P,. Often the force line
is so long that it reaches to the edge of the paper, and although
only a portion may be ultimately inked in, still a little left at
each margin may be indicated. The line is fine and long for the
practical purpose of setting the rollers or T square accurately
parallel to it. A plane set of forces all acting on one body, which
may at first be assumed to be the sheet of paper, are numbered
in any order ; but for a successful issue they must be numbered
in cyclic order. The thick lines drawn parallel to the thin, each
to each, form a polygon round which the corresponding numerals,
printed heavier or larger, run consecutively ; for some purposes
it is convenient to put half barbs on the thick lines. This is
the " Polygon of Forces." When the forces are all vertical the
sides would lap on each other, but this is avoided by slightly
displacing some, and drawing a polygon which often looks like
a gridiron pendulum ; the " eyes," however, should all be in
one vertical line. In this case, the downward forces being all
vertical, the polygon is often called " the load line " ; while the
closing upward sides constitute the reactions or supports.
Given, graphically, fig. 10, a plane set of forces to construct
the balancing force or resultant. On the upper lefthand corner
is shown an analytical definition of five plane forces in terms
of intercepts and angles and lbs., agreeable to which the force
polygon and lines of action are drawn to a pair of scales.
82 APPLIED MECHANICS. [CHAP. IV.
Construction. — Draw a line closing the " force polygon," and
scale off its length in lbs. on the force scale for the magnitude
of Bs. Eeckon its " sense " in the same order round the
force polygon as the other forces. Choose any pole (not on
the closing side of the force polygon), and, to that pole, construct
a " link polygon " A B C D E F amoDg the lines of action.
Through F, the closing point of the link polygon, draw the line
of action of B^. parallel to the closing side of the force polygon.
When scaled off, the values of B(„ x^, and A^ should be nearly
the same as the values given on fig. 10 which were calculated
by trigonometry.
Pkoof. — Consider the set of six forces, 1, 2, 3, 4, 5, 6 ; they
are a balanced set of forces. For, add a pair of balanced forces
7 and 7' equal and opposite to each other, having the link A B
for their common line of action, and having the vector, which
comes to the junction of 1 and 2, for their common magnitude.
Add another pair of balanced forces of 8 and 8', with BC for
their common line of action, and the vector, coming to the
junction of 2 and 3, for their common magnitude. Add in the
same way the pairs of forces 9 and 9', 10 and 10', 11 and 11',
12 and 12'. We have now altogether a set of eighteen forces
1,7,12'; 2,8,7'; 3,9,8'; 4,10,9'; 5, 10', 11 ; 6,12,11';
which are exactly an equivalent set to the set of six forces with
which we began, since all the forces we added balanced in pairs.
Now, of the set of eighteen, the first group of three, 1, 7, 12',
act at the same point A and have magnitudes proportional to
the homologous sides of a triangle; they are, therefore, a balanced
set of three; similarly the second group of three, 2, 8, 7', act at
one point B and have magnitudes proportional to the three
homologous sides of the triangle. In the same way each group
of three is balanced ; hence the set of eighteen is balanced ;
hence the original set of six is balanced.
Cor. — The graphical conditions of equilibrium of a plane
set of forces are two in number. The force polygon must close.
A link polygon must close.
The proof given here is very important, for, in the first place,
conceive the rigid body to be the paper, say, a sheet of brass,
and let the six balanced forces be attached to it at the points
A, B, C, D, E, and F. Suppose the brass all cut away except
the narrow strips under the thick lines A B C D E F, which
may further be supposed to be pinjointed at these points.
These strips of brass actually apply to the pins, the pairs of
forces given by the vectors, viz., 7, 8, 9, 10, 11, and 12, and
CHAP. IV.] RETAINING WALLS. 83
we have designed an articulated structure in equilibrium under
the given load system.
This is a balanced polygonal lineal frame or rib, and if
each strip of brass be sectioned for the load on it, we have
then an actual balanced frame. A model of this frame sits
on a horizontal table in the Engineering Laboratory of
Trinity ' College. It is in unstable equilibrium, for a sharp
blow on the table causes it to distort.
The pair of diagrams, figs. 12 and 13, show designs for
retaining walls. The load on the back of the wall is constructed
by the method of the ellipse of stress and its auxiliary figure.
This load is reduced so as to be expressed in terms of the
\\ eight w of a cubic foot of masonry.
Triangular or rectangular blocks of masonry are added one
after another, and the partial resultant constructed graphically
till the last resultant, that with the greatest number of barbs,
at last passes through a centre of stress deemed to be sufficiently
far in from the face of the wall.
The data and construction are sufficiently given on the face
of the diagrams, which, however, are small, having been reduced
half the lineal size of a set of graphical exercises, published by
Macmillan & Co. for the authors.
It wiU be seen that the slopes of the earth, both actual and
limiting, are given 3 to 1 and 2 to 1 which are nearly the same
as 7 = 20°, and <j) = 30°.
The steps of the calculations corresponding to those graphical
solutions are as follows. The fourth is only approximate so that
one auxiliary figure may serve for two walls.
In the process of designing either by the equations
(a), (b), etc., and especially by the graphic method shown
on figs. 12 and 13, it is sufficient to get the centre of stress
to pass through the toe of the wall and then adding a slice
to the face of the wall to throw this centre of stress in a
suitable distance from the new toe. This addition will not
sensibly shift the centre of stress, as may be shown thus : —
Minimum distance of the centre of stress from the heel of the
wall.
For a very thin wall (fig. 11a) the weight to be compounded
with the load is so small that the centre of stress lies far out
beyond the face and such a wall might be of metal and sunk
firmly in the earth. As slice after slice is added to the wall,
the centre of stress comes nearer and nearer to the heel till
G 2
84
APPLIED MECHANICS.
[chap. IV.
(fig. lib) it passes through the toe ; but for further slices
added begins to move away from the heel, at first very slowly,
as is always the case in the neighbourhood of a minimum.
t
A
■■'"}
I
\>
f
/
w " / 1
3 / ,
^
■„ ,Jbs. /
/
rieei.
~aneai
^
Minimum.
Increasing.
Fig. 11 — rt, b, c.
For the same reason wedges can be removed from the face of
the rectangular wall to give suitable batters without disturbing
the centre of stress.
Examples.
1. For the first wall on figure 12, the corresponding analytical solution is
tanii = 26° 34' ; k= ! ^'" "^ = 382.
1 + sin ip
p = \5tv' ; q = kp = b'lZw'.
E = Q = y X 30= 1719!<.''= 1375t<^^, or 138 times the weight of a
cubic foot of masonry.
M = \Q H = Wibw; M=ZOtci x^t = U'2bwf.
Equating, t = 122, and ^ = 1 1 feet nearly.
2. Second wall on figure 12. Approximate prnciical solution. In the last
suppose an upward force S to act through the centre of gravity of the wedge
removed from the hack of the wall equal to the excess weight of masonry over
earth.
5 = ^x30x5x
I5w,
and its lever about the centre of stress is (it — f) = 8 nearly as t is almost 11.
Correcting the equation of moments
ll>5icl = \37bw + low X 8, or <*= 133 and <= 115 feet.
3. Second wall on figure 12, solution exa<tly corresponding to the graphic
solution,
p + g iv'
■ ^ = — (15+ .573)= 10365et7'.
0M =
MR = '
4635M7'.
2
2
OM"^ = 107.'). MK^ = 21 0 ; dropping it'.
OR = OM + MR^ +20M . MR cos 20.
= _(15.V73)
E*RTH.
TIm wclihl of on*
CuMc Foot Is (aur
nttto that ol inuon
,«»
'^ DESIGNS
RETAINING WALLS.
30 J'oof ■ Ca/umn of JCarfh <
AUXILIARY FiaURE tor First ELLIPSE
Fig. 12.
86 APPLIED MECHANICS. [CHAP. IV.
Now e = P = cot^l = 80° 33', 80 that 20 = 161° 06', the sup. of which
is 18° 54'.
OiJ2 = 129  9607 X 946 = 3812.
r= 0R = 6174U'',
and CC the hack of the wall is 30 cosec fi = 304 square feet.
R = 304> = 1877^' = 150w.
sin 7 MR . 4635
^^e = OR^ .•.sm7 = ^:^x324=245, , =. 14° 11'.
To 7 add 9° 27', the complement of 6, and we obtain 23° 38' as the inclination of R
to the horizon.
Resolving R into horizontal and vertical components
H= R . COB 23° 38' = 150m> x 916 = 1374m.
r= 7? • sin 23° 38' = 150m; x 401 = 6015mj.
The equation of moments about F, the centre of stress, is
1374 = 30 (!!  5) (f (!  f ) + 75 (^<  iJt) + 60 (.<  ),
<2 _ 1I6i; = I'^O, or <= 115 feet.
4. The first m all on figure 13 has the additional datum 7 = tan' i = 18° 26'.
r = low' cos 7 = \bw' X 9487 = 1423w'.
, _ cos7V(cos^7cos''(^) _ 9487  v(9000  8000) _ 9487  3162 _ .
~ cos 7 + V(cos^7  cos*^) ~ 9487 + \J{^mQ  8000) ~ 9487 + 3162 ~
/•' = k'r = 7115if', also 7' = 7 = 18" 26'.
R'  30/ = 214m;' = 171m; ; E' = R' cosy' = 162m;.
r' = iE'sin7' = 54t<;. Also JF = SOwt.
Equation of moments about F, the centre of stress, at ^th of t in from the face
of the wall.
lOJy = 7Fx< 4 F'x<,
1620 = 1125i!2 + 471^ )/, and < = 93 feet.
5. Approximately for second wall on figure 13, by removing a wedge at back.
1620 = 1125^ + 4725< 15 X 8, and t = 105 feet.
6. Detailed calculations for second wall on figure 13 corresponding to the
graphical solution shown on it.
p 1 + 8in^ 14 4472 , ,^^
— ~ = 1"144.
r cos 7 4 v'Ccos'27  cos2(^) 9487 4 3162
p = l144r = 1144 X 1423m;' = 1627m'.
q = kp = 382 X 1617m;' = 621m'.
OM = ^(p + q) = 1124M; ; 3IR = Up  q) = 503«;'.
ain 20 OR r 1423
sin y ~ MR~ ^ (/?  y) " 503 '
. , 1423
sin 2a = — X 3162 = 8956; 26 = 63° 34' : 6 = 31° 47'.
j^ _ aoFnof C»Tumn of £(vth tj
AUXILIARY FiaURE tor Second ELLIPSE
' ' ' ""
Cubiv net of Masohhy.
too POO 300
TTTTTt
fiv/
I L' :UIUU.i:
/cii ^oo 3c^ 4cn
.ill. . ^...Li jij^ii >. I ,:i=r:
40 feet
Fig. 13.
88 APPLIED MECHANICS. [CHAP. V
Now the angle between the normals ON' and ON" can be expressed in two
ways.
e + e" = 7 + /3, or 6" = 18° 26' + 80^ 33' 31° 47' 67° 12'.
•lef' = 134° 24', the sup. of which is 45° 36'.
[OR'Y = OM 4 iMR^ + 20M. MR cos 6".
•24^ + O03  2 X 1124 x 503 x 6997 = 7253.
(^)'
r" = 8516«f''; R' = >•" x 304 = 259tt' = 2072w.
siny" MR" 503 . „ 5030 „, , ,„, , „ „ ,„,
■ ; „ = — „ = —,—7; .. sin 7" = ;— X 7145 = 422, and 7" = 24° 58'.
sm 26" OR" 8516 ' 8516 '
Obliquity of OK' to the vertical is
;8  7" = 80° 33'  24° 58' = 55° 35.
R" = K sin 55° 35' = 171m^, and F" = R" cos 55° 35' = l\lw.
lOH" = 30,^ {to)(ftA) + nw (it  la) + V" (it  %],
or i + 327t = 1582, and t = ll'OSfeet.
CHAPTER V.
TKANSVERSE STRESS.
In the preceding chapters we have considered the internal
stress at any point within a solid, and have shown that it can
be expressed by means of three principal stresses. We began
with one principal stress, the other two being zero ; this was
illustrated by pieces strained under one direct simple stress,
such as tie rods and struts ; and at each point in these pieces
the strain was similar in every respect. We next considered
two principal stresses, the third being zero or identical with
one of those two ; this was illustrated by small rectangular
prisms of earth under foundations, or loaded with the weight
of superincumbent earth, the prism being strained by two
(or three) direct simple stresses upon its pairs of opposite faces.
'I'here we saw that the strain at all points, in certain parallel
planes, was similar in every respect; varying, however, as we
passed from points in one to points in another of those parallel
planes. It was pointed out that earth might have the stress
in one horizontal direction artificially increased by a direct
external stress, in which case there would be three principal
stresses at each point, the intensities of which might be difterent
at different points.
CHAP, v.]
TRANSVERSE STRESS.
89
In all such examples, the internal stresses were due to strain
produced in the simplest manner possible, viz , by direct external
stresses ; and in many the stresses at internal points were given,
without specifying what the solid was, or in what manner it was
strained. These exercises served to illustrate methods, but it
will afterwards appear that the data specifying the stress at
such points were obtained by supposing that the body was
strained by external stresses, definite though by no means either
simple or direct.
We now come to consider the stresses at points within solids,
due to strains produced in the next simplest manner, viz., by
external stresses which are all parallel. Pieces under such
stresses are called beams, and the stress is called transverse stress.
The case in which both ends of the beam are simply supported
will be primarily considered. For simplicity, the external
stresses, as shown on the diagrams, are all vertical ; they
consist of the two upward thrusts concentrated at the extremi
ties, and the loads concentrated on intermediate portions and
acting downwards. These external stresses are uniform in the
direction normal to the paper ; and whatever be the breadth of
the beam, they may be replaced by forces all in one plane, the
plane of the paper.
On fig. 1, AA' B' B is the longitudinal section of a beam
of length 2c, depth A, and breadth h, and OX is any line chosen
as axis. W^ is a force in the plane of the paper, replacing a
stress spread uniformly over the breadth of the beam, as shown
w,
1"
413.
w,
._ll
.Q
Fitr. 1.
on the crosssection below it. Similarly P and Q are forces at
the extremities and in the plane of the paper. In order to have
these forces specified, it is necessary to know their amounts, and
the distances measured from some origin O, say at one end of
90 APPLIED MECHANICS. [CHAP. V.
the beam, to the points where their lines of action cross OX.
Such distances are called the abscisste of the places of applica
tion of the loads. Thus P acts at 0, W^ at rci, and Q at 2c.
C is the centre of span ; its abscissa is c.
The varieties of load to be considered are : —
V. Loads concentrated at one or more points of the span
as Wu
2°. Loads uniformly spread over the whole or parts of the
span, as w lbs. per running foot. Such a load is represented on
fig. 1 by a set of arrows, each equal to v), and consequently
they are one foot apart ; to save trouble, it is more convenient,
as on fig. 2, to represent .such a load by means of a parallelo
gram surrounding all the arrows.
3°. Combinations of such loads.
The loads concentrated at points might be the ends of cross
beams resting on such points, or the wheels of carriages, &c.
The weight of the beam itself is often to be considered as a load
spread uniformly over the span.
To find the relations among the external forces, we consider
the equilibrium of the beam as a whole. The beam is to be
consideied as perfectly rigid and indefinitely strong. In order
to find the supporting forces P and Q, we require to know the
amounts and positions of the loads, and the length of the beam.
Since the forces are all parallel and in one plane, there are
two conditions of equilibrium : —
I. The algebraic sum of the forces is zero.
II. The algebraic sum of the moments of the forces about
any point is zero.
From the first condition we have
P+Q= W,+ W, +W, + &c. =. ^(W) (1)
where 2 ( W) represents the sum of all the quantities
W„ W,, W„ &c.
If we take moments about 0, then P has no moment, and Q
tends to turn the beam in one direction about 0, while the loads
all tend to turn it in the other direction. By the second con
dition the sum of these moments is zero, and we may, if we
CHAP, v.] TRANSVEKSE STRESS. 91
choose, put the moment of Q equal to the sum of the moments
of the loads, thus
Q X leverage = sum of the products got by multiplying each
load by its leverage,
or ' Q.2c= Jr,.r. + W^x^ + W^^ + &c. = 2 ( Wx) ;
hence Q = ^—\ ,2)
where 2 ( Wx) represents the sum of all the quantities
W]Xu WzXi, &c.
P may be found in a similar manner by taking moments
about the other end ; or it may be found at once, since we know
P + ^ by equation (1).
An uniform load, such as tv lbs. per running foot, spread over
a portion of span, is to be treated as one force equal to the
amount, and concentrated at the middle of that portion.
Examples.
1. The span of a beam is 20 feet, iind there is a load of 80 tons at 5 feet from
the left end. Find the supporting foi ces.
2c =20; JFi = 80, and a;i = 5 ; Q . 2c = JFiXi.
Othei wise,
IFiZi 400
Q = — = — = 20 tons, /* + e = 80 tons, P= 60 tons.
load 80
Q = X segment remote from Q = — x 5 = 20 tons ;
span 20
load , ^ 80
and 1' = X segment remote from T— — x 15 = 60 tons.
span 20
2. A beam of span 24 feet supports loads of 20, 30, and 40 tons concentrated,
in order, at points which divide its length into four equal parts. Find the sup
porting lorces.
»^i = 20 ; 7r2 = 30 ; TFs = iO ; zi = 6 ; x^ = 12 ; 2:3 = 18 ; 2c = 24.
^ :S,(TFx) 20 X 6 + 30 X 12 + 40 X 18
Q = ^ — ' = = 50 tons.
2c 24
P = 2 ( ;r)  Q = (20 + 30 + 40)  50 = 40 tons.
3. A beam 30 feet span supports three wheels of a locomotive v hich transmit
each 6, 14, and 8 tons ; the distances measured from the left end of the beam to
the wheels are 8, 18, and 24 feet respectively. Find the supporting forces.
JFi = 6; Jr2 = U; JF3 = 8; 2 ( ^) = 28 tons ;
an = 8 ; 2:2 = 18 ; xs = 24 ; 2c = 30 feet.
Atis. P= 11*6 tons; Q = 164 tons.
92
APPLIED MECHANICS.
[chap. V.
4. The span of a beam is 60 feet ; an uniform load of 2000 lbs. per running foot
is spread over the portion of the span beginning at 40 and ending at 50 feet from
the left end. Find the supporting forces.
See fig. 1, and suppose the spread load alone on the beam.
Replacing ww ... by
W2 = w (50  40) = 2000 X 10 = 20000 lbs.,
concentrated at
a;2 = i (40 + 50) = 45 feet, F = 5000, and Q = 15000 lbs.
5. A beam 60 feet spnn, and weighing 100 tons, supports an uniform load of
2 tons per running foot, which extends from the left end of the span to a point
20 feet therefrom. Find the supporting forces.
M? = 2 ; JFi = w X 20 = 40 tons ; x\ = 10, the middle point of uniform load.
"Weight of beam, TF^ = 100 tons ; 3:2 = c = 30.
Aus. P= 83^ tons ; Q = 56 tons.
6. A beam 60 feet span, and weighing 100 tons, supports a locomotive as in
example 3, and an uniform load of 2 tons per running foot which extends from the
middle to the right end of the span. Find the supporting forces.
Q X 60 = 6 X 8 + 14 X 18 + 8 X 24 + 60 X 45 + 100 X 30.
Aus. 0=1032, and P = 848 tons.
14 8
vy,vy
100
Of— ;^<— ^V^^'b X^
I M
60
1*
iwibsprft
X
iQ
Fig. 2.
7. A beam 36 feet span weighs one ton per lineal foot. Tlie first half is
loaded uniformly with 2 tons, and the second half with 3 tons per running foot.
Find Pand Q.
Alls. P= 585, and Q = 675 tons.
8. A beam 42 feet spun supports five wheels of a heavy locomotive. The tore
wheel is one foot from the left end, and the distances between the wheels, in order,
are 5, 8, 10, and 7 feet, and the loads transmitted, in order, are 5, 5, 11, 12, and
9 tons. Find the supporting forces. See fig. 3.
lFi = 5; ir2=o', 7r3=ll; ;r4=12; 7/'5= 9; 2(7r) = 42.
.ri = 1 ; a:2 = 6; ars = 14 ; J* = 24; 3:5 = 31; 2c = 42.
.. Q X 42 = 5 X I + 6 X 6 J 11 X 14 + 12 X 24 + 9 X 31.
Ans. P=24, and Q= 18 tons.
CHAP, v.]
TRANSVERSE STRESS.
93
For a system of loads such as Wi, Wi, &c., there is a point at which, if they
were all concentrated, the supports would share the load as they do for the acttnal
distrihutiDH at different points. This point is called the centre ot gravitij of the
load system ; its position will he marked G, ami its ahscissa 00 will be denoted
by X. Hente, snpposing the total force 2 ( ^F) concentrated at O, we have
Q .'Ic^liW) .X for the single force 2 ( 7F).
Q . 2(; = 2 ( Wx) for the actual distribution ; see equation 2 on page 91.
Hence :S.{W) . x = :Z{Wx), or x = ^^ ,
which gives the position of G.
Having calculated .r, we can now find P and Q as for the single load 2 ( JF)
at X from the left end ; see example 1, setond method.
, . , , totiil load
The supporting force at either end = x remote segment, (3)
span
P =
2c
(2c  X) ;
2c
9. Solve exercise 102 by the method just described.
We have
_ _Z{Wx) _ 5 X 1 + 5 X 6 + 11 X 14 f 12 X 24 + 9 X 31
^^ ■S.[W) ~ 5 + 5 + 11 + 12+9
and the other segment (2c  i) = 24 feet.
= 18 feet,
and
total load
span
42
X remote segment = — x 24 = 24 tons,
total load 42 „ „
X remote segment = — x 18 = 18 tons.
span 42
The position of G relative to the loads W\, W^, . . . can be found, although
the span of the beam and the position of tlie loads upon it be unknown, provided
the amounts of the loads and their distances apart be given. On fig. 3 let S be
111
42 tons
is.
9 tons
' i^iy^ffj;^^f~\^^rr^jj^
ov
s>
■V* G C
o v^
Fig. 3.
the point where the first load W\ is situated, and let a\ (= 0), a2,ra3, a, Ui, 05 be
the distances from »S to W\, Wt, JF3 G, JF4, W^, respectively. Taking moments
about S, the moment of 2 (fF) ucting at a will equal the sum of the moments of
^1, JFi . . acting at m, rto . . . .
94 APPLIED MECHANICS. [CHAP. V.
That is
2(?r)x« = 2(^«); .. « = ,^
gives position of G measured from S the left end of the load.
10. In example 9, find G the position of the centre of gravity of the load
measured from tlie fore wheel.
Wi = 5, ^2 = 5, TTs = 11, Wi = 12, JTs = 9, 2 (R^) = 42.
«i = 0, «2 = 5, rt3 = 13, (74=23, as = 30.
■S,{Wa) + 5x5 + 11x13+12x23 + 9x30
Then
a =
= 17 feet.
2(Jr) 6 + 5 + 11 + 12 + 9
11. In example 3, find a the distance of G from the fore wheel.
TTi = 6, W2=U, TF3 = 8. m = 0, ^2 = 10, as = 16 ;
.. « = 957 feet.
It is convenient to calculate a, if it be required to find values of P and Q, as in
examples 3 and 8, corresponding to the given load system shifted into some new
position upon the beam ; thus
12. In example 8, find F and Q when the locomotive shifts till its fore wheel
is 6 feet from the left end.
(7=17: hence adding six feet we have x = 23, so that Pand Q will now be the
same as for a single load 2(^r) = 42 tons concentrated at G, a point dividing the
span into the segments 23 and 19.
^ load 42 „„
.. Q = X remote segment = — x 23 = 23 tons.
span 42
Similarly
42
P= — X 19 = 19 tons.
42
In like manner P and Q may be found with great convenience for other
positions of the locomotive, all of whose wheels must, however, be on the beam,
because, if one wheel goes off, the beam is under a different load system altogether.
13. Find P and Q in example 3 when the locomotive is shifted so that its
fore wheel is 10 feet from the left end of the beam.
Here
« = 957 feet, and 2 ( ^T) = 28.
^«.9. P=973, Q = 1827 tons.
Neutral Plane and Neutral Axis.
The phenomena which accompany transverse stress are : —
Every horizontal straight line parallel to the axis of the
beam becomes a curve, one line on the diagram showing the
curved condition of all lines lying in the same horizontal layer.
All points in the beam, except those over the supports,
arrive at a lower level.
CHAP, v.]
TRANSVEKSE STRESS.
95
The
shorter
consequence is, that some
and others are longer than
top
bottom
horizontal
they were
layers
before
are
the
stress was applied. The  j^q^j^^jj^ layer is that which is
^ f compressed 1
«^^«t) I extended J'
and one nearer the
r top 1
\ bottom /
IS more
than one not so near. Since this condition of
( compressed 
{ extended ]
being extended diminishes gradually as you pass upwards
from layer to layer, and passes into a condition of being
compressed, there must be one intermediate layer which is
neither extended nor compressed. This layer is indefinitely
thin, is in fact a plane, and is called the neutral j^^f^ne of the
beam, and the line which is its trace upon the diagram is
called the neutral axis of the beam.
On fig. 4, the straight line OX, the neutral axis, while the
beam is unstrained, is chosen as an axis of reference. Let S be
any point on the neutral axis after the beam has been strained,
5 its distance from O measured along the curve ; let S' be the
point on OX directly above
S, and X its distance from v,
measured along OX. It
is to be observed that
the curvature, although ex
aggerated on the diagrams,
is really in practice so slight
that X and s will be sensibly
equal to each other, and x
may be put for the amount
of either unless where it is
absolutely necessary to dis
tinguish between them. S'S is called the deflection of the point
S ; the greatest value of this is called the deflection of the beam,
and when the beam is symmetrically loaded, it occurs at the
centre of span.
Let T be another point on the curve. Draw 8H and TR
normals to the curve at 8 and T meeting each other at H ;
then H will be the centre of a circle which will coincide with
the arc ST. Draw also 8K a tangent at 8 (fig. 5), meeting any
horizontal line KL at K. Then
ds = arc ST is the small difference between the values of
s for the two points T and 8.
96
APPLIED MECHANICS.
[chap.
ds
P
dx = S' T' is the small difference of the values of x, the
abscissae of T and S.
= dx so far as value is concerned.
= SH is called the radius of curvahire at aS ;
, its reciprocal, is called the curvature at S\
P
and H is called the centre of curvature at ;S'.
= the angle SKL is called the slope at S\ its greatest
value is at one end, and is called the slope of the
beam.
= the difference of the slopes at T and S
di
arc ST ds dx ...
_^ =  = — sensibly.
oil p p
These angles are in circular measure.
Let ya be the height of the top layer A A above, yi, the
depth of the bottom layer BB below the neutral axis, and
let { + ) yhe the distance of any layer CC \ ^qIq^ j the neutral
axis.
Fig. :,.
The portions of these layers intercepted between the two
radii ffS and HT before being strained were all equal to ds
in length. Let (ds  a), (ds + 3), and (ds ± y) be their lengths
respectively when strained ; then by similar triangles
arc A A arc ST ds  a ds
CHAP, v.] TRANSVERSE
STRESS.
1
.. ds  a = ds  y„
^^^'"^ BR = SR '
ds
' p'
or
ds
or a= y„;
P
ds + ft ds
P + l/b p
' ds p
>
.. ds ^ B ^ ds + ijb . — ,
P
or
P
_/3_ ^ yb
ds p
, . arcCC arc^T
^^^^^ CR = SR '
or
ds ± y ds
p±y P '
.•. ds ± y = ds ± y . — ,
P
or
ds
y = y  \ •••
p
JL = 1..
ds p
97
Now the intensity of the longitudinal strain on the layer
AA at the point A is
augmentation of arc A A a
original length of arc A A ds \ t' •/
Similarly the intensity of the longitudinal strain on the
layer BB at the point ^ is ^ , and that on CC at any point
± y
C is —7^ . Hence, from the above equation, we have
Ts''ds'l''^"'^'''^' ""' ^^ ^°^^^
The intensity of the longitudinal strain on each layer of the
point vjhere it crosses a section AB, is proportional to the distance
of the layer from the neutral axis.
Elements of the Stress at an Internal Point
OF A Beam.
To specify the stress at a point within a beam, it is
necessary and sufficient to find the intensity and obliquity
of the stress at that point upon any two planes through, it ;
for convenience we take two planes at right angles to the
plane of the paper. In fig. 6, (>X and dY are rectangular
axes ; OX coincides with the neutral axis of the beam, and
the origin is at the middle of its length. Let distances
measured along OX to the j^.^^^ , along OY ^""^l^^^] ^
H
98
APPLIED MECHANICS.
[chap. V.
be \ P°^^ J:^® i Then c and  c will be the abscissae of the two
(^negative J
ends of the beam ; x^, x^, &c., the abscissae of the weights. Let
H be any point in the beam, its coordinates being x and y ;
that is, on the diagram, x is the distance of H to the right or
left of 0, and y its distance above or below the neutral axis.
Of the planes at right angles to the paper and passing through
H, choose two, viz., AB and CD, vertical and horizontal.
According to custom, CD may be called the ^j^a^i through H, and
AB, the crosssection, or shortly the section ; further, it is called
the section at x, meaning that the abscissa of every point in the
section is x. H may be any of the points on the crosssection
at the distance y from the neutral plane ; and as all these points
are exactly under the same stress, it is unnecessary to say which
of them H is ; or, in other words, it is not necessary to give the
W,, Wi W3
A
C,D
I H y
V V* *f
u
^
'v*
B
♦y;
^Neutral Axis of
Cross Section
Q
Fig. 6.
third or Z coordinate of H required to specify its distance from
the plane of the paper. The trace of the neutral plane upon
the crosssection is a horizontal straight line, dividing the
crosssection into an upper and under portion, and this line is
called the neutral axis of the crosssection. In order to specify
the stress at H, we find the intensities and obliquities of the
stresses at that point upon the two rectangular planes through it.
The stresses upon these two planes are due to the strain upon the
beam; that is, to the fact of its being bent at the section AB,
leaving out of account the particular forces which actually cause
the beam to be bent. We may suppose that the beam is bent by
these particular forces, and surrounded by an envelope of some
rigid material, and then that these particular forces are removed.
CHAP, v.]
THANSVERSE STRESS.
99
Upon this consideration it is evident that the stress on CD
will have no normal comjjonent. Certainly, if one of the weights
of the actual load happened to be at 4, then a normal stress on
ri) would be directly transmitted to it; such a stress, however,
being accidental is left out of the present investigation. Having
remarked this about the plane CD, we leave its further con
sideration for some time, and give our attention to the section
A£. At the point // on the section AB,yfe see there will be a
normal component stress ; and we know, further, that it is a
thrust, since the horizontal fibre through // is compressed. If,
however, H be on the neutral axis of the crosssection, there is
no normal component stress, since a horizontal fibre through
such a point is unstrained. On the other hand, if H be below
the neutral axis, that is, if its ordinate y be positive, there is a
normal component stress ; and we know, further, that it would
be a tension, since the horizontal fibre through such a point is
stretched. Generally, then, at a point JI on a, section A B, there
will be a normal component stress of opposite signs for points
situated on opposite sides of the neutral axis of the cross
section ; and generally, also, there will be a tangential com
ponent stress acting in the same direction for all points on the
section, as will afterwards be seen.
To ascertain the stresses at points on the section AB ;
suppose the beam cut into two portions at that section, and
consider the equilibrium of one, say the left portion, as a rigid
body (fig. 7). The forces acting on it are P, W^, Wz, shown on
H 2
100 APPLIED MECHANICS. [CHAP. V.
the figure by strong arrows, together with the stress upon the
cut surface, shown by fine arrows. Let P be greater than the
sum of W\ and Wn \ then the vertical components of the fine
arrows must act downwards to conspire with W^ and W^ in
balancing P. For some positions of the section, M\ + Wo + &c.,
the sum of the loads on the portion of the beam to the left of the
section AB, may exceed P ; and on such sections all the vertical
components of the fine arrows must act upwards ; while for
other positions of the section, the sum of W^, W^, &c., may
equal P, and on these sections the fine arrows will be normal to
AB. Let ',' be the intensity of the stress on the section at the
point H, and y its obliquity. Eesolve the fine arrows into
vertical and horizontal components — that is, resolve the stress
at each point into a tangential and normal component stress.
On figs. 8 and 9, p and q are the components of r shown
separately, one set on each diagram ; pa and Pb are the values
of ^ at the highest and lowest points of the crosssection, and
the value of p at the neutral axis is zero ; qo is the value of q
at the neutral axis, and, as will afterwards appear, the value of q
at the highest and lowest points is zero.
The equilibrium of these forces gives the three following
conditions : —
I. The algebraic sum of the arrows p is zero.
II. The algebraic sum of the external forces (strong arrows)
together with the arrows q is zero.
III. The algebraic sum of the moments of all the forces
about 0' is zero.
Condition I. is equivalent to — The sum of the arrows p
which are thrusts acting on the portion of the section above
the neutral axis, equals the sum of the arrows p) which are
tensions acting on the portion of the section below the neutral
axis. Hence the resultant of all the arrows^ is a pair of equal
and opposite forces not in one straight line ; or, in other words,
a couple in the plane of the paper. Since this couple, by con
dition IIL, balances the sum of the moments of the external
forces about (>', therefore the moment of the couple is equal to
that sum, and acts in the opposite direction.
Definitions. — Fx, the algebraic sum of the external forces
acting on a portion of a beam included between one end and
the cross section at x, and comprising the supporting force
(if any) at that end and the loads on that portion, is called
the shear in/f force at that crosssection.
CHAP, v.]
TRANSVERSE STRESS.
101
Ml, the moment of these external forces about any point
on the cross section at .r, is called the bending moment at that
crosssection.
F„ the amount of the tangential component stress on the
crosssection at x, is called the resistance to shearing of that
crosssection.
Mj, the couple which is the moment of the total stress on
the crosssection at x about any point of the crosssection, or,
Iw,
p'^
r 1^
— ■(CX) ■?
:....{X.X) ,
Result ant
of tensions
JletulUnt
of thrusts
M the moment
of couple is th»
Resultant of nil
arrows *j^
F Is the
^'Resultant of
all arrows g.
Fig. 10.
what is the same thing, the moment of the normal component
stress on the crosssection about any point in the plane of the
paper, is called the moment of resistance to hending of that cross
section.
On fig. 10 these four quantities are —
F, = P  W, W..
M, = P[cx) W,{x, x) W,'x,  x).
F ^ = Eesultant of arrows q.
M,= Resultant of the pair of equal and opposite forces, one
of which is the resultant of all the thrusts p, the
other of all the tensions p.
The conditions of equilibrium for any portion of the beam
comprised between one end anrl a crosssection at x are —
F. = F.. (1)
M, = M,. (2)
Since F^ and 3fx are calculated from external forces alone.
102 APPLIED MECHANICS. [CHAP. V,
they are independent of the size or form of the crosssection,
and depend only upon the amount and distribution of the load ;
and, when calculated at a sufficient number of crosssections,
they form the data from which to design the beam. On the
other hand, F^ and M^ depend only upon the size and form of
the cross section at x, and upon the material of which the beam
is to be made. Having fixed upon a material for which we
know the working strengths to resist tension, thrust, and shear
ing, the two equations above enable us to design the different
dimensions of the crosssection at x in any required form. The
form to be adopted in any particular case depends in some
degree upon the shapes in which the material is naturally
obtained, or in which it can be manufactured cheaply ; and
when it can with equal facility be manufactured in several
forms, that one is to be preferred which takes greatest advan
tage of any difference in the resistance of the material to the
various kinds of stress ; since, by doing so, we require the least
quantity of material. The form of crosssection chosen must
be suitable for the particular nature of the load, and locality of
the beam. Having thus designed the crosssections at a suffi
cient number of places, we are said to have designed the beam
so as to have sufficient strength to resist the given loads.
Besides the above qualities of suitability, cheapness, and
economy of material, a beam must also have sufficient stiffness.
We will, further on, derive equations to enable us to find the
form of economy mentioned above, and also to select the ratio
of the dimensions of the crosssection to fulfil the condition of
stiffness.
The crosssection at x either being given or having been
designed, equations d) and (2) enable us to calculate the
elements of the stress at any internal point ff.
The Cantilever.
A beam may be supported by one prop placed exactly below
the centre of gravity of the load, as shown on fig. 11. It is
evident from the definitions that, in this case, the shearing force
and bending moment are maxima at the point of support ;
because the external forces upon one of the portions into which
the section through this point divides the beam, say upon the
left portion, are the loads on that portion, and they all act in
one direction. The bending moment at that section is the
arithmrtical sum of cacli of these loads into its leverage ; and
for any other section to the left, the number of loads to be
reckoned in calculating the bending moment may be fewer, and
CHAP, v.]
TRANSVKKSE STRESS.
103
at the same time all the leverages are shorter. In considering
the left portion alone, it is unnecessary to draw the other ; and
instead, the left portion may be supposed, as on the figure, to
terminate at its right extremity in a vertical plane ; this plane
is supposed to give the necessary resistance to balance the
shearing force and bending moment at the point of support.
T
atfliec"
end
Fig. 11.
Such a piece is called a Cantilever; as shown on fig. 11, the
right is itsjixed and the left its/ree end ; its length is c. When
the cantilever is strained, that is bent, the tangent to the axis
at the fixed end remains horizontal ; the upper layer is stretched
while the lower is compressed, and the lettering will be accord
ingly. The deflection v of the cantilever is the difference of
level of the two ends, and its slope i is the inclination of the
tangent to the axis at its free end with the horizontal.
We shall now proceed to calculate the bending moments on,
and construct bending moment diagrams for, beams and canti
levers under various loadings ; in such a diagram, the horizontal
and vertical dimensions are respectively the span and bending
moments at the different points of the span drawn to scale.
Each of these diagrams requires two scales — a scale, say, of feet
for horizontal, and a scale such as foottons fcr vertical, measure
ments ; there may be required, also, a scale, say, of tons for loads,
if the loads are drawn to scale. Such diagrams must be drawn
upon a large scale if intended to be used as graphical solutions,
in which case only the construction requires to be known. The
diagrams of this treatise are too small for such purposes, but
they serve to show clearly the constructions, and are principally
useful as maps upon which to note the analytical results. In
every case, these diagrams are constructed so that one of their
104
APPLIED MECHANICS.
[chap. VI.
boundaries is straight — is, in fact, the span. This is a matter of
importance, as the diagram so constructed assumes a suitable
form for a practical purpose which will be afterwards pointed out.
One boundary of such diagrams is generally a parabolic right
segment in certain simple positions ; and we will now give a
short chapter on the equations to, and the construction of, this
curve.
CHAPTER VI.
THE PARABOLIC SEGMENT.
THE parabola.
On tig. 1, Xand Tare two rectangidar axes passing through A,
any point of reference. Let distances measured from A to the
( left ) J J. ^^ . p ^ (downwards) ^{ positive 
( right j '
and from the axis of X
upwards
be
negative
Fig. 1.
The points Pi, P., &c., have their ordinates IPj, 2P2, &c.,
proportional to their abscissae Al, A2, &c. ; that is, if JTand Y
be the coordinates of any point P, then
Y = mX,
where /» is any number, whole or fractional, and the locus of P
is a straight line passing through A. In the figure m = \, and
for any point P, X, and Y are both positive or both negative.
The coordinates of points Pi, P2, &c., are marked ; e.g. for P3,
X=3, and Y= f.
On fig. 2, Pi, F2, &c., are points coiTesponding to the points maiked similarly
on fig. 1, bnt in this case the points P], Pi, &(;., have their ordinales \P\, 'IPz, &c.,
proportional to the squares of their ahscissse ; that is
'I'his is the principal equal inn to the pMrabola, and the quantity m is the itwdulus.
The locus of this equation, P, is the parabola ; and it is altogether on one side of
the axis of X, since, although X, the abscissa of any point P, is positive or negative,
the quantity X"^ is always positive.
CHAP. VI.]
THE PARABOLIC SEGMENT,
105
In the figure m = ^, and for any point 2\ while X may he positive or negative,
J" is always positive. The coordinates of tlie points Pi, T2, &c., arc marked;
e.ff. for F3, X = 3, and Y = 2^. The point A is called the vertex, and the line A Y
+ X
Fig. 2.
the axis of the parabola ; the curve is symmetrical about this axis. The points
P\, P2, &c.. thus found are points on the curve ; and if a sufficient number of such
points be found and a fair curve drawn through them, the curve will be sensibly
a parabola.
Fig. 3.
Whenever ■\e have an equation of the above form, we conclude that the locus
a parabola vbose vertex is at the origin, and whose axis lies along the axis of Y.
106 applied mechanics. [chap. vi.
Parabolic Segment.
Any line as BC, fig. 3, meeting the curve in two points, and drawn parallel to
the axis of X, may he called a right chord ; and the tigure enclosed between thi&
chord and the curve may he called a parabolic right segment. This chord is the
base of thf! segment, and AO, the distance of the vertex A from the hase, is the
height of the segment.
The following is a convenient construction for drawing a parabolic right
segment of a given height and on a given base £C, fig. 3. Plot A at the required
height above the nii<idle of the base, and complete the rectangle BOAS. Let it
he required to construct accurately twelve points at equal horizontal intervals.
Divide AH and SJi each into six equal intervals, and number them as on the
figure. From each number on AH draw vertical lines, and from the point A draw
a. ray to each number on HB. Then Pi, the point of intersection of the vertical
through 1 and tlie ray Al, is a point on the parabola , so also is F2, the point of
intersection of the vertical through 2 and the ray A2 ; &c.
It is evident that, for instance, H/'i, is piopoitional to the square of AD ; for,
calling IPi unity, then JjJ = 4, and T)Pi = 4 DJ = 16 = 4, while the abscissae of
the points Pi and P4 are proportional to 1 and 4 ; similarly with the other points.
Hence onefourth is the common ratio of the depth of each point below AH to the
square of its distance laterally from A 0. We will return to the subject of drawing
a parabolic segment, and will now show how to draw a
TANGENT AT ANY I'OINT OF A PARABOLA,
say P4 ; on Pi A and P4D construct a parallelogram and draw its diagonal P4£:
this line will he the tangent required. P4^ is the same fraction of F^D that P4L
is of Pi A, and ZP3 is parallel to P4fir. If KP3 were parallel to P4i, then would
KL be a parallelogram similar to DA, and so P3 would be on the diagonal PiE.
But since KP3 is not parallel to P4X, but coTiverges towards it, the point P3 where
EP3 meets LG lies to the right of the line PiE. Produce ^P4 through P4. PjiV
is tlie same fraction of /'il) that PiM is of P4^ ; if NPt were parallel to PiM,
then P5 would he on the diagonal PP4 produced ; but, since NP^ diverges from
PiM, itie point Ps where iVPs meets MP5 lies t<> the right of the line PiE produced
through P4. In the same manner every point on the curve, either above or below
P4, lies to the right of I'iE; that is, PiJE is a tangent at the jioint P4. Now the
diagonal PfE bisects the other diagonal AT) in P, and the most convenient way of
dra>ving a tangent at anj point P4 remote from the apex is to project P4 on the
hoiizontal tlirough the apex ^ ; bisect AD in F, and draw P4P; this is the tangent
required. It is readily shown that PiE is parallel to any chord as PsPa with its
ends equidistant horizontally from P4.
To plot the locus of an equation of the form F= »«X* ; in other words, to draw
the parabola whose modulus is m. For instance, let ni = ^. Draw any lii\e liC
parallel to the axis of X, fig. 3, as hase; lay off OA equal to ^05; then dmw
the segment by plotting accurately a number of points Pi P2, &c., by the con
struction already given, and draw a fair curve through them. Usually we fix only
a few points on the curve accurately, and from these the rest of the curve is
sketched in. By making the number of these points sufficiently great, we can.
draw the curve as accurately as we please.
A parabolic segment might be constructed on, and cut out
of, a piece of cardboard, and used exactly like a set square.
The parabola could then be quickly drawn on our diagrams
thus : — If the apex A be given, place the parallel rollers to
the axis of X, shift the rollers, slide the cardboard segment
CHAP. VI.] THE PARABOLIC SEGMENT. 107
along them till the apex is at A, and draw the curve. Again,
suppose we are given (fig. 3) the axis OJi and a point B on the
curve, and that we wish to draw the curve ; place the rollers at
right angles to the given axis OE, slide the cardboard segment
with its apex on this axis till the curved edge passes through B,
and then draw the curve. Instead of cardboard we may use a
parabolic segment cut out of a slip of peartree or brass, and
with one such segment we can, if we choose, draw all parabolas.
Thus, suppose our peartree segment to have the modulus , and
that we require to draw a segment whose modulus is 1. We
may choose, as suitable for horizontals, a scale of 10 parts to an
inch ; lay off the base with this scale and draw the curve with
the peartree segment ; if we now measure the verticals on the
scale, we find for every point Y = ^X"^. If we draw a new scale
of 40 parts to an inch for vertical measurements, then for every
point on the curve we will have, as required,
F= 1 X X^ = X\
In like manner the curve drawn by this peartree segment
may represent any jiarabola, if the verticals be measured on
a suitable scale ; this is similar to the common practice of
exaggerating the vertical scale for sections. All the diagrams
which immediately follow may be very conveniently drawn
with one such seijment, since it is much easier to draw an
additional scale than to construct a new parabola; for example,
suppose the segment, fig. 90, so drawn, and that we wish it to
represent the parabola Y = \X'^.
Using a horizontal scale of four parts to an inch, we have by
measurement OB = 6, and OA or HB = 9 ; that is, HB = {OB' ;
but we wish HB to measure ^OB", that is 12. It is only neces
sary then to divide HB into 12 equal parts, and lay off a scale
of such parts to be used for verticals ; this scale is evidently
that of 3 inches divided into 16 equal parts.
If the new scale be only slightly finer than the old, then the
old scale plotted sloping to the vertical at an angle can be used
by ruling horizontals across to it.
Equations to the Parabola.
In figs. 4 and 5, lei ihe middle of the span be the origin of rectangular
coordinates, the span BC being taken as the axis of X ; let distances measured
from to the  ^ ■,. \ , and from the axis of A^  "P^^*" ^ i be  ' „^„,;,.^ } •
I right I ' I downwards ) ' ( negative )
In fig. 4, the axis of the curve passes through 0, whereas in fig. 6 it passes to
108
APPLIED MECHANICS.
[chap. VI.
one side. Let x, y be the coordinates of any point P\ and for the apex let x = K,
and y = H.
In fig. i K ^ 0, since Kis on the axis of Y; and in order to find the equation
to the curve with the origin at 0, we have
r =  mX'
Instead of X and Y, we substitute their values, thus
(origin at A).
H = — mx^ ;
y
In what immediately follows, let Co, Ci, and Ca be co«s<flM< quantities; then,
if we have an equation of tlie form y = Co + (hx^, consisting of a term not con
taining X. and a term in x"^, we conclu<ie ihat the locus is a i>arabola, with its axis
veitical, its apex on the axis of Fand at the distance Co from the origin ; and that
the modulus of the parabola is C2, the coefficient of x". so that the principal equa
tion is y = CiX'^.
For example, suppose we wish to plot a number of points above the span BC,
such that the coordinates of each point may fulfil the equation
y = ;i (36  a:2) = 9  \x' ;
we conclude that all the points are on a parabola whose axis is the vertical through
0; that the apex A is at the height IT = 9, the term not containing x ; that the'
modulus is — ^, the coefficient of x ; and tlierefore that the principal equation is
r=iX2; .. S=9 = ^0D\ or OD = 6.
Fig. 4.
To draw the locus, lay off OA = 9, OD = OE = 6, aud con
struct points on the curve, as already shown on fig. 3 ; or more
quickly, by means of the peartree segment we may draw a
curve through the points D and E just found, aud construct a
scale for verticals upon which OA measures 9.
Again on fig. o, tlie equation to the curve with the oriain at is derived
from the principal equation by ."substituting x  K nr\A y  H for X and Y, thus
y ~ H= m {x A' )■,
or y = H~ mx + 2mKx  m A'*, or y ^{H  niK^) + Im Kx  mx^,
which is an equation of the form ;/ = Co + Cix + dx, consisting of a term not
CHAP. VI.]
THE PARABOLIC SEGMENT.
109
containing x, a term in x, iind a term in x : and we conclude that the locus ia a
parabola, with its axis vertical. In this equation, if C1C2 is { ** *" '^^ i , the
•^ ' H » ' i J neg:it:ve ) '
apex is to the  .^^ I of the axis of T; if Co  — r is  \ ^ { , the apex is
I above ) . . f ,
{ 1 > the axis of A.
[ under I
To find the value of A' and of H, arrange the equation thus: —
y = m (2K X)x ^ H— mlf^.
Now when y = JT, its value is a maximum, and the corresponding value of .»• is
K; so that to ascertain A', we huve onl\' to tind that value of x which makes
y = m ('2K — x) x + S — mK^ = maximum.
This is a maximum when {IK a;) ^is a maximum, since the rest of the expression
is constant for all values of x.
Suppose that 2 A is the length of a line, which is divided at a point into two
segments ; let x be t)ie length of one of them ; then 'IK  a: is the lengtli of the
other, and (2A— x)x is the rectangle contained by the two. "We kno.v by Euclid
that this rectangle is greatest when the segments are equal, each being half of the
line 2A'. So that, wlien x = K, \^1K — x)x is a maximum, m{2K — x)x + H  iiiK'^
is also a maximum, aiid y = S.
Fig. 5.
For example, suppose we wish to plot a numher of points whose coordiiuites
X, y fulfil the equation
y = i (4  a:)a: + 3 = 3 + a;  ix2
From the form of this equation, we conclude tliat all the points are on a parabola,
whose axis is vertical and to the left of ; that the ap»x A is above BC ; that the
modulus is  \, the coefficient of x"^, and therefore that the principal equation is
The distance K at which the axis lies to the left of is found thus : — the value
of X which makes
y = ^ (4 — x)x + 3 = a maximum,
is that in which \ — r = x, or x = 2 ; that is, A'= 2.
Again, to find the height of the segment ; when
x = K, y^H, and IT = i (4  2) 2 4 3 = 4.
Substituting in the principal equation, we have
i = ir= \SI) ; • therefore haKbase = 6'i) = + 4.
110
APPLIED MECHANICS.
[chap. VI.
To draw the locus : — lay off OS = 2, and draw a vertical
through S ; from JS lay off SD = SB = 4, and SA = 4 ; and
construct points on the curve, as in fig. 3 ; or more quickly
by means of the peartree segment, we may draw a curve
through the points I) and U just found, and construct a scale
for verticals upon which SA measures 4.
Theorem A. — Fig. 6. Tlie quadrant of a parabola and a riglitangled triangle
stand on a common liorizontal base, witli the right angle of each at one end (the
right end in the figure) ; let a be the lieight of the apex of the parabola, and b ihe
height of the vertex of the triangle, above that end. If, ai each point cf the base,
the ordinate of the parabola be added to that of the hypotenuse of the triangle, and
a. new curve be plotted ; it will be the sawj^ parabola with its axis still vertical, and
having its apex shifted beyond that end (the right in the figure) of the base .ibove
which the apex was, and the curve will still pass through the other end of the base.
In like manner, if the ordinate of the parabola be deducted from that of the
liypotenuse, a similar result is obtained ; the apex of the new figure, however, is
filiifted to the other side. The horizontal distance through which the apex shifts is
(oC=£base)
of 4 Ord
Fis
the same fraction of the base that the height of the vertex of the triangle is of
twice the lieight of the apex of the parabola, ; or if o be the lateral distance through
which the apex shifts, then
a = — X base of the quadrant.
2a
The addition does not afi'ect the coefficient of x"^, so that the modulus is
■unchanged. In short, the algebraic addition of the ordinates of the straight slope
makes the parabolic base segment shift on the rollers from the position on tig. 4
to that on fig. 5.
Theorem Ji. — If to the ordinates of a paiahola, axis vertical and ape.x to left or
right of origin, the ordinates of another parabola, with axis vertical and passing
through the origin, be added ; then the new locus is a parabt)la, its modulus is the
sum of their moduli, and its axis is vertical ; its npex lies on the same side of the
origin as the apex of the first parabola ; and the abscissa of its apex is the same
fraction of the abscissa ol the first parabola's apex that the modulus of the first
parabola is of the sum of the moduli. (Fig. 7.)
CHAP. VI.]
TUE I'AHABOLIC SEGMENT.
Ill
For the parabolas whose apexes are Ai and Ao, respectively,
y'l = yo + yi = fiTfl  "»»^ + « (iTi  mKi^) + 2inKix  mx',
or y 1 = (flo + fiTi  mi'r') + {mo + ♦«) ( "'"' ^  z\ x, (1)
which is a parabola of modulus (>w„ + m), its axis is vertical, its apex lies on the
same side of origin as the apex of the tirst parabola ; and if K'l is the value of x
... , / ImKi \
which makes  — x \ x greatest, then
\}>io + m /
ir\ =
i)li) + in
A',
(2)
Q. E. T).
Fig. 7.
Theorem C. — If to the ordinates of a locus consisting of two parabolas, with
axes vertical, wiih a common modulus m, and intersecting at F, there be added the
ordinates of a parabola, axis vertical, apex above origin, and modulus nia ; then
the new locus consists of a pair of parabolas with axes vertical, having the common
modulus (;«o + m), and intersecting at F^ on the same vertical as F. (Fig. 7.)
For the parabolas whose apexes are Ai and A2, respectively,
yi = {H\  niKi'^) + 2mKix  mx'^ ; 1/2 = (If^  mK^'^) + iniEiX  mx'i.
If f be the abscissa of the point of intersection, then where yi = yz, x = f ;
subtracting, we have
= (.S"i  Hz)  m(Ei  Ki) + 2m[Ki  K^Y,
which gives the value of/. '
(3)
112
APPLIED MECHANICS.
[chap. VI.
Again, by the previous tlieorem, equation (1),
y 1 = [Ho ^ Si  mKr) + 2mK\X  (wq + m)x ;
similarly
y'2 = (Ho + ff2  mKi^) + 2mE2X  (wo + tn^x^.
Let/A be the abscissa of the point of intersection, then where y'^\ = y^i, x =/* ;
subtracting. Me have
= (£"i  Hi)  m{Ki^  jfo^) + 2m{Ki  Ki)/^ , (4)
hence
r =/.
Q. E. B.
(5)
Theorem D. — If two parabolas with axes vertical and a common modulus »«
intersect at a point, then the horizontal projection of any double chord drawn
through that point is constant and is equal to the double horizontal chord drawn
through it. And conversely, if a line whose horizontal projection is equiil to the
double horizontal chord through the point of intersection be placed with one
extremity on each parabola, either the line or the line produced passes through the
point of intersection. (Fig. 8.)
vn and vb
are constant.
Fig. 8.
The equation to the parabola (fig. 5) is
y = (H mK^) + 2mEx  mx^ ;
choosing the origin at the point of intersection, then
Hi = mKi\ and Hi = mK>^ ; hence [Hi  )nKi^) = (Hi  mEz') = ;
and considering Ei and E2 positive, we have
yi = 2mEiX  mx^, (6)
2/2 =  2mE2X  mx. (7)
The equation to any straight line thiough may be written
.'/ = M^ ; (8)
equating (6) and (8), we find for Fthe point of intersection x = 2E\ ; sinii
m
larly for N, x =  2Ei  ^.
Hence vn = 2{Ei >t E>^ = ST. (9)
Further, if VN be such that vn = 2 (E\ + Ei), and T moves on parabola No. 1,
while iV moves on parabola No. 2, it is evident that J'iV always passes through 0.
When V arrives at 0, then VN is a tangent at to parabola No. 1; and if V
moves to F", then NV produced passes through 0. Q, E. D.
CHAP, VI,]
THE PARABOLIC SEGMENT.
113
Theorem E. — While the liouble chord 7'OiV turns, if for each position there be
taken on it ti point li whose horizontal projection b divides rTt in a constant ratio ;
th»'n this point traces out a parabdla whose axis is vertical, which has the same
modulus m as the parabolas Nos. 1 and 2, and which passes through the point of
intersection 0. (Fig 8.)
X, tj being the coordinates of F; let X, Y be the coordinates of B \ and
vb = C a, constant ; then
x = X+ C, and Y = ^ X.
Now 1/ = 2mKix  mx^, or — = m {2Ki  x) ;
•' X
therefore Y=m {2Ki  C  X) X,
(10)
the locus of B ; and it is a parabola with axis vertical of modulus m, and passing
through 0. The abscissa of the apex A3 is equal to JS"]  iC. Q. E. I>.
O^pe*
A pro
«»Vy
y
14
30
.^
12
18
&)
A
6 X
Fig, 9.
Fig. 10,
Degradation of a Locus.
Definition. — If a locus be drawn whose ordinates are proportional to the square
roots of those of a given locus, the new locus is the given locus degraded.
2'heorem F. — Fig. 9. If a locus which is a straight slope be degraded, the new
locus is a parabola with its axis horizontal, and its apex at the point where the
locus crosses the horizontal base.
For the straight slope, y = ax, if we take the origin at the point where it cuts
the base ; if A be the corresponding ordinate of the new locus,
h^ oc y, or ax ; that is, x oc h^,
a parabola with axis parallel to OX, and apex at the origin.
Theorem G. — Fis;. 10. If a locus, which is a parabola with axis vertical and
apex on the base, be degraded, the new locus is a straight slope cro.'ssing the base
at the apex.
For a parabola with origin at apex, y = mx ; but A oc y; therefore
h oc \/ mr,
a straight line passing through the oiigin.
114
APPLIED MECHANICS.
[chap. VI.
Theorem H. — Fig. 11. If a locus which is a parabolic segment be degraded,
the new locus is an ellipse with its centre at the middle of the base, the base being
a major or minor diameter.
O \0 <iO 2,0 ^0 ^0 ^oX
2 OA ' 1 "*
Fig. 11.
Taking the origin at the centre of base, we have for the equation to a parabola,
y = m{c'^ — x^), where 2c is the base. Let h' = fih, where /x is a quantity such
that h'^ = y; then since h^ oc y, we have
— = c — x'^
/i'2
+ a;2
^2^2
= 1;
the central equation to an ellipse whose semidiameters are c and . The ratio
of the semidiameters depends on m.
On fig. 11, the ordinates of the degraded figure have been multiplied by 4 ;
thus OA is 24, or four times the square root of 36. So that BAF i?, the degraded
figure if the vertical scale be four times as coarse as the horizontal. If the multi
plier were 10, then BAF would be a semicircle.
Hence a figure consisting entirely of arcs of the same parabolic segment can,
when degraded, be represented by a series of arcs of circles, provided a suitable
vertical scale be constructed.
The false semiellipse, fig 11, is readily struck from five centres. The first at
a distance rz = 60 ^ 24 = 150 below A. With this radius AD is swept out, and
BC with n = 24* H 60 = 10 nearly. While DC is drawn from the centre E
defined as shown on the diagram.
CHAP. VII,] BENDING MOMENTS AND SHEARING FORCES. 115
CHAPTER VII.
BENDING MOMENTS, AND SHEARING FORCES FOR FIXED LOADS,
Definition. — The Beriding Moment at any crosssection of a
beam, or, as we may more conveniently say, at any point of
the span, is — The sum of the moments about that point of all
the external forces, acting upon the portion of the span on either
side of the point.
For convenience in the case of beams supported at both
ends, we calculate this bending moment from the forces acting
upon the portion to the left of the point. These forces comprise
{vide figs. 8 and 10, Ch. Y) the supporting force P acting
upwards at the left end, and the loads acting downwards
between that end and the point. Taking the centre of span as
origin, the abscissa of P is c ; and, if x be the abscissa of the
point about which moments are taken, then (c  x) is the
leverage of P, and P tends to break the beam at the point
by bending the left portion upwards with a moment P{c  x) ;
the abscissa of ir, is x^, its leverage is {x^,  x), and it tends
to break the beam at the point by bending it dovmwards with
a moment Wx{xi  x) ; all the other loads to the left of the
point have an effect on the beam similar to that of W^ ; and
since, in this case, the left portion of the beam is bent upwards
at the point, the moment P{c  x) exceeds the sum of all these
moments W^{xx  x) + W^ixz  x), &c. ; and if Mj: represent the
bending moment at any point x, then
M^ = P{c  x)  JV,{x,  x)  W,(x2  x), &c,
all the loads on the portion of the beam to the left of the point
being taken into account,
A Bending Moment Diagram is a figure having a horizontal
straight line for its base, equal in length to the span on a scale
for horizontals which should accompany the diagram. Above
this base is an outline or locus consisting of a curve, a polygon
with straight sides, or a polygon with curved sides, and such
that the height of any point on the outline gives the bending
moment at the point of the span over which it stands, measured
on a scale (say) of ft.lbs. or of ft.tons for verticals which also
should accompany the drawing. It is evident that this outline
always meets the horizontal base at both ends, since the bending
i2
116 APPLIED MECHANICS. [CHAP. VII.
moment at each end is zero. It will be seen that x and Mjc
are respectively the abscissa and ordinate of a point on this
locus or outline ; an equation between those two quantities,
such that, when you substitute into it any value for r, it gives
you the corresponding value of Mx, is called the equation to the
lending moment.
An approximate method of drawing a bending moment
diagram is, to calculate the bending moments at a number of
points of the span, say at equal short intervals ; plot these to
scale, and then join the tops of the verticals with straight lines,
or draw through these points a fair curve. Such a diagram will
give the bending moments accurately at the points which were
plotted, and approximately at intermediate points ; and its
principal use is to mark the calculated results thereon. On
the other hand, if upon investigation we find the locus to be
of a form which we can draw readily, then, drawing the diagram
first, we may afterwards, by measurement from it, find the value
of the bending moment at any point of tlie span. Such a method
of proceeding is called a graphical solution.
The MAXIMUM BENDING MOMENT is that value of the bending
moment, than which no other value is greater ; if this value be
at one particular point of the span, that point is called the
jwint of maximum bending moment ; sometimes this value ex
tends iDetween two points of the span, in which case any point
intermediate may be so called. The determination of the
maximum bending moment, and of the point at which it
occurs, is of great importance. A graphical method is pecu
liarly successful in giving these, as we know or readily find
the highest point on the diagram ; the height or ordinate of
this point is, of course, the maximum bending moment, and
its abscissa is the point at which it occurs. In all the cases
which follow, that point on the diagram is either the angular
point or side of a polygon, or the apex of a parabolic right
segment.
Definition. — The Shearing Force at any crosssection of a
beam, or, as we may more conveniently say, at any point of
the span, is — 2'he algebraic sum of the external forces acting
upon the portion of the beam to either side of the point.
We will denote this shearing force by Fx, and calculate its
amount systematically from the forces acting on the portion of
the beam to the left of the point; thus for a beam
Fx = P W,  W,  &c., (1)
the external forces on the left portion.
CHAP. VII.] BENDING ilOMENTS AND SHEARING FORCES. 117
Similarly, for a cantilever
t\ =  W,  W,  &e., (2)
On t!ie beam then, i^^. is P , •  according as P
' I negative) *
IS
1^ , I than the sum of the weights to the left of the
section at .r. The beam may be divided at a certain point
into two segments, such that for every point in the I • , ,
o ' J f jnght)
segment, the shearing force is j^ ^ • I ; at this point the
shearing force changes sign. The shearing force on either
segment may be considered as positive ; for convenience we
have chosen as just stated. Having fixed the sign thus, on
a cantilever as shown on the diagrams, that is with its fixed
end to the right, the shearing force is everywhere negative ;
and in considerations regarding beams atid cantilevers, it is
necessary to take this into account ; when, however, the
cantilever alone is considered, it will be better to reckon the
shearing forces as positive.
It may be observed that the symbolical expression for F^
does not de]>end for its form upon the position of the origin ;
in this respect it is unlike J/^
A Shearing Force Diagram is a figure having a horizontal
base representing the span on a convenient scale, and an outline
or locus. For a cantilever this locus lies wholly under the base ;
for a beam, it lies above the base for the segment to the left,
under the base for the segment to the right, and crosses the
base at an intermediate point. The heiglit of any point on the
locus, measured on a scale for forces, say tons or lbs., gives the
shearing force at the point of the span over which it stands.
Both these scales should accompany the drawing.
Mcu:imum Shearing Force. — On a cantilever it is evident
that the greatest value is at the fixed end. On a beam there
are two values each greater than the others near it, one at the
left end and positive, one at the right and negative ; the value of
the greater of these is the greatest for the whole span.
This locus, in the diagrams for all the cases of fixed loads
which we consider, consists of straight lines. For a portion of
the span between any two adjacent loads, the straight line is
evidently horizontal ; for portions uniformly loaded it slopes at
118 APPLIED MECHANICS. [CHAP. VIL
a rate given by the number which indicates the intensity of the
load; thus if w lbs. per foot be the intensity of the uniform load,
then w vertical to one horizontal is the slope. Where a weight
is concentrated at a point, the line is vertical ; that is, at such
a point the locus makes a sudden change of level, the change
being equal to the weight.
On a cantilever, the shearing force at the fixed end is equal
to the load, and at the free end it is zero ; we can readily draw
the straight lines as above described to suit the nature of the
load, and so complete the diagram.
On a beam the shearing force at the left end is P, at the
right end it is  Q, and at some intermediate point the locus
intersects the base and changes sign ; the manner of fixing the
position of this point will be explained immediately. The whole
locus is then readily completed by drawing the lines as above to
suit the nature of the load.
Theorem. — The Shearing Force at any point of a beam or
cantilever is the rate of variation of the bending moment at
that point; and on the beam the shearing force changes sign at
the point of maximum bending moment.
As before
i^. = PS(IF),
where ^{W) means the sum of the loads to the left of x ; and
M, = P{cx) S(ir..r),
where '^{W.x) means the sum of the products got by
multiplying each load to the left of x by its leverage about x.
Hence if we suppose F^ to be positive, and take the bending
moment at any interval d further to the right, the second
bending moment will exceed the first by F^ . d, if there be
no load on the portion d ; and by Fx . d, minus the load over
the portion d into its leverage about x, if there be a load
on the portion d.
Since the leverage of the load which is on the portion d is
less than d, we can by taking d small enough make this
product as small as we please ; that is, i^^ • ^ is the change of the
bending moment in passing from x through a small interval d.
Now the rate of change of M^ means the change in passing
from ./■ through an unit interval, say one foot, if the change
continued uniform throughout that interval, and at the same
rate as at x; in otlier words, the change in Mj^ for d reckoned
CHAP. VII.] BENDING MOMENTS AND SHEARING FORCES. 119
equal to unity, without taking into consideration any additional
loads which may be over that interval ; hence the
rate of change of M^ = Fx ;
also, as we pass from the left to the right end of the span,
'^ ^^ ^^ {neglti^e}' ^^^ ^« { rc^eashl ) ' ^"^^ ^^ ^^^^ P°^^
where Fx changes sign M^ is a maximum.
For fixed loads tlie shearing force diagram greatly assists the
construction of the bending moment diagram, besides readily
locating the maximum bending moment. It is convenient to
consider and draw these two diagrams conjointly.
Beam with Unequal Weights at Irregular Intervals.
Let a beam 42 feet span (fig. 1) support weights, viz.,
W, = 5, W., = 5, IF3 =11, W,= 12, and W^ = 9 tons at points
whose abscissce, reckoned from the centre, are x^ = 20,
X2 = 15, .rj = 7, Xi =  3, and x^ =  10 feet. This is the
problem of example 8, fig. 3, Ch. V, and we have P = 24 tons,
at c = 21 feet, and Q = 18 tons, at  c =  21 feet.
To calculate the bending moment at any point x, x =  S,
for instance, we may take the forces upon the right portion, and
i/_3 = ^ X 18  1^5 X 7 = 18 X 18  9 X 7 = 261 ft.tons.
We will now calculate the bending moments at the points
where the weights stand, in each case systematically from the
forces upon the left portion.
Ft.tons
J/,.  P{cx,);
il/20 = 24 X 1 = 24
M^^ = P(cX2) W, (ai  x^) ;
i!/,5 = 24 X 6  5 X 5 = . . . . . . .119
M^z = P{cX3)  }F, (x,  X3)  W2 (X2  x^) ;
iJ/, = 24 X 14  5 X 13  5 X 8 = 231
Jil.^ = P{c~x,) W,{x,^x,) W^(x,x,) W,{x,x,);
iT/.3 = 24 X 24  .5 X 23  5 X 18  11 X 10 = . . . 261
^Z5 = P{c  X5) W^ (Xi  X,)  Wi {x2  x^)  &c. ;
M_,, = 24 X 31  5 X 30  5 X 25  11 X 17  12 X 7 = 198
120
APPLIED MECHANICS.
[chap. VII.
In the interval between W3 and W^, consider any two
sections A and B, the latter being situated further to the
right; let the distance between them be d. The leverage of
F is greater at B than at A by the quantity d, and the
wpward bending moment of P is greater at B than at A by
the quantity F . d ; again, the leverages of W^, W^, and W^,
respectively, are greater at B than at A by the quantity d,
and the dovmward bending moment due to W^, W^, and IFj
"V ;d< O <V udJ **
W=12 W=9fon»
1
18= Q
^
W'9
«='Q
Forces on Right
Portion of Span
W,= 5 W.= 6
24= P
24 = P
....
I Sect/on af)
■ r I,.
U—.^XXJ^13
^^*T^I '^"'■::ri
(CX) = 24
24Pri( L.2..6'. (X!^X)^30 —
tion at)
,= 70 i
Fig. I.
is greater at B than at A by the quantity (Wi + W^ + W3) d.
Hence M^ is to be derived from M^ by adding F . d and sub
tracting ( I'T, + W2 + Ws) d ; in our example F>( W^ + W^ + W^),
so that the quantity to be added exceeds the quantity to be
subtracted ; that is
Ms > M,, , by (FW, W,  W,) d,
a quantity proportional to d.
CHA?. VII. J BENDING MOMENTS AND SlIEAUING FORCES. 121
Again, for any two points C and D in the interval between
W^ and Wi, Md is to be derived from M c by adding P . d and
subtracting ( W^ + W. + W^ + W^) d ; in this case, however,
P < {Wi + Wi + Wi + }Vi\ so that the quantity to be added
is smaller than the quantity to be subtracted ; that is
Jin < Mc, by ( Wi + IF, + TV, + IV,  P) d,
a quantity proportional to d.
In both cases, as we pass towards the right from one point
to another in the interval between two weights, the change
in the bending moment is uniform, and is proportional to the
I"' I"'
I
Fiof. 2.
horizontal distance passed over — uniformly increasing or uni
formly decreasing according as the supporting force P is greater
or less than the sum of all the weights to the left of the points
being considered ; and if for some interval, P is equal to the
sum of the weights to the left, the bending moment at points
in that interval is constant.
Bending Moment Diagram. — With a scale of feet for hori
zontals, lay off the span, fig. 2, and plot the positions of the
loads ; at each of these points erect a vertical equal to the
bending moment thereat upon a suitable scale of, say, ft.tons
122 APPLIED MECHANICS. [CHAP. VII.
for verticals ; join the tops of these ordinates by straight lines,
and join each end of span to the top of the nearest ordinate.
The lines just drawn give ordinates which vary uniformly in
each interval, so that at any point whatever the ordinate gives
the bending moment at that point in ft.tons when measured
on the vertical scale. The bending moment at each load having
been calculated analytically, a diagram thus constructed is a
graphical solution for every other point.
Maximum Bending Jloment. — In our example, since P > Wi,
the line AB slopes ujj towards the right, that is, the bending
moment increases in this interval ; BC also slopes icp towards
the right since P > (W^ + W^), and Cl> slopes up towards the
right since F > {Wi + W2 + W3) ; BU slopes doivn towards the
right since P < ( ^1 + W2 + JV3 + Wt), and lastly, BF slopes
down towards the right since F < {Wi + W2 + W^ 1 W^ + W^).
That is, beginning at the left end and passing towards the right,
we find that the sides of the polygon slope , ^ towards the
right, while P is p .  than the sum of the weights we have
passed. For some loads there is one interval where P is equal
to the sum of the weights passed, and in such cases the side of
the polygon above that interval is horizontal. Evidently the
highest ordinate is that of the angle of the polygon made by
the last side which slopes up to the right, either with the first
side which slopes down, or with the horizontal side if there is
one. On fig. 2, that angle is D, and its ordinate is the maxi
m\mi bending moment ; so that, in our example, the maximum
bending moment occurs at W^, that is at the point a; =  3, and
its value is 261 ft.tons. If one side of the polygon be hori
zontal, the ordinate to any point of this line is constant, and
is a maximum.
The Point of Maximimi Bending Moment is found thus —
From P subtract the quantities Wi, JV2, W3, &c., in succession
until tlie remainder becomes zero, or first negative ; when the
remainder becomes zero, the maximum bending moment occurs
at the weight last subtracted, at the weight next in order, and
at every point between them : when the remainder becomes
negative for tlie first time, the maximum occurs at the weight
last subtracted, and at that point only. In our example, from
P = 24, subtract Wx = 5, Wn = 5, W^ =11, and the remainder
is + 3 ; subtract W^ = 12, and the remainder is negative for the
first time ; at this weight, that is, at Xi =  3, the maximum
bending moment 261 ft.tons occurs.
CHAP. VII.] BENDING MOMENTS AND SHEARING FORCES. 123
Graphical Solution for Bending Moment Diagram. — The
followiug purely graphical solution for the same problem
requires no analysis, but must be drawn with accurate in
struments and upon a large scale. Draw the vertical lines
Scale for Bending Momemts (Verticals)
2o°Ft.Tons
' pO 60
I I ' I I I [ I nr
ScALE FOR Forces Tons
Scale for Dimensions (Horizontals)
■I'll'
FiiT. 3.
P, W„ W^, Ws, Wi, TFs, and Q, fig. 3, at the given horizontal
distances apart upon a scale for dimensions ; draw ab, be, cd,
de, ef equal respectively to the forces Wi, W2, W3, W^, W^,
upon a scale for forces, that is equal to 5, 5, 11, 12, and 9 tons
124 APPLIED MECHANICS. [CHAP. VII.
in our example. Choose any pole 0' , and join it to a, &, c, d, e,
and /. From H' any point on P draw H'A parallel to O'a,
and meeting W^ at A' ; draw A'B', H'C, CD', D'E', and E'K
parallel respectively to O'h, Oc, O'd, (7e, and O'f. Join H'K' ;
and draw O'g parallel to K'K', meeting af in g. Then, upon
the scale for forces, /^ = Q, and ga = P. In the example,/^ = 18
tons, and ga = 24 tons ; so far this is a graphical solution for
finding the supporting forces. The polygon HA' B' C D' E' K'
is a bending moment diagram ; but as it is inconvenient to have
^'iiT' sloping, we draw another polygon thus: — Choose a new
pole on the horizontal line passing through g, and such that
the distance Og is some convenient integral number upon the
scale for dimensions, on the diagram Og = 10 ; draw Oa, Oh, Oc,
Od, Oe, Of, and Og. From H any point on P draw HA parallel
to Oa, and meeting W^ at A ; similarly, draw AB, BC, CD, DE,
and EK parallel respectively to Oh, Oc, Od, Oe, and Of
We have a new polygon, HABGDEK, which is the bending
moment diagram, and HKis horizontal as desired. The vertical
ordinates give the bending moments at each point of the span,
upon a scale for verticals obtained by subdividing the divisions
on the scale for forces by the number which Og measures on
the scale for dimensions; thus on the figure this new scale is
made by subdividing each division on the scale for forces into
10 equal parts, because we made 0^ = 10 on the scale for
dimensions. We can now find the bending moment at z, any
point of the span, by measuring the ordinate zs upon the proper
scale ; for instance, the ordinate at D measures 261, the bending
moment in foottons at that point.
Produce the two sides HA and ICE to meet at h ; then G the
point below h is the centre of gravity of the loads JF,, W2, W3,
Wi, and W5 ; on fig. 3, HG = 18 on the scale for dimensions :
see example 9, fig. 3, Ch. V, p. 93.
Proof — Draw the vertical hn, and produce all the sides of
the polygon to meet it ; at z, any point of span, draw the vertical
zq, and let the sides of the polygon that lie to the left, or those
sides produced, meet it in the points r and s ; then sr is called
the intercept on the vertical through z, made by the two sides of
the polygon which meet at the angle B. Other intercepts on
this vertical are zq and rq, made by the pairs of sides from the
angles H and A respectively. Again, on the vertical through h
we have the intercepts Gh, kh, mk, and nm made respectively by
the pairs of sides from the angles H, A, B, and C ; and also the
intercepts Gh, ph, and np made respectively by the pairs of sides
from the angles K, E, and D.
CHAP. VII.] BENDING MOMENTS AND SHEARING FORCES. 125
The triangle Bsr on the base sr, and of height zv, is similar
to the triangle Ocb on the base cb, and of height [/(), because
their sides are respectively parallel ; the bases of these triangles
measured on one scale will be in the same j)roportion to each
other as the heights measured on any other scale ; hence
~ , ~j on scale . . ~~ . /^ on scale of ,
&/ . CO of forces » ' ZV ' yU dimensions J
~T on scale >^ ZZ. on scale of
on scale of forces ^ ^" dimensions
or sr of forces — — „n j^ale of
y^ dimensions
_ Wi tons X ZV feet _ moment of W^ about z
~ 10 ^ 10
But sr measures 10 times as much upon the scale for
verticals as it does upon the scale for forces, since we made
the scale for verticals by .subdividing each division on the
scale for forces into ten parts ; therefore
sr on vertical scale = moment of Wt about z.
That is, measuring on the vertical scale, the intercept on the
vertical through any point z, made by the pairs of sides from
any angle of the polygon, equals the moment about z of W,
the force at that angle.
To show that if P = ga, and Q = fy, they will balance
JF, + n\ + W^ ^ W,+ W,. It is evident that P + (> = 2 fr ;
and if you produce Hli to meet the vertical through K, the
intercept by the pair of sides from H equals the moment of
ga, that is of P, about K; the five intercepts made on the
vertical through K by the pairs of sides from A, B, C, D, and E,
are the moments about K of the weights Wi, W^, W3, Wi, and
W^ respectively, and it is evident that the first of these inter
cepts is identically equal to the sum of the other five. Hence
the moment of F equals the sum of the moments of JF,, JV^,
^Vh Wi, and Wi about K', that is, ga exactly represents the
value of P.
To show that G is the centre of gravity of the weights
Wx, W2, W3, Wi, and PF5. The sum of the moments about G
of all the weights to the left of G, that is of W^, W^, and JFj,
is hk + km + mn = hn ; again, the sum of the moments about
G of all the weights to the right of G, that is of W^ and W^,
is np + ph = hn; hence the sum of the moments about G of all
the weights to the left equals the sum of the moments about G
of all the weights to the right ; and since these moments tend
126
APPLIED MECHANICS.
[chap. VII.
to turn the beam in opposite directions, the one sum destroys
the other, or the sum of the moments of all the weights W
about G is zero ; that is, G is the centre of gravity of the
weights.
To show that zs, the ordinate of the polygon, measured on
the vertical scale equals the bending moment at z,
zs = zq qr  rs, all measured on the scale for verticals
= Mom. of P  Mom. of TFi  Mom. of W^, all about z
= Bending Moment at z.
This is Culman's Theorem. It will be seen that this is a
special case of fig. 10, Chap. IV. Also HABCDUK is a balanced
frame, the vectors from giving the thrust on the bars, &c.
w, iw,
w.
m
w.
a'^Q
Fisr. 4.
In all cases of fixed loads the maximum bending moment
occurs where the locus of the shearing force diagram crosses
its base.
Graphical Solution for Shearing Force Diagram. — Fig 4.
P and Q are determined graphically by finding the point g,
as on fig. 3, or by moments. Through g, fig. 4, draw a
horizontal line and produce the lines of action of P, W\, TFj, . . .Q
to cut it; join these lines of action in pairs by horizontals
CHAP. VII.] BENDING MOMENTS AND SHEARING FORCES. 127
through a, b, c, d, e,f, respectively. The scale used for forces
is also the scale for shearing forces.
The construction is cyclical ; through each of the points
a, h, e, d, e, f, g, a horizontal line being drawn to join the lines
of action of the forces in pairs, viz. : — from a joining P and JF,,
from h joining TF, and W^, from c joining W^ and W^, from d
joining W^ and W^, ho\\\ e joining W^ and W^, from / joining
Wi and Q, add from g joining Q and F.
Cantilever with Unequal "Weights at .Intervals.
A cantilever (fig. 5) 12 feet long supports three weights,
ir, = 8, W. = 6, and W^ = 4 tons at points whose abscissae
reckoned from the fixed end are Xi = 12, x^ = 10, and Xz = 4^ feet.
This loading is shown also on fig. 11, p. 103, but on that figure
the cantilever projects beyond Wi, and so c is greater than a;, ;
it is evident, however, that the part which so projects is not
strained, so that although a cantilever does so project, yet, for
purposes of calculation, its length may be considered as the
distance from the fixed end to the most remote load; this is
shown on fig. 5, where c = a?,.
We will now calculate the bending moments at the fixed end,
and at points where the weights stand, systematically from the
forces upon the lefthand portion.
Ft. tons
Mx,,,,^
Mx^ = JV,(x,x,); 3f,o = Sx2= . . . . 16
Mxs = W, {x,  x^) ^ W, (x,  ^3) ; i^i = 8 x 8 + 6 x 6 = 100
Mo = JV,x, + W,x, + JV,Xs ; = 8 y 12 + 6 X 10 + 4 X 4 = 172
The Bending Moment at any point is the sum of the
products got by multiplying each weight to the left of that
point by its distance therefrom; the above is an arithmetical sum,
since all the weights tend to bend the left portion downwards.
Having found the bending moment at one point, we may derive
the bending moment at another point nearer the fixed end and
such that no weight intervenes, by adding the product of the
sum of all the weights to the left into the distance between the
two points, since no neiv weights have to be considered, and all
the leverages have increased by the distance betv/een the two
points. Hence the bending moments increase uniformly in
each interval as you move towards the fixed end.
128
APPLIED MECHA^'ICS.
[CHAP. VII.
W, Wj
jW,W,
I i—(XX>6
^...(XXJ~8
4 tons
ii
•p''
Bending Moment Diagram. — With a scale of feet for
horizontals, lay off the length (fig. 5)
and plot the positions of the loads ;
at each of these points, and at the
fixed end, draw a vertical ordinate
doivmvards, equal to the bending
moment thereat, upon a suitable scale
of foottons for verticals ; join the
ends of the ordinates by straight lines.
The ordinates are drawn doivmvards
to signify that the moments on a
cantilever are of a different sign,
as compared to those on a beam.
In either case the moments are all
of one sign, which will always be
reckoned as positive.
Maximum Bending Moment. — It is
evident that the maximum bending
moment is at the fixed end, and is
Jfo = S ( Wr).
Fig. 5.
W,
w.
(i;
«a
Graphical Solution for Shearing Force Diagram. — Pig. 6.
From a, b, c, d, joints of loadline, draw horizontals joining
the lines of action of the forces in pairs, the vertical through
the fixed end being reckoned as a line of action. Thus draw
horizontals, from a joining the ver
tical through the fixed end and W^,
from h joining W^ and W^, from c
joining W^ and W^, and from d
joining fFs and the vertical through
the fixed end. The scale for forces
is also the scale for shearing forces.
Graphical Solution for Bending
Moment Diagram. — Draw a vertical
line through K, the fixed end, and
draw the vertical lines W^ W^, W^
at the given horizontal distances
apart upon a scale for dimensions (fig. 7). Draw ah, he, cd,
equal respectively to the forces Jr,, W^, W^ upon a scale for
forces; that is, equal to 8, 6, and 4 tons in our example.
Choose a pole on the horizontal line passing through a, and
such that the distance Oa is some convenient integral number
upon the scale for dimensions ; on the diagram Oa = 10 ; and
draw Oh, Oc, and Od. From A any point on IF", draw AB
.W,
ih
<.c
id
Fig. 6.
CHAP. VII.] BENDING MOMENTS AND SHEAUING FORCES. 129
parallel to Oh, and meeting Wz at B ; similarly, draw BC, CD,
and A K parallel, respectively, to Oc, Od, and Oa. Then AB('1)K
is the bending moment diagram ; its vertical ordinates give
tlie bending moment at each point of the span upon a scale
for verticals obtained by subdividing the divisions on the scale
for forces by the number which Oa measures on the scale for
dimensions ; thus, on the figure, this new scale is made by
subdividing each division on the scale for forces into 10 equal
parts, since Oa = 10 on the scale for dimensions.
Proof. — At any point e draw the ordinate eh, and produce
those sides of the polygon that lie to the left till they meet it.
On the figure, they meet it at e, /, g, h. As in the proof to
fig. 3, the intercept on the vertical through e made by the
pair of sides from any angle of the polygon will, when measured
Fig 7.
upon the scale for verticals, give the moment about e of the
force at that angle. Now the ordinate at e is the sum of the
intercepts made by the pairs of sides from each angle to the
left, and so will give on the vertical scale the sum of the
moments of the forces to the left of e; that is, the bending
moment at e. On fig. 7, ef, the intercept by the pair of sides
from A, gives the moment of Wy about e ; fg, the intercept by
the pair of sides from B, gives the moment of W^ about e, and
gh gives the moment of JFs about e. Hence eh gives the sum of
these three moments, that is the bending moment at e.
Shearing Force Diagram due to One Load.
For a beam with W at the centre, the shearing force
diagram consists of two rectangles of height i W, one standing
above the left half and the other below the right half of span.
For a beam loaded with ^ at a point dividing the span
into any two segments, the shearing force diagram consists of
two rectangles, one standing above the left segment, the other
130
APPLIED MECHANICS.
[chap. VII.
below the right segment. The height of each is inversely pro
portional to the length of the segment on which it stands, and
the sum of their heights is IF.
In these two cases it is evident, and it can easily be proved
for a general case, fig. 4, that the area of the part of the
/
>
W
1
^W
• V
/
(
"1
Fig. 8.
shearing force diagram above the base is equal to the area of
the part below the base.
For a cantilever with W at its free end, the shearing force
diagram is evidently a rectangle of height W, standing below
the span.
Beam Loaded at the Centre.
Fig. 9. Let W be the load at the centre. By symmetry
P ^ Q = I W. For a section distant from the centre x towards
the left, consider the forces on the left
hand portion. The only force is P, and i^
its leverage is {c  x): hence,
pjw /\ wiQ
M,^P{cx) =— (c X),
the equation to the bending moment.
The value of Mx is zero at the end,
that is where x = c; it increases uni Fig. 9.
formly as x decreases, that is, as you
approach the centre, and it is greatest where x = Q, that is at
the centre ; by symmetry for the other half of the span, the
value will decrease uniformly till it is again zero at the right
hand end ; hence the maximum bending moment
1/ ^
Wl.
In Rankine's " Applied Mechanics," the maximum bending
moment in each case is given in the above form, viz., maximum
CHAP. VII.] BENDING MOMENTS AND SHEAKING FORCES. 131
bending moment = constant x total load x span, or Mg = m. W.l,
Throughout a great portion of that work, vi stands for this
constant, which he calls the numerical coefficient of the maximum
hendiiyj moment exjnrssed in terms of the load and span. The
value of m depends upon the manner of loading and of support.
In the case we have just solved, we specify the mode of support
by calling the piece a beam, and the manner of loading when
we say that W is at the centre, and we find m = \.
Bendin;] Moment Diagram. — Upon a convenient scale of feet,
lay off BC, fig. 9, equal to the span ; construct a triangle
with its apex above the centre 0. Draw a scale of ft.lbs. to
measure verticals upon, such that OA shall measure upon it one
fourth of the product of the load in lbs. into the span in feet.
Graphical Solution. — The simplest graphical solution is to
draw the bending moment diagram as above ; one which is
purely graphical may be made as in fig. 3, and it will not
be necessary to use the first pole 0', &c., as we know that
P = Q = h^^' Draw ah vertical and equal to the load W ;
from its middle point draw a horizontal line; choose at a
distance from ah equal to some convenient integral number
on the scale for dimensions, and draw Oa and Oh. Then
fig. 9 is constructed l)y drawing BA parallel to Oa, and
AC parallel to OB; and a scale for verticals is obtained by
subdividing the scale for forces by the number chosen for the
distance of from ah.
Cantilever loaded at the end. — Fig. 10. To find the bending
moment at a section distant x from the fixed
end K, consider the loads to the left of that jw
section. The only force is fF, and its leverage r
about the section at x is (c  x), and we have ^^^ * —
M^ = W{c  x), \\^ I
the equation to the bending moment. ^<,
The value of Mjc is zero when x equals c,
that is at the free end; it uniformly in fw. lo.
creases as x decreases, and is a maximum
when .r = 0, that is at the fixed end ; the maximum bending
moment is
Mo^ Wc= W.l,
and the value of the constant is m = 1.
Bending Moment D.agram. — Upon a convenient scale of
fe^et, lay off KA (Hg 10) equal to the length; draw below
KA the rightangled triangle FAK, with the right angle at
k2
132
APPLIED MECHANICS,
[chap. VII.
the fixed end, and construct a scale of ft.lbs. for verticals,
such that KF may measure upon it the product of the load
in lbs. into the length in feet.
Beam Uniformly Loaded.
Fig. 11. Let w be the intensity of the uniform load in
lbs. per foot of span. This is represented by a load area,
consisting of a rectangle of height w feet standing on the
span, and weighing one lb. per square foot. The total load
W^is the area of this rectangle,
so that
W = 2wc, or w = —
W
2c
LOAD AREA
P ' =iyc
i'
luc =Q
toe
rr,i
lw[cx)
IOC*
•— {caJ>
To find the bending moment
at a section distant x from the
centre 0. Consider the load area
standing upon the portion of the
span to the left of that section ;
it consists of a rectangle of length
(c  x) feet, its area is w{c  x)
square feet, and its weight is
w{c  x) lbs. This weight may
be considered to be concentrated
at the centre of gravity of the area — that is at its middle point ;
this gives a bending moment about the section equal to that
for the actual distribution. We have two forces to the left of
the section, P = wc lbs., half the total load acting upwards with
a leverage of (c  x) ft., and w{c  x) lbs. acting downwards with
a leverage of 2" (c  x) ft. ; hence
Fig. 11.
Mjt = P{c  x) w{cx)x — — = wc (c  x)  —
xY
. ^ (c x) {2cc^x)=^(cx) {c + x) ='(f= .r)i;
W
= — (c  x^), the equation to the bending moment.
4c
The value of J/^ is zero where x ^ ± c, that is at the two
ends ; it increases as x decreases numfically from c and from
 c that is from both ends towards the centre; and it is
CHAP.
.1.]
BENDING MOMENTS AND SHEARING FORCP^S.
133
greatest where .r = 0, that is at the centre. By making .v = 0,
we have .1/,, = \ Wc ; or hy considering the section at O, the
maximum hendinar moment is
Mn =^ F . c  wc .  = wc .c 
IOC
= \Wc, or ^Wl,
In this case, the value of the constant is m = \.
Bending Moment Diar/ ram. — Examining the equation to the
bending moment, we see that the ordinate M^ equals a constant
term, minus a term in .r ; hence we know that the locus is a
parabola, with its axis vertical and with its apex A (tig 12)
exactly above the origin O at the height  Wc = I Wl ; and that
. W
the modulus of the parabola is — , the coefficient of .r^ so that
the principal equation to the parabola, that is, taking A as
origin, is
W
4c
Graphical Solution. — With
a scale of feet for horizontals,
lay off the span BC (fig. 12j
and draw a vertical OA up
wards through () ; apply the
parallel rollers to HC ; place
any parabolic segment cut upon
peartree or cardboard against
the rollers (see fig. 4. Ch. VI),
with its apex on the vertical
through ; shift the rollers till
the curved edge passes through
B and C, which it will do simul
taneously, and draw the curve
BA C ; construct a scale of f t.
Ibs. for verticals, such that
OA = i jr . /.
Fig. 12.
Shearing Force Diagram. — Fig. 13. Draw verticals, one
upwards from the left end, another downwards from the right
end of span, each equal to  JV; join their extremities with a
straight line, which will cut the base at the centre.
134 APPLIED MECHANICS. [CHAP. VIL
The shearing force diagram consists of two rightangled
triangles, whose common height is i W; one is above the left
half, the other is below the right half, of span, and the right
angle of each is at the end of span.
Cantilever Uniformly Loaded.
Fig. 14. — Consider a section at the distance w from the
fixed end. The load area standing on the portion to the left
of that section is a rectangle of length (c  x) ft. and height
w ft. ; its area represents a weight of tv (c  x) lbs., which
may be considered to be concentrated at the centre of gravity
of the rectangle ; and so to the
left of the section, there is to be ^^.^.
taken into account only one force \4j!.^^ k
IV [c  x) lbs., having a leverage y  f '■q^q area h
about the section of  (c  x) ft. ; \n=\ivc °P
hence ^ 1___ .^k
Mr. = w{c
w W
2 («  ^)' = 9^ (^  ^y
Fi. 14.
the equation to the bending moment.
The value of M^. is zero wherp x = c, that is at the freer
end ; it increases a 5 x decreases ; and it is greatest where
X = 0, that is at the fixed end. Putting a; = we have the
greatest value, or directly from the figure we find the maximum
bending moment
In this case, the value of the constant is m = 5.
Bending Moment Diagram. — For the sake of comparison with
the diagrams for beams, we may consider the bending moments
on a cantilever to be negative, when the equation becomes
W W TFc
the locus is a parabola (see fig. 5, Ch. VI) with its axis vertical
CHAP. VII.] BENDING MOMENTS AND SHEAKINrx FORCES. 135
and to the left of O, and with its apex on the axis of X. The
vahie X = c makes (2c  x)x great(\st, and therefore makes Mc =
ii positive maximum ; so that the ap(>x
lies to the left of () at a distance
c^ — that is, the apex is at the free
enil. Since the coefficient of x is
— , the principal equation
to the
parabola is Y
2c ^
X..,^^
Graphical Solution. — With a con
venient scale of feet, lay off KA,
tig. 16, equal to the length ; place the
parallel rollers to KA ; set any para
bolic segment against the rollers,
with its apex at A ; draw the curve
AG, till it meets the vertical through
K at G. Construct a scale of f t.lbs.
for verticals, such that KG may
measure upon it onehalf the pro
duct of the total load in lbs. into
the span in feet.
Shearing Force Diagram. — Fig.
15. Draw a vertical downwards
from the fixed end of the base equal to W, and join its
extremity to the free end of the base.
The shearing force diagram is a rightangled triangle of
height ir, standing below the span, and with the right angle at
the fixed end.
Fig. 16.
Beams Subjected to Equal Loads at Equal Intervals.
The actual loading on a girder is often of this kind hecause the dead weight of
the platform is transmitted to it by equidistant crossgirders. This again may
have to be combined with the uniform weight of the girder itself : see such a com
bination following. Al!^o the wheels of rolling stock riding over a girder or stringer
while at rest are sensibly uniform loads of uniform di.~:tances. If the rolling stuck
move, the problem is changed and will be treated of in a chapter to follow.
On tigs. 17ff, 174. Let JF, the total load, be distributed over the span /, in « parts
each equal to w, and at equal intervals a apart ; then ntv = W, (n + \)a = 2c — I,
P=knw. Take S, a point directly under one of the weights, and let BS = ra,
then »• will be a whole number ; and if j bi the distance of S from the centre,
then m = (c — x). On tlie portion of the beam to the left of S, there are in all
r forces, viz., P= \mv acting upwards with a leverage about 5 of ra, and (r  1)
forces each equal to tv and acting downwards; the nearest to S has a leverage a,
136
APPLIED MECHANICS.
[chap. VII.
the next a leverage '2a, the next a leverage 3«, &c., and the last a leverage
(r  1) a ; hence
Mx = F .ra — w . a — w .'la — w .Za . . , —tv .{r — \)a
nw
= — . »•«  wa (1 + 2 + 3 ... + >• 1)
nw (r  Y)r w
= y . »•«  wa — ^ = 2 • '■" (« + 1  '■)•
But
W
2c
{n + 1) (c  x)
substituting, we have
M,
iV
2«
W
2w
{c x) \n+l 
w + 1
(« + 1 ) (c  X)
2c
n
(..)(„+l)''_lf=^^±I (,._,.),
2c
4(7 n
which'is the equation to the locus of T.
This equation gives the bending moment only at points where weisrhts are,
that is for values of a; which are multiples of a, but not at intermediate points ; it
n equal loads at equal intervals. W= total load,
if W be the same in both so also is M„ the ma>
Fig. 17a.
Fiff. 176.
only'differs from the equation we had for an uniform load, by the constant factor
t^__2J^« locus of T, the tops of the ordinates at the points where the weights
are situated, is a parabola with its axis vertical, and its apex above O. Putting i/
instead of Mx for the ordinate to the curve at any point, we have
y = (c^  x^),
4c n
80 that y is a maximum where a: = 0, and
W n + I
.Vn = — .
ic V
«+ 1
W . I
a maximum, and ;he height of the apex A in every case.
On fig. 17 it will he seen that if n be, odd, a weight comes exactly at the
CHAP. VII.] BENDING MOMENTS AND SHEARING FORCES. 137
centre ; so that j/o, tlie ordinate of the parahola, is tlie maximum bending moment ;
hence the maximum bendinu: moment
Mo = * '^i^ W. I, (« odd),
n
and the value of the constant is
« 4 1
m =
_ 1
ff
When n is even, the maximum bending moment \Kq is le»s than yo» and equals
the ordinate of the parabola at the weight on either side of the centre ; so that to
find the value of .l/o, it is only necessary to substitute tor x half of an interval,
that is ha, or , ; hence the maximum benditiar moment
Mo = ij  — \c'  I I =5 W.l, (n Qven),
(_c_\ ic n \ \«+l/ ) S»+l ' ^ ^'
\n * 1 /
and the value of the constant is
i '»±±^
'" ~ ^ m + 1*
The cords of the parabola give tlie bending moments at points intermediate
between the weights, since the bending moment varies uniformly in these
intervals.
Examples.
1. A beam 20 feet span supports a load of 10 tons uniformly distributed.
Find the shearing forces at intervals of 5 feet.
Am. Fio = 5 tons ; Fi = 25 tons ; Fo = 0. On right half of span, the
values are the same, but negative.
2. A betim 20 feet span supports a load of 10 tons concentrated at the centre.
Find the shearing forces.
Ans. i^iotoo = 5 tons, i^otoio =  o tons, and the sign changes at
the centre.
3. In the example given on page 119 and with the data shown in fig. 1,
find the shearing forces.
At2s. Fii to 20 = i* =24 tons.
i^2otoi5 =24 jr, = 19 „'{
i^l5t0 7 = 19  ^^2 ■= 14' ,,
i^7to3 = 14  JTs = 3 „
Faioio = 3  F'l = 9 „
/'ioto21 =  9  TFs =  18 tons = (  Q).
See figs. 2 and 4, Ch. VII, and note that the shearing force changes sign,
and that the bending moment is a maximum at the point a; = — 3.
4. A lieam 24 feet span is loaded with 20, 30, and 40 tons at points dividing
it into equal intervals. Find the maximum bending moment and the point where
it occurs.
As in example No. 2, p. 91, vee have P = 40 tons ; deduct 20 and it leaves 20 :
138 APPLIED MECHANICS. [CHAP. VII.
deduct 30 and the remainder is negative for the first time : hence the maximum
occurs under the load 30, that is at 12 feet from the left end, and maximum
JJfi2 = P X 12  TTi X 6 = 40 X 12  20 X 6 = 360 ft.tons.
0. In example No, 4, find the bending moments at the other two weights.
j{/6 = P X 6 = 40 X 6 = 240 ft.tons.
Jlfis =Px 18 ^Tix 12 TTz X 6 = 40x 18 20 X 12  30x 6= 300 „
or il/i8 = Q X 6 = 50 X 6 »= 300 „
6. Find the maximum bending moment in example No. 12, Ch. V.
An^. Mi9 = 256 ft.tons, maximum.
Or, taking the centre as origin, Mz = 256 ft.tons, maximum, and is at the
wheel transmitting W3 = 11 tons.
7. A beam 40 feet span supports four weights TFi = 50, TF2 = 10, W3 = 20,
and TFi = 30 cwts. at points whose abscissaj, measuring from the centre to left and
right, are xi = 10, X2 = 2, .rs =  12, and Xi — — 16 feet. Find the supporting
force at the left end, the maximum bending moment, and the place where
that maximum occurs.
Ans. P = 50 cwts. : and since P — TFi = 0, the maximum bending
moment occurs at zi, at X2, and at every intermediate point, and
^10 to 2 = 500 ft. cwts., maximum.
8. A beam 50 feet span has ^'eights o, 8, 9, 12, 9, 8, and 5 cwts. placed at
equal intervals of 6 feet, in order, upon the span, and with the hiad 12 cwts. at
the centre. Find the maximum bending moment.
Ans. Mo = 500 ft. cwts., maximum.
9. A cantilever 12 feet long bears four loads Wi = 8, Wi= 6, Wz = 9, and
Wi =12 tons at distances from the fixed end of a;i = 12, X2 = 8, Xi = 6, and Xi = 2 ft.
Find the bending moment at each weight, and also the maximum bending
moment.
A71S. J/,2 = 0, Ms = 32, J/e = 60, M^ = 152, and M^ = 222 ft.tons.
10. A cantilever is loaded with Aveights of 8, 6, and 4 tons at distances
of 12, 10, and 4 feet from the fixed end. Find the bending moment at each
weight, and draw a bending moment diagram upon a large scale: see fig. 5.
11. Draw also a bending moment diagram by the graphical construction
(fig. 7) to large scales. From either diagram, by measurement, find the bending
moments at intervals of two feet.
Ans. Mo = 172, M2 = 136, J/4 = 100, Me = 72, Mg = 44, Mio = 16,
Mi2= ft.tons.
12. A cantilever 12 feet long is uniformly loaded with 3 cwts. per footrun.
Find the equation to the bending moment.
Ans. .l/x= I (12 .jpft.cwts.
13. In the previous example, find the bending moments at intervals of two
feet by substituting for x into the equation.
Ans. Mn = » (12  0)= =216 ft.cwts. 3^ = # (12  2)^ = 150 ft. cwts.
Jfi = 96, Ma = 54, Mg = 24, Mio = 6, M12 = 0.
14. A beam 40 feet span supports seven biads, each two tons, and placed
symmetrically on tlie span at interviils of five feet. Calculate the maximum
CHAr, VIII.] B. M. AND S. F. FOR FIXED LOADS. 139
bending monient by substituting in the proper equation, and calculate at each
load the height of the parabola which gives the bending moments.
Here
;r= 14 tons, c=20 feet, I = 40 feet, and w = 7.
Tlie maximum bending moment is
J/o = yo = i ^^i TT . / = J X f y 14 X 40 = 80 ft. tons.
The equation to the parabola is
y = ^ . "^ {c x) = U Xf (400  ^) = i (400  X') ;
therefore, at the weights,
^ViOT.o = i (400  25) = 75 ft. tons ; Jfio or  io = 60, Jfisor 15 = 35ft.tons.
CHAPTER Vni.
BENDING MOMENTS AND SHEARING FORCES FOR COMBINED
FIXED LOADS.
Beam uniformly loaded and with a load oi its centre. — Fig. 1.
Let U be the amount of the uniform load, then the bending
moment at x due to it alone is r (c"  a:") ; let W be the load
4c
at the centre, then the bending moment at .r due to it alone
IS ^ [c  X) ; summing these, we have
•^ = Tc ^''  ^^) + — (^  ^) = 4^ (^  ^) (^^ + ^ + ir}
the equation to the bending monient for positive values of r,
that is for the left half of the span. Putting y instead of J/^,
we have
U. ,/ 2Wc\
a curve, the ordinates of which are the bending moments for
the left half of span ; this curve is a parabola with its axis
vertical, and its apex above BC the span. To find the
position of the apex ^,, it is only necessary to find that value
of x which makes y greatest ; now y is greatest when
. ./ 2Wc
(c  ;r) f c + ;r + —
140 APPLIED MECHANICS. [CHAP. Vni.
is a maximum; and since the sum of these two factors is constant,
their product is greatest when they are equal ; putting then
2JFc ^ JF
c + X + —jr= = c  a*, we have x =  jfC
as the value of x which makes y greatest ; the negative sign
denotes that Ai lies to the right of 0, so that OSi is to be laid
W
off towards the right and equal to jy^ Thejheight of A,, is
the value of y when we substitute this value for x ; that is,
byAi = 7 C + ^^C C  ~rC +
4c V U ) \ U U
] ^{\ui) =
U J ^^^^' \ u
times the height of A^, the apex of the] parabola for the
uniform load alone.
A^DG is the same parabola with its apex at the symmetrical
point Ai and the portion DC gives the bending moments for
the right half of the span. Since the coefficient of o:^ is r ,
° 4c
the principal equation to the parabolas AiDB and A^DC,
each referred to its own apex as origin, is
Ac
this is also the principal equation to the parabola BAJJ for tlie
uniform load alone, so that all three parabolas are identical.
AVe might suppose the diagram for the uniform load alone
to consist of two parabolas lying on the top of each other ;
and that upon the addition of the load W at the centre,
they both move upwards, while the one moves towards the
right and the other towards the left. This follows at once
from the Thcorein A, Ch. VI.
The Bendiivf Moment Diagram is BDC. It can be shown
that the tangent at 1) to the parabola A^DC cuts of!" CE equal
to the height of A„. ODBC is an approximate bending moment
diagram made with straight lines ; and it is safe, since the
ordinate of any point on DE is greater than the ordinate for
the corresponding point on DC.
CHAP. VIII.] B. .M. AND S. F. FOR FIXED LOADS.
141
Graphical Solution. — With a scale of feet for horizontals,
lay off BC equal to the span, fig. 1. From the centre O lay otl"
W
OSi and OSz to the right and left, each equal to —  c, and draw
verticals upwards from Si, 0, and S^. Apply the parallel rollers
to the span £C ; place ani/ parabolic segment against the rollers^
as in fig. 5, p. 109, with its apex on the vertical through 6', ; shift
the rollers till the curved edge passes through i/, the end of the
span on the opposite side of the centre from Si, and draw the
curve BDAi. Again, place the segment with its apex on the
V^^IOHd at centr
Fig. 1.
vertical through So ; shift the rollers till the curved edge passes
through the end C, and draw the curve CD A.; then BDC is the
bending moment diagram. The scale, of say ft.lbs., for verticals
is to be constructed such that OD measures (Z7+ 2W)l, where
U and W are in lbs. and I is in feet. The same scale may be
constructed as follows : — Place the parabolic segment with its
apex on the vertical through ; shift the rollers till the curved
edge passes through B and C; draw the curve BAoC, and make
142
APPLIED MECHANICS.
[chap. VIII.
a scale of ft. lbs. for verticals, such that O.I,, measures upon
it I Ul, where U is in lbs. and I is in feet.
Cor. — For the same uniform load, although different loads
be put at the centre, A^DB is always the same parabola; as the
load at the centre increases, the apex A^ moves from the centre,
and the are DB is a part of the wing of that parabola further
from the apex. Now the
wing of a parabola gets \ r „ ^mt.ot un.form.o^d. ~\ n
flatter as its distance from
the apex increases; hence,
if U be constant and W be
increased, BD becomes flatter
and flatter ; and if W De
very great compared to U,
DB is sensibly a straight
line.
Beam uniformly loaded atid
supported on three props. — Fig. 2.
Let BC be a beam with a load U
uniformly distributed on it and
supported on three props, one at
each end and one at the centre ;
let P, JF, aud Q be the forces with
which they press upwards, so that
/'+ W +Q= (7 always, and P= Q
by symmetry. If W=Q, the cen
tral prop bears no share, P= Q = \U,
and the bending niiiment diagram
is the parabola HAK, as on fig. 2.
If the central prop bears a share,
then the beam is loaded with a
uniform load U, and a negative load
W at the centre; and the bending
moment diagram is two parabolas,
each tlie same as HAK, but with
its apex away from the centre, and
in a direction opposite to that on
fig. 1. The horizontal distance
through which each apex moves is
given by the equation o — — c.
Suppose RAK to be two parabolas
lying one on the top of the other,
and let the prop press up with a
greater and greater force, then the
two parabolas shift away from each
other. When W=^U, Ai is over
S\, and Ai is over iSj, the middle
points of OH and OK respectively,
and the bending moment at is
zero ; this is the best value of JF, if the beam may only be bent so that its convex
side shall be down. It is evident that t^vi ./i  iff A : and tliat the beam might be
CHAP. VIII.] B. M. AND S. F. FOR FIXED LOADS. 143
sawn through at 0. when it would be two beams of span c, each uniformly loaded
and supported at the two ends.
If the heara may be bent both upwards and downwards, let the central prop
press upwards till W > \U, then 0S\ is greater than \0H, or in symbols o > ^c ;
and it can be seen troin the figure that the ordinales from H' to K' are negative,
from H" to ^positive, and from K' to jf positive, while at M, R', K, and A" the
bending moments are zero. Hence the beam will be bent with the convex side
downwards from JTto H', and from K' to K, and upwards from H' to K' . A hinge
might I'O put on the beam at H" and E? , since there is no lendencj' to bend at these
points, which are called points of contrary flexure. There are three maxima
bending uuiments, two equal positive ones at ^i and S^, and a negative one at 0.
If the material of the beam may be bent upwards and downwards equally well, then
the best value of W is that which makes the positive and negative maxima equal,
as their common value in this case is less than the greatest value would be in any
other; for, suppose them equal, then if /F be increased the parabolas move outwards,
and the ordinate at will increase ; while, if W be made smaller, the parabolas
will approach and the ordinates at S\ atid S2 will increase. Let the three maxima
be equal to each otiier, then OX = SiAi, or
JiSi : AiB : : 1 : 2 ; hence SiE' : liX : : 1 : V2
since the curve is a parabola ; or in symbols
3 : a : : 1 : V2.
Practically this could be accomplished by causing the props to press upwards till
OSi a V2
7r= — i U= U= 4 F= 586 Z7;
OH o + y3 V2 + 1
or by fixing hinges at the two points IT and K' at the proper distances from 0,
the central prop may be made to bear the above share of U.
If the prop at the centre press upwards so that TF = U, then P= Q= 0,
and 0S\ = OH; A\ coincides with H, and Ai with K; the bending moment is
everywhere negative, and its niaximiun value ON equals OA. OS and OK are
cantilevers as on fig. 16, p. 135.
The Continuous Beajnl
Beam uniformly loaded and supported on many props. — Fig. 3. Let BE be a
beam bearing an uniform load and supported on many (5 in the figure) props ;
this is only an extension of the previous case. The end C, fig. 2, instead of resting
2^^ .^J.1^^(°</3J
r . *^ w w w
A
H.
Fig. 3.
on a prop, might be hinged to the end of a cantilever, which in turn might have
its other edge hinged to abeam, «fcc. ; see fig. 3. Here a double cantilever, as B'C,
over each intermediate prop is hinged on each side to the end of a beam ;
144 APPLIED MECHANICS. [CHAP. VIII.
each cantilever bears an uniform load over itself, as well ns half the load on the
beam concentrated at its end, and is therefore in the condition of the cantilever
shown on fig. 2. E<)ih intermediate beam as C' is uniformly loaded, and is
supported by a hinge at each end; tlie two end beams B' li and K' E are supported
at one end iiy a iiinge, and at the other by one of the extreme props ; each central
span L consists of a beam and tsvo cantilevers; the end span / of a beam and a
cantilever.
If ihe tiinges be put in the positions indicated by H' and K' on fig. "2, then
the negative maxima bending moments over the props at 0, T, &c., are equal to
the positive maximii bending moments at .S'l, Si, &c., the centres of the beams ;
the common value of all these maxima will be less than the greatest for the hinges
in any other position, and
I : L : : a+ fi : 2a : : I + V'2 : 2^2 : : 854 : 1.
Cantilever unifoimly loaded and with a had at its free end. — Fig. 4. As in
the previous case, add the bending moments at the section distant x from the
fixed end, due to the loads separately ; thus
 M^ = — {c  xf +W{c x).
We consider the bending moments on a cantilever negative as compared with those
on a biara, and so they will be represented by ordinates drawn down from the span
instead of up as in the case of beams. It we further put y instead of Mx for the
ordinate at the point on the curve corresponding to any value of x, then y will
be Mr only for values of x from to c ; and
U U I 2Wc\
U ^ . / 2Wc\
This curve is a parabola with its axis vertical and its apex above the span.
To find the position of the apex A, it is only necessary to find that value of
x which makes y greatest ; now y is greatest when
{cx) \xc —^
is a maximum ; and since the sum of these two factors is constant, their product
is greatest when they are equal; putting then
2Wc / , ^ \
is the distance of A to the kft of 0, or —  c is the distance of A to the left of E\
W
that is, the apex of the parabola is beyond the free end by the distance ^ c, or the
same fraction of the span that W the concentrated load is of U the distributed loiad.
CHAP. VIII.] B. M. AND S. F. FOR FIXED LOADS.
145
The height of A above OH is the value of y when we substitute the above
value for x; or height of ^ is
U / U+W \ (11+ W 2WC\ I^V.yrrn f^V
times the maximum bending moment for the uniform load alone. Since the
coeflficient of jr is — , the principal equation lo the parabola AEF is
r =
u
X;
and it is therefore the snme parabola as that for the ui.iform load alone, which
Mouhi follow at once from the Theorem A, Ch. VI.
'I'he BeudiKg Moment Buigram is EFO, formed by the parabola EG, whose
apex is at E. and which makes the diagram for the uniform load alone, shifted
without turninsj till its apex is at A.
Graphical Solution. — With a scale of
feet for horizontals, lay off 0£ (fig. 4)
equal to the length, and produce it t" <S,
W
so that ES = jjrC, or is the same fraction
of the lengtli as the load at tlie end is
of the uniform loa(i, and draw a vertical
upwards through 6' ; apply the parallel
rollers to the line 0£ ; place any
parabolic segment against the rollers
with its apex on the vertical through 6' ;
sliift the rollers till the curved edge
passes through the free end E, and
draw the curve AEF ; OEF is tiie
hending moment diagram. The scale
of, say, ft. lbs. for verticals is to be
constructed such that OF measures
(^)'•
where V and W are in
height of A is('^)a= SA
Fig. 4.
lbs., and I is in feet. The same scale
may be constructed as follows : — Place
the parabolic segment with its apex on
the vertical through the free end E, and move the rollers till the apex comes to E;
draw the doited curve EG, and make a scale of ft.lhs. for verticals such that OG
measures upon it  Ul, where U is in lbs., and I is in feet.
Beam Loaded both Uniformly and with Unequal Weights
FIXED AT IrREGULAE INTERVALS.
On fig. 5 the uniform load is U = 36 tons or  a ton per foot
on a 72foot beam, which bears also four concentrated loads
dividing the span into five bays. BA^C the parabolic segment
giving the bending moments for the uniform load alone may be
supposed to be fivefold. Each moves up and away into the
position shown by dotted lines when the slopes due to the
dead load diagram are added; see Theorem A, Ch. VI.
£ £
Jl_£
Fig. 6.
CHAP. VIII.] 15. M. AND S F. FOR FIXED LOADS. 147
Graphical Solution. — The concentrated loads Wi = 8, W^ =10,
Wi = 18, and Wi  13 tons in the example on fig. 5 are drawn
downwards in cyclic order to a scale of tons. Then h, the joint
between the reactions at the supports, is to be found either by
drawing a link polygon or by taking moments about one support.
The supporting forces are P = 20 and (^ = 29 tons. The uniform
load is f a ton per foot, so that tlie supporting forces due to it
are 18 tons at each end. The dotted lines show the shearing
force diagram for the loads separately standing on the common
base BC. The thick lines show the two compounded, and
crossing the base at dz the position of the maximum bending
moment. The other sloping lines produced to meet the base
at f?i, ^2, di, and d^ determine the points over which must stand
the vertices of the parabolic arcs for the other " fields."
Now draw BC for the base of the bending moment diagram,
produce it each way, and from did,. . . . project down on it the
points D^D., . . ., &c. Apply the parallel rollers to BC ; place
the parabolic segment against them with its apex on the
vertical through D^, move the rollers till the curved edge passes
through B, and draw BE meeting the vertical through W^ in E,
Shift the segment till the apex is on the vertical through Di ;
move the rollers till the curved edge passes through E, and
draw EF. Shift the segment till the apex is on the vertical
through A ; move the rollers till the curved edge passes
through F, and draw EG meeting the vertical through Wz
in G, &c. The accuracy of the drawing is checked by obser\dng
whether the last curve passes through C, the other end of the
span. Lastly, with the same parabolic segment draw BA^C
standing on the span, and construct a scale so that the height
of its apex A^, shall measure on it  U7, that is, oneeighth of
the product of the uniform load and the span.
A Diagram of the Square Roots of Bending Moments is shown
at the bottom of fig. 5. The parabolic arcs are replaced by
arcs of circles, the centres being the points Z>i, D^, &c., where
the slopes of the shearing force diagram when produced meet
the base. See the Tlicorem G, fig. 9, p. 113, as to the parabolic
segment degraded.
Beam Uniformly Loaded on a Portion of the Span.
On figs. 6 and 7, let 2c be the length of the span, and its
centre; let G be the centre of the load area, and Ik the
extent of the load. The intensity of the uniform load is w lbs.
l2
148
APPLIED MECHANICS.
[chap. VIII.
per footrun ; and, as in the previous case, we take w as the
height of the load area in feet, so that every square foot of
load area represents one Ih. The total load area is a rectangle
of height vj ft., and length 2k ft. ; its area is 2wk square feet,
so that the total load on the span is 2v:k lbs., which may be
supposed to be concentrated at G, the centre of gravity of the
load area ; this gives the supporting forces P and Q as for the
actual distribution. Let x be the distance of the centre of
gravity of the load from the centre of the span. Then
2wk wk
P = ^(c + x), Q = — {c x),
are the supports at the left and right ends of the span. They
are also the height and depth of the rectangles on the right
and left unload fields of the shearing force diagram, fig. 6, so
that KM, the sloping locus for the central loaded field, crosses
the base at s. Hence *S^ (fig. 7) divides K'F' inversely as P
w
\ i
G S
i
\n
'■Sw
!
i
f
"""^;)
Fig. 6.
rig.
to Q. But G divides the span directly at P to Q. Putting
h = GS the distance to the point of maximum bendingmoment
from the centre of the load, we have
Q
K'S
C[G
B'G
CHAP. VIII.] B, M. AND S. F. FOR FIXED LOADS. 149
or
/r + ^ c, + X » hx
k  B c  X c '
and the maximum bending moment is
wk f . kx\ wk"
JI,^(c^x)[cx + j^(c:xy
'^(c^'x)(2ck)
(^^)(2^
W f^ k
The locus of the bending moment diagram (fig. 7) is
parabolic on the central loaded segment and tangents to the
parabola on the end unloaded segments. The modulus of the
parabolic segment is — ; for suppose the whole span loaded
and two upward forces at the centres of the end fields to be
introduced, then the negative locus for those two forces would
only introduce sloping loci to be subtracted mechanically from
the right parabolic segment due to the whole span loaded.
That segment would thus be shifted into the position HAE
without change of modulus. Theorem A, Chapter VI.
Cor. If k becomes very small, the load is sensibly at the
point X = x, and
Zc
That is for a load, assumed at a point, but really spread over
an inch or two, the scalene triangle is really rounded at the
apex.
Beam Loaded on one Segment.
Note. — This case is of importance, with the half span, only, loaded; itis required
in the treatment <>f the Iron Arched Girder. Also th6 load covering a segment
varying in length is required in the treatment of the girder with its ends fixed
horizontally.
Graphical Construction of the SJiearing Force Diagram and
Bending J\Ionient Diagrum (figs. 8, 9). Calculate or construct
the supporting forces P and Q. To a suitable scale, prick A'
150
APPLIED MECHANICS.
[chap. VIIL
at the height P above the base of the shearing force diagram
at left end, and J/ at the depth Q below it at the right end
of the load. Then KsM is the locus for the loaded segment,
and a horizontal through J/ is locus for the unloaded segment.
Place a parabolic template, with its axis vertical and it&
lower edge against the Tsquare. Push up the Tsquare and
shift along the parabolic set square till its curved edge passes
through B, the vertex being on the vertical through s, then
W
"A'
G
S
— 4vA
1'
s"~~ — ^
>
Fiar. 8.
J/
draw the arc BAE, and join B to C with a straight line.
Construct a scale of foottons to suit the calculated height of
any point, say that of ^ or of ^. As before, s divides the load
inversely as G divides the span.
Beam uniformly loaded on its Two Segments with loads of
different intensities (fig. 10). — This case is not of much practical
importance, but serves to illustrate the perfectly general case
of fixed loadings which is to follow.
To construct the shearing force diagram, calculate P and Q, the left and right
supporting forces, by taking moments about one end. Prick Ji' at Ptons above the
If ft end of the base, and /* at Q tons below tlieriglil end. Draw the str;iif:ht locus
A'M falling Wi tons vertically per font horizontal so that 31 shall bo lf'\ tons lower
than K. Draw Mh falling ta tons per foot. The maximum bending moment
must occur at t where this locus crosses the base.
To construct the bonding moment diagiam (fig. 1 1). If U'l to the right of Z
CHAP. VIII.] B. M. AND S. F. FOR FIXED LOADS.
151
were the only load on the s}«in, tlien the locus of the bending moment diagiam
on left segment would be a straight slope Ji'E.
Suppose, again, that tlie whole span was loaded with u'\ lbs. pfr foot when the
bonding; moment diagram would be a parabolic locus standing on h'C, with its vertex
under 0, the middle point of the span, the modulus of the parabola being ^«'i. Now
remove completely the load to the right of Z, which is the same as adding an
upward load ; from B' to /this would require the ordinates of a straight slope to be
W,
Wi
G,
)W,
Ti?+Tr
Fig. 10.
subtracted from those of the parabolic arc, which then, by Theorem A, Ch. VI,
would still be an arc B'ae of the same parabola of modulus \ w\, but with its vertex
ti shifted to the left under Si.
Fig. 11
Lastly, adding the ordinates of the loci B'ae and B'K, we have for the left
"field" BF the parabolic locus BAiE of modulus ^tvi with its vertex over Ti
defined by i where the slope KM of the shearing force diagram crossed the base.
152 APPLIED MECHANICS. [CHAP. VIII.
In the same way the locus of the bending moment diagram for the " field " FC
is a parabolic locus AiKC oi moiiwlus \wi, and with its vertex over Tj defined by
the point on the shearing force diagram where the slope hM produced meets the base.
Mnxxmum Bending Moment. — It is easy on any numerical example to find BT\ so
that the load on it shall equal the supporting force Pat B. Then M = T\A\ is
a maximum and is to be calculated directly at that section.
Graphical Construction of Beudini/ Moment Diagram. — Having first constructed
the shearing for(;e diagram as explained above, construct two parabolic segments
standing on a common base, but with the height of their vertexes in the ratio of Wi
to W2. These are to be out out in ciirdboard. The first is to be placed against the
T square and pushed up with its vertex im the vertical through T\ (where the slope
of the shearing force diagram crossed the base), and shifted till the curved edge
passes through B, and the arc^^i^drawn. This same paraholic template is then
to be pushed up with its vertex on the verticiil through till its curved edge
passes through B and C simnltaneouslv. and its vertex An pricked. Then a scale
is to be constructed so that OAo shall read on it \^w\P where I — span, when the
maximum T\A\ can he scaled off. To finish the locus for the right segment the
second parabolic template must be used, and pushed up with its vertex on the vertical
through T2 (where the corresponding slope hM produced on t'ne shearing force
diagram meets the base) when its curved edge will simultaneously pass th tough E
and C, and the arc EC is then to be drawn.
Beam Loaded in any manner with Fixed Loads.
Graphical Construction of the Shearing Force Diagram. — The supporting forces
P and Q at the left and right supports are to be calculated by taking moments about
one end of the span. Or all the distributed loads may temporarily be concen
trated at their centres of gravity, and Pand Q constructed hy drawing a load line
and link polygon as on fig. 3, p. 123. To a suitable scale of feet for horizontals, draw
a base line for the shearing force diagram, and w ith a suitable scale of tons lay P
upwards from the left end, and Q downwards from the riglit end. Starting from the
top of P, draw a locus which will cross the base once and end at llie bottom of Q.
This locus will cover "field" after "field" of the span thus. ItwiW cvo&s unloaded
Jields horizontally ; at a concentrated loud it will drop the amount of that load
vertically, and will cross uniformly loaded Jields sloping downwards with a str«ight
slope inclined at the given rate of loading.
Graphical Construciion of the Bending Moment Diagram. — If there be "fields"
of the span, some loaded at one uniform rate, and some at others, then as many
parabolic cardboard templates should be prepared as there are varieties of uniform
load. These templates should have a common base, and their heights be in a con
tinued proportion the same as tliat of the difiierent intensities of uniform loads (see
fig 11> . . . . ....
The base for the bending moment diagram being drawn, it is to be divided into
" fields " ; each concentrated load is the boundary bet ween two " fields " ; so also is
a point at wliicli tlie intensity of the uniform load suddenly changes, including the
ends of unloaded fields. The locus of the bending moment di;igrain will bcirin at one
end of the base and end at the other; being a maximum at the same point ar
which the shearing force changes sign. This locus will cross tlie fields thus : it
will cross unloaded Jields with a straight locus sloping up at a rate given by the
absolute height of the shearing force horizontal locus in that field, or doivn at the
depth of it. It will cross uniformly loaded fields in a parabolic arc, the vertical
axis of the parabola being the verticnl through the point where the sloping locus of
the shearing force diagram for that field crosses the base when produced. Also the
modulus of the pambola shall be half the intensity of the uniform load in the field.
Lastly, the arc shall at its emls liave a cotnmon tangent with the arc in adjoining
field, or, if the locus in the adjoining field he straight, it shall be a tangent. At
points where a load is concentrated, the loci meet at an angle whose tangent is the
CHAP. VIII.J B. M. AND S. F. KOR FIXED LOADS.
153
sudden fall on the shearing force diagram. The locus is readily drawn over tield
after tield, using the template for any loaded field already prepared to correspnrul
to it.
The snale is to be prepared to give the value at some point as calculated th^re,
sav at the point where the shearing force diagram crossed tlie liase ; oi with one of
the prepared parabolic segments, a parabolic segment may be drawn standing on the
span as base when a scale is to be prepared upon which the height of its veifex
shall measure ^ivl, where tv is the intensity corresponding to the template
used.
Examples.
1. A beam is uniformly loaded, and a prop in the centre bears onethird of the
load. Find the maximum bending moment (fig. 2).
Let U = the amount of the uniform load,
IF = the upward thrust of central prop = ^ U.
Here a = ^c ; and since a < le, the two parabolas intersect above the span, and
there are no negative bending moments ; there
is a positive minimum at the centre, and
a positive maximum at a on each side of the
centre.
The base of the segment
SJiE = 2)3 = ic = il.
Height SiAi : Height OA : : (§)2 : V :
therefore
SiAi = iOA,
Fi. 12.
Ma = ^ X i^V.l = x^U.l, maximum.
2. A cantilever 18 feet long is loaded
uniformly for twothirds of its length from
the free end, svitli 10 cwt. per footrun. Find
the bending moments at intervals of two feet.
(See Hg. 13.)
Look upon the loaded part as a cantilever
uniformly luaded, then
l=\.wk
w lbs pern^i(cky^
fe— — ;» ,
M. = (c
x)^ = 5 (18 xy;
therefore
Mi8 = 0, and Ih = 720 =. BE. '^k
Now consider BK?i cantilever load at B with 120 cwt. ; then
k
Mr
W
80 that
icx\ = 120(12 a;);
ifti = 720 = D^, and J/o = 1440 ft.cwts.
Ans. 0, 20, 80, 180, 320, 500, 720 ; 960, 1200, 1440 ft.cwts.
13 upon a large scale, and
A graphical solution is obtained by drawing fi
measuring the ordinates at intervals of two feet.
154 APPLIED MECHANICS. [CHAP. VIII.
3. A beam 72 feet span is loaded with 8 and 10 tons at points 18 and 6 feet to
the left of the centre, and with 18 and 13 tons at points 12 and 24 feet to the
riiiht of the centre ; there is also an uniform load of half a ton per foot of span.
Find the position and value of the maximum bending moment. These data are
drawn to scale on tig. 5.
Taking moments about the right support
/ P X 72 = 8 X 54 + 10 X 42 + 18 X 24 + 13 X 12 + 36 x 36.
So that P= 38 tons, including the reactions for spread load. At the central point
the shearing force 13^0 = ^81018 = 2 tons ; and to bring this to zero,
we must pass to a point 4 feet further to the right : hence F.\ = 0. The maximum
bending moment will be at a point 4 feet to the right of the centre, and
3f.i = 38 X 40  8 X 22  10 X 10  20 X 20 = 844 ft.tons.
The shearing force changes sign at the point a; = — 4, and at this point the
bending moment is a maximum.
4. A beam 54 feet span is loaded uniformly for twothirds of its length from
the left end with 10 cwt. per footrun. Find the position and magnitude of the
maximum bending moment. (See tig. 9.)
In this case, c = 27 ft., and OG = 9 ft., measured to the left of 0.
From G lay off GS = 6 ft. = §G0, since the load extends over twothirds of the
span, and the maximum moment occurs at S, that is, at 3 feet to the left of the
centre. Suppose the whole load TF= 360 cwt. is concentrated at G ; then
360
P= — X 36 = 240 cwt.
54
This, of course, is equal to load up to S, that is, 24 x 10.
Taking a section at S, the portion of the span to the left is 24 ft., so that the
load upon it is 240 cwt. acting downwards, and if supposed to be concentrated at
its centre, its leverage about the section is 12 ft. ; at the same time F acts upwards
with a leverage of 24 ft., and
Ms = 240 X 24  240 x 12 = 2880 ft.ewt. max.
5. The left half of a beam 32 feet span is uniformly loaded with 1 ton per foot
run. Find the position and magnitude of the maximum bending moment.
^7is. The maximum occurs at the section 4 feet to the left of the centre,
and its value is .Ui = 72 ft.tons.
6. A beam 36 feet span is loaded uniformly from the middle point towards
the left to an extent of 12 feet, with 2 tons per footrun. Find the position and
magnitude of the maximum bending moment. (See tig. 7.)
In this case, OG = 6 ft., and GS = ^OG = 2 feet, since the extent of load is
onethird of span ; the maximum bending moment is at S, 4 ft. to the left of the
centre. Suppose the whole load Jf, 24 tons, concentrated at G, we have
24
P= — X 24= 16 tons.
36
Taking a section at S, the extent of the load to the left is 8 ft., and is equivalent
to 16 tons acting downwards with a leverage of 4 feet, while Pacts upwards with
a leverage of 14 feet ; hence
JA = 16 X 14  16 X 4 = 160 ft.tons.
7. A beam 72 feet span is loaded in two segments ; the left segment is two
thirds of the S[)an, and is loaded uniformly at the rate of two tons per foot. The
right segment is uniformly loaded at tlie rate of one ton per foot. Find the position
and amount of the maximum bending moment.
CHAP. VIII.] B. M. AND S. F. FOR FIXED LOADS,
155
Fig. 11 is this example drawn to scale.
The data in symbols are
1e = 72, 2/t = 48, 1ki = 24, w\ = 2, and xci = 1 :
/» X 72 = 96 X 48 + 24 X 12 and P = 68 tons.
In order that the shearing force at T\ maybe zero, it is necessary that the load
between the left end and T\ should equal F, that is 68 tons ; hence T\ is 34 feet
from the left end or 2 leet left of the centre. And
ifs = 68 X 34  68 X 17 = 1156 ft. ton maximum.
8. Draw the shearing force diagrams for the beam and cantilever where*^a
uniform load is combined with concentrated loads (figs. 14 and 15).
i>/ 1 in I
Fig. 15.
Fig. 14 combines figs. 4 and 12, Ch. VII ; fis. 15 combines figs. 6 and 15,
Ch. VII.
9. Find, by analysis, the positions of the apexes in the example solved
graphically on fig. 5.
On the shearing force diagram the locus begins at height 38 ; it then slopes at
5 to 1, so that Bd\ = 76, or Od\ = 40. The second slope being 8 lower, then d\di
is 16 and Od% = 24. See the numbers engraved at the centres of the circles at toot
of fig. 5.
Just as
OJo = 324 = i (36)2 = i (^0)2,
so also
Di^i = i (Bd,f = i(76)2 = 1444,
measured on the scale of ft. tons. So also the height of Ai above E is ^th the
square of their known horizontal distance apart. Adding this height to the
calculated height of E itself, we have BiAi = 1044.
156 APPLIED MECHANICS. [CHAP. IX.
/
CHAPTER IX.
BENDI1<G MOMENTS AND SHEARING FOKCES FOR MOVING LOADS.
In the first Chapter the action of a live load wlien applied to
a tie or strut is described ; the action is somewhat similar when
a live load is applied to a beam. Thus for a beam loaded at the
centre, the load W may at one instant be in contact with the
central point of the beam, and yet not be resting any of its
weight on the beam ; the next instant its whole weight may be
resting on the beam. It does not follow directly from Hooke's
Law, but is a matter for demonstration, that, for an instant, the
strain thus produced is douUe that which the dead load pro
duces, provided the greatest strain does not exceed the proof
strain.
One way of applying the actual weight W to the centre as
a dead load is, as in the case of a tie, to put it on bit by bit ;
another way is to put the whole weight W on the end of the
beam, when the strain is zero, and then push it very slowly
towards the centre, when the strain gradually increases to the
full intensity due to JF as a dead load. If W, on the other
hand, be pushed from the end to the centre in an indefinitely
short time, it will be the same as if it had been applied suddenly
at the centre ; in this case, then, W is applied as a live load.
Definition. — A load which passes along a beam, and which
thus occupies at different instants every possible position upon
the span, is called a moving or travelling load.
A moving load may be dead or live or of intermediate
importance, but not of greater importance than a live load.
A travelling crane, which moves very slowly, and so as not
to set the suspended weight swinging, is practically a dead
moving load. The action of a moving load on a railway bridge
is of intermediate importance; when the span of the bridge is
short, say less than 20 feet, this importance is about equal to
that of a live load ; and when the span is long, say more than
200 feet, it may be considered as about midway between a dead
load and a live load.
The Commissioners on the Application of Iron to Eailway
Structures at p. xviii of their report say : — " That as it has
appeared that the effect of velocity communicated to a load is
CHAP. IX.] 15. .M. AND S. F, KOIJ MOVING LOADS. 157
to increase the deflection that it would produce if set at rest
upon the bridge ; also that the dynamical increase in bridges of
less than 40 feet in length is of sufficient importance to demand
attention, and may, even for lengths of 20 feet, become more
than onehalf of the statical deflection at high velocities, but
can be diminished by increasing the stiffness of the bridge ; it
is advisable that, for short bridges especially, the increased
deflection should be calculated from the greatest load and
highest velocity to which the bridge may be liable ; and that
a weight which would statically produce the same deflection
should, in estimating the strength of the structure, be considered
as the greatest load to which the bridge is subject."
In the same way the shearing strain produced by a moving
load is greater than that produced by the same load when fixed.
In the cases which follow, it is to be understood that the loads
as given are dead loads, or the equivalent reduced dead loads.
When the load is partly fixed and partly moving, the equivalent
dead load is the sum of the actual dead load and the dead load
equivalent to the actual moving load.
Definition. — For any point oc, the Mange of Shearing Force
due to a moving load is its extent ; and the limits of this extent
are the maximum positive and maximum negative values which
Fx assumes during the transit of the moving load.
Classes of Moving Loads. — An uniform load coming on at
one end of the span, covering an increasing segment till it is
all on, then moving to a central position on the span, and
passing off at the other end, is called an advancing load. A
train of trucks, shorter than the span of a bridge, coming on at
one end, travelling across and going off at the other end of the
bridge, is an approximate example of, and is generally to be
reckoned as, an advancing load. The reason that it is called
approximate, is that although the weight of the trucks may be
uniform per foot of length, yet they are not continuovisly in
contact with the bridge, but transmit the load thereto by means
of wheels at a number of points. An advancing load may be
equal in length to the span ; in which case, in passing across, it
covers the whole span for an instant. If the load be longer
than the span, it will continue to cover it for a definite time
v^hile passing, but as time does not come into our consideration,
it will be included in the advancing load equal in length to the
span.
A load concentrated at a point, and which moves backwards
and forwards on the span, is called a rolling load ; a wheel
which rolls along a beam is a practical example of this. In.
158 APPLIED MECHANICS. [CHAP. IX.
reality the load is distributed over a small area, and if
now the load be taken to be uniformly distributed over this
small area, it may be considered as an advancing load of
small extent ; on the diagrams it is represented by a wheel or
circle.
A TravelliTig Load System is a load transmitted to the beam
in definite amounts at points fixed relatively to each other, the
whole load moving into all possible positions on the span; a
^xlocomotive engine is a practical example of such a system, and
a rolling load is its simplest form. On the diagrams, the load
is represented by a number of circles or wheels with their
centres fixed on a frame (see fig. 3, Ch. IV), or for ease in
drawing by a number of vertical arrows connected by a thick
horizontal line (see fig. 7).
It will not be necessary to consider moving loads upon
cantilevers, as in practice there is seldom such a thing. It is
only necessary to suppose the load fixed in the position most
remote from the fixed end ; this, it is evident, gives the greatest
bending moment at each point, the maximum being at the
fixed end.
Bending Moments on a beam under an advancing load equal
in length to the span. — Suppose the load to come on from
the left end and cover a segment of the span, the bending
moment diagram is shown on fig. 8, Ch. VIII ; when the whole
span is covered, on fig. 12, Ch. VII ; and when the load is passing
off, by fig. 8, Ch. VIII, reversed. Since the parabolas in these
two figures are the same, it is evident that the apex A is
higher on that where the base of the parabolic segment is
the whole span ; the ordinate, not only for A, but also for
every point. Hence the maximum bending moment at each
point of the span occurs when the whole span is loaded ; of
these maxima, the maximum is at the centre, and this case
resolves into that of a beam uniformly loaded.
Shearing Forces on a beam under an advancing load of
uniform intensity (fig. 1).— At any point x, the positive
maximum shearing force occurs when the front of the load
is at the point. Suppose the load to be in the position shown
at the top of the diagram, then the shearing force is positive
and equal to P. Jf the load move towards the right, P will
decrease, and Fx will equal that decreased value. If the load
move towards the left, and if the total load be not yet on the
span, then the new load is exactly the same as the first with
an additional load to the left of r ; this additional load is
shared in some manner between the two supports, so that
CHAP. IX.] B. M. AND S. F. FOR MOVING LOADS.
159
the increase of P is only a fraction of that added load ;
in reckoning Fx, however, we subtract from this increased
value of P the whole of the added load, that is, we subtract
more than we add. Again, if the load be shorter than
the span and be wholly on the span, then, after the advance,
the new load is the same as the tirst with a portion added
to the left of x and an equal portion taken off at the tail
of the load ; the portion added increases by a fraction of itself
the value of P, while the portion taken off decreases by a
smaller fraction of itself the value of P ; and in reckoning F^,
we subtract the whole •
of the added portion
from this increased
value of P. Hence,
whatever be the
length of the load,
Fx is a positive
maximum when the
front of the load is
at X. Similarly, by
calculating the nega
tive shearing forces
from the supporting
force Q, it can be
shown that the shear
ing force is a negative
maximum when the
tail of the load is at
the point. At each
point, during the transit of the load the shearing force
assumes all values between the two maxima, and passes
gradually through the whole range in the same time that
the load takes to make a transit.
Sliearing Force Diagram. — Length of load equal to, or greater
than, span (fig. 1). When the front of the load is at the left end
of span, the whole span is covered, and P = ^W, where W is
the load which covers span ; hence the positive maximum at
the left end is h W. When the front of the load is at the
right end of span, P is zero, as no load is on the span ; hence
the positive maximum at the right end is zero. When the
front of load is at any intermediate point, the right segment
is loaded, and Fx = P di positive maximum ; the value of this
maximum increases as the point approaches the left end of span,
because the length of the loaded segment is increasing, and
iW
160
APPLIED MECHANICS.
[chap. IX.
because its centre of gravity is nearer the left end; or in symbols,
this positive maximum
load on span length of loaded segment _ ^' / y
span
4c
Hence, the locus giving F^ the positive maximum shearing
force at each point is a parabola with its axis vertical and
its apex at the right end of the span, and whose ordinate
at the left end is ^W; the locus giving the negative maximum
is a similar parabola below the base ajjd with its apex at the
left end of span : the range at each point is given by the
double ordinate.
Graphical Solution. — Length of load equal to, or greater than,
span. With any parabolic segment draw a curve with its
apex at the right end
meeting the vertical
through the left end
at L (fig. 1) ; draw
another below the
base, and construct a
scale for verticals upon
which
where W is the load
which covers the span.
Shearing Force Diagram. — Length of load less than span
(fig. 2). Let 2k be the length of the load. While the load
advances from the right end, and so long as it completely
covers the right segment, that is up to a distance 2k from G,
the diagram is a portion of the parabola (fig. 1); for the further
advance of the load, the maximum shearing force is increasing,
because the centre of gravity of the load is approaching the
left end, and the remainder of the locus is therefore a straight
line. The straight portion of this locus and the locus for an
equal rolling load (see fig. 6) are parallel, and are separated
from each other by a distance k measured horizontally ; this
straight portion, when produced, cuts the base at a point F such
that GF = k, and it is therefore a tangent to the parabola.
Graphical Solution. — Length of load less than span (fig. 2).
Draw the parabola GD as in the previous case ; construct a
vertical scale such that BD = wc, that is equal to half the load
CHAP. IX.] B. M. AND S. F. FOR MOVING LOADS.
161
which would be on the span supposing the whole span covered ;
ink in CE the portion of this parabola extending from the
apex C through a horizontal distance equal to 2k; and draw
for the remainder of the span, a tangent to the parabola at
the point E ; this tangent is drawn by laying off CF = k,
drawing FE, and producing it to L.
Bendiug Moment for a Beam under an advancing load less in length than the Span
(fig. 3). — Let 2c = spati ; 'Ik = extent of load ; iv = intensity of load ; C the origin,
and centre of span; G the centre of load; and let the load be upon the span in any
position.
Here
k < c, and JF= 'Iwk = total load.
To find P, we may suppose the whole load concentrated at 0, and we have
where x is the abscissa of any point of the span reckoned positive to the left of C,
and y is the distance of the same point reckoned positive to the left of G. Taking
a section at the point x, we have two forces acting on the portion of the span to the
=1 2 (t'fc
 J< (c + xx
W=i 2 wk
(cxty) J< (c + xyj s)
Moving Uniform load of intensity to and extent 21c
f^wt::::?_:iri.__jc
V X 1 >^
Section at x
(Span =2 c)
Fig. 3.
left of the section, viz. P acting upwards with a leverage (c — x), and a load area
equivalent to a force w (k  y) acting downwards with a leverage  {k — y); hence
the bending moment at this section is
Mx = P{c  x)  IV {k  y) y
k 
wk
{c + xy){cx) ^ {k yY
(wk „ „, ivk'^\ w llkx \
— (^^)^) + y( y)y.
(1)
As the load moves about, y varies and the bending moment Mr at the section x
depends upon the position of the load, that is, upon the value of y. To find the
position of the load which gives the greatest bending moment at the point .( , it is
only necessary to find the value of y which makes Mx a maximum ; now Mx is
M
162
APPLIED MECHANICS.
[chap. IX,
I2kx \
greatest when the product ( y\ y is greatest ; and as the sum of the two
factors of this product is constant, the product is greatest when the factors are
equal to each other, that is when
Ikx
— y = y\
so that w • have
kz
(2)
or
y : k : : x : e. (3)
This proportion expressed in words gives the following : —
KuLE. — The greatest bending moment at any point of the span occurs when
there is directly over it, that point in the load which is situated in the extent of
the load in a position similar to that in which the point is situated in the e> tent of
the span.
Substituting in (1) the value of y in (2), we have
This is the equation to the maxima bending moments, and may be written thus
where Co is a constant quantity ; the locus is therefore a parabola with its apex
above the centre of span, and the maximum of these maxima — that is, the maximum
for the whole span — is at the centre, or where x = 0\
*^T(^:)^4^')"" = i'^('*>
(5)
=2ivk
By the preceding rule the maximum bending moment at the centre of the span
occurs when the centre of the load is over the centre of the span.
Graphical Solution. — Fig. 4. With a scale of feet lay off the span and draw
a vertical upwards through the centre ;
apply the rollers to the span; place any
parabolic segment against the rollers with
its apex on the vertical through 0; shift
the rollers till the curved edge passes
through the ends of the span, which it
will do simultaneously, and draw the curve.
Construct a scale of' ft. lbs. for verticals
such that OA = \ W{1 k), where W is
in lbs., and I and k are in feet.
Note that to give the maximum bending
moment at any point, the load assumes a
different position tor each point according to
the above r\ile, and that it is possible to ful
fil the condition of the rule for every point
without any of the load going off the span.
Co;. 1. — Suppose the extent of load equal to the span
k = c, (2')=l,
then
CHAP. IX.] B. M. AND S. F. FOR MOVING LOADS. 163
and we have
max.3/r = —(c  X), and max.'l^o = ^^.^
ic
the same as for the span uniformly loaded (6g. 12, Ch. VII) and as shown in the
preceding case. Note further, that the rule for fixing the position of the load so
as to give the maximum bending nionieni at any point is fulfilled simultaneously
for a/l points of the span ; as it is evident that when the centre of load is over the
centre of span, every point of the load is over the corresponding point of the span.
Cor. 2. — Suppose the extent of load to be zero ; then
0, (.4) = .,
and the load is a rolling load for which
TF
max.iWx = ^i'''  ^')> a°<i max. 1^0 = ^ ^■^^
When the rule for finding the position of the load which gives the maximum
bending moment at any point is applied to this case, it is found that the maximum
occurs at any point oi span when the rolling load is at that point.
As this is an important case, and leads to cases still more important, we will
give a separate investigation.
BendtTig Moments for a beam under a rolling load (fig. 5). —
Consider any point of the span at the distance x from the
centre, distances to the left being reckoned positive. Let B
be the amount of the rolling load, and suppose it over the
point in consideration.
We may calculate the bending moment Mx from either of
the two equations
Mx = P {c  x), or Mx = Q{c + x).
If the load moves to the right, then the upward supporting
force F' is less than P, and
M\ =^P' {cx)< Mx ;
if now the load moves to the left, the supporting force Q' is
less than Q, and
M"x= Q' {c + x) < M^;
thus Mx decreases whether B moves to the right or left, that is,
Mx, the bending moment at any point x, is greatest when B, the
rolling load, is over the point.
Let B be over the point x, then
7? P
max. ^4 = P{C X)  —{C+X)[CX)= ~{c^  X"').
This is the equation to the maxima bending moments ; the
bending moment diagram is a parabola, with its axis vertical
m2
164
APPLIED MECHANICS.
[chap. IX
, t4j'
and its apex above the centre of span ; and the maximum of
these maxima^ that is the maxi
mum bending moment, for the
whole span occurs at the centre
when the load is over the centre;
putting X = 0, we have
the value of the constant m = \,
and the principal equation to
the parabola is
'2c
Bending Moment fcnr a beam, under both a roiling and an
uniform dead load, — Let R be the dead load equivalent to
the actual rolling load, and U the uniform load. For each
load separately the bending moment diagram is a parabola
with its apex over the centre. For the combined load, the
diagram is a parabola whose apex is also over the centre, and
whose modulus is the sum of their moduli ; hence
jc = (^ + 27) ^'^ ~ ^^^ " — ir~^ ^^ ~ ^^' ™*™^
and
M, =  {U r IB) = ^ {U ^ 2R) .1, max. of maxima,
the bending moments being in terms of a dead load throughout.
Shearing Forces in a beam under a rolling load (fig. 6). —
At any point of the span, F, the shearing force is positive and
equal to P, so long as ^ is to the right of the point : since P
increases as R moves towards the left support, F, is evidently a
positive maximum when R is indefinitdy close to, and on the
right side of, the points ^^
When R passes to the
left of the point, F^ = P
 R=  Q: since Q in
creases as R comes closer
to the right support, it is
again evident that F, is a
negative maximum when
R is indefinitely close to,
and on the left side of,
the point. When R is
Fig. 6.
inde6nitelj close to the point, P and Q have sensibly the same
CHAP. IX,] B. M. AND S. F. FOR MOVING LOADS. 165
values as for R exactly at the point, and we have the follow
ing :—
To find the maximum shearing force at any point x, place R
over the point, and calculate P and Q for that position of the
load ; these are respectively the maximum positive and maxi
mum negative values of Fj^.
During the passage of the load, the shearing force assumes
all values lying between these maxima ; and it is important to
observe that the shearing force not only changes sign at any
point as R passes over the point, but that it changes from its
greatest positive to its greatest negative value, or vice versa,
according as R is moving to left or right, and does so even
although the load be moving slowly. If moving quickly, this
sudden change produces what is called a hammer blow.
The positive maximum at each point is the value of P as R
comes to the point ; and since P is proportional to the remote
segment, it follows that the positive maximum at each point is
proportional to the distance of the point from the right end of
span ; it is zero for the right end, and increases uniformly till it
is R for the left end ; for, when R is just to the right of the
right end of span, P is zero, no load being on the span ; and
again, when R is just to the right of the left end of span, P is
sensibly equal to R.
For a rolling load the range at each point is constant, and is
equal to R.
Graphical Solution. — From the left end of the base draw
upwards a vertical equal to R, and join its extremity to the
right end of the base ; similarly from the right end, draw down
wards a vertical equal to R, and join its extremity to the left
end of the base ; at each point the ordinate upwards gives the
maximum positive, and the ordinate downwards the maximum
negative, shearing force ; while the double ordinate gives the
range.
Two Wheeled Trolly confined to a Girder like a
Travelling Crane.
Bending Moments for a. beam, vender a travelling load system
of tv)o equal weifjhts at a fixed interval aj)drt (fig. 7). — Let R be
the total load, and W^ = W^ the weights numbered from the loft
end ; let 4.s be their distance apart, so that if G be the origin for
loads, the abscissie of W^ and W^, are 2s and  2s, respectively ;
the origin for the span is the centre, and x is the distance
from to G.
166
APPLIED MECHANICS.
[chap. IX.
First, let Wi be over the point x, the whole load being on
the span ; then F may be calculated as if the whole load E
were at G, that is,
B 7?
P=(c:r), and ^M^ = P{c  x) =  (c  x){c  x). (1)
If the load travels a little to the right, P diminishes, and
therefore the bending moment at x diminishes ; if the load
travels until Wj is at a small distance a to the left of the
section, then (fig. 8)
^ = 0"^ (c  •* + a),
and, if x is positive,
M'^ = P'{cx) W,a = ^ (c  .r + a){c  x)  ^ a
2c 2
{cx){c x)  ^ ao: = ,M^
Pa
"2c
t..£+j: 2S — 4
'^ ^£ ,
f/?
J5£
„....€ X.
2S K.2S
^?H
** (?
. X < ,ilf.
•CXtet X
',1V>#72
Xi XH
"h:
rin
— cx — V Xi
Fig 7.
Fis. 8.
n7j^sgs ir.
\Fy) — <''' — ^jf
(4S=C)
}
Hence ^M^ is the maximum bending moment for values of
X from to c; that is, for any
point in the left half of the span
the maximum occurs when W^
is over it, provided the whole
load be then on the span. By
symmetiy the maximum for any
point in the right half of the span
occurs when W2, the right weight,
is over it.
Bending Moment Diagram (fig.
9). — Substituting for x its value
(2s  x), we have
p
i^/x = 2p (c  2s + X) {cx), (2)
the equation to the maxima bend
ing moments for first half of span. The locus is a parabola with
CHAP. IX.]
B. M. AND S. F. FOR MOVING LOADS.
167
its axis vertical, and the principal equation is Y = — X^ ; it is
therefore the same parabola as for a rolling load H (see fig. 5,
p. 164). To find the horizontal distance to the apex Aj, find
that value of x which makes the product (c  2.s + x) (c  x)
greatest ; since the sum of the factors is constant, this occurs
when they are equal ; thus
c2s + x = cx, or X = s. (.3)
That is, the apex ^i lies to the left of at a distance s, one
quarter of the distance between the two weights. To find the
height of Ai, put
x = s, and ,if, = ^ (c  sf, (4)
the maximum of maxima for first half of span.
It is evident that Az will lie to the right at a distance s,
and that the two parabolas will intersect at I) on the vertical
'JP7
iR . ir,^sf2sJTT^ =^j;
(4S>C).
(4S<C)
W,
^L
f S W— S );
4S»{2W2).2C
"f.^ .t/„ is max:
A,hE fA,
OR = 0F={4SC)
Fiff. 10«.
Fio. \0b.
through the centre. It is convenient to call the first half of
the span Jielcl 1, and to say that this field is governed by W^ ;
and we observe that the maximum in field 1 occurs when Wi,
being in its own field, lies as far to one side of the centre, as
G lies to the other.
If it be possible for Wi to occupy every point in its field
without W^ going off the span, we say that W^ can overtake
its field. In the present problem it is necessary that 4s, the
distance between the weights, be not greater than c th^ half
span, in order that each weight may be able to overtake its
168
APPLIED MECHANICS.
[chap. IX.
field. Two cases are shown on figs. lOo and lOj ; in the second
the distance between the wheels being greater than the half
span, one wheel can stand at the centre alone on the span
giving the height E, a maximum and a rival to the maxima at
Ai and A,. The condition that all three maxima may be equal
is engraved on the figure.
w,
f'
A,
P=Ii
Fig. 11.
The complete interpretation of the locus BAiDAtC is shown in fig. 11.
Suppose the beam to extend beyond the supports at B and C, and to be fixed at
these supports so that P and Q may act upwards or downwards. In the figure the
travelling load is standing with G over B, so that /*= li, and Q is zero ; hence at
L\, the point under W^, the bending moment is zero, and this is the point at which
CAiD meets BC. Let the load move towards the left until TFz is over any point
as JT; Q now acts downwards ; at the point K, the beam is bent upwards and the
bending moment is negative ; the value of this negative moment is Q .KC, and it
is given by the downward ordinate Xb "When IFz arrives over B, the bending
moment a.t B is negative; and its value, Q.BC'= Wi .As, is given by the
ordinate Bd.
As the load moves farther to the left, the bending moment at each point, as W^
comes over it, is of the constant value Bd ; BC is now u cantilever under the down
ward load Q, and the bending moment at each jioint of BC is now negative,
increasing definitely as the load moves towards the left. For all positions of the
load, with no restriction on the value of 4.*, .B^i7)^2C' gives the maximum positive
bending moment at each point; and the height of .i4i is the greatest positive
bending moment tliat can possibly be produced by the load system.
Bending Moments for a beam U7ider a travelling load system
of two unequal weights at a fixed interval apart (fig. 12). — Let
li be the total load ; PFi and JFj the weights numbered from
the lef^ end ; let G their centre of gravity be the origin for
CHAP. IX.] B. M. AND S. F. FOR MOVING LOADS.
169
loads, 2Ai and  2/t2 being the abscissae of Wi and Wj, so that
the distance between the weights is 2Ai + 2h>. Let Wi be over
any point of the span whose abscissa is x measured (positive to
i(<;»a2/(^N
m
}g Zhr^2h,^^^
— <;aj— i x^— X — ^
ii:
J2i
cxid.;
ih;
'yr ■>;
Q'
Fis. 12.
Fig. 13.
left) from the centre as origin, and let it be understood all
through that the whole load is on the span, or rides out on a
continuation of it as on fig. 11.
As in the previous case,
i^^x ^^(c ^)(cx)
(1)
is the equation to the bending moment at any point x when Wi
is over it.
If the load travels a little to the right, P diminishes, and
therefore 3f^. diminishes ; if the load travels until Wi is at a
small distance a to the left of the section, then (fig. 13)
M'x = n {c  X + a){c  x)  Wia
That is, the bending moment at x any point of the span, when
Wi is over it, is greater than when the load is in any other
position, provided that the point itself is situated between B
the left end, and F a point whose distance from the centre is
or from the left end is
BF=^^.2c.
Jx
170 APPLIED MECHANICS. [CHAP. IX.
Further,
01'=^. 2c,
and the bending moment is greatest at any point of CF when
li\ is over it. BF and CF are fields 1 and 2, and they are
commanded by W'l and W2, respectively.
Bending Moment Diagram (fig. 14). — For x substitute
2hi  X in equation (1), and we have
^M, = ^^(c2h,^z)(cx), (2)
the equation to the maxima bending moments for field 1,
The locus is a parabola whose axis is vertical, and principal
equation is
2c
it is therefore the same parabola as for a rolling load B.
The abscissa of the apex A^, that is OSi, is found as before
by equating the factors of equation (2), thus
c  2hi + X = c  X, or x = hi.
That is, the apex Ai lies to the left of (? at a distance OSi = Jh ;
similarly Ai lies to the right of at a distance OS2 = ho, each
being half the distance between W and G. Putting x = h^
we have
SiAi = ~{c h,)\ (3)
zc
Similarly,
SU. = § (^  ^Y' (4)
If the point /S, does not lie in field 1, the ordinates which
are the bending moments for field 1 continually increase from
zero at B the left end to their greatest value at F the other
end of the field ; if S^ lies in field 1, then
.Mni = SiA, j^{c Ihf (5)
the maximum of maxima for field 1,
CHAP. IX.] B. M. AND S. F. FOR MOVING LOADS.
171
Similarly, if aS^o be situated in field 2, the height of A2 will
be the maximum bending moment for field 2.
Suppose Wz > Wi ; then, since both parabolas are the same
as that for the rolling load JR, and are therefore the same as
each other, A. is higher than A^ because the quadrant CAnS,
stands on a longer base than BAiSi ; and
(6)
the maximum of maxima for field 2, and maximum for whole span
Now F is both in field 1 and field 2 ; and when Wi arrives
at F, the ordinate of the first parabola gives the maximum
bending moment at F; again, when W., arrives at F, the
ordinate of the second parabola also gives the maximum bend
ing moment at F\ that is, the ordinates at i^are equal, or the
two parabolas intersect at D, a point on the vertical through F.
It is well to observe
that the maximum in either
field occurs when the weight
commanding the field, while
lying in its own field, is as
far from the centre of the
span upon one side as G is
upon the other. If it be
impossible in one of the
fields for the weight so to
lie, then for that field the
bending moment contin
uously increases towards the
end of the field not coincid
ing with the end of the span.
! ''^
i '' /
''Tk
?>;
~\!eld2
1
"1
j
1 // _.i
// •'' '^
i '
1 N
%.
i
1/ ' H
i 1
i i
\
\
B
S,FO S:
Fk. 14.
C'a^ie l,2ki + 2/(3
smaller weight
X span, or distance between the weights
— shorter field.
< total weight
In this ease it is evident that each weight can overtake its field.
Case 2, 2hi + 2^2 >
smaller weight
X span, or distance between the weights >
total weight
horter field (fig. 15). — The two apexes Ai and A2 are further apart than in the
previous case ; D occupies a lower position, and is no longer on BEC, of which a
portion kEh is above BAiBAiC ; and the diagram showing the maximum bending
moment at each point is now BIcEhAzC. The positions of h and h are indicated on
the fig. lo. That E may be the same height as Ai we have, equation (4)
SiAi = — ie  hi) ; and OE == \W2.l=hW2 c;
2c
equating these gives distance between weights = — ? 2c.
C)
172
APPLIED MECHANICS.
[chap. IX.
Graphical iSo/M<toM. — Construct G, the centre of gravity as indicated on fig. 15.
Mark S\ and ^2 at distances half of those at which the wheels are from G. With
any paiaboiic riglit segment, cut ont of cardboard or celluloid and used like a set
square, draw the three parabolic arcs with their apexes Ai, A2, and An on the
verticals through 'Si, ^2, and 0. The arc of the Icit parabola to pass through Ji,
and of the right to pass through C, while the central one passes through both.
Then construct a scale fur verticals so that the height of Ag shall read ^ Jil where
Ji = IFi + TF2, and I is the span. Note thai the trolly on fig. 15 is standing with
fVi in its most commanding position, which is neatly expressed by saying that
fo^''
2hr
— i,
2h,
f
)
' \g
'v )
,''''
■fh,
•f
h,',
i
/
y'^
h
^y
4^
K
S',"F OW S.,
%. Oist. between Wta »J
I
( Oiit. between Wta increased In ratio of Wts'
Fig. lr>.
the wheel is as far from the one support as the common centre of gravity is from
the other. A convenient way of drawing the scale for verticals is to select one on
which OAq reads a little less than it should read, and slope the scale over till a
horizontal from Aq reads correctly, when the required values of SiA\, &c., can
be lead off by horizontal lines ruled over to the sloping scale. If the trolly can
ride out on the beam prolonged past the supports, fiA\ DA2C is the locus of
maximum bending moments at each point: see fig. 11. If the beam be not
prolonged, and if the two wheels he fnr apart, as in fig 15, then the remaining arc
kEh is to he drawn by a second iiarabolic segment whose height shall bear to
OAo the ratio of JFz to ( fF^ + Wi) or Ji. Note that F below the junction D
divides the span into Jields proportional to the loads.
On fig. 1, Ch. VIII, we had a parabolic locus which we
supposed to be duplicate. A load at the centre made them rise
and move towards each other. It can readily be shown that a
load at i^on fig. 15 of magnitude
will in a like manner make the two side parabolas approach
CHAP. IX.] B. M. AND S. F. FOR MOVING LOADS. 173
when At, An, and A, will all coincide. Such a fixed load at F
and the trolly give as maxima BAoC, just as a single rolling load
Ii = {W,+ W2) would do.
Moving model illustrating the bending moments on a girder
bridge dne to a trolly passing over it. — A model appeals to some
minds to which analysis and descriptive geometry are tiresome.
The model consists of two boards about 18 inches square,
with a girder a foot long at the top of each. The rails are
directly over the girders, so that the boards are spaced about
2 inches apart, the gauge of the rails (hg. 16). The trolly is
supposed to weigh 36 tons, and it has three wheels ; the two
wheels on the near side, axles 12 feet apart, ride on the near
girder, and transmit to it a total load of 18 tons, 12 by the
leading and 6 by the trailing wheel. The axle of the single
wheel on the remote side is at the centre of gravity, being
4 feet from the leading and 8 feet from the trailing axle, and
rides on the remote girder, transmitting 18 tons directly to it.
On the face of the model are three hands which turn upon
pivots as the trolly is pushed along the bridge. The side hands
are pivoted at the ends of a horizontal line representing the
span of the girder, 36 feet to a scale of 3 feet to an inch. This
line is the base of the bending moment diagram. The central
hand is pivoted directly over the point lying 6 feet to the right
of the centre of the span, the point dividing the span in the
ratio 2 to 1, just as the two wheels share the load. The height
of the pivot is 5^ inches.
As the trolly is shoved along the bridge, the hands turn so
that their intersections are always directly under the wheels.
At the same time the pole, at the extreme left of the model,
moves up and down, keeping the three threads attached to it
always parallel to the three hands, one to each. The other
ends of the threads pass through three eyelet holes in the face
of the model, and little weights are attached to them to take up
the slack out of sight. The eyelet holes are ranged on a vertical
line, which is the load line to a scale of 3 tons to an inch. The
polar distance is 6 feet ; the scale for bending moments is conse
quently 18 foottons to an inch.
While the trolly stands still, the three hands form a polygon
of three sides standing upon the base. This polygon is the instan
taneous bending moment diagram. It also is a balanced frame
for the loads fixed in that position, and the threads from the
pole form the reciprocal figure giving the thrusts along its sides.
Between the boards a pantagraph is pivoted loose on the
pivot of the central hand. One point of the pantagraph is
174 APPLIED MECHANICS. [CHAP. IX.
driven by the trolly, and therefore the middle point of the
pantagraph describes a horizontal path, with half the travel of
the trolly. From it a thread passing over a pulley makes the
pole move up and down with half the speed of the trolly. From
the central point of the pantagraph three ivires radiate out — the
middle one slides in a tube fixed at right angles on the pivot of
the central hand, and rocks the pivot about ; the other two
wires in the same way rock two pivots (not seen on the wood
cut), which are placed horizontally right and left of the central
one, just as the end eyeletholes are placed above and below
the central one. These end pivots transmit the rocking motion
to the pivots of the end hands by means of cranks and
connecting rods.
The mechanism will be readily understood by supposing the
threads from the poles to be wires transpiercing journals at the
eyeletholes, and causing them to rock as the pole goes up and
down ; and then supposing these geared to the pivots of the
hands, one to each, by pulleys and endless bands between the
boards, thus compelling each hand to be always parallel to the
corresponding thread.
The intersection of the hands below the leading wheel
sweeps out a parabola which is painted on the face of the
model. It is at the vertex for the position of the locomotive
shown on the woodcut. The leading wheel being 16 feet from
the left abutment, the centre of gravity is 16 feet from the
right abutment. Now the supporting force at the left end is
proportional to the secondmentioned 16, being in fact omha/f
of it, this being the ratio of total load to span. And the lever of
the left supporting force to cause bending on the girder under
the leading wheel is the firstmentioned 16 feet. So that
the bending moment there, for the position of trolly shown, is
M_i = i X 16 X 16 = 128 foottons, a max.
(Compare Example No. 8, p. 180.)
Should the trolly move one foot right or left, one of the
factors 16 becomes 15, and the other 17, giving a lesser product.
Should it move two feet right or left of the position shown on
woodcut, the factors become 14 and 18, with a still smaller
product by a proposition of Euclid. Hence the locus of the
intersection of left and central iiands is a parabolic right seg
ment, half base 16 feet, and passing through the pivots of those
two hands.
When the trailing wheel is 14 feet from the right abutment,
CHAP. IX.] B. M. AND S, F. FOR M0VIN<1 LOADS.
175
the centre of gravity is 1 4 feet from the left abutment, and the
intersection of the central and right hands is at a height,
3fi = I X 14 X 14 = 98 foottons, a max.,
and is at the vertex of its path, which is again a parabolic
right segment, half base 14 feet, modulus onehalf, and passing
through the pivots of those two hands. The position of the
pivot of the central hand is then on the intersection of those
two parabolas. The height of that pivot is readily found thus :
Observe that the perpendicular dropped from the pole upon the
load line divides it into two segments which are the supporting
forces. If the locomotive be moved till the central hand is
horizontal, then the central thread being horizontal is perpen
dicular to the load line, and so we have tlie left supporting
force 12 tons, twothirds of the whole load ; hence the centre
of gravity of the trolly must now trisect the span, and be
176 APPLIED MECHANICS. [CHAP. IX.
12 feet from the left abutment. The leading wheel then will
be 8 feet from left abutment, and the bending moment under it
is 12x8, or 96 foottons. But the central hand being hori
zontal, this also is the height of its pivot, namely, 5i inches.
On the other face of the model not shown there is no
central hand nor thread, and the triangle formed over the base
with the two end hands is the instantaneous bending moment
diagram for the remote girder, upon which only a single load,
18 tons, rolls. This intersection sweeps out a parabolic right
segment, having 18 feet for its half base, with the same modulus
as before, namely, onehalf ; it passes through the pivots of the
end hands, and its vertex is over the centre of span.
The fact that the polar distance remains constant shows that
the central parabola (giving the bending moments on the remote
girder due to the 18 tons rolling on one wheel) and the pair of
parabolas (giving the bending moments for the near girder)
constitute a pair of diagrams to one common scale. Of course
the scale for the single central parabola, which alone appears on
the back face of the model, is readily constructed. The scale
must be such that the height of the vertex of the central
parabola shall measure on it \WL= 162 foottons. Otherwise,
as already stated, the polar distance being 6 feet makes the
scale for bending moment 18 foottons to the inch, that is,
6 times finer than the scale for tons.
The height to the central parabola at a point 6 feet to one
side of the centre is onehalf oi the product of the two segments
into which that point divides the span, or it is  x 24 x 12 = 144
foottons. But the height of the central pivot is 96 foottons,
so that the depth of the central pivot from the central parabola
is 48 foottons. Now 48 foottons is the statical moment or
product of 12 tons, the leading wheel into 4 feet its distance
from the common centre of gravity.
Another way then of stating the position of the central
pivot is to say that the vertical through it must di\TLde the
span in the ratio 2 to 1 in which the leading and trailing
wheels share the total load, and that it must lie on this
vertical at a depth below the central parabola by an amount
given by the statical moment of either wheel about their
common centre of gravity.
The model is made by Messrs. Dixon & Hempenstal, Dublin,
and may be seen in Dublin, at the Engineering School, Trinity
College, or at the College of Science, Poena.
The description of this model in our second edition was
original, except that the figure appeared on the correspondence
CHAP, IX.] B. M. AND S. ¥. FOR MOVING LOADS.
177
in Mr.Farr's paper on "Moving Loads on KailwayUnderBridges"
in the Transactions of the Institute of Civil Engineers, 1900.
A model of ruder construction, with sliding frames, and
called " A Bending Moment Delineator," is described in the
Transactions of the Engineers and Ship Builders of Scotland,
November. 1889, and was exhibited at the Munich Exhibition in
1893, and described in the /Catalog.
We add here a photograph of the " Delineator " from a large
blackboard model (fig. 17). It will be seen that the three
arms are constrained bv slots to move exactlv as the hands on
Fig. 17.
the model 'fig. 16). The loads on the wheels are equal, being
9 tons each, and are due to a uniform load of 1 tons per foot
spread on the wheel base. The bending moments on the wheel
base itself are shown by the parabolic segment painted on a
" Distorting Table," which consists of a number of vertical
strips of wood slightly spaced, and connected to each other by
two horizontal lazytongs across their backs, each strip being
pinned to a joint of the upper and under lazytongs. This
N
178 APPLIED MECHANICS. [CHAP. IX.
distorting table has holes, one at each coiner, which slip over
four pegs connected to the crossarm which bears the two knob
handles at the ends. If the distorting table be pulled forward
off the pins, it can then be stretched like an accorciion, but, when
on the pins, it can only be subjected to sliding displacement.
In the middle position the figure painted on the distorting
table is a right parabolic segment ; but when the trolly is shoved
to one side as shown on the photograph, it is oblique, but of course
the lengths of the white strips are unaltered. The two oblique
slotted bars are always tangents at the ends of the parabolic
segment.
If the uniform load drop directly down upon the girder, and
move about on it without the intervention of the trolly, then
the instantaneous bending moment, as shown on the photograph,
is the parabolic locus for the loaded central segment of the span
and the tangents to it for the two side unloaded segments.
Compare fig. 7, Ch. VIII, which this model illustrates when the
load is at rest.
In the photograph the load is so disposed on the girder that
the lefthand quartering point of the load is directly over the
lefthand quartering point of the span. Consider the bending
moment at the left quartering point of the span for this position
of the load. It is given by the height above the base of the top
of the left quartering white stripe painted on the distorting table.
We know that this is the maximum bending moment at this
point of the span (see figs. 3 and 4, Ch. IX). Now the model
exhibits this beautifully ; for if we move the load a little to the
right, the " white strip " is replaced by one shorter than itself,
for which reason the top is Imver, but the base of the parabolic
segment has risen^ for the crossbar on which it stands has
rotated clock wise about a pivot under the centre of the girder
which exactly neutralises the efi'ect of the shorter " white strip "
replacing the original one. That is, for a small motion of the
load to the righthand the bending moment at the quartering
point of the span (as given by the new instantaneous diagram)
is unaltered. Similarly it is unaltered for a slight motion of
the load in the other direction. At that point, then, the
variation of the bending moment is zero for a small movement
which is the criterion for a maximum.
This " distorting table " also serves to illustrate the addition
of the parabolic slope to the straight slope (see theorems in
Chapter XI).
It also serves to show how the shearing force diagram
painted on it for any moving load is to be distorted and
superimposed upon the shearing force diagram for the dead load.
CHAP. IX.] 1$. M. AND S. F. FOR MOVING LOADS. 179
EXAMPLKS.
I. An advancing load, as long as or longer than the span, and of intensity I.V
tons per foot, conies upon a beam 36 feel long. Find the maximum bending
moment for the whole span for all positions of the load
mas.^^^o = iJr. I = ^54 x 36 = 243 ft. tons.
Find the maxima at intervals of six feet
4c
therefore
^f±^8 = 0, .l/±i2 = 135, .¥±6 = 216, and Mq = 243 ft. tons,
all maxima.
2. For the same beam of 36 feet span, with the same intensity of load Ih tons
per foot, the whole length of the load being now only 12 feet, find the maximum
for the whole span
max..l/o = ^TF{lk) = i X 18 X (36  6) = 135 ft. tons.
Find now the maxima at intervals of 9 feet
m^xMx = — (c  x^){2  ^] = i%(324  z^) ■
and MiB = 0, Mg = lOli, Mn = 135 futons.
all maxima.
3. A beam 42 feet span is subject to an advancing load of 3 tons per foot and
12 feet long. Find the maximum bending moment at 7 feet on either side of the
centre. Where is the centre of the load situated when this maximum is produced ?
Atis. max..ilf±7 = 288 ft. tons. Five feet from centre of span.
4. In the previous example, find the maximum bending moment for the whole
span, for all positions ot the above load.
Alls, njax.il/o = 324 ft.tons.
5. A beam 30 feet span is subject to a rolling load of 40 tons : find the
maximum bending moment for M'hole span. At what point does it occur, and
how is the load then situated ?
Aus. maLiVo = jff . I = 300 ft.tons. Ji is at centre.
6. In the previous example, find the maxima bending moments at intervals of
5 feet. How must the load be situated in each case P
Am. max^/r = ^ (c  ar^) = 1(225  x^) ;
therefore "
M±is = 0, M±io = I665, J/±5 = 266§, and M^ = 300 ft.tons,
aU maxima. The above load of 40 tons in each case is over the point.
7. If the load of 40 tons in the above examples be spread uniformly over
3 inches, instead of being concentrated at a point, how much are the above
results in eiTOr ?
Ji , „ „ /, extent of load\
raax.i»/^r=  (c  X) 2 ) :
4c \ extent ot span/
N 2
180 APPLIED MECHAiNICS. [CHAP. IX.
this differs from the above expression by the factor
/ 2yt\ , / •25\ 239
I (2  ) = ^2  ) = ,
hence the results above would be in excess by a ^Jgth part, or by i^jths percent.
8. A beam 36 feet span bears a travelling load of 18 tons concentrated on
two wheels 12 feet apart, there being 12 tons on the left and 6 tons on the right
wheel. Find the maximum bending moment; the equaiiona to, and amounts at
intervals of 6 feel of, the maxima bending moments.
Data: jr, = 12, W2 = &, if =18 tons, c = 18, 2/(1 = 4, 2A2 = 8feet.
Dividing 36 directly as 1 and 2, we have JiF and FC, fields 1 and 2, equal
respectively to 24 and 12. Since the distance between the wheels is not greater
than tlie shorter field, the eXMmjde comes under Case I. (see fig. 14), and the
maximum bending moment at each point is the locus liDC, M'hether the load
is confined to the span or makes a transit.
The maximum bending moment for whole span occurs at x = h\ = — 2 ;
its value is to be found by supposing the load standing with the left wheel
two feet to the left of the centre as on the model (fig. 16), to which the text
now refers, and then calculating the moment at that point,
P = 8 tons, and \M.2 =128 ft. tons max.
The equations to the maximum bending moment at each point are.
For field 1, iM^ = ^ (14  x) (18 + x), for values of x from  18 to 6.
For field 2, sJ/^ = ^{IQ + x) (18  x), for values of x from 6 to 18.
And evaluating at intervals of six feet, we have
J/.18 = 0, Jf.i2 = 78, i/.6 = 120, J/o = 126, J/e = 96, Mn = 66 ft.tons, &c.
Note. — The height of Az may be calculated from fig. 14 thus : — OSi = 2,
the base of quadiant then is 16 ; the moiiulus is the load divided by span,
that is A ; hence the height i>f apex, or modulus into base squared, is 128.
To inake the graphical solution, lay off 0S\ = 2, 08^=4, and dr&w JJIjC as
previou.ly desciilied ; make a vertical scale upon which OAo = 2 x 18 x 18 = 162
(see model, fig. 16).
9. A beam 36 feet span bears a travelling load of 18 tons concentrated in.
equal portions on two wheels 12 feet apart. Find the maximum bending moment.
In this case, Wi = Wz, and 4s < c, so that (fig. \0a) UHx = OS2 = s.
The maximum occurs at 3 feet on either side of the centre, whether the load is
confined to the span like a travelling crane or makes a transit like a truck ; its
amount is found by assuming the load to be standing with the left wheel 3 feet to
left of centre.
Aug. 1' = 7^ tons ; iMa = zM.s = 112j ft.tons, maximum for whole
span. That is, half of the square of fifteen (see the model, fig. 17).
10. If the load of Ex. 9 shift till the left wheel is 6 feet from the left end
of the beam, calculate the bending moment 9 feet from the left end.
P = 12 tons, J/9 = 12 X 9  9 X 3 = 81 ft.tons.
This is the position of the load on the model (fig. 17), and is the bending
moment on the girder under a point 3 feet to the right of the left wheel. Also
J (6*  S"^) or 20J is the bending moment at the same point on the wheel
base or 12 foot beam joining the two wheels, due to the uniform load of 1^ tons
spread over it. These two added together give 81 4 20^ = 101; ft.tons. This
CHAP. IX.] B. M. AND S. F. FOR MOVINOx LOADS.
181
is the maxiniinn bending niomoiit on the girder at the point 9 feet from the left
end, if the uiiiform load rested directly upon it without the intervention of the
twowheeled trolly.
Compare this Example with the preceding, and with Exercises 2 and 3.
11. A beam 56 feet span bears a travelling load of 16 tons concentrated on
two wheels 32 feet apart, 7 tons being on the left wheel and 9 tons on the right.
Find the maximum bending moment during the transit.
From equation 7, p. 171,
iz.v^:z::5o, = 32feet,
which liappens to be the distance between the weights ; hence there will be two
equal maxima, one at 7 feet to the right of the centre when the greater load is
over it ; the other, at the centre also when the greater load is over it, the smaller
load not being then on the span. Place the load in those positions respectively,
and calculate the moments.
Ans. 2.U.1 = iM'o ■■= 126 ft.tons.
12. A travelling load of 5 tons concentrated on two wheels 10 feet apart,
1 ton being on the left wheel and 4 tons on the right, passes over a beam of
•10 feet span. Find the maxima bending moments at intervals of 4 feet, and the
maximum for the whole span.
Distance between weights x ratio of weights = 40 feet = span, so that k
coincides with B (fig. 15), ;ind therefore arc BhE lies everywhere above arc
BB ; that is, the maximum at each point of span occurs when the greater weight
is over it. Further, the height of A2 is greater than that of E, since the distance
between the weights does not exceed ^„lhs of the span. Placing the greater
load over points at intervals of 4 feet and calculating the bending moments, or
substituting into the equations to the loci Bh and hAoC, we have
zM'io = 0, zM'iG = 144, 2M'i2 = 256; 2M^ = 35, Mi = 42, zMo = 45,
2^4 = 44, oiWg = 39, a.Vx. = 30, 2.I/I6 = 17, 2iIf2o = ft.tons.
Since the greater weight lies 2 feet to the right of the centre of gravity of the
load, the maximum lies at 1 foot to the right of the centre of span, and
2Jlfi = 45^ ft.tons maximum for whole span during transit.
The equations to the loci from which these may be calculated are for Bk,
,M'. = A(c  x^) = i, (400  x%
for values of x from 20 to 10 ; and for hAiC
.Mr = /o (.0 + 2/i2  x){c + X) = ^ (18  ^) (20 + x)
for values of x from 10 to — 20.
182
APPLIED MECHANICS.
[chap. X.
CHAPTER X.
BENDING MOMENTS AND SHEARING FORCES DUE TO A TRAVELLING
LOAD SYSTEM.
Bending Moments for a beam under a travelling load system of
unequal weighU fixed at irregular intervals the load being confined
to the span so that no veight passes off (fig. 1). — Let R be the
total load ; W^, Wi, ... Wr . . . IVn, the weights numbered in
order from the left end ; G the centre of gravity and origin
for the weights ; 2Ai, 2^2 . . . 2h,. . . . 2h„, the abscissse. of the
V.'> Vil ^n,'^ >< ' %x<»"
]T7 l4 ^4  ^trl ; llly.^,.^^
\s \' <S.b ^<
w; ' 'fVr'^^
p'
C*X*Ot y\
nv '■ ■ Wr
.f
Fis. 1.
weights, those to the right of G including their negative sign ;
the origin for, and 2c the length of, the span, distances
towards the left being positive; and let cc be the abscissa^of
any section.
First. — Let the load be in a position such that the i'"" weight
is over the section ; cr^, x^ . . . Xr.\, the abscissae of the (r  1)
weights, measured from ; it the abscissa of G, and c that of the
left end of span ; then
and
R
rM^ = {c^x){cx) W,{x, x) W,{x,  x)
.. . Wr.^iXr,  X)  Wr{Xrx); (1)
the last term is zero, since Xr = x, and it may either be expressed
as above or omitted.
CHAP. X.] B. M. AND S. F. — LOCOMOTIVE. 183
Second. — Let the load be situated at a short distance a to
the light of its former position ; then
^' = 2^ (c + ^  ") ;
and
M\ =  (c + X a)(c X)  W,{x,  a  x)  Wi{x2 a x)
2c
 ...  Wr.i{Xr.i  a X)
2c
= ,.M, ■^(cx)a + (W,+ W, + ...+ Wr.,)a. (2)
<,.!/„ a ^(c,r)>^rW^^ or if c  ^^ ^ 2c> a; ;
2c Ji
that is, the bending moment for the first position is the greater,
if the distance of the section to the left of the centre is less
than c '—^ — ^ . 2c ; or, what is the same thing, if the distance
of the section from the left end is greater than ^ — f, . 2c.
Third. — Let the load be situated at a short distance « to the
left of the first position ; then
P' = — (c + X + a),
and
M"x ^ ^ (c + X + a) (c  x)  Wi{xi + a  x)  Wzix, + a  x)
2c
 ...  Wr{Xr + a  X)
= ,J/, + ^ (c  ,r)o ~{W, i W,^...+ Wr)a. (3)
R "^ ^{ W^
< rM.„ if  {cx)< Sx'X W), or if c ^^^j^ .2c<x;
that is, the bending moment for the first position is greater than
for the third, if the distance of the section to the left of the
centre is greater than c „ . 2c ; or, what is the same
thing, if the distance of the section from the left end is less
than M!Q . 2«.
184 APPLIED MECHANICS. [CHAP. X.
Hence there is a portion of the span, lying between the
point whose distance from the left end is — ^^ — . 2c, and the
'2i'"^( TV)
point whose distance from the left end is p —  . 2c, such
that the bending moment at any section in that portion is
greater when the r^^ weight is over it than for any other
position of the load, if for each position all the weights are
on the span as premised. This portion of the span we call
the r''^ field, and we say it is commanded by the r"" weight.
The extent of this r*'' field is
^^J.2.^p.2c.I^.2c, (4)
.he same fraction of the span as the r"* weight is of the total
load ; hence, in order to mark the fields, the span is to be
divided into as many portions as there are weights, these
portions being proportional to the weights and in the same
order. The maximum bending moment at any point occurs
when the weight, which commands the field in which the point
lies, comes over that point. Since no weight is to go off the
span, sometimes there is a part of a field which the commanding
weight cannot occupy, and then the weight is said not to be
able to overtake that part of its field ; as will be proved when
we come to the graphical solution, the maximum bending
moment for such points occurs when the commanding weight
is as close thereto as it can be brought.
Into the expression for ,J/, the maximum at any point of
the r'* field, substitute as follows : —
x = {x 2hr), {x, x) = (2h,  2hr), [x,  .r) = {2h,  2h,.), &c.,
. . . !>,._, x) = {2h,..,  2//,,) ;
and we have
rM,= ^^{c + x 2K) {cx) W, (2A, 2/^)  ]V, (2A,  2A,)
 ...  Wr_, (2A,.,  2/^.)
= ^ (c + '•  2hr) (c  X) + 2hr Sr' {W) Sr\ W.2h) ; (5)
this is the equation to the maxima bending moments for the
r'* field. The locus is a paral)ola ; its axis is vertical ; its apex
is above the span, and may lie to either side of 0, the centre of
CHAP. X.] r.. M. AND S. V. — LOCOMOTIVK. 185
span ; its modulus is ^, a quantity which is the same for all
fields, and is the same as the modulus of the parabola due to E
as a rolling load. The abscissa of A,, the apex of the /•''' para
bola, is that value of .r which makes ,J\^x or (c { x  2h,.) (c  x)
greatest, that is where
a; = //,.; (6)
hence the apex of each parabola lies on the same side of, and
horizontally half as far from the centre of the span as the
commanding weight is from G, the centre of gravity of the load.
The apexes for some of the fields may lie in their own fields ;
and in such fields the maximum of the maxima is given by the
ordinate of the apex ; for other fields, the apexes may lie out
side of their own fields, and in these there is no maximum of
maxima, but the maxima increase continuously from one end of
the field to the other.
Bending Moment Diagram (fig. 2). — The locus is the polygon
BDtDn, &c., formed with parabolic arcs £Di, Z>iA, &c. ; each
parabola being the same as BA^G that for the rolling load B,
but lying with their apexes at the distances OSi = h^, OS. = hz,
&c., where hi, lu, &c., are half the respective distances of W^,
Wo, &c., from G the centre of gravity of the load. The para
bolas intersect in pairs on the verticals through the junctions of
the fields ; that is, through F^, Fo,, &c., points such that
BFi : F,F,, &c. : BC : : IF, : W,, &c. : B;
for, if F,. be the junction between the r"' field and the (?• + If'
field, then Fr is the last point in the '/'' field ; the maximum at
Fr occurs when W, is over it, and is given by the ordinate at F,
to the r''' parabola. Again, Fr is the first point in the (r + 1)'''
field ; the maximum at F,. occurs when W,. + , is over it and is
given by the ordinate at F,. to the (r + 1)^^ parabola. But the
maximum at Fr is some one definite quantity ; hence the two
parabolas have a common ordinate at Fr.
Maximum Be7uling Moment for whole spaji. — In fig. 2, A^
and Ai lie respectively in their own fields ; and the one which
has the greater ordinate gives the maximum for whole span.
In the general case one or more apexes will lie in their own
fields ; one at least, as we cannot conceive of such a series of
curves lying so that every apex is inside of another of the
curves. The ordinate of the apex that lies in its own field, if
only one is so situated, is maximum for the whole span ; the
.ordinate of the highest apex, if more than one be so situated.
186 APPLIED MECHANICS. [CHAP. X.
Suppose the first curve continued to the right, past i>, ; the
second past D2, &c. ; then each parabola is the locus of the
bending moment at each point as the corresponding weight
comes over it, the whole load being on the span. If the load
stands still in any position, as, for instance, that in which the
load is drawn in fig. 2, the bending moment at the point where
Wi stands is the ordinate there of the first parabola, at the
point where W^ stands the ordinate of the second parabola, &c.
if then the ordinates of these points be drawn each to the
proper parabola, and the tops of the ordinates be joined, we
will have the bending moment diagram for that set of fixed
loads ; we will have in fact the diagram shown in fig. 2,
Ch. VII, because the load is now the fixed load shown in
fig. 3, Ch. V.
Further, for the position of the load shown in fig. 2, the
straight line joining the tops of the two ordinates, one drawn to
parabola 2 from the point where Wn, stands, and the other to
parabola 3 from the point where W3 stands, will pass through
D^ the intersection of these parabolas, because the horizontal
projection of that joining line is constant, being equal to the
distance between W^ and W^ ; and we know that one end of
this joining line coincides with Do when either of the weights
W2 or JF3 is over F^ ; hence by the theorem (D, Ch. VI} it will
always pass through D2. Now, for the position of the load
shown, the joining line gives the bending moments at all inter
mediate points, so that the ordinate of D^ is the bending
moment at F,. for that position of the load ; and similarly for
any other position for which F2 lies between the weights W2
and JF^j. In other words, the bending moment at F the
junction of two fields is the same, whether the weight com
manding the field on either side is over it, or whether the load
stands in any position for which F lies between those weights.
When the load stands in a position where those two weights
are both to one side of F, then the joining line prochtccd still
passes through Z), but the ordinates of the produced line are not
bending moments. For instance, the line joining the tops of
the ordinates on fig. 2, one drawn to parabola 4 from the point
where W^ stands, and the other to parabola 5 from the point
where T[\ stands, will, when produced, pass through Dx, the
junction of those parabolas ; however, not the ordinate at Fx to
that produced line, but the ordinate to the line wliich termin
ates at C, gives the bending moment there.
On fig. 2, observe that if the load moves more than one
foot to the left, W^ goes oil' the s[)an ; and there is a portion
CHAP. X.J B. M. AND S. F. — LOCOMOTIVE. 187
of field 3 at its left end, which W:^ cannot overtake, and for
which the corresponding portion of parabola 3 is dotted, being
inadmissible. For the position of the load shown, we saw that
the bending moment at each point of that portion of field 3
is given by the ordinate to a straight line from Do to the top
of the ordinate of parabola 3, at the point where W3 is stand
ing ; that is, by the chord of parabola 3, from D^ to the top
of tliat ordinate. Now, the closer JFs comes to Fo the steeper
will that chord be, and consequently the greater the bending
moments at all these points. Hence the chord of the dotted
or inadmissible part of parabola 3 gives the maximum bending
moment at each point of the portion of field 3 which W3 cannot
overtake. Similarly for any portion of any field which the
commanding weight cannot overtake, the chord of the parabola,
instead of the arc, gives the maxima bending moments for the
whole load on the span. Fig. 15, p. 172, has already been quoted
as an example of this, when we assumed the arc Bk, chord kD,
chord Dh, and arc hC to be the diagram of maxima bending
moments for load confined to the span.
Graphical Solution for Bcndinq Moment Diagram (fig. 2). —
Lay off the wheel base and find G the centre of gravity of the load
by analysis as at fig. 3, p. 93, or graphically as in fig. 3, p. 123,
and indicated in fig. 2. Lay off BC equal to the span ; divide
it at the points F into fields proportional to the weights, either
by arithmetic or as indicated on the diagram, and draw vertical
lines through them to separate the fields. Lay o&OSi equal to
half the distance of W^ from G, OS2 equal to half the distance
of Wo from G, &c. — each point S being on that side of the
centre on which the corresponding weight lies with respect
to G ; draw verticals through the points ;S' ; apply the parallel
rollers to BC; place any parabolic segment against the rollers
with its apex on the vertical through S^ ; shift the rollers till
the curved edge passes through B, and draw the arc BDi,
stopping at the vertical through F^ ; shift the segment till the
apex is on the vertical through So, move the rollers till the
curved edge passes through A, and draw the arc DJ)>, stopping
at the vertical through F2. Similarly draw arc after arc in
succession for each field, and if the arc for the last field passes
through C the extremity of the span, it checks the accuracy of
the drawing. Lastly, shift the segment till the apex is on the
vertical through the centre ; move the rollers till the curved
edge passes through the two extremities, mark A^, and con
struct a scale for verticals and bending moments such that
OAq = \R.l. Find by inspection the portions of the fields
I I t '' 'm
■ ; I ■» \' i
\t. field l'jLfield 2jh field 3 jl^ field 4 » field 5 i
VERTICALS.
HORIZONTALS
_2?0FT TONS.
I I I I I I I I I I ^
CHAP. X.] H. M. AND S. F. — LOCOMOTIVE. 189
which the commanding weights cannot overtake, and over such
portions replace the arcs by chords. Then the locus gives tlie
maximum bending moment at each point for all possible posi
tions of the load, the load being confined to the span.
Graphical Solution for a Diagram of the Square Hoots of
Bending Monieiits. — Divide the span into " fields " proportional
to the loads on the wheels. From the middle point of the span
prick points at distances one half of those at which the wheels
lie from the centre of gravity of the load. From those points
as centres draw circular arcs one over each fiekl, and we have
a diagram of the square roots of the maximum bending moments.
It only remains to construct a scale. On the span draw a
semicircle and construct a scale such that ««, the height of the
crown of the semicircle, shall measure on it the square root of a
fourth of the product of the total load and total span.
The height of ffi may be scaled off the drawing, and its
value as compared to the known height of a^ found directly by
the rule of three without constructing a scale.*
Shearing Force Diagram (figs. 3 and 6). — At any point the
shearing force increases as each weight in succession approaches
from the right; and when a weight passes the point, it
suddenly diminishes by an amount equal to that weight. At
each point there is a maximum when a weight is just to the
right of the point, and a minimum when it is just to the
left.
In order to find for any point the maximum and minimum
corresponding to a particular weight : — Place the load system
so that this weight is over the point ; from P subtract the
weights to the left of that weight for the maximum, and
further subtract that weight for the" minimum. In figs. 3 and
6, a locus giving the maximum at each point due to a particular
weight approaching, is shown by a full line ; the locus giving
the minimum due to the same weight is evidently parallel to
the first, and below it at a constant distance equal to that
weight ; this parallel locus is shown by a dotted line. In
each of these diagrams there are five loci drawn in full lines,
and giving the maximum at each point due to the approach
from the right of each of the five weights respectively ; of the
five dotted loci giving the minimum at each point due to the
* This simple and elegant graphical solution of so complex a problem by means
of circular arcs only was published in a note in the Transactions of the Institute of
Civil Engineers for 1900.
190 APPLIED MECHANICS. [CHAP. X.
receding of each weight respectively, only one is shown, and
it is drawn parallel to the locus shown by the lowest full line,
and at a constant depth below it equal to the weight at the
right end of the load. Having determined the full lines or
positive loci, it is easy to draw the others.
The first locus Aa is the value of F when JFj is over any
point ; so that as the load comes on from the right end, and
so long as Wi alone is on the span, Aa is a portion of the
diagram due to TFj as a rolling load ; that is, it slopes at an
angle whose tangent is W^ f 2c, or, in other words, at a rate
proportional to Wi ; further, Aa extends over a horizontal
distance equal to that between IV^ and TF2. When IV. comes
over any point, W, and W2 alone being on the span, then F^ is
calculated by finding F and subtracting the constant quantity
TVt ; F is now increasing as for a rolling load ( Wi + W^) con
centrated at its centre of gravity, so that the second link aa
slopes at an angle whose tangent is ( W^ + Wo)  2c, or, in other
words, at a rate proportional to Wi + W^ ; further, aa extends
from the point where the preceding link ended, and continues
through a horizontal distance equal to that between W2 and W3.
Thus, the locus Aaaa . . . begins at the right end of the span,
each link sloping more and more at rates in direct proportion
to the sum of the weights on the span, and extending respec
tively over horizontal distances equal to those between the
weights; the last link extends over a distance which is the
excess of the span over the extent of the load. On fig. 6, the
first four links are short and equal, the last one is long. The
other loci Bbh . . . , Ccc . . . , &c., consist of links sloping more
and more as weights come on at the right end, and less and
less as they go off at the left ; for instance, Ccc . . . (fig. 6)
consists of two equal short links and a long one increasing in
slope, and two equal short ones decreasing in slope ; the rale
of slope of each link is directly proportional to the sum of the
weights on the span at the time corresponding.
After having drawn the first locus Aaa . . ., the initial
points B, C, D, &c., of the other loci are found as follows : —
The ordinate of B represents the value of the shearing force
at the right end of span when W^, is just to the right of that
point ; if the load be placed in this position, the shearing
force at the right end of span, and in the interval between
that end and the point where W^ stands, is constant; the
value of the shearing force just to the left of Jr, is given by
the ordinate of a, the left extremity of the first link Aa, and
this quantity diminished by JF, is the value required. Hence
CHAP. X.]
U. M. AND S. F. — LOCOMOTIVE.
191
B is on the vertical through the right end of span, and at a
depth W^ below the level of the first joint a on the locus
Aaa . . .; similarly G is on the same vertical and at a depth
Wi below the level of the first joint h on the locus Bhb ....
Graphical Solution for Shearing Force Dia<jram (fig. 3) —
The example shown in the figure is the same as that for which
ON THE SPAN
Fig. 3.
the bending moment diagram is given in fig. 2. Lay the
weights up, in order, on the vertical through the left end of
span, and then down again in order ; draw a ray from the
right end of span to each junction ; rays to joints in ascending
order are drawn in full lines, and to joints in descending order
are drawn in dotted lines. Draw another line Ae equal to
the span, so as not to complicate the figure. Draw the locus
Aaa . . . link after link parallel to the slopes in ascending order,
each link extending in order for a horizontal distance equal to
that between the weights in pairs, the last link completing the
locus ; in the figure, the first link Aa extends for a distance
equal to that between W^ and W^ ; the second link aa to that
between W^ and Wz, and so on ; the last link extends for a
distance equal to the ditt'erence between the span and the
length of the load. Plot B on the vertical through the right
192 APPLIED MECHANICS. [CHAP. X.
end, at a depth equal to Wi below the level of the first johit
a in the locus already drawn; draw the locus ^&& ... parallel
to the respective slopes in ascending order ; the first link Bb
is parallel to the slope for (ir, + W,), each link extends till
a new weight comes on, the second last link extends till Wi
goes off, and the last link is drawn parallel to the slope for
(W2 + W3+ W\ + Wi). Plot C at a depth equal to W^ below
the level of the first joint b, and draw the locus Ccc . . . ; the
first link Cc is parallel to the slope for {Wi + Wz+ W^); the
other links are drawn parallel respectively to the slopes in
ascending order, and when these are exhausted the remaining
links are drawn parallel to the slopes in descending order ; the
extent of each link is determined as each weight after JF3.
comes on, and then as weight after weight goes off. In the
same way, for each weight, a locus is drawn in full lines,
consisting of as many links as there are weights ; the highest
ordinate at any point gives the maximum positive shearing
force thereat, for transit of load. Dotted loci are drawn, one
parallel to each locus shown by a full line, and below it at a
distance equal to the weight to which it corresponds ; the
deepest ordinate at each point gives the negative maximum.
On the diagram, ee'...E' is drawn parallel to ee...U; and
since, in this case, the locus e'e'...E' gives the maximum
shearing force for every point of span, the other four dotted
loci are not shown.
Bending Moments for a Beam under a travelling load si/stem of equal iceighta
fixed at equal xnlervals and confined to the span (figs. 4, 5). — The locus BD\I)iDzO
drawn as in the general case, will be symmetrical about the centre. If the
number of weights be n, and their distance apart be ith of the span, then it is
evident that the span will be divided into « equal fields whose common extent is
the same as the distance between two weights, and that each weight will just be
able to overtake its field ; it the distance between two weights be less than
ith of the span, each weight is still able to overtake its own field. There
fore, for the common distance between the n weights equal to or less than ^^th of
the span, the bending moment diagram is the locus BD\D%Di . . . C everywhere
following the curves ; the maximum bending moment is at the centre, or at one
quarter of the common interval on either side of the centre, according as w is odd
or even. On the otlier hand, when the common interval between the n weights
is greater than ith of the span, each weight will always be in its own fieM, and will
only be able to overtake a portion of its field ; the bending moment diagram is
the locus BDyDiBz . . . C following the arcs for portions of fields overtaken by
the weights commanding, and the chords for the remainder. If « be odd, the
middle weight can always be placed at the centre of the span, and at that point
the maximum bending moment will occur; if n be even, and it be possible for
a wei^ihl to come as close to the centre as or closer than a quarter of the conimon
interval, then the inaximun) bending moment will be on both sides of the centre,
and at one quarter of the common interval therefrom ; if a weight cannot come so
close, the maximum will still be on both sides of the centre, and nt points as near
thereto as the weight on either side of it may approach.
CHAP. X.]
B. M. AND S. F. — LOCOMOTIVE.
193
For n even, the proof that the maxinmm is at the point, a quarter of the
cominon interval <m either side of the centre may be shown thus :— Let 6 be
H'/ =
\w.= \m =
Im
k
D..
^
4
,_ Span
No.of wts.
Fig. 4.
over the centre : then the weight nearest to the centre will he distant onehalf of
an interval ; if the end weight be distant from the end at least a quarter of an
interval, it will be possible for a weisht to approach the centre, a quarter interval ;
hencH, the span must equal the w — 1 equal intervals, and at least two quarter
intervals more, that is
= _ 2«
2c > (»  1) s + s ; 2c > — 
 1
— s.
It is readily shown for a system of n equal weights at a common interval
not greater than Hh of the span and confined to the span, that the locus] of. the
maximum bending moment at each point will entirely include the loci due to any
fTn
{ ir..
__J'I
i
1 i
iw; =
1 11', =
H
h>:
m
a" O
Fig. .5.
smaller number of the same equal weights at the same common interval. Each
passes through the points T)\ of the last, and if the common interval he less each
figure is inside the last without touching. For intervals greater than Ath of
the span the figures cut each other as shown on fig. 5.
O
194
APPLIED MECHANICS.
[chap. X.
Also fig. 6 shows tl;e shearing force (iiajrram already described conjointlv with
fig. 3.
4^
Fig. 6.
Examples.
1. A beam of 42 feet span supports the five wheels of the locomotive shown
on fig. 2 : find the locus of the maximum bending moment at each point for all
positions of the load, it being understood that no wheel moves off the span (fig. 2).
Loads 5, ■'), 11, 12, 9 tons = 42 tons.
Intervals o, 8, 10, 7 feet = 30 feet.
Distances from G, 17, 12, 4, 6,  13 feet.
Dividing the span in the ratio of the weights, we have the extent of the fields as
follows : —
1st field from 21 to 16 ; 2nd field from 1(5 to 11 ; 3rd field from II to ;
4th fiehl from to  12 ; .5th field from  12 to  21.
Substituting in the general equation
,J/, = £ (c + ,■  2f<r) (cx)\ •IKS.rK ^n  21*' (/'■ ■ 2/0,
we have
I Jf^ = _ (c + a  2Ai) (c  a:) = (4 + x) (21  .r),
2f
for values of x from 21 to 16;
R
iM^ = _ (c + X  2/12) {c  x) + ihi . U'l  }}\ . Ihx
'Ic
= (9 + z)(21 x)  26,
CHAP. X.] B. M, AND S. F. — LOCOMOTIVE.
195
for values of x from 16 to 11
R
3,V, ^ — {c + x 2/13) (c x) + 2/(3 ( JFi + TTo)  ( JFi . 2hi + Wn . 2h)
2 c
= (17 +x){2lx) lor.,
admissible for values of x from 8 to 0.
lif. =  (c + a;  2/(4) {c  r) + 2/m ( Wi + W^ + W3)
{JFi . 2hi + TFi . 2hz + /Fs . 2hz)
= (27 + a;) (21 _ a)  315,
admissible for values of x from — 2 to — 12 ;
sMr = — {c\x 2/(5) (c  X) + 2/t5 ( Wv + JTo + 7F3 + Wi)
(TFi .2hi+ Wi . 2A2 + W3 . 2/t3 + Wi . 2hv
= (34 +a;)(21  x)  546.
for values of x from — 12 to  21.
To find the ordinates of the apexes ; substitute x = In = S'o in the first equa
tion, a; = A2 = 6 in the second, and so on, and we have
1^8.5=15625, 2^6 = 200 3iI/2 = 256 4^1/":. = 261, 5y6. 5= 21025.
Only two are expressed by the letter M, as they alone of the five are bending
moments ; the others do not lie in their own fields. Hence the maximum bending
moment for the whole span is
4^_3 = 261 ft. tons.
2. Calculate the maxima positive and negative shearing forces at intervals
during the transit of the load (fig. 3).
The slopes of the links are 5, 10,' 21, 33, 42 ; 37, 32, 21, and 9 tons per 42 feet ;
42 feet heing the span.
The coordinates of the joints of the links are : —
Links.
Ft. Tons.
Ft. Tons.
Ft.
Tons.
Ft. Tons.
Ft. Tons
Ft.Tons
Aaa .
— 21 — ooo
— 16 06
8
25
2 75
9 130
21 25
Bbb ..
—21 — 44
—13— 25
3
25
4 80
16 20
21 244
Ccc .
—21 — 75
— II — 25
4
30
8 150
13 194
21 255
DM ..
—21 —135
— 14 — 80
— 2
40
3 84
II I4'5
21 195
E^e . .
— 21 290
— 9—17
4
— 126
4 65
14—15
21 OQO
For po.sitive maxima, the locus Ccc ... is to be taken between the left end of
span and the point a; = — 1, and the locus Aaa
of span ; for negative maxima, the locus ^"ee
span.
o 2
from that !)oint to the right end
. is 10 be taken for the whole
196 APPLIED MECHANICS. [CHAP. X.
3. Find tlie equation to the maxima bending moments in the third field
(Ex. 1) directly, and without using the general formula.
Choose liny point whose abscissa is x, such that when the load is placed with
the 3rd weight over it, the whole loud will be on the span. The distance of G the
centre of gravity from the right end will be (c + a; — 2A3), or (17 + a;), and
P=ff X (17tx) = (17 +a;).
Taking a section at x and considering the forces on the left side of it, we have
3^/, = P{2\ x) Wi{b + 8)  ^2 X 8
= (17 + a:)(21 .r)  105.
4. Find the maximum bending moment at any given points, say z = 8, and
X = 10, in Ex. 1 ; and find the maximum for the wliole span without using the
general equations (fig. 2).*
1°. To find tlie maximum bending moment jit any given point, divide the span
into fields proportional to the weights ; consider which field the point lies in, an<l
place the load so that the corresponding weight is over the point. Then if the
whole load be on the spar, the bending moment calculated at the point for the
load fixed in this position is a maximum. If when the load is so placed, some
weights are off the span, move the load the least distance vhich will bring them
all on the span, and calculate the bending moment at the point for the load fixed
in that position.
2^. To find the miiximuni bending moment for the whole span. — By inspection,
plai e the load so that any paiticular weight and the centre of gravity may be upon
different sides of the centre of span and at equal distances therefrom ; then if the
whole load be on the span and the weight be in its <iwri field, calculate the bending
moment at the point where the weight stands tor the load fixed in this position,
and it will be the maximum at the point. For each weight which can be placed
so as to fulfil tliese conditions, there is a maximum, and having calculated these
maxima, the greatest is the maximum for the whole span.
In Ex. 1, the point x = 8 lies in the 3rd field ; place the load so that Wz is
over it, and it will be tbund that all the wheels are on the span; calculating the
bending moment at a; = 8 with the load in this position, we have
3M8 = 220 ft. tons maximum at point.
The same result may he obtained by substituting a: = 8 into zMx.
Again, the jjoint 10 lies in field 3, but when Wz is placed over it, W\ is not on
the span ; moving tlie load two feet to the right brings W\ just <in the span, while
Wz is at 8 ; that is, the load is in the veiy same position as previously. Calculat
ing the bending moment at a; = 10, with the load in this position, we have
J/'io = 190 ft. tons maximum at point.
The same result may lie obtained by substituting a; = 8, a!id x = 11, into zMx,
and then adding onethird of their difference to the smaller.
In Ex. 1, we find that hy placing Wz at the point x — 2, it is in its own field
while G is at the point .r = — 2, and the whole load is on the span ; calculating tlie
bending moment at 2 for the load in that position we have zMi = 256. Again, by
placirg Jf'i at the point x = — 3, it is in its own field while <? is at a = 3, and the
whole loa<i is on the span ; calculating the bending moment at  3 for the load in
that position, we have (.see example, fig. 1, p. 109) 4JI/.3 = 261. Now since lu) other
weifilit can be i)laced according; to the rule, it follows tliat 4J/.3 = 261 ft. tons is
the maximum for span not only for the position on that figure, but for all possible
positions of the load on tlie span.
• These Kiiles were first piibli.hed in " En<;iiicering,'' Jan. 10th and July 26th,
1879, previous to the conception of our grapliical constructions.
CHAP. X.] B. M. AND S. V. — r,OCOM()TIVE. 197
0. Find the maxima bending moments at tlie junctions of the " fields."
That is, tind the heights of tlie points Di, 1)2, &c., on tig. 2.
FiBi = i.lfie = 100 ft.tons ; FiDz = •.Vu = 175 ft. tons ;
F3D3 = i^fo = 252 ft.tons ; F^Ih = iM_io = 180 ft.tons.
Where 1.1/^16 means the hending moment at the point 16 feet left of centre when
the first wheel is ovei* it ; 2/'Ai tlie bending moment at 11 feet left of centre when
the second wheel is over it, &e.
The eciuation to the dotted parabola BAoO is
^^ {c"  X) = C2V  x^'),
so that the ordinates at Fi, Fz, F3, and F4 to this dotted parabola are to be found
by substituting for x the values 16, 11, 0, and  12, respeciively. They are there
fore 185, 320,^441, and 297 ft.tons.
Hence the depths of the points 7), !)■•, T)^, and D^ (which are the junctions of
the " fields '') below the dotted parabola got by subtracting are 85, 145, 189, and
117 ft.tons.
It will now be seen that the depth of T)\ below the dotted parabola is
Wi . 2hi = 5x17= 85 ;
of Da is JFi . 2/(1 + IF. . 2hi = 5 x 17 + 5 x 12 = 145 ;
of 7)3 is /Fi . 2/«i + lF> . 2/(0 + JFz . 2/(3 = 145 + 11 x 4 = 189 ;
and of 7)4 it is Jr, . 2/h . . . TFi . 2/(4 = 189  12 x 6 = 117.
A general proof can readily be made of this remarkable theorem. That the
junctions of the parabolic arcs (fig. 2) lie on vertiial lines dividing the span in
segments which are proportional to the individual load on the wheels, and that the
depths of any one of these junctions I) below the parabola which is t*ie bending
moment diagram for the wliole load rolling on one wheel are given by 'S,[W . 2A)i'",
that is by the geometrical moment of the weights about their centre of gravity from
the first up to and including all those that command fields left of that junction.
A beam 20 feet span is subject to the transit of 5 weights, each 2 tons and
fixed at intervals of 3 feet. Find the maximum bending moment at each point
during the transit.
All the loads may be on the span at once, therefore n — b\ since the common
interval is less than a fifth of the span, the locus is that for the whole load on the
span (fig. 4). For the left half of span
biv
iM^= — (c + x  2s){c x) = h(i: + a;)(10  x),
2c
for values of x from 10 to 6.
2.1/";, = i (10 + a;  3)(10 a)2x3 = i(7+ x){lO  x)  &,
for values of x from 6 to 2.
iMx=h (10 + a;  0) (10  a.)  2 x 6  2 X 3 ^\{10  x)  IS,
for values of x from 2 to 0.
By symmetry the values for the right half of the span may be obtained, and the
maximum for span during transit is zMq — 32 ft.tons.
198 APPLIED MECHANICS. [CHAP. XI.
CHAPTER XI.
ON THE GRAPHICAL CONSTRUCTION OF MAXIMUM BENDING
MOMENTS ON SHORT GIRDERS DUE TO A LOCOMOTIVE.
Much attention has recently been given to the bending effects
upon bridges of short span, due to the concentration of the
loads on the wheels of locomotives. The subject assumes
special importance on account of the everincreasing weight of
the rollingstock. In a paper by W. B. Farr, read before the
Institute of Civil Engineers in 1900, the subject was discussed
at great length, as at that time the Board of Trade had
required all the railway companies to strengthen their bridges.
Farr, in his paper, contends that a period has arrived when the
weight concentrated on any wheel cannot further be increased,
so that any further increase of weight must be spread over a
longer wheelbase and a greater number of wheels. This
would, he held, narrow the problem to that of finding the
maximum bending moment for a span accommodating a loco
motive and tender, and a like moment for a series of decreasing
spans accommodating portions of the locomotive. Then, for
each span, an equivalent uniform load was calculated and the
rate of this uniform load tabulated for use in designing. Much
of the discussion on this papti turned upon the question as
to what was the best method of estimating such an equivalent
uniform load.
Each type of locomotive and tender gave a special table of
its own ; and these tables were given in the paper for a large
series of locomotives, and were amplified to allow for shock and
other important practical considerations, again leading to much
important discussion.
It was suggested, too, that each railway engineer might use
a table derived from a hypothetical locomotive, which was an
average of their actual passenger engines.
In calculating the bending moments for each span of one
series, Farr used Culman's original method of drawing a link
polygon for the locomotive standing still, then moving the span
about within the polygon, and when in a promising position
projecting its ends up to the polygon, which is then closed by
CHAP. XI.] LOCOMOTIVE ON SHORT GIRDERS.
199
an oblique chord, having the span for its horizontal projection.
The highest ordinate of this closed polygon is scaled off, and
gives a maxinmm bending moment for that particular position
of the span placed under the locomotive or under some portion
of it. Taking a number of those positions, by a sort of trial and
error, an approximate value of the maximum of maxima is
obtained. As this method is laborious and not quite certain,
Farr thought of trying the authors' method, which required the
use of a parabolic setsquare, but found a difficulty in adapting
it to the continual change of span.
In the correspondence on Farr's paper the authors proposed
their method of using circular arcs, and received inquiries as to
that method from engineers both at home and in the colonies.
For the purposes of this chapter the ideal locomotive weigh
ing 42 tons is adopted, having its weight divided among five
wheels as shown on the under line, while the spacing of the
wheels is shown on the upper line, thus —
5
5 5
10
11 12
= 30 feet.
9 = 42 tons.
In the first instance a span of 42 feet is chosen, as it greatly
simplifies the description of the graphical constructions to have
the span and total load given by a common number. The centre
of gravity of the locomotive falls in the 10foot space, and is
6 feet from the 12ton wheel and 4 feet from the 11ton wheel.
The locomotive is shown both on fig. 1 and fig. 2 standing in
200 APPLIED MECHANICS. [CHAP. XI.
the same particular position. On fig. 1, the positions of the
wheels are given left and right of 0, the centre of the span.
An instantaneous bending moment diagram ABCDEF is drawn
to scale for that particular position of the locomotive.
On the other hand, fig. 2 gives the position of the wheels
left and right of G, the centre of gravity of the load. By
inspecting these two figures it will be seen that the wheel
Wx= 12 tons, which we will call the rulinrjvjhed, since under
it, as will be shown subsequently, the greatest bending moment
occurs, lies ^4 = 3 feet to the right of the middle of the span,
while G lies .3 feet to its left. On fig. 2 the bending moment
under the ruling wheel is given to scale by Sx At, so that A^ is
identical with the apex D on fig. 1. We now contemplate
moving the locomotive left and right of the position on the
figures, so as to find the locus of the apex D on fig. 1. It is well
at this stage to conceive the girder as extending some distance
beyond each support as shown on the model, fig. 10, at the end
of the chapter.
On fig. 2 the locomotive stands with the rulingwheel at a
distance (21  3) = 18 feet from the right abutment, while the
centre of gravity G^ is 18 feet from the left abutment. Since
the load and span have the common value of 42, therefore the
pushup at the right abutment is 18 tons, because G is 18 feet
from the other abutment. The bending moment under the
rulingvjheel is 18 tons multiplied by 18 feet less 63 foottons,
the product of the next wheel load 9 tons by its distance 7 feet
from the rulingwheel. The height of D, fig. 1, and of Ai, fig. 2,
is
tons. ft.
4if_3 = 18 X 18  63 = 261 foottons.
The subscript figures meaning that the fourth wheel stands
3 feet to the right of the middle point of the span. If the
locomotive moves one foot either to the left or right, one of the
" eighteens " in the above product becomes seventeen and the
other nineteen, and
tons. ft. tons. ft.
M.2 = 19 X 17  63 or ,]\L, = 17 x 19  63.
These are the common height of the locus of D, fig. l,at one
foot left and right of its figureil position according as the ruling
wheel Wi arrives at the one or other point. Hence Ai, fig. 2, is
the vertex of a parabolic right segment, its halfbase being 18,
and its height, 18 x 18 ; but placed with its vertical axis 3 feet
CHAP. XL]
LOCOMOTIVE ON SHOUT GIRDKR.S.
201
to the right of the middle of the span and with its base lowered
63 units below BC, the base of the bending moment diagram.
It will be seen that the point 6 feet to the right of the middle
and the middle point of the span itself are equidistant about
the axis of the parabola, so that the bending moment is the
VERTICALS.
1 1C0
HORIZONTALS
Fij)
same at the two points when the wheel Wi arrives at them
respectively.
To find the locus of the apex C, fig. 1, the locomotive should
be shifted till the wheel W^= 11 tons is two feet to the left of
202 APPLIED MECHANICS. [CHAP. XI.
the centre, when C will coincide with A^, fig. 2 ; and the height
S^Ai will be calculating from the left end,
tons ft.
3il/. = 19 X 19  (5 X 8 + 5 X 13) = 256 foottons.
For a movement of one foot left or right the moment has
the common value
tons. ft. tons. ft.
,J/3 = 20 X 18  105 and 3J/, = 18 x 20  105.
So that ^3 is the vertex of a right parabolic segment ; half
base, 19 ; height, 19 x 19 ; placed with its vertical axis 2 feet
to the left of the middle of the span, and with its base lowered
105 units below the base of the diagram.
The apexes A, B, C, D, and E of the instantaneous bending
moment diagram, fig. 1, describe the five parabolic loci whose
vertexes are Ai, Ao, A3, A^, and A^, shown on fig. 2. These have
their five vertical axes placed about the middle of the span in
the same way that the five wheels are placed about the centre
of gra\'ity of the locomotive, but at half the distance in each
case thus —
0S\ OSn OS3 OSi OS5
85 6 2  3  6i feet.
The dotted parabola BA„C standing on the span is the locus
of the apex of the triangle which is the instantaneous bending
moment diagram for a single load of 42 tons rolling on the
42foot span. This triangle is scalene for every position of the
single load, except when the single load is at the middle of the
span, when the triangle is isosceles, and its height is then
OA^ = \ load X .span = 21 x 21 = 441 foottons.
We now see that all six parabolic loci have this in common
that each parabolic right segment has its height numerically
equal to the square of its half base. If a teynplate of the
parabolic right segment BA^C be made, then the other five
segments are drawn from portions of that template. If the
template be only a quadrant OAX', it is then called i\ pambolic
setsqnarc. A parabolic template, such as BAX', whose height
OAo is 441, the square of its half base, which is OC = 21, is
said to be of morJulvs unity ; and, further, the lieight of the
template at any other point is the product of the two segments
of the base BC into which it is divided at that point. Such a
CHAI'. XI. I LOCOMOTIVE ON SHOKT GIRDERS. 203
parabolic template or setsquare would be far too lofty for actual
use ; but any template or setsquare of convenient proportions
can be used by employing two scales — one for horizontal or feet
measurements, and another for vertical or footton measure
ments. The modulus of the template for drawing the six
parabolic loci in tig. 2 is unity, because the ratio of the load to
the span is unity. Had the locomotive only weighed half as
much, say 21 tons, then the modulus must have been onehalf;
still the same template would serve by employing a new
vertical scale for foottons — in that case twice as coarse.
We may now consider the point Di on fig. 2 where the
fourth and fifth parabolas intersect. At the point F^ the
bending moment will have a common value F^D^ whether the
locomotive be moved so as to bring Wi or W„ to that point.
If we put Z for BFx in the first case, the distance from B to G
will be (Z  13) feet so that the pushup of the abutment at C
is (Z  13) tons and the bending moment at Ft is iZ  13)
(42  Z) foottons. In the second case the distance from £ to
G is {Z  6), and the bending moment at F^ is (Z  6) (42  Z)
 9 X 7. Equating these, gives ^ = 33 feet, which is numerically
equal to the sum of the loads on the first four wheels. In this
way (Ex. 1, p. 194) it is shown that the first parabola is above
all the others for the first 5 feet of the span, the second
parabola for the next 5 feet, the third parabola for the next
11 feet : that is, each wheel commands a portion of the span or
a " field," which is the same fraction of the span that the load
on that wheel is of the total load. As the load and span have
been taken numerically equal, the five fields into which the
span is divided are 5, 5, 11, 12, and 9 feet respectively. At
any point in any field the maximum bending moment occurs
when the commanding weight is over that point. In the two
fields, one on each side of the middle of the span, there is a
maximum of the maxima SzA^, 2 feet to the left of the middle
of the span, and SiA^, 3 feet to its right. Or in symbols
■iM, = 256 foottons and iM_^ = 261 foottons.
As the second of these is the greater, we have called the
wheel Wi = 12 tons the rulingivheel of the locomotive when
riding on a girder or bridge of span 42 feet.
To find an uniformly distributed load which will give the
same bending moment, 261 foottons, at the same point, 3 feet to
the right of the middle of the span, we can assume a parabolic
locus like BAqC standing on the span, but passing through Ai.
204
APPLIED MECHANICS.
[chap. XI.
This locus representing the bending moments due to an uniformly
distributed load of intensity v.' tons per foot, gives if we put
c = 21 the half span,
SU* = ^^xBF, xCF,= '^(c + 3)(c  3);
and equating this to 261, we get vj = 1209 tons per foot of
span.
On tig. 3 we now show a method of drawing the five loci
without considering the junctions of the fields. It does not
give so good a definition, but is instructive. Thus, the locomo
tive is to be fixed with its centre of gravity over the middle
point of the span, and the instantaneous bending moment
polygon BahcdeC drawn to the vertical scale, upon which the
Fig. 3.
height of A» reads 441 foottons. Now on page 201 it was pointed
out that the locus whose vertex is Ai has equal heights at
6 feet to the right of the middle point of the span and at the
middle point itself. But on fig. 3 the wheel JFi is 6 feet to the
right of the middle of the span. So that the perpendicular
from d on fig. 3, dropped on the vertical line through the
middle of the span, is the base of a small part of the locus
whose vertex is At, which can be drawn with the template. In
the same way perpendiculars on the middle vertical line from
a, b, c, and e give bases to guide the template to draw parts of
CHAl'. XI.] LOCOMOTIVE ON SHORT GIRDERS. 205
the other four loci. Note that to the common height of caiul d
it is only necessary to add the square of the half of the small
bases, thus : 252 + 4 = 256, and 252 + 9 = 261 are the heiglits
of A3 and Ai respectively.
The height at F^, the junction of the fourth and fifth
fields, to the dotted parabola BA^fi, fig. 2, is the product
BFi X CFi = 33 X 9. In a preceding paragraph we had
FiDi = (33  13) X 9 ; hence the height from A to the
dotted parabola is (9 x 13), that is, the load on the wheel
J/'j = 9 tons multiplied by its distance from G, the centre of
gravity of the locomotive. In the same way it is found that
Os^o is the sum of the moments of the two weights W^ and
Wi about G. Generally the depths of the junctions of the
parabolic arcs in pairs below the dotted parabola are given by
the moments of the weights of the wheels about their common
centre of gravity. The numerical value for i?i is 5 x 17, for
Di it is greater by 5 x 12, for A it is still greater by 11 x 4,
while for Di it is lesser by 12 x 6, calculating them in order
from the left end. Depths from the dotted parabola BAf^C,
due to the 42 tons rolling on one wheel, to the junction of the
arcs of parabolas giving the locus due to the 42ton locomotive
crossing the 42foot span, are (Ex., p. 197) : —
At Z>, n. D; D,
85 145 189 117 foottons.
On fig. 4 are shown the two parabolic right segments
BAoC standing on the span, and c'Ajj', the arc BjAiB^ of
which is the locus of the maximum bending moments for the
fourth field commanded by Wi. Let us suppose (see the dis
tortingtable, fig. 17, p. 177) c'mnh' to be the end of a pack
of cards stacked vertically into a rectangle, and having the
right parabolic segment painted upon it, each vertical hatch
ment being on the edge of a card. The pack is then to be
distorted and packed into the paiallelogram cmnh, and lifted
up till Dz coincides with A^, and D^ coincides with di. We
shall then have the arc of the oblique parabolic segment
CA^ttidJ) coinciding at every point with the arc of the segment
BA^C. Also F will be above at a height DsAq =189 foot
tons, and F will be above F^ at a height Didi =117 foot
tons. Hence FF is an oblique base whose horizontal range
is OFi =12 feet, the extent of the fourth field, and the
vertical ordinates from that oblique base measured up to the
segment BAqC give the maximum bending moments at each
206
APPLIED MECHANICS.
[chap. XI.
point of the fourth field individually as W^ comes over it.
The maximum of these maxima is given by StC/i = SiAt = 261.
foottons.
If each of the five ritrht parabolic segments on fig. 2 be
distorted and lifted up in a like manner, we then have, on
fig. 5, only one parabolic right segment of the height 441 foot
tons standing on the span as a base, and a polygon ACDEFB
standing on the same base, and on the same side of it, having
SA^Sa =26] futons.
its four apexes on the lines dividing the five fields from each
other. The heights of these apexes are : —
G
85
D
145
E
189
F
117 foottons,
being the same as the depths of the junctions of the arcs i),,
D^, 1)^, and D^ below the dotted parabola on fig. 2.
Fig. 4 furnishes a neat geometrical proof of the heights of
the apexes G, D, E, and F. Consider the quadrant Aiiu ; its
base is 3 feet, being half the distance of the wheel W^^ 12
tons from the centre of gravity of the locomotive. The height
oi the segment is 1» foottons, being the square of its lialf base.
OHAP. XI.] LOCOMOTIVE ON SHORT GIRDERS. 207
Now, the vertical through A^) meets the tangent m^i at a point
twice as high above the base. Hence nui slopes at an angle to
the horizontal, whose tangent is twice the square of the half
base of the segment A^a^ divided by its half l)ase. So that
the tangent of the slope of EF to the horizontal is given
numerically by the distance of Wi, the 12ton wheel from
tiie centre of gravity of the locomotive. To get the amount
that E is higher than F it is only necessary to multiply OFt
by this tangent, when we get the product 12 x 6 = 72 foot
tons. In the same way the height of F above BC is the
tangent of the slope of FC, that is, the distance of the wheel
TVs = 9 tons from the centre of gravity multiplied by FiC
when we have the product ofl3x9 = 117 foot tons.
Then the maximum bending moment at each point of the
span is given by the vertical height from the polygon to the
parabola. The maximum of maxima in any field — say, the
fourth field — is to be found by producing FF to meet the
parabola at b and c, then he is to be bisected at the black spot
where the height to the parabola gives 261 foottons, the
maximum of maxima ; provided that the bisecting point falls,
as it does, on the side J^JF, and all the wheels are on the span.
In the same way for the third field, the height at the centre
of the chord of the parabola given by 1)F produced (fig. 5) when
measured vertically to the parabola gives 256 foottons. For
the other three sides of the polygon produced to give chords of
the parabola, the bisecting points do not fall on the sides.
Observe, too, that the black spot bisecting the chord cFFb falls
3 feet, measured horizontally, to the right of the vertical
through the middle of the span, so that the graphical diagram,
fig. 5, gives the maximum of maxima for the 42 ton loco.,
crossing the 42foot girder, to be
4if_3 =261 foottons.
that is, when the fourth or 12ton wheel stands 3 feet to the
right of the middle point of the girder.
The polygon ACDEFB, which is mechanically subtracted
from the parabolic locus on fig. 5, is the bending moment
diagram for four fixed forces acting upwards at the junctions
of the fields ; 5 tons at C, 8 tons at D, 10 tons at E, and
7 tons at F, as these forces would give the moments 85, 145,
189, and 117 foottons of negative bending at those points.
Generally, then, the polygon is that due to a set of upward
fictitious forces at the junctions of the fields whose magnitudes
208 APPLIED MECHANICS. [CHAP. XI.
are numerically the same as the distances between the wheels
muUiplied by the ratio of the iveight of the locomotive to the
length of the span.
First Graphical Method with one Parabola only.
For the 42ton locomotive on fig. 5, after finding the centre
of gravity G, choose the smallest span which will accommodate
the loco, standing with G and the rulingwheel W^ equidistant
from the two abutments. That span is 40 feet ; but we have
taken 42 feet, which is only slightly greater, and makes the steps
of the construction more evident, and simplifies any arithmetical
checks.
Gonstructio7i, fig. 5. — Lay down the span AB = 42 feet on a
suitable scale of feet, and divide it into f\\& fields proportional to
the weights on the wheels of the locomotive. On the left side
draw the loadline equal to 42 tons, the weight of the locomotive
to a scale of tons judiciously chosen, so that with a polar
distance of 10 feet a wellconditioned isosceles triangle is
formed. Join the pole to the ends of the loadline. Construct
the apex of an isosceles triangle standing on the base AB, and
having its sides parallel to the extreme vectors from the pole.
The height of this apex above the middle of the base should
measure 441 foottons, being a fourth of the product of 42 tons
aud 42 feet on a scale ten times finer than the ton scale, the
polar distance being ten. Construct the scale of foottons by
renumbering the ton scale accordingly. Taking the apex of
the isosceles triangle as a vertex, construct the parabolic right
segment standing on the span AB. Construct the points close
together near the vertex, but more sparsely well out from it.
A construction is shown, fig. 3, Ch. VI. It will be seen at this
stage that the shape of this parabolic segment depends on the
choice of scales.
To construct the polygon ACDEFB, draw vertically npnards
the lines of action of iowv fictitious forces through the junctions
of the five fields. On the right side draw upwards tlieir load
line 7, 10, 8, and 5 tons, being numerically the distances between
the wheels multiplied by unity, the ratio of the weight of the
locomotive to the length of the span. To a trial pole draw the
dotted linkpolygon among the four upward forces and the two
holdingdown forces at A and B. A vector from the trial pole
parallel to the closing sides gives the junction between the
magnitudes of the two holdingdown forces. From this junction
lay oil" horizontally 10 feet for the true pole, and construct the
CHAP. XI.] LOCOMOTIVE ON SHOKT GIRDERS.
209
210 APPLIED MECHANICS. [CHAP. XI.
linkpolygon ACDEFB. Otherwise the height of F may be laid
up directly to scale as 9 tons x 13 feet = 117 foottons, being the
product of the 9ton wheelload and its distance from G. Then
the height of ^ is 189 foottons, being 12 x 6 greater than the
last, while the height of D is to be 145 or 11 x 4 less than the
last, also the height of C is 85, or 5 x 12 lesser again. In this
way by taking the moments of the weights of the wheels about
G, their common centre of gravity, the polygon ACDEFB may
be plotted, or, being drawn as first described, it can be checked
in this way. Still otherwise, if the parabola be engraved on the
paper or drawn first of all with a celluloid template. The span
AB is, io be placed on it and divided into fields. The fictitious
forces on the right are now drawn upwards to a scale for which
the height of the vertex is 441. Now from the junction of the two
holdingdown forces lay up onehalf of the span and lay down
the other half, and from tlie two ends draw vectors parallel to
the two sides of the isosceles triangle on the base AB and with
its apex at the vertex. These will meet and determine the polar
distance and the pole.
To scale off the Maximum of Maxima..
Produce EF, the oblique base of the fourth field, to meet the
parabola at c and d, bisect cd at the black spot, and scale off the
vertical height to the parabola, which find to be 261 foottons;
also find the horizontal distance of the black spot from the
vertical through the middle of the span to be 3 feet, and we
have
iMz = 261 foottons.
A rival maximum, to be found by producing BE the oblique
base of the third field, gives
3M2 = 256 foottons.
For the other three chords AC, CD, and BF, the bisecting
points do not fall in the fields. The fourth weight W^ = 12 is
the ridingicheel, and the symbol ^M.3 means the bending moment
at 3 feet to the right of the middle of the girder when the loco
motive stands with its fourth wheel at that point, provided all
the wheels of the locomotive are actually on the girder.
To deal with shorter spans which only accommodate a part
of the locomotive.
Diop ofl' the 0ton wheel at the left end of the locomotive,,
and drop ott' 5 feet from the left end of the span. Joining B to C
CHAP. XI.] LOCOMOTIVE ON SHORT GIRDERS. 211
gives CDEFBC, the polygon for the reduced span of o7 feet.
But G is to be projected up to 6" and tlie polygon completed on
the oblique base C'B. Next the side corresponding to EF is to
be produced both ways to h' and d, then h'd is bisected at the
black spot, and the vertical height to the parabola scaled ofi"
214"o foottons. Also the horizontal distance of the black spot
from the open ring at the middle of the oblique base C'Z? should
scale 1"85 feet, being half the distance of the rulingwheel,
12 tons from the centre of gravity of the group of wheels
5, 11, 12, 9 tons.
In like manner another wheel 5 tons is dropped off the loco
motive, and 5 feet is taken off the span. Joining D to B gives
DEFBD, the polygon for the reduced span of 32 feet. Project
D up to D' on the parabola, and D'B is the oblique base. Com
plete the polygon ; produce the side corresponding to EF both
ways to meet the parabola. Bisect this chord at the black spot,
and scale off' the height to the parabola 170 foottons. Also find
the horizontal distance of the black spot from the ring at the
middle of the oblique base to be "734 of a foot.
Further, removing the trailing wheel 9 tons, we have
remaining the two drivingwheels 11 and 12 tons. Removing
9 feet from the right end of the span, then DEED is the polygon
for the span, 23 feet. Project both D and F up to the parabola
at D' and F', then D'F' is the oblique base. Complete the
polygon, and produce the side corresponding to EF to df ; bisect
F'd' at the black spot, and scale off the vertical height to the
parabola as 83 foottons. Scale off' the horizontal distance of
the black spot from the ring at the middle of the oblique span
F'd', and find it to be 24 feet.
To find the equivalent uniform rate of loading to give the
maximum bending moment on each span at the same point of
the span, we must double the moment, and divide by the
difference of the squares of the halfspan and the displacement
from the middle of the span of the point at which the maximum
bending occurs.
Span. Moment. Displacf.ment. Rate of Load.
Feet. Foottous.
42 261
37 2143
32 170
23 83
It will be seen that the rate of loading on the 23 foot span
p2
Feet.
Tons per foot.
300 .
1209
185
1265
•73
1331
240
1312
212 APPLIED MECHANICS. [CHAP. XI.
is less than that on the 32foot span. But the rate should
constantly increase as the span decreases. By inspection it
will be found that the two drivingwheels, 11 and 12 tons,
can be accommodated on a span of 15 feet, instead of
23 feet.
To interpolate spaiis, slightly smaller or larger than those given
hy dropjnng ojf parts of the span proportional to the loads mi the
wheels dropped off the locomotive.
Thus to determine the max. bending moment for a span of
20 feet loaded in the most trying way by the group of wheels
11, 12, 1> tons. Consider the parabolic segment and polygon
standing on the oblique base B'B, the horizontal projection of
which is 32 feet. By adopting a coarser scale for feet we can
make this horizontal projection measure 20 feet instead of 32.
Now, however, the height of the parabolic segment must also
be measured on a coarser vertical scale in the like ratio assumed
to save the trouble of redrawing the parabola. But the
polygon must be redrawn. On fig. 5, then, the rulingside of
that polygon is shown by a hatched line with its two ends set
up higher from the oblique base D'B' in the ratio of 32 to 20,
and produced each way to h and Iz ; then hh is bisected by a
black spot the height from which to the parabola measures 119
on the old scale. This is to be decreased in the ratio 20 to 32,
when we have 74*4 foottons, and the horizontal distance of the
black spot from the ring at the middle of the oblique base D'B
measures on the old scale 1 19, but when altered in the ratio of
20 to 32, we get 734 feet.
Again, if the original locomotive is to ride on a span of
50 feet, instead of 42 feet, it is only necessary to lower the side
of the polygon EF in the ratio of 42 to 50, shown by a hatched
line ; produce it each way to m and n ; then bisecting mn in the
black spot, and reading the height to the parabola, we have
2886, which is to be increased in the ratio of 50 to 42, giving
34356 foottons.
Also the horizontal distance of the black spot from the
middle of the span measures 252, which, increased in the ratio
50 to 42, brings it to 3 feet.
In this way the shortest span to accommodate any group
of wheels may be interpolated. It will require in general two
trials, for at the first trial we may find when the deviation of
the point of max. bending from the middle point of the span is
measured, that placing tlie ruling wheel, there sets an end
CHAl'. XI.] LOCOMOTIVE ON SllOKT GIKDKKS.
213
wheel oH' the span altogether. We have then to slightly
increase the span and proceed again.
Span.
MOMKNT.
D
SPLACEMENT.
Rate of T,oad.
Feet.
Foottons.
Feet.
Tons per foot.
20
744
•734
1495
50
34356
3
1115
Second Graphical Method with Circular Arcs only.
We have defined a parabolic right segment of modulus unity
as having the height at any point nvimerically equal to the
product of the two segments into which the point divides the
^c
iiTz!!.
Fig. 6.
base. And we have also pointed out that it represents the
maximum bending moments for the transit of a single rolling
load, the height of the segment being made to scale in foot tons
an amount equal to onefourth the product of the load and the
span. Now, by Euclid II, 14, a semicircle will serve as a
diagram of the square roots of the maximum bending moments
in the same case to a vertical scale upon which the height of
the crown of the semicircle shall measure the square root of the
above product. Compare fig. 2, Ch. X, and Theorem H, Ch. VI.
If we suppose every vertical height on tig. 2 to be replaced
by a height equal numerically to its square root, we will have
the diagram fig. 6, all the parabolas becoming circular arcs with
centres at the points S^, S^, &c., as shown on fig. 6, each arc
beginning on the vertical line through the junction of the fields
214 APPLIED MECHANICS. [CHAP. XI.
where the last arc ended. On fig. 6 the vertical scale is the
same as the horizontal scale because of the load and span being
numerically equal. The height of oti will scale 16"16 parts on
the scale upon which the height of the crown of the semicircle
scales 21. Then
iM.^ = (1616) = 261 foottons.
On fig. 7 is shown the 42ton locomotive standing on the
42foot girder. On j&g. 8 the span is divided into five fields
proportional to the loads.
Then the centres for the arcs are set off about the middle of
the span at half the distances at which the wheels stand from
the centre of gravity of the locomotive.
The centre of gravity of the locomotive is defined by Gn on
fig. 7, where the end links of the linkpolygon meet, the link
polygon being drawn to a polar distance of 10 feet. The five
circular arcs are then drawn on fig. 8, each beginning at the
junction of a field where the last arc ended. The vertical scale
is made so that the height of the crown of the semicircle shall
scale 21. Next the crown of the circle in the fourth field is
ruled over to the scale and reads 16*16, and this when squared
gives
,M_^ =261 foottons.
The front wheel, 5 tons, is now dropped off when Gy„ fig. 7,
defines the centre of gravity of the remaining four wheels ; from
Ga a perpendicular is dropped on the vertical from the ruling
wheel 12 tons ; this perpendicular is bisected, and the half scales
1*85 feet. On fig. 8 the new span, with the 5 feet at the left
end left off, is bisected at the ring, and the centre for the arc
corresponding to the wheel, 12 tons, is laid at 1"85 feet to the
right of the ring, and the centres for the other three arcs spaced
relative to it. The arcs are then drawn for the second, third,
fourth, and fifth fields, and the crown of the arc on the fourth
field is ruled over to the scale where its height reads 14*64 ;
and squaring this, we have M = 214 foottons, a maximum, at
1'85 feet to the right of the centre of the 37foot span when
the 12 ton wheel stands over it.
Dropping off the second 5ton wheel we have the centre of
gravity of the remaining three wheels defined at Gxi ; and the
perpendicular from this point upon the vertical from the
rulingwheel, 12 tons, gives, when bisected, 0*734 feet. This
distance is laid off to the right of the ring marking the middle
of the 32foot span and the two remaining centres placed about
CHAP. XI.] LOCOMOTIVK ON SHORT GIRDERS. 215
it as before. The three arcs are then drawn, the highest crown
ruled over to the scale, where it leads 13"04. Squaring this we
get M = 170 foottons, a maximum, at a point 734 feet to the
right of the centre of the 32foot span w[ien the 12ton wheel
stands over it.
In the same way fig. 8 shows the solution for various spans
loaded with various groups of the wheels of the locomotive ;
and a table is shown giving the equivalent uniform loading.
Third Graphical Method, being Culman's Method
rendered precise.
Only our original construction, fig. 2, drawn with a para
bolic segment, and the diagram, fig. 8, of circles derived from it
show which is the ruling ivheel. For this purpose it would be
well to draw the diagram, fig. 8. in conjunction with Culman's
method. Hence we have placed fig. 8 under fig. 7, which is
Culman's method rendered precise, so that there is no searching
about by trial and error.
Lay down on fig. 7 the loadline 5, 5, 11, 12, 9 tons. Choose
a polar distance, some round number — say, 10 feet. The two
scales, one for tons and the other for feet, should be so related that
the two endvectors may meet at the pole at a wellconditioned
angle. Draw the vertical lines of action through the wheels of
the lf)Comotive, and among them draw the linkpolygon.
Produce the two endlinks to meet at G^i, which determines
the position of the vertical through the centre of gravity of the
locomotive. Either the 11 or the 12ton wheel will be the
rulingwlieel. By drawing the semicircles on fig. 8 and the locus
of circular arcs, we at once find the drivingwheel carrying
12 tons to be the rulingwheel.
From Gil drop a perpendicular on the 12 ton load. Bisect
the perpendicular, and from its middle point lay off 21 feet
horizontally on each side, that is, half the span. Project the
two ends down on to the two endlinks, and draw the oblique
base closing the polygon. This closed polygon is the instan
taneous bending moment diagram when the locomotive is
standing in the most trying position, that is, with the 12 ton
wheel 3 feet to the right of the middle of the span. Scale
off the depth from the apex on the 12ton load down to the
oblique base, and find it to read on the ton scale 261 ; multiply
by 10 feet the polar distance, and we have ^M,^ = 261 foot tons.
Drop off the 5 ton wheel from the left end of the locomotive,
and leave off 5 feet from the left end of the span. For the
o oof oos ooe oo* suoxiJ'
a or OS OP ot tiuoj,
n I I M I I I I n I I I H I I I I I I I I H I I I 1 M I I » I I TTTR
'""IV }""3 "I'fU ^S "■'/" ' '"J
.£1 QL SL
■S 4
^
5 "^
s
a.
5S
•o
S 5
•n
fl
_g>
• II
\
II
II
•J o
1
s
s
:
:
~
•\
•§. °
:S
~
'


fci 5
• i
fe
>.^
^
>
*
»
i
*
0
i «
f.
«
e
^
1
i5
3
^
ui S
(A <
■yf;MDJ£)
yb >Jiu?Q uowiuo} jpyi viojf si; sj^sy^H Smpuods^JMj
inoqD p>3Djd }jlv sij y jpjnpjij it/t J'o suiusj »y_^
218 APPLIED MECHANICS. [CHAP. XI.
remainder of the locomotive the two endlinks meet at G3,^.
From this point drop a perpendicular upon the 12ton load, and
bisect it ; from the bisecting point lay off horizontally 18*5 feet
on each side. Project the ends down upon the two endlinks,
and draw the oblique base, closing a polygon which is the
instantaneous bending moment diagram for the remainder of
the locomotive standing on the 37foot span in the most trying
position, that is, with the 12ton wheel at a point 1"85 feet to
the right of the middle of the span. Scale ofi' the depth from the
apex on the 12ton load to the oblique base. Measure it 2142
on the tonscale, and multiply by 10 feet the polar distance,
and we get 211'2 foottons. In the same way fig. 7 shows the
maximum of maxima constructed for other spans and loads
obtained by dropping off some wheels from one or other end.
or from both ends of the locomotive, and taking off corresponding
segments from the original span. In some cases observe that
the 11ton drivingwheel becomes the rulingwlieel.
To find the maxmucm of maxima bending movients where there
is a given uniform load and tlie locomotive.
Eeturning to fig. 5, suppose that on the 42foot span AB
there are 16 tons uniformly spread, which would give a parabolic
locus the same as that for 8 tons rolling on one wheel, the
height of the locus being ^ x 8 tons x 42 feet or 84 foottons.
To allow for this it is only necessary to adopt a finer scale so
that the height of the parabolic segment in fig. 5 shall measure
441 + 84 foottons instead of 441. Adopting this finer scale
for verticals, it becomes necessary to lower the chord EF down
to the position shown hatched, that is, lower in the ratio of 42
to 50, just as we already noted when we passed from the
42foot span to a 50foot span. So that by bisecting mn at the
black spot, measuring the vertical height to the parabola on
the original vertical scale, we get 2886, but multiplying by the
ratio 50 to 42 it becomes 34356 foottons. Rut in this case
the original horizontal scale still obtains, so that the liorizontal
distance of the black spot from the middle point of AB is only
252 feet. In the former case, where mn was the oblique base
for a 50foot span bearing the locomotive only, this '52 had to
be multiplied by the ratio 50 to 42, which made a product
3 feet, as it must, for on any span that bears the locomotive
only, 3 feet to the right of the middle is the most critical pouit,
for then the rulingwheel, 12 tons, and the centre of gravity of
CHAP. XI.] LOCOMOTIVE ON SHOUT GIRDERS. 219
the locomotive are equidistant from the two abutments, or are
each 3 feet from the middle of tlie span. But in the case we
are now considering, the rulingwheel, 12 tons, must stand 2"52
feet to the right of the centre ; and it will be found that the
centre of gravity, not of the locomotive but of the load made
up of the 16 tons spread uniformly, together with that of the
locomotive, is 252 feet to the left of the middle of the span.
Mr. J. T. Jackson has pointed out that the similarity
between the above construction and that described on p. 208 at
once suggests that it must be possible to devise an ideal
locomotive which shall produce the same maximum bending
moments at every point of the span as are actually due to the
combined effects of the real locomotive and the uniform load.
To see how this may be done let us compare the modification
of the bending moment diagram in fig. 5 to allow for change of
span with that for the addition of an uniform load.
(1.) Change of Span. — The ordinates of the ^olygonACDEFB,
the unit of length and the unit of bending moment are all
altered inversely as the span, while the unit of load is
unaltered.
(2.) Addition of Uniform, Load. — The ordinates of the poly
gon AODEFB, the unit of load, and the unit of bending moment
are all altered inversely as the sum of the rolling load and half
the uniform load, while the unit of length is unaltered.
The effect on the fortn of the diagram in fig. 5 of an increase
of span from, say, 42 to 56 feet, would then be the same as that
due to the addition of an uniform load of 2 x (56  42) or 28
tons ; i.e. to a load of twothirds of a ton per footrun on a
girder of 42feet span. The effect on the scales would, however,
be different : in the case of the increase of span the scales both
of length and bending moments would be made finer in the
ratio of 42 to 56, or 3 to 4, while in the case of the addition of
the uniform load the lengthscale would be unchanged, while
the load and bending moment scales are made finer in the above
ratio.
It is evident that the diagram of maximum bending moments
for the 30foot 42ton locomotive of fig. 5 running on a 42foot
span which already carries an uniformly distributed load of 28
tons and drawn to scales of, say, 3 feet, 4 tons, and 48 foottons
to the inch, is identical in form and size with the diagram for
the same locomotive crossing a span of 56 feet, and drawn to
scales of 4 feet, 3 tons, and 48 foot tons to the inch. Now, on
changing the scales of the latter figure to 3 feet and 4 tons to
220
APPLIED MECHANICS.
[chap. XI.
the inch, it is seen to represent equally well the diagram of
maximum bending moment for a 22^foot 56ton locomotive
crossing a 42 foot span. So that the effect on the diagram of
maximum bending moments of the addition of an uniform load is
A SHORTER, HEAVIER IDEAL LOCOMOTIVE, EQUIVALENT TO
REAL LOCOMOTIVE
H AND THE DEAD LOAD
S 6 Tons.
\& 3^' rector cecc CMS 9~A\
7 >< . 3oFf. ^'"^
(^^^2 Tons,
Decffi Model
'^'TON
^^'PER FOOT.
Ol4)
Fig. 9.
the same as that prcjduced by nhorteniifKi the intervals between
the wheels in the ratio of the loco, weight to the sum of the
loco, weight and half the uniform load, and making the several
wheel loads Iteavier in the reciprocal ratio. This is illustrated
in fig. 9, where the derivation of the diagram of square roots of
CHAP. XI.] LOCOMOTIVE ON SHORT GIRDERS. 221
luaxiiuuui beudiug moments for the loco, of figs. 15 is effected
when the loco, is supposed to cross a bridge of span 42 feet
already carrying an uniform dead load of twothirds of a ton
per foot.
The method is useful as illustrating clearly the eftect on the
bending moment diagram of the addition of an uniform load ;
but it requires to be applied with some degree of caution, as it
must be remembered that the change in the character of the
diagram which takes place when a wheel goes off the span
occurs when a wheel of the real locomotive iroes off, and as the
ideal locomotive is shorter than the real, it might readily be
forgotten in examining a particular portion of the locomotive
that one wheel was off, since the corresponding wheel of the
ideal locomotive might be well within the span.
Kifiematical Model, demonstrating the variatioTis in bending
moment at all points of a girderbridge as a locomotive
comes across the bridge.
This model is specially designed to exhibit the manner in
which the diagram of maximum bending moments, consisting
of arcs of parabolas (as shown in fig. 2), is traced out by the
vertices of the linkpolygon corresponding to the loads on the
wheels of the locomotive, as the locomotive moves across the
span. The model (see fig. 10) consists of two parallel plates of
mahogany each 18^ in. x 15 in., the front or faceplate being
f in. and the back plate \ in. thick ; these plates are set I5 in.
apart. On the top of the plates is the model girder constructed
to a span of 42 feet on a scale of 4 feet to an inch.
Running on top of the girder is the locomotive, with a wheel
base of 1\ in. (to represent 30 feet on the assumed scale). The
girder does not terminate at the supports, but overhangs each
support by an amount sufficient to prevent any wheel of the
locomotive from running off the girder so long as the centre of
gravity of the locomotive remains between the supports. The
object of this is to avoid the change in the character of the
bending moment dia2;ram which would ensue if an^• wheel were
regarded as passing completely off the girder.
Passing round the frame and tightly stretched over four
small pulleys set on the back plate near the corners is an
endless chain. A stud projecting downwards from the loco
motive is attached to the upper horizontal side of this chain, so
that the chain carries the locomotive with it as it moves.
A detailed description, suitable for the modelmaker, is given
0')9
APPLIED MECHANICS.
[chap. XI.
in a paper to the Royal Irish Academy hy Mr. Jackson, assistant
to the professor of engineering in Trinity College, Dublin. He
Fig. 10.
designed tliis model modified from the Rim]dei' model described
at fig. 16, Chap. IX. The load and si)an being numerically
CHAP. XI.] LOCOMOTIVE ON SHORT (JIUDKRS. 223
equal, the span divided into fields servos as the loadline, turned
through ninety degrees. The rocking pivots are at the junctions
of the ^fields. Tlie chain moves the pole at the same rate as the
locomotive, but in an opposite direction. The wires (vectors)
from the pole rock the pivots which again turn the hands, one
directly above each pivot, at the junctions 2),, Z>2, &c., of the
parabolic arcs. In this way each hand is always at right angles
to the rocking wire, and so parallel to the thread (vector) in the
load diagram on the left.
These models were made by Messrs. Dixon and Hempenstall,
Dublin, through the liberality of the Board of Trinity College,
Dublin.
Examples.
1. Suppose with the weights in Example 1, Ch. X, and shown at fig. 9, we
combine an uniform load of rds of a ton per foot of span, that is, a total load
U = 28 tons, and find the equations to the maxima bending moments for the
various fields. See Ex. 4, Ch. Xyil.
Summing the moduli for R rolling and for ZT" spread uniformly, we have
M I7'_jr+2i?_28 + S4_4
2c "*" 4c ~ 4c ~ ~84 ~ 3'
the modulus of the parabolas for combined load.
The modulus for R rolling, divided by the sum of the moduli, is
\2c/ " \2c'^ ic)
2E 84 3
U+2R 28 + 84 4
Substituting into the general equation,
lifx = 1(21 +xly 17)(21 x) = f (825 + /)(21  x)
for values of x from 21 to 16.
oilfx = 1(21 +x^x r2)(21  a;) + 12 X 5  5 X 17 = (12 + .r)(21  x)  25,,
for values of x from 16 to 11.
sMx = 1(21 + a:  I X 4)(21  a:) + 4(5 + 5)  (5 X 17 + 5 X 12)
= A(18 t a:)(21  x)  105,
for values of x from 8 to 0, the chord being taken for values from 11 to 8.
4ylfx= A(21 +a; +  X 6)(21a;)6(5+ 5 + 11)  (5 x 17 4 5 x 12 + 11 x 4)
= A(225 + a;)(21  z)  315,
for values of a: from — 2 to — 12, the chord being taken for values from to — 2.
sifx  1(21 + a; + I X 13)(21  x)  13(5 + 5 + 11 + 12)
 (5 x 17 + 5 X 12+ 11 X 4 12 X 6)
= 1(3075 + a:)(21  x)  546, for values of x from  12 to  21.
224 APPLIED MECHANICS. [CHAP. XII.
The abscissae of the apexes are
a X 8.) = 6375 ; ^ x 6 = 45 ;  x 2 = 15, lies in field 3 ; J x ( 3) =  225,
lies in field 4 ;  x ( 65) =  4875.
Here only two apexes lie in the corresponding fields ; substituting for these,
we have
3M1.S = 402 maximum in field 3 ; and
4i^/.2.25= 40575 maximum in field 4 ; and maximum for span ;
the greatest bending moment occurs at 2^ ft. to the right of the centre, when the
fourth weight is over it. The joint centre of gravity is 2^ ft. left of centre.
CHAPTER XII.
COMBINED LIVE AND DEAD LOADS WITH APPROXIMATION BY MEANS
OF AN EQUIVALENT UNIFORM LIVE LOAD.
We already found that when the loads were all dead or fixed
the joint construction of the shearing force diagram and the
bending moment diagram recommended itself, the one aiding
in the construction of the other. The shearing force diagram
in itself, easy to draw, helped to construct the more difficult
bending moment diagram.
For moving loads we continued to discuss the two diagrams
together, although they no longer had much in common, owing
to the fact that the i^osition of the load for a maximum bending
moment and for a maximum shearing force at the same point
was not the same position.
In the combination of live with dead loads to be dealt with
in this Chapter matters are reversed, for the bending moment
diagram is easily dealt with, while the shearing force diagram
demands thorough investigation.
For practical purposes the dead load is to be reckoned as
of uniform intensity per foot of span. The live load is also
reckoned as uniform per foot of length, and as of sufficient
length to cover the whole span.
The actual live load, no matter how it may be disposed, in
positions concentrated on wheels regularly or irregularly spaced
is to be replaced by an equivalent uniform load. That is, an
uniform advancing load (at least as long as the span) which,
in its transit, produces the same maximum bending moment
CHAP. XII.] COMBINED LIVE AND DEAD LOADS. 225
at the central point of the span as the actual load produces at
or near the centre. It is to be noted that although irregular
load systems produce their maximum effect at a section a few
feet from the centre, yet this is covered by the fact that in
the practical design of girders the crosssections at the centre
and for a few feet on each side of it are identical. The
equivalent load also produces the same shearing forces at the
abutments as the actual load.
Again, it is necessary to augment the actual moving load
to allow for the impulsive effect due to high velocities in a
large proportion for short spans, and in a lesser proportion for
longer spans. This we will afterwards illustrate by tables from
a recent paper in the Transactions of the Institute of Civil
Engineers.
The moving load when thus augmented and equivalated to
an uniform intensity may now be called the effective equivalent
uniform live load, but for brevity it will now be called simply
the live load.
Seeing that both the dead and live loads are now uniform,
the maximum bending moment will simply be a right parabolic
segment standing on the span, for the live load must be put
covering the span for maxima.
The bending moment diagram is now a parabola, the height
of its vertex being oneeighth of the product of the total load
and the span ; where the total load means the sum of the dead
load and of the equivalent live load covering the span. If u be
the dead load per foot of span, and ui the live load per foot, the
equation to the bending moment diagram is
u + w ,
M, = —^ (c*  *2),
where 2c = ^ is the span, and x the distances of any cross
section from the centre. Now we shall investigate the
important practical case of the
Shearing Force for a beam under a fixed uniform load together
with an advancing load of uniform intensity (fig. 1). — Let u be
the intensity of the fixed uniform load, and w the intensity of
the advancing load in terms of its equivalent dead load ; as in
fig. 13, Ch. VII, draw DE for the uniform dead load, and from
DE as a sloping base plot the ordinates of the two parabolas
(fig. 1, Ch. IX) up and down respectively ; the two loci FLEqmA
DKG give the maximum and minimum, or the positive and
negative maximum, shearing force at each point.
Q
226
APPLIED MECHANICS.
[chap. XII.
In fig. 1, Ch. TX, the parabola DC may be supposed to be
drawn on the " Distorting Table " described at fig. 17, Ch. IX.
The table is then to be distorted till its base is at the slope DE
(fig. 1), when we have FLE as the locus for the joint loads ;
this is still an arc of the same parabola with its vertex at Az,
for which see Theorem A, Ch. VI.
The ordinates to DKN, the portion of the curve which
extends over the left half of span, are derived by taking the
ordinates of the slope 01) due to the dead load alone, and
subtracting the ordinates of DN; or, what is the same thing, by
adding the ordinates of the slope OD, and of the parabolic
segment JDN, and then subtracting the constant quantity ON.
When the ordinates of the slope and of the segment are added
ii^^'^
UC or i V
wc or i W
Fig. 1.
by the I'heorem A, Ch. YI, the resulting locus is a portion of
the same parabola as DN, but with its apex to the left of B at
a distance
BD \U
^'^=^" = 2^^^==2Tllr
2c.
w
(1)
When the constant quantity ON is subtracted, the locus
unaltered in form moves vertically downwards through that
distance ; the apex still remains to the left of B at the
distance a ; the curve passes tlirough the point U whose
height is BD = ?<c, and through the point G whose depth is
GO = (w + w)c. Similarly ELF is symmetrical, left and right,
up and down.
CHAP. XII.] COMBINED LIVE AND DEAD LOADS. 227
Taking the centre of span as origin, the equation to the
parabola AxT>G is
wc '2u {■'WW. , 
,, = ^^xx\ (2)
The point K is found by making y = 0, and finding the corre
sponding value of X ; thus
OK.cojf^'^) (3)
X^y w^ w v)j
The ordinate to the apex A^ is found by substituting (c + a)
for X in equation (2) ; and is
^^1 = [ + 1^ c. (4)
\w J
Graiihical Solution (fig. 1). — Lay up BD and lay down GE,
each equal to half the dead load. Then further lay down EG
equal to half the live load. To construct as many points on the
parabolic arc DKG as may be required, divide GB in any manner.
In fig. 3 it is divided into twelve equal parts corresponding to
the l3ays or panels in the girder. More correctly, however, it
should have been divided by points over the centres of the hays,
that is, it should have been divided into eleven equal central
divisions and two half end divisions. Now divide EG similarly
to the manner in which BG has been divided ; then vectors drawn
from D to the points of division of EG meet verticals drawn
through the points of division of BC, and determine the corre
sponding points of the curve DKG. ELF is the same curve
transferred by the dividers.
To the left of K the shearing force is always positive, to the
right of L it is always negative, and between K and L there are
both positive and negative maxima. The range at the centre is
\ W as for W alone ; if u be great compared to w, the range
between K and L is nearly constant and equal to \ W, since the
apex is then far out and the portion of the parabola over KL is
very flat.
The critical points K and L, between which the shearing
stress sometimes produces distortion of the girder in one direc
tion and at other times in the opposite direction, according to
the position of the live load, are of great importance. Between
these points the girder requires counterbracing. When the fig. 1
is drawn graphically, K and L are at once fixed ; while, by
Q 2
228 APPLIED MECHANICS. [CHAP. XII.
calculation, the exact position of K is given by a modification of
equation (3) ; thus
BK2c(E.^l). (5)
This equation is so awkward and the position of K so
important that we will determine its position by ascertaining
the position of the live load which shall bring the common
centre of gravity of the girder and the load on its left segment
as near to the left abutment as possible.
The centre of gravity of the girder itself is at the middle.
As the live load is pushed on from B, the common centre of
gravity moves from the middle point nearer and nearer the
left abutment B for some time ; then it begins to move back
towards the centre 0, where it arrives when the whole girder is
covered with the live load. For some segment such as BK
loaded, the common centre of gravity of girder and load will be
nearer the abutment B than for any other. This is the case
when the common centre of gravity is at K the end of the
loaded segment. For adding a load to the right of ^will move
the centre of gravity to the right of K ; and removing a load to
the left of X will have the like effect. To find K then it is
only necessary to assume the shorter segment BK to be covered
with the live load and at the same time the common centre of
gravity to be at K. This is readily remembered, and can be
applied at once to any numerical example.
That it leads to equation (5) is proved at once, for the sup
porting force at the right end can be calculated in two ways.
First, we may suppose the dead load w . 2c concentrated at the
centre 0, and the live load vj . BK concentrated at the middle
of BK, when the right reaction is
Q = uc ■¥ v: . BK^ — = ^ic + — BK •
^ 2c 4c
Again, we may suppose the whole load {u , 2c + w . BK)
concentrated at K, when
Q={2iui + w. BK) ^ = uBK + '^BK\
^ 2c 2c
Equating these, and solving the quadratic equation, we have
the same value for BK as tliat in equation (5).
Another way of considering the live load — that which Levy
CHAF. XII.] COMBINED LIVP: AND DEAD LOADS.
229
uses in his Graphiquc Statique, and which we follow in a later
chapter — is to consider an equivalent rolliiifj load on one wheel
which will produce the same bending moment at the centre,
and the same shearing force at the abutment as the maxima
produced by the actual load. If R be the rolling load and U
the dead load, then the bending moment diagram is a parabola,
the height of the vertex being {{R + \ U)l. By superposition
of the diagrams, figs. 13, Ch. VII, and 6, Ch. IX, first distorting
fig. 6, Oh. IX, so that its base will coincide with the slope DE,
we have the shearing force diagram, fig. 2.
Girder bridges are tested by passing back and forth over
them two or three locomotives of the heaviest type and fully
loaded, during which the deflections are observed. The eflt'ect
is the more marked the shorter the span. With spans above
100 feet the effect is nearly the same as that calculated for a
uniform load. That is the fact of the load being concentrated
Fig. 2.
in parcels at the wheels may be overlooked almost. Compare
figs. 4 and 5, Ch. X, and the formulas referring to them.
In 1890 the heaviest locomotive and tender used on the
Caledonian Railway corresponded very nearly to the following: —
Loads on wheels 14, 15, 15, 12, 13, 11 tons = 80 tons.
Intervals, 7, 7, 9, 8, 6, 7, 6 feet = 50 feet.
Two such locomotives, the leading one running backwards,
serve to test all girder bridges from the shortest spans up to a
span of 100 feet. That wheel or set of wheels adjacent to each
230
APPLIED MECHANICS.
[chap. XII.
Other which gives the greatest bending moment is taken for
each span, and the greatest bending moment near the centre is
calculated in this Chapter by the preceding rules or graphic
constructions. Then the equivalent uniform load which would
give a like maximum at the centre is calculated.
Thus on a span of 5 feet only, one pair of wheels could
be on the bridge at one time, that is, the greatest load is the
heaviest pair, namely, 15 tons of a rolling load. The equivalent
uniform load is 30 tons spread or 6 tons per foot. For a span
of 10 feet the 15 tons above the span and at its centre is
the same as 30 tons spread or as 3 tons per foot ; but when
the 12 tons and 13 tons are both on the span, and placed so
that the 13 tons is as far from the right abutment as their
common centre of gravity is from the left abutment (see fig. 14,
Ch. IX), in which reckon Wi = 12 tons, JFo = 13 tons, and
2h^ + 2^2 = 6 feet ; then the maximum bending moment under
Wi is to be calculated, and it will be found to be 4 per cent,
greater than with the 15 tons alone at the centre.
In each case if vi be the intensity of the equivalent uniform
load, we must equate ^wV' to the maximum bending moment
for the severest position of the load.
The results are given in the following table* : —
Span in
feet.
Maximum
in foottons.
Equivalent
uniform load in
tons per foot.
Span in
feet.
Maximum
in foottons
Equivalent
uniform load in
tons per foot.
5
10
20
30
40
50
i8f
3«
104
214
37'
563
600
304
208
191
I 86
I 80
60
70
So
90
100
774
lOIO
1303
1633
2020
172
165
163
162
162
Within the last few years the weight of rolling stock has
greatly increased, and results similar to the above, but on the
most elaborate scale, are given in an excellent paper by
Mr, W. B. Farr, called " Moving Loads on Eailway Under
Bridges," and published in the Transactions of the Institute of
• These tabulated results were given by Dr. Thomson in a paper to the
students of the Institute of Engineers and Sliiphuilders of Scotland in tlie year 1890.
CHAP. XII.] COMBINED LIVE AND DEAD LOADS.
231
Civil Engineers, 1900, No. 3195. Mr. Farr has included the
rolling stock of all the British and Belgian Railways. His
argument, too, seems excellent, namely, that there is now no
further possible increase of the weight on the wheels of rolling
stock, so that increased weight of the rolling stock in future
must be accomplished by increasing the number of wheels
wliich would not affect the equivalent uniform load.
Another excellent feature of Mr. Farr's paper is that he
gives a suitable augmentation of the equivalent uniform load
on a sliding scale to allow for the impulse or impact of the
load, so that his tables furnish at once reliable working values
for the effective equivalent uniform load which are not likely
to alter much in the immediate future.
We quote here the summary of his results.
Main Gikders.
X = maximum uniform load per foot run (40 types of
locomotives).
y = suggested per cent, increase for impact.
Spans
X
y*
x + y
50 feet
760 tons
30"oo per cent.
988 tons
75 ..
555 ..
2750 ,,
707 .,
100 ,,
485 ..
2500 „
606 „
150 „
3*74 >.
2250 ,,
458 „
20'0 ,,
3 20 ,,
20*00 „
384 „
30*0 ,,
263 ,,
i5"oo ,,
301 „
400 ,,
240 ,,
1450
275 ..
6oo ,,
217 ,,
1350
246 „
800 ,.
206 ,,
I20O „
230 „
1000 ,,
197 ..
lOOO „
2i6 „
* "We think Mr. Farr's addition too low, and suggest that 1/ he increased
througliout. See Fidler's Bridge Construction, page 241.
232
applied mechanics.
CrossGirders.
[chap. XII.
X = maximum concentrated loads in tons on each single line.
y = suggested per cent, allowance for impact, &c.
Distance apart
X
y
x+y
3 feet
igoo tons
50 per cent.
2850 tons
5 ..
igoo ,,
45
2755 „
7 ..
1900 „
40
2660 „
8 „
2150 ,,
35
2902 „
9 „
2350 .,
30
3055 ,.
lO „
2511
25
3139 «
Examples.
1. A bridge 32 feet in span is subject to the transit of a loton road roller .
find the equivalent uniform load.
Here H = \o tons and I = 32 feet, so that Mq = {R( = 120 ft. tons. Equating
this to ^wP, we have Atis. to = if th of a ton per foot run.
2. A beam 36 feet in span is subject to the transit of a load on two wheels,
viz., 12 tons on one and 6 tons on the other, spaced 12 feet apart. Find the equi
valent uniform live load (see the model, fig. 16, Oh. IX).
1.V2 = if X 16 X 16 = 128 ft.tons maximum.
Put lu>P = ^ X 36 X 36 = 128.
And to is a ^a part of a ton.
3. The beam 42 feet in span, shown on fig. 2, Oh. X, is subject to the transit
of the locomotive shown on that figure. Find the equivalent uniform live load.
From the figure iM.a = 261 ft.tons is the maximum bending moment.
Equating this to ^w x 42 x 42, we find w equal to 1184 tons. Compare p. 204.
4. A beam 30 feet span is subject to the transit of five weights, each 3 tons,
fixed at intervals of 7 feet. Find the equivalent uniform live load.
We have iMo = 495 ft.tons. Equating ^wF to this, we find tv  44 ton per
foot.
5. An advancing load in length not less than the span, and of uniform intensity
3 tons per foot, passes over a beam 42 feet span. Find the maxima shearing fones
positive and negative at the points .r = 21, 14, 7, and 0.
Positive tons. Negative tons. Range tons.
Ans. F21 .. 630 .. 00 .. 630.
Fu .. 4375 .. 175 .. 455.
F. .. 280 .. 70 .. 360.
/■n ., 1575 .. 1576 .. 315.
6. In the previous example, find the maximum value of the positive and
negative shearing forces at the point z = 7, directly.
CHAP. XII.] COMBINED LIVE AND DEAD LOADS. 233
Let the segment to the riglit of the point be loaded ; the amount of load will
then be 28 X 3 = 84 tons; suppose this concentrated at the centre of the 1 laded
segment and + Ft — P = (84 ^ 4) x 14 = 28 tons ; asiaiii let the segment to the left
of the point be loaded, and  Ft = Q= (42 4^ 42) x 7 = 7 tons.
7. Find, for the previous example, the points between which the shearing
force is sometimes positive and sometimes negative ; and tind the coordinates of
the apex A\. See tig. 1.
Ans. 0K= 21  42 {^(^ + i)  j} = 7 feet ; 0L = 1 feet.
For the apex Ai,
OS = z = c \ '2c = 35 feet on horizontal scale,
w
SAi = 1/ = ( h " ) e = 28 tons on vertical scale.
By using the coordinates of ^i, the graphical construction can be made with
greater accuracy, more especially if the poiiiis Ai and D aie situated near each
other.
8. A beam 24 feet span bears a load of 6 tons uniformly distributed, and is
subject to a rolling load of 6 tons. Find the amounts and the range of the shearing
forces at intervals of 2 feet.
Since the span is a little greater than 20 feet, the dead rolling load equivalent
to the actual roUing load of 6 tons will be, say, 12 tons.
For the uniform loiid, F varies uniformly from 3 tons at the left end of span,
to — 3 tons at the right ; for the rolling load, i^ varies uniformly from 12 tons to 0,
positive from left to right, and negative from right to left; and we have for both
loads
Fin = 15,  Fi2 = ; Fw = 135,  J'lo = ;
Fb = 12,  i^s =
J?4 = 9,  Ft = 3
Fs = 105, Fe = 15
Fo = 75,  Fi = 45
Fu = 6,  Fo = 6 tons.
The results for the right half of the span are similar to the above, the signs
alone being changed.
The range at intervals of 2 feet for the left half of the span, beginning at the
point of support, is 15, 13*5, 12, 12, 12, 12, and 12 tons.
9. A beam 24 feet span bears an uniform dead load of 100 lbs. per foot, and
is subject to an advancing load as long as the span, and of intensity 400 lbs. pei
foot. Find the shearing forces, and their range, on the left halt of span, at
intervals of 2 feet. Find the critical point K, and the coordinates of the apex Ai.
See tig. 1, Ch. XIII.
Since the span is a little greater than 20 feet, the dead advancing load equiva
lent to the actual advancing load of 400 lbs. per foot will be, say, 800 lbs. per foot.
Taking this value, we have
u = 100, and w = 800 lbs. : c = 12 feet.
Substituting in equation (2), p. 227,
Fvi = 10800,  Fi2 = ; Fio = 9067,  Fig = ; Fs = 7467,  Fs = I
Fc = COOO,  Fe =0 ; Fi = 4667,  Ft = 667 ; Ft = 3467,  F2 = 1467 ;
Fo = 2400,  Fo = 2400 lbs.
The range at intervals of 2 feet beginning at left end of span is 10800, 9067,
7467, 6000, 5334, 4934, and 4800 lbs.
OK = 6 feet ; OS = 15 feet ; 6'^i = 1350 lbs.
234 APPLIED MECHANICS. [CHAP. XIII.
CHAPTER XIII.
RESISTANCE, IN GENERAL, TO BENDING AND SHEARING AT THE
VARIOUS CROSSSECTIONS OF FRAMED GIRDERS AND OF
SOLID BEAMS.
Framed beams are built of pieces either freely jointed, or so
slightly connected at their joints that they may be considered
as freely jointed. If the load be applied at the joints, each
piece is either a strut or a tie ; and if the load be applied at
intermediate points on, or distributed over a portion of, a piece,
it (the load) is to be replaced by a pair of equivalent forces at
the ends, evidently equal and opposite to the supporting forces
of that piece looked upon as a short beam. Such a piece has
two duties to perform, viz., to resist the bending moments and
shearing forces as a small beam, and to act as a strut or tie in
the built beam considered as loaded at the joints only; by
making such pieces short enough, the duty they have to dis
charge as beams can be made so small compared to what they
have to perform as struts or ties, that when designed to fulfil
the latter duty they will also be able to fulfil the former ; by
making the pieces continuous at their joints, they are greatly
strengthened for their duties as short beams.
If a crosssection cuts not more than three pieces, the
unknown stresses are not more than three in number, and the
three conditions of equilibrium (fig. 10, Ch. V) enable us to
calculate the stress on each ; or using the bending moment and
shearing force diagrams, we can readily apply the conditions to
as many sections as necessary, and so design the whole beam.
The sections chosen for this purpose should lie just to one side
of the joints for bending moment diagram, but for the shearing
force diagram they should more correctly be at the centre of
the bays, as the stress on a diagonal is constant and equal to
the shearing force at centre of the bay multiplied by the secant
of its slope. The variations of the shearing force for the half
bay on each side of the centre are borne by the boom or stringer
bridging across from joint to joint.
For any given system of loading a beam is said to be of
CHAT. XIII.] RESISTANCE TO BKNDING AND SHEARING.
2'M
unifonn .strength when the crosssection of each piece is such
that the ratio of the ultimate or proof resistance of the material
Advancing Loart
equivalent to 800 lbs. p. ft
Uniform Load
100 lbs. p. ft.
Bending Moment divided by dipth
gives amount of stress on Horizontal
boom.
The Shearing Force is the vertical
component of the load on diagonal. ^
Two distinct diagonals may resist the positive ''
and negative shearing force. If one resists both,
St be able to act as a strut, and the load is
to be reckoned as the sum of the two loads.
Vf3, hft.
COUNTERBRACING
KMn 11000 iir.'ijo igsoo sjooo iioooG
„tlO Jl'IMO/^i
r,<;(j(i
*c.
V
'
/
%
/ '^\
%
w.
V.
w^
50.oooff(.;6s. I '•'""" "■"" """' Warren Girder
Load on flange is B. M. divided by depth.
fLoad on Web is the L «
fiange of S.F. reck (»
oned uniformly ) i>
distributed.
Fig. 1.
and the tension or compression which it has to bear is constant.
236 APPLIED xMECHAKICS. [CHAP. XIII.
an allowance being made, if necessary, for pieces which also act
as small beams.
The term " flanged girder " is employed to denote all girders
consisting of a web and one flange, or of two flanges connected
together either by a continuous web or by open lattice work ;
in bridges, one at least of the flanges is usually straight, and
also horizontal.
In figs. 8 and 9, Ch. V, the stress at A is horizontal and is
denoted by 2^a ', if the flange is thin, as is often the case in iron
bridges, this stress is sensibly constant ; and since the intensity
of the stress to which a piece is exposed should not exceed the
strength of its material, we have
Pa ^ ./, (1)
where / is the working or proof stress as may be desired ; and
if < = amount of stress on the horizontal flange, and S = the
crosssectional area of that flange, then
t ^ S.f. (2)
When there is only one set of triangles, as is shown in the
lowest of the three systems of bracing (figure 1), and shown also
in the two upper systems if we neglect the counterbracing, the
amount of stress on the straight boom may be found thus : —
Take a crosssection at a point, say F, just on either side of a
joint in the boom opposite the straight boom (in the figure both
booms are straight) ; take moments round the point F, and
since we neglect the counterbrace, the only member not passing
through the point F is the upper boom ; the product t . h gives
the moment where h is the depth of the beam at the point ; this
is equal to the l:)endiiig moment, and we have by substituting
the value of t given in equation (2)
t.h = S.f.h = M. (3)
The quantity M is different at difi'erent points of the beam,
and / is a constant quantity ; if the above equation is to be
fulfilled for every point, we make aS . h vary as 31 ; in practice
one of these two factors is generally kept constant, and the
other is made to vary.
If we make //. constant, then both booms are liori/.ontal, and
S varies as the bending moment ; hence the bending moment
diagram gives, upon a suitable scale, the area of the boom at
each point. The vertical component of the stress on any
CHAP. XIII.] RESISTANCE TO ItKNDING AND SHEAHING.
237
diagonal is the amount of the shearing force at that end of the
diagonal where the shearing force is greatest ; hence if t be
(jhe P
I rectftngu
Load on two nvheets.
' The curves are arcs of
circles with their centres
on the verticals through the
apexes of p^trabolas.
ELEVATIONS OF THIN FLANGED
GIRDERS OF UNIFORM STRENGTH,
derived from Bending Moment Diagrams.
Fig.
the stress on a diagonal, F the shearing force at the end of the
diagonal where it is greatest and as given by the shearing
238 APPLIED MECHANICS. [CHAP. XIII.
force diagram, and 6 the angle made by the diagonal with the
vertical, then
r = i^ sec e. (4)
The depth h is chosen from ^th to xr ^^ ^^ ^^^ ^P^'^ ^^ ensure
stiffness ; in fig. 1 the depth h is taken at 3 feet, that is ^th of
span, and that figure shows how the stresses on the booms and
on the diagonals are found. Between the points K and L,
counterbracing is required ; this is accomplished in the two
upper girders by introducing ^mrs of diagonals between the.se
points, both being ties or both struts as the case may be ; the
one diagonal resists the positive, the other resists the negative
shearing force. In the Warren girder, the third in the figure,
one diagonal resists both the positive and negative shearing
force, the one being applied suddenly after the other; each
diagonal should be designed to bear the stress due to the sum
of these stresses. The lowest girder shown in the figure is one
with thin flanges and a thin continuous web ; part of the bend
ing moment is resisted by the web and part of the shearing is
resisted by the flanges : these parts, however, are small ; prac
tically the flanges are considered, as in the case of open beams, to
resist the whole of the bending moment, and the web is con
sidered to resist the whole of the shearing force : an approximate
result is obtained if we consider that the shearing force is
uniformly distributed over the crosssectional area of the web.
If we make S constant, then h varies as M \ and (fig. 2)
the elevation of the beam will correspond with the bending
moment diagram ; h, the depth of the beam at the centre is,
as in the previous case, to be taken sufficient to ensure stiffness.
The curved boom will bear a share of the shearing force ; this
compounded with the stress on the horizontal boom at the same
section will give the resultant stress on the curved boom. It
is usual in practice to make the area of the curved and of the
straight booms uniform, and to make the diagonals sufficientl)'
strong to resist the whole of the shearing force as in the previous
case. Where the curved member slopes considerably, as in the
bowstring girder, it is made sufficiently strong to bear the
whole shearing force, the diagonals being intended for another
purpose, viz., to distribute partial loads in a sensibly uniform
manner.
The theoretical elevations reduce to a height zero at the
ends, and so give no material to resist the shearing force at
the point where it is greatest ; sufficient material is generally
CHAP. XIII.] RESISTANCE TO BENDING AND SHEARING, 239
allowed at the ends either by making the span of the girder
exceed the clear span, or by departing from the theoretical
form along a tangent near the end. Further, whatever the
curves may be — and, as we have seen, they are generally
parabolas — they are usually replaced by circles which nearly
coincide therewith ; when the iigure passes from one curve
to another, the passage is made along a tangent, as will be
seen on some of the figures. Approximate forms, consisting
entirely of straight lines enveloping the bending moment
diagram, are sometimes adopted.
Moment of Eesistance to Bending of Eectangular and
Triangular CrossSections.
The moment of resistance to bending we have defined as
the moment of the total stress upon the crosssection about
any point in it ; and this we have shown (figs. 7, 8, 9, Ch. V) to
be equal to the couple which is the moment of the normal
stress on the crosssection.
The stress (fig. 8, Ch. V) might be artificially produced by
building on the portion O'A, columns of a material tending to
gravitate towards the left ; and on the portion O'B, columns
gravitating towards the right. These columns standing on very
small bases, being of uniform density, and of the proper height
to produce the intensity at each point, will, if we suppose them
to become one solid, form a wedge with a stepped or notched
sloping surface ; the more slender the columns are, the more
accurately do they give the stress at each point, and the smaller
are the notches on the wedge; hence tvjo right wedges exactly
represent the normal stress on the crosssection. Such a stress
is called an uniformly varying stress. Taking the density of the
wedges as unity, the height of one will be expressed hj 2Ja, and
the other by jib ', the volumes of the wedges will give the two
normal forces ffig. 10, Ch. Y) the resultants of the thrusts and
tensions respectively ; these forces are equal ; the volumes of
the two wedges must therefore be equal, and the position of the
neutral axis of the crosssection is thus determined. Further,
each wedge, instead of distributing its weight over its base, may
be supposed to stand on the point below its centre of gravity ;
this enables us to find the positions of the normal forces (fig, 10,
Ch. V), and gives us the arm of the couple ; if we multiply the
volume of either wedge by this arm, we have M the moment of
resistance to bending.
240 APPLIED MECHANICS. [CHAP. XIII,
For a rectangular crosssection, the neutral axis is at the
centre since the wedges are equal, and pa equals ])}, ; and if the
common volumes of the right wedges be represented by V, then
V = ^Pa >i h bh = Ipabh ;
where h is the breadth and h is the depth of the beam, as shown
in fig, 1, Ch. V. Each wedge stands on a rectangular base, so
that the point on the crosssection below its centre of gravity
is distant from 0' by twothirds of O'A, or ^h ; hence the arm
of the couple is §/«, and
M = iPabh X 1^ = iPabh\
By increasing the bending moment we can increase M till
Pu becomes equal to /, the resistance of the material to direct
tension or thrust, but no further ; because if Pa becomes greater
than/, the fibres at the skin will be injured ; hence
M = iph' (1)
is the ultimate, proof, or working resistance to bending,
according as / is the ultimate, proof, or working strength of
the material. Since /is generally expressed in tons or lbs., per
square inch, it is necessary to express b and h also in inches ; in
which case M will be in mc7ttons or lbs. ; on the other hand,
M \s> generally expressed in footion^ or lbs. ; and it is well to
observe that, if such be the case, M requires to be multiplied
by 12 to reduce it to inchton^ or lbs. before equating to M.
For all crosssections, if b and h are the dimensions of the
circumscribing rectangle, the equation giving the moment of
resistance is of the same form, but the numerical coefficient
assumes values other than ^, as will be proved hereafter.
Rankine uses n for the value of this constant, which he calls
the numerical coefficient of the moment of resistance to bending
of any crosssection, and we may put
M = nfbh\
We shall now verify this for a triangular crosssection ; an
isosceles triangle is taken, but exactly the same result would
be obtained ff)r any triangle; the angle A, however, must not
be too acute.
CHAP. XIIL] resistance TO BENDING AND SHEARING 241
Triaiiffular cross section (figs. 2 and 3). — Assume that the
neutral axis passes through the centre of gravity of the
Fi. 3.
triangle, and divide the part below into three areas as shown ;
put aS' and V with suffixes for the areas and volumes of the
242 APPLIED MECHANICS. [CHAP. XIII.
different parts, put I for the leverages about the neutral axis of
their centres of gravity, and, for convenience, suppose h = 36a%
and h = 6;/.
Vi is a pyramid on S^ as a base, and let/ be its height; F,
is a right prism of height / ; F3 is a pyramid on S:, as base
and of height 12r. Now the centre of gravity of a pyramid is
on the line joining the apex with the centre of base, and at
threequarters of the length of that line from the apex ; and
;S', = i . 24a; . 4y = 48>;?/ ; S^ = 12.r . 4y = 48xy ; Ss ='^ y.
2'
V, = \fS, = Wxy V, = },.^.S, = Ufxy;
V,= \.12x.S, = 2fxy= F,;
hence F, = F, + F3 + F4, and therefore the assumption as to
the position of the neutral axis is correct.
Now
Zi = 12a:, I2 = Sir, h = li = ^x;
and
M = V,h + V,h + V,h + F«Z4
= (192 + 96 + 18 + 18)/7/.r = 324/?/a;^
Substituting y = g, and a:' =f — j = ^^^,
we have M = iJhh\ (2)
so that for a triangular section, n = ^.
Solid beams are sometimes made of uniform section ; that is,
at the point of maximum bending moment the section is made
sufficient to resist the bending moment, and this section is
adopted along the entire length ; at every other section there
fore, the beain is too strong. This is frequently done with
small timber beams cut out of one piece, because the material
in excess, even if cut away, would be lost ; the weight of this
excess is very little, since timber is light, and probably *he
cutting of it away would add to the expense of the beam.
When the section of an uniform beam is designed to resist
the maximum Ijending moment, it is generally many times more
than sufficient to resist the maximum shearing force.
Solid rectangular beams are designed of uniform strength to
economize material, and to reduce the weiglit when tlie material
itself is heavy.
CHAP. XIII.] RESISTANCE TO BENDING AND SHEARING.
243
Bectangidar beam of uniform strenrfth and uniforin d^pth. —
In this case the breadth is varied so as to make tlie cross
section at every point just sutficient to resist the bending
moment. It is evident that the elevation is a rectangle, while
the plan will vary according to the load ; the plan, however, is
to be symmetrical about a centre line. The theoretical shape
of the plan so designed has to be departed from near the ends,
so as to make the end sections large enough to resist the
shearing force ; and some additional material is required at the
€nds of the beam to give it lateral stability ; now
nfbh^ = M; therefore h = ^rr„,
nfh
or h is proportional to M, since n, f. and li are constant quantities.
Quad, of Par., but circular arc ia close approx.
Lo»as/m.ontwoWhe»lt Ip
Plans of Rect. Beams of Uniform Strength i Uniform
Depth same asB.M Diagrams.
Fig. 4.
Hence the half plan is the bending moment diagram reduced so
that its highest ordinate is equal to ^^o, the half breadth of the
crosssection which can resist the maximum bending moment
(see figs. 1 and i). As has previously been explained, the
parabolas being very flat are readily replaced by arcs of circles
with their centres on the axes of the parabolas.
In designing such beams, then, the uniform depth is fixed
as a fraction of the span from an eighth to a fourteenth to ensure
the required degree of stiffness ; &,„ the breadth of the cross
section where the bending moment is a maximum, is calculated
to make that section sufficiently strong, exactly as in the pre
ceding examples ; then the bending moment diagram is reduced
till its highest ordinate is ^^o, and drawn on both sides of a
R 2
244 APPLIED MECHANICS. [CHAP. XIII.
central line. All the curves may now be replaced by circular
arcs, and all sudden changes of breadth bridged over by tan
gents ; the curves are to be departed from, near the ends, to
make the sections there strong enough to resist the shearing
force. Otherwise, tlie l)readths may be calculated at a number
of sections and plotted, and a fair curve drawn through them, &c.
Bedcnr/ular beam of uniform strength and nniform hracUh. —
It is evident that the plan is a rectangle ; and since ufbh"^ = M,
we have h proportional to VtI/. Hence the elevation of the beam
is obtained by degrading the bending moment diagram, so that
the derived figure has its highest ordinate equal to Ao, the depth
required to ensure stifl'ness ; h is then to be made sufficient to
resist the maximum bending moment (see Theorems F, G, and IT,
Ch. VI).
The figures in brackets refer to the corresponding bending
moment diagrams.
Fig. 5 shows what the bending moment diagrams (figs. 9,
10, lo, and 16, Ch. Vll) become when degraded by the preceding
theorems. The elliptical elevation of a beam for an uniform
load is readily struck from three centres, as shown in fig. 11,
Ch VI ; AD from a centre on AO and with a radius
,.. = ^■' = 60'^24 = 150;
OA
BC from a centre K with radius
OA
ri = ^ = 24^ f 60 = 10 nearly ;
OB ^
E is found by drawing OE from the first centre, and a circle
about K with a radius
7, = A.„/, = 2410 = 14;
further, if we choose we may retain the single circular arc AD,
and depart along the tangent at D. For a beam with the load
at the centre, the two parabolas may be replaced by their
tangents ; this gives an approximate form, and the beam will
now consist of two straight portions tapering so that the depth
at each side is \K, K being the depth at the middle (see fig. 3,
Ch. VI) ; the area of the theoretical elevation is twothirds,
while that of the approxinuite elevation is threequariers, of a
rectangle of height h„; these areas are as 8 to 9; hence the volume
•CHAP. XIII.] ItESISTANCE TO BENDING AND SHEARING,
245
of the approximate form is only oneeiglitli in excess of the
theoretical one, and the additional material is well placed to
resist shearing ; the approximate form is in some cases preferable,
since it has the great advantage of straight boundaries. The
same remarks apply to the elevation of a cantilever with the
load at the end ; on this principle spokes of wheels, when of
uniform thickness, taper to half the depth from boss to tyre.
More particularly in timber beams {bcor,unt)
and cantilevers are the straight ..—"X b.m.d
boundaries desirable. The material '(''
is light, and the extra wood at
the ends holds the bolts and^'.
fastenings securely. ^'
ELEVATIONS
UNIF. STRENOTH,UNIF. BREADTH
B.M. DIAGRAM DEGRADED.
Fk
Fig. 6.
In fig. 6 are shown the effects of degrading such bending
moment diagrams as figs. 2, 5, 17, Ch. Vll, diagrams consisting
of straight slopes. Producing the slopes to meet the base as at
A, B, &c., the elevation for beams of uniform strength and
uniform breadth will consist of a series of parabolas with their
apexes at A, B, &c., and intersecting on the lines of action of
the concentrated weights.
An approximate elevation is to be derived thus : — Lay up
ho at the point of greatest bending moment, the proper fraction
of the span to ensure stiffness, and calculate the breadth of that
crosssection as in the preceding examples, so as to give the
proper resistance to bending there ; from the top of h^, draw a
tangent to the parabola whose apex is A, that is, draw the line
which intercepts Via on the vertical through A, and it will cut
246
APPLIED MECHANICS.
[CHAl'. Xill,
off hi on the line of nearest weight; from the top of A, draw a
slope to cut off /'i on the vertical through ]i, the apex of second
parabola and the point where the second slope of the bending
moment diagram meets the base; this will be parallel to the
tangent to that parabola, &c., &c.
It is evident then that these approximate elevations for
concentrated loads consist of straight lines, each sloping at
half (Jk: rate of the corresponding side of the bending moment
diagram ; from this fact they are readily drawn thus : — From
the highest point in the bending moment diagram draw a line
at half the slope of the adjacent side till it cuts the line of the
weic'ht nearest to that point ; from the point thus found draw a
line to cut the next w eight at half the slope of the corresponding
side of the bending moment diagram, &c. ; lastly, reduce the
ordinates so that the highest is Ji„.
(bconatant)
W^ o
When fig. 13, Ch. VIII, is degraded, from K to D it will he
a portion of the parabola whose apex is at B the centre of load,
and from Z> to 6' a straight slope ; the approximate form is a
straight slope to D the end of load, and which, when continued,
tapers to Ih^ at B ; then from D a straight slope tapering to
zero at the free end. On degrading the two parts of fig 4,
Ch. VIII, separately, that for £/" will be a taper from a at the
fixed end to zero at the free end ; while the approximate form
for W will be a taper from b at the fixed end to i at the free
end ; hence the approximate elevation for the combined load is
a taper from (a + J) at the fixed end to ^h at the free end ; sub
stituting for a and b their values, we have H : Ji ■.: 2JV + U : TV.
Fig. 7 shows figs. 1, Ch. VIII, and 10, Ch. IX, degraded, the
quadrants of parabolas becoming quadrants of ellipses.
Ikctawjvlar beam of uniform strength and similar cross
section. — In this case both b and // vary, but they bear to each
CHAP. XIII.] RESISTANCE TO BENDING AND SHEAUING.
247
Other a constant ratio ; it is evident that the plan and elevation
will ha\e the same form. The plan is always to be symmetrical
abont a centre line ; the elevation may either have one straight
boundary, or be symmetrical about a centre line.
Since
now
b oc h, then b/r oc b'\ or h^ ;
nfbh = 31, bh' oc M:
therefore
b and h
oc
J^l^
That is, both plan and elevation are derived by drawing a locus
whose ordinates are proportional to the cube roots of those of
the bending: moment diagram.
Approx. Uniform Strength
Similar Cross Sections.
Fig. 8.
For bending moment diagrams with straight slopes, that is,
for concentrated loads, the degraded figures were parabolas with
axes horizontal, &c. ; in the same way, when the new figure is
made with its ordinates proportional to the cube roots of the
ordinates of the straight slopes, it becomes what is called a
cubic parabola ; a property of this curve is that the tangent
cuts off ticothirds of the ordinate upon the vertical through the
apex, instead of oneJialf as in the case of the common parabola.
Hence figs. 5 and 6 give approximate elevations and half plans
248 APPLIED MECHANICS. [CHAP. XIII.
of beams of uniform strength and similar crosssections, if
we use f^n instead of hhn in making the construction ; or if we
draw the tapers at twothirds of the slopes of the bending
moment diagram instead of onehalf. Observe that in this
case there is a double taper in planes at right angles to each
other.
On this principle the crosshead of a pistonrod may have a
conical taper, so that the diameter at each end may be two
thirds of the diameter at the centre ; and the spokes of wheels
may have a conical taper from the boss to twothirds (linear
dimensions) at the tyre.
ExAMPLKS.
1. Find the working moment of resistance to bending of a rectangular section,
10 inches deep and 3 inches broad, the woiking strength of the material being
4 tons per square inch.
M = n/b/r = ^ . 4 . 3 . 100 = 200 inchtons.
Find the same for an isosceles crosssection inscribed in the above rectangle,
and with the base horizontal.
Ans. M = «/M2 ^ ^\. . 4 . 3 . 100 = .50 inchtons.
2. Find suitable dimensions for a castiron Deam 20 feet span, of uniform and
rectangular crosssection, and subject to a load of 10 tons at the centre.
Taking h = 20 inches, a twelfth of the span, to ensure stiffness ; / = 2 tons per
square inch ; and M = Mo = m Wl, the maximum bending moment.
Ans. i . 2 . « . 20^= i . 10 . (12 X 20); .. i = 4.5 inches.
3. If the breadth be taken at 6 inches, what depth would give sufficient strength ?
Am. i . 2 . 6 . A2 = i . 10 . (12 X 20) ; .. h = 173 inches.
4. If the load is uniformly distributed, and the crosssection is a triangle whose
base is horizontal and 8 inches broad, find the height of the triangle.
Ans. ^^ . 2 . 8 , //' = 1 . 10 . (12 x 20) ; .. h = 212 inches.
5. Find the greatest crosssection for u wroughtiron beam of rectangular section
and 15 feet span, to bear a load of 20 tons uniformly distributed, together with a
load of 5 tons at the centre. Take /= 4 tons per square inch, and /» = lo inches
to give sufficient stiffness.
A7i>. 1.4.*. 1.5 = (i . 20 . 15 + I . o . 15) X 12 ; .. * = 45 inches.
6. Taking the depth onetwelfth of tlie span, and /" 4 tons per square inch,
find the breadth for a wroughiiron beam of rectangular section, to resist the
maximum bending moment in Ex. 9, Ch. IX.
Ans. ^ X 4 X 4 X 36» = 112^ x 12 ; . . b = 156 inches.
OHAl'. XIII.] RESISTANCE TO BENDING AND SHEAHINO. 249
7. Design a rectangular cantilever 1 feet long of approximately uniform strength
and of uniform breadtli, of timber whose working strength is 1 ton per square inch ;
the load is 2 tons at the free end, and ho = 15 inches, an eighth of the length.
Ans. nfbhu = m . W . I; i . 1 . i . 15^ = (1 . 2 . 10) 12,
gives b = 6*4 inches for uniform breadth ; the depth tapers from
15 inches at tixed end to 7*5 inclies at free end.
8. If an additional uniform load of 4 tons be added, and /(o be still retained
15 inches, find the value now of b, and of h at the free end for Ex. 7.
Ans. At the fixed end the bending moment will be double its
former amount, so that b will be doubled ; that is, b = 12*8 inches,
and 7(0 :/*:: 2 JT + r : Zr :: 15 I/":: 2 X 2 + 4 : 2 :
7(10 = 37o inches.
9. A wooden cantilever 12 feet long bears 3 tons uniformly distributed on the
half next the free end. Design an approximate elevation, supposing the breadth to
be uniform, and/= 1 ton per square inch ; take depth at fixed end as 18 inches.
Ans. J/o = 27 foottons = 324 inchtons; i . 1 . i . 18^= 324 ;
.*. b = 6 inches constant.
Between the fixed end and the end of the load next thereto, the elevation
will taper at the rate of 1 inch per foot, and thence to zero at the free end :
that is, /io = 18, ht = 12, and /ns = inches.
10. Design a cantilever for Ex. 7, supposing its section to be a square, taking
dimensions to the nearest whole number in inches.
Ans. Put ^0 = 7(0 = side of square at fixed end, and equating
n/b^ = in . W . I, gives bo = ho = 113, say 12 inches. The side
of the squai'e at the free end is 612 = hn = 8 inches.
Area, Geometrical Moment, and Moment of Inertia.
The functions of a plane surface which we require for our
investigations regarding the moment of resistance to bending
of a crosssection in general, are the area, geometrical moment,
and moment of inertia.
The area of a rectangle is the product of its two adjacent
sides ; the area of any other surface is the sum of all the
elementary rectangles into which it may be divided. We take
it for granted that the area of the triangle, the circle, the
ellipse, and parabolic quadrant are respectively half the product
of the height into the l)ase, tt into radius squared, tt into the
product of the semimajor and semiminor diameters, and two
thirds of the product of the circumscribed rectangle.
Definition. — The geometrical moment of a surface about any line in its plane
as axis, is the sum of the products of each elemental area into its leverage or
perpendicular distance from that axis ; the leverages which lie to one side of the
axis being reckoned positive, and those to the other side negative.
250
APPLIED MECHANICS.
[chap. XI I L
It will be convenient for us always to choose a horizontal axis; and if we
consider leverages up to be positive, then, when the axis is below the aiea. the
geoniftrical moment is positive ; when ahove it, negative; when the axis cuts the
area, it will he positive ornetrative accordinji as the axis is near one or other edge;
and for one position of the axis, cutting the area, the positive and negative products
will destroy each other, and the geometrical moment will he zero.
An axis about which the geometrical moment of an area is zero passes through
a point called the geo»ietrical centre of the area. From this it appears that the
geometrical centre of an area corresponds with the centre of gravity of a thin plate
of uniform thickness and of that an a ; and for this reason the geometrical centre
of an area is often called its centre of gravity.
Thenrem A. — The geometrical moment of a surface about any axis in its plane
is equal to the area multiplied by the distance of the geometrical centre from the
axis (fig. {')• Suppose G the geometrical centre of
the surface to he known ; through G draw 00'
parallel to the axis A A ; let s and «' be a pair of
elementnl areas, one on each side of 00', such that
the sum of their geometrical moments is zero ; that
is, i' . a = s .b. It is evident that the whole area
can he divided into such pairs from the definition of
the geometrical centre. The geometric al moment of
*' about AA is s(d + a), that of s is 8[d  b), and
their joint moment is («' + s)d + (*'«  sh) = («' + .s)rf,
since the second term is zero. In the same way the
moment of each pair is their sum multiplied by rf ; \ \ \
hence the geometrical moment of the area about AA j, i i^j— 4
is the sum of all the pairs into d, that is, the area
multipiitd by the distance of the geometrical centre Fig. 9.
from the axis.
Cor. — The geometrical moment of an area which can be divided into simpler
figures whose geometrical centres are known, may be found by multiplying the
area of each sucli figure by the distance of the axis from its geometrical centre and
summing algebraically.
T'neorem Ji. — The geometrical moment of an area about an axis in its plane is
expressed by the number which denotes the volume of that portion of a right
angled isosceles wedge whose sloping side passes
through the axis, and which stands on the area as
base.
Let AEKFA' (fig. 10) he the wedge, and AF its
sloping side passing through the axis A A' ; the angle
at A is 4.5° and that at E is 90' ; let BCBE be the
area, then the geometrical moment of BCDE rela
tively to the axis of A A' is represented by the volume
of l'bckl.
If s be an elemental poition of the area BCBE, its
geometrical moment about AA' is s multiplied by its
distance from AA' \ but the column of the wedge
standing on « is sensibly a parallelepiped whose height
is the same as the distance of s from AA', and the
volume of that column expresses the geometrical
moment of s; hence the volume oiUBCKL, a portion
of the isosceles wedge, is the geometrical moment of
BCDE about A A'.
When the axis cuts the area, the plane sloping at
45° will form a wedge above one portion and bclotv the
other; by considering these of different signs, their algebraic sum is still the
geometiical moment. For example, if we wish to find the geometriial moment of
the triangle (fig. 3) about the axis JSA ; let /= J//, th.n the plane will slope at
45° ; in the figure, the volume on one side of the axis is equal to the volume on the
•O^
Fijr. 10.
CHAr. XIII.] RESISTANCE TO BENDING AND SHEARING, 251
other, tliiit is, the geonietrifiil monieiit is zero; the axis lY^I must pass therefore
through the geometrical centre.
To find an axis passing through the geometrical centre of a plane area then, it
is only necessarj' to draw a plane sloping at 45° which will cut off an equal volume
of the wedge above and below ; the intersection of this plane with the area will be
the axis re(uired.
By similar triangles a plane through that axis at any slope will cut off equal
volumes above and below ; the wedges which represent tlie normal stress (fig. 8,
Ch. V) require to be equal ; hence the neiiiral axis of a crosssection passes through
the geometrical centre (or centre of gravity) of the crosssection.
The distance of the neutral axis from the furtiiest away skin is, in each cross
section, a definite fraction of h the deptli of the circumscribing rectangle ; for
instance, for a trianguhir crosssection (figs. 2, 3)
OA = %h.
Bankine expresses this generally thus
OA = m'h.
where m' is the fraction rvhich the distance from the neutral axis to the farthest
atray skin is of the depth ; for all crosssections, symmetrical above and below, as
a rectangle, ellipse, hollow rectangle, &c., «»' = ^.
Definition. — The moment of inertia of a surface, about a line in its plane as
axis, is the sum of the products of each elemental area into the square of its
distance from the axis.
Theorem C. — The moment of inertia of a surface about any axis in its plane
equals that about a parallel axis through its geometrical centre, together with the
product of the area into the square of the deviation of the axis from the centre.
Let s and s' (fig. 9) be a pair of elemental areas, the sum of whose geometrical
moments about 00" is zero; and let Xi and X2 be their leverages about A A
respectively. The sum of their moments of inertia about A A is
s X2 + sxr = s'{d + a)'^ + s{d — b)
= (.«' f s)d^ + 2(s'«  sb)d + (.■>■'« 4 sb)
= (s' + s)cP + (s'a + sb'), since (s'a — sb) = 0.
Summing the leftside for all pairs we have the moment of inertia of the area about
AA. The first term on the right side is the area of each pair into the square of
the deviation ; and the sum of these for all pairs is the area into the square of the
deviation ; the second term on the right side is the moment of inertia of s and 4'
about 00", the sum will be the moment of inertia of the whole area about 00'. If
I^i and lo represent the moments of inertia i ound the axes A and respectively,
then
I A = S . cP f lo,
where S is the area of the figure.
Cor. — For any set of parallel axes, the moment of inertia about that axis which
passes through the geometrical centre is a minimum ; and those axes which give
minima values for the moment of inertia intersect at a point.
For every crosssection, lo will be of the same form, a constant multiplied by
the breadth and multiplied by the cube of the depth of the circumscribing rectangle.
Rankine puts generally lo = n'bh'^ where n' is the numerical coefficient of the
moment of inertia of the crosssection about its neutral axis, the other factors
being the breadth and cube of the depth of the circumscribing rectangle.
If through the neutral axis of a crosssection we draw a plane sloping at 45°,
it wUl form two isosceles wedges, or two portions, one on each side of the plane ;
the sum of the products got by multiplying the volume of each by the distance of
the point under its centre of gravity from the neutral axis gives lo for the ciosa
section ; for this purpose we may take each portion as a whole or subdivide it into
a number of parts if such is more convenient.
252 APPLIED MECHANICS. [CHAP. XIII.
For a rectangular wedge ^EKFBA' (fig. 10) let S = area of hase AEBA',
f = height EK, V — volume, xa = distance from A A' to the point which is under
the centre of gravity, then
V = ^S./; xq = fZt'.
For an isoscelestriangular wedge ACCS (6g. 3) let S = area of base ACC,
/■= height AS, Xo = distance from CC of the point which is under the centre of
gravity, then
r= ^S .f; xo = \0A.
If any doping plane be drawn through the neutral axis, it will cut off two
wedges ; and since the volumes of all such wedges are proportional to their heights,
we have
F: V::f :f::f .mh,
where V is ilie volume corresponding to/' the new value of EKin fig. 10, and of
AS in fig. 3 ; and 7n'h is the height of the isosceles wedge, that is the distance of
the skin from the neutral axis. Since the leverages are the same as before, the
statical moments of the wedges are also in the above ratio.
CHAPTER XIV.
c;rosssectio>'s : their resistance to bending and shearing,
AND distribution OF STRESS THEREON.
PtESISTANCE TO BENDING OF C ROSS SECTIONS.
We see then that Iq the moment of inertia of the crosssection
about its neutral axis is represented by the statical moment of
the isosceles wedges made by a plane sloping at 45^ and passing
through the neutral axis ; that the highest point of these wedges
is ya or 6 = 'ni'h, the distance from the neutral axis to the further
skin ; while M , the moment of resistance to bending, is repre
sented by the statical moment of the wedges made by a plane
passing through the neutral axis, and sloping so that the height
of the highest point of these wedges is^„„r6 =/, the stress on
the skin A or B, whichever is further from the neutral axis ;
hence
1^ normal stress on skin furthest from neutral axis j
distance of skin from neutral axis °
mh ma m k
is the proof, or working moment of resistance according as / is
the proof, or working strength of tlie material supposed to be
the same for both skins, that is. for thrust and tension.
Rectangular Section (fig. 1).— Let BCDE be a rectangle. Its
moment of inertia about DC one side is found thus : — The
CHAP. XIV.] RESISTANCE TO BENDING AND SHEAKING,
253
height of the isosceles wedge is zero at C aiul BC at /> ; the
average height is therefore \BC ; its vohmie is ^EB . BC, and
since the point below the centre of gravity of the wedge is two
thirds of BC from DC, we have
I DC = h^B . BC^ X ^BC= \EB X BC
as the moment of inertia of a rectangle about a side.
Now, if BE'' be a rectangular crosssection of which DC is
the neutral axis, then its moment of inertia about that axis will
be double the above ; and thus the moment of inertia of a rect
angle about an axis through the centre and parallel to a side is
In this case m' =
M
/o = 2 X I . EB . BC = ^b/i\
h, so that
f_
X J^bh^ = i/7>/r.
Thus, for a rectangle, m' = i, n' = yV, and u = i.
Hollow Rectangular Section. — For a hollow rectangular section
symmetrical above and below the neutral axis, that is, when the
whole rectangle and rectangle removed have their centres on the
12 k h » B .,,
^^Mkj^^^l T . +o— M~
io
Fijr. 1.
Fig. 2
same horizontal axis, the moment of inertia is the difference of
the moments of the two rectangles.
Let H, h, and B, b be the outer and inner dimension.^
respectively, so that the area is Bff  bh ; then
bh'
BH
fBH\
BR'
Hence, for a hollow rectangle,
bh' \
BH'j
and for the dimensions given in fig. 3, viz.,
H=30, B= 10, 7t = 24, J = 6 inches,
m = ^, n = jL ( 1  ^, ), and n = l(l ^^
254
APPLIED MECHANICS.
[chap. XIV.
we obtain
lo = 15588, and M = 10392/ inchlbs., if/ be in lbs.
Tabular Mdliod. — The following is a convenient form for
expressing the area, geometrical moment, and moment of inertia
of a rectangle: — Let BCDE (fig. 2) be a rectangle, and A A'
an axis parallel to DC, and let Y and y be the distances of its
sides EB and DC from AA'. The area is
S=h{Yy\ (1)
the breadth into the difference of the ordinates. For EL, part
of the isosceles wedge (fig. 10, Ch. XIII) EK = Y, CL = ?/; and
the geometrical moment of the rectangle BCDE about ^^' is
G , = volume of part of isosceles wedge
= h{Yy) X i(r+ y) = h.h{Y'  f), (2)
onehalf the breadth into the diflerence of the squares of the
ordinates of the sides parallel to the axis. By theorem C,
Fig. 3.
Ch. XIII, we have the moment of inertia of the rectangle BCDE
(fig. 2) about AA'
I^ = Io + Sd' = j'jEB . BC + EB . BC (^^)'
= ^^h(YyY + h{Yy)^^^=!j.b.(Y^^y^), (:))
onethird of the breadtli into the diflerence of the cubes of the
ordinates of the sitles paraUel to the axis.
CHAP. XIV.] RESISTANCE TO BENDING AND SHEAUING.
255
To apply this to the case of a hollow rectangle and a sym
metrical doubleT section, which give similar results (tig. ?>).
Choose 00 the neutral axis in the centre from symmetry ; and
if only the upper half of the section be considered, it consists of
two rectangles, the ordinates to the edges being 0, 12, 15 ; for
one rectangle Y = 15, ;/ = 12 ; for the other Y = 12, and y = '■
and for each (equation (3), /« = F>(^'^'  y^).
h
Y j ya
ya_v3
WVy')
10
4
15 3?75
12 1728
1647
1728
5490
2304
 1  1  *,.=.:«
if
X::
This tabular method applies to any section made up of rect
angles and which is symmetrical above and
below the neutral axis. As we require Gr
the geometrical moment of the semisec
tion relatively to the neutral axis when
we come to resistance to shearing, it is
convenient in making the table to find o* j=
Gr ; thus, for the symmetrical section,
onehalf of which is shown in fig. 4, we
have the following results : —
Fio;. 4.
b
Y
yi
Y^f
\h{Y'—f)
ys
y3_^,3
^A(y3_y3)
10
100
1000
12
8
64
36
216
512
488
1952
9
6
36
28
126
216
296
888
3
36
54
216
216
—


G, = 396

—
^^0=3056
Where (r is for semisection onlv.
256
APPLIED MECHANICS.
[chap. XIV.
Suppose the working strength / = 4 tons per square inch,
then the distance from 00 the neutral axis to the skin is
10 inches ; hence
M = 4, /., = tV X 6112 = 24448 inchtons.
m h
The use of G" will be shown at the proper place.
Symmetrical sections are suitable for materials for which
the strengths to resist thrust and tension are equal, as ^^and^*
become /'simultaneously (tig. 8, Ch. V). For materials whose
strengths are unequal, /„ is put for the greatest value oi ])„ and
fb for that of pb ; for wrought iron, for instance, the working
resistance to thrust is ./„ = 4 tons per square inch, while to
tension it is /j = 5 tons per sq. f^"^
inch. For symmetrical sections
in such materials the value of f
employed must obviously be the
smaller. Because of this property
of materials, crosssections are
made unsymmetrical above and
below.
Unsym met rical Sect ion. — Let
the unsymmetrical section (fig. 5)
be given, and let/a = 4, and/j =
5 tons per square inch. In order
to determine its resistance to
bending we find Iq '. we will also
?/^«
i^//////m^/AA
m.
v;r/*o—
ys'^
Jfj 5 Ions pe r aq. in,
Fiir. a
m o.'._
Axis
Orainatet.
A
find the area of the section,
and G' the geometrical moment of the portion of the section lying
to one side of the neutral axis whose position is not yet known.
Choose an// line as axis, say, the upper skin ; the diagram and
table show the breadths, and ordinates laid down from this line.
b
Y
Yy
Area
y^
y2 _ j2
t'^y y^)
ya
Y3.y3
i3(y3>3)
12
3
3
36
9
9
54
2
27
108
6
6
3
18
36
27
81
216
189
378
15
i6
lO
15
256
220
165
4096
3880
1940
4
i8
2
8
324
68
136
5832
1736
2315
8
2
16
76
304
2l6S
5781
20
400
8000
~
Ia =10522
S=93

Ga = 740
CHAP. XIV.] RESISTANCE TO BENDING AND SHEARING.
257
Q
Now y„= ^ = W = 8 "iches (by theorem A, Ch. XIII) ;
o
.. //6 = 20  8 = 12 inches.
Agcain, I, = /.,  Sija' = 10522  93 x (Vj")" = 4634.
The neutral axis divides the depth almost exactly as 2 to 3 ;
and if we had/a r/j : : 2 : 3, then both skins would come to their
working stress simultaneously, and we might obtain M by
f f
multiplying /« either by — or by  ; but since fa'.fb ' '■ ^ ■ o,
Va Vb
it is evident that both skins cannot come to their working
strength at once. In this example, when the skin B comes to
fb = o tons, the skin A will be at /j or 3^ tons per square inch,
and so is not at its full strength 4, yet the stresses are now at
their greatest ; on the other hand, the skin A cannot come to
its strength /„ = 4, because then the skin B would be at f x 4,
or 6 tons, and so be overtaxed ; hence M is to be obtained by
f f
multiplying /„ by the ratio —, not by — , which would give too
yb Va
much. Of the two ratios
•^ = 4 = 5, and *^ = ^ = 42,
select the less, and
M
Vb
X 7o = 42 X 4634 = 1931 inch tons.
Taking the neutral axis now as origin, the geometrical
moment for the part of the section lying above or below, is
(?o = 296 ; a quantity to be used in calculating the resistance
to shearing.
b
Y
Y2
Y2y^
hHY^y^)
i'5
8
§4
64
48
4
10
lOO
36
72
8
44
176
12
144

—
—
—
Gn 0=296
258 APPLIED MECHANICS. [CHAP. XIV.
Graphical Solution (fig. 6 : for the same data as fig. 5), —
Eeplace the areas of the rectangles by the forces (1), (2), (3),
(4), (5) acting at their centres of gravity, the amount of each
force being the same as the number of units in the area
corresponding ; draw the first link polygon as in fig. 3, Ch. VII,
and Z, the intersection of the end links, gives the centre of the
forces and therefore the centre of gravity and neutral axis of
the crosssection. The intercepts on 00 are the geometrical
moments of the areas respectively, that is, the product of each
area into its leverage about 00 \ the scale is found by sub
dividing the scale for areas by 10, since 10 on the scale for
dimensions was taken as the polar distance.
Considering these intercepts as magnitudes of forces acting
in the same lines as before, and drawing a second link polygon,
its intercepts on 00 are the products of these forces each into
its leverage about 00; or these intercepts are the areas each
into the square of its leverage about 00 ; the scale is derived
from the previous one by again subdividing by the polar dis
tance, 10 on the scale for dimensions. The proof is given at
fig. 3, Ch. VII.
The sum of the intercepts made by the second link polygon
is nearly /o ; being deficient by onetwelfth of the sum of the
products of the breadth of each rectangle into the cube of its
depth.
Also KL the geometrical moment of areas (1) and (2) is
nearly G^n, the deficiency being that of the rectangle lying
between the area (2) and the neutral axis ; this rectangle,
however, is small and has a short leverage.
There is also shown in fig. 7 a link polygon drawn for the
three areas (1), (2), (3"), which constitute the portion of the
section lying above the neutral axis ; the sum of the intercepts
is (?o. ^
Corrections. — The manner of correcting is shown in figs. G
and 7. Take the lines of action of the areas first along their
(say) n2')'per edges, and construct the first link polygon ; treating
the intercepts on 00 as forces acting along the unthr edges of
the rectangles, construct a second link polygon ; then to the
sum of the intercepts made by the second link polygon in
fig. 6, add, as a correction, onethird of its excess above the
sum of the intercepts made by the second link polygon in
fiu. 7.
CHAP. XIV.] RESISTANCE TO BENDING AND SHEARING. 259
The proof of the correction on Z,, which is exact, is shown
thus : — For any rectangle
Ia = \KY'  f) = hK^ y)ir'+ Yy + 1)
= S\d' + ^ {cP  Yy) } = ScP + ;1 (ScP  SYy) ;
where S is the area of the rectangle under consideration.
The smaller h becomes, the more nearly does the value of
Yy approach d, and the smaller is the correction. The
greater the number of rectangles into which the crosssection
is divided, the more nearly accurate will be the approximation
given by fig. 6 alone.
In the example, the correction on /o is about 4 per cent. ;
in the rectangle EBFE' (fig, 1), the areas of the two halves
each into the square of its leverage about 00 is less than I^ by
onetwelfth, or about 8 per cent.
Any crosssection can be blocked out into rectangles, and
/o and G\ easily calculated for it by the tabular method, if we
have a table of squares and cubes. If the crosssection has a
very irregular outline, it may require to be blocked into a great
many rectangles, and the construction (fig. 6) will probably
give a sufficiently close approximation.
Definition. — A crosssection for which the neutral axis
divides the depth in the same ratio as the strengths of a given
material to resist tension and thrust is called a crosssection of
uniform strength for that material.
The crosssection (fig. 5, for instance) would be of uniform
strength for a beam of a material whose resistance to tension
and thrust is as 3 to 2, A A being the compressed skin, and BB
the stretched skin ; for a cantilever of the same material it
would be turned upside down.
It is readily seen that the area between the first link
polygon and its end vectors produced to meet at L is propor
tional to the moment of inertia.
s 2
260
APPLIED MECHANICS.
[chap. XIV,
Fig. G.
CHAP. XIV.] RESISTANCE TO BENDING AND SHEARING. 2G1
Fi. 7.
262
APPLIED MECHANICS.
[chap. XIV,
Triangular crossseclion and sectiotis which can be divided into triangles. — They
are of no practical importance, but lead »ip to others which are.
Triangular Section. — On making /= f A, the wedges (figs. 2 and 3,;Ch. XIII)
become isosceles, and we have by substituting in the expressions given thereat
and
C = vol. of wedge on triangular portion above NA = ^^hk.
In the following way. /p may be derived from the rectangle (fig. 8)., The
moment of inertia of the shaded triangle about the central axis CC is half that of
the rectangle ; lor the triangle, then
/o = Jc  Sd' = i\bh  — X f  j = h
■bh^.
M = ^ /o = ^'ffbh\
Hence for a triangular section
».'/ = 3 , «'  "ST) " = "5 4 •
Hexagonal Section. — As an example of a built figure, we will t.ike a hexagon
about a diameter as axis (fig. 9}.
"
.
/
— \
' ^
/ '•
' / \
/ *'■»■*
sH
^mm^^
— r— V — '■ — ■
1'
\ »
/. ■*!■» /
\ ■ *
Jr. _ /
^^^^
jt
Fig. 8.
Fig. 9.
Taking a quadrant, we can divide it into three equal triaiigles;*the breadth of
each is \b, and the height \h ; hence the area of ench is "iVMj »"<! tl'e moment of
inertia of each about its own neutral axis is aV x \b x \h'^  \\ rribh'^ . If 7o be
the moment of inertia of the hexagon about 00, we have for a quadrant
j/ri = moment of each of the three triangles abotit its own neutral axis,
together with the area of each into the square of the distance of
its centre from 00,
hh^ bh i /h\ /h\' /h\ »i
bh\
CHAP. XIV.] EESISTANCE TO BENDING AND SIIEAUING.
263
Dividing the semihexagon above the axis into any set of convenient figures,
and multiplying the area of each by the deviation of its centre of gravity, we have
Also
Go = iV*^"'
m h ik ■' ® * "
Hence for a hexagonal section about a diameter as axis,
«,' = ^, n' = A = 05208, H = A = 10417.
Rhomhoidal Section. — As another example, we will take a section in the form
of a rhombus, with diagonal as neutral axis (fig. 10).
^•6>k B —
7b m&ka
Fig. 10.
Fig. 11.
bh
For upper half section, the area is — , the moment of inertia about its own
neutra
hence
1 axis is s^h f  j = ;^„, and the deviation of its centre of gravity is  ;
,^ bh^ /bh\/hy
therefore
Jo = Mh\ and M = £ /o = rrfbhK
Hence for a rhomhoidal section with diagonal as neutral axis,
til = ^, n = 4 S". " = "£"4".
Square Section. — A square section lying with its diagonal horizontal is a par
ticular case of this.
264 APPLIED MECHANICS, [CHAP. XIV.
In order to compare the strengths of a square section when lying with a side
horizontal, and when lying with a diagonal horizontal; let n be the side of the
square, then d = aV2 is the diagonal ; for b and /t. substitute a in the one case, and
rtV2 in the other ; then
M = i >^ M' = ^:V/(« y/2V = Y^/«' ;
so that M : M' : : \/2 : 1, being stronger when the side is horizontal.
Trapezoidal Section (fig. 11). — Observe that a triangular section is of uniform
strength for a material whose strengths to resist tension and thrust are in the ratio
1 : 2, or 2 : 1. It is evident, then, that for a material, the ratio of whose strengths
is between 1 and 2, a crosssection of uniform strength can be made by selecting
the proper frustum of a triangle ; that is, a trapezoid with the parallel sides
horizontal. Suppose the strengths are as 6' : s, not greater than 2:1; then to find
the ratio of the parallel sides B and b (fig. 11) ; a wellknown construction for
finding G is to lay off .B in a continuation of b, and 4 in a continuation of B, one
in each direction ; the line joining the extremities cuts the medial line at G ;
hence
h Tt
: S:s;
therefore
b 2s S
B 2SS
For instance,
if the strengths are as 3 : 2, the
in the ratio
i _ 2 X 2  3 ,
B ~ 2 x~3'~2 ~ *
when the neutral axis will divide the depth as 3 : 2.
Though such sections are thus far economical, still they are not the most
economical, as too much of tlie material lies near the centre of the section, where it
cannot act effectively in resisting bending.
Circular Section. — In the circle (fig. 12) inscribe a hexagon
with a diameter horizontal ; for the hexagon
b = d, h = dj~, and 7, = ^^ {d) ld^\ = OSSSd*,
an approximation on the small side to that for the circle about
a diameter.
Again, circumscribe a hexagon with a diameter horizontal ;
for this hexagon
h^d, b = d.~=jJ3, ami I, = ^\(^^^3ydy = 0602d\
an approximation on the large side.
CHAP. XIV,] ItESISTANCE TO BENDING AND SHEARING.
265
Taking the average of these, we have for the ciicle
where d is the diameter of the circle, ami the breadth and depth
•of the circumscribing rectangle.
It will be seen tliat n' = '047 is a close approximation, being
correct to 2 decimal places.
In the same way an approximation to (?'o may l)e found.
The exact value of /„ is found thus : — From the centre
of the circle (fig. 13) draw three axes OX, OY diameters
.at right angles, and OZ normal
to the plane of the paper. Let
for Circle lh^ie^
FiK. 12.
Fis. 1.3.
/, J, K be the moments of inertia of the circle about these
axes respectively, and let s be any small elementary area.
Now OA, OB, and OS are its leverages about the three axes
respectively, and by definition
/ = (.s X 0A) summed for each element,
J ={sx OW)
K=(sx OS^)
but, by Euclid i 47, OS ^ OA' + OB for each element, hence
K = I + J \ and further, since / and J are equal, each being the
moment of inertia about a diameter, K = 2/, so that if we find
K, the value of / is at once obtained.
On the element 5 build a column of material of unit density,
and whose height is OS; suppose this column to gravitate
tangentially, that is, at right angles to OS, then its statical
moment about OZ, that is its volume into OS, gives the moment
of inertia of the element about OZ ; all these columns will form
266
APPLIED MECHANICS.
[chap. XIV.
a solid standing on the circle as base, whose height at the cir
cumference is r, and whose upper surface is a conical surface,
apex at 0, and sloping at 45° to the plane of the circle. The
volume of this solid is that of a cylinder of height r standing
on the circle, minus a cone of equal height and base ; its volume
is therefore twothirds of the cylinder, viz.
F = "I X irr X r = ^rrr
= •?■;
every particle gravitating tangentially and in the same direction.
If we cut this solid (fig. 14) into slices by planes at right angles
to the plane of the circle, and passing through consecutive radii,
the slice between two adjacent planes will be a pyramid with
its apex at 0, and having its base on the cylindrical surface ;
one dimension of the base, there
fore, is r, and the other an arc
of the circle. Now by taking
the slices thin enough, the base
/ Plane sloping
mt 45°
Height at K=LK
r^Yal.of Cyl.Vol of Cone
Fis. 14.
Fis. 15.
of each pyramid becomes in the limit a rectangle, and the points
in the circle below the centres of gravity of these pyramids will
form a circle of radius r ; the whole weight may be supposed
to be applied by a cord on a pulley of radius ^r, and hence
K = volume of solid x radius of pulley = f tt?^ x ir= ^wrK
/o =
k"
or
64
hjr
M
yr<^ 32^ '
or
i/».
Hence for a circular section.
m' = i
,q'««.
« = 32 = 098.
To find G\ : — Suppose a wedge standing on tlie quadrant
(fig. 15) formed by a 45° sloping plane passing througli OX;.
CHAP. XIV.J RESISTANCE TO BENDING AND SlIEAlilNG.
267
the geometrical moment of a semicircle about a diameter will
be twice this volume. Cut the wedge into slices by planes at
right angles to the plane of the circle, and passing through
consecutive radii. Each slice is a pyramid with apex at O,
and of height ?•; its base is part of a cylindrical surface, and
the upper edge of the base is sloping. Let « be the short arc
as shown in the figure, and 8 its projection on OX; then in the
limit when the slices are thin, the base becomes a plane rect
angle whose dimensions are a and the height of the sloping
plane at K ; but height of ^ = LK = r cos B ; hence
volume of pyramid = ^ x r y. a . r cos 6* = ;^r^ . 8,
since
S = a cos 0.
For each pyramid \r is constant : so that the volume of
wedge on the quadrant is ^r"^ multiplied by
the sum of the quantities 2 ; this sum is r,
whether the slices be thick or indefinitely
thin ; hence the
volume of wedge = jr' ;
and doubling, we have for semicircle
CP,
2 ,^
3' '
If we divide G\ by the area of the semicircle, / • ; }'
we obtain the distance from the centre of ' *
the circle to the centre of graA'ity of the
semicircle ; k. — r — i
< rb i
 =^. ' '
Sir Fi?. 16.
Elliptic Section. — The ellipse is immediately derived from
the circle thus : — Let the ellipse (fig. 16) have the same minor
diameter as the circle above it, so that & = 2r ; all the vertical
dimensions of the circle are, however, to be altered in the
constant ratio a : r, since the vertical radius of the circle is
altered to a. Let the circle be divided into rectangular ele
ments, each with an edge lying on the neutral axis ; when the
circle is changed into the ellipse, each element remains of the
same breadth h' as before, but its vertical dimension is changed
from y to Y, where Y : y : : a : r.
268 APPLIED MECHANICS. [CHAP. XIV.
For each element of circle and ellipse respectively
/ = i7/(?/3 _ o»), and / = ^b\ Y'  0').
Por each element, and therefore for the whole figure, the
moment of inertia has changed in the ratio
For circle
hence, for ellipse.
y3 = ^3 . ^.3^
^'^Ul''')>'r^^'''
putting ^b for r, and ^h for a,
"i' li^'"
Hence, for elliptic section,
64' '' " 32'
For a seraiellipse about a diameter as axis, we have
4ff
3^"
G^' = f ?'«* ; and // = — .
Hollow Circular Section. — For a hollow circle or ellipse
the reduction is the same as for the hollow rectangle only
— replaces \ ; hence
hh'
«ii^.V^'
Eesistance of CrossSections to Shearing, and Distribution
OF Shearing Stress on a CrossSection.
In fig. 17 let AB be the crosssection shown in figs. 6, 7,
8, 9, and 10, Ch. V, and let A'B' be another section lying at a
small distance S.v to the left thereof ; in figs. 6 and 7, we
CHAP. XIV.J RESISTANCE TO BENDING AND SHEARING.
260
assume AB to be in a position such that F is greater than
Wi + IF>, that is, AB lies to the left of the section of maximum
bending moment; hence the bending moment on A'B' will be
less than that on AB.
Consider the horizontal equilibrium of AHH'A' (fig. 17)
part of the slice of the beam between these sections, and
bounded below by the horizontal face HH' a portion of the
plane CD (fig. 6, Ch. V). There is no stress on the free
surface. On AH the horizontal stress is indicated by arrows
Pay &c. ; and on A'H' by the shorter arrows p'a, &c. (see fig. 8,
Ch. V), shorter because the bending moment on AB' is less ;
in so far as these affect the horizontal equilibrium of AHH'A',
they may be replaced by arrows {pa  p'a), &c., on the face AH
alone. For horizontal equilibrium, there must be a tangential
•V
Pa ^ A' A Pa
\
/
/
\ *
< /
\ '
; /
\h'
X.— \
8r,
/ t
B'
""
B
^ 'A ji]:i'a) A
yuJi'^9^
B
0'
A_.A
B
Fiff. 17.
stress q acting towards the right on the remaining face H'H ;
and since H'H is very small, the stress on it will be of constant
intensity q ; therefore
q X area of plane H'H = sum of the arrows (pa p'a), &c.,
on plane AH]
or q . z .})X = volume of frustum of wedge standing on HAA
the shaded part of section as base, and of height
(Pa  Pa)
Now if G stands for the geometrical moment of the shaded part
of the crosssection relatively to the neutral axis, that is, for the
270 APPLIED MECHANICS. [CHAP. XIV.
volume of the frusLum standing on HAA of the isosceles wedge
made Ijy a plane sloping at 45° and passing through the neutral
axis, then the frustum of the wedge above is the same fraction
of G, that iiJap'a) its height is of y,, the height of the isosceles
wedge, and
^ , !> , VaPa fi
(1 . Z . CJO = Cr.
Va
Xow on the crosssection AB, the bending moment equals the
moment of resistance, that is
M^Ui'^L; .: ''" f. (1)
For the sections AB and A'B', ?/„ and the moments of inertia
are the same, because in the limit when ^x is indefinitely small
the sections coincide ; so that if M' be the bending moment at
A'B', then
Ji = — In, . .  — — —^, anu — — = — ,
Va Va A lla ^o ^0
where ^M stands for the small difference of the bending
moments on the sections AB and A'B' at the small distance
h' apart ; therefore
/MA
„ ■BM \&v .G
q .z.hx = y G, or q = — = —  •
In the limit when Sa; becomes indefinitely small, so does IM;
but their ratio ^ becomes F the shearing force at the section
AB, by theorem, page 118, and therefore
F G
since H' now coincides with H, we are warranted in assuming
q constant over IfH.
Now q is the intensity of the shearing or tangential stress
at the point H in the horizontal plane CD (fig. G, Ch. V) ; but
OHAP. XIV.] RESISTANCE TO BENDING AND SHEARING. 271
it is also the intensity of the shearing stress at the point H in
the vertical plane AB (see fig 14, Ch. II) ; hence (tig. 9, Ch. Y)
F a
^ = 7 • T (2)
is tlie intensity of the shearing stress at H, any point of the
crosssection AB. "Where F is the shearing force on the
crosssection ; /o is the moment of inertia of the crosssection
about the neutral axis ; G is the geometrical moment of the
portion of the crosssection beyond the point, about the neutral
axis ; and z is the breadth of the cross section at the point.
On the crosssection AB, it will be seen that q acts down
wards at any point 1I\ if we choose K oX an indefinitely small
distance ahoxe H, the tangential stress on ICK will also be q
since it is indefinitely near H'H ; and on ICK t\\Q lower surface
of AKK'A', q will act to the right just as on H'H, but on K'K
the uirpcr surface of the small prism KHH'K\ q acts to the left.
The horizontal stresses q form a couple tending to turn the
small prism in the lefthanded direction ; hence for equilibrium,
the vertical stress q on the faces KH and K'H' tends to turn
it in the opposite direction ; so that q on the face KH acts
downwards, an assumption made in fig. 9, Ch. V, which is now
proved.
On the other hand, if P were less than W^ + W^, &c., then
the bending moment on A'B' would be the greater ; q on face
H'H would act towards the left, the arrows q all round the
prism would be reversed, and so q on the face HK would act
upwards ; that is, q at any point H of the crosssection would
act upwards.
Observe that if q be evaluated for two different points of a
crosssection of a beam loaded in any manner, F and /o will be
the same in both, and the two values of q will therefore be to
each other directly as the geometrical moments of the parts of
the section beyond the points, and inversely as the breadths at
the points respectively. But for any given form of crosssection,
as rectangular, circular, elliptical, &c., the ratio of the geometrical
moments of the portions beyond two points definitely situated
in the section with respect to each other, is the same whether
the section be large or small ; whether, for instance, it be a
large circle or a small one ; so also is the ratio of the breadths ;
and hence the distribution of the shearing stress on a cross
section, or the manner in which jq varies from point to point in
the section, depends only upon the form of the crosssection.
272
APPLIED MECHANICS.
[chap. XIV.
It is instructive to know how the shearing stress is distributed
on crosssections of various forms employed in practice ; and
it is of the greatest practical importance to know ichere the
intensity is a maximum, the amount of that intensity, and
the ratio of its maximum and average values. This ratio is
an abstract quantity, and depends only upon the form of the
crosssection.
In the graphical solutions it is inconvenient to draw the
tangential arrows q as in fig. 9, Ch. V, since they interfere with
each other ; we will therefore draw them at right angles to
AB, when their extremities will give a locus which specifies
the distribution. For such a locus the origin will always be
at the neutral axis, the abscissae y are measured on the
vertical axis, and the ordinates q are
measured on the horizontal axis; and \y
at any point H whose abscissa // is
given, the ordinate q gives the inten
sity of the shearing stress.
Rectangular crosssection (fig. 18). —
Since z = h = constant, q will vary as G
the geometrical moment of the shaded
portion of the section, that is as the
isosceles ^frustum) wedge standing on
that portion ; now this wedge will be
greatest when y = 0, for then the shaded
portion will be the whole isosceles
wedge on the semisection. If ?/ be Fig. 18.
negative, we have more than the semi
section shaded, but the portion lying below the neutral axis
gives a negative geometrical moment, and q is again less
than ^o; hence g is a maximum at the neutral axis, it has
equal values for equal values of y above and below 0, and is
zero at each skin. The maximum value is thus
^^"M")
\a
q^n.»J ^J
y^=f^
V 5.=M
^o =
/o
(r'
F
u{(,o=
w^'
3F
2bh'
Graphical Solution. — Lay off OA = q„ ; and if we take A
as origin, qo  q the ordinate of any point will be proportional
to the volume of the isosceles wedge on the semisection minus
tliat on the shaded part, that is to the isosceles wedge on the
])art of the section from to // ; but the breadth being constant,
the volume of that wedge is proportional to Oy ; so that from
CHAP. XIV.] RESISTANCE TO BENDING AND SHEARING.
27J
A as origin, the ordinate of any point on the curve is propor
tional to .?/, and the curve is a parabola. The modulus of the
parabola is
go ^ 4 go
and the equation, with as origin, is
4 go fh
h \ 4
 r
Let g'aver. represent the average value of the intensity of the
shearing force on the crosssection, then
g aver. \ ~
shearinsT force F
area bh '
and we have for the ratio of the maximum and average intensity
(7 " 2'
1 aver. ■
Hollolo rectaw/idar crosssection, or symmetrical donhleJ sec
tion (fig. 19). — For values of y from ^H to hh, q varies as^ G
(, — J5=6...J
,3
20 =H
q„nnax. 3 (BirbhXRI{br >
q'^ver. ^(BHhK'j(Bb)
\
^o max .
:A,
S:s:KL::B:Bb
Figr. 19.
exactly as in the previous case, so that the locus from Y to L is
a parabola whose apex is ^i ; the modulus of this parabola is
greater than that for the solid rectangle since the constant
274 APPLIED MECHANICS. [CHAP. XIV.
divisor /„ is now less. For values of // between ^h and 0,
consider the eifect upon each of the items that make up q.
If the hollow were filled up, SA2 would be part of the para
bola in fig. 18 ; IqIS now less, however, and allowing for that
fact alone SA2 would coincide with LA^. The removal of the
centre decreases z from B to Zi; this increases each ordinate in
the same proportion, so that SA2 is still a parabola. Lastly,
for values of y between ^h and 0, the removal of the centre
alters the value of G from what it would be for the solid figure,
by the geometrical moment of the part of the hollow beyond y ;
now the geometrical moment of the part of the hollow is
§& I (ihy  i/'\ ', this leaves the equation consisting of a term
in y"^ and a constant term, so that SAo is finally a parabola
with apex at A2, but with a modulus different from that of
ZAi ; and hence q^ = OA2 is the maximum.
For semisection,
G\, = l(Bff'  W) ;
for total section,
Area
,S' = (BH  hh), and z, = {B  h)
is the breadth at the neutral axis ;
?'>= T
F G\ SF Bff'hJr
5
lo' z, 2 {BH'hh'){Bb/
F F
aver.
,S (BR  bh) '
. 2oma^ J,{BR'hh'){BHhh)
• • q\,„, 2{BH'  hh') {B  h)
For the dimensions given in fig. 19,
%^ = 24, and KS : KL : : G : 2.
2 aver.
The locus (fig. 10) gives tlie shearing stress on the horizontal
layers as well as on the crosssection. Now, as you pass from
the horizontal layer K cutting the web to another cutting
the liange, there is a sudden change of intensity from KS to
CHAP. XIV.] KESISTANCE TO BENDING AND SHEARING. 275
KL ; this change cannot be supposed to take place altogether
at K, as the free overlianging surface of the section at that
point does not bear any share of the shearing at all, and for
a small distance just above K the portion overhanging does
not bear its proper share ; in the vicinity of K, therefore, the
stress is not constant on the horizontal on the crosssection.
The consequence of this is to introduce shearing stress on the
vertical plane through K. In order that the intensity of the
stress should change from KL to KS absolutely at K, there
would require to be an infinite amount of shearing force on
the horizontal plane at K ; and since the intensity changes
from KL to KS in passing through a small vertical distance
at K, there must be a great amount of shearing force on the
horizontal plane at K. Hence sections of this shape very
readily give way by shearing at K ; castiron sections being
specially liable to do so under the shearing force developed
by irregularities in cooling. Eeentrant angles, as that at K,
are to be rounded off in castings and rolled plates, and filled
in with angle irons in built sections ; this allows the breadth
to change gradually from B to (B  b), and the intensity of
shearing stress to change gradually from KL to KS.
It follows as a corollary from tig. 19 that, for symmetrical
sections made up of rectangles with breadths diminishing
towards the centre, ju is the maximum.
Approximate mctJiod. — For a crosssection, such as is shown
in fig. 19, the web bears the greater share of the shearing
stress; and, moreover, the stress is nearly uniform in its
distribution. A close approximation to the resistance to
shearing for such a crosssection will therefore be obtained
by multiplying the area of the web into qo =f, f being the
shearing strength of the material. This is equivalent to
considering that the web bears all the shearing stress uniformly
distributed over it ; or that the central parabola, in such a
diagram as fig. 19, is replaced by a rectangle of height h and
breadth q„, and that the upper and lower portions of the
diagram are left out of consideration. The area of the web
required for a double T section is readily found by the converse
of the above, and is given by the equation
F
'?■
where aS* is the area of web in square inches ; F is the amount of
the greatest shearing force in lbs. at the section ; and /is the
resistance of the material to shearing in lbs. per square inch.
T 2
276
APPLIED MECHANICS.
[chap. XIV.
Circular Crosssection (figs. 20 and 21].— On the shaded
sector (fig. 20) suppose an isosceles wedge standing, made by
a plane sloping at 45° and intersecting the horizontal radius,
and cut up into pyramids as in fig. 15. In this case the sum
of the small quantities S is OC = r sin d, and the volume of the
wedge is \r x r sin 6/ ; for the sector, the geometrical moment
about OX, G = Jr^ sin B ; and deducting for triangle ODB, we
have for the geometrical moment of ADB about OX as axis
G = \r^ sin B  {I . r sin B . r cos 6) x /' cos B = ^ sin^ft.
Hence for the shaded part of the circle in fig. 21
G = f/' sin'0 ; : = 2/ sin B ;
and — = ^ sm* B,
z 6
where B is half the angle subtended at centre.
But y  r cos B ;
cos = — ;
sin^e =
r'  W
and
6^„
= Wy')\
A
D
^^m
\d
^
ram 9 ^
^ \
K
"e>^^
.se\
so that Ci OC i(r*  y"'), and the locus is a para ^
bola with its axis on OQ. Fig. 20.
Hence 5,, is the maximum ; and since
<t\. = i^ r^ ; Zi = 7 r^ ; aS' = ttf ; and c, = 2>' ;
Vomax.
^0
Lz
\F
F
<2
Elliittical section (fig. 21).— Let the ellipse shown in the
figure be derived from the circle by altering in a constant ratio
the breadth ~ of the circle at any point, to z' the breadth of the
ellipse at the point corresponding, that is, let z = nz .
Suppose the shaded part of the circle to be divided by
vertical lines into elementary rectangles, and let the ellipse
be correspondingly divided ; by considering each of these ele
mentary rectangles, it is easily seen that I, G, S, and z for the
CHAP. XIV.] KESISTANCK TO BENDING AND SHEARING.
277
ellipse are derived from the corresponding quantities for the
circle by multiplying by n ; for the circle we had
F a
^= 77'
Fig. 21.
so that for the ellipse
F iiG F G
4 F
111 nz 1 z 6 rnv
9_ aver. ~ / '■> ~~? ~ q '
77?'/' q aver. ^
the same as for the circle.
St/mmetricnl section of three rectangles. — When the middle rectangle is the
narrowest, qo is the maximum, as we saw in the previous case.
go„,ax ._ 3 {bB:UBb)h'\{hII*(Bbjk]
q'a.er. 2 { bJli (BbJ h'l B
\ KS KL ::b : s
3hmax.y f >
/ Umax.^ 3 (SLh){bS^(B h)h\
quaver. 2 bSU(Bb)h'
Fig;. 22.
278
APPLIED MECHANICS.
[chap. XIV.
In fig. 22 the middle rectangle is the broadest ; and in this case, the intensity
of the shearing force has two maxima values, one at y = 0, the other at y = .
Their values are shown on the figure. Numerically they are
qo = QUF, and qn = 013^; ^^j = 13 ; 4 = 12.5.
? aTer.
'i aver.
Triangular section (fig. 23). — Let ij, which includes its sign, be the ordinate of
tlie base of any portion such as is shaded in the diagram ; then for that portion
(?o = ^Y y). \!/+l[^y)} = ^ (2A  3y)(A + 3y).
Now
F Go ^ . F .
g = f — , and since — is constant.
Go
9 « — , or q x (2A  3y)(A + 3y).
The sum of these factors is constant, and their product is therefore greatest when
they are equal ; that is when
h
{2h  3y) = (/* + 3y), or y = g ;
hence $ is a maximum at middle of depth h, and the locus is a parabola with its
axis horizontal.
K I, X
Fig. 23.
For the portion above this central point
^'=^•2 (6^*2) ='24'
For the whole section /o = s^JA* ; heme
■ 24
FiK. 24.
Jo • 2
and :
3F
bh^ b bh '
^ ^ 2
CHAP, XIV.] KESISTANXE TO BENDING AND SH BAKING. 279
2^'
q aver.
iS \bh bk ^'aver. 
The equation to the parabola is easily found.
Rhomboidal section (fig. 24). — For a portion such as is shaded, in the ui)per
half of section, that is for positive values of y,
<^.=i(^.)^[..i(^,)j=ijr.^^,).
Now
G
q X — IX h'^ + S>/
(:')•
so that the locus is a parabola with its axis horizontal. The sum of the factors
.'/ ( >/) is constant, and the product is greatest when these are equal, so that
when y = , ^ is a maximum. The locus for the upper half is a parabola whose
8
axis is A above the neutral axis, and for the under half a similar parabola
symmetrical below.
For that part of the section above y = ,
o
Zh 3b
For the crosssection
1
isection
Ko = aV*y +**2" (^.j =96= ••• ^» = I8'
?A
FG 9F
•IF
qh
4bh' ^^''^' ^bk bh '
In . z i bh " ifbh bh. 5aver.
Regular hexagonal section, diameter vertical (fig. 25). — If we suppose the sloping
Base of each
Par.Seg.s ih.
Heights as 1'i:9.
Fig. 25.
sides above and below produced to meet the neutral axis, we have a form like
280
APPLIED MECHANICS.
[chap. XIV.
fig. 24; and for values of y from \h to \h, everything so far as regards q is the
same as if those sides were produced, except the constant /„ which is less ; hence
for those values of y, the locus is a parabola with its apex A\ on the horizontal \h
above the centre as in the previous case. For values of y from to \h, suppose
the vertical sides produced and tlie rectangle completed ; the four triangles which
require to be added on to complete that rectangle increase the constant /q, hut so
far as that affects q the locus is still a parabola with its apex on the neutral axis ;
they also increase G^q for every value of y, augmenting both the constant term and
the'term in y^ in the expression for q ; the locus, however, is still a parabola with
its axis coinciding with the neuti'al iixis, but the modulus is altered.
Now we know that this parabola intersects the other pair at K and i, points
beyond their apexes, hence OAi. is greater than ordinate of 4i ; and ?o = 0A% is
the maximum. We have readily
G\ = hhli^ ; /o = yfri/t^ ; =1 = * ; and
FG\, 28i^ , _ ^ _ 4 f
?omax: = T;;^ = X^/' 9 aver: " p " 3 j^ 5
90 max .
Examples.
1. Find the working moment of resistance to bending, and the working
resistance to shearing of the section, of which the upper half is shown in fig. 26.
In wroughtiion built sections, the
piercing of holes for rivets reduces the
effective area to resist tension but not
to resist thrust ; the area as thus reduced ,«j [t— 
is to the original area in the ratio of "^
about 4:5, which is the same as the
ratio of the strengths ; hence it is usual
to make such built sections symmetrical
above and below, to consider the woridng
resistance to tension and thrust as each
eqtial to 4 tons per sq. inch, and then to
neglect tlie fact that the holes diminish
the effective sectional area; the working
resistance to shearing may also be taken
at 4 tons per square inch. Fig. 26.
3\
b Old.
Dif.
Dif X b
Ord.
Dif. Dif.xt
Otd.
JJ,/.
Dif.xi
3
12
3
I
10
S
35
2
45
35
24
I3'5
35
100
64
123
36
52
12
216
78
6
1 1000
1
1 512
43
4SS
469
43
'952
469
14
41 X 2=82=S.
Gvo=3oo.
2435 X 2=487O=/0.
•CHAr. XIV.] KESISTANCE TO BENDING AND SHEAIUNG.
281
M = r, 1» = ^^r X 4S70 = 1948 inchtons.
F0, qohzx 4x4870 x 1
^»=:^= •• ^ = lF7^ — 300— =*^^^'^"^
2. If the rivets a, a, fig. 26, be pitched at 4 inches apart, find the diameter
necessary for each rivet.
For part of section beyond SS, (^o = ^ x 12 (10=  9) = 114 ;
?o =
FGq _ 67 X 114
~hz ~ 4870 X 12
= 13 tons per square inch
is the intensity of the sliearing stress on the horizontal plane SS at the cross
section, and it will be sensibly constant on SS for a few inches on either side of
cross section. There is one rivet for each 24 sq. inches of SS; lience a rivet has
to bear 13 x 24 = 3*12 tons of shearing force. If the rivet be very tight, this
shearing force would be uniformly distributed on its section, and the area required
would be found by dinding by /= 4. If the rivet be not perfectly tight, there
will be a bending moment on it, and we must consider the shearing stress
distributed as in fig. 21 ; in which case
</y = 4; .•. (?' = fx4 = 3 tons per square inch average intensity.
area = ^ = 1'04, which gives a diameter d = \2 inches.
Taking 1*2 inches as the diameter of the rivets, the six holes reduce the area
of the crosssection by 168 sq. inches, almost exactly a fifth as we supposed it
would do.
This is on the supposition that the section bears the full shearing force that it
can resist ; at other crosssections where the shearing force is less, the rivets might
either be made smaller in diameter or be more widely pitched.
3. Find the resistance to shearing of a crossformed section ; width of each
pair of wings 5 inches, thickness of metal 1 "5 inches ;
/= 4 tons per sq. inch (fig. 27) ; see also fig. 22.
For semisection
G'o=lx 10 (25 75) + ^ X 5(7o20) = 57
G' Q 5"7
and — = — ;r = 1'14 IS a maximum.
^1
i/o = i X 15 (25»  753) + i X (753  0^) ;
therefore 7u = 16'6.
For portion beyond K
G^o = ^ X 15 (25  lb) = 427 ;
Gq 427
and — = = 2"84 is therefore the maximum.
: 10
F G^, F
^ = ZT" ^=16:6^ ^^^^
F = 234 tons.
SI
dS
T
vm
(
:)^
/:iy
\S
C7
:)
,IZ
—.075
.Ln
Fie
4. If the section (fig. 27) be built of halfinch plates as shown, and fixed with
^rivets a, a, oneinch diameter, find what should be the pitch near the crosssection
282
APPLIED MECHANICS.
[chap. XIV
which is under ^ of the full working sheaving force F, if /= 5 tons per square
inch.
For the part beyond SS, Go = ^ x lo (2.5'  lb) + J x 5 {lb 252) = oo ;
therefore
?.25 =
IF Go _ ^ X 234 X 55
; ~ 166 X 5
= '5 tons per square inch
is the intensity of the shearing stress on the horizontal plane SS.
If the working strength of the rivets be 5 tons per square inch, the average
resistance to shearing will be  x S = 4 tons per square inch nearly : hence b ^ 4,
that is ^t*" of the area along ST, and normal to the paper, must be pierced with
holes. But ST= 2 inches; so that for every four inches measured on iST normal
to the paper, there should be one square inch pierced ; that is, rivets about one inch
in diameter should be pitched four inches apart.
\f.... 4 *2..J
9 I U. 9
Semi. Symmetrical Section,
Fig. 28.
FiK. 29.
5. Find the moment of resistance to bending of the symmetrical section, the
upper half of which is shown in fig. 28 ; the material is wroughtiron, for which
the weaker working strength is fa = 4 tons per square inch.
Consider the two triangles as one of breadth 8 inches ; for a triangle about its
own neulral axis, / = Jg dh^ = 48 ; hence for semisection
7o = (48 4 24 X 82) for triangle + ^ x 2(10' x O^) for rectangle ;
therefore /„ = •^501 ; and M = ,— Jq = ,*,r x 4501 = 1800 inchtons.
6. For the crosssection shown in fig. 29, and which is the same as fig. 28,
find the working value of M, if the working strengths of the material be f„ = 4,
and fb = 5 tons per square inch.
Choose ££ as an axis of reference ; then
5 = 24 + 20 = 44 ; Gs = {24 >. S) + ^ \ 2 (10  0=) = 292.
/^= A/u of Ex. 5 = 2251.
Gb 292
yb= Zr = — = 664, and >ia = 10  664 = 336 ;
o 44
/« =//?«. 'Jb = 315 ; ^" = ^ = 119, and ^ '
y„ 336 ijb
taking the smaller value, we have
M = "752 X /„ = 237 inchtons.
(3G4
= 752:
CHAP. XIV.] RESISTANCE TO BENDING AND SHEARING. 283
7. Find M for the section, the upper half of which is shown in fig. 30; the
material is cast iron, for which /j = 2 tons per square inch.
For the circle about its own neutral axis
1= ^bh^ = 049 X 8< = 201 ; and 6' = 5026.
64
For the crosssection,
Uo = (201 + 5026 X 8=) + A X 3 (4^  0*) ;
.. Jo = 6964 ; and M = jy^ /q = A x 6964 = 1161 inchtons.
Fi?. 30.
Fig. 31.
8. Find M for the section shown in fig. 31, the working strengths /« and/*
being 4 and 5 tons per square inch respectively.
It is convenient to choose CC the diameter of the ellipse as the axis of
reference, and we have
Ellipse.
5 = 7rx4x3 +
Gc =
Middle rect. Lower rect.
3x6 + 10 x2 = 757;
t Jx 3(9 32) + i X 10 (1129=) = 308;
ic = — X 8 X 63 + 1 x 3 (9»  33 + J x 10 (IP  93) = 2793 ;
308
y =
= inches sensiblj ; so that i/a = 7, and tjbl inches ;
S 757
J^ = IcSr = 1582.
The neutral axis being sensibly in the middle, take ft = 4 the smaller strength,
and M = f/o = 904 inchtons.
9. Find the resistance to bending for the section shown in fig. 32 ; the
dimensions are in inches, and the strengths of the material are /« = 4 tons (thrust),
and /ft = 5 tons (tension) per square inch.
Choose CC the diameter of the ellipse as an axis of reference ; and for semi
ellipse 5 = 23 ; 6 G^c =  30 ; Ic = 53.
Alls. M = 519 Jo = 11G7 inchtons.
284
APPLIED MECHAKICS.
[chap. XIV,
10. Find the resistance to bending of the wroughtiron section, fig. 33 ; the
dimensions are in inches, and the metal everywhere is 1 inch thick.
1 ; '_\ ji.. (gi. )
0012
y.^as.r.
Fig. 33.
Choose CCthe diameter of the hollow semicircle as an axis of reference.
Ans. M = •512/n = 668 inchtons.
1
S
<= \'^
Hollow semicircle . .
Rectangle above CC,
First rectangle below CC,
Second rectangle below CC,
14
4
10
10
 41 144
— 8 21
50 333
105 1 1 103 j
Whole section .. 38 106 1 1601
11. For a wheel, design an elliptical spoke of approximately uniform strength,
of a material whose smaller strength is 2 tons per square inch ; length of spoke
3 feet, load at end due to a force applied to the circumference of wheel oV*^ ^^ ^
ton, and at each crosssection the depth is to the breadth in the ratio 3:1.
M„
Wl = _,^, ft. tons = 18 inchtons ;
M = nfbli^ = —x2 X b {3b) = llGb^ inchtons.
Equating these values of M and M, we have
17643 = 18 ; therefore b^ = 1.
Therefore b = I, and A = 3 inches, are the diameters of the elliptic section at boss,
while twothirds of these are the diameters at tyre (see tig. 8, Ch. XIII).
CHAP. XIV.] RESISTANCE TO BENDING AND SHEARING. 285
12. Find the thickness of a castiron pipe whose external diameter is 2 feet,
that it may have a working moment of resistance to bending of 800 inchtons; the
smaller working strength of castiron is 2 tons per square inch.
Put D = 24 inches, and let d be the inside diameter; then
/, = ^(2)*rf*), and M = ^/,:
therefore 800 = OOSIS [B^  d*) or {D'  d*) = 97800.
Ans. d = 22 inches ; and for the thickness of the metal, we have t = \ inch.
13. The section of a beam is a rectangle 2 inches broad by 6 inches deep ; the
material is wroughtiron whose working resistance to shearing is /= 5 tons per
square inch. Find the working resistance to shearing of the crosssection.
F will be such that qo the maximum intensity shall attain to /= o tons per
square inch ;
<?o = 5 ; but <?'„er. = f?o = V ^^^s.
F = average intensity x area = q' . bh = ^ x 12 = 40 tons.
14. Find the resistance to shearing of the crosssection, fig. 5, taking/= 4^tons
per square inch, ^q is the maximum value.
From the tabular form or graphical solution we had
h
= 4634, (?'o=296,
and ri = 15 ;
F&'o
Fx 296
or 4 —
F = 94 tons,
vi'
4634 X 15'
but ^0
15. Find the resistance to shearing of the section shown in fig. 4, taking y the
resistance of the material to shearing at 4 tons per square inch.
F(?'o , Fx396 _ _.
?o = y^; •^ =fiTT^^5 ••• F = lSotons.
loZ\ 6112 X 3
16. In section (fig. 4), find the ratio of the maximum and average intensity of
the shearing stress.
FG'o ^ , F
9o = f — , and q = x,;
to^i o
., . q^ma^. SG'o 120 X 396 „ ^
therefore = ^^ = FTT, o =26.
9 aver. 'o^i 6112 x 3
17. Find F for fig. 19, taking/= 4 tons per square inch ; and find the ratiojof
the maximum and average intensity of the shearing force.
i7o = i X 6(10363) +i X 2 (630^); .. /o = 3424 ;
(?'o = ^ X 6 (102  62) + 1 X 2 (62  02) = 228.
^5 = 6(106) + 2(60); .. .S=72.
Ff"
21 = 2, and we wish to have qo = 4 ; but qo = — — ; .•. F = 120 tons.
Joi
F go max 4
? aver. = r„ = f tons ; and = v = 24.
72 q aver. f
286
APPLIED MECHANICS.
[chap, XIX.
18. Find the ratio of the maximum and the average intensity of the shearing
stress for fig. 31.
Take the under half which consists entirely of rectangles, and find G'o = 157'5.
Show /o = 1582, jS = 757, and ci = 3. ^^^^ <?omax. _ <^.^
Q aver.
19. Compare the resistance to shearing for the crosssection shown in fig. 3
as obtained by the exact and approximate formulae respectively, taking /= 5 ton
per square inch.
loziqo 15588 X 4 X 5
F = —^ = g^2 = ^^*^ tons (exact).
F =/<S' = 5 X 9fi = 480 tons (approx.).
20. Suppose the Meb of the crosssection shown in fig. 4 to extend to the
under side of the upper plate, take /= 4 tons per square inch, and find approxi
mately the resistance to shearing. Compare the approximate result with that
obtained for Ex. 15.
F = 4 X 48 = 192 tons.
CHAPTER XV.
STKESS AT AK INTERNAL POINT OF A KEAM.
Keturning to fig. 6, Ch. V, we have now found the intensity
and obliquity of the stresses at the point H, on AB and CI)
the pair of rectangular planes through it, viz. :— On CD the
liL = ( R' = q, in figs. 198 to 201 = r,,
in fig. 47.
N = tangential component stress on the
rectangular planes AB, CD.
OL = p, in fig. 84 = >„, in fig. 47.
= normal component stress on plane
AB.
Fig. 1.
total stress tangential and given by q in such diagrams as
fig. 18, Ch. XIV, and on AB the total stress of intensity r at
obliquity y (fig. 1) given in terms of its normal component;;,
and its tangential component q on such diagrams as fig 18,
CHAP. XV.]
LINES OF STRESS ON A BEAM.
287
Ch. XIV. The planes of principal stress at H are to be found
as on fig. 10, Ch. IX; making that construction, and noting that
y is a right angle, we have (fig. 1)
OM = \p, and 3IR = jt^ + g^.
These are readily calculated, and become known quantities ; we
also have
tan 26 = ^^ = ^^^Q^^ g (fig 18, Ch. X IV)
ML halfarrow p (fig. 3)
giving d the angle which the plane of greater principal stress
makes with the plane of crosssection AB^ both planes being
normal to the paper. Further
OM + MR = intensity of greater principal stress (of same
kind as arrowy;, fig. 3) ;
OM  MR = intensity of smaller principal stress (of oppo
site kind from arrow p, fig. 3) ;
and MR = intensity of greatest tangential stress, being
on the planes inclined at 45° to the planes
of principal stress ;
that is, we have the maximum value of the thrust, tension, and
shearing stress at the point H.
288 APPLIED MECHANICS. [CHAP. XV.
That the two principal stresses are of opposite kind, follows
from the fact that the stress on the plane CD is wholly tan
gential (fig. 14, Ch. II).
If the principal planes be drawn through H for a short
distance on each side, that upon which the stress is thrust by
a full line, and that upon which it is tension by a dotted line,
and if this be done for a number of points H on the elevation,
then the full lines and the dotted lines will each form a series
of polygons with short sides ; and if we take the points U close
enough, each polygon becomes a curve. These curves are called
lines of principal stress ; and the tangents thereto at any point H,
where a dotted line and a full line curve intersect, are the planes
of principal stress at H. See Kankine's " Applied Mechanics,"
sect. 310, and his " Treatise on Shipbuilding."
These curves have the following properties : — They cut the
neutral axis at ■45'^ ; for, considering a point there, ? = 0, and
in figure 1 R will fall on the line R'O produced upsvards,
26* = 90°, and 6 = 0' = 45°. Similarly all lines that meet the
end section, that is the section over the point of support, meet
it at an inclination of 45° ; for, since the bending moment is
zero, we again have p = 0. All lines meet the upper and under
skin at right angles, since q = a,{, A and B (Hg. 4).
In the elevation of tlie same beam, these curves will be
different for different loads, except when the loads are kept in
the same positions on the beam and altered in a fixed ratio ;
thus, for a rectangular beam — if the load be uniform, then for
positions of H ranged on a horizontal line, the arrows p and q,
figs. 3 and 4, will both vary, since the bending moment and
shearing force alter for each crosssection ; if the load be at the
middle of span, p will vary, but q will not, since the shearing
force is the same at each crosssection.
The planes of principal stress will differ in the elevations of
two beams which are loaded alike, and whose crosssections are
different but uniform throughout in each case ; for instance, if
the crosssections be a rectangle and a triangle respectively, we
have (/max. ft the centre of the crosssection in both cases ; the
neutral axis, however, is at the centre in one case but not in the
other ; so that, though both be loaded alike, the planes of prin
cipal stress will differ in the two elevations.
In designing beams, it will be seen that we have followed
the usual practice of considering p„ and pb at the section of
maximum bending moment to be the greatest value of thrust
and tensile stress respectively, and //„ for the section over the
greater supporting force to be the greatest intensity of shearing
CHAP. XV.] LINES OF STRESS ON A BEAM. 289
stress. Strictly the points at which these maxima occur are to
be defined thus : — Let .r, ;// (fig. 6, Ch. V) be the coordinates of
any point H referred to the centre of the neutral axis as
origin ; then for the crosssection at x
M^ = npaW, and pa = rf. ;
also p,:v^::y:r,a ,. Vy = ^^^^,
We also have (f = ^, and q^ = — ^^ ;
k being the ratio of the maximum and the average intensity of
the shearing stress for such crosssection ; q^ is readily derived
by considering the manner in which the stress is distributed.
Then (fig. 1)
ME^ = q\ + ap,y,
and by finding the values of x and y which make this a maxi
mum, we get the point at which the intensity of the shearing
stress is greatest ; also
mi ± MR) = ip, ± Jq\ + [^^py)\
and by finding the values of x and y which make this sum and
difference a maximum respectively, we get the maximum value
of the intensity of the thrust and of the tension.
To find these maxima is difficult even in the easiest cases ;
for instance, for a beam of uniform rectangular crosssection,
loaded at the centre, we have
_ ^ W{c  x) _ 3 W{c  x) _ y_ _ 6 W{c  x)y
^'^ ~ nbh' ~ hh^ ' ^^ " p;^" ~ Ih^ " '
and ME^ = (i^)^/^^  ^I/J + 16r(c  xf] ;
u
290 APPLIED MECHANICS. [CHAP. XV,
for every value of y, MBr is greatest when x = 0, and then
the greatest vahie of this is obtained by putting
h ^ oWc
2/ = ±  , then MR = ^^
at the surface of the beam at centre of span.
The quantity Py is greatest when
h ^ 6W{cx)y SWc
X = 0, and y = ^l then p, = j^ = ^,
and since p^y ^^'^^ ^^J^ thus have their greatest values at the
same point, viz., at the surface at the middle of the beam, the
greatest principal stress is there situated, and its amount is
OM+MR = ~j'.
oh'
Determination of internal stress by the ])olariscope. — On fig. 5
is shown a plate of glass strained in a frame like a beam
supported at the ends and loaded
in the middle. The plate is then ^^WF
placed in a lantern, when a cone , ^ ^,j^ ~:~ , "^
of polarized light throws an ' '"~
image on a cardboard screen.
The whole image will appear
luminous, except along certain
lines, which are the loci of points
at which the axis of greatest ^ i^' •'•
or least elasticity is parallel to the vibrations of the rays leaving
the polarizer, which lines will appear black. If the plate be
E
""i'.iifiiiimM'™'"^"'''^''" — ^'"W I
^•^
C
%
* Extract from " How to use the Poliiiiscope in tlie Practical Determination
of Internal Stress aiul Strain." By tlie late Prtifes.or Peter Alexander. .\ paper
read liefore the Matheniaiic.il and Pliy&ital Section ot tlie Philo^opllical Society of
Glasgow, 3rd February, 18S7, und rejMinti'd in full in our liist edition by the kind
ptrmissioa of the Society.
CHAP. XV.]
LINES OF STRESS ON A BEA.M.
291
now rotated about a line through its centre and perpendicular
to its faces, the black lines will assume new positions in tlie
plate.
Fig. 6 shows the forms and positions of the black line for nine
positions of the bar, from the horizontal position to the vertical
position and corresponding to the first eight points of the
compass.
Fig. 6.
It will be observed that the central point of the bar is
fdivays black ; consequently we conclude that there is 7io stress
there. At every other point of the black loci lines of stress
cross them horizontally and vertically, that is to say, parallel
and perpendicular to the direction of vibration of the polarized
ray.
u2
292
APPLIED MECHANICS.
[chap. XV.
A practical way of drawing the lines of stress is to rotate
the card in front of the lantern simultaneously with the
rotation of the glass bar, so as always to have the
image of the glass bar occupying the same position
on the card, and let the tracing point, while carried
round by the card, at the same time move horizon
tally, so as always to be on the black loais projected
on the card. The line drawn in this way v/ill be
an accurate delineation of a line of stress. This motion
of the tracing point may be accomplished as follows : —
Let the tracer be a small toothed wheel, T (fig. 7),
capable of rotation about a fixed axis, AA', at the end of
a tracing pen, F, and let a heavy load W, be attached
rigidly to the pen at the end of a rod, JD, so as to
compel the axis, A A', of the tracing wheel to be vertical
when in use.
The tracing point will thus be incapable of any motion
but that communicated to it by the rotation of the card,
and the horizontal motion given to it by the rotation of the
tracing wheel as the tracing point is made to keep on the
black locifs.
The network of dotted lines (fig. 8) shows what is the state
of internal stress of the glass bar, and consequently that of the
loaded beam.
The black loci, i, ii, Sec, correspond to the angular positions,
1, II, &c., of the card. The dotted lines are the lines of internal
stress.
Example.
318. A beam of constant rectangular section 9 ins. broad, 20 ins. deep, and 20 ft.
span, bears a load of 96 tons uniformly distributed. At a point in the section half
way between the centre and left end, and halfway between the neutral axis and
CHAP. XVI.] CURVATURE, SI.Ol'E, AND UEFLECTION. 293
upper skin, tind the greatest intensities of thrust, of tension, and of sheuriug
stress; and find the inclinations of the planes of principal stress at the point.
.Vf. = 2160 inchtons ; irPabh = M^; ■■ . pa = — — = 3'6,
and py = ^pa =1*8 tons per square inch.
2? 24 2
/^ = 24tons; ,' = _ = ^^ = _ ; ,.= ./= '^ ;
and qy = ?o = '15 ton per square inch ; ME = lb + '9^
Therefore MR = "91 ton per square inch max. intensity of shearing stress.
OM + MR = "9 + gi = rSl ton per square inch max. (thrust) ;
OM — MR =  01 ton per square inch max. (tension).
Tan 2d = ~= — = 016 ; .. 26 = 9° 28' ;
ML 90
6=4° 44', the inclination of the plane of greater principal stress to the cross
section.
CHAPTER XVI.
CURVATURE, SLOPE, AND DEFLECTION.
At any point in a plane curve, the direction of the curve is that
of the tangent at the point ; and the curvature is the rate of
change of direction at the point.
If two points be taken on a circle, the change of direction
as you pass along the arc from one to the other is the angle
which the tangent at one of the points makes with the tangent
at the other ; this angle is equal to the angle at the centre
subtended by the arc between the points ; since this change of
direction takes place uniformly, the rate of change is found by
dividing the total change by the arc ; the total change as just
stated is the angle at the centre, and this angle when expressed
in circular measure is the ratio of the arc to the radius ;
dividing this ratio by the arc, we then have for every point of
a circle,
1
curvature =  • (1)
r
If we take a point in any plane curve, we can find a circle
■which coincides with the curve for a short arc in the vicinity
294 APPLIED MECHANICS. [CHAP. XVI.
of the point ; and if p be the radius of that circle, then the
curvature of the circle everywhere being 1 ^ 0, it is clear that
for that particular point of the plane curve,
1
curvature = • (:)
P
For any point in the neutral axis of a beam (or cantilever), we
have (fig. 5, Ch. V),
1 \ds) strain on any horizontal layer
curvature =  = = .. ^ „ :— 7 — ,
p y dist. irom neut, axis to layer
\dsj \dsj
strain on either skin
y„ yj dist. of skin from neut. axis
Now the strain on a horizontal layer is equal to the longi
tudinal stress on the layer divided by U the modulus of
elasticity of the material (see page 7) ; hence
1 stress on any horizontal layer 1 _p 1 Pa 1 ,
p dist. of layer from neut. axis £ y' E ya' E'
Pa being the normal stress at the skin on the crosssection,
and 7/„ the distance of the skin from the neutral axis as shown
in fig. 3, Ch. XV, we have
Pa ^^
hence at any point of the neutral axis, the curvature due to any
load which induces the bending moment M on the crosssection
passing through the point, is
/ being the moment of inertia of the crosssection, and U the
modulus of elasticity of the material. Choosing as origin that
point where the neutral axis crosses the section of greatest
bending moment, we have the curvature at that point for a
beam or a cantilever,
/o<i ti io
and if we wish this to correspond to the proof or working load.
1 _ 1 ,/■ 1 / .
p„ E ' v/o E ' m'h '
1
E
tK°=i_ /. or ^ ~'~
= 1.
CHAP. XVI.] CURVATURE, SLOPE, AND DEFLECTION. 295
it is only necessary to make M^ the bending moment due to
the one or the other. These vahies of J/„ can be easily obtained
from equation 3 by substituting for^;„ the value /corresponding
to the proof or working strength of the material. For this
crosssection of greatest bending moment, putting t/o instead
of Ua,
so that
Multiplying the value for by this quantity, which being
unity will not alter the value, we have
(5)
The slope of a beam or cantilever at any point whose abscissa
is Xi, the origin being at the point where the neutral axis is
horizontal, is found as follows : — Let S be any point between
the origin and that point ; then at fig. 5, Ch. V, we had for the
increment of slope between T and S, supposing these points to
be indefinitely close,
di =  X clx ; (6)
P
if we add these increments di from point to point between the
origin where the slope is zero and the point Xi, it gives us the
slope at that point ; and
1
\E ll\i/, mJ
f ^n.
p
' Ey,MJ
I;©
dx ; (7)
the righthand side expresses, in the language of the Integral
Calculus, the summation of the products equal respectively to
these increments.
In the case of a beam symmetrically loaded, the origin will
be at the centre of the span, and in practice, for any load, the
origin will be sensibly in the same position ; in the case of a
cantilever, the origin will be at the point of support.
296 APPLIED MECHANICS. [CHAP, XYI.
At the point of support of a beam, and at the free end of a
cantilever, we thus have the slojx,
^()d.v. (7a)
V/
that is the integral with respect to ./; of the general expression
for the curvature between the limits c and 0.
Dejiection of the neutral axis of a heam or cantilever at any
point wliose abscissa is Xz. Let Xi be the abscissa of tS (figs. 4 and
5, Ch. V) ; in the case of a beam let 7i be the height of *S^ above
the lowest point of the neutral axis, in the case of a cantilever
let 2C be the depth below the highest point, and let du be the
small difference of level between >S and T, two points indefinitely
close ; then
J— = tan ijr = ii (the angle being small),
or the difference of height of S and T is
du = 2xj . dx^ ; (8)
and summing all these increments from point to point, between
the point a; = (the lowest point) and the point .Tj, we have the
height of Oh above 0,
'"{i.)dx,. (9)
The height of the end above the centre of a beam, or above
the free end of a cantilever, is given by the equation
[' {i.) dx, ■ (9«)
J
this in the case of a beam is the deflection of the centre below
the points of support, and in the case of a cantilever it is the
deflection of the free end below the fixed end ; so that the
dejiection
v.) =
(t.,) dr.. (10)
Again the deflection at any point x. is
V^ = Vo  Vr,^ = t^„  I ' (ir) dx,. (10(/)
CHAP, XVI.J CURVATURE, SLOPE, AND DEFLECTION.
297
For ije we may substitute the value in equation 7, and
— ) dx . iU\ ;
v^ = ?',. 
/I
^7
dx . dr,.
(11)
(11a)
Beatn of v.niform section, load at centre. — For all cases of
^uniform section / is constant ; and for load at centre, we have,
fig.'9, Ch.YII,
M W ^ ( x\_ 1 / / X
M.
^'^0
(12)
\P) %0
1  x\. do:
c '2
Ic
Eyo\ c 2
Ut^'^^
In every case the expression for ic the slope can be arranged
into these three factors ; a numerical factor depending on the
form of the crosssection, in this case \, and for which Eankine
/ " c
puts 7ft" ; a factor — depending on the material, and a factor —
depending on the dimensions. The general expression is
%c = m"i^. — : (13a)
and 7/t" is called the numerical coefficient for the slope.
The deflection
lA
/
x/>/ %0
J
"' (^1 _ l.r ) dx . dxr
\ ^
 \ \x  1 %~\' dx. = I [ (x, ^% O) dx,
oJoL c 2J„ %oJoV c 2 )
= Z(^?_l^'_o
'Xx' 1 ^1
T~2't!"3 Jo %oV2 2c 3
(14)
In every case the expression for the deflection can be arranged
into these three factors : a numerical factor depending on the
298
• PPLIED MECHANICS.
[chap. XVI.
form of the section, in this case ^, and for which Eankine puts
f . c'
n" ; a factor V, depending on the material ; and a factor —
depending on the dimensions. The general expression is
(14a)
and w" is called the numerical coefficient for the deflection.
In this case, the numerical coefficients are
m" = h ; n" = ^s
Observe that / is the working or proof strength of the
material, according as you may want the working or proof values
of ic and v, If / be given, as is generally the case, in lbs.
per square inch, then £ is to be in lbs. per square inch, and y^
and c in inches ; if we then calculate p or p,,, v or v^^, they will be
in inches also.
From equation (12), if we put x = 0, we get
P."f" (15)
This will be the expression for p^ in every case, and we shall
not repeat it.
The inclinations i, when calculated from these formuht, are
in circular measure.
Beam of uniform section, uniform load. — From fig. 12, Ch. VII^
(16)
p £//, ' MJ Uf/o \ <f
Ic =
Jo \P
dx = ■^—
1  , x^ \ dx
__f_( _ 1 i' _ a\ ? / ^
■•"\[r.'e)""=i..[j;^?")^^"^
(17)
'c
r 1 x''
f
%0
x'
2
1 x''
~37'4_
Ui/q\2 3c' 4 J VlE'ij^'
(18)
CHAP. XVI.] CURVATURE, SLOPE, AND DEFLECTION.
299
The numerical coefficients are
111
= 2. .
3 >
TJ*
Cantilever of uniform section, load at end. — From tig. 10,
Ch. VII,
.,r(..).ire=(i'j)
exactly as in the case of a beam loaded at the centre ; and the
numerical coefficients are
m'' = \ ; n" = i.
Cantilever of uniform section, imiform load. — / cancels /^ ;
from fig. 16, Ch. VII.
M.
•J.C
(19)
i. =
)c^»=77 [lJl> +X'' \dx
pj Mjo J V C C"
2 x^ 1 x^'Y _ / / _ 2 c* 1 63
(20)
Vn = I I ^ I  ) f^^ • f^^i = ¥r
The numerical coefficients are
^ ( 1   X i  x'^ \dx , dxi
c c
2 X I x^ \xi ,
X ^ % ^ • dxi
c 'Z c 3Jo
_ _ 1 ^^ l^x,'
2 c y "*■ 3? T
(21)
 1 .
~ 3 5
500
APPLIED MECHANICS.
[chap. XVI.
Beam of uniform section, bending moment constant. — If a beam
be symmetrically placed on, and extend beyond its two points
of support ; and if the two projecting parts be loaded symme
trically, while the intermediate portion is unloaded ; the bending
moment on the portion of the beam between the points of
support is constant, and we have /and /„, J/and .1/,,, cancelling
each other.
p ^>o '
a constant quantity, so that the neutral axis is circular.
(22)
Ic =
■y,. =
]d.r
\PJ
\ (23)
^1 /1\ f [" [^
\dx , dxi = ir—
d^: . dxi
^•^^^•"=^1/^^'^^'
%o 12 Jo
(24)
The numerical coefficients are
m" = 1 ; n" = \.
Beam of uniform section, loaded with tujo equal weights at
equal distances from the centre of span. — Let Whe the total load,
Xr and  Xr the abscissa of the loads. P= ^W; i^/o = i IV (c  Xr),
and the bending moment is constant along the central portion
of span; for values of x between x, and c, M^ = ^W{c  x).
H T
For the central portion of span, ^ = ^\ for the end portion
(25)
J/^ C  X
il/o ~ c  a,. '
1 / M
P Eyo Mo
/
For values of x from to Xr,
1 ./■ 1
■(c 
f) Fl/n C  Xr
For values of x from x,. to c.
(25a)
CHAP. XVI.] CURVATURE, SLOPE, AND DEFLECTION.
1
301
/
dx +
c  X,.
c  X,.\xr
2C
(c  .r) dx
tJy =  .dx . d,i\ = \ d.v .dx^ + \  dx. dx^ ;
J J p J J 0 J iV J /O
Ey,, [Xr XI [" i 1 IT, )
(26)
[.*■,  0] f/.r, + {ex,  I Ci»  ;r,.2) f?.>;i
c  X, J .r^
.[ir.'fH ^
c  ^,.
^  l.ri^  Wx,
9''i 6'M
= ^r +
c .r,.
2 6 2 '^' ' 2 ^'' "^ ' ' "*" ^'^''^ I '
2(;^  3c.r,^ + .g.^ ) / _ jl £: _ i ^ 1 / ^
''" ^ 6(c^,.) ]Eyr l*"'3c * cM^"2/o
■i'*('Jli,r
(27)
Thus, for an unifovm beam, loaded at two points which are equi
distant from the centre, and so that the stress on the outer fibres
of the middle crosssection is/, we have
m" = i(i + :^); i^'=\l(l''
Beam or cantilever of uniform strength and uniform depth. —
For the proof or working load, the bending moment at each
section equals the proof or working moment of resistance to
bending there ; since, however, the depth is constant, the moment
302 APPLIED MECHANICS. [CHAP. XVI.
of resistance to bending at each crosssection is proportional to
the breadth, that is the breadth at such sections is proportional
to M; and since the depth is constant, / for each section is
M .
proportional to the breadth ; hence j is constant at every
section. fj.l, and \I^, (2S)
as in the preceding case of a beam of uniform section and
bending moment constant ; so that for beams or cantilevers to
resist (my load, and made of uniform strength by varying the
breadth only, the numerical coefficients for the slope and
deflection are
7/i" = 1 ; n" = i.
Proportion of the greatest depth of a beam to the span. —
Putting the working strength of the material as the value of/,
we have, as a general formula,
and since 3/0 = m'h^, we have
2c 4r/<,' a h^
Now ^ is the ratio of the deflection to the span, and its recipro
2c
cal — represents the stiffness of the beam ; ^ is the ratio of
depth of beam at centre to span, and we have
2c 4m' E h, ^
v,~ n" f2c'
or, the stiffness of a beam is proportional to the ratio of the
depth at centre to span.
For instance, to give a working stiffness 1000 to a wrought
iron beam of uniform symmetrical section uniformly loaded;
we have the ratio of depth at centre to span (Eankine, C.F. § 1 70)
2c ^'ni"E'v^
(iV) 10000 lbs. p er square inch _^^ 1_
" Ti) 30000000 l¥sT per square inch ~ 144
CHAP. XVI.] CURVATURE, SLOPP], AND DEFLECTION. 303
That is to secure the degree of stiffness 1000 required for the
wroughtiron beam, the depth must not be less than a fourteenth
of the span. In the same way it may be shown that, in general,
to give to beams the degree of stiffness that practice shows to be
necessary, and that is usually proscribed, the depth at centre
must bear to the span a ratio varying from ^th to ,^4th, accord
ing to the material, form, and manner of loading.
Slope and deflection under any load less tlian the proof load. —
For the proof load, the stress on the skin furthest from the
neutral axis at the crosssection of maximum bending moment
is/' and for a smaller load it is pa (fig. 8, Ch. Y) ; now, though
the load is less than the proof load, yet being distributed in
the same way, the ratio M : liL^ is in no way changed ; / is the
same as before, and we have E and y^ constant ; so that
But since
we have
,,Pa c , „ p^ e
E y. E y.
mWl . „
no^fi Q
mm" Wlc , inn" Wlc^
I . =
m'n UbohJ'' " m' n EbJiJ
^O't'o
For a beam / = 2c, and for a cantilever I = c\ so that for any
load for a beam
.. _ 2mm" Wc , _ 2mn" W(^
^''~ m'n Ehji/' ^'~^i^EbJJ"
'O'^O
and for a cantilever
mm" Wc' , mn" Wc^
v„ =
m'n EhXy ' m'n EbJiJ>
O''O
The coefficients
mm . mn
— T and —r
m n m n
assume values which depend on the crosssection and the
system of loading.
For similar beams similarly loaded we thus have v\, the
deflection under any load less than the proof load, propor
tional to W the total load, and to c' or the cube of the lenoth ;
304
APPLIED MECHANICS.
[chap. XVI.
and inversely as h the breadth at centre of span, and h„'' the
cube of depth at centre of span.
That v\ is proportional to W does not follow from Hooke's
Law, but has been established upon the supposition that the
slope is so small that the tangent and circular measure of the
slope are sensibly equal. In bending small pieces of wood in
a machine which registers the load and the deHection as the
coordinates of a line, it is found that the line is straight even
when the piece of wood is bent to a considerable amount ; this
proves that the formula above is a close approximation, even
when the slope is considerable ; it must not be forgotten, how
ever, that the formula is only approximate when the slope is
great. It has been stated that where such a machine's register
ceases to be straight, the elastic limit has been passed ; no such
conclusion can be drawn ; the only just conclusion to draw is
that since the slope is visibly great (say 8° or 10°), the formula
above has ceased to be a close approximation. Were a second
approximation made, it would be found that v\, was not exactly
proportional to W; and so long as the register did not depart
from the curve which is the locus of that new equation, it would
be inaccurate to infer that the elastic limit had been passed.
Defiedion of heam suf ported on three props. — Let HK be a
beam of uniform crosssection, bearing an uniform load of
amount U; and let it be supported on three props, one at
each end and one at the centre. Let W be the reaction of
the central prop, and P = Q, the reaction of each end prop ;
then P+Q+ W=U.
Fig. 1.
Fig. 2. U.
In fig. 1, let W = U, then F = Q = 0, and OFT is a canti
lever of uniform crosssection fixed at and uniformly loaded,
for which ?i" = i ; in fig. 2, let W = 0, then P = Q^ ^U, and
CHAP. XVI.] CUrxVATURE, SLOPE, AND DEFLECTION.
305
HK is a beam of uniform crosssection uniformly loaded, for
which n" = fV ; hence
BH{'[ig. l):DO{^g. 2) ::3 : 5.
Putting BH = Zz, then DO = 5z.
In fig. 1, if we suppose the end props to be pushed up till
they support all the load, then 7/0^ will assume the form LOB,
where HL = 8z ; and in fig. 2, if we suppose the central prop
to be pushed up till it just supports all the load, then HO K will
assume the form HTK, where OT = 8z. Hence we have the
following theorems.
Theorem. — If to an uniform cantilever {OH, fig. 1) loaded
uniformly, we apply at the end a load [P = ^U) equal to that
uniform load, it will produce an additional deflection in the
direction of the applied load {HL = ^BH) equal to eightthirds
of that due to the uniform load alone.
In the figure, P = lU \s applied upwards at H, but the
theorem holds for P applied downwards, since deflection is
sensibly proportional to load ; it being understood that the
total deflection in no case exceeds the proof deflection.
Cor. — A load P', applied to the end of an uniform cantilever
loaded uniformly, will produce an additional deflection in the
direction of P' ; and the amount of this additional deflection is
proportional to the load P'.
Theorem. — If to an uniform beam {HOK, fig. 2) loaded
uniformly we apply at the centre a load ( IF = U) equal to
the uniform load, it will produce an additional deflection in the
direction of the applied load {OT = ^DO) equal to eightfifths
of that due to the uniform load alone.
In the figure, W = U\& applied upwards at 0, but the theorem
holds for W applied downwards, as for the previous theorem.
Cor. — A load IF', applied at the centre of an uniform beam
loaded uniformly, will produce an additional deflection in the
direction of W ; and the amount of this additional deflection
is proportional to the load W. iSee Thomson and Tait's
" Natural Philosophy," first edition, §§ 618, 619.
Uniform beam uniformly loaded and supported on three
props at the same level, one at each end and one at the centre
Fig. 3.
(fig. 3). — This figure is obtained by pushing up the central prop
X
306
APPLIED MECHA^■ICS.
[chap. XVI.
in fig. 2 till coincides with D. Then since the reaction of
the central prop in fifj. 3 has produced the deflection 01), which
is fiveeighths of OT, it follows that W = ^U, and therefure
From fig. 2, Ch. YIII,
_ W ^
OS, = jj OH = ^OH;
hence OH, = \0l}.
At centre 0,
Mq =  ^^Ul, max. ; F =  lVf^ max.
At Si and /S'2 the maximum deflection occurs, and, as proved
in the next article,
n" =■ \ X ^. = 043.
1024
The central prop increases the strength four times and the
stiffness nearly ten times.
Uniform beam tmifoi'vily loaded fixed at one end mid sup
ported at the other (fig. 4).— The left
half of fig. 3 represents such a beam.
Let W = ^U the uniform load, and
/ = Off the span ; then P = }W ; the
shearing force at is the remainder
of load, viz. fTF; the shearing force
changes sign at S where the bending
moment has a maximum value ; and
the locus of tlie shearing force diagram
is a straight line, since the load is '
uniform.
From fig. 2, Ch. VIII, the bending
moment diagram is AFBE a parabohx
with axis vertical and apex on vertical through S.
'dW S[_ S_W 3^ _ _9_
^~ ■ ¥ ~ ~8~ • 16 " 128
a positive maximum ; and since AFBE is a parabohi
FD : Fa : : DFT : O'R : : 25 : 9 ;
therefore FD = ^FO',
and CE = O'D = FI)  Fa =  ^O'F ^ 111,
that is Mo =  ^^'^/. tlie greatest value; this quantity may also
be calculated directly by taking moments about 0.
0'F=M,=
Fis. 4.
Wl.
CHAP. XVI.] CUKVATUUE, SLOPE, AND DEFLECTION. 307
This solution is exact, and all the results are readily got by
remembering that OH' is a quarter of the span. The approxi
mate solution indicated in Rankine's "Applied Mechanics,"
sec. 308, assumes H' to be sensibly on the same level as H.
To find the deflection at S ; considering the beam HW alone,
we have for the deflection at aS^, since SH = f/ = c;
= AZ* ^'= 5_A ^i= ^A t.
^' 12 E ' ?A, 1'^ %> ■ 16 64 E ' y, '
where /, is the intensity of stress at the skin at S.
Next consider the cantilever OR', taking into account its
uniform load alone. If the cantilever were of length SH, its
deflection would be Iv, ; and since the deflection of a cantilever
is proportional to the cube of its length and to the load we have
for the deflection of OH for uniform load alone.
^ ~ ^T? • 3' 1  U • 8^ •
\SH
:V, =
_ I'l 6
T3 T^l >
further, if a load equal to that on OH' be put at the end, it will
produce an additional deflection frj ; the load at end of OH,
viz., that on SH, will produce a proportionate deflection, and we
have for the deflection due to load at end of OH',
and the total deflection of OH' = 5vi = Ifvi.
The total deflection of the point S is therefore
1 16 _ 35 _ 35 15/, c _ 175 /,c
^^'i^ 2 27"^ ~ 27''^ ~ 27'64:E'i,'"576Ei:
Let/o be the proof strength of the material; when the beam
is loaded with the proof load, then /^ = intensity of stress on
skin at the fixed end, and
/.:/,::i)/,:ilf, ::9:16;
hence/, = t%/,; substituting this, we have
= ns 9^^ c' _ 175 /, r
^ 576*16 ^'y," 1024*^ >/
1 175
so that m = , and n' = = 171.
o 1(J24
x2
308
APPLIED MECHANICS.
[chap. XVI.
Uniform beam uniformly loaded and fixed at both ends
(fig. 6). — Suppose the central of the three props that support
UK, fig. 5, to push up till is above the level of H and K
such a distance that H' and K\ the points of contrary tiexure,
shall be on the same level as H and K. Let x = QIC the dis
tance of the point of contrary flexure from 0, and let u be the
intensity of the uniform load.
Consider OK' alone. It is a cantilever fixed at 0, of length x,
loaded uniformly with intensity v, and loaded at X' the free
end with 7^ (c  x), half the load spread on K'K. For the
uniform load alone, compare OK' with OK (fig, 1), whose
deflection is 3z ; their deflections are proportional to the cubes
Fig. 5.
of their lengths and to the loads ; hence for uniform load alone
the deflection of OK (fig. 5) is
fOK'^
\0K)
3s =
3.
OK being shown in fig. 5, and OK in fig. 1. By theorem
at fig. ], a load at the end of this cantilever, and of amount
ux, would produce an additional deflection 8 (   z; the load
\u{c  x) will produce a proportional deflection ; and we have
for the deflection due to the load at end of OK,
and for the total deflection of the point K,
OD = (4c  x) z.
CHAr. XVI,] CURVATURE, SLOPE, AND DEFLECTION. 309
To pass from fig. 2 to fig. 5, the central prop has been
pushed up through OD (fig. 2) together with DO (fig. 5) ;
that is through
and by the converse's of corollary to theorem at fig. 2, the
reaction on the central prop will be
1 / (4c  x)x^\
8s I c^ )
Now, OS, = \(c + x); and since (fig. 2, Ch. VIII) OS, =^c, we
have i(c + a;)=i(5+^l^i^
Uc,x)=l[5+^±^l^y;
solving this equation we find
OK' = x = c(2  v/3),
which determines the position of the points of contrary flexure.
Substituting this for x in the expression for W as given above,
we have the reaction of central prop
If we take OK'K, half of this beam, and suppose the end at
fixed, and the end K to be supported by the cantilever O'K
similar to OK' in all respects ; we have (fig. 6) a beam fixed
at both ends and uniformly loaded ; its semispan is
n^ 3 v/3
c = OS,= —  — c ;
but OK' = {2 ^S)c = (l yojc';
hence the distance from the centre to a point of contraiy
flexure is
SK' =^,0= 289/',
where /' is the span.
310
APPLIED MECHANICS.
[chap. XVI.
Let W be the total load on the beam (fig. 6) ; then the
shearing force diagram will
vary from ^W to  ^W
from left to right end.
The bending moment dia
gram, is the same parabola
as 'for the beam supported m. ^^
at the ends, except that itn" = j
■passes through the points
of contrary flexure ;
FO" :FD':: 0"B : D'L\
FiK. 6.
or 0"F is onethird, while
0"D is twothirds of the
maximum bending moment for the beam supported only ; hence
at fixed end
'"max. ~ S a" "^ ' TU
To find the deflection; consider first the part KK\ for it
5/, SK^
5 /; c'
5/, c'
12E y.
12 ■ Ey, 3
36 E ' y„
where /, is the stress on the skin at S. Consider next the
cantilever OK', taking into account the uniform load alone ; a
cantilever of length SIC would deflect j^v^, and taking account
of its length, we have for the deflection of OK' for uniform load
alone
fOKy 3
A load at end of OK', equal to the uniform load on it, would
produce an additional deflection of ^v^ ; but the load at end of
OK' is that on K'S, and it will produce a proportionate deflec
tion ; so that we have for the deflection of OK' for load at end
fSK'\ 8 fOK'\ 8
'' = \0K')Z'' = \SK')V^'
Hence the total deflection of ;S' is
L ZfOK'\ SfOK'Vi
CHAP. XVI.] CURVATURE, SLOPE, AND DEFLECTION. 311
OK' / 1 \ 1
fOK'y (0K'\^
(^ =(2816 v/3), and (^^,j = (6^3  10) ;
o j 5 4. E y, ^ E y,
since J/ at the end and at the centre are in the ratio of 2 and 1
We thus have m =  tV 5 ^^^ ''^" = s
From the above, we see that fixing the ends of an uniform
beam which is loaded uniformly increases the
strength in the ratio ^ : ■^, or 3:2;
stiffness in the ratio j^2" • ¥» or 10:3.
As already shown at fig. 2, Ch. VIII, the strength is econo
mised to the greatest extent if by means of hinges, or in some
other way, we shift the point of contrary flexure to a distance
c f v^2 from the centre ; we then increase the
strength in the ratio 2:1;
stiffness in the ratio r^ : ^^, or 10 : 5*66 (see below).
Maximum Stiffness. — For some position of the hinges, the
stiffness will be a maximum ; in order to find this point, let /3
be the distance of the hinges from the centre, and/, the stress
on the skin at centre ; then as above we have
{5 j '\ {i j )''E y.
_ 3c^  4c''/3  6c/3' + 12/3" ^ c^ ^
12cj3^ E'iJ
312 APPLIED MECHANICS. [CHAP. XVI.
where/, is the proof strength so long as 3fniax is at the centre,
c
that is so long as /3 > — ; putting
/3 = c, we have ?/' = /'j. (Hinges at ends) ; /3 = rr, we have
J2
n" = ^ • (Hinges at "TOTc from centre.)
When j3 < ^, then iT/max is at the end, and it is necessary
to substitute for /« in terms of /„ the proof stress on skin at
end ; thus, if
/3 = put f, = \fn,
and the above expression gives
or ?i
_ 1
~ 8
The stiffness is a maximum when v is a minimum ; this
occurs when /3 = ^c, sensibly giving
1 J'^ 1 yo'' //I Ml • •
f = ^ ^;— = 4 t;— : or 'Ji = TT sensibly a minimum.
Ut/o ' %„ '^ ^
Fixing the ends and placing the two hinges at the quarter
points of span increases the stiffness in the ratio ■f'j : ^, or 4 : 1
nearly.
Some economy of a material whose strength to resist tension
is, say less than that to resist thrust, can be secured by making
the constant crosssection of such a form that the upper skin at
end section, and under skin at central section, shall simul
taneously come to their working stress. Thus for a material
half as strong to resist tension as thrust, and with the hinges
fixed so that /3 = — , a section whose neutral axis is half as
far from the upper as it is from the under skin, would give
considerable economy ; as the upper and under skin at end
sections, and under skin at central section would all come to
their working strength at one time.
CHAP. XVI.] CURVATURE, SLOl'E, AND DEFLECTION. 313
Beam of uniform strength, uniform depth, fixed at the ends. —
Suppose that OK'SKO', fig. 6, is the beam. From equation 28,
p. 302, the curvature is constant so that OK' is part of a circle
whose centre is on tlie vertical through 0, and SK' is part
of a circle of the same radius whose centre is on the vertical
through S; by symmetry SK' = O/v"', and /iTand K' the points
of contrary liexure are midway between the centre and the ends
of the span. Since the beam is of uniform strength and depth,
the breadth varies as the bending moment ; hence the plan
should correspond with the bendingmoment diagram. For the
case of an uniform load, the plan will correspond with the
bendingmoment diagram in fig. 6, the curve being drawn on
both sides of a centre line, the two parabolas passing through the
quarter points of the span ; the breadths are then to be reduced
till that at any point (say the end) is just sufficient to give the
necessary resistance to bending, and we have the plan of the
beam ; see Eankine's "Applied Mechanics," fig. 144. Since 0"B,
fig. 6, is half of D'L, and since the load is uniform, 0"F is
onequarter, and 0"D' the maximum bending moment is three
quarters, of FD' the amount of the maximum bending moment
for the same beam not fixed at the ends. The plan must allow
sufficient breadth at the points of contrary flexure in order to
be able to resist the shearing force at these points ; on K', for
instance, the shearing force will be the load on SK', and the
breadth at K' must be sufficient to resist this amount.
EXAMPLKS.
1. A beam 24 feet span, of uniform symmetrical section as shown in fig. 26,
Ch. XIV, is made of wrought iron whose working strength /= 4 tons per square
inch, and whose modulus of elasticity E = 11600 tons per square inch. Find the
radii of curvature at intervals of 4 feet, when loaded uniformly with the working
load.
Taking equation 16, we have
p Eijo \ c' J 11600 X 10 \ 144/
where x and c are to be in one name,
144
p = 29000 — ; , X being in feet.
144 — X'
The radii of curvature are
po = 29000 ; Pi = 3262.5 ; p^ = .52200 inches : pn = infinity.
= 2420 ;• = 2720 ; = 4350 feet.
The reason that p is in inches is because vo is in inches, and the proportions
•derived at fig. .5, Oh. V, show clearly that p and y are in one name. "We had to put
j/oin inches, because for the material/ and E are given in tons on the square inch.
314
APPLIED MECHANICS.
[chap. XVI.
2. Calculate the slope and deflection in the previous example
4 144
5 / c
11600 10
■5 4 (H4)'
•0033; .. i, = 0° 11' nearly;
= 030 inch = 002.5 foot.
12 2; yo 12 11600 10
Tliis is nearly onethousandth of tlie span, which is ahout the extreme ratio of
working deflection to span allowable in practice
(Rankine's "Applied Mechanics," sect. 302) ; and
this degree of stiff"ncss is secured by making the
depth of this wroughtiron girder a jj — j^ part
of tlie span.
3. Find the slope and deflection in example 7
by means of a graphical solution.
As in fig. 7, which is diawn to a scale of
20 feet to an inch, draw verticals ; one to represent
the vertical through the centre of span ; the others
to be drawn on each side of the centre, and at 2, 6,
10, and 12 feet therefrom ; the last two will be
through the extremities of span.
From c\ any point in the central vertical, with
radius 24*2 feet, that is a hundredth part of po,
describe the arc B'AB between the verticals
through B and B' ; produce Bc\ to C2, so that
CiB = 27"2, a hundredth part of p4, and about c^
describe the arc BO: similarly (■36'= 435; and
from D, DE is drawn at right angles to Dc^. On
the other side of the point v/ construct AB'E';
draw the horizontal chord EOE', and the tangent
E'K. Then EAE' represents the curve assumed Y'l".
by the neutral axis of the beam when under the °
proof load, but with its vertical dimensions exaggerated 100 fold.
The deflection
^■■'=iTo^^^
1
100
2'n(by scale) = 025 foot,
OK . 40
t, = tant<:=  f OE' = —  124 = 003 = 0' 11'.
ICO 100
4. If the beam in example 1 be loaded with one ton per foot of span, tiiul the
deflfction.
Let W" be the working load, then mW'l equals the working value of M,. : that
is \W'{1\ y 12) = 1948 (see fig. 26, Ch. XIV) ; therefore W = ')4l tons.
Now the load in this example is 24 tons, ^""^ of W the working load : and
since deflection is proportional to load, we have
Deflection = v'o = J x 30 = 013 inch = 0011 foot.
Oihcrwise, if the working deflection had not been already calculated, we have
Wc^ 2inn" Wc^ 2init" Wc^
'jV '~E~
I2mn"\
Deflection = Vo = I — p I
\ mn I
Ehhn^ n'boho^ E
2(i)(A) 24x(144in.)3
/o = 4870
4870
11600
= 13 in. = (HI foot.
CHAP. XVI.] CURVATURE, SLOPE, AND DEFLECTION. 315
5. Find the working deflection for a wrouglitiron beam of uniform sticngth
and uniform breadth, and loaded uniformly ; the span is 24 feet, and the upper
half of the crosssection at centre of beam is shown in fig. 26, Cb. XIV ;
/= 4 tons, and 2.'= 11600 tons per square inch.
vo = »" . 4 .  = I „  1 1 ^ .  = '5" X :r • = '41 m. = '034 feel.
A' I/O \2 / ii Vo 11600 10
6. A Mroughtiron rectangular beam, '20 feet span, 16 inches deep, and
4 inches broad, is loaded at the centre. Calculate the proof deflection if the proof
strength of the iron be 7 tons, and its modulus of elasticity 12000 tons per square
inch.
f c" 7 120
Vo = n" . 4; . — = i . — . = 35 inch = 029 feet.
7. Find the resilience of the above beam (see Chap. I).
Let JF be the proof load,
.Vq = Mo, or m Wl = nfbh^ ;  TT. 240 = i x 7 x 4 x 16= :
therefore 7F = 20 tons nearly.
Resilience = \ proof load x proof deflection = J x 20 tons x "029 feet
= 29 ft. tons = 650 ft. lbs.
8. Find a general expression for the resilience of a rectangular beam loaded at
the centre.
Let W be the proof load, and Vq the proof of deflection, then
_ /2mn"\
\ m'n )
Ebh>'
('})l)i \ c
—r ^rrr:; x W^ ; but m . TF . 2c = nfbh ;
m n I Eoh^
••• ^= o ; •■• resilience = ; . ■—  . ^ = ; .•'— . Icbh.
The first factor depends on the form of crosssection and the distribution of load,
and the second upon the material ; the third is the volume of the beam.
Hence for a rectangular beam of a given material loaded at the centre with the
proof load, the resilience is directly proportional to its volume ; a result correspond
ing with that obtained for direct stress, Chap. I. Suppose for any given material
we take a oneinch cube ; then, when loaded at the centre as beam we have
resilience per cubic inch = ( , ) "—  ;
for a rectangular section n = J, tn' = ^ ; for load at centre
1 ;i " 1 "''" ^
'" =  "^"^ " ='' s;;^' = T8'
.1/1
and resilience per cubic inch = — ■ / • "f ~ To ^ P'oof stress x proof strain.
IS £ 18
'^ = ^V^,L.^ = il£ ^ . 600 = 156.
316 APPLIED MECHANICS. [CHAP. XVI.
In the case of Mroughtiron, for which the proofstress /= 7, and E= 12000
tons per square inch, and remembering that the deflection is in inches,
resilience per cubic inch = 77: x = 000227 inchtons = 0423 ft. lbs.
For the previous example, the volume of the beam is 15300. cubic inches;
multiplying this quantity by '0423, we find the result given there.
It. For a rectangular timber beam of uniform section and uniformly loaded,
find the ratio of depth to span, so that the working deflection may be a si.x
hundreuth part of the span. The working strength of the MOod is one ton, and
its modulus of elasticity is 800 tons per square inch.
2c * in ' £' Vo * i 800
That is, the depth is to be between a sixth and a seventh of the span.
10. What stiffness will be secured for an uniform rectangular beam of the
same timber, loaded at centre, by making the depth an eighth of the span ?
15 = 1!!:' L 1 = ^ ^ i 8 = —.
2c * m' ' E' liQ •* ■ ^ ■ 800 ' 600
That is, the working deflection will be a 600*'' part of the span.
11. Find w" and w", the numerical coefiBcients for slope and deflection, for a
beam of uniform section loaded with two equal weights at points which trisect the
span.
In the expressions following equation (27) substitute c for Xr ; and
12. A beam of uniform crosssection is loaded with two equal weights symme
trically placed on the span ; the amount of each weight is for each position such
that tlie outer fibres of the middle crosssection bear the working stress/. Compare
the amount of deflection when the abscissae of the left weight are as follows : —
Xr = 0, Xr = he, and Xr = c.
Ahs. For Xr = 0, n" = J ; for Xr = h, n'' = %% ; for a, = %c, n" = ?§.
CHAPTER XVII.
fixed and moving load.s on an uniform girder with
ends fixed horizontally.
Unsvmmkti;1(Al Fixed Loads.
We now consider the detleclioii of an uniform beam and tlie
slopes or tip.svji at the freely hingetl ends due to loads placed
unsymmetrically ui)on it.
CHAP. XVII.] GIRDERS FIXED AT THE ENDS.
317
On tij,'. 1 let a load W be concentrated at D a distance z
from the centre of the span. D divides the span into two seg
ments III and n. The bending moment diagram is ACB. The
height of C is h, and of any other point is y. The origin is at
Q the lowest point of the beam which divides the span into
segments a and 6. The deflection is u^ = «3, and the tipsup
are o and /3. The span is / = 2c.
In Chapter XV f. equation (5), / cancels /o, and all the
factors are constant except J/ the bending moment, y may be
supposed to include this constant in every case.
h
111 ' h
b  Xo
•ll  ^'' ~
h III
The slope at D and its height above Q are
ill }o in L 2 Jo 2
Ub 71 = — I (« + '^'i) dxdx =
6 m
(m  «)■ (??i + 2a).
fi i { ri\ x^ '/
And h =  \ (h  Xj) dxdx =\ hx  — . dx= 
(a  X3) dxdx =
oJo
Also Ua =
3 w
318 APPLIED MECHANICS. [CHAP. XVII.
Equating w„ and v.b we have (fig. 1),
Ua = «6 = t'hn + 'ilOh.n + A',
2ft" = (m  ay {m + 2a) + I6n {m  a^) + 2nhn,
(m + n) (m^ + 2mn  3f') = 0,
3a' = w (/ + n) = {c + z) (3c  =). (1)
Substituting we have the deflection of the beam
l\l = Ua = Uh =
ha^ hyS .
3m
27
(3c  s) ^y{3e  z)(c + z).
(2)
(i = 6b = dhn +  f Q> ^2) dx=^ {m  a) +
hx 
2jt.n
hf a^ \ h( a?
2 V m / 2 V ?n
But by equation (1) (7^ ^ 7n = (/ + ?i) H 3, so that
/3 = ^(2^70=^^ (3c + .).
ha h ,. . /« ,.,
(ft  .r^) rfr» =  = (/ + w) = F^^c  z).
2 ?n 6
And
a _ ZcZ
]3~ Sc + z
C — X
C + X
(3)
(•i)
(5)
where .^ is the distance horizontally of G the centre of gravity
of the bending moment diagram from the centre of the span.
Also
(« + /3) = he = area ACB. (6)
The following are perfectly general : —
Theorems. — If a girder of uniform section and hinged at the
ends be loaded in any manner between them — (ft) The sum of
the tij)!iv2^ at tlie ends is proportional to the area of the bending
moment din^ram, and (//) their ratio is the same as that of tiie
segments into wliich the perpendicular from the centre of
gravity of the bending moment diagram divides the span, but
in tlie inverse order.
Ecjuations (G) and (5) prove these theorems for one concen
trated load.
"CHAW'. XVII.] (HKDKKS FIXKD AT THK ENDS. 319
Let there be two sucli concentrated loads so that a = a, + a,
a.n(A /3 = /3i t /Sj ; then, from eqnation (5),
or 2t'oi = AiC  AiJ = AiC  G^
a, + i3, 2c
whc'^re Ai is the area of the bending moment diagram for JJ\
aloiu^, while G^ is its geometrical moment about the centre of
the sf :>an. For IF, alone on the girder, we have 2e(r. = A.c  Gn,
and auUling 2ca = Ac G for the joint bending moment diagram.
Similai 'ly, for any number of loads.
Pu ttiiig  = 0, in equations (4) and (2), gives us
ip = ^hc and u^ = \hr, (7)
whicni aie just the same as equations (13) and (14) of Ch. XVII
for t he Uniform beam with load at centre if we put
/Again put z = c, and from equations (3) and (4), we have
/3 = /ic and a = Ihc. (8)
Ttte bending moment diagram is now rightangled at B ; and as
^J^is at B, and over the abutment, there is no strain on the beam,
ajad a and /:• are both zero, but still just before W went ofi* they
v/ere in that ratio.
There is tinother way of loading the beam so that the bend
ing moment diagram shall be rightangled at B, that is, by
applying a couple at the end B.
Suppose then a righthanded couple Mq applied at B, the
hinge A will hold down with a force F, and the hinge B will
push up with Q =  P. In this manner the two hinges will
constitute a reacting couple, with the span as an arm, and of
oppiosite sense, but equal in moment to Mg ; so that
The ends of the beam will drooj) by a and /3 at the ends A and
B, the droop /3 bt'iug double the droop a by equation (8). Their
sum will be the a I'ea of bending moment diagram, viz. the right
angled triangle of height Mg and base 2c. In the same way let
a lefthanded couple 3[phe applied at the end A. The joint
bending moment diagram due to both couples will be of height
3fp and Mq at A and B, with a straight locus MK, fig. 2,
between them. If these two end couples be the only loads on
320 APPLIED MECHANICS. [CHAP. XVII.
the beam, Mn being bigger than Mp, then the hinge A must hold
down and the hinge B hold up with equal forces P=  Q, form ing
a reacting couple with the span as arm to balance Mq  Mp, so
that P =  Q == (^^<j ^^p) ^ 2c. The beam will be every whc^re
convex upwards, its two ends drooping so that the droops a ai id
j3 shall have their sum equal to the area of the bending monif^nt
diagram AHKB, and bear to each other the same ratio "mto
which the perpendicular from g divides the span only ii'i the
inverse order. /
Suppose now that the uniform girder hinged at the pnds is
loaded between them in any way, as, for instance, merely with
the weight W (tig. 2), trisecting the span. The two e^c Is will
tip up. By applying suitable end couples Mp and M^, whihh alone
would produce droops exactly equal to those tipsup, iwe! may
destroy the tipsv/p and make the ends level. \
The conditions are that Mp and Mq are given by A H and
BK when HK is drawn across the bending moment diagram
ABC, in such a way that the areas AHKB and ACb shall be
equal, and have their centres of gravity ij and Q in one
vertical line.
We have the following graphical solution for a uniform
girder with its ends held horizontal and loaded between ihvni
in any way. First draw the bending moment diagram for the
extern loads as if the girder were hinged at the ends ; next draw
a straight line cutting across it so that the foursiderl figure shall
include the same area over the span as the bendino moment
diagram, and have its centre of gravity on the st^ue vertical.
This line, HK on fig. 2, is the new base cutting the locus in
two points S and T between which the bending njoments make
the girder convex downwards, beyond which they inake it convex
upwards. From S to T ordinates are to be measured iip to SCT
for positive bending moments, and from S to J£ and T to K
downwards to SA and TB for negative bending i.noments.
Beam of Uniform Section fixed Hohizontai.lv at the Ends
AND subject to a PiOLLING LOAd.
Let W be at any distance z (fig. 2) froi.n the centre 0,
then ACB, with C directly under W, is the, bending moment
diagram for the ends hinged. Let HK n^ap out the area
AHKB = ACB, and let <j and 6', their cent/res of gravity, be
in a vertical line.
For shortness, put h ■= AH, and /.: = BK\ also put m for the
lieight of C from AB, and OA = OB = c, t/he half span. For
CENTRE OF HYPS. IS
OVER Vat 4 times
ITS HEIGHT AND
2 t/JC = {3XC){CX) W
■ ® ®.®
r^4)
X2
2C'
FEET 10
I I I I I I I I I
s
POSITION
OF THE LOAD
Fig. 2.
Y
322 APPLIED MECHANICS. [CHAP. XVII.
areas equal we have h + Jc = m. For g under G take moments
about A, first dividing AH KB into two triangles by the
diameter RB ; their areas are he and kc ; their levers about A
are ^ of 2c and § of 2c, wliile the area of ACB is 9«c, and its
lever c + ^z, so that
mc(c + As) = he X fc + ke x *c ; ?;2(3c + z) = 2c(h + 2/i) ;
T c + z ^ ^ c  z
therefore k = —rr— m, and h = tn.
2c 2c
The equation to HK is
2/,  h h  h
X + c 2c
Jc  k k + h zm m
y'^2^'''2r = 2^^'''2' "
The equation to BC is
i/z  m  c  X
i/i = m ,
X  c z  c
Let us take the negative bending moment at any point x to
the right oi T. It will be the intercept from TK down to TB,
and
M
• " '^' ~ ''''' " 2& ic  z) r^ ~ *^ ^^' "^ "^^^ " ^^' ^^ ~ "^^i •
Now, Oil, the bending moment under W, is the height of C
above ^^', or
and substituting ancrsiniplifying
^^neg =^(0 + 2)X(2c2).rC^). (9)
On the girder above T and <S' are shown virtual hinges or
points of contrary Hexure. To find , the abscissa of T, sub
stitute Mneg. = zero, and .r = ? into (9), and change sign of z for
r, then
CHAP. XVII.] GIRDERS FIXED AT THE ENDS. 323
Now t, is a miiiiuium when z =  c, that is, when W i.« just
at left end, and substituting this vahie of z, we get the niininiuni
vahie of ^ to be Jc.
Hence, for the middle third of the span, the bending moment
cannot be negative at any point for any position of the load.
For negative values of M, then, we require x >^. Inequa
tion (9) let X be any constant value between ^c and c, while
z varies,
Ac' dM
W^ ^^"^^^^ ^ (^  ^)  ^^"")>
4c' dm _ . „
— 7r =  6^^  2c.
W dz
±1 irst, z =  c makes —, = 0, and rrr —rir =  be { .r ~
dz W dz \ 6
positive quantity, as x is between ^ and c. Hence z =  c makes
J/jjeg a minimum.
^ . .,, 2c' dM _ , 4c^ dW f c
Second, with s = c  — , ^r = 0, and jr 3—7 =  be [x  —
S.r d.r ' IF dx' \ 3
a negative quantity, since x > — . Substituting this value of z in
equation (9), we have
c \»
3fx = JV ~— , a negative maximum. (11)
When the load is at a point.
2c'
= = .  3. (12)
On fig. 2 there are laid doivn to scale the values of Mx at
1, II, III, IV, and v, for the values of x, namely, 2, 3, 4, 5, and 6
twelfths of the span. The corresponding positions of the rolling
load are shown with circles round the Eoman numerals agreeable
to equation (12). The maximum of negative maxima is at the
end when x = c, and z = Jc. At the top of fig. 2, the load W
is shown in this position, namely, trisecting the span, and the
maximum of negative maxima bending moments at the end is
8 4
3fc= ^r Wc= ^rr IVl maxlmum. (13)
27 27
Y 2
324 APPLIED MECHANICS. [CHAP. XVII.
By substituting for m in the preceding expression for h and k
we have
W
3Iq=BK=^Xc^z')(c^z). (14)
M^ = AH= ]^(c^  z') (c  z). (14a)
We come now to consider the positive bending moments.
Changing the sign of (9) we have
M', = i^(c + zf (r  (2c  ) x) from C to T,
or as long as x lies between z and  = ^ _ _ . Now Jf j; is only
admissible while ^ :^ ^ ; and as it continually increases with z,
M'x is clearly greatest for z = x, that is, when the load is at the
section. As the diagram is of the like form right and left of
W or C, it follows that the positive maximum bending moment
occurs at any point when the load W is over it. Substituting
2 = a; we have
W
Mj, = —i{c^  •«')■> a positive maximum, (15)
Cor. — If the girder were only hinged at the ends, the
maximum bending moment is 3Io = } Wl ; when fixed horizontally
at the ends by equation (13) it is ^S^^^^ > hence fixing an uniform
girder at the ends increases its strength in the ratio 27 to 16 to
resist a rolling load. This increase is 69 per cent. With the
load confined to the central point the increase would be 100 per
cent,, for HK would then be horizontal, and ACB isosceles, and
for the two bending moment diagrams of equal area we would
then have AH = CD = BK, and all three maxima half of that
for the ends hinged.
It can readily be proved that the loci of ^S'and T are the pair
of hyperbolas shown on fig, 2, The equation to the locus of
T is 2i/x = (Sx  c){c  x) W. The principal axes of the pair
meet on the vertical OK produced four times that amount.
It is interesting to notice from fig. 2, that at a point ^ of
the span from the right end, the positive and negative bending
moments 25 and 27 are nearly equal. That is, for the end sixth
CHAP. XVII.] GIRDERS FIXED AT THE ENDS. 325
of the span the negative niaximuin bending moment is greater
than the positive. This point might be found from the equation
4c^ x'
2
It will be found that by trial and error x = c gives
25 1
j ^r^ = 7 nearly.
4 X 27 4 "^
Levy gives this point as '171 from the end, a trifle more than
I" of span, as the diagram (fig. 2) shov^^s it to be.*
Shearing Force. — For the position of the loads shown on
figs. 2, 3, the right supporting force Q is composed of two parts,
one a major part of W, namely W {c + z) ^ 2c, just as if the
girder were simply hinged at the ends. The other part is
{Mq  Mp) ^ 2c due to the couple at one end being greater than
the other, so that
^= ^2r^4^3^(^'^^)
Now Q is the instantaneous shearing force at every point in the
shorter segment of the girder from JV to the right abutment B.
It is clear, then, for a section x = z, that is, for a section inde
finitely close to W on its right side, that Q is the maximum
shearing force. Hence
c + X W' . . ,.
i\ = yy—^ — + TV ^{(^ ^') a maximum,
2c 4c^ ^
the load being just to the same side of the section as the more
remote abutment
IF
Fa: = "ttC^ + ^)"(2c  x) maximum. (16)
* In Za Statique Graphique, second edition, vol. ii., p. 124, Levy gives equa
tions (20) and (19) derived by constructing graphicully "lines of influence." He
finds that (19) supersedes (20) in magnitude for '17 part of the span from each
end. In Levy's equations (20) and (19) the origin is at the end of the girder;
substituting (c  x) for his x, and 2c for his I, these equations (20) and (19) are
identical with our equations (11) and (15) given above in their symmetrical form,
the origin at the centre, and the loci of which are plotted on the lower part of our
fig . .
The solution in the same form as Levy's is given in Du Bois on The Stresses in
Framed Structures, second edition.
326
APPLIED MECHANICS.
[chap. XVII.
Now Fc = TV at the ends just as if the girder had been hinged
at the ends. So also at the centre Fo= h W, as indeed is evident,
for with IV at the end there are no end couples ; while with W
at the centre the end couples are equal and do not modify the
supporting forces. Hence fig. 6, p. 164, is an approximation to
the shearing force diagram.
On fig. 3 is shown the diagram of maximum shearing forces
plotted from equation (16).
F(XED
Sct
F=j5(c*x)tk<s.tO
2F
\^' X*
Fisr. 3.
Levy shows this diagram in his fig. 25, p. 125 (see footnote).
From his equation (21 his), identical with our equation (16), he
infers that the locus starts horizontally at each end. It is readily
shown by differentiating (16) when the differential coefficient of
the variable part is 3 (c*  x'^) , then x = ± c makes this zero, so
3 IF
that the slope at end is zero, and at centre is tan"'— r— •
It is important to observe that at any section the positive
maximum shearing force is instantly succeeded by the negative
as the load passes it, so that their arithmetical sum or the rcoigc
of the shearing force, as we luive called it, is IF, and is in no
way altered by the fixing of the ends of the girders.
Fig. 4.
328 applied mechanics. [chap. xvii.
Beam of Uniform Section fixed hokizontally at the ends
AND subject to AN UNIFORM ADVANCING LOAD.
For convenience the uniform load may be taken as 1 ton
per footrun.
Consider first the negative bending moments. For the
middle third of the span they are zero at each point ; for the
load, however it be disposed, may be looked upon as a number
of concentrated loads, no one of which can produce a negative
bending moment on the middle third by the last problem
(fig 2).
Let X be the abscissa of any section on the third next the
right abutment, measured from the centre as origin, so that x
lies between the values \c and c, where c is the semispan. Let
the load advance from the left abutment till \hQ front is at the
section z from the centre, where z is measured in the same
direction as x, but is less than x.
On fig. 4, A VGB is the instantaneous bending moment
diagram for the ends hinged ; compare fig. 9, Ch. VIII. A VC
is parabolic, but CB is a straight locus. By theorems {a), 'h),
HR' is drawn across so that the areas AHH'B and AVCB are
equal and have their centres of gravity on one vertical line.
As z is less than x, the new base B.H\ for ends fixed, cuts
at S on the straight part CB. For ends hinged
the height of C above AB is
^ (c + z)\c  z),
while the height at x to SB is
y = 4^ (<^ + )" (c  ^)
The bending moment diagram AUCB for the extern load
divides into two parts a triangle ACB of area \v:{c>r zf {('  s),
and lever about A of AD = (c + ^z), and a parabolic part AUG
of area Yl'U{c + z), and lever about A of AE = \{c + z), where
TU = ^w(c + zf, just as for a span (r + z) uniformly loaded.
CHAP. XVII.] GIRDEKS FIXED AT THE ENDS. o29
Agaiu, the bending moment diagram AHirB, for the end
couples m and n that destroy the tips up, divides into tv^o
triangles ; AIIH' of area mc and lever about A of fc, and
AH B of urea nc and lever ^e.
Equating the areas of these two Ijending moment diagrams
and also their moments about A, satisfies the two conditions
laid tlown by preceding Theorems [a) and (7;) ; we have the two
equations
(m + n]c =.(c + zy{cz) + — {c + z)\
;So that
m = ^_{c + zf (llr  lOcs + .V), (17)
n = ~{c^zf{^c^z). (18)
The height to HH' at the section x is
c — X
y = 71 + —^ — (m  n),
So that at .r
M.,,. = 2/'  2/ = 4^3 (^ + ^)' (3(3c  z).v  ^0% (19)
If iT = ^, when M is zero, we have for S the virtual hinge
e _ ^^*
^~ 3(3cs)'
Putting it for the part of equation (19) that varies with z, then
^^ = 12{c + z) i2cx  d"  o:z),
dz
~^=12{c + z) [^cx  2c*  2>xz).
330 APPLIED MECHANICS. [CHAP. XVII.
z = 2c — , ( 20)
X
which, observe, is less than x as premised, makes
di.t fli( i c \
— = 0, and — = 36c(c + z){ x],
a negative quantity, as we are only dealing with values of x
which are greater than \c, in that we have disposed of the
middle third of the span already as far as negative bending
moments are concerned. Hence this value of z makes il/ucK. ^
maximum for all points in the third of the span towards the
right abutment. Substituting and reducing, we have
^^^neg. = j^ ^ max., (21)
when the front of the load is at z, equation 20.
We come now to consider the positive bending moments.
Let the load cover the span, and simultaneously at all sections
the bending moment is given by the dotted parabola on fig. 4,
as already explained, measuring up and down from the hori
zontal base across it. The equation to this parabola with the
centre of that base as origin is
y = ~ (c^  Sx'). (22)
See iig. 4, Ch. VI, for such equations, and fig. 6, Ch. XYI, for
the conditions.
Now y is positive in the middle third and even further out
from the middle, but we shall only consider the middle third.
Xow removing any part of the load is the same as if we had
left the load all over the span and applied a loail of equal
intensity n2waTcl at a part, and this upward load being itself
negative cannot cause any positive bending moment in the
middle third of the span. Hence the dotted parabola itself
gives the maximum bending moment at each section in the
middle third, the load being all over the span
'l/pc. = "^ (c^  3.r) max. (23)
for values of x from to ^c only.
For the third of the span next the right abutment, let x be
the abscissa of any section as before. Now we have shown that
CHAP. XVII.]
GIRDERS FIXED AT THE ENDS.
001
•joi
equation (20) il/„eg. is greatest for the load extending from the
left abutment to z = 2c — . If this load were acting upwards,
the bending moment at x would still be the maximum, but of
course positive. But before applying this upward load, let the
whole span be loaded, and the upward load removes the load
from left abutment to z and leaves it from z, to the right abut
ment. Hence the maximum positive bending moment is the
F=:^s(cf^yQc% ^cvjcy
SCALE
16 TONS TOAN INCH
Fis. 5.
same as J/neg. only measured upwards from the dotted parabola
(fig. 4), and occurs when the short segment from z is loaded.
Hence for values of x from jc to c
ir ^^ / 2 o ox 27WC
X  ■
(24)
a maximum when the tail of the load is at z, equation 20.
The same argument, on the last page, shows the negative or
downward intercepts between IIH' and the arc A U, all to be
maxima for the whole span loaded. These maxima only reach
down to the dotted parabola, and are superseded by the locus
of equation 21.
332 APPLIED MECHANICS. [CHAP. XVII.
Sliearing Force. — For the ends fixed (fig. 5}, we have
^nd this is the shearing force at x for the position of the load
shown. Xow the differential coefficient of Q with respect to z
is positive, or Q continually increases with z till ;; = a. Again
z> X means z = .r, and an additional load between x and the right
abutment, a share of which is thrown on the left abutment, so
that the increase of Q is more than cancelled when from Q we
subtract this whole extra load, as we must for the shearing force ;
hence the shearing force at any section x has maximum positive
and negative values when the segment from x to one or other
abutment is loaded, exactly as was the case for the ends hinged
(see fig. 1, Ch. IX), only now
V)
(c + x)\U + 2cx  x^). (25)
Examples.
1. Grapliical construction of the bending moment and shearing force diagrams
for a uniform girder of 36 feet span, fixed horizontally at the ends, and loaded
with 12 tons at 9 feet from the left end, and 6 tons at 21 feet from the left end
(fig 6)
The load line is laid down to a scale of 10 tons to an inch. From a trial pole a
link polygon constructed among the four forces P, W\, JFj, and Q, laid down at
10 feet to an inch, and a vector from the pole parallel to the closing side determines
the joint between the reactions, making /"scale \\\ and Q scale 6^ tons, being the
values for hinged ends. They may also be found by taking moments about the
left hinge. Through this joint the base of the shearing force diagram is drawn
horizontally, and the diagram completed for hinged ends.
A new pole on the horizontal through tlie joint is chosen at a distance 6 feet,
a horizontal base laid off for tlie bending moment diagram (with hinged ends),
which is then constructed as a link polygon to the new pole.
Thus (ar it will be seen that the load and girder correspond to the moving
model (fig. 16, Ch. IX), with the locomotive standing so that its heavy wheel is
9 feet from the left abutment.
To return to fig. 6, it will be found that the scale for bending moments is
60 foottons to an inch, tliat is, six times as fine as the ton scale.
For the ends fixed we must draw a base, cutting across the bending moment
diagram, so as to include the same area over the base, and have the centres of
gravity in one verti<:al (see Theorems («) and (A)).
To do this graphically the bending moment diagram is divided into thru
triangles and one rectangle. The first triangle has a base '•• feet and a height
103'o foottons, being the bending moment under 7/'i, either as scaled oflF or as
calculated. The last triangle has a base Ift feet and a heiglit 97'5 foottons. The
dimensions of the remaining triangle and the rectangle are readily seen.
CHAP. XVII.] GIRDERS FIXED AT THE ENDS. 333
It is convenient to reckon 18 feet the half span as a horizontal unit for these
areas, which will then be found to be Ai = 26, A^ = 2, Ai = 65, and Ai = \\.
The lines (if action are drawn doivh through the repei;tive centres of gravity.
Next the lines of action A^ and A,, are drawn upwards through the trisecting points
of the span to represent the areas of the two triangles, into which the bending
moment diagram for the pair of end couples can always be divided. A load line
of the areas is laid down to the foolton scale, a link polygon drawn among the
lines of action when a vector from the pole drawn parallel to its closing side divides
the areas At and Aa from each other. Since the halfspan is unity, it will be seen
that Aj and A(, are M^ and Ma on the footton scale. Their values scale 51 and
83 foottons, and are pricked up at the ends and the base drawn across.
To modify the shearing force diagram M^  ,1/5 = 31 foottons, dividing this by
36 feet, the lever between the supports, we get gths of a ton as the common force
by which the right end must hold down and left hold up to resist the lefthanded
couple J/e  l^o Drawing a new base for the shearing force diagram ths of a ton
lower down decreases the one and increases the other support by this amount.
2. To find the righting end couple M^ (fig. 6) by calculation. For equal areas
and equal moments about 0,
M& 4 3h = 26 + 2 4 65 + 41 = 134,
\1Mz + 24ilf5 = 26 X 6 + 2 X 13 + 65 X 15 + 41 X 26,
and Mi = r>\, M(. = 83 ft. tons.
3. A uniform girder is loaded on the left half only with a uniform load : find
the righting end couples to hold the ends horizontal.
Substituting for m and n in equations (17) and (18), we get
«j = ^^ IV c and n = ^^ wc'.
Now the bending moment at centre for hinged ends is Q a quarter of total load,
multiplied by c half span: that is, the height of C (fig. 4) is now \icc or \%wc.
Hence, if on fig. 4, we suppose the load to be on left half only then m and n
are \\ and ^4, respectively, of the height of Cthe middle point of the bending
moment diagram.
4. The case of a uniform girder, uniformly loaded on the left half only, is so
important that we shall calculate the righting end couples directly.
First since Q is a quarter and F threequarters of the load, it follows that the
shearing force diagram crosses the base at a point th of the span from A (fig. 4).
Also AE is onehalf of span so that U and C are the same height and ET= TU
are each half the height of C. With c the half span as unity for horizontal areas,
the area A VC is %TU or ird height of C. Putting /.• for the height of C, area A UG
is \k and its lever about A is +c while area ACB is k, and its lever about A is c.
Again for AHH'C divided into two triangles by AH', the area of AER' is m, its
lever 3 x 2c; the other AH'B has an area n and lever j x '2c. Equating areas
and moments of areas about A
k , 2c 4c k c , 11 , , 5
~ + k; >n^ + n.~= ^.~ + kc; m = ~~ k and « = 
The graphical construction is shown at top of fig. 3, Ch. XX.
Fi 6.
CHAP. XVIII. 1 COMBINED STRESSES. ;J35
CHAPTER XVIII.
skeleton sections for beams and struts ; combined thi'.ust
with bending and twisting with bending ; long struts.
Thin Hollow Crosssections.
Let t be the uniform thickness of a thin hollow crosssection of
any form, and let n and n be the numerical coefficients respec
tively of the moment of inertia and the moment of resistance to
bending of a solid crosssection of the same form ; let B, H be
the breadth and depth of the rectangle circumscribing the
section, and h, h the breadth and depth of the rectangle cir
cumscribing the hollow ; then
/o = n [BW  hh') = 71 B (IP  h'), since B = h nearly
= n' B (Hh) {H + Hh + h')
= n' B.2t. 3H' since H=h nearly = n' . 6BHf.
M = 4^I. = {]Qf.B.H.t
m H \m J
= n . QfBHt (1st approx.) (a)
Thin hollow rectangle, M = i . QfBHf =fBHt.
Thin hollow circle, M = ^ ^BSt = Qfrf, {d = diam.)
A closer approximation for any form is obtained thus : —
/„ = n' \BH'  hh'\ = n'{ {B  h) m + h {H'  h') 
= n {{Bh)H' + h[H  h){W + Hli + h'] 
= n' \2t.H' + b. It: . SHh] (approx.)
= 2n'H\tH + '6ht'h\ (approx.) ;
.. M = 2??/ (^^^ + 3^570 (approx.),  [h)
where t is the thickness of each side, and /' the thickness of the
top or bottom.
To design a thin hollow crostssection — choose the depth H
the proper fraction of the span to give the required degree of
336 APPLIED MECHANICS. [CHAP. XVIII.
stiffness ; assume B a suitable fraction of U to give sufficient
lateral stiffness, and the above is a simple equation from which
to find t; or t may be assumed a multiple of the thickness that
the metal plates are usually manufactured, and B found from
the formulre. Xovv find the moment of resistance to bending,
and the resistance to shearing, by the accurate formulae ; and if
this differs from M and F, alter t or B for further approxima
tion.
Crosssections of Equal Strength.
When a beam is made of a material whose strengths to resist
tension and thrust are different, the area of the upper fiauge is
made different from that of the lower (fig. 1), in order that
both flanges may be brought to the proof or working stress at
the same time. In the case of cast iron, the strengths to resist
tension and thrust are as 1 and 6 ; and on this account the area
of the upper flange (compressed) is about onesixth that of the
lower flange (extended). This form of crosssection was first
proposed by Mr. Hodgkinson. On account of the liability of
cast iron to crack if unequally cooled, sudden changes of thick
ness of metal are to be avoided ; on this account the top of the
web may be made of the same thickness as the top flange, and
the bottom of the web of the same thickness as tlie bottom
flange.
Double T crosssectio7i (fig. 1) — The position of the neutral
axis, and the moment of inertia about that axis, in terms of the
areas and depths of the three rectangles, are expressed as
follows, the notation being :
.\reas. Depths.
Upper flange, . . . A^, hi.
Web, A,, 1u_.
Lower flange, . . . A^, h^. q^
Totals, . A
Fig. 1.
Exact Solution. — The height of the neutral axis above the
lower side of the crosssection is obtained thus —
h {hi + hi) As  [hi + hi) Ax  ijh  hi) A^ , .
II » =2 2 A " ^^^
CHAP. XVIII.] COMBINED STRESSES. 337
The moment of inertia for each rectangle is shown ofi page
215 ; and letting 2 as before denote the " sum," the moment of
inertia for the crosssection is
+ A,A, {Ju + 2h + IhY + A,Az {K + h)'}. (2)
The moment of resistance to bending is as before
llh
Approximate Solution. — (Eankine's "Civil Engineering,"
§§ 163, 164). When 1h and A3 are small compared with lu,, we
may leave them out of the exact formulae ; and we obtain
_k {A.a;)1u
yi>2 2 A — ' ^^
lo =^' + ^ (^1^2 + 4:A,A, + A,A.). (4)
Put h' for hi + ' ^ ^ that is for the distance between the
centres of gravity of the flanges ; and let Az be the area of the
crosssection of the vertical web measured from centre to centre
of the top and bottom flanges ; then, nearly
A.JjA.vi^A.^. mJ^A.Z^A... (5,6)
Substituting this value of Az in equation (4), we obtain
J/o = h' j/Us + (2/6 /.) ^ j = K /„^, + (24  /,) ^^ j . (7)
In designing a beam to resist a given bending moment, the
depth li is taken at a fraction, say ^th to j^th of the span so
as to ensure stiffness ; the thickness of the web is then fixed by
considerations of practical convenience, and so as to give suffi
cient resistance to shearing ; and the area of the upper and
lower flange can then be found by equations 5 and 6 ; having
thus fixed the value of Ax and A^ we can then choose breadths
and depths suitable for the flanges.
338
APPLIED MECHANICS.
[chap. XVIII.
Fof the section as thus fixed calculate M and F by the
tabular method shown for figs. 5, 26, Ch. XIV ; if ^i and h^, the
depths chosen for A\ and A^, are very small, it will be found that
IVI and F are sensibly what is required, and that the neutral
axis sensibly divides the depth of section as/o and/^, so that no
further calculation is necessary. If, however, one or both of
the depths /;,, hi, be not very small, the solution by the tabular
method will differ considerably from the data, and further
approximation will be necessary. When one of the flanges, as
in the case of cast iron, is comparatively deep, the inaccuracy of
the results will be considerable, and one or more further ap
proximations may be required. For different examples, the
error will be diiferent in amount, and we have no simple means
of judging how great this error will be in any particular case.
In the equations given above, the results for M and I are
close when hi and hs are small ; the result for 7/a will be close
although hs is not small, and that for yb will be close although
hi is not small; thus, suppose that Aj is small, we have
h (hi + h>)Ai
2/4 = 9
hiAi + hiAi
2A
(8)
putting h'  Ai = /ij, and h' = h,
,24l + ^2
yb = h'
2A
, nearly ;
(9)
a result similar to that found previously in equation (3), but ya
cannot now be found by interchanging Ai and A^.
Common forms for castiron beams are shown in fig. 2 ; the
corresponding equations for these Tshaped sections are derived
from the above by putting Ax = 0, or A^ = 0, according as/„ or fh
is the greater. Thus for a section of this form, when/,, is greater
than fb, the flange will be required on the extended side ; when
/i is greater than /„, the flange will be required on the com
pressed side ; and we have
A.J'^^A,; (10)
ovA,J^A,; (11) [
J
as the case may be.
Fit,'. '.2.
CHAP. XVIII.] COMBINED STRESSES. 339
Similarly for resistance to bending, we have
M=h'i2^,f,)^; (12a)
or =h'i2/,fa)~'; (Ub)
as the case may require.
In the case of trough shaped beams the same formulae are
applicable, if we consider the web A2 to consist of the two
vertical ribs.
Examples.
1. Find the moment of resistance to bending of a castiron pipe 18" external
diameter, metal 1" thick, and/ = 2 tons per square inch, by means of the exact
formula ; and compare the result with that obtained by the approximate formulae
(a) and (b).
M = 430 inchtons by exact formula = 390 and 429 by approximate formulce.
2. Find the thickness of metal required for a castiron pipe 24" external diameter,
so that its moment of resistance to bending may be 50 foottons, and its resistance
to shearing 10 tons ; taking/= 2 tons per square inch,
600 = 6 X 2 X 244 X t; .. t = 0"87 nearly.
■On checking the calculation by the accurate formula, and taking t = 0"87,
M = 705 inchtons ; so that i = 0"87 is more than sufficient ; and for shearing
this thickness will be several times too great. On account of the difficulty of
casting so as to have the thickness quite uniform, an allowance has to be made,
and t would probably be taken at about 1""25.
3. Find, by the approximate formula, the moment of resistance to bending of
the crosssection shown on fig. 3, p. 254, and compare the result with that given
in the text at that figure,
M = /i(2 . 900 + 3 . 3 . 6 . 24) = 1032/(2 x 7794/^ 15).
4. Design the central crosssection for a wrought iron beam, for which /„ = 4,
and/j = 5 tons per square inch, suitable for example 1 and fig. 9, Ch. XI.
Since the span is 42 feet, we may take h' = 40 inches ; M= 40575 ft. tons
= 4863 inchtons ; the web may be taken as f " thick.
From equation 7 we have
4863 = 40 {4^1 + (8 5) V} ; .. ^1 = 285 sq. in.
4863 = 40 {5^3+ (104) ifi^} ; .. ^3 = 213 „
Adopting as a first approximation — breadth of flanges, 21 inches ; thickness of top
flange, 136 in., of bottom do., 1 in. ; thickness of web,  in., and depth of girder
(outside to outside), 412 in. ; we get by applving the exact method shown at
p. 256,
4 X 21340 „ . ,
M = — rj^T — = 4652 inchtons,
a result differing from what is required by only 4 per cent.
z2
340 APPLIED MECHANICS. [CHAP. XVIII,
The upper flange may therefore be taken 1 in. thick, the lower flange 1 in.
thick, and the breadth of each 22 in. This does not take into account the angle
irons, and the loss by rivetholes.
Solve this example by another approximate formula.
Jf = A'{/„^i  i(A/a).:f2}. (7a),
4863 = 40 {4^1 i (5 4)15} ;
Ai = 31, and similarly A3 = 24 square inches for a first approximation ; we may
fix on t„ = 125 and th = I inch, and from these obtain suitable breadths,
y'„ = 40 =^ 1778 in., and y'j = 2222 in., y, = 184 in., and yj = 2272 in. ;.
.. /'„ = 4 (1  — ^j = 386, and /'& = 489 tons.
We have therefore
4863 = 40 {386^1  ^(489  386) 15} ;
^1 = 305 square inches, and ^3 = 256 square inches.
Adopting the following dimensions, upper flange 244 in. x 125 in., lower flange
25*6 in. X 1 in., and h = 4112 in., and solving by the exact method, we obtain
4 X 24250 . ,
M = — f^.T," = ^^^^ inchtons,
a quantity differing from the required result by less than 4 per cent.
Allowance for Weight of Beam.
After having designed a beam which is sufficient to bear sn
given external load, it is necessary to make an allowance for
the weight of the beam itself; especially is this the case for
beams of long span, as then the weight of the beam bears a
considerable proportion to the amount of the external load.
This allowance is readily made by increasing the breadth of
the provisional beam sufficient for the external load alone ;
since the breadth is a dimension which appears in the first
power in the expression for the resistance to bending.
Consider the weight of the beam, and the external load
reduced to its equivalent dead load, as uniformly distributed, a
supposition sufficiently exact for our present purpose ; let b'
denote the breadth, and B' the weight of the provisional beam,
computed for IV the external load alone ; let b, B, and W
CHAP. XVIII.] COMBINED STRESSES. 341
denote the same quantities for the actual beam sufi&cient to
bear the external load and its own weight ; then
V~ B'~ W' ~ ]r~ IF'
W W  B W
W ~ W  F W  E
Therefore the breadth of beam required,
The weight of beam required,
The gross load
Eesistance to Twisting and Weenching.
One end of a cylindrical bar is rigidly fixed, and to the other
•end a couple is applied in a plane at right angles to the axis of
the bar ; or, what is the same thing, as shown in fig. 3, a pair
•of equal and opposite couples are applied to the ends of the bar ;
the tendency of these couples is to make the bar rotate about
its axis ; and if we suppose the bar to consist of fibres originally
straight and parallel to the axis, each of these fibres will now
have assumed a spiral form. The moment of each couple is
called the twisting moment or moment of torsion applied to the
bar, and it is constant for each crosssection ; on account of the
bar being uniform, the stress will be similarly distributed on
each crosssection ; and since the bar is circular in section, the
stress at all points equidistant from the axis will be the same.
Suppose two crosssections to be taken at the distance dx
apart ; the twisting moment causes the one section to move
relatively to the other through an angle di ; and if we consider
two points originally opposite to each other, that is in the same
fibre, one in each section and at a distance r from the axis, then
these points, relatively to each other," move laterally through a
342 APPLIED MECHANICS. [CHAP. XVIII.
distance r .di; and since the two sections are dx apart, the rate
of twist is
a quantity directly proportional to the distance of the points
under consideration from the axis.
We have thus at any point in a crosssection, a shearing
stress at right angles to the radius drawn to the point, and pro
portional to that radius in intensity ; this may be expressed
thus
,Crl^; (17)
where C is the coefficient of transverse elasticity for the material
of the cylinder under consideration ; Eankine gives
For cast iron, C = 3,000,000 lbs. per square inch (approx.).
For wrought iron, C = 9,000.000
The greatest value of q occurs at the surface of the cylinder ;
and if / represent the resistance of the material to shearing,
and ?'! the radius of the cylinder, then we have
9=/} (18)
If we consider s a small portion of the ring of the cross
section, with its middle point at a distance r from the axis,
then r . di will be its mean length., and we may denote its
Fig. 3.
breadth by dr ; its area then is r . di . dr ; the intensity of the
shearing stress at s is q; the amount of shearing stress on
the small area is therefore
/
g . r . di , dr = — r"^ . do . dr;
r,
CHAl'. XVIII.] COMBINED STRESSES. 343
and its moment round the axis, found by multiplying this
quantity by r, is therefore
^~ r" ir .di . dr).
The quantity within brackets is the small area, and r is its
distance from the centre; r^{r . di . dr) is therefore the moment
of inertia of the small area about the centre ; summing for
every such small area, we have the moment of resistance to
torsion for the cylinder
M = 'K=^2I,; (19)
where K is the moment of inertia of the surface about the
centre, and I^ is the moment of iner
about a diameter. We therefore have
centre, and I^ is the moment of inertia of the same surface
where di is the diameter of the cylinder.
For a hollow cylinder, let r^ and Vi, d^ and di, be the
internal and external radii or diameters, as the case may be ;
let /'o and Iq be the moment of inertia about a diameter of a
cylinder equal in radius to Tq and ?'i respectively, then I^  I\
is the moment of inertia about a diameter of the rinar under
consideration ; we have therefore
M = ?/ (/„  /;) = ?^(i7rr,*  iTrn*)
' 1 ^1
= 5/!il!i' = . 196/1^^'. (21)
2 ri d^ ^ '
Comparing : these equations with those on page 266, we find
that for equal values of the limiting stress /, the resistance
of a cylinder, solid or hollow, to wrenching is double its
resistance to breaking across.
The working values of the limiting stress /, suitable for
shafts, as given by Eankine, are
cast iron, / = 5000 lbs. per square inch ;
wrought iron, /= 9000 lbs. per square inch.
344 APPLIED MECHANICS. [CHAP, XVIII.
For a crosssection whicli is not circular, the above formulae
are inapplicable, since the ratio  is no longer constant. For a
square shaft M. de St. Venant gives as the moment of resis
tance to torsion
M = 028 1/^^ (22)
Angle of torsion of an uniform cylindrical shaft. — Let x be
the length of the shaft, and i the angle in circular measure
through which the one end has turned relatively to the other ;
then, since the angle of torsion per unit length is constant, we
have, from equation (17),
— = i = i=/ A^tl.J^ (C)'X\
dx X Cr Or,' Cr^ Cd\' ^^^
If/ be the working resistance of the material to shearing, we
have the same angle, whether the shaft be solid or hollow ;
the values of /and of C for cast and wrought iron have already
been stated, and for
cast iron, ^'=3^0^' ^'^^^
1 X
wrought iron, *=^vm"r' (^^Z*)
where i is the angle in circular measure through which the one
end of a shaft of length x and diameter d^ has turned relatively
to the other end, when the working strain has been produced ;
the coefficient for cast iron is somewhat uncertain.
When subjected to J/ any twisting moment not greater than
the proof moment, we liave for a solid shaft (equation (20))
r, 2 ' r 2
TT/'i
and from equation (23)
i ^
*" Cr
For a liollow shaft, similarly
. q^x 'IMx _ , Mx ,^_.
' ' "2 CW ,:..')• ^''>
CHAP. XVIII.] COMBINED STRESSES. 345
If we make x=l, or, what is the same thins^, if the distance
between the two crosssections which we consider is unity, the
stiffness of the shaft will be measured by the I'eciprocal of i,
i being the angle in circular measure through which the two
crosssections have turned relatively to each other, when the
skin has been brought to the proof strain.
For two shafts of the same length and material, but of
different diameters, we see from equation (25) that the twisting
moment to be applied to each, in order that both may be turned
through the same angle of torsion, is proportional to the fourth
power of the diameter, the proof stress not being in any case
exceeded ; thus
»= 102^= 102^, or 3^, = ^, (27)
Bending and Torsion Combined.
Let the shaft shown in fig. 4 be acted upon by a bending
load and a pair of equal twisting couples ; and at the point H
let Ml be the moment of the first, and Mo the moment of the
second ; then in order to find the amount and direction of the
greatest principal stress, we require to combine the greatest
direct stress due to bending with the greatest shearing stress
due to twisting ; this is done by the method of the ellipse of
stress.
At the point H, let ]> be the intensity of thrust (or tension)
due to the bending moment il/i, and q the intensity of shearing
stress due to the twisting couple M^ ; then we have
4.1/, 2J/.
P = — J 1 = — i (^^)
Let pi be the greatest intensity of stress (thrust) at the
point ; then fig. 1, Ch XV, shows the construction required to find
its amount ; in that figure OL =^:», OR' = q, and we have
:P =0M ^ MrJ^^ +k^+,/ (29)
the greatest intensity of shearing stress is represented by
MR=J^+q'; (30)
346
APPLIED MECHANICS,
[chap. XVIII.
and the angle made by the greatest stress pi with the axis of
the shaft is given by the equation
tan 26 =
JRL
ML
2q
P
(31)
By substituting for 2^ and q the vahies given in equation 28,
we have
and for the greatest intensity of shearing stress
2
ME
(32)
(33)
A very important example of this principle is that of a shaft
with a crank attached ; in this case we have a force applied to
the centre of the crank pin, and resisted by the equal and
opposite force at the bearing ^S'. If F represent the force, then
the moment of the couple is (fig. 4)
M=P .SP;
(34)
this couple may be resolved into two couples, one a bending
couple
M, = P .N8=McoQj^
t\\e other a twisting couple,
M,^P .NP = M sin /.
t if,
(35)
(36>
Cl
S)
Fi. 4.
The greatest intensity of stress is found by equation 32,
2
Pi =^3 (M cos J + ^3P cos'j + M^ sin^y)
2 2
= — ; {M cos j + M)  —  M (I + cos ;■)
TTVi
= ^ M (1 + cos J)
^r.
(37>
CHAP. XVIII.] COMBINED STRESSES, 847
If instead of px we put / the resistance to torsion or thrust
(the smaller), we get
5*1
d^ = — M{1 + cos j) ; (38)
which enables us to calculate the diameter required for the
shaft. If we put /, for the greatest intensity of the shearing
stress, we have
d^=^^M; (39)
which also enables us to calculate the diameter required ; and
the greater of the two, one got from equation 138, the other from
equation 39, is to be adopted.
The angle made by the principal stress with the axis of the
shaft is given by equation 31,
ta„2S?'' = ^.?^^ = tani; ..  i. (40)
p il/i COS 7 "^ 2 ^ '
Thrust or Tension combined with Torsion.
Let the shaft shown in fig. 5 be acted upon by a thrust
(or tension) P and a pair of twisting couples of moment M\ the
stress due to P is uniformly distributed, and that due to M is
greatest at the skin ; the greatest intensity of stress
will therefore be at the skin. If under thrust, the \p
length of the shaft is to be so short compared with its fr^
diameter, that the bending action need not be taken '^^^
into account. At the point H we have a thrust (or
P 2M
tension) p = — ^, and a shearing stress q = • — ^\ pro
ceeding as at fig. 1, Ch. XY, we have
X
0^/ = i2~ («) $^
^^J? = J(a^(iJ ^)
Fig.
The greatest intensity of thrust (or tension) is
lh = OM+ MR^ (43)
the angle B made by p^. with the axis of the shaft is given by the
equation
tan20=S=?^. (44)
ML p
The greatest intensity of shearing stress is MB.
348
AITLIED MECHANICS.
[chap. XVIII.
Thrust and Induced Bending.
When thrust is applied to a pillar or strut whose length is
great compared with its diameter, it will collapse not by direct
crushing but by bending and breaking across.
Let a long thin vertical bar, originally straight, be deflected
to an extent not greater than the proof deflection by the appli
cation of a horizontal external force applied, say
at its middle, while the ends are guided so that
they cannot move laterally, and let it be held in
that position ; it will then have a form such as
is shown in fig. 6 ; let the load P he now applied,
then when the restraint is withdrawn, the bar will
tend to assume its original vertical form, it will
remain neutral, or it will collapse according to the
amount of P ; that is to say, if the moment of P
relatively to the centre of the bar, viz. P . v, is less
than the moment of resistance of the bar to bend
ing, the bar will tend to right itself.
The stress on the crosssection AB consists of
one part 2^' due to the load P, and another part p'
due to the bending which takes place in the direction in which
the pillar is most flexible ; since P is uniformly distributed,
we have
P
P
S
(45)
where S represents the sectional area of the bar ; by equation
given on page 240, we have M = ivp'hh', and since M = Pv
(46)
where h is the smaller, and h is the larger diameter, when these
are unequal ; the proof deflection v, Ch. XVI, equation (14rt), is
directly proportional to the square of the length and inversely
proportional to the depth ; that is to say
that is
/ Pl^ Pt^
v:r.~, and /' x ^^ ac ^, ;
,, , /* , p p" (^ N'
» ccp —■. therefore —i cc
(47)
that is, the additional stress due to bending is to the stress due
CHAP. XVIII.]
COMBINED STRESSES.
349
to the direct thrust, as the square of the proportion in which
the length of the pillar exceeds the least diameter.
The total intensity of stress p + p" must not exceed the
strength of the material ; equating that intensity to /, the
strength of the material, we have
f = p + 1i" = p + o.}/
P
F
" S
1 + a
P =
fs .
i'^'T
(48)
1'
Fig. 7.
where a is a constant coefficient to be determined by experi
ment. The above investigation is due to Tredgold, and values
of the coefficient a, as determined by Professor Gordon, are
given below.
For a strut or pillar securely fixed at the ends : —
Wroughtiron solid rectangular section, . . . a = yi^Vir'
*Angle, channel, cruciform and Tiron (see fig. 8), a = ^55.
Castiron, hollow, cylinder, a = tItt*
fFor timber struts, oak and pine, « = t^tt*
A pillar rounded at both ends is as flexible as one of double
the length fixed at the ends ; so that
fS
fS
1 + a
21V
h
1 + 4«
(49)
The strength of a pillar fixed at one end and rounded at the
other is a mean between that of a strut fixed at both ends and
one rounded at both ends.
* As deduced by Unwin.
t As deduced by "Weisbach.
350
APPLIED MECHANICS.
[chap. xvin.
The values of / in lbs. per square inch are given by Kankine
as follows : —
Breaking load.
Proof load.
Working load.
Wroughtiron solid >
rectangular section/
36,000
18,000
6,000 to 9,000
Wroughtiron cell
27,000
13.500
4.500 ,, 7,000
Castiron cylinders
80,000
26,700
13,300 ,, 20,000
British oak, dry
10,000
—
1,000
American oak, dry
6,000
—
600
Red pine and larch, dry
5.400
—
550
For green timber the values of/ should be halved.
A pillar or strut securely fixed at both ends corresponds
with a beam fixed at the ends (fig. 6, Ch. XVI) ; the points of
fracture are the points where in that figure the bending moment
is greatest, viz., at the centre and at the ends. A pillar fixed
at one end corresponds with the beam shown in fig. 4, Ch. XVI ;
the points of fracture being at the fixed end, and at about one
third of the length from the rounded end. A pillar rounded at
both ends corresponds with the beam shown in fig. 12, Ch. VI,
the point of fracture being at the middle of the length.
The following table gives the results of the above formulae
for pillars of wrought and cast iron, whose diameter and lengths
are in different proportions, and whose ends are securely fixed : —
h
p
Breaking load, lbs. per sq. in.==^
h
p
Breaking load, lbs. per sq.in. = ^
Wroughtiron
solid rectangular
section.
Castiron hol
low Cylinder
Wroughtiron
solid rectangular
section.
Castiron hol
low Cylinder.
10
15
20
26 4
34,840
33.490
31.765
29,230
64,000
51,200
40,000
29,230
30
35
40
50
27,700
25.56c
23,480
19,640
24,620
19,700
16.000
11.030
€HAP. XVIII.] COMBINED STRESSES. 351
From this table it is seen that, so far as the ultimate strength
of such pillars is concerned, cast iron is stronger tlian, as strong
as, or less strong than, wrought iron when the proporlion , is
less than, equal to, or greater than 264 ; this result was first
pointed out by Professor Gordon.
For struts in wroughtiron framework, such crosssections as
are shown in fig. 8 may be chosen on account of their stiffness ;
these are called T, angle,
channel, and cruciform ra p^i^ tp^
iron, respectively ; in fix
I
ing the proportion 7 for Fig 8.
such sections, the value of h is to be taken as the least diameter ;
this is marked in the diagram.
Since, however, such crosssections are not made of very
large dimensions, struts above a certain size require to be built ;
in this case they usually consist of four thin plates forming a
square and connected at the corners by angle irons ; of two thin
plates held parallel and at a fixed distance apart, by means of
a web and angle irons, or by a lattice work of small diagonal
bars ; or of two f irons or channelirons, with the ribs turned
towards each other, and held by a lattice work of small diagonal
bars as in the case just stated.
Examples.
•5. A lattice girder 80 feet span bears a load of 100 tons uniformly distributed;
depth from centre to centre of flanges 6 feet, and f — 4 tons per square inch ;
the breadth of the flange is 1 ft. 9 in., and is constant ; the thickness, however,
varies.
Taking the provisional breadth as 1ft. 9 in., find how much this has to be
increased so as to allow for the weight of the beam itself.
Jfo= 12,000 inchtons; J/io= 11,250; Jfao = 9000 ; Jfso = 5250 ; Mm = ;
since the flange is thin, we have M = fhh't = 4 x 72 x 21 k (!, and we obtain
<o=2in., <io = 2 in., U,^ = 15 in., <3o = 1 in., and <4o = 05 in. say; the
average being 1*2 in. nearly. Taking this as the thickness of the flanges, and
allowing for bracing, we obtain 10 tons nearlj'^ as the weight of the provisional
girder ; so that V = 21 inches, B' = 10 tons, and W = 100 tons ; from which
we readily find J = 24 inches, B =\\ tons, and W  \\\ tons for the actual
girder.
6. A water wheel of 20 horse power makes 5 devolutions per minute : find
the diameter suitable for the malleable iron shaft which transmits this force.
For each revolution 132,000 ft. lbs. of work are performed ; this is equivalent to
21,008 lbs. acting on a wheel of radius one foot, and we have iL = 21,008 ft.lbs.
= 252,096 inchlbs.
252,096 = 196 x 9000 rfi^ = 1764 «^i=; .. d = 523 inches.
352 APPLIED MECHANICS. [CHAP. XYIII.
If this shaft be 12 feet long, what is its angle of torsion when the working
moment as above is applied? Take /= 9000, and C = 9,000,000; then
1 144
» = TTT^ Ti^ = "05507 = 3" 9'.
500 523
7. The crank shaft of an engine is 5 inches diameter; the distance from
the centre of the bearing to the point opposite the centre of the crank pin, NS
in fig. 4, is 12 inches; the half stroke, NP in figure, is 16 inches; and the
pressure applied to the crank pin is 5000 lbs. Find the greatest intensity of
thrust, tension, and shearing stress ; and 6 the angle made by the line of
principal stress with the axis of the shaft,
FS = 20 inches ; .•. M= 100000 inchlbs. ; J/i = 60000, and Mi = 80000.
51
pi = — 100000 (1 + f ) = 6530 lbs. per square inch,
the greatest intensity of thrust and of tension, at the bearing, the one being at the
one side and the other being at the other side of the shaft. The greatest intensity
of shearing stress is ytt 100000 = 4080 lbs. per square inch. The angle
e = ^ = 27°.
8. A shaft 8 inches diameter is subjected to a thrust of 100 tons uniformly
distributed over its two ends, and a twisting moment of 30 foottons. Find
the greatest intensity of thrust and shearing stress, and the angle made by the
line of principal stress with the axis of the shaft,
100 « 720 „ ,„
» = = 199 tons ; q = = 3'o8 tons per square inch ;
^ 5026 ^ 20104 ^ ^
/ji = 10 + \/99 + 12*81 = 471 tons per square inch ;
the greatest intensity of thrust ; the greatest intensity of shearing stress is
371 tons per square inch ; and
tan 20 = ^ = 3*6 ; 6 = 37°.
9. The diameter of one shaft is double that of another of the same material ;
the smaller gave way when subjected to a twisting moment of 2 ft. tons. What
twisting moment will be required to wrench the other? Aus. M = 16 ft. tons.
10. A shaft 12 feet long and 6 inches diameter is subjected to a twisting
moment of 16 ft. tons, and the two ends are thus twisted through a certain
angle; a second shaft of the same material, 16 feet long and 9 inches diameter,
is twisted so that its angle of torsion is exactly the same as that of the first :
find the twisting moment required to do this,
„ 192 X 144 ,^ ^ M X 192
therefore
192j<J44 ^ ^_xJ92 ^jjgj.^.f^j^,j ^ ^ 29 i„..tons = 6075 ft.tons.
1296 6561
CHAP. XVIII.] COMBINED STRESSES. 353
11. What thickness of metal is required for a castiron hollow sliaft, 10 inches
outer diameter, so as to resist a twisting moment of 10 ft. tons?
10*  ^/„'
A>is. M = 120 X 2240 = 196 x 5000
rfu* = 7258 ; therefore d^ = 923, and t'ne
thickness required is 0'4 inch.
12. A malleable iron shaft 20 feet long and 6 inches diameter is subjected
to a moment which twists the ends through an angle of 2' ; taking C the
coefficient of transverse ehisticity as 9,000,000, find /, the stress at the skin,
2/* 240
i = 0349 = — — • : therefore / = 3926 lbs. per square inch.
9 X 10« X 6 ^ f H
13. The inner and outer diameters of a hollow steel shaft are 10 and 12 inches,
and / = 6 tons per square inch is the working value of the resistance to shearing.
What is tiie twisting moment this shaft is capable of transmitting?
12*  10'
M  196 X 6 — — ,^ = 10.')2 inchtons.
14. A castiron column, securely fixed at the ends, external diameter 8 inches,
length 20 feet, is to bear a steady load of 30 tons. Find the thickness of metal
required.
Here  = 30, and for that proportion the breaking load is 24,620 lbs. per
//.
square inch ; taking g for a factor of safety, we get the working stress
/= 4100 lbs. = 1*8 tons per square inch; the area of metal required is there
fore 17 inches; this gives 67 inches for the internal diameter, or finch nearly
for the thickness of metal. Since an allowance has to be made for slight
irregularities in casting, the thickness of metal should be 1 inch.
15. Find the working load for a castiron pillar 12 inches diameter, 40 feet
long, metal 1 inch thick, taking 6 as the factor of safety.
Here  = 40 ; 6' = 34'6 square inches ; /" = 1'2 tons per square inch ;
h
and the steady working load is 42 tons nearly, both ends being securely fixed ;
but if the load is such as to cause considerable vibration, from onehalf to
twothirds of this amount may be taken.
16. What is the crusliing load for a malleable iron bar 6 in. x Sin. x 10 feet
long, securely fixed at one end ?
I . .
S = 18 square inches ;  = 40, and the ultimate stress corresponding is
n
10"5 tons per square inch, wlien fixed at both ends; when rounded at both
ends, this reduces to ol tons; when fixed at one end and rounded at the
other, the result is the mean of these quantities, viz. 7*8 tons per square inch.
The crushing load is therefore 140 tons steadily applied.
2 A
.^54 APPLIED MECHANICS. [CHAP. XIX.
17. Find the working strength of a strut formed of channel irons  inch
thick. 6 inches broad, width of each flange (outside) 25 inches, and length
6 feet, fixed securely at both ends.
Here 5=5 square inches ;  = 29 nearly, and the working strength is
h
\\ tons per square inch, or 7 tons nearly.
18. What is the working load for a strut of seasoned American oak firmly
fixed at the ends, 20 feet long, and 1 foot square ?
Ans. P = 33,000 lbs. nearly, or Hi tons.
19. A strut of red pine whose ends are to be well fixed, is to be 4 inches
thick and 6 feet long ; the thrust applied to its ends is calculated to be 4 tons.
Find the breadth required.
Iiy 324
\h) ^ "^ 27o
S = F = 4 = 37 square inches nearly.
and the breadth required is therefore 9^ inches nearly.
CHAPTER XIX.
DESIGN OF LONG STEEL STRUTS.
For long struts in wrought iron and steel, the double Tsec
tion is of great importance, as most struts, where merely booms
of girder or themselves girders, can be hlockcdout into such a
form. The extreme shapes where the flanges are equal, and
where one flange is zero, that is, double T sections equal above
and below, and the Tiron sections are especially important. The
angleiron too is often constrained to bend like a Tiron by being
braced in other directions.
We will establish a general expression from which to select
an approximate one suitable for designing struts. The sizes are
shown clearly on flg. 1. E<j. 9, p. 338.
25 + qc h' 2a + qc h'
a + b + qc 2 a + o + qc 2
CHAr. XIX.] LONG STEEL STRUTS. 355
and taking a mean of the term* in ^ repeated twice, once with
each of the evident values of 8, for symmetry, we have
/„ = afm^ + hM + J^ {aP + hP + qtc^)
_thf^ \{2a + qc) (2b + gcf + i(2& + qc){2a + qcf
4 (rt + 6 + qcy
 Iqdh'' + ^\qcH + JL (a + &) ^'
4 a + + qc
= *' (2« *,c)&.gc) _ _ , ^ ^^ _^ ^ ^
4 a + h + qc
/o = .4 ( //iyi + t;^ )  \qctli'{li' + 0 (1)
/o = ^('m7i + ^^^M'c. (2)
/„ = ^r/z,?i  ^^'A,'^ nearly. (3)
Here we have (3) a remarkable expression for the minimum
moment of inertia of the double Tsection, the degree of approxi
mation clearly shown by comparison with the exact expression
(2). The approximate expression may be given in words thus : —
The moment of inertia of a double Tsection is sensibly
equal to its area multiplied by the product of the two segments into
which the centre of gravity divides the depth bctiueen the middle
points of its flanges, from which is to be deducted twice the
moment of iTiertia of the web considered separately, and as reaching
from centre to centre of the flanges.
Cor. — For a given area the value of /o and therefore of the
stiffness is a maximum when m = 71, that is, with equal Hanges.
2a 2
356
APPLIED MECHANICS.
[chap. XIX.
By a further reduction we have
th'^ fl2ah + 4(« + b)qc + qr „ .A t^ , , ^ , ,,
a + b + qc
th' Vlnh + 4 (« + h) qc + q^c
12'
nearly.
(4a)
+ 2i!' + &c.
a + b + qc
By an exactly similar reduction
t]ifl2a fi+4(a + ^)qh + qVi'
~ 12 V, a + ^ + qh
Or, for the double Tsection thickness of Metal constant,
, th" 12aZ/ + 4(« + b)h + 1r
^" = 12 • — ^rrm . ' ^^^^
is a sufficient approximate value for designincr struts, even with
h for h'.
Economical double "Xscction of VMifwm drenffth to resist
bending. In fig. 1 suppose the two
skins b and a to simultaneously
come to their working or proof
strengths /j and /„, and putting
ft = fb ■ fa their ratio, then p wQl
sensibly be equal to the ratio
./"ft :/'„, that is, 7i : on.
Just as for f/en em I stiffness h or h'
must be a suitable fraction of the
span, so also for loccd stiffness must
also t\ the thickness of the web, be pi„ i
a suitable fraction r of h or h'.
The resistance to bending is, by equation (3),
t'h''l
6 n
< r / >
: Kf
±._.L.;_.Lo'
SA
n
or 7 = Ani 
./ 6
But
h'
and
pA'
il/
1 + p' 1 + p'
so that
rh'\\ + p)
i + jO Op 1 + p 6p
Equating zero to the differential coefficient with respect to li\
A _ rh'\\ + p)
1 + p "
= ^^
dK /'ft
:2/>
CHAr. XIX.] LONG STEEL STKUTS. 357
Or the area of the web measured from centre to centre of the
flanges, namely,
''"'iil'.r' <5)
makes the strength a maximum. Its vakie is
J/ _ 2 Ah' _ (1 + p)rh'' _ (1 ^ p) t'h'^
f\ ~3"(lT7)~ Zp ~ 3p
(6)
For homogeneous strength p = 1, and the section has then
equal flanges. For wrought iron p = 4 : 5 commonly, and is
never less than 1 : 2 for steel or steelyiron. For values of p
from i to 1 the coetflcient 2p f (1 + p)* is \ sensibly constant.
Hence the condition given by (5) that tlie duuhle "X section of
uniform strength shall he the most economical for a given degree of
local stiffness is that the area of the nrb measured from centre to
centre of the flanges shall he half of the toted area.
Substituting p = 1 in (5) and (6) we get for the most
economical section of equal flanges for a given degree of local
stiffness the two conditions
th' = A and ~ = —  = /h. (7a, 1h)
2 f I o 6
The second may be written M = 4 {^fht'h') where the
expression in brackets is the share of the resistance to bending
exerted by the web measured as before from centre to centre of
the flanges. Hence the two conditions (7) for designing the
economical crosssection with equal flanges for a given degree of
local stiffness can readily be remembered thus — Design the section
so that the iveh measuring from centre to centre of the flanges shcdl
he half the area and take I'.p a fourth of tJie resistance to hending.
For the purpose of direct design, however, (7j is to be
modified by putting
f'b = —fb ; h' = {h  t); t = sh, and f = rh,
when we lm\e, where fb is the lesser strength,
2 2 1/
rh''=^,r{l.s)%'=^ (8)
bo fb
•358 APPLIED MECHANICS. [CHAP. XIX.
For comparison put Mi in (6) and M, in (7/^), when
Ml  3f, 1p 31 1  M, lp ,. ^
is the fractional excess of strength of the one section over the
other where both have the same area A, the same depth li, or
same r/eneral stiffness, and the same vahie of r = t' ~ A', or same
luca/ stiffness. In each section the *' web " is half the total area
and each is the most economical of its kind. For the ordinary
value /» = 4. the expression (9) is 10 per cent.
To design the economical double Tsection of uniform
strength for a given degree of local stiffness. From the required
strength, deduct the percentage indicated by (9), and design
the economical section with flanges equal, of the required degree
of local stiffness by (8) ; then borrow from the flange a a por
tion, and add it to the flange b, till (a + ^h) : (b + gA) is in
the ratio p : 1.
Care must be taken that the weaker strength fb in (8; is
used always for the compressive strength of wrought iron and
steel. Then, too, it must be the weaker or thrust flange that
is increased at the expense of the other.
Also the beam is to be placed with the proper flange up, and
there must be no possibility of the bending reversing. For
beams a saving of 10 per cent, may be effected. Such sections
are not suited for struts for which the flanges are best to be equal,
the bending being equally like to be one way or the other.
The Tsection and angleiron constrained to bend like a
Tsection are often used as struts, as they suit constructional
purposes.
The angleiron if not braced so as to bend like a Tiron is
very weak as a strut, as it tends to open flat and bend over.
These points are best illustrated by the numerical examples
to follow.
Double Tsection vnth equal flanges, fig. 2. — In equation (3)
/y = Aran  jit'h'^;
we must now put
a = b, VI = n = }Ji', t' = qt,
and we have
A = t{qh' + lb), (10a)
^0 = jV ('/''' + t)6), (10&)
CHAr. XIX. J LONG STEEL STRUTS. 359
and the radius sq. of gyration
A 12 ^/t^ + 2i ' ^ ^
where ^ is the ratio of the thickness of the web to the common
thickness of the flanges.
J section and angleiron constrained to bend like it (fig. 2, iv).
— In equation (4«) put « = 0, and now c = h\
A=t(qc + h). (lla)
_ t}i qc" + 'khqc _ t'Ti^ qc + 4h ^ ty q c + 4h
" " 12 qc + b ~ "12" qc + b ~ 12 qc + b' ^ ■
/o qc^ qc + 45
A 12 qc + b
(lie)
Angleiron itnsupported or a cruxsection (fig. 2, v.). — Let
t' = t, for angleirons are always of one thickness being rolled
from a plate of one thickness. The section is readily identified
with a rectangle of breadth B and depth H, where
H=hiimd = bh^ yh + &'
/I , 1
and B = t (sec 6 + cosec d) = t ^/h + ^'^ [J"^l,
so that putting Iq = yj^'^^ 3,nd reducing, we get the values, of
little practical use, given in line V. of the accompanying table.
In the Table the struts are constrained to bend only about
the dotanddash lines.
Line /. is box or double channel section.
Line //. is double T section of uniform metal.
Line ///. has the web half as thick as the flanges ; these
lines are got by putting 2, 1, and  for q in equation 10.
Line IV. is the uniform metal T or angleiron constrained
to bend like a Jiron. Unity is put for q in equation 11. The
approximation is very close, as the terms in t^ and t^ vanish
together, in the exact expression expanded in powers of t. It is
of great practical use, and is not given in the Table, Eankine's
Civil Enyiiutring, p. 523, where there only appears the value in
•our Table at
Line V. for unconstrained angleiron a section quite unsuited
for struts.
360
APPLIED MECHANICS.
[CHAl*. XIX.
Table of Akeas, Moments of Inertia, and Uadh Square
OF Gyration.
SECTIONS.
h
7.
oL
^ ./;....
jj
T ■■■
.. ,
'h
n
V 
 ^
■ir
cz:
...i5....
! M',
ft (■
2 n
K.— Tf—" U6
^
_.H.Ht; I ^(^'^*)!
■n;
f=<? A <•
<(2A' + 26) :
t(/t'+2i):
<(iA'+2i);
^■1 ',
t{h + i) ;
'A"(2A + 6J)
'£'(/*■ + 6i);
j'7(K+6A)
M'^ A' + 46
72" ■ ~h'~Vb '
thh^ h\h
12 ' Ti^b
h for A'
h h+U
12 A + A
A' A + 66
12 ■ A r 27/"
12 ' T+ 4'/ ■
A' A + 4*
12 ■ (A + bf
A2 _J2_
12 ' A' + r"
Fio. 2.
CHAP. XIX.] LONG STEEL STUUTS. 361
KankineCJordon Fokmula.
Eankiiie first proved that Gordon's formula, equation 49,
Ch. XVIII, was rational (see Eankine, C. E., p. 523). He then
substituted v for .^ , because that is the relation of the radius
square of gyration to the depth for the solid rectangular sections
upon which the actual experiments were made. He then says
the formula is general, only for skeleton sections the proper
value of i must he used. To be of practical use the values of
^ like those in the above table must not contain t, the thickness
of the metal, otherwise there would result a quadratic equation
to determine its value, since t is generally the unknown
quantity in this formula.
We prefer to present the formula suitable to a strut with
hinged ends, because, as Fidler justly says, in wroughtiron and
steel struts, the ends, at best, can only be imperfectly fixed in
direction, as the other portions of the steel, in a bridge, for in
stance, to which such struts are fixed, are themselves so elastic,
that, in yielding to the loads imposed on them, the end of the
struts do deflect. Fidler allows for this imperfect fixing by
supposing the virtual length of such a strut to be yV^hs of its
actual length. But every strut imperfectly fixed at the ends
can be judged on its merits, and a deduction made from the
length never more than ^Vths. As the length in the Rankine
formula is in inches, it is always wise in making the above
deduction to so do it as to leave the virtual length expressed in
a round number of inches. The reason that I is in inches is
because h and i are always in inches.
The Rankine formula then for ends hinged is
T „ , 4 tons X A /io \
Iron, F tons = ,, (12a)
1 + ^— 
^ ^ 9 i
Mild Steel, p tons = ^^^^^^^ ■ (12h)
1 + — '— 
^ ^ 7 ^2
These, of course, are only good average values.
362 APPLIED MECHANICS. [CIIAP. XIX.
For a practical knowledge of this formula for designing long
struts Fidler's Treatise on Bridge Design should be studied.
The formula for many years was thought to be too general,
and, in America, most eminent engineers made formulae of their
own, of like form, but varied to suit the different crosssections,
in a way that led to great confusion. Fidler has, in the most
skilful way, arranged the Kankine formula, and tabulated results
which agree, with great exactness, with the results of the more
recent exhaustive and elaborate experiments made on struts
with the skeleton sections. His tabulated results agree also
with the results of the special formuke for special sections.
In Britain again many writers put forward a modification
of Euler's formula. In our opinion Euler's formuka is quite
unsuited for struts in engineering steelwork, Euler contem
plates a load which will just not bend the strut. Now the
whole duty of a steel strut in a structure is to resist bending,
and bend it will, no matter how it is designed, for it will take a
set to one side or other with its own weight if with nothing else,
for the weight of the immense struts in modern structures is
very considerable. Also heat will distort a strut ; and in riveted
work the straining of the adjacent members sends bending
strains along a strut, which must be designed with an ample
radius of gyration, and it is undesirable to talk about a load
which will just not bend it.
Fidler has done great service in vindicating the supreme
position of the EankineGordon formula as a master formula for
tentative design, though, of course, experiments on the actual
strut of any special section will supersede all formula^, provided
the experiments are on a proper scale, and in this matter no
perfunctory experiment in special struts can compare in im
portance with the lifehistory of such a strut in some large
structure.
We quote a part of two of the tables given in Fidler's
Practical Treatise on Bridrje Construction. These tables gi\e
the coetticient for stiffening long struts, that is, a multiplier
for converting the net area of the strut, where quite short, into
tlie gross area to resist both thrust and induced bending. In a
good design, this multiplier should not exceed 1'5, that is, the
extra volume of steel for stiffening the strut sliould not be more
tban 50 per cent, of the net volume. Otlierwise the main cross
dimension h should be increased to give a bigger radius of
gyration.
CHAP. XIX.]
LONG STEEL STRUTS.
363
For section (fig. 2 n.) with h = 2I> = IGt and ft, = 4 tons
per square inch.
Ends fixed, . .
5 feet.
10 feet.
20 feet.
30 feet.
40 feet.
Ends hinged.
3 
6 .,
12 ,,
18 ,.
24 ..
/' tons
8
/in sq ill
I 04
II4
I '45
I So
2'l8
l6
4
I'02
107
125
I 49
174
32
8
I 01
I 04
ri4
128
144
48
12
lOI
103
iio
120
132
64
16
I 01
102
ioS
ii6
I 26
80
20
I 00
IQI
I "06
113
I 21
For section (fig. 2 L) with h = 7; = 16;! and /j = 4 tons
per S(}uare inch.
Ends fixed, . .
5 feet.
10 feet.
20 feet.
30 feet.
40 feet.
Ends hingeci,
3 .
6 „
12 „
18 „
24 .,
P tons.
16
A" sq.in,
4
I 04
II3
142
176
213
32
8
102
I 07
124
146
171
48
12
roi
105
117
134
152
64
16
I'OI
104
II3
127
142
So
20
I 01
103
I'll
122
135
q6
24
ioi
I "02
iog
iig
131
Examples.
1 . Compare the strength to resist bendinf,' of four double T crosssections, all
having the common area of 18 square inches, and all having the common local
stiffness afforded by the constant thickness of the metal being a tenth part of
the depth of the section. The four sections are respectively to have deptlis of
134, 12, 10, and 8 inches.
The fig. 3 shows the four sections drawn to scale, and the calculations made by
the exact tabular method of Chapter XIV, fig. 3.
The section iii. has ^ = 492 a maximum, and it will be seen that the web is
ialf of the area. It is the strongest section with area 18 square inches, and the local
stiffness due to ( = ,\,/'
364
APPLIED MECHANICS.
[chap. XIX.
Table of
COMPAPw'vTIVE h^TJtENGinS OF FOUR DOUBLE T CKU^SSECTIONS,
WITH Thickness of Metal iVth of the depth.
IIai.fSections.
5y^ X } X 6
M^f
Area.
^
> 6 7
2.
o
Area
/v=_</j^4:
re
•^
^
1^^
Aroa_
/i=/o
.."^o
J^
1
vm
L^o
[ZIZU^EZISj
/8.
IK
_._ o
noo8
00
2160
1106
00
250
640
00
C40
328
00
3008 X i X 134 = 13433
i /= 13433
f
I_
6^7
= 401
1054 X j^ X 270 = 9486 j Jfj /
1106 X ^ X 120 = 4424 j 7"" "T
= 464
\1= 13910
610 X ^ X 500  10167
640 X ^ X 100= 2133
\1= 12300
312 X i X 805= 8380
328 X ^ X 0SO = 880
\I= 9260
Mi
I
f b
= 49 2
~y "4
= 463
Fig. 3.
CHAP. XIX.] LONG STEEL STRUTS. 365
The section ii. has the web tliieefoiirths of the area if it be measured from
<entre to centre of the flanges, and is the stijf'est section of area 18 square inches,
and local stiffness due to t = y^h.
For substituting from (lOrt) into (lOi), we have 7o = "iVA' (3^  2l/i'), which,
with A constant and t' a constant r"' fraction of h', give the variable factor
/It = SAh' — 2rh'^. The ditt'erential coefficient with respect to li' equated to
zero gives QAh' — 8r/i'^ = 0, or the area of web t'h' = ^^A. See fig. 3, ii.
2. Design the economical double T section with equal flanges in wrought iron
lor M'hich fb = 4 tons per square inch. Its strength to be M = 196'8 ft. tons,
and to have the local stift'ness afforded by the thickness of the web, t' being
a tenth of its depth h' . The thickness of the flanges to be t = t', that is,
" unifoi'm metal.''
We must plan the web measured from centre to centre of the flanges to have
its area t'h' a half of the total area J., and its share in the strength ^f'bt'h' to be a
fourth of IVl This is done bj putting in (8) M r fh = 492, and s = r = iV
when
I . TV(nrA? = 492 ; A» = 1013 ; // = 10.
So that t = t' = 1, h' = 9, and c = 8, while t'h' = 9, and so A = 18 square
inches. The upper half of this section is shown at fig. 3, iii., and the whole
sectkm at fig. 2, ii.
3. Design the economical double T section of uniform strength for wrought
iron with y's = 4 and /„ = 5 tons per square inch. For local stiffness the web
is to have t' — xiyh , and the thickness of the flanges t = t' . The required
moment of resistance to bending is M = 2I8'6 inch tons.
In (9) put p = f , and we tind that the section will be 10 per cent, stronger
than if its flanges were equal. From 218*6 deduct 10 per cent., and it leaves
1968 ft. tons. In Ex. 2 we have already designed the section with equal
flanges and found the values there quoted. It only remains to borrow from
one flange and add to the other, so that, with a + l> stUl equal to 10, we get
a {■ he : b + ^c equal to 4 : 5. And
« + 4 4
a + b=lO; ——■ = ^ ;
+ 4 r>
or ff = 4 and b = 6 inches.
The halfsection is seen at fig. 3, iii., if an inch of breadth be taken off the
upper flange and added to the lower. The section may now be checked by the
tabular method at fig. 5, Ch. XIV, when, with the axis at the lower skin of b, the
results are S = 18, Gi, = 81, I„ = 606, and y = 4*5, so that h' is exactly divided
as 4 to 5, although h is only nearly divided in that ratio ; further /o = 242, and
multiplying this by the smaller of the quantities 4 tons^ 4i in. or 5 tons i 5j in.,
namely, the first, we get the exact value of M = 215, almost exactly what was
prescribed.
4. Design a double T section for a wroughtiron strut 24 feet long and hinged
at the ends, the load being 64 tons. The depth should be A = 12 inches for stift'
ness, and with the breadth of the flanges each 6 inches, and flanges and web of one
thickness t as in fig. 2, ii., we iiave
., A h + 6b
'l2h^2b^''
I = 288 in. ; /= 4 tons per square inch.
A
= {2b + h)t = 24< nearly.
then
being the same degree of approximation as i.
366 APPLIED MECHANICS. [CHAP. XIX.
By (12fl)
fA 4 X 24«
9000 i*
< = 1 inch nearly, but A' = 2it nearly, or A = 22t = 22 square inches.
5. Design the same strut, using Fidler's table, corresponding to fig. 2, ii.
^0 = 64 f 4 = 16 square inches is the net area. The multiplier is the last in
the second last line, that is, 1*26. The gross area A  1*26 x 16 = 20 square
inches. As the relative dimensions on the table are
^ = 2* = 16<, so that A = {h 2t)t + 2bt = 30(=,
and therefore
<* = 20 ^ 30 = 67, or (=816, and A = 16< = 13 inches.
This is the nearest that can be done with the table. It is evident that with
/i = 12 inches instead of 13 inchts, then A must be greater, say, 22 square inches.
6. A square box steel strut, virtual length 30 feet, is to take a thrust of 80 tons.
Design the section. For stiffness b = h = \& inches might be suitable (fig. 2, i.).
„ h h  3A
1=; r = 44 ; A =2t(b ^ h) = %U ; / = 360 ;
12 A + i
f^ o. 7 X 64i!
TTTyr " ^^'
7000 i
As the divisor 142 indicates that the excess metal to stiffen the strut is not over
50 per cent., we can feel assured that h = 16 was ample enough. Proceeding
t = a lo = \ inch, and A = \& square inches.
7. Design a T iron or angle iron constrained to bend like one. The strut is
!2 feet long, the load 30 tons. Taking A = 8 inches and J = 4 inches, then
(fig. 2, IV.)
, A3 A + 4A
A = t[h^b) = \2t ■ i^ =  — — = 7 ;
12 (/♦ + *)
4 V \2t 48 t
" 9000 7
t = S, and A = 9*6 square inches.
8. If the above strut be an angle iron not constrained in any way, design it.
Taking h = b = 10 inches, say, we have by fig. 2, v,
^ = 20^ .42 nearly; .30 . — Lfi^, = i^ ;
_1_ 144 1 00
9000 42
and t = •58 = g inches, and A = 201 = 125 square inches.
CHAP. XX.] STEEL ARCHED GIRDER. 367
CHAPTER XX.
THE STEEL ARCHED GIRDER.
For roofs, the steel arched girder is of great importance. It is
used as a roof principal to cover the multiple platforms of great
railway stations, a gigantic example being the St. Pancras Eail
way Station, London. It is struck in arcs of circles with a slight
cusp or peak at the crown, span 240 ft., rise 96 ft. The ribs,
wliich are spaced about 30 ft. apart, are each 6 ft. deep, and uni
form except a part of the top where the moment of inertia is of
double the value elsewhere. The rib is without hinges either at
the crown or springings, and a tiegirder holds the ends together.
These tiegirders are underneath the platforms which they
support. We shall quote from Walmisley on Iron Eoofs, 2nd
edition, p. 83, the test prescribed in the specification of this
roof: — "Two main principals of the roof shall be erected at the
works upon proper abutments erected for the purpose upon tlie
flooring. The principals shall be erected with purlins complete
in every respect, and shall be tested in presence of the engineer
or assistant appointed by him. First a weight of 200 tons or
100 tons on each principal shall be evenly distributed. Xext
25 tons shall be taken from one side of each principal, leaving
50 tons upon the other side, and making the relative loads
25 tons and 50 tons on the two sides of each principal." We
would note, too, that in Mills' Railwai/ Construction, 1898, p. 271,
he gives the weight of iron work per square (of 100 square feet)
as 2'07 tons for the Central Station, Manchester roof, only
30 feet less in span than the St. Pancras. In Mills' book
there are four consecutive pages showing arched roofs, most of
them like the St. Pancras without hinges. One, however, of the
roof of the Anhalt Railway, Berlin, he shows with a hinge at
each end, parabolic in outline, uniform in depth, which is greater
in proportion to the span 205 feet than that of the St. Pancras.
The Anhalt principals suddenly taper at the ends towards the
hinges.
Again, arched girder roofs for buildings, often of great span,
are generally made in two halves affording three hinges, one at
368 APl'LIED MECHANICS. [CHAP. XX.
the crown, as well as the two at springing. The third hinge
eliminates to a great extent the temperature effects, and facili
tates the erection. Besides, the two halves can, with economy,
be swelled out in the middle, and shaped with a curved lower or
inner member, and yet have a polygonal upper member to suit
the covering. Two of the finest examples can be seen super
imposed on each other in the World's Columhidn Exjjosition, re
printed from Engineering of April 21st, 1893. They are, one
the roof of the Manufacturers' and Liberal Arts Building of the
Exposition, and the other the roof of the Machinery Hall,
Paris, 1889. The spans have the common value of 368 feet.
The Columbian roof has a rise 206 feet, being about 50 per cent,
higher than the Paris one, the principals being spaced 50 feet
apart. The girders are 8 feet deep at the top, and at least
twice as deep as the haunches. Two rival modes were employed
in erecting the Columbian roof by two contractors : figures illus
trating them are shown. They are called the Fives Lille mode
and the Cail mode.
AVe mention those examples, as the drawings can readily
be seen by British students. A complete list of roofs and
bridges is given by Howe in his Treatise on ArcJies, Wiley
and Son, 1897, an excellent and exhaustive treatise on the
subject.
For bridges, the steel arched girder seems likely to entirely
supersede the suspension bridge. A comprehensive view can be
had on one page of Engineering, of April 28th, 1899. AVith the
exception of the top one, the spans shown are all between 500 and
600 feet, the rises varying from, judging by the eye, onefifth to
twothirds of the span. Some are circular, some parabolic, some
have no hinges, when the girders have a small depth at crown
and increase till they are from twice to thrice as deep at the
spriugings.
Others have a hinge at each springing, and are either uniform
with a sudden taper at the hinges or are very deep at the crown
and taper gradually to the hinges.
Again, some have abutments, others tierods suspended from
the arch. Also in some of the examples there shown, tlie railway
runs over the crown on straight girders supported l»y "Eiffel"
towers rising from the arch; in other exampk^s these straight
girders are suspended or are partly suspended from crowns and
partly supported over the haunches.
The figure at the top of the group is the great Niagara Falls
bridge, 1898, span 840 feet, rise 150 feet, depth of parallel
booms 26 feet. The shape of archedril) is parabolic, and it
CHAP. XX.] STEEL ARCHED GIRDER. 369
rests on steel pins 12 inches in diameter one at each end. Two
arched girders support two railway tracks and side walks. The
horizontal girders ride over the arches on which they rest by
vertical and horizontal spandril bracing. The live load is pre
scribed in one form as a string of the Ely Tram and Trailer,
spaced 200 feet clear apart per track. Such a tram is roughly
40 tons in a 60foot length.
Disposition of loads. — It is evident that the live load all over
the span of an arch produces the greatest thrusts on any section
of the rib as a whole. But the live load, sa}' half over, produces
bending moments which at some sections greatly increase the
thrust on one boom, at the same time decreasing it on the other,
it may be to such an extent as to change it to tension. The
unsymmetrical position of the live load causes the thrust at
some sections not to be truly along the rib but oblique. This
oblique thrust is then equivalent to a direct component thrust,
and a tangential component which distorts the panel or bay
of the rib, which must accordingly be braced by one or two
diagonal members.
These effects would be most pronounced for the middle panel
or bay with the live load half over the span. And with this
position of the live load these effects are sufficiently pronounced
at all bays for practical purposes, the span being very great,
although with short spans it might be well to put the load three
fourths over for some bays.
From these remarks on bridge arches, and from the specifi
cation of the test on the St. Pancras' roof arch, it will be seen
that for large span lofty steel arched girders, either for roof or
bridge work, it is practically sufficient to consider only two
dispositions of the load — (1) an uniform load all over the span,
and (2) an uniform load on each half, but greater on one half of
the span than on the other.
Levy has given his solution of this problem in the happiest
way. He draws the stress diagram for a unit uniform load on
left half of the span only, from which by suitable multipliers the
stresses on the members cut by any section can be computed
for — (1) the live load as well as the dead load all over the
span; (2) the live load on the half in which the section lies,
but not on the other half ; and (3) the live load on the other
half, but not on the half of the span in which the section itself
lies.
The curve of henders and curve of flatteners. — If an arched rib
stand on its ends on a horizontal plane, and if we suppose each
2b
370 APPLIED MECHANICS. [CHAP. XX.
end to be a pin furnished with wheels, then if any load be
imposed on the back of the arch, it will straddle out and be
everywhere flatter than it was before, and the flattening moment
at each section can be shown by the diagram which, for a straight
girder hinged at the ends bearing the same load, is called the
bending moment diagram. We shall now call that figure the
curve of flattcners.
If we measure the amount by which the arch has straddled
or increased the span and remove the load, and if we now put a
tierod across between the pins and screw it tight, till they are
brought nearer to each other by the same amount that they
straddled before, the arch will be like a bow a little more bent
at each section than it was at first. That is, there is now at
each section a bending moment the amount of which is tlie pull
on the tierod multiplied by the height of the middle point of
the section above it as a lever. Hence the curve drawn up the
middle of the arched rib is itself the curve of lenders due to the
pull on the tierod. If the pull on the tierod be ascertained in
any way, a scale could readily be constructed for this curve of
benders.
Suppose that the free girder as at first is loaded with the
external load which straddles it out, and that then the tierod
is put across and screwed till the straddle is destroyed, and the
hinges brought to the original span. We have now two diagrams,
one the curve of flatteners due to the extern load drawn arbi
trarily to any scale, and the other, the middle line up the rib
itself, a curve of benders due to the pull on the tierod, the scale
for which is unknown, unless we should happen to know, say by
experiment, the pull on the tierod. Suppose for a moment that
we did know the pull on the tierod and so were enabled to
construct the scale for it. We propose to call this scale the
normal scale. On looking now at our curve of flatteners, which
is hardly likely by accident to be to this very scale, we could
reduce it or draw it over again to this normal scale, and in doing
so, we could draw it on the chord of the rib as base so as to have
both curves, the one of flatteners and the other of benders, on the
chord as a common base, and both to the normal scale.
We have already shown that a bending moment diagram is
itself a balanced linear rib for the load, so that the two curves
thus superimposed on the chord as base and to the same scale,
may now be distinguislied with propriety as the real rib and the
ideal balanced rib.
It will be seen that if tlic free unstrained rib be first pro
vided with rigid aliutments that will not yield appreciably, or
CHAP. XX.] STEEL AKCHED GIRDER. 371
with a lierod which will not stretch an a^jpreciable amount, the
load placed on its back produces simultaneously the flatteners (a
curve of which we can draw to an arbitrary scale independent
of everything but the extern loads) and the benders due to the
induced pull on the tierod, of which the central line of the rib
is itself the curve to the normal scale unknown. This normal
unknown scale depends only on the induced pull on the tierod,
that is, on the straddle of the girder without a tierod under the
extern load. Now this straddle depends on the variations of the
crosssections of the rib from section to section, so that the
normal scale is statically indeterminate as a rule.
The exception is the presence of a third hinge, say at the
crown. Suppose there had all along been a hinge at the crown,
but that it had been securely clamped. The minute it is
undamped the normal scale is known, for the ideal rib must
pass through this third hinge in order that the flattener and
the bender may at that point neutralize each other ; hence the
mere reduction of the curve of flatteners so that it may pass
through the third hinge changes its arbitrary scale at once into
the normal scale which is then known.
Eankine advised the adoption of a third hinge at the crown
of castiron arched girders, and says the assumption of such a
hinge is always a good approximation.
Levy illustrates his graphical construction for an arch with
hinged ends by an application to the Oporto bridge over the
Douro. In this bridge the rib is crescentshaped, and is hinged
at the spriugings where it rests on abutments. Span 160,
mean rise 425, and depth of rib at crown 10 metres. The
moments of inertia of the crosssections from crown out to a
hinge vary at intervals as 10, 8, 6, 5, 4. It is described in the
Mem. et Compte Rendu dcs Trav. dc la Soc. des Ing. Civils,
Sept., 1878, par T. Seyrig. The line up the middle of the rib
is an arc of a circle.
Levy primarily assumes the rib to be of uniform cross
section, and indicates a correction for the variation.
The stresses on the members of this bridge are calculated by
Howe in his treatise already referred to, and his results com
pared with those of Seyrig. Howe quotes the moving load
covering the entire roadway as 277 tonne or 270 tons, which is
roughly half a ton per foot, or half a share of a single row of
locomotives.
Levy illustrates his graphical construction for an arch with
fixed ends by a rib with the same arc of a circle as the other,
but with the section increasing from the crown outwards so that
2 b2
372 APPLIED MECHANICS. [CHAP. XX,
its moment of inerlia at each section is proportional to the secant
of the slope of the rib to the horizon. Such an arched rib is of
uniform di/fhcss. This is proved quite simply by Rankine {see
" The Sloping Beam," Civil Eng., p. 292).
It is evident that this arch should deepen in section as it
goes outward, for two reasons. The bending moments at the
ends of the straight boom fixed at ends are greatest. Also a
broad foot is required to securely fix it to the skewbacks.
Lastly, before illustrating Levy's construction, the relative
weight of steel on the three types are given by Howe in the
following proportion : —
(1) Three hinges, ... 1.
(2) Two hinges, . . . .12.
(3) No hinges, .... I'o.
The economy of (2) over (3) would, we think, be increased'
by the extra material in the skewbacks to supply the reacting
couples at the ends.
Observe that the type (o) cannot, when circular, subtend
much more than 120° at the centre, as the secant of 60° is 2,
and beyond 60° it increases enormously. Eoof girders of this
type (3) are of ten the complete semicircle with a vertical part in
addition at each end. Here only the portion out to 60'^ on each
side of the crown should be treated as the actual elastic rib, the
lower part on each side being reckoned as part of the vertical
post, and is generally along with the post buttressed or braced
to a smaller rib rising on the opposite side of the post.
Other references are The Engineer, June 30 and July 14,
1899, for a Danish bridge ; April 26, 1907, for a road bridge
over the Usk ; also Engineering, March 30, 1900, for the Ehine
bridge at Bonn. See, too, a Mem. cle Viaduct Garaht, par G. Eiflel,
1880.
Levy's Graphical Solution of the Arched Bib of Uniform
Section Hinged at each end.
On fig. 1, 2 is shown to scale a segment of a circle, span 240
feet, rise 80 feet, radius 130 feet. It is the lineal real rib
up the centre of the iron arch which consists of two parallel
booms 10 feet apart. This, then, is the curve of benders to
the unknown normal scale. It is divided into 16 equal parts
along the curve, and the ordinates are numbered from to 8 to
the right hand, and from to 8' to the left.
The fig. 1, 5 shows the load line W = 120 tons drawn to a
•CHAP. XX.] STKKL AHCIIKl) GIRUEIt. MTo
scale of tons, being one ton per foot on left half of span only.
The reactions are Q = ^W and F => f W. A pole Q is chosen
arbitrarily, and the curve of f,attciicrs (fig. 1, 1) drawn to it,
and consisting of a straight tail over the unloaded half parallel
to the vector Qa. A parabolic arc must be drawn over tlie
loaded half. The vertex is over the point where the shearing
force diagram (dotted) crosses the base, that is, at a point divid
ing the span in the ratio o to 5. The curve of flatteners when
tinished is like a great fish or " Jon Dore," and it is best to
construct the points on the parabolic arc corresponding to the
ordinates of the figure below ruled up. This is done by producing
the straight tail till it makes an intercept on the vertical through
the left end (see fig. 3, 1), dividing this vertical intercept pro
portional to the way the left half of span is divided when vectors
from the top point of the straight tail will cut ofi' the proper
heights on the ordinates. Number these ordinates from 1 to 8
right, and from 1 to 8' left (see fig. 3, Ch. VI).
The next step is best done with proportional compasses.
Double all the ordinates of fig. 1, 2, drawing right and left a
series of fictitious parallel forces 1 to 8 and 1 to 8'. Set the
compasses to the convenient ratio 4 to 1, construct the load line
(fig. 1, 3) with equal to a quarter of the middle ordinate of
the curve of benders, with 1 equal to a quarter of the sum of its
ordinates 1 and 1', &c., and to a convenient pole draw the link
polygon among the horizontal fictitious forces going left, and
produce the two end links to make the first intercept on the
chord joining the hinges of the real rib.
In the same way lay off the load line (fig. 1, 4) with equal
to a quarter of the middle ordinate of the " Jon Dore," with 1
equal to a quarter sum of its ordinates 1 and 1', &c. With the
same polar distance as last draw the link polygon among the
horizontal fictitious forces going right, and produce the end
links to make the second intercept on the chord joining the
hinges of the real rib.
On fig. 1 this second intercept is about twothirds the length
of the first. Now we know that these two intercepts should be
equal, for their algebraic sum equated to zero means that the
moment of all the fictitious forces, with respect to the chord
joining the hinges, is zero.
In this problem three definite integrals are involved, namely,
^Mclx = 0, IxMdx = 0, lyMdx = 0,
where x and y are the coordinates of the real rib, and the limits
•of the integration being from hinge to hinge.
374 APPLIED MECHANICS. [CHAP. XX.
The first two integrals we have fully discussed as regards
the straight beam and illustrated by examples in Chapter XVII,
at the beginning of which are theorems {a) and (6), corre
sponding to these first two integrals, and for an illustration
see fig. 2, Ch. XVII, in which AK is drawn that its area may
equal that of ACB, and that its geometrical moment about the
vertical through the centre shall equal that of ACB. That
is, the joint areas are zero, and the joint geometrical moments
are zero.
So with the third integral ; 3Idx is the joint algebraical area
of the two corresponding strips of the two curves, one of benders
and one of flatteners, and by drawing a fictitious force for this
partly right and partly left, and giving them the common lever
about the chord, namely, the height of the rib itself (only
doubled for good definition), we obtain the product Mdz x y
for that finite part of the rib by producing the two links that
concur on the fictitious force till they make an intercept on the
chord. The total intercept must be zero, and it matters not
that we have grouped them in two categories according to
whether they tend to flatten or further bend the rib.
The reason that the two intercepts are not equal is that the
arbitrarily chosen pole Q, to which we drew the " Jon Dore,"
does not make that figure be to the normal scale.
It is now necessary to draw the " Jon Dore " or curve of
flatteners over again to a vertical scale increased in the ratio in
which the first intercept exceeds the second. It is best by trial
and error to set the proportional compasses till one end spans
one intercept, and the other spans the other intercept. Take
the height of the middle ordinate of the " Jon Dore " on the
smaller end, and, with the large end, set it up on the middle
ordinate of the real rib. On fig. 1, 2 it will be seen to reach
a little above the crown of the real rib ; joining this with a
straight line to the right hinge we have the tail of the new
curve of flatteners, and with the proportional compasses the
seven intermediate points of the left half may be set up, or the
new parabola may be constructed at those ordinates as at first.
Next from L the Ijottom of the load line fig. 1, 5 a line
parallel to the tail of the new curve of flatteners svill meet the
horizontal from d, the junction of the reactions P and Q, at the
new pole. Or dO is shorter than dQ in the ratio of the intercepts.
This polar distance, which is 130 feet, determines the
normal scale of the joint curves of flatteners and benders, or,
as they may now be called, the ideal and real ribs. For it is
evident that if the righthand link polygon were redrawn for
CHAP. XX.] STEEL ARCHED GIRDER. 375
the new curve of flatteners, the new intercept 2 would now
equal the first intercept.
The virtual hinge or point at which there is no bending
moment is a little to the right of the crown of the real rib
where the tail of the ideal rib crosses it. Had there actually
been a hinge at the crown of the rib, the curve of flatteners
could have been drawn at once with its tail drawn from the
crown to the right end, and the normal scale determined atonce.
The elevation, to the double scale, of the steel arched girder
is shown on fig. 2, 6 ; it is sketched as if for a roof. It has
radial stiffenevH with double diagonal ties in tach square.
Counting the crown square as zero, we will now take out
the stress at a radial crosssection through the middle of the
third square to the right hand.
This crosssection is shown at A on fig. 1, 2, and A' is th&
corresponding crosssection in the loaded half of the girder.
A portion of the arch reaching from A' to A is shown above,
furnished with arrows to indicate the stresses on the upper and
under booms and the radial or shearing stress.
The thrust on the ideal rib at the point below A is given by
the vector 10 of fig 1, 5, which is resolved into two components,
a tangential thrust 50 tons, and an outward radial force 19 tons.
The 50 tons is divided into 25 tons each (of thrust), on the upper
and under boom. At ^ the height of the curve of benders or
real rib is greater than that of the curve of flatteners or ideal
rib. The intercept at A between the ribs is the residual
bender ; it measures 5 tons and, when multiplied by the polar
distance 130 feet, we have the bender ilf^ equal to 650 ft.tons.
The booms have a lever of 10 feet to resist it, and accordingly
there is a pull of 65 tons on the upper and thrust of 65 tons
on the lower boom at A due to this bender. This is the
most businesslike way of taking out the effect of the bending
couple at A.
But there is another and instructive way of looking at it
and of scaling it off. Suppose the thrust hO tons acting on the
real rib at the point below A. Then to shift it parallel to itself
so as to act at A instead, we must compound with that force
10, a lefthanded couple, the common force, of which shall be
hO tons and whose arm is the perpendicular distance between
the old and new positions of the force before and after the shift.
Now this arm is shorter than the vertical intercept at A. It is
equal to the vertical intercept multiplied by the cos 0, if d be
the slope of the tail of the ideal rib to the horizon. So that if
we measure the vertical intercept at A mfeet, we have the arm
376
APPLIED MECHANICS.
[chap. XX.
Jon OORE.
^Cue tirErtern'Xoizd.' draxvTi/ to a.TZ/
arinjrary Scale. J'clc £.
tkc"^?! OJiarTia^ic coiled
e^ "the J'u^Mcd, tu/ /i" a^v ujOc. —
Fig 3.
^'g^. bearing thosamoTUijnicr: cft}i6Su^7vdfa,pcurcf07din/J/ea
Fii;. 1.
CHAP. XX.] STEEL AECHED GIRDER.
377
1
MA
_!.
^
Ton
PrR
F0or
1
1
TiaZf tx^ ttm .
h
!jr
1
TuOy^ ci. ten, 1
Stresses due, te thcXoeiA c
cA^* ha^ pf the. Girder:
AleidSc Thorrtson.
4f? ^/7 ^a 7& 80 sa Tffc 710 7V/iet^.
Fis. 2.
378 APPLIED MECHANICS. [CHAP. XX.
of the shifting couple multiplied by sec. 61. But if instead of
measuring hO in tons we take instead dO in tons, we have the
force of the shifting couple decreased in the exact proportion
that the arm was increased. Hence the moment of the shifting
couple equals the product of the vertical intercept measured
in feet, and of dO measured in tons. But the "product is in no
way altered if we interchange the scales upon which we scale off
the two factors. Hence what we measured oft before, namely,
the intercept at A in tons, and the polar distance dO in feet,
give, when multiplied together, the moment of the couple
which shifts the thrust on the ideal rib up to act at A on the
real rib. "When the thrust arrives at ^, it brings with it this
lefthanded couple, so that the two booms have not only to
oppose the thrust, but have also to exert a righthanded couple
to neutralize the shifting couple.
This is tlie penalty, as it were, that the real rib pays at A
for not coinciding with the ideal one. Again, now that the
thrust is at A it is not directly along the rib, and when resolved
into a direct thrust along the rib there is also a radial outward
or shearing force as v/ell, tending to distort the square at A, and
this distorting force is the penalty, as it were, that the real rib
pays at A for not being parallel to the ideal one.
Note that dO plays two parts ; it is 130 feet as polar distance,
but is also 44 tons as the thrust at crown of ideal rib or pull on
the tierod.
At A' the vertical intercept measures 5 tons, and when
multiplied by 130 feet we have J/.^i = 650 foottons a residual
flattener, giving a thrust of 65 tons on upper boom, and a like
pull on the lower. A tangent drawn to the parabola at A
happens to be horizontal : it resolves into a direct thrust of
43 tons along the real rib at A', or 22 on each boom, together
with an inward tangential force of 10 tons.
The like quantities are constructed for the pair of points
B and B' corresponding to the twelfth square from crown. In
drawing the tangent to the parabolic part of the real rib at B' it
is best to construct the tangent at B" by ruling it horizontally
to meet the vertical tlirough the vertex, setting up a like height
above the vertex whicii gives a point to join B" to ; thus Qh' is
drawn parallel to this tangent, and, joining h' to 0, we have the
tangent to the real rib at B'. When the point B" lies near the
vertex, this construction is bad, and a better one in such a case
is shown for the point S" on fig. 3, I, namely, to draw a
chord from c (the most remote given point on the parabola
to the right of *S'") to d, a point lying at the same distance
CHAl'. W'.] STEEL AIJCHED (illlDKR. 319
horizontally from S" as c does, but on the left side. The tangent
at S" is parallel to dc. The other tangent at K" is drawn from
K" to /where Vf = e/'as already described.
On fig. 2, 6 three varieties of load are shown. Fig. 2, 7
gives the stress at the third square selected from values at A and
A' as follows : the pull on the upper boom at 4 is 65  25 = 40,
due to a load of 1 ton per foot on left half of span ; but at
fig. 2, 7 only half of this or 20 tons is set on the upper boom
beside a half barb or arrowhead, because there is only half a
ton per foot on the left half of the arch. In the same way we
have set 45 as thrust on low boom, and 9 as outward radial
stress. These are set beside half barbs to indicate that the
stresses are due to the load on left half of arch only.
Now if we suppose the figs. 1, 1 and 2 turned over so as to
be seen through the paper at the back, A' will have taken the place
of A, so that by halving the stresses at A' and writing them on
fig. 2, 7 with full barbs, we have the stress due to the load on
the right half of the arch. These barbed and half barbed results,
are summed algebraically and written with a black figure a little
to the left of the section (fig. 2, 7).
On fig. 2, 8 the quantities for half barbs are unaltered, as
the load on the left lialf for the second variety is unaltered, but
the quantities at the full barbs are halved, as the load on right
half is reduced.
On the other hand, fig. 2, 9 shows the stresses for the third
variety of load. Here the quantities at the full barbs are
unaltered and the others halved.
On the third square to right hand of the arch itself 33 tons
is written on the upper boom, being the greatest of the three
varieties. On the lower boom it is 35 tons. On one of the
diagonals in the third square is written 10 tons : this is a pull.
The other diagonal is dotted to indicate that there is no pull on
it for any of the three varieties of load.
It will be seen that the gross radial stresses for the three
varieties of load are 4, 7, and 0, all outurird. As the cross
section cuts the diagonal at about 45°, and as 7 tons is only one
component of the pull on the diagonal, it has to be multiplied by
sec. 45° or by 1^ nearly, giving the pull of 10 tons. To make the
roof stiff the two diagonals in the third square might both be
tightened by a thimble, or one only, till they had an initial pull
of 2 tons each, giving 12 and 2 as the crreatest pulls on the two
diagonals respectively.
"The figs. 1 to 4 are reduced half lineal size from a set of
graphical exercises advertised near the titlepage.
380
APPLIED MECHANICS.
[chap. XX.
AtBX.'i
y^^
■^c? f/vr((^r\'ii Sief\f,i(>e/ injrjaf^s.
Fig.
CHAP. XX.]
STEKL ARGUED GlKDP^Il.
381
• ot^r
znzmrrrtvr. i
' ' ' ' ^' ' ' 1 ' ^ I ' I I M I
the Stresses ciae to ^
I mol,on6lo/i the hft if
Fig. 4.
382 applied mechanics. [chap. xx.
Levy's Graphical Solution of the Arched Rib of Uniform
Stiffness fixed at the ends and without hinge any
where.
The preceding description must serve in detail, and we shall
point out the marked differences in this case.
We have the same central line for the linearrib. The booms
are now 7 feet apart at crown, and 17 feet apart at springing, and
the moment of inertia of the crosssection is everywhere pro
portional to the secant of the slope. The stiffeners are vertical
and are prolonged as suspending rods to support the horizontal
girders that carry a railway.
The first marked difference in the construction is that the
real rib is divided into equal parts along the span or horizontal.
There are more parts, being 12 in each half. The curve of
flatteners is shown at fig. 3, 1, drawn as before to a short load
line with a polar distance of 50 feet. Then for theorems (a), (b),
Ch. XVII, as applied generally at fig. 4, Ch. XVII (and
particularly at Exs. 3, 4, Ch. XVII), we must draw a new base
ran across the curve to include the same area an that a Vh
contains, and with its centre of gravity on the same vertical
line. This is accomplished by niaking am = Woe, and hn = f^oc.
The graphical construction is shown to left of fig. 3, 1. The
half span being reckoned as the horizontal unit, a force is laid
down equal to oc to represent the area of the isosceles triangle
acb, followed by a force fa^ to represent the area of the para
bolic segment aac. The lines of action of those two areas are
drawn downward ; also two lines of action trisecting the span
are drawn upwards to represent the two triangles into \vhich
the unknown figure an can be divided. A link polygon drawn
among these four lines of action is shown, its closing side dotted ;
a vector from the pole parallel to this divides the load line into
two segments which are the areas of the two triangles amii and
(nil). But the half span being unity, those segments are am
and hn respectively.
MNis drawn across the real rib so that JTmay have the
.same area as ACB. To draw MN its height ^ J/ will be a little
more than twothirds of the height of G. For if ACB had been
parabolic, it would have been exactly two thirds. The area
ACB can be measured with a planimeter, or calculated from
three ordinates in a half of it, using Simpson's multipliers.
CHAP. XX.] STEEL ARCHED GIRDEK. 363
In drawing the load line, fig. 3, III., 0, 1, ... 7 go left, as
long as the ordinates of the real rib in pairs are above MN, but
8, 9, ... 12 go right, and only onclialf of ordinates 12 are to
be taken. This load line should be a closed polygon, but it
will only nearly close. This must be averaged so that the
two end vectors may coincide. When the link polygon to left of
fig. 3, II is drawn, its two end links are parallel, and make
an intercept on the horizontal, the load line being horizontal ;
the two ends links of the right polygon make another intercept
on the horizontal.
This second intercept, which should be equal to the first, shows
that the curve of flatteners is to a scale too small. The propor
tional compasses may be set so that one end shall span the one
intercept and the other end the other. The curve of fiatteners
is then to be increased and superimposed on the real rib with
mn coinciding with MX. To do this from the middle of MN
set up oC equal to the increased value of oc. Set downwards
NE equal to the increased value of nh, and MF equal to the
increased value of nia. Join CE with a straight line and it is
the tail of the ideal rib, and the points on the curved half FS'G
■can have their values set up from the chord EC increased from
those of aS"c from its chord ac.
The load line to a bold scale is drawn on fig, 4, V, the poles
Q and determined, and the stresses at four sections ;S', *S", K, K',
scaled off as in the last, and then these are manipulated with
the proper multipliers as on the hinged rib.
The reason that equiintervals are taken along the span is,
that / is not constant, being the integral J (if ^ I) y dx = ;
but / cos 6, the projection on the chord, is constant, since
I is proportional to the sec 0, so that in taking finite intervals
it is necessary to make them have a constant horizontal
projection.
Numbere engraved along the upper and under booms of both
of our girders show how the plates must vary, and it is interest
ing to note the difference in the variation for the two cases.
Some booms sustain a pull instead of a thrust for the live
load half over, and we note that eighttenths of this reverse stress
is prescribed on the figure of the Niagara bridge as a quantity
to be added to the direct stress.
384
APPLIED MECHANICS.
[chap, XXI,
CHAPTER XXI.
ANALYSIS OF TRIANGULAR TRUSSING IN GIRDERS WITH
HORIZONTAL PARALLEL BOOMS.
The lower member A,A,n (fig. 1) is supported at each end
and is divided into 2n equal parts. Upon the first n of these
booms as bases similar triangles are erected, the upper booms
joining the vertices of the triangles. This constitutes the left
half of the girder with which we have principally to deal. The
right half of the girder extending from An^ to Ain is the image
/^(l_^\ Reaction due to uniform dead load W.
xt,(i^\ Reaction due to rolling load W at A2y. g,
A3 Ajii
2 1+ 1 ^(ztT'^z^"^^
« is No. of parts in half span.
r is ratio h to L, depth to span. ;^
c is segment of base of triangle remote from
centre of girder.
Triangular Trussing.— (Fig. 1.)
of the left half, that is, it has the pattern reversed just as an
image of the left half would appear in a Aertical mirror forming
a right crosssection at the centre.
It will be seen that we are dealing with a girder divided
into an even number of baysy This is partly because of the
comparison to be drawn with the Fink system of trussing,
which only admits of even subdivision, and partly because an
odd number of bays is not so desirable from an economical point
of view. Neither could the notation be conveniently adapted
to both the odd and even subdivision of the span.
Eeturning to lig. 1, it will be seen that there are 2n  1
upper booms all equal in/ length with the exception of the
central one. At each lowet^ joint or apex the letter A is placed
CHAP
XXI.]
ANALYSIS OF TRUSSES.
385
witli an even suffix, and at the upper apexes with an odd suffix.
The depth of the girder is h, and the span is X = A^Ai„, and
the common length of all the booms, with the exception of the
upper central one, is X ^ 2n.
A variable c to regulate the shape or form of the triangle is
now introduced in the following manner. Fig. .2 is one of the
triangles, singled out from those n triangles forming the left
half of the girder. Its base is A.i.iAii, and from its vertex
A.j.i a perpendicular h is dropped dividing the base into two
segments ; c is the segment of the base rcrrote from the centre
This side of the triangle, and
the segment c ot its base, are
said to be remote from the centre
of girder.
Form of the tri
angle varies with c.
This side and segment are
adjacent to centre of girder.
Position of tri
angle varies with
The i"' Tkiaxgle from left end of Girder. — (Fig. 2.)
of the span (the shorter in the diagram), while {L ^ 2n)  c is the
segment adjacent to the centre. The surd ^y {h + c^) gives the
length of the diagonal Aii^iA^io, which, in the same way, is
said to be remote from the centre of the span, and a similar surd
gives the length of the diagonal adjacent to the centre. For all
symmetrical loads, such as the dead weight of the girder itself,
the diagonals remote from the centre are struts, and those
adjacent to it are ties. It will now be seen that the upper cen
tral boom is twice the length of the segment adjacent to the
centre. To the variable i are to be assigned the integer values
1, 2, 3, &c., . . . n, when the triangle (fig. 2) will become in
succession the triangle at the left support, the second, the
third, &c., each and all of the triangles forming the left half of
the girder.
2c
386 APPLIED MECHANICS. [CHAP. XXI.
Dealing only with the left half of the span, we have
Lower apexes, A^A^Ai . . . A^.^A^i . . . Atn
Stress on upper boom opposite, ?'2 7\ . . . r^.i r>i . . . r2„.
Upper apexes, AiAi ^211 ^2ni.
Stress on lower boom opposite, Vi Ts r;,_i rjni
The stresses on diagonals remote from the centre are
tAh • • • Uii . . . <2ni, and on the adjacent diagonals they are
*2'4f6 • • • ^2i • • • ^2»1
The bending moment at a section through any apex is
represented by the letter 3f with the same subscript which the
A at the apex bears, and the shearing force, constant for any
interval of the span, is represented by the letter F with two
subscripts separated by a comma, which subscripts are those
borne by the ^'s at the two apexes bounding that interval.
If we confine ourselves to the representative triangle
(fig. 2), we have generally
Bending moment at upper apex, . JAfi.
Stress on a lower boom, . . rji^i = Mai f h.
Bending moment at under apex, . Mu.
Stress on an upper boom, . . rji = M^i ^ h.
Shearing force in the i"' bay, . . i^2.2, 2.'
Stress on a diafjonal remote from , ,,, ,,
centre, . . . . . yO'll^
h
F,i.
,2)2i
Stress on a diagonal adjacent to yM
■In '
centre, > \" ' w .
The values are given irrespective of sign ; the first two are
obtained by dividing the bending moment at the section through
the opposite apex by the depth of the girder. The second pair
of quantities are obtained by multiplying the shearing stress in
the bay by the secant of the slope of the diagonal brace to the
vertical. It is to be remarked here that all loads, even the
dead loads, are to be looked upon as something apart altogether
from the girder itself, and tliat the loads are all redncal to the
joints of the lower boom. For this reason i^is taken as constant
from ^2,.2 U) Aii.
CHAP. XXI.] ANALYSIS OF TRUSSES. .'387
Dexd Load. — Let the dead load be w tons per incli of span
uniforndy distributed on the lower boom, then the total load is
W = irL tons where all lineal dimensions as L and li are to be
taken in inches. This load when reduced to the joints of the
lower boom gives a share oi W  2n concentrated at each of the
lower joints except the two end ones, which have only a half
share each, and those are directly over the abutments, so that
the reaction of the left abutment on the left end of the truss is
W uL _W f 1
T ^n ~ 2 \ 2n
The shearing forces are
W2n\ W
_ W%i\ W
^ ^ W2n1 ^. ^^ W W(2n+1 i\
2 2n 2n 2 \ 2n n)
The bending moments are
,, W2n1
W2n\(L \ W
2 2% \2n y 2
zn
W2nlf^ Z \ WfL ^
M.= 2. — ^ c] — + 2c
2 2n V 2n 7 27^1 2^^
,^ W2nlf^ L \ Wf L \
M.= ^^— 4.~ + c 1 + 2 + 3— + 4c),
2 In \ In / 2n \ 2n J
2 271 ( ^ ^ 2n ) 2nl 2 2n ^ ' )
2 c 2
388 APPUED MECHANICS. [CHAP. XXI.
The numerical coefficient of the second last term being the
sum of the natural numbers from 1 do [i  2). Then by
reduction,
WL ,. ,, [In +1 i\ Wc [2n + 1 i
Mn, = r (*  1) — 7r] +
'*" \ Zn in)
M,
4?t \ '^11 2nJ 2 \ '2n n
W 2n \ Z
2 2n 2n '
'~ 2 2n ■ ^ ■ 2^ ~2n'2n'
2 2n 2n 2n ^ ■ 2n'
"■~T ~ll^'^'2ii ~2^x 2 2n'
The numerical coefficient of the last term being the sum of
the natural numbers from 1 to (t  1). Then by reduction,
M^i = — 2 1  —
471 \ In
Substituting these values in the preceding expression for
the stresses on the booms and diagonals, and arranging in
ascending powers of i, we have
JVXf 2n_h2 n+ 1 . i
4:nh \ 2n n 2n
Wc (2n + 1
+
2h \ 2n 11 J '
WL I . i'
''''~4:nhV 2n
W/2n+l i\ ,,,
The strength of the material is/ tons per square inch. It
is this unit which demands that the lineal dimensions shall be
CHAP. XXI.] ANALYSIS OF TRUSSES. 389
in inches. It is reckoned to be of the same vahie with respect
to tension and thrust. It will be shown afterwards how a cor
rection can be made if the material l)e not homogeneous in
strength.
Dividing r..,_i by / we have, in square inches, the theoretical
area of the crosssection of the i'^ lower boom ^.,_2 A^i, reckoning
from the left abutment, that is, of the boom which is the base of
the /'^ or specimen triangle (fig. 2). Multiplying by L ^ 2//, its
length, we have a general expression, containing the integer
variable i, for the volume in cubic inches of that boom. To i in
that general expression is to be assigned the values 1, 2, . . . n
and the whole summed up, giving then tlie volume of the left
half of the lower boom. Doubling so as to include the right
half we have the theoretical volume of the lower boom in cubic
inches to resist the dead load
t=»i ~ T ^, . ■
V o^ _ Jill
' under booms — Z^ o J!
dead load ,v, L^" /
+ 7^77 2
^'hi + 1 n
2nfli ^ \ 2n n\
WD (_ 2w + 1 n + 1 n (n + 1) »(» + 1)(2// + 1)
4n»//i ( 2~ ^ ~7i 2 12^1 :
WLc\2n + l M(n + l)
(n l)(in + 1) +
lu summing, terms not containing i are simply to be multi
plied by n ; those containing i involve the summation of the
natural numbers from 1 to n, while the term containing t*
involves the summation of the squares of the natural numbers
from 1 to n.
The theoretical volume of the upper booms from the left
end to the centre of the span is obtained in the same way, only
in summing for values of i from 1 to n we include a portion of
the upper central boom extending a distance c past the centre
of the girder. Hence a deduction equal to c multiplied by the
390
APPLIED MECHANICS.
[chap. XXI.
crosssection of the middle upper boom is to be subtracted,
before doubling to include the right half of the girder.
' upper booms ~ ■* ^
dead load ,_,
_ WU \ n{u + 1) )t{n+ 1)(2» + 1)
WD . .... ^. WLc
 L
_2)t
 2r —
n
2
1
^ in
WLc
r • i"'!
I 
2>i
2nf/i
V ~ 2;^\
WLc
Infh
'^^nfh
4/A
It is remarkable that the sum of the volumes of the booms
both upper and under is independent of c, that is, of the shape
or form of the triangle. It amounts to
WL^
(1)
a quantity continually increasing with n, but sensibly constant
for values of n above 10. It has its greatest value when n is
infinite, that is, for the ideal plategirder with areas of flanges
everywhere proportional to the bending moment. It is verified
by considering that particular case in which the load is uniformly
distributed, the maximum bending moment oneeighth of the
product of the load and the span, and the average bending
moment twothirds of that maximum. Dividing that average
bending moment by h and by/, and multiplying by 2Z, the joint
WL"
length of the flanges, we get •
For the diagonals remote from the centre of the girder,
'211 /rrz r. I
remote diagonal
dead load.
= 21
.=. L ./'
w
w
w
(AM r)2
yii' + r j
2n+ 1
^ •■)(. ^^0
1 ("')■
CHAP. XXI.] ANALYSIS OF TRUSSES.
And for the adjacent diagonals similarly,
391
adjacent diagonal _
dead load.
Wn
/r +
2«
Adding these four quantities, we find the theoretical volume
of the truss in cubic inches, to resist the dead load of W tons
uniformly spread on the lower boom, to be, after reduction,
dead load
Consider now the effect upon this volume of varying only
the shape or form of the triangle. The only variable is c, and
Fig. 3.
it will be seen that the volume is a minimum when the product
c r c] is a maximum which occurs when the factors are
\2n J
equal, that is, when the perpendicular from the vertex upon the
base bisects it. Hence the isosceles, or, as it is sometimes called,
the Warren truss (fig. 3), is the most economical form of
FiK. 4.
triangular trussing to resist a dead load. Substitute, then, in
(2), c = half a bay = Z ^ 4n, and for convenience put h = rZ,
where r is the fractional depth of the span, or ratio of depth to
span, and we have, after reduction,
V;
isosceles
dead load
WL
nr +
Sn +
48^:
2 1)
(3)
392
APPLIED MECHANICS.
[chap. XXI.
A form of bracing that rivals the isosceles is the rectangular,
of which there are two types ; in one the struts are vertical ; in
the other, the ties are vertical (figs. 4. 5). Their volumes are
obtained from equation (2) by substituting c = 0, and c = L ^ 2n
respectively, and, what is remarkable, both have the same value,
so that one expression serves for the volume of either. It is
y rectangular
dead load.
WL
nr +
(» + l)(4»l) 1
24n* • r
(4)
To resist a dead load the Warren truss is considerably more
economical than the rectangular truss when the number of bays
is small. Thus if both have a depth onetenth of the span and
a common span, and if there be only two bays in the span, then
Fig. 5.
the Warren has an economy of 33 per cent., as may be seen from
the tabulated results which follow, or directly by substituting
71= 1 and r = ^ff"* into equations (3) and (4). When, however,
w = 4 or 8, between which it will usually lie in bridge work, the
percentage of economy is reduced to 7 and 3. Now the rect
angular joints will probably recover this economy.
The economical values of n, and especially of r, the ratio of
depth to span, will be discussed after the live loads have been
dealt with.
Live load. — We shall now consider a live load consisting of
an uniform load of u tons per inch per girder which comes on
at the right end of tlie deck or platform, passes across the
platform, completely covering it, and then passes oft' at the
left end.
The bending moments everywhere are simultaneously a
maximum when the platform is covered. If we put W = v.L,
we have the same Vdhime for the upper and under booms
conjointly as we liad with the dead load.
Vol. of l)ooms
=w('4y (i*)
CHAP. XXI.]
ANALYSIS OF TRUSSES.
393
The maximum sheaving force in any triangle of the left
half of tlie girder occurs when all the triangles to its right are
loaded and a portion z of its own base. This is illustrated by
fig. 6, which sliows the third triauijle BCD resisting a shearing
force I\ due to the load standing with its crest at a distance z
from D. If we put the base BJD = a, we will see that z = fa
makes F3 a maximum, while Fi is a max. for 2 = fa. That is,
to get the maxima values of F for each of the eight hays on
lig. G, we must lay up from A half the total load W = uL,
draw the parabolic concave locus beginning at the right end of
the girder, with the span as a horizontal tangent and ending at
A with the height huL, which is the max. for the end bay.
The span must now be divided into seven equal parts where the
heights to the locus give the maxima 2^ositive shear on each
bay.
Fig. G.
On fig. 6, let BCD be the i"' bay from the left end. Let m
be the number of equal bays on the span, each bay being a
inches in length. Then a {m  i) + z is the length of the
loaded segment, while half of this is the distance of the centre
of the load from the right end of the girder. And
F =
P
load
span
X dist. of c of ^ from other end.
u {a (m  i) + z) a (m  i) + z
ma
u
2ma
P = ^r^^ {{m  i)a + zf.
V I'll '» ^ ^ ' '
When the loads are apportioned to the joints, we have at
each joint to the right of D a load iv = ua. Let the load uz
spread on the segment z divide into x at B and y at D. Then
X uz , IIZ''
t— = — and X = ^
Iz a 2a
394 APPLIED MECHANICS. [CHAP. XXI.
The shearing force in the i'* bay is
Ima 2a
when
and
2; = — a
m  1
„ uaCm  iy uu (2ni  i)^ ,_.
F = — ^ — = — ^ (5>
2 m  1 2 2n  1 ^ ^
a maximum, where n is the number of parts in the half span.
Now Fi sec a ^ / is the sectional area of BC in square inches,
while its length is h sec a, and substituting Ji^ h [h^ + cy for
sec. a we have
Vol. of 5^="^'^^^^'^^'^^"^'
2/ 2n  1 h
Summing for values of i from 1 to n, and doubling so as to
include both half girders, the volume of struts is
^struts = 77 7^ T S (4?i^ + 4?^^ + i^)
But
S (4w*) = 4?^^ 2 (47it) = 4w . — ^^^ — ^
and substituting and reducing
12/' V ^''
CHAP. XXI.] ANALYSIS OF TRUSSES. 395
For the volume of the ties k is to be substituted for c, and
adding, the volume of the diagonals is
^^^^\^'^^'^) <«)
Since the sum of c and k is constant, the sum of their squares
is a minimum when
so that
a L
The economical shape of triangular trussing is the isosceles
triangle. Adding this to the expression (1) for the volume of
the l)ooms, we get the volume of the isosceles truss to resist the
transite of a unifoim load W which can just cover the span
Wj'inl ) WL^ nn  1 In 1 \1
6/ ' ^ '^ "^J^\i^^ Un Jh
T^ WL \7n  1 16?r + 7n  5 1j
r = r + — — • — > (7)
.isosceles / ( 6 9671" r\
moT. load
having substituted h  rL.
Again in the expression (6) if we put c = ^ and k = 0, or
conversely we obtain the common value of rectangular trussing
(figs. 5 and 4) to resist the live load, and
V =^fi^z} . 8?r + 77t  3 1
rectangular / ( 6 4872' ?' I
mov. load
In the expressions for the theoretical volumes of trusses
given in equations (2), (3), (4), (7j, and (8), if all the quantities
be constant except n, the volume will be seen to vary directly
with n. Practical considerations fix the values of n between
very narrow limits, the chief being the necessity for stiffness,
which is of paramount importance, especially with moving
loads. The stiffness of a girder is of two kinds — stiffness of the
girder as a whole and local stiffness.
396 APPLIED MECHANICS. [CHAP. XXI.
Xow these two affect the question of economy in opposite
ways, for increasing the depth of the girder increases the stiff
ness of the girder as a whole, but makes the struts so long that
to be effective they must be greatly stiffened individually.
Considering these jointly, a depth of girder from oneeighth
to onetenth of the span is most suitable for iron and steel
girders. Where the booms aie timber which demands a large
scantling for strength, and so affording at the same time local
stiffness, the depths may be greater in proportion to the span.
In the American early practice with timber, depths of even one
fourth the span were used. In the British early practice, in
passing from cast iron to solid plate girders, depths as low as
onefourteenth the span were used. The modern practice in
both countries is now from oneeighth to onetenth of the span.
Econcyinicul Depth. — Having discussed the economical shape
and subdivision, it now remains to consider the variable r in
WL
the six expressions above. They are all of the form —r x it
where n = ar + Q, (9)
r
and, what is more, the theoretical volume of all trasses without
redundant bars reduces to this form. There is a value of r
which necessarily makes u a minimum, because with very large
values of r, u is sensibly «/•, incrcasiivj with r, wliile with small
values of v, n is j3 ^ /", increasing as r decreases. Put
making u = 2 Jn ./3 a minimum. (11)
Tablks uf Theoketical Volumes of Trusses.
The quantities n (9) are given in the tables, both for isos
celes or Warren bracing, and for rightangled bracing to resist
a rolling load IV, an uniform dead load W, and an uniform
moving load. The five rows are for 2, 4, 8, 16, and o2 subdi
visions or bays on the span, and in each pane of the table tliere
appear four values of u correspomling to the four depths
expressed as fractions of the span, viz., r = p'^, yV k' o^' T'
respectively. In each ' pane ' there appears also a heavyfaced
CHAP. XXI.] ANALYSIS OF TRUSSES 397
number, being the niininiuni value of u by equation (11), wliile
the corresponding vahie of r by equation (10, appears with an
asterisk (*) at it. These heavyfaced numbers are put in their
proper sequence among the plainfaced ones, so that one can
see at a glance whether this most economical depth is feasible,
that is, whether it falls within those demanded by stifihess.
This, of course, decides whether the economy of material can be
realized or not. For instance, on the firsfe bottom ' pane,' p. 398,
*r = 103 while on the third, *r = '096, so that jointly for live
and dead load the economical depth lies in the neighbourhood
of r = y'o' J^^s^ ^s demanded by stiffness. On the other hand,
consider the second last bottom ' pane ' of the second table,
p. 399, *r = '258, so that the economical depths for the Bollman
trusses are quite out of the question, and for the Fink trusses
too they can hardly be realized, if we may use such a word.
The tables give the numerical values oiu, see expression (9),
WL
the other factors being — —  where / being usually expressed in
tons per square inch, then W must be in tons and L in inches.
Having obtained this theoretical volume of the truss to resist
the load, it will be in cubic inches, and we must add 20 to 30
per cent, to stiffen the struts.
The tables deal with the isosceles and rectangular trusses
resisting an uniform dead load, an uniform moving load as long
as the span, and a rolling load. The Fink trusses for a dead
and rolling load are also shown for comparison.
The values for the rolling load are quoted from the last
edition, and are Levy's values.
For the isosceles truss with vertical members to transmit
part of the load to the upper joints, the formulas are shown
and the results tabulated on a separate table along with the
Bollman truss.
Girder with an Odd Number of Bays.
For the span divided into an odd number of parts (2m + 1)
we find as in equation (1) that the joint volume of the upper
and under booms is
6 A \ (27U + l)'^
And the volumes are as under
UNIFORM DEAD LOAD.
UNIFOEM MOVING LOAD.
Ratio
of
1
X ^ 1 ■»
^ 1 V.
V)
< 1 i.
'/
— 1 ~
05
depth
to
c
CO
■S
C5
•X
Si
1
1^
a
span,
6 ^
Z
do
•■^
w 3
•< +
w .
<
1
h
■1.
u *
H a
u 7
r
w 1
»■ = V
"^
X +
U ^
)i< T
U
L
^*
o
u: h 1
9 1 1
u
P<
'■r. ^
a :. 1
a:
~
1
t
CO
r + vVi
r + ii
f 4 2 J.
» + 1 8 r
r+ 1 J
Is
■h
^ 2879
^ 3817
_ 2879
o 3817
Oi
i^.r
II
^ 1975
c 2600
w 1975
f2 2600
"S
1
J*
1 625
■„ 2125
1625
•„ 2125
g^
i
1 (100
." 12.50
." 1000
I 1250
C4
* 866
'' 1000
* 866
* 1000
09
2r + Ai
•>r J. J" i
Vr f ./.V ^
Vr + A\ \
^
T^/f
.^ 2946
_ 3415
_ 2996
., 3 504
P4
ce
II
^ 2075
1750
^ 2388
■, 2000
^ 2118
• 1792
Ji 2456
•„ 2()63
J.
^
." 1250
.: 1.375
." 1302
I 1437
Tt<
* 1225
'' 1323
* 1 283
* 1393
4r + .vVJr
4; ^ aV i
Sr + ;\V i
f r + AV ^
.\
^ 2961
^ 3196
3025
„ 3288
Pi
h
Tt<
t; 2197
a 2353
o 2266
;:j 2442
X
II
19.37
• 2<;63
2016
■, 2157
•1 1895
;3
s
i 1696
i 1768
." 1808
1
* 1719
' 1 781
\ * 1852
* 1922
GO
8/ » .%V i
8r + v'A i
Vr + niii
Vr + 3W4 i
S
i\
:5141
3258
3236
3370
1— 1
W
00
':; 25;j8
2 ^"616
« 2660
:; 2750
0«
7i
J.
II
II 'J.. '5 9!
II 2453
II 2516
11 2617
f3*
:, 2359
i 2 411
^ 2532
i 2592
CO
\
2 695
2727
2991
3028
16r+,v;'H i
Ifir + ^AVni
Vr + itSii
3 '+ 4nt€ r
;2
3624
3682
3800
3866
P4
lt
CO
o
r .'5:304
o
r 3343
g 3552
1 c 3560
S 3 584
9 3605
a
II
II 3302
II 3341
M
II
&
X
s
:^' 336.3
i 3395
^ 3681
» 3717
^
\
4682
4697
1
5309
5327
• The value of r (or ininiriniin volume.
ILOLLINe LOAD.
FINK TEU88.
1
Ratio
— ' ^
« 1 i.
■A
— 1 ■_
^ ei
r/; »0
1
of
a J. •:.
•/; 1
„
; y
< ^'
deptli
r^ «;?
ai o
4<
13
C ,7.
O ^'i^
oT
Se]
1
'c
u
O
to
span,
2
c
CJ i^
H 3^
3
^ ^ "^
~ _
A
•yj H
O 1
a 1
' 
s =
O j
'■=1
Si2 8
s: s
^
^
■^ £i.
CO
CO
5
CI
2r + ' Ji
2r + J i
r4 i I
2/ + ^ J;
OB
^ 5758
o 7633
II
o 3817
_, 7633
iV
5 39A0
g 5.200
~
5 2 600
.9 5200
i^r
O.
".. 3'50
;, 4250
II  '^
•■ 4250
i
g
;^ 2000
^ 25O0
II
^ 1251)
v^ 250()
X
4
o*
* 1732
* 2000
1 ~l
* 1000
"^ 2 000
9
or + ^ i
5r + U J
I
■2r + iV i
6r T ^ ^
OS
^ 5958
^ 7365
II
.. 4821
„ 9775
tV
»4
 4J50
o 5188
r
?; 3 325
2J 6850
1
At
. 3 125
o
II
•,; 2750
:1 1750
■; 5750
^ 4000
i
* 2 739
* 3062
c^
* 1581
* 3873
""^
llr+Ui:
llr + ,VKi
CO
II
3r+fi I
14r+ K^ i
^ 6124
^ 6944
5122
11246
T^>
2 4H94
2 5241
^
;: 3582
a5 8275
iV
04
; 4250
i 3977
• 4688
I 4268
o
II
t= 3000
II
f 7J50
1 6 205
i
"3
1*
* 4188
* 4406
e
*^ 2063
1984 i
*"" 6250
t
CO
^3r + ^\2g J
23r + Hii^
4r + ^",V >
30r+ 5} J
6748
7188
II
5 247 1
12781
^ 1
^ 5777
?, 6070
5g 3720 !
:g 10188
tV !
Is
P4
r 5655
T
^
?* 1
7"
II 5656
11 5 891
2
11 3156
ii 9500
"3
*■
*^ 5 889
II
^T
*^ 9287
7141
7258
fi
2:? 28
2 305
10 375
1
CO
47r + ,W*^
47r + /„V),i
II
o/^tV.'5 J ■
1
8246
8458
5328
15149
1 6 f
S 8005
SS 8180
^
30
C3
cd
9 8108
9 8250
o
?» 3 830
r 13544
iV
II
11
C>1
CO
II
" 13495
"3
,> 8590
*" 8715 ,
II
% 3289 1
*~ 13625
»
49
13113
13 170
5<
25SJ
2581
18438
i
eo
The hetivy figures give the minimutk volume.
400 APPLIED MECHANICS. [cHAP. XXI.
Warren Triiss. — Dead and rolling load, odd number of bays.
^ _WL m(»t + 1) /^ 8m + 7 1\
j^ _ WZ m
f 2m,
(13)
. 2(3w + 2) r + ^ o/^ — T^ •  • (14)
1 \ ^ ^ 3 (2m + 1)^ ry ^ ^
Bectangular Trusses. — Dead and rolling loads.
„ WL m(m + 1) /_ 4m + 5 1\ . _.
^d = —r ^ T^ 2r + ,^j^ .  • (15)
IFZ m /^,^ ^, Sm^ + 2lm + 10 1
^r ,^ A 2(3W + 2) ?• + ^T^r r— — . 
/ 2?n + 1\ ^ ^ ^{2m + lf r
For (13) and (15) there is no shearing stress in the odd
middle bay. In (16) it is assumed that there is only one
diagonal in the middle bay. If there are a pair acting only
one at a time, say both ties, and incapable of acting as struts,
then (8??i2 + 21?^ + 10) is to be replaced by (8m^ + 21?;? + 13).
We have not tabulated these quantities, as values sufficiently
correct can be found by proportional parts from those which
are tabulated for the even division of the span.
For the uniform moving load in both trusses the coefficient
of r is (14?;^' + 12??^ + 1) f 6 {2m + 1) ; but for the Warren or
isosceles truss the coefficient of  is
r
32m.'' + 62m' + 28m + 1 ^
24(2m. + 1)' ' ^^
For the rectan'nilar truss, assumiuLf two diagonals on the
middle rectangle acting fine at a time, the coefficient of  is
16;»=' + 38m' + 23m + 1 .^.
Vl{2)v + iy ^ '
CHAP. XXI.]
ANALYSIS OF TKUSSES.
401
The Fink Truss.
Here the beam A A' (fig. 7) is divided into a number of
equal parts by a series of repeated bisections. There are thus
1 primary truss, 2 secondary trusses, 2'^ tertiary trusses, &c., so
that the number of trusses of the i*'^ variety is 2""* where n is
the number of bisections. Hence summing 2"'' for integer
vahies of i from n to 1, we have
1 + 2 + 2 . . . 2"' = 2"  1,
the total number of apexes of all orders, and therefore also 2"
is the number of equal parts into which the boom is divided.
The Dead Load. — If W be the dead load uniformly spread
on the boom AA', then the load directly over each vertical
strut is
W ^ L
v'+^, and a=^
is the common length of each segment.
A A3 /
2 ^
3 Ai
Ao
a
a'
1
V,
^
^
^
J
^^
^
\
vtx
/I
S\
^
1
\
7^
1 \
A
\L
\
w\
i^
X
A
B^_, B^
I ''
n is N'o. of bisections.
« is No. ot varieties of Trusses.
2'» is No. of parts in span.
FixK Tiiuss.— (Fig. 7.)
Consider an apex of the order i?„,, the stress on the vertical
strut is 10 the load on top of it, and this is all for i = 1; but for
1 = 2 there is besides the downward component pull of a pair of
tie rods attached to the top of the strut, being between them
another v:, or in all 2u\ Now a pair of the steepest ties
transmit loads with a vertical component w per pair, the next
steepest a vertical load of 2iv per pair, and the third steepest, a
vertical component load 2hv per pair, &c.
The load on the strut ^„_,, B„^i (fig. 8) then is 2''?r. The
pull on each tie from Bn.i is  x 2'"'w sec 0j.
The thrust on all segments of the boom is constant. At the
end A a tie rod from one of each variety of apex comes in. The
2d
402
APPLIED MECHANICS.
[chap, XXI.
horizontal component load on the tie from Bn.i is ^ 2''hi: tan Bu
Substituting for tan 0., and summing for all values of i from
1 to 11 so as to include one of each variety, we have
thrust on boom = '^ 2" 2^('i) = ^"^ (1 + 2^ + 2'
111 1 2li
7m 2*" 1 2"u'.2'Va 1(^)'"
WL 1  (1)^
Dividing this by / gives the constant sectional area of the
boom, and further multiplying by its length, we have
volume of boom =
Wl ilY' D
f
6
h
(A)
B77, 2
i z w
or i w
The i'*' Vakiety oi' Thlss. — (Fig. 8.)
The volume of the strut A,,., B,,, is 2'"'?'.7i ^f, and multiply
ing by 2""' gives us 2"'^ ich f ,/'as the volume of all the struts
of the same variety. But the expression being independent of i
shows that one strut of the primary truss has the same volume
as the Uoo struts of the secondary trusses, as the fi>vr struts of
the tertiary trusses, &c. We have to simply multiply the last
expression by n, the number of varieties, and
volume of struts =
(2"K >) nh _ W nh
(B)
CHAP. XXI.] ANALYSIS OF TRUSSES. 403
Again, tlie volume of the pairs of ties from all apexes />„_,• is
stress , , . i2^%sec0i , „
2 X number x ~—y x length = 2 x 2"^ x x A sec %
= 2" > — SeC^/ = 2" ' — r r:;
/ / ^^'
Summing for values of i from 1 to n,
volume of ties = 2"^ '^^ + 2"' ^ t 2M'i)
/ ¥ I
1''w nh 2"w a ,, ^, ^
= + Q + 2 + 2* . f 2")
/ 2 f 2h^ ^
_ Wiih PT «^ 2^"  1
/ 2 f 2h 3
/ I 2 '^ 6 77
(C)
Summing (A), (B), and (C), and putting h = rL, then the
volume of the Fink truss to resist a dead load is
^ jvLf 1ar 1\
Xr'"' — 3 r)' ^^^^
TA€ Rolling Load.. — When W the rolling load is at the apex
Ani, the load on the strut Ani Bn.i is W, which also is the
vertical component load on the pair of ties supporting it at Bni
The stresses in all pieces from the other apexes of the same
order, or of higher orders, is i^ero ; while those on struts from
apexes of lower order are fractions of W. Multiplying W, the
greatest load on a strut, by h its length, and dividing by/, we
have its volume, and as they are all alike, we must multiply by
2"  1 their number.
W
Volume of struts = (2"  1) h ■ (D)
The greatest pull on a tie from J5„.j is hW sec Oi, the area
of this tie W sec b ^ 2/" and its length h sec 0„ and there are
2d 2
404 APPUED MECHANICS. [CIIAP. XXI.
twice as many as there are apexes of order Bnj, that is, there
are 2 x 2""*. The volume then is
r 2"' sec Qi = — 2"' ^ ,
which has to be summed for vahies of i from 1 to n, and
Wh
vohime of ties =— r (1 + 2 + 2= . . . + 2»^)
Wa
fh^ . . . + ^ ;
= — (2"])/^ + (2"l)(i)'^J
= _(^(2"l)/;f (2"l)a)"J. (E)
The greatest thrust on the boom occurs when TV is at 4„
the centre of the span. It is ^W ^ \L ^ h, dividing by /and
multiplying by L,
volume of boom = . r 7 • (F)
f i h
Summing (D), (E), and (F), and putting h = vL, the volume
of the Fink truss to resist a rolling load W is
^^.J^(2(,..l)...iL4ill',i). (20,
If the dead load be heloiv the truss, each vertical strut is
relieved of a load ic, but is not altogether relieved of load, as
Levy assumes. Diminishing then the volume of each strut by
w/i ^ f, and multiplying by 2"  1 the number of struts, then
we have the proper deduction to be made from V ^ to be
2"  1 Wh
which caves the volume of the Fink truss to resist a dead load
hcloio the truss
r. = '*^^{o.l + (4)")'^^.'{ (19')
CHAP. XXI.]
ANALYSIS OF TRUSSES.
40;
And again for the rolling load below the Fink truss, we must
not leave out the whole volume of the vertical struts, as Levy
inadvertently does. For tlie rolling load JV l)elow the truss, the
greatest stress on a strut occurs when W is at the foot of the
next adjoining strut. The amount is then
1
Iv ' — I — +  . . —■ —
V2 4 8 2*V
= JV I
iW)
Hence the volume of all such struts is
2.Mr(iQj
HC
Summing for all the n varieties,
volume of struts =
^<Q
=
/ f 3
1
(D*)
Summing (D*), (E), and (F), and putting L = rh, we find
the volume of the Fink truss to resist a load PF rolling below it
to be
WL (/ 2"^^ + (I)"
F,.=
 2
3  [\Y^ V)
4 "r)
(20*)
We have not thought it necessary to tabulate the values for
(19*) and (20*), as the Fink and Bollman trusses are only given
here for comparison, and for an exercise in analysis. The trusses
themselves are out of date and far from economical for the
II = 2w is No. of parts in span.
Bollman Truss. — (Fig. 9.)
rolling load. The economy for dead load, too, is only apparent,
as it could only be realized with a depth of truss which is im
practicable, as it would involve a stiffening of long struts that
would eat up the economy.
406
.APPLIED MECHANICS.
[chap. XXI.
The Bollman Truss.
In this system of trussing (fig. 9) the upper boom AA' is
divided into m equal parts ; at each joint is a vertical strut
whose foot is tied back to the two abutments.
The i"' Truss from Left Abutment. — (Fig. 10.)
Dead load. — Consider the i^^ truss from the left abutment
(fig. 10).
__ , . , ^ length
Volume of struts = number x common stress x ° —
,,W h m1 W ,
= {m  1) —  = . . .h;
m f m f
sin ABjA' = sin (6 + cf)) = sin 9 cos </> + cos sin 6,
(G)
and using the surds in fig. 10, which are derived by the forty
seventh proposition of Euclid I, we have
sin(^ + ^) =
mha
y/h'^ + ia^ y/li + (w  ifa^
For equilibrium at Bi we have
W ,
i^i : i!2 : — : : sin AiBiA : sin AiB[A : sin ABiA
m
: : cos B : cos 9 : sin {Q * (j>).
Using the surds in the fig. 10, we have
^ ^ W (m  i) y/rTlv
in mh
and multiplying by the length ABt and dividing by/ we have
the volume of
ABi = '' .(m z)(/i^ + i'a'),
in nj
CHAP. XXI.] ANATA'SIS OF TRUSSES. 407
and also the volume of
IFi
A'Bi = ^777 ('''" + (^'^  *')•«■)•
mlif
Their sum is the volume of the pair of ties from the apex Bi
and is
W
—., (h^ + i(riii)(ir).
mlif
To sum the first term, which is independent of i, it must be
multiplied by {m  1) the number of apexes ; the other two
terms are to be summed for values of i from 1 to (m  1).
Putting L for a m finally
jy / »'i '"1
vol. of ties = — rz, (in l)h + ma^ V i a V v
^V (, ,,,, o [m  l)m , {m  l)?n{2m  1)
' (m l]/r + mar —^ a
mh/y ' 2 6
Wmlf^ m + lL'\
/ m \ Qm h
Part of the thrust on AA' is the horizontal component of ti,
which is
W m  i . ■ W . . .. ((■
^1 cos f/» = — . — 7" . ta = —  I [m %) —•
m mil mil L
Summing for values of i from 1 to {in  1) includes the horizontal
pull of one rod from each apex, and
thrust on boom = — r 7 //^ T ?  2 i' )
mh X V 1 1 /
W a}(m^{mV) ( ml)w(2ml)
6
mil L \
2
WL m 
1
6/t nt"
Multiplying by L :/, we get
volume of boom = — ; — — — ^ • (I)
f D??r li
•iOS APPLIED MECHANICS. [CHAP. XXI.
Summing 'G), fH), and (I), and putting m = 2n so that n may
be the number of parts in half span as in the other trusses, also
putting h = rL, we have volume of Bellman truss to resist a
dead load W over the truss to be
WL 2nlf 2n+l 1\ ^.^^.
/ n \ I2n rj
Lire load. — Let the live load W^come to the point A,, and
it will be seen that the load is now m times what it was for the
dead load. Now as W comes to every point in succession, the
volume of the struts and of the ties will be obtained by multi
plying the expressions (G) and (H) by m, when we have
TF
volume of struts = (w  1) — 7i ; (J)
volume of ties = {jii  1) y{h+ — — ) • (K)
/ V 6)/^ h
With the boom it is otherwise. When W is at Bi , we get
the thrust on the boom by multiplying the former value of the
horizontal component of ^i by m, which gives
W.. .cr
^.(m^);
but this is thrust on the boom only as long as IF remains at
Ai, but i(m  i) is greatest when the factors are equal, so that
the maximum thrust on the boom occurs when i = m  i or
i = hn, that is, when W is at the middle point of the boom.
Substituting this we have
max. thrust on boom = ; — r v
li i L
Dividing this by / to get the sectional area, and multiplying
by L, we get
, „, JF m'cc' WD
volume of boom = — — = —  — • (L)
jli 4 f/l 4:
Summing (J), (K), and (L), and putting m = 2)i, and 7/ = rL,
CHAP. XXI.]
ANALYSIS OF Tl'JJSSES,
400
we have the vohune of the Bollman truss to resist a load W
rolling over it to be
F. = y (2 {2n  l)r + ^^^ • " ) • (22)
BOLLMAN
ISOSCELES TRUSS.
TRUSS.
Part of Load sent to vertices by vertical
rods.
i.

^^T^
: "
o r
Kstio
LOAD
12n
s "
11 e*
,
i!
1
9 1
^
of
depth
A
a +
o S^
a =■
O e
to
M
ir
oi +
2 S
K 00
epaD,
1
S 7 .
° ^
o 7
A
» i *
« i. .
b 1
X J_ c«
'"Z
(S
rti i
2r + i i
' r + i i
2r + i i
o 3817
^ 7633
3817
. 7 633
iV
o 2600
o 6200
o a600
5 5200
T^r
—
•„ 2125
■ 4250
t 2125
 4250
i V
II
: 1250
^ 2500
" 1250
" 2500
1
* 1000
• 2000
* 1000
<■ 2000
f (» + A i)
er + i ^
ir*^i i
ir*H i
=» ^'^f
^ 13525
3382
ra ''332
i^r
« 3275
g 9350
=» 2338
£i 5138
i^.
(M
•„ 2688
• 7>75a
" 1937
" 4313
i
^
1, 1625
* 1369
i 500O
• 4683
:: 1250
* 1146
° 3000
*" 2655
i
5039
14r + fJ >
24371
.!r+,V. i
^»r + ,y. i
?? 3456
;5 17025
3096
6844
•^h
T 2844
^ 14250
2 2203
§ 5091
l^r
f*
T^ 1876
« 4500
i
I]
.,> 1750
»> 9750
9354
11
11 3967
»
1516
»i~ 1406
»> 4031
1
1398
v(»■^ii y
30/+ J} i
5105
45594
!^ + .V. i
Vr + iv* i
_ 3508
_^ 32063
3025
6955
iV
CO
5 2266
g 5720
■>';,
00
• 2891
„ 27000
?*
T
11
i
II
11 2016
t: 5454
1.
^
» 1797
•'' 19125
,^ 1808
»' 5422
1578
18675
1852
6383
ilir + ii i)
62r+V.'^ J
\hr 4. ^aiSjtf 1
^%hA% i
5124
87805
3182
7973
iV
11^
o
CO
o 7500
CO
:;; 3524
o 61981
:>: 2597
g 7509
<^
11
." 2906
" 52375
" 2458
11 7786
*■
*" r816
• 37813
* 2,436
•
1607
37190
2822
11280
i
If the load be below the truss instead of above it, the volume
of the struts (G) and (J) can be left out, as there is no load on
them theoretically, so that Vd and V, are obtained from (21) and
(22) by halving the first term.
410 APPLIED MECHANICS. [CHAP. XXI.
Examples.
1. A Warren girder has a span of 160 feet and a depth of 20 feet, and is
subdivided into 16 eqnal parts of 10 feet each. Find the theoretical volume of
wrought iron for which /= 4 tons per square inch, that the girder may resist a
rolling load of 80 tons. Here
« = 8, r = ^, iJ = 80 tons, /= 4 tons per sq. in. and Z = 1920 inches.
From Levy's table, second column, fourth line, we get 56o6, and
HZ
I', = .5656 — = 217190 cubic inches.
2. Determine a girder of similar type to bear the dead weight of the last, there
being 8000 cubic inches of iron to the ton, that is
TFi = 217190 ! 8000 = 272 tons,
and also ^' = 160 x  = 60 tons,
a share of the deck or platform which is threeeighths of a ton per foot.
Vi = 2391 — TT = 28167 cubic inches, TTj = 352 tons,
TF'L
Yz = 2391 — r = 68866 cubic inches, Wi = 861 tons.
3. Find now the theoretical weight of the girder to resist the rolling load E,
its own weight, and a share of the deck or pl.itform
W= ZFi + W2 + TFs = 3933 ions nearly.
But no provision has been made to resist W2 = 3"52 tons. For this
V = 2391 X IFiL 4 /= 4040 cubit; inches,
or its weight is half a ton, so that TF= 40 tons. Adding 20 per cent, to stiffen
the long struts and allow for rivet heads, &c., the probable weight is 48 tons.
4. Find the weight of the girder if it were braced on the Fink system. The
tabular numbers are now 9500 for the rolling load, and 31.56 for the dead load.
^"1 = 456, Jr2 = 86, 7r3=ll4, and «• = 067,
JF= Wx + W% + 7F3 + w = 663 tons.
Adding 20 per cent, we have SO tons.
5. Find the theoretical volume of the isosceles truss shown on fig. 1 1 .
Here n = 3 and r = i. As w = 3 is not given in the table, these values of n
and r are to be substituted into the general formula at the top of columns one an<l
two, giving 1709 for the dead load, and 3703 for the rolling load, as piinted on
tig. 1 1, which see.
CHAP. XXI.]
ANALYSIS OF TRUSSES.
411
The upper figure shows the "Warren or isosceles girder. The number of bays
in the half span is n (3 on the figure). The span is L inches, so that r = L ^ 4m
is half a bay. The total uniform dead load is IF tons, giving a parcel of hv over
each abutment, and of w= W~'2n at each intermediate joint between them.
The left supporting force is P = \W  ho = tv {n — §). The depth of the girder
is A inches, and h = rL, when expressed as a fraction of the span. "We will find
ISOSCELKS AND RECTAXOfLAU BkACINO.
4»
V = 1.709 — jr~ cub. in.
r=// ^ I. = 1. V = 3.03:
i — r~ ■ l>
/
y
/
\
\
\
< — c > 1
^ 2n <
I
1
f
\
\
\/
/
1
c = o
r
554 — Y cub.
WL
cub. in.
11.
the bending moment at each of the sections through the upper joints in left half of
span, and add them. This sum divided by h gives the sum of the tensions on the
n lower booms, and dividing this by/, we have the joint crosssectional area in
square inches of those n booms as if packed in one bundle. Lastly, multiplying by
their common length "ic, we have their volume in cubic inches.
Using the odd digits for the upper joints, beginning at the left end, the bending
moments are —
.1/, = F.c.
Mz = F.2civic).
Ms = P.!icu>(c + 3c).
Mt = F.lc w(c+ 3c + 5c).
412 APPLIED MECHANICS. [CHAP. XXI.
5 J/ = Fc {I r 'i + b to II terms)  my (1 + 4 4 to h  1 terms)
= w {n  ^) or  ur . i (n  1) u (in  1)
= lwcn{in'^  1).
The sum of the odd digits being the square of the number of terms, while the
sum of the squares of the natural numbers up to n is g n (n + 1) (2« + !)• Multi
plying the above sum by 2c r/%, and doubling to include the other half of the
girder, we have for the volume of the lower boom —
2 If
 — »<(4w^ — l)c cubic inches.
3 jh
With zero at the left abutment, and the even digits at the lower joints —
Mz = F. 2c.
Ml = F.4c w.'Ic.
Ms = F.6i u{2c + 4c).
Me = F. Scw(2e + 4c + 6c).
Mi,, = F. 2>ic  2uc (1 + 2 to «  1 terras).
This last line is the bending moment at the centre, half of which must be
subtracted, so as to exclude half the upper central boom.
2Jf  ^if2„ = 2Fc(l + 2ton terms)  2t*c (I + 3 + 6 to u  1 terms)
= 2w («  i) c . in (« 4 I)  2tvc . ^ {n  1) n (« + 1)
— tv {« — 5)mc 4 uc , ^ («  1) il.
= iwc«(4«« 1),
which gives the same volume as for the lower boom. The only additional series
being 1I3 + 6 + 10 + &c., the sum of which is ^» {n + 1) {n + 2), where u is
the number of terms.
The length of each diagonal brace is ^(c^ 4 h), and if d be its slope to the
vertic il, sec d = V(c* + A*) f //.
The shearing forces in the bays, beginning at the left abutment, are —
K„2 = F\ F2,i = P w ; Fu a = F 2k,
2J'= F. n  :o{l + 2 + 3 to «  1 terms),
This sum, wlien multiplied by sec 0, becomes the joint thrtist on all the
diagonal struts in left half of girder; dividing by /gives us their sectional area as
if all in one bundle ; multiplying by their common length, and doubling, so as to
include both halves ot girder, we have vulume of diagonal struts —
uw* sec X . yc' + h = — »j2 (c* + A).
CHAP. XXII.] DESIGN OF MASONRY AKCHES. 413
The same expression is the volume of the diagonal ties. Adding, we get the
volume of the Warren or isosceles girder to resist a uniform dead load.
^' = 3^^' ('"'')'•' + 7^' "'(^' + ''')
= —  1 nh + „ c
W ( , 8w" + 3«  2 L
h +
3/t 16/i=
WL I 8« + 3>(  2 1
+ — ' \ nr +
/ ( iSn r)
6. A giriier is braced as shown in fig. 1. Span Z = 168 feet, h = 12 feet, and
« = 6. The struts are shorter than the ties somewhat in the manner of the Post
truss. The shape of each triangle is such that a perpendicular from the vertex
divides the base into a short segment c = 6 feet remote from the centre, and a
longer one of 9 feet adjacent to it. It will he seen that the struts are 13 feet long,
while the ties are 15 feet long. Find the theoretical volume of tlie truss to resist
an uniform load of foursevenths of a ton per toor. That is, 7r= 96 tons or 8 tons
concentrated at each lower joint of fig. 1. The jtudent should work this example
by taking out the volumes in detail as on last example, and cheek his result with
that obtained by substituting
JF= 96, / = 4, i = 12 X 168, r = ^^, c = 12 x 5 in equation (2).
Atis. 148480 cubic inches.
CHAPTER XXII.
THE SCIENTIFIC DESIGN OF MASONRY ARCHES,
In this chapter we have chiefly in view the design of important
Masonry Kailway Bridges and Viaducts, having arches of long
span, with as small a rise as is consistent with moderately heavy
abutments, and in which the surcharge, over the top of the key
stone, is no more than sufficient for the proper formation of the
bed of the railway, which may be taken at about eighteen
inches. To successfully resist incessant, swift, heavy traffic, the
archring, or at least some central portion of it, must not be
restrained by rigid backing.
An archring well built of wedges of stone, or voussoirs, as
they are called, and more especially one built of brick masonry
in concentric rings, as is the most modern practice, has ample
elasticity to accommodate itself to all the vagaries of the
live load, provided it be not overmuch constrained by a too
rigid superstructure. The two joints at equal distances from
414
APPLIED MECHANICS.
[chap. XXII.
the keystone, which map out this centra elastic part of the
•archring, are practically the joints of rupture ; and a horizon
tal line, through the upper edges of these joints, is the level of
heavy backing. The backing above this level should consist of
light spandril walls bonded into the archring by through
headers, or simply riding on the archring. Their principal
function is to give a horizontal passive reaction on the arch
ring when it bulges out due to the live load on the other half.
They stand upon heavy spandril walls below them, and stretch
horizontally such a distance that the friction on their Ijase shall
at least be equal to the excess of the horizontal thrust of the
loaded half arch over that of the unloaded half.
The light spandrils are best built of rubble masonry, or of
brickwork with plenty of mortar in the vertical joints, to allow
the archring sufficient play while the passive horizontal re
action is being called out. The spandrils serve also to lighten
Fis. I.
the superstructure and reduce its average weight per cubic foot,
as the spaces can be left void and the walls covered across.
The heavy backing consists of thicker walls with vertical
dressed joints, and having the stones or bricks packed close
horizontally, so as to yield but little when the centre is struck.
The voussoirs, below the point of rupture, have their backs
•dressed in a flight of steps, so as to take the horizontal thrust
truly. The heavy spandrils must stretch horizontally a sufficient
distance to give friction on their base equal to from twothirds
to onehalf of the horizontal thrust of the arch at the crown.
The earth tilled in behintl tlio spandrils can be relied upon to
give promptly the remaining tliird or half, according as it is
tilled loose, as in an embankment, or is rammed in layers
between the masonry and the face of the excavation, as in some
cuttings in old consolidated earth or rock.
CHAP. XXII.]
DESIGN OF MASONRY AUCIIES.
415
Observe, before the earth is filled, tliat the heavy backing
can only take a large fraction of the horizontal load. But the
centre of an arch is always struck when the superstructure is
only partly built, else the superstructure would crack, due to
the aftersubsidence, so that only a part of the horizontal load
is required. Of course the earth must be filled behind the
spandrils before the remainder of the superstructure is built.
Two rival designs present themselves for longspan light
surcharged masonry arches — the Segmental Arch and the Semi
elliptic Arch.
In the one, the soffit is a segment of a circle, with a rise
about a quarter of the span. The joints of rupture should be
at the springings, which demands that the total headroom, from
springing level to level of rails, should bear to the span the
ratio, almost constant, of one to three. In this design, too, it
^z:^^
'2— o,< ' I on
^^rCA''^' \
Fk
will be found that the archring must thicken outwards from
the crown to the springing, so that a skewback shall be from
once and a half to twice as thick as the keystone. This may
be disguised on the front of the bridge by an uniform moulding,
of the thickness of the keystone, cut upon the face of the pend
ring of stones.
In the other design, the soffit is a false semiellipse struck
from five centres. At the ends of a quadrant the radii are
those of the ellipse, being the square of one semidiameter
divided by the first power of the other. Between them a
centre is interpolated, as on fig. 2 (see Eank., C.U., p. 420), at
the intersection of two arcs, one from each of the main centres,
with a radius equal to the difference of that main radius and the
rise. It will require three templates to work in these three
varieties of groups of voussoirs. In this design the rise of the
arch may be about a fifth of the span, and the ring in this case
is of uniform thickness. As the thrust at crown will not exceed
twice that at springing, the ring can accommodate the line of
stress in its middle third.
416
APPLIED MECHANICS.
[chap. XXII.
It is well to observe that this soffit, although called elliptic
by courtesy, approximates in an equal degree to one of the
elastic curves, called by Rankine the Hydrostatic Arch, or to
that elastic curve moditied and then called by him the Geostatic
Arch. (Eank., C.E., pp. 208, 419 ; 212, 420.)
We have already given a practical definition of l\\& joint of
rupture, and, as Eankine says {ihid., p. 428), the more rigid part
of the ring, below it, is really part of the pier or abutment. It
will appear as we proceed that this practical definition corre
sponds to Eankine's^wm^ of rupture for a linear rih loaded in any
way. By locating these joints as far out from the crown as is
consistent with equilibrium, two advantages are gained : one,
economy of heavy backing ; the other, a long elastic field.
Line of Steess.
In calculations connected with the arch, it is convenient to
consider one foot of the ring in the direction of its axis, just as
if the bridge were only a foot broad, and the same symbol t
gives the thickness of the ring in feet and the area in square
feet of the joint on which two adjacent voussoirs abut. It is
Fig. 3.
assumed that their mutual pressure on each otlior is either
uniformly distril)uted over that joint, or varies unifi)rmly from
a maximum value at the one edge to a minimum at the other,
Such a stress is indicated on a drawing by a sheaf of arrows
varying in length, and with their feathers ranged in a straight
slope, or more simply by a trapezium mapping them out.
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 417
Three varieties of this distribution of the stress on the left
cheek of the keystone are shown upon fig. o. Each has the
same maximum value ./'at the lower edge of the joint, and the
minimum values are Ci = 0, c^ = \/, and es = ./' at the upper
edges respectively, while the average values at the middle point
of the joints are respectively 2h, Ih, and ih, equal to 4, 5, and 6
eighths of the maximum. The whole stress may be supposed
concentrated at a point called the centre of stress. To find the
centre of stress it is only necessary to construct the centre of
gravity of the wedge whose face is the trapezium of which c
and /'are the parallel sides. A construction shown on fig. 11,
Ch. XIY, and fig. 3, Ch. XXII, is— Produce e to the left a
distance equal to./', and./' to the right a distance equal to e, and
the line joining the extremities cuts the line joining the middle
points of e and ./ in the point g. If now S stands for the
deviation of the centre of stress from the middle of the joint, we
have, by similar triangles,
U + S:htS::y+/:e+ hf :: p + hf : '2p  If.
Since the average value ^j = hf + he, then the ratio —
Maximum stress /' 6S
Average stress p ' t '
So then, on fig. 3, we have
^1 = g , c\ = — , and ^a = jg , and e^ = 0, £■, = If, and e^ = \f.
Let the ring on fig. 3 be that of a small model, the blocks
resting on a centre and separate from each other, with square
pins to preserve the spacing. The ring being loaded in a
suitable manner, the pins are to be moved up and down in the
joints till it is found that the centre can be lowered, leaving the
ring in equilibrium. Then the curve joining the pins is the
line of stress for that manner of loading. The equilibrium is
unstable : any slight disturbance would make each pair of blocks
to rotate about the pin between them till either their upper or
lower edges came in contact. Xote, too, that the thrust between
some of the blocks being oblique to the joint, there is a tendency
for the pin between them to roll over, if it were too small, or
2e
I (1)
418 APPLIED MECHANICS. [CHAP. XXII.
had its corners rounded. In the one case the equilibrium is
said to be destroyed by heitdinr/, which is only stopped by the
edges coming in contact, while in the other the equilibrium is
destroyed by slylin;/, which is only stayed by the roughness and
squareness of the pins.
Equilibkium and Stability.
For every alteration or rearrangement of the load, the pins
or fulcrums would require to be shifted into new positions to
maintain equilibrium, so that the line of stress changes its form
to suit the load. A small extra load laid on the centre of the
part of the arch shown on fig. o would slightly flatten the arch,
and tend to make the central joints shut at their upper edges,
so that the pins there would have to be moved up, and the new
line of stress would then be a sharper curve.
The same thing liappens in the masonry arch, where the
mortar fills the spaces between the joints. Every slight alteration
of the load alters the distribution of the stress over the joints
in the elastic untrammelled part of the arch. The centres of
stress move up and down in the joints, just as we moved the
pins in the model, and the line of stress freely changes its
shape, and accommodates itself to the change of load. And
wherever the ring is it?,Q\i flattened and depressed, there the line
of stress becomes sliarper in curvature, and rues into a hif/hcr
position in the ring. Also, the friction prevents the stones
sliding where the obliquity of the stress is only slicrht, say less
than 17°.
To render the equilibrium secure it is only necessary to
have deep joints, and to limit the displacement of the centre of
stress to some central field of the joints. This displacement
should not be greater than a sixth of the thickness of the joint,
for then the minimum stress at one edge is zero (see fig. 3) ;
to force the stress further would mean opening the joints, so
that as the live load crossed and recrossed, the joints would
' work,' the mortar crumble and drop out, and the whole be
destroyed.
The conditions of statical and dynamical equilibrium of an
archring practically reduce to this — That the line of stress
shall always be in the middle third of the archring, and the
joints be normal to the soffit ; it being then practically im
possible to draw a curve, in the middle third or ' kerne/ ' of the
ring, which shall cross any of those joints at an angle greater
than 17°.
chap. xxii.] design of imasonry arches. 419
Strength.
"We will take the ultimate resistance of sandstone to crushing
at / = 576,000 lbs. per square foot, and its weight at 140 lbs.
per cubic foot. And the least factor of safety shall be ten.
The ./ shown in fig. o, indicating the working intensity of
the thrust at the keystone, is not to exceed 57,600 lbs. per
square foot, if the ring be sandstone. For strong brick we
take/= 154,000 lbs. per square foot, and vj = 112 lbs. per cubic
foot. Now, as the loads are to be reckoned in cubic feet or
columns of the actual substance of the bridges, then, speaking
directly as to the strength, but inversely as to the mass, strong
brick is onethird as efficient as sandstone, while granite is twice.
as efficient. For granite, /= 1,350,000 and vj = 164.
For large bridges the question of strength is entirely domi
nated by that of stability ; but for small arches where the s%ir
cliavfje is rclativeli/ larger, the question of strength, on the other
hand, dominates that of stability, and it becomes convenient to
confine the line of stress to closer limits, and we then take as
a ' kernel,' a middle yZ/i^/t of the ring, and as a ' kernel,' a middle
ninth (see fig. 3). In this last, the line of stress may be looked
upon as practically up the centre of the ring, furnishing no
reliable clue to the proper thickness of the ring. But even in
this case, the maximum and average stress on the keystone
differ by 25 per cent. In designing arches (usually short in
span) with heavy surcharges, one is really designing a bent
strut — a ^'ery difficult matter in itself. The line of stress is
practically up the centre, but with a probable variation of
25 per cent. The difficulty in the one case is, that the factor
of safety to be used is left entirely to the judgment, whereas,
in the long span lightly surcharged arches, the stability
demands so deep an archring, that the factor of safety
against direct crushing is beyond what prudence would
otherwise dictate.
Balanced Linear Rib or Chain.
The Linear Eib and Chain are ideal structures, usually
curved or polygonal. They are designed to suit or resist a
given external set of loads, with the intention of afterwards
clothing the sides of arcs in timber, metal, or masonry. They
are then designs for actual engineering structures. The pins
in the model (fig. 3) map out such a linear rib, and its clothing
is masonry. Eib and chain are to distinguish whether the
2e 2
420 APPLIED MECHA^'ICS. [CHAP. XXII.
members of the structure are in compression or tension. The
rib is of far greater practical importance, but its equilibrium is
unstable, until a measure of security is given to it by the
material clothing. It is hard for the mind to conceive of the
linear rib witliout its clothing ; and still harder to write about
its e(iuiliV)rium, clearly and accurately, in words. On the other
hand, the linear chain may be regarded as having a slight
amount of material, of too little depth to interfere with its
flexibility, and of little weight compared to the external loads.
It may also be supposed to be furnished with hinges at short
intervals, and so be perfectly free to assume the proper shape
for equilibriiim, whatever may be the external loads, and to
return again to that shape if disturbed. Consider the equili
brium of the pin of one hinge. The two links which it joins
must be pulling the pin in opposite directions with equal force.
This can only be the case if the mutual interaction of the two
links, through the medium of the pin, be along the tangent to
the chain at that point. We must assume a hinge at every
point, and so — The rjcncral condition of equilibrium is, that the
2ndlor thrust along a linear chain or rib, at each point, is alone/ the
tangent. Having designed a linear chain, for given loads, it is
only necessary to invert it, or suppose the loads reversed, and
we have the corresponding linear rib. The funicular polygon
of older writers, and the link polygon (see tig. 10, Ch. IV) of
graphic statics, are examples of linear structures. All the
ribs we are about to treat of shall be horizontal at the central
highest point or croini, and shall spring from two points on the
same level.
The suspension bridge is the only practical example of a
linear chain. In its primitive form, with vertical rods, and free
from diagonal bracing or stiffening girders, part of the load
was nearly uniform along the chain, and another part uniform
along the span or platform. It formed a good example of a
balanced chain under vertical loads alone. The two theorems,
that the linear chain is in form a catenary when loaded uniformly
along the chain, and a 2}<^'>'(t'l'ola when loaded uniforndy along
the span, seemed to have an important application to the
suspension bridge, and much has been written about them
under that misapprehension. It is impossible to liave eitlier of
these loads alone, in a useful structure. Still, the two theorems
inform us that tlie chain of tlie bridge is, in form, something
between the catenary and the parabola. As both of these curves
are sharpest at the vertex, they show that this also is the char
acter of the bridgecliain. But the sharpness at the centre of
CHAP. XXII.] DESIGN OF MASONRY AllClIES.
421
the Inidoechain is not very marked, because the dip is always
a mere fraction of the span, so that the corresponding ares near
the vertex of both catenary and parabola are sensibly circular.
In the modern suspension bridge, what with oblique rods,
stiffening girders, and bracing, the load is no longer wholly
vertical, and the chain may have a variety of forms.
These two theorems, although of little practical use, are
exceedingly convenient as a startoff in the study of the
balanced semicircular rib or chain. An elegant proof that the
linear chain hangs in the catenary when loaded uniformly along
the chain, and that that curve is sharpest at its lowest point,
may be found in Williamson's Iniajral Calculus.
^ * ^ ' ' I 'Earih (loose or gunned).
Fis. 4.
The parabolic chain, balanced under a load uniform along
the span, is figured in Kankine's Civil Engineering at page 188.
The proof is simple and instructive, and may be given here by
supposing fig. 4 modified for that purpose. Let the quadrant
ABS be the edge of a thin steel rib or curved ribbon, one
foot broad normal to the paper, and supported by a horizontal
thrust T^ at the crown hinge A, and a tangential thrust T^
at the springing hinge >S'. The only load on it is to be the
load of 2? lbs. per foot, spread uniformly on the horizontal
platform ah. This load is to be transmitted down to the
rib by vertical struts without weight themselves. The plat
form ah is also one foot broad normal to the paper, and is
to be quite flexible, so as to allow the load to be transmitted
422 APPLIED MECHANICS. [CHAP. XXII
to the rib without constraint. The parallelogram standing
on the platform ah, and having the height p, is called the
verticalloadarea. Consider the equilibrium of a portion of the
rib from the crown hinge A, out to any other hinge B, and
suppose it to be rigid, having all hinges between A. and B
clamped. The corresponding part of the verticalloadarea,
which is sliaded on fig. 4, may now be supposed to be con
centrated into one force P, acting down through the centre of
gravity of that shaded area. AB is balanced by the three
forces 1\, T, and P. They must therefore meet at one point.
The general condition of equilibrium requires T to act along
the tangent at B. If we lower the shaded area till a coincides
with the crown A, it becomes apparent that the tangent at B
bisects its base ae. The rib is parabolic, for that is the only
curve whose tangent at any point behaves in this manner.
Now, as B is any point on the rib, it can be in succession each
joint from the crown outwards, so every joint can be undamped,
still preserving equilibrium.
Hence, the particular condition of equilibrium of a linear
rib under a verticalloadarea alone is — That the tangent at any
point shall meet the crovni tangent on the vertical through the centre
of gravity of the 'portion of the loadarea from the croicn out to that
point.
ConjugateLoad Areas.
We have already discussed a linear rib to resist a vertical
loadarea alone. The vertical load was spread on a horizontal
platform, equal in length to the span of the rib, and one foot
broad normal to the paper. The shape of the area was a
parallelogram of height a The rib itself was parabolic, and
so peaked at the crown A. On fig. 4 this rib is shown : ABS
is a quadrant of it. It is no longer parabolic, but has been
deliberately pulled out horizontally at each pair of points,
except the springing hinges, until it is a semicircle. It will
no longer be in equilibrium ; the vertical load will tend to
flatten it at the crown, and spread it out further at each pair
of points. To prevent it spreading horizontally, we may suppose
a pair of vertical platforms, of which rd is one, each in length
equal to a radius, and a foot broad normal to the paper, and
furnished with loads gravitating horizontally inwards. One of
these loads is indicated on fig. 4 by a parallelogram of height
q = /), standing on the platform cd as its base. It is called the
conjugate horizuntalloadarea. If we consider the whole circular
CHAl'. XXII. J DESIGN OF MASONRY ARCHES. 423
rib, it is almost axiomatic, that the two conjugateloadareas
shouhl be alike, from symmetry. The proof is as follows : —
Let the quadrant ABS be rigid, with hinges only at its ends
A and S ; then for horizontal component equilibrium, we have
the thrust at the crown Tq equal to the whole of the horizontal
loadarea on cd. Again, let AB be rigid, with hinges at its
ends, and resolve 2\ the thrust along the tangent at B, into
its vertical and horizontal components V and H. Then, for
horizontal component equilibrium we have T^ equal to H,
together with the unshaded part of the horizontalload area. It
follows then that H must be equal to Q, the shaded part of that
area. Also, V is equal to P, the shaded parts of the vertical
loadarea. If 6 be the slope of the rib at B to the horizontal,
then V must bear to H the ratio of sin Q to cos B, that T may
be along the tangent at B, as required by the general condition
of equilibriunr. But it is evident that P and Q are also in this
ratio, as the bases ac and cf are sin i) and cos B, respectively,
when the radius is taken as unity. Hence, with q constant and
equal to p, we have the condition of equilibrium satisfied at each
point B of the quadrant.
For a right circular or other quadrantal linear rib, that is, a
quadrant horizontal at the crown and vertical at the springing,
which has to resist a pair of conjugate loads, one vertical and
the other horizontal, the particular condition of equilibrium is —
That the part of the verticalloadarea from the crown out to any
point, ami the part of the horizontalloadarea from the springing
up to that point, shall hear the same ratio to each other as that of
the sine and cosine of the slope of the rib, at that point, to the
horizon.
On fig. 5 is shown a manner of loading the linear quadrant
ABS of a circle with another pair of conjugate loads which
balance it. The horizontal conjugate load, being the simpler,
may be described first. It is what is called a uniformly varying
load. The horizontalloadarea Gj stands on the base Hj and
is mapped out by the 45"^, or 1 to 1 sloping line drawn from X,
where the bases of the two areas meet. The vertical load con
sists of two parts. The first part is the substance between the
horizontal base DCL and ABS the rib itself. The second part
is a load distributed itniformly along the rib. The rate of loading
along the rib is such that the amount on an arc AB would
fill up the area OAB with the same substance as the first part
is made of. On the arc AB the first part of the vertical load
is a slab ABCD, one foot wide normal to the paper. The
second part might be a uniform ring of voussoirs, pinned to the
424
APPLIED MECHANICS.
[chap. XXIL
arc AB at close intervals by their middle points, as on fig. 3.
If the ring of vonssoirs be a foot wide normal to the paper, and
be made of the same substance as the slab ABCD (fig. 5), it
is clear that the thickness of the roussoirs (t, on fig. 3) must
be half the radius, that their volume might complete the slab
OBCD. Il will now be seen that P, equal to the shaded area
OBCD, is the vertical load from the crown out to B, while the
shaded area EFGH equals Q, the horizontal load from the
springing up to B. The two shaded areas have their parallel
sides equal, each to each, BC = EF, and OD = HG. Their
areas are proportional to the distances between the parallel
sides, respectively. Hence P : Q : : CD : HE : : sin 6 : cos 0,
and the rib is balanced.
Sur/a^e L of C FluU
Fig. 5.
It is to be observed that only the upper part ABCD of the
verticalloadarea truly represents the distribution of the first
part of the vertical load. The area GAB merely gives the
amount of the second part of the vertical load on AB, but in no
■way represents its distribution ; the arrow P should be further
to the left on fig. 5, instead of going through the centre of
buoyancy. Just as in Graphic Statics a line is used in the
forcepolygon to represent a force in the limited degree of
magnitude only, so also is the area OBCD used here.
Superposition of Loads.
It is evident that if a rib be balanced under two systems of
loading separately, it will balance under the systems conjointly.
CHAR XXII.] DESIGN OF MASONRY ARCHES. 425
The whole load will then be represented by a pair of conjugate
loadareas described by geometrically adding or subtracting the
two given pairs of areas. The proof of the balancing of the
circular quadrant ABS for the manner of loading shown on
fig. 5 is quite independent of the height of the horizontal
platform BCL above A, the crown of the rib. Suppose it to be
lowered till J) coincides with A, then AJ, the horizontal platform
in its new position, cuts off a parallelogram JD from the
verticalloadarea. At the same time the new 45° sloping
boundary jn cuts off a parallelogram jG, of exactly the same
area, from the horizontal conjugateloadarea. This is just the
same thing as if we had removed the load shown on fig. 4. So
that equilibrium of a circular quadrant ABS for the load on
fig. 4 follows as a corollary once the equilibrium for the load
fig. 5 has been established. In the same way the load on fig. 4
may be added to that on fig. 5. The rectangle P will sit on the
top of CD, while the rectangle Q may either be placed against
JIB on its right side, or Q may be distorted till its sides
slope at 45° and placed against GF. See Distortingtable
<fig. 17, Ch. IX).
The practical importance of the load shown on fig. 4 is now
evident, and the assumption of weightless struts transmitting
the load to the rib is got rid of. The vertical load F is the
share of the live load, which falls to a slice of a bridge one foot
broad, due to two rows of locomotives covering a thirtyfoot
broad platform. The height of P is ^j ^ 1"5 feet when reduced
to a column of the same material as the superstructure of the
bridge. If iv be weight per cubic foot of the superstructure,
then 2^ = I'ow. A good average value of ic is about one cwt.
The other rectangle Q is the passive additional horizontal reaction
with which the backing muse promptly oppose the extra effort
of the arch to spread when the live load comes upon it.
On the other hand, fig. 5 is a picture of the dead load on the
bridge. The level of rails is DCL, while ABS is the line of
stress up the middle of the voussoirs. These two lines form the
upper and lower boundaries of the great hulk of the dead load.
It is convenient to have names for such boundaries. They are
called the extrados and the intrados of the load. The other
portion of the vertical load, spread uniformly along the line of
stress, is the weight of the half of the arch ring itself lying
below ABS, and the excess iveight of the half lying above ABS,
the material of the ring being always heavier than that of the
superstructure by about 50 per cent. The conjugatearea y^'G^
maps out from the cer// croivn down to the springing, the passive
426 APPLIED MECHANICS. [CHAP. XXII.
horizontal resistance with which the backing must oppose the
tendency of the linear rib to spread due to the dead weight.
As far as the masonry arch is concerned, fig. 5 is, as yet, a very
imperfect picture of the dead load, the load spread along the
rib being out of due proportion. We saw that it would require
the thickness of the archring to be half a radius, whereas the
actual thickness of the keystone is only about onefifteenth of
the radius. It is only for the convenience of being able to draw
the horizontalloadarea with the 45° setsquare, that we
assumed this enormous load along the ring, the intention being
to afterwards remove the greater part of it, making at the same
time the corresponding correction upon the horizontalloadarea.
It is better to remove the whole of the load assumed along the
ring, as a small proportion can readily be restored again to suit
the practical requirement.
The next step will be to find the shape of the horizontal
loadarea for the circular quadrant ABS bearing an uniform load
along the rib of an intensity half its radius. We will calculate
its breadths at ten equidistant points of the platform jH, and
map out its shape with three straight lines or hatters sufficiently
correct for practical purposes. The area is then to be drawn
standing on the left side of jH, so as to subtract it from the
area already drawn. The treblebatter then furnishes, once and
for all, one boundary of the conjugate horizontalloadarea for
the linear quadrant ABS bearing the vertical load between
itself and the straight line DCL. The other boundary of the
horizontalloadarea is the 1 to 1 batter LFG, drawn with the
45° setsquare ; this boundary, and this one only, changes its
position, as you vary the position of the rails DCL to different
heights above the crown, or as you add or take oft' the live
load.
Fluid Load Uniform or Varying Potential.
The pair of conjugate loads shown on fig. 4 can be produced
simultaneously by placing the circular ring horizontally in
water, the water being excluded from its inside. The platforms
now form two pairs of barricades running north to south and
east to west. It is evident the water will attack the barricades
in the manner shown on the figure. The stresses ^; and q are
now both passive. The active stress is the column of water
from the surface down to level at which the ring of the cylinder
lies. The height of this active column is called the potential,
and the load shown on fig. 4 is shortly called a Jiuul load of
uMiforni potential.
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 427
The stress as depicted on tig. 4 is called imiplauar stress,
because the equilibiiuin of bodies subjected to it is in no way
aftected by iheiwtnitial strrss vvliich is iionnal to the paper. The
whole stress at this depth in the fluid is represented by a sphere.
From this it will be seen that the resultant stress on the
back of the rib due to the two conjugate equal loads p and q,
fig. 4, is a normal stress r =^ p = q as indicated on the S. W.
quadrant. For if we remove the four barricades, we know that
is how the water will attack the back of the rib.
The formal proof is given in Chapter II.
In the same way the load on fig. 5 is described as a jiuid
load of varjiing jJotcntial together with a uniform load along
the rib of an intensity such that the load on any arc equals the
weight of the fluid which would fill up the hollow subtended
by the arc at the centre.
Thrust at the Crown of a Rib.
To get the thrust along the rib at any point B of the circular
rib, loaded with the normal load of constant intensity r, as shown
on fig. 4, we have T = V^ + H = ^j^ . ar + q^ . cp = r"W for
!p = q = i\ and the radius R is the hypotenuse of the rightangled
triangle with sides equal to ae and cf. The thrust at any point
is the product of the radius and the intensity of the external
normal stress on the back of the rib at the point. At the crown
A the constant value of the thrust is readily found by inspection
of the figure, for T(, = q.cd = rR. Now this would be the thrust
at the crown, although the rib only extended for a little arc of
the circle on each side of it, and would still be the thrust at
the crown, if the rib, after that arc, continued on, of any form
or shape whatever. At any point of a rib, if we are sure the
external load is normal, and if we know the radius of a circle
which j^^.'? the rib for a little arc there, we can, by taking their
product, get the thrust along the rib. Now the ribs we are
considering are all horizontal at their crown, and suffer only a
vertical load there, so we have this rule : — The thrust T,, at tJie
crown of the rib is the product i^p,^ of the depth of the load at the
ci'oimi and the radius of curvatiirc there.
If we apply the rule for the load shown on fig. 5, and
remember that the part of the verticalloadarea inside the
rib is due to roussoirs of thickness h\ we get H^^ = {DA + \r)r,
where the weight of a cubic foot of the material is taken as
unity. This is exactly the value of the horizontal conjugate
loadarea y^6', which hns to be balanced by Hq, and so the rule
is verified.
428 applied mechanics. [chap. xxii.
Buried Arches, Shafts, and Sewers.
An application to the design of tlie above, chiefly a question
of strength, is given in the examples.
EXAMPLKS,
1. a biiuk shaft or well, inside diameter 6 feet, is sunk vertically in water
which is kept on the outside. h\ a onefoot length, see the east half of fig. 4, at
a depth of S2 feet, t = IS" being made up of four rings of common bricks. Find
the factor of safety against crushing.
JR = 375 ft. and q = 32ic, where w is 62 lbs., the weight of a cubic foot of
water. To = q . dc = 'S2w . R = 7440 lbs., and as this is uniiormly distributed over
To sq. feet of brick, we have then for the intensity of the crushing load 4960 lbs.
per sq. foot. The crushing strength of common brick is only half of what we gave
for strong brick, that is 77,000 lbs. per sq. foot. The factor of safety against
direct crushing is 77,000 4 4960 = 15, which is a prudent value on the lace of so
treacherous a load as an actual fluid load.
2. If the same shaft be loaded with earth spread all round it in hoiizontal
layers, the earth being filled quite loosely, find the factor of safety against crushing
at the same ring.
The potential is doubled, as earth is twice as heavy as water. But the passive
horizontal stresses p = q are no longer equal to the potential, but are now only
onetliird of it. Hence p = q is now twothirds of what it was before. Hence
the apparent factor of safely, 50 per cent, greater, is 23. Distortion may halve it.
3. If the earth be rammed in 9" layers as it is filled outside the shaft, or punned
as it is called, find the factor of safety now.
The punning can only make the passive horizontal conjugate stres>es p = q,
fig. 4, equal to the potential, so the earth may now be assumed to be pressing like^
a fluid, but like an imaginary fluid twice as heavy as water. Hence the factor of
safety is now of half the value in Ex. 1, namely 8.
Notwithstanding the fact that tlie factor of safety in this example is only 8,
while in Ex. 2 it is 23, yet the absolute element of safety is just as satisfactory,
for the punned earth is a steady load.
Practically, then, the answers to Exs. 1, 2, and 3, are 15, 11, and 8.
4. If the shaft be bored and lined, discuss the factor of safety.
If the shaft be bored in old consolidated earth, and then be lined with brick,
the three rings would suflice at the depth of 32 feet.
At first, the load on the lining is nominal, but increases gradually as cracks run
out into the earth, and suddenly when slips occur. In this way, the potential comes
into play, and crowds the earth in behind the shell. At depths beyond 32 feet,
the cracks due to this small bore hole would not run out sufficiently far to bring
the whole potential into play, so that the three rings of brick might practically
suffice down to 64 feet, double the depth. The mere arithmetical calculation
would infer that the factor of safety was 4, the half of 8, though, in all probability,
it is still 8. Thus it is that with buried arches, the factor of safety may seem to be
as low as 4 : see Rankine's Civil Enyiuccring, p. 437.
5. The upper right quadrant of fig. 5 shows, to a scale of 10 feet to one inch,
lialf of a semicircular arch ring, 14 feet in span. The ring is 2 feet thick, and
made of rou.ssoirs which are 2^ times as heavy as water. Show that the line of
stress A.U up the centre of the vnussoirs is balanced under a water load uj) to any
level BK, together with the weight of the voussoirs themselves.
CHAl'. XXII.] DESIGN OF MASONRY ARCHES. 429
If «e suppose the water to reach from DK down to AM, the excess weight of
the upper half of the ring is I J times tliat of water. That is, u share of the weight
of that upper half of the ring is already reckoned ; it is, as it were, Inioyed up.
The lower half of the ring is, in weight, 2^ times that of water. That is the excess
weight of the 2 foot thick ring, as a whole, is twice that of water. That is the
same as if the roiissoirs were 4 feet thick, and of the same weight as water. But
4 feet is half the radius of AM, and tliis is exactly the required load along the
circular rib which is required in conjunction with llie fluid load of varying potential
to give absolute equilibrium. Such an archring requires slight spandrils to balance
it where the centre is struck and before the water load is tilled in.
6. A culvert siphon or sewer, of 7 feet inside diameter, with two circled
rings of brick, is laid horizontally along the bottom of a sheet of water. Show that
the equilibrium is perfect when the inside is empty, if the bricks be '1\ times as
dense as water.
The section is shown on fig. 5, to a scale of five feet to an incli, >' = 4 and <= 1,
both in feet. The problem is that already discussed in Ex. 5, only with the
dimensions halved and the circle completed. The equilibrium, as a whole, must
be considered. If the cylinder be one foot long, and if iv be the weiglit of a cubic
foot of water, the weight of the empty ring is '2irrl x 'I'bw, and its buoyancy is
ir(j' + ^tyic, which are in the ratio 20 to 20^, so that the empty cylinder would
float upwards. Increasing the thickness of the ring by a couple of inches the
proportion would now be 23 to 21, and the cylinder would sit on its bed, or if the
brickwork were a trifle more than twice and a half the density of water, the same
end would be accomplished.
7. If the axis of the cylinder be at a depth of 20 feet, find the thrust at the
highest and lowest points A and a, fig. 5, for a ring of the cylinder one foot
long.
Eo =jG = ^^LH  ^Lf = 1 (20=  IS^) iv = 72k = 72 x 62 lbs. = 4464 lbs.,
Si = GN= ZiV^  ^LH = h{2i  202) j^ = gSjt; = 88 x 62 lbs. = 5456 lbs.,
and this is spread on one square foot, so that, for ordinary brick of crushing strength
77,000 lbs. per sq. foot, the factor of safety is 14.
8. An empty circular cylinder, 7 feet inside diameter, with a shell 12 to 13 inches
thick of brickwork 2i times as dense as water, will, if completely submerged,
remain at rest in neutral buoyancy at any depth and in any position. If the axis
he vertiial, the surface of stress is the circuLir cylinder up the middle of the brick
work (fig. 4). But what is most remarkable is that this is still the surface of stress
if the axis be horizontal (fig. 5). It follows, by induction, that it is the surface
of stress for all positions, so that there is no tendency for the tube to collapse under
the external water pressure, though it float in any position whatever, completely
submerged.
9 If the cylinder be made of a metal very much denser than water, it would
be a thin shell, and there would be no necessity to distinguish between the inside
and outside diameters. Thus a platinum circular shell need only have a thickness
onetwenty second part of half its radius, so as to displace, when empty, its own
weight of water. Such a shell might be S8 inches in diameter, and 01 inches
thick. Let it be submerged with its axis horizontal, the water being kept out by
face plates at the ends of the same density themselves as water. There is no
tendency for the cylinder to collapse even if it were made in staves like a barrel.
The staves might be hinjied together, and the whole covered witii a thin coat of
watertiffht material, the hinges to have a little stifiness to give slight stability to
the equilibrium.
10. If this hinged cylinder were split into two semicircular troughs, one of
them would float with the water up to its lip, and remain semicircular in form in
430 APPLIED MECHANICS. [CHAP. XXII.
perfect equilibrium if only the hinges have a little stiffness to give stability. In
the same way two vertical plates, submerged in and e.xcluding water from between
them, might be arched over and inverted under with the two semicircular troughs.*
The practical importance of Exs. 3, 4, and 5, where the
shell by its very thickness in proportion to its radius, and by
the very density of the brickwork, coirccts the tendency of the
extern fluidload to distort its circular form, will appear when
the conditions in which culverts, sewers, and inverts are usually
built are considered. The circular form recommends itself in
situations where great strength is required; the simplicity
of this form, too, lends itself to good workmanship. The
surrounding load is due to earth, loose or more or less consoli
dated and in a transitional unsettled state. With the gathering
of water sapping the earth close to the brickwork, the external
load becomes more and more like a fluid load ; and it is well that
the very increased thickness of shell required in such exposed
situations, to give extra strength, should at the same time render
the equilibrium more secure.
Stekeostatic Eib.
This is the name by which Eankine calls the most general
€ase of a rib balanced by two conjugate loads : see his Applied
Mechanics, p. 198. Three quantities are involved, the shape of
the rib and the shapes of the two component conjugateload
areas equivalent to the actual distribution of the uniplanar load ;
these two may be either oblique or rectangular. The theorem
is that it is sufticient and necessary to give, either implicitly or
explicitly, two of these shapes to find the third. In most cases
the solution ends in integrals which can only be approximated
to very roughly and with great labour.
We are confiuing ourselves to the particular case where the
ribs are complete, that is they are horizontal at the crown, and
vertical at the springings. The conjugate loads are vertical and
horizontal, and the two quadrants are symmetrical. For one
quadrant the loadareas stand on flnite bases, the base or platform
* This theorem of the equilibrium of the thin horizontal empty circular cylinder,
displacing its own weight of fluid, was given by the authors in a letter to Nature,
of 18th February, 1897. From some private correspondence with scientific men
interested in it, it appears as if the theorem were new to hydrostatics. FVom
criticism of the method of conjugatcloadareas employed, it seems as if this
elegant method, especially lending itself to graphical construction, were little
known or understood. The theorem can be proved by the strict but laborious
methods of the integral calculus. See letter on I'olygonal Shells, Nature,
Ist May, 1902.
I
CHAP. XXII.J DESIGN OF MASONKY ARCHES. 431
of the verticalloadarea being the half span of the rib and the
rise of the rib being the base or platform of the conjugate
horizontalloadarea. We ha\e already solved two important
examples (figs. 4 and 5) by the geometry of the areas. One
was the direct problem : given the shapes of the rib and the
verticalloadarea, to find the shape of the horizontalloadarea.
In the other, the shape of the rib is given and the shape of
the horizontalloadarea assumed, and a suitable vertical load
built up in two portions : tliis is the inverse problem. We
established a rule for building up these areas, namely, that the
shaded portions of the areas should be in the same projiortion
as the sine and cosine of the slope of the rib.
Given the rectangular equation to the curve, with the axis
of X horizontal, and that of Y vertical. Given also the vertical
loadarea so that P is known, and how it changes as we shift
from point to point. Let q be the unknown breadth of the
horizontalloadarea at the level of B. Let B shift out a little
and the decrement of Q is the small parallelogram dQ = q . d//.
Also Q: P\ : cos : sin or $ = P cot d. So that
horizontalloadarea for semicircular elb, loaded
Uniformly along the Eib.
A quadrant ACB is shown on fig. 6. It is convenient to
take the intensity of the uniform load along the rib as half
a radius. We may suppose it to consist of a ring of material
of unit density, with a uniform thickness ^r, so that the area of
the part of the ring loading any arc AC is, equal to the area
OAC, subtended by the arc of the rib at its centre. By the rule
for the thrust at crown, we have H^, = ^r . r = \r. This must
be the area of the total horizontalloadarea ahM, standing on
the platform ad = r. On fig. 4, if the intensity of the load
spread on the horizontal platform were p = ^R, then q = ^R, and
the total horizontalloadarea would be ^R', the same in each
case. In this new case, however, the boundary hhk no longer
maps out a parallelogram.
Suppose, at first, the rib is balanced under the uniform load
along the rib alone. Its shape, the catenary, would be peaked
at the crown A. If it were deliberately pulled into a semicircle,
each pair of points on one level, except the springing pair, have
been pulled further apart, giving the load an advantage to
spread the rib. The horizontal load must, at each level, 2>?'ess
432 APPLIED MECHANICS. [CIIAP. XXII.
imvards, to prevent the spreading, most violently at the springing
level, and decreasing gradually till it is least at the crown level.
To find/y, the breadth of the horizontalloadarea at the level
of any point C on the rib, we have for the verticalloadarea
from the crown A outwards to 0, a ring of breadth h', and
whose length is s = AC the arc. With i for the slope of the
curve at C.
p r r . r .
r"^ d r
P cot i = X i cot i ;  (P cot i) =  (cot i  i cosec i).
^ Ct% d
,, . di/ . . , di  1
y = jd = r cos i; j.=  r sin i ; and
di dij r sin i
So that irrespective of sign the value of /r is
d ,^ . d ,^ . di
q = (P cot ^) =  (^ cot ^) —
df/ di di/
72 1
=  (cot %i cosec* %)
2 r sin i
sin i cos ii
= r
2 sin^ i
This can be modified into a form more convenient for calcu
lation, as it is engraved on fig, 6,
. 2i  sin 2i ,.,,
« > ' 3sin»Bin3» ''■ ^^^
To find q^ = ha = ^r, the breadth of the area at the level of
the crown A, we must substitute i = 0, when the coettieient of r
takes the indeterminate form of the ratio of zero to zero. The
numerator and denominator are to be difierentiated separately
and i = substituted, which has to be repeated three times when
the ratio becomes determinate. 1^'or i = 0, we have
21  sin 2i 22 cos 2i 4 sin 2i
3 sin i  sin 3i 3 cos i  S cos '3i  3 sin i + 9 sin Si
_ 8 cos '«^* _ 1
 3 cos i + 27 cos '3i 3'
or tliis may be found by substituting the series for the sines.
CHAP. XXII.] DESIGN OF MASONRY ARCHES.
433
y
1 v
i>N
>
<s
• 5^
%
1
>C5» '.
•iNsr'
~JS
v<
^ 
^ K vi y= p$ Ni ^
'1 F
434 APPLIED MECHAJNICS. [CHAP. XXII.
The value of a^ = kd = r, the breadth of the area at the
4
TT
springing level B, is found by substituting i = ^. Since kd acts
normal at B, and is the total external load on the back of the
rib there, by the rule, the vertical thrust along the rib at B
must equal the product of kd and the radius. But the thrust at
B is the area of the quadrant of the ringarea whose thickness
TT
is hr, that is, an area r x r, and from this the xaXwe of kd is
found by dividing out the radius.
The breadths of the horizontalloadarea at ten equidistant
points of ad are found by substituting in the expression for/v,
ten values of the angle i, whose cosines are 0, 1, 2, 3, 4, 5, 6, 7, 8,
and 9 ninths of r the radius, respectively. They are marked
on fig. 6.
The exact boundary is a gentle curve, but a second boundary
maps out the area with three straight lines hg, gl, and Ik. The
first extends downwards for 4 ninths of the radius, giving a
breadth of 3 and 4 ninths of the radius at h and g ; the second
extends downwards for 4 ninths of the radius, gi^'ing a breadth
of 6 ninths at /, and the last extends downwards the remaining
ninth of the radius, and gives the final breadth of 7 ninths at
k. This approximate boundary begins with the proper breadth
onethird of the radius at the top, and encloses the exact area
;'^, which may be found by counting the number of squares on
the paper, squareruled with the sides a ninth of the radius. Thus
hg encloses 14, gl encloses 20, and Ik encloses 6h squares, in
all 40 1 squares ; but each square is ?•■ ^ 81, so the whole area
included is ?■^ It is called shortly the trchlehaffcrhoundarg
hglk, and may be seen clearly near the centre of fig. 6, drawn
in dotted lines with the hatters written upon them. Beginning
at h on the crown level, bg is drawn with a hatter of 1 in If, for
the first 4 ninths, then gl with a hatter of 1 in 2 for the next
4 ninths, and lastly, Ik is drawn with a hatter of 1 in 1 for the
remaining ninth of the radius.
Horizontalloadarea for Semicircular Eib. loaded with
THE Area between Itself and a Horizontal Straight
Line over it.
Let 4 Ci? be the circular quadrant, and D^ the horizontal
extrados (fig. 6). Suppose, for a little, that besides the shaded
CHAP. XXII.] DESIGN OF MASONRY AKCHES. 435
verticalloadarea bounded thus, there were also, as on fig. 4, a
tinifonn load along the rib of intensity half a radius. Then
the 45^ slope, or 1 to 1 batter no, gives the horizontalloadarea
for the jointloads. Now remove, on fig. 6, the load along the
rib, and at the same time remove the corresponding horizontal
loadarea ahfilkd, it is to be drawn on the same base ad, and to
be supposed to gravitate horizontally outwards from the rib,
while the area already drawn gravitates horizontally imoards
towards the rib. Hence, for the given verticalloadarea between
the rib and the straight extrados, the horizontalloadarea lies
between the 1 to 1 batter uo, and the curve approximated to by
the treble batter hglk. In form it is like a figure 8, as these
two boundaries generally cross each other at a pointy. The
part below p gravitates horizontally invmrds, and is to be
reckoned as positive', the part above p gravitates horizontally
oidv:ards, and is negative. They are to have the same density as
the given verticalloadarea.
Allourincefor the JExccssiveiyht of a Masonryriiuj of Uniform
thickness. — It is Eankine's practice in the masonry arch to
assume as a first approximation, that the line of stress is along
the soffit. In the meantime we follow this practice in fig. 6,
so that the quadrant of the circle is at once the soffit of the
masonryring, the intrados of the verticalloadarea, and the
linearrib or line of stress. It will be seen that the weight of
the masonryring has been included in the verticalload area,
just as if it were of the same density la as the superstructure.
It is usually fifty per cent, denser, and its uniform thickness,
see t^ = jir, on the figure, may be taken as a fifteenth of the
radius of the soffit, and although its area lies wholly above the
.soffit it may be assumed to act along it in the meantime. We
have then to consider an additional vertical load uniformly
spread along the rib, and mapped out by two concentric circles
at a distance apart ^r, and of density hvj. But this is the
same as if the two concentric circles were at a distance apart
■onefifteenth of a halfradius, and the area of the full density w.
This is exactly onefifteenth of the load which for convenience
we assumed along the rib, and the removal of which introduced
the treblebatter hglk. It is only necessary, then, to restore a
fifteenth part of the area ahglM by pricking back bh' = j^ha and
kk' = yi^kd, and similarly at g and I, and drawing the dotted
treblebatter instead. Or, shortly, with the line of stress assumed
to be along the soffit, an allowance is to be made for a masonry
ring of an excess density of 50 per cent., and of a uniform
thickness a fifteenth of the radius by receding the treblebatter
boundarg onefifteenth.
2f 2
436 APPLIED MECHANICS. [CHAP. XXII.
Allowance for the Excess vciijht of a Masonnjring, Thickeniiiff
outivards from Keystone to Skevjhcichs. — In the best French and
English practice, the skewbacks of the segmental circular arch
are from oiice and a half to twice as thick as the keystone. It
depends in part on the size of the segment, and it will answer
all purposes to suppose the masonryring included between two
slightly excentric circles, with their centres on one vertical line,
with their crowns at a distance apart t^^ and the pair of points
60° out from the crown It^ apart nearly. They are shown
on fig. 6, by the softit and the upper dotted circle. The lower
dotted circle is concentric with the soffit, and we have already
considered the excessweight of that part. Consider now the
two dotted circles. Their crowns toucli, and if completed, their
greatest distance apart is at their lowest points, and equals the
difference of their diameters. At the points left of the centre,
their distance apart is half as much, being nearly the difference
of the radii, slightly more or less, according as we measure
along a horizontal line through the one or other centre. It will
be seen, then, that the distance apart of the two dotted circles
on fig. 6, measured along the radius through 0, at any angle i
out from the crown is given closely by z = 2tQ (1  cos i), as
at the three cardinal points for ■i = 0, 90°, and 180^ we have
2=0, 2^0' ^"^ '^'^0' t^® proportions indicated, while with i = 60°
we have 2: = ^^ as required.
An element of the area between these dotted circles \b z .ds
or zrdi, and the verticalloadarea from the crown out to i is the
definite integral of this between the limits and i. Hence
P = r\zdi = 2r/'„J(l  cos i)di
= 2rt^ [i  sin t'j = 2^„ [ri  r sin ?']
= 2^, [s  .r]. (4>
Interpreting the two terms separately : — The first is a uni
form load along the arc equivalent to the area l)etween two
concentric circles 2t„ apart; the second term is a negative
load uniform along the span. That is a parallelogram of height
2^0 to be tdhii 0//' the span. But since the excess weight is only
^10, each of the uniform loads is to be an area of unifttrm
breadth t^^, and of the normal density v, the one added along
the rib and the other taken oil" the span. The corresponding
conjugateloads we already know.
On fig. 6, suppose the load only to reach up to the hori
zontal thrDUgh the crown A, that is, the pair of parallelograms
•CHAP. XXII.] DESIGN OK MASONRY AllCIIES. 437
AH and ((0 simultaneously removed. The conju,i!;ate horizontal
loadarea is drawn inside the arch in dotted lines. One
boundary is the dotted 45° line AgB, and the other the dotted
treblebatter boundary h(/lJc. This last is to recede fV^lis of the
area hO, by pricking back hh' = fjfiA and kk' = ^kO, and
similarly at g and I. Onefifteenth part is due to the uniform
part of the ring, and the other twofifteenths to the first term
of the spreadout part. At the same time the boundary AB is
to recede p^th of a radius into the position an, due to the
second term. Lastly, a'o is to be added on when we restore
AE. The conjugate horizontalloadarea as modified for the
thickeningout ring is shown shaded, Q^ is the portion of
it pushing in below ^>", while ^2 is the part pulling out
above ;p".
Note that at the point r/, where the two dotted boundaries
cross, the effect of the weight of the thickeningout part of the
archring makes both boundaries move to the right the same
distance. For the one moves a J^th part of r, and the other
fjths of the breadth of the figure hO, which at g is almost half
of r. Hence j/ and p" are practically at one level, whether the
ring be uniform or thicken outwards.
Note, however, that the assumption that the area between
the dotted circles acts along the soffit makes the approximation
much coarser than in the case of the uniform ring.
liANKiNE's Point and Joint of Ruptuke.
We are now in a position to define Rankine's point of
rupture, and to exhibit it to the eye by our graphical con
struction on fig. 6. The point j/ or ;/' projected horizontally
on to the circle gives P the point of rupture, that is, the point
■on the rib where the conjugate horizontalload changes sign.
Below P there is required a thrust Qi on the back of the arch,
but above P an outward pull Q,. The angle of rupture is the
slope which the tangent to the rib at P makes with the horizon,
which in the circle is the same as AOP. For the shaded load,
fig. 6, between the circle itself and an extrados ^r over the
•crown, together with a slight additional load due to the excess
density of the voussoirs, the shaded conjugate horizontalload
area may be readily drawn on a large scale on squareruled
paper, taking nine sides of the square to represent the radius.
P is to be projected from p", and the angle of rupture measured
Avith a protractor, when it will be found to be AOP = 4;')" 15'.
The joint at P, between two voussoirs there, is the joint of
488 APPLIED MECHANICS. [CHAP. XXII.
rupture, and the point where this joint meets the back of the
masonryring gives Eankine's level of lieavy backing.
In the simpler case, with no excess load along the linear rib,
p determines the point of rupture : see fig. 6. This is where the
45° line no meets the treblebatter hjlk, or more strictly the gentle
curve to which that treblebatter closely approximates.
It is instructive to observe how y behaves as the extrados
moves lower or higher over the crown. First let the extrados
touch the crown A, then the 45° line an meets the treblebatter
exactly at the joint //, the positive and negative parts of the
conjugate horizontalloadarea are glkn and gob, respectively.
They each contain exactly six squares or
cjnk=gah = ^i^^=()Ur
Their algebraic sum is zero, and JZ, the thrust at the crown is
zero. As the extrados cannot come lower down, g projected over
to G gives the lowest possible position of the joint of rupture,
and putting % for the corresponding biggest value of the angle
of rupture we have the cosine of i, given by the ratio of the
height of g above dk to the radius ; but the height of g is five
sides of the square, that is, r, so that cos % = '5 and % = 56° 15'
As the extrados DiJu is placed higher, ? moves up and P
approaches the crown and % is decreasing. It will be seen, by
inspection of fig. 6, that the positive part of the conjugate
load area is increasing and the negative part decreasing. Their
algebraic sum expresses H„ the thrust at crown, the unit being
the weight of a cubic foot of the superstructure. That sum must
be equal to the product of the height of the load over the crown,
and the radius there. When the extrados reaches a height one
third of r above the crown A, the conjugateloadarea bglks is
wholly positive, the point of rupture has reached the crown,
Zo is zero, and counting the squares
/rV r
Us ^ „. ^ 27 [^)  .y
The area remains wholly positive for all higher positions of the
extrados.
Since the triangle peg has one side sloping at 1 in 1, and the
other at 1 in 4, it follows that the height of the triangle is ^rds
of its base eg, that is irds of AD, tlie height of the load over the
crown of the rib. It is convenient to express the height of the
load over the crown as a fraction of the radius ; if .s is this
important ratio, then <g = AD = sr.
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 439
For the height of i) the point of rupture, we have the sum
of the heights of fi above OB, and of p above <j. It is
r cos t = ^/' + ^$r,
and cos i =  + ^s (5)
is a close approximation to the angle of rupture.
For the positive part of the conjugate horizontalloadarea,
we have
pok = (jnh + (JO + peg
= '•= (tr*¥ + ¥')■ (6)
The negative portion equals in area
fjah  ea + peg or ?•• (^^  f « + f «'), (7)
giving as their algebraic sum sr, which is the product of the
height of the load over the crown and the radius there.
Semicircular jVIasonry Arch.
In the application to the semicircular arch, the left quadrant
of which is shown on fig, 7, the assumed line of stress is the
quadrant cm with the centre at d, and having a radius of 48 feet.
It is taken up the middle of the masonryring, which has a
thickness AD = to = S feet at the keystone, and a thickness
twice as great, ^o = 6 feet at the joint near C which is 60^ out
from the crown a. So far the ring is mapped out by two
excentric circles as already explained : one is the soffit AECB
drawn from the centre S with radius SA = 43'5 feet, and the
other from the centre c with radius eD = 52"5 feet. Beyond the
joint C the back of the ring is stepped to receive the heavy
backing, and the ring ceases to widen out any further.
From a the middle of the keystone ah is laid oft' horizontally
and equal to onethird of 48 the mean radius ; then hg is drawn
at the batter 1 in 4 till ^r is at a level ths of 48 lower than h ;
next gl is drawn at the batter 1 in 2 till I is ths of 48 lower in
level than g, when U: is drawn at the batter 1 in 1, Also agn
is drawn through a the crown of the assumed line of stress at
the batter 1 in 1. These dotted lines are readily drawn on
squaredpaper, making da nine sides ; they include the conjugate
440 APPLIED MECHANICS. [CHAP. XXII.
horizontalloadarea, neglecting altogether the weight of the
half of the ring below the assumed line of stress, and the excess
weight of the other half over the average weight of the super
structure, which is v: lbs. per cubic foot. The load between a
horizontal through the crown a and the formation is in the
meantime also neglected. Now the archring is to be built of
masonry half again as heavy as the superstructure, so that the
excess density of the lower half is ^v:, and of the upper half it
is he, or conjointly the excess weight of the archring is w, and
as this is twice as great as it was in fig. 6, the boundaries must
now recede twice as much. So that, on fig. 7, the treble
batterboundary hfjlk recedes pths of the area into the position
h'k', while the 45' boundary an recedes y^ths of 48 when there
is added to it a parallelogram, by drawing uo. equal to the
parallelogram between the horizontal through the crown a
and the formation through 0, which is at a height aO = 8 feet
above it. Xote ^ is still sensibly JU.
The area of ph'o, the positive part, is readily found by
measurement or by counting the squares, or it may lie calcu
lated from the manner in which it was constructed. It is
almost 400 square feet, so that Qx = 400?r is the inward thrust
with which the' solid backing must resist the tendency of
the arch to spread at the haunches. For this purpose the
squaredressed heavy backing must stretch out at the springing
joint B, a distance z = 12 feet. For the weight pressing the
backing down on its base is then 12 x 48/r; and taking the
coefficient of friction of masonry on masonry at "7, the f rictional
stability of the backing at the joint ^ is i^ = y% x 12 x 48?r
= 400/f nearly. The level of this heavy backing might be up
to the pointy?. Above this level tlie conjugate horizontalload
required is Q^ = 112?'' outward, and if it were practicable to
apply this in any way, for instance, by the centering which
might still be inside the arch, then the line of stress would be
the circle of radius 48 feet as assumed, and the thrust at the
crown ffo = Qi  Q. = 288«\ This is exactly the value of H^,
got by the rule for the thrust at tlie crown of the circular
linear rib of 48 feet radius with the normal load at a oi AO
made up of the two parts IJO = l5w and AD = 3 x 4"', in all
equivalent to Qw, so that ^, = 48 x 6?/' = 288 /r.
In practice the negative part Qi of the horizontalloadarea
has to be left out, and for a masonryring of uniform thickness
the heavy backing may come up to the level of p, as recom
■ffSi'
442 APPLIED MECHANICS. [CHAP. XXII.
mended by Eankine. He gives the joint at 45^, as being
always on the safe side. With the archring as in fig. 7, we
need only bring the backing up to the level of g, as the very
thickening of the ring outwards is sufhcient backing down to
that point. It will be seen that we are leaving out a part of
the positive area from p down to g, as well as the negative
part Qi. This positive part, shown shaded across from p down
to g, is not completely left out ; it is only applied lower down, so
that the whole inward thrust §, = 400^:' of the squaredressed
heavy backing is applied to the back of the arch between the
joints B and C. A portion of the negative area from p upwards,
20« is also shaded across : equal in area to the shaded part
below p. In this way we determine fico joints of rupture E and
C, one on each side of Eankine's joint. And just as the average
position of his joint is 45°, so the average positions of our pair
are 30° and 60° out from the crown.
The leaving out of the horizontal load above C requires the
line of stress from that joint to the same joint on the other side
of the crown to be modified. For want of the outward pull Qi,
the arch tends to shrink horizontally and the crown to go up.
This brings the centre of stress at the crown joint down from
the bisecting point to the lower trisecting point. As the thrust
^0 at the crown is now greatly increased, we must divide 400?'/
by AB + BO or %w, when we have the raflius of curvature of
the modified line of stress at its crown, yo,, = 67 feet. A circle
swept out with this radius, having its centre on the vertical
through the crown, and beginning at the lower trisecting point
of the crown joint A, crosses the assumed line of stress, and
approaches the upper boundary of the middle third of tlie arch
ring. Consider now the thrust at E, the upper end of the
portion of the archring CE. At E, its upper end, a thrust a
little greater than 400?^ has shifted from the bisecting point to
nearly the upper trisecting point of the joint, a distance nearly
half a foot. This is the same as applying at E, the upper
end of the block CE, a lefthanded couple 400?^? x ^ = 2007r,
and takes the place of the couple 20u' x 9 feet, constituted
by the shadedacross parts of the horizontalloadarea, above
and below p, which ought to act on the back of the block
CE, but which have been left out as far as that block is
concerned.
Three points on the left half of the modified line of stress,
fig. 7, are the lower trisecting point of the crovn joint A, the
upper trisecting point of the first joint of rupture E, and a point
slightly above the middle of the second joint of rupture C. This
CHAP. XXII.] DESIGN OF MASONKV ARCHES. 443
is to be verified as follows : — Consider the structure rigid from
A to C, and examine whether the three forces acting on it meet
at a point. Namely, the weight of the structure from C to
680h', acting vertically downwards through the centre of gravity,
the horizontal force i/o = 400iy acting horizontally through the
centre of the stress at the crown joint assumed to be its lower
trisecting point and the oblique thrust at the centre of stress of
the joint assumed to be sensibly at its middle point. Taking
moments about this last point, we have G80«' x 13 feet and
400?/' X 22 feet sensibly equal. It is only necessary then to show
that the whole weight from A down to G is 6807/', and the centre
of gravity 15"4 feet measured horizontally from the right vertical
boundary. The weights and leverages of four parts into which
the area AC can be divided are printed on the right half of
fig. 9.
Again, consider the structure rigid from A to U when the
three forces must again riieet at a point : namely, 287?/' vertical
through the centre of gravity, 40026' horizontal through the
lower trisecting point of the joint A, and an oblique force
parallel to the tangent of the line of stress acting through the
upper trisecting point of F. Taking moments about this last
point, we have AOOiv x 7*5 ft. and 287'?/^ x 105 ft. sensibly equal.
The areas and levers of the various regular figures into
which the area under consideration can be divided are printed
on the lower part of the right half of fig. 7.
In this way we have indirectly designed a Segmental Arch,
the half span of soffit being 37*5 feet, the rise of the soffit 21*5
feet, the formation level being 45 feet above the crown of
soffit, or otherwise the total rise of the formation above the
springing level is 26 feet. (See the right half of fig. 7.) We
know five points on the line of stress : the lower trisecting point
of the crown joint A, the upper trisecting points of the first
pair of joints of rupture U and U at about 30° out from the
crown one on each side, and a pair of points slightly above the
central points of the two springing joints C and C. These five
points being just comfortably located in the "Kernel," con
sisting of the middle third of the archring, we will prove
that the line of stress lies wholly inside that ■' Kernel."
It is not sufficient, as Eankine does in his C. E., p. 421, to
ensure the centres of stress to be in the middle third of crown
joint and his (45") joint of rupture, as there is a point of
maximum curvature on line of stress near 30°, of which he
was unaware.
Light elastic rubble spandiils are shown riding out to a
444 APPLIED MECHANICS. [CHAP. XXII.
stretch of 5 = 7 feet on the heavy backing or abutments, but
they only exert a liorizontal reaction when the live load conies
on the opposite half of the arch. They can offer a horizontal
resistance 7 x 23?6' x 7 = 113h\ Xow the heicrht of superstruc
ture representing the live load is I'o?/; this when multiplied
by /oo or 67 feet, gives 100 w, the excess of the horizontal thrust
•of the loaded half over the unloaded half.
Equilibrium Eib or Tkansformed Catenary.
It will be seen on fig. 7 that the modified line of stress in
the segmental archring CC bears only a vertical load, the bulk
of which is given by the area included between itself and the
formation level. The remainder of the load is spread in a very
similar manner because of the thickening of the archring from
the crown outwards.
The curve which is a balanced rib for the vertical loadarea
alone included between itself and a horizontal straight extrados
is known as the Equilibrium Curve. Let AB, fig. 8, be a finite
arc of the curve balanced under the verticalloadarea NOAB,
so that ^and T meet at the same point .^ on W, drawn vertically
through the centre of gravity of that area.
H= T cos e, W = T sin ^, W = H tan Q. Expressing tbe
area by an integral, and the tangent of the slope at B by a
differential coefficient
w f ydx = !{—,
^ dx
and differentiating each side
wy = H /^, or // =  [,
d.ar 1IJ d.r
TT
or with 7>i^ as a short name f(n — , the differential equation to
>r
the equilibrium curve is
'/ = »' % (8)
dx
There are only two functions of x which differ from their
original values by a constant m^, when differentiated twice
successively, so that their sum is the general solution of this
equation. It is
r r
y = Ai"' + Be''^. (9)
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 445
When ^ = 0, y = y,, = OA, giving j/,, = A + B.
By differentiating (9), we have
n (^y 1  
tan e =  =  {Ai'"  Bi "•). (10)
Now when x = 0, tan ^ = 0, giving ^ = A  B, so that
A = B = 1^05 ^^^
m  —
y = r ^ (e'" + e '"),)
where ?/o = OA is put equal to rm.
If '/• = 1, the curve is the common catenary. It is shown
inverted by a thick line on fig. 8. Any other of the curves
can be described by dividing the ordinates of the catenary in a
constant ratio r. The catenary might be drawn on a sheet of
indiarubber fixed along its edge OX, and previously stretched
in the direction OY only. If the sheet were now to contract
in that direction only, the catenary would become in succession
the Transformed Catenaries AB, ac, hd, &e. In this way a whole
family of transformed catenaries is derived from the common
catenary by assigning a succession of graduated values to r, the
ratio of transformation. There is only one such family, for the
catenary itself, like the circle, admits of no variety ; you may
make ni larger or smaller, but you only draw the catenary to a
coarser or finer scale. It is therefore convenient in tabulating
numerical values of the different dimensions of the family of
transformed catenaries to assign to m the value unity, and to r
a series of fractional values graduated at close intervals.
If the curve AB, with its tangents Ag and B(j, be trans
formed towards OX by the shrinking of the stretched sheet of
indiarubber, it is e^'ident that g travels on a vertical locus. It
follows that the centres of gravity of the areas from ON down
to hd, to ac, to AB, &c., all lie on one vertical, and the same is
true of the areas included between any pair of the curves.
Hence any one of the curves is a balanced rib for the vertical
loadarea alone between any other pair.
As a matter of fact, hd is the modified line of stress,
in the segmental arch, iig. 7. Its soffit is a circle coiiicident
446
APPLIED ■MECHANICS.
[chap. XXII.
nearly with ac, while ONX is ihe formation level. The portion
of extra density lies between another pair, of which ac is one ;
the other being a curve a little higher than hd, but not shown
on fig. 8, Of course for this comparison between the figs. 7
and 8, the scale of fig. 8 must be such that Oa shall measure
absolutely 4*5 feet. We then have Oh = 35 feet = ,?/.,, while
py is 67 feet, so that for the curve hd we have the ratio
2/o : Po = '05 nearly.
ThcWireefrJX .
Mo of f/ie
^raUsforfV(d(07i,
Ik
Now, we only used iir as a convenient abbreviation of
H^ f w, that is the thrust at the crown of the equilibrium
curve fexpressed in square units of the verticalloadarea, so
that ^0 = ^^^y ^^^t from the general rule for the thrust at the
crown of a rib we have H^ = ■2t'//„p,„ so that p^ = ??r ^ 2/o = ^" ^ '".
but ?/,, = rm, so that
= ?•' = N.
(12)
That is — The important ratio .9, namely, that of the depth
CHAP. XXII.] DESIGN OF MASONRY AKCIIKS. 447
of the load at the crown to the radius of curvature there, is
in the transformed catenary, exactly the square of the ratio of
transformation. On fig. 8, AB is the transformed catenary
for which s = \. All the members of the family of curves
below this are like the catenary or parabola in this, that their
curvature is sharpest at the vertex. They are of little
importance in their application to arches.
On the other hand, all the curves for which s is less than
onethird, such as ac, bd, differ in form completely from that of
the catenary or parabola. They have a flat curvature at the
vertex a or b. The curvature becomes sharper and sliarper as
you travel out from the crown on both sides, till it is sharpest
of all at a pair of points on each curve, one on each side of the
vertex, one of the pair being shown with a little ring on fig. 8.
From this, as you go outward, the curvature begins to flatten
till at d or c it is again exactly the same as it was at the crown.
This finite part of the curve, having the same curvature at its
middle point and at its two ends, must now engage our attention.
Beyond d or c the curves flatten indefinitely outwards.
Observe that the value s = J, which divides the transformed
catenaries into the two sets, those sharpest at the crown and
those flattest at the crown, corresponds exactly to the same
value of s in fig. 6. On that figure s ^ J makes the conjugate
horizontalloadarea wholly positive. So that in an example
like fig. 7, but with the formation level above the crown of
the assumed line of stress by a third or more of its radius, the
horizontalloadarea would be wholly positive or inwards. If
then an upper part were left out, that would decrease the thrust
at the crown, so that the modified line of stress would sharpen
at the crown instead of flattening. In bridge practice the
formation would never be at such great heights.
The TwoNosed Catenapjes.
We have ventured to use this name for the important part
of the family of transformed catenaries having their ratio of
transformation less than the square root of onethird. We
shall only give the part of the treatment necessary for the
construction of the tables. That their importance is appreciated
is shown by their adoption by Professor Howe in the chapters
of his elegant treatise which deal with the masonry arch.* We
are indebted to the courtesy of the Eoyal Irish Academy for
* "A Treatise on Arches," by Malverd A. Howe, C.E., Professor of Civil
Engineering, Kose Polytechnic Institute, TeiTe Haute, Indiana. New York:
John Wiley & Son. London : Chapman & Hall, Ltd., 1897.
448
APPLIED MECHANICS.
[chap. XXII.
the use of their blocks, and especially for a photoblock from
the design of an arch by this method engraved on a grand scale
Ijy the splendid liberality of the Academy.*
For simplicity, put m equal to unity, and
% = !^(t'60 = tan« = y(y^r^).
dx
vw
sec^O (if + 1  /•)
or
P =
(y + (ty
y
Fiff. 9.
if (1} = 1 r = 1  6\
dy ~ //'
dp ^ v/(?/ + (r)
df ~
y
(V)^^^
* " On TwoNosed Catenaries tind their Application to the Design of Segmental
Arclies," by T. Alexander, M.A.I., Profissor of Engineering, 'i'rinitj College,
Dublin, ami A. W. Thomson, B.Sc, Lecturer in the Glasgow nnd "West of Scotland
Technical College, Transactions of the Jioi/al Jiuh Acadtmij, Vol. xxix.. Part iii.,
1888.
CHAP. XXII.] DESIGN OF MASONRY ARCHES.
449
Now (2}/ a) = makes ^ ^ 0, and as a first step makes
(^f' y
a positive quantity upon a further substitution from %f  a = 0.
Hence 2y'  a = gives the value of y, which makes p a
minimum, namely, a pair of points symmetrical about the crown
or vertex. They are the noses, and are shown at Bi and B'^ on
fig. 9. The abscissa, ordinate, slope, and radius of curvature of
the " nose " B^ are designated by ^i, //i, d^, and p,. If p^ be
produced to meet the vertical through and Q,, and if now a
Directrix
Fig. 10.
circle be drawn from Q^ with a radius QiBi = ^i, it will touch
the curve externally at its two " noses." This is the described
circle, and is outside of the finite arc B^, B'^ with which we are
concerned.
On the other hand, fig. 10 shows a circle drawn from the
centre Q. with a radius Q2B2 = B2 passing through the three
points B:, A, and B'. of common curvature. It is the three
point circle. It touches the curve internally at A, and sensibly
touches it at B\ and B., as the slope of the curve 62 and that
of the circle /3 at B2 never differ by more than two and a half
degrees. For all practical purposes this circle is inscribed in
the curve. There is evidently a mathematically inscribed circle,
but its equations are very troublesome.
2g
450 APPLIED MECHANICS. [CHAP. XXII.
For the describedcircle, we have
2.
3;, = Pi = '' —  — ' = ^yAy, = ^ (1  s), (14)
tan Q, = ^(i/; s) = J^^ (15)
So that when s = ^ then ^i = 0, and the crown is the sharpest
point.
Again, by adding the expressions for // and for its first
differential coefficient, we get
n' = y + tan 0,
. 7/ + tan , v/(l  .s) + v^(l  os).
X = log '^ , X, = log ^^ '^z^^ — — ''
''' ■/•2s
El = Zi cosec ^1, Fq = //, + Bi cos ^i  B^
For the threepoint circle, we have
sec^ d = py = fj y (r^ + tan^ 6), sec" 9 = p (r + tan 9),
bnt p2 == p., =  =   and ?/o = '" = /^ ',
T ^ S
.'. r sec^ 9i = r'^ + tan' 0,,
9, + 1 =
1 sec^ 021 1 , ,. ../>, Po
or  = — —,, , or   sec* 9. + sec 0. + 1 = — • (16)
r^ sec 021 s Vn
Solving this like a quadratic equation
tan* 02
je^)'
(17)
So that when s = L, then 6i = 0, and B. comes to the crown just
as Bi did. Also
y2 = ys sec^ 02,
QO2 —
fVi tan 02\ , / ,  tan 02\
log  + = log sec* 02 +
As 62 = ^ nearly
1 J?.
7 = sec* 03 + sec 02 + sec 0, = — . ( I8)
CHAP. XXII.]
By Euc. iii. 35.
DESIGN OF MASONRY ARCHES.
X\ = (^2  !/,) \2Ii.,  (l/.  %)],
451
■1/2 ^S
^'  2(y. v's) ■ 2 '
8o = 3/o  Yo,
^2 = Bi  Bn + A cos 02 sensibly,
where A = Hi  Bi + ^o (See fig. 11.)
Table A, for each assigned value of s, the following ten
quantities are calculated by the above equations in the order y^,
pi, 0u Ru Yo, p2, or /3„, 02, Bn, do, ^2. The values assigned to s
are at equi distances apart.
Table A.
1 ^
c . =
11
ei
e>
^1
i?2
"' 1
11
Yo
60
Si
vi
•333
a
173
173
173
173
■577
■577
•000
000
i
•250
19
28
197
I 98
I 94
200
•500
499
•000
•000
—
•180
25
39
224
225
213
235
■424
•419
•004
•002
—
•170
26
41
229
230
215
242
•412
■406
•005
003
—
•160
27
42
234
235
2i8
250
•400
•392
•007
004
—
•150
27
44
2 39
2*40
220
258
•387
■378
•009
•005
—
•140
28
45
244
245
223
267
•374
•363
on
•006
—
•130
28
46
250
251
226
277
•360
•347
•013
•008
—
•120
29
48
257
258
228
288
•346
■330
•016
■010
i
•III
30
49
263
263
230
3 00
■333
•313
•019
•012
—
•100
30
51
272
272
233
316
■316
•292
024
•oi5
—
•090
31
52
2 So
2 80
236
3'33
•300
•270
029
019
^
—
■080
31
54
289
288
239
3 53
■282
•247
035
•025
—
•070
32
55
3'oo
2.99
241
377
•264
•220
•043
•032
J
i
063
32
57
309
307
243
400
•250
•199
•051
039
I
—
•050
33
59
32S
324
246
447
•223
•156
•o65
•054
i
•040
33
61
346
340
249
5 '00
•200
•115
•084
073
^
—
•000
35
90
00
00
2 "60
00
•000
—
—
—
2 G2
452
APPLIED MECHANICS.
[chap. XXII.
Fig. 11 shows the finite q.xc BzBiAB'iB\ sandwiched between
the arcs of these two circles, and So = 2/0  Yo is their distance
apart at the crown, while ^2 is their distance apart at each
pesc
Qj Pole
Fig. 11.
end. These distances are so small, that as far as being ^the
houndari/ of a load area the arc of the threepointcircle is prac
tically the same as the finite arc of the twonosed catenary
itself. The dotted curve should pass through A.
Tables for the Immediate Deskjn of Se(;mental Arches.
Table B (p. 457, et seq.) is derived from Table A (p. 451)
line by line by dividing the linear quantities in I'able A by
the value of ^,. Ifatios r, s, di, and 62 are unaltered. The
description of the table is fully given on the face of it.
In the Supplementary Table B, the thickness of the key
stone ^0 is made equal to 3^0 (See DA on fig. 12.) This
confines the line of stress to a " kernel," which is the middle
third of the archring. The upper limit of this " kernel " is the
describedcircle of the line of stress. It is still the described
circle of the new line of stress due to a uniform live load all
over the span, which is equivalent to raising the directrix.
The deptli of the crown of the sojit from the formationlevel or
directrix is cl = y^ + Sq. (See OA on fi<4S. 7 and 12, and
0(1 on fig. 8.) The soffit is the threepointcircle of a lower
down member of the same family as the line of stress itself.
This is accomplished as follows. In any line of Table Bi, take
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 453
the value of d, which divide by the vahie of m in that line
produced into Table B. Look for the quotient in Table A,
under the heading ?/o ; take the corresponding value of B. and
place it in Table Bi, under the heading R, but, of course, first
multiplying by m. The rise and span of the soffit are /.' = R
vers Q,, ^md 2e = 2R sin B,.
The quantities engraved on fig. 12 correspond to those set
down on the fourth line from the bottom of Tables Bi and B,
only multiplied each by 5007. This line will serve to explai)i
the construction of the Table B,, while inspection of fig. 12
may help to make it clear. The position of the line fourth
from the bottom is designated as the line on Table B, with
s = 05. In the third column, 104 is the maximum multiplier
which may be used on the linear quantities in this line that there
may be a factor of safety 10 against crushing a sandstonekey.
First, the thrust at the crown wd x p„ = 140 x 089 x 1*3634 =
16987, and dividing this by ^o = "061, the number of square
feet exposed to it, we get the average thrust on the crownjoint
to be 278 lbs. per square foot. Multiplying by 2, as the maxi
mum is double the average (see fig. 3), the maximum stress on
the cheek of the sandstonekey is 556 lbs. per square foot. Now
this increases directly with the multiplier used ; hence dividing
556 into 57,600 lbs. per square foot, the resistance to crushing
of sandstone allowing a factor of safety of 10, we get 104.
Any smaller multiplier gives a factor of safety proportionately
greater. Thus on fig. 12, the multiplier being 50*07, gives the
factor of safety 20. In the fourth column 155 feet is the max
mum span between the points of rupture for a sandstone bridge
with its proportions as on that line; it is 2c = 1*498 multiplied
by 104. The first two columns are the same numbers divided
by three for strong brick, while the fifth and sixth columns are
the same numbers doubled for granite. Ptefer to p. 419 under
" strength." In column seven {cl  Q = 028 is the surcharge ;
it is DO on fig. 12, and should be only sufficient to make a soft
bed for the railway that the sleepers may not hammer the top
of the keystone.
It only remains to explain how t. = "123, the thickness of
the voussoir at the joint of rupture sloping at B^., that is at the
skewback or springer of the segmental arch, is obtained. The
radius of the describedcircle (fig. 12), as given on the fourth
lowest line of Table B is 1. Also QA = R = "869 is the radius
of the sojfit. If now a circle were drawn from the centre Q,
concentric with the soffit and touching the described circle at the
crown, its radius would be {R + 2go) = '869 + 2 x 0204 = 9098.
tlO
CHAl'. XXII. ] DESIGN OF MASONRY ARCHES, 455
Here, then, we have two circles touching at the crown, the
describedcircle itself of radius l,and the circle concentric with
the soffit of radius "9098, just like the two dotted circles on
fig. 6. By the approximation, page 436, the distance apart
of these two excentric circles at the springing (fig. 12) is
the difi'erence of their radii multiplied by (1  cos 6^2) where
0, = 59 ol'. Putting Z for this, we have
Z=(l 9098) (1  5078) = 044.
Add to this the distance between the soffit and the circle
we drew concentric to it, namely 28„ = 2 x 0204, we have "084
the distance between the soffit and the described circle at the
springing joint on fig. 14. Increasing this by 50 per cent., we
get ti = '126 ; and by a closer approximation we make U = •123.
The thickening of the arch ring from t^, at the crown out
wards to /; at the springing (lower joint of rupture) by this
tabular method ensures that the line of stress, beginning at the
loicer trisectingpoint of the crownjoint A (fig. 12), just reaches
the upper trisectingpoint of the joint Ji" (higher joint of rupture
at 33° 6') ; then it comes back into the whitekernel.
This thickening makes the extra load due to the excess
density of the ring suited to the line of stress, since the two
circles embracing the voussoirs are, approximately, for boundary
purposes, two members of the same family of transformed
catenaries as the line of stress itself.
The Tables Bo and B3 are only wanted for small arches in
which the surcharge is relatively larger and the economy of
material in the archring has to give place to the consideration
of strength.
Apparent Factor of Safety. — In looking at fig. 7, it will
appear that the design there made by assuming sizes and
drawing the conjugate loadarea is the same as that made
immediately from Table Bj on fig. 12.
In fig. 7, the apparent thrust at the crown increased from
288?c to the actual thrust 400?r from the leaving out of the
negative part of the conjugate load, and the radius at crown of
line of stress changed from 48 to p^^ = 37 feet. Dividing
equation (16) by (18), we have
Pq = i^o(sec 02  1 + cos 021, (19)
which may serve instead of drawing the conjugate loadarea or
using the tables. Thus taking R^ = 43*5, the radius of soffit
roughly, and 9. = 60° ; then p = I0R2 = 65 feet. As 02 is never
greater than 60"", it follows that the real factor of safety against
crushing the keystone will not be less than tvjothirds the
apparent factor.
SUPPLEMENTAllY TABLE B,. i
StrongBrick.
Sandstone.
Granite.
"o
U ■!
.5 rt =^ '
 1
"c** "o as'
sf
a»=112
/= 164000
7c = 140
/= 576000
71. = 164
/= 1350000
o
li
th of load fi
trados (direct
crown of sotTi
1°
J5
=e =
= o f ;
o'n"o
^ 1
o '
"o '
4.1 ^ 4., hfl <
E^ E"^ '
° II
■~ B
So*,
c •
H
5; J " ■"
3
MJ3
^
"o Z
O V
PS ° 1
U O u
•ji 2'o
^ t
t 9
rt
^. =
"3
c
"3
c
■s.
q"*
'^ c u
_^_.^
a 1
^ 1
5 ""
a."
■j:
E
".
E Z 1
>•?
E
M
rt u
dtu
d.
/u. h.
R.
k.
2c.
s.
hH
t^i
<r.
/^■S.
rr. f',
^
^
w is
in lb. I
pr cub
ft.
i
1
/i
s in lb.
per sq.
ft.
• Kernel" Middle Third of Archring.
1
Intensity
of thr
ust on
Sandstone
The Soffit is the Threepoint Circle of a lower
£ j
Key — • mu
St not e
^ 576000
''"«''r0xT40
1*
• 2 '
curve transformed from same Catenary as Line of
Stress, the directrix being: surface of Kails.
2
its average \
jeing i
its maximum.
!! 1
1 1
a i
Feet.
'
1
j
1
1
1
1
50
72
•122
•141
•019
•026
•973
•325
1451
•120 i
55
81
•116
•137
■021
•029
•970
•333
1462
•115
61
90
•110
•133
■023
•033
•966
•341
1473
•110
Feet.
68
101
•104
•129
•025
•037
•962
•3l9
1^482
■105
38
56
76
112
•098
•125
•027
•041
•957
•357
1491
•100
42
63
84
126
•092
•121
•029
•045
■952
•364
1497
•095 ;
46
70
92
140
•085
•117
•032
•019
•946
•372
1504
•090
61
78
102
156
•07'J
■113
•034
•055
•940
•380
1509
•085
Feet.
19
29
67
86
114
172
•073
•110
•037
•062
•933
•388
1514
■080
21
32
63
95
126
190
•006
•106
■040
•071
•925
•395
1516
■075
23
35
69
105
138
210
•059
•103
■044
•081
•917
•403
1518
■070
•25
39
77
116
1.54
232
•051
•099
■048
■089
•907
•409
1516
•065
28
43
85
128
170
256
•043
•095
•052
•097
•896
•416
1514
•060
31
47
94
141
_
_
■036
•092
•056
•110
■8S3
•421
1506
•055
35
52
104
155


•028
•089
■061
•123
■869
•427
1498
•050
38
67




•018
•085
•067
•141
■850
■431
1480
•045
42
62




•009
•082
•073
•159
•831
•436
1462
•040
46
67
—


•000
■078
•078
•177
•812
•441
1^444
•036
a.'
•1
S
Subject to a mult., loss than irivcn max., to give
i
«
•(5
a
factor of safety greater than 10 in inverse ratio.
""
B
O
o
o
•
Q
TABLE B.
\
A
Srn'ex o/" " Twoxoskd CATKXAiaics" inscribed
n the Circle
of Iiadiiis
(ii, ) Unihf, and hariur/
Parallel Directrices a
t Graduated
Distances
{h\ + Yu) from its Centre
from (1 t. 187) to (1
+ 026).
Tl
is Table forms a continuation of the Supplementary
lables Bi, B,,
and B3, and
lias
then for its purpose the designing of archrings, so as
to secure the
condition of
the
ine of stress Iving within a A
•#>•««/ forming the middle third, fifth,
or ninth of
tlie an lirin
, respt
ttively.
s.
Ou
^3.
m.
lil.
It..
pi.
pd = P2
yo.
To.
5o.
5.2.
C "1
^ «
5 w
S ^
V
J) 1
u.
«
— >.
u
c •—
I. 0,
C u
3
„
>y
> .
3 3
u
u rt
3
V u
u
a
k.T3
c
rt
"rt
s
j= "
? «.
•3
U M «
= .1
•; >
'"' 5
u
t
= e
III
c—
=
= :=£
.0
rt
"H
5
t ^
c =
•£ '
5^
ill
3 V "
^:5 B
rt
V
3
rt J2
i. '^
tn
"0 °
in
.2 5
C .
c S
•=1'=
" X
S « a
13
:=.a
u
i:
U
■5
Cm
5
^ r
— »
7 '^H
a.
3
u
'p
a c
•a. 5
• a
C/3
£^
^^
T°
0
Q"
fi
•180
25°37
39°46
•4455
1^
r0024
9491
1^0501
1890
•1869
•0021 •ooio
•175
25°58
40°2S
•4410
r0024
•9452
1 0543
1845 1 1821
•0023 0011
•170
26°20
41°11
•4365
1^0024
9413
roose
•ISOO 1774
0026 0013
•165
26°40
4r53
•4319
Hg
1^0024
9370
1^0635
1754
•1726
0028 001.5
■160
27°01
42°3G
•4273
^■"
1^0024
9327
1^0684
•1709
•1678
0031 0017
•155
27°20
43°18
•4226
3 ^
1^0023
•9279
1^0738
•1664
•1629
•0034 j 0019
•150
27°40
44°00
•41S0
n ~
1^0022
•9232
r0793
•1619 ^1581
0038 : 0021
•145
27°59
44°42
•4132
=
10021
•9180
10856
•1574 ^1532
0042 ; 0023
•140
28°1S
45°24
•4085
r
** ^
1^0020
•9129
10919
•1529 ^1484
0045 0026
435
28°36
46°06
•4036
.2 y
rt
1^0019
•9071
ru990
•1483
•1434
0049 0029
•130
28°55
46°49
•3988
1^
rooi7
•9014
M061
•1438
•1384
•0054 ^0032
•125
29°r2
47''31
•o9.SS
rooi4
•8952
11142
1392 1333
•0059 0035
•rio
29°30
48°14
•3888
10011
•SS90
11224
1347 •1'283
0064 0039
•115
29°47
4S°o7
•3836
'§^
10007
•8821
M318
1301 1231
•0070 0043
•110
30°(i4
49°41
•3785
10004
•8752
1^1412
1255
•1180
0076 ; 0048
•105
30°20
50°25
•3731
c ;;
tr
09998
•8676
M521
•1209
•1127
0082 0053
•100
30°37
51°09
•3678
1
09993
•8600
1^1631
•1163
•1074
0089 0058
•095
30°52
51°54
•3622
09987
•8517
1^1760
•1116
•1019
•0097 0064
•090
31°08
o2°40
•3567
0^9981
•8433
1 1889
1070
•0965
0105 [ 0071
•085
31°23
53°27
•3508
^
09972
•8340
1^2043
1023
•0908
•0114
0079
•080
31°39
54°14
•3450
0^9963
•8246
1^2197
0976
•0852
•0124
0087
•075
31°54
55°03
•33S8
0^9951
•8141
r2384
0928
•0793
0134 0097
•070
32°09
o5°53
•3326
09940
•8036
P2o71
0880
•0735
0145 1 0108
•065
32°23
56°45
•32o9
09925
•7917
1^2803
0831
0673
0158 0121
•060
32°38
57°38
•3193
09910
•7798
1^3036
0782
•0611
•0172 ^0134
•055
32°52
58°34
•3121
0^9890
•7661
13335
0732
•0544
0188 0150
•050
33°06
59^31
•3049
09870
•7524
13634
•06S2
0478
0204 0167
•045
33°20
60°34
•2968
09842
•7363
14037
0630
0406
0223 , 0190
•040
33°33
6r37
•2888
09815
•7202
14440
0578
0334
•0243 ' 0213
•035
33°46
62°49
•2808
.
09788
•7040
14843
•0526
•0260
0263 ! 0236
i
Independent of /i^.
Direct
y proportioned to Ji^, and s
ubject to any mu
Itiplier.
SUPPLEMEXTAllY TATTLE
B,.
—   il.„ 1 tC :«
*— .*?
— «
,
Strons Brick
M. = 112
/= 154000
Sandstone.
7,= 140
/ = 576000
Granite.
7,. = 164
/= 1350000
O CM
a 1 £■" 
.S 3
11
1°
£ 2 c
lii
=e =
O =
iEc
o
*. so
c =
11
3
si
I' 5
^ 3
« 2
a"
al
° II
o c
• o
"a
c
C3
lilt, to give
>f safety lo.
ilueofspnn
that mult.
"3
n in
•/3
a M 3
Q
s
^X. >.s
C
X
7<
:1
5 1 2S
rs
5
dl„. d.
<0
<2.
R.
k.
2c.
$.
s.
f.
3^ 3 <=i
f;
«5
w is in lb. per cub. ft.
/is in lb. per sq. ft.
"Kernkl"' Middle P'ifth ok Archring.
Intensity of thrust on Sandstone The Soffit is the Threepoint Circle of a loTOer
_, curve transformed from same Catenary as Line of
Stress, the directrix bein? surf.ice of Rails.
p^d , 576000
Key y must not exceed jQ^^j^ . g
its average being % its maximum.
18
20
22
24
27
30
32
35
38
42
46
50
Feet.
17 24
26
29
32
35
39
43
46
50
54
59
64
69
40
45
50
55
61
67
73
81
89
97
106
Feet,
56
63
71
79
87
95
104
116
128
139
161
1
Feet.
52 j
70
•150
59
80
•145
66
90
•139
73
101
•134
80
112
•128
90
126
•122
100
142
•115
110
158
•109
122
174
•103
134
190
•096
146
208
•089
162
232
•082
178
256
•075


•068
_
_
•060
_
_
•053
_
_
•044
_
_
•036
•027
•169
•166
•162
•159
■165
•151
•147
•144
•141
•137
•134
•131
•128
•126
•122
•120
•117
•115
•113
019
021
023
025
027
029
032
035
038
041
045
049
053
057
062
067
073
079
086
026
030
034
038
042
046
050
056
063
070
077
086
095
105
115
128
142
159
176
•972
•968
•964
■960
•956
•951
•945
•939
•932
•925
•917
•908
•899
•877
•862
•848
•831
•814
273
280
287
294
302
309
316
322
329
335
342
34S
354
359
365
368
372
375
378
351
362
373
383
394
402
410
415
420
424
429
4 '29
429
426
423
414
404
390
376
•150
•145
•140
■135
•
•130
•125
•120
•115
•110
•105
•100
•095
•090
•085
•080
•07'.
•070
•065
•060
Subject to a mult., less than eiven max., to Rive
factor of safety ureater than 10 in inverse ratio.
TAT.LE 1
1.
A
Series o/ " Twoxosed Catenaries" inscribed
■« the Circle of Radius
(i?,^ Utiittj, and hari)ig Parallel Direc
trices n
t Graduated Distances
(ill ^ Yt) ff'om its
Centre from (1 > •IS
7) to (1
+ 026). i
Tt
is Tahle forms a continuation of the Supplementary
Tables Bi. Bj, and Ba, and
has
then for its purpose th
B designing of arcliringrs, so as
to secure the condition of
the
ine of stress lying within a kernel formiu!? the middle third, tifth, or ninth of 
the
s.
irchring, respectively
fli.
02.
m.
JRi.
Hi.
pi.
P0 = P2
yo.
To.
5o.
52.
S
u
a
•5
3
a
.S~
» »
o u
O O
Ji VI
a) w u
»2
c rt
. o
U o
3 e
u 01 5
u. 2 2
0— o
5.
S.2
o _
ih CO
«
ill
04
•3
3
O
•O
w
o. =
o u
•5 °
JO "2
li.
= ■5,2
3 j.t;
o
o
a
o
3
Is
.= •=
li*
i'^ ■
3 ■" 2
U 3 u
m
o
.5 «
a u
° o
H
J3 CO
sd • lit
o
3
a.
3
Ui
"o
'5
« c
dl
•180
25°37
39°46
•4455
u3
10024
•9491
l^OoOl
1890 1869 0021
•0010
•175
25°58
40°2S
•4410
U jj
1 0024
•9452
1 0543
1845 1821 . 0023 0011
•170
26°20
41°11
•4365
IHa
10024
9413
r0586
1800 1774 0026 0013
•165
26°40
41°53
•4319
^"a
10024
•9370
1^0635
•1754 1 •1726 ^0028 ' •OOlo
•160
27''01
42°36
•4273
>v
r0024
9327
10684
•1709
•1678 •OOSl ^0017
•155
27°20
43°18
•4226
'S °
10023
•9279
10738
•1664
•1629 ^0034 •0019
•150
27''40
44°00
•4180
c iJ
O 3
10022
•9232
1^0793
•1619
•1581 0038 i 0021
•145
27°59
44°42 4132
10021
'9180
10856
•1574 1532 ^0042 ' 0023 1
•140
28°1S
45°24 4085
10020
•9129
10919
1529 1484
0045 0026 1
•135
28°36
46°06 •4036
10019
■9071
10990
1483 1434
W49
•00^29
•130
1 '
2S°55 46°49 3988
•3 u
10017
•9014
11001
1438 1384
•0054
•0032
•125
29°12 47=31 • 3938
10014
•8952
11142
•1392 1 1333
•0059 ' 0035
•120
29°30 48°14 i •3888
S a
10011
•S890
11224
•1347 ^1283
0064 i 0039
•115
29°47 48°57 ; •3836
u{
rooo7
•8821
M318
•1301 1231
0070 0043
•110
30°(i4
49°41 1 3785
Q u
1^0004
•8752
11412
•1255 11180
0076
(X)48
•105
30°20
50°25 •3731
■^5
en o
0^9998
•8676
11521
•1209
•1127
0082
0053
•100
30''37
51°09 ^3678
3
0^9993
•8600
11631
•1163
1074 1 WS9 1 0058 
•095
30°52
51^54
•3622
09987
•8517
11760
•1116
1019
0097 j 0064
•090
31°08
52°40
•3567
09981
8433
1 1889
•1070
0965
0105 ' 0071
•085
31°23
53°27
•3508
j5
09972
•8340
12043
•1023
0908
•0114 ^0079
•080
31°39
54°14
•3450
1
09963
•8246
r2197
0976
0852
0124 OCKT
•075
31°54
55°03
•3388
1
09951
•8141
12384
0928
0793
0134 ' (DO?
•070
32°09
55°53
•3326
1
09940
8036
1^2571
0880
0735
0145 ; 0108
•065
32°23
56°45
•3259
I
09925
•7917
r2803
0831
0673 0158 i 0121 
•060
32°38
57°38
•3193
09910
•7798
13036
•0782
0611
•0172 ^0134
•055
32°52
58°34
•3121
I
09890
•7661
13335
0732
0544
0188 0150
•050
33°06
59=31
•3049
1
09870
•7524
13634
0682
•0478
0204 1 0167
•045
33''20
60°34
•2968
1
09842
•7363
14037
•0630
0406
0223 i 0190
•040
33°33
61°37
•2888
1
09815
•7202
14440
•0578
•0334
■0243 ^0213
•035
33''46
62°49
•2808
09788
•7040
14843
•0526
•0260
•0263 •0236
Indep
endent of R^.
Directly proportioned to
R,, and
subject to any multiplier.
SUPPLEMEXTAllY TABLE B,.
strong Brick.
w=lI2
/= 164000
e
e
M
M
K
S
rt
s
^
Sandstone.
T< = 140
y = 576000
Granite.
tc = 164
/= 1350000
S 2
0.3
.5 rt
o ■ ^ — '
 ?
° £
c B ■
Bl
=> II
I 2
 E
°3
dlo.
£.
■ur is in lb. per cub ft.
/is in lb. per sq. ft.
Intensity of thrust on Sandstone
p„rf ^ 576000 3*
Key y must not exceed ^q ^ ^^q . ^•
its average being J its maximum.
20
22
25
27
30
33
36
39
42
46
50
55
Feet.
27
30
33
37
40
44
48
52
56
61
66
71
40
44
49
55
61
67
75
Feet,
50
57
64
72
80
89
83 110
91 • 121
99
108
118
132
145
157
58
64
70
80
98
110
122
134
150
166
182
198
Feet.
72
80
88
101
114
129
144
161
178
199
220
242
264
"Kernel" Middle Ninth of Archring.
The Soffit is the Threepoint Circle of a lower
curve transformed from same Catenary as Line of
Stress, the directrix being surface of Rails.
•178
•197
.
•173
•194
•167
•190
•162
•187
•155
•183
•149
•180
•143
•177
•137
•174
•130
•171
•123
•168
•116
•165
•109
•162
•102
•160
•0G5
•158
•088
•156
•080
•154
•072
•152
•064
•1.51
•054
•149
019
1
•028
021
•031
023
•034
025
•037
028
•041
031
•047
034
•054
037
•060
041
•067
045
•073
049
•080
053
•089
058
•098
063
•108
068
•119
074
•131
080
•144
087
•1.58
095
•173 '
964
•223
963
•230
962
•238
958
•245
953
•252
948
■258
942
•265
936
•269
929
•274
921
•281
•913
•288
903
•293
893
•298
882
•302
•870
•307
857
•311
844
•315
828
•317
S12
•320
1234
1250
1266
1278
1290
1299
1309
1316
1323
1327
1331
1331
1332
1334
1326
1320
1315
1303
1291
C4
180
175
•170
165
•160
•165
150
•145
•140
136
•130
125
120
115
110
•105
•100
09.5
090
. •
• 1
o
CO
o
CO
•
c
Subject to a mule, less tli.m eiven max., to g^ve
tactor uf saluly greater than 10 in inverse ratio.
TABLE B.
A Sen'es o/" " Twoxosfd Catexakies" inscribed in the Cirde of liadiuH
{R\) Unity, and having Parallel Directrices at Graduated Distances
(lii + To) from its Ce7itre from (1 + 187) ^o (1 + 020).
This Table forms a continuation of the Supplementary Tables Bi, Bj, and 0:^. and
has then for its purpose the designiiifj; of nrclirings, so as to secuie the condition of
the line of stress lying within a kernel forming the middle third, fifth, or ninth of
the archring, respectively.
Independent of A'j.
s.
0i.
6i.
m.
Hx.
■ § c
pi.
P0 = P2
yo.
To.
5o.
8>.
i
u
a
rt
"5
3
c.ti
c5
<J o
O O
u ■"
». «
o » .
t) ■« «
C ,1
3
. U
¥j
ill
E.2
^
.O
•r
o
J
0. c
■^ c
•£ °
o
a
3
> £
Si
O "1
3.5
;5^
t .
u ^
~ 3
«^
Ill
i i
a
.£ H
si
^ u
S
u
'o
"o
■51
?•■"
1^
JJ,
5 = o
o
3
a.
3
'o
'5
d c
d ~
•180
25°37
39°46
•4455
I'S
10024
•9491
10501
•1890
•1869
•0021
•0010
•175
25°o8
40°2S
•4410
1^0024
•9452
1 0543
•1845
•1821
•0023
•0011
•170
26°20
41°11
•4365
10024
•9413
r0586
•1800
•1774
•0026
•0013
•165
26°40
41°53
•4319
^'b
10024
•9370
1^0635
1754
•1726
•0028
•0015
•160
27°01
42°36
•4273
^■~
1^0024
•9327
10684
•1709
•1678
•0031
•0017
■155
27''20
43°18
•4226
r0023
•9279
1^0738
•1664
•1629
•0034
•0019
•150
27°40
44°00
•4 ISO
rt ~
c tB
o =
1^0022
•9232
1^0793
•1619
•1581
0038
•0021
•145
27°59
44°42
•4132
10021
•9180
10856
•1574
•1532
•0042
•0023
•140
28°1S
45°24
•4085
10020
•9129
1^0919
•1529
•1484
•0045
•0026
•135
28°36
46°06
•4036
M O
rooi9
•9071
1^0990
•1483
•1434
0049
•0029
•130
28°55
46<'49
•3988
10017
•9014
M061
•1438
•1384
•0054
•0032
•125
29°12
47°31
•3938
10014
•8952
ril42
•1392
•1333
•0059
•0035
•120
29°30
48°14
•3888
■^ %
10011
•f.890
M224
•1347
•1283
•0064 ;
•0039
•115
29°47
48°57
•3836
'iM
10007
•8821
M318
•1301
•1231
•0070 1
•0043
•110
30°()4
49°41
•3785
1^0004
•8752
1^1412
•1255
•1180
•0076
•0048
•105
30°20
50°25
•3731
M O
09998
•8676
M521
•1209
•1127
•0082
0053
•100
30°37
51°09
•3678
3
•g
o
0^9993 1
•8600
11631
•1163
•1074
•0089
•0058
•095
30°52
51°54
•3622
0^9987
•8517
11760
1116
•1019
•0097 j
•0064
■090
31°08
52°40
•3567
09981
•8433
1 1889
•1070
•0965
•0105 '
•0071
■085
3r23
53°27
•3508
.2
H
0^9972 ;
•8340
12043
•1023
•0908
•0114
•0079
•080
31°39
54°14
•3450
09963 ;
•8246
12197
0976
•0852
•0124
•0087
•075
31°54
55°03
•3388
0^995 1
•8141
12384
•0928
•0793
•0134 1
•0097
•070
32°09
o5°53
•3326
0^9940
•8036
12571
0880
•0735
•0145 !
•0108
•065
32°23
56°45
•32o9
1
09925
•7917
12803
0831
•0673
•0158 1
•0121
•060
32°38
57°38
•3193
1
09910
•7798
P3036
•0782
•0611
•0172 i
1
•0134
•055
32°.72
58°34
•3121
1
09890
•7661
1^3335
•0732
•0544
•0188 •
•0150
•050
33''06
59^31
•3049
09870
•7524
13634
•06«2
•0478
•0204 ;
•0167
•045
33°20
60°34
•2968
09842 1
•7363
14037
0630
0406
•0223 1
•0190
•040
33°33
61°37
•2888
09815 1
•7202
14440
•0578
0334
•0243
•0213
•035
33°46
62°49
•2808
09788 1
•7040
14843
1
•0526
•0260
•0263
•0236
Directly proportioned to /?,, and subject to any multiplier.
402 APPLIED MECHANICS. [CHAP. XXII.
EXAMPLKS.
11. Design of a sandstone segmeiitul arch with vertical load, directly from the
tables, span, 75 feet, and depth of surcharge at crown about 1' 4" : the springing
to be the joint of rupture. (See fig. 12, left half.)
Here '2c = 75, and d — to = I'S ; their ratio is 56"25. "We find by trials on
Table Bi, that 2c ^ {d  to) = 53 occurs on the line where »■ = "05, and the multiplier
required on that line to make 2c into 75, is 50'07, about half of the max. mult,
given under " sandstone " ; so we shall have a factor of safety of about twice ten,
and need not consult the other supplementary Tables, £2, £3.
That line gives the following relative and absolute values : —
Mult.
dto
d
^0
h
It
k
2c
5007
j 028
( 141
•089
446
•061
305
•123
6^15
•869
435
•427
214
1498
75 ft
Rx
Po
Yo
5o
62
1
1^3634
•0478
•0204
•0167
5007
08265
2393
102
•836 ft.
•050
The radius and rise of soffit are 43^5 and 21*4 feet ; the thickness of archring
at crown and springing 3' 0" and 6' 2" ; the surcharge being 1'41 feet, or about
1' 4", as required.
On the same line, continued on Table B, we have —
s Bi Mult.
•050 59= 31' 5007
At tlie crown the thrust on the archring per foot of the breadth is
S = wpud = 140 X 68265 x 446 = 42624 lbs. ; the average intensity of the stress
is 42624 ^ 305 = 13975 ; and double of this 27950 lbs. per sq. ft., is the maximum
intensity, giving a factor of safety of 576000 4 27950 = 20. Otherwise, the
multiplier being half the maximum in the Table, the factor of safety is double ten.
At the springing, 2'= jff sec ^2 = 8402;j ; the average stress" 84023 ^ 615
= 13062 ; and since the deviation of the centre of stress is ^^2 — 5i = 1^025 — 836
= 189 above the centre of the joint ; we can substitute this for 5, the deviation of
the centre of stress from the middle of the joint, equation (1), at page 417,
Max. stress 65
= 1 + _ = 1184.
Aver, stress ^2
Hence the maximum intensity of the stress on the springing joint is 16180 lbs. per
sq. ft. Dividing this out of 576000, the crushing strength of sandstone, we get
the factor of safetj* 35.
A simpler way to proceed to get the factor of safely against crushing, sufficiently
close for all purposes, is to divide the crushing strength 576000 by 13062, the
average, and the quotient 42 is the apparent factor of safety against crushing the
sandsioneskewback. For the three deviations of the centres of stress shown on
fig. 3, expressed as fractions of t the thickness of the joint, the average bears to
the maximum the ratios below.
Deviations of c. of stress, iVth, "iVth, ith.
llatios of aver, to max., *> «, ^•
Now tlie deviation of the centre of stress on the skewhack (fig. 12) is 180 in
6 15 ; as this is less than jf , then the factor of safety is at least ths of 42, that
is, there is a real factor of safely greater than 31.
lil
Pu
To
«o
52
\ .5007
12384
62007
•0793
3970
•0134
•671
•0097
•486 ft,
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 463
12. If a live load of 2"20 lbs. per sq. ft. of tlie ])lutfoiiii be all over the span of
the bridge (Ex. 11), find the new line of stress in the archring and the intensities
of the stresses at crown and springing. (Fig. 12, right half.)
The height of superstructure equivalent to this live load is
A = 220 f 140 = 1571 ft.
of sandstone. Here we have to find a new twonosed catenary still inscribed
in the same circle, ii'i = J0'07, forming the upper boundary of "kernel"
(middle third) of the archring, as already designed (Ex. 11), but to a directrix,
h higher than before. Adding h to the old value of Yq, we get 2393 + 1'571 or
3904, which, divided by the multiplier 5007, gives us 0793 as a nctc relative
value of I'o, which is found in the Table B, at the line —
.y ei Mult.
•075 55° 3' 5007
This is a new twonosed catenary, of a different modulus and of a different
family, so that the soffit already designed will not be mathematically the three
point circle of another member of the family of this line of stress, but it "wiU
sensibly be so. The joints of rupture have gone up to 55° 3' ; but this is
immaterial, as the line of stress is now closer to the upper "kernel," and will
therefore be wholly in the " kernel," down to 59° 31' the springing joint. In
designing the abutment, the tangent at the joint 55° 3' should lie in its middle
third.
At tbe crown, now we have the thrust, 5"' = M;poC«^ 4 A) = 140 x 62007
(446 + 157) = 52346 lbs. ; average intensity 52346 f 3^05 = 17162 ; the devia
tion of centre of stress is \to — 5,. = 509  671 = •162 ft. hdow centre of the
joint. The apparent factor of safety is 576000 f 17162 = 33. The fractional
deviation of the centre of stress from the middle of joint being 162 in 3"05 or
1 in 18 nearly, the factor of safety is of ths its apparent value, that is, 27 is the
factor of safety at crown when the live load is on the bridge.
At the springing joint, T = if ' sec 59° 31' = 103188 lbs., nearly ; the
average intensity is 103188 ^ 6^15 = 16778 ; the deviation of centre of stress
is J<2  52 = 1"025 — •486 = 539 ft. above the centre of joint. The fractional
deviation of centre of stress is 539 in 615 or less than 1 iu 10. The apparent
factor of safety is 576000 f 16778 = 34, and the factor of safety not less than
fths of this or 21.
13. Let the live load (Ex. 12) cover only onehalf of the span: find the
horizontal thiust to be balanced by the backing of the voussoirs.
The horizontal thrust due to dead load is (Ex. 11) 42624 lbs. ; and due to dead
and live loads combined it is (Ex. 12) 52346 lbs. ; the difference is 9722 lbs. per ft.
of breadth.
14. Suppose the archiing, spandiils, &c., of Ex. 11, have, by means of voids
in the superstiucture, an average density of 100 lbs. per cubic ft. Find results
corresponding to those of Ex. 11.
For stability, and to give the required surcharge, the dimensions in Ex. 11 are
required, just as before, but the stresses will be altered in the ratio 140 : 100.
H = 30500 lbs., nearly ; T= 60020 lbs., giving factors of safety, 28 at crown and
49 at springing. These values aie engraved on fig. 12.
The voids in the superstructure should be so arranged that their boundary
may be roughly a member of the same family as line of stress, by making the
ordinates of their boundary a constant fraction of those of the soffit.
464 APPLIED MECHANICS. [CHAP. XXII.
15. Let a live load of 157 lbs. per sq. ft. of platform be over the \rhoIe span
of bridge (Ex. 14) : find the line of stress and the intensities of stress at crown
and springing.
The equivalent height of structure is l'o7 ft., taking the new density into
account ; so that the solution is the same as Ex. 12, only we must alter the
quantities in the ratio 140 : 100.
E' = 37500 lbs., nearly; T'  73700, nearly, and the factors of safety are
increased to 35 at crown and 31 at springing. These values are engraved on
fig. 12.
16. Let the live load in Ex. 15 be only over onehalf the span: find the
amount of horizontal thrust to be balancedby the frictional stability of vault covers
butting against the higher voussoirs. Find also the distance back to which the
vault covers must extend to balance it. (See tig. 12.)
The thrust is, P= E"  H= 37500  30500 = 7000 lbs. per foot of breadth. If
the underside of the vault covers come up to the level of the crown of the soffit,
then the weight per foot of breadth of bridge on the spandrils due to the vault
covers, and dead load over them alone, is wdl=^ 140 x 446/ = 624'. Taking the
coefficient of friction at "7 ; then 7 x 624/ = 7000 or / = 16 ft. to the neare.'^t foot.
The voussoirs near the keystone should have squaredressed sidejoints until the
sum of their vertical projection is to, the depth of keystone, so as to receive the
horizontal longitudinal resistant thrust of the vault covers truly; and the covers
are to be built with squaredressed side joints set closely so as to yield as little as
possible. Light spandril walls may be built up to level of crown of soffit, and
extend hack so as to give 16 feet longitudinally of vault covers ; then the spandrils
may step rapidly to lower levels.
17. Design of a semicircular archring of common sandstone, the span to i.e
100 ft., and a surcharge of at least Ijft. being required for the formation of the
roadway, laying of gaspipes, &c. The data are R = 50, and .T( f (rf — <y) not
to be greater than 33. On Table Bi, the lines above that with .< = OS (in order
to make li into 50) require a multiplier greater than the maximum given for
sandstone ; these lines are tlierefore excluded on the question of strength, while
the lines below that with s = '05 give Ji i {d — t^) greater than 33, and are
excluded by requirements of the roadway. These two limiting lines give —
■08
•05
The upper gives greatest economy of material in archriiiff, wliich is only 2 ft.
at crown, but kss economy of material in superstructure, as d is larger, and also
less economy of solid backing, which has to be built to a joint 5° higher. Hence
the line midway between them would be most suitable all round. For a single
arch a line a little nearer the upper may be adopted ; and for a series of arches a
line nearer the lower, that is, in favour of a heavier archring to withstand the
shocks transmitted from arch to arch. The best lines, then, are for a Single
Arch, or a Series, respectively —
Mult.
R
d
tn
03
Factor of Safety.
536
50
5'9
2
54° 14'
57 X 10
575
50
01
35
59° 31'
ol'b
Mult. R d (n h ei Factor of Safety.
) X 1
d4"5
5 X 1
55^8"
69 X 10
•07 54526 50 o^6 24 44 55° 53' = 12^7
d4^5
•06 55804 50 6^3 29 54 .'57° 38' ^— = l.j 2
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 465
Compare Kunkine's empirical rule, Civ. Eng., Art. 290, giving
t^ = v/^12~>r50 and \/T7 x 50
245 ,, 292 respectively.
The solid backinK must be brought up to the point where the joint at Q2 meets
the back of the archring, and below that joint the archring may be of the
uniform thickness /■.>. The su!)erstriicture may readily be reduced by voids and
the employment of material of less density than sandstone, till the average
density of the whole is a fifth less than that of sandstone, which would raise
the factors of safety at crown to 16 and 19. The factors of safety at joint of
rupture are even greater as the centre of stress is nearer the centre of the joint,
and is r <i > sec 6n. By means of the values obtained for w, p^, d, 5j the thrust
at crown and joint of rupture, and the centre of stress at joint of rupture are
calculated, as in preceding example. A tangent from this last point enables a
suitable abutment to be designed.
18. Design of a segmental brick arch, having the joints of rupture at the
springing, a span of 30 ft., and a surcharge of 2 ft., slightly more or less.
Here the ratio '2c  (d  i^) is 12. But this ratio (nearly) occurs in the
Tables Bi, Bj, and B3, as noted under —
d  /,j
s
Max. Span
Brick.
Factor of
Safety.
Table Bi
119
•120
12 ft.
10
„ B2
Wh
•125
21ft.
10
)> B3
115
•130
37 ft.
10
^2
Mult.
dt^
d (0 h
R
46° 49'
22^54
( 116
t261
•165 05 08
372 12 18
913
206
po
1106
2492
k
•29
4^5
2c
1331
30 ft.
Factor of Safety.
37x10 ,„ ,
= 125.
30
It is clear, then, that we cannot adopt the proportions from Table Bi or Bo ;
for althougli the surcharge and span would be in the required proportion, the
multiplier to make 2c equal to 30 ft. would only give in Table Bi a factor of
safety \% x 10 = 4, and in Table B2 a factor ^ x 10 = 7, which are not
sufficient. From Table B3, then,
■130
The arch is to spring at about 45°, have a rise of \\ feet, the radius of soffit
being 20' 8", while the thickness of the crown is to be 14 inches, and at the
springing 50 per cent, greater. These dimensions would suffice for the archring
built of the strongest red bricks moulded into the proper wedgeshaped forms.
The line of stress may be exactly located, as in previous examples ; but it would
be sufficiently near, in this case, to take it as being up the very centre of the
archring : and a line drawn normal to the springingjoint from its middle point
is sensibly the line of thrust on the abutment. The superstructure could easily be
arranged with voids to make the average mass of the whole m; = 90 lbs. per
cubic ft., or ths that of brickwork, which would increase the factor of safety to
about 20. The thrust then would be —
At crown, B. = xcp^d = 9262 lbs. per ft. of breadth.
At springing, jr= ZT sec 62 = 13532 lbs. ,, ,,
For ordinaryshaped bricks the archiing ought to be built of a uniform thick
ness, the line of stress and thrust remaining sensibly unchanged, as the extra part
2h
466 APPLIED MECHANICS. [CUAP. XXII.
of the archring at crown will really form part of the superstructure. Fcir a single
arch, an average of to and Cj could he taken ; but for a series of arches the greater
had hetter be chosen, giving respectively
t= 15 and 18 feet.
Compare Rankine's empirical formula, Civ. Eiig., Art. 290 —
t = v/206 X ^ and \/206 x 17
— 157 ,, I '87 respectively.
19. If the bridge, lefthalf of fig. 12, is to carry a road, approaching with a
slope 1 in 100, and passing over with a curved profile sensibly circular; to find
position of profile.
To make the profile sensibly a member of the same family as line of stress, it is
sufficient to derive it from the soffit, which is sensibly a member of the family.
Let Mo and ui be depth to profile at ciown and springing. The tangent to profile,
being like that to a paiabola, gives slope of half chord of profile 1 in 200; hence
?<2  Mo : 370 : : 1 : 200, and ii2 : tio : : 4G + 214 : 446. Hence >/„ = 04 ft.,
and «2 = "23 ft. The profile is to be below the directrix about i an inch at crown,
and about 3 indies at springing. The load left out between directrix and profile
increases the factors of safety as 4*46 ; (440 — 004).
20. Compare the segmental masonry arch of 75 feet span as already designed in
Example 10, for the moderate surcharge of I' 4" with others with the surcharge
heavy and light iespectively.
Deriving the dimensions from the three lines of Table Bi where s = 'OS, s = '05,
and s = •035, the multipliers are 496, 5007, and 51"94, that in each case the span
may be 75 feet. The corresponding maximutn spans for sandstone in the fourth
cidumn are 86, 155, and 201 feet; and 10 increased in the ratios which these bear
to 75 gives the factors of safety against crushing the keystone it' the ring be sand
stone. The values for strong brick are onethird part, while those for granite are
double that for sandstone. The three designs are sketched du fig. 13, and we have
the following dimensions and factors of safetj' against the crushing of the key
stone: —
Surcharge,
. 3' 8"
r 4"
0' 0"
Span,
. 75' 0"
75' 0"
75' 0"
Keystone,
. r ID"
3' 0"
4' 0"
Ske whack,
. 3' 1"
6' 2"
9' 0"
Rise,
. 19' 3"
21' 5"
22' 10"
Total rise,
. 24' 8"
25' 10"
26' 10"
Strong brick,
4
7
9
Sandstone,
12
20
27
Granite,
24
40
54
It is to be observed of Table B, that it gives designs of the
utmost economy of the material in the archring itself con
sistent with equilibrium under the verticalload only. But as
the surcharge gets heavier, as, for instance, on the first part of
fig. 13, there is not sutiicient strengtli with biick or sandstone,
tliough granite would suffice. With still heavier surcharges
(see Example 8), we are forced to sacrifice economy of material
and use tlie other Tables Bz and B3. The problem is passing
CHAP. XXII.] DESIGN OF MASONRY AKCHES.
46^
from one of stabiliti/ to one of mere streivjth. See page 419,
under " strength." As the surcharge further increases relative
to the span, the line of stress becomes practically up to the very
centre of the ring, and we have buried arches as in examples to
follow, and in tunnel roofs for the deep lengths.
Heavy Surcharge.
Directrix
Moderate Surcharge.
Inspection of the Tables B,, Ba, and B3 shows clearly that
for balanced segmental masonry arches strong brick is economi
cal for spans from 20 to 50 feet, sandstone from 50 to 120 feet,
and granite from 60 to 250 feet.
2 H 2
468 APPLIED MECHANICS. [CHAP. XXII.
Linear Transformation of Balanced Eib.
We had an example of this in fig. 8, wliere we supposed
one of the curves AB to be painted on a sheet of indiarubber
pinned down along its edge OJ^, and already stretched in the
vertical direction. The sheet is, of course, made of a purely
imaginary kind of indiarubber which can be stretched in one
direction without changing in the direction at right angles to it.
At the ends of the arc A and B are to be painted the forces
which balance it, H and 2' to scale ; while the veiticalload
area, the only load oil the arc AB, is mapped out by the four
painted lines, the arc itself and the three straight boundaries
OA, NB, and ON. Then, by further stretching the sheet
vertically or by allowing it to contract, we obtain a new
balanced rib with new forces at its ends and a new loadarea.
Otherwise the lines may be painted or traced through upon a
sheet of glass lying flat on tlie paper and himjed to the paper
along its upper edge OX. The glass is then to be tilted up on
its edge OX, and the work projected normally on the paper.
The sheet of glass, although hinged to the plane of the paper
along OX, may extend beyond the hinge. It may also have
been originally tilted up when the work was projected normally
from the paper on to it, then flattened down before projecting
the work back on the paper. This is the simplest case of
the orthogonal or linear transformation to obtain, from a rib
balanced under a verticalloadarea alone, another rib balanced
under the transformed verticalloadarea. In the work which
follows, it is more convenient to transform horizontally.
In dealing with a rib balanced by a verticalload area and
horizontalloadarea conjointly, both of these areas themselves
may not be painted on the stretchingsheet or tiltingplate.
For convenience, the verticalloadarea may be supposed to be
painted on the tiltingplate, but after the transformation the
density or weight per square foot of area is to be altered.
That is, the superstructure, if transformed linearly, must then
be reckoned to be of a new material of different density or
weight per cubic foot.
All forces, thrusts along the rib or their components,
cross the hinge at the same points after the transformation
as before. Forces or component forces in the direction of
the transformation alter in the ratio of the transformation ;
tho.se at right angles remain unaltered.
Ilesultant stresses or arrows representing the e.xternload.
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 46!»
such as r, r, &c., on the southwest quadrantaliib, tig. 4,
also cross the hinge at the same points after the transfonnatiou
and before. Tlie new arrows obtained this way, altliough correct
in direction, are not of the proper length. Consider one of the
arrows r (fig. 4) ; it represents to scale the extern load normal
on a unit tangentplane to the back of the rib with a certain
southern exposure to stress and a certain western exposure.
After the transformation the tangent plane is no longer a
unit plane, and one exposure to stress has altered, while the
other has remained unaltered. But the length of the new
arrow must give the externload on a unit plane, allowing
for every alteration.
Let the northwest quadrant, fig. 4, in equilibrium under
t\\Q fluid load represented by the two equal rectangles, be pulled
out horizontally till it is twice as broad as before. It is
now shown on fig. 14. The quadrant of the circle AB'S' has
become ABS, the quadrant of an ellipse twice as long east and
west as it is north and south. Since Tq has doubled in value,
so must also the area of the east load, and as this load stands
on the same base or platform as before, so the height of the
rectangle must have doubled. On the other hand, T, is
unaltered by the transformation, so that the southload is
unaltered in area ; but as it stands now on a platform or base
doubled in length, the height of the rectangle is hcdf what
it was at first. The ratio oi q to p is now 4, which is the
duplicate of the ratio of transformation. The shaded part of
the eastloadarea before and after the elongation is the
horizontal component of T' and of T, respectively. The
shaded part has therefore doubled in area. By starting
with B at «S' and moving it up to A, the shaded area has
added to it strip after strip, each strip being stretched to
double what it was at first. AVe see that the new eastload
area is of the same for/n as at first, only with the height
at each point of its base increased in the ratio of transformation.
In the same way it will be seen that the southloadarea is
also of the same original form, but with the height at each
corresponding point of its base decreased in the ratio of
transformation.
The thrusts T and T' along the ribs at B and B' meet
at the same point j on the hi7igc Oj. The extern resultant
stresses on the backs of the ribs r and r'l also meet at the
common point on the hinge.
The whole stress at B' for a unit plane, its edge towards
you and its face turned in all directions, is given by a circle
470
APPLIED MECHANICS.
[chap. XXII.
of which r' is a radius normal to the circular rib. The whole
stress at B for a unit plane, its edge towards you and its face
turned in all directions, is given by an ellipse of which r is
tlie radiusvector parallel to BO, and whose major and minor
semiaxes are q and p, the heights of the east and south load
areas, respectively, for the point B. For clearness, a quadrant
of this ellipse is shown with its centre at D, the point on the
base of the eastloadarea in a line with B. It is called the
Ellipse of Stress. It completely determines r in direction as
well as in length. Thus through D draw CC, the unit plane,
parallel to Bj the tangent to the rib at B; draw JDN normal
to GO laying off DM equal to the half sum of g and p,
f/onsiant J^otentzaZ,
^JRatto ofCan/uqate
9.P
\
/7
/ /
/ / ^
^Pu C^ Xoad! on ' ^'iffpi^
Fig. 14.
make Md isosceles with MB, and lay off MR equal to the
half difference of q and p. This gives RD or ?•, which is to
be shifted parallel to itself to act at B. It is not necessary
to know the point O on the lihifjc through which r must pass ;
it is only necessary to know the shape of the rib itself so
as to be able to draw a tangent to it at B, and further to
know q and p tlie semiaxes of the ellipse of stress at that
level.
In the example, fig. 14, both of the rectangles, giving
the south and eastloads on tlie elliptic(]ua(lrant, may be
doubled in height, and their density per square foot taken
CHAP. XXII. J I»ES1GN UK MASONKY AKCIIE8, 471
as half that of the rectangles giving the loads on the circular
(juadrant. The new southloadarea is then obtained simply
by projection, while the new eastloadarea has been obtained
by projection in tlie duplicate ratio.
Generally then a balanced quadrantal rib and its vertical
loadarea, that is its superstructure, can be projected by a pull
out horizontally, provided that the conjugate horizontalload area
he pulled out in the duplicate ratio. After which the density of
the new loadareas, that is, the weight per cvMc foot of the nev:
superstructure, is to he reckoned as less than that of the old super
structure in the ratio of the transformation. (See figs. 14 and 15.)
Geostatic Load Unifoem or Vaeying Potential.
The pair of conjugateloads shown on fig. 14 can be produced
simultaneously by placing the elliptic ring horizontally in a fric
tional granular mass with a free level surface, the mass being
excluded from the inside. The east and westload q on the pair
of short barricades being uniform, and four times as intense as p>
the north and southload on the pair of long barricades. Com
pared to the greater, the lesser load is passive. If the shaft or
well were a thin elastic cylinder erected first and stayed inside
by an elliptic centre, which could be gradually shrunk upon
itself, and if now the granular mass were spread loosely around
it in horizontal layers, the genesis of the horizontal stresses
would be as follows. The potential is the weight of the column
of grains from the surface down to ihe ring of the cylinder under
consideration ; it is the active force. Every cube of grains around
the elliptic ring would try to spread equally in all horizontal
directions. The horizontal load on all the four barricades around
the elliptic ring would be p, just as on the circular ring, fig. 4.
But/? would only be a, fraction of the potential, for the friction
among the grains assists in supporting the overheadload on a
cube of grains. Let us suppose that p is about a fourth part of
the potential. This would be the proportion if the granular mass
were crushed stone or, in a model, smallshot. Any cube of the
smallshot would cease spreading from the load overhead, when
the horizontal load applied to its four vertical faces was a fourth
of the overheadcolumn.
Here we have the elliptic ring, fig. 14, with a fluidload p
onefourth of the potential outside as in fig. 4, and supported
inside by an elliptic centre. If now the centre were gradually
shrunk, the elliptic ring would gradually collapse or flatten,
472 APPLIED MECHANICS. [CHAP. XXII.
«Uie to the advantage which the tiuidload has on tlie long
l>arricades. In elongating, the elliptic ring would consolidate
the granular mass opposite the short V)arricades, till it pressed
back with a stress ^ = 4^ = the potential, when the ring would
cease elongating and the centre could be removed. The ellipse
would now be more than twice as long as it was broad. To
remedy this, the granular mass is to be spread around the
shaft in thin layers and punned or rammed hard, layer after
layer opposite the ends of the ellipse, but filled loose opposite
the sides. The granular mass will be earth or shivers in the
case of masonry cylinders or wells, and must be filled as
described, as the distortion would he fatal to an elliptic brick
cylinder long before the earth at its ends would be sufficiently
consolidated. The load on the elliptic rib, fig. 14, is called
shortly an earth or (jeostatic load of unifoi^m jjotcntial. One
ellipse of stress, a quadrant being shown with its centre at
D, serves to give completely the stress on the vertical faces of
every unitcubeof grains all around the rib. If the unitcube
lie north and south, p and (y = 4p) are the loads on its vertical
pairs of faces, while the potential, also q, is the load on the
horizontal faces. The whole stress at D is given by an oblate
sjyJieroid. If the unitcube be turned till a pair of faces is
parallel to CC, then the vector r gives the load on that pair of
faces, and is therefore the mutual pressure between the face of
the unitcube and the back of the rib at B which are in contact.
In the same way the ribs shown on fig. 15 are subjected to
the one a fluidload, and the other a grostaticload of uniformly
varying 'potential. The vector r' gives the load on the face of a
unitcube parallel to jj at the depth B' below the surface of the
earth, that is the stress on the back of the rib at B.' One
ellipse of stress serves for all cubes at depth B', :rF being the
semimajor axis which is itself the potential. The load on the
face of the unitcube parallel to the paper is equal to the minor
semidiameter, and the whole stress at the depth B' is given by
a prolate si^heroid. As the depth of B' increases, this spheroid's
axes increase at the same rate ; but tlie ratio of any pair of its
axes remains the same.
EyuiLiuiauM OK A FuiuTioxAL Ghanilah Mass.
If the loads on the elliptic rib, fig. 14, were actually due to
a granular mass of crushed stone, filled around the cylinder in
the manner described, then, in order that the equilibrium may
l)e pernuinent it is necessary that the unitgranularcube at 7>
CHAP. XXII.] DESIGN OF MASONKY ARCHES. 473
should itself be in ec[iulibriuni with the load 7 on its east and
west faces 4 times as great as the load j) on its north and south
faces. These loads are called the 2^f'^ncipal stresses at D because
they are normal to the faces of the cube lying north and south.
If we map out a unitcube at I) with a pair of faces parallel to
CO, the load on these faces is r which is no longer normal, but
" makes an angle BDM. The manner of constructing r is to lay
off Z)il/ equal to the halfsum of the principal stresses along the
normal DN, and then to lay off MR equal to their halfdifference
along Md isosceles to MD. Hence ■?' is most oblique when the
unitcubeofgrains is so mapped out that MBD is a right angle.
It is only necessary then to see that grains, in the cube mapped
out in this most critical position, shall not slide on each other
due to the extreme obliquity RDM of the load on its faces.
But when i? is a right angle, then sine RDM equals R.M divided
by MD. If we put RDM = ^, the angle of friction between the
grains, then for the cquilihrium of the granular mass at D it is
necessary that sine (ft shall not he greater than the ratio of the difference
of the principal stirsscs to their sum. Or what is the same thing,
if the friction, among the particles of the unitcubeofgrains
mapped out by the principal planes (northandsouth and
east and west on tig. 14) is to be sufficient to enable the
minor principal stress to resist the tendency of the major to
flatten out the cube, then the ratio of the minor to major stress
must not he less than (1  sin ^) : (1 + sin (p).
The value of the angle of friction <f> for a granular mass
may be found by shooting the mass out on a level plane, and
when the surface of the conical mass ceases to run, its slope to
the horizon is to be measured. For crushed stone (p = 37°, and
sin = "6, so that
(1  sin 1^) : (1 + sin 0) = 1 : 4
which would exactly suit the conditions required by the rib,
fig 14.
For dry granular earth spread in layers horizontally or
gently sloping = 30°, so that (1  sin 0) : (1 + sin 0) = 1 : 3.
Compare the preceding chapter on Rankine's Method of
the Ellipse of Stress, Chapter II.
Approximate Elliptic Pjb.
Consider the circular quadrantal rib ACB, fig. 6, loaded
with the verticalloadarea between itself and the straight
extrados ^r above its crown A. The conjugate horizontal
loadarea hglks is wholly positive or inwards, and its area is
474
APPLIED MECHANICS.
[chap. XXII.
i/*. On fig. 15 the same quadrant of a circle ARS' is shown,
having the same directrix OE at a height AO = \BA = \r as
before. The conjugateloadarea is hglke, only it has iDcen
turned round so that hglk the treblebatterboundary is outside,
while the other boundary he is now vertical instead of sloping
at 45°. The area is still ^r^ and the breadth at each point
Hydrostatic ctrtf/ Ceos ta t/c Ribs /o.
l/ju forint y YmyfiKr TofoT(/laL
Y
CoTfjimai^e
S<n/'rrc^"^f
Z L\ '^^.^^^^ ' ,  .>^^.a i
Fig. 15.
exactly the same as before. The slopes of the treblebatter are
now, h(j battering at 3 in 4, (jl at 1 in 2, while Ik is vertical.
Of course the exact outer boundary is a gentle curve to which
that treblebatter is a close approximation. Let now a sinrjle
houndarij f'li be drawn so that when produced it will pass
through F, and such that the enclosed area hf'h'e shall equal
CHAP. XXII.] DESIGN OF MASONRY AUCHES. 475'
h(jlh\ that is ^r. It is only necessary to draw Ff'h' at the
l)atter 4 in 10. We have then hf'li'e as a rough approxi
mation to the conjugate horizontalloadarea for the circular
([iiadrantal rib AB'IS' loaded up to a straight extrados at a
height OA, onethird of the rise DA above the crown.
Suppose now that the quadrant be slightly modified in form
so as to exactly balance under these two loadareas. We have
A B'S' the quadrant of some curve differing but slightly from
a circle, and balancing under the verticalload area between
itself and a horizontal extrados at a heighi over the crown
OA which is onethird of the rise DA, together with the
horizontalloadarea standing on the platform he equal to DA
the rise, and mapped out by the straight slope Ffli battering^
at 4 to 10.
The modified quadrant AB'S' is called a geostatic rib,
because both loads can be simultaneously imposed on the rib
by immersing it in a frictional granular mass of which OEY
is the free horizontal surface, and the friction among the
particles such that the unitcubes mapped out horizontally
and vertically are themselves in equilibrium with the horizontal
load on each fourtentlu of the vertical. Dry granular earth
spread outside the arch in horizontal layers can, as already
explained, produce such a load which is shortly called a
geostatic load, the potential uniformly increasing with the
depth below the free surface. At x, a point at the same
depth as B , the ellipse of stress is shown on j&g. 15, FW is
an arc, and the semidiameters are in the ratio of 4 to 10.
By transforming horizontally the rib AB'S', in any ratio,
we obtain a new rib ABS. If AB'S' be the quadrant of a
circle, then ABS is the quadrant of an ellipse, the new hori
zontal load area being bg'l'k'e derived from h[/lke by a horizontal
transformation in the diqjlicate ratio. But if AB'S' be a true
geostatic quadrantal rib, differing little from a circular quadrant,
then the new rib ABS is also the quadrant of a true geostatic
rib differing but little from the quadrant of an ellipse, and the
horizontalloadarea is bfhe derived from bf'h'e by a horizontal
transformation in the duplicate ratio.
By choosing a suitable ratio for the horizontal transforma
tion, namely, one for which the duplicate ratio is 10 to 4, we
have Efh sloping at the batter 1 to 1, the load is then o, fluid
load, and the rib ABS is called a liydrostatic rib. Xow the
ratio whose duplicate is 10 to 4 is approximately 3 to 2 ; hence
the hydrostatic rib, of which ABS is the left half, has its span
three times the rise, and the depth of the load over the crown
476 APPLIED MECHANICS. [CHAP. XXII.
onethird ^)a?*^ of the rise. The ellipse of stress has become a
circle of stress.
In this way we see that a semielliptic rib, or one which in
courtesy (see fig. 2) is called semielliptic, may with like pro
priety be called a complete hydrostatic rib. That if the span be
treble the rise the surface of the fluid must be over the crown at
a height oncthircl the rise.
It is better now to suppose that, on fig. l.l, we start with
this liydrostatic rib ABS loaded outside up to the fluid surface
OFF, so that the 45°, or 1 to 1 line Ffh, defines the horizontal
loadarea. Tlien to suppose the geostatic rib AB'S' to be
derived from ABS, by a crushin in the ratio 2 to 3, the hori
zontalload area crushingin in the ratio 4 to 9, or 4 to 10
nearly, and defined by the 4 to 10 battering line Ffli' : the
circle of stress at x becoming the ellipse of stress with diameters
in this duplicate ratio. The particular geostatic rib AB'S'
derived by this crushingin of the particular hydrostatic rib
ABS is sensibly the quadrant of a circle. One might readily
fall into the error of supposing that all hydrostatic ribs might
l)e crushedin so as to be sensibly quadrants of circles by
choosing a fractional ratio of transformation to make the semi
span equal to the rise. But by considering the genesis we have
employed, it will appear that only those hydrostatic ribs, with the
surface of the fluid at a height over the crown at least onethird
of the rise, can be crushedin to give derived geostaticribs with
half span equal to rise, so that such derived ribs may be, in
any other respect, like the quadrant of a circle. Because
starting as we do on tig. 15 with a horizontalloadarea hilkc
wholly positive or inwards, the rougher approximation hf'h'e is
reasonable, but with a lower directrix to start with, the hori
zontalloadarea would be partly positive and partly negative,
to which one sloping boundary Ffh' could, in no sense, give an
approximation.
We have this important practical distinction, that all sensibly
semiellipticribs may be treated as sensibly geostatic whether
derived from a hydrostatic rib by a 2^Mout or di pushin, pro
vided the height of the load over the crown be at least onethird
of the rise. For loads over the crown less than onethird of the
rise tlie hydrostatic rib itself, and geostatic ribs got by a pull
out, are sensibly elliptic, but not those got by a pushin.
Hydrostatic and Geostatic Ribs.
Let AliS be the quadrant of a hydrostatic rib. fig. 15.
The origin is the point over the crown, while the axis OF is the
CHAP. XXII.J DESIGN OF MASONRY AKCIIES. 477
surface of the fluid, the vertical is the axis OX. The depths
of A and S the crown and springing are a\, and Xi. The rise
of the arch is a = (^r,  Xo), and the halfspan is //i. The
verticalloadarea is OESA, while the horizontalloadarea,
mapped out by the 45° or 1 to 1 line is hflie; it is the difference
of two half squares, namely h (^i"  x^), and this multiplied
by IV, the weight of a cubic foot of the fluid, gives us the thrust
at the crown 7",, equal to }iW{jr^  x^^). Since the extern load
is normal at each point on the back of the rib, the thrust along
the rib is constant. So that we have, at *S the springing point,
at B any intermediate point, and at the crown A,
iij
1\ =T = 1\ = — (a,'  ;r;). (20)
The rule, p. 427, for the thrust at the crown of a rib applies
alike to every point of the hydrostatic rib. To get the radius of
curvature at any point of the rib it is only necessary to divide
the constant thrust T by the intensity of the extern normal load^
which itself is the same as the mass of the fluid vj multiplied
by the depth of the point on the rib below the surface.
As w is common to both expressions, the same quotient is
obtained by dividing the horizontalloadarea by the depth of the
point. The radii of curvature of the points >S', B, and A, are —
X\' — .^0 2Ji" Xf, Xi' —
pi = ^ » P = ~~ o^ ' l°" " 7,—^ ' (21)
The central of these expressions is an equation to the
hydrostatic rib, and is to be expressed simply in words thus —
at any point of the rib the radius of curvature and the depth
of the •point below the surface have a constant product ; — the value
of the constant being the horizontal conjugateloadarea for the
quadrant. The constant involves two linear quantities called
parameters; they are x^ and x^, or, what is the same, a = {x^ x^
the rise of the rib, and x^ the depth of the load over the crown.
Any two values assigned to those determine a particular
hydrostatic rib, when ijx the halfspan can be approximated to.
In having two parameters this rib is like the ellipse. Further,
as the depth to the rib increases continuously from the
crown A down to the springing S, it follows that the radius of
curvature decreases continuously between those points just as
in the quadrant of an ellipse.
478 APl'LIEl* MECHANICS. [CHAP. XXII.
Kankine approximates to the halfspan ^, as follows :
see the lowest part of fig. 17, which shows the quadrant
of the hydrostatic rib, and the quadrant of an auxiliary
ellipse having tlie same absolute rise a = (.Tj  r„), and having
its extreme radii of curvature in the same ratio to each other
that they have on the other quadrant. If h is the halfspan
•of this ellipse, it will be a fair approximation to y^. Now,
the extreme radii of curvature of the elliptic quadrant are
Jy^ f a, and « ^ h, so that their ratio is J^ : a'. For the
quadrant of the hydrostatic rib their ratio is /og : p,, or Xx : x^.
Equating and arranging we have
= a 'j^ . (22)
\ «0
That is, an approximation to the half span of the hydrostatic rib
is the rise multiplied l^y the cube root of the ratio of the depth
of the springing point to that of the crown.
That is always about 5 per cent, too great, or, more definitely,
from h is to be deducted onethirtieth of the radius of curvature
at the crown of either the auxiliary ellipse or of the hydrostatic
rib as it may be convenient. The closer approximation to the
half span of the hydrostatic rib is
a
(23)
In this way we shall find the span of the hydrostatic rib
having the depth of the springing points four times as great as
that of the crown, or, what is the same thing, having the rise of
the rib treble the depth of load over crown. Putting unity for
the depth of the load at the crown A, fig. 15, then x^ = OA = 1,
and the rise is f = DA = 3, while Xi = BS=4:. By equation (22)
J = 3 ^J4 = 476.
Squaring this and dividing by a we get 7*55, the radius of
curvature of crown of auxiliary ellii)se. Or for the hydrostatic
rib, the horizontalloadarea being ^ (4  1), we divide it by x^,
and get p„ = 7*5. Also a thirtieth part of these radii is "252,
which is the same as 5 per cent, of b. Keducing h by this
amount, we get y^ = 4*51, and doubling, the span of the rib
is 002.
Hence for the hydrostatic rib with the load over tlie crown
u third part of the rise the span is three ti/iu\s the rise,
CHAP. XXII.] DKSIGN OF MASONRY ARCHES. 479
a result which we arrived at in the last section by a ([uite
independent approximation of our own.
For this particular hydrostatic rib the approximation given
by (Rankine's) equations (22) and (23) are verified graphically on
fig. 17, the curve being drawn in small arcs beginning at the
crown A. It is now more convenient to have OA = 3 feet,
and OD = 12 feet. The scale was double before reduction
by photography. Horizontal lines were ruled at different
sou)idin;/s below OJkJ the surface of the fluid. The conjugate
horizontalloadarea h (144  9), or 67'5 square feet, being
divided by 3, gives po = 22'5. With this radius the first arc
is drawn beginning at A, and ending when it cuts the
horizontal at the sounding 3'75. Along the old radius at this
point a new shorter radius is laid off, with which a second
arc is drawn till it cuts the horizontal through the next
sounding 4o. In this way arc after arc is drawn till the last
one has a vertical tangent at (script) s, which completes the
approximate quadrant. As each arc has been drawn of constant
radius, while the radius ought to have decreased, it follows
that the semidiameters of this approximate quadrant are
too large. By measurement Ad is found to be 5 per cent,
greater than the given rise AD, and we conclude that ds is
5 per cent, too big also ; hence BS is laid off at that reduced
value when ABS, the false quadrant, is struck from two main
centres, and an intermediate one as already explained at fig. 2,
the whole rib being struck from five centres. By measure
ment the halfspans ds and BS verify the values already
calculated by the equations (22) and (23).
The soundings and radii used in the graphical construction
of the curve are given below in two rows. Each radius is
found in turn by dividing the sounding out of the area Q7\).
These radii begin w4th the largest p,^ = 22*5, and end with
jOi = 56.
300, 375, 45, 60, 75, 90, 105, 1125, 12.
225, 180, 15, 11, 90, 7*5, 65, 600, 56.
Observe that in this particular rib, 5 per cent, of the half
span and joth of the crown radius are the same, but in other
ribs the fraction of the radius is to be preferred.
In the same way the spans of two other particular hydro
static ribs may be calculated from the equations (22j and (23),
The depth of load over crown (jto) being taken as unity, then
for the rise (a = Xi  .r^) three times the load, the span (2_?/i)
was found to be 3*01 times the rise. With the rise four
480
APPLIED MECHANICS.
[chap. XXII.
times the load, the span is 3*22 times the rise, and with the rise
five times the load the span is 3"41 times the rise. Exact
values, by elliptic functions, are 3"06, 3'28, 3"47.*
As already pointed out, on page 476, only the first of
these hydrostatic ribs may be transformed by pushingin. All
three may be transformed by pulling them out horizontally,
and the resulting geostatic ribs are sensibly semielliptic. Let
the pullout, in the first place, be such that the 1 to 1 sloping
boundary of the horizontalloadarea becomes a 1^ to 1 sloping
boundary. That is, the duplicate of the ratio of transformation
is 1*5, so that the ratio of transformation is 1"225, which is the
multiplier to give the three new spans. In the second place,
the 1 to 1 sloping boundary is to become a 2 to 1 sloping
boundary, so that the ratio of transformation is now 1414,
which again is the multiplier to give the three new spans.
We have then the following nine geostatic ribs, that is, ribs
sensibly semi elliptic, able to be struck out from five centres as
on ficj. 2, and such that, when loaded between the ribs themselves
and a straight extrados at the given height above the crown,
require for equilibrium a horizontalloadarea mapped out by
a sloping line at the batter given in the table. In the following
table the depth of the load over the crown is given as a
fraction of the rise of the rib, and the span as a multiple of
the rise : —
Spans in Terms of the Rise of Semi Elliptic Bibs which are
sensibly Geostatic.
Boundary of hori
zontalloadarea.
Load at Crown
rd of Rise.
Load at Crown
jth of Rise.
Load at Crown
th of Rise.
1 to 1
301
322
341
U to 1
368
394
418
2 to 1
425
455
482
With the semielliptic rib, it being desirable to liave the
span between four and five times the rise, it appears from the
above table that the depth of the load over the crown should
be between th and J^th of the rise.
•Notes on the Hydrostatic Curve, and its Application to the Stability of
Elliptic Miisonrv Arches, by J. T. Jackson, m.a.i., Assistant Professor of Civil
Engineering, Trinity College, Dublin. Trahs. of the Inst, of Civil Engineers in
Ireland, vol. xli, 1915.
CHAl'. Wll.] DKSJIGN OF MASONRY AKGIIE8. 481
SemiElliptic Masoxrv Auch,
111 this arch tlie soffit is a false semiellipse struck from five
centres as on fig. 2. The span being cleterniined upon, the rise
of the sotftt may be any fraction of the span from a fourth to a
fifth. The radius at the crown is now determined, being the
square of the halfspan divided by the rise. Then a mean pro
portional between this radius and '12 gives liankine's empirical
thickness of the keystone ; to this a suitable sr.rcharge over the
keystone being added, we have the depth of the load over the
crown, which will be from a third to a fifth of the rise. For
example, take the lower part of fig. 16, showing a quadrant of
an elliptic masonry arch, halfspan of sotiit 38 feet and rise
19 feet, the radius of crown will be 38 ^ 19 = 76 feet. By
Eankine's formuke the thickness of the keystone is ^7Q x '12
= 3*02 for a single arch, or for one of a series it is v/76 x 17
= o*6. Now, it is usual to have the ring of stone of uniform
thickness, instead of thickening out as in the segmental arch,
so that 4 feet had better be allowed for the uniform thickness
of the ring. To this a surcharge of not less than a foot and
a half is to be added. With this minimum surcharge the
depth of load at crown of soffit is sensibly a fourth of the rise.
But we have on fig. 16 adopted a surcharge of 3 feet, so that
the depth of the load at crown of soffit is 7 feet. This we
have chosen, that the total rise may be 26 feet, so that this
semielliptic arch may have the same span and total rise
or headroom as the segmental arch, figs. 12 and 5, the two
patterns being rival designs for the same railway viaduct of
many arches.
The line of stress is to be supposed to be along the soffit in
the first place. The soffit is now, instead of elliptical, to be con
sidered as sensibly a geostatic rib. The quadrant is shown on
the upper part of fig. 16. The three data are OA = ^Tq = 7,
OD = Xi = 26 feet, and the half span c = sy, = 38 feet, where x^
and .Ti are the extreme depths of the hydrostatic rib from which
the geostatic rib is derived by a horizontal transformation in
the ratio s. The thrust Tq at A the crown of the geostatic rib is
the mass of the superstructure to multiplied by the horizontal
loadarea hf'h'e, while Ti the thrust at the springing is an sth
part of T„, for before the transformation they were equal, and
only T^ has increased. We shall return to the strict numerical
solution of the arch on fig. 16 among the examples following.
2 I
0, Q,
CHAP. XX II. J DESIGN OF MASONRY ARCHES, 483
In the nieautiiue fig. 16 may serve to illustrate tlie quite general
treatjiient.
Since j„ = 7 is sensibly ^rtl of a = 19, we may look in the
second column of the table, and since the span 76 is exactly 4
expressed in t^rms of the rise a, the slope of the boundary of
the horizontalloadarea will lie between 1 to 1 and 2 to 1, just
as 4 lies between 3*68 and 4"25. By proportional parts this
slope is If to 1. Quite strictly it is 1"853 to 1 : see the numerical
examples following. On fig. 16, s to 1 is taken as sensibly 2 to 1.
At A the thrust is T„ = w x area hf'h'c = 627'?t', y.nd dividing by .s,
Ti = 420u' nearly.
Consider the quadrant of the masonry ring as rigid, and
apply a pair of equal and opposite couples, one at the crown,
and the other at the springing. Let the couple at the crown
be of a moment such that it shifts T^ just into the middle third
of the keystone as shown on the lower part of the figure ; then
the other couple will shift Ti well into the middle third. In
no ease will 1\ be shifted past the middle third, as 7\ is never
twice as great as Tj. The modified line of stress begins and
ends in the middle third, and as its curvature is continuously
varying, the whole curve is confined to the middle third of the
masonry ring, shown white on the figure.
For the middle elastic part of the ring, shown dense black,
the heavy backing is left out. It stretches between the pair of
joints where the tangents to the soffit are inclined at 30° to the
horizon. For one thing T,, is now slightly reduced. The soffit
for this middle part of the ring is actually all struck from the
main centre, and is really a segmental circular masonry ring
springing at 30° to the horizon. The modified line of stress of
this central part, for the most perfect economy of masonry, would
start at the lower limit of the middle third at the keystone
and would reach the upper limit of the middle third at the
joint of rupture, the springing joint for this black part of the
ring, that is at the joint which, when produced, goes through
the main centre, and makes 60° with the vertical. This will
now determine whether the thickness adopted for the archring
is sufficient, for the horizontal through the lower limit of the
middle third of the crown joint, and the line at SO'' to the
horizon drawn through the upper limit of the joint of rupture
must meet on the vertical through the centre of gravity of the
corresponding superstructure. The construction will be similar
to that shown at the lower righthand corner of fig. 7. In this
case the thickness of the ring is the least for permanent equili
brium, and at both crown and joint of rupture the maximum
2l2
484 APPLIED MECHANICS. [CHAP. XXII.
stress is double the average. With a more liberal thickness of
ring, different pairs of tangents may be drawn to meet on the
proper vertical, and the actual line of stress will be determined
by the ceasing of the subsidence of the keystone when the
centre is slowly removed. It is always safe to assume that the
maximum intensity of the stress at the crown joint is double
the average, oc that the real factor of safety is only half the
apparent factor against crushing.
If a uniform live load be added all over the span equivalent
to an additional height of the superstructure, 1 feet on fig. 16,
the calculations could be repeated, making 3;,, and Xi greater l)y
this amount. The increase of Ty is simply 7v multiplied by the
area of the rectangle of base 38 feet, and height 1 feet, or
(T/  Ti) = 48w". And for the increase of To we have (by the
rule for the thrust at the crown of a rib, p. 427) merely to
multiply l'2ovj the extra normal load at the crown by 76, the
radius of curvature, when we have {TJ  To) = 95?'.'. Light
spandrils to resist this load are shown on fig. 16. The depth
from the formation to the top of the hea\y spandril walls upon
which the light spandrils ride measures 8 feet on the figure.
If I be the distance they stretch out longitudinally, then 8ivl is
the overhead load pressing the spandrils down on their base.
Then with '7 for the coefficient of friction of stone on stone we
have the frictional resistance of the light spandrils 56 wl equal
to 95i/', so that / = 16"9 feet.
Examples.
21. In a semielliptic masonry arch the span of the soffit is 76 feet and the
rise is 19 feet, and the level of rails is 7 feet above the crown of the soffit. The
soffit which is to be struck from •') centres is to be taken in tlie fiisi place as the
line of stress. Assuming it to be sensibly a gcostatic rib, find the thrust at the
crown and springing.
The extreme depths of a quadrant of tlie hydrostatic rib from which the
gcostatic rib is derived are .To = 7 and x\ = 26. A rough and a close approxima
tion to the halfspan of this ky/rostatic rib are by equations (22) and (23),
=;/!=
29424,
b
y, =29424 ,,Vj,j
== 29424  1Jil8 =27906.
The halfspan of the gcostatic rib being given s>j\ = 38, we have then the ratio
of transformation or pullout by which the halfspan 27906 has become 38 to be
38
, = — — = 13617
27900
and .(2 = 1853.
Hence the 1 to 1 boundary of the horizontalloadarca for the hydrostatic rib
CHAP. XXII.] DESIGN OF MASONRY ARCHES. 485
ha\'ing been puUedout in the dupUcale ratio l8o3 to I, this furnishes the slope of
the boiiiuJiirv mapping out the horizontalloadaiea of the geoiflatic rib.
Multiplying w the weight ot a cubic foot of superstructure into this area the
thrust at the crown is
r„= 18.53 X ^(26 7) = 5Shr,
and Ti c=To^s= ^— = 427^.
l"oo2
Note. — These are the more exact values which should have appeared on
fig. 16, only that the slope of the boundary of the geostatic rib was taken 2 to 1
instead of 1853 to l^for clearness of explanation.
22. If a liveload equivalent to an additional height of superstructure of 125
feet be added all over the masonry arch in last example, find now the thrusts at
crown and springing. 
We have now Xo = 825 and .ri = 27'25,
6= 193/^1^=28297,
\ 82o
yi = 28297  3V ^ = 26892,
og
 = 1414 and * = 2,
26892
w
To = 2 (27252  8252) ^ 67o?c,
Ti' = To ^s = il8w.
Note that now tlie boundary of the conjugate horizontalloadarea slopes at
2 to 1. Also To'  To = Glow  oSliv = Sitv, about the product of \2r)w and
38 ^ 19 already used for an approximation : see fig. 17. Again Ti'  Ti = blw,
a little more than the, simple product of l2bw and 38 the halfspan.
23. The pier between two arches is 12 feet thick at the top, and the faces
batter 1 in 20 : find the centre of stress at a joint 45 feet below the springing level.
That joint is 16 o feet thick, the weight of tiie pier (;41(i'. The half arches
rest 427ct' and 478u on its top ; the whole weight above the joint is 1546«. The
excess of the horizontal load on one side over the other is 95«. If ; be the devia
tion of the centre of stress from the midiileof the joint 45 feet down, the equation
of moments is
loiGtcz = 45 X 9.5ir.
;: = 276 feet.
This is onesixth\oi 165, so that the centre of stress is just within the middle
third of the joint.
24. The abutment pier is 20 feet thick at top, and its faces batter at 1 in 20.
At*a joint 45 below the springing, find z the deviation of the centre of stress from
the middle, when the pier bears a halffinished arch on one side only.
The joint is 245 feet thick ; and if we take the mass ot the pier 2m and reckon
a halffinished arch on one side only, then weight of pier is 2002«, and the
equation of moments is
2002it" r 427« (0 f 8) = 581m x 45,
so that z = 93 or threeeiyhths of 245 ; the thickness of the joint and the line of
stress does not come closer to the face of pier than an eighth of its thickness.
486 APPLIED MECHANICS. [CHAl'. XXIL
25. If. when the viaduct is finished, an arch fiiUs, it leaves a whole arch on
one side only of the abutment pier.
The first term in the equation above must be halved, ae we cannot now consider
the pier of double density ; at the same time we have now a mass of masonry
20 feet by 26 feet over the pier, so we must add 520Msand deduct lOOlw; from the
first term and
152hf: + 427!f(; + Sj = oSlw x 4o.
And now z= 11 0 less than half of 245, so that the centre of stress is still
within the masonry.
Masonry Abutments.
In the case of the segmental arch (fig. 12), the abutment
is readily designed by building the masonry in benches stepped
at the back, the oblique thrust of the arch, shown by a
dotanddash line, to lie in the middle third of the abutment
if it be no deeper than the rise of the arch. The masonry may
be entirely in square dressed course.«^, but the central part may.
with advantage, be in radial courses. The abutment is best
honeycombed at the back with thick walls for the thin spandrils
to ride out upon. The spaces between the thick walls to be
thoroughly drained and packed with heavy ballast and broken
shivers of stones. The spaces between the thin spandrils to
be coated with waterproof material, arched or bridged over,
and preferably left void, drainpipes and ventilating holes
being provided.
In the semielliptic arch, lower part of fig. 16, the heavy
spandrils, with square dressed joints, are shown in three
benches, from tlie springing level up to the joint of rupture.
The thickness z, at springing level, is calculated in the first place,
as if the backing required were like that of a fiuid load, or as if
the horizontalloadarea were mapped out by a 1 to 1 boundary.
In the example 7 x 26//'' = ^?' (26  8), or / = 17 feet,
taking v/ and ii' as e(iual. A rightangled triangle, with the
base z, maps out the backing. The reentrant angles of the
benches are to lie on the hypotenuse of this triangle, shown with
a dotanddash line ; then the middle and top benches are slid
along inwards till they bult against the squaredressed backs of
the voussoirs. As v' is likely to be 20 per cent, greater than w,
that is, the mass of tlie backing greater than that of the
superstructure, we will have the reduced value ~ = 14 feet.
We have sliown that the conjugate horizontalloadarea
for elliptic arclies has its boundary sloping at 2 to 1 at most,
that is double the horizontal action of a Huid load. l»ut if the
CUAI'. XXII.] DKSICN OF MASONRY ARCHES. 487
abutment, as on the lower part of fig. IG, is built in an excavation
in old consolidated earth, then the earth filled between the
abutment and the face of the excavation can be matle to press
horizontally, like a Huid, by punning it hard in thin horizontal
layers. This reinforcing the frictional stability of the masonry
abutment, as already designed, supplies the necessary resistance
to tlie spreading of the arch. The centre of the arch is to be
struck when about half of the superstructure is built, and
before the earth is rammed behind the abutment. When the
masonry has settled and consolidated, the earth is to be first
rammed behind the abutment, and then the superstructure
finished. On the other hand, if the abutment is built and a
loose embankment built over it, then even by filling the
best stuff behind the abutment, it will only press horizontally
with half a tiuid load, so that the masonry abutment must
now be equal to a 1 to 1 bounded load area. That is, z must
be increased by 50 per cent, from 14 feet to 21 feet. Adding
to each of these, 4 feet for the thickness of the springing stone,
we have the total thickness of an abutment for the semielliptical
masonry arch (fig. 16) to be 18 feet and 25 feet in the best and
worst circimistances respectively. These are almost exactly
^th and ird of 76, the radius of curvature of the crown of
the soffit. And so we have verified Eankine's proportion :
see his Civil Eng., p. 514: The thickness of the abutments of a
masonry bridge are from a third to a fifth of the radius of curvature
at the croivn.
Masonry Piers and Abutment Piers.
The design for a common pier is shown on the left half of
fig. 17. It is for a viaduct of many arches, of which half an
arch is shown on fig. 16. Every fourth or fifth pier is to be
an abutment pier, the design of which is shown on the right
half of fig. 17. Both piers are 45 feet high, and are divided
into three blocks of 15 feet each. The common pier is 12 feet
thick at the top, and the abutment pier is 20 feet thick.
Their faces batter at 1 in 20. The arch to the left of the
common pier bears a live load e(uivalent to 1^ feet of extra
height of its superstructure. Hence the excess horizontal
thrust of this arch above the one on the right side of the pier
is 1252/; X 76, or 9oiy. It is shown on fig. 17, acting on the
horizontal line, through the centre of gravity of the horizontal
loadarea, as drawn on the figure above. Then 468w and
420i^; are the vertical loads on the springing stones at the
€HAP. XXII.] DESIGN OF MASONIJY ARCHES. 489
•conieis of tlie pier, due to the halt' arch loaded, and the half
Arch unloaded. They are quoted from fig. 16. There are also
shown four barbs on the vertical line down the middle of
the pier, which are the weights of the column of masonry
between the two half arches, and of the three blocks of
masonry into which we divided the pier itself. The mass of the
masonry f is taken as unit. The vertical scales, both for
•dimensions and loads, are three times as fine as the horizontal.
By drawing a force and a link polygon, the centres of stress
at each of the three joints below the springing are defined by
the barbs of oblique arrows. Each centre of stress is well
witliin the wliite "kernel," which is the middle tliird of the
masonry of the pier.
It is absolutely necessary, for the permanent stability of
the pier, that the line of stress be confined to the middle third,
otherwise the joints would open, as the pier was rocked by
the live load shifting from one arch to the other, the mortar
drop out, and the whole l)e destroyed.
The everyday work of the abutment pier is to reflect vibra
tions passing from arch to arch along the viaduct. Once in its
history it has to act as an abutment, the arches on one side
only being finished, or partly built, Now the arch resting on
the abutment pier will have its centre struck when itself half
complrterh On fig. 17 the horizontal and vertical loads, 280^(7
and 210^r, at the left corner of the abutment pier, are due
to the half finished arch on that side, and are quoted from
fig. 16, being slightly modified because the pier is of dense
masonry. The smaller force polygon, and its corresponding
link polygon, define the centres of stress to be within the white
kernel. That is, the centres of stress do not approach so close
to the face as onccigldli of the thickness, which is a suitable
limit, as the load is steady, and only in one direction, just as in
retaining walls.
If by an accident, such as the failure of a foundation, an
arch of the viaduct should fall, the damage should only reach
to the nearest abutment pier. The larger polygon shows the
full load of an arch on one side, and the oblique dotanddash
arrows show the centres of stress to be still within the masonry,
so that the abutment pier could be expected to sustain the
complete arch for a short time till it could be shored up. A
study of the construction shows that the second last full line
oblique arrow produced to the lowest joint defines the same
centre as the last oblique dotanddash line. This saves the
trouble of drawing the larger polygon.
4ftO APPLIED MECHANICS. [CHAP. XXII.
Tunnel Shell.
On fig. IS is shown a tunnel shell ; it has nearly the same
profile as the Blechingley tunnel (see Plate 1 of Simms'
Practical TnniuUinfj). In other respects the shell is designed
to illustrate the mutual stability established between the shell
and the surrounding cubes of earth.
The left part is a half crosssection of {\\e front louith of the
tunnel, where a deep cutting is first made in old consolidated
earth, the sides as nearly vertical as may be. The thick side
walls are first built, 4 feet thick, and the roof turned on a
centre. It is 2 feet thick at the crown, thickening by a ring of
bricks at intervals outwards to the haunches. The notches are
covered with a band of asphalt to render the roof waterproof.
From A, the crown of the soffit, the invert is swept out truly
with a suitable versine, to accommodate a covered drain
between the rails. The earth is now filled between the back
of the walls and the face of the cutting. Up to the level of h
it is punned in thin layers, that the cubes may press horizon
tally like a fluid ; above that it is filled more loosely in thick
layers, so as to be pressing horizontally, with little more than a
third of column overhead. The lowest level of the filling when
completed is to be SP, so that Sa may be a third part of 14' 5,
the radius of the circular quadrant ah assumed to be the line
of stress. This quadrant, then, is a geostatic rib, whose hori
zontalloadarea is bounded by the 4 to 10 sloping line lying
close to the ^rd to 1 slope, indicating the minimum horizontal
load necessary for the equilibrium of the cubes of earth. The
depths of a and h, below SP, are 5 and 20 feet, so that the
horizontalloadarea measures ^^ . ^ (20'  5') = 75. The liorizon
tal thrust at a, the centre of tlie crown joint, is lov:. At h the
vertical thrust is Ibv), multiplied by ^, the ratio of transformation
by which the quadrant ah is converted into the corresponding
hydrostatic quadrant with the same extreme depths (see fig.
15). To this vertical thrust of the rib at h there falls to be
added a column of earth, 20 feet by 5 feet, making in all 2V2ii\
acting down through h. The total upward tlirust, acting partly
on CD, the base of the tb.ick wall, and partly on DE, the half
invert, must also be 212/'. Now, from the manner of con
struction, most of this load is thrown on VI), the invert only
ottering the minimum reaction to the swelling of the earth
Itoneath it being built later and l^eing elastic and yielding.
TUN'NEL SHELL
Cut ajxd Core jr.
TLo soffit lias neariy the samo jitoGle as tho Bletchingloy Tnnnel :
SP is tho (jnidrant o£ a falso ellipso irith gA = 10 and gF 12 feet. Tho
fonnalion is 22 feet belo^r^, and the invert 13 struck from A 03 centre.
In the shaUowJength tho roof ia 2 feet thick at tho crown, the sido iralk 4,
and lUo'IuTcrl li feot thick. Tho shcU has a uniform thickness of 3 feot
on the deep length.
^OfL eselra/'tilina, ^
CeyeL of'J^illtna,
2iay
7i Si 00 oJlsp. SaJir,
EARTH PRESSINgC
LIKE A FLUID.
^ % k
dJXXXXXuXVIJ
Kclow tlio juiut li the earth
is rammed in 9iucU layers
hetween tlio face cf tho ex
cavation and tho wall, above
that it is spread, in 12'iiic!>
lajers.
lun;. Cot. Ben. ISDJ.
30 ziTGach
i J ^cLofCoOanjr lilllUIII
souieaeh
EARTH ON THE POINT OFHEAVING UP.
Zfxapei'.
8
W — J20 lbs.
r£ET lO
aoorr. ^ ^ * «
3 J20nre^
2 g/^cois\
Fig. IS.
492 APPLIED MECHANICS. [CHAP. XXII.
Suppose all the unitcubes of earth at the level CE, 30 feet
below the free surface PS, to be pressed horizontally with 30w,
a load equal to the overhead column. Then the least vertical
pressure on the 12 unitcubes (white) under the invert DE,
consistent with the stability of those cubes themselves, is lOw,
or onethird of column. From the ranmiing of the earth at C,
tlie vertical load, on eacli of the unitcubes (black) under CD
the base of the wall, may be taken as column, or 30?/', more or
less. This furnishes an equation to verify if the thickness of
the walls' foundation is sufficient. For we must have
30?/ X t + 10?/; X 12 = 212w.
This gives t nearly 4 feet, and taking t exactly 4 feet, and
modifying the equation, we liave the vertical load on the
four (black) unitcubes below the wall to be IZw, and
•2Zw X 4 + lOif X 12 = 212^6'.
The radius of the line of stress in the invert is 25 feet,
and it has the upward load of lOvj uniform along the span,
just as in fig. 14, so that the thrust along it is constant,
and equal to IQiv x 25 = 250«.
Next to join together these two lines of stress, in roof and
invert, by a line of stress confined to the middle third of the
lower part of the thick wall, we must consider the horizontal
load applied to the liack of the wall by the punned earth
l^ehind it. This load is given by an area mapped out by any
boundary lying between the 1 to 1 and ^ to 1 boundaries.
On fig. 18 these areas are measured up to the 1 to 1 boundary,
and at intervals of 2 feet of depth are o^v\ 46?r, b^w. and 54?'".
Beginning with 21 2?/', at h, and compounding those in order by
means of the force polygon, pole 0, and the link polygon LM,
the centres of stresses are (lefined — c by the arrow with two
feathers ; d by the arrow with three feathers ; c by the arrow
with four feathers; and lastly, B, by the arrow with five feathers.
In another example the points c, d, c, &c., are to be laid down
one after another, in suitable position, each in its own joint,
and the horizontal areas necessary to make them so lie
determined in succession, and provided the boundary of those
load areas lies between the two practical limits, the 1 to 1
and J to 1 boundaries ; then hcdeB is a possible line of stress,
consistent with tlie equilibrium of the punned earth itself.
CilAT. XXII.] DESIGN OF MASONItV AUCHES. 49.'>
Addiug 70 /r for the horizontal load below B, we liave
the sum of 30h', 46?^', 50?r, 547r, and ^Qiv, equal to 250?t',
the thrust at crown of invert.
Both the top and bottom quadrant, then, of the left half of
tig. 18 are in horizontal and vertical equilibrium, consistent
with the equilibrium of the surrounding cubes of earth, and
throwing as light a load on the invert as may be. There still
remains the equilibriunt of moments of the shell as a whole.
This is satisfied by the fact that the principal axes of stress for
the cubes of earth are vertical and horizontal, the free surface
being horizontal, and as well the axes of the shell are also
vertical and horizontal. With the free surface of the earth,
at a heavy sidelong slope, it may become necessary to find the
cant which the principal axis of the earthpressure makes with
the vertical, which is easily done by the formulae on the model
(fig. 31, Ch. II), and then to prevent the shell of the tunnel
from twisting like a stick of barleysugar under the unbalanced
couples, one of two things has to be done. One is to build the
shell of the tunnel at tlie same cant, an example of which is
the Dove Tunnel, a section of which can be seen on the
plate xxii of the fourth edition of Simms' Practical lunnellinfj.
But the more modern way is to load the shell by an extra mass
of masonry at one foot, so as to cant the virtual axes where it
may be required.
As the filling, over the tunnel shell, gets deeper and deeper^
the problem becomes less and less one of stability, and more
and more one of strength. The right half of fig. 18 shows an
extra 30 feet of filling. The total depth is now QB = 60 feet,
the usual maximum for tunnels built in cut and cover. Even
for a tunnel built in a cutting greater than 60 feet in depth,
and covered, it is likely that only to this height will the load
overhead affect the shell, as the earth itself forms a relieving arch.
And, again, in a bored tunnel, the disturbance is only likely to
affect the earth to a limited height above the hole. If on the
diameter of the hole as base, an isosceles triangle be constructed,
with its sides sloping at 75° to the horizon, the vertex gives the
probable extent of the disturbance due to the loosening of
consolidated earth due to the borehole. For cracks will run
out on the plane on which the direction of the thrust is most
oblique. Thus, in all cases, the height of load shown on the
left half of fig. 18 is the probable maximum (see quotations,
p. 497, from Simms' treatise).
Consider the right half of fig. 18 : the additional vertical
load is a column, 30 feet by 16 feet, or 480;'. The upward push
494 APPLIED MECHAKICS. [CHAP. XXII.
of the unitcubes under the right half of invert is doubled,
as the depth is doubled; and taking the upward load on the
four unitcubes under the thick wall at li as 1207/, the total
extra upward push on the right half of the shell is 10?' x 12,
together with 90 tr x 4, or, in all. 480i'j'. The horizontal load
among the cubes, to the right of JS, must be equal to the
column QR, so that no one of these sixteen cubes shall have
the rectangular pair of loads on it in a ratio exceeding 3 to 1.
This requires that the horizontal thrust of the punned earth,
at the depth B, shall equal the column, but at the depth
35 feet below Q, it may be less, say, instead of Sou, it be only
20w, while at the depth 50 below q, it is 50w. Then the
loadarea standing on the base, from ?>5 to 50, has a height
half the sum of 20^/^ and bOw, that is 35?r, and multiplying
by the base 15, then we have 525?r as the probable horizontal
■extern load on the upper rigiit quadrant of the shell. With
15 feet as the radius of the new line of stress at the crown,
and the load over it at 35a6', we have, by the rule for the thrust
at the crown, 35?y x 15 = 525?'^ This establishes the horizontal
•equilil)rium of the upper right quadrant of shell. For the
lower right quadrant, the horizontal load due to the punned
w
«arth is — (58  50^) = 432?6 nearly, and adding Q8ir for the
load below 58 at B, we have in all oOO?'*, the same as the new
thrust at crown of invert. Thus the thrust at both crowns
is the same, and the shell is now of a uniform thickness
of 3 feet.
The thrust at crown is o'2dv: ^ 3 = 21000 lbs. per sq. foot
of brick, taking w =120 lbs. per cubic foot. The crushing
strength of strong brick is 154000 lbs. per square foot, so that
the apparent factor of safety against crushing is 8, and the
real factor cannot be less than 4, onehalf of it. (See p. 427,
and Kankine's Civil Enfjincerinff, p. 514, on buried arches.)
Allowance for Excessload alon<j the Ellijjtic Masonry Ming. —
Rankine's assumption of the line of stress along the sofht of the
•elliptic arch makes allowance for the excess load of hw along
tlie ring, because the softit, not being so elongated as the
line up the middle of the ring, the ratio of transformation is
greater.
Compare the two quadrants ABS (fig. 15) and ABS (fig. 5).
In the first the half span is 50 per cent, greater than the rise,
and it balances under the fluid load alone. In the second
the halfspan equals the rise, and it balances under the
fluid load together with a uniform load along the rib.
CHAP. XXII.] DESIGN OF MASOXUY AUCIIKS. 495
equivalent to voussoirs of the same mass as the lluid, and of a
depth half (the radius) of the rise. By proportional parts the
halfspan of a quadrant of given rise, to Ijalance under a
fluid load, and an excess uniform load along the rib, can be
found by reducing the halfspan for fluid load above by a
proportionate reduction.
Thus, in the numerical example (fig. 10), our second
approximation is as follows : —
Assume the line of stress to be up the middle of the
masonry ring: then Xo = 5, x^ = 26, and c^i = 40. For the
corresponding hydrorib : then h = 36382, and y^ = 34*28.
If the ring be 50 per cent, heavier than the superstructure,
the excessmass is u', for the upper half having been included,
its excess is hw, and the under half having been excluded, its
excess is I'ovj where vj is the weight of a cubic foot of the
superstructure.
Now, with no load along the rib, in addition to the fluid
load, pi = 34'28, with voussoirs 10^ feet deep, that is half of the
rise 21, //i = the rise 21, being a reduction of 13*28 feet.
Hence with the actual voussoirs of 4 feet in depth, the propor
tionate reduction is 13*28 x 4 ; 10 = 5*06. 8o that we have
the reduced value, i/i = 29*22, giving s = 1*37, and s' = 1*88.
The horizontalloadarea is, therefore, 1*88 x  (26  6") = 612.
Then Tq = 612m?. This is practically the same as the results of
Eankine's approximation (see fig. 16).
In an aqueduct, when the masonry ring has a mass greatly
denser than water, our second approximation becomes essential
to accuracy (see numerical example which follows).
ExAMrr.Ks.
26. Design of a semielliptic masonry arch of sandstone. Span 80 feet, and
with a surcharge of I'o feet.
Taking the rise a fifth of the span 16 feet, the radius of crown of soffit is
40 f 16 = 100 feet. For depth of keystone, <o = v(*17 x 100) = 4 feet, say.
First approximation. — Take line of stress along the soffit and Xq = vb, x\ = 21'5,
a = x\ — Xf) = 16, and 47/1 = 40. The horizontalloadarea is \ {xi — aro*) = 216
for the corresponding hydroarch. By equation (22) and (23),
3/ 21 5
h = 16 / — = 25204.
V 50
y: = 2.5204 3! f^ = 2388i:
40
s = —■■ = 1676, s2 = 280S.
JiO*00
Hence the thrust at the crown is 2808 x 216, or To = GOGiv.
49 G APPLIED MECHANICS. [CHAP. XXII,
27. The extreme depths of a quadrant of a hydrostatic rib uie xn = 7 anck
= 9.ri fpf>h below tiie siirf.ii;« <it tbe fluid ffiir. 161. At !i noint R' Kvo laar
27. The extreme depths of a quadrant of a hydrostatic nb uie xn
x\ = 26 feet below the surface ot the fluid (fig. 16). At a point £',
lower than the crown, find the slope of the rib to the horizon.
five Ceet
and if the
28. In a complete geostatic rib, span 76 feet, rise 19 feet, height of enrth load
over crown 7 feet (tig. 16), find the sloite of tlie rib at B, a point five feet below
the level of the crown. So that xo = 7, ^i = 26, and st/i = 38 feet.
As in last example, we must find the slope 8 of the corresponding hydro
arch, iind also s = 1'362, as in example 21, and cot 6' = stotd determines the
slope at Ji.
29. Design of an aqueduct with a semielliptic masonry ring twice as dense as
water. The clear span is to be 40 feet, and the depth of water over the keystone
2*5 feet.
We proceed by trial and error, assuming values for the rise, and calculating
the span. If there were no excess load along the ring, a first trinl value foi the
rise wouM be a tliird part of the span, but a greater vuiue is indicated, because of
the large excess load along tlie rib. Try the rise of the soffit 15 feet. This gives
the radius of crown of soffit 20 ^ 15 = 27. As an actual fluid is a troublesome
load, the greater of Raukine's values is preferable for the thickness of the keystone,
or ^0 = V('17 X 27) = 2^ feet nearly. Adopting the value 2^, and assuming tlie
line of stress up the centre of the ring, we have x„ = 3'75 and xi = 20, and
for the horizontal load area ^(20  3'75) = li)2"97. Also the rise of the line
of stress is a = 16' 25, so that
A = 1625^ (xi fa:o) = 2839, and i/i = b  s^jh ^ a = 2G7i.
Now the excess load of the upper part included is w, and of the lower half
excluded is 2w, or, on an average, l5if. This is the same as if the voK.siuiis were
of a thickness 15 ^o = 37t feet. "With no load along nb >/i = 2674, with
imaginary voiissoirs half the rise, that is Sy feet, yi = the rise = a = 1625, or is
reduced by lO'o feet. The proportionate reduction for vous.soirs 3"7o feet is 4'84,
so that the reduced value of the half sp.m of line of stress is y'l = 2674  484
= 219. The half span of soffit would then be 219  125 = 206. This is near
enough to 20 feet, the required halfspan of soflEt, so that further trial values of
the rise of soffit are not required.
For the line of stress, then, the horizontalloadarea is 192 '97 square feet,
multiplied by w = 64 lbs., gives the thrust at crown ^'o = 12350 lbs., and the
factor of safety can be evaluated.
30. A semicircular rib of radius 27 feet is loaded between itself and a hori
zontal straight line 3 feet over the crown. Find the angle of rupture, and the
inward and outward parts of the horizontalloadaiea.
By equations (5), (6), and (7), p. 440,
r,  5 4 1
C«s, = f.._^=.04.
/ 2 o 1 2 1 \ ,,.,,
^'= (27^99^38l)(')^'^=^'^"
/ 2 4 1 2 1 \
^ V27 9 9 3 8iy ^ *
Their algebraic sum is 81w, the product of the radius and load over crown.
CHAP, XXII.] DESIGN OF MASONRY AKCHES. 497
31. One of tlie shafts in the Hoosac Tunnel, Western Massachusetts, is elliptical
in form, the axes being 27 by 15 feet (Simms* Tunnelling, 6th edition, p. 527).
Show that 2 feet of brick lining is strong enough for such a shaft, surrounded by
earth, even at great depths.
The major diameter of the bore hole is M feet, so that 50 feet is about the
height of the isosceles triansile, with base 31 feet, and vertical angle 30°. This
is the deepest column of earth likely to be disturbed by the bore hole. Consider
a ring of briekwork .'jO feet bolow tlie surface of the earth, the ring itself 2 feet
thick and 1 foot deep. Assuming the line of stress, in the first place, to be up the
middle of the ring, its diameters are 29 and i? feet, their ratio is 17, almost
e.\aetly %JZ. Hence the diameters of the ellipse of stress for the extern cubes of
eai th may be in the ratio 1 to 3. Now the lesser horizontal stnss may be a third
of the potential oOw, so that the greater horizontal stress is equal to 50w. The
horizontalloadarea for a quadrant, in the direction of the major axis, is 50(f x 85,
or42oM7. This also is the thrust 7'„ isee fig. 14), at the ei.d of the minor axis of
the ring. But the area of the section of the ring is 2 square feet, so that the thrust
per square foot of brick is 2r2ow, or 25500 lbs. For strong brick the strength is
154000 lbs. per square foot. Hence the apparent factor of safety is six.
QuOXAXtONS FROM RaNKIKe's CiVIL ENGINEERING.
Page 425. — To determine, with precision, the depth required for the keystone
of an arch by direct deduction from the principles of stability and strength would
be an almost impracticable problem from its complexity. That depth is always
many times greater than the depth necessary to resist the direct crushing action "of
the thrust. The proportion in which it is so in some of the best existing examples
has been calculated and found to range from 3 to 70 ... good medium values are
those ranging from 20 to 40.*
An empirical rule, founded on dimensions of good existing examples of bridges,
is — depth of keystone, a mean proportional between the radius of curvature of the
intrados at the crown, and a constant "12 or 17 for a single arch, or one of a
series.
Page 428. — In some of the best examples of bridges, the thickness of the
abutments ranges from onethird to onefifth of the radius of curvature of the arch
at its crown.
Page 429. — The thickness adopted for piers in practice ranges from onetenth to
onefourth of the span of the arches ; the latter thickness, and those approaching
to it, being suitable for "abutmentpiers." The most common thickness for
ordinary piers is from onesixth to oneseventh of the span of the arches.
Page 435. — It appears that, in the brickwork of various existing tunnels, the
factor of safety is as low as four. This is sufficient, because of the steadiness of
the load; but in buried archways, exposed to shocics, like those of culverts under
high embankments, the factor of safety should be greater ; say, from eight to ten.
QuoxAiiONS FROM SiMMs' Pracxical Tunnelling (4th edition,
pp. 202, 203).
It is known, too, that the pressure on tunnels in comparatively shallow ground,
say, less than 40 feet below the surface, may be localised and concentrated upon
the crown of the arch with peculiar severity. Mr. Simms accounts for the greater
* Factors of safety calculated from existing bridges are likely to be what we
have called apparent ones, so that the actual factors will range from 10 to 20, the
10 being the lowest assumed in our tables.
2 K
498 APPLIED MECHANICS. [CHAP. XXIII.
pressure upon the work in shallow ground by the supposition that the whole
superincumbent mass acts vertically downwards, whilst at greater depths it is more
or less sustained as an arch over the tunnel, which is proportionally relieved of the
pressure. In the building of the Stapleton Tunnel the arc)i was first built with
only four rings of brickwork in morlar. "When the autumnal rains set in, the
ground began to press heavily ; so much so, that the line of the tunnel could be
traced as a hollow on the surfiace of the ground, which was not more than 40 feet
above the tunnel. A portion of the tunnel fell in, and was rebuilt with five and
six rings of brick.
The pressure of clay and shale, when disturbed by excavation, is, in some
situations, something almost immeasurable. The phenomena of disturbance
supply powerful examples of the flow of solids. There is the familiar phenomenon
of the bending and snapping of huge poles, and strongly timbered frames, by the
moving pressure of clay. In the construction of the Primrose Hill Tunnel,
through the London Clay, the lengths were limited to 9 feet, and strongly
timbered, till the arching was completed. In virtue of its mobility, however,
the moist clay exerted so great a pressure on the brickwork, as to squeeze the
mortar from the joints, to bring the itiner edges of the bricks into contact, to grind
them to dust by degrees, and to reduce the dimensions of the tunnel slowly but
irresistibly. The evil was counteracted by using very hard bricks, laid in lloman
cement, which, setting before the back ))ressure accumulated, hardened and
resisted the pressure, and so saved the bricks. The thickness of the brickwork
was augmented to 27 inches.
A similar accident occurred to a portion of the invert of the Netherton Tunnel,
built on a foundation of "bluebind," or marl. Some weeks after it was built,
the invert was forced up in several places by the swelling of the ground, and at
one point the bricks were crushed almost to powder. The invert was rebuilt with
a greater versed sine.
CHAPTER XXIII.
THE METHOD OF KECIPROCAL FIGURES.
This method is a rival to Putter's method of sections. It is
especially applicable to irregularly shaped frames such as roof
principals. Fig. 1 shows the simplest pair of reciprocal figures,
each consisting of the six lines joining four points. The
(three) lines forming a (triangle) closed polygon on any one of
the figures meet at a point on the other. Also the lines
as numbered are parallel. We have chosen the case where the
fourth 'point is not inside the triangle joining the other three,
so that a pair of the six lines pass each other at E. If the
second figure be drawn, then each side bisected, and the dotted
lines drawn at right angles to each side through the bisecting
points, we have the dotted figure. If it be turned through a right
angle, it is similar to the first of the pair of reciprocal figures.
Again, drawing the first figure, and looking upon it as a
quadrilateral frame with a pair of diagonals passing each
other at E, it is evidently able to be self strained by tightening
the diagonal (5). Now draw 1* and Q on the section figure
CHAP. XXIII.] METHOD OF RECIPROCAL FIGURES.
499
equal and opposite to each other, and giving to scale the pull
on the diagonal (5). Next, a triangle of forces is drawn for the
apexes B, C, and I), as shown in fig. 1. In this way the second
figure is a stress diagram for the selfstrained frame. The
frame with the bar (5) removed is said to be i/ideformaUe.
In general for an indeformable frame two conditions are
required to balance each pin, giving in all twice as many
conditions as there are pins. But three of these conditions
are first used to balance tlie external loads acting on the frame
as a rigid body. Hence we have a rule that the frame may be
indeformable, viz. : l^wice the number of joints less three should
equal the number of bars.
A Pair Of j^Recjprocal Figures.
c
^he Of/terttaned
(ipsitle (/otY/7 .
Jitress
JJlagj^ocyn.
FiK. 1.
The stress diagram for a roofframe is usually only a 'part
of a reciprocal figure, the frame itself, being only a part of the
first figure. We now describe at great length the process of
drawing it, using the scientific notation of Levy.
On figure 2 is shown a skeleton elevation of an iron
roofframe, span 32 feet, rise 12 feet. The rafter is trisected
by two isosceles struts PN and FO, also PQU is 60°. The
frames or principals are 20 feet apart, so that a fully loaded
joint such as R has apportioned to it all the load on an oblong
•of the roofcover 20 feet x 7 feet nearly, since MQ, is 20 feet,
being the hypotenuse of a triangle of which the sides are
16 and 12 feet. Taking the weight of the cover at 4 lbs.
2k2
500 APPLIED MECHANICS. [CHAP. XXIIL
per square foot, and of one foot deep of snow at 12 lbs.
per square foot, the vertical load (5) concentrated at the
joint H is 140 ^ 16 = 2240 lbs. = 20 cwts. The vertical loads,
then, are 20 cwts. at the joints iV, 0, Q, R, and S, but only
10 cwts. at J)!f and T. The windload is on the left side only,
and its direction is at 45° to the rafter MQ. We take the
normal component windloads as 25 per cent, greater than
the vertical load. That is, the two loads at N and normal
to the rafter are each 25 cwts., those at Q and M being half
as much.
The Reactions at the Supports. — The frame is securely
anchored to the wall at the stormend M, while the other end T
has a free horizontal motion on expansion rollers on the top of
the wall there. This is only the first case ; later on we will
develop the procedure if both ends be anchored.
The Forcepolygon, or load line, is now drawn to a scale of
cwts. Beginning at M, and going round the frame in right
cyclicorder, the loads at the joints have the numbers 1 to 7
allotted to them. Thus at M there is 1„ and 1^ for the normal
and vertical component loads there. Because of the trolly, the
reaction at T is necessarily vertical, and 8 is its number.
The reactions at the anchored end M are 9t, and 9^, being
vertical and horizontal component reactions, respectively. The
sides of the forcepolygon are now drawn in the same cyclic order.
It will be found to be complete, with one exception, namely,
the joint x between the vertical reactions 8 and 9„ is unknown.
Because the halftruss is a 3, 4, 5 rightangled triangle, it will
be found that all the joints in the forcepolygon are horizontally
opposite multiples of 10 cwts. on the scale.
The Li7i]cpolygo7i. — By consulting fig. 10, Ch. IV, this part
of the construction will be more readily followed. On fig. 2 a
pole is chosen just opposite the last bend in the forcepolygon,
and so as, roughly, to be the vertex of an equilateral triangle
standing on the forcepolygon. Vectors are to be drawn (in
pencil only) from the pole to each joint of the forcepolygon
except X, which is not yet determined.
Since the joint x is at the junction of 9, and 8, the vector
Ox must be parallel to the link of the linkpolygon joining up
those two vertical forces. For this reason we should begin the
linkpolygon at a point on the vertical line of action 9,,, and end
it at a point on the line of action of 8. Choose any point i on the
line of action of 9^. On fig. 2 it is chosen on 9,„ produced about
an inch above 3f, and is marked with a little square, as a memo
randum that i is the beginning point. The link ii' is drawn
/CHAP. XXIII.] METHOD OF KECIPUOCAL FIGURES. 501
between the lines of action of 9„ and 9/„ parallel to the vector
from tlie pole to the junction of those two forces. In the same
way the other links, i'n, aa, ah, he, c'd, dd', d'r, ef,fq, and gh, are
■drawn consecutively between the lines of action of the forces in
pairs, each link parallel to that vector which comes from the
pole to the junction of the corresponding pair of forcCvS. The
line (fh has no finite length, the lines of action of 7 and 8
•coinciding ; hence g and h coincide too, and have a little circle
with a little square round their common position. Joining the
two little squares i and li with a dotanddash line, we have
found the closing side of this linkpolygon, while a vector (or
searchlight) drawn from the pole parallel to it determines x the
■closing point of the forcepolygon. The judicious choice of the
pole, and of the startingpoint i, ensure that the linkpolygon
will 'rainbow' over the truss, out of the way of the further
construction, and also that the links will go on continuously
without looping back on each other.
The Stress Diagram. — Put halfbarbs on the sides of the
forcepolygon, and fullbarbs on the lines of action of the forces
external to the frame. It will then be seen that at the apex M
six forces concur. Four of them are known, that is, they are
already drawn to scale on the forcepolygon. In naming these
four they are to be taken in righthand cyclic order round M,
thus, 9,„ 9a, 1,„ It,. It is convenient to call the remaining two
forces exerted on the joint M by the rafter and tie 10 and 11,
keeping to the same cyclic order round M.
Now, the foursided open polygon 9„, 9^, 1„, and 1„ is to be
completed as a closed sixsided polygon, by drawing a side 10
parallel to the rafter to meet a side 11 drawn backwards from
the initial point x. In this way the magnitudes of 10 and 11
are now determined, and by going round the sixsided closed
polygon in the direction indicated by the halfbarbs, it will be
seen that 10 acts along the rafter toicards M, while 11 acts
along the tie away from M. On 10 and 11, in the neighbour
hood of M, place half barbs pointing accordingly ; but at the
other end of 10 put a full barb in the opposite direction from
that of the halfbarb, so as to indicate the reaction of the rafter
there, and a fidlho^vh to indicate that it is already drawn to scale
on the forcepolygons. In the same way put a/w//barb reversed
at the other end of the tie 11.
It will now be seen that of the five forces that concur at N,
all are furnished with barbs except 12 and 13, which reminds
us that all the sides of the fivesided closed polygon for the
apex N are already drawn on the forcepolygon, and the lines
502 APPLIED MECHANICS. [CHAP. XXIII.
drawn parallel to 12 and 13 close it, when the senses of 12 and 13
are then found to be both towards N, so that they are to be
furnished with halfbarbs in the neighbourhood of N, and with
fullbarbs reversed at their other ends. The next apex that can
be solved is 0, for here the group have all barbs but two. In
this way the apexes are solved in the order Q, R, and S, the
known and found parts being duly registered in the table for
that purpose on fig. 2, while the letters w, n, &c., give the
corresponding closing points. The next apex T is abnormal :
there all the forces are known but one, namely, xi, which
cannot be compelled to close the corresponding polygon. When
XI is drawn from ?;, if it only fails to go through s by a little
deviation, that is the allowable error of drawing. In this way
we have a check upon the construction. If it utterly fails to go
through s, we must look back to the construction of the
polygons for some point at which we did not keep to strict
cyclic order. If with every care to draw exactly it still quite
fails to close, that would mean that the truss is deficient in
members — that it is not rigid. Of course this could not happen
with the truss on fig. 2, which is necessarily rigid, being a set
of triangles.
Anchor on Leesule. — If we suppose the anchor shifted to ^f,
that will make no difference in the magnitudes of the vertical
reactions of the walls. For one way of calculating the magni
tude of 8 is to take moments about M. Now, the horizontal
reaction of the anchor passes through M, whether the anchor
be at M or 2\ and so has no moment, and cannot affect the
result. There is no need to draw the linkpolygon anew, but
only to shift or into the new position ;:, and the forcepolygon
is now 1„, Ip, 2„, 2e„ 3„, 3^, 4„, 4j„ 5, 6, 7, \iUh, viiir, 9. Nor is
there any need to redraw the stress diagram, polygon by polygon,
for we may suppose the original stress diagram as having been
constructed in two steps, as follows : — In the first step the six
legs are drawn of indefinite length, one from each joint of the
forcepolygon. In the second step we might suppose a billiard
ball to start from x, describing the route 11, 13, 15, 16,
rebounding from the legs in order at m, n, and o, and arriving
on the horizontal at ^), then completing its route xvi, xv, xiii :
rebounding from the legs at q, r, and s, and returning along xi
to the pocket at x. The dotted route is mapped out at once in
this way, only beginning and ending at z.
i<calin(j off the results. — The full lines are scaled off' and the
results are written on a small elevation of the roof on tig. 2,
having the anchor on the left, or storm side. Another reverse
Fig. 2.
504 APPLIED MECHANICS. [CHAPs XXIII.
elevation is shown with the anchor shifted to the right, or lee
side, and on it are written the loads as sealed off the dotted
diagram. Xow this is called a reverse elevation, because aft«r
all it is not the oMchor which has really shifted, but the wind
has, and we have gone round about to look at the back of the
roofframe. Hence this reverse elevation requires to be read
through on the back of the paper when comparing the stresses
with those marked on the first direct elevation. That is, the
quantities scaled off for the bars bearing arable numbers on
the first elevation are to be compared with those scaled off for
the bars bearing roman numbers on the reverse elevation.
Both ends fixed. — If both ends of a truss be fixed, it is quite
indeterminate in what proportions the two anchors will take
up the horizontal load due to the wind. Suppose for a moment
that both ends of the roof, fig. 2, are fixed, and that we assume
that the two anchors share equally the horizontal load. It
would be solved on this assumption, by simply placing the
closing point of the forcepolygon at a point midway between
X and z, and of course the results would just be an average of
those for the full and dotted stress diagrams shown there.
These are marked on a halfelevation at bottom of the figure.
In every ease, then, it is best to assume an anchor at the
storm side for the reactions ; then shift .r horizontally in
accordance with any assumed proportion between the holding
power of the anchors.
With the anchor at the storm side, the stresses are marirna,
so that it is a safe assumption to make in all cases. The dotted
stress diagram, fig. 2 gives minima stresses; but in some
examples these minima are very important. For a member
which was a tie for a maximum may have, in the minima
diagram, decreases through zero, and become a strut.
Figs. 3 and 4 show the Kingpost frame and its stress diagram
drawn as already described, but the loose ends of the loads
at K, L, &c., are removed. Fig. 3 is now to be looked upon
as a selfstrained frame. It is composed of two frames, the
link polygon, an open balanced frame (fig. 10, Ch. lY) and
the iiuleformahle Kingpost, each straining the other through
the bars 1, 2, 3, &c., joining them point to point, and all idea
of external load is dismissed.
Fig. 4 is the one and only one reeiprocal fjijnre drawn to
fig. 3. The reciprocity in the other direction is not so complete,
as fig. 3 is only one of the many reciprocal figures to fig. 4,
for we might have begun the link polygon at a different point.
/^ Jb A Anchor
£ L EVA T /ON.
Jfeal and Jc^eczl Frame^ (coTTTZmecf)'
ft 10 o
LULX ' M I I ^
40 SO ft.
I J
Fig. 3.
Stress JDia^rcuny.
Fig. i.
Fig. 5.
CIIAr. XXIII.] METHOD OF KKCU'ltOCAL FICfURES. 507
Exercise I.
Here the Queenpost Eoof, having a span 56 feet, is shown on
fig. 5 with the central rectangle doublebraced with ties which
can only act one at a time. There is an additional load 9 at
the joint I. The truss is anchored at both walls, the strong
wall (square anchor) taking twothirds of the horizontal
displacing ellbrt of the windload.
In the first place assume that the left (square) anchor takes
all the horizontal load, and leave out the load 9 at the joint I.
After drawing the link polygon we find the point x separating
the two vertical reactions at 8* and 10".
Next draw a vertical line through the right trisecting point
of vy. Here is the true position of 8* and 10". They are to be
slightly displaced left and right of each other, and 8" is to be
made longer than z.r by twothirds of 40 cwts. (the value of the
extra force 9) and 10" made longer than .ry by onethird of
40 cwts. The force 9 will now fill the gap between the two
augmented vertical supports.
After completing the solution of apex after apex as shown
on the table, the dotted reactions are placed in the vertical
through the otlier trisecting point of vij, and augment with
the portions of the 40 cwts. in the reverse order, as 9 will now
be at the joint U and be called ix, also the counterbrace has
replaced the brace, for we suppose the truss turned end for end
and the wind to remain as shown by the storm arrow. The
dotted stress diagram is now to be drawn.
The half roof is a 3, 4, 5 triangle, as the rise 7 is 21 feet.
Assume the roof to be rigid, and the whole vertical load
120 cwts. may be concentrated at C, and the whole normal
windload 90 cwts. at the centre of the rafter AC, and then
further decomposed into a vertical component of 72 cwts. and
a horizontal component of 54 cwts. Taking the moment of
this 54 cwts. about A and U alternately adds 10^ cwts. to 8'
and subtracts it from 10". Hence yv = 54, 20: = 60 + 18 + 10,
and xy = 60 + 54  10^ cwts. Scale 50 cwts. to an inch.
C U R 1/ E D
ROOT.
ft /O O
/o
ZO
Jo
■*o
So
OoJ^eef.
1 i 1 1 :) , 1 1 1 1 ! 1 I 1 ! 1
'O o C,t/^
so
/oo
/SO
SOO C'U/CS
._tj .::.■!■!•:■'■ i; 1 1 ■'■ !
Fig. 6.
CHAP. XXIII.] METHOD OF RECIPROCAL FIGURES. 509*
Exercise II.
This is a great cur\ed roofframe (fig. 6). In drawing the
legs 10, 14, 18, they should be drawn at right angles to Mj,
Mk, Ml, and the vectors 11, 13, 17 at right angles to Np, Nq,
Nr. The legs may be firm lines drawn indefinitely outwards,
and numbered boldly near the forcepolygon. The vectors are
then light lines numbered on an arc of a circle. The indefinite
lines may be temporarily numbered again in pencil at their far
ends. It will then be easier to work out the contour 12, 15, 16,
being verticals and obliques alternately, and joining a leg and
a vector alternately.
On the dotted figure for the truss turned round, taking the
anchor to the other end, the contour XII, XV, XVI, is put in
by inspection. Note that 16 is drawn down instead of upwards
on the first contour. Hence the stress on the bar HI has
changed from a pull to a thrust.
The frame is primarily of the Kingpost pattern. On the
first auxiliary figure the scaled results are marked for wind on
each side alternately. The ^•ertical bars are all ties except
that HI changes with the wind. On the second half auxiliary
figure the vertical bars are compelled to be struts by double
bracing as on King's Cross Eailway station roof. A thimble
acting on a right and left screw on the counterbrace BE puts
upon it an initial pull of 80 cwts., which induces a pull of
61 cwts. on the rod CD and thrusts on all four sides of the
trapezium. Choose a scale on which BE measures 80 and the
lengths of the other bars on this scale give the induced strains,
those on the parallel sides being interchanged, see fig. 1. If
both BE and CD be furnished with tightening thimbles, they
assist in adjusting the shape of the frame.
COMPOUND ROOr TRUSS
'iind on Anchored Sida,
Mp,
knon
fcunrl
i
R
1? ?„ ;>»
r
LI n
c n
F
JSI6H.ln M
X H
r
22 23
n
lO fl PI
ti
Mti Sa. m
1
!SK
J
X 2i>
It 30
A
I. .»
31 3i.
/.
3i IS t? 7
33 a
3i. 3r.
.V
J7. 3* .K
nCJtrdl
n
free Motion
HomontaJly
o
10
^
^ncAerrd on Lrr Sidr.
fret O
T.AIwmT Trin. Coll. Hu^Ha
' ' " ' ^^
4» reft
Fig. 7.
CHAP. XXIIl.] MKTHOL) OF KECIPROCAL FIGURES. 511
Exercise III.
Here is a compound roofframe (fig. 7). A vertical section
through the crown cuts only one bar, the main tie. Hence the
pull on tlie main tie is virtually given when the loads are
determined upon.
In this example both x and y have to be found by the
link polygon. It is to be drawn passing through the crown H,
and may be drawn Hd'd backwards, and Hfg forwards.
A vector from the pole parallel to the closing side of the
linkpolygon gives x, and a vector from the pole parallel to Hz
cuts off xy on the horizontal through x, where z is the point
where jk cuts the main tie (produced).
The span being 56 feet, and the rise 21 feet, then the half
frame is a 3, 4, 5 rightangled triangle, which makes an
arithmetical check on the values of the supports and component
loads easy. Observe the joints of the forcepolygon ticked
over on to the vertical scale. As on Ex I, with the frame
assumed to be rigid, the loads reduce to a vertical load of
80 cwts. at the vertex H, a vertical load of 48 tons at E the
middle of the storm rafter AH and also a horizontal load there
of 36 cwts. The calculated values of the supporting forces are
(lift) = 36, (10) = 40 + 12 + 6, and (11„) = 40 + 36  6 cwts.
In this particular pattern the slightest error in the vertical
position of x causes a greatly increasing error as you pass from
joint to joint. This is to be eliminated by drawing the four
parallels 12, 14, 20, 22, with one setting of the parallel rollers
or sliding setsquare and of an ample indefinite length, and also
the four 25, 27, 33, 35. Then, while solving apex after apex,
the fact, due to symmetry, that 15 and 23 are in one line, and
as well 24 and 35, enabled any slight error in the position of x
and of y to be corrected.
512 APPLIED MECHANICS.
Note to Chapter X.
In this edition we have omitted the transite of the locomotive
over the constant span. It is purely of academic interest, and
had to give place to the matter in Ch. XI, which is of practical
importance.
The transite was illustrated by figs. 158 to 166. The
fig. 158 showed a number of parabolic segments on a common
base equal to the span of the girder but of varying heights,
proportional to — 1st wheel only on span, 1st two wheels only
on the span, &c. With templates made from these a set of
diagrams gave the maximum bending moment at each point
of the span, with only the set of wheels on the span corre
sponding to the particular template. Each of these diagrams
consisted of arcs for such points as the wheels could actually
reach without any other wheel coming on or off, and tangents
to the arcs for the other points.
3AyONARA^
Printed by Ponsonby & GiBBS, University Press, Dublin.
/,
8i3i
TA
350
A5
1916
Alexander, Thomas
iilementaiy applied raechanicE
c3d ed.,
Eof
irxrrmt
PLEASE DO NOT REMOVE
CARDS OR SLIPS FROM THIS POCKET
UNIVERSITY OF TORONTO LIBRARY
III!'
HiHilulKitUiI
i:!iiH!i