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ELEMEXTAKY APPLIED MECHANICS
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MACMILLAN AND CO., LiMiTEr.
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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 non-seulement 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 title-page, 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 full-page 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 set-square 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 bent-glass 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 double-tee 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 tee-section, or angle-iron constrained to bend like it, as it usually is. This is an important addition to Rankine's list, as these sections are of every-day 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 heavy-faced 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 semi-elliptic 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 two-nosed 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 arch-ring 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 Square-roots 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 Square-roots 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 CROSS-SECTIONS 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.
CROSS-SECTIONS : 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 Cross-Section and Sections partly built of Triangles, Hexagon, Diamond, Trapezoid — Circular and Elliptic Cross- Sections — Resistance of Cross-Sections to Shearing and the Dis- tribution of the Shearing Stress thereon, including the Hollow Rectangle, Doul)le T-Section, 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 Cross-Section 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 cross-girder — 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 Bending-Moment 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 Shearing-Force 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 Cross-Sections of Beams — Cross-Sections of equal strength, Rankine's approximate formula for their design in cast-iron, 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 cast-iron and wrought-iron 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 — Rankine-Gordon 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 full-page 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 full-page 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 Aich-ring — 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 Arch-ring — 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 Load-area for semi-circular 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 Arch-ring — 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 two-nosed 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 — Semi-elliptic Masomy Arch — Exs. 21 to 25 — Abutments and Piers — The Tunnel Shell — Allowance for excess-density of Elliptic Arch-ring — Exs. 26 to 31 — Quotations from Rankine's C-E.y and from Simms' Practical Tunndling, .... 413
CHAPTEE XXIII.
THE METHOD OF llECIPKOCAL FIGURES.
A rival of Ritter's Method of Sections — Suitable for roof-frames 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 self-strained frame and the other as its stress diagram, the reciprocity being perfect — General application to an iron roof-frame — 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 pre-determining 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 self-strained — 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 self-strained frame is one of its many reciprocal figures— Ex. I, Queen-post roof double braced in the centre and with an extra load between the supports — Ex. II, Great curved iron King-post roof — do., double braced like King's Cross Ry. station roof— Ex. III. Compound roof-frame, 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 0 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 right-hand 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 tie-rod in a roof, M-hose 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 tie-rod 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 cast-iron 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 2-75 inches. What strains do they sustain ? Which sustains the greater ? Give the ratio of the strains.
For the 100-fathom cable
augmentation 3 in.
strain = , = — = -000417.
length 7200 m.
APPLIED MECHANICS. [CHAP. I
For the 90-fathom cable
augmentation 2*75 in. „„^ „ strain = ^. ,, - = ^.^^ ■ = -000424. length 6480 in.
The 90-fathom cable is the more strained. Eatio of these strains is 417 to 424.
5. A 30-feet 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 produce-a 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 right-hand 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 one-millionth part. Pliability is a term applied to the property which indiarubber possesses in a higher degree than steel.
Examples.
7. A wrought-iron tie-rod 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 cast-iron 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 = — = 31111-1 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 tie-rod 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 inch-lbs. of work, which will be stored up as a potential energy in the stretched rod.
Suppose a scale-pan 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 scale-pan. 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 tie-rod, 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 one-fourth 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 wrought-iron 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 wrought-iron 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 cast-iron pillar has a sectional area of 12 sq. in. It is advisable only to strain cast-iron 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 one-half 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 one-half 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 foot-lbs. 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 -j-V-K 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 iii-staiitaneous 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.
= 35-75 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 35-75 x 120 in. x -5 sq. in.
= 178-75 foot-lbs. 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 — , = 178-6 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 1-44 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 = 1-613 ; 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. = 314-8 lbs. 1-44
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 34-8 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 =11-7 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.' = 13-3 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 = 24-05 ft. length of rod.
Equating for second case,
— a; = 720 + ^x, 738a: = 144000, 4 oO
X = 195-12 ft. length of rod.
For the first case length is 288-6 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 = 585-3 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 0 on the plane normal to OX is
P =
total stress area of plane
P
AA
Through 0 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 5-6 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 0 (fig. 2) represents the stress at the point 0 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 0 clef.
:Pn = p cos' e ; (fig. 3)
also
Pt
p cos 0
= 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 0 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 0 cos 0.
On DJD p'n = p cos^ (j), p't = p sin ^ cos (j>.
But sin 0 = cos <p, and cos ^ = sin 0 ;
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 0 (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 0 on the - /' one side of OX, and the^^ --, other at 0 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 0 cos 0 = ^ sin 20.
Therefore pt is greatest when sin '2B is greatest,
when sin 20 = 1, when 2B = 90°, or 0 = 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 1000077^^ 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 0 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. |
0 |
- |
— |
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
/ruerhfU-ces of UmgcnuaL ccmnonenCa
Fig. 7.
Force. Leverage. Force. Leverage.
qt • OA X OB = p, . OB ^ OA.
«nHBatf:i<«HBn<aiit;aHia!iaiinas.THiswii
26
ATPUED MBGHA2flCS.
[chap n.
Now die area OA nmltiplied by the length OB gives the volxEme (rf the parallelepiped, and the area OB mnltiplied by the length OA also gives ihe volnme.
tT^:-'-. '.' i i/oint within a solid in a state of strain, the
nente of the stresses npon any two planes
:... . -s.. - ; 41 . . " rles to each other are of an eqnal intensity.
Oyr. — If : apon any plane throngh a point be wholly
normaL then ts„. -e wholly normal upon another
plane at right ani.. .^ne.
Snch a pair of 3trt»g*as are called principal -' nd the
planes trpon which they act are planes of i ^ , - stress. When these two stresses are given, the nniplanar stress at a
point i= c " — "t given. There is ^so a third principal stress
a/iting or -^ at right angles to the other two.
Jjua/jj Peoelem, the PursaPAL .Stkesses givek.
Ef/tfjcd-iibs prineipaJ stre^^es (fig. 8;. — If the pair of principal stresses at a point be like (Tx»th thrusts or both tensions), and
beof- " ■ - ■ '
on ^:
'tj%
J^
'-.fi
any third plane inclined at H to
PPPP
iiir
Fi:
Let A A' and Blf be the planes of principal stress at the point 0, and let the intensities of the pri '
g, be er, thmsts>.
CC is throngh 0. AA'/
OAB is a small triangular prism at 0, having its faces in those planes, Tr is in equiliVjrium under the three
forces — the total ijion OA, OB, and AB.
Tixn three total pressures F, Q, B 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 O.
Lay off 0// - total stress parallel to OJC = p . OA,
and OB = UjW parallel Uj OY = ^/ . OB.
O-onplete the para. ...
CIIAV. U.
KANKINKS KUirSK tH' STKKSS.
TluMi .'!(> r«'pn«stM\ts [\\v total stross o\\ A l^ '\\\ vlii«vtii»i\ .tiul anuniMt.
.-. KOn - OAli - 0: .. Olx is u{H>u O.V. lltMui' f\0 iv*« normal to .•(/»'.
Now
a»»ioui\i otstrossou A l^ HO
aioa of .1 />'
. I />
|) ov
V
('(•;•. - Kvon- piano tlM\nigh (> is a piano oi \nn\v\\^\\ stivss. K^u'h point, iti a tlnid is in this sUto of strain.
K(ttutl-intlikr prim^fxtl sttyaars (tig. 0\ — If tht^ pair of prinoipal strossos at a point l>(> unliko (^ont> a tlnwst aiul tlu" otiior a tonsion^ a»\tl Ih» of (Mpial intiMisit V. tho st nvssonany I l)inipl-m(^ tliron«;h tl\»^ point is of tl\atsatnt>ii\tonsitv,an»l is itu'linodat ananghMo llvo iwMtual to tho plaiu' of principal st\t\><s,onual to that vvluolt tho >ior
nial to thistlnnl plant> v- .-. >'^ "*" irnniA «)*
nialuv** thon^wilh. l>nt upon tho »•/»/>( IX »V«' siiit .
Lot .'(.•l'an«l />/»'l>o Kij; ;>,
tho pianos of principal
stross at tho poitU (>; jn and ./ tho mtonsitios yA iho pvinoijvvl stnv-^sos of o*pial valno. /• hotnj; a (hriist ju\il </ a tonsion ; ;nnl (V a>\y thinl piano thronivl* (> inolinod at ti to AA \
(>,</»', a stnall tnat»gnlar prisni at O bo\>n»lo*l l\v thoso iluo(^ pl'uios. is in tvptilihriiun \nulor tho th»ro foivos. vi«., Iho amount of st«t»ss «m it.s far<>s (>.^. (>/>\ an*! .-( />
l,ay olV <>/> l«'tal stross parallel to O.V - y. (>.•<. ii\.l OK - t«>lalst>vss paralh^l to (M u\ thoilinvtionof «/
i^.Oll
m
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.
Eqiud-like 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 0 is a plane of principal stress. Each point in a fluid is in this state of strain.
Equal-unlike 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 -4-4' and .5^ be Fig. 9.
the planes of principal
stress at the point 0 ; 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 0 inclined at 6 to AA'.
OAB, a small triangular prism at 0 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 0 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—n-A =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 0 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 0 ; 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 p-q
, 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
jp-q
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 equal-like 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 0 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 0 (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 =
p-q
= 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 p-q
= ?;
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
p-q
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 0 ; 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=OM-MT
= 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^ 0 ( — ^ + ^—^ ) + 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- 0 + g' cos'' d. Now, r„ = JO cos^ 0 + q sin^ 0,
and adding, we get
Sn + I'n = p (sin- B + cos- B) + q (sin- 0 + 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 . P-fL
0, 9.
p;
and the minor semi- ,' axis is )
OR^ = OM, - M,R,
^p+q P-1^ 2 2^'
The moving model, fig. 31 following, shows these positions nicely.
This is called the ellipse of stress for the point 0 within a solid in a state of strain. Its principal axes are the normals to the planes of principal stress, the principal semi-axes 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 0 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 semi-axes 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 0 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 semi-axes 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=OM-MB; 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 thi-oiigh 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^y-i'^)
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 ;
p-q 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' ; 0 = 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 0 > 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 0 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 0 > 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=P-J,
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 0 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 0 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 0 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, 0 = 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,
P-1
= MR = ^ (03/2 + 7-2 - 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 -jp-q
and cos 20 = — ~ (C)
p-q
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 ;
p-q
tan 20
(r cos y - r cos y) 4 RL
+ r- sm-7 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
P-1 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 0 (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 head-stock of the T square, represents the trace on the blackboard of any plane through 0 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 crank-pin.
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
p-q
= 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
0 = r^ - j-'2 _ 20M{r cos 7 - r' cos 7') ; r^ - r"^ p+q 2016
OM = - , „.
2 (»• cos 7 - r' cos 7') 2 52-43
MR^ = (38-45)2 + (54)2 _ 2 x 38-45 x 54 cos 10° = 1478-4 + 291G - 4088-8 = 305-6 ;
^-^ = \/305^ = 17-48.
= 38-45.
The principal stresses are P
p + q p — q
+ -;r— = 55"93 lbs. per square inch thrust, like r,
2
2
p-q
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
p-q
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='-
.-. ^^ = -/(1263-8 - 1200) = 8.
Adding and subtracting (A) and (B),
I? = 43*5 a thrust, and j = 27 5 a thrust,
78-8 - 71 cos 20 = ~ = -49.
16
(B)
(C)
E 2
52
APPUED MECHANICS.
[chap. II.
Therefore 28 = 60° 40', and 0 = 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^ = b0bb-2\ bnt(B), 4j09 = 4800 ; .-.jt?'- 2;?^ + j^ = 255-2; .-. j»-j=16:
adding to and subtracting from (A),
.-. 2;? = 71-1 + 16, and 2^ = 71-1-16: .-. jo = 43-55, and ? = 27-55.
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= 38-14, = MR = 150-3 ;
and
p= 188-4 (thrust),
.-. q = - 112-2 (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' ;
= 237-6 =190 =33-4
iLJ-1 = OM =
213-8,
P - <1 2"
= MR = l\ (''" ^''"\ +r,'j = v/(566+ 1115) = 41
j9 = 254-8, and 9=172-8; tan 26 26 = 54° 32'; .-. d = 27° 16'.
-l'V = 1-4034; 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 surface-soil, 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 right-hand 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 0
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 0
1 - sin
? =
1 - sin 0
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 0
1 + sin </»
w . D a.t the bottom,
and therefore
1 - sin 1 + sin 0
0 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 0 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 bed-joint 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 foot-lbs. 6 1 + Sin 0 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 bed-joint be stable : but for some forms of wall it is necessary to go through all calculations for each bed-joint 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 bed-joint 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 <^'
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 =
p-q_
(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^(cos-y - 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 0 = ^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 Ij-q
(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 right-hand 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-</)) ,,
= -7r-cos7 '- 77 r- ttIds.
2 ' C0S7+ v/(cos-7-cos-0)
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 bed-joint must be greater than R.
H tends to overturn the wall with a moment
= h(^\ -
M = H
V3
wD^ J cos 7 - -v/(cos-7 - 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 0 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> 2-y, .•. 0 > 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 = 100-5 lbs. ; M = 100-5 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 = 1225-5 lbs. per lineal foot.
Thrust per strut = 6127-5 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 = 1-21 ft. 1440
Note.— The height of wall above ground is 10 - d = 8-79 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 — COS-rf))
- = ^7^ — ,r ^ = '62 ; .-. r = 299 x -62 = 185-4,
r COS 7 + v{cos''7 — cos^<p)
and average intensity of stress = 92*7 lbs. per square foot, r' 2
M = H X - = 231-6 ft.-lbs. o
6. A cutting having 3-foot 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/(cos-7 - C08^(^) _ 'oTG _ r cos 7 + V(tos''7 — cos-<f)) 1-304
.-. >•' = 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. = 70-2.
Total do. = average intensity x area = 70-2 x 3 = 210*6 lbs. per lineal foot of wall.
Moment = 210-6 x - = 210-6 ft.-lbs. 3
I. j\ 9 ^ ~ ^^^ 'P '^
r cos 7 + V(cos-7 - co.s2(f>) 1-304
.-. y = 338 x -383 = 129-3 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 cross-section ; 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 0 3 ■ ^ ^
That is, the horizontal pressure at any point of AC, the back of the wall, is one-third 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 semi-diameters 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 bed-joints 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 = 891-2 lbs. per square foot at base of wall ; AC = 20-3 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- 3-52) X 20 X 140 X from which
t = 82 ft. = thickness at base of wall, t - 3-52 X 2 = 1-2 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 u-d cos ^ ; and thus the intensity of the two equal conjugate pressures r and /, for a point 10 feet deep, is 1039-2 lbs. per square foot. In drawing the triangle OMR, proceed as in first case :
<RMX= '29 = 120°; <ROM=r^ = 30°; OR = 1039-2 ;
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=2078-41bs. 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 = CBcos-SO° - -St sin 30^ W^UOtd, Ll^-'M
Fig. 3«.
Fig. 3*.
20784 (5-77 - -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^=(^-3-52)x 20 xl40,
^ = 8-9ft. = thickness at base,
t - 3-52 X 2 = 1-8 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 bed-joints 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 cross-section 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 = 3-2 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 0
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 .
l-sin0 l-sin'/< , , /l-sin0
^ ^ — 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 3-52 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 - 15-4 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^ - 15-4 X -03^, t = 5-78 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)
+ 0 (t,, - U) (140 + 120) (-3^,0 + -bh)
from which ^i,, = 4'1 ft.; similarly ^15 = 6"3 ft.; and tw = 8-4 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
/ = 1-7 + ^' + 6 = ^' + 7-7 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 S6-77 - "4 {f + 7-7) ) ,
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 bed-joints 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 heaving-up. And for the unit-cube at heel of trench, it is as a practical lower limit, one-third 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 = 4-32 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--: (4-32 + | x -96) : (i x 4-32 + -96), and ya + yb = 1-45^,
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
l-J <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
t-H
W
H
O
w pq
H
«- II
« "5
«N •<»• O C^
t-« CO
H
* c^
^ S
-ja -e-
W
w
<M 00 lO CI
r-i 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 hand-made 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 arrow-head 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 left-hand 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 0 (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 pin-jointed 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
tan-ii = 26° 34' ; k= ! ^'" "^ = -382.
1 + sin ip
p = \5tv' ; q = kp = b'lZw'.
E = Q = y X 30= 171-9!<.''= 137-5t<^^, 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 <= 11-5 feet.
3. Second wall on figure 12, solution exa<tly corresponding to the graphic
solution,
p + g iv' ■ ^ = — (15+ .5-73)= 10-365et7'.
0M = MR = '
4-635M7'.
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 - 96-07 X -946 = 38-12. r= 0R = 6-174U'', and CC the hack of the wall is 30 cosec fi = 30-4 square feet. R = 30-4>- = 187-7^' = 150w.
sin 7 MR . 4-635
^^e = -OR-^ .•.sm7 = ^:^x-324=-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 = 137-4m-.
r= 7? • sin 23° 38' = 150m; x -401 = 60-15mj.
The equation of moments about F, the centre of stress, is
1374 = 30 (!! - 5) (f (! - f ) + 75 (-^< - iJt) + 60 (|.< - -|),
<2 _ 1-I6i; = I'^O, or <= 11-5 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 = 14-23w'.
, _ cos7-V(cos^7-cos''(^) _ -9487 - v(-9000 - -8000) _ 9487 - 3162 _ .
~ cos 7 + V(cos^7 - cos*^) ~ -9487 + \J{-^mQ - -8000) ~ 9487 + 3162 ~
/•' = k'r = 7-115if', 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 = 11-25i!2 + 47-1^ )/, and < = 9-3 feet.
5. Approximately for second wall on figure 13, by removing a wedge at back.
1620 = 11-25^- + 47-25<- 15 X 8, and t = 10-5 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 = l-144r = 1-144 X 14-23m;' = 16-27m'.
q = kp = -382 X 16-17m;' = 6-21m'.
OM = ^(p + q) = 11-24M; ; 3IR = Up - q) = 5-03«;'.
ain 20 OR r 1423
sin y ~ MR~ ^ (/? - y) " 5-03 '
. , 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 j-i-j^i-i >-. 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^ + O-03- - 2 X 11-24 x 5-03 x -6997 = 72-53.
(^)'--
r" = 8-516«f''; R' = >•" x 304 = 259tt' = 207-2w.
sin-y" MR" 5-03 . „ 5030 „, , ,„, , „ „ ,„,
■ ; „ = — -„- = —,—7; .-. sin 7" = -;— X -7145 = -422, and 7" = 24° 58'. sm 26" OR" 8-516 ' 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,^ {t-o)(ft-A) + nw (it - la) + V" (it - %],
or i- + 3-27t = 158-2, 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"
4-13.
w,
._ll
-.Q
Fitr. 1.
on the cross-section 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