3-1 ERSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CALIFORNIA /fa ERSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CALIFORNIA i | SITY OF CUIFOOU LISfllRY OF THE UNIVERSITY OF CUIFORNU III i 9 = SfTY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CALIFORNIA Llf THE AET OF CALCULATION BY DEAWING BY SrOTTISWOODE AND CO., XKU'-STKEliT LOXDOX GRAPHICS OR THE ART OF CALCULATION BY DEAWING APPLIED ESPECIALLY TO MECHANICAL ENGINEEBING WITH AN ATLAS OF DIAGRAMS BY EOBEET H. SMITH <\ PROFESSOR OF ENGINEERING, MASON COLLEGE, BIRMINGHAM; FORMERLY PROFESSOR OF ENGINEERING, IMPERIAL UNIVERSITY OF JAPAN : AUTHOR OF ' CUTTING TOOLS WORKED BY HAND AND MACHINE '; M.I.M.E. ; ASSOC. M.I.C.E. 5 MEM. ORDER OF MEIJI, JAPAN PART I. LONDON LONGMANS, GEEEN, AND CO. AND NEW YOKE: 15 EAST 16 th STKEET 1889 All rights reserved PREFACE. THIS book will not enable the student of Practical Mechanics to dispense with the use of other books treating mechanics in the ordinary manner. If it did, it might be entitled ' Engineer- ing Mechanics developed Graphically ; ' but in this case it would have been necessary to include in the text a very great deal that would be merely tedious repetition of what is to be found in other well-known books, and the author would have been open to the accusation of ' mere book-making.' On the other hand, if he assumed on the part of the reader a familiar acquaintance with all the higher mathematical developments of mechanical science, and merely showed how to use graphical constructions in applying these results to practice, the book would lose much of its most important and legitimate utility. It is intended to enable those who have a knowledge of ele- mentary mechanics to advance that knowledge to any degree of thoroughness they may find useful, and to apply that know- ledge to the every-day problems of engineering science, with- out the aid of the more complicated portions of algebraic and trigonometrical mathematics or of the differential and integral calculus. Many have no taste or faculty for this latter sort of mathematics ; others have not the time needed to keep themselves aufait in its use ; and, again, it is undeniably true that the solution of many a problem becomes practicable in point of time and ease by the graphic method which would be intolerably tedious and difficult without its aid. yi GRAPHICS It seems to be one advantage of the graphic method that it requires a more intimate knowledge of the physical natures of the quantities dealt with than does the algebraic method, whose essence is that it proves certain propositions regarding x, y, and z, without the smallest reference to what is meant by x, y, and z. This thorough understanding of fundamen- tals is particularly insisted on throughout this book ; and the reading of Chapter VII., on Vector and Eotor Addition, is intended to assist in the clear comprehension of some mechanical principles which are too much slurred over in ordinary teaching. In Chapter VIII., too, the explanations that are given regarding Moments and the limited sense in which Eesultants may be taken as equivalents for their Com- ponents, should help in the same direction. On reference to Chapter II., it will be seen that the whole subject is divided into eight heads. Of these, only two namely, Arithmetic and Statics have been dealt with in previous books on Graphics. The introduction of Algebra and Trigonometry treated graphically is a novelty. It is hoped these chapters will be found interesting and useful, but it has not been deemed desirable to devote very much space to them. A good deal also of Chapter III., on Graph-Arithmetic, is believed to be now for the first time published. The Kinematics of Bigid- Bar Mechanisms was first dealt with on the present system in a paper by the author, written in 1884, and read before the Royal Society of Edinburgh in January 1885. In the application of Graphics to the Statics of engineering structures, the stress-diagrams for linkages containing beams and those for linkages built in three dimensions i.e. for solid structures are both new. The latter especially seems to be an important development of stress- diagram construction, seeing that all structures, if rationally investigated, must of necessity be treated as solid. Again, the author has not found in other books on Graphics any directions as to how to deal with structures containing no PREFACE Vli joints where only two links meet ; with structures on which the loads act at internal joints ; or with others offering similar difficulties. Nor does he find that the Cyclic Order of pro- cedure in stress- diagram construction is anywhere else ex- plained, much less insisted on, as it ought to be in view of the fact that this is the whole and sole key to the unravelment of difficulties in construction and in interpretation. Indeterminate Abutment-thrusts is another subject fully explained in this volume, although unfortunately lost sight of in most text-books. Several shorthand symbols have been introduced, and, it is hoped, will be found acceptable in offices where graphic work is done. The symbols i^> c; a b ; and a $ b enable one to write out stress- diagram constructions in one- tenth of the space that would be occupied if uncontracted language were used, and the descriptions gain greatly in definiteness and ready comprehensibility as well as in mere conciseness. The words Locor and Kotor have been adopted and freely used, the latter in a somewhat different sense to that em- ployed by Professor Clifford. The words were very greatly needed, and were only finally chosen after very lengthy con- sideration and experiment with other possible expressions. The word Pen is another new one, the choice of which involved much deliberation. At one time the author inclined towards the use of ' cell ; ' but, this having an already established appli- cation different from that intended here, the preference was finally given to * pen.' Again, the Field in which a motion or other vector occurs has proved to be a convenient and expres- sive phrase. These and the symbols =$= and ^ of vector and locor equality enabled the author, as he hopes, to avoid much diffuseness in, and give greater definiteness to, his diction. A Glossary of the meanings of all the terms and symbols used that are either new or not quite commonly understood, is given a prominent place at the beginning of the book. Vlll GKAPHICS It is hoped that the Index to the Engravings, giving page references to the text for each figure, will be found specially convenient. It may be useful to insert here a word of warning to ordi- nary readers and of appeal for consideration to critics, to the effect that it is not intended that the text should everywhere be comprehensible without reference to the diagrams, or that the diagrams should be always capable of interpretation without reference to the text. If the present volume meet with a favourable reception, and no unforeseen obstacle arise, Part II. will be issued at an early date. In this second part it is intended that the subjects dealt with should include ' The Distribution of Stress and Strain ; ' * The Strength, Stiffness, and Design of Beams and of Struts ; ' * Economy of Weight in Structures ; ' ' Stresses in Eedundant Structures ; ' * The Statics and Dynamics of Ma- chines ; ' * Frictional Efficiency ; ' * Governors ; ' ' Fly-wheels ; ' 1 Valve Gears ; ' ' The Practical Thermodynamics of Furnaces, Boilers, and Engines,' including series of curves facilitating the calculation and design of boilers and steam and gas engines ; * The Hydrostatics and Hydrokinetics of Ships and Hydraulic Machines.' Generally speaking, Part II. will deal mainly with synthetic problems, and aim more at the design of structures and machines than does Part I., which is chiefly analytic. MASON COLLEGE : October, 1888. CONTENTS. PAGK INDEX TO ATLAS OF PLATES xv GLOSSARY OF SPECIAL TERMS AND SYMBOLS . xix INTRODUCTION. Origin of Graphics. Clerk Maxwell, Cremona, Culmann, Bow, Henrici. Advantages .1 1. Continuous quantity. 2. Artificial scales. 3. Degree of approximation. 4. Vectors. 5. Business utility. 6. Training to accurate draughts- manship. 7. No large error possible. 8. Conditions necessary for superior rapidity by graphic methods. 9. Less mental fatigue . 1 G CHAPTER I. INSTRUMENTS. 1. Need of accurate instruments. 2. Scales. 3. T-square. 4. Straight- edges. Set-squares. 5. Drawing parallel lines. 6. Needle-pricker. 7. Compasses and pencils. 8. Curve templates. 9. Splines . 714 CHAPTER II. DIVISION OF THE SUBJECT. 1. Graph-arithmetic. 2. Graph -algebra. 3. Grapho-trigonometry. 4. Grapho-dynamics. 5. Grapho-statics. 6. Grapho-kinematics. 7. Tabulation of Experimental Besults. 8. Tabulation of Mathe- matical Besults 1517 CHAPTER III. GRAPH-ARITHMETIC. 1. Addition and subtraction. 2. Integration. 3. Multiplication. 4. Con- struction. 5. Well- and ill-conditioned intersections. 6. Possible error of intersection due to thickness of line. 7. Possible error of inter- section due to error in direction of line. 8. Scale of the product. 9. Twelve multiplication constructions. 10. Division. Combined X GKAPHICS PAGE multiplication and division. 11. Sectional Tablet for multiplication and division. 12. Use of Tablet. 13. Positive and negative integral powers. 14. Fractional powers ....... 18 31 CHAPTER IV. GEAPH-ALGEBEA. 1. Five graphic methods of representing equations connecting two variables. 2. Linear co-ordinates. 3. Polar radius-angle co-ordinates. 4. Focal angle co-ordinates. 5. Focal radius co-ordinates. 0. Focus and directrix linear co-ordinates. 7. Equations represented by a straight line, Dz + E?/ + F = 0. 8. r=. or f sm(a + 0) cos(a-0) 9. a cot 6 4- b cot 4> c. 10. Equations represented by conies, Ax- + ~Bxy + Cy- + Dx + ET/ + F = O. Ellipse, parabola, hyperbola. 11. r- = a- cos -0+b- sin -6. 12. r = acos0+ 6 sin 0. 13. a cot (B a) + 6 cot (4>~)=c. 14. r A + r K = c r ^ r^-c\ r = mp. 15. Equations in general. 1(3. Solution of equations. D x + F = 0. 17. jD.s + E.y + F^O) i 8 . ^ectional Tablet.' 19. ACC- + DOM F=-0. 20. Simultaneous quadratics. 21. Gz 3 + Dz + F = 0. 22. A/ (#) + D a + F = 0. 23. /(a?) = 0. 24. A Vr^ti- + D x-+ F = O. 25. F (x) +f (x) = 0. 26. |F (xy) = and f(xy) = 0}. 27. General method of procedure, first with tangents, second without tangents. 28. Miscellaneous methods ........... 3253 CHAPTER V. GEAPHO-TEIGONOMETEY AND MEXSUEATION. 1. Protractors. 2. Plotting angles by tables of chords. 3. Solution of triangles and other plane rectilinear figures. Datum circle. Examples. 4. Areas of triangles. 5. Parallelograms. 6. Irregular quadrilaterals. 7. Irregular polygons. 8. Curvilinear figures. 9. Mean height of in- dicator diagrams by gridiron parallel ruler. 10. Graphic correction at point of cut-off. 11. Area by Planimeter. 12. Graphic progressive Integration of Areas. 13. Proof of Integration Construction. Scale of Integral Area curve. 14. Application to Water Storage in reser- voirs. 15. Other applications. 16. Curvature, slope, points of inflexion of integral area curve. 17. Correction by versine of chord of integral area curve ........ 54 70 CHAPTER VI. COMBINED MULTIPLICATION AND SUMMATION. MOMENTS OF PAEALLEL VECTORS. 1. Combined multiplication and summation of products of any kind. 2. Moments of parallel forces and loads on beams. 3. Construction. Pole distance, and scale of moment-diagram. Eight or Left Pole. CONTENTS XI Locus of Axis indefinite. 4. Moment round any axis. 5. Position of axis of zero moment. 6. Locors. 7. Centre line of parallel locors or resultant locor. Partial moments. 8. Cyclical lettering of diagram. 9. One balancing locor. 10. Closed moment-diagram. 11. Two balancing locors. 12. Eule of signs for sectional moments. 13. Inde- terminate Abutment-thrusts 4 . . .- . 7183 CHAPTER VII. VECTOR AND ROTOR ADDITION. General definition of a vector. 2. Coincident and opposite directions. 3. Illustrations of different kinds of vectors. 4. The signs =, |k 4J r> =|=. =^=L ) zjjjr, =j|j=. 5. List of Locors occurring in mechanics. 6. Kotors. 7. Rotor representation of surfaces. 8. Kelativity of all vectors and rotors. 9. The ' field ' of a motion. 10. The meanings . of the words ' through,' ' past,' ' over,' and ' round.' 11. Keciprocal duality of all vectors and rotors. 12. Graphic process of Vector Addi- tion. 13. Distributive law. 14. Physical meanings of vector addition in cases of successive displacements of one body, and simultaneous dis- placements of different bodies. 15. Mean position. 16. Displacement of Centre of Volume or of Mass. 17. Integral mass displacement. 18. Zero sum of displacements from centre. 19. Co-planar displace- ment of a rigid body. 20. Axis of rotational displacement. 21. Non- planar displacement of a rigid body. 22. Axis of screw displacement. 23. Addition of simultaneous co-planar displacements of one body in different fields. 24. Adaptation of Hamilton's nomenclature to de- scribe this addition. 25. Similar addition of non-planar displace- ments. 26. Physical meaning of addition of simultaneous velocities and other time-rate vectors of different parts of one body. 27. Differ- ences of simultaneous velocities of parts of one body. 28. Sum and difference of simultaneous velocities in different fields. 29. Change of velocity. 30. Instantaneous velocities and hodograph. 31. Addition of accelerations. Eadial and Tangential Accelerations. 32. Intro- duction of the mass factor. 33. Addition of parallel displacement rotors ; the axes fixed in the field, and fixed in the body, and one axis fixed in the field with the other fixed in the body. 34. Similar addition when the axes are not parallel and intersect. 35. Coincidence of rotor with vector addition when the rotors are small. 36. Addition of angular velocities and other instantaneous time-rate rotors . . 84 120 CHAPTER VIII. LOCOR ADDITION AND MOMENTS OF LOCORS AND OF ROTORS. 1. Meaning of a locor moment. 2. Central position of a group of locors. 3. Sum of two intersecting locors. 4. Resolution of a locor into com- ponents. 5. Zero moment of axial component. Resolution into any number of Components. 6. Rectangular Components of forces, velo- Xll GRAPHICS cities, &c. 7. Sum or resultant of any number of co-planar locors. The Single-pen and Vector-pencil. 8. Sum or resultant of any number of locors not parallel to one plane. Two methods. 9. The balancing locor. 10. The locor Moment-diagram. 11. Variation of the Single- pen for one fixed set of locors. 12. Proportional displacement of the Single-pen. 13. Determination of two supporting forces at two given points. 14. Determination of three supporting forces at three given points. 15. Sum of parallel locors not in same plane. 16. Moment of a rotor. 17. Addition of angular velocities round parallel axes. 18. Diagram of linear velocities due to sum of angular velocities round different axes. 19. Diagram for angular velocities round non- parallel intersecting axes. 20. Moments of angular momenta. Pro- duct of Parallel Eotors. 21. Locor and rotor couples. 22. Substitution of ' equivalent ' couples. Limitation of the applicability of the ' result- ant ' as a substitute for the group of units. 23. Eesolution and composition of couples. 24. General case of reduction of a group of locors or of rotors . j . . . . ' . ~ . . 121 143 CHAPTER IX. THE KINEMATICS OF MECHANISMS. 1. General definition of a mechanism. Bed-plate. 2. Eelation between numbers of joints and bars in a mechanism. 3. Specialties of rigid- bar mechanism. 4. Motion-paths. 5. Displacement diagrams. 6. Velocity diagrams by differences and by Beuleaux's method of centroids. 7. Polar Velocity Diagram for a single rigid bar. Velocity Image. 8. Acceleration Diagram. Constructions for Centripetal Acceleration. 9. Scale of acceleration diagram. 10. Acceleration Image. 11. Integral momentum and acceleration of momentum. 12. Nomenclature of diagrams. 13. Four-bar mechanism. 14. Ordi- nary steam-engine mechanism with straight guides. 15. Six-bar mechanism ; direct solution. 16. Six-bar mechanism ; solution by two trials. 17. Four-bar mechanism with straight or curved sliding- joint. 18. Toothed gear 144162 CHAPTER X. FLAT STATIC STBUCTUBES, FBAMES, OB LINKAGES WITHOUT BEAM LINKS. 1. Conditions of balance in a set of forces and graphic test of such balance. 2. Meaning of moment diagram. 3. Closed chain of links. 4. Defi- nition of a link. 5. Pin-and-eye joints. 6. Practical deviation from theoretical idea of frictionless pin-joint linkage. 7. Distribution of load between joints. 8. Bending moments at joints. 9. Joint stiff- ness and flexibility. 10. Two-joint links. Simplicity of two-joint- link structures. 11. Identity of moment-diagram and diagram of balanced one-pen linkage of two-joint links. 12. How the link stresses arise. 13. Equilibrating shapes either stable or unstable. 14. Three CONTENTS xiii PAGE elements of arbitrary choice in balanced one-pen linkage. 15. Two- pen linkages with six elements of arbitrary choice. 16. Division of the linkage into any number of pens and introduction of three ele- ments of arbitrary choice with each new pen. 17. Limitation of this rule. 18. Degree of flexibility or stiffness dependent on number of pens. 19. Geometrical relations between linkage and force-diagram. 20. Nomenclature of diagrams. 21. Outside and inside links. 22. Method of Sections. 23. Definition of stiffness. 24. Criterion of stiffness in plane linkages without and with beam-links. 25. Criterion of stiff- ness in solid linkages. 26. Definition of redundancy. 27. Bedun- dancy due to earth connections. 28. Closer analysis of the criterion of Stiffness and non -redundancy. 29. Cyclic order and method of marking ditto. 30. Beading stress-diagrams for tension and compres- sion without the use of arrow-heads. 31. Objection to arrow-heads. 32. Use of + and . 33. Two colours, or thick and thin lines, for struts and ties. 34. Same cyclic order for whole structure, for each pen, for each link, for each joint-pin, and for any fractional part of whole structure. 35. Loads at internal joints and imaginary links. 36. Be- petition of stress lines. 37. Crossing links and imaginary joints. 38. Lettering for imaginary links and joints. 39. Two-link joints with- out applied load. 40. Three-link joints with two links in same line. 41. Four-link joints with two links in line. 42. Order of commence- ment and procedure in drawing stress-diagrams. 43. Distribution of links at the joints of a stiff structure. 44. Structures with no two- link joints. Commencement by ' Method of Sections,' or by ' Method of Two Trials and Two Errors.' 45. Detailed examples ; Arch with stiffening girder and loads at inside joints. 46. Girder with diagonals crossing. 47. Problem ; to draw a line through an intersection lying beyond the limits of the paper. 48. Contracted method of writing out Stress-diagram Constructions. 49. Symmetrical Boof truss. 50. Symmetrical Bridge truss with a series of loadings. 51. Example of Method of Sections. 52. Suitability of method for symmetrical and unsymmetrical loading. 53. Special case of ditto with parallel links in section. 54. Example of Boof truss illustrating method of sections and other specialties. 55. Cantilever Bridge structure illus- trating method of ' Two Trials and Errors,' and a number of peculiar difficulties. Interpretation of Stress-diagram for Coincident Actual and Imaginary Links 163 214 CHAPTEK XI. FLAT STATIC STRUCTUBES CONTAINING BEAM LINKS. 1. Besolution of stresses on beam sections into Axial, Shear, and Bending Stresses. 2. Various methods. 3. Simple cases. 4. Boof truss with four beams by method of sections. 5. Braced Pier transverse section by method of sections. 6. Three-beam Boof truss by method of ' Two Trials and Two Errors.' 7. Two-beam Crane truss by sub- titution of triangulated truss for beam . . . . 215230 XIV GRAPHICS CHAPTER XII. SOLID STATIC STEUCTUEES. PAGE 1. Necessity of solving solid structures in practice. 2. Fundamental dis- tinctions between flat and solid structures as regards (a) stiffness, (6) stability, (c) solubility of joints. 3. Simple tetrahedral linkage lettering of the diagram and cyclic order of construction. 4. General statement of method of solution. 5. Construction for the tetrahedron. 6. Find the whole stresses from their plans and elevations. 7. General case of three supporting forces found by help of an imaginary penta- hedral frame. Composition diagram. 8. Shorthand symbols for writing out the construction of the diagrams. 9. Braced Pier with four sup- ports. Simplification. Construction of the diagram in plan and elevation. Composition diagram. 10. Method of solid section cutting six bars . . . . . , 231 GENERAL INDEX . 253 XV INDEX TO ATLAS OF ILLUSTKATIONS. N.B. The 'Page' is that of the Text where reference is made to the Figure. Subject. No. of Plate. 1 Section of Scale 2 Division of Scale 3 Positions of Set- Squares 4 Divided Curve-Templates of Uniform Shape 5 Principle of Graphic Multiplication ... V I. 6 Error of Intersection due to Thickness of Line 7 Error of Intersection due to Error in Direc- tion of Line 8 Twelve Multiplication Constructions ... II. 9 Combined Multiplication and Division . . 10 Continued Multiplication 11 Multiplication Construction on Sectional Tablet 12 Division Construction on Sectional Tablet . 13 Positive and Negative Integral Powers . 14 Fractional Powers . 1& Signs of Polar Co-ordinates , 16 Straight Line Eectangular Equation . . . j 17 Straight Line Polar Equation 18 Straight Line Focal Angular Equation . . 19 Elliptic Quadratic Equation 20 Incomplete Hyperbolic Quadratic Equation |- jy. 21 Complete Hyperbolic Quadratic Equation . 22 Parabolic Quadratic Equation 23 Circle Polar Equation 24 Conic Focal Angular Equation 25 Sectional Tablet for the Solution of Linear, Quadratic, and Cubic Equations ... VI. 26 Solution of Special Equation 27 Plotting Angles 28 Height of a Spire from Theodolite Measure- f ments > 29 Theodolite Survey across a River . . . 30 Areas of Triangles 31 Areas of Parallelograms VII. 32 Areas of Irregular Quadrilaterals . . . 33 Ditto, another construction . . . . * 34 Areas of Irregular Polygons 1 VIII 35 Ditto, another construction > XVI GKAPIIICS "o 2 *! Subject. No. of Plate. Page of Text. 36 Areas of Curvilinear Firires 59 37 38 Two Constructions for Correction at Point of Cut-off in Steam Engine Diagram . Progressive Area Integration I VIII. 62 64 68 39 Water Storage Integration 67 40 41 42 Correction by Versine of Integration Curve. Addition of Moments of Parallel Locors . Cyclical Lettering of Moment and Vector Diagrams . . . . IX. 69 72 76 43 44 Indeterminate Abutment-Thrusts .... Vector Addition 80, 132 92 45 46 Co-planar Displacement of Rigid Body . . Addition of Translatory Displacements in Different Fields . 98 102 47 Addition of General Displacements in Dif- ferent Fields ... X. 103 48 49 Vector Differences or Vector Subtraction Addition of Simultaneous Velocities in Dif- 108 108 50 51 52 53 Differences of Simultaneous Velocities . Change of Velocity and Hodograph . . . Addition of Simultaneous Accelerations . Addition of Successive Parallel Displace- ment Kotors with Axes fixed in moving . XI. 109 109 110 112 54 55 56 57 58 a 586 59 Ditto with Axes fixed in the Field .... Addition of Inclined Displacement Rotors . Addition of Locors meeting in One Point . Addition of Co-planar Locors, General Case and Construction of Single-Pen Linkage Addition of Non-planar Locors, General Case Addition of Parallel Non-planar Locors . . Linear Velocity Diagram of Rotating Rigid Body 1 \ XII. 114 116 122 124, 129, 163, 169 127 134 137 60 Locor and Rotor Couples 139 61 62 63 64 65 66 Resolution and Composition of Couples . General Locor and Rotor Reduction . . Displacement Diagram of a Mechanism . Velocity Diagram of a Rigid Bar . . . Acceleration Diagram of a Rigid Bar . . Three Constructions for Centripetal Acce \ XIII. 141 143 146 148 150 111, 151 67 Velocity and Acceleration Diagrams for XIV. 153 68 69 Ditto for Steam-engine Mechanism . . Ditto for 6-bar Mechanism .... 1 154 155 70 lla lib 72 73 Ditto for 6-bar Mechanism by Trial and Error 4-bar Mechanism with Straight Sliding Joint Ditto with Curved Sliding Joint .... Mechanism with Two Straight Slides . . . Velocity Diagram for Toothed Gear . . . XV. XVI. 156 159 159 159 160 INDEX TO ATLAS OF ILLUSTRATIONS XVll ss =.l ^En Subject. No. of Plate. Page of Text. 74 Variability of Shape of Linkage acted on by a Set of Balancin^ Forces XVI. 173 75 7r> Cyclical Lettering of Joints, Pens, Poles, and Pencils in Linkage and Stress-Diagram . Outside Links . [ XVII. 179, 187 182 77 J 184 78 79 Stiffened Arch loaded at Internal Joints Lettering of Lattice Girder with Crossing I XVIII. 190, 197 193 80 Non-redundant Lattice Girder . 194, 200 81 82 Hoof Truss wrongly designed Symmetrical Eoof with Symmetrical Load- ing I XIX. 195 202 83 Symmetrical Bridge with Distributed E oil- I XX. 203 84 Method of Sections 1 206 85 Hoof without 2-link Joints solved by Method of Sections XXI. 207 86 87 Cantilever Bridge solved by Trial and Error Simple Beam Linkage XXII. 210 216 88 Inclined Beam Linkage | XXIII. 218 89 Roof with Four Beams XXIV 219 90 Pier of Two Columns connected by Trans- verse Top Bracin " . XXV 221 91 Eoof with Three Beams solved by Trial and Error XXVI. 222 92 Crane with Two Beams solved by Substituted XXV. 227 93 94 Tetrahedral Linkage with One Loading Force and Three Supporting Forces . . Pentahedral Construction to find the Sup- porting Forces of a Solid Linkage with any number of Loads . . XXVII. XXVIII. 233 134, 239 95 Solid Braced Pier with Vertical Loads, Wind Forces, and Four Supports XXIX. 243 XIX GLOSSAEY OF "SPECIAL TEEMS AND SYMBOLS. 1. Sectional Tablet'. Square plate of cardboard, porcelain, or slate ruled with small squares, used for Multiplication and Division. 2. Direction indicates not only the lie of a line, but also the sense along that lie. A line has two possible exactly opposite directions, unless it be specified to have one only of these two. 3. Vector : a quantity having the two properties, Magnitude and Direction. 4. Locor : A quantity having the three properties, Magnitude, Direction, and Position. 5. Rotor : A quantity which, not being a Vector or Locor, is yet capable of accurate graphic representation by a Vector or Locor line ; such as a rotation or a force-moment. 6. = means Quantitative or Numerical Equality. 7. J^= means Quantitative or Numerical Equality, combined with Parallelism without reference to Direction or Sense. 8. r^jr means Vector Equality. 9. =j means Quantitative or Numerical Equality and Parallelism, but of opposite directions. 10. 4|4; means Locor or Rotor Equality. 11. jp indicates Locor Oppositcness ; i.e. quantitative equality, identity of position and of opposite directions. 12. Centre of Volume, Area or Mass : The point whose position is the average of the positions of all the individual or part volumes, areas, or masses considered, the average being taken with regard to their magni- tudes. 13. Centre-line of Locors : The line whose position is the average of the positions of all the locors considered, taken with regard to their magni- tudes. 14. Positive sign of Rotation is used when the rotation is right-handed when viewed in the direction counted positive along the axis of rota- tion. If the motion of a screw through its nut be positive, its rotation is positive if the thread be right handed, and vice versa. XX GRAPHICS 15. Field of a body, or of a group of bodies maintaining their relative positions among themselves invariable, means the space geometrically attached to the body or bodies, or the space defined, and measured in, relatively to the body or bodies. The field of a body moves along with the body when this latter is displaced through the field of any other body. The field includes the space both inside and outside the body itself. 16. ' Through ' a field means relatively to the field specified. ' Past,' 1 Over,' or ' Round ' a body or point means relatively to the body or point specified. 17. Integral Area, Volume, or Mass Displacement, means the area, volume, or mass, multiplied by the displacement of the centre of area or of volume, or of mass. 18. Axis of Rotational Displacement is the axis round which the actual displacement of a body may be produced by pure rotation without translation. 19. Axis of Screw Displacement is the axis round and along which the actual displacement of a body may be produced by pure uniform screw motion. 20. A space completely surrounded, or ' closed,' by lines or surfaces, by bars or plates, is termed a Pen. 21. Two contiguous superficial spaces, being called A and B, the dividing line between them is called A B. This line is common to both spaces, and is in this book frequently called the 'joint between the spaces. 1 22. Similarly, the straight line joining two points is called the 'joint between the points', ' the intersection of two lines is called the ' joint of the two lines ; ' the surface between two volume -spaces, and common to both, is the 'joint of the volumes.' 23. Any space P being bounded by, and separated from other spaces by, the straight lines P A, P B, PC, P D, &c., and being open on one side to infinite space, the polygon so formed is called (P) ABCD &c. oo. "When the polygon is closed it is termed a ' pen ' and the sign oo at the end of the name is omitted, thereby indicating that the space has no side of it open to infinite space ; thus (P) A B C D a pen of four sides. In this name the letters must be placed in the order of the sides consecu- tively, taken either right- or left-handedly ; thus the above may be called (P) A D C B. The pen may often for simplicity's sake be called simply P i.e. by the letter placed in the space when no ambiguity arises through thus naming it. 24. The joint of two lines A B, B C is called (A B C). When several lines A B, B C, C D, D E have one common joint, i.e. meet in one point, their joint is called (ABODE). This is the joint of four lines ; but if the spaces E and A have a common boundary, it is the joint of five lines, the fifth line being E A. The letters must be placed consecutively in this name in the order in which the spaces are arranged round the joint, either right- or left-handedly ; thus the above joint may be also called (E D C B A). The joint of more than three lines may, however, GLOSSARY OF SPECIAL TERMS AND SYMBOLS XXI be more shortly and quite explicitly expressed by naming three only of the spaces, as (ABC). 25. When the joints of the spaces A B C D E, &c., have one common joint (A B C D E), this latter is a point common to all the spaces and may be called the 'point-joint ' of the spaces. When the spaces A B C D E completely surround a pen P, the pen (P) A B C D E may be called the 'pen-joint ' of these spaces. 20. When a number of lines p a, p 6, p c, &c., radiate from one point p, the pointy is called the pole, and the pencil of lines is symbolised collec- tively by (p) a b c, &c. 27. The perpendicular distance of a point p from a straight line a b is in this book written^? (ab). Here a and b may be points. If A and B be spaces, the perpendicular distance of the point p from line A B may still be called p (A B). Sometimes, for sake of clearness, this will be written p (a b) or p (A B). The moment of a locor A B round an axis P is written A B P (A B). 28. A couple formed of two equal and opposite locors or rotors A B and C D is written (A B ^= CD). 29. * means right-handed cyclical order in stress-diagrams, arid Q means left-handed cyclical order. 30. The tension stress is counted + ; the compressive stress . 31. The elongation strain is counted + ; the contraction strain . 32. Shear stresses and strains, when requiring to be indicated by a sign analogous to + and , are given the signature L. 33. ^^> c -^ g means ' draw from the known points a and b parallel to the known directions P Q and R S two lines, and mark their inter- section c.' Usually ac is understood to be drawn parallel to the simi- larly named known line A C, and b c to the similarly named line B C ; and in this case the complete symbol ^/-c J^Q is contracted to >c simply. 34. ' a $ b || P Q ' means ' find point b by drawing from the known point a in plan a line parallel to the known line P Q, and by projecting verti- cally downwards on to this line from the known point b in elevation. 1 Usually the line a b is understood to be drawn parallel to the similarly named known line A B, and in this case the complete symbol is con- tracted to ' a $ &.' The converse process of finding a point in elevation by upward projection from its already known plan is indicated by **4&. 1 35. '0-6 = 0' means that, the length a b being found to be zero, the point b is to be marked coincident with the already found point a. 36. In linkages for which stress-diagrams are drawn, the thin lines repre- sent tie-bars and thick lines represent struts. GEAPHICS INTRODUCTION. THE principles according to which stress -diagrams are drawn out are stated in a very complete and general manner in a paper entitled ' Reciprocal Figures and Stiff Frames,' by Clerk Maxwell, in vol. xxvii. of the Philosophical Magazine, 1864 ; but this paper is quite useless to the ordinary engineer who wishes to know how to apply these principles. The theoreti- cal development of the subject is chiefly due to Cremona, an Italian, and to Culmann, a German. A very great advance in adapting the method to practical requirements, and in sim- plifying the comprehension and construction of the figures, was made by the introduction of the elegant system of mark- ing the diagrams, commonly known as Bow's Notation, the earlier invention of which is, however, also ascribed to Pro- fessor Henrici. Believing that none of the books published on the subject in the English language is well adapted to the needs of the engineering student who wishes to fit himself for the use of the method in business practice, it has been thought well to attempt to supply this deficiency. To succeed in doing so it is absolutely necessary to illustrate every new statement of general applicability by examples worked out in accurately drawn diagrams; and this will accordingly be done. It is just as essential for the student who wishes to profit by the B 2 GKAPHICS study of this book to draw out for himself similar examples as accurately and carefully as he can. There is no subject in which it is more easy to imagine that one understands a problem or theorem, and to find, when one comes to the practical application of it, unexpected difficulties cropping up which bar progress even at the outset. These difficulties are found out only by experience in the art. The way out of such difficulties will as far as possible be pointed out in the course of the present volume. 1. Advantages and Disadvantages of the Method. The great importance and convenience to engineers of this method of calculation have become very generally recognised of late years, but it may be well to make clear what are the grounds upon which, and in what respects, it can claim superiority to the more time-honoured algebraic, trigonometric, and arithmetric methods. The following advantages may be easily recognised : All the things that engineers deal with have magnitudes, or quantities, which vary ' continuously '- that is, one magnitude may differ from another by an indefi- nitely small quantity. Designers who never mark on their Continu- drawings dimensions involving a fraction of T ^ in. may be apt quantity ^ ^ or g e ^ this, but the workman who has to file down the piece to the exact T ^ to which it has been designed is in no danger of forgetting it. But numerical representation of continuous increase of magnitude is impossible ; to represent a very small difference a very long decimal fraction must be used, and to represent an excessively minute difference is practically im- possible by numbers. A line, however, is continuous in mag- nitude in the same way as all the quantities to be calculated are the difference between two lines may be indefinitely small. It is, therefore, a very suitable vehicle upon which to mark off the magnitudes to be dealt with. 2. Again, the graphic method brings into strong promi- Scaies nence before the mind the important fact that all * scales ' are essentially artificial, and that no two scales for quantities of INTRODUCTION 3 different kinds can be alike in any respect. The magnitudes of all sorts of things of different kinds are plotted off along lines. Sometimes an inch length along the line means so many tons, or pounds, or kilogrammes; sometimes so many cubic feet ; sometimes a quantity of energy, or of momentum, or a linear velocity ; it may even mean so many degrees of angular space, or, perhaps, an angular velocity. No confusion of mind is more common than that of thinking that a quantity of one of these kinds may be in some sense equal to a quantity of another kind, and that both may be represented to the same scale. At first sight the habit of marking off all such quantities as Scales lengths along lines might seem to tend to confirm this confu- sion of mind. No commoner mistake is made by learners on the threshold of the subject than to ask if they should mark off moments, for example, to the same scale as they have em- ployed for forces, or forces to the sa^ne scale as has been used for feet. But they cannot work out thoroughly to the end a single problem in graphic calculation without having this absurd idea very much weakened in its hold on their minds, and it will be driven entirely away before they have completed half a dozen examples. 3. Also, the accuracy of the measurements that can be made in this method is in a certain sense proportioned to that actually attainable in quantitative knowledge of facts. It becomes at once apparent to the calculator that absolute accuracy in his results is impossible, and he becomes familiar with the true and important idea that all engineering calcula- tions and, indeed, all physical calculations possess only a Degree of certain degree of accuracy, and are burdened with a certain amount of unavoidable probable error. Nothing is more silly than the statements we often unfortunately meet with in reports from engineers, who ought to know better, of results calculated to a nicety sometimes a thousand, or even a hundred thousand, times greater than is at all possible in fact. The tensile strength of a material may be stated to five B 2 GRAPHICS Degree of approxi- mation Vectors or six figures, when actually it is not known within 5 per cent. The breaking strength of a girder may be stated by three or four figures, when it really cannot be certainly known within 10 per cent. The horse-power of an engine may be stated to within T Vth per cent, when it has been cal- culated from half a dozen measurements, each of which may be subject to a probable error of from ^ per cent, to 5 per cent. Such mistaken exactitude in calculation is in the first place purely deceptive, and thereby injurious, and in the second place it most surely reveals the calculator in a most ridiculous light to the eyes of those who know better. 4. Fourthly, a very large proportion of the quantities engineers have to calculate are quantities having direction, and if these are represented by lines which have correspond- ing directions as well as magnitudes, the comprehension of the relations among the different elements of the problem becomes much more natural and easy. The word * graphic ' etymologically means ' byline drawing,' and it is in this sense that it is used in the title that has been given by common consent to this art ; but in its popular sense of * peculiarly representative and calculated to assist the imagination ' the word ' graphic ' may, for the above reason also, be considered particularly appropriate and descriptive. To such directed quantities the general name 'vector' has been given. Ex- amples are : Distances between two positions, i.e. ordinary dimensions, motions, velocities, momenta, forces. 5. These are general considerations in favour of the graphic method, but so far as business utility is concerned they might easily, and in fact are frequently, overbalanced by more practical necessities, such as, for example, rapidity and accuracy. Absolute accuracy, as has been already said, is never attainable, and the pretence of it is in most cases ridiculous ; but in many special circumstances it may be essential to obtain as high a degree of accuracy as is by any means possible. In such special circumstances it not infre- INTRODUCTION 5 quently happens that the capability of ordinary arithmetic for the attainment of exactitude quite outdoes that of any graphic method. It is, therefore, a mistake to suppose that in every case the graphic method is necessarily the best. With skilful Minute draughtsmanship, however the special points to be attended unattain- to will be mentioned hereafter an accuracy of -fa per cent. a ( in nearly all cases, and very frequently of l ^ per cent., may be obtained. 6. To get at all good results exact drawing is needful, and Training to exacti- the endeavour after this exactitude is in itself a good training tude to every engineer of whatever class. 7. Kegarding accuracy, also, there is one very important advantage the graphic method possesses namely, that in a large variety of most important problems the drawing always Automatic furnishes a very easy test of its own accuracy, so that it be- comes absolutely impossible to make a large mistake without its being discovered. 8. As regards rapidity, graphic calculation is in a great number of cases the most rapid method that can be adopted, provided certain obvious conditions are fulfilled. These are that the calculator has his drawing instruments, pen, paper, pencil, and ink always at hand and in good condition ready for immediate use. If he takes half an hour to get his instru- ments ready in order to make a small calculation which when these preparations have been completed takes ten minutes more to finish, he must not be surprised if his arithmetical or Limitin ff J conditions trigonometrical competitor has beaten him as regards time, and he must not blame the method for his want of success. As paper, drawing-board, and instruments cannot always be ready at hand, this points to a class of calculations for which the graphic method offers no sort of advantage namely, short, easy calculations. We have heard of the product 4x5 being found with the help of logarithms. We do not admire the method, although we remain fully convinced of the advantages obtained from using logarithms. If one has to 6 GRAPHICS make a single multiplication, say (347) 2 x -7854, there is no utility in putting down a piece of paper on a board, sharpening a pencil, and getting out one's scale, in order to find the product. It may be found by ordinary arithmetic, or Limiting by reference to a table of areas of circles, in the time that 8 must be spent in sharpening the pencil. If, however, fifty or a hundred multiplications of odd factors have to be performed, it may be worth while to spend the time necessary for the above preparations. 9. In these last hundred calculations it is also an impor- tant consideration that they involve much less mental fatigue when performed graphically than by any other method. This is one of the greatest, if not the greatest, advantage offered by graphics. In scientific designing, and, indeed, in nearly all engineers' head-work, a large amount of wear and tear of brain power occurs through the need of performing lengthy and laborious calculations that are, to a large extent, mere mental drudgery. Nothing is more fatiguing to the eyes, and through the eyes to the brain, than dealing with interminable Mental arrays of figures. Graphic calculation does away with such fatigue to a very great extent, and in fact a moderate amount of such exercise is felt to be pleasant and enjoyable merits which few will venture to ascribe to numerical arithmetic. No universal superiority, then, can be ascribed to this art over its rivals. It should be used in such circumstances as bring into prominence its peculiar advantages. Adherence to it under all conditions, and to the entire exclusion of assist- ance from other modes of reckoning, is to be avoided as dogmatic and irrational, while a denial to it of very great superiority in innumerable practical problems would show equally unreasonable prejudice. INSTRUMENTS CHAPTEK I. INSTRUMENTS. 1. IT is only with scrupulously exact drawing that the generally useful degree of accuracy can be obtained in graphic calculations. It is, therefore, of the highest importance that all the instruments used should be true. 2. The Scale is of the first importance. It should be engine-divided. No better material can be chosen for the scale than boxwood. Cardboard scales are not sufficiently exact for this class of work. The oval section, useful as it may be for ordinary working drawings of details of machinery and structures, is unsuitable for graphic calculation because of the unsteadiness with which a scale of this form rests on the paper. Long lines have to be marked off, and measured with exactitude ; and as one cannot look fairly at both ends of a long line at once, one must be able to hold the scale firmly on the paper, secure against the slightest slipping, Scales while one moves from end to end. The flat section shown in Fig. 1 is therefore the best, offering as it does the fullest amount of frictional resistance to slipping over the paper. The flat underside should be plain, without markings of any sort. The two bevel edges may conveniently be divided in inches and in millimetres. It is important that the division should be decimal. Fig. 2 shows a small portion of the length of the scale used in the engineering classes of the Mason Science College. The numbering of the divisions proceeds in one direction only, from left to right. For each edge there are different lines of numbers. The first series on GRAPHICS CHAP. I Scales the inch edge reads inches, or ^ , or - x J , or y^Vo of an inch, or any submultiple of tenths of an inch. The second series is for reading to Jin., or yV^ 18 * or sin^hs, &c., of an inch. The third is for -J-th, or -g^ths, or si^hs, & c -> of an inch. The millimetre edge has two similar series of numberings. The minute division to half millimetres and to fiftieth-inches is placed at both ends beyond the end line of the scale, and also on each side of the middle line of the scale. The length of the scale should not be less than 20 in. Long lines, such as frequently occur in the diagrams, ought not to be measured in sections. The scale should stretch the whole length of the line, otherwise inaccuracies will accumulate, and much unnecessary loss of time will occur. A scale 30 in. long is often useful. 3. T-squares are used in the same manner as in ordinary T-squares mechanical drawing. The straightness of the edge is of special importance. 4. Next to the scale the straight-edges and set-squares are of greatest importance. Two straight-edges one about 18 in. long, the other 3 ft. 6 in. or 4 ft. long are convenient. The longer one cannot be dispensed with. Mahogany is a good material, giving a good edge without special edging. If these are edged with ebony, the exactness and smoothness of the edge are greatly improved. It is essential that the straight- edges be kept dead true. They should, therefore, be fre- quently examined, and trued-up with scrupulous care, if they be found to have bent or warped in the least degree. The same remarks apply to the set-squares their edges must be kept accurately straight. The right angle of the set-square should be exactly correct. The accuracy of the 60, 30, or 45 angle is seldom of any importance, but they should never be used to set off these angles with unless they are known to be correct. Six set-squares should be provided: One GO , of 12 in. or 14 in. length of side; one 45, of about 9 in. or 10 in. side; and one 75, of similar size. Straight- edges Set- squares INSTRUMENTS ^^L CALIFORH\V^ 9 Three small set- squares about 4 in. to 5 in. side, and with the above angles, viz. 60, 45, and 75, are very useful. Frame set- squares of mahogany edged with ebony are the best, but Set- simple pear-wood set-squares keep their shape and straight- ness of edge very well if a number of round or oval holes be cut out of the central portion. 5. In graphic calculation lines have constantly to be drawn accurately parallel to each other, and frequently at a consider- able distance apart. Parallel rulers are very inaccurate in doing this work. The only accurate and convenient method is to slide a set-square along a straight-edge which may be an edge of another set-square, if this be of sufficient length. In performing the transfer of the direction across the paper it is important to make the operation as simple as possible. In nearly all cases it is possible to complete the transfer in one sliding operation. It is at first sometimes rather con- fusing to see in what way the set-square and straight-edge are to be arranged for this purpose, and a beginner has frequent recourse to a tortuous policy of many successive slidings of one set-square over the other. This is seldom necessary, even for the longest transfers. The long straight-edge should be Drawing laid on the paper always in the direction of the desired para transfer, and as close as convenient to the positions of the line to be drawn and of the already drawn line to which it is to be parallel. Then this last line is found to make with the straight-edge some angle lying between and 90. The set- square angle is now to be chosen which most closely approxi- mates to this angle between the straight-edge and the line to be drawn i.e. between the desired direction of transfer and the direction to be transferred. With the two 60 and 45 set- squares there is a choice of six set- square angles. The 60 set-square gives 30, 60, and 90; the 45 set-square gives 45 and 90. By laying the side of one set-square against the straight-edge, and the side of the other set- square against the hypotenuse of the first set-square, there are 10 GRAPHICS CHAP. i. obtained the two angles 15 and 75; because 30 + 45 = 75, and 45 - 30 = 15 ; or 60-45 = 15. These last arrange- ments are shown in Fig. 3. The single set- square with angles 75 and 15 is, however, very greatly to be preferred to this combination of the two 60 and 45 squares. Putting a set- square edge along the line to be transferred, we may thus place the straight-edge inclined to it by any of the six angles 15, 30, 45, 60, 75, and 90, which angles increase by a differ- ence of 15. Thus the setting of the straight-edge need never deviate from the exact desired direction of transfer across the paper by an angle of more than 7J at the most. This will throw the position of the line to be drawn further from or nearer to the straight-edge than that of the line to which it is to be drawn parallel. Since the sine of 7^ is about J, the maximum amount by which it may be thus thrown to one side is about one-eighth the distance of transfer. The excess Drawing of the length of the edge of the set- square over that of the line to be drawn will generally cover this deviation of the straight-edge from the exact desired direction of transfer. If it does not do so the line of the edge of the set-square may be extended by laying another set-square against it. Sometimes the direction of the parallel lines nearly co- incides with that of the transfer. In this case the straight- edge is simply laid along the given line ; then near the posi- tion of the desired parallel, a side of a set- square is laid against the straight-edge. Finally, a second set-square is laid against this first, and the first is slid along the second into the required position. When a long line has to be drawn through a given point and parallel to another given long line, and at no great distance from this latter, the best procedure is to take with the pencil compass the distance of the point from the given line ; with this as radius strike an arc with centre in the given line as far away as possible from the given point, and to draw a tangent to this arc from the given point. 6. In marking off lengths upon lines in the diagram, it is CHAP. i. INSTEUMENTS 1 1 necessary, in order to secure exactitude, to prick them off with a needle point. An ordinary fine sewing needle, stuck in a small piece of wood to serve as handle, is a very much better instrument for this purpose than the prickers usually sold with mathematical instruments. If intersections of lines Needle- are pricked off with this needle point they become much more P ncker sharply defined. The dividers should never be used to take dimensions from the scale with ; nor should the points of bows or compasses ever be placed on the scale. 7. The pencil points of the bows and compasses used should be filed flat, with a rounded profile. This rounded profile is best obtained by filing one side of the pencil quite flat, and the other to a rounded conical form. The flat must be per- pendicular to the line between the two points of the compasses. If this is not attended to, and if the profile be not well rounded, the compasses or bows will draw circles of slightly different Pencils radii according as they lean to the paper on one side or the other. These instruments should have sharp points and stiff, inflexible legs. Two pencils are desirable one for drawing lines, the other for marking and lettering the diagrams. The latter may be a No. 4 Faber's, and is given a round point. The former may best be a No. 5, and its point should be filed flat and broad, and kept always perfectly sharp. 8. In many of the more complex graphic calculations curves need to be drawn. A set of pear-wood curves assists greatly in this work. Of these, what are called Trench* curves are frequently useful, but ' ship ' curves are more generally applicable. A good curve for this kind of work possesses the same character that a * ship ' curve ought to Curves have namely, the curvature should change gradually and continuously from point to point. It is surprising what a variety of curves may be fairly drawn out with the help of only three or four wooden templates of this sort if these have been skilfully shaped. The characteristic shape of any portion of a curve depends upon the rate at which its curvature varies 12 GKAP1IICS CHAP. i. from point to point. It may be very conveniently stated by specifying the change in the length of the radius of curvature per unit length of arc. Thus the rate of change of radius of curvature might be i in. per 1 in. length of arc. If one purve be derived from another by diminishing the lengths of suc- cessive small arcs, all in the same ratio, and at the same time diminishing in the same ratio the radii of curvature of these successive small arcs, the two curves will differ simply in the second being drawn to a smaller scale than the first they will have the same general shape. It would be very convenient if the radii of curvature were marked in figures at the different points of the edges of wooden template-curves. A very useful description of curve is obtained by keeping the above rate of change of radius of curvature per unit length of arc uniform throughout the length of the curve. Take, for instance, the above given example of ^ in. change of radius per 1 in. length of arc, and suppose this rate maintained uniform from point Caives to point. Let the radius of curvature at point A be 12 in. ; let B C D E F be points in the curve whose distances from A measured along the arc are in., J in., f in., 1 in., and 1 in. Call the radius of curvature p ; for instance, at A, p a = 12 in. ; then /> fc =12|- in.; p c =12J in. ; /o d =12f in.; p e =12J in.; p r =12| in., &c. Let the point K be distant from A 24 in. measured along the arc. Here p k =p a + V = 12 + 12 = 24 in. = 2/v Take now on the curve points L M N P distant from K by arcs of lengths ^ in., 1 in., 1 in., 2 in., and 2 J in. ; that is, at distances apart double those between BCD, &c. We then have p=p k + i x i = 24 in. = 2/> & ; /> m =24 in. = 2/> c ; p n = 24| in. = 2p d ; /> = 25 in. = 2p e ; and /o p = 25J in. = 2/> / . Thus this latter portion of the curve differs from the first only in that it is drawn double size. It is clear, therefore, that the different portions of the curve have all the same ' shape ' as defined above, and differ only in size or scale ; each portion is simply a repetition of any other portion drawn to a larger or smaller scale. With a set of curve-templates designed on this CHAP. I. INSTRUMENTS 13 principle and drawn to a suitably graduated series of ' rates of change of radius of curvature per unit length of arc,' a practically inexhaustible variety of curves may be fairly drawn, whether they be required to be drawn to a large or a small scale. For instance, the series may be the following : Tem- plate No. 1, rate of change of radius of curvature \ in. per 1 in. length of arc ; template No. 2, J in. per 1 in. length of arc ; template No. 3, | in. per 1 in. length of arc ; template No. 4, 1 in. per 1 in. length of arc ; template No. 5, 1^ in. per 1 in. length of arc ; template No. 6, 2 in. per 1 in. length of arc ; template No. 7, 2J in. per 1 in. length of arc ; template No. 8, 3 in. per 1 in. length of arc ; template No. 9, 3^ in. per 1 in. curves length of arc ; template No. 10, 4 in. per 1 in. length of arc, &c. Fig. 4 shows three examples of these curves accurately drawn. A scale of inches is divided off on the edge, and the radius of curvature figured at the chief points. The scale should be marked on both sides for convenience in drawing two-sided symmetrical figures. To ensure perfect symmetry of the second side with the first, it is only necessary to plot off two points, and lay between them the reverse side of exactly the same portion of curve-template as has been used between the corresponding two points of the first side. 1 9. Curves are frequently drawn with the help of splines of lance wood. If these are well made and in good condition they answer the purpose very well, but if by warping or other- Splines wise the spline has become irregularly bent, then it is very difficult to bend it so as to draw fair curves, the local natural bend or twist in the rod always being reproduced to a greater 1 The equation representing the above kind of curve is -]dy Eadius of curvature p = e c * ~ c ' tan ax c where e - is the radius at the point where the curve has its maximum height above the axis of x (c 2 , therefore, merely showing the position in which the curve lies on the paper), and c } depends on the ' shape ' as defined above. 14 GRAPHICS CHAP. i. or less extent in its artificially bent condition. It is difficult to preserve splines, even if made of the best wood and perfectly straight and regular at first, so as to avoid comparatively speedy injury by local bending. This is partly produced by the use of the spline, whereby, owing to the imperfect elas- ticity of the wood, severe permanent set is produced in special Splines places. Carefully made splines of slightly hardened steel of small section would not be subject to this disadvantage, but it would be difficult to harden them as much as would be necessary to avoid permanent set, arising through their con- tinued use, without twisting and bending them in the harden- ing. Pieces of watch and clock spring of different strengths are often very useful. CHAP. ii. DIVISION OF THE SUBJECT 15 CHAPTEK II. DIVISION OF THE SUBJECT. GRAPHICS may be divided in correspondence with the ordi- narily recognised different methods and subjects of calcula- tion. These are : (a) Arithmetic, (b) Algebra, (c) Trigono- metry, (d) Dynamics, (e) Tabulation and Analysis of Experi- mental and Mathematical Eesults. A few words of explanation regarding each of these sections of the subject may be useful before proceeding to the detailed treatment. 1. Graph-arithmetic.- Arithmetic shows how to find in- creased or decreased quantities when they are altered by given amounts or in given ratios. As the solution of every practical problem involves, and, in fact, to a great extent consists, in a more or less complex series of such operations, the rules of Arithme- arithmetic are applied continually throughout all graphic con- tlc structions. It is thus of great importance to be thoroughly familiar with them, and with the special suitability of each rule for the circumstances to which it is most adapted. 2. Graph-algebra consists in the solution of equations by drawing straight lines and curves. The usefulness of the method in solving equations, which would be very difficult or Algebra impossible to solve by other means, will be illustrated by examples. 3. Grapho-trigonometry is the * solution ' of triangles and other rectilinear figures, that is, the calculation of unmeasured Trigono- sides, angles, and areas from the sides and angles that have metry 16 GRAPHICS CHAP. II. been measured. Applications to surveying measurements will be given. 4. Grapho-dynamics. Dynamics may be considered under three heads : Kinematics, or the pure geometry of motion ; kinetics, or the laws of motion as dependent on the masses of Dynamics the bodies moving ; and statics, that special branch of kinetics dealing with cases in which the motions are zero, and the forces in equilibrium. Some simple constructions which apply equally to all three sections of dynamics will first be illus- trated. 5. There is great practical convenience in treating statics separately from kinetics ; and since the bulk of the interesting engineering problems to which the graphic method has been statics applied belongs to statics e.g. applications to bridge and roofwork this portion of the subject will be taken before the more difficult problems of the kinetics of moving masses. 6. The plan of separating kinematics from kinetics has been followed in many modern text-books of high authority. This plan is specially convenient for engineering students Kinema- because it is frequently desired to study the relative displace- ments, velocities, and accelerations of velocity of the various parts of a mechanism without references to the masses of the moving parts or to the driving and resisting forces. A special chapter will therefore be devoted to the Kine- matics of Kigid Bar Mechanisms. 7. Experimental Tabulation. The results of a series of experiments for example, on the relation between the speed of a vessel and the horse-power indicated by its engine, or on the relation between the pressure and temperature of steam are best made clear by plotting them graphically, i.e. draw- Co-ordi. ing a curve, the rectangular ordinates to which are the values lation " of the quantities whose relation is to be investigated. This assists in the elimination of experimental errors ; it shows the relation found in a very clear manner to the eye, and through it to the mind ; and if a formula is desired to represent the CHAP. ii. DIVISION OF THE SUBJECT 17 variation, the curve can be analysed as to its geometrical pro- perties. 8. Mathematical Tabulation. The best and most accurate formula by which to design some dimension is often compli- cated and tedious in its application to each special case. This prevents its use in practical life where men are busy and have to economise time. Its use is also prevented by the difficulty of understanding the general meaning or the effect of so com- Mathema- plex a rule. These difficulties are entirely done away with if JJ-^Jte the results of the formula are represented by a curve, and the tabulation application of a difficult and cumbrous formula becomes abso- lutely as easy as that of the most simple. These curves ought to be drawn on square sectional paper, the divisions of which ought usually to be decimal. This plotting out of experi- mental and mathematical results may be called graphic tabu- lation. c 18 GRAPHICS CHAP. HI. CHAPTEK III. GRAPH-ARITHMETIC. 1. SIMPLE addition and subtraction can seldom be performed by graphic means with any advantage, so far as ease and rapidity are concerned. Suppose two or more quantities known, and that they are to be added together. The sum can be found by ordinary numerical addition much more easily and quickly than can be completed the process of plotting off the magni- tudes to a certain scale along a straight line, each successive Addition length plotted having its left-hand end at the right-hand end of the preceding one, and then reading off to scale the length of the line made up of these separate parts. This is evidently the only possible graphic process of arithmetical addition. If any of the magnitudes are to be subtracted, they are to be measured off in the opposite direction to that of the others that is, backwards along the line on which these others have already been plotted off. This graphic method of addition is, nevertheless, often convenient as a step in a more lengthy and complex graphic calculation. Suppose that by graphic means we have obtained lines the lengths of which represent to a Subtrac- certain scale certain magnitudes. These magnitudes taken separately may be of no interest, but their sum may be the final object of the calculation, or may be needed in order to continue the calculation to its completion. It would cause more trouble, use more time, and be less accurate to read off each of these parts to scale and add the scaled lengths numerically than to add them graphically by careful use of the dividers, or otherwise, and to read off to scale only the CHAP. in. GRAPH- ARITHMETIC 19 resulting sum of the lengths. The scale cannot be read to such minuteness and accuracy as the dividers can be set to, and the sum of the errors in reading the different quantities to scale is therefore always probably greater than that of the errors due to inexact setting of the dividers. Moreover, the error in reading to scale is nearly always in the same direc- tion either always a little too much or else always slightly Addition too small, the direction of the error depending on the pecu- traction" liarity of the eyesight of the draughtsman. The error in setting the dividers has not the same invariable character ; it is as often positive as negative, and the chances are that numerous errors of this sort will not accumulate, but will more probably neutralise each other to so great an extent that the sum of a large number of errors will be by no means correspondingly large. 2. Sometimes a quantity can only be found by adding up a very long series of very small parts. The magnitude of each small part in the series may be determined beforehand, but not infrequently it cannot be found until the sum of all the previous parts in the series has been calculated. This kind of addition is called integration. Sometimes, when the law determining the successive values of the small parts is a simple mathematical one, the process of integration is very much simplified by mathematical calculation, as explained in the Integral and Differential Calculus. To attain a moderate Integra- approximation to accuracy, the parts require to be taken very tlon small and correspondingly numerous. Thus to integrate by ordinary numerical addition is an immensely tedious opera- tion. The same process, however, may be carried out much more rapidly, and with much less fatigue, by graphic means. In the later parts of this book, when we deal with somewhat complicated constructions, we shall have many illustrations of this graphic integration. As illustrations of the beneficial employment of graphic integration occur only in these some- what difficult problems, we may pass by the subject for the c 2 20 GRAPHICS CHAP. III. Integra- tion present, promising to return to it when its utility will have become more evident and its interest, therefore, greater. The graphic addition and integration of areas will be dealt with under Trigonometry, Chap. V. 3. Graphic Multiplication. The problem is to find the pro- duct of two or more known quantities. Let a and b be the quantities ; x = a b is to be found. This may be thrown into one of the two forms Multipli- cation Choice of unit and x _ a b~ I' The graphic construction is to draw two similar triangles, in one of which two sides are made 1 and b (or 1 and a), and in the other of which the two similar sides are a and x (or b and x). This will give us a line x, the length of which to the proper scale represents the product a b. The pair of triangles may be formed in the two ways represented in Fig. 5. 4. In the first of the Figs. 5, b is associated with 1 in the one triangle and x with a in the other, 1 and a being marked off along the same straight line or parallels and x and b lying along the other side of the angle or parallel sides. The two lines on which b x a I are marked may stand inclined to each other at any angle. In the figures the heavy lines indicate the data. The light full lines indicate the lines that require to be actually drawn on the paper. The dotted lines (?/i, and m 2 ) do not need to be drawn, except at the extremity of m 2 , where the intersection has to be marked by drawing a short portion of m 2 across the other line. After marking off b a and 1 the edge of a set-square is laid across the end points of b and 1 in the first figure, or of a and 1 in the second figure ; then the set-square is slid on another straight-edge until its edge passes through the end point of a in the first figure, or of b in the second figure ; finally, in this position the intersection CHAP. in. GRAPH-ARITHMETIC 21 of m. 2 and b produced, giving the extremity of x, is marked with pencil. 5. These two diagrams of Fig. 5 of course give the same length for x ; but the first is a better construction than the second. In the first the intersection of m. 2 determining the length of # is a more sharply defined point than in the second figure, because the angle between m 2 and x is greater in the first than in the second. This arises from the fact that b is more nearly equal to 1 than is a. There results the rule for the above construction that 1 should be marked off on the same line with that one of the two quantities a and b which sharp differs most from 1. The difference between the two cases is sections expressed by saying that the triangles in the first are ' better conditioned ' than in the second. If in the second the ratio between a and 1 were considerably greater than it is in the drawing, the triangles would be ' ill-conditioned.' As the angle at which two lines cut each other becomes smaller, their intersection becomes less well defined, and the reading of a length to it becomes liable to a greater possible error. This arises in two ways, illustrated in Figs. 6 and 7. 6. In Fig. 6 the thickness of the intersecting lines is magnified, each line being shown by a double line. The in- tersection of the two has really the length marked e on the diagram. The angle of intersection being 0, and the thick- ness of the line t, it is easily shown that the length of the Possible intersection is 1 + cos 6 e = t sin 6 which becomes very rapidly larger as 6 becomes smaller. The reading of a length to this intersection must be indefinite within this range e. 7. In Fig. 7 is shown the error in the position of the inter- section resulting from drawing one of the lines in a slightly Error incorrect direction. If the incorrectly drawn line is drawn from a point distant H from the other line, it is easy to prove 22 GRAPHICS CHAP. in. that the error e resulting from an angular error s in the direction is equal to "~"Sn*P Error which for the same error e increases still more rapidly as decreases than does the error shown in Fig. 6. 8. In Fig. 5, of course, a and b may represent any quan- tities either of the same kind or of different kinds. For in- stance, they may both be lengths, and then the desired pro- duct is an area. If they are both of the same kind, they can be marked off to the same scale. If they are of different kinds, they must of necessity be represented on different scales. In either case their product cannot be measured to the same scale as either of the factors. What, then, are the relations between the different scales employed in the construction ? This is explained at once by observing that a length to repre- sent unity (1) has been marked off. This unit length has been, of course, measured to a certain scale. In the first diagram of Fig. 5 it is set off on the same line as a. Suppose it has been measured to the same scale as has been used for a, then x must be read to the same scale as b, in order that Scale of its length read to that scale may numerically equal the pro- ttct duct of a and b, the geometrical ratio equation being x __ a b I' But if x is to represent, not only the numerical magnitude of a b, but the real product a b itself to a scale of its own, that scale cannot be the same as that of b. The unit of that scale represents unit quantity of the particular kind resulting from the multiplication of a and b. For instance, if a and b are lengths, say in feet, the unit of the x scale is unit area, say 1 square foot ; or if a is a length in feet, and b a weight in pounds-weight, the scale of x is one of quantities of work, or of force moments, in foot-pounds. This may be more clearly CHAP. in. GKAPII-ARITHMETIC 23 understood, perhaps, by considering the equation in the form xx 1 = a x b. This equation may be expressed in words thus : ' x to the scale of a multiplied by 1 to the scale of b equals a multiplied by b,' or ' x to the scale of b multiplied by 1 to the scale of a equals b multiplied by a.' It is plain that the scale of x depends on that to which 1 has been marked off, and that any scale may be adopted for 1, provided a corresponding in- verse change is made in the scale to which x is read off. For example, in the third diagram of Fig. 5, 1 is marked off to double the previously used scale. The length in inches obtained for x is just half that obtained in the first two dia- grams, but when read to half the scale used for x in these first two diagrams, the same result is obtained as before. To illustrate further : suppose a is a number of pounds weight, say 160 lb., measured to the scale of yj^- in. = 1 lb., and b is Scale of a number of feet, say 11 ft., measured to the scale of yVin. = 1 1ft. Suppose, now, that unity is marked off to 'the same scale as that of b that is, -^ in. is marked off as 1. Then x must be read to numerically the same scale as a that is, to the scale yJ-Q in. = 1 foot-pound. It would be found to be 17'6in. long, and to this scale would mean 1,760 foot-pounds. Suppose, however, that 1 is marked off to double the 6-scale, that is, *2 in. is taken as unity ; then the length obtained for x would be 8*8 in., and this length must be read numerically to half the a-scale, namely, -^ in. = 1 foot-pound ; and to this scale it will mean as before 1,760 foot-pounds. Once more, suppose 1 in. taken as unity (1) that is, ten times as much as represents 1 ft. on the b -scale. Then the length obtained for x will be 1'76 in., and the unit of the x scale must be yV of the length that represents 1 lb. on the a scale that is, yoV in. To this scale x measures as before 1,760 foot- pounds. If 2 in. is taken as unity, the length got for x will 2-1 GRAPHICS CHAP. IIT. be -88 in., which, read to the scale TF Vo in. = 1 foot-pound, means again 1,760 Ib. In the first two constructions of Fig. 5, 1 = 1 in. ; in the third, 1 = 2 in. If y 1 ^ in. had been taken as unity, the lines T/I! and m 2 would have been inclined to b and at a very small angle, and the intersection of ra 2 denning the end of x would have been an ill-conditioned one. It is true that a long length can be read with a smaller percentage of error than a short one. If lengths can be read to -01 in., an error of -005 in. is ten times more serious in a length of 2 in. than it is in one of 20 in. But the error that may arise from a small inac- curacy in the direction of the line m 2 due either to inexact setting of the set-square to the line m lf or else occurring in the sliding of the set square from the position m { to the posi- tion m 2 increases much faster than does the length of x nearly in the ratio of the square of its length. Also it must Scale of be remembered that, if it is possible to read the long x with act greater proportionate exactitude, to obtain the long x a short 1 must be used. As the proportionate error in reading x de- creases, the proportionate error in marking off 1 increases. It is evident, therefore, that such a length should be adopted for 1 as will make the intersection of m 2 with x as well-con- ditioned as possible. This result is obtained by adopting for 1 a convenient length as nearly equal either a or b as possible. But it must not be chosen so as to give an awkward scale by which to read x. Thus, if the scales used are parts of inches, 1 may be chosen 10 in., or 5 in., or 2 in., or 1 in., or in. If millimetre scales are used, 200, 100, 50, 20, or 10 mm. may be used as 1. This rule of arranging the units so as to get well-condi- tioned triangles cannot always be attended to through long complicated graphic constructions involving series of successive multiplication of a variety of quantities of greatly different magnitudes, because in order to follow it it would be necessary to change the unit and the scales from time to time. This CHAP. in. GRAPH-ARITHMETIC 25 would lead to hopeless confusion, and in such circumstances it is frequently necessary to work with ill-conditioned tri- angles. The above considerations are, however, of the great- est possible importance throughout the whole of graphic calculation, and they have, therefore, been presented here very fully. Whenever it leads to no confusion or other in- convenience, the unit should be chosen according to the above explained principle. Whenever it is impracticable to do so, it is well to remember that increased care and exactitude in drawing is necessary whenever intersections at acute angles have to be used. In all cases it is necessary to have a clear conception of the true meanings of the different scales used throughout the diagram, to understand the relations between the scales, and to avoid the confusion of imagining that scales which are essentially different in kind can be in any sense the ' same scale ' that is, for example, that 1 in. = 1 lb., and 1 in. = 1 ft., and 1 in. = 1 foot-pound, and 1 in. = 1 square scale of foot area, are in any sense the same scales, or that they are pro uct equal in any way, except that they are to be read numerically in the same manner. While the difference of the scale of x from those of a and of b should be remembered, its relation to these should be clearly comprehended, and the manner in which it is to be deduced from these and from the value taken as 1 should never be lost sight of. The construction of Fig. 5 can be modified in a great variety of ways according to convenience in special circum- stances. The special circumstances result chiefly from the different relative positions on the drawing paper that are found to be occupied by the factors a and b in the course of an extensive graphic calculation. The factors generally result from previous portions of the calculations as lines in certain parts of the drawing. It is not desired to draw them over again in order to perform the multiplication. They are to be used in whatever positions they may happen to have been placed in already. They may be near or distant, parallel, 26 GKAPHICS CHAP. 111. Scale of product Multipli- cation methods perpendicular, or oblique to each other. They may both radiate from one point ; the extremity of one may lie in some intermediate point of the other; or they may cross each other. 9. The diagrams given in Fig. 8 show the most usually useful variations of the construction. The geometrical proof of each is so simple that it need not be given here. The same dimensions for a and b as are used in Fig. 5 are used through- out, so that the different constructions may be more readily comparable. The circumstances in which each should be used are explained in the attached notes. Throughout, thick lines have been used to indicate the data ; thin full lines, the con- struction lines, that require to be actually drawn ; and dotted lines, those that do not need to be pencilled or inked, but only obtained by laying the set square or other edge along them. The construction lines are in each case lettered m lt m v m 3 , &c., in the order in which they are drawn. Of course, in the actual constructions none of the letters attached to the dia- grams of Fig. 8 are needed, but the numerical value of x as read to scale should be written on the line. DIVISION. 10. In division we have the converse problem to that of multiplication, and very slight alterations of the constructions given for multiplication serve for division. In multiplication we had x = a b, or ~ = ?. In division we have x = ^, or b 1 b If, therefore, in each construction given for multipli- Division cation we make b and 1 change places, the construction is converted to that for the division of a by b. This subject, therefore, requires no further explanation. Very frequently there occurs a problem in combined mul- tiplication and division, say, to find the quantity x = . CHAP. in. GRAPH-ARITHMETIC 27 This could, of course, be found by first finding a line to repre- sent a I) by a multiplication construction, and then operating on this line with c by a division construction. But it may be found more directly by a single construction, for it may be written X -= -, so that if in the multiplication construction c is b c marked off in place of 1, the line marked x will be the quantity ( - - J desired. The diagram as thus modified is shown in Fig. 9, in which a special example is worked out. Example, Fig. 9. An engine piston has an area of 343 square inches. The steam pressure on it is 85 Ib. per square inch. The whole force is transmitted along the piston-rod, which has an area of 5*23 square inches. What is the stress on the piston-rod section ? It is in pounds per square inch - , and this is calculated graphically, in the diagram Division 5*Zd and Mul (Fig. 9), to be equal to 5,580. The number is easily found, by ordinary arithmetic, to be 5,574-5 Ib. per square inch, but the degree of accuracy of the data in a problem of this kind does not justify the answer being stated to a pound per square inch. Similarly, continued multiplication of several quantities can be readily completed in one diagram. Fig. 10 shows an example of the calculation of the weight of water filling a rect- angular tank, whose sides are 4'75 ft. and 6*25 ft., and whose depth is 5*15 ft. The weight of water per cubic foot is taken as 62-4 Ib., and the product to be found is 62-4 x 6-25 x 5*15 x 4*75 = x. Here we have set off 62'4 to the scale T V in. = 1 Ib. per cubic foot, and in this first multiplication, instead of unity to this scale, viz. y 1 ^ in., we have used 100 to this scale, or 10 in. We have set off the sides to the scale 1 in. = 1 ft., and in the second multiplication used 10 to this scale 10 in. instead of unity ; and in the third multiplica- tion have used simply unity 1 in. The scale to which x is 28 GRAPHICS CHAP. III. to be read is, therefore, in. = 1 Ib. The Continued Multipli- cation Sectional tablet Multipli- cation Division 100 x 10 1,000 result is obtained correct to four figures. It is read to scale 9540, whereas the exact product is 9540-375. The object of changing the unit from 10 in. to 1 in. is to avoid ill-conditioned intersections. Still better intersections would have resulted from taking 5 in. as the unit for all three multiplications, but this would have involved reading x to the awkward scale of = _ in. = 1 Ib. ; x would be read first in inches, 50 x 5 x 5 1,250 then multiplied by 10,000, and finally divided by 8. The combined operation of first calculating what the scale is, and then reading x to that scale, would be altogether too cumbrous. 11. By the help of a piece of finely and accurately divided sectional paper and a straight-edge, multiplication and division can be performed very rapidly and with fair accuracy. The paper should be divided in millimetres or in twentieths of inches, and should be 200 of these small divisions square. A piece of fine cardboard or a flat plate of ground glass or por- celain ruled in this way is better than paper, which shrinks and expands somewhat irregularly as the air becomes dry or damp. The straight-edge may be an edge of a set square, or better, a very fine phosphor-bronze wire held stretched. The readiest constructions for use with this 'sectional tablet ' are those shown in Figs. 11 and 12. 12. In Fig. 11, for multiplication, at the horizontal distance 1 from zero on the scale of the tablet and at the vertical height a place the straight-edge, letting it also pass through zero. Then at horizontal distance b the vertical height of the line measures the product a b x. In Fig. 12, for division, place the straight-edge from zero to the point at horizontal distance b (the divisor) and vertical height a (the dividend). Then the quotient, ~ = x, is the height CHAP. in. GRAPH- ARITHMETIC 29 of this line at the horizontal distance 1. Here, however, it conduces to accuracy to read off 10- at the hor zontal dis- tance 10. 13. Similar graphic processes of continued multiplication may be applied to finding the positive and negative integral powers of numerical values, such as a 3 , a 6 , - , &c. The a' a 4 neatest construction for this purpose is shown in Fig. 13, in which two examples are given of series of positive and negative powers of, first, a value greater than unity ; and, second, a value less than unity. Evidently, for numbers much larger than unity, the construction is of no use for high powers, because these increase so rapidly that the lines representing Integral them soon become so long as to extend outside any convenient * limits of the paper if the unit chosen be not extremely small. Between two cross lines at right angles a series of lines is drawn, starting from a point on one of the cross lines distant unity from their intersection, and each diagonal line of the series being perpendicular to those immediately behind and in front of it. The length a or 6, whose powers are wanted, is measured from the centre perpendicular to 1, and the first line of the series is drawn to this point. The construction depends on the similarity of all the triangles in the four angles of the cross. This gives all the integral positive and negative powers. 14. The inverse problem of finding fractional powers, such as --, ^, -f , f , &c., has never been solved by any direct graphic process, although an easy and well-known construction gives the square root. In Fig. 14 is illustrated a graphic or me- Fractional chanical method by which this problem may be very easily P ^ solved with great accuracy and rapidity. Suppose we wish to extract the cube root of a quantity a. From the construction of Fig. 13 it is evident that if we lay off a on the line D in Fig. 14, and from its extremity manage 30 GRAPHICS CHAP. III. to draw between D C, C B, and B A three lines at right angles to each other, and so that the last will cut A at unit distance from the centre of the cross, then the distance of the intersec- tion on B of the oblique lines will be equal to V^ and the similar distance on C will be a*. This could be done on ordi- nary drawing paper only by a tedious process of trial and error. But with the help of some accurately and finely divided square sectional transparent tracing paper, the trials may be completed in a small fraction of a minute. Any line of the sectional tracing paper is laid on the extremity of &, as marked off on the line D. This line is followed by the eye to its intersection with C. This intersection does not lie exactly on any tracing-paper line passing from C to B, but an inter- polated intermediate line is easily followed by the eye without actually drawing it in. This is done along to its intersection with B, and from this a similar line followed from B to A. Fractional This cuts A either further or nearer the centre than unity. If the intersection is further than unity from the centre, the sectional paper is turned round right-handedly a small dis- tance, and the line from the end of a again followed through the three angles of the cross. If it cut A still too far out, the tracing paper is again shifted in the same direction, and the process repeated. By means of a very few of such repeated shiftings the paper is got in the right position, and then the ^rd, rds, rds, f rds, &c., powers of a can at once be read off. To facilitate this reading off, the cross lines A B C D are finely graduated, the marks being made clearly visible through the tracing paper. The setting is facilitated by fixing a line of the tracing paper at the extremity of a by a needle-point pricker. In Fig. 14 the tracing paper is set so as to read the fifth root and all integral powers of the fifth root of 0*7. It reads V 0-7 = '93. The accurate value, as found by loga- rithms, is -931. The method is thus capable of great accuracy, and is, in fact, of much practical utility. To deal with large numbers in this way is not inconvenient in the same way as CHAP. in. GRAPH- ARITHMETIC 31 in the graphic construction for integral powers. Thus if, for example, it is desired to find V7,000, we first divide 7,000 by 10 5 , and deal with the quotient, viz. *07, instead of with 7,000. Graphically, we find V -07 = "585, and this gives 10 x -585 = 5-85= V7,000. The correct number is 5*875. Fractional In Chaps. VI. and VIII., in treating of Linear Velocities l due to Eotations and of Force Moments, graphic processes for the multiplication of a series of factors, taken in pairs, and the simultaneous addition of the products, will be explained. 32 GKAPHICS CHAP. iv. CHAPTEK IV. GRAPH-ALGEBRA. 1. IN ordinary algebra the relation between two quantities varying continuously together is expressed by a letter-symbol Five equation. The same may be expressed graphically by a line, methods straight or curved, in one or other of several ways. Generally, one of the following five methods will be convenient. 2. (1st.) The two quantities whose continuous variation together is to be expressed are represented by the lengths of the two ordinates to a curve. The scales to which they are so represented are necessarily different if the two quantities are of different kinds ; if they be of the same kind, they are conveniently plotted to the same scale. The ordinates are in most cases conveniently taken rectangular, but are not so necessarily. Positive ordinates are measured upwards and to the right hand of the axes ; negative ordinates downwards and to the left hand from same axes. For this first graphic Linear mode of algebraic representation, ' square sectional paper ' is nates* 1 " ^ ver y 8 rea * va l ue > especially if it be accurately ruled. The division of the paper should be decimal. Centimetre or half- inch squares each divided decimally are convenient, but inch squares divided decimally are also useful. Every tenth line being heavy, every fifth line should be of medium thickness. Light blue is the best colour for the sectional lines. If the ' slope,' or tangent of inclination, be readily found by differen- tial calculus, it is of very great assistance in drawing these curves to calculate and make use of this slope. 3. (2nd.) The two are represented by the length of a line CHAP. iv. GRAPH-ALGEBRA 33 and the direction of that line, or the angle defining that direc- tion. The line is usually a radius vector from a fixed pole ; the angle is measured from a fixed radius vector from the same pole. A right-handed rotation from the fixed zero radius should represent a positive angle; a negative angle should be plotted off left-handedly from this zero direction. The line along which the length is to be plotted being thus determined, outwards from the pole along the side of the angle is to be reckoned positive ; the opposite direction, or backwards from the pole along the side of the supplementary angle, should be counted negative. Fig. 15 explains these conventions regarding signs. The datum or zero radius is P 0. The angle 6 A is positive, and with this angle positive lengths are measured from P towards 4- A and negative lengths towards A. The angle 6 B is negative, and with this angle lengths from P towards + B are positive; those towards B being negative. Evidently Polar co- a positive length with the angle 7r-f or 180 + # will lie in the same position as a negative length with the angle 0. Again, no distinction can be made between the angles 0, 6 -i 2 TT, 64 4-7T, &c. These ambiguities make this mode of represen- tation inconvenient except when one of the quantities, viz. that represented by the angle, is of a recurrent character ; in fact, it is not often suitable except when the one variable is actually an angle. In this latter case, of course, it is eminently ad- vantageous in some respects, but not always even then. Thus, for instance, when the driving moment on an engine crank is to be co-ordinated with the angular position of the crank, this method will give a polar diagram than which nothing could be more intelligible to the eye ; but if the work done during any given period of rotation has to be calculated from this diagram, the first method explained above, or the ' square diagram,' is preferable, because in it the area underneath the curve will represent the work. Similar remarks will apply to the representation of the relation between the electro-motive D 34 GRAPHICS CHAP. iv. force of a dynamo and the angular position of the armature. Circular sectional paper is convenient for this kind of dia- gram, but is not made as a regular article of sale. The Polar co- drawing of the tangent of a polar curve at each point ordinates pi ^ e( j j[ g O f even greater help in drawing the curve fairly than in the case of curves with rectangular co-ordinates. The ' slope ' from the radius, or the tangent of the angle of inclination to the radius, is f-4 , and can often be found simply by the ordinary rules of the differential calculus. 4. (3rd.) The two variables may be made the two angles Angniar formed by two lines drawn from a point in the curve to two nates " fixed foci with the line joining the foci. Radial co- 5. (4th.) The two variables may be made the distances of a point in the curve from two fixed foci. 6. (5th.) The two variables may be two distances from a Focus and point in the curve ; the one distance being from a fixed focus, the other being the perpendicular distance from a fixed straight line called the ' directrix.' The 3rd and 4th modes of graphic representation are not of general applicability, but are suitable for special problems. THE STRAIGHT LINE. 7. Rectangular Equation. Using the first diagram with rectangular ordinates, the straight line is the graphic repre- sentation of the linear equation y + Dx The simplest method of plotting the line from the given data D and F is to plot y vertically at any two of the three values of the horizontal ordinate x, 1,0 and -f 1. For these . points y takes the values (D F), - F and (- D - F). To obtain accuracy the two points plotted should be as far as possible apart, and, therefore, in general x = 10 and x = + 10, giving y = 10 D F and y = 10 D F, are the best points to use. If the diagram stretch so far it is still better CHAP. iv. GEAPH- ALGEBRA 35 to use x = 100 and x = + 100. Of course, careful atten- tion must be paid to the signs of D and of F, either of which may be + or . If F be , the line crosses the vertical axis above the origin, and vice versa. If D be , the line has a right-hand upward slope ; if D be + , it has a right-hand downward slope. If the equation be given originally in the DX + E/ * T 1 form D# + E?/ + F = 0, the easiest pair of points to plot out and ~ .,, F y = with x =-. Fig. 16 is given to illustrate the most simple graphic means of obtaining these points. Very careful attention must be paid in this construction to plotting F, E, and D in the proper directions according to their signs + or . 8. Polar Equation. C r = sin (a + ey where C and a are the two constants and r and the two variables, is represented by a straight line. For in Fig. 17 where P is the pole and P the datum direction, and o is the distance at which the line K crosses P at the inclination sm (), we have r _ sin a o sin ( + 6} ' The constant C, therefore, corresponds to o sin . To plot the line representing = sin ( + 6) from the data C and , we have, therefore, to measure from the pole along P 0, the datum direction, the length -(when 6 = o, r = ), a \ sm / 36 GEAPHICS CHAP. IV. and from the point so obtained draw a line inclined ( a) to P 0. If an arc with radius C be first drawn from centre P, and tangent to this arc a line be drawn making the negative or left-hand inclination a with P 0, then this line is the desired line K, the radius vector to which from P at the angle 6 is r. The equation may also appear in the form cos a 1 _ C COS (a 1 - 0)~COS (*>-&)' c Here a 1 is the complement of a in the above. From P cos (a 1 - 0) pi ^ O ff (j a i on g a ii ne> making the right-handed angle a 1 with P 0, and from its extremity draw a second line perpendicular to this first. The radius from P to this second line at the + angle is r. 9. Focal Angular Equation. Let and be the two variable angles, and c the distance between the foci. Then the equation a cot 6 + b cot cf> = c is represented by a straight line (see Fig. 18). To obtain this line plot the foci A and B at a distance c apart. From A and B, and perpendicular to A B, plot b and a. The line is to be drawn through the two points so obtained. This is evident because when 6 = 90, cot 6 = 0, and, therefore, cot = ^; and again, when = 90, cot < = 0, and, therefore, cot 6 = - . a a cot e + To prove that the intersections of lines drawn from A and B at the angles 6 and , connected by the above equation, all lie on a straight line, draw parallels to A B through the tops of a and b. From A draw any line at any angle 6 meeting the first parallel at a distance z from the line b. From this intersection draw a line perpendicular to A B to meet the second parallel. The line from B inclined at to A B must pass through this last intersection, because z = a cot 0, and, therefore, we must have b cot c z. Let P be the inter- CHAP. iv. GRAPH-ALGEBRA 37 section of the two lines at and < from A and B. Let x and y be the rectangular co-ordinates of P referred to b and c as axes. We find x- z y Jj 16 - and , ^y Eliminating z from these equations, we obtain a b ,7 y = _ ^x + b, c which is the equation of a straight line passing the axis of y a cot + i b cot at a height b above the origin, and lying at a slope c from the axis of x. If 1 and (j> 1 be the complements of 6 and of the above figure, then the same equation may also take any of the three other forms : a tan 6 l + b tan } = c ; or, a tan O l + b cot = c ; or, a cot 6 + b tan 1 = c. CONICS. 10. The general equation of the second degree between two variables x and y, A# 2 + Bxy + Ci/ 2 -f D# + Ey + F = 0, can be represented graphically by one or other of the conies namely, the ellipse, parabola, or hyperbola, x and y being taken Conies as the co-ordinates, most conveniently at right angles to each other. The conic is an ellipse if 4 A C > B 2 ; a parabola if 4 A C = B 2 ; and an hyperbola if 4 A C < B 2 . Choosing any two axes for the curve to represent the above equation, the centre of the conic is to be plotted off with the co-ordinates 38 GRAPHICS ^ B 2 -4AC' CHAP. iv. = -BD = B 2 -4AC ' If B 2 = 4 A C, the divisor in these fractions becomes zero and the co-ordinates of the centre become infinite ; that is, there is no actual centre, the curve being a parabola. If the divisor be positive, the curve is an hyperbola ; if negative, an ellipse. The centre being thus obtained, it is more convenient to proceed with the construction from a new set of rectangular axes through the centre as origin and inclined to the previous axes by the angle 2 tan ~ l -^, that is, by an angle 6 given tan 2 = B A-C' If we then take f= AE 2 -BDE + CD 2 B 2 -4AC Conies and use co-ordinates (called, say, x l and y l ) from these new axes through the centre as origin, the curve will be expressed by the equation This will be an ellipse if both A and C be positive, and an hyperbola if either A or C be negative ; it cannot be a parabola, because in that case/ becomes infinite. The semi-major and semi-minor axes of the ellipse are A / L and A/ L. v A ^ C To construct the ellipse draw round the centre two circles of radii, A/ /and */_, as in Fig. 19. Draw any radius ^ A ^ C to cut the two circles, and from its intersections with these two circles draw parallels to the axes. These parallels will meet in a point in the curve. For accurate drawing the points so constructed must be closer together near the major CHAP. iv. GEAPII-ALGEBRA 39 axis than near the minor axis because of the sharper curva- ture at these places. The ratio of the two circle-radii being A / J- x - = \/ - the correspondence of the construe- Ellipse V A / V A tion with the equation is evident if the equation be written Fig. 20 is introduced here only to show clearly the relation between the ellipse and the hyperbola, and not as a desirable construction for the latter, y' is increased in the ratio ^ by means of a pair of circles similar to those used for the ellipse, and this increased length is taken as the height of a right-angled triangle whose horizontal side is the constant Z_. The hypotenuse of this triangle is taken as the A horizontal x' for the curve corresponding to y'. The construc- tion is inapplicable to any value of y' greater than which would be the maximum value of y' in the ellipse if G had been positive. Fig. 21 shows an easy construction applicable to the whole range of the hyperbola. From the centre along axis of y' are plotted */ f. and \/A >' v A and horizontally from same axis is plotted \/ are the two variable angles and a, b, c, a and ft are five constants, may be represented by a conic curve. It determines only the shape, not the size of the conic, because six data are required for the complete determination of a conic. Thus our curve may be drawn to any scale, or, in other words, it may be constructed on any base. Take any base AB (see Fig. 24), and use A and B as foci, measuring 6 at A from A B to the radius from A to a point in the curve, and < at B from B A to the radius from B to the same point acot(e-a) of the curve. At A and B draw perpendiculars to AB, and "i_m* plot off from them outwards, i.e. away from AB, the angles a and ft. At distances a and b from A and B draw lines A' A' and B' B' perpendicular to these last ; these will be inclined and to AB. Along these lines A' A' and B' B' from the feet of the per- pendiculars a and b measure off c. Call the points so obtained 6 on A' A' and on B' B', and divide the two lengths c c similarly and reversely. In the figure each is divided into six equal parts, numbered from to 6 on A' A' away from the foot of a, and from to 6 on B' B' toicards the foot of b. Thus 2 on B' B' equals 2 on A' A' ; and 2 on A' A' plus 26 on B' B 7 equals the constant c. Also, for the point P on the curve corresponding to the points 22 on the lines A' A' CHAP. IV. GHAPII-ALGEBEA 43 and B' B', the angle 2 A B being and A' 2 A being, therefore, (0 - ), evidently 2 on A' A' equals a cot (0 - ). Similarly, the angle PB A being and B 26 being (< /3), we have 2 6 on B' B' equal to & cot (< ). Thus the construction gives c = a- cot (0 ) + & cot (_0) have been obtained. Evidently the curve passes through A and B. As the points taken on A' A' and B'B' recede indefinitely to the right hand, the lines drawn through A and B become in the limit parallel to A' A' and B 7 B'. The cor- responding point on the curve is marked oo. The lines joining it with A and B make the angles and ft with A B ; thus the conic is one circumscribing the triangle A B oo. If a or /3 or both are negative, attention must be paid to this change of sign in plotting them off from the perpendiculars to A B ; they should be plotted towards A B if they be negative. If both are negative, the curve is always an ellipse. CONIC FOCAL KADIAL EQUATIONS. 14. The equation r A + r B = c, as is well known, can be represented by an ellipse, r A and r u being the variable distances of a point in the curve from the ( r A * r u foci A and B. The distance between the foci may be taken at |r=mp pleasure, so that there is an infinite number of ellipses which will represent the equation with a given value of c. Simi- larly, an hyperbola represents FA - r B = c. If r and p be two variables, the equation r = mp 44 GRAPHICS I - .1 I? -*B mp General equations m, being a constant ratio, can be represented by a conic if r be taken the distance of a point in the curve from the focus and p its perpendicular distance from the directrix. If m < 1, the curve is an ellipse. If m = 1, it is a parabola ; if m > 1, it is an hyperbola. Of course, the same equation between the two variables r and p may be graphically expressed in a simpler manner by a straight line through the origin, taking r and p as rectan- gular or oblique co-ordinates. 15. Any equation connecting two variables, no matter how complicated it be, whether it involve high or fractional powers, trigonometrical or exponential functions, is capable of graphic representation in one or other of the five methods mentioned at the beginning of this chapter. As one method is usually much preferable to others as regards ease of construction and applicability to the special problem in hand, the draughtsman will always be repaid for the bestowal of very careful considera- tion upon the choice of method. The choice should, of course, depend on the use to be made of the curve. A familiar know- ledge of miscellaneous curves as explained in books on the more advanced modern geometry will be found of great assist- ance. Eagle's ' Constructive Geometry ' is a useful book of this class. Many curves of special utility in engineering are dealt with in various subsequent chapters of this book. SOLUTION OF EQUATIONS. LINEAR EQUATIONS. 16. The solution of the simple equation Vx -f F = TGI means no more than the finding of the quotient ^ = x. Simple This can be done by any of the graphic methods of division equations a i rea( jy explained in the chapter on Graph -Arithmetic. It is only in certain circumstances that any advantage is obtained CHAP. iv. GRAPH-ALGEBKA 45 by the adoption of the graphic method. But it is instructive to view the graphic solution in another way. Choose rect- angular axes and draw the straight line D x + F = y. Simple equations The point where this line cuts the axis of x, i.e. where it makes y = o, gives the value of x desired. 17. This immediately leads us to the solution of two simultaneous equations involving two unknown quantities. F 2 = 0. Choose rectangular axes ; draw the two lines representing Simple the ahove two equations. The co-ordinates of the inter sec- talneous tion of the two lines give the values of x and y, which are the e< l Tiat;ion8 roots of the equations. 18. This method, with the help of a piece of finely and accurately divided sectional paper, is not only rapid (when one has practised it sufficiently to have become familiar with the method of constructive procedure), but is also accurate enough for most practical purposes. For instance, with a piece of paper 200 mm. square that is, with its square edges each 100 mm. distant from the centre, which is taken as origin the roots of the equations can be obtained with an accuracy of 3 places of figures. A piece of paper 10 inches square divided in half-inches and twentieths will do equally Sectional well, being slightly larger in scale and therefore to most per- ta et sons more readable. Ordinary sectional paper, however, is not accurate enough in its division, and deforms somewhat irregularly with variations in the hygrometric condition of the air. It is better, therefore, to prepare a piece specially for the solution of equations. Fine hand-made cardboard may be used for this purpose. Finely ground glass, porcelain, or any of the white materials used for memorandum tablets, would be still more suitable, not only because they are more permanent in shape under change of temperature and damp- 46 GRAPHICS CHAP. IV. Sectional tablet ness of the air, but chiefly because fine pencil lines may be drawn on them and afterwards rubbed out. If paper or cardboard, which it is desired to preserve permanently for this purpose, be used, pencil lines should not be drawn on it, but the lines should be drawn by stretching across the card fine wires ; phosphor-bronze wire being very suitable. Since two wires have to be handled, one must be held stretched by weights after being placed in position. It will facilitate the manipulation if the wire be fastened to the two weights either permanently by soldering, or by being twisted round pegs screwed in the weights for this purpose. If a few simple curves (see Fig. 25) be drawn permanently on a sectional tablet of this sort, quadratic and other more difficult equations may be rapidly solved by its help. QUADRATICS. 19. The equation Quadra- tics j)2 has the roots x == - D 2 Of course each item of this result might be calculated graphic- ally, and the items combined by graphic construction so as to find the two roots. This process is, however, too clumsy and tedious to offer any practical advantages. 1 Since, however, the equation may be written - - A A' a rapid graphic solution is possible. Choose axes and draw a curve of squares i.e. the curve y=x*. Note that this curve, being once drawn carefully in a permanent manner on the 1 In Eagle's Constructive Geometry a solution on this principle is given which looks simple ; but it starts with the equation in the reduced form Really fully half the work of the solution of any quadratic consists in finding the quantities a and 6 to reduce it to this form. CHAP. IV. GRAPH- ALGEBRA 47 sectional tablet, is sufficient for the solution of all quadratic equations. Now draw the line Vx + Ky + F = 0, which is the same as D F "--i*-'r Evidently the two intersections of the line with the curve of squares give the two solutions of the quadratic. The line may be ' formed ' without drawing it by simply laying a set- square edge or by stretching a fine wire along its true posi- tion. If the line does not cut the curve the two roots are ' impossible ' or ' imaginary,' and cannot, therefore, be found, or represented in any way, by graphic means. Since the curve of squares rises very rapidly, it is more Quadra- 2 tics convenient to use the curve y = -, as in Fig. 25. The ver- tical scale of the diagram being now -jV the horizontal scale, the straight line that must be drawn is that is, the line has to be drawn through the two points p and y = with x = =-. In the example given in Fig. 25 the two roots are x l = 4- 3-56 and x 2 = - 8'20. 20. The solution of simultaneous quadratics involving two unknowns consists in drawing the two conies representing the simulta- two equations. Their two intersections give the two root pairs of values of x and y. If the conies do not intersect, the tics roots are imaginary, and the equations are incapable of graphic solution. 48 GRAPHICS CHAP. iv. 21. The cubic equation Gx 3 + Vx + F = is of not infrequent occurrence. It can be solved by drawing the curve and drawing the straight line C^ij, Their intersections give the real roots of the equation, there being either three or only one intersection. The curve once drawn will serve for the solution of all such equations. The vertical scale is taken y^- the horizontal scale in order to get the cubes from 10 to + 10 into the sectional tablet. 22. Similarly, if f(x) be any function of x, and we have to determine the value of x from the equation A/(*) f Dx + F = 0, the graphic solution consists in drawing the curve **/<*) and the straight line Dx + ky -f F = 0. The intersection or intersections of the line and curve give the real roots of the equation. General For instance, an important equation insoluble otherwise except by approximation is A log x + D x + F = 0. Draw the curve y = log T, and the above-mentioned straight line ; their intersection gives the desired value of x. Similarly, the equations A sin + D 6 + F = 0, A tan 3 + D0 + F = 0, Aa x + D# + F = 0, &c., may be dealt with. 23. Generally, if the equation be f(x) = 0, it is solved CHAP. iv. GEATII- ALGEBRA 49 graphically by drawing the curve f(x) = y down to its inter- section with the axis of x. Only that portion need be drawn accurately near the intersection. The proper mode of pro- cedure will be understood from what follows regarding simul- f (x) taneous equations for two unknown quantities. If the tangent to the curve be easily constructed, it should be used in con- junction with calculation of the points. 24. The function may, however, sometimes be considered as the sum of two functions which are easier to deal with graphically separately than combined. The last equations mentioned above are examples, one part of the whole function being very simply represented by a straight line. Or, again, for instance, the equation A V r 2 - x* + D x + F = 0, in which r is a constant, is most rapidly dealt with by drawing Special .. . , equation the circle y = yV x 1 and the straight line Ky + Vx + F = 0, the intersections giving the two roots. 25. In such cases, the equation taking the form F(x) +f(x) =0, General the two curves Y = F (x) and y= -/ are drawn, and their intersections give the roots of the equa- tion. 26. Similarly, if we have for the determination of two un- General known quantities x and y, the two equations F (x, y) = and f(*,y) = 0, the two curves represented by these two equations have to be 50 GEAPHICS CHAP. iv. drawn, and the values of x and y satisfying the equations measured at their intersections. 27. These various processes must, however, be carried out systematically in order to economise labour. For distinction's sake let the vertical ordinates of F (xy) = be called Y, and those of/ (xy) = be called y. First, note that it is waste of time to complete the calcu- lation of several points or to draw a considerable stretch of one curve before proceeding with the other. Pairs of points for the two curves should be calculated and plotted for each value of x taken before proceeding to another value of x. Secondly, note that, although the calculation and drawing of the tangents at the points found assist greatly the good plotting of the curves, there is practically little use of doing this for one of the curves unless it can be easily done also for. the other. Method of Thirdly, the plotting of one point and the tangent at that point is nearly, but not quite, so useful as the plotting of two points without their tangents. Whether tangents should be used or not, therefore, depends upon whether the nature of the curve allows the drawing of a tangent with considerably greater ease and less expenditure of time than would be occupied in the plotting of an additional point. Fourthly, after plotting a pair of points in the two curves corresponding to a certain x, the next x should be chosen as near the intersection as can be estimated. Thus the process should be carried out in the following manner : Choose any value of x, say x r Find the corresponding Y lf and y l9 and plot the two points. If the tangents are to be used, draw them and produce them to their intersection. Choose x 2 in the neighbourhood of this intersection, but rounding off to an easy value to deal with if the finding of Y 2 and t/ 2 involve any complicated arithmetic. If these latter are to be found entirely by graphic construction, then take x 2 exactly at the CHAP. iv. GRAPH-ALGEBRA 51 intersection of the first two tangents. Plot now Y 2 and ?/ 2 , and draw the two new tangents. Next, through these last two points draw arcs of circles to touch the two pairs of tangents at the two pairs of points 1 and 2. (This can be very quickly done by anyone well practised in circle drawing.) Take x 3 exactly at or close to the intersection of these arcs, and repeat the last process, finding two points 3 and their tangents, and drawing circular arcs through them to touch the tangents at 3 and 2. The fourth chosen # 4 at the intersection of these arcs will in all probability be found to give Y 4 and y 4 so nearly equal, that a practically exact solution will be obtained by drawing arcs as before through these fourth points. But if not, the process must be repeated once more. In most cases, indeed, it will be found unnecessary to go beyond the third pair of points. " If, on the other hand, the tangents are not to be used, Method of after finding the first pair of points choose any horizontal pro ordinate, x 2 . Find Y 2 and y 2 and plot the points. Draw the chords between the pairs of points 1 and 2, producing them to their intersection. Take x 3 near or exactly at this intersection, and proceeding as before find and plot the pairs of points 3. Now draw arcs of circles through the two sets of three points, 1, 2, 3. Take x 4 at the intersection of these arcs, and plot Y 4 and y 4 . Draw the arcs passing through the two sets of three points, 2, 3, 4. Compare the radii of curvature of these latter arcs with the former arcs passing through 1, 2, 3, and estimate a change of radius of curvature from the averages through 2, 3, 4, up to those at the points 4. In making this estimate consider the average radius through 2, 3, 4, to be the actual radius mid- way between 2 and 4, and the average through 1, 2, 3, to be the acual radius midway between 1 and 3. Using these estimated radii at the points 4, draw arcs through these points and tangen- tial at these points to the last drawn arcs through 2, 3, 4. The intersection thus found will probably be a very close approxi- E 2 52 GRAPHICS CHAP. iv. mation to the solution sought for ; but if it be not found to be so, the same process must be repeated to a fifth or possibly a sixth pair of points. In carrying out these processes it must be remembered that each new estimate of x may just as probably as not overstep the mark aimed at that is, lie on the opposite side of the intersection of the curves to that on which lay the last Method of taken x. Thus the successive #'s taken may be alternately procedure g rea ter and less or vice versa. Similar general methods of procedure should be followed when the curves chosen to represent the equations are polar or focal. In these cases the drawing of tangents through the points plotted is even more useful than when rectangular co-ordinates are used provided they can be obtained without much labour. 28. Sometimes special graphic methods of trial and error are much more direct and rapid than the above general method. A good example is furnished by the equation A sin (20 + ) = B sin (0 + ). Draw two concentric circles of radii A and B (see Fig. 26). Draw any datum radius P 0. From P set off negatively, i.e. to left hand, the angles a = P a and = P b. If either a or y8 be negative it must be set off to right hand of P 0, i.e. in positive direction. Then take any trial point 1 on circle b. With this as centre, and with radius 1 strike the arc 1', cutting circle a in I'. Then if angle P 1 were 0, angle P I/, would be 2 ; a P 1' would be (2 + ) and b P 1 would be (6 -t- /3). Also the perpendicular distance of V from a P would be A sin (2 6 + ), and the perpendicular distance of 1 from b P would be B sin (6 + /3). By trial with the dividers we find the first of these distances much greater than the second. Take a second trial point 2 and find the corresponding point 2' by striking an arc from 2 with radius 2 0. We now find distance of 2' from a P less than distance of 2 from b P . CHAP. iv. GEAPH-ALGEBKA 53 Assume a third point, 3, and find the corresponding point 3'. We find 3' more distant from a P than 3 is from b P. The fourth point, 4, chosen makes these distances equal, and, there- fore, P 4 is the angle 6 satisfying the equation. In a very short time this angle 6 can be found with a degree of accuracy such that the possible error is less than can be measured on the paper, the error, therefore, being in inverse proportion to the scale of the diagram drawn. 54 GRAPHICS CHAP. v. CHAPTEK V. GRAPHO-TRIGONOMETRY. 1. WE will deal here only with plane trigonometry. We have to make calculations regarding plane figures bounded by straight lines. In doing so we must frequently plot off angles. The instruments called ' protractors ' are nearly all of them very rough devices at the best, and are far too untrustworthy for accurate work. The vernier protractor made by Stanley Protrac- with a silvered divided circle and two opposite arms is a re- liable and accurate instrument, but it is costly. The card- board protractor of 12 in. diameter made by the same maker is also useful, although not so reliable as the other. But as any angle can be set off very easily with ordinary instruments with any desired degree of accuracy, the use of protractors is best wholly avoided. 2. The method is the following, and requires the use of a table of chords, such as one finds in Chambers's or in most other mathematical tables. The table found in Molesworth is not sufficient for the plotting of angles taken in surveys, because it gives the chords for every degree only ; whereas the angle is read in the field to minutes. First draw from the Plotting centre from which the angle is to be plotted a circle of, say, 10 in. radius. On this circle the chord of the desired angle is evidently 10 in. multiplied by the tabulated chord for the given angle. This quantity in inches, taken in the dividers, is set off as the chord of the desired angle. For example, suppose the angle to be plotted is 69 22'. We find 10 chord 69 22' = 11-38, and set this off as the chord on a circle CHAP. v. GRAPHO-TKIGONOMETRY 55 of 10 in. radius. Now, from the table we find that 11-39 = 10 chord 69 27' and 11-37 = 10 chord 69 17'. Now, it is possible with ordinary care to set the compasses to y-J-g-in., but much greater accuracy than this is not easily possible. Thus we cannot pretend to set off very accurately as a chord anything between 11-37 and 11-38 or between 11-38 and 11-39. These, as we have seen, correspond on a circle of 10 in. radius to angles differing by 5'. With this size of circle, then, we cannot pretend to plot angles to any greater accuracy than 5'. With very small angles, indeed, the accuracy is increased to 3^, but with angles larger than the above it is considerably Plotting reduced. Thus it is advisable never to plot off by this method ang e angles greater than 45. The complement of the angle for instance, 90 - 69 22' = 20 38' should be plotted off in- stead. This is also the degree of accuracy obtainable with a circular protractor without vernier, with a divided circle of 20 in. diameter. If a circle of 20 in. radius is used, an accu- racy of 2J' is obtainable in plotting. To make the construc- tion on this large scale requires beam-compasses, and, of course, to maintain this accuracy throughout the diagram it requires to be drawn to a correspondingly large scale. 3. The ' solution ' of any triangle or other plane rectilinear figure is accomplished graphically by plotting it off accurately to scale, and measuring the quantities desired. If it is a Triangles length that is to be found, it is measured in the ordinary way ; an angle is to be measured by a reversal of the above ex- * inear plained process ; and the measurement of areas we will im- mediately proceed to explain. But it should first be observed that, in plotting off, all the angles required should be set off in the first place upon one and the same circle, and the direc- tions so obtained then transferred by sliding set squares upon straight-edges into the positions required in the drawing. That is, we are not to draw a new 10 in. circle at each new Datum point of the drawing where an angle is to be set off, because circle such a proceeding would involve the waste of time and labour, 56 GRAPHICS CHAP. V. because it would cover the paper with unnecessary and con- fusing lines, and because at several of these points it will usually be found that there is not room inside the limits of the paper to use a good-sized circle. The centre of the mark- ing-off circle is conveniently chosen near the middle of the paper. Thus in Fig. 27 let it be required to mark off from the line A downwards the angle 140. Instead of doing this directly, we mark off upwards 180 140 = 40 from A to B, and thus get B as the desired direction, which can now be transferred to any part of the drawing. Again, let it be required to mark off 70 downwards from A, the angle A V 1 being less than 70. We first mark off 20 to C 1 from H _L V, then take the chord A C 1 in the dividers and set it off from V to C. We thus obtain C as the direction wanted. We now give two examples of plotting off. In Fig. 28 is shown the calculation of the height of a church spire from Height of theodolite measurements. The theodolite is first placed at a station A. The height of its axis from the ground is measured 4*12 ft. The angle of elevation to top of spire is measured 43 25'. Another station B is chosen in line with A and top of spire. Distance A B on ground is measured 120 ft. Dif- ference of level of ground at B and A is measured by reading with theodolite at A placed with telescope level on surveyor's levelling staff held at B, 2-34 ft. Theodolite is now placed at B, and height of its axis above ground measured 4*47 ft. The angle of elevation to top of spire is measured 25 15'. The construction is so simple and so easily understood from the figure that no explanation is needed. The marking off circle is struck from 0, the position of the theodolite axis at station B with radius H = 5 in. The line a 1 is drawn parallel to a, the angle H a being made 43 25'. From x the intersection of a 1 with b is measured X down to the ground line through station A. This is the height required. The distance Y from station A to centre of base of spire may also be directly measured from the diagram. The calculation I UNIVERSITY CHAP. v. GRAPHO-TRIGONOMETRY 57 of X by ordinary trigonometry involves the solution of the two following equations : (120 + Y) tan 25 15' + 4-47 - 2-34 = Ytan 43 25' X = Ytan 43 25' + 4'12. The solution of these equations gives X = 121-207 and Y = 123-744. In Fig. 29 is shown the plotting of a piece of ground surveyed with the theodolite, and through which flows a river/ preventing the direct measurement of the two .sides C D and A G. All the other five sides are measured in the field, and the external angles between each contiguous pair of sides are also measured. The sum of these external angles measured in semicircles is two more than the number of sides, so that, except as a check on the accuracy of the work, the measure- ment of one of these angles may be omitted. The sides C D Survey and AG are measured on the plot and stated below the dia- rivr S gram. In plotting off the angles at the left-hand of the dia- gram, always the difference between the given angle and the nearest greater or less multiple of one right angle is made use of. After plotting D E F G, from G is drawn G b parallel to the direction of A B and equal to the known length of B A, namely, 230 ft. Then b c is drawn parallel to the direction of, and equal in length to, B C. Then c C is drawn || G A, and D C in the known direction to meet c C in C. Finally, C B is drawn || c b to meet b B || A G, and from the intersection B there is drawn B A || to the known direction to meet G A in A. To find CD and AG by ordinary trigonometry involves a considerable amount of tedious preliminary trigonometrical calculation, and the solution of two not very easily formed equations. 4. Areas. The areas of rectilinear figures already plotted can easily be calculated by taking one side as a base and Areas multiplying it by the mean height of the area above that base. The following are amongst the most convenient constructions 58 GEAPHICS CIIAP. V. Parallelo- grams Quadri- laterals for this purpose. The proofs of the constructions are easily recognised and do not need formal demonstration here. Fig. 30. Area of Triangle = A = base x Height=B x ~ or ~ = 5. To find A, mark off 2 or 20 or 200 or 2,000 along Triangle B to scale of figure. Draw m { and m 2 \\m r The height of intersection of ra 2 with opposite side measures the area ; that is, A x 1, or A x 10, or A x 100, or A x 1,000 = area of tri- angle = i H B. 5. Fig. 31. Area of Parallelogram = A = H B. Mark off 1 or 10 or 100, &c., along one side, A = height of inter- section of m l and w 2 , or of m 2 and w 3 . Area = A, or 10 A, or 100 A, &c. 6. Fig. 32. Area of Quadrilateral with Unequal Sides = A. The given quadrilateral is indicated by heavy black lines. It is split into two triangles by the diagonal m p along which is marked off 2, or 20, or 200, &c. m 2 and m 4 are drawn from end of 2 on w, ; w 3 is drawn || m 2 and w 5 1| ?tt 4 to meet the two sides of the quadrilateral meeting at corner from which 2 was marked off; w 6 is drawn || m r A is measured perpendicularly to m l or w 6 . The construction is equally good whether the diagonal m l is > or < 2. The area = A, or 10 A, or 100 A, &c. Another good construction is the following. The quadri- lateral is divided into two triangles by the diagonal m p from one end of which a circular arc is struck with radius 2, or 20, or 200, &c. (see Fig. 33). A tangent ra 3 to this arc is drawn from the other end of m p and m 4 and w 5 are drawn || m^ They cut off A on w 3 . The radius 2 in this construction must be less than the diagonal m r The area = A, or 10 A, or 100 A, &c. 7. Fig. 34. Areas of Irregular Polygons. Eeduce to a Polygons triangle, and proceed as in Fig. 30, or to a quadrilateral, and proceed as in Fig. 32 or Fig. 33. The given polygon in Fig. 34 is indicated by the heavy lines. The extension of the side CHAP. v. GRAPHO-TRIGONOMETKY 59 09 is chosen as the base because it is the longest. Draw 12'||20. Then the triangular area 92'29 equals the quadri- lateral area 90129. Similarly draw 23 / ||32 / and 34 / ||43 / . The triangular area 94/49 equals the area of the polygon 9012349. Also draw 87'||79, and 76'||67', and 65' || 56', and 54 // ||45 / . Then the triangular area 4" 4' 4 4" equals the polygonal area to be calculatednamely, 90123456789. From 4' mark off 2, or 20 or 2,000, &c., and from the end of this to 4 draw m lt and draw w 2 from 4" || to m l to intersect 4' 4 produced. The length of ra 2 from 4" to this intersection measures the area that is, the area equals A or 10 A or 100 A, &c. Fig. 35 shows the method of applying the construction of Polygons Fig. 33 to the polygon. It is somewhat shorter and neater than that of Fig. 34. It consists essentially in reducing the area to a triangle whose height is 2, the base of which there- fore, of course, measures numerically the area. From any corner 6 describe arc m l with radius 2 or 20 or 200, &c. To this arc draw a tangent w 2 from any other corner, such as 0. Use m z as a base on which to reduce the polygon, in same manner as in Fig. 34. The polygonal area equals the tri- angular area 7', 6', 6 T ', of which the height is 2. Therefore the area = A or 10 A or 100 A, &c. 8. Areas with Curvilinear Boundaries. These are best calcu- lated by dividing them into parallel strips, each of a width easy to deal with as a multiplier of the mean length or height of the strip. If there is no inconvenience in making them so, all the strips are made of equal width. In this case all the mean heights are first added together and their sum then multiplied by the common width. But trouble and time are often saved Curved by taking the widths unequal. The sharper the curvature of gur the boundary the narrower ought the strips to be taken. The mean height is obtained in all cases with sufficient accuracy upon the supposition that each small portion of the curve belonging to one strip is parabolic. The approximation shown in Fig. 36 does not imply the supposition that the whole curve 60 GKAPHICS Curved figures Indicator diagrams from end to end is a part of the same parabola. The different small portions are supposed to be parts of different parabolas, such as most nearly coincide with them throughout the small length. In Fig. 36 the strips are ruled off in the direction of the greatest length of the figure in order to have as few strips as possible, and in order to make the small areas at the ends of the strips between the boundary and the chords as small as possible in proportion to the areas of the strips. The full lines indicate the dividing lines between the strips taken, and the dotted lines are midway between the full lines i.e. they are the centre lines on which the mean lengths of the strips are to be measured. The first two strips are taken T 2 ^ in. wide, the next two -fo in. wide, the next two -f^- in. wide, and this leaves a portion of width x X which is an odd fraction of an inch. The mean length of this last portion is xy, and it is reduced to a strip of width nj-in. ( = 67), and of length X Y by the construction shown i.e. by drawing the line x Y. The mean lengths taken on the dotted lines are not measured to the curve itself, but are taken as the lengths between the chords plus two-thirds the distances between the chords and the curve. This is shown on the small diagram to the right hand. Instead of measuring to n on the curve, or to m on the chord, the mean length is measured to r, m r being two- thirds of m n. The division of m n into three parts can be accurately enough performed by the eye, and the points r need not actually be marked as in Figure 36 ; the length can be read off on applying the scale to the line without the point r being visibly marked on the paper ; in fact, the small divisions on the scale assist in the accurate taking of the f of the small length m n. The construction lines need not be drawn in fully as shown in the diagram ; it is sufficient to mark their intersections with the curve and with the chords. 9. One of the most useful and interesting examples of the CHAP. v. GRAPHO-TRIGONOMETRY 61 calculation of areas with curvilinear boundaries is that of steam-engine indicator diagrams. Here, however, it is not the area but the mean height that is really wanted. The mean height might be arrived at by calculating the area and dividing .by the extreme horizontal length of the diagram ; but this is not the most direct way of proceeding. The ordinary method is to divide this extreme length into an exact number of equal parts, to measure the mean heights of the vertical strips of equal breadth corresponding to these equal parts, to sum up these heights and divide the total by the number of strips. For simplicity's sake the number of parts is usually taken as 10. To facilitate the exact division of the length into 10 equal parts there is used what is called a gridiron parallel ruler that is, a set of 11 small steel straight- edges all linked together by two cross-bars so that all the 11 edges must always remain parallel and equidistant, but may be placed with larger or smaller spaces between them ; in indicator fact, constructed in exactly the same manner as an ordinary ia rams parallel ruler, only with 11 instead of 2 bars. It is very im- portant to observe that the heights of the strips have to be measured each on its middle line. Thus, if the corrections for curvature of the top and bottom boundaries of the strips be neglected, as is usual, that is, if the height be measured simply to these boundaries and not to the two-thirds point between chord and curve it is really unnecessary to draw the vertical lines dividing the strips, but, instead of these, it is better to draw only the middle lines. The gridiron parallel ruler is first set on the diagram as if to draw the dividing lines between the ten strips. With it so set the widths of the two end strips only are marked with pencil. These are then bisected either with dividers, or, as is usually sufficiently accurate, by the eye simply. Between the centres of the two end strips thus obtained we now set the edges of the parallel ruler so as to enclose nine strips between these centres that is, not using the end one of the series of 11 straight-edges of 62 GKAPHICS CHAP. V the gridiron. We then draw in the 10 centre-lines on which the heights have to be measured. 10. Although to the small scale to which steam-engine indicator diagrams are drawn it is nearly always sufficiently accurate to measure the height of each strip to the curve, yet it must usually happen that the point of cut-off falls some- where within the boundary of one of the strips, and in this case it would evidently be very inaccurate to measure simply to the upper boundary without correction, especially if the cut-off be sharp. In Fig. 37 are shown two constructions by which the necessary correction may always be made with suf- ficient accuracy for practical purposes by freehand sketching or by ruling with the help of the gridiron parallel ruler, c d is the last part of the ' admission ' portion of the diagram, and d Cut-off is the point of cut-off, a c and ft/are the dividing verticals enclosing one of the strips. It is sufficiently accurate to take d/as a straight line, and, therefore, c df as a triangle. In construction (A) m l is drawn from c to /and wjjw,. Then ??i 3 is drawn from c to the intersection of w 2 with bf. The height of the strip is to be measured on the centre line to m y In construction (B) the height g f is bisected in e, and m l is drawn through e horizontally. From the intersection of a c with m l the line w 2 is drawn to g, and the height of the strip is measured to the intersection of ra 2 with the vertical drawn through the point of cut-off d. It is evidently unnecessary to actually draw the lines m l and w 2 on the paper ; it is enough to mark their intersections with the other lines. 11. Indicator diagrams are sometimes measured with the aid of an instrument called a planimeter, which measures areas enclosed by re-entrant curves. The instrument carries Piani- a tracing point which is carried round the complete curve, and at the beginning and end of the circuit a difference of readings on a small dial-plate is obtained which measures the area. The most commonly used is Amsler's Planimeter, which, when carefully used, gives very fairly accurate readings. It is CHAP. v. GKAPHO-TEIGONOMETRY 63 , no part of the work of this treatise to describe or demonstrate the action of these instruments. They are, however, very valu- able aids in many graphic calculations, and, therefore, they deserve notice here. Wherever an enclosed area bounded by an irregular curve has to be measured, they may be usefully employed. But where, as in the case of indicator diagrams, the real object aimed at is not the area, but the mean height, their use is clumsy, slow, and not to be recommended unless in conjunction with logarithmic calculation. The method em- Piani- ployed is, of course, to measure the area by the instrument and divide arithmetically by the length. In dealing with indicator diagrams this division must always be an awkward one, because the length is never any exact simple number or fraction of an inch. If, however, the division be performed with the help of a logarithmic slide rule (of which Fuller's Spiral Slide Eule is the most accurate pattern), and if there be a large number of diagrams to work, the use of the plani- meter saves much time and labour. 12. By the above processes the value of a complete enclosed area is found or 'integrated.' It is frequently required to integrate an area progressively that is, to find the progressive value of the portion of the area lying behind a straight line cutting through the area, which straight line is gradually moved (or progresses) perpendicularly to itself, thus always keeping the same direction or parallel to its original direction. If the line progress towards the right hand, the portion of the area to be found lies to its left hand, and vice versa. To Integra- represent the result of this progressive calculation, usually the lon best way is to represent the gradually increasing area by the gradually increasing height of a curve measured from a straight base line drawn anywhere perpendicular to the moving line. Such a curve may be called the integral area curve. If y be the height between the boundaries of the curve enclosing the area, measured perpendicularly to the above base line, and corresponding to an abscissa x measured parallelly to the same 64 GRAPHICS CHAP. V, base, the ordinate to the above integral curve corresponds (is rather identical) with the integral of y with respect to x as determined by the integral calculus. Thus the graphic con- struction of this integral curve, which is about to be explained, is really a method of performing graphically any desired piece of integration. The construction is shown in Fig. 38. The irregular curve bounds the closed area to be integrated by strips perpendicular to the line 00'. 0' is any line drawn in the required direction (i.e. the direction of progression of the integration). The dotted lines separate off the strips into which the whole has been divided. The portions of the strip- areas lying above the line 0' have been called A B C D, &c., and those lying below the line (X are called A' B' C' D'. Each line separating these strips is conveniently named by the two letters which name the two strips it separates. Thus the line between strip A and strip B is called A B ; that between strip Integra- E and strip F is called E F. The vertical centre lines of the strips, on which their mean heights are measured, are called a' a, b' 6, c' c, &c. Through any point p in the line 0' pro- duced a perpendicular is drawn, and on this perpendicular all the points a b c, &c., and a' b' c', &c., are projected. Thus on this perpendicular we have measured upwards and downwards from the point p all the mean heights of the strips A A', B B', C C', &c. On the line p is measured off the length p P, which we may take in the meantime as unity. Then from is drawn 1 across the space A and parallel to the line from P to the projected a. Then across B is drawn 12 parallel to P b (pro- jected) ; then 23 across space C parallel to P c ; then 3 4 across space D parallel to P d ; and so on with the series of succes- sive lines 4 5, 5 6, 6 7, 7 8, 8 9 across the spaces E F G H I. Simi- larly, starting again from 0, the lines 1', V 2', 2' 3', 3' 4', &c., are drawn across the spaces A' B' C' D', &c., parallel to P a', P 6', P c', P d'. If now fair curves are drawn through the points 1 2 3... 7 8 9 and V V 3'... 7' 8' 9', the vertical height between these curves will, to a certain scale to be explained CHAP. v. GRAPH 0- TRIGONOMETRY 65 immediately, measure the integral area to the left hand of the line at which this height is measured. 13. To prove this we notice that the triangle 1 1' is similar integra- te P a a', because 1 is parallel to P a, 1' parallel to P a', and n of 1 1 / 1 V parallel to a a'. Therefore the ratio - . is the same as a a that of the horizontal breadth of the strip A to the distance PP. Thus, since a a' is the mean height of the strips A and A 7 taken together, we have the rectangular area 1 V x Pp area of strip (A + A'). That is, 1 V measures to a scale numerically equal to the reciprocal of P p the area of the com- bined strips A and A . To this scale evidently the height of 1 above line 0' measures area A, and the depth of 1' below 0' measures area A'. Again, suppose the line 1 drawn parallel to 0'. The triangle 1 2 is similar to P bp, and, since 1 is the breadth and bp the height of the strip B lying above 0', we have the rectangular area 2 a x ~Pp = area B. Thus, since the height of is the same as that of 1 above (X, we have the height of 2 above 0' measuring to the above-mentioned s cale the sum of the areas A and B. Similarly, 3 /3 measures area C, and 4 y measures D ; so that the height of 3 measures A + B + C, and the height of 4 measures A + B -f C + D. Thus the height of each of the points 1 to 9 measures the . . portion of the area to its left hand lying above 0', and the depth of each of the points V to 9' measures the portion to its left hand lying below 0'. The point P is called the pole of the projection diagram, and I* p the pole distance. The scale to which those heights measure the corresponding areas is inversely as the length scale of chosen for the pole distance, Pp. As this is a most important areacurve point often leading to confusion in engineering problems if not clearly understood, it must be fully explained. If the diagram be drawn full size and the dimensions of the area be measured in inches ; and if, further, the pole dis- 66 GRAPHICS CHAP. V. tance P p be taken the natural unit, namely 1 in.; then our equation becomes Height of integral curve x P_p = rectangular area of breadth 1 in. and of height equal to height of integral curve = the integral area within the given irregular boundary* The breadth of this rectangular area being 1 in., its height namely, that of the integral curve if read off in inches, will give the number of square inches in the desired area. But in most practical problems the integral curve would be inconveniently steep, and would soon run out of any ordi- nary size of paper if P_p were taken so small as 1 in. Suppose P p be taken 10 in. Then all the heights of the integral curve will become just one-tenth of what they were before. Thus -jJg- in. = 1 sq. in. is now the scale to which the heights measure the areas. Another way of putting it is, the height of the integral curve to the scale of the given figure (that is, in Scale of inches) multiplied by the pole distance to the scale of the pro- jection diagram (namely, 10 in.) equals the desired area in square inches. If Pp were taken 5 in., then the scale would be ^ in. = 1 sq. in. ; if it were taken 20 in., the scale would be ^ in. = 1 sq. in. Now suppose the figure has been measured, say, in feet, and has been plotted off to a scale of, say, in. = 1 ft. It is desired to measure the area in square feet. The projected heights are to the scale = 1 ft. Suppose we find it convenient to take the pole distance equal to 10 in. This to the scale of the pro- jection diagram means 80 ft. Then the height of the integral curve measured in feet to the scale in. = 1 ft. multiplied by 80 ft. equals the area in square feet ; or the height of the integral curve measured to the scale - = - - in. = 80 x 8 640 1 sq. ft. equals the area in square feet. Now suppose that for some reason or other the figure has been plotted with different vertical and horizontal scales. For CHAP. v. GRAPIIO-TRIGONOMETRY 67 example, suppose it to be a plot of a longitudinal section for a railway cutting, in which the horizontal scale is invariably smaller than the vertical scale. Suppose the vertical scale to be ^ in. = 1 ft., and the horizontal scale $ in. = 1 ft. The projection diagram has the same scale as that of the vertical ordinates namely, | in. = 1 ft. Let 12 inches be the distance found convenient for the pole distance that is, Pp measures 60 ft. on the scale of the projection diagram. If the horizontal were the same as the vertical scale in our figure, we would now read the heights of the integral area- curve to the scale _ = , in. = 1 so. ft. But since the horizontal scale is 5 x 60 300 only one-tenth the vertical scale, the actual areas are evidently ten times greater than we would find if we supposed the hori- zontal scale equal to the vertical. Therefore, the scale to which the heights of the integral area-curve must be read is Scale of 1 ^ area curve = 1 sq. ft. To take a numerical ex- 3000 " > ample, and referring to the strip B and the triangle 1 2 of Fig. 38 ; suppose the mean height p b measures 6" = 30 ft. to scale |" = 1 ft., and breadth 1 a measures 4" == 200 ft. to scale 3-y = 1 ft. Then the area is, of course, 6,000 sq. ft. The pole distance being 12 in., we have P = --=-^-=-^, Pp 12 1 4 whence 2 * = 4 * 6 ' = 2 V . This height to the scale -^ ff in. La = 1 sq. ft. means 3,000 x 2 = 6,000 sq. ft. Otherwise ex- pressed : the pole distance is 60 ft. to the scale of the projected mean heights, and the horizontal scale of the area is $" = 1 ft. Then the scale to which areas are to be read off on the integral area-curve is 1 sq. ft. = -$ x -fc in. _ fraction of inch representing 1 ft. on horizontal scale of area pole distance in feet read to vertical scale of area 14. One useful application of this area- integration is the Water storage of water in reservoirs. A curve is plotted, giving the F 2 68 GRAPHICS CHAP. V rate of supply from day to day, which, of course, varies with the season. Another is plotted to represent the rate at which water is drawn off. The integral of the area between these two curves is the total excess or deficiency of supply over or below discharge. An example is shown in Fig. 39. The Water excess curve is alternately positive and negative, and its maxima and minima points occur simultaneously with the crossings of the supply and drain curves. The points at which the excess curve reverses the direction of its curvature (points of inflexion) are simultaneous with those at which the supply and drain curves run parallel to each other, i.e. where these two are at maximum or minimum distance apart. 15. Another extremely important application of the inte- gration construction that has just bean explained is the calcu- lation of the excess of energy delivered by an engine to the crank-shaft during one portion of the revolution and the defi- Appiica- ciency during another. Examples are worked out in subse- quent chapters. Still another use of the same construction is the calcula- tion of the sheer force on different sections of a beam or girder from the known manner in which the load is distributed over the span. This also is treated of in a future chapter. 16. Keferring again to Fig. 38, it is to be noted that we have there determined only ten points in each of the two inte- gral curves. It must not be supposed that the diagram is composed of straight lines drawn between these points. To get the intermediate points more accurately than can be done by simply drawing a ' fair curve ' through the points already integral determined, it would only be necessary to divide the area into a larger number of narrower strips, and thus obtain more points. In most practical calculations, however, no great re- finement is needed, and the curve may be drawn in with enough accuracy from a few points only, provided the draughts- man is guided in doing so by an intelligent and careful com- prehension of the true character of the curve at its various CHAP. v. GRAPITO-TRIGONOMETRY 69 points. Thus it must be noted that, so long as the mean height of the strips is increasing, the integral curve bends upwards, i.e. its centre of curvature lies above. If the mean height remain constant throughout a certain length, the integral curve becomes a straight line throughout that length : it bends neither upwards nor downwards, but maintains a uniform inclination. If the mean height decreases, the curve bends downward*. In the ordinary case the mean height increases Charac- to a maximum, and then immediately begins to decrease. ofYnte- With this maximum or minimum of the mean height coincides & ral c a reversal of curvature of the integral area curve. If the mean height becomes zero at any point, then at this point the curve becomes horizontal. It slopes upwards so long as the mean height is positive. If the mean height ever become negative, the curve slopes downwards. 17. Although the upper boundary of a strip may be practi- cally a straight line, it does not follow that the integral curve also approximates to a straight line throughout the corre- sponding stretch. Let (in Fig. 40) A a B b be one of the strips, and let a be the corresponding stretch of the integral curve. Since the mean height of the strip increases at a uniform rate from a to b, therefore the inclination of the curve a JB increases also at a uniform rate. It is easily recognised that the curve is one of the second degree. It is, in fact, parabolic. Its curvature depends on the steepness of the line a b. The dis- Correction tance of the curve on the middle line p q from the chord a /3 may be calculated graphically without trouble if desired. Let the letters a b ft be taken to represent the heights of the corresponding points. Then, tv being drawn horizontally, av= i (/3 a). It can be shown that the distance from chord to curve ty = \ (B a) 7 ~ . Now, a r being drawn t (o + a) horizontally and r q being bisected in s, we ha.ve evidently q s = \ (b a) and p q = J (b -f a) . If, then, we draw the lines q ex. and p v, and through their intersection draw the line s x t 70 GRAPHICS CHAP. v. then evidently ^~x = ^v . qS = ( ) f-yr^ ?i and is therefore, the distance ?/ required. It is seldom that this calculation of the versine is needed in practical work, and it must be remembered that the same result namely, to de- terminethe point y of the curve independently can be just as Correction well, and probably more quickly, obtained by taking the strip in two parts instead of in one in the general construction for deriving the curve. This latter course is, in fact, the more satisfactory one, because the correction above explained is founded on the assumption that a b is a straight line, which in most cases is only an approximation. It must be understood that in all the constructions from Fig. 30 to Fig. 40 the dotted lines represent lines which are not to be actually drawn but only ' formed ' by laying the edge of the set-square along them and making the required intersections. The practical draughtsman avoids drawing in as many lines as possible, because unnecessary lines not only dirty the paper when in pencil, but they also make the diagram confused and obscure, and render the useful result less readily perceived by the eye. Frequently the number of measurements taken, from which the figure has to be plotted, is in excess of the number absolutely required. Such is the case always, for example, in a survey. There is then an embarras de richesses of data for the calculation, and the results calculated from different sets of data may not exactly agree. The disagreement shows that small errors have been made in the measurements. Later on we will show how these errors are most equally distributed in the graphic constructions. CHAP. vi. COMBINED MULTIPLICATION AND SUMMATION CHAPTEK VI. COMBINED MULTIPLICATION AND SUMMATION. MOMENTS OF PARALLEL VECTORS. 1. IN the last chapter the graphic summation of area strips was explained. Each area strip was the product of its mean height and its breadth, and the strips stood side by side. Any product of two quantities may he represented by a rectangular area whose sides are the two quantities measured as line magnitudes to suitable scales. A number of such areas can be added together only if they all represent things of the same kind and if the scales for the different areas are the same, gum of We have now to consider the frequently occurring case in P roducts which the areas representing the products overlap each other, each of them having one edge lying on the same straight line common to all. In the diagrams necessary for the calcula- tion it is seldom necessary to actually draw in the actual sides inclosing the rectangular area ; it is sufficient to draw in their correct directions the two perpendicular dimensions of the rectangle. 2. This problem is of very frequent practical occurrence. The most familiar example is that of the total moment round some axis of several parallel forces ; for instance, the total turning moment round the fulcrum of a lever of a number of weights hung on the lever, or the total bending moment on Parallel any section of a beam due to a number of different loads. The constructions given in this chapter for parallel forces are capable of extension to the case of non-parallel forces, and this development will be dealt with in Chap. VIII. .72 GRAPHICS CHAP. vi. The general construction applicable to the summation of products of any kind of quantities is shown in Fig. 41. Let the products be A a, B fc, C c, &c., the factors ABC being all of one kind, and a b c being all of one kind. From any point X along any straight line lay off the factors a b c to any suitable Parallel scale, each factor being measured from X. Then from X draw a line perpendicular to this first. Also draw another perpen- dicular C' in any other position on the paper, and from any point mark off along this line to any suitable scale the factors ABC successively. These factors are not measured all from one point so as to overlap, as the factors a be' are, but are added along the line so that C' is the sum of the three. 3. From any point p in this line draw a line p P perpen- dicular to C' and, therefore, parallel to X a, and make p P equal to 1, or 10, or 100, &c., to the scale of the factors ABC. Then from a draw the line a o \\ P and a || P A'. The tri- angle o a a is similar to the triangle P A', and therefore similar dimensions in the two triangles have all the same ratios ; thus the bases A' a,nd o a have the same ratios as the heights Pp and &X. But A' = A and &X = a ; there- fore, we have P p o = A a. If Pp be taken unity to scale of A, then o a to the scale of a will numerically equal the pro- Pole duct A a. If Pp be taken 10 to scale of A, then ten times o a to scale of a will equal A a. Whatever length it be convenient to take for P p, we have o a to scale of a multiplied by P p to scale of A equal to the product A a. Now draw b b' \\ o X to meet a in 6',- and draw b' ft || P B'. The two triangles I)' a. ft and P A' B' are similar ; therefore, the heights b X and Pp have the same ratio as the bases a /3 and A' B'. Thus Pp a fi = B b ; that is, a ft measures the second product B b to same scale as o * measures the first A a. Also we have o ft the sum of the two products. Again draw c c' || o X to meet b'fi in c' and draw c' y \\ P C'. We have now /3 7 measuring the product C c to the same scale CHAP. vi. COMBINED MULTIPLICATION AND SUMMATION 73 as above. Tims we can find the sum of any number of pro- ducts by this construction. The scale to which the result is read is inversely propor- tional to Pp. The larger the factors we have to deal with are, the larger must we take P p in order to get the result to a sufficiently small scale to allow of the sum being represented by a line within the limits of a convenient space on the paper. Pis called the 'pole' and Pp the 'pole distance' of the diagram. The pole distance should be taken a convenient number (to the scale of ABC) by which to multiply the quantity read to the scale of a I c. In different practical pro- blems it is convenient to use 1, 5, 10, 50, 100, 500, 1,000, 10,000, and even sometimes 100,000. The pole may be either to left or right of the vertical C'. If it is at P' to the right, the diagram a V c y would slope up- Right or wards intead of downwards, and a o would slope downwards ; the diagram would be simply reversed on the line a X. Suppose all the factors ab c to be diminished by the same amount. This is equivalent to shifting X to X', a distance equal to the diminution of each factor. The rest of the diagram remains unaltered : o'y' now measures the sum of the new products. If the diminution exceed c ; for example, let it be X X", then c becomes negative and the product C c is to be subtracted instead of added. This is still effected without Change of alteration of the diagram, o" y" measured to the line c' y produced backwards gives the algebraic sum, because o" l>" measures (A a X" -f B 6X 77 ) and V y" measures the nega- tive product C . c X". If X be shifted to X'", to the left hand of the crossing I of the lines ao and c'y, then the negative products overbalance the positive ones, and the sum o'" y'" is to be read as a nega- tive quantity. 4. It is on account of these last peculiarities that this construction is most useful for calculation of the moments of Moment parallel forces or loads on beams or levers, a X represents the 74 GRAPHICS CHAP. VI. line of the lever or beam : X the fulcrum or section round which the moments of the forces are to be taken. The points ab c are the positions at which these forces act perpendicularly to the line a X that is, parallelly to the line C' and the quantities a b c are their leverages around X. ABC marked and added on this line represent the forces. The diagram composed of the two straight lines 7 7"'' and and o o'" parallel to the two lines Moment P and P C' form a complete graphic calculation and record of all the moments this set of forces ABC exert on any and every point or section of the lever or beam, because X may be taken anywhere. The moment on any section is read off between these two lines 77"' and oo'" on the vertical drawn through the section. To the right of the intersec- tion I of these two lines the moment may be regarded as positive ; to the left of the same crossing it has the opposite sign. 5. Immediately under the crossing the moment is zero ; that is, round a fulcrum under this crossing the given set of forces would have zero turning moment or would balance so far as moments are concerned. Also it is evident that if a single force equal in magnitude to C', the sum of the given forces, were applied to a point of the beam immediately under this same intersection, this single force would produce on all sections the same moment as the given set of forces actually produce. For example, round the section X this Axis of single force C' applied through I would produce the moment moment 7> because the two triangles 107 and P C' are similar, and, therefore, Pp-oy equals C' multiplied by the hori- zontal distance of I from o 7. Any of the forces may be negative, i.e. directed upwards. They are then marked off upwards on the line C', and from the beginning to the end of this line still represent the alge- braic sum of the forces. Thus the left-hand diagram gives the ' resultant force ' and the right-hand diagram gives the ' resultant moments.' The CHAP. vi. COMBINED MULTIPLICATION AND SUMMATION 75 first has, therefore, been called the Force Diagram, and the Axis of second, the Moment Diagram. It is to be noted that the pole moment distance P p represents a force. 6. All quantities which have, like forces, directions and magnitudes are called 'vectors.' These will be explained fully Vectors in next chapter. Examples are motions, velocities, momenta, accelerations of momentum or of velocity. Vectors whose pro- perties or effects depend upon the definite position in space in which they occur, have been called ' rotors' In this book, Locors however, they will be called ' locors.' The moments of all locors are calculated in the same way as those of forces, and, therefore, the above construction is applicable to the moments of all kinds of parallel locors. It will be extensively used throughout the rest of this work. In all such diagrams the pole distance represents a vector. The extension of the method to vectors that are not parallel will be explained in Chap. VIII. 7. By varying the pole distance the inclination of the two final lines oo'" and 77'" is also varied; what may be called their combined ' slope ' or the sum of their slopes to the horizontal line a X varies inversely as the number of units taken for the pole distance measured to the vector scale. It follows that, although a shifting of P parallel to C' alters the inclination or slope of each of the lines o o"' and 7 7'", it Position does not alter the sum of their slopes. Also any shifting of ant e! P either vertically or horizontally does not alter the horizontal distance of the intersection I from any line o X ; the point I shifts vertically as P is moved, but it always remains in the same vertical line. This line" parallel to the forces in which I lies is called the centre line or resultant line of the locors. In the case of bending moments or. beams and in other similar cases, we wish to determine for each' section the resul- tant moment of all the forces lying to one side of that section, ex- cluding the moments of those lying to the other side. The above diagram, Fig. 41, gives this at once. For instance, if the section 76 GRAPHICS CHAP. VI. Partial Moment Cyclical lettering to be investigated be at X", we have A and B acting to the left of this section, and their resultant moment round this section is o" V : while we have C only to the right hand and its moment (of opposite sign to the former) is measured by It' 7". Thus for the resultant moment of all forces to the left of any section we measure vertically between a o and the diagram outline a b' c f 7, where a o is the final line of the diagram outline going from force to force towards the left. For the resultant moment of all forces lying to the right hand of any section we measure vertically between c' 7 (or the same produced) and the outline c' b' a o, where c' 7 is the last line of the diagram going from force to force towards the right. 8. The lettering of Fig. 41, although almost necessary in explaining and proving the mathematics of the construction, is by no means the best for practical use. Fig. 42 shows the more convenient method of lettering. The utility of this method increases rapidly with the com- plication of the figure. The spaces between the force lines are called by the names ABC, &c., and the lines themselves separating these spaces are called by the two letter-names of the spaces they separate; thus, AB, BC, E F, &c., are the lines separating spaces A and B, B and C, E and F, and so on. These lines are of indefinite length. The vectors along these lines are called ab, be, c d, &c. They are plotted off successively along a line parallel to the given vector direction. The small letters on this line in- dicate points, and the magnitudes of the vectors are measured between these points. Thus the line between the points d e in the vector diagram give the magnitude and direction of the vector whose position and direction are shown by the line D E on the moment diagram. The pole p is placed anywhere at a convenient pole distance. Then parallel to the radii (which need not be actually drawn), from p to a b c d t &c., are drawn lines across the corresponding spaces ABCD, &c., CHAT. vi. COMBINED MULTIPLICATION AND SUMMATION 77 these being drawn so as to form a chain of successive links that is, the two lines across adjacent spaces meeting in the line separating these two spaces. The whole of the space on one side of this diagram outline or chain is named P, the names ABC, c., being confined to those portions of the spaces between the vector position-lines that lie on the other side from P of the chain. The lines or links of the chain are called P A, P B, PC, &c. Thus, corresponding and parallel lines in the two diagrams are called by the same large and cyclical small letters. The series of lines or * chain ' may be referred letterm K to as (P) A B C D E F G H, and the pencil of radii from pole p may be called (p) a b c d, &c. This makes a complicated dia- gram easy to comprehend and easy to read, and it makes it almost difficult to make an error in its construction. The two final lines or links in Fig. 42 are P A and P H, and their intersection I is on the centre line or resultant of the seven locors whose resultant magnitude is a h. The resultant moment round any axis is measured between P A and P H. 9. This resultant moment is, therefore, not zero except round axes situated in the line drawn through I in the direc- tion of the locors. This set of locors, therefore, does not balance generally with respect to moment. In order that there should be such balance, there must be zero moment round all possible axes. This evidently can occur only if the two final lines of the chain coincide. If an extra vector h r along a line H K were added to the existing set, the line P H would cross the space H to its intersection with this new locor H E, Balancing and from this intersection would be drawn a line P R parallel locor to p r in the pole diagram, in which h r must first be plotted off. This new line P E can only coincide with P A if the new locor line H E cut P H in the intersection I. In order that the extra locor should produce moment balance, its line, there- fore, must pass through I. It is further necessary that P E should have the same direction as P A, which can only be if r coincide with a, because P E is to be drawn parallel to p r. 78 GRAPHICS CHAV. \i. Thus the new balancing locor must have the magnitude h a that is, must be equal and opposite in sign to the sum of the other vectors. The sum of the magnitudes of the complete Balancing set then becomes zero, and the moment round every possible locor . . axis is also zero. Thus, if the first given set is to be balanced by a single locor, both its magnitude, its direction, and its position are determined by the condition that it must balance the others. 10. Evidently in the diagram there will be no need for either of the letters K or r ; r coincides with a and is not wanted. The letter P will now refer only to the enclosed space inside the chain. The chain is now a closed chain, its Closed final links meeting in I. The space A extends from line B A diagram ^ o ^ e resultant line through I. The space H extends simi- larly from line G H to same line through I. Therefore, this last line is now to be called HA. The closed chain. or outline of the enclosed space P is called (P) A B C D E F G H. 11. If, however, the set is to be balanced by two other locors, these may have positions assigned to them indepen- dent of the condition of balance, which condition may then be fulfilled by assigning proper magnitudes to the two. A familiar example is that of the two vertical supporting forces at the ends of a beam which keep the loads in balance. Let B! A and H B 2 be the assigned positions of the two balancing locors. Then to complete the diagram there have Two to be drawn a line P B! across the space B t from the intersec- tion of P A and E, A, and a line P B 2 across the space B 2 from the intersection of P H and H B 2 , these lines being parallel to corresponding lines drawn from the pole p. But if these two new locors are to balance the set completely, these two lines P B, and P B 2 must coincide with each other, and must there- fore lie from the first-mentioned intersection of P A and Bj A to the second- mentioned intersection of P H and H B 2 . P B t and P B 2 being thus one and the same line, they correspond to one line only in the pole diagram. This line p r 2 is to be CHAP. vi. COMBINED MULTIPLICATION AND SUMMATION 79 drawn from p parallel to the line P K joining the above two intersections. The point r 2 thus found gives h r 2 as the mag- nitude of the balancing locor along H K 2 , and r 2 a as that along K, A. Also the chain is once more a closed chain, and the letter P should be applied to the space enclosed by it. It must further be noticed that there is no need of both letters Ej E 2 and r l r 2 . The space K may be supposed to stretch from line A E t to line H E 2 on the other side of the Two space P enclosed by the chain to that occupied by AB C, &c. Also, r simply may be substituted for r 2 . The new closed chain is now to be called (P) ABCDEFGHE, the space inside it being P and the whole space outside it being divided by the nine locors into nine spaces, each of which is definitely bounded on three sides by three lines but has no definite outer boundary. 12. This closed chain is still used as a moment diagram ; not, of course, of moments of the whole set of nine locors, because the set taken as a whole has zero moment round any and every axis, but of the total moments of all the locors lying on one side only of any line parallel to the locors. The total moment of those lying to one side of any such line round any axis lying in this same line is the length of this line intercepted inside the closed chain, measured to the scale determined by the reciprocal of the pole distance. This moment must be taken (say) positive for the set of locors n u i e O f lying to the left hand and negative for those to the right signs hand, the sum of the two being zero or the moment of the whole set. If the line pass altogether outside the closed chain, the intercepted length is zero, which means zero moment because we have now the whole set lying to one side. There is an extremely useful convention to notice with regard to the signs of these locor moments. It is the fol- lowing : The pole being taken at any convenient position may stand on the paper either right or left of the line on which 80 GRAPHICS CHAP. VT the vectors are marked off successively and added. If the vector line stand to the right of the pole, call the figure a right- handed diagram ; if to the left, call it left-handed. Now observe the cyclical order in which the adding of the vectors leads from letter to letter i.e. the order in which the letters must be read in the pole diagram to give the true directions of the vectors. Now, taking the sides of the chain surround- ing the closed space, P, in the same cyclical order, pass from one end to the other of the intercept measuring the moment inside the closed chain, following round that portion of the chain lying to the same side of the intercept as do the locors whose total moment is to be measured. This will give the intercept measured in a definite direction, either up or down ; up for the locors lying on one side and down for those lying Eule of on the other side of the intercept. The rule is now that an upward intercept means a positive or right-handed moment ij the diagram be right-handed, and if it be a left-handed diagram an upivard intercept means a negative or left-handed moment. Vice versa, a downward intercept means a left-handed moment on a right-handed diagram, and a right-handed moment on a left-handed diagram. This rule is of universal application, and may save considerable confusion of mind if the diagram be a complicated one with the sides of the chain crossing each other several times. A precisely analogous convention will be shown hereafter to be of the greatest utility in stress diagrams for complicated frameworks. In this construction we have assumed the positions and directions of the two balancing locors as known, and that these directions are parallel to the other known locors. 13. In the cases of such structures as roofs and bridges resting on walls, piers, or abutments, it is seldom known in Abutment which directions the supporting forces are exerted ; the * thrusts ' or horizontal components of these forces are un- known. All that is known is that they are exerted through CHAP. vi. COMBINED MULTIPLICATION AND SUMMATION 81 certain comparatively small surfaces which we may assume in our diagram as known points. If this condition is all that is known regarding the action of these forces, it leaves the problem of calculating their amount and direction to a certain extent insoluble. But only to a certain extent. Although they cannot be completely determined, these forces have cer- tain determinable relations which tell the engineer a great deal about them that is useful. Later we will find how these rela- tions enable the forces to be completely determined as soon as certain coefficients of pliability of the abutments and of the structure are known. Meanwhile the graphic calculation of these relations is the following in Fig. 43. The diagram shows in heavy lines the outline of a girder which is taken as sloping, that is, with supports not at the same level, for the sake of greater generality, the method being equally applicable and easy whether these sup- ports are on the same level or not. The loads applied are Abutment supposed to be four only, their lines being A B, B C, C D, D E, thrusts and their magnitudes, a b, b c, c d, d e. The supporting points or foundations are marked Fj and F 2 . Through these are drawn lines (dotted in the figure) parallel to the loads. A pole, p, is chosen at a convenient distance and the chain (P) ABODE drawn, the sides being drawn parallel to the radii from p. The intersections of PA and PE with the lines through F, and F 2 are joined, and parallel to the line (PK) joining them is drawn p r. Then we know that if the forces at F! and F 2 were parallel, their magnitudes would be er at F 2 and ra at F,. But if they are not parallel, then each of them may be resolved into two components, one parallel to the loads and the other along the line F t F 2 . These latter components parallel to F! F 2 may again be resolved each into a component parallel to the load and one perpendicular to the load. These perpendicular components must have equal and opposite magnitudes in order that the whole system may balance, because all the other forces in the system are at right G 82 GEAPHICS CHAP. VI. angles to them. Since the two primary components along F, F 2 and exerted at F, and F 2 are in the same direction, and since their secondary horizontal components have the same magnitude, therefore these two primary components have also equal and opposite magnitudes. Thus, each of the supporting forces being resolved into a component parallel to the loads and a component parallel to Y l F 2 , which we may call the * abutment line,' we find that the latter components parallel to the ' abutment line ' are equal and opposite. Since they also act along the same straight line, their moments round any and every axis exactly neutralize each other. Therefore, the rest of the system, exclusive of these two, must independently balance with respect to moments ; that is, the two remaining components parallel to the loads of the forces at F 2 and F, must be e r and r a, the same as if the whole sup- porting forces were vertical. Abutment Through r draw the line r p \\ F, F 2 , the abutment line. We now see that the force at F 2 is represented to scale by a line from e to some undetermined point in the line r p, for it is compounded of e r and a force parallel to r p ; and that the force at F, is similarly represented by a line from the same point in r p to a. The indeterminateness of the problem is now expressed by the indenniteness of the point p at which the two forces e p and p a meet in the force diagram. If any one other particular concerning either of the two supporting forces be known, then the problem becomes capable of solution. For instance, if it be known that the support at Fj be capable of exerting only a horizontal force, then the point p can be at once found by drawing from a a horizontal line to intersect r p. If either force be known to be wholly vertical, then the point p will coincide with r and the horizontal thrusts be zero. If the direction of the force at F 2 be known, a line can be drawn from e parallel to that direction to inter- sect r p. Thus, in the special example shown, if the force at F 2 were horizontal, or had a less inclination to the horizontal CHAP. vi. COMBINED MULTIPLICATION AND SUMMATION 83 than r p, the intersection would fall to the right hand of a e and the force would be a tension. It will, however, be after- wards shown that, if the direction of one of these forces be known, their magnitudes can be found by a slightly more direct and expeditious construction than the one here ex- plained. As a further example of the utility of the above Abutment construction, suppose that the horizontal component of either of the forces exerted at F t and F 2 be known. This horizontal component is to be plotted off in the proper direction to the right or left hand of the line a e, and at' the distance so ob- tained a vertical line is to be drawn to intersect r p. This intersection will then be the true position of p. G 2 84 GRAPHICS CHAP. VII. CHAPTEK VII. VECTOR SUMMATION. 1. THERE is a large series of different kinds of things many of the useful calculations regarding all of which are made by exactly the same methods because the various members of this group have two characteristics in common. Vectors They all have magnitude or quantity and they all have direction. This circumstance has led to its being convenient to give a distinctive name to this class namely, vector. Everything that has both direction and quantity is a vector, whatever be the other qualities it may possess. A vector may, therefore, be shortly and completely denned as a directed quantity. 2. Here the word direction is used in its strictest sense. A given line in space does not necessarily have a definite direction, or rather it has two possible directions exactly opposite to each other. It is not a vector unless it is specified to have one and one only of these two opposite directions that is, to be directed from one specified end of the line tmcards the other end. When two lines have the same lie Direction in space, if they have opposite directions or if one or both have no specified direction, their coincidence in lie is expressed by the word parallelism ; they are said to be parallel. The word ' direction ' has often been used to mean the same as * lie ' in the last sentence, and to avoid ambiguity the word ' sense ' has been employed to convey the stricter meaning ascribed above to ' direction.' But, although we have no other word meaning exactly what has been expressed above by ' lie,' there seems to be no practical need of using * sense.' To say CHAP. vn. VECTOR SUMMATION 85 that two lines have the same ' lie ' would, apart from the con- text, be awkward and ambiguous, but the same meaning can be exactly expressed by saying that they * are parallel.' In this book the word direction will be used to indicate one only of the two possible opposite directions along a straight line. Different kinds of vectors have, of course, other properties Direction besides direction and magnitude which distinguish them one from another. But when dealing with them as vectors we assume that the consideration of these other properties is abstracted from ; we deal with them as having only these two properties, magnitude and direction. 3. Thus the motions of the different soldiers of a regiment marching straight forward through a field take place in differ- ent parts of the space of the field. These different motions are distinguished from each other by having different positions in the field. But abstracting from their property of position , they are vectors all having the same direction and the same magnitude, and are, therefore, all one and the same vector. Two coaches running along different but parallel roads and in the same direction, and with the same speeds along these roads, have velocity vectors which are equal or rather are the same vector, although the actual motions occur in different parts of the country. Again, simple changes of position, velocities, and momenta niustra- are all vectors, but of different kinds, being distinguished one from the other by the first having position only as a property besides direction and magnitude, while besides these the second involves both position and time, and the third position, time, and mass. Again, if two masses have the same momenta both quan- titatively and as regards direction, these two momenta are one and the same vector although they have different positions and occur in different masses. It is to be noted that vectors of different kinds cannot possibly be added together. 86 GRAPHICS CHAP. VII. Symbols of Equiva- lence Locors 4. The sign = indicates quantitative or numerical equiva- lence. The sign || ordinarily indicates simple parallelism, but we will, when applying it to vectors, use it invariably to indicate identity of direction. These two signs are combined into 4p to symbolise vector equality or combined quantitative equality and directional identity. This sign is to be used in all vector equations. The symbol ^ is used to indicate equality in magnitude and in parallel but opposite directions. The symbol =^= indicates quantitative equality combined with parallelism without reference to direction or ' sense.' When it is desired to express equality, identity, or coinci- dence of the other properties of the vectors compared besides those of direction and magnitude, the sign ^ will be used. This sign will be more particularly used to express coincidence of position or situation as well as of direction and magnitude of the vectors compared. When we consider the results or effects of the combination of the three properties, position, direction and magnitude, the things dealt with will be called locors, and we may term =||j= the symbol of locor equality. Nearly all vectors are actually locors ; a vector is, usually if not always, simply a locor considered without regard to its definite position. The sign ^. will indicate exact locor oppositeness ; thus a 4( & means that a and b are locors with equal magnitudes and exactly opposite directions lying along coincident lines. 5. The locors or vectors that are chiefly dealt with in mechanics are the following : Difference of Position. Change of Position or Displacement. Velocity and Change or Difference of Velocity. Acceleration (i.e. Time Bate of Change) of Velocity. Momentum and Acceleration of Momentum. Force and the one-directional aspect of a Stress, including the stresses called Gravitation, Cohesion, Electric and Mag- netic Induction. CHAP. vii. VECTOR SUMMATION 87 Flow of Fluid or Flow of any quantity of any kind what- ever, and this being either the Integral Flow throughout a given time or the time-rate of flow, and either the Integral Flow through a given area or the flow per unit of area. The Locors flow may either be of one or other kind of substance or matter, or it may be of a condition for example, of a dynamic condition such as momentum, stress, strain, kinetic energy. Electric current may be taken as coming under the latter head. 6. Eotations and Angular Velocities may also be repre- sented by vectors or locors, the direction and position being taken along the axis of rotation and the quantity (represented to a convenient scale by a length measured along the axis) being an angle or angle per second. The length along the axis is to be directed + or according to the direction of rotation ; the most convenient convention being that it is to have that direction in which the rotation will appear right- handed. Viewed in the opposite direction, the rotation would, of course, appear left-handed. But there is a distinction between this kind of quantity and those previously mentioned which should not be lost sight of. While all the others are Rotors vectors in themselves and in reality, rotations are evidently not vectors at all ; it is only the graphic representation of the rotation by a directed line that is a vector. The vector is the conventional symbol, not the real thing itself. This essential distinction should be preserved in the language used, and this kind of quantity will be called a ' rotor,' a name applied by Professor Clifford to all locors as well as to what is here called a * rotor.' When the position of the axis of the rotor is con- cerned in the problem dealt with, as well as the direction of that axis and the magnitude of the rotation, the rotor may be called a ' locorotor.' Angular momenta and force- couples are rotors of a kind very similar to rotations and angular velocities. 7. In some circumstances it is convenient to represent surface plane surfaces similarly by directed lines. These may also rotors 88 GRAPHICS CHAP. vn. be called rotors. The line is taken of a length measuring to the scale adopted the area of the surface. It is taken normal to the surface, and directed in such a way as to indicate whether the area is to be counted positive or nega- tive. An area is frequently to be regarded as generated by the motion of a line i.e. as being swept out by a line, and the sign of the area then indicates the general direction in which it has been generated. Thus it may be a ' polar area ' swept out by the radius vector from a fixed pole to a moving body. In this case the sign given to the area is naturally Surface made to coincide with that of the rotation of the radius vector round the pole i.e. the line representing the area is to be drawn in that direction in which the sweeping out of the area round the pole appears to be right-handed. There are several other conventions with regard to the signs of areas used under various circumstances. This rotor representation of a surface indicates nothing as regards the shape of the boundary of the surface, and, therefore, is not a complete description of the surface any more than the vector representation of a mo- mentum is a complete description of the momentum ; this latter not affording any information regarding the mass or the velocity separately of the moving body, but only regarding the product of the mass and of its velocity. 8. We habitually talk of all these things having position and direction in space. It should be held clearly in view that in doing so we think only of relative positions ; that the space referred to is that occupied by and geometrically attached to, Relativity so to speak, and surrounding, some material object or set of objects. Careful consideration will convince anyone that we have, and can have, no other idea of space than this relative one ; that talk of absolute space and absolute position is talk only and does not correspond to any real consciousness or any real thought, probably because it quite certainly corresponds to no real experience. Clerk Maxwell says with keen and delicate sarcasm, * Anyone who will try to imagine the state CHAP. TIL VECTOR SUMMATION 89 of a mind conscious of knowing the absolute position of a point will ever after be content with our relative knowledge.' ' 9. The positions of things in space can only be recognised and defined in relation to other things. Thus the position of a picture is thought of in reference to the walls and furniture of the room in which it hangs ; the position of a cricket-ball at any instant is thought of in relation to the wickets, bat, fielders, and the field on which' the game is being played ; the motion of an engine-crank is regarded as taking place in the space defined by reference to the engine bed-plate ; the motion of a pinion in a watch is thought of in relation to the space defined by reference to, that is ' geometrically attached to,' the watch-case; the motion of a ship propeller-blade The may be viewed as taking place in the space attached to the ship itself, or in the space attached to the water, or in the space attached to the solid earth, this latter being different from the water-space if there be a water current giving motion as between the water and the earth. The space geometrically attached to the body in relation to which a vector, locor, or rotor is defined is conveniently called the * field ' of that body. Thus, the stellar field is that in which the motion of the sun is astronomically calculated ; the field of the sun is that in which the planetary orbits are mapped 1 Matter and Motion, p. 20. ' Absolute space is conceived as remaining always similar to itself and immovable. The arrangement of the parts of space can no more be altered than the order of the portions of time. To conceive them to move from their places is to conceive a place to move away from itself- ' But as there is nothing to distinguish one portion of time from another except the different events which occur in them, so there is nothing to distin- _, ..., A guish one part of space from another except its relation to the place of mate- rial bodies. We cannot describe the time of an event except by reference to some other event, or the place of a body except by reference to some other body. All our knowledge, both of time and space, is essentially relative. When a man has acquired the habit of putting words together without troubling him- self to form the thoughts which ought to correspond to them, it is easy for him to frame an antithesis between this relative knowledge and a so-called absolute knowledge, and to point our ignorance of the absolute position of a point as an instance of the limitation of our faculties. Anyone, however, who will try to imagine the state of a mind conscious of knowing the absolute posi- tion of a point will ever after be content with our relative knowledge. 1 90 GRAPHICS CHAP. VII. out ; the field of the earth is that in which wind and ocean currents and railway trains, &c., are regarded as moving ; the field of the locomotive frame is that in which most com- monly the momenta, force?, &c., of the various parts of the machine are considered ; the water flowing through a turbine may be considered to move either through the field of the blades, which field is rotating along with the blades in the field of the fixed casing, or it may be regarded as flowing through this field of the fixed casing i.e. the field of the earth the word * fixed ' here meaning motionless in the field of the The earth. We hardly ever think of the motion of a watch-pinion 'Field' ag re i a tive to any other field but that of the watch-case, because it is that motion alone that has any great importance as regards the construction and action of the watch. It was probably only after the discovery that the rate of a watch depends slightly on the position as regards the force of gravity and as regards the force of terrestrial magnetism, that any one ever thought of the motions of its wheels relatively to the earth field. On the other hand, the motions of a ship propeller reckoned relatively to the ship and relatively to the earth are equally important, and, therefore, the recognition of the relativity of the motion in this case has always been familiar to engineers. It is the same in the case of turbines. 10. Since all vectors are thus relative to one or other field, there will be no occasion to use the phrase ' relative to.' If Through,' the vector is said to be ' through * a specified field, that will over',' mean that it is to be regarded relatively to that field. If it is said to be * past/ ' over,' or ' round ' a specified material body, such as the bed-plate or frame of a machine, that will mean that it is reckoned relatively to the ' field of that body.' 11. It is now evident that every possible motion or other Recipro- vector has a dual aspect ; or rather that the phenemenon of duality which the given vector is one aspect is also capable of being viewed in an exactly opposite aspect. It is necessarily a dual phenomenon, capable of interpretation by one or other of two CHAP. MI. VECTOR SUMMATION 91 equal and opposite vectors. Tims, while the propeller-blade has a certain motion through the field of the water, the water has a simultaneous exactly equal hut opposite motion through the field of the propeller-blade. While the pump-bucket is moving upwards in the field of the pump cylinder, the pump cylinder may be considered to be, and actually is, moving downwards in the field of the bucket. While the water of a river is flowing downwards through the field of the river bank, the bank is travelling upwards through the field of the water. While a crank is rotating right-handedly through the field of the engine bed-plate, the bed-plate is rotating with an equal left-handed velocity through the field of the crank. The ancient belief that the sun moved round the earth daily from west to east has exactly the same truth in it as the more common modern notion that the earth rotates daily about its own axis ; these are only the two opposite aspects of one and the same phenomenon ; the sun moves round the earth in the Eeci- field of the earth, while the earth revolves in the opposite direction in the field of the sun, the two axes of rotation being coincident. The two experiments that will produce the greatest personal conviction of the truth of this doctrine of the essentially dual and relative nature of motion are first, riding in a train side by side with another train moving through the field of the earth with only a slightly different speed from our own, and, secondly, going down and up a deep mine shaft with the cage moving at a considerable velocity. The un- familiar sensations that are experienced when these experi- ments are first made are not illusions, as they are often thought to be ; the} 7 are, on the contrary, the awakening of the mind to a consciousness of the fact that it has hitherto been resting in the habitual delusion that all motions necessarily take place in the field of the solid earth. These two experiments should be re- peated until the sense of strangeness wears off and the mind becomes habituated to, grasps familiarly, and makes part of its permanent instinct the true notion of the dual relativity of all motion. 92 GRAPHICS CHAP. VII. All other vectors are similarly the one-sided aspects of dual phenomena, the complete vector description of each of which consists of a pair of equal and opposite vectors through Becipro- two different fields. The student is left to trace out for him- auaiity se ^ tn ^ s principle of dual relativity or reciprocal duality through the series of phenomena corresponding to the different kinds of vectors mentioned above. 12. The graphic process of summation of vectors is almost self-evident. Since the vector properties do not depend on position, we may place the vectors where we please in order to effect their addition, provided only we keep their directions and magnitudes unchanged. They may, therefore, be placed so as to form a continuous chain, the beginning of each vector being placed so as to coincide with the end of the one preceding it in the chain. vector Thus in Fig. 44 the five vectors AB, B C, CD, D E, and E F, or , /3, 7, , and e, are arranged in this successive or tandem fashion. Here the vectors are added one on to the other, and the total result of this addition is the same as, or equivalent to, a vector from A to F i.e. with direction A F and magnitude measured by the length A F to same scale as that employed in plotting the separate vectors. This being the total result, the vector A F may be called the Vector Sum, or in symbolic language calling A F by the letter and its rotation in this earth- field round the same point A through the angle B^ A 2 B 2 Hamii- (call this angle 6} ; and, second, in the displacement of the n0 men- earth past the ship. In terms of the data of the last problem clature we have and 6 = The displacement of the earth past the ship might be given as the movement of the point of the earth originally at A 2 to A 1 combined with a rotation s round A 1 as axis. But the original position of A of the rod in the earth-field is not A 2 but Aj. Therefore, in order to follow out the exact con- verse of the procedure in the last problem we must have the earth displacement past the ship defined by the movement of its point originally at A r Call this movement 7. Now this point would move a e through the ship-field, if there were no rotation of the earth in this field ; but, seeing that there is the rotation s, the displacement is a e turned through 106 GRAPHICS CHAP. VII. the angle s in the ship-field. Thus, according to our pre- vious nomenclature, 7^ ( 1) v ( aj. Then according to the analogy of the result in the previous problem, namely, a displacement \a e + ( 1)* a s I along with a rotation (s + a) , we obtain from the present problem a dis- )- \ 7 + ( 1) $1 com- bined with a rotation (6 s). Writing these as above in terms of the data of the first problem, the displacement becomes Hamil- =H=( !) " I <*e ton's ( nomen- 4j_ a clature because a e -\-a e =o and ( 1) '"* = ( 1) = 1 ; and the rotation becomes s + a s = cr. Thus the whole displacement of the rod through the ship- field consists in the movement a & of the point A of the rod and the rotation cr round the same point A of the rod as axis. It will be a useful exercise for the student to draw out for the last problem the diagram corresponding to Fig. 47 for the former problem. Evidently this solution is equally true, whether the dis- placement of the body in the one field and the displacement of that field in the other field be simultaneous or successive in point of time. 25. It has been explained here only for co -planar motions. An exactly similar result can be obtained for non-planar motions in different fields. This case, however, can only be understood after closer consideration of the laws of the addi- CHAP. TIT. VECTOR SUMMATION 107 tion of rotors. It need not be given in this book because the graphic construction is not a convenient one ; the problem is best solved by other than graphic methods. What has been said regarding the addition of vector trans- Simni latory displacements of points or rigid bodies applies equally to the vectors we have called ' Integral Volume Displacement ' and ' Integral Mass Displacement.' menta 26. In none of these cases is there involved any consider- ation of time-rates. The changes occurring concern only the differences between the initial and final conditions of the quantities changed ; the consideration of the intermediate conditions of position, &c., is entirely eliminated from the problem. Let us now see what the addition of time-rate vectors means. The simplest of these is linear velocity. The average velocity during the interval occupied by any displace- ment is simply the quantitative comparison between, or the quantitative measure of the physical relation between, the dis- Time- placement and the time occupied. By analogy with purely numerical ratios between quantities of the same kind, it may be called the physical ratio between the displacement and the time, but it is not a pure ratio in the ordinary and strict sense of the word. When we talk of the velocity of a body at a certain instant we mean the average velocity during the very small interval of time referred to as that instant. When the simultaneous vector displacements of the different parts of a body or assemblage of bodies are added together as previously explained, no difference in the process will be effected if each displacement is divided by the interval of time in which all have occurred. This interval of time being common to all, all are affected by this division in the same ratio. The summation of the vector velocities now gives the velocity of the ' centre ' of the system multiplied by the number of equal parts into which it has been divided, and simul- this vector velocity of the centre may be obtained by dividing the vector- sum by this number of parts. If the parts be of 108 GRAPHICS CHAP. VII. Integral momen- tum Difference of velo- cities Velocities in differ- ent fields equal mass, the velocity of the centre of mass multiplied by the sum of the masses will be the integral vector momentum of the system. No physical meaning can be given to the vector summa- tion of the velocities of the individual parts of a system if these do not occur simultaneously, but at different periods of time. 27. The vector differences of the simultaneous velocities of the parts are the velocities of these parts relative to each other. Thus in Fig. 48, if a /3 and 7 be the velocities of the parts A B and C, and if from any point p chosen as pole, these vectors, a ft 7, be drawn to any convenient scale ; then the line marked (a /3), directed as shown by the arrow, is the velocity of A relatively to B ; the velocity of B relatively to A being the exact reverse of this. Similarly, (7 ft] and (7 ) with the directions shown by the arrow-heads are the velocities of C relatively to B and of C relatively to A ; the velocities of B and A relatively to C being the exact opposites of these. 28. When a body has a translatory velocity in one field, and that field has a simultaneous translatory velocity through a second field ; as, for example, when a portion of water flows with a certain velocity along a pipe in a locomotive, and that pipe at the same time moves with the locomotive with a certain velocity over the earth ; then the velocity of the body through the second field is the vector-sum of these two velocities. For if in Fig. 49 the two velocities be a and ft, then, calculating the displacement over the earth from the position in any time t, by means of oblique co-ordinates parallel to and ft re- spectively, and calling these co-ordinates a and b, we have a = t , and b = ftt, and therefore . = ^ = a constant ratio for different ', so long as the velocities and ft are main- tained constant, that is, during the interval of time, long or short, during which we reckon and ft to be the velocities of water through pipe and of pipe over earth. This ratio being constant, the locus of the displaced position is a straight line CHAP. vii. VECTOR SUMMATION 109 coinciding with the vector (a + fi) drawn through 0, that is, the water moves along this line in the field of the earth ; and as the length of this displacement along this line bears, by similar triangles, the constant ratio magnitude of ( 4- ft) magnitude of a to the length of the simultaneous displacement along the pipe (in the field of the pipe), therefore the velocity over the earth equals both in magnitude and direction the vector- sum (a + ft). Similarly, differences of simultaneous velocities through the Velocities same or different fields may be obtained graphically. Thus, en t fields if we know (see Fig. 50) a to be the velocity over the earth of the water entering a turbine, and ft to be the velocity over the earth of the part of the turbine which the water is entering, then ( ft) is the velocity of the water through the turbine, that is, relatively to the turbine. There is no physical meaning to be attached to the addi- tion or subtraction of successive velocities of one body in different fields. 29. But the successive velocities of one body in the same field may be compared by means of their vector differences. These velocity changes may be called time differences to dis- tinguish them from the simultaneous differences previously explained. If these differences be divided by the time during which they occur, the result of this comparison will be the average time-rate of change of velocity during that interval of Change of time. velocity Thus if (see Fig. 51) any point A of a body have at one time a position Aj and a velocity j ; if it move from here along the curved path A, A 1 A 2 to the position A 2 in the interval f, and have then the velocity 2 ; then, drawing from any pole p and to any convenient scale the vectors et l and 2 , the line marked ( 2 j) in the figure, directed as indicated by the arrow-head, is the change of velocity that has occurred 110 GRAPHICS Hodo- graph Sum of accelera- tions in the interval t. If this vector (* 2 * { ) be divided by t we obtain the average time-rate at which the velocity has changed during this interval. This time-rate is called the average velocity acceleration. 30. If the interval of time be taken small the change in the direction and magnitude of the velocity will be small in a corresponding degree, because there occur in nature no absolutely sudden changes of velocity. If a large number of velocity vectors occurring at successive small intervals of time be drawn from p, a continuous curve can be drawn through their extremities. This curve is called the ' hodograph ' of the motion. During any instant the velocity acceleration is evidently the linear velocity along the hodograph of the extremity of the velocity vector radiating from p. Thus the acceleration is continually tangential to the hodograph. Velocities reckoned from averages taken over only a minutely small interval of time may be called * instantaneous ' velo- cities. 31. Velocity accelerations are thus vectors, and can be added in the usual vector way. The addition of simultaneous accelerations in different fields being very important, we give here Fig. 52 in order to show clearly the truth of the state- ment that ' the acceleration of the resultant velocity is the resultant or vector-sum of the accelerations of the component velocities.' Let there be three velocities a b and c added together in one body, which in a given time change from 0, b l and c l to a 2 b 2 and c 2 . Call a 2 a l =(j= and b 2 b l ^j= /3 and c 2 - 0,4^7. From the end of b { placed as in the figure draw a. This leads to a point to which the vector from the end of a 2 is evidently b { . From this point draw/3; this will lead to a point to which the vector from the end of a 2 is 6 2 , because 6 2 r^jr 6, -f /?. From the end of c l as drawn in full line in the figure draw (a + /9). This leads to a point to which the vector from end of b 2 is o p because the end of 2 is (a + y9) CHAP. vii. VECTOR SUMMATION 111 away from the beginning of c { in full line. To the last ( + ft) plotted we now add 7, which leads, therefore, to a point to which the vector from the end of b 2 is c 2 , since c 2 4p c l -f 7. This point, therefore, gives a 2 + & 2 + c 2 from the point p, and it differs from a^ + b l -f c, by -f /3 4- 7. This is the proof of the proposition stated. The acceleration of velocity of a point may often be con- veniently split into two components : one along the line of motion, termed ' tangential ; ' the other normal to it, and called ' centripetal ' or ' radial.' Referring to section 30 for the meaning of the ' hodograph,' and calling the velocity along the hodograph diagrammatic (as being only that of an imaginary point in a constructed diagram) to distinguish it from that of the actual motion of a real point now spoken of, the tangential acceleration in the real motion is clearly the component of dia- grammatic velocity in the hodograph directed away from the fixed centre of the hodograph ; and the radial acceleration is Radial the component of the same diagrammatic velocity perpendicular g en tiai " to the radius of the hodograph. Let the linear velocity in the real motion be v, this being also the radius in the hodograph ; and let the radius of curvature of the path of motion be R. The angular velocity round the centre of curvature is then .= . K This is also the rate at which the direction of v changes, and is, therefore, the angular velocity of the hodograph radius. The linear velocity of the end of the hodograph radius resolved perpendicularly to the radius is, therefore, v x ?- = ^- = &> 2 R, if a) = - be the angular velocity. K This centripetal acceleration may be graphically calculated by any of the multiplication and division constructions shown in Fig. 8, Chapter III. Three special constructions for this purpose are shown in Fig. 66, Chapter IX. 32. The introduction of a mass factor into any of the vectors we have dealt with alters none of the additive pro- 112 GRAPHICS CHAP. VII. perties we have considered. Thus momenta, accelerations of momenta, forces, stresses, rates of flow of water mass, sand mass, or any other kind of mass are all subject to the same graphic laws of vector addition and subtraction as have already been explained and illustrated. These laws are so simple that the student runs little risk of making mistake. In applying these processes, however, to any new kind of vector Mass not previously dealt with, it is very necessary to inquire care- fully whether vector addition or subtraction has any real physical meaning in this new connection, and, if there be such a meaning, what it is exactly. The student should in no case be content to employ graphic processes without clearly and accurately understanding the physical interpretation to be put upon them. 33. Eotors are in some cases added in the same way as vectors, but only in special cases. Displacement rotors can- not be added in this way ; velocity rotors can. The difference arises from the fact that a velocity may be taken as a function of an extremely minute interval of time, or of an extremely minute displacement. Displacements are not in general minute. The vector mode of addition may be applied to dis- placement rotors provided they are extremely small. On the other hand, this process of addition cannot be applied to velocity rotors if these are average velocities through an Displace- interval of time or through a displacement not extremely tors small. In fact, the general graphic formula for rotor addi- tion reduces to coincidence with that for vector addition when the rotors to be added are minutely small. Thus the velocity rotors which may be added as vectors may be called ' instantaneous ' as distinguished from those that are obtained by taking averages over an interval of time longer than an instant. Suppose that, in Fig. 53, A B represents a rod which suffers first a rotation round an axis (supposed perpen- dicular to the paper) through A, this rotation displacing the CHAP. vir. VECTOR SUMMATION 113 point B of the rod to B', and which subsequently is rotated ft round a parallel axis drawn through B in its new position B'. Bisect the angles a = B A B' by A b and ft = A B' A' by B'a, and let these bisecting lines intersect in P. Then A P = A' P and B P = B' P, and, therefore, the actual double rotational displacement might be produced by a single dis- placement round a parallel axis through P. The angle A' P A = B' P B because the first is double a P A and the second is double b P B', while these two last equal each other, being formed by the crossing lines A b and a B'. But each of these last equals the sum P A B' + P B' A. Now P A B' = J a and P B' A = 0. Therefore A P A' = B P B' = * + ft. Thus a single rotation (a + ft) round the axis through P would produce the same whole displacement as the two suc- cessive rotations, and this, therefore, may be called the re- sultant or the rotor sum of the two component rotors. The position of P is most simply defined as the intersection to two lines through A and B making the angles \cx. and Dispiace- -J /3 with A B and B A. Considering the exact reversal of the whole process bringing the rod back from A' B' to AB, the same axis P can be obtained by drawing through B' and A' two lines making the angles ft and + J a with B' A' and A' B'. In this reverse process the rotation ft must first be performed round B' and then round A. If the rotation ft had been performed round B first, this would have brought A to A", and if the rotation a had then been performed round A" the new position of A B would have been brought to B". The rotor sum would then be a rotation ft + round P" instead of round P ; P" being found by drawing from B and A two lines making the angles + ^ ft and -iawithBAand AB. Thus in this case the axis of the rotor sum a + ft (a being first in point of time and ft subsequent to ) has not the same position as the axis of the rotor sum ft -f- . Notice that here the axes are defined as lying in given positions in the moving i 114 GEAPHICS CHAP. VII. body and are carried along with the body through the field in which the rotation takes place. In this case the magnitude (a + /3) = (J3 + ) of either of these rotor sums (a + /3) or (ft + ) equals the arithmetic sum of the magnitudes of the component rotors. This results from the parallelism of the two axes. Now suppose that the two axes of rotation are given fixed in position in the field through which the rotations take place, not in the field of the moving body. In Fig. 54, let A be the axis in the field of rotation round which the first rotation takes place. This brings point B of the body to B', B being the position in the field round which the second rotation /5 takes place. This second rotation round B brings point A of the body to A' and the point originally at B but now at B' from B' to B" The axis of resultant rotation is to be found by bisecting line A A' and drawing through the bisect- ing point a perpendicular to A A' ; drawing a perpendicular to Displace- B B" through the middle of B B", and taking the intersection P rotors of these two perpendiculars. But this last perpendicular passes through A', because the triangle B A' B" is simply the triangle BAB' turned round through /3, and in this latter triangle BAB' the sides B A and B' A are equal, so that the perpendicular to B B' through the middle of B B' passes through A. The per- pendicular through A on B B' makes the angle J a with A B. Therefore the line A' P makes the angle a with A'B. Also, since P B bisects angle A' B A, and since A' B = A B, therefore angle P A B = P A'B = *. Thus B AP = ~ i . There- fore, the position of the axis P is found by drawing through A and B lines making angles i and + i /? with A B and B A. Similarly, if /3 round B were performed first and <* round A second, the axis P" of the rotor sum (fi + ) would be found by drawing through B and A lines making angles ft and + \ a with B A and A B. The two axes P and P" are placed sym- metrically on opposite sides of the line A B at equal distances CHAP. vii. VECTOR SUMMATION 115 from that line. The angle of the rotor sum in either case is (* + P). It is evident that the axis of the rotor sum (a + 0) with the axes of and fixed in the field of the rotating body coincides with that of the rotor sum (/3 + ) with the axes of and /3 fixed in the field through which the rotation takes place coincidently with the initial positions of the previously mentioned axes in the body. Again, the axis of (/8 + ) with axes of /3 and a fixed in the body coincides with that of (a -f 0) with the axes of and fixed in the field of rotation. In all these cases the distances of the resultant axis P from the component axes A and B have the ratio AP _ sin j 13 B P sin i a* If the angles and are of opposite signs and nearly equal in magnitude, the resultant axis is very far off. If a. and are exactly equal and of opposite signs, the displace- Displace- inent becomes one of translation alone. rotors There is still a third case really more important than either of the others namely, that in which one axis is fixed in the field through which the motion occurs, and the other axis fixed in the moving body. For instance, if a crank revolves on an axis fixed in the field of the earth, and carries with it on a crank-pin a wheel or a rod which revolves round this pin through a given angle relatively to the crank, then the motion of the wheel over the earth corresponds with the third case of the problem now being stated. Referring to Fig. 53, it will be seen that the position of the resultant axis in this case is precisely the same as in that figure, with the proviso that the axis fixed in the field must be considered as the first taken in the problem of that figure. There is, in fact, in this third case no question of the order of the rota- tions ; the rotor sum of the component rotors is the same in whichever order they be taken. If A be the axis fixed in the field and the rotation round it, and if B be the axis fixed in i 2 116 GRAPHICS CHAP. VII- the displaced body and ft be the rotation round B, then the place of the resultant axis is found by drawing through A and B two lines making angles J a and ft with A B and B A. This position is the same whether the rotations be performed in one or the reverse order, or if they be simultaneous. The rotor sum is a rotor round this resultant axis equal to (a + ft) in magnitude. 34. Suppose now the axes to be inclined to each other, and to intersect. We have here also the three cases distin- guished as above with precisely similar variations in the result. In the first two cases the position of the resultant axis depends on the order in which the rotors are taken ; in the third case it is independent of that order ; the magni- tude or angle of the rotor sum is the same in all three cases, and does not depend on the order of rotation. In Fig. 55 let C A and C B be the two axes intersecting in C. Diagrams I. and II. are two views at right angles to each other, the first (I.) being taken backwards along the axis of , Displace- that is, from A towards C, or in the negative direction along rotors ^is axis ; the second (II.) being taken in the direction per- pendicular to both CA and CB, that is, the plane of (II). coincides with that containing the two axes of a and ft. Imagine a sphere of unit radius described about C, and suppose A and B be the points where the axes of a and ft intersect the surface of this sphere. The great circle per- pendicular to C A is marked in both views a a a. It appears a straight line in II. and a circle in I. The great circle per- pendicular to C B is marked bib, being a straight line in II. and an ellipse in I. and ft are taken both positive i.e. right- handed, as viewed in the positive directions along their axes. In (I.), a and ft appear left-handed because the view is taken in the negative direction as regards both axes. Through C A draw two planes making angles on the opposite sides of plane A C B. Through C B draw two planes making angles ft on opposite sides of the same plane B C A CHAP. vn. VECTOR SUMMATION 117 Let these intersect in C P and C P'. These lines C P and C P' are the resultant axes occurring in the three cases of the problem. For a line in the body originally coincident with C P would be shifted to C P' by the rotation a round C A, and would be shifted back again from the position C P' to C P by a subsequent rotation /3 round C B. These two rotations taken in this order, therefore, leave finally the line C P without dis- placement, and C P is therefore the resultant axis of a followed by /3, the axes being fixed in the field. C P' is the resultant axis of followed by /3 if the axes be fixed in the displaced body. Similarly, the rotations j3 and a taken in this order round C B and C A would shift a line originally coincident with C P' first to C P and then back again to C P'. This last is, there- fore, the resultant axis if the component axes be fixed in the field. Suppose the plane C P B to be fixed in, and moved along with, the body. The first rotation a round C A would bring this plane into the position C P' B' where the angle B' P' B Displace- nient equals double the angle BPa. After the second rotation rotors which swings B P' round C B back into the position B P, the above plane will make with B P C (that is, its original position) this same angle B P' B' or double a P B. This angle is, there- fore, the rotor magnitude of the rotor sum ( + /3), and its direction is right-handed or positive round the resultant axis C P. The angle a P B is the supplement of B P A, and double a P B is, therefore, one whole revolution minus double B P A. Since one whole revolution leaves the resultant displacement zero, a revolution less 2 x B P A taken right-handedly gives the displacement the same as would be effected by a left- handed rotation 2 x B P A. The total rotation may thus be considered 2 x B P A, but for many purposes it seems better to write it {2 TT 2 x BPA}, which indicates more unam- biguously the real magnitude and direction of the angle through which the body has been turned. 1 1 8 GRAPHICS CHAP. Til. The arc or angular distances of the resultant axis from the component axes have the ratio sinACP Sin 2 sinBCP" . a sin - This result can be graphically represented in another way, which shows a more evident analogy with vector summation. Draw Cp' perpendicular to plane B C A. Draw C b l perpen- dicular to plane AGP. Thus angle b { Cp' = \ . Draw C a 2 perpendicular to plane P C B, thus making angle p' C 2 = J ft. Make the angles b l Cp l = b' Cp f = p' C b l = \ a in the plane perpendicular to axis CA. Make angles a 2 Cp 2 = a' Cp' = p' C 2 , and the arc p } a^ = p' a' = p 2 a 2 = \ ft. Draw a great circle arc from a { to b { . Suppose the spherical triangle a l p l b { Displace- to be fixed in and moved along with the rotated body. The rotors fi rs t rotation a round C A will bring this triangle into the position of p'V because angle p { Q>p r = . The second rota- tion ft round CB will bring the side a' p' of this triangle to position a 2 p 2 because angle p' Cp 2 = ft. Suppose the triangle now to stand in the position a 2 p 2 6 2 . Draw a great circle arc from b l to a 2 . Comparing the triangles a l p l b l and a 2 p' 6,, since the arc sides a l p l = a 2 p' and p l b l = p' b l and the angle a, p l 6, = a' p' b' = a^p' b l ; therefore, the arc b { a 2 = b l a l and the angle a { b l p^ == a 2 b l p f f and b l a 2 p l = b { ci l p l = b 2 a 2 j> 2 . Therefore, the three arcs a { b lf b l a 2 , and a 2 b 2 all lie on the same great circle. Thus the triangle could be brought from its first to its final position by a single rotation through the angle a { C a 2 = 2 x ^ C a 2 round an axis perpendicular to the plane b { C a r It can be shown that this last axis coincides with C P previously found. Thus if the component rotations be represented to half CHAP. TII. VECTOR SUMMATION 1 1 9 size (i.e. J a = ) by arcs on great circles perpendicular to their axes, and these arcs be placed joining each other so as to form two sides of a spherical triangle, taken in the order in which the rotations occur, the great circle arc forming the third side of this triangle, taken in the direction from starting point of the first rotation arc to end of second rotation arc, will represent the true direction and magnitude (this also to Displace- half size) of the resultant rotation or rotor sum, the resultant ^^s axis being perpendicular to this last arc. By this last construction any number of rotors round axes intersecting in one point can be added by a precisely similar process to that of vector summation, there being sub- stituted for the straight line representations of the vectors great circle arcs on a sphere of unit radius to represent the rotors. 85. When the rotors have minutely small displacement- magnitudes, their arc representations in the last construction become practically straight lines. In this case the law of rotor summation becomes identical with that of the addition of vectors. The vector method of addition can be applied to the lines taken along the axes of the rotors to represent both Small magnitude and direction of the rotors ; because, in the case of adding two rotors of minute magnitude, evidently the axis of their sum lies in the same plane as the component axes, the three arcs to which these axes are perpendicular all lying on a minutely small spherical surface, i.e. practically in one plane. 36. Angular velocities are rotors referring to minutely small displacements, and are therefore subject to this graphic law. Angular momenta, accelerations of angular velocity or of angular momenta, and force couples are other examples of Angular rotors all falling in this class and all to be dealt with gra- phically by this vector method of summation. When the rotor magnitudes refer to minutely small dis- 120 GRAPHICS CHAP. VII. placements the angular distances of the resultant axis C P from the component axes C A and C B have the relation Angular BID A C P = velocities sin B C P a * If the axes be also parallel, the linear distances of these axes have the ratio . BP CHAP. viii. LOCOR ADDITION AND MOMENTS OF LOCOES 121 CHAPTER VIII. LOCOE ADDITION AND MOMENTS OF LOCOES AND OF ROTORS. 1. WE have already seen how to find the average, mean or * central ' position of a number of points, mass-particles, or volume-particles. The process consists in adding the vectors from an origin to the different units and dividing the vector sum by the number of units. We have also dealt with the problem to find the mean or central position of a number of parallel locors. The solu- tion consists in multiplying each locor by its perpendicular distance from a datum axis, adding these products, which are called the moments round this axis, and dividing the sum of the moments by the sum of the vector magnitudes. The graphic execution of this process has been fully explained in Locor Chapter VI., and from what follows it will be seen that it is moment only a special case of a general construction applicable to all locors, whether parallel or not. The moment of a locor round an axis perpendicular to it is the physical product of the locor and of its perpendicular distance from that axis. The magnitude of the moment is the arithmetical product of the magnitudes of the two factors in the product. The unit of the product is a quantity having a definite rotational direction round an axis which has definite direction and position. A locor-moment is, therefore, a rotor. But in the moment-product all indication of the definite position or direction of the locor is lost. From the moment 122 GRAPHICS CHAP vin. one can only deduce that the locor has one of the infinite number of possible directions and positions perpendicular to the axis of the moment. 2. While locor moments thus afford no indication as to the positions of the component locor s, still they afford means of finding the average or central position of a number of locors. If the moments round any axis chosen as a datum be summed up, and the moment sum be divided by the vector sum of the locors, a position is obtained defined by a distance from the Central axis and by parallelism to the vector sum. A locor taken at position this distance from the axis, parallel and equal to the vector sum, will have a moment round the axis equal to the above sum of moments. This locor may be looked on as the sum of the given locors both with regard to moments and with regard to vector sum. This equivalence with regard to moment and to vector sum is expressed by the sign ^fc. 3. The locor sum of two locors whose lines intersect lies in the line passing through this intersection. Thus in Fig. 56 let AB and BC be the two locors, and P the projection of the axis. From C draw C D ^ A B. Then BD^ AB + BC; and since B D considered as a locor lies through the inter- section of the lines of A B and B C, the proposition is that BD^AB + BC. The magnitude of the moment of AB round P is double the area of triangle A B P ; that of B C intersect- round P is double the area of triangle B C P ; and that of B D added 00 " round P is double the area of triangle B D P. The two triangles ABC and C D P having their bases A B and C D parallel, the sum of the heights of their vertices C and P above these bases equals the height of P above A B. Their bases being also equal to each other and to that of A B P, the area A B P = area A B C + area C D P. = area C D B + area GDP. Add area B C P to each side of this equation and note that BCP -I- CDB + CDP = BDP. We find area ABP + area B C P = area B D P. Therefore, taking moments round P, CHAP. vin. LOCOR ADDITION AND MOMENTS OF LOCORS 123 moment of A B + moment of B C = moment of B D, or The proof of the proposition is independent of the position of P ; it is true for every possible axis. Thus B D is the locor sum independently of, and without reference to, any special axis. 4. From this it becomes evident that any locor may be split into component locors at any point of its line, the Resoiu- magnitudes and directions of the components having the same relations to those of the resultant as in the case of vectors, and the component and resultant locors having their lines all intersecting in one point. 5. The above refers only to the case of components and resultant lying all in one plane perpendicular to the axis. If a locor be parallel to an axis, it has no rotational direction round that axis, and it is, therefore, said to have no moment, or zero moment, round that axis. If now a locor be oblique to an axis, the locor may at any point of its length be split into two com- ponents, one parallel to the axis and the other perpendicular to it. The former component has zero moment, so that the whole moment of the locor equals that of its latter component perpen- dicular to the axis. This convention or system of calculating moments corresponds exactly" with physical facts, as, for ex- ample, the total turning power or influence round an axis of a Axial number of different forces. It also enables the above proposition regarding locor summation, or composition and resolution, to be stated perfectly generally, no matter into how many components the resultant may be split up, or whether they all lie in one or different planes. If S be the sum, and ABODE be five com- ponents into which it is split up, such that the vector equation S=H=A + B + C + D-f E is satisfied, and if A, B, C, D and E be taken as lying along lines all passing through one point in the line of S, then, looking on these as locors, the equation Sf=A+ B + C + D + E 124 GRAPHICS CITAP. Till. Moments of forces, &c Co-planar locors added will be also satisfied, no matter which point on the line of S be chosen as the common intersection ; the meaning of this last equation being that the moment of S round any axis whatever (perpendicular, parallel, or oblique to S) will equal the sum of the moments of A, B, C, D and E round the same axis. 6. Thus a force F may be resolved into three rectangular components F x F y F 2 . If these three be taken as all acting along lines through any point in the line of F, the sum of their three moments round any axis whatever equals that of F round the same axis. Similarly with a velocity V, or an acceleration of velocity ; with a momentum or an acceleration of momentum, or any other kind of locor. 7. If there be a number of locors not meeting in one point, the following graphic process gives their sum or resultant very easily. Using the notation explained for parallel locors at end of Chapter VI., let in Fig. 57 the spaces A, B, C, D, E be separated by the lines of the locors AB, BC, CD, DE, whose magnitudes and directions are given in the vector diagram abed e; the vector sum being a e. At any point in the line of A B resolve this locor into two component locors A P and P B chosen in any two directions. In the vector diagram draw through a and b the two lines a p || A P and bp \\ B P. Then the magnitudes of these two components in the chosen directions AP and PB must be ap and p b. These two A P and P B may be taken as a locor substitute for the single locor A B. Now combine the locor P B with B C. The resultant is P C $=p c and taken through the intersection of P B and B C. Since A P + P B $= A B and therefore, Next add locor PC to C D. The sum is ^= p d ; and if P D be drawn || p d through the intersection of P C and C D, this sum will lie along the line P D. Then we find CHAP. vin. LOCOR ADDITION AND MOMENTS OF LOCOES 125 Similarly, from the intersection of P D and D E draw the line PE||pe, the vector sum of pd and de. Take along this line a locoY^=pe. If this locor be called P E, we have Therefore, We have now got AP + PE as a locor substitute for AB + BC + CD + DE. But we can combine A P + P E into one locor A E 4p a e and taken along the line A E drawn || a e through the intersection I of A P and PE. Thus the single locor AE, whose magnitude and direction are those of the vector sum a e, is the locor equivalent or sum of AB, B C, C D, D E. By this statement it is meant that the moment of A E round any axis whatever is equal to the sum co-planar of the moments round the same axis of A B, B C, C D and D E, JjJ^S the locor sum having also the same magnitude and direction as the vector sum of these same four. The space P is called the chain or the ' pen,' or * single pen ' (P) AB C D E, and the lines drawn from the pole p are called the ' pencil ' (p) abcde. The whole process may be thus indicated : ' draw the single-pen (P) A B C D E parallel to the pencil (p) a b c d e.' The construction as explained involves the assumption that all the locors lie in one plane. But if they do not do so, but are still all parallel to one plane, it will still apply so far as concerns moments about any axis perpendicular to that plane, because the moment of any locor round such an axis is not affected by its distance above or below the plane. 8. The construction is, however, not directly applicable to the summation of locors not all parallel to one plane. In this case the locors may be dealt with in either of two modes. In Non- the one construction the locors are each to be resolved into anTnon- two component locors. The resolution is to take place in each P arallel locors case at the point in which the locor line cuts any plane that 126 GRAPHICS CHAP. VIII. may be chosen as convenient for the purpose of this resolution, the same plane being adhered to for all the locors. The one component is to lie wholly in this plane ; the other is to be perpendicular to it. This furnishes two sets of locors. The first of these sets lie all in one plane and can be summed up by the construction already shown. The second set are all parallel but are not co-planar. They are to be reduced to a single resultant as explained later in this chapter (v. section 15, Fig. 58 b). If the lines of these two resultants intersect, the two can be added so as to give a single locor as the complete locor sum of the whole given set. If they do not intersect, it is impossible to reduce the system to any single equivalent locor. Keduced to its simplest form, its equivalent is either the above-mentioned pair of non-parallel and non-intersecting locors, or else a single locor combined with a rotor. The method of reduction to this last form is given at the end of the present chapter. u on . The second mode of dealing with this problem in locor anTnon- summation is to form three orthogonal projections of the parallel given set of locors. Projections on any three different planes locors will serve, but, especially in engineering practice, three planes at right angles to each other are most convenient, and these may be called the 'front elevation,' the 'side elevation,' and the ' plan.' The plan projections of the locors form a set all parallel to one plane, and by the construction already given can be reduced to a single locor parallel to the same plane, which single locor will be their equivalent so far as vector sum and so far as total moment round any axis normal to that plane are concerned. This plan resultant is not the total resultant of the whole set of locors. The whole is equivalent to this plan resultant taken at a definite distance from the plane of the plan combined with a locor perpendicular to the plan taken in a definite position. This latter locor will be represented by a point in plan, and the position of the point depends on the CHAP. vin. LOCOR ADDITION AND MOMENTS OF LOCOES 127 height above the plane of the plan at which the plan resultant is supposed to act. The front elevations and side elevations of the locors are treated similarly. Each is by the ordinary process reduced to a single resultant. In the front elevation the whole system of locors is represented by the front elevation resultant and a point ; and similarly in the side elevation. It remains to determine the proper positions of the points in each of the three views. In what follows the three views are referred to by the letters (TT), plan; (s) front elevation; and (cr) side elevation. In Fig. 58 a the three partial resultants are p v p e and p fft supposed to have been obtained from the projection components of the several locors by means of suitably chosen poles and single pens drawn with sides parallel to the pencils radiating from the poles. The components of p n and p e measured horizontally on the paper are equal; those of p e and p ff measured vertically on the paper are equal ; and that of p v measured on the paper vertically equals that of p v measured horizontally. The ends of p e are placed directly above those of p v and on parallel the same level as those of p o merely for convenience ; they are not necessarily so. OE, S and S' may be called ground lines; V the plumb line. S and S' represent the same line, and it may be sup- posed to be a level south line. E may be thought of as a level east line ; and V as a vertical line. The plan resultant p n may be looked upon as having been obtained by resolving into a horizontal and a vertical com- ponent each of the group of locors at that point of its line of action where it cuts a given horizontal plane, say at the level a b. All the horizontal components then lie in one plane and their resultant p n lies in the same plane, viz. at the level a b. The whole group of locors is now equivalent to p w at the level b a and a single vertical locor, the position of which is to be determined. In elevation (s) this system of two locors may 128 GEAPHICS CHAP. VIII. be considered as three by resolving p n , which lies along b a, into two; one along b a parallel to plane of (e), one perpen- dicular to same plane (i.e. really || S) through some point in b a, and a third vertical || Y. Since the second is a point only in (s) it contributes nothing towards the formation of p e . Therefore, since the vertical component together with that along b a gives p e , the former must act through a where b a meets p e ; i.e. in (e) it lies along line a c and in (TT) it is represented by some point in the same line c a. Now in (. vnr. LOCOR ADDITION AND MOMENTS OP LOCORS 139 20. Angular momenta are simply angular velocities mul- tiplied by masses. They form rotors which are to be dealt with precisely as angular velocities by the foregoing methods. The product of an angular momentum and a radius perpen- dicular to it is evidently a linear momentum. These are the only rotors whose moments have to be in- vestigated in ordinary mechanics. The product of two rotors with parallel axes is generally Moments an undirected quantity. For instance, that of an angular momfnt^ velocity and of a force-moment is a rate of doing work (horse- power). That of an angular velocity and the moment of an acceleration of linear momentum is a rate of increase of kinetic energy. That of an angular rotation and of the moment of an acceleration of momentum is a quantity of kinetic energy. That of an angular rotation and of a force- moment is a quantity of mechanical work done. 21. Suppose we have two locors along parallel but different lines, oppositely directed and equal in magnitude. The con- struction given above in Fig. 57 fails in this case. In Fig. 60 let A B, B C, a b, b c represent the two locors. Choose any pole p and draw A P || a p, and B P || I p, and C P || c p \\ a p. The lines A P and C P being parallel, do not intersect, and it becomes impossible to find a position for the locor sum or resultant. Evidently, however, the two final lines PA, PC still form the correct moment diagram for the pair of locors that is, the moment of the pair round any axis whatever is ipcor the intercept between these two parallel lines on the line drawn conpies r through the axis parallel to a b ; the scale being determined by the pole distance p (a I}. This moment has the same magnitude round all possible axes perpendicular to the plane containing the two locors. The resultant vector being zero, it has no direction. The system has zero vector magnitude and has lost the two other characteristics of locors viz. direc- tion and position. It retains, however, its character of giving a moment, this moment .being a rotor and having a definite 140 GRAPHICS CHAP. Till. Locor and rotor couples Equi- valent couples direction for its axis. The resultant of the two locors cannot be said to be a rotor, but it has lost all locor character except this of giving a moment which is a rotor. When the two equal and opposite locors are forces, the pair is called a ' couple ' or ' force-couple.' If the two equal and opposite forces be A B and C D, the couple is written (AB=(=CD). Again, if in the above figure A B, a b, and B C, I c, repre- sent equal and opposite angular velocities round parallel axes, the magnitude of the resultant angular velocity is zero and its axis has neither position nor direction. The two parallel lines P A and P C, however, continue to form a diagram of linear velocities of all points of the body that has these two simul- taneous angular velocities superimposed upon it. These linear velocities are all equal that is, the body is moving with a velocity of translation only. The angular velocity has become infinitely small and the distance of the axis of rotation infi- nitely great ; the product of the two, or the linear velocity, being in the limit the distance between the parallels A P and C P taken || a I. Here the combination of two rotors has lost all rotor cha- racter except that of giving a moment which is a vector. It cannot, however, be correctly said that the resultant of the two angular velocities is a translatory linear velocity, this latter really being the moment of the resultant zero angular velocity. The resultant of the two angular velocities may be called a ' rotor-couple.' 22. Neither a locor-couple nor a rotor-couple having any but moment properties, all couples giving the same moment must be considered as equivalent ; any one of an equivalent set of couples may be substituted for any other so far as vector or moment phenomena are concerned. Thus for the two equal and opposite forces at a given distance apart may be substi- tuted two other equal and opposite forces of different magni- tudes and directions from those of the original couple, provided CHAP. vin. LOCOR ADDITION AND MOMENTS OF LOCOES 141 the distance apart of the new pair is greater or less than that of the first pair in the inverse proportion in which the mag- nitudes are less or greater, and provided also that the plane of the second is parallel to the plane of the first pair. This latter condition ensures that the axes of the two couples have the same direction. A similar substitution may he made in the case of rotor couples. When students observe how freely this substitution is made use of in text-books of mechanics they are apt to imagine that the effects of these equivalent couples as applied to pieces of material are in every respect the same. They should care- fully avoid this error. The distributions of stress and strain throughout a piece of material are among the most important conditions that engineers have to inquire into. The states of Equi- s tress and strain in bodies due to the action of different but JJapfcg ' equivalent ' couples are wholly different. If the question to be investigated involves stress and strain, the above sort of substitution cannot be made. If only the bending moment on a certain section of material has to be found, then this substitution is permissible; but no problem of strength or stiffness is completely solved by the calculation of bending moments alone, and this viz. the case of bending moments is the only item of stress and strain calculation in which an ' equivalent ' couple may be substituted for the actually exist- ing couple. A similar warning may be given regarding the substitution of a * resultant ' of any kind such as are explained in this chapter. The resultant can only be substituted for the group of units for purposes relating to the ' vector sum ' or to the moments. The resultant applied instead of the actual forces would not have even approximately the same effect in stressing and straining the material. 23. In Fig. 61 let the plane of a couple be perpendicular to Resoiu- the paper ; let its forces be represented by a & and b a ; and let its moment be represented to a convenient scale by a m. am ti(m of couples may be taken as the axis of the couple ; it is parallel to the 142 GRAPHICS CHAP. VIII. Resolu- tion and composi- tion of couples General locor and rotor re- daction paper, and perpendicular to the plane of (ab^zba). Each of these two equal forces may he resolved in any two rect- angular (or ohlique) directions a c and c b. Then (a c ^ c a) will he equal and opposite forces, and will form a couple whose moment is a n obtained by drawing a n J_ ac and m n j_ c b. Also (c b ^ b c) are equal and opposite forces forming a couple whose moment is measured by n m. Again, a n and m n are perpendicular to the planes of these two couples (ac^c a] and (c b =$E b c) , and these may, therefore, be taken as their axes. It is now evident that couples may be resolved and compounded, subtracted and added in exactly the same way as vectors or instantaneous rotors, and that these operations may be effected by treating as vectors or instantaneous rotors lines measured along their axes (such as a m, a n, n m) to such lengths as will represent the moment magnitudes to any con- venient scale. These axial lengths are complete graphic de- scriptions of the couples, and the term * axis of a couple ' is usually employed to indicate this definite length perpendicular to the plane of the couple. 24. A locor-couple combined with a single locor R per- pendicular to the axis of the couple evidently reduces to a bingle locor, which is equal and parallel to R but along a different line. If a set of locors reduce to a couple and a single locor parallel to the axis of the couple, these cannot be added so as to give any simpler result. If the set reduce to a couple and a single locor B oblique to the couple axis, the couple may be resolved into two com- ponents, the axis of one component being parallel and that of the other perpendicular to R. The latter component combines with R to give a single locor 4j= R, and the whole set is now reduced to the above last-mentioned case. If the set reduce to two non-parallel and non-intersecting locors, Q and R, either of these, say Q, may be resolved into two components, one component being equal and opposite to R (i.e. ^R). This component forms with R a couple, and CHAP. vin. LOCOR ADDITION AND MOMENTS OF LOCOKS 143 the whole set is now reduced to a couple and a single locor, which is in general oblique to the axis of the couple, but by the previous method can be further reduced to a couple and a locor parallel to the couple axis. It is, however, questionable whether there are many cases in which there is any practical advantage to be gained by substituting a couple and locor (which together really mean three locors) for a system of two single non-intersecting locors. Two oblique non-intersecting locors may always be reduced to two other perpendicular non-intersecting locors. Thus, in Fig. 62, let a b be the one locor in the plane of the paper and General c d be the projection on the same plane of the other. This ro tor latter can be resolved at the point c where it cuts the reductloa plane of the paper into two components, of which the pro- jection c d is one and the other is perpendicular to this plane. c d and a b can be added, the sum being a locor efm the plane of the paper. The system is now reduced to ef in this plane and the vertical component of the second locor perpendicular to same plane and passing through c. One is at liberty to choose any plane that may be most convenient for this reduc- tion, and for this reason among others this solution may in many instances be more convenient and useful than the reduction to a couple and a perpendicular locor, for which latter there is only one definite plane possible for the couple. GRAPHICS CHAP. IX. CHAPTEK IX. KINEMATICS OF MECHANISMS. 1 . IN this chapter there are dealt with the relations between the velocities and accelerations of velocity of the various parts of a Mechanism. A ' mechanism ' may be denned as a combination of frames, plates, bars, or flexible members jointed together in such manner as to ensure that, while the parts may move relatively to each other, the relative positions of all parts are determi- nate for each given possible relative position of any two parts. The different rigid members of a mechanism may be termed ' links ' or ' bars.' The flexible members may be called * bands ' or * belts.' That bar relatively to which the motions of the other parts Bed-plate are measured is here termed the * base-plate ' or * bed-plate ' or ' frame.' The motions of these other parts are thus con- sidered as taking place in the ' field of the bed-plate ; ' and in this field the parts of the bed-plate itself have, of course, no displacement, velocity, or acceleration of velocity. 2. In the next chapter (Chapter X.) there is explained the relation that must subsist between the number of joints and that of links in order that a flat structure may be ' stiff ' i.e. non-deformable without stretching, contracting, or bending the links. The relation is Definition of me- chanism ' Joints and bars where I = number of links and j = number of joints. When a link has more than two joints in it, only two of them must be included in the count ofj; and such a link, which is called a beam-link, is to count as one only. CHAP. ix. KINEMATICS OF MECHANISMS 145 If a flat pin-and-eye linkage have a number of links less than the above, it is no longer stiff and can be moved into different shapes. If the number of links be one less than the above, or I = 2j - 4, there is one degree of freedom in the possible motion. This means that, although there is an infinite number of possible shapes, still for each definite position of two parts relatively to each other there is a definite relative position of every other pair of parts. This agrees with our definition of a mechanism. The definition of a machine will be given in Chapter XIII. A machine has two more bars than a mechanism with the same number of joints, and two of its bars are not rigid, but are capable of resistant deformation. In a pin-and-eye joint mechanism the relation between the numbers of links and of joints is, therefore, In a quadrilateral, for instance (see Fig. 67), which is the joints and simplest mechanism, there are 4 joints and 4 links, and 4=2x4-4. In applying this rule only two joints are to be counted to each bar, no matter how many may actually exist in it. Thus in Fig. 70 there are 6 bars ; there are also 7 joints, but two of the bars have in each 3 joints, so that the number of joints to be counted in order to apply the criterion is 5. Now 6 = 2x5-4, and the mechanism has, therefore, definite motions. A pin-and-eye, or full and hollow cylindric, joint is called by Eeuleaux a * lower pair,' as distinguished from the joint between two spur-wheels in gear with each other, which latter is called a ' higher pair.' The above criterion-equation re- quires modification for mechanisms including higher pairs. For each * higher pair ' joint existing in the mechanism, the number of bars becomes one more than that given by the L 146 GRAPHICS Rigid-bar mechan- isms Motion, paths Displace- ment diagrams above rule. This is evident because a higher-pair joint may be looked upon as a substitute for two lower-pair joints with a link between them. The suppression of one of these joints would correspond to the suppression of two links, but actually only one link disappears in the conversion of the link and two joints into a single higher-pair joint. 3. The motion of any one point in a bar relatively to any other point in the same bar may be conveniently resolved into two components : one along the line joining the two points, which will either lengthen or shorten the dimensions of the bar along this line ; the other at right angles to this same line. If the bar be * rigid,' that is, if its dimensions be incapable of altera- tion, the former component is always zero. Eigidity does not interfere with the latter component taking any value whatever, great or small. Thus, in a rigid bar, the instantaneous dis- placements and velocities of the various pairs of points rela- tive to each other are all normal to the lines joining the pairs of points, and all parts have the same angular velocity. 4. The graphic determination of the simultaneous positions of the various bars of ordinary mechanisms is easy. When difficulties occur, as in some reversing link-motions, the methods here explained enable them to be readily overcome. The loci of the successive positions of the various parts may be called the 'motion curves,' or more simply the 'paths,' of these parts. These paths are drawn in on the ' mechanism diagram.' 5. From these ' paths ' the displacements from any assumed initial configuration can be directly measured. It may, how- ever, be often advantageous to have a separate ' displacement diagram,' consisting of a series of curves showing the succes- sive simultaneous displacements of all important points of the mechanism as vector-radii from one and the same pole. These curves in the ' displacement diagram ' are, of course, exact copies of the ' paths ' in the ' mechanism diagram.' Let ABC and BDE in Fig. 63 be two bars jointed together at B. Let P' be the pole of the displacement dia- CHAP. ix. KINEMATICS OF MECHANISMS 147 gram, and let the curves A' A', B'B', D'D' be the displace- ment curves of the three points A B D. All these curves, of course, pass through the pole F. P'A'p P'B' P and P'D^, being simultaneous displacements of AB and D, draw on A.\ B^ and B^ T>\ as bases, the triangles A^ B' t C\ and B^D'jE^ similar to the triangles ABC and BDE in the mechanism. It can easily be shown that P'C'j and P'E'j are the simultaneous displacements of C and E if the two bars be rigid. By joining all the points C' and E' found by such constructions, the displacement curves of C and E for a rigid-bar mechanism can be drawn in. Numerous simul- taneous points on the various displacement curves should be marked, and numbered 1, 2, 3, 4, &c. The advantage of such a displacement diagram over the set of ' paths ' dispersed over the mechanism diagram consists in the greater facility of comparison between the dis- Diepiace- placements of the various parts of the mechanism that it digrams affords. Thus, comparing simultaneous points A' and C' belonging to the same bar, the vector A' C' is the displacement of C past (or relatively to) A in the base-plate field. Con- versely, C'A' is the displacement of A past C in the same field. The same holds for points in different bars ; thus, D' C' is the displacement of C past D in the base-plate field. It may be noted that any line, such as A' C', belonging to one bar is perpendicular to the line bisecting the angle between the simultaneous and ' initial ' positions of the line A C in the mechanism. This does not, however, apply to a line, such as E 7 C', joining points belonging to different bars. This latter vector may be looked on as the sum (E'B' + B'C'), to each of which components the above rule applies, although it does not apply to the resultant. 6. The method of obtaining the velocities by taking the Differ- small differences of the displacements, which method is the velocity basis of kinematic analysis developed by means of the differen- tial calculus, has often been adopted as a graphic process for L2 148 GRAPHICS CHAP. IX. the solution of specially complicated problems. After con- structing the velocity hodographs, the same method may be followed to find the velocity accelerations. As a graphic process, however, this method is capable of no accuracy ; it s, in fact, practically useless. Ken- Professor Keuleaux's method of centroids, more properly centrdids can " e d axoids, has now become famous. These axoids are, how- ever, very tedious of construction, and when constructed furnish no direct easy means of measurement of useful quantities. 7. The method now given furnishes velocity and accele- ration diagrams, somewhat similar in appearance to stress diagrams, which show the true directions and magnitudes to scale of the velocities and velocity accelerations of all points in the mechanism ; there being one pole only for each diagram from which all vectors radiate, so that the velocities or accele- rations of all parts and at all times of the complete cyclic period can be compared with maximum facility. Fig. 64. Let A B C D be a rigid bar. Suppose the velocity of A over the base-plate P to be known. Choose any pole p, and draw p a parallel to the velocity of A, and of a length to represent its magnitude to any scale considered convenient Polar for the velocity diagram. If now the angular velocity &> of the diagram ^ ar ^ e a ^ so known, a & mav ^ e plotted perpendicular to AB, and equal in length to o> A B to the above velocity scale. Then, p b is the velocity of B over the base-plate. If, instead of o> being known, we know the direction of the velocity of B, then p b may be drawn from p in this known direction to intersect in b the line a b drawn _L A B from the point a. This gives p b the velocity of B ; and the angular velocity of the bar may, if desired, be calculated by dividing a b by AB. Since the (relative) velocity of C round A is perpendicular to AC, and its relative velocity round B is perpendicular to B C ; if a c and b c are drawn perpendicular to A C and B C, their intersection c gives pc the velocity of C through the base- plate field. Similarly, p d is found to measure the velocity of CHAP. ix. KINEMATICS OF MECHANISMS 149 D. The diagram gives not only the velocities over the base- plate P, but also all the velocities of pairs of points relative to each other. For instance, bd is that of.D round B, and d b is that of B round D, these relative velocities being through the field of the base-plate P. It is clear that the figure b a c d forms a diagram of the bar B A C D to a diminished scale and turned through a right angle in the direction of . It may be called the ' velocity image ' of the bar. Further, on this new diagram of the bar, altered in scale and rotated through 90, the pole p represents the position of the instantaneous axis of rotation. Theoreti- cally, the original diagram A B C D, with the position P of the instantaneous axis added, would serve equally well as a dia- gram of velocities, the scale being chosen suitably so that P A would represent the velocity of A. In fact, this construction is really the gist and basis of Keuleaux's method above mentioned, various modes of finding the instantaneous axis Velocity being adopted according to circumstances. But for practical image graphic construction it cannot be so used, for several reasons. First, the usual variation of the position of the instan- taneous axis is extremely inconvenient, and in almost all mechanisms this axis periodically recedes to an infinite dis- tance. Secondly, the scale to which the diagram P A B C D could represent the velocities is always varying throughout the periodic motion of the mechanism ; it is always necessarily an awkward scale to measure to, and it periodically becomes in most cases an impossible scale by becoming infinitely large. Thirdly, the various bars of a mechanism have all different instantaneous axes, and the scales of the velocities would be entirely different for the different bars. As will be shown presently, in the method explained in this chapter the velocity diagrams of all the bars of even the most complicated mechan- isms are grouped together so as all to radiate from one pole, and so as to be to the same scale for all the bars and at all times throughout the periodic motion of the mechanism. 150 GKAPHICS CHAP. ix. 8. A similar construction is applicable to accelerations of velocity. In Fig. 65 let A B C D be one rigid bar. Let the acceleration of velocity of point A through the field of the base- plate be known, and represented in direction and magnitude by p' a' drawn from any convenient pole p' to any convenient acceleration scale. The acceleration of B can be obtained by adding to the vector p' of the vector acceleration of B in its relative motion round A. If &> be the angular velocity of the bar in the base-plate field, and if a/ be the acceleration of an- gular velocity, the radial or centripetal component of velo- city acceleration is w 2 A B and the tangential component is d A B. Thus plot A 7 = 1, and from 7 plot backwards to- wards A the magnitude 2 to any convenient scale. From the point so obtained plot to same scale o>' JL A B to the point . From B draw B ft || 7 a to meet A a in ft. Then Eft equals the acceleration of the velocity of B round A. The whole Acceiera- acceleration of relative velocity is, therefore, AB'Vw 4 + , and, therefore, also the same &>' ; and, since the magnitude of the acceleration of one point round any other is proportionate to the distance between them (e.g. that of B round A is oc A B), therefore, if a' V be drawn inclined to B A at the angle tan"" 1 2 and of length A B */ a>* + a/^, and if the figure a' b f c' d' be made similar to that of the bar A B C D, then p' b r , p' c', p' d', will be the accelerations of the points B C D in the base-plate field. Further, a' c r , for instance, is the acceleration of C round A. In the graphic construction it is simplest to plot a' ft' = AB o> 2 , and parallel to B A (not A B), and ft' b' = AB CD' perpendicular to A B and in the direc- tion given by the sign of o>'. The radial component is usually obtainabfe from the already constructed velocity diagram, CHAP. IX. KINEMATICS OF MECHANISMS 151 where the velocity of B round A is called a b, and the radial acceleration is therefore ^r-n-- Fig. 66 gives the two most ready graphic constructions for calculating ^? U In Figs. (1) A. -D and (2) the velocity a b is plotted along B A from B to #, to- wards A in (1), and away from A in (2). From B as centre with B /3 = a b as radius, a circular arc is struck intersecting Acceiera- in , any other radius from A. From /3 is drawn fi fi', diagram parallel to that other radius, and intersecting B /3 L in ft'. Then B/3' is the radial acceleration feSL In (3), (ab) is A 15 plotted from B as B y3 perpendicular to A B, and a circular arc with centre in B A is struck through A and /3. This arc cuts the diameter A B in ft' giving B /3' the desired radial acceleration. 9. If the bar be plotted to the scale m" = 1 inch, m being a fraction ; and if the velocity be plotted to the scale n" = 1 inch per second; then such constructions give the acceleration to the scale ( ) inch = 1 inch per second per second. If it Scale of \m / accelera- be desired to diminish the size of the acceleration diagram Diagram from this scale in the ratio q, it is only necessary to make the constructions of Fig. 66 with the radius A B increased in the same ratio q. To increase the size of the acceleration diagram, in any ratio, diminish A B in Fig. 66 in the same ratio. 10. It is evident that the figure a' b' c f df of the accelera- tion diagram is simply a reproduction of the figure A B C D of the bar altered in scale and rotated through an angle I 180 tan ~ l } in the direction of o>, where, in tan" 1 . Acceie- ft) 2 ) o> 2 ration the sign of a/ is to be taken positive or negative according as L it is in the same or the opposite direction to that of a. The figure a'b'c'd' may be called the ' acceleration image ' of A B C D. In this image p' is the point of the bar if the bar extends so far, or of its field if it does not extend so far, which suffers no 152 GEAPHICS CHAP. IX. Accelera- tion image Integral momen- tum. Accelera- tion of momen- tum Nomen- clature acceleration or is moving uniformly in a straight line. This point does not in general coincide with the instantaneous axis of rotation. If the velocity diagram were rotated and altered in scale, and placed on top of the acceleration diagram so that a'U c' d' coincided with abed, then p' p would represent in direction and magnitude the acceleration of that line in the bar or in the field of the bar which is at any time the instan- taneous axis. Both the velocity image and the acceleration image of a rigid bar moving without rotation, i.e. with a motion of trans- lation only, reduce to a single point. The smaller the rota- tional velocity the smaller will these images be. 11. If G be the centre of inertia of the bar and the similar points g and g f be plotted in the velocity and acceleration diagrams, then the products of the bar-mass by pg and p'g' are respectively the integral momentum and the integral acceleration of momentum of the whole bar. 12. In what follows the capital P will denote the base-plate through whose field the velocities, &c., are reckoned. The pole of the velocity diagram will be called p. The pole of the acceleration diagram will be called p'. Points in the mechanism will be named by capital letters, ABC, &c. The corresponding points in the velocity diagram will be named by the same letters in small type, a be, &c. ; so that p c will denote the velocity of C over the base-plate and b c the velocity of C round B, and c b that of B round C. The corresponding points in the acceleration diagram will be named by accented small letters ; this being in accordance with the common mathematical convention, whereby x f repre- dx sents ^. In finding, for instance, the point 6, it is sometimes neces- sary to find other points which are not afterwards required in the completed diagram. When such construction points CHAP. ix. KINEMATICS OF MECHANISMS 153 need to be named, they will be called ft l /3 2 , &c., if used to Nomen- find b in the velocity diagram, and ft\ /3' 2 , &c., if used to find l) f in the acceleration diagram. In the displacement diagrams described above, the accented capitals A'B'C', &c., are suitable. 13. The simplest mechanism is that with four rigid bars and with two joints, P^, in the base-plate, and two joints A B coupling the other three bars together. An example is shown in Fig. 67, the calculations being made for five dif- ferent phases of the periodic motion. The velocity of the crank-pin A is supposed known at each phase. From any pole p, and to any convenient scale, this velocity p a is plotted perpendicularly to P t A. From p a line is drawn perpendicularly to P 2 B. Evidently the extremity b of p b, the velocity of B, must lie in this line. But also p b ^= p a plus a velocity perpendicular to A B. Therefore, from a a line is drawn perpendicular to A B to meet the above line in b. Thus p b is determined. In the example p a is taken of the same magnitude at all the five phases. To obtain the acceleration diagram we assume the accele- Four-bars ration of A. On the supposition that pa is constant in magnitude, this acceleration is also constant in magnitude (v d) 2 = ^-{ , and is wholly radial. "i A From any pole p r this acceleration ~-^ = p'a' is plotted * i A parallel to APj (not to P! A). The calculation of the magnitude is performed by the graphic construction previously explained. By the same construction the magnitudes \^ ^ and Y-_; f * ne radial com- Jr 2 > A.D ponents of the accelerations of B round P 2 and round A are found and plotted off from p' (as p' ft') and from a' (as a' ft') parallel to B P 2 and to B A. From these two points ft' thus obtained, lines are drawn perpendicular to B P 2 and to B A. The point b' sought for must lie on both of these last lines, 154 GRAPHICS CHAP. IX. and is, therefore, at their intersection. The acceleration p' b' of the joint B through the field of P is thus obtained for the five different phases of the motion. The method of procedure is plain. Each joint of the mechanism is a point in two different bars, and therefore the calculation for that joint may be approached, as it were, from two different sides. In each of the two calculations there is one element missing, and the Four-bars last stage of the calculation cannot be completed directly ; for example, approaching the calculation of the acceleration of B through A, we can calculate the radial component (parallel to B A) of the acceleration that has to be added to that of A ; while of the tangential component the direction only is known, but this gives a line in which the desired point must lie. Another conditioning line being similarly found by approach- ing the calculation in another way, the point is found at the intersection of these two lines. 14. In the ordinary steam-engine with guide bars, the radius bar B P 2 swinging in the base-plate bearing at P 2 is re- placed by the cross-head sliding in straight guides which form part of the base-plate. The effect is the same as if B P 2 were infinitely long. On account of the cross-head joint being guided in a straight line passing through the crank journal centre, a symmetry is given to the motion which materially lightens the labour of drawing complete velocity and accelera- tion diagrams. Fig. 68 illustrates this. The angular velocity of the crank is taken as uniform. Therefore, the linear velocity of A is uniform and the points a all lie in one circle whose centre is p. Here the four positions 1, 2, 3, and 4 of the crank-pin A are taken equidistant from the dead-points and 0'. There- fore, the two cross-head positions B t and B 4 coincide, as do also B 2 and B 3 . Therefore also the four velocities pa lf pa 2 , p a 3 , and p a 4 are equally inclined to the velocity line p b, and the four points a p a 2 , a z , a 4 are equidistant from the line pb. Also at 1 and 2 the connecting rod has the same inclination Steam- engine mechan- ism CHAP. IX. KINEMATICS OF MECHANISMS 155 to the centre line, which inclination is equal and opposite to that at 3 and 4. Thus the lines a^ 6 P a 2 6 2 , a 3 b 3 , and a 4 Z> 4 are equally inclined to p b, the inclinations of the former two being opposite to those of the latter pair ; and, therefore, the velocities p & 4 and p b l have equal magnitudes, as also have p& 3 and_p& 2 . Therefore also the radial accelerations a\ft' lt a\ /3' 4 have equal magnitudes, as also a\ /3' 2 and a' 3 /3' 3 , and are equally inclined to p' b f on opposite sides ; while also the tan- gential accelerations, /3\b f } , &c., are equally inclined to the same line. Therefore, finally, b\ coincides with b\, and Z/ 2 with b' 3 , but p r &' 2 =y b' 3 differs in magnitude as well as direc- tion from p' b' 4 =p' b' r This symmetry is, of course, destroyed by want of unifor- steam- rnity in the rotation of the crank. mechan- The joint lines of the bars of a mechanism, the velocity ism lines, and the acceleration lines need be drawn in full for one position only. The results for the other positions are indi- cated by numbered points on the three sets of curves, which are the loci of the corresponding points or extremities of lines. The first set of curves are the paths of motion of the joints. The second series of curves are the hodographs of the velocities of these same joints. The third series are the loci of the ex- tremities of the lines representing the velocity accelerations. In Fig. 68 the velocity and acceleration diagrams marked (C) (D) are constructed in this way, and show the velocities and accelerations of the joints A and B for thirty-two equidistant positions of the crank-pin. 15. Six-bar motion is nearly equally easy to deal with by positions this method. The first example given in Fig. 69 is quite simple, because the velocity p a of the joint A is assumed as known, the bar Six-bars P! A being one of the quadrilateral Pj A B P 2 . The determin- ation of the velocity p b is, therefore, the same as that given already. Thus p b and a b are drawn perpendicular to P 2 B and A B, and their intersection gives 6. Then the triangle 156 GRAPHICS CHAP. IX. abc is made similar to ABC. p d is then drawn per- pendicular to P 3 D, and cd to CD, the intersection giving pd the velocity of D. To find the velocity of E, there are drawn p e and d e perpendicular to P 3 E and D E. The con- struction of the acceleration diagram here offers no special difficulty. 16. The solution of the next example in Fig. 70 is not quite so direct, because here the velocity assumed as known, namely, p a that of A, is that of a joint in the pentagon P l A B C P 2 . First, p a is drawn of the known magnitude and perpendicular to P! A ; and then a ft of indefinite length perpendicular to A B. Then p 8 and p c are drawn of indefinite length perpen- dicular to P 3 D and P 2 C that is, in the directions of the velocities of D and C. The points b c and d now sought for are known to lie on the lines a ft, p c, and p S, and also it is known that the line joining b and d is perpendicular to B D. Any point ft on a ft is chosen, and from it ft 8 drawn perpen- Six-bars dicular to B D ; and then the triangle ft S 7 is constructed similar to BDC, corresponding sides being perpendicular. The triangle bdc that is sought for must evidently be similarly placed to ft B 7 between the lines p B and a ft. Other- wise stated, the locus of 7 for various positions of ft and S is a straight line passing through the intersection of the line aft and p 8. Therefore, 7 is joined with the intersection of p 8 and a /3, and this line is produced to intersect p c me. This gives the true position of c, and the triangle d c b is then com- pleted by drawing c d and c b perpendicular to C D and C B to meet p B and a ft. The point e is obtained by drawing p e and d e perpendicular to P 3 E and D E. The acceleration diagram has in this case to be con- structed according to a similar indirect method. The accelera- tion of A being supposed known can be plotted at once. Then the radial components of the accelerations of B round A, of C round P 2 , and of D round P 3 , are calculated and plotted off from a' and p' in their proper directions namely, parallel to CHAP. ix. KINEMATICS OF MECHANISMS 157 BA, CP 2 , and DP 3 ; their magnitudes being ^*, ^, JJ A JT 2 L> y^-yl. From the three points so obtained, three lines, which r 3 D we may call fi', 7', and ', are drawn of indefinite length per- pendicular to BA, P 2 C, and P 3 D. The tangential com- ponents of the above three accelerations lie along these lines, which, therefore, contain the three points 6', c', and d f sought for. On the line &' any two points, B\ and S' 2 , are chosen, and from each the centripetal acceleration of C round D C .D is plotted parallel to C D ; and from the two points thus ob- tained are drawn two lines perpendicular to C D to meet the line 7' in two points, say 7/ and y 2 '. On the two bases S/ 7/ and S 2 ' 7 2 ' are constructed two triangles similar to D C B, whose two vertices may be called /3/ and /?/. Neither of these points /3/ fa' will be found to lie on the line fi', and their distances from this line may be taken as measures of the six-bars errors involved in the two guesses S,' / at the position of df. The error thus found in the resulting position of ft' is a linear function of the error in the guessed position of ' ; and, there- fore, the interpolation between these two errors in order to reduce them to zero is to be performed by simple proportion. This linear interpolation is at once effected graphically by drawing a line through fi^ and /3 2 ', and producing it until it meets the line fi'. The intersection thus found will be the true position of b'. Or, otherwise, the two error-distances of /3/ and fi 2 r from the line fi' may be plotted off from the points / and &/ perpendicularly to the line &' (or both in any one direc- tion inclined to S'), and through the two points thus obtained a straight line is drawn to cut the line S' in the true position of d'. From d' or b' thus determined, the other points are constructed as usual. This construction of the acceleration diagram of Fig. 70 will be easily followed without further explanation. From 6', found as above, is plotted off '- CB 158 GRAPHICS CHAP. IX. parallel to C B, and from the point so obtained is drawn a line perpendicular to C B to meet the line 7' in c'. Then on V c' as base is constructed a triangle similar to BCD, the other corner of which is marked d' and should fall on the line 8' if the drawing has been accurate. six-bars This indirect method of ' two trials and two errors ' and linear interpolation between them is adopted in drawing the velocity and acceleration diagrams for the ordinary forms of steam-engine reversing link motion. These diagrams could not be obtained except by this method. 17. In the common steam-engine mechanism we have already had a case of one bar sliding on another namely, the cross-head sliding in the guide-bars of the bed-plate. A cir- cular slot in which sliding takes place may, of course, be looked upon as simply an incomplete pin -joint of large size, the radius of the pin becoming infinite when the slot is straight. But when the radius of the slot is large, this manner of regarding the joint is not practically useful owing to the very distant position of the centre. A more direct application of the present graphic method to sliding joints is effected thus : If B be a bar sliding over the bar A, the differ- Four-bars e nce of the velocities of any two touching points in B and A sliding- is a velocity parallel to the slide- surface, or * guide-surface.' Thus, the velocity of the ba.r A being known, the velocity of any point in B can be obtained by adding to the velocity of any touching point in A a velocity parallel to the guide- surface, and further adding a velocity perpendicular to the line joining this touching point with the point in B whose velocity is to be found. This last added component is that due to the rotation of B in the field of the base-plate. If the touching surface of B has the same shape as that of A, so that B always ' fits ' on to all parts of A into contact with which it comes, and if during the sliding these fitting surfaces are forced always to lie close together, then the angular velocity of B is different from that CHAP. ix. KINEMATICS OF MECHANISMS 159 of A only in consequence of the sliding that occurs in the slot. In this case, if the velocity of A be completely known, the linear velocity of any point in B can be calculated by adding to the velocity which the point would have if B were rigidly attached to A, a velocity parallel to the nearest part of the guide-surface that is, perpendicular to a normal let fall from the point on the guide-surface. In the illustration (Fig. 7 la) the velocity of point A round P, is supposed to be known, and it is plotted as p a. The slot is a straight one. Then p /3 and a /? are drawn perpen- dicular to P! B and A B. This gives p ft the velocity that point B would have if the cross-head were rigidly attached to the guide-bars, and if ft b be drawn parallel to the slot the point b must lie in this last line. But B is guided by the radius rod P 2 B. Therefore, pb is drawn perpendicular to P 2 B to meet y3 b in b ; then p b is the velocity of B in the P field, and @ b is the velocity of sliding in the slot. In Fig. 716 Four-bars the slot is curved. There is first found p fi in same manner ^iai ng as in Fig. 7 la from the known velocity of A, and p b is drawn J oint perpendicular to ~P 2 B. From ft is drawn ft b parallel to the guide-surface at C, the foot of the perpendicular from B on the guide surface. Note that C is an actual point of the guide- surface only if the curvature of this latter be uniform. If it be not uniform, then C must be taken in the extension of the circular arc at D which is coincident with the small portion of the guide-surface over which actual contact takes place. Such guide-surfaces in mechanisms are commonly of uniform curvature, because this is an evident condition of continuous close fit between the two sliding bars. If a block C (see Fig. 72) slide in two slotted bars A and B, the first of which has a translatory velocity p a, and the second a translatory velocity p b, evidently the method of finding the velocity p c of the block is to draw from a and b two lines parallel to the two slots in A and B. If these lines meet in c, then p c is the velocity required. 160 GRAPHICS CHAP. IX. If the slotted bars have rotational instead of purely trans- latory velocities, then precisely the same construction is to be followed, making p a and p b the linear velocities of the touch- Four-bars ing points of the guide-surfaces in A and B. Now, however, it sliding- is evident that one and the same block cannot constantly fit joint close to both slotted guide-surfaces. But if two fitting blocks, one fitting the one slot and the other the other, be pinned together, then the above construction may be applied to find the velocity of the centre of the joining pin, and from the velocity of this centre it is easy to deduce by methods already explained the velocities of all other points in the two sliding blocks. 18. The following application (see Fig. 73) of the con- struction for sliding motion to toothed-wheel gear well illus- trates the complete generality of the method. The sketch represents four wheels, P A A, P B B P P B B 2 , and P c C, pinned to the base-plate at P A , P B , and P c . The point A of the first touches the point E { in the second, the two surfaces having here a common tangent to which the line (A BL) T AB is drawn normal. The third wheel being mounted on the same shaft as the second, these two are to be looked upon as forming, along with the shaft, one bar of the Toothed mechanism. The third and fourth wheels touch at the common point (B 2 C), and the line (B 2 C) T BC is drawn normal to the common touching surface. The points T AB and T BC are in the centre lines P A P B and P B P c . The velocity of the wheel A, and therefore of its touching point A, is supposed known, and this velocity is marked off as p a from any pole p, the line p a being drawn perpendicular to P A A. Then^! and ab l intersecting in b { are drawn perpen- dicular to P B Bj and to Bj T AB . This gives p b { the velocity of B! and a b { the velocity of sliding of one tooth over the other. Then p b 2 and b { b 2 intersecting in 5 2 are drawn perpen- dicular to P B B 2 and to B, B 2 ; p 6 2 is the velocity of the point B 2 . Finally, p c and & 2 c intersecting in c are drawn perpen- CHAP. ix. KINEMATICS OF MECHANISMS 1G1 dicular to P c C and to the normal C T BC . This gives p c the velocity of C, and b. 2 c the sliding velocity of this second pair of teeth over each other. The process may be carried on in- definitely through a whole train of wheel work, however com- plicated. As a method of finding the velocities throughout such a train, however, it is not a practically useful one, be- cause the directions of the normals to the touching surfaces cannot be very accurately obtained on the drawing unless the ' pitch points ' T AB , T BC , &c., are known, and if these are known to start with the various velocities can most simply be determined from them directly without reference to the touching points. The point T AB maybe looked on as indicating two points one in the first wheel, which may be called T A , and the other in the second, which will be called T B . To obtain the velocity of T A , the triangle p a t a is constructed similar to the triangle P A AT A . ' In this triangle at a coincides with the line ab lt and Toothed p t a is perpendicular to P A P B . Making a similar construction 8 for the velocity of the point T B , we find that the point t b coin- cides with the point t tl . Thus the points T A and T B in the two wheels have the same velocity p t ab , and the point T AB is therefore called the ' pitch point.' The angular velocities of the two wheels are therefore inversely as the distances P A T A and P B T B , this being a familiar theorem proved in the ordinary treatment of toothed gearing. Similarly, if pt bc be drawn perpendicular to the centre line P B P c to its intersection with 1 2 c, this p t bc is the velocity of the pitch point T BC of the pair of wheels (B C) . If the teeth be so shaped as to give constant angular velocity ratios between the wheels, and the angular velocity of any one wheel of the train be kept constant, the points T AB , T BC , &c., in the diagram of the mechanism and the points t^, t bc , &c., in the velocity diagram remain fixed throughout the periodic motion of the train. It may also be noticed that since 162 GRAPHICS _ AT A , 6, 6 _B,T CHAP. IX. therefore n t -n / n t AT PI 1 PT * "ta .P * 6 -"-- L A - L B J -B__- L B J -B M; ' "M^ == M^~ := IV^A ' B^TH "" PIT? so that - also measures the ratio of the angular velocity of wheel A to that of wheel B. The condition that the angular velocity ratio should remain constant may thus also be ex- pressed by the condition that the line a b in the velocity diagram, representing the velocity of sliding of tooth over tooth, should be divided in a constant ratio by the fixed point t ab . [This point tat is only fixed if the angular velocities themselves, as well as their ratio, remain constant, the magnitudes of these Toothed angular velocities being proportional to p t ab .~] Whether this proposition can be utilised in simplifying the practical drawing out of the teeth-profiles, so as to secure a constant velocity ratio, the author has not yet investigated. This kinematic method has been applied to many actual mechanisms. The diagrams when fully drawn are to a certain degree complicated, although not difficult to produce ; and in order that a large number of points should be distinctly marked and lettered on each of the various curves, the dia- grams should be drawn to a moderately large scale. This fact prevents any of these diagrams being usefully reproduced in the atlas to this book ; but the student should not fail to work out for himself a sufficient number of practical examples to familiarise him with the method. CHAP, x. STATIC LINKAGES 163 CHAPTEE X. STATIC STBUCTUEES, FRAMES, OB LINKAGES. 1. WE have already described the construction of the * Moment Diagram' for parallel forces, and in Chapter VIII., Fig. 57, this construction is extended to non-parallel co-planar forces. The construction consists in drawing a connected series of lines across the spaces lying between the force-lines, the joint-points of the lines lying on the joint-lines of the spaces, and the lines being drawn parallel to the radii drawn from a * pole ' to the joints of the vector-addition diagram. This series of lines across these spaces forms a closed polygon if the group of forces ' balance ' both with regard to vector sum and with regard to moment. This state of * balance ' involves two distinct conditions : firstly, that the vector sum Conditions is zero, the graphic test of which is that the vector-addition c diagram forms a closed polygon, and the physical meaning of which is that this group of forces when applied to any portion of material has no influence in changing the velocity of the centre of inertia of this material ; secondly, that the sum of the moments of the forces is zero round any and every axis, the graphic test of which is that the moment-diagram forms a closed polygon, and the physical meaning of which is that the application of this group of forces to any mass does not result in any change of the angular velocity of this mass round any axis whatever. Ill-taught students are apt to think that when the vector sum is zero the moment must necessarily also be zero round every axis, because a resultant of zero magnitude can have, it is supposed, no moment. M 2 164 GRAPHICS CHAP. X. Conditions of balance Moment- diagram Closed chain Definition of link In doing so, they forget the case of force-couples, which have a moment although the vector resultant has zero magnitude. The locor resultant in this case is not zero ; it is two equal and opposite forces. If the vector resultant is zero, and the moment diagram does not close, then the locor resultant is a couple. This is indicated by the first and last lines of the moment diagram being parallel. 2. Confusion also sometimes arises from the fact that a group of forces which balances and has no moment round any axis still has a ' moment-diagram ' from which one measures bending moments produced by this group of forces. It is to be remembered that these bending moments are due to partial sets only of the whole balancing group. Thus the height of the diagram at any point measures the moment round an axis lying in the line on which the height is measured, of all those forces of the group that lie on one side, either right or left, of that line. When the forces are not parallel, it was explained that this polygon was inconvenient when used directly as a ' mo- ment-diagram,' although it always remains of the greatest utility in finding the moments because it supplies the re- sultants of the various partial groups of forces. 3. We must now look upon this closed polygon in another light, namely, as a skeleton drawing of a jointed linkage which would keep its shape and position i.e. be generally in equili- brium, under the action of the forces. Looked on in this light, it is sometimes called a 'chain/ sometimes a 'link- work,' or a ' linkage.' The latter term is preferred by the author. 4. A stiff 'link' may for our purpose be defined as an approximately rigid piece of material forming one part of a structure or machine and jointed to the other parts in such a way that the forces exerted between it and those other parts act in lines passing through definite and easily ascertainable points in the link, which points are called the ' joints ' of the CHAP. x. STATIC LINKAGES 165 link. 1 The word ' point ' must not here be understood in its mathematical sense. No forces act through mathematical points ; they are all more or less distributed through certain volumes or masses and act always through sectional areas of finite magnitude. When one speaks of a force acting through Definition a point, one either refers to the point through which passes the centre-line, or line of the resultant (as previously defined), of a group of distributed forces ; or else one means that the distributed force is concentrated through a very small mass and acts on a very small sectional area. 5. If two links be connected by a cylindrical pin-and-eye joint in which the pin fits the eye loosely, each link acts on the pin over a very small portion of ita circumference ; and if there be very little friction between the pin-and-eye surfaces, then the force, being nearly exactly normal to the surface, passes very nearly through the centre of the pin. Whether the pin be loose or tight, provided there be very little fric- tion, the pressure must be almost symmetrically distributed on either side of a line passing through the pin-centre, and, therefore, the centre-line of the pressure passes approximately through the pin-centre. Thus a rigid mass connected to other pieces of material Flexible only by pin-and-eye joints is a ' link ' according to the above J01 definition, because it is definitely known that it is acted on only by forces whose lines pass through the centres of the pins ; pass through those centres, at any rate, with a degree of approximation quite equal to that of the other data assumed in the course of engineering calculations. For example, our knowledge of the loads on the structure or machine is always far from accurate, as is also our knowledge of the strength and the modulus of elasticity of the materials out of which the structure or machine is built. 1 We will need afterwards to admit into our investigations flexible, ex- tensible, and compressible links; but in the meantime, and except when specially mentioned as flexible, &c., a ' link ' will here be understood ot mean an ' approximately rigid ' link. 166 GRAPHICS CHAP. X. 6. As applied to actual structures, the accuracy of this idea is interfered with in the first place hy the friction between pin and eye. In the case of machine -joints the coefficient of fric- tion is always small and may usually be known with consider- able exactitude. In this case it is not difficult to take into account the effect of friction in the graphic calculations of the forces ; the method of doing so will be explained in a subse- quent chapter devoted to the dynamics of machines. In the pin-joints of roofs, bridges, and similar static structures, the coefficient of friction may often be very large, and the friction, moreover, may be increased very much beyond the product of the effective force through the joint and the coefficient of fric- . tion in consequence of the pin being driven in tight. Also, since there is no relative motion between the links, it is im- possible to say in which direction the friction at each joint is active. By laborious calculation of the stretching and con- stiff traction of the various links under stress, it would be possible to discover in which direction the friction at each joint acts during the process of erection of the structure, that is, when the stresses are first produced. But the friction during the permanent life of the structure is not necessarily the same as that during erection, and it may evidently be reversed at various joints many times in consequence of to-and-fro strain- ing due to rise and fall of temperature and variation of load. Thus it becomes practically impossible to take account of joint friction in these structures ; it may be theoretically possible to do so, but, as a matter of fact, it is never done, and the labour of the operation would be out of proportion to the utility of the result. It is useful to remember, nevertheless, that, however great the friction may be, it can never shift the centre-line of the force through the joint away from the centre of the pin so far as the radius of the pin, except in the case of a pin driven in extremely tight. Probably in actual structures the deviation seldom exceeds two-thirds the radius of the pin. 7. Secondly, it must be noticed that the forces applied to CHAP. x. STATIC LINKAGES 167 certain members of these structures are, instead of being con- centrated, distributed over nearly their whole length. If the pieces be so large that their own weights must be taken into account, then these weight- forces are distributed throughout the whole mass of each piece. Again, the weight of a road- way borne by the booms of the girders is usually, and should always be, applied to these booms as a well-distributed load. Similarly, the rafters of a roof carry the weight of the roof covering as a distributed load. In constructing the force- diagrams for such structures, these distributed loads are usually supposed replaced by con- Distri- centrated loads applied at the joints, each whole load being divided into parts in proportion to the distances of the joints from the centre-line or resultant of the load. This substitu- tion, although it is usually employed in the applications of graphic statics, is by no means necessary, and in what follows there will be shown the method of constructing the diagram without using this artifice. The disadvantage of using this method of substitution is that, in following it, one loses sight of the bending moments produced by the distributed loads. The method hereafter explained will give in the diagram these bending moments as well as the direct forces along the joint- lines. 8. Thirdly, it may be here remarked that the graphic method applicable to pin -joint structures is very commonly applied to structures built up of members riveted together, there being many rivets in each joint. Even if there were Moments only one rivet at each joint, the method would not be strictly a J01 applicable, because such a joint is stiff that is, is capable of resisting relative rotation of the jointed links, or, in other words, capable of transmitting a bending moment. 9. If there are many rivets in the joint, the magnitude of the bending moment the joint is capable of transmitting stiffness is proportionately great, and the application of the method J becomes proportionately objectionable. A link pin-joint may 168 GRAPHICS CHAP. X. be distinguished from other joints (otherwise than as in the above definition) by saying that it is incapable of transmitting a force-couple. Any set of co-planar forces may be reduced to a force and a couple, and by varying the magnitude of the couple the position of the force-locor can be changed at will ; because the resultant of a force and a couple is an equal and parallel force shifted in position a distance dependent on the magnitude of the couple. The sum of the forces across any section of a bar may thus be reduced to a force through the geometrical centre of the section and a force-couple. To pro- duce a force-couple there must be opposite stress-forces exerted at different parts of the section, and the magnitude of the couple will be proportionate to the distance apart of those parts of the section where the opposite stresses are exerted. Thus, if the cross-dimensions of the section or of the joint be small, it is incapable of transmitting a large bending moment, stiffness and the resultant force through it can have its line passing only at a proportionately small distance from the geometrical centre of the section or joint. Otherwise expressed, the section or joint has an amount of flexibility or rotational freedom which diminishes as its cross-dimensions increase. The greater the flexibility or rotational freedom of the joints of a structure, the greater is the accuracy of the force calculations made by the graphic method of this chapter on the assumption of approximately frictionless pin-joints. The degree of error at any joint involved in the adoption of this method is measured by the magnitude of the force-couple the joint is capable of exerting divided by the vector-force-sum actually transmitted. This quotient is the maximum possible deviation of the line of this force-resultant from the centre of the joint, through which centre the resultant is assumed to act. Some examples of the possible errors involved in applying the method to riveted-joint structures will be given subsequently. Evidently the error occurring at strut joints is apt to be greater than that at tie-bar joints, because struts are necessarily made stiff CHAP. X. STATIC LINKAGES 169 to avoid buckling, while flexibility is rather a merit than a dis- advantage in a tie-bar. 10. When a pin-joint link has only two joints in it, then the link is acted on by two forces only. If the link be kept in balance under these two forces, they must be equal in magni- tude and opposite in direction in order that they should give zero vector-sum, and they must act along one and, the same line in order that they should not form a force- couple which would give the link an increasing angular velocity. Since each Two-joint of them acts through one of the joint-centres, this common line of action must be that joining the two joints in the link. This line is called the 'joint-line' or the ' centre-line' of the link. The statics of structures built up of two-joint links only is a great deal simpler than that of structures in which occur links with more than two joints. These simpler struc- tures will, therefore, be exclusively studied in the first place. Subsequently the method will be extended to more complicated cases. 11. In Fig. 57, Chapter VIII., looking on the series of lines (P) ABODE as the centre lines of the links of a closed chain of two-joint links, the links P A and P B will push the pin at the joint (PAB) in the directions p a and bp, and if these latter represent the magnitudes as well as the directions of the pushes of these links, then the pin will be kept in balance by the three forces pa, ab, and bp, since these three form a triangle. Both these links are in compression. Now if bp be single- the magnitude of the push of the link B P on this pin, this j^ ka same link will push the pin at its other joint (P B C) in the opposite direction p b with an equal force, provided the link is subject to no acceleration of velocity. This latter pin will then be in balance if it also receive a thrust from the link C P in the direction and of the magnitude c p, the three balancing forces being pb, be, and cp. Similarly, the pin (P CD) will then be in balance if dp measure the pull exerted on it by link D P, this link being in tension and pulling the next pin (P D E) 170 GRAPHICS CHAP. X. in the direction p d opposite to that of its pull on pin (PCD). e p must now represent the thrust of the compression link E P in order that the pin (P D E) should he in equilibrium, and on this supposition this same link will exert the opposite force Single- p e on the pin (P E A) at the point I. But the link A P has Linkage been assumed to be in compression to the amount dp, and thus the three forces ap,pe, and ea will keep this last pin balanced. The pins being originally motionless, and being kept motionless by the balance of these sets of forces, the whole linkage is also kept in balance i.e. keeps its shape and position unchanged, because the positions of the joints define that of the linkage as a whole. 12. So far the reasoning shows that the shape of the linkage is consistent with the supposition that it may be kept in equilibrium under the given group of forces. It will so rest in equilibrium if the above stress-forces arise in the links. It remains to be examined how and why these stresses do actu- ally arise when the external forces are applied. Suppose the linkage to be given this shape before these external forces are applied, the links being all unstressed. Since there are no stresses in the links at the instant of application of the forces, it is evident that at this first instant the above balance cannot exist, and the first effect of the application of the external Origin of forces at the joints is that the pins begin to move through at least minute distances in the directions of the applied forces. Since these forces form a balancing system, and, therefore, do not move the centre of mass of the linkage, the small motions of the joints are in different directions and produce changes of length of the joint-lines of the links. The material of the links is resistant, and these changes of length, or strains, cause stresses to arise in the links of such character as to resist the further motion of the joints. The further the joints move, the greater necessarily become the strains in the links and the resisting stresses increase in magnitude towards the above limits which produce balance at the joints, and which CHAP. x. STATIC LINKAGES 171 limits are actually reached provided they are not beyond the ultimate breaking strengths of the links. The extent of the movements of the joints and of the strains of the links that actually occur during the application of the load before this equilibriated condition is reached that is, the general defor- mation or strain of the linkage as a whole that is necessary before balance can be attained, depends on the extensibility and compressibility of the various links. These latter depend on the modulus of elasticity of the material and upon the sec- tions of the links. In linkages containing three-joint links it depends also on the flexibility of the links. The greater the extensibility, compressibility, and flexibility of the links, the greater is the general strain needed before balance is attained, and balance is never attained if the links do not possess sufficient strength to exert the stress-forces shown by the diagram. In the above it has been attempted only to explain the origin of general way in which the stresses arise. It has not been 8tresBes attempted to show that the small motions of the joints due to the application of the external forces are exactly such as will produce the precise strains needed to develop the stress-forces that will produce balance. It has not been even attempted to show that these stresses as they develop bear to each other the correct proportions as shown on the balancing stress- diagram whose pole is p. In such an investigation one would need to take into account the distribution of mass throughout the linkage as well as the distribution of sectional stiffness ; because during this adjustment to the condition of equilibrium one has to do with unbalanced forces producing acceleration of momentum. This difficult and complicated investigation is unnecessary, firstly, because our graphic calculations will be limited to such frames the links of which are sectionally so stiff as to prevent the actually occurring strains being large enough to produce any sensible alteration in size or shape of the structure ; and, secondly, because in actual engineering 1 72 GEAPHICS CHAP. X. structures the loads are applied almost invariably very gradu- ally, so that at no instant of the erection and loading does there exist at any part any considerable unbalanced force. Again, in all practical cases in which the linkwork is in Origin of itself flexible i.e. capable of taking different forms under the action of different sets of forces, as in the case of a simple chain such as that of Fig. 57 we do not find it set out in the shape suitable for equilibrium before the loads are applied. It has at the outset some other shape, and, when loaded, not being in balance, it is drawn into a new, possibly an entirely different, shape such as may (or may not) produce balance. 13. Now there are two cases to be distinguished carefully from each other. The distinction depends on the shape possessed by the linkage in relation to the particular forces applied to it. If the shape be such that, when the forces are applied, the linkage is drawn by the action of the forces further and further away from the shape which would har- monise with balance under these forces, then the loading will result in the complete collapse of the linkwork. If, on the other hand, the forces draw the linkwork nearer and nearer the configuration in which it would balance under these forces, then, conversely, stability will be the result. A similar dis- stabiiity tinction is to be made between two kinds of equilibriating shapes. If the balancing forces be slightly altered or dis- turbed in any way and the linkage have in consequence its shape slightly changed, and if then the external forces be exactly restored to their former magnitudes and directions ; then, if the shape be of the one kind, the linkage will come back to its original position and shape, but will deviate still further from these if it be of the other kind. The first kind of shape is said to be one of ' stable ' equilibrium, and the second of ' unstable ' equilibrium. Thus, if the links be arranged in the form of an arch and the loads be directed in the general direction more or less towards the centre of the arch, then collapse will take place under the disturbing con- CHAP. x. STATIC LINKAGES 173 ditions described. But if the form be that of a suspension bridge or inverted arch, and the loads act away from the centre of curvature, then after the disturbing influence has stability passed away the linkage will tend to draw back to its original shape, and is, therefore, in stable equilibrium. 14. As there can be drawn an infinite number of moment or locor-summation diagrams for the same set of balancing forces, so there is an infinite variety of shapes and of sizes of linkages which would keep in balance under the same set of forces. The joints of all these possible linkages lie, of course, on the same force-lines. The possible variety of shape may be expressed as before by saying that the pole of the force- diagram may have any position', and the size may be varied without shifting this pole. If the order be settled in which Three the spaces between the force-lines are to be crossed by the variable links, then the possible variety of linkage may be otherwise expressed by saying that three consecutive joints may be chosen anywhere on their respective force-lines ; or that the direction and position of one link and the direction of one con- tiguous link may be arbitrarily selected. There are evidently three elements of arbitrary choice : as soon as these are chosen the external forces determine the rest of the lines of the struc- ture, and at the same time the stress-forces in all the links are fixed. 15. Let the full lines of Fig. 74 represent a set of balancing forces ac.ing on such a balanced linkage and the correspond- ing force-diagram with its pole p corresponding with the inside space or pen P. At the joint (P 71) or E we have the external force 7 1 balanced by the forces i p + p 7 exerted by the links 1 P and P 7, the first of which is in tension and the second in six points variable compression. On the two links P 7 and P 1 choose any two positions B and A, and suppose the linkage altered by the abolition of the remaining lengths, BE and AE, of these links and the inser- tion of three new links A B, B D, and D A, jointed at B and A 174 GRAPHICS CHAP. X. to the links P 7 and P 1, and also jointed at any point D on the force-line 7 1. We will now prove that the linkage so altered will still be in equilibrium under the same set of forces. From 7 and p in the force-diagram draw ip' and pp r parallel to B D and A B. Then at the joint B we have the force p 7 exerted on the pin by the link P7. This link may be kept in balance under this force and the forces exerted by links B D and A B. These latter are in the directions of these lines, and, therefore, must equal 7p r and p' p in order to pre- serve the balance. The pin at D is now acted on by the forces p r 7 (exerted by link D B and opposite to the thrust of this link at its other end B) and 7 1. It can only be kept in balance by a force ={j=i_p f , and this force can only be exerted by the link D A. It has to be proved, therefore, that i p f \\ D A. If this be proved, then the pin at joint A will also be in balance, Six points because it will be acted on by the three forces \p (already determined as the stress in the link 1 P of the original un- changed linkage) and pp r and p' i, which three form a triangle in the force-diagram. The intersection of i p and p' 7 we will call q, and that of P 1 and B D we will call C. The triangle pp'q has its sides parallel to those of the triangle ABC; therefore, The triangle q i 7 has its sides parallel to those of the triangle C E D ; therefore, . 97 CD The triangle pq7 has its sides parallel to those of the triangle E C B ; therefore, ?9_EC pq CB ' < "- CIIAP. x. STATIC LINKAGES Therefore, But since A.C\\iq and C D j| q p f , therefore angle A C D = iqp'. The triangles ACD and i qp' t having these angles equal and the sides containing them in the same proportion, are, therefore, similar ; and these sides being parallel, the remain- ing side A D in the one triangle is parallel to the corresponding side ip' in the other. This is what had to be proved. For the simple linkage of one pen P only we have now substituted a linkage of two pens. The second pen may be called P' ; and the links A B, B D, and D A may now be called P'P, P'7, andP'l. The first pen is now (P) 1 2 3 4 5 6 7 P', and corresponding six points to it we have in the force-diagram the pole p from which variable radiates the pencil (p) 1234567J/, the radii of which are parallel to the sides of the above pen. The second pen is (P') 1 P 7, and corresponding to it there is the pencil (p f ) i p 7, whose radii from the pole p' are parallel to the sides of the pen. In commencing the construction of the original one-pen linkage, or say that of the pen P of the new two-pen linkage, we saw that there were three elements left to the arbitrary choice of the designer that is, rather, left to be chosen from considerations unconnected with the problem of obtaining equilibrium in the linkage under the given external forces. In constructing the second pen P' we find that three more elements of arbitrary choice have been supplied. Thus all three points A B D may be chosen arbitrarily anywhere on the three lines P 7, 7 1, and 1 P ; or the direction and position of one of the three new links and the direction of a second may be so chosen. 1 76 GRAPHICS CHAP. x. A linkage of one pen for a given set of loads thus leaves the designer with a free choice of three elements. A linkage divided into two pens leaves him free to choose six elements. It can easily be shown that the link dividing the whole into two pens can be introduced at any place. Thus in Fig. 74 it might stretch from a joint such as A in PI to a joint at any position in link P 5. With the link A B as taken above, the forces through links P 7 and P 1 added together as locors equal the locor resultant of all the external forces applied to the linkage to the one or the other side of the dividing link A B. These two forces taken each in one direction namely, in the directions in which they act on the joints A and B equal the locor resultant of the forces 12, 23, Six points 34, 45, 56, and 6 7 ; taken in the opposite direction- that is, as they act on the joints (P 6 7) and (P 1 2) they equal the single force 7 1, which is the only force to the right of the dividing link AB. But in investigating the balance of the pins at the new joints A and B we had only to consider the substitution of the forces through the links A B, B D, and D A, for the stress-forces along B E and A E ; and it is evidently of no consequence to the proof whether these last two added as vectors equal a single actual force 7 1 or the resultant of a number of forces. It is thus clear that the dividing link can be placed in any position in the linkage ; and indeed this can be proved directly following for any special case exactly the model of the proof given above, the number of steps, however, and the complication of the proof increasing directly in proportion to the number of forces acting on, or to the number of joints in, the new pen. 16. We can again operate similarly on the pen P of the two-pen linkage, splitting it once more by another dividing Varia- link, say from a joint in P2 to a joint in P 3, or in fact bll i ty stretching between any two points in the sides of the pen. piexity The whole linkage will then become a three-pen linkage, and, for a given set of loads, there will be nine elements in the CHAP. x. STATIC LINKAGES 177 construction of this three-pen linkage at the arbitrary choice of the designer. There will be a third pole in the force-dia- gram, from which will radiate force-lines parallel to the links bounding this third pen. 17. The introduction of each new dividing link, which means the splitting up the whole linkage into one additional pen, gives, as already said, three new elements of arbitrary choice in the construction of the linkage. But it must be noted that if the new dividing link be inserted between two already existing joints, its insertion in that particular position Limita- amounts to the choice of two of these three elements namely, tlon the positions of the joints of the new link and there remains only one element to be chosen capable of giving a new outline to the linkage as a whole. Thus, if A and B were placed at the joints (P 1 2) and (P 6 7), the change of form effected by the creation of the new pen resolves itself into the single choice of the position of D on the line 7 1. 18. A most important point to notice is that the change of shape of the linkage made on one side of the newly intro- duced link does not necessitate (i.e. in order still to maintain the condition of equilibrium under the given set of forces) any alteration of shape on the other side of this link. It is important to note this, because it means a first step towards stiffness in the linkwork. Either portion of the linkage to the right or to the left of the dividing link that is, either of the pens P and P' can be changed in shape at will in a mul- Degree of titude of various ways. Any change in any part of one pen sl affects the shape of the whole or most of that pen, but does not affect that of the other pen ; except, of course, in the case of the change being effected by displacing the dividing link, which is common to both. Thus, the shape of each pen is in- dependent of that of the other. The shape of the linkage at any joint is denned by the angles between the links meeting at that joint. Complete stiffness is attained when the shape at the joint between each pair of jointed links is independent of N 178 GRAPHICS CHAP. X. Degree of stiffness Relations of locor- to vector- diagram the shape at every other part. So long as the shape at one part depends on that at other parts (the loading forces remain- ing the same), the linkage is to a certain degree flexible. The flexibility is diminished and the stiffness increased the more the whole is split up into numerous pens. At the same time, a larger number of elements of free choice remain open that is, the designer of the structure has a freer hand in shaping it as he thinks fit. At a certain limit this multiplication of the pens produces complete stiffness, and at the same limit the shape of the structure comes completely within the power of the constructor to make it what he likes. That is, the general shape and outline of the structure will no longer depend in any degree on the external forces applied to it ; if these forces form a balancing group of forces, then the structure will be in equilibrium whatever shape be given to it ; while, if these forces do not form a balancing system, then, of course, the structure cannot be brought into equilibrium under their action by giving it any particular shape. This limit will be explained presently. The effect of introducing more dividing links than are needed to produce this limit of subdivision is to cause what is called * redundancy,' a term the meaning of which will be afterwards fully expounded. 19. The relations between the linkage and the force- diagram are extremely interesting from a purely geometrical point of view. It was Clerk Maxwell who first investigated them in the fullest and most general manner. For instance, the point p may be chosen anywhere, as may also one of the three points ABC, say A on its line P 1. Then the lines A B, A D, and B D being drawn parallel to the three radii from p' to p, i and 7, the locus of the intersection of the two latter lines is the line 7 1 in the linkage diagram. On account of a certain reciprocity in the geometrical relations between the two diagrams, Clerk Maxwell called them ' Eeciprocal Figures.' But a purely mathematical view of- any matter is the last thing an engineering student should allow himself to take. CHAP. x. STATIC LINKAGES 179 Fig. 75 will serve to illustrate the properties most important in connection with the use of these diagrams for the investi- . gation and design of engineering structures. 20. Each of the pens A B C D E F has a corresponding pole in the force-diagram, which poles are given the similar names abed ef. From each pole radiates the same number of force-lines as there are links bounding the corresponding pen. These lines are parallel to these bounding links, and measure to scale the stress-forces exerted by the links. Each link separating two pens is common to both pens, and the corresponding line in the force-diagram, therefore, joins the poles of these two pens. The separating link may be called Nomen- the joint between the two pens. The joint between the two poles, therefore, measures the force exerted by the joint between the pens. Throughout this book the pens will be named by capital letters, ABC, &c., and the corresponding poles by similar small letters. For instance, from the pole b radiates a pencil of 5 lines (b) a, 3, 4, c, 10, parallel to the 5 links bounding the pen (B) A, 3, 4, C, 10 and measuring the forces through these links. The spaces outside the linkage and separating the lines of external load will be named by large figures, 123, &c., and the corresponding points in the force-diagram by similar small figures. These spaces are not pens proper, because they are not completely surrounded or closed by definite boundaries ; towards one side they are open and of indefinite extent. But Outside the force-diagram contains a point corresponding to each of pen these external spaces, from which point radiate the same number of lines as the space has definite bounding lines, and these radii are parallel to these bounding lines and measure the forces exerted along them. Therefore, these points may consistently be termed the poles corresponding to the ex- ternal spaces, and the external spaces may be called ' outside pens ' or ' open pens ' or ' unclosed pens.' 21. The outside pens are separated by the locor lines of N 2 1 80 GEAPHICS CHAP. X. the external loads. The distinction between these joints (or dividing lines) of the outside pens and those of the internal pens is not very simple to explain, but ought to be thoroughly grasped. It is still more important to understand the essential similarity between these two kinds of joints, and this can be best done by recognising clearly their one point of difference. Each stress line corresponding to a link of the structure is evidently used as a force line twice over in considering the balance of the joint-pins. At one end of the link it represents a force acting on the pin in a certain direction ; in considering the balance of the pin at the other end it represents an equal but opposite force. If it be in compression, the link uses the pin at its one end as its abutment for the thrust it exerts on the pin at its other end. If it be in tension, each pin acts as the anchorage of the link for the pull exerted on the other pin. But the line of the stress- diagram representing the Outside action of an external force, and corresponding to the joint * 16 5 between two external pens, is used only once and in one direction only. It is so, at any rate, so far as the stress- diagram is used for investigating the forces throughout the structure. This then is the distinction : the one line is a stress line representing two opposite forces ; the other is a force line, to be used only in one direction, since it represents only one force. But this external load is simply one aspect of a stress action between the structure under investigation and its ma- terial surroundings. Suppose it to be the supporting force exerted by a pier on a bridge. Then this is only one aspect of the compression stress in the pier. The pier acts as a strut between the bridge and the earth ; it is a link connecting the bridge with the rest of the material world outside the bridge. At one end this link has its abutment on the earth ; at the other end it abuts against the bridge. It is, therefore, to be properly called an external or outside link i.e. outside the structure under investigation. As the line of pressure through CHAP. x. STATIC LINKAGES 181 a pier does not always, or generally, coincide with the geome- trical axis of the pier, it may frequently be preferable to call it an * outside link line ' rather than ' outside link,' meaning thereby that it is the force-centre-line, or the centre-line of the force through the outside link; but when no ambiguity arises, the shorter phrase may be legitimately used. The links belonging to the structure itself will be called, in contra- distinction to these, ' inside ' or ' internal links,' this designa- tion including the peripheral or boundary links which separate the inside from the outside pens. When the ' external force ' is due simply to the weight of a mass resting on or hanging from the structure, there is no material link, such as a pier, communicating the stress between the structure and the outside material world. The stress bond in this case is the attractive force of gravity between the load- Outside ing mass and the earth. But whether we consider this ^ k g ns attractive force to be exerted across absolute empty space, or to be a real active stress existent in that portion of the sub- stance of the radiant-energy- carry ing ether lying between the attracting masses, 1 we may take this space intervening between the earth and the heavy load as representing an outside link along which the external stress is transmitted, the line in which the weight acts being the centre-line of the external stress. When this weight-load is applied to the under side of a structure, this imaginary or representative external link wil lie wholly outside the structure, and will be a tie-bar. When the load is applied to the upper boundary of the structure, the direct external weight-link to the earth would pass down- wards through the structure, thereby confusing the spaces into which the link-lines divide the whole diagram. It will 1 This latter is the belief of nearly all modern physicists. To those who believe the doctrine of conservation of energy it would appear impossible to account for the conversion of active into so-called ' potential ' (better called ' latent ') energy, except by supposing the latter to exist in some form of stress- energy filling the space intervening between or surrounding the attracting bodies. 182 GKAPHICS CHAP. X. be found later that it is important to have all the external link-lines lying clear of, and outside the structure. In this case, therefore, it is best to adopt an artificial convention, and imagine the weight to be due to a compressive stress through an outside link (a strut) lying above the structure. This outside strut must be supposed to have its abutment at its upper end against an arch built up from the earth and over the structure. Fig. 76 will help to explain these conventions. The outside links 11, 1, and 6, 7, are compression links con- Outside sisting of the end piers or abutments of the bridge. 7,8; 8,9; links 9 9, 10; and 10, 11 are tension links due to loads resting on the bottom boom. 1, 2 ; 2, 3 ; 3, 4 ; 4, 5 ; 5, 6 are external compression links due to loads on the top chord. No essential distinction, therefore, exists between internal links and external links. The forces of all the external links taken as acting upon the structure must form a balancing group of forces if the structure is to remain in situ. "These forces, taken as acting on the earth i.e. on the world external to the structure also form a balancing set of forces. 22. Take any section /3 through the structure. We may now consider the portion to the left of this section as a struc- ture by itself, held in balance by a set of stresses between it and the external world. The outside links are now 9, 10; 10, 11 ; 11, 1 ; 1, 2 ; 2, 3 ; 3, F ; F, G ; and G, 9. The latter three are internal links of the whole bridge, but are now to be considered as external links of the structure consisting of that part of the Method bridge to the left of the section line aft. This shows still 8 more clearly how non-essential is the distinction between in- ternal and external links. This latter set of external stresses, including 3 F, F G, and G 9, must form a balancing system. From this fact the stresses 3 F, F G, and G 9 can be deduced from a knowledge of the rest of the group 9, 10 ; 10, 11 ; 11, 1 ; 1,2; 2, 3. The two con- ditions of vector and locor balance are sufficient to determine the unknown magnitudes of those three forces, whose directions CHAP. x. STATIC LINKAGES 183 are known. This method of finding the stresses in members of structures is called the 'Method of Sections,' and will here- Method of sections after be shown to be of great utility. The graphic construc- tion by which it is effected has already been explained in Section 14 of Chapter VIII. 23. If the pens of a structure be all triangular, the struc- ture is evidently * stiff,' in the sense that its shape cannot be altered except by lengthening or shortening the individual links. The members of engineering structures either actually having, or being intended to have, such sectional sizes as pre- stiffness vent the tensive and compressive forces coming upon them from altering their lengths in any degree enough to change appreciably the general shape of the structure, the above is the sense in which the term ' stiff ' is technically applied to such structures. 24. In order to effect complete stiffness it is not necessary that all the pens should be triangular. In building up a plane structure, in order to locate ' stiffly ' each new joint as it is added to the design or in the actual erection, it is evidently necessary and sufficient to add ttvo new two-joint links. Thus, in adding any number of joints, double the same number of links must be added. Since in the simplest possible structure viz. a triangular Criterion one-pen truss there are three joints and three links, that is, frames there are three less links than double the number of joints, therefore the general formula connecting the number of joints with the number of links in a completely stiff but non-redun- dant plane linkage is I = 2j - 3 where,/ = number of joints and I = number of links. 1 1 This condition is sometimes expressed as a relation between the number of pens and that of joints namely, the number of pens is two less than the number of joints. But in some structures the pens overlap each other and the counting of their number becomes confusing. Again, in ' solid ' as distinguished from plane structures, the corresponding rule connecting the numbers of links and of joints is of easy application, whereas that connecting the numbers of joints and of enclosed volumetric pens is confusing and difficult to apply. 184 GRAPHICS CHAP. X. Criterion in plane frames Criterion in solid frames Thus in Fig. 76 the number of joints is 11, and 2x11 3 = 22 3 = 19 = number of links. In Fig. 78 the number of joints is 41, and 2 x 41 - 3 = 82 - 3 = 79 = number of links. In Fig. 80 there are 18 joints, and 2 x 18 - 3 = 36 - 3 = 33 links. This only applies to flat structures built of two-joint links, If the linkage contain ' beam-links 'i.e. links capable of resisting cross bending and containing more than two joints then evidently in order to locate stiffly in a beam-link a third, fourth, fifth, &c., joint, no further links need be added to the structure. Thus the above formula can be applied unaltered to structures containing beam-links, provided that in counting j only the end joints (or any two joints) of each beam link be included in the count. In the diagrams in this book beam-links will be distinguished from others by being indicated by double lines ; one line being thin and representing the centre-line of the link, and the other being thick. 1 Thus in Fig. 77, counting only two joints to the horizontal beam, the number of joints is 7, and 2x7 3 = 11 is the number of links, counting the beam as one link only. This structure is, therefore, stiff in the sense of the word we are now using. In the ordinary meaning of the word, of course, its degree of stiffness will depend on the rigidity of the beam. 25. In a ' solid ' structure i.e. one of three dimensions as distinguished from a flat one in order to locate each new joint stiffly it is necessary to add 3 new links. The simplest pos- sible solid linkage is a pyramid with a triangular base. This has 4 joints and 6 links i.e. its number of links is 6 less than 3 times the number of its joints. Therefore, the general formula for a stiff and non-redundant solid linkage is I = 3 j - 6 where j is the number of joints and I the number of links. 1 This system is consistent with the representation of ties and struts by thin and thick lines respectively, because one side of a beam is in tension i.e. acting as a tie while the other is in compression. CHAP. x. STATIC LINKAGES 185 As in the previous case, in counting j only two of the joints of each beam- link are to be included in the count, and the beam is reckoned as one link only. 26. If the number of links be greater than that given by the above formulas, the linkage is ' redundant,' and it becomes Redun- impossible to determine all the stresses in it either by algebraic dancy or graphic means unless account be taken of the elastic extensions, contractions, and bending of the ties, struts, and beams. 27. If the structure rest on the earth by means of fixed connections, so that the joints between structure and earth are approximately immovable, then evidently the earth forms one link that goes towards the stiffening of the structure ; and a number of links in the structure itself (exclusive of the earth link) one less than given by the above rules is sufficient for stiffness. Thus, an arch hinged at centre and at abutments becomes a stiff structure only when considered in connection with the earth, which performs the function of a tie-bar whose tension balances the horizontal thrust through the arch. If Earth a structure stiff in itself be connected with the earth by fixed ^ O n n n s ec " joints, by this connection it is converted into a redundant structure, the stresses in whose members can only be deter- mined by taking account of the elastic deformation of the structure and of the earth abutments. If, however, the joint with the earth have a certain degree of freedom; if, for example, a stiff non-redundant girder rest at one end on hori- zontal roller-plates that leave its one end free to move hori- zontally without any retarding horizontal force worth taking account of ; then the directions of the supporting forces at the earth joints become determinate, and the problem of finding the stresses throughout the structure can be always completely solved. 28. To ensure stiffness and non-redundancy the structure taken as a whole must fulfil the conditions expressed by these closer rules ; but their fulfilment for the whole structure is not an anal y sis 18G GRAPHICS CHAP. x. infallible criterion of stiffness and non-redundancy, nor is it necessary that a stiff and non-redundant structure should fulfil these conditions in each of its separate parts. Thus, a structure may be flexible in one part and redundant in another, in such a way as to leave it flexible as a whole and yet fulfil as a whole the rules. For example, a quadrilateral with braces across both diagonals is redundant, and to one side of this may be joined another quadrilateral with braces across neither diagonal. The second quadrilateral is flexible and the whole linkage is flexible, but it fulfils the conditions, there being 6 joints and 2x6 3 = 9 links. To test completely, therefore, the stiffness of a proposed design of a complicated character some additional criterion is necessary. This consists in taking a number of sections, each dividing the structure in two parts. The parts must then fulfil certain conditions wherever the section be taken if the Closer structure be stiff and non-redundant. These conditions are analysis the follow ing : If the section pass through a joint and one bar ; or if with- out passing through a joint it cuts through three bars ; then in each of the two parts the usual rule (I = %j 3) must be fulfilled, the joint cut through in the former case being in- cluded in the count for each part. The section cannot pass through two bars only if the linkage be stiff, unless the two bars are actually jointed to- gether. If the section pass through a joint and two bars, or through no joint and through four bars ; then in one of the two parts the usual rule (I = Zj 3) must be fulfilled, while in the other part there must be one degree of flexibility (i.e. I = 2J 4). This rule is exemplified in Figs. 78, 80, and 86. If the section be through a joint, the joint has to be included in the count for each part. In no case are the bars cut through by the section to be included in the count of links for either part. CHVP. X. STATIC LINKAGES 187 These conditions must be fulfilled for every possible section. In these tests, in counting the number of links cut through Closer by a section a beam-link is to count as two. Thus, a section of a stiff structure cannot pass through two links only if both be two-joint links, but it may do so if one of them be a beam- link, as may be seen from Figs. 77, 87, 88, 89, and 90. 29. The method of finding the stresses by constructing at each joint a graphic diagram showing the balance of the forces acting on the pin at that joint has been in use for a long time. But this method is clumsy ; it necessitates the drawing of each line twice over, because the stress in each link acts on two pins, and, therefore, enters into two of the above diagrams. The modern improved method of stress diagrams consists in combining all these into one diagram, so that no line, or as few as possible, need be drawn twice over. It is not always possible to avoid repeating the same stress in the diagram, but in a good diagram this repetition is minimised, and in almost the majority of cases it does not need to occur at all. Its avoidance depends altogether upon taking the stress lines in the correct order in building them Cyclic into the diagram. Eeferring to Fig. 75, take for example the order stress in the the link 5 C. This stress gives one of the forces acting on the joint-pin (5 C B 4) . It, therefore, appears in the closed vector polygon for this joint ; and thus it must appear in connection with the stress 4 5, i.e. in the same polygon. But it must also appear in similar connection with 5 6, because it forms one of the balancing forces on the joint- pin (5 6 DC). It acts in opposite directions on these two pins, and, therefore, in the one diagram it will be read 5 c, and in the other c5, and it must radiate from the joint of stress lines 45 and 56, because these latter have to be in similar connection with each other since they form. part of the polygon showing the vector balance of the group of external forces. Similarly, 6 d must radiate from the joint of 5 6 and 188 GRAPHICS CHAP. X. 6 7. Therefore, in ths diagram for joint (C 5 6 D) the stresses along C 5, 5 6, 6 D, must follow each other either in this order or the exact reverse. Since the three follow in this manner and there are only four altogether in the polygon for this joint, it results that the same must hold for the four. That is, that in the stress- diagram the four must follow in the same order Cyclic as the locor lines are cut in the linkage in circulating round order the joint. The circuit of the joint may be made either right- handedly or left-handedly that is, either as (C 5 6 D) or as (CD 6 5). Such order is called 'cyclic' order, and there are two possible cyclic orders that may be followed in building up the stress-diagram, the right-handed or the left-handed. But the same cyclic order must be preserved throughout the stress- diagram for the whole structure. 30. The cyclic order that has been adopted in building up the stress diagram must be marked on it either in words under the title, thus, 'Left-handed' or 'Right-handed,' or very much better graphically by a circled arrow-head thus *J or Q'. By so marking it, it becomes quite unneces- sary to place any arrow-heads on any of the lines of the diagram (neither on the external nor on the internal force lines) to indicate force directions. In the link diagram to Tension distinguish compressions from tensions it becomes unnecessary * use e ^her arrow-heads or -f and . Thus, in Fig. 75, which is marked *, in order to find whether the link D E is in tension or compression, it is only necessary to notice that on the lower joint (9 C D E F) the stress on this link will be read d e, and on referring to the stress-diagram we find that this is downwards, or towards the joint that is, the link is thrusting downwards on this pin, and is, therefore, in com- pression. Similarly, link C D is in tension, because at the lower joint its force is to be read c d, and this we find to be directed upwards, or away from the joint. 31. Arrow-heads placed on the lines in the linkage are very objectionable because they are ambiguous. Thus, a link CHAP. x. STATIC LINKAGES 189 marked 3 *^ >- (T would by some people be read as being in tension, the arrows being taken to indicate that the ends of the link tend to separate ; while others would under- Arrow- stand that the arrows indicate the directions of the forces exerted by the bar on the pins at its opposite ends, and therefore read the bar as being in compression. 32. The signs + and affixed to the link-lines are not ambiguous, and are in fact very commonly and usefully em- Use of + ployed. When the stresses are read off in figures, these figures should be written along the links, and a plus may be prefixed for tensions and a minus for compressions. 33. But a readier and more quickly perceived distinction between tension and compression members may be made in inking in the linkage by employing two colours ; for example, by inking all the tie-bar lines in black and the strut lines in Two red. In this book all tie -bars will be shown by thin black lines, and strut lines by thick black lines. The stress-diagram * nd thin may very suitably be inked in blue. In this book, however, fine black lines are used for the stress-diagrams. 34. The cyclic order of the stress-diagram applies to the vector summation diagram of the outside forces. The same order as holds for each individual joint is that in which these external forces must be arranged to follow each other in the stress-diagram. The order here refers to that in which the outside links are cut through in making the circuit outside the whole structure, either right-handedly or left-handedly. Thus, in Fig. 75, they will be found in the stress-diagram in Constancy the order 123456789 10, the same as that in which the outside pens are traversed in making a right-handed circuit outside the linkage. The same cyclic order will be found to be preserved in the vector- summation diagram showing the balance of the forces acting on any individual pen. Thus, the pen (C) 9 B 5 D in Fig. 75, consisting of 4 links, is kept in balance under the action of the links 9, 10; 10 B; B4; 45; 56; 6D; DE; 190 GRAPHICS CHAP. X. E F ; and F 9. This is the order in which these links are cut in making the right-handed circuit of the pen, and in the stress-diagram it will be found that the corresponding forces follow each other in this order and form a closed vector polygon. Here C is the ' pen-joint ' of the above links, and we see that the same cyclic order holds for the balance of ' pen -joints ' and for that of ' point-joints.' Taking two pens together, for example C and D of Fig. 75, and considering them as a compound or complex pen, we find the same law holds ; thus, these two are held in balance by the forces in the stress-diagram 9, 10 b 4 5 6 7 e/o, these being arranged according to the right-handed cyclic order in which Constancy the corresponding spaces are traversed in making the external order* 5 * circuit of these two pens. The same will be found to hold true of any combination of pens one can take. Thus, the application to the group of external forces under which the whole structure is in balance is only a special application of a general law. The same law again holds good for the balance of individual links. Thus, the link CD is in balance under the forces shown in the stress-diagram by 9 c 5 6 d e/9, this order showing the correct direction of these forces. Thus pins, links, simple pens and complex pens are all subject to this one law of the same cyclic order in the stress- diagrams exhibiting their balance. The importance and utility of this law in the art of constructing graphic stress-diagrams cannot be overrated. 35. There is an important exception to the rule that in the stress-diagram no stress-line is to be repeated twice. In the examples hitherto given the external forces are assumed loads at to act at external joints, and there will be found at once a difficulty in following the method explained if they do not do so. But the loads are actually sometimes applied at internal joints. For example, Fig. 78 shows a 'stiff' arched linkage in which the load, which is the weight of a roadway, rests on the bars B M, D 0, F Q, &c., &c., and is supposed to be con- CHAP. x. STATIC LINKAGES 191 centrated at the joints along this line. In order to get over the difficulty arising in this way, the best and, so far as the author knows, the only correct method is to suppose added a number of extra links lying along the lines of the external forces and terminating at joints in the outer periphery of the linkage. In Fig. 78 these imaginary links are shown by dotted lines running from the joints where the loads are applied to the upper boundary or chord. They might as simply have been drawn downwards to the lower chord, in which case the stress-diagram would have taken a slightly different shape from that shown but would have given exactly the same stress-magnitudes. It is generally a matter of theo- retic indifference towards which boundary the imaginary links are inserted, but it is of practical importance to place them so that they shall intersect as few actual links as possible. As in Fig. 78, it is generally unnecessary to allow each to inter- sect more than one actual link namely, that at the outer Loads at boundary ; but in Fig. 86 there are examples of them crossing J^ two actual links. At this intersection a new joint must be imagined to be inserted, which joint will divide the actual boundary link into two. The supposition of the insertion of one imaginary link will thus increase by two the total number of links in the linkage and by one the total number of joints. 1 Thus, the question of the flexibility, stiffness, or redundancy of the structure will not be affected by the introduction of these imaginary links. This is still the case whether or not the imaginary links intersect more" than the one peripheral actual link. At each ' internal ' intersection of imaginary and actual links there must be supposed introduced a new joint. Each such joint will split the imaginary and the actual links each into two parts, thus increasing the number of links by two and the number of joints by one. Thus in Fig. 78 the load actually applied at the joint (0 P Q F E D) is supposed transferred to the upper chord by 1 The whole of this argument refers only to plane linkages. 192 GRAPHICS CHAP. X. the imaginary link E E'. This divides the space E into two spaces E and E'. The link joining joint (CD Ell) with (E F G 10) is divided by the imaginary joint (E E' 10, 11.) into two links, E' 10 and HE. The supposed insertion of the imaginary link leaves the stresses through these latter two links entirely unaffected that is, it leaves them exactly the same as if the imaginary link and joint were not inserted ; because at this joint we have jointed together four links lying in two straight lines. The vector diagram showing the balance of this joint-pin must, therefore, be a parallelogram, and the stresses through the two links lying along the same line are, therefore, equal and quite independent of the magnitude of Loads at the stress in the other pair of links lying along the second joints"* l me - I* 1 the Present case the parallelogram is a rectangle and is seen on the stress- diagram to be e e' 10, n. Thus the stress condition of the link (11, 10) E is unaffected by the suppo- sition of the insertion of the extra link. Furthermore, che stress on the imaginary link E E' being necessarily equal to the external load, the thrust of this link on the joint-pin (0 P Q F E D) is equal to that actually exerted by the load ; and the supposition made, therefore, leaves the stress condi- tion at this joint entirely unaffected. This artifice is, therefore, legitimate, since it alters no actual stress conditions^ and its adoption eliminates all difficulty in constructing the stress- diagram according to the ordinary simple procedure already explained. 36, The result, however, is that a number of lines in the stress-diagram appear twice over. Thus all the ' external stresses ' applied at internal joints are repeated ; each appear- stress- ing once in the external- force vector-balance polygon and repeated once as an internal stress, namely, that through the imaginary link. E very actual link that is divided into two by imaginary joints has also the line representing the stress through it repeated twice in the diagram. This repetition is unavoid- able. Although it increases the number of lines and thus CHAP. x. STATIC LINKAGES 193 decreases the ' readability ' of the diagram, it can never lead to any ambiguity or real difficulty in reading off results. 37. When two actual links cross each other without being actually jointed at their intersection, the same artifice must be adopted in order to make it possible to draw out the stress- diagram without confusion and without break of con- tinuity. An imaginary joint must be supposed inserted at their intersection. This will divide each of the links into two. The stress-diagram for the joint will be a parallelogram, and the stress on each of the two links will be repeated twice. Crossing The spaces separated by the links should be lettered so as to give similar names to the two stress-lines representing one and the same link stress. Thus, in Fig. 79, we have a lattice girder of common design except that vertical struts across the quadrilateral spaces are usually introduced. In this design the diagonals that run from top to bottom chord are not actually jointed at their crossings, but at these points joints must be imagined for the purpose of enabling one to draw out the stress-diagram in an orderly and continuous fashion. 38. If the spaces are lettered after a system similar to that shown in Fig. 79, the identical stresses in the stress- diagram will appear with names, not the same, but with a distinctive similarity to each other. Thus, C 3 C 4 and D 3 D 4 are really parts of one and the same link, but they are treated as if they were two separate links. Their names have, how- ever, this in common, that the suffices 3 4 are identical in both. These suffices in the name of each differ from each other Lettering that is, 3 differs from 4 ; while the letters appearing in each do not differ, as for instance in C 3 C 4 only the one letter C appears. The stresses of these two links will appear as c 3 c 4 and r/ 3 $ 4 ; and these will be opposite sides of a parallelogram, and are identical in name so far as the suffices 3 4 are con- cerned. The other pair of equal sides of this same parallelo- gram will be c 3 d 3 and c 4 d 4 , whose names are identical so far as the letters go, although the suffices differ in the two stresses 194 GRAPHICS CHAP. X, while in each stress only one suffix number appears. These are the stresses in the links C 3 D 3 and C 4 D 4 , whose names have the same sort of identity, and which are imagined to be separate links, although in reality they are one and the same. Using this system of lettering, no two bars which are really different bars will be found to have names that are not dis- tinctive. In each name either the two letters are the same while the suffices differ, in which case the suffices are to be Lettering taken as the part of the name individualising the real link or the corresponding stress ; or else the two suffices are the same number while the letters differ, in which case the letters are to be taken as the distinctive part of the name. The linkage in Fig. 79 has 7 joints in each chord, or 14 in all. In order to be stiff it requires, therefore, 2 x 14 3 = 25 links. It has actually 26, and is, therefore, redundant in one degree, and its stress-diagram cannot be drawn without taking account of its elasticity. 39. Fig. 80 shows a non-redundant stiff linkage of similar design. There are here at the ends 4 joints at each of which only two links meet. At each such joint no stresses can exist in the links unless an external force acts at the joint. Thus at the joint (9 J 2 ) no outside link acts, and therefore the stresses in the two internal links, J 2 9 and 9 J 2 , are zero. 2-iink It must be understood that this rule applies only to 2-joint without links. If one of the pair of links be a beam, then force may load b e transmitted through the joint. If, again, the two links meeting at such a joint lie in the same direction, then it follows, not that their stresses are zero, but that they are equal, both being subjected to equal pulls or to equal thrusts. 40. Another case, frequently of use in surmounting special difficulties in complicated structures, is that of a joint at 3-link which three links meet, two of which links lie along the same nne: Here the force-component exerted normally to the links in line by the link not in line must be equal and opposite to CHAP. x. STATIC LINKAGES 195 the component of external force acting on the joint resolved in the same direction. This enables one to find the stress on the oblique link at once without working up to this joint from other parts of the structure; in other words, it may be possible to begin drawing the stress-diagram from this joint. In many roofs as actually put up such 3-link joints occur 3-link on the lower chord where no load is applied. The joints 2 in line ABCD in Fig. 81 are examples of this. The stay-bars at these joints cannot be stressed except by the weight of the tie-bars of the lower chord, and are, therefore, practically of no use whatever. If the lower chord were composed of two stiff beams, then the stays would be of use. 41. In the case of an external force acting on such a 3-link joint, if we look on this force as that of an external link, the joint is a 4-link one. Thus, at any 4-link joint two of whose links lie in one line, the force-components of the other two per- pendicular to the first two must be equal and opposite. It not infrequently helps one out of a special difficulty in pro- ceeding with the stress- diagram to remember this obvious fact. Again, if a known external force acts at such a 4-link joint, 4-link the component of external force perpendicular to the two links 2^ J^e in alignment must balance the algebraic sum of the similar components of the forces along the two other links. Thus in Fig. 80 all except the end joints are of this character. There- fore, the algebraic sum of the vertical components of the forces of any pair of diagonal braces meeting at a joint must equal the vertical component of the load at that joint. The student should inspect the stress-diagram to trace out the truth of this proposition for each of the joints of this structure. 42. In drawing the stress-diagram of a plane structure, the general rule is that one must begin at a 2-link joint i.e. one of two internal links. The outside link makes a third, Order of and, its stress being known, the triangle of balancing forces can pr at once be drawn. From this one must proceed to a con- 02 196 GRAPHICS CHAP. X. Order of procedure Himber of links at a joint tiguous 3-link joint ; a 4-link joint would leave three unknown forces which it would be impossible to determine. The third joint taken in building up the diagram may be a 4-link one, but must not be 5-linked. At each successive step the new joint attacked must be one at which all the forces except two are either known as data or have already been found by the stress-diagram construction. It is here that the beginner gene- rally goes wrong ; he goes prematurely to a joint where more than two forces remain undetermined, and, being unable to solve this, becomes confused and perhaps disheartened. It is only in special circumstances, however, which are dealt with below, that he has not at each step a choice of one or more joints fulfilling the condition necessary for solubility. Nearly all structures have at least two joints at each of which meet only two inside links. It is a matter of theoretic indifference at which of these two the stress-diagram is begun. Sometimes one will be preferable as a matter of convenience to the draughtsman. Usually, the commencement having been made at one of these two joints, the other is the last to be dealt with at the completion of the diagram, and the last two joints usually furnish the graphic test of accuracy of draughtsman- ship. Sometimes, however, it is convenient to work part of the diagram from one of the 2-link joints, and another part from the other similar joint, the two parts being eventually joined somewhere near the centre of the diagram, and this junction then affording the test of accuracy. 43. The normal distribution of links at the various joints is the following : two 2-link joints, two 3-link joints, all the others 4-linked. This may be recognised from the formula (2/i 3) links for the n joints of a stiff non-redundant plane frame. Each link goes to two joints, and, therefore, there are (t/t e) link-ends at the n joints. The above distribution accounts for {2 x 2 + 2 x 3 + (n 4) 4} = 4 + 6 + 4 n 16 = 4 n 6, the same as just mentioned. The two 2-link joints can never be found contiguous to CHAP. x. STATIC LINKAGES 197 each other in a stiff linkage. The two 3-link joints are gene- rally contiguous to the two 2-link ones. If there be one 5 -link joint, then there are three 3-link ones. If there be several 5-link joints, there is a corresponding extra number of 3- or 2-link ones. Each 6-link joint leads either Number to an extra 2-link instead of a 4-link joint, or to two extra 3- at a joint link instead of 4-link joints. Each extra degree of complexity in any one part of the structure beyond the normal arrange- ment in which the most complex joint has four links leads to a corresponding amount of extra simplicity in another part. 44. There may also be no 2-link joints along with six 3- link ones, all the rest having four ; in this case the count of the link-ends being 6 X 3+ (n 6) 4 = 18 -f 471 24 = 4W 6 as before. Figs. 85 and 86 give examples of this distribution. NO 2-link It is here impossible to start drawing the stress-diagram in J01 the usual way. One method of surmounting the difficulty is to start by help of the ' Method of Sections.' Another is to use the * Method of Two Trials and Two Errors.' Explana- tions and examples of both methods are given below. 45. These principles of stress-diagram construction will now be illustrated by going through in detail a few examples. First, take the arched linkage of Fig. 78. The structure is symmetrically built, but the load is not symmetrically dis- tributed. The loads are 2, 3 = 12, 13 = 2 tons each ; 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 4 tons each ; stiffened and 9, 10 = 10, 11 = 11, 12 = 7 tons each ; all these loads being vertical and parallel to the axis of sym- metry of the linkage. The supporting force 13, 1 is inclined at 45 to the horizontal. These are the load data. A left-handed stress-diagram is adopted : the outside pens are numbered in left-hand cyclic order ; the loads applied at internal joints are transferred to peripheral joints by means of imaginary links ; and the inside pens are lettered as shown. 198 GKAPHICS CHAP. X. Stiffened Arch Next, the loads are plotted to scale consecutively from 2, 3 at the right-hand end to 12, 13 at the left-hand along a vertical straight line lying near the left-hand end of the paper. The scale actually adopted is J" = 1 ton ; but this is made small for the sake of engraving. In practice " = 1 ton would be better. From 13 in stress-diagram a line is drawn at 45 to horizontal. Any pole is chosen, and parallel to the radii from it are drawn the lines of a simple link polygon through the spaces 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 of the linkage. 1 The lines through spaces 2 and 13 are produced to meet, and through their intersection a vertical line is drawn, this being the centre-line of the loads. The line 13, 1 of the support- ing force is drawn through the joint (13, 1 K) at 45 to meet this centre-line. The line to this last intersection from joint (1,2K) gives the direction of the other supporting force; and this force is now drawn from 2 in the stress -diagram to meet 13, i, thus determining the point i. The following table now shows the order in which the stress-diagram is built up. In the force-diagram for each joint certain force lines (all except two) have been drawn already before the completion or closure of the diagram. The last two, or those requiring to be drawn when the diagram is completed for the joint, are indicated in the table by a bar being placed over them : Name of joint in linkage Balanced force- diagram Name of joint in linkage Balanced force- diagram KIM M1N N10 O1P P1Q Q1E R1S SIT 13,1, A;, 13 fci mk T1U 12, 13, K MB 12BC 11,1200' C'CBMNOD 110'DE 10, 11 E E' E'EDOPQF lOE'FG 1 1 U t 12,13, Jcmfrw 12 6 C 12 11, 12, CC'll c' cbmno dc f nc' de 11 10, 11, ee' 10 e' edopqfe' m mm nion o ]po qijrq r iar a \ta 1 The simple linkage to determine the supporting force 12 is omitted from the engraving for clearness' sake, but was, of course, drawn in the original diagram. CHAP. X. STATIC LINKAGES 199 Name of joint in linkage Balanced force- diagram Name of joint in linkage Balanced force- diagram 9, 10 G G' G'GFQBSH 9G / HI 89IF TIHSTUJ 81'JV 78VV 12X 1 KIVl 1 1 Iff llffO lOP 1P<* 1QR imp 9, wgg'v g'gfqrsWg* 9g'JM9 ssii's i'iTistujI' 6 i'j V 8 78VV f 7 12~k7 1ST ITU V'VJUllTJ 7V'I 67X1' I I JUTS H r.i'HC :.I-,CG' G G H S R Q F >G FE 45EE' E E F Q P I> 1 E DC 34CC' C'CDOXTMB liTFi 1 tU 1 v' vj u i u j i>' 7v'jT7 6 7 ii^T i r i j u t shi r ei'hgre 5 6 gT %' 5 g'ghsrqfg' sgr'fes 4 5 e e' 4 e'efqpotle' 4 e' d c 4 3 4 C <*' 3 c' c d o n in b c' 1 km [ imn i 1 no i lop i ipqi iqrl irsl This completes the drawing of the diagram with the excep- tion of the line b 3. But both the points b and 3 are already located, and the line can therefore be drawn by joining them without reference to the direction of the line B 3 in the link- stiffened age. The test of the accuracy of the drawing consists in find- ing whether these two lines in linkage and stress-diagram be parallel as they ought to be. Another, form of test (although not an independent test) is to locate point b once more by drawing the force- diagram for the joint 3 C' B. If the drawing be accurate, the two b's thus found will coincide. A third mode of locating it is to draw the diagram for the joint MK23B. If the inaccuracy be found to be much, it shows that some downright mistake has occurred in forming the diagrams ; if it be small, it shows that no such mistake has been committed, but that strict parallelism between the lines has not been maintained, or that the lines have not been drawn strictly through the right points. There being no external loads at the joints along the upper chord where the diagonals meet, and the diagonals having equal, although contrary, inclinations, the stresses in 200 GKAPHICS CHAP. X. the two members of each pair of diagonals meeting at an upper joint are equal in magnitude and of contrary sign, stiffened Also the stresses in the pair of horizontal members meeting at each upper joint differ by the arithmetical sum of the hori- zontal components of the stresses in the two diagonals i.e. differ by double the horizontal component of one of these diagonal stresses. 46. The next example taken will be Fig, 80. Here we need to use a number of imaginary joints on account of the crossing of the bracing-bars. The structure is symmetrical, and is supported by abutments at the four points Q P S T. At the latter point T the supporting force is known to be zero, and, since no load acts here, the stresses on the links J 2 9 and 9 J 2 are zero. The loads 1 2, 2 3, 3 4, 4 5, 5 6, 6 7, 7 8, 8 9 are equal and inclined 30 to the horizontal. The loads 10, 11; 11,12; 12,13; 13,14; 14,15; 15,16; 16, 17 ; 17, 18 ; 18, 19 are equal and vertical. The whole load on the upper chord 1, 9 is in magnitude half the whole load on the lower chord 10, 19. The supporting force 20, 1 is supposed known in direction as at 45 to the horizontal, and Compound also in magnitude as f that of 1, 9. The supporting force girder 19, 20 is known to be inclined at 60 to the horizontal. These are the data. The right-hand cyclic order has been adopted in construct- ing this diagram. The stress-diagram is begun by plotting to a suitable scale 20, i and 123456789; then vertically downwards from 9 to 19' the loads on the lower chord. The location of these latter is only temporary, as the un- known force 0, 10 has been omitted from its proper cyclic position, and has to be inserted subsequently after being found. A pole is chosen, and parallel to radii from it are drawn the links of a simple one-pen linkage through the out- side pens 20, 1 2 3 4 5 6 7 8 (9, 10) 11, 12, 13, 14, 15, 16, 17, 18, 19 : 9 and 10 being taken as one space because of the omission of the force 9, 10 ; and the latter links being drawn CHAP. x. STATIC LINKAGES 201 parallel to the radii from the pole to the points n' 12' 13'. . . . 19'. This simple polygon is omitted from the engraving. The first and last link through spaces 20 and 19 are produced to meet, and through their joint is drawn B the line of the resultant || 20, 19'. The known supporting force line 19, 20 through P is produced to meet B, and from their joint is drawn a line to Q. This latter line gives the direction of the supporting force 9, 10, and parallel to it is drawn from 19' compound the line 19', 19 in the stress-diagram. In same diagram from si *H* n 20 is drawn 20, 19 || 20, 19 in linkage i.e. at 60 to horizontal. This gives the final position of 19 in the stress-diagram. The parallelogram 19, 19', 9, 10 is completed by drawing 9, 10 and 19, 10. The line 9, 10 is now properly located in the stress- diagram, giving the magnitude of the supporting force at k Q. The lines 9, 19' and 19' 1 9 may now be rubbed out for clear- ness' sake, and the lower chord loads are plotted off along the line 10, 19. 47. It is frequently the case, as here, that the joint of the load resultant line with one supporting force line lies far out- side the paper. The difficulty of drawing from this joint to another point such as Q is got over by a construction such as that shown in the figure. The point r is the joint of B and line P Q. From any other point r" in B directions to P and Q are taken, and lines rp f and r q' are drawn parallel to them. p' is the joint of rp' and 19, 20. From p' is drawn p f q' || P Q, Joints and its joint cf with r q' is a point in the desired line 9, 10 t he n through Q. The proof is obvious. Owing to similarity of tri- drawm angles, Pr _rr" __ rQ / ~j """ 7 ' """"" ~~7 *"> P r r r r q and therefore the three lines P p', r r' and Q q' converge to the same point. The difficulty may be otherwise avoided by the construction explained in Section 13 of Chapter VIII. 48. In the following table, describing the further construc- tion of the stress-diagram, a shorter method of writing out 202 GRAPHICS CHAP. x. the process is adopted. In each new joint- diagram we here leave unnamed the lines already drawn, and mention only the two new lines to be drawn. The joint is also left unnamed. The new point or pole determined by the joint of the two new stress-lines is placed last in the triangular name given to the process of finding it. Thus ^ > b 3 means that two lines are to be drawn from the already known points 8 and 6 2 , and that their intersection gives the point & 3 . It is, of course, to Con- be understood that these two lines are to be drawn parallel descrip- to the lines 3 B 3 and B 2 B 3 in the linkage. By help of this constrac- s J s ^ em the whole process for Fig. 80 is accurately expressed tion in the following table. The consecutive order in which the table is to be read is along the horizontal lines from left to right as in ordinary reading. 18 ^ 20 ^ ^iSfc 17 N &2\7 C 2\ 16 3^3 b^ C * C3 d *^J * N , 9 S; *a (coincident ) y s. II X 5 / -5 A'i n "i and ? - with and . 8 >i 3 * 2 >/l 2 * 3 >/* 3 ] >h, 6 \ a. \. ~ a, ;> g 5 ^ 4 /> j. y ? 94 h The points e 5 and / 5 are now determined. The line join- ing them should be || E 5 F 5 , that is, vertical. This is the test of the accuracy of the drawing. 49. When the structure is symmetrical and the distribu- tion of load also symmetrical, the two axes of symmetry coin- Symme- ciding, the stress-diagram is also symmetrical, its axis of tncaiEoof g y mme t r y being perpendicular to that of the linkage and of the loading. This leads to much simplification in the con- struction of the diagram. This is illustrated by Fig. 82, which shows a simple roof CHAP. X. STATIC LINKAGES 203 truss with vertical loading and vertical supporting forces. From the symmetry of the load distribution it is evident that the two supporting forces are equal, and, therefore, it is un- necessary to draw in a simple linkage in order to determine them. Each is equal to half the whole downward load. The lines 5 6, G G, and 14, 15 lie on the axis of symmetry. Eight- hand cyclic order is used. The loads are 1,2 = 9, 10 = 2tons; 2,3 = 8,9 = 3tons; 3, 4 = 7, 8 = 4 tons ; 4, 5 = 5, 6 = 6, 7 = 6 tons ; 11, 12 = 12, 13 =13, 14 = 14, 15 = 15, 16 = 16, 17 = 17, 18 = 3 tons. The whole downward load is, therefore, 57 tons, and each supporting force is 28 tons. These outside stresses are plotted to a suitable scale along the vertical line i, 10, n, is, i. Symme- The horizontal dotted line at the middle of this vertical line is the axis of symmetry of the stress-diagram. The construc- tion proceeds in the following order : There is no need of completing the other half, as the results would only be a repetition of those already obtained. The only test of accuracy is that the line g g should be bisected by the axis of symmetry. If the diagram were completed each pair of points with similar letter names (i.e. poles of similar right- and left-hand pens) should be equidistant from the same axis and should be vertically one over the other. Each pair would thus give a test of accuracy similar to that given by#g. 50. In Fig. 83 we have again a symmetrical structure, namely, a railway bridge with roadway along the bottom boom. Symme- The first stress-diagram is for a symmetrical load viz. a Bridge. uniformly distributed load over the whole length. As pens jjj**"~ 1, 2, 11, and 12 have only } the breath of the others, the loads rolling 12 and 11, 12 are each f , and 2, 3, and 10, 11 are each f , of 204 GRAPHICS CHAP. X. the loads at the other joints. The depth of the girder is f of the span. The loads are 1.2 = 11,12 = 6 tons; 2.3 = 10,11 = 7 tons; 3, 4 = 4, 5 = 5, 6 = 6, 7 = 7, 8 = 8, 9 = 9, 10 = 8 tons. Total load = 82 tons. Each vertical supporting force = 41 tons. Left-hand cyclic order is chosen, and the load line i, 12 is, therefore, plotted near the left edge of the paper. The three bars A 1, A 12, and L L have no stress in them. The two former bars have stresses developed in them when an end wind acts on the bridge ; when the girder expands by rise of temperature and thus thrusts horizontally on the abutments ; and when the head of a train comes on the bridge, thus com- municating to it a horizontal shock. The bar L L serves to support part of the weight of the long compression member Symme- (L L) and to lessen its buckling by keeping its middle point Bridge. in line with its end points. None of these three bars are use- buted 1 " ^ ess > therefore, although the stress-diagram shown indicates roiling zero stress. L L must be constructed as a strut, but requires load only a small section. The order of the construction is as follows : Mark a coincident with i. Then draw Mark 1 coincident with I, Then J>k, &c. The lower half of the diagram is symmetrical with the upper half ; j h and / should coincide with the corresponding heavy type letters, and kigecdb should be vertically over the corresponding heavy letters. These latter coincidences are the tests of accuracy in drawing. In the stress- diagram No. 2 the pen 12 is supposed to be unloaded, the other portions being loaded as before. All the CHAP. x. STATIC LINKAGES 205 joint loads remain the same except 11,12, which becomes 3 tons, and the supporting forces 1 becoming 41 4 ? T = 40f tons, and 12, = 41 - 2|f = 38 T \ tons. The order of proceeding in drawing the diagram is the same as above. This diagram is shown in the plate drawn on top of No. 1 diagram, and distinguished by being drawn in dotted lines, thus ..... . The third diagram, drawn in same place with ----- lines, is for the load covering all but the last two pens 11 and 12. The load 11, 12 is now zero, and 10, 11 = 4 tons, the others remaining unchanged. The total load is now 82 9 = 73 tons, of which 40ff - \\ 40 T 2 T tons is balanced by 1, and the remaining 32^ tons by 12, 0. The fourth diagram shows the results with all but the last three pens loaded as before ; the fifth, those with all but the last four pens loaded ; and so on, there being 12 stress-dia- grams drawn, the last of which corresponds to the case of the first pen only being covered with a load of 6 tons, and Symme- the only external forces considered being, therefore, a load of Bridge. 3 tons in line 1, 2, and the supporting forces, which are, in this case, 1 = 3 x || = 2f| tons, and 12, = 3 x -fa = T \ rolling ton. The student should carefully trace through all twelve diagrams and compare the results, which are very instructive. It will be noticed that the maximum stress in any bar of the upper or lower chord occurs when the bridge is fully loaded ; while the maximum for any diagonal or vertical bracing bar occurs when the larger only of the two portions of the bridge to one side of the lower end of the bar is covered by the load. This is not an invariable law, as will be seen presently, but it commonly holds good for the shapes employed in engineering structures. As under these various systems of loadings the individual links are sometimes in tension and sometimes in compression, it is impossible to distinguish between struts and ties by thick and thin lines in the frame- diagram. All those bars that are at any time hi compression have been marked by thick lines ; those that are always in tension by thin lines. 206 GRAPHICS CHAP. X. 51. The ' Method of Sections ' has already been men- tioned. One of its uses is to investigate the stresses in part of a structure without taking the trouble to draw out a stress- diagram for the whole. The dotted-line section P Q in Fig. 83 may be taken as an example. It crosses the pens 2, C, D, and 0, two of which are outside and two inside. The method is applicable only to a section crossing not more than 2 inside pens i.e. cutting not more than 3 inside links. The locor sum of all the outside forces (in this case 01 + 1 2 4 2 ^ E) is first found either by a one-pen linkage, or, as in the present very simple example, by numerical calculation. From the joint between E and the link 2 C is drawn a line to the joint between C D and D 0. The portion of the frame to the left hand of the section line is in balance under the four forces E, 2 C, C D, and D 0. The locor sum of the first two must pass through their joint ; the locor sum of the last two must pass Method of through their joint ; these two locor sums must be equal and opposite along the same line that is, E + 2C^CD + DO; and therefore each of them lies along the line last drawn. Parallel to this line, therefore, is drawn the line c o in Fig. 84, where o 2 4= E and 2 c \\ 2 C. This gives 2 c =|= stress in link 2 C, and c o =|= vector sum of stresses in C D and D 0. This latter is now resolved into its two components by drawing cd \\ C D and o.i || OD that is, by the operation J>d. We have thus obtained the stresses in the three links cut by the section plane P Q. The construction of Fig. 84 may be made in either of two forms as shown one corresponding to the left-hand cyclic order, in which the locors are cut in making left-handedly the circuit of the portion of the frame on the one side of the section, and the other being similarly right-handed. The former is coincident in every respect with the diagram o 2 c d o of the first stress- diagram of Fig. 83, except that it is drawn to a larger scale. CHAP. x. STATIC LINKAGES 207 52. For symmetrical loading the utility of this * method of sections ' is not great, because in this case one does not need to draw a one-pen linkage in order to determine the supporting forces ; and, therefore, it must be drawn specially for the sake of finding the position of the locor E in solving Symme- the ' section.' But one can draw the complete stress-diagram * n up to the section -line practically as quickly as the one-pen linkage can be drawn. But for unsymmetrical loading the supporting forces cannot be found without drawing the one- pen linkage, and a portion of this already drawn can be used to fix the position of E. Its use in starting the diagrams of peculiarly formed structures has already been mentioned. 53. Suppose the two boundary links cut by the section- line be parallel to each other and E parallel to both. The joint of E with either of these links is .at infinity, and the above construction fails. But in this case evidently the trans- verse bracing bar has no stress in it, because the resultant external force has no transverse component to be balanced by Parallel the thrust or pull of this bar across the plane of section. The stresses in the two boundary links will, therefore, be found by dividing E between them in proportion to their distance from E. This, of course, is easily done in the ordinary graphic mode. This case arises in taking sections of vertical and non-tapered piers with vertical loading. 54. This ' method of sections ' is not applicable to a section cutting three links which meet at one joint. As an example take Fig. 85. This is not put forward as an economical form of roof truss ; it is inserted here simply as illustrating several of the special features that have been spoken of. It contains Roof no joint where only two links meet, and yet it is non-redun- dant and stiff. A section through the pens 2 B A is not ;PJ, cia J dinicul- amenable to treatment by the above ' method of sections,' ties because the three bars 2 B, B A, and A all meet in one joint. The stress- diagram can, however, be commenced by help of 208 GRAPHICS CHAP. X. a section through the pens 4 E F 0, or of one through 5 G F 0. The former is used in the following construction. The load data are the following. The loads 1 2, 2 3, and 3 4 are vertical, the first 5 tons and 2 3 = 3 4 = 7 tons. The load 45 = 9 tons and is inclined at 30 to the vertical. The load 5 6 = 6 7 = 10 tons, each being inclined at 20 to the vertical, while 78 = 7 tons and is at 15 to the vertical. The horizontal component of the supporting force 8 is known to be -J- of the horizontal component of 1, both components having the same direction (viz. of course, from left towards right hand). With these data proceed as follows. Plot out the outside stresses i 2, 2 3 . . . . 78 in consecutive right-hand cyclic order. Join 8 i of stress-diagram, and parallel to it draw two lines from the supporting points P and P r Proceed first on the supposition that the supporting forces are parallel, and, there- Roofiiius- fore, both parallel to 81. Choose any pole p, and draw in a special corresponding one-pen polygon, drawing first the links through difficul- the spaces 12345678, and carrying the first and last of ties these up to intersect with the lines drawn through P and Pj parallel tosi. Join these two intersections. This gives the ' closing line ' of the pen, and parallel to it draw from p a line intersecting 8 i in o'. The actual supporting forces are 8 o' and o' i compounded with a pair of equal and opposite forces along the line P P r Therefore, from o' is drawn a line parallel to PP P and o is known to lie on this line. The horizontal projection of i 8 is divided in the proportion of 3 to 1, and o is found vertically over the point \ of this horizontal projec- tion from 8. The supporting forces 8 o and o i are now known, and lines parallel to them through P and Pj are to be drawn. That through P meets the link of the one-pen linkage through space 1 in a point from which must now be drawn a line \\p o. The joint of this line with that through pen 4 is to be found. It is called K in the diagram. Through this is drawn a line Q K || o 4, and the joint Q of this with link F is CHAP. x. STATIC LINKAGES 209 found. This joint Q is joined with joint (FE45G), and parallel to this line is drawn 4/ in the stress-diagram, / being the joint of this and of\\ F. The correct position of/ being thus found, the diagram proceeds in the following order : The line between b and a must now be found parallel to B A, and, therefore, coincident with da> if the drawing has been accurate. Continuing {>a >l >ft >i ^>/L A second & J 9 * test of accuracy is now found in that the line joining the points I and k so found should be parallel to L K. It may be noticed that since 2 B and 3 C are in the same line, we might have commenced by finding b c. Thus draw through points 2 and 3 lines 2 b and 3 c parallel to 2 B and 3 C ; any line such as b c drawn between these two and parallel to B C gives the stress on B C. Having found this, cd can be Koofiiius found by drawing b d \\ B A and cd\\CT>, because B A and gpeciaf A D are in the same line. In a similar manner e d may next difficul " ties be found. But this beginning will carry us no further than this, because we now find at each of the joints (A D E F) and (0 1 2 B A) that three unknown forces remain to be 'found. A similar remark might be made regarding the other side of the roof. It should also be observed that, in spite of the dissymmetry of the load distribution, the stresses in A and L are equal, as are also those in AF and FL. This results from the symmetry of the structure and from the fact that, while the joints (0 A F) and (0 F L) are 3-link only and are directly joined by a link F, no external forces act at these. They would still have been equal in pairs if equal and symmetrically disposed loads acted upon these joints ; for instance, equal portions of the weights of the bars themselves form actually occurring loads which still leave the above equalities un- disturbed. 210 GKAPHICS CHAP. X. If the links OA, OF, and OL were in line (along PPj), then A F and F L would be useless, there being no stress in them. Their only utility would be in suspending the long tie-bar (P P^ and preventing it drooping by its own weight. 55. The other method previously referred to as an alter- native to the method of sections in starting a diagram, when there exist no 2-link joints, is the ' Method of Two Trials and Two Errors.' In the present example one would assume a value of o a, and from it find the stresses in the neighbouring bars namely, the following stresses : 2 b, a fe, 3 c, b c, c d, a d, de, ef. The point e having now been determined, a line from it parallel to E 4 should cut the load line at point 4 in the Two trials stress-diagram. It does not do so, however, because of the 5 error in the assumption made regarding o a. The distance of the intersection from 4 is a measure of the error made. Next assume a new value of o a and repeat the process, obtaining a new error or deviation from the point 4. By linear inter- polation between these two errors the true points are at once obtained. It does not matter in the least whether the two guesses are near or far away from the truth. If the two errors come out -f and , so much the better ; and if one error be small and the other large a good accurate interpola- tion will be obtained. This process is illustrated in detail in Fig. 86, in which is Cantilever represented a bridge structure without any 2-link joints. The Bridge f ra me is of the cantilever type which has become common in recent years. One way in which this problem might be commenced would be to take a section through the joint at the top of pen A and through the bar OA. The moment of the force in OA round the above joint must balance the moment round the same joint of the resultant of the outside forces 01 89. The locor sum of these would be found, and its intersection with A joined with the joint ABU. This joining line gives CHAP. x. STATIC LINKAGES 211 the direction of the third side of a stress-triangle of which one side is the stress in A. A second mode of commencing would be to take a section through pens 9, U, B, 0, or one through 10, TJ, B, 0, proceeding as previously explained. In either of these methods the intersection of the locor-sum of external forces with the line of the bar A or B would be an ill-conditioned intersection, and although either solution, would probably be more rapid than the following, neither would be so accurate. The load data are the following : The downward loads are applied to the joints along the horizontal straight line K Q, E M, S L, S I, &c., all which loads are vertical, and are transferred to the upper boundary by the introduction of imaginary links ; and the two supporting Cantilever forces are applied at joints (01HG) and (18, OGH), that at TWO trials the latter joint being known to be inclined at 30 to the vertical, and the direction of the former having to be found by help of a one-pen frame. (This latter is shown in the figure in lines thus , and its stress-pole is marked .) The loads at the joints (Q 12), (P Q R), (L M R), (K L S) = each 13 tons. Those at joints (F I S), (E F T), .(C D T), (B C U) = each 20 tons. Those at joints (ABU), (CBTJ), (DCU), (EDT), (I FT), (K I S) = each 25 tons. Those at joints (MLS), (N M R), (Q R, 17) = each 12 tons. Total downward load = 318 tons. Using the method of two trials and errors, we may begin (after having drawn in the closed polygon of outside forces including is, o, and o i) by choosing for ' trial ' any arbitrary p 2 212 GKAPHICS CHAP. X. value i q f for the stress in 1 Q. From this point <( chosen arbitrarily on the line i q drawn from i || 1 Q, we proceed to find the following other points, in which the accent ' indicates that their positions correspond with the supposition that q' is the correct position of q : ?Sr' r 'Sr' r S*' *iS.' 2 /i 3/2 4/1 5^2 in' ^ 7, i\ T / i \ 7/ o \ / o - L/>* 7/>^7 V. h' d' Having now found the point 6', we notice that it is not on the line o b drawn from o \\ B as it ought to be. A second trial value of 1 q viz. 1 q'' is now taken, and exactly the same process as above repeated, resulting in a second wrong position for b, namely 6", which is still out of Cantilever the line o b but not at the same distance from it as b'. Next TWO trials through b' and b" a straight line is drawn to cut ofc||OB. and errors mj^ intersection b is the true position of 6, and the true diagram can now be drawn backwards from b towards q by reversing the series of operations indicated above ; thus, 9\ 8\ 7\ 6\ W Q \ 7 S UA > W > It., > U, 7 3 > C. Zr 1^4 u^ 2 %2 & Here we are stopped in this backward course because no proximate joint leaves only two stresses undetermined. This will not be generally the case, this special example having been chosen because it presents an unupual number of pecu- liar difficulties. Next draw through cf and q" two parallel lines in any direction different from o q, and plot off along these from ./' and q" to p' and p" the distances b' b and b"b. These are measures of the errors resulting from the two trials. In this example the errors are of opposite signs, and are therefore CHAP. x. STATIC LINKAGES 213 plotted in opposite directions from q' and q". Join p f and p" y and let the joining line meet o q in q. This intersection q is the correct position of q, and from it the diagram may be proceeded with up to c in the same order as that used for the trials. That this forward series of operations gives the same position for c as the above-mentioned backward series, affords a proof of the accuracy of the drawing. Evidently the backward operations were not really necessary, because, having found q correctly, the forward series could be continued as in the trials up to b. The proof of accuracy of drawing would then consist in finding the b given to lie on o b \\ B and also in the line b' b". The rest of the diagram may now be drawn as follows : At this point the same difficulty reappears ; there is no cantilever proximate joint at which only two as yet undetermined JjjJ^ais stresses act. Since q is not yet known, we have once more and errors to fall back on the method of trial and error. Draw from u^ a line || II 2 T 2 and select at random any point t/ upon it. Then proceed as follows : This s 2 ' is found not to be on the line 14 s drawn from u || 14 S, as it ought to be. Another point t 2 " is now selected on u 2 1 2 , and the process repeated whereby a point s 2 " is found, still not on the proper line 14 s but not at the same distance from it as s/. Then s 2 ' s 2 " are joined and the line produced to meet 14 s> the intersection s 2 being the true position of s 2 . The two errors s 2 ' s 2 and s 2 " s 2 are now plotted from t,/ and t 2 " along parallel lines in any direction to the points TT' and TT". The ine joining IT' and IT" cuts u 2 1 in t 2 , the correct position of t 2 . 214 GRAPHICS CHAP. X. This point having been obtained, the drawing of the true diagram proceeds as in the trials up to the point s 2 . It is then drawn as follows : Finally the points m and n are joined, and the final test of accuracy of drawing is that this line m n should be parallel toMN. It is to be noticed that the diagram shows a stress r 2 s l different from the outside force 3 4, which is really applied at joint (L M R S). Now of this stress r 2 s { a part = 34 belongs to the imaginary link supposed to lie along this line. The stress on the actual link R S is, therefore, the algebraic differ- ence between r 2 s l and 3 4, the stress in the imaginary link being, of course, a compression. Since r 2 s t indicates a resul- Cantiiever tant tension, the real tension on the actual link is the arith- metic sum (T^SI + 84), taking both these lengths positively. This is a most important rule to attend to in using imaginary links ; if it be not very carefully adhered to, in the case of imaginary and actual links lying along the same line, the stress-diagram will be altogether wrongly interpreted as re- gards these actual links. The same remark applies in the present example to the links S T, T S, and S R. The stresses on these four actual links are on R S + 65 + 13 = + 78 tons tension ; S T - 355 + 20 = - 335 compression ; TS 343 + 25 = 318 compression; S R + 61 + 12 = + 73 tension. Here the imaginary stresses 13, 20, 25, and 12 are added (i.e. given a + sign), because they are compressions (or negative stresses) to be subtracted. CHAP. xi. BEAM LINKAGES 215 CHAPTER XI. FLAT STATIC STRUCTURES CONTAINING BEAM-LINKS. 1. WHEN the structure contains beam-links that is, links exposed to and capable of resisting bending moments it *is necessary to determine these bending moments for all the. sections, as well as the shear force and the direct pull or thrust along the axis of the beam. The total or resultant force across any section is always capable of being looked on Beam as a single force through the centre of figure of the section 8 combined with a force-couple. The latter furnishes the bend- ing moment on the section. The former is usually resolved into two components : one, viz. the shear force, parallel to the section plane, and the other normal to it and producing either compression or tension. 2. There are various methods by which these beam stresses may be dealt with graphically. One is to suppose substituted for the beam a stiff non-redundant triangulated frame. When properly interpreted the stress- diagram for this substituted frame gives all the required forces acting on the actual beam. This method, although legitimate and Various elegant, is seldom necessarily followed. A somewhat similar device was employed in his paper on ' Frictional Efficiency of Mechanisms,' read by Professor Fleeming Jenkin before the Royal Society of Edinburgh in April 1877 and February 1878. Another mode of solution is the successive use of the ' method of sections ' at various parts of the frame ; or else of the method of ' two trials and two errors.' 3. In some simple cases, however, it is not necessary to 216 GRAPHICS CHAP. XI. resort either to this substitution or to the latter method. The diagram on account of its simplicity becomes easily intelligible without the artificial aid of the supposititious triangulation. Fig. 87 illustrates the method to be adopted for such simple linkages. The beam is indicated by a double line viz. a fine line indicating the position of its axis, underneath which is drawn a thick line. The loads are given in direction and magnitude. The supporting force 8, 9 is given in direction ; it is vertical. By the construction of a single-pen linkage the direction of 14, 1 and the magnitudes of 14, 1 and 8, 9 are found. Beginning at the joint 4 5 C, the stress-diagram is then constructed as follows : This supplies the poles of all the inside pens. The points abde are then joined with u, 13, 10, 9; and c with both 12 Sim le an ^ 1 1 * Next dotted lines from abcde are drawn parallel cases to the beam axis, and their intersections with the load line 9, 14 are marked a' b' c' d' e'. The stresses on the non-beam links require no comment. The force actions of the various parts of the beam, exclusive of couples or pure bending moments, on the pins of the differ- ent joints are as follows : That of part A, 14 on pin 14, 1, 2 A is a H and on pin 14 A B 13 is 14 a ; B 13 14 A B 13,, 6 13 13BC12 1.36; C 12 13BC12,, ci2 12,0,11, i2c; C 11 12C11 en 11 CD 10,, nc; D 10 11CD10,, dio 10DE9 ,,iod; E 9 10DE9 69 9E78 9 e. Each of these except the first and last is combined with a bending moment. Each of them is conveniently resolved by the dotted lines into a shear force normal to the beam-axis (which direction here coincides with that of the lower external CHAP. xi. BEAM LINKAGES 217 loads) and an axial force producing direct tension or com- pression. For instance, the force 13 b with which the part 13 B acts on the joint (13 B C 12) is resolved into the upward shear 13 &' and the axial pull b' b. The force of part 10 D on joint (10 D E 9) is resolved into the downward shear 10 d' and the pull d f d. With the particular data of this example the beam is in tension throughout its whole length, but in general it may be in tension at one part and compression at another. At the joint of the force-line 11, 12 with the beam viz. joint (11, 12, C) the shear force of the left-hand portion is 12 c', and that of the right-hand portion is c' n, the shear in both these portions being of the same sign viz. such as to give a right-handed twist to the material of the beam viewed as in the figure. The sign of the shear changes at joint (11 C D 10), the actions n c' and d' 10 on the two sides of this joint being both upwards. A B and B C are tie-bars ; C D and D E are struts, as also Simple 2 A and E 7. The forces acting on the beam are those exerted by these bars and the external loads applied at the beam-joints. The components of these forces parallel to the beam-axis have no bending moments on the sections. To make this statement generally true, it must be carefully noted that the joints where the forces are resolved into their com- ponents must be taken as the geometrical joints between the force-lines and the beam-axis. The actual joint-pin centres that is, the actual points of application of the forces to the beam may lie out of the beam-axis, in which case the draughts- man must be careful to avoid using these actual joints instead of the geometrical joints of the axial lines. At the right-hand end of the beam the forces e 7, 78, and 8 9 have the resultant e 9, whose vertical component is e 9. At the left-hand end the forces are u, i + 12 + 2 a=jj=ua, whose vertical com- ponent is 14 a. The vertical component of a b, the force exerted by link A B on the beam, is a' b' ; and so on. The 218 GRAPHICS CHAP. XI normal, or bending-moinent-producing, forces acting on the beam are, therefore, e' 9; 9, 10; 10, n ; n, 12 ; 12, 13 ; 13, 14; ua' ; a'b'; 6V; c' d' ; d' e f . We find these arranged in consecutive order in the stress- . diagram that is, forming a continuous and closed chain. The pole p is chosen, making the pole-distance p (e' 9) a con- venient multiplier. In this case twenty tons is used. Then parallel to the pencil (p) e' 9, 10,11, 12, 13, H, a' V c' d' is drawn the one-pen linkage (P) E' 9, 10, 11, 12, 13, 14 A' B' C' D'. It is to be observed that the spaces A' B' C' D' E' do not coin- cide with ABODE, the dividing lines A'B', B'C', &c., being vertical i.e. normal to the beam-axis. If the work has been accurate this will be a closed linkage. Its vertical depth directly under any section multiplied by the pole- Simple distance p (e' 9) .gives the bending moment on that section. Since the frame has been drawn to the scale " = 1 ft., there- fore, one inch depth on moment-diagram = 5 ft. x 20 tons = 100 foot-tons. We have thus obtained a diagram giving the stresses on all the two-joint links, and also the direct thrust or pull, the shear force, and the bending moment on every section of the beam. In Fig. 88, a structure of similar design, but with the loads on the lower joints oblique to the beam axis, is shown. The points a b c d e are found as in last case. Then from 14 is drawn a line Ha' perpendicular to the beam axis, and upon it are projected at right angles to it the points a b c d e 9, 10, n, 12, 13. The projections are marked a' b' c' d' e' 9' 10' n' 12' 13'. Then u a', 13' &', 12' c', 11' c', 10' d', $' e' are the shears on the transverse sections of the portions 14 A, 13 B, 12 C, 11 C, 10 D, 9 E of the beam. The shear changes sign at joint (13BC12). The axial thrust CHAP. xi. BEAM LINKAGES 219 along 14 A is a' a; and that along 13 B is (b b f is, is'). The rest of the beam is in tension, the tensive stresses being (12, 12' + c c) on 12 C ; (n, n' + c c) on 11 C ; (10, 10' + df d) on 10 D ; and (9 9' + e e) on 9 E. The set of transverse forces producing pure bending are : Simple The set of transverse iorces producing pure bending are : H a', a' b', V c', c f d', d' e', e' 9', 9' 10', 10' 11', n' 12', 12' 13', is' H'. These being arranged consecutively along the line H a', a pole p at a suitable pole-distance is chosen, and from it the moment diagram or one-pen linkage (P) 14' A' B' C' V E' 9' 10', &c., is drawn through the spaces divided by the dotted lines drawn from the beam-joints normal to its axis. 4. In these cases the easiness of the problem arises from the fact that there exists in the frame a joint at which only two links meet, these two links being each a two-joint link i.e. not a beam. The diagram can be at once begun at this joint. In Fig. 89 at the right and left hand extremities there are joints where only two links meet, but in each case one of the two links is a beam. The reaction of the beam on the pin is not in the direction of the beam axis ; it is in an unknown direction, and, therefore, the diagram cannot be commenced at either of these points. A section is taken through pens 4B1. The resultant locor K4fM2 + 23 + 34is found and its joint with B 1 de- cases termined. This joint is joined with joint (B 4 5 C), and Roof with parallel to the joining line is drawn 4 b in the stress-diagram to meet i b \\ I B at the point b. b 4 gives in direction and magnitude the force exerted by the beam on the joint-pin B 4 5 C ; also b i gives the true stress on B 1. The diagram now proceeds as follows : P>a; join as; take section through spaces IE and joint (ED67F); find resultant locor B / ^(12 + 234 34 + 45 + 56) and its joint with E 1; join this joint with joint (E D 6 7 F), 220 GRAPHICS CHAP. XI. and parallel to the joining line draw 6 e in stress-diagram to meet i e drawn || 1 E ; this gives e ; then f]> c; e ^> d ; join c 5 U C and d 6 ; take section through spaces 9 H 1 ; find K" 4f|= (9, 10 + 10, 11 + 11, 1) and its joint with H 1 ; join this joint with (H G 8 9) and parallel to the joining line draw 9 h in stress-diagram to meet i/*||lH; then JX/; ^9> ^X/J joints and/7. The beam (3, 4) (A, B) is kept in balance by the following forces : 12, 23, 34, 45, 5C, cb, bd, dl. These resolved transversely across the beam are : l'2', 2' 3', 3' 4', 4' 5', d' C, cb, ba,ai', where it must be noted that the struts A B, B C, C D are all Roof with normal to the beam-axis. The shears upon the two parts 3 A and 4B of the beam are 3' a and 4' 6, these being of opposite signs. The axial stresses are 3 3' on 3 A and 4 4' on 4B, both being thrusts. The bending moment-diagram is drawn for the above set of balancing transverse forces namely, as', 3'*', *'b, and ba, the three forces ai', i'z', 2'$' being equal to the single force as', and similarly for the other end. The pole-distance p l (z'b) is chosen so as to give a convenient scale for the moment-diagram. The force-diagram having been drawn to the scale \ inch = 1 ton, the pole-distance has been taken 2 tons. The moment- diagram consists of four lines only, and is called P t in the truss-diagram. The beam (5, 6) (C, D) is treated in the same way. The shears are s' c and 6'd in the two portions of the length, and the axial forces, both thrusts, are 55' and 66'. The moment- diagram is drawn with the pole p 2 with same pole-distance 2 tons from the pencil (p 2 ) c56d, and is marked P 2 in the diagram. The beams (7 8) (F G) and (9, 10) (H J) are treated simi- CHAP. XT. BEAM LINKAGES 221 Itirly, poles p 3 and p being chosen for them with the same pole-distance namely, 2 tons as for the two beams at the left-hand side of the the truss. Here the substitution of triangulated frames for the beams has been avoided by making free use of the ' method of sections,' no less than three different sections having been requisite for the completion of the diagram. 5. In Fig. 90 is shown a simple illustration of common occurrence of a beam structure which can be dealt with by the method of sections. It is a pier composed of two upright columns capable of resisting bending and braced together by three two-joint links. The upper joints are exposed to vertical loads, and horizontal wind forces act on the two joints of one column. The structure is supported at PL and P 2 , where it is pin-jointed to the foundation. The force exerted by the foundation at P t is known to be vertical ; that at P 2 (or 61) is found by the construction of a single-pen linkage, the pole p being used. Then the line 6 1 in the frame is produced to meet 1 B, and its joint with this line is joined with joint (6 B A 5) . Parallel to this last line is .drawn 6 b in the stress- diagram to meet in b the line i b drawn parallel to 1 B. This gives the true position of b as is recognised by considering a section through spaces 1 and B and through joint (6 B A 5) . Then proceed with In the diagram it is an accident that the point a falls on the line 62. Next join as and project a vertically to a! on line 2 4, and project c to e' on i b. The axial and shear stresses on the beams are now : Axial Transverse or shear Lower part of left-hand column . . Upper part of left-hand column . . Lower part of right-hand column . . Upper part of right-hand column . . 12 tension a of compression 6 6' compression 6 6' compression 3 a' = 16' 66' Braced Pier 222 GRAPHICS CHAP. XI The transverse forces on the left-hand column are 2 3, and the horizontal component of ai, applied at joint (AB123), and (34 + 5 a), applied at joint (3 4 5 A). Since these com- bine into zero forces applied at these points, the moment- diagram reduces to zero, which means that the stress along this column is wholly axial. The transverse bending forces on the right-hand column are b 6' applied at the top joint, and which is the horizontal Braced component of the resultant (b a + a 5 + 5 o) ; G' i applied at the bottom joint, and i b applied at the middle joint. The pole distance p l (i b) is chosen equal to 10 tons, and from the pencil (PI) b G' i is drawn the moment-diagram P. The scale of this diagram is ^ inch = 1 foot-ton. The left-hand column is a pure strut, not exposed to trans- verse bending forces. This is the result of our having taken as one of the data that the force at P t is vertical that is, in the direction of the axis of the column. If it had been in any other direction there would have been a moment-diagram for this column as well as one for the other. 6. In the example taken in Fig. 89 it was possible to start by taking a section. The section was through one tension-bar and a beam ; but the part of the beam cut by it was an end part i.e. no joint intervened in the beam between the section and the end of the beam. Thus a point was known namely, the end joint of the beam through which the resultant force- Koof with action of this part of the beam was known to act. It was beams. only for this reason that the section was directly soluble, and error "** m ^ ac * ^ m ^^ ^ e simpler to consider the section as taken through the end beam-joint and the tension-bar. A section through a two-joint link and a beam joint which is not an end joint (and at which, therefore, the bending moment is not necessarily zero) is not directly soluble. In the following ex- ample (Fig. 91) many such sections could be taken, but they are useless as aids in commencing the diagram. There is also CHAP. xi. BEAM LINKAGES 223 no section possible through a beam end-joint and a simple non-beam link. The substitution of a triangulated truss for the main lower beam would furnish no direct means of commencing the dia- gram. If such triangulated trusses were substituted for all the three beams, then the diagram could be wholly deduced by the method of trial and error already explained. This method of trial and error is the only available one for solving this problem, but we will here carry it out without the aid of substituted triangulated frames. At first sight it might be supposed that this was a pe- culiarly easy case to deal with because there are no less than three joints namely (L 4 5), (16 V 15), and (Q 9, 10), at each of which only two links meet ; and there are other two namely, (4 D B 3) and (1 7 J H 16), where two links are in line and where, therefore, the stress in the third can be found at once. But a little examination shows that the beginnings Roof with that can be made at these points do not lead far towards the J^ms building up of the diagram. Trial and Not counting beam-joints intermediate between the ends, the whole number of joints in the structure is 22. There should, therefore, be 44 3 = 41 links ; and this is the actual number, namely, 3 beams and 38 two-joint links. All the loads on the right-hand side are vertical ; those on the left-hand and the top joint are inclined 15 from the ver- tical, except 12 and 45, which are at 45. The loads 11, 12; 12, 13; 13, 14; and 15, 16 are each 3 tons; 56, 67, 78, 8 9, and 9, 10 are each 5 tons ; 16, 17 and 17, 18 are each 6 tons ; 1 2 and 2 3 and 3 4 are each 10 tons ; 18, 19 is 4 tons, and 4 5 is 7 tons. The load V V is 2 tons and Q Q' is 4 tons; these are applied at internal joints, but are transferred to the boundary by introducing the imaginary links V V and Q Q' and imaginary joints at the upper ends of these links. The vertical direction is given for the supporting force 19, 20 as one of the data. 224 GKAPHICS CHAP. X). After plotting consecutively the above loads, the single- pen P is drawn with help of the pole p, chosen in any con- venient position, and the pencil of rays drawn from p to the corners of the load-line. The resultant E ltl9 of these loads being found by help of this pen, the given line of 19, 20 is produced to intersect this resultant, and the line from this intersection to the left-hand end joint of the lower beam is that of the supporting force at this end. In this way the magnitudes of 19, 20 and 20, i are obtained in the stress- diagram. We now begin by solving the joint Q 9, 10. As the stress on the imaginary link equals the imaginary external force applied at this joint, the balancing diagram can be drawn omitting these in the first place, thus giving the stresses on Hoof with the bars Q 9 and 11 Q'. These being obtained, the known beams. imaginary forces at this joint are next introduced in their error *** P r P er cyclical order, and the diagram for the joint is thus completed. The diagram now proceeds thus : 5S. *^\s SN. 7 \ 771s. 15\ / I4v 8 v , > U >. V U So far no trial and error has been used. But we can proceed no further in this direct manner, and the beams F (L N 0) and F (S T V) are still unsolved. The former of these beams along with the portion of the roof lying above it is acted on by a set of forces 46 + 56 + 67 + 7 8H 89 + 9 10+10 11 + 11 12 + 12S, all of which are known. Their resultant is found in the usual way, its magnitude and direction being given in the stress- diagram by 4 s, and its line in the /rawe-diagram being ob- tained by help of the link polygon and being marked E 4 ,. This set of forces is to be balanced by a force S F exerted by the right-hand beam at its upper joint and by two forces F D CHAP. xi. BEAM LINKAGES 225 and D 4, which may meantime be considered as compounded into a single force F 4 acting through the lower joint of the left-hand beam. The lines of these two forces S F and F 4 must intersect in some point of the line E 4S . Guess this point to be /3 P and parallel to the two lines from /3j to the two ends of the beam draw in the stress-diagram the two lines 4/j ands/j from the two already known points 4 and s. Mark their intersection / r This is a position of / result- ing from guess ft^ on line E 4S . Next find also by the link-polygon method the resultants B** #45+56+ 67 + 78 + 89 + 9 10+10 11 + 11 12 + 12 18 + 1314 + 1415 + 1516, and Ko,i6 #08 + 8 9-!- 9 10 + 1011 + 11 12 + 12 13 + 18 14-; 1415 + 1516. The former E 4> 16 is the sum of the known forces acting on the Roof with part resting on the two beams, and has to be balanced by beams. 16F^16H + HF through joint (16 H F) erroJ and and + D4 through joint (F D 4), this last being the same force as already mentioned and guessed to be in the direction from (F D 4) to r The other resultant K , 16 is the sum of the known forces acting on the part resting on the right-hand beam, and has to be balanced by 16 F, the same force as has just been cited, along with F 0, the force exerted by the left-hand beam at its upper joint. Produce the line (F D 4) & { to meet E 4>16 in p and let the line ! (16 H F) meet E 0>16 in y r If the guess /3 l were correct, then 7j would be the point in E 0> 16 where the two forces 16 F and F would meet. Parallel to the two lines from y { to the two ends of the beam F (S T V) draw in the stress-diagram from the known points o and 16 the two lines o//and ie// intersect- ing in//. This is another determination of the point / resulting from the guess y^ ; and the discrepancy between this position Q 226 GRAPHICS CHAP. xi. and the first found / is a measure of the error involved in this guess. Next guess a second point /3 2 on line K 4S , and repeating ex- actly the same process, find points 2 72/2 an d//- The latter two are the two discrepant positions of/ resulting from the suppo- sition that /3 2 is the correct point, and the distance / 2 / 2 ' is a measure of the error involved in that supposition. From the properties of the ' parallel displacement of the links of the single pen' or the 'proportional displacement of the joints of the single pen,' explained in Section 12 of Chapter VIII., it follows directly that all possible positions of/ found from the first step in this process must lie on the line /!/ 2 , while the locus of those found from the second step is////. The intersection of these two lines is, therefore, the true position of/. The position thus found should be checked before proceeding with the diagram by drawing from the three joints at the ends of the beams four lines parallel to /4, fs, /o, /16. The first pair Roof with should meet in E 4S ; the latter in E 0) 16 ; and the first and the beams. last of the four should meet in E 4> 16 . In the example shown 'rial and ^ nese intersections came right without visible error. error Having thus found /, the diagram proceeds as follows : s t> />* All the required points are now found, and all the forces acting at the various joints of the three beams are known. These forces are next resolved, as in the last examples into axial and transverse components, and from the latter are drawn in moment-diagrams. The stress-poles chosen for these moment- diagrams are marked p l p z and p y The pole-distance taken for the long lower beam is four times as great as that taken for the two short upper beams, so that its moment-diagram appears to a scale only one-quarter as large as that of the short beam diagrams. The accurate closing of these three inonient-dia- CHAP. xi. BEAM LINKAGES 227 grams furnishes additional tests of the correctness of the drawing. 7. In Fig. 92 is shown a stiff non-redundant structure with two beam links. This might be dealt with by means of a section taken through the two-joint link 1 A and the joint (2 1 A) namely, the joint between the upright beam and the nearly horizontal beam. The section must not pass through the upright beam, but only through the spaces 1 A 2. The joint is a flexible one, so that the force exerted by the upright beam on the other passes through a known point namely, this joint. The resultant of the forces A 1 and 1 2 must, there- fore, pass through this joint. The lines A 1 and 1 2 being produced to meet, their joint would be joined to joint (2 1 A), and parallel to the joining line would be drawn 2 a in the stress-diagram from the point 2 to meet in a the line i a drawn from i parallel to 1 A. The problem is, however, solved here by the aid of a tri- Crane v angular truss substituted for the upright beam as an example two of this latter method. l^ti. The data include, besides the dimensions of the structure, tuted . triangu- the magnitude and direction of the loading force 1 2 applied lated at the extremity of the jib beam ; the position and direction of the supporting force 2 3 ; and the point of application of the supporting force 3 1. These three being the only external forces, their lines must meet in one point. Producing 1 2 and 2 3 to meet, their joint is joined to joint (313). This line gives the direction of 3 1, and this being known the triangle in the stress -diagram is completed J>3. Any point to the left of the upright beam in space 1 is taken as the position of an imaginary pin, and this joined to the four joints of the beam by the imaginary links 1 G, G F, F E, and El. In the diagram these links and their stresses are indicated by dotted lines in order to distinguish them clearly from the real links and actually occurring stresses. It Q 2 228 GRAPHICS CHAP. XI. Crane with two beams. Substi- tuted triangu- lated truss must be distinctly understood that they are only construction lines ; they are not used in the final reading off of the stresses throughout the structure, and in practical designing they might be drawn in pencil only, and rubbed out after inking the other lines. The stress- diagram is now built up as follows : Then 2 is joined with ab c and d\ 11" and 22" are drawn _L 3 e ; through a is drawn i' d r the beam 2 (A B C D), and n', 2 2', b b', c c' and d d' are drawn 1 i' d r . All the forces acting on the beam 2 (A B C D) are now found, so that it is unnecessary to suppose substituted for this beam an imaginary triangulated truss. The interpretation of the diagram so far as the beam stresses are concerned is the following : Upright Beam. In part Axial stresses Shear stresses 31or3E 2 1 or 2 F AlorAG 3 i" compression 2"i" compression Zero l"l Zero ai The axial stress on the top part A 1 of the beam is zero because the two-joint link A 1 is placed at right angles to the beam : if it were inclined otherwise this axial stress would not be zero. The shear stress is zero in the middle part of this beam because the load 1 2 acts parallelly to this beam, so that the components of 2 3 and 3 1 perpendicular to the beam must balance each other that is, the horizontal component of 23 + 314^2118 Z ero. The shears on top and bottom parts are of opposite signs. CHAP. XI. BEAM LINKAGES 229 Inclined Beam. Iii part Axial stresses Shear stresses 2A 2 i f compression 2 ' a 2B (2 2' b b^) compression 2'b' 2C (2 2' c c') compression 2'c' 2D (22' ^ df} compression 2'd' The shear in this beam changes sign at joint (2 C D), and it is here, of course, that the maximum moment occurs. In the upright beam the bending moment is constant throughout the middle part 2 F where the shear is zero. 1 Two poles p l and p 2 are chosen at a convenient (and each at the same) distance from the lines i" a and ad', and from the pencils (p,) i"i a \" and (p z ) a b' c f d' iV a are drawn the two moment- diagrams Pj and P 2 for the upright and inclined beams. The force a i which the vertical beam exerts on the pin at Crane its top joint is the resultant (a g + g i) . The force 2 i with |, eams . which the middle part of this beam acts on the pin at top of middle part is the resultant &f + fg + g i) ; the stiff beam triangu- section h here equivalent in its action to the three two-joint truss links 2 F, F G, and G 1 of the substituted triangulated truss. On the same pin the upper part of the beam acts with a force i a, the resultant (i g + go). The vertical beam has, there- fore, a total force-action on this same pin, and therefore also upon the left-hand end of the other beam, equal to the resul- tant (zf + fg + gi + i g + 9 o) =jj= 2 a. The reaction of the inclined beam on the vertical beam is a 2. The positions of the points efg depend altogether upon that arbitrarily chosen for the joint (1 G F E) ; but the actual stresses in the diagram are not dependent in any degree upon the positions of efg. The student should satisfy himself that the positions obtained for the points abed do not vary with 1 The bending moment on a beam always reaches its maximum at the section where the shear is either zero or changes sign. 230 GEAPHICS CHAP. xi. change of position of joint (1 G F E) by taking several differ- ent points for this joint and working out the diagrams for all of them. If the drawing be accurate, the different diagrams will all lead to the same points a b c d, and the two moment- diagrams will both close. CHAP. xii. SOLID STATIC LINKAGES 231 CHAPTER XII. SOLID STATIC STRUCTURES. 1. STRUCTURES built in three dimensions offer naturally much more difficulty in the investigation of their stresses than do plane structures. Notwithstanding this superior difficulty, it is of the greatest engineering importance to be able to deal with them so as to arrive at a more or less exact knowledge of the forces through the various sections. The importance of this part of the subject becomes evident when it is re- membered that as a matter of fact all structures are solid. A AH stmc- girder or roof-truss may be treated as a plane structure with S oiid S * a certain approximation to truth so long as the forces acting on it are parallel to its central plane; but it must not be for- gotten that the girder, viewed in this light, is in unstable equilibrium transversely to its plane, and that it would not stand up when side winds act upon it except for the wind- bracing. This wind-bracing does not lie in the above plane, and when it is included as part of the girder the structure becomes a solid one. 2. It will be well at the outset to tabulate the fundamental distinctions between plane and solid link structures as regards stiffness, equilibrium, and solubility of the stress problem for an individual joint. Stiffness. It has already been explained that to ensure Flat and stiffness and avoid redundancy, the number of two-joint links f^ mea in a plane structure must be three less than double the number com P ared of joints ; and that in a solid structure it must be six less than thrice the number of joints. 232 GRAPHICS CHAP. XTI. Equilibrium. In a plane frame, when the known loads are balanced by a single force, as in a steel-yard, the condition of balance determines the magnitude, the direction, and the posi- tion of the necessary balancing force. In a solid frame, similarly, when the loads are balanced by a single force, this is determined in magnitude, direction, and position by the condition of balance ; but it is not gene- rally possible to produce the balance by a single force. When the loads on a plane frame are balanced by two forces, the two points of application and the direction of one of them are independent of the condition of balance ; and, these being otherwise fixed, the direction of the second and the magnitudes of both are determined by this condition. A further limitation is that the two supporting forces must be in the plane of the frame and of the loads. If the loads on a solid structure be balanced by two forces, Flat and the supporting forces must lie in one plane with the resultant frames of the loads, and they are subject to the last set of conditions ; compared ^ u ^ a g a j n> j^ w ^ no j n g enera i b e possible to obtain balance with two forces only. If the loads on a solid structure be balanced by three forces, then the three points of application, the direction of one absolutely, and the direction in a given determined plane of a second supporting force, are independent of the balance- condition ; but, these being otherwise fixed, this balance-con- dition determines the magnitudes of all the three supporting forces, the plane of the second, and the direction absolutely of the third that is, as regards the second, the condition of balance fixes a plane in which it must lie, but not its direction in that plane ; and as regards the third it fixes the direction completely. In general the supporting forces of a solid structure, what- ever their number be, are determined in six elements by the condition of vector and locor balance. Solubility of Joints. In a plane frame the stress-diagram CHAV. xii. SOLID STATIC LINKAGES .233 for any particular joint-pin cannot be drawn out until all the forces except two acting on it have been found. In a solid structure the stress-diagram for any particular joint-pin can be drawn as soon as all the forces acting on it except three have been found. Thus in the plane problems the stress-diagram must be commenced at a joint where only two links meet, unless one Flat and of the special methods already explained be employed. But J2jj es in a solid structure there are no such joints. The stress- com P ared diagram is begun at a joint where three links meet, of which joints there are generally two in a stiff, non-redundant, three- dimensional linkage. These fundamental laws of solid linkages will have their truth made evident in the course of the following illustration, showing how the stress-diagram may be obtained. 3. Fig. 93 shows the simplest possible solid linkage namely, a tetrahedron. It has four joints and six links, thus fulfilling the condition of stiffness I = sj 6, or 6 = 3 x 4 6. ABC and A' B' C' are two projections of the structure on two Tetra- planes at right angles to each other, the dimensions of either of which projections parallel to line G G may be derived from the other by projection perpendicular to G G. The lower view ABC will be called the plan, the upper the elevation, and the line G G the ground line. G G is taken as the plan of the vertical plane of projection of the elevation ; and G G is also the elevation of the horizontal plane of projection of the plan. It is impossible to carry out completely the system of cyclic nomenclature found so convenient in plane diagrams Lettering. namely, that of lettering the surface-pens lying between the order 5 link-lines. The corresponding system would be to give a letter- name to each volume, or solid pen, and if this were practicable it would give a perfectly complete and consistent nomenclature for all parts of the frame and stress-diagrams. In the frame of Fig. 93 there is one solid inside closed pen, and there are four solid outside unclosed pens namely, one corresponding to 234 GRAPHICS CHAP. XII. each face of the tetrahedron. At each joint, if the four solid pens meeting at it be taken in a certain cyclic order, the link- line common to any two successive solid pens may be called by the two letters which are the names of these pens ; and in this case the two corresponding letters in the stress-diagram would Tetra- be found at the two ends of the stress-line for this link. But frame as ^ ne drawing is on a flat sheet it is impossible to place the letters on the paper in such a way that they unambiguously indicate the volumes which they are intended to name. More- over, the choice of the correct cyclic order is difficult and con- fusing. Therefore, the ordinary system of naming the pens of flat structures is here applied to one of the projections, and the forces arranged in the proper cyclic order for this one view, the lettering of which serves as a complete guide throughout in building up and in reading the stress-diagram. In Fig. 93 it is the plan that is thus lettered. The names ABC are given to the plans of the three upper faces of the tetrahedron, lettering. Notice that these letters indicate not the faces themselves, but order 6 ^ ne plan* of the faces. 1234 are the names given to the plans of the outside spaces bounded by the plans of the frame and of the outside load-lines. No name is given to the lower face of the frame. This plan now gives a name to each inside and outside link, the left-handed cyclic order being preserved in the particular example now under consideration. The two letters of this name for each link are now written along the elevation of the same link, the letters, however, being accented in order to distinguish readily plan from elevation. This accentuation, although it is here employed in this first ex- ample for the sake of clear description, will be found (vide subsequent examples) to be unnecessary in practice when the operator has become familiar with the method of procedure. It is to be noticed that, although the outside force 1 2, 1' 2' acts at an outside joint, still in the plan this joint is inside ; and, therefore, the pen B in the plan has to be split CHAP. xir. SOLID STATIC LINKAGES 235 in two, Bj and B 2 , and an imaginary link E l B 2 and joint (1 2B 2 B,) introduced. The data are the dimensions of the structure ; the point Tetra- of application, the direction, and the magnitude of the load f r e a me 1 2, I' 2' ; the direction of supporting force 3 4, 3' 4' ; and the direction of the elevation of the supporting force 2' 3'. At each soluble joint in a solid frame there act a number of known forces that can be compounded into one known re- sultant, and three unknown forces acting along the lines of Typical three links whose stresses have not yet been determined. The top joint of Fig. 93 at which the known force 1 2, V 2' acts is, therefore, typical of the general problem. 4. The mode of solution is to resolve the known force into two components one perpendicular to the plane of two of the unknown forces, and the other parallel to this plane. The third unknown force being imagined similarly resolved into two components perpendicular and parallel to the same plane, it is clear that its perpendicular component must be equal and opposite to the perpendicular component of the known load. General Because the resultant of the first two unknown forces must be " in the plane of these two, and therefore this resultant cannot help to balance the perpendicular component of the known load. Thus if from the joint in question along the line of the known force a length be plotted off representing to a con- venient scale that force-magnitude, and if from the end of this line be drawn another line parallel to one of the three ' unsolved ' links ; and if the intersection of this last line with the plane of the other two unsolved links be determined ; then the length of this last line down to its intersection with the plane mentioned measures the stress along the parallel link. The scale, of course, is the same as that to which the known force has been plotted. This second force being thus deter- mined, the remaining two forces can be found at once by closing the polygon by lines parallel to the links. 236 GRAPHICS CHAP. XII. 5. In Fig. 93 the force 1 2 is plotted to scale from the joint to the point ff, and from this point is drawn a line parallel to the link A B to intersect the plane of A C, B C. The plans and elevations of the lines A C, B C are produced to cut the ground line G G. The intersections with G G of the plans AC, B C give the plans of the points where these lines cut the vertical plane ; the elevations of these points are found directly under- neath their plans on the elevations A' C', B' C' of the lines ; Typical and these elevations being joined, the joining line is the vertical Joint * race f the plane of AC, B C - -that is, the line in which this plane intersects the vertical plane. It is marked V AC) BC . The horizontal trace H AC>BC is similarly found by taking on the lines A C, B C the plans of the points where the lines A' C' and B 7 G' cut G G and joining these plan-points. Only one of these latter need be found, however, because the vertical and horizontal traces necessarily intersect in G G ; and thus can t> e found by joining one of them with the inter- Construe- section, already found, of V AC>BC and GG. The finding of diagram both points, however, involves very little extra work, and supplies a most useful and much-needed check on the accu- racy of the work. Through / and /' are now drawn lines parallel to A B and A' B'. These are plan and elevation of the line through point //' parallel to AB, A'B'. The plan /ran is also the plan of the line of intersection of a vertical plane through this same line and the plane AC, B C. The plan cuts H AC>BC in a point m whose elevation is ??z', which point is a point in the above-mentioned line of inter- section of plane A C, B C and the vertical plane passing through the line through / parallel to link A B ; it is the point where this intersecting line cuts the horizontal plane. This same intersecting line meets the vertical plane in a point whose plan is n where the line || A B through / meets G G, this point having the elevation n' on V AC) BC . Thus ra' ri is the elevation of this intersecting line, while /ra n is its horizontal projection. CHAP. xii. SOLID STATIC LINKAGES 237 The line || A B, A' B' through ff lies in this vertica plane, whose plan is/n, and it will therefore meet the plane A C, B C in a point in the line in which this latter plane intersects the former. Thus n' m' meets the line / q' through / || A' B' in the intersection sought for namely, q'. q' is now projected to q on fm. The projections of the stress on bar A B, A' B' are q'f and qf. Then from q is drawn a line || A C to meet the prolongation of B C ; and from q' is drawn a line || A' C' to meet B' C' produced. These give the projections of the stresses on the two bars A C, A' C' and B C, B' C'. But it is better to remove this last part of the construc- tion to another part of the paper, where the stress-diagram Typical can be built up in plan and elevation free from confusion and j}|^ without each stress-line being repeated. Thus in the plan stress-diagram b l b 2 \\ B 1 B 2 represents to scale the force 1 2 in horizontal projection, and to same scale &/ &/ represents its elevation. When q' is obtained by the preceding con- struction, there is plotted &/ a' ^f q'. Then there is drawn Construe- b l a || B A, and the point a is obtained by vertical projection Diagram from a'. The two stress-quadrilaterals are now completed by drawing 6 2 \ IIB C . , a ' c A C m P a b ' IIB' C/ ' elevation. Next in the plan from the joint where 3 4 acts is plotted off along AC to p a length = a c. Through the point p so found, a line || 3 4 is drawn, and through its vertical projec- tion p' on A' B' is drawn a line || 3' 4'. The elevation of the intersection of this line with plane A 4 and C 3 is next found. This involves the finding of the traces of this plane ; they are found as previously explained for the other plane and are marked H A4 , C (horizontal trace) and V A4 , C3 (vertical trace). The intersection is marked r' in elevation and r in plan. Then r' p' and r p are the projections of the outside force 3 4. 238 GRAPHICS CHAP. XII. Tetra- hedral frame Construc- tion of diagram Flan and elevation Stresses From r is drawn a line parallel to A 4 to meet C 3, and these give the other two stresses acting on this joint in plan. These are inserted in the plan stress- diagram in their proper cyclic order ; and from them is deduced the vertical projections, the lines in the vertical stress-diagram being drawn parallel to the elevations of the corresponding links. There is next taken the joint at which force 2' 3' acts. Here the stresses 3 c, 3' c' and c b. 2 , c' b. 2 f are already known. The plane containing their resultant and the link B 2 2 must also contain the outside force 2 3, 2' 3'. This outside force is, therefore, already partly determined, but the direction in this fixed plane is at the free choice of the designer. In this problem this element of choice is represented by the direction of one projection namely, the vertical 2' 3' being included in the data. This being so, the stress quadrilateral for the joint 3' can be immediately completed in elevation; thus 7 /^>2', the ^2 line & 2 7 2' || (!' 2') B'. This gives 2' in elevation ; its plan is next found by projection downwards from 2' to 2 on line drawn through Z> 2 || B 2 2. The point 2 so found is now joined to 3, and 2 3 gives the plan of this force in direction and magni- tude. Next the plan and elevation parallelograms for the ima- ginary joint 12 B 2 B, are drawn in ; and finally the points 4, i, and 4', i' are joined. These latter give the direction and magnitude in plan and elevation of the remaining sup- porting force. 6. This completes the drawing of the stress-diagram. The stresses are known by their plans and elevations. Their un- resolved magnitudes are to be found by compounding the plan with the vertical component measured from the elevation. A convenient mode of effecting this composition graphically for all the stresses in one ' composition diagram ' is shown in Fig. 95 (see Section 8). 7. When the stresses in a large and complicated structure CHAP. xn. SOLID STATIC LINKAGES 239 have to be investigated, generally the known loads are more or less numerous. The frame is supported by at least three forces acting at known points, and the first part of the problem is to calculate these supporting forces. Of these the condition of equilibrium is sufficient to determine six elements, as in the last problem, where the three magnitudes, the plan and elevation directions of one force, and the plan direction of another force are calculated. To find these six elements it is necessary to reduce the known loads to as simple a resultant General as possible ; but, as has been explained in Chapter VIII., tSree Section 8, they cannot generally be reduced to a single resul- tant force. Their simplest equivalent is a pair of non-inter- secting forces at right angles to each other. The mode of reducing them to this simplest form is fully given in Chapter VIII., Section 8, Fig. 58 a. In Fig. 94, in the three orthogonal rectangular projections (TT) (s) and ( e and project it hori- zontally to 8" on p ff in (a). Measure the horizontal distance of 8" from V and plot it vertically downwards in (TT) to 8 from line E on the vertical projection line through 8'. In the elevation stress -diagram d l and d 2 are vertically over d l d 2 in (TT), and in (s) d l d 2 is horizontal. d 2 e r 4j= /o v , the vertical component of p ; and d { e of known forces has now its equivalent in (TT) as p along line 12 and at level 8', and p v through 8. Next draw 8 X || p n in (TT) to meet in X the line 7, and project X upwards to X' on p e in (s) . The same set of forces is also equivalent to p n acting horizontally in the line 12 at the level of X' and p v acting vertically through X. Next mark 6 in (TT) where p n meets a 7, and project 6 up- wards to 6' in (s) on the horizontal line through X'. Now suppose erected a 9-bar linkage, or pentahedron, on the base ay^with its two other joints at (XX') and (66'). At joint (XX') suppose p v to act, and the force p n to act at joint (6 6[) . Draw in the bars of this linkage in (IT) and (s) and letter the spaces in plan as shown. The right-hand cyclic order has been chosen. In plan the bars D E and E (1 2 3) lie coincidently with the bars E A, D B, and D C. In order to obtain places for the letters E and D the former two bars are represented by curved dotted lines ; but, of course, in order to draw lines parallel to these bars in the plan stress-diagram, the directions are taken from the straight line ^7. It is CHAP. xn. SOLID STATIC LINKAGES 241 especially important to have these spaces clearly marked, be- cause D is divided by the external force line 1 2 into two, viz. D! and D 2 ; and E is divided into three, E p E 2 , E 3 , by the external forces 1 2 and 2 3. This last 23 is p v > which is really vertical and represented in the plan stress* diagram by a point only, but in (TT) it is represented by a finite line for the same reason that curved lines are drawn to represent the above- mentioned bars. Note that the bar DB between the two joints X is horizontal. The linkage is taken of this shape with the faces a X y and X 6 7 in the same vertical plane because so doing very much simplifies the resolution of the forces at joints X and 0. Taking first joint 0, the horizontal component of stress in B C normal to 7 must balance that of p u ^= d^ d 2 in the same direction* Therefore draw d 2 c' || B C and d l c' \ &. 7. This gives d 2 c f the plan of stress in BC. In (e) draw d 2 c'||BC and obtain c' by vertical upward projection from c' in (TT). Next in (e) draw c' c || D B and d l c \\ D C, and then c b \\ C B to Pentahe- meet d^d 2 in b. These points, c and b in (e), give d 2 b, be, struction and c d l the (e) projections of the stresses in D B, B C, and C D. Next in (TT) draw d 2 b and d^ c, both || 7, and on these obtain b and c by projection downwards from b and c in (e)* Checl* the accuracy of drawing by finding b c \\ d 2 c' || B C. Take next joint X. The force E 2 E 3 and all the others, except that along A B, lie in one vertical plane, and the plan- projection of their resultant must therefore lie along line 7. Therefore, there is zero stress on A B, and a must be marked along with b in the stress -diagram both in (TT) and in (s). The known forces acting at this joint are a d 2 and d 2 e' in (e). Draw a e 3 \\ A E and e' e 3 \\ D E, and mark the joint of these e y Draw d 2 e 2 ^= e' e 3 and e^ e 2 ^ e' d. 2 * Next in (TT) draw d 2 e 2 \\ 7, and find on it e 2 and e 3 (coinciding) by projection from e 2 e 3 in (e). Next find e l in (TT) and (s) by completing the parallelograms e 2 d 2 d l e l . For the joint at in (TT) draw 4' 4 || y and a 4 || A 4, meet- 242 GRAPHICS CHAP. xn. Pentahe- dral Con- struction Composi- tion Diagram Short- hand symbols ing in 4. Then draw 4 3 ^= 4' d 2 , the point 3 falling in line e 3 a produced. The plan of the stress- diagram for this joint is now complete, viz. a e z 3 4 a. Complete it in (s) by drawing e s 3 |! E 3 3, i.e. horizontal, and finding in this line the point 3 by projecting upwards from 3 in (TT) ; also finding 4 in the horizontal line through a by projecting upwards from 4 in (TT) ; and finally joining 3 4. Next for joint /3 there are already obtained in (TT) the lines c b a 4. Draw 4 5 ^= 4' 5', and join 5 c. The accuracy of the drawing is checked by finding 5 c || 5 C. In (s) draw c 5 || C 5, i.e. horizontal, and find on this 5 by projection from 5 in (TT) . Also join 4 5 in (s). The diagram for this joint in (s) is c b a < 4 5 c. Next, in both (TT) and (e) complete the parallelogram 3 e. 3 e f . 2, giving 2 coincident with 3 in (TT) and making 2 3^= p v m (s) ; and also the parallelogram 2e 2 e l i, which in (e) gives i on the horizontal line e l e 2 , and makes i 2 4J= e l e 2 4|= d l d 2 . Join 5 i in both (s) and (TT), and check by finding in (TT) 5 i || & f d { . The diagram for joint 7 will now be found complete in both (TT) and (e), namely, 5ie l d l c5. The elevations of the three supporting forces have now been found, as well as their plans, and their resultant magni- tudes are found on the auxiliary * composition diagram ' by plotting from 0' horizontally and vertically the plans and the vertical components of the elevations. On this diagram is also seen the resultant magnitude of the datum- load, viz. 13. 8. In these constructions, in which the stresses are shown by elevation and plan, a point is very frequently to be found partly by help of projection from elevation to plan, or vice versa. In order to write out the construction for a compli- cated frame in a form that is not excessively tedious, it is advisable to adopt some shorthand symbol to represent this operation. In what follows the symbols used are 1 1 t and t. The phrase * (TT) d $ e ' means ' find the point e in plan (TT) by drawing from d, already found in plan, a line parallel to D E in plan, and projecting vertically downwards on to this line CHAP. xn. SOLID STATIC LINKAGES 243 from the point c already found in elevation.' Again ' (s) g $/' means ' find/ in (s) by drawing from g in (e) parallel to G F in (s) and projecting upwards from / in (TT).' Not infre- quently the diagram for a joint cannot in the first place be drawn in correct cyclical order ; it has to be otherwise drawn at first, and after finding the closing lines the polygon has to be partially redrawn to put it into proper cyclic order. In that case temporary lines have to be drawn in the stress-diagram Short- parallel to frame-lines named differently. To express this symbols clearly the foregoing symbols are, when needful, expanded thus : * (TT) d t e \\ G E ' means ' in (TT) draw from d a line || G E and find on it the point e by downward projection from e in (e).' Also ' > ^ I r vj means ' draw from c and/ two lines parallel to D E and L M, and mark their intersection k.' Again, ' p q = o ' means ' mark q coincident with already found point p.' 9. In Fig. 95 is shown a more complicated solid structure. It represents a braced pier for a bridge, although, 'for the sake of greater clearness in the illustration, proportions have been chosen that deviate somewhat from those commonly found in such piers. The three views, (TT), (e). and (