THE ELEMENTS OF GRAPHIC STATICS THE BROADWAY SERIES OF ENGINEERING HANDBOOKS VOLUME XXVI THE ELEMENTS OF GRAPHIC STATICS A TEXT-BOOK FOR STUDENTS, ENGINEERS AND ARCHITECTS BY ERNEST H. SPRAGUE, A.M.lNST.C.E. ASSISTANT AT UNIVERSITY COLLEGE, LONDON ; LECTURER ON HYDRAULICS AND STRENGTH OP MATERIALS; LECTURER AT THE WESTMINSTER TECHNICAL INSTITUTE; FORMERLY PROFESSOR OF ENGINEERING AT THE IMPERIAL CHINESE RAILWAY COLLEGE, SHAN-HAI-KUAN WITH WORKED EXAMPLES AND ONE HUNDRED AND SIXTY-THREE ILLUSTRATIONS LONDON SCOTT, GREENWOOD. & SON (E. GREENWOOD) 8 BROADWAY, LUDGATE, E.C. 1917 [All rights reserved] PREFACE THE graphical treatment of problems in En- gineering and allied subjects is often much less laborious and much more lucid than the corres- ponding mathematical analysis. Large mistakes are less likely to be overlooked, whilst the accuracy obtained is all that is required in most cases. Also, the more difficult the problem the more necessary becomes the use of graphical methods, the procedure in the more difficult cases being usually similar to that in the more simple ones, and therefore readily understood though perhaps complicated in appearance when finished ; whilst in the mathematical treatment the use of the higher branches of Mathematics is frequently necessary even in comparatively simple cases, and the labour involved and the consequent liability to error is so great as to render a solution in many cases impracticable. Experience in the use of graphical methods and comparison of the results with those ob- tained by calculation will soon convince the sceptical that the accuracy obtained with reason- able care in draughtsmanship is not only suffi- cient, but is greater than might be anticipated. The Author believes that the graphical an- alysis of Engineering problems, which is of comparative recent development will grow in favour as the advantages of the method are 370661 VI PREFACE better appreciated, and hopes that this element- ary exposition of the fundamental principles as applied to Statical problems may be of assist- ance to the beginner. Further applications will be found in Vol. xvii., "The Stability of Masonry Structures " ; Vol. xx., " The Stability of Arches"; and in "Moving Loads by In- fluence Lines and other Methods," to which the present book may be regarded as an intro- duction. ERNEST SPRAGUE, A.M.lNST.C.E. UNIVKKSITY COLL HI IB, LONDON, March, 1917. CONTENTS I'AGES PREFACE ... v-vi CHAPTER T COMPOSITION AND RESOLUTION OP FORCES . . . 1-15 Advantages of Graphic Statics Scalar and Vector Quantities Definitions Vector Addition Ap- plication to the Composition and Resolution of Forces Equilibrium of Concurrent Forces Equilibrium of Three Forces Triangle and Poly- gon of Forces Parallelogram of Forces Prac- tical Illustrations Examples. CHAPTER II COMPOSITION AND RESOLUTION OF FORCES MATHEMATIC- ALLY CONSIDERED 16-25 Composition and Resolution of Forces Rectangular % Components-^Resultant of Two Forces Result- ant of a System of Concurrent Coplanar Forces Equations of Equilibrium Three Forces in Equilibrium Illustrations and Examples. CHAPTER III FRAMED STRUCTURES 26-14 Assumptions Bow's Notation Sense of the Internal Forces Application to Roof Trusses, Framed Girders, etc. Structures in Equilibrium under the Action of Three External Forces Roof under Wind-pressure Resolution of a Force into Three Non-concurrent Components Illustrations and Examples. b vii Vlll CONTENTS I'AGKS CHAPTER IV SUPERPOSED AND SPACE FRAMES . . . . 45-58 Perfect, Imperfect, and Redundant Frames Con- ditions for a Perfect Frame Finck Truss Boll- man Truss Lattice Girder Braced Tower Space Frames Sheer Legs Belfry Tower Tripod Examples Resultant of Concurrent Forces in Space of Three Dimensions Resolution in Space of Three Dimensions. CHAPTER V MOMENTS 59-77 Definition Resultant Moment Equations of Equili- brium Examples Method of Section- Appli- cation to French Roof Truss Graphical Treat- ment of Moments Couples Composition of Couples Resultant of Non-Concurrent Forces Link-polygon Graphical Conditions of Equili- brium Parallel Forces Graphical Treatment of Moments Application to a System of Parallel Forces. CHAPTER VI ROOFS 78-94 Dead and Live Loads Weight of Roof Coverings, Rafters, etc. Slopes for Roof Coverings Weight of Snow Wind-Pressure Duchemin's Formula Graphical Construction for Duchemin's Formula Increase of Wind-pressure with Height Wind-pressure on Various Surfaces Distribution of Load on Roof French Roof Truss Station Roof Tru^s with Fixed Supports Crescent Roof. CHAPTER VII SHEARING FORCE AND BENDING MOMENT DIAGRAMS 95-111 Shearing Force and Bending Moment Diagrams for Point-Loading and for Uniformly Distributed Loading Shearing Force at a Point Special Cases of Shearing Force and Bending Moment Graphical Construction of Parabola (i) by Points, (ii) by Tangents Combined Diagrams Indirect Loading Determination of Forces in a Framed CONTENTS IX PAGES Girder from the Diagrams of Shearing Force and Bending Moment Parallel and Non-parallel Booms. CHAPTER VIII RELATION BETWEEN THE CURVES OP LOAD, SHEAR, AND BENDING MOMENT . . ... . . 112-119 Load-Curve Sum-Curve or Curve of Integration Shear-Curve as the Sum-Curve of the Load-Curve Bending Moment-Curve as the Sum-Curve of the Shear-Curve Scales Application to a Con- crete Raft. CHAPTER IX SHEARING FORCE AND BENDING MOMENT FOU A MOVING LOAD 120-185 Direct Loading Single Concentrated Load Two Concentrated Loads Train of Concentrated Loads Maximum Values of Shearing Force and Bending Moment Schlotke's Method Indirect Loading Maximum Values of Shearing Force and Bending Moment for a Train of Concentrated Loads Distributed Loading. CHAPTER X MOMENTS OF AREAS, ETC 136-162 Graphical* Determination of the First and Second Moments of a System of Parallel Forces Centres of Gravity Two Point Loads System of Point Loads Triangle Trapezium Radius of Gyra- tion Equimomental Systems Triangle Rect- angle Relation between Second Moments about Parallel Axes Polar Second Moment Circular Areas Graphical Treatment of Moments of In- ertia of Areas Momental Ellipse Symmetric and Asymmetric Sections Unequal-legged Angle Formulae for Centroids of Lines, Areas, and Solids Formulas for Moments of Inertia of Com- mon Cross-Sections. CHAPTER XI STRESS DISTRIBUTION ON CROSS-SECTIONS . . 163-183 Load Point or Centre of Pressure Neutral Line and Axis of Flotation Resultant Pressure on Sub- X CONTENTS PAGES merged Area Relation between Neutral Axis and Load Point Position of the Load Point for a Submerged Area Rectangular Section Ex- amples Stress Intensity at any Point -Examples Critical Distance Core of a Section Rule of the Middle-Third Asymmetric Sections When the Material is Incapable of Resisting Tension Mohr's Method Numerical Example. CHAPTER XII THE LINE OF PRESSURE 184-191 Thrust, Shear, and Bending Moment on a Structure Crane Structure Three-Hinged Roof or Arch Diagrams of Shear and Bending Moment Stability of a Masonry Structure Conditions of Stability Flying Buttress. INDEX 192-19G CHAPTER I. COMPOSITION AND RESOLUTION OF FORCES. 1. Statics is that branch of Mechanics which deals with bodies in equilibrium, and since Structures are usually in equilibrium as a whole, or in the case of a moving structure the parts which compose it are in equilibrium amongst themselves, a knowledge of the principles of Statics is essential to the science of structural engineering. Also, as will soon appear, the action of a system of forces is usually much more conveniently investigated by graphical than by purely mathematical processes, and therefore the application of drawing to statical problems under the name of Graphic Statics has come to be regarded as an important branch of engineering study. The results so obtained are usually more than sufficiently accurate to satisfy all practical requirements, the error in careful draughtsmanship being usually within about 1 per cent of the truth, and therefore within the limits of accuracy warranted by the data ; whilst the economy effected in time and the ease with which mistakes exhibit themselves, recommend its use not only to the less mathematical designer, but also to those who are skilful in analytical methods ; whilst the one process may often be used* as a valu- able check upon the other. i5LEMENtS O^ GRAPHIC STATICS The student cannot be too strongly advised to make himself familiar with the fundamental prin- ciples upon which the graphical constructions are based. Many mistakes arise from lack of this know- ledge, and it is impossible for anyone to think a way out of a difficulty unless the principles involved are understood. The mechanical solution of a problem without an intelligent appreciation of the modus operandi, not only burdens the memory with numberless variations of construction, but leads to the adoption of wrong methods. The most profitable course of study will always be found to consist in understanding the reason of every process used. The fundamental principles are few- in number, and their application to new problems not only develops an interest in the work, but also a constant improvement in a knowledge of the theory of the subject. Festina lente (make haste slowly) is therefore a motto, which, though always to be recommended, deserves particular attention in the present subject, where the applications of a few important constructions are so varied and extensive. 2, Scalar and Vector Quantities, The quantities with which we have to deal in everyday calculations are not, as is often assumed, amenable to the same mathematical laws, and it will be a great simplifica- tion if at the outset we realize the important fact that they may be divided into two groups, (1) Scalar quantities or Scalars ; (2) Vector quantities or Vectors. Scalar quantities are such as may be represented completely by numbers, such as intervals of time, lengths, speeds, temperatures sums of money, and COMPOSITION AND RESOLUTION OF FOKCES 3 so on. All these quantities have size which can be numerically expressed as in ordinary arithmetical operations ; but since size may vary by becoming larger or smaller, it is desirable to indicate the sense of growth by the algebraic device of affixing a posi- tive or negative sign, according as the size is increas- ing or decreasing. Scalar quantities may, of course, be represented graphically by lines drawn to scale, and their sense may be indicated by affixing an arrow- head to the line. In this case the addition or sub- traction of such quantities will be performed by joining thes'j lines end to end in their proper sense and in the same straight line. Thus if a person have AT- *T| ^c * K--2--M? 3---*** 5 H FIG. 1. assets of 2, 3, and 5 say, and debts of 4 and these amounts may be represented by lines drawn to any convenient scale, the assets being distinguished from the debts by arrowheads pointing in opposite senses. If we choose a scale of 1 inch = 1 say, and plot AB = 2", BC = 3", CD = 5" (fig. 1), AD represents the total assets ; and if we plot DE = 4" and EF = li" in the opposite sense, then DF repre- sents the total debts, and AF represents to scale the excess of assets over debts. All this is quite simple, but at the same time there is usually no advantage gained by performing the operations of addition and subtraction of scalar magnitudes in this way, and the graphical method is therefore of little use except in special cases, such as its application to the Slide-Eule, 4 THE ELEMENTS OF GRAPHIC STATICS where lengths representing the logarithms of num- bers are added and subtracted mechanically, in order to perform the operations of multiplication, division, etc. (see "Mathematics for Engineers"). Vector quantities, on the other hand, not only have magnitude and sense, but also direction, and it is because the direction of a line may be so readily expressed on the drawing-board, that the composition or addition of vectors may be so much more conveniently effected by graphical methods ; for whilst a line may repre- sent magnitude by its length, and sense by an arrowhead, the slope of the line will represent its direction, whereas in expressing this property as an angle in degrees we introduce the processes of trigono- metry and the comparative difficulties attending it. Any vector magnitude therefore, such as a dis- placement, a velocity, an acceleration, force, etc., in fact any quantity which has magnitude, sense, and direction, may be completely represented by a line or vector. Such vectors have in general no definite position ; a vector having a definite position is called a rotor or a localized vector. 3, Law of Vector Addition, Vector magnitudes do not obey the ordinary laws of arithmetic and algebra. If by the sum of two quantities we are to understand the result of their combination, then it is clear that in general this combination must obey a law which involves the idea of direction. Thus, for example, the sum of two forces will only be their arithmetic or algebraic sum if the forces have the same direction ; but when they are inclined to one another they can no longer be added or subtracted in the ordinary way. Suppose a body to move from A to B (fig. 2). Then the line joining A to B represents the displace- .ment of the body, and this no matter what the path along which it moved in passing from one point to the other. Let the body be further displaced from B to C, and again from C to D. Then if the lines AB, BC, CD represent to some scale the magnitude, sense, and direction of the several displacements, these lines or vectors when joined end to entl in the same sense, that is to say, with the arrowheads all pointing the same way, will represent the combina- tion of the displacements, and the line AD which joins the beginning of the first to the end of the last will represent the total or resultant displacement ; and since this law is not confined to displacements, but evidently applies equally to all vector quantities, it is known as the law of vector addition. It is ob- vious that this law is quite different to the law which o THE ELEMENTS OF GKAPHIC STATICS governs the addition of scalar magnitudes, except in the special case when the vectors have one and the same direction. Order of Vector Addition Immaterial. Since a body which moves through given distances in given directions and in given senses will arrive at the same spot in whatever order these displacements occur, it is evident that the vectors which represent them may be added or compounded in any order. Thus if a, b, c represent the magnitudes of the vectors in fig. 2, and s represent their vector sum or resultant, we may write s = o.-H-&-ff-cors = a -[[- c 4f 6 or again s = b -ff a -jf c where the symbol -ff is used to denote the operation of vector addition. 4. Application of the Law of Vector Addition to the Composition of Concurrent Forces. Forces, being affected by direction, are vector magnitudes, and must therefore obey the law of vector addition enunciated above. Consequently the vector sum or resultant of any given system of forces acting along given directions which pass through a common point, may be found by com- pounding the vectors which represent them. For if Pj, ?<>, P 3 , in fig. 3, be three given forces acting on a body at the points A, B, and C, in such directions that their lines of action when produced pass through a common point O, then the effect of any one of them will be the same, if we suppose it applied at the ' point O. This is often spoken of as the "Principle of the Transmissibility of Force," according to which a force may be supposed to act at any point in its line of action. Hence P lt P 2 , and COMPOSITION AND RESOLUTION OF FOBCES 7 P 3 may be supposed to act at O instead of at A, B, and C respectively, without in any way affecting their influence upon the body as a whole. Conse- quently, as there is no tendency to produce rotation about 0, the force which is equivalent to the given forces, or their resultant, must also pass through O, and it is only necessary therefore to find its magni- tude, sense, and direction in order to determine it completely. This is accomplished at once by adding the vectors 01, 12, 23 which represent the forces, as .shown; then the vector 03, which closes the vector figure, represents the required resultant ; and a line drawn through O parallel to it will represent the line along which it acts. 5, Equilibrium of a System of Concurrent Forces in a Plane (Coplanar Forces). Let P p P^, P 3 (fig. 3) be any given coplanar forces whose directions pass through a common point O. These we have shown above may be reduced to an equivalent force E = 3 acting as shown in the figure. Now the only single force that can keep another in equilibrium is one which is equal and opposite to it in the same straight line. Hence a force E, equal and opposite to E, will keep E in equilibrium, and will, of course, be represented by the vector 3 reversed, that is by 3 0. Such a force is known as the equilibrating force, or Equilibrant, and since it keeps the resultant E at rest, it will also keep at rest the forces P lf P 2 , P 3 , of which E is the equivalent. The forces P x , P 2 , P 3 , E will therefore be a system in equilibrium. Hence we see, that when a system of concur- rent forces is in equilibrium, their vectors added in the same sense form a closed polygon ; and con- 8 THE ELEMENTS OF GRAPHIC STATICS versely, when the vectors form a closed polygon the system of forces they represent, if concurrent, will be in equilibrium. FIG. 3. 6, Equilibrium of Three Non-parallel Forces. Let Pj, P._, (fig. 4) be any two non -parallel forces whose FIG. 4. vectors are 01, 12. Then the line 2 which closes the figure in the same, sense represents their equili- COMPOSITION AND RESOLUTION OF FORCES 9 brant, because it is equal and opposite to the vector 2 which represents their resultant. Hence, when three forces are in equilibrium, (1) these forces are represented by the sides of a triangle drawn parallel to them and taken in order, and (2) the forces must be concurrent, because the resultant of P l and P 2 must pass through the point 0, and the only force which can keep it in equilibrium is one equal and opposite to it and acting in the same straight line. Hence if one of the forces is known, the other two FIG. 5. can be found. For, suppose three forces Pj, P. 2 , P 3 (fig. 5) to pass through O and to be in equilibrium. Then if any one of them, say P x , is known, and a vector 1 be drawn to represent it to scale, lines 1 2, 2 parallel to the other two forces will represent them also, and not only in magnitude, but in sense also, by the arrowheads. These indicate whether the forces are acting towards or away from the point O, and therefore whether the forces are tensile or compressive. It is of no consequence on which side of the given vector the triangle is constructed, for if we draw 1 3, 10 THE ELEMENTS OF GRAPHIC STATICS 3 parallel to P 2 , P 3 respectively to meet on the opposite side of 01, it will be seen that we get vectors having the same magnitude and sense as in the first case, and this is sometimes convenient. The vector figure obtained in this way is com- monly spoken of as the " Triangle of Forces," and it will readily be seen to be only a special case of the " Polygon of Forces " which we obtain as the vector figure when more than three forces are concerned. Also the Triangle of Forces is identical in principle with the "Parallelogram of . Forces," which states that if two forces are represented by the sides of a FIG. 6. triangle, their resultant is represented by a diagonal. For if OP X (fig. 6) represent a given force P p and OP., a given force P 2 and we complete the parallelo- gram, the diagonal OR will represent the resultant R ; for it will be seen that if we set off OP 2 to repre- sent P 2 and add to it a vector PR 2 = OPj, the line OR which joins O to R is the resultant, by the triangle of forces ; and in fact it is not necessary to refer at all to the parallelogram of forces which is only another way of obtaining the same result as that derived by the triangle of forces. 7. Application of the Triangle of Forces, Let W (fig. 7) be a given load supported by cords passing over pulleys at A and B, and let Pj, P 2 be the ten- COMPOSITION AND KESOLUTION OF FORCES 11 sions in the cords. Set down 1 to represent the load W to scale, and construct the triangle of forces 012 for the forces acting at C. Then the vectors 1 2, 2 parallel to CB, CA respectively represent the tensions in these cords, a fact which may be tested experimentally with flexible cords and freely moving pulleys. If the tensions P l and P 2 are increased, the point C will rise to some point C', and the tri- angle of forces will now be 1 2' ; so that as we A B FIG. 7. increase the tensions, the more nearly do the cords approach the horizontal, and the longer become the vectors which represent the tensions, so that it is evident that a force such as 1 may produce other forces much greater than itself. This fact is indeed utilized in mechanical engineering in the form of the " toggle-joint," used for hand-brakes, stone- crushers, etc., a small force being used to develop very large forces when the arms of the toggle are nearly in line. 12 THE ELEMENTS OF GRAPHIC STATICS As, however, the development of large forces is, in structural work, to be avoided as far as possible, the members of a structure should as far as con- venient make fairly large angles with one another. Thus in the roof truss (fig. 8) the force in the bar AD will increase as the angle CAD gets smaller, as may be seen by the vector triangle 012, and will decrease as the angle gets larger, the smallest stress FIG. 8. possible, if it were feasible, occurring when the bar AD is perpendicular to AC, as shown by AD', in which case 103 would be the triangle of forces for the forces at A, arid the force in AD' would then be that represented by the vector 1 3. When AD, DB are horizontal, the tension in them is represented by the length 1 4, drawn parallel to AB. It will be seen from the foregoing that when not more than two unknown forces act at a point, these forces can always be found, for however many others there may be, so long as these are known, a COMPOSITION AND EESOLUTION OF FOBCES 13 vector figure can be drawn for them and their resul- tant determined. This resultant being in equilibrium with the two unknown forces, the latter can be found by constructing the vector triangle. Thus if P lf P.,, P 3 (fig. 9) be three known forces acting at O, and X, Y be two unknown forces, the vector figure 0123 being drawn for the known forces, 3 is their resultant K, and if we construct on it a triangle 043 having its sides parallel to X, Y, then the forces FIG. 9. 04, 43, which are in equilibrium with K, are also in equilibrium with the forces of which E is the re- sultant. Bufr it will be seen that there is no neces- sity to draw in the line 3 at all, because the same result is obtained by drawing lines from the ex- tremities 0, 3 of the vector figure which represents the known forces, parallel to X and Y, to close the polygon. Examples. 1. Find the resultant of two equal forces of 10 Ib. inclined to one another at angles of 0, 30", 60", 90, 14 THE ELEMENTS OF GRAPHIC STATICS 120, 150, and 180. Plot the results on a base of degrees and construct a curve to show the resultant for any other angle. Ans. 20; 19'3 ; 17'3 ; 14-1; 10; 5-18; Ih. 2. Find the resultant of four forces of 8, 12, 15, and 20 lb., making angles of 30, 70, 120, and 155 with a fixed line ; and check the result by compound- ing the forces in a different order. Ans. 39-5 lb. at 111 46' to the line. 3. A weight of 6 lb. is suspended by two strings, of lengths 9 and 12 ft. from a horizontal beam at points 15 ft. apart. Find the tensions in the strings. Ans. 48 and 36 lb. 4. A bar 10 ft. long, hinged at its lower end against a wall, and attached to the wall by a chain 6 ft. long at a point 8 ft. above the hinge, carries a load of 2000 lb. Find the forces in the bar and chain. Ans. 2500 and 1500 11). 5. Two rafters AB, BC of equal length are hinged at B, and loaded there with 50 lb. vertically. Find the compressive force in each bar when the angle ABC is (i) 30; (ii) 90; (iii) 120; (iv).150 . Ans. 96-59 ; 70'71 ; 50 ; 25'88 lb. 6. A body weighing 20 lb. rests on a smooth plane inclined at an angle of 30. It is acted on by a pull inclined upward at 20 from the plane. Find the magnitude of the pull necessary to keep the body at rest. Ans. 13-4, 25, 50, 186-5 lb. 7. A string ABCD hangs in a vertical plane, the ends A and D being fixed. A weight of 1Q lb, is COMPOSITION AND EESOLUTION OF FORCES 15 hung from B and an unknown weight from the point C. The middle portion BC is horizontal and the portions AB and CD are inclined at 30 and 45 to the horizontal respectively. Determine the unknown weight and the tensions in the three portions of the string. (I.C.E. Exam.) Ans. Weight = 17-32 Ib. ; in AB, 20 ; in BG, 17-32 ; in CD, 24-49 Ib. FIG. 10. 8. A weight of 5 tons hangs from a chain as shown in fig. 10. Find the forces in the bars AC and BC. Ans. 15 and 8'66 tons. The above examples may with advantage be worked analytically also as explained in the follow- ing chapter. CHAPTER II. COMPOSITION AND RESOLUTION OF FORCES MATHEMATICALLY CONSIDERED. 8. Although many classes of problems are very conveniently solved by graphical methods, it is equally true that others are more adapted to mathe- matical treatment, and it is essential to a clear grasp of the subject to understand both methods, the one being often not only useful as a check upon the other, but supplementary to it. The following chapter will therefore traverse the ground of the previous chapter from the mathematical standpoint. A slight acquaintance with trigonometry is necessary, and this is desirable in any case. 1 9, Composition of Two Forces at Right Angles, Let P p P 2 (fig. 11) be two such forces, whose lines See " Mathematics for Engineers," vol. xxi. iu this Scries. (16) COMPOSITION AND RESOLUTION OF FORCES 17 of action intersect in O. Then if we draw Oa to represent P lf and add to it ab = P. 2 , the line Ob represents their resultant K, as we have shown in the previous chapter. Consequently since the triangle Oab is right-angled we have R 2 = PS + P, 2 or k= vpTTiy. Also, if 6 be its inclination to the force P 1? tan = Q = p 2 ; from which the magnitude of the angle 6 can be found. Example 1. Let Pj = 10 Ib. and P 2 = 15 Ib. Then E = ^/fdOTWd = 18-03 Ib. and tan = -}-J = 1-5, whence = 56 18'. 10, Resolution of a Given Force into Two Rect- angular Components, Let P (fig. 12) be any given force, which it is desired to resolve into components FIG. 12. along two given axes OX, OY perpendicular to one another. Then if Ob represent the given force P to scale, its projections Oa, ab parallel to the axes OX, OY will be the required components, by reversal of 18 THE ELEMENTS OF GRAPHIC STATICS the above reasoning. If we call these Px, PY we have P x = P cos 0, and P v == P sin respectively. Example 2. Let P = 100 Ib. and = 30. Then P x = 100 cos 30 = 86'6 Ib. and P Y = 100 sin 30 = 501b. 11. Composition of Two Forces Inclined at any Angle to One Another. Let fche given forces P l and P 2 (fig. 13) intersect in O. Take two perpen- dicular axes OX, OY, and let OX (for simplicity) FIG. 18. coincide with the direction of P t . Then if we re- place the force P 2 by its components P., cos and P., sin 0, we have a total force of P, + P., cos acting along the X axis, and a force P.. sin along the Y axis ; and since these are at right angles to one another they may be compounded as in the pre- ceding paragraph, and \va get R- == (?! + P., cos 0)- + (P., sin 0) 2 = PJ- + 2P,Po cos + P,, 2 cos 2 + P. 2 sin 2 = P, 2 + 2?^ cos + P,, 2 (cos 2 + sin 2 0) = P t 2 -f P 2 2 + ^PjPo cos since sin 2 + cos 2 0=1. .-. K = yP/ 2 + P, 2 + 2P T P 2 cos 0, COMPOSITION AND RESOLUTION OF FORCES 19 and its inclination a to the X axis is given by the formula _?2 sin B \ + P 2 cos & Example 3. Let P! = 10 lb., P 2 = 201b., and = 35. Then E = ^/lOO + 400 + 400 cos 35 = v /827'68 = 28-77 lb. 20 sin 35 11-472 tan a and tan a = .-. a = 27 56'. 12, To Find the Resultant of a Given System of Concurrent Forces, Let P,, P,, etc. (fig. 14) be the given forces which pass through O and make angles O lt 6.,, etc., with the axes OX, OY. Let each of the forces be resolved into components along OX 20 THE ELEMENTS OF GRAPHIC STATICS and OY. Then taking into account the sense in which the forces act, and denoting by 2x an( l 2y the algebraic sum of their components along the X and Y axes, we get 2x = PI cos #, - P.> cos 0., - P., cos 0., + P 4 cos 0,1 1 ~M Z 2. 3 .1 4 \ 2v = PI sin O l +. P 2 sin #._> - P 3 sin 3 - P 4 sin 0J whence we get R =^ V^x 1 + Sv 2 M and the direction of R will be given by tan = - Example. 4. Let P l = 10 lb., P 2 = 15 lb., P 3 = 20 lb., P 4 = 10 lb., and let the angles be B l = 32, 0, = 48, 3 = 50, 4 = 35. FIG. 15. Then 2 X 10 cos 32 ' - 15 cos 48 - 20 cos 50 + 10 cos 35 U \ v v = 10 sin 32 + 15 sin 48 - 20 sin 50 - 10 sin 35 J .-. ^ x = 8-480 - 10-036- 12-856+ 8-192 = - 6'221 and S Y =5-299 + 11-146- 15-32 - 5-736= -4-61J 7-743 lb. and .-. R == 76-222 + 4-61 2 - 4-61 tan = -- - '7411. .-. 6 = 36 32' (fig. 15). COMPOSITION AND RESOLUTION OF FORCES 21 13, Equations of Equilibrium for a System of Concurrent Coplanar Forces, When a system of forces is in equilibrium their vector figure must close, or in other words the resultant must be zero. Therefore E = ^x 2 + 2y 2 = 0, and this can only occur when both 2x = and 5y = 0, whether we consider the matter from the graphical or the mathematical standpoint. Whenever a system of non-parallel forces is in FIG. 16. equilibrium therefore, it is always possible to form two independent equations by resolving the forces into components along a pair of axes and equating the sums in each case to zero. Consequently if all the forces except two are known in magnitude these two can be found, provided their directions are given. Example 5. Let the forces shown in fig. 16 act at a point O, and let P x , P 3 be unknown in magnitude. 22 THE ELEMENTS OF GRAPHIC STATICS Then for equilibrium 2x = 100 + P! cos 40 - 150 cos 30 \ - P 3 cos 65 0. and or and 2 Y = PI sin 40 + 150 sin 30 - P 3 sin 65 = 0. 100 + -7660 P! - 129-9 - -4226 P 3 = 0\ 6428 P! + 75 - -9063 P 3 = 1 7660 P! - -4226 P a = 29'9 ) 6428 P, - -9063 P 3 = - 75 j P! - -552 P 3 = 39-0 ) P, - 1-41 P :5 = - 116-5) whence P., - 181-2 Ib. and P, - lOO'O Ib. 14, Special Case, Three Forces in Equili- brium. Lanic's Theorem. Although, of course, the general method explained ab'ove will apply to this FIG. 17. particular case also, the following method is more convenient when only three forces are concerned. Let P (fig. 17) be a known force and Q, S two unknown forces acting at the given angles a, y with P and in equilibrium with it. COMPOSITION AND RESOLUTION OF FORCES 2 Then if the vector triangle abc be drawn P : Q : S : : cb : ba : ac by the triangle of forces ( 6). : : sin (180 - ft) : sin (180 - a) : sin (180 - y) : : sin ft : sin a : sin y, 1 that is to say, the forces are proportional to the sines of the angles opposite them. "Example 6. Two cords attached to a beam carry a vertical load of 100 Ib. The cords are inclined at 120 and 1651 to the load. Find the tensions in them (fig. 18). FIG. 18. Here =-^ = p^ 100 sin lf>r> sin 75 sin 15 C Also 100 sin 75 .-. P = 26-8 Ib. sin 12Q _ sin 6Q C * sin 75 " sin 75 C /. Q = 89-5 Ib. 2588 9659* 8660 Example 1. A. body of weight 20 Ib. rests on a smooth plane inclined at 25 to the horizontal, being supported by a force P acting upward at an angle See " Mathematics for Engineers," p. 54, Form (8). 24 THE ELEMENTS OF GRAPHIC STATICS of 35 to the plane. Find the value of P and the pressure on the plane (fig. 19). If, for convenience, we take the X axis parallel to the plane and the.Y axis normal to it, then 2 X = P cos a - W sin 6 (I) and 2v = N + P sin a - W cos (2) FIG. 19. and since for equilibrium these must 1x3 zero, we get Pcos a Wsin0orP = W COS .1 and and N ,. p. N == W cos 6 - P sin a. 8in 25 = 8-45 cos d5 -819 20 cos 25 -- 10-3 sin 35 == 18-1 - 5-9 = 12-2 Ib. Example 8. A body whose weight is 130 Ib. is suspended from a horizontal beam by strings whose lengths are 2 ft. and 4 ft., the strings being attached to the beam at points 5 ft. apart. Find the tensions. Ans. 88-9 and 126-6 Ib. Example 9. A simple triangular roof truss 24 ft. span and 3 ft. deep is supported at its ends and COMPOSITION AND RESOLUTION OF FORCES 25 carries a load of 3 tons at its apex. Find the forces in the rafters and in the tie-rod. Ans. 6'2 and 6 tons. Example 10. A wall crane consists of a strut 10 ft. long, hinged at the bottom end and supported by a horizontal rod 6 ft. long attached to its upper end, and carries a load of 2000 Ib. Find the forces in the bars. Ans. 1500 Ib. ; 2500 Ib. Example 11. Five bars of a structure meet at a joint. The bars make angles of 0, 30, 90, 120, and 210 with the horizontal measured contra-clock- Jfr FIG. 120. wise, and the forces in them are 10, 8, 6 tons, P and Q, the first two being tensile, the third com- pressive, and the last two unknown. Find the values P and Q. Ans. P = 10-22 ; Q = 13'65 tons. A cord supported at A and B (fig. 20) carries a load of 2 T at C, and an unknown load W at D. Find the tensions in the cord and the value of W when CD is horizontal. Ans. AC = 2-83; GD = 2; DB = 4; W = 3-46 tons. The above examples may with advantage be worked graphically also, as explained in .the pre- ceding chapter. CHAPTER IIL FRAMED STRUCTURES. 15. In dealing with the forces which occur in framed structures, it is usual to suppose that the loads which come upon the frame are transmitted to the joints. When this is not so, the memher will he sub- ject to hending and the stresses induced in this way will he considerable unless the loads are very small or act near the joints. In any case it is desirable to avoid bending stresses. Moreover, it is usual to neglect the rigidity of the joints and to assume that these are free to turn. Actually of course this is frequently not the case, the principals of a roof truss being continuous from end to end, and the bars being frequently, and in fact usually, riveted at their ex- tremities both in roof trusses and girders ; but it is necessary to make the assumption of pin-joints for the sake of simplifying the problem. 16, Bow's Notation, The system of notation commonly known as Bow's is very convenient in dealing with forces and their vectors. If P lf P 2 , P 3 (fig. 21) be any forces acting at a point O, let figures or letters, e.g. a, 6, c, be written in the spaces be- tween them, and let the vector which represents the force lying between the spaces a and b be denoted by the same letters placed at its extremities, and ('26) FRAMED STRUCTURES 27 similarly for the other forces and their vectors. Then since the equilibrant is represented by the closing vector da, this force P 4 must be parallel to da, and have the magnitude and sense which this denotes, a and d lying on either side of it. The convenience of this notation will best be understood in its applica- tion. Example 1. Suppose a load W (fig. 22) to act at the vertex of a triangular frame ABC, pin-jointed at the apices. Then if the direction of one of the re- actions at A or'B be known, the values of these re- actions can be found, as well as the forces acting in the frame. For first, since the external forces at A, B, and C are in equilibrium, if the direction of the reaction at A, say, be given, it will intersect the vertical force W in a point D ; and since three forces in equilibrium are concurrent the direction of the reaction at B must pass through D. A triangle of forces abd for these forces which intersect in D will then determine their magnitudes. 28 THE ELEMENTS OF GRAPHIC STATICS It is to be noticed that if the direction of neither of the reactions at A or B is known, the problem of finding these reactions is impossible, because there are any number of solutions ; the only condition for equilibrium being that their directions shall intersect on the line of action of W, and for each pair of forces satisfying this condition the magnitudes will be different. If the bearing surfaces are horizontal then FIG. 22. the reactions will be vertical for vertical loading, and in this case their values can be found. Lettering the spaces of the frame and the vectors by Bow's notation, as explained above, the reaction KA at A will be given by the vector oc, and the re- action KB at B by the vector cb. Also if we draw cd in the vector figure parallel to AB, the forces at A will be represented by the sides of the triangle acd, the force cd in the frame being represented by cd in the triangle both in magnitude and sense, and as the sense is away from the point A, it indicates FRAMED STRUCTURES 29 pull. Similarly the vector triangle cbd represents the forces acting at B ; the sense of dc in the vector figure being now opposite to what it was before, and indicating a force acting away from B, and therefore a pull ; which checks with what we got when con- sidering the force in the same bar acting on A. It will be seen therefore that the vectors referring to the bars of the frame represent the internal forces acting in the bars, which balance the external forces ; in fact they represent the elastic forces which the material exerts to prevent being deformed, and which act in both senses. Thus when considering the forces at C, the force in AC acts towards C, but when considering the forces at A, it acts towards A, the elastic forces at every section being equal and opposite on either side of the section. For this reason it is convenient to indicate the sense of the internal force by an arrowhead in the frame diagram directed towards the point whose equilibrium is under consideration, when the force is a thrust ; and away from it, when the force is a pull. It will be under- stood therefore that after having found the force in any bar AC by consideration of the forces at A say, we must reverse the arrowhead when considering its action at the opposite end C. Example 2. Let fig. 23 represent a roof truss symmetrically loaded as shown. Then if the loads on the joints are set down along a vector line from e to e' (e f not shown in the figure), and ee be bi- sected in k, e'k, ke are the reactions. The resultant vertical force at A is therefore ke - ef = kf, from which it will be seen that any load acting at the supports may be disregarded, so far as any effect MU THE ELEMENTS OF GRAPHIC STATICS on the forces in the frame are concerned. The resultant reaction at A, viz. kf, being known, the triangle of forces kfa may be drawn. Then fa being known, and /r* - I0f- - H*- - -/O'- - if* - 10'- FIQ. 38. hinged and the right end resting on rollers. Tabulate the stress in the members. (B.Sc.) FRAMED STRUCTURES 43 Ans. Bars: 1234567 Forces: +9-1 +1-9 +1'9 +8-4 -5-2 -58 -8-4 tons. Bars: 8 9 10 11 12 13 Forces: -2-8 + 5'62 -4-Q5 +2-Q -4-12 +2-2 tons. 8. In fig. 39 is shown the outline of a roof truss, span 28 ft. and height 7 ft. The full panel loads are 800 FIG. 39. each 1600 Ih. Find graphically the stress in each member. Ans. Stresses marked on bars. (B.Sc.) 9. Part of a pin- jointed frame shown in fig. 40 is loaded with a vertical dead load of 10,000 Ib. and a FIG. 40. normal wind pressure of 15,000 Ib., both being taken as uniformly distributed along AB. The supporting 44 THE ELEMENTS OF GRAPHIC STATICS forces P, Q, and E are shown by dotted lines. Find these forces and the forces in the bars which meet in C, indicating the struts and ties. Ans. P = 13,000 ; Q = 1950 ; R = 14,500. Other forces are marked on bars. CHAPTEB IV. SUPERPOSED AND SPACE FRAMES. 19. A framed structure is said to be (1) Perfect, (2) Imperfect, (3) Bedundant, according as the members which compose it are just sufficient, less than or more so, to keep the frame in equilibrium. The simplest case of a perfect frame is the triangle, which is indeformable by its geometrical property that the lengths of the three sides completely determine its shape. Thus ABC (fig. 41) is a perfect frame. If FIG. 41. another triangle ACD be built upon it, two additional bars will be necessary to fix the new joint D. Any less number of bars would render the frame unstable and therefore imperfect, and any greater number would make it redundant ; in which case the forces in the bars would be statically indeterminate, that is to say, their magnitudes could not be found except (45) 46 THE ELEMENTS OF GRAPHIC STATICS by considering the elasticity of the members. Since every new joint requires two and only two additional bars to constitute the whole a perfect frame, we get for a triangle 3 bars and 3 joints for a quadrilateral 5 ,, ,, 4 ,, for a five-sided frame 7 ,,5 and so on. Hence if b be the number of bars and j the number of joints, the condition for a perfect frame is b = 2/ - 3. Thus in fig. 42, b = 8 and FIG. 42. j = 6. .-. 2; - 3 = 9. The frame is therefore im- perfect, and unless another bar be added, its stability must depend upon the rigidity of its joints. In the case of the ventilator frame in fig. 43, b = 8 and j = 5. Fiu. 43. .'. 2/ - 3 = 7. Hence the frame has a redundant member. In the case of the Fink truss (fig. 44), b = 13, j = 8. .-. 2j - 3 == 13 and the frame is therefore perfect. A framework of this description may be described as a superposed frame, for it will be seen SUPEEPOSED AND SPACE THAMES 47 that the bars overlap one another, and in this respect differ from those which form an open triangulation. 20, Superposed Frames, In dealing with the Fink truss of fig. 44, if W lf W 2 , W 3 be loads acting at the upper joints, we may begin at joint b. The thrust in bf will be equal to Wj and may be repre- sented by 1 in the vector figure. Construct the triangle of forces 012 for the joint /. Then 2 1 FIG. 44. represents the tension in cf, and 23, 31 are its vertical and horizontal components at c. Similarly if 4 5 represent the load W 3 at d, and therefore the thrust in dh, the line 4 6 will represent the tension in ch and 4 7, 7 6 are its horizontal and vertical components at c. The load on the centre strut eg is therefore equal to 2 3 + 7 6 + W 2 , and therefore if we add these as shown and construct the triangle of forces for the joint tf t viz. 268, the lines 6 8, 8 2 represent the stresses in the bars ge and ga respec- tively. 21. Bollman Truss, The Bollman Truss, which 48 THE ELEMENTS OF GRAPHIC STATICS is a common American type, consists of superposed triangles as shown in fig. 45. Set down 01 = Wj, and construct triangle of forces for joint g. Set down 23 = W 2 , and draw triangle of forces for joint h, and so on. Then if these are set off as shown, the line 9 8 will be the sum of the vertical components of the tensions in the bars, and will therefore be equal to the reaction at a, and 9 will be the sum a FIG. 45. of the horizontal components of the tensions and will therefore be equal and opposite to the com- pressive force in the upper chord of the girder. 22, Lattice Girder, Girders of this type (fig. 46) are statically indeterminate in theory, but for prac- tical purposes a quite sufficiently accurate result may be obtained by resolving the framework into its com- ponent systems, finding the forces which occur in each in the ordinary way, and then compounding SUPERPOSED AND SPACE FRAMES 49 the results by adding the forces which occur in bars of the upper and lower chords which are common to the separate systems. Thus the compressive forces which act in ab and cd must be added to get the force in CD ; likewise the compressive forces in ab, cd, and ef to get the force in DE, and so on. *" s ' tf f f f f ~~ /0/ ~ Sr - ''- gy*- - |o '- -y- - l0 '- *r - /0 '~ -4"^"* FIG. 46. Example. The double Warren girder (fig. 47) is supported at the ends and carries the loads shown. Find the forces in the members. State the assump- tions made. (B.Sc.) Ans. Forces shown on bars. In other cases an approximation to the stresses in the bars of a redundant frame may be obtained 4 -50 THE ELEMENTS OF GRAPHIC STATICS by neglecting the effect of the compression mem- bers in cases where these are designed only to take up tension. Thus in the case of the trestle shown in fig. 48, the diagonal members may be put in tension or compression according to the direction of the applied pull, but as the bars are usually weak *- "b ' Fia. 48. to resist compression they may be assumed to buckle sufficiently under thrust to throw the load upon the tension members alone. In the present SUPERPOSED AND SPACE FRAMES 51 case, for example, the bars ab, be, bd, ce, fy may be neglected, and then a reciprocal figure may readily be constructed for the remaining bars, from which the forces in them may be found. These are shown on the corresponding bars in the figure. 23. Space Frames. Space frames are those whose members act in space of three dimensions, as in the case of domes, towers, etc. The stresses in these may be very complex, but a few simple cases may be here considered. In resolving a force into two components by means of the vector triangle, the FIG. 49. force and its components necessarily lie in a plane. A force may however be resolved into definite com- ponents at a point along three given directions, pro- vided that these directions are not in one plane. Let ABC (fig. 49) represent a sheer-legs carrying a load W, and supported by a guy-rope AD. Take a plane containing the load and the guy-rope and let it cut the plane of the sheer-legs in AE. The load W may now be resolved into components along AD and AE as before, and the component along AE may then be resolved along AB and AC in the same way. 52 THE ELEMENTS OF GRAPHIC STATICS 24, Belfry Tower, Fig. 50 represents a belfry tower carrying a load of 10 tons at A, This load may be assumed to act half along a vertical plane containing AjH^Jj and half along a vertical plane con- taining AjIjLj. Having resolved the half load of 5 tons along AB and AD, the component along AB may be resolved into components along BE and BD, and the last may be again resolved in plan along B^, BjC.,. Similarly the force along BE may be resolved along EH and EF, and the force along EF into components along EjF,, E^ in plan, and so on. The results obtained are shown on the bars in the figure. 25. A more general case would be that of three bars of unequal length meeting at a point. Such cases occur in the construction of framed towers and domes. Let OA, OB, OC (fig. 51) re- present three such bars whose plans are OA lf OB P OC 1 resting on the ground at A x , B 1? C l and carrying a vertical load W. Then if this load be first resolved FIG. 50. SUPERPOSED AND SPACE FRAMES 53 into components along OC, OD in the plane OCD, and the force along OD is then resolved along OA and FIG. 51. OB, the forces in the three legs will be found. In order to carry out this idea a plan of the points A, FIG. 5U. B, C is drawn to scale, viz. AjBjCj, and the lengths of the legs being known the triangles OjAjBj, OjBjCj equal to the triangles OAB, OBC are drawn. If we 54 THE ELEMENTS OF GRAPHIC STATICS suppose these triangles turned about their bases AjBj, BiCp the plans of their vertices will move along lines perpendicular to these bases, and these lines will intersect in a point F which is the plan of the vertex of the tripod. Hence FA P FB lf FC X will be the plan of the three legs. Produce CjF to Dj. Then FD X is the plan of OD. It is now only necessary to construct the true shape of the triangle ODC, and this may readily be done, for if a line FO 2 be drawn perpendicularly to the base DjCj, and the length of the leg OC be taken and turned about C x to cut F0 2 in O 2 , the triangle CjCXD, will be the true form of the triangle ODC. If therefore we suppose the load W to act along O._,F, it may be resolved along O..C,, O 2 Dj. The first force will be the compressive force along the leg OC, and the second may then be set off along the line OjD, and resolved along OjAj, O,Bj, by which we obtain the forces along OA and OB. Examples, 1. A pair of sheer-legs are 60 ft. high when up- right, their feet being 30 ft. apart. The guy-rope is 90 ft. long. Find the forces acting in the legs when lifting a load of 20 tons which overhangs 25 ft. Ans. 17'9 tons in the guy-rope, and 17'5 tons in each leg. 2. A load of 7 tons is suspended from a tripod, the legs of which are of equal length and inclined 60 to the horizontal. Find the thrust on the legs. Ans. 2'75 tons. 3. In an unequal-legged tripod, the lengths of SUPEKPOSED AND SPACE FEAMES 55 the legs are OA = 11 ft., OB = 9 ft., and OC = 10 ft., the length AB = 8 ft., BC = 9 ft., CA = 10 ft. Find the stresses when carrying a load of 10 tons at O. Ans. 4*5 tons ; 6*15 tons ; -87 tons. 4. A rectangular frame 3 ft. x 8 ft. is supported on castors and carries bars 7 ft., 7 ft., 4 ft., 4 ft. long at its corners meeting in a point O. At O is hung a weight of 400 Ib. Find the forces in all the members. Ans. In bars : 192, 192, 110, 110 Ib. ; in frame : 25, 72, 94*5, 94-5 Ib. 5. A weight of 1000 Ib. is carried by three con- vergent ropes 10, 10, and 15 ft. long, hanging from points A, B, and C in a horizontal ceiling. AB is 16 ft., and BC, CA are each 20 ft. long. Find the pull on each rope. Ans. 736'5 ; 881 ; 881 Ib. 26, General Treatment of the Forces in a Tripod, When a given force acts in any direction FIG. 52. in space, it may be completely represented by the plan and elevation of its vector. Thus if P (fig. 52) represent the plan of a vector E, and-P' its elevation, the magnitude and direction of the vector may be 56 THE ELEMENTS OF GRAPHIC STATICS found by setting off the plan along XX = AB, draw- ing a vertical at B, and a horizontal through D to cut it in C. Then CA is the resultant vector and a is its inclination to the horizontal. 27, To Find the Resultant of a Given System of Concurrent Forces in Space. Let OP lf OP 2 , OP 3 (fig. 53) be the plans and OF 1} OP' 2 , OP' 3 be the elevations of a system of forces whose resultant is required. Fia. 53. Construct a vector figure with the given plans of the forces, viz. abc. Then the closing line ad will be the plan of the resultant. Similarly, if we construct a vector figure a'b'c with the given elevations of the forces, the closing line a'd' will represent the elevation of the resultant. The plan and elevation of the resultant being thus known, its actual magni- tude and inclination can easily be found by 26 : or if we draw a'e' horizontal and set up de = d'e at right angles to ad, ae will represent the resultant and a will be its inclination to the horizontal. SUPERPOSED AND SPACE FRAMES 57 28, Resolution of a Given Force into Compon- ents along Three Given Directions not in the same Plane, Let P (fig. 54) be a given force kept in equilibrium by three other forces acting along the lines A, B, and C, whose projections are a, a' ; b, b' ; c, c', and let p, p' be the projections of the given force P. We FIG. 54. first resolve P into two components, one acting along the line of one of the unknown forces, say A, and the other along the line in which the plane containing P and A intersects that containing the other two un- known forces B and C. First, if we join the feet of a and b we get the horizontal trace of the plane con- taining the legs A and B. Next find the horizontal trace ; of the force P by producing p' to meet the line XX, and erecting a perpendicular to cut p pro- 58 THE ELEMENTS OF GKAPHIC STATICS duced in /,. Join this to the foot of c, viz. t.-,. Then t^j is the horizontal trace of the plane containing P and the leg C, and the intersection of this plane with the plane containing the legs A and B is the line DE, whose projections are de and d'e'. The plan and elevation of P, viz. ab, a'b' are now drawn in another place for convenience, as shown, and these are re- solved along ed, et z and along e'd', e't\, respectively, giving ac k a'c' as the plan and elevation of the force along ED, and be, b'c' as the plan and elevation of the force along the bar C. The first of them, viz. ac, may be easily resolved along directions parallel to the plans of the legs A, B ; and the second, viz. a'c', along directions parallel to the elevations of A, B. In this way we obtain the plans and elevations of the required forces along the legs A, B, and C, from which the forces themselves are easily deduced, as in $ 16. CHAPTER V. MOMENTS. 29. When the line of action of a force does not pass through a point, it tends to produce rotation about it, and this tendency increases both as the force in- creases and also as its distance from the point or its lever -arm increases. In other words, the tendency to cause rotation is jointly proportional to the magni- tude of the force and its perpendicular distance from the point. The value of this product is known as the moment of the force about the point. Thus if Pj_ (fig. 55) be a force and a^ be its distance from a given point Q, then P l ci l is the moment of the force about the point. Similarly, if P 2 be another force (59) 60 THE ELEMENTS OF GRAPHIC STATICS whose distance from Q is a 2 , then P,ao is its moment about Q, and the resultant moment due to both forces will be the algebraic sum of the individual moments, viz. P^ + P 2 a 2 . Further, if no rotation takes place, then this resultant must be zero, and therefore for equilibrium we get the equation P^ + P 2 a 2 = or P 2 a 2 = - P^, that is to say, the moments must be equal in magni- tude and opposite in sign. This is the fundamental condition that there shall be no rotation. Suppose P! = 10 Ib. and a } = 2 ft., and we re- quire to find what force P 2 acting at li ft. from Q will maintain equilibrium. Then since 10 x 2 + P 2 x 1^ = 0, we get P 2 = - 13J Ib. the negative sign indicating that the moment of P 2 must be opposite to that of P,. 30. We have seen formerly that when a body is acted upon by a system of concurrent forces in equilibrium, two independent equations can be ob- tained, viz. 2x = 0(1) and 2y = (2) ( 13), and when the system of forces is non-concurrent we have the additional equation 2 (moments) = (3). The first two are the conditions that there shall be no translation, and the third is the condition that there shall be no rotation. These three equations are known as the Three Fundamental Equations of Equilibrium for forces in a plane, and it is there- fore possible to determine three unknown quantities when such a system of forces is in equilibrium and the forces are non-concurrent ; a very common case MOMENTS 61 being to find the magnitudes of the two reactions and the direction of one of them, when a system of forces acts on a structure supported at two given points, the direction of one reaction being given (see Example 3, p. 40). Examples, 1. Suppose a load of 2 cwt. to act upon a horizontal beam supported at A and B (fig. 56), as shown, and whose weight is neglected. Then as the beam is at rest, whatever point we choose to consider, the sum of the moments about it must be zero, because there is |2cwt Jh B FIG. 56. no rotation. To avoid unnecessary labour therefore, consider either the point A or B, because by doing so we get rid of one of the unknown forces. If e.g. we consider B as our moment-centre, then taking moments about it we have BA x7-2x4=0 where the clockwise sense of rotation has been adopted as positive. Hence BA = ? = 1? cwt., and since KA + BB = 2 cwt. BB = ? cwt. 2. A horizontal pull of 200 Ib. is applied to a vertical post which is supported by a stay, as shown in fig. 57. Find the pull in the stay if the foot of the pole is assumed free to turn. I THE ELEMENTS OF GRAPHIC STATICS Taking moments about the foot we have 200 x 15 + P x a = 0. Now a = 5-656 ft. ''*>. Tof---** . I POl FIG. 57. 3. A beam, whose weight is neglected, carries the loads shown in fig. 58. Find the values of the re- actions at A and B, where B is a pin-joint. Problems of this kind are more conveniently dealt with by graphical methods, as will be shown FIG. T>8. later, but it is desirable to be able to proceed by cal- culation also, and in this case it is best to resolve the loads into their horizontal and vertical com- ponents. This may of course be done graphically or mathematically. MOMENTS 63 Also let X and Y be the components of R B . Taking moments about B we get R A x 10 - 2 x8 - 3sin60 x 6 - 1 sin 45 x3 = 0, or R A x 10 = 16 + 1-5 x 6 + -707 x 3 = 27-1. .-. R A = 2-71 tons. Also since the horizontal forces must have a zero resultant, we get X - - 3 cos 60 - 1 cos 45 = or X = 1-5 + -707 = 2 '207 tons, and since the vertical forces must have a zero resultant, we get Y + R A - 2 - 3 sin 60 - 1 sin 45 = 0. .-. Y = 2 + 1-5 + -707 - 2-71 = 1-5 approx. .-. R B == VX* +~^= V7-12 = 2-67' tons and tan = |= ^=-678. .-. B = 34 8'. 4. Three loads of 2, 1, and 3 cwt. rest upon a beam at 2, 5, and 9 ft. respectively from the L.H. end. Find the vertical reactions. Ans. 2-4 and 3'6 cwt. 5. A flap-door, whose weight is 100 lb., is raised by a pull applied at an angle of 45 and at 6 ft. from the hinges. Find the pull required, if the resultant weight of the door acts at 3 ft. from the hinges. Ans. 70-7 lb. 6. Loads of 100 lb., 150 lb., 100 lb., 100 lb. act at angles of 45, 90, 60, and 90 at intervals of 2 ft. from the left along a beam 10 ft. long, the left end of which is hinged, the reaction at the other end being vertical. Find the reactions. Ans. Vert, reaction 210 lb. ; inclined reaction 233 lb. 64 THE ELEMENTS OF GRAPHIC STATICS 7. A platform in the shape of an equilateral triangle, with sides 6 ft. long, supported at its three corners, carries a weight 3 ft. from one corner and 4 ft. from another. Find the fraction of the weight borne by each of the three supports. (I.C.E. Exam.) Ans. -232W ; -336W ; -432W. 8. Find the stresses in the frame sketched in fig. 59, the members AB and CD being continuous. Find the maximum bending moment in CD, and show how to calculate the direct stress in it. (B.Sc.) Ans. Stress in AD = - -416 T ; maximum bending moment in CD = 2-5 tons ft. Stress in EF= + 1-59 T . 31, The Method of Sections, The ".Method of Sections," as it is usually called, is an application of the method of moments to framework structures which are statically determinate with the object of finding the internal force in any particular bar, when the dimensions of the structure and the loads acting on it are known. This method consists in taking an MOMENTS 65 imaginary section through the structure, so that the forces in all the bars cut by it are known except three at most. If then the force in one particular ^ar be required, and we take as moment-centre the point in which the other two intersect, the equation of equilibrium so obtained will involve only one un- known quantity, which can therefore be found. Example 1. In order to make this important method clear, consider the roof truss in fig. 60, FIG. 60. where we require to find the stress T in the tension bar. Suppose an imaginary section XX taken, as shown. Then the structure on the left of this section may be regarded as in equilibrium under the action of the known external forces together with the three internal forces in the bars cut by the section, and which may now be regarded as external forces applied along them at the points where the bars are cut through. If then we choose the point O as centre of moments, we get 1-1000 x 50 - 4000 (37-i 4- 25 + 12$) + T x 30 = 5 66 THE ELEMENTS OF GRAPHIC STATICS whence T = 20000 lb., the negative sign indicat- ing that the force acting on the extremity of the bar tends to produce contra-clockwise, rotation about O, and is therefore a tensile force. As a further illustration, suppose we require the force S. In this case the point O' in which the other two bars meet must be taken as centre of moments, and we get for equilibrium 4000 (12-| + 25 + 37^) + S x s = 0. 300000 , .. S = - where s = 20 ft. - s .'. S = - 15000 lb., the negative sign representing contra-clockwise rotation about O' and therefore a tensile force. Similarly, if we take moments about O'' to find the force P in the principal, we get 14000 x 29 - 4000(16-5 + 4) + 4000 x 8-5 + P x 10-8 = whence P = - 33200 lb., and contra -clock wise rota- tion about O" indicates a compressive force. Example 2. Determine by the Method of Sections the forces X, Y, Z in the three bars of the roof truss in fig. 61. Ans. X = 32-2 tons comp. ; Y = 11*7 tons tension ; Z = 37-Q tons tension. 32. Graphical Treatment of Moments, Let P (fig. 62) be any given force acting in a given direc- tion AB, and let Q be a given point at distance a from AB. Then P x a is the moment of the force P about the point Q. Let be be a vector representing the force P to scale, and let O be any pole whose distance from be is p. Join Ob, Oc, and through any MOMENTS 67 point P in AB draw lines P6', PC' parallel to 06, Oc, making an intercept b'c' = i on a line through Q parallel to AB. Then by similar triangles be p . , - or be x a = p x i. .'. P x a = p x i ; that is to say, the moment of the force, viz. P x a, is equal to the product of the polar FIG. 61. FIG. 62. distance measured in the scale of forces by the inter- cept i measured in the scale of lengths. At first sight the graphical representation of a moment in this way will appear to be less convenient than its mathematical expression, but this disadvan- tage will disappear as the subject develops. 33, Couples, Two equal and opposite forces not in the same line are known as a couple, and the moment of a couple is measured by the product of one of the forces into the distance between them. For if P and - P (fig. 63) be two equal and opposite 68 THE ELEMENTS OF GEAPHIC STATICS forces distant d from one another, and if we consider their moments about any given point Q whose dis- tance from the one force is x and from the other x + d, then their resultant moment about Q will be P(x + d) - Px = Pd. HeDce, the resultant moment being independent of the value of x, is the same about all points whatever, being equal to P x d. 34. Effect of a Couple, If P (fig. 64) be any force acting through A, and if we suppose at any point B a pair of equal and opposite forces applied along a line parallel to the original force P, then FIG. 63. FIG. 64. these three forces produce the same effect as the original one, and may therefore be considered as replacing it. But these forces consist of a single force P acting at B in the same sense as the original one, together with the couple P x d ; so that the force P acting at B is equivalent to the same force P actmg at A, together with a couple whose moment is equal to the product of the force into the distance through which it has been shifted. Conversely, the introduc- tion of a couple is equivalent to shifting a force parallel to itself, so that the moment due to its shift is equal to tha.t of the couple, MOMENTS 69 35. Composition of Couples, Since couples have magnitude, sense, and direction, they are vector magnitudes and may be compounded by the law of vector addition. In order to effect this the moment of a couple is represented by a line perpendicular to its plane, the sense of its moment being indicated by the sense of the arrowhead attached to it. The vectors representing two or more couples may then be compounded in the usual way, and the resultant vector will represent by its magnitude and sense the FIGS. 66, 65A. resultant moment and will be perpendicular to the plane of the resultant couple. Thus if P^, P 2 a 2 , P a fl 3 (fig. 65) are the moments of a system of forces tending to rotate a shaft AB, and if ab, be, cd represent these moments, the resultant vector ad will measure the resultant moment E, which is obvi- ously in the same sense as that of the first two moments. 36, Resultant of a System of Non-Concurrent Forces. In the preceding paragraph we have seen that if a force P l (fig. 66) act upon a body at a point A, and another force P 2 at a point B, the effect of 70 THE ELEMENTS OF GRAPHIC STATICS the latter is the same as if it were to act at A, together with the couple whose moment is P 2 x d. But since a couple consists of two equal and opposite forces, having a zero resultant, the magnitude and direction of the resultant force cannot be affected by it, but only its position. Hence, when a system of non-concurrent forces acts upon a body, the magni- tude and direction of the resultant may be found exactly as if the forces were all transferred to one point ; but its position will depend upon the resultant moment -of all the couples due to this transference FIG. 66. of the forces. The position of the resultant force is however more conveniently determined by means of the following construction : Let P lf P.,, P 3 (fig. 67) be any system of non-con- current forces. Then if 1, 1 2, 2 3 be the vectors representing them, 3 is their resultant in magnitude, sense, and direction, as we have just shown. In order to find its position, take any convenient pole O and join OO, Ol, O2, O3. Through any point a in the force P 1 draw lines xa parallel to OO, ab parallel to Ol, be parallel to O2, and cy parallel to O3. Then P! may be replaced by its two components acting along j'rt, ba and represented by 00, Ol in the vector MOMENTS 71 figure. Similarly, P 2 may be replaced by its two components acting along ab, cb and represented by 1O, O2 ; and again P 3 by forces along be, yc repre- sented by 2O, O3. Now the forces along ab and also along be are equal and opposite forces, being represented by the same vectors 1O and 2O respec- tively in opposite senses ; hence they cancel one another, and there remain only the forces along xa, FIG. 67. ya as the equivalent of the original system of forces P lf P 2 , and P 3 . The resultant of these forces along xa, ya must pass through the point of intersection d, and therefore if a line be drawn through d parallel to 3, it will represent the line of action of the re- sultant force E ; and since the same reasoning may be extended to any number of forces, we conclude that if a polygon be drawn as above its first and last links will intersect in a point which lies on the re- 72 THE ELEMENTS OF GRAPHIC STATICS sultant force and therefore fixes the position of its line of action. Such a polygon is known as a Link- or Funicular- Polygon. 37. Further, if the first and last links xa, yc, along which act the two forces represented by OO and O3, coincide along the line ac, then OO and O3 will coincide also. But as we have shown above these two forces are equivalent to the original system, and they will now form a pair of equal and opposite forces acting along ac, from which it is evident that when the link-polygon is closed by the lines xa, yc produced being in coincidence, the given force system is in equilibrium ; but if the first and last links are parallel, the points 3 in the vector figure will coincide, and this figure will close, but the link- polygon will not. In this case the forces acting along the first and last links form a couple whose moment is equal to that of either of the forces repre- sented by its corresponding vector multiplied into the distance between the first and last links of the link-polygon. Hence, 1. When the vector and link-polygons both close, we have complete equilibrium. 2. When the vector-polygon closes, but the link- polygon does not, we have a couple. 3. When neither polygon closes, we have a definite resultant represented by the line which is required to close the vector-polygon and acting through the point in which the first and last links of the link- polygon intersect. 38. Special Case, When the load system con- sists of vertical forces, (a) When all the loads lie between the reactions. This case is of very common MOMENTS 73 occurrence in practice, because the loads which act on structures are usually vertical forces. The determination of the resultant of the given loads is exactly the same as for that of non -parallel loads explained above. Thus if W lt W 2 , W 3 (fig. 68) be any given vertical loads, and 1, 1 2, 2 3 be the vectors which represent them, 3 will be the re- sultant vector. To find the position of the resultant force take any pole O and construct the link-polygon FIG. 68. abcde as before. The reactions are usually assumed vertical (though of course this really depends upon whether the supporting surfaces are horizontal or not), and therefore the first and last links cut them in a and e. Join ae and draw through the pole O a line O4 parallel to ae. Then as before we may show that the resultant load B acting at / may be replaced first by two forces along af and ef, and then again by forces along ae and Aa, and along ea and Be respectively, of which those along ae cancel out, leaving forces represented by 4 and 4 3 as the 74 THE ELEMENTS OF GRAPHIC STATICS components of R along the verticals through A and B. Consequently the line O4 drawn parallel to the closing line ae of the link-polygon divides the vector line into two parts which represent the vertical pressures on the supports, or the vertical reactions, if reversed in sense. This method of finding graphic- ally the magnitude of the reactions at A and B is particularly useful for reasons which will appear FIG. 69. later ; but if the reactions are alone required then they may he obtained by calculation with less trouble than by drawing, as in 29. (b) When all the loads do not lie between the re- actions. Let W p Wo, W 3 (fig. 69) be the loads and A, B the supports. As before set down vectors 1, 1 2, and 2 3 to represent the loads, and with any convenient pole O construct a link-polygon abode, the first and last links being produced to cut the support reactions in a and c.. Join ae, and draw a line 4 MOMENTS 75 parallel to ae. Then 4 0, 3 4 are the reactions at A and B respectively, the reason being as in the previ- ous case. By calculation, taking moments about B, we have R A x9-2x6-4x2 + 2x4 = 0.- ... K = 12 + * ~ IS = IJand R B = 9 - 1J = 6 tons. 39, Graphical Determination of the Moment of any Given System of Forces about a Given Point, FIG." 70. Let P lf P 2 , P 3 be given forces and Q a given point. Construct the vector figure 0123 for the given forces. Then 3 is their resultant. With any pole O draw a link-polygon abcde. Then a line drawn through the point / in which the first and last links 76 THE ELEMENTS OF GRAPHIC STATICS meet, parallel to 0-3, is the line of action of the re- sultant force. Now the moment of this force, which will also be the moment of the given forces about Q, can be found as described in 32, viz. by drawing through a point / on the line of action of the force E two lines parallel to the lines O , O 3 in the vector figure. Now these lines correspond to the first and last links of the link-polygon produced and they cut off an intercept ea = i, which measured in the linear scale and multiplied by the polar distance p measured in the force scale gives the moment of the resultant K about Q, and therefore also the moment of its components. \ \ F.G. 71. Hence, if a link-polygon be drawn for a given system of forces and its first and last links be produced to cut a line through the given point parallel to the resultant of the given forces, then the intercept on this line multiplied by the polar distance gives the moment of the forces about the given point. 40, Application to a System of Parallel Forces, Let E, Pj, P 2 , be any given parallel forces whose MOMENTS 77 moment about Q is required. Draw the vector figure 0123 and the link-polygon abode. Then if a line be drawn through Q parallel to the resultant 3, the intercept ae on this line between the first and last links produced to cut it measures the moment of the forces about Q. CHAPTEE VI. ROOFS. 41. In calculating the probable maximum stresses that are liable to occur in the members of a roof truss, the first step is to estimate the maximum loads which are likely to come upon the joints. These loads are due to the weight of 'the truss itself and of the roof covering and any other loads which the truss may support. These constitute the permanent or dead load, and in addition we have to consider the possibility of snow and wind. The following Tables I, II, and III will be useful in estimating the probable dead load that may be expected in any given case. (78) HOOFS 79 - gj M 'o 1 3 o is - 1 1 1 1 CO 00 O5 O <* CO 1 1 1 CO CO O5 O <* CO 1 1 1 O * 00 ' CO 00 05 >0 n CO 1 I I 0 ^ CO | I IH IH < o o * n e- I " H t I 1 I 1 I >O >OOOCO OOCOCO. 00 ^ t- 05 TH CD 4h CO 6> I >0 ' of sloping surface. All other roofs, 56 lb./ft. 2 on a horizontal plane. This last condition for slightly sloping roofs is to allow for the possibility of a crowd of people collecting on the roof. At the same time the working stresses for iron and steel are given as follows : ROOFS 85 Tension. Compression. Shear. Bending. Cast Iron 1-5 8 1-5 10 tons/in. 2 Wrought Iron 5 5 4 7 Mild Steel 7'5 7-5 j 5-5 1 11 i 44. Increase of Wind Pressure with Height. For high structures, like factory chimneys, etc., it is desirable to recognize the fact that the pressure and velocity of wind also increases very greatly as the height above ground increases, as shown by the fol- lowing figures : t above ground 5 9 15 25 52 miles per hour. 4 6 6 6-5 7'5 > 7 17 18 21 23 13 23 25 30 32 } 19 28 31 35 40 26 32 34 37 43 , These figures indicate that the pressure varies roughly as the square root of the height above ground. At the Firth of Forth Bridge, on 1-5 sq. ft. gauges, the pressure ranged from a maximum of 65 lb./ft. 2 at 378 ft. elevatjion to 20 lb./ft. 2 at 50 ft. 45, Wind Pressure on Inclined and Curved Sur- faces,' The following are experimental coefficients showing the relative pressures on various shaped bodies ; taking a flat plate normal to the wind as unity : Flat plate, 1. Hexagon, 0*65. Octagon, 0'75. Cubs (normal to face), 0'80. 86 THE ELEMENTS OF GRAPHIC STATICS . Cube (parallel to diagonal), 0'66. Lattice girders, about O8. Wedge (vertex angle = 90), 0'6 to O7. Sphere, 0'3. Elongated projectile, 0'5. Cylinder, -54 to '57. (height = diam.), 0'47 normal to axis. Cone (vertex angle = 90), 0'69 to 0'72. ( = 60), 0-54. ,, (height = diam.), 0'38 parallel to base. 46, Distribution of Load on Roof, The loads are estimated on the assumption that each truss FIG. 73. carries half the load on either side of it, and that this load is transmitted to the joints of the frame by means of the rafters and purlins. Thus if w be the load per sq. foot of roof surface due to the dead or live load, as the case may be, I be the length of the rafters, and d the distance between the trusses, ivld will be the total load on one principal such as DE (fig. 73), and as each joint, such as E or F, e.g., is likewise assumed to carry half the load on the panels to the right and left of it, the full panel loads will act on all the joints except those at the walls (in the case of the dead load), which carry one-half of a full EOOFS 87 panel load ; but in the case of the wind pressure we have to consider it as being fully loaded on one side or the other in turn, and in this case we have full loads on the one side at all the joints except at the top and bottom. Special calculations must of course be made for lanterns or other variations. The end trusses of the roof carry only half the load which comes on the intermediate trusses, but they are made to the same dimensions as the rest for the sake of uniformity. 47. Example 1. French Roof Truss. One of the most economical types of truss, and therefore one FIG. 74. of very frequent occurrence, is that particular type shown in fig. 74, and commonly known as the French truss. If we set down the joint-loads along a vector 88 THE ELEMENTS OF GRAPHIC STATICS line, we get ab as the reaction at one end, and if we proceed in the usual way we get the forces in the bars without difficulty up to the joints A and B. At each of these points we are met with three unknown forces, and these cannot therefore be found unless we can determine one of the unknown forces in- directly. This may be done in several ways : I. By calculating the stress in the bar a7, as ex- plained in 31, and plotting it on the force diagram. II. By the method of bar substitution, which con- sists in removing the bars 4 5 and ' 5 6 and putting in a new bar BC. As the force in the bar al may readily be seen by the reasoning of method I to be independent of the system of bracing adopted, this force may be found graphically from the new ar- rangement in the ordinary way, instead of by calcula- tion as in method I. The old bars are then replaced and the force diagram may be continued in the usual way. III. Suppose the bars 12, 23, 45, and 5 6 re- moved and the half-load on the principal, viz. ce, con- centrated at A. If this force ce is then resolved parallel and normal to the principal,' as shown, we get 3 4 as the force acting in the corresponding bar, and when this is known, all difficulty is removed, and we can proceed in the usual way. 48. Example 2. A station roof of the form shown in fig. 75 being given, find the stresses in the members, if the dead load works out at 200 Ib. per foot of rafter and the wind pressure at 300 Ib. per foot normal to the roof, assuming that the support reaction is vertical on the side opposite that on which the wind acts. Width 30 ft. ; height 10 ft. Base EOOFS 89 and principals supported at equal intervals. Since the length of each principal is 18 ft., the dead load on 1 truss = 2 x 18 x 200 = 7200 lb., which gives 1200 lb. at each joint except the extreme ones, on which the load is 600 lb. Also the wind load on FIG. 75. one side is 18 x 300 = 5400 lb., or 1800 lb. on the intermediate joints and 900 lb. on the extreme ones. Method I. A diagram may now be drawn for the dead load, and another for the live load, and the re- sults compounded. Method II. The dead load and the live load may be compounded at each joint, and one diagram drawn 90 THE ELEMENTS OF GRAPHIC STATICS for the whole. A link- and vector-polygon may then be drawn, by means of which the resultant of the loads may be found in magnitude and position ; or for the sake of greater accuracy the resultant of the whole dead load acting at the centre of the roof may be compounded with the resultant wind pressure acting midway along the principal. This course has been adopted here. The two resultants intersect in C, and a line drawn through C parallel to the re- sultant of the total load, viz. ah, gives the line of action of this resultant, and this cuts the vertical reaction at A in D. Hence DB will be the reaction at B, and the magnitudes of the reactions can be found by constructing ahj on all. We can now pro- ceed to construct the force diagram in the usual way, as shown. In order to complete the investigation, the wind should also be assumed to act on the other side of the roof and the stresses again determined. The maximum value in either case must then be taken as the load liable to come on any member. This will be illustrated in 49. The forces in the bars will be found to be as follows, where + signifies compression and tension. Ib. Ib. Ib. bk - 1100 ak + 350 kl + 1100 cl - 2200 jn + 700 Im + 1200 dm - 2200 hq + 2550 mn + 2400 eo - 3100 no + 2000 fP - 4250 op + 3250 f JP - 2450 pq + 2950 Truss with Fixed Ends, When the truss is fixed, that is to say pin-jointed at its ends (because it is ROOFS 91 always assumed that the joints have no rigidity), the problems of finding the forces in the members is indeterminate, and the method often given in the textbooks of resolving the resultant load into two parallel components through the points of support is incorrect and may easily be made to give contra- dictory results. As no determinate solution is pos- sible, the best course to pursue is to make an assumption which will introduce the worst con- ditions likely to arise, and this will be to suppose (i) that one of the supports takes all the horizon- tal thrust and (ii) that the other takes it all, the wind acting on the same side in both cases. This of course means that we assume each of the reactions in turn to be vertical, the other being regarded as a pin-joint, exactly as is done in the case when one of the ends is supported on rollers. 49, Crescent Roof. In order to illustrate the above, we will now consider a crescent roof with fixed supports (fig. 76). The span was taken as 50 ft., with a rise of 8-J ft. from the chord AB to the top of the roof, and a depth of 5 ft. from the upper to the lower chord members, the. outer radius being 30 ft. and the inner 41'3 ft. The resultant wind pres- sures at the joints make angles of 56, 42, 28, and 14, with the tangents at those points, and the wind pressure being taken at 56 lb./ft. 2 normal to its direc- tion, the values of the normal pressures by calcula- tion or graphically will be found to be 55, 51*8, 43*2, 25*6 Ib. respectively. The lengths of the panels measures 7'3 ft., and assuming the distance between the trusses to be 12 ft., the wind pressure on the joints will be as follows : 2410, 4540, 3780, 2240 Ib. 92 THE ELEMENTS OF GRAPHIC STATICS FIG. 76. KOOFS 93 Also assuming the dead load of the roof and truss to be 10 Ib./ft.^, the dead loads on the end joints will amount to 438 Ib. each and on the inter- mediate joints to 876 Ib. each. As explained above we shall first assume the wind to act on one side and consider the whole of the horizontal thrust to be taken alternately first by the one support and afterwards by the other. Taking the wind to act from the left, and assuming (i) that the whole horizontal thrust is taken by the left-hand support, then the right-hand reaction will be vertical. We now compound the live and dead loads on the joints, and by means of a link- and vector-polygon the resultant B is determined in magnitude, direc- tion, and position in the usual way. Producing the line of action of this resultant to cut the vertical support reaction at B in Q, we get QA as the direc- tion of the reaction at A. A triangle of forces bb'a (i) may now be drawn which determines the re- actions at A and B. After this the force diagram may be constructed in the usual way as shown in (i), and their magnitudes scaled off, whilst the sense may be found from the direction of the vectors in diagram (i), as explained in 6. These results should now be tabulated. Next, the whole of the horizontal wind thrust will be assumed to be taken up by the right-hand support B. In this case the left-hand support will be vertical. Producing this to cut the resultant load E and joining to B, we get the direction of the reaction at B, and a triangle of forces bb'a (ii) determines the magnitudes of the reactions. A new force diagram is now drawn (ii) and the forces in the bars are again found from it 94 THE ELEMENTS OF GRAPHIC STATICS aiid entered in the table. If we now compare the forces in the members which are symmetrically situated, and take the larger values, these will be the maximum forces which will occur whether the wind blow from the right or from the left. CHAPTEE VII. SHEARING FORCE AND BENDING MOMENT DIAGRAMS. 50. Definition, The shearing force at any section of a structure is the force parallel to the section which tends to cause sliding between it and the adjacent section. Let ABCD (fig. 77) represent a cantilever which FIG. 77. is cut through at XX, the portion BCEF being kept in position by a spring-balance attached at F which prevents vertical motion, and by a wire FG attached to a spring balance at G, and a small strut XE, the tension in the wire together with the thrust in (95) 90 THE ELEMENTS OF GRAPHIC STATICS the strut preventing rotation. The spring-balance FH will register the weight of the portion BCEF which tends to cause vertical movement, and which is therefore equal in magnitude to the shearing force at the section EF. If any loads W x , W 2 be applied to the free portion, the spring-balance will register their sum as the additional shearing force. At the same time the moment of the couple due to the pull in the wire GF and the push of the strut will be measured by the pull P in the spring-balance multiplied by the arm a of the couple. This moment is known as the Bending Moment at the section, being the moment of all the forces on the one side of the section which cause bending in the beam. Hence we see that the effect of a load on any section of a beam may be analysed into a shear- ing force tending to produce a slide and a couple tending to produce rotation. The first is resisted by the shear stress in the material, and the second by the moment of resistance of the tensile and compres- sive stresses induced. In order that a beam or girder may be strong enough at all sections to resist the joint action of the shearing forces and bending moments, it is important to be able to construct diagrams showing the magni- tudes of these at all sections. So far as the shearing force is concerned, this merely amounts to finding the algebraic sum of all the forces parallel to the section on either side of it, because at the section these form a pair of equal and opposite forces which tend to produce a slide either upward on the one side or downward on the other. Thus at any section XX (fig. 78) the shearing force on the left of it will SHEARING FORCE AND BENDING MOMENT 97 be equal to the reaction KA upward, and on the right of it will be equal to W - E B downward, and since W - RB = RA we see that the shearing forces are equal and opposite on either side of the section. This we indicate by plotting their values on opposite sides of the base line AjB^ and for the sake of uni- formity we will make the convention that a shearing force which tends to make the left-hand segment move upward relatively to the right shall be con- sidered positive, whilst one which tends to make the right-hand segment move upward relatively to the left shall be considered negative, the first being c RO j & ;. | "", i ; FIG. 78. plotted above the base-line A^ and the second below it. 51. Diagrams of Shearing Force and Bending Moment, It will now be shown that the dia- grams of shearing' force and bending moment may be easily drawn at one and the same time by the same construction which has been explained in the last chapter for finding the reactions. For suppose loads of 2, 3, and 4 tons to act as shown in fig. 79 on a beam simply supported at A and B. Then if we construct the link- and vector-polygons, and draw through the pole O a line O4 parallel to the closing line bg, we get 3 4 as the reaction at A, and 4 as the reaction at B. Draw a horizontal 4C through 4. 7 98 THE ELEMENTS OF GEAPHIC STATICS Then if we project the point to D the ordinate of the rectangle CDDjD 2 represents the constant posi- tive shearing force at any section of the beam between CD 2 , this shearing force being equal to the reaction &t the left-hand support. After passing D.> however, the resultant of the forces on the left of a section becomes equal to the reaction 4 the load 01 = 4 1. Hence we project the point 1 as shown to EEj. Similarly, on passing the next load the resultant force becomes negative and equal to 4 2, and finally l FIG. 79. on passing the last load it becomes 4 3 downward, being equal in magnitude but opposite in sign to the reaction at the right-hand support. The diagram thus obtained is a shearing force diagram, and shows at a glance the values of the shearing forces at any section. A difficulty frequently occurs to the student in contemplating such a shearing force diagram, in that we get apparently two. different values for the shearing force at one and the same section. Thus at D 2 we get D 2 E and D 2 D l . This ambiguity arises from the assumption tacitly made that a load may SHEAEING FOEGE AND BENDING MOMENT 99 act at a point. This is, of course, impossible, because the shearing stress which is - would then be area infinitely great. Thus if W (fig. 80) be a load this load must rest upon a base of some appreciable width, however small. If we suppose this width to be CD, then the shearing force on the left of C will be represented by AA 1? and the shearing force on the right of D will be represented by BB 1? and the value will change more or less uniformly from the one value to the other over the length CD on which the load rests. If therefore we assume a load to be 6 PIG. 80. concentrated at a point, as is often done for con- venience, the base CD shrinks to a point and the line EF becomes- vertical. It must, however, be understood that this is only a conventional representa- tion of the facts. Actually the upper ordinate will represent the shearing force on the left of the load, and the lower ordinate will represent that on the right of it, however small the base may be, and the two values do not occur at the same section, although so represented for convenience. Not only, however, do we in this way obtain a diagram of shearing forces, but at the same time the figure abcde by which the reactions were found is a 100 THE ELEMENTS OF GRAPHIC STATICS diagram of beading moments. For if we refer to fig. 71, p. 76, we have shown that the bending moment about any point Q is represented by the intercept be- tween the first and last links of the link-polygon on the vertical through Q, and if we suppose Q to be on the axis of the beam, then it will be seen that the intercept ae in fig. 79 corresponds exactly to the inter- cept ae in fig. 71, which we have shown to represent the moment about Q. Hence, the bending moment at any section of the beam under vertical loading will be represented by the vertical intercept of the link-polygon which is made by the section produced, this intercept being measured in the linear scale and multiplied by the polar distance in the load scale. 52. When some of the Loads Lie Outside the Supports. The procedure in this case is similar to I FIG. 81. 'that in the previous case. The loads are set down along a vector line (fig. 81) and a link-polygon ab . . . f is drawn with any pole O for the given forces Wj . . . W 4 , the first and last links being produced back SHEARING FORCE ANb BENDING 'MOMENT 101 to cut the support verticals at a and /. The line af is then the closing line and the hatched figure is the bending moment diagram. In constructing the shearing force diagram the load W x = 1 is first set down below the base line of shear drawn through 5. At D the vertical reaction E A = 5 is set up from Dj to D 2 . The points 2 and 3 are then pro- jected across as before, until we come to B when the vertical reaction RB rnust be set up from G to G!- The figure so found is the shearing force dia- gram, and it will be seen that it merely represents the algebraic addition of the loads from left to right. Note. In constructing the bending moment dia- gram it is advisable to use a polar distance which is a round number of units, in order to simplify the numerical evaluation of the bending moments from the diagrams. 53, Shearing Force and Bending Moment Dia- grams for Special Cases, Case I. Cantilever with concentrated load. When a load W (fig. 82) acts at the end of canti- lever the shearing force at all points is constant and IQ'J THK ELEMENTS. OP GRAPHIC STATICS equal to W. Hence the diagram will be a rectangle of constant height representing W. The bending moment at a distance x from the free end will be W#. This represents a value which increases uniformly from zero when x = to a maximum of WJ when x = I as indicated in the figure. Case II. Cantilever with uniform load w per foot run (fig. 83). In this case if we consider a section at any distance FIG. 83. x from the free end, the total load on the right of it will be wx, and therefore the shearing force = wx ; that is to say, it will be represented by a straight line BC which increases from zero to a maximum at the fixed end, where its value is the whole load W = wl. Also the bending moment M r at x will be the load wx x the distance of its centre of gravity from the section, viz. \x. .'. M. x = wx x \x = ^wx 2 . If the values of M x be calculated for different values of x we get a curve DE whose ordinates vary SHEARING FORCE AND BEMDlNG MOMENT 103 as the squares of the abscissae, viz. a parabola, the maximum bending moment being -^- = JWZ where W is the total load. Case III. Beam with concentrated load W (fig. 84). Taking moments about B, we get B A x I = W x b. ' >: E A = W.^ a and similarly E= - B FIG. 84. Hence the shearing force diagram will be as shown in fig. 84. Also the bending moment will increase uniformly from the supports up to the load, where its value will be b EA x a = W . -j .a = I or the maximum bending moment product of segments = load x - . span If the load is central, then a = b = -j-/ and the W7 bending moment at the centre is -j- . 104 THE ELEMENTS OF GRAPHIC STATICS Case IV. Beam with distributed load W = wl (fig. 85). In this case the reaction at A being ^ , the shear- ing force will have this value at A and will decrease at a uniform rate, until it becomes zero at the mid- span, and then increases to - ^ at B. Also the FIG. 85. bending moment at a section x from the centre will be wl /I \ (I \ /I Here again the bending moments depend upon the square of x and the curve is therefore of para- bolic form, the maximum ordinate being obviously wl- W/ at the centre when x = and its value is - o o or half the value if the same load were concentrated at the centre. (Compare with Case III.) 54. It will be seen that for uniform loading the bending moment diagram under the load is always SHEARING FORCE AND BENDING MOMENT 105 of parabolic form, and therefore it is necessary to be able to construct a parabola for the following conditions : Case I. When the base, height, and axes of the parabola are given. Let AB (fig. 86) be the given base and CD the a c Fia. 8G. axis and height. Join AD, and draw any vertical ab to cut it in c. Project c to d and join dD, cutting the line ab in e. Then e is a point on the parabola, and other points can be found in the same way. Case II. When the tangents to the parabola are given and its end points. FIG. 87. Let AC, BC be the given tangents and let A and B be the points of contact. Divide AC into any number of equal parts, and BC into the same number of equal parts as shown in fig. 87. Join 1 1 106 THE ELEMENTS OF GRAPHIC STATICS 22, 33. These lines will be tangents to the para- bola and will sufficiently define it, if enough tangents are drawn. 55, Example 1. Construct combined diagrams of shearing force and bending moment for a beam carrying a point load W and a distributed load w per ft. Set up CC X = W^- to any scale, and join B Fro. 88. j, BC r Then ACjB is the bending moment dia- Calculate the maximum bending moment WJ 8 ' Set it up at DDj (fig. 88) to the same scale as before and construct the parabola ADjB. Then the whole figure represents by its ordinates the combined dia- gram of bending moment. SHEARING FORCE AND BENDING MOMENT 107 Next draw a vector OK = W, and OO, KO parallel to A 1 C 1 , G l B l meeting in O. Then O is the pole of the vector diagram, and a line OP parallel to AB, i.e. horizontal, will give the two reactions. The shearing force diagram may then be drawn as usual, via: EFGHKP. If we now plot EL and PM = %wl and join LM we get the shearing force diagram for the distributed load, and the whole figure will repre- sent the combination of the two diagrams on a base FIG. 89. LM, any areas which are positive in one diagram and negative in another cancelling out. Example 2. A beam AB carries the load shown in fig. 89. The distributed load W 2 is first con- sidered as a concentrated load acting through its centre of gravity, and a link-polygon AKHLC is drawn as usual ; and the closing line AC. O 4 parallel to this determines the reactions at A and B. Since, however, the load on DE is uniformly distributed instead of concentrated, as assumed, the link-polygon between F and G should be a parabolic one. 108 THE ELEMENTS OF fJRAPHTC STATICS 56, Indirect Loading, When a load does not bear directly upon a beam or girder supporting it, but is transmitted to it through the medium of cross-girders or otherwise, so that the load bears on the beam at intervals, the loading is said to be indirect. This is the commonest case in practice, because in all large girders, whether triangulated or plate girders, the load is usually transmitted to the main girders through the cross-girders. In this case the bending moment will be modified, as will be seen from the following example : Let Wj, Wo, W 3 , W 4 be any system of point loads on a girder, as shown in fig. 90, the loads resting on the platform, which is carried by the cross-girders at C, D, E and P. Construct the bend- ing moment diagram in the usual way, viz. aHJKLB. Then neglecting the continuity of the longitudinal members, if we join cd, d\\, K/', the new diagram obtained in this way will be the actual bending moment diagram, the parts cHd, dJeK, KL/ being the bending moment diagrams for the panels CD, DE, EF for the loads upon them, and it is only SHEARING FORCE AND BENDING MOMENT 109 therefore at the panel joints that the bending moment is the same as for direct loading, being less between the joints than if the load rested directly on the main girder. Similarly, for a uniformly distributed load, if we draw the parabolic diagram in the usual way, and drop verticals from the joints to cut it, the chords connect- ing the joints will be the true bending moment dia- gram ; that is to say, the true bending moment diagram is a polygon inscribed in the parabola. 57. Determination of the Forces in the Bars of a Braced Structure from the Shearing Force and Bending Moment Diagrams, Case I. Parallel Booms, Let ACDEFGB (fig. 91) be a triangulated girder, carrying loads W p W 2 G FIG. 91. at its joints, as shown, and let the shearing force and bending moment diagrams be constructed in the 110 THE ELEMENTS OF GRAPHIC STATICS usual way. Then if the force in a diagonal CD is required, this force can be found at once from the shearing force diagram when the chord members CE and AD are horizontal, because as the forces in these will then have no vertical components, the whole of the shearing force will be equal and opposite to the vertical component of the force in CD. There- fore if we take a vertical section XX cutting the shear- ing force diagram in a and b and draw cd parallel to CD, cd will represent the force in CD ; because it is evident that the vertical component of cd is cc and is therefore equal to the shearing force ab. Similarly, the stress in the bar DE will be given by the line ef drawn parallel to it, and the stress in FG by gh. In order to find the force in one of the boom members such as CE, we suppose the girder cut by a section XX through the three bars CE, CD, and AD. Now it will be evident that if the bar CE be supposed cut, the girder will collapse due to the bending moment about D, and it is the force acting along EC whose moment is in equilibrium with this bending moment. Hence the force in EC x the depth of the girder d must equal the bending moment about D. But this may be measured off the bend- ing moment diagram, being represented by the ordin- ate Kl x p, and therefore the force in EC x d= Kl x p or the force in EC = K/ x ^, and if p had been drawn equal to d, then force in EC = K/ measured in the scale of loads. In the same way the force in the bar DF may be found by writing force in DF x d = moment about E = mn x p, and so on. 58. Case II. Non-parallel Booms, Let ACB SHEARING FORCE AND BENDING MOMENT 111 (fig. 92.) be a braced 'girder, with curved upper boom, loaded as shown, and let AjLBj be the bending mo- ment diagram. Then the force in any bar such as CD may be found from the fact that this force multi- plied by its distance from O is in equilibrium with FIG. 92. the bending moment about O, which is given by the ordinate EF. For force in CD x d = FL. . ^ FL bending moment at O /. force in CD = =- = - ~d~ ~' If we set off cd = force in CD and draw de parallel to DO, then de will be the force in the diagonal DO, because the vertical components of cd and dc together must balance the shearing force at the section. CHAPTER VIII. RELATION BETWEEN THE CURVES OF LOAD, SHEAR, AND BENDING MOMENT. 59, Load-Curve, When a load is supposed to act at a point it may be represented by a straight line, but when the load is distributed, it will be represented by an area. Actually as we have ex- _ plained it should always be represented af\^ in this way, because the load is distri- [ y buted over some area, however small, in all cases. If the load is uniform and equal to w per unit length over any given distance d, then the total load over that distance will be wd t and . . ,, will be represented by a rectangle of ' t i base d and height w. When the load FIG. 93. is variable, the average intensity w lt over a short length rf,, may be taken, and the load on d l will then be represented by w^ or graphically by the area of a rectangle whose base is cl l in the scale of lengths, and whose height is w } in the scale of load intensity (fig.- 93). Similarly, the load on d 2 will be represented by a rectangle of height w<>, representing the mean value of the load intensity over do, and so on ; and the sum of these areas will represent the total load on the length which is the (112) RELATION BETWEEN CURVES OF LOAD, ETC. 113 sum of their bases. The smaller the lengths d lt d z , etc., are taken, the more numerous are the rect- angles, and the more nearly do they represent the actual load distribution AB, but in practice unless the mathematical expression for the load-curve is known, we are compelled to proceed graphically by dividing the load area into small strips and to assume that these areas are equal to their widths multiplied by their middle ordinates. The summation of the load may then be effected graphically as follows Fio. 94. (fig. 94). Suppose AC to be the load-curve for the load upon a beam AB, the ordinates of the curve representing to scale the varying load intensity. Let the area under AC be divided into vertical strips by the ordinates aa lt bb lt cc l9 etc., and let a^j, %<$<>, etc., be the mid-ordinates. Project y v y 2 , etc., upon a vertical axis OY to y\, y' 2 , etc. Take a pole O on BA produced and join Oy' ly Oy' 2 , etc. Next, start- ing from O draw a line Oa 2 parallel to Qy\, and from a 2 draw a. 2 b> 2 parallel to Oy'^ and so on. Then if we consider the area of any one of the strips, say 8 114 THE ELEMENTS OF GRAPHIC STATICS that on the base be, and compare the similar triangles 6 2 6 3 ~ OA p ' .-. b. 3 c. 2 x p = b 2 b s x # 3 2/ 3 = base x mean height. Hence we see that the increase in the ordinate in going from b to c, viz. 6 3 c 2 multiplied by the polar dis- tance p, measures the increase in area. Similarly, C -A x P represents the area on the base cd, and so on ; so that the ordinate of the curve at any point x d represents the sum of the areas up to that point. For this reason the curve Aa. 2 b. 2 c. 2 d. 2 etc. is known as a Sum-Curve or curve of integration. 60, Shear-Curve, The foregoing construction will enable us to draw a curve of shearing force for Frc. <>f>. irregular loading. Thus, suppose the sloping line to be the load-curve for a beam AB, and let Aa . . . e be the sum-curve of this load-curve drawn with a polar EELATION BETWEEN CURVES OF LOAD, ETC. 115 distance p. Then since the increments of ordinate of the sum-curve represent the increments of area, if we project the points a, 6, etc., to a 2 , 6 2 , etc., the lengths Ba 2 , a 2 6 2 , etc., will represent the areas or loads on the bases Aa 1? a-J)^ etc. These loads are sup- posed to act along their mid-ordinates, and a link- polygon ghk . . .pis drawn whose closing line is gp. Through the pole O draw Or parallel to gp. Then Br represents the reaction at A, and re that at B. Draw a horizontal rs through r. Then rs is the base-line of shear and the diagram Asre is a diagram of shearing force. For consider any section, say b lt of the beam ; the shearing force at this section will be the difference between the upward reaction at A and the sum of the load on A6 X acting downward. Now b^ = As represents the first and bb-^ represents the second, so that bt represents their difference ; that is to say, the ordinate at any point between the line sr and the sum-curve Aa . . . & gives the shearing force at that point. 61. Bending Moment Curve, The shearing force at any section of a beam is by definition the summation of the loads on one side or the other of the section, and is represented by the area of the load-curve ; and it may be shown that the bending moment at any sec- tion is in a similar way represented by the area of the shear-curve. For let the shearing force diagram for the loads in fig. 96 be drawn. Then the bending moment at any section XX will be equal to E A x A X X - W x x D 2 X - W a x E 2 X = A X C x AjX - DD X x D.N - EE; x EN = area AjCDD^EEjPX = area of the shearing force diagram. 116 THE ELEMENTS OF ftRAPHIC STATICS Hence in the above problem ($ 60) the bending moment diagram might be obtained by constructing a sum-curve of the shear-curve on the base sr, but as J ID - i yjl f 1 qrtj C *vJa N; i i Q | Rp R fl| Pl E l^ { r. j .. Fir . 96. ci i' the figure gh . . . p is already a bending moment diagram, it is unnecessary in this case, at any rate, to draw a second sum-curve. 62, Scales, In order that the numerical values of the shearing forces and bending moments may be known, it is necessary to find the scales of the dia- grams. If the scale of load intensity be 1 inch = a tons or pounds per unit length, then the ordinates of the shearing force diagram will have a scale of 1 in. = p^ where p l is the polar distance used, measured in the linear scale of the drawing. Thus if the linear scale be 1 in. = I feet and p l is in inches, the scale of forces will be 1 in. = p } al, and this will be the scale of the shearing force diagram. Again, if the sum-curve of the shear-curve be drawn with a polar distance of p., inches, then the scale of bending moment will be 1 in. measured in the force scale multiplied by the polar distance p in the linear scale or 1 in. = p^al x p. 2 l, or if the polar distances are taken the same in each KELATION BETWEEN CUKVES OF LOAD, ETC. 117 case, and equal to p, then 1 in. = ap~l 2 tons/ft. 2 or lb./ft.- as the case may be. / \ i / H f i T ' ua's* 2 s < 16' > ^ 4 T .^.r~ t^ \ FIG 97. 63, Concrete Raft, Numerical Example. Fig. 97 represents a concrete rafft or distributing the load due to a building or other structure over a soft foun- dation. The raft was designed to be 3 ft. thick, and its weight may be taken as 125 lb./ft. 3 , the super- posed load consisting of the weights transmitted to the three piers, which are estimated at 14 tons, 12 tons, and 14 tons per foot length of walls respectively, symmetrically situated, and which may be taken to 118 THE ELEMENTS OF GEAPHIC STATICS act at the centres of the walls where they rest upon the raft. Total weight of concrete raft per foot length = 44-5 x 3 x 125 lb./ft. 3 = 7'45 tons. Load on raft per foot length = 14 + 14 + 12 = 40 Total load = 47 '45 The load on the soil is therefore ^ = 1-067 44-5 ton per sq. foot. This load was plotted = ba and a horizontal line through a represents the reaction of the soil. The 125 x 3 weight of the concrete = -g- = 0-167 ton per foot was deducted, as represented by the dotted line. The wall whose weight is 14 tons causes a pressure 14 of ^TK = 5'6 tons per foot, leaving a resultant down- ward pressure of 4'7 tons per foot, which was plotted as shown below the base line. In this way we get a load diagram abcdefghij. A sum-curve of this diagram was next drawn with a polar distance of 1 in. = 5 ft. This diagram is bklmnop, and represents the shearing force to a scale of 1 in. = 2 tons per ft. x 5 ft. = 10 tons. Next a sum-curve of the shear- ing force diagram was drawn, as shown, with the same polar distance. This will therefore be the bending moment diagram to a scale of 1 in. = 10 tons x 5 ft. = 50 tons/ft. The maximum ordinate was found to measure 1-3 in. on the drawing = 65 tons/ft. = 780 x 2240 lb,/in. RELATION BETWEEN CURVES OF LOAD, ETC. 119 .. the stress per sq. inch M 780 x 2240 x 6 = z = -isirssr =18>71b -/ m - 2 The concrete may be relied upon to withstand this working stress, and therefore no reinforcement will be required. CHAPTER IX. SHEARING FOECE AND BENDING MOMENT FOR A MOVING LOAD. 64, I, Single Concentrated Load Crossing a Simple Beam, Maximum Shearing Force. Let AB (fig. 98) be the beam, and let C be any section distant x from B. Then as long as the load W lies FIG. 98. on the right of C, the shearing force at C will de- pend iipon the value of the reaction at A, and as this will increase as W approaches C, the maximum shearing force at C will occur when the load has (120) SHEARING FORCE AND BENDING MOMENT 121 just reached C from the right. Now in this position EA can be found from the equation W E A x I = W x x, or EA = -y . x, I which shows that E A increases uniformly with x, and attains its greatest value when x = I, when it has the value W. Hence if we set up AA X = W and join AjB, the triangle AA^B will be a diagram of maximum shearing force for every section, and if we agree to call the shearing force positive which tends to cause the left-hand segment to move up- ward relatively to the right, then this diagram will represent the maximum positive shearing force at all sections. On the other hand, if we set down BBj = W and join AB lf the triangle AB X B will be a diagram of maximum negative shearing force. So that, for example, when the load stands at C the drdinate CC l will represent the maximum positive shearing force at the section, that is, on the left of the load, and the ordinate CC 2 will represent the maximum negative shearing force on the right of the load. At any other section such as D, DD X and DD 2 will represent the maximum positive and nega- tive values. Maximum Bending Moment. Since the bending moment at a given section D (fig. 99) is equal to BA (I x), and since EA increases as the load ap- proaches the section, the bending moment will be greatest when the load reaches D, and since EA is then equal to W -, the bending moment is I W6 x a 122 THE ELEMENTS OF GRAPHIC STATICS which being an expression of the second degree in x represents a parabola (see " Mathematics for Engineers "), whose central ordinate will be found by making x = iJ, when we get for the WZ central bending moment MC = -j-. If therefore we set up CC { = - and construct the parabola ACjB, the diagram so obtained will be the diagram of maximum bending moments for all sections of B FIG. 99. the beam. This parabola is the locus of the vertices of all the triangles such as ADjB which are the bend- ing moment diagrams for the loads at various points of the span. 65, II. Two Concentrated Loads, Distant d from One Another, Let W x (fig. 100) be the leading load and \V 2 the following load, and sup- pose them to move from right to left. Set up AAj = Wj and AA 2 = W 2 and join AjB, A 2 B. Then as in Case 1, A X B, A 2 B will be the diagrams of maximum shearing force for the separate loads. Now when Wj stands at C, the maximum shearing force at C is CC lf and as \V 2 has not yet come SHEARING FORCE AND BENDING MOMENT 123 upon the beam, BC l is the diagram of maximum positive shearing force for the segment BC. When the load W x reaches A, then AA X is the shearing force at A due to it. At this instant W 2 stands at D, and the additional pressure it causes at A is re- presented by DD r If therefore we make A X E = DD P AE will be the total shear at A, and if we join EC^ the diagram ECjB will be the diagram of maximum positive shearing force for both loads. The diagram ?r ^ B of maximum negative shearing force will be a similar diagram reversed below the base-line AB. Maximum Bending Moment Diagram. Set up CCj = _1- (fig. 101) and construct a parabola on 4 AB as base. Similarly, set up CC 2 = -^- and con- struct another parabola. These represent the maxi- mum bending moment diagrams for the individual loads. Then as in the case of shearing force, the maximum bending moment for the segment BD will 124 THE ELEMENTS OF GRAPHIC STATICS be represented by the curve BD lt because the load W 2 has only just reached B. If we now advance the front load to E, EEj will be the bending moment that it causes at E. But at this instant W 2 is at D and causes bending moments represented by the tri- angle AD 2 B. Consequently its effect at E is repre- sented by EE 4 . Make EE 4 . Then EE will be the maximum bending moment at E, and Fio. 101. so on. In this way we get the diagram AC.X^B as the diagram of maximum bending moment. 66. Ill, Train of Concentrated Loads. Maxi- mum Shearing Force. Suppose a train of loads Wj, W 2 , etc. (fig. 102) to cross a simple beam AB from right to left. Then as long as the leading load Wj lies to the right of a given section C, the shearing force at C will be the same as the reaction at A and will therefore go on increasing as the loads approach C. Let the loads be set up from A in order to scale, as shown, and let these loads be also set off in re- versed order with the leading load W x at B. Using B as a pole, construct a link-polygon ABDEFG. Then the ordinate y between the first link AB and SHEARING FORCE AND BENDING MOMENT 125 the link cut by the.ordinate at any section of the beam will represent the moment at that section for the assumed position of the loads reversed. Thus at the point C the moment of the reversed loads is measured by the product of y l in the length scale, by / in the force scale. But this is the same as the FIG. 102. sum of the moments of the actual loads about B which is equal to RA x /. RA x I = y 1 x I, or R A = y r Hence we see that the ordinate at C measures the reaction at A when the leading load stands at C ; and when the leading load is not small relatively to the loads which follow it, this value of RA will be the maximum shearing force at the section ; but when the leading load is relatively small, this may not be the case, because the increase in the reaction 126 THE ELEMENTS OF GRAPHIC STATICS due to advancing the loads may be greater than the decrease due to subtracting the front load which has crossed the section C. In order to test this, let the second load be advanced to C and let H be the corre- sponding position of W r Then .y. 2 MH will be the new reaction at A ; but since Wj is now on the left of the section the actual shearing force will be 2/ 2 - W lf so that if we make HK = W lf MK will be the new value of the shearing force at C, and we can therefore now see whether the shearing force is greater than before or not by comparing the values of T/J and MK. Maximum Bending Moment. Let AB (fig. 103) be the beam and W,, W 2 , etc., the given train of loads crossing it. This case is the most difficult to B Fio. 103. find a direct solution for,[ because, in the first place we do not know which combination of loads will cause the maximum bending moment, neither do we know under which load it occurs or at which section of the beam. The usual procedure therefore SHEARING FORCE AND BENDING MOMENT 127 is to draw link- and vector-polygons for the given loads, as shown in fig. 103, and to extend the first and last links indefinitely. Now if the beam AB be brought into any position as indicated, and verti- cals be drawn through A and B to cut the link- polygon in Aj and B p the closing line A. 1 B 1 will evidently give the bending moment diagram for the loads which are on the beam, and the maximum bending moment for the assumed position may be scaled off. In order to arrive at the absolute maximum bending moment, however, it will be necessary to shift the beam into various positions, and to find which of these gives the greatest vertical ordinate. This method is usually carried out by plotting curves of bending moment for selected sections of the beam, as it is moved by equal in- tervals under the loads. The highest point of each curve is then the maximum bending moment for the section to which it relates. If this is done for a series of sections, a curve drawn through the maxi- mum values at each section will be a curve of ab- solute maximum bending moment. Schlotke (" Lehrbuch der graphischen Statik ") has given an ingenious solution for the case when the span is longer than the load, so that no loads enter upon or leave the beam whilst these are shifted a little to one side or the other of their maximum position. Let \V t , W 2 . . . W 6 be any given loads (fig. 104) represented by the vectors 01, 12, etc. With any convenient polar O construct a link-polygon abc . . . g and produce the first and last links to meet in Q. Then a vertical through Q re- presents the position of the resultant. Draw in the 128 THE ELEMENTS OF GRAPHIC STATICS beam in any position A B, and at A set up A C = to the total load = 05 and join CB , cutting the resultant in D, and draw in the stepped figure or load-line E6FGHIJKLMNP, where CF, GH, etc., FIG. 104. are the loads 01, 12, etc., set down in order, and starting from b where the horizontal through D cuts the first load. Bisect A S, B P in X and Y and join XY. Then the maximum bending moment will occur when the beam is drawn through the point or points in which the line XY cuts the vertical SHEARING FORCE AND BENDING MOMENT 129 steps of the load-line. In the present case it cuts the steps KL and MN. Therefore A l B l or A 2 B 2 is the position of the beaih for maximum bending moment. In order to tell which of these gives the absolute maximum value, it is shown that the area of the triangle TLV represents the diminution in bending moment which occurs in moving the beam from the position AjB x to a position coinciding with the base LV, after which the bending moment in- creases again, until when it reaches the position A 2 B 2 it has increased again by the area of the tri- angle VMU. The position AjBj or A 2 B 2 will there- fore be the position of absolute maximum bending moment according as the triangle TLV is greater or less than the triangle VMU. In this case there- fore A^j is the position required, and if we now erect verticals through A lt B x to cut the link-polygon in A 3 , B 3 , the line* A 3 B 3 will be the closing line which determines the bending moment diagram, and the maximum value will be given by the ordinate which passes through T. 67, Indirect Loading, In the previous cases the loading has been .direct, but in practice it is more usually indirect, that is to say, the load acts on the main girders through the medium of cross-girders. The load on any panel is then distributed so that it acts on the cross-girders which support the ends of the panel. I. Maximum Shearing Force for a Train of Loads, Let AB be an N girder (fig. 105) and let Wj, W 2 , W 3 , W 4 be a train of point loads crossing it from right to left. Proceeding exactly as in 66 for direct loading, we reverse the loads and construct a 9 130 THE ELEMENTS OF GEAPHIC STATICS link-polygon with the span for polar distance. Sup- pose now we investigate the maximum shearing force in the second panel from the left. When the leading load arrives at the panel-point C, the reaction at A will be C^CX, which will therefore be the maximum shearing force so far, and it only remains to investi- gate whether it will become greater than this value. As the maximum shearing force will always be greatest when a load arrives at the panel-point, let o' Wa FIG. 105. W 2 be moved up to C ; then Wj will be at Dj and DjDjj will be the reaction at A. From this, however, must be subtracted the downward pressure at E due to the load W x standing at D r Set up EjE 2 = W x and join C^. Then the ordinates of this line represent the pressure at E as the load Wj rolls across the panel CE, and therefore D^g represents the pressure at E when the load is at D r Sub- tracting this from the reaction at A 1? viz. DjDg, the remainder D 2 D 3 is the shearing force at C when the load W 2 has reached this point, and since D 2 D 3 is SHEARING FORCE AND BENDING MOMENT 131 found to be less than C^, the latter value is the maximum shearing force in the panel, and if we draw C 2 F parallel to the diagonal, then C. 2 F is the maximum tensile force which will occur in it due to the moving load. In this way the maximum positive and negative stresses in the diagonals may be deter- mined. The maximum tensile stresses in the dia- gonals due to the dead load may be found in the ordinary way by calculation or by means of a force diagram. The forces due to the live and dead loads should then be tabulated, and if the compressive or negative stress due to the live load exceeds the ten- sile stress due to the dead load in any diagonal, then the panel containing that diagonal must be counter- braced, because it is assumed that the diagonal members are incapable of resisting a compressive force. When compression occurs therefore in the one diagonal the other is in tension and prevents deformation. ' II. Maximum Bending Moment. Let any sys- tem of point loads W lf W 2 , W 3 (fig. 106) say cross a span /, the loading being applied to cross-girders or otherwise transmitted at intervals to the main girders. In the present case let 1, 2, 3, 4, 5 be the cross-girders. Then we require to know first of all the position of the loads which will cause maximum bending moment at each of the points, 1, 2, etc. This may be very easily found as follows (for proof see " Moving Loads by Influence Lines " in this series) : Set up at B vectors 1, 1 2, 2 3, as shown, to represent the loads W Jt W 2 , W 3 . Join AO, and through the points 1, 2, etc., draw lines la, 26, etc., 132 THE ELEMENTS OF GBAPHIC STATICS parallel to AO. Then the loads which are cut by these lines are the loads which must act at the corresponding points to cause maximum bending moment there. Thus the lines from 1 and 2 cut the load Wj. Hence Wj will stand both at 1 and 2 when maximum bending moment occurs at those points. FIG. 100. Similarly, W 2 will stand at 3 when there is maximum bending moment at that point, and so on. In order to find the numerical values of these maxima, place the girder first in position AjBj so as to bring the point 2 under the load W,. With any convenient pole O (and making the polar distance an even number of tons on the force scale for the purpose of simplifying the arithmetic), construct a link-polygon for the SHEAEING FORCE AND BENDING MOMENT 133 given loads. Project A p B t to a lt b r Then a-J)\ will determine the bending moment diagram and the ordinate y l in linear units multiplied by the polar distance in force units will be the maximum bending moment at the point 1. Similarly, if we place the girder in position A 2 B 2 , so that the point 2* lies beneath load 1 and drop perpendiculars A 2 a.,, B 2 6 2 from its ends to cut the link-polygon in a 2 , 6 2 , the line a 2 b. 2 will determine the bending moment diagram, and the ordinate y., will give the maximum bending moment which can occur at the point 2. Also 2/ 3 in the same way will give the maximum bending moment at 3 when the point 3 is brought under load 2, and so on. , 68, Uniformly Distributed Load. The determi- nation of the maximum values of shearing force and bending moment is much more readily effected for a uniformly distributed load, and for this reason it has been a common practice in this country to work out the values of distributed loads for various spans and various assumed axle loads which will produce maxi- mum shearing forces and bending moments at least as great as those which the concentrated loads will cause. There seems to be no particular reason why this indirect and more or less indefinite procedure should be adopted in these days when the use of influence lines leads to a reasonably simple and direct solution, but it has been widely adopted and is simple in application, especially for girders with parallel booms. Maximum Shearing Force. In the case of a girder divided into equal panel lengths, it may be shown (see "Moving; Loads by Influence Lines") 134 THE ELEMENTS OF GRAPHIC STATICS that the position of the load which causes maximum shearing force in any panel can be at once found by dividing the span into a number of equal parts one less than the number of panels. Thus in fig. 107 where there are given equal intervals we divide the span into four equal parts at a, 6, and c. Then when the head of the load reaches a maximum shear will occur in that panel. When it reaches /; this position FIG. 107. of the load will cause maximum shear in the corre- sponding panel, and so on. Then in order to avoid the trouble of calculating the correct value of the maximum shearing force, the following graphical method may be used, the proof of which is given in the book referred to above. Let AB be the girder and let w be the load per foot run which passes over it. On a base-line A 1 B 1 set up AjA 2 = \wl, that is to say, one- half of the load on the whole span when covered, and join AoB r Divide the span into n - 1 equal SHEARING FORCE AND BENDING MOMENT 135 parts where n is the number of intervals into which the span is divided. Here n = 5, therefore n - 1 = 4. Suppose now, for example, we require the maximum shearing force in the interval 1 2. Drop a perpen- dicular from a, cutting A^ in C, and project C horizontally to A 3 . Join A 3 B X cutting a perpendicular from the point 2 in D, then DD 1 is the maximum possible shearing force which will occur in the inter- val 1 2. A similar construction will give the maxi- mum values for the other intervals, and also the maximum negative values. In the latter case set down B X B 2 = $wl and join AjBg cutting the vertical through a in b. Project b to B 3 and join AjBg. Then a perpendicular through the point 1 cuts this line in E and the ordinate EE X measures the maxi- mum negative value of the shearing force in the interval 1 2, and similarly for the other intervals. Maximum Bending Moment. As regards the maximum bending moment nothing need be said in addition to what has already been said for a uni- formly distributed load covering the span, because the maximum bending moment will occur at all sections of the span when the load completely covers it, and therefore apart from considerations of dyna- mic action, there is no difference between the case of a moving load and that of a static load. This case is therefore dealt with as in 57. For further treatment of moving loads, the volume on " Moving Loads by Influence Lines and Other Methods," already mentioned, may be referred to with advantage, where the subject is more fully dealt with. CHAPTEK X. MOMENTS OF AREAS, ETC. 69. Graphical Determination of the First and Second Moments of a System of Parallel Forces, In later developments of the subject it will be necessary to find the first and second moments of a system of quantities about an axis, and occasionally also the third moment where by the first, second, or third moment is meant the product of the quantity into the first, second, or third power of its distance from the axis. Thus suppose, for example, that a small area or mass be supposed to act at its centroid, and that a number of such quantities whose magni- tudes are ap a 2 , etc., lie at the points A lf A 2 , etc., dis- tant y lt ?/ 2 , etc., from a given axis XX (fig. 108). Let (136) MOMENTS OF AREAS, ETC. 137 vectors 1, 1 2, 2 3, etc., be drawn parallel to XX, and a pole O be taken with polar distance 'p. Also let lines be drawn through A x , A 2 , A 3 , etc., parallel to XX, and a link-polygon abode constructed, and its sides produced to cut the axis XX in b lt c lt d^ Then the intercept b^ on the axis between the first and last points measured in the scale of lengths and multiplied by the polar distance p in the scale of the magnitudes of a lt a 2 , etc., will give the resultant of the first moments of these quantities about the axis. Proof. For the resultant moment is etc - Now in the similar triangles bc l b l9 001, c.b, 1 -V* : - - .-. Vi x P = 1 x y L = a, x y r Similarly, c^ x p = 1.2 x y. 2 = a 2 x y., and dfa x p = 2 3 x y. 3 = a 3 x y z . .-. (bfr + c^i + d^Jp = a' 1 y l + a. 2 y z + a z y 3 = sum of the moments about XX. This result was proved also in chapter v. in a somewhat different manner. Further, the second moment of the system of loaded points about the axis will be represented by the area edcbb 1 between the first and last links multiplied by twice the polar distance. Proof. For the second, moment of the quantities is i2/i 2 + a s 2 2 + a 32/3 2 + etc - Now area of the triangle 66 1 c 1 is %b l c l x y l and if 6 1 c 1 = -. Also the area of the triangle cc^ is i . c l d l x 7/ 2 where c l d l = ^- , and so on, so that 138 THE ELEMENTS OF GEAPHIC STATICS the whole area = bb^ + cc l d l + dd^ flit/, 2 a.,v. 2 = 4. 321-4. i -ML P P P .'. 2p x whole area = a-flf 4- a 2 2/ 2 2 + a a 2/ 3 2 = the sum of the second moments of the quantities a,, a lM etc., about the axis XX, the area being measured in linear units on the scale of the drawing and the polar distance p in the scale of the magni- tudes of the quantities. The second moment of a system of areas or masses about any axis is commonly called their moment of inertia about the axis, but the expression is not a good one. If we find the distance from the axis, of an area or mass which is equal to the sum of the areas or masses, so that the second moment of this area or mass about the axis is identically the same as the resultant of the given areas or masses, then this distance is commonly known as the radius of gyration of the system. In other words, if A represent the sum of all the quantities a, and k x be the value of their radius of gyration, then A x k? = a!?//-' + a. 2 y.r + a 3 ?/ 3 - + etc. = S[ay 2 ]. Hence *, = 2[< f 1 The radius of gyration of a system of quantities serves the same purpose in regard to second mo- ments that the centroid or centre of gravity does for first moments, by enabling the whole system of points to be regarded as concentrated at a single point, and so simplifying the subsequent treatment, but the distance of the centroid of a system of areas MOMENTS OF ABEAS, ETC. 139 or masses from an axis is always less than the radius of gyration. Example. To find the centroid distance and the radius of gyration of a system of points loaded with quantities whose magnitudes are 2, 3, and 4 and lying at distances of 1, 3, and 2 units re'spectively from a given axis. Then since the moment, of the resultant = the sum of the moments of its components, (2 + 3 + 4) x their centroid distance -2x1+3x3+4x2 = 2 + 9 + 8 = 19. .-. the centroid distance is y = 2 units = 2 -11. N.B. It will be observed that if the sum of the moments about the axis is zero, the centroid distance is zero also, or the axis passes through the centroid. Further, since (2 + 3 + 4) x &, 2 = 2 x P + 3 x 3* + 4 x 2* = 45, fc,s = .45_ = 5 or ^ = v/5 = 2-24. 70. Graphically, Let XX (fig. 109) be the axis and let 2, 3, 4 be the loads. Set off vectors parallel to the axis to represent these loads and take any polar distance (but preferably a length = sum of the vectors). Construct the link-polygon aocde. Then ae x p is the first moment about XX. But the moment about XX = sum of the loads x the centroid distance x from the axis. Therefore sum of loads x x = ae x p. But if we take p = sum of the loads, then x = ae where x is the centroid distance = 2'11 as before. 140 THE ELEMENTS OF GRAPHIC STATICS Further, area abode x 2j> = second moment of the loads about XX = sum of the loads x A;, 2 . /. if p is the sum of the loads, k,- = 2 area abode. - in H Now the area measured 2-5 sq. inch. .-. k* = 5 as by calculation. 71, Centres of Gravity. Case I. Both Loads Positive. Centroid of Two Loaded Points. Let A and B (fig. 110) be any two points at which we may suppose areas or masses to be concentrated, and let LA, LR be the magnitude of these loads. Then the resultant of these must lie on the line joining A to B, and since the moment of their resultant acting at the centroid C is equal to the moments of its com- ponents, if we take C as moment centre, MOMENTS OF AKEAS, ETC. 141 L A x AC - L B x BC = O since the moment of the resultant = O. inversely as FIG. 110. ' ' L BC ' or fc k ^ me l the loads. Graphical Construction. Set up AD = LA and BE = LB where AD and BE are parallel lines. Join DE cutting AB in C. AD BE Then by similar triangles -r^ = gp '-gn = jg = j- B . Hence AB is divided in C inversely as the loads, and is therefore their centroid. B FIG. 111. Case II. One Load Positive and the Other Nega- tive. In this case, referring to fig. Ill, 142 THE ELEMENTS OF GRAPHIC STATICS L A x AC + L B x BC = O. .-.. ^ = ^TH i , and since the moments about C - L A BC are of opposite sign, the point C must lie on AB pro- duced, as shown in the figure. Set up AD = LB and BE = LA parallel to 'one another. Join DE and AC LR produce to cut AB produced in C. Then = ^ * as required. Hence the resultant lies on AB pro- duced, and on the side of the greater load. 72, Centroid of a System of Loaded Points on a Straight Line, This is identically the same FIG. 112. problem as that of finding the position of the re- sultant of a system of parallel forces. Hence if we MOMENTS OF AREAS, ETC. 143 construct a link- and vector-polygon for the loads, the intersection of the first and last links will de- termine a line passing through the centroid. Thus if ABCD (fig. 112) be a rectangle with two circular areas cut out of it, the actual area may be regarded as consisting of a positive part ABCD and two negative parts, each acting through their centres. Hence if we set off 1, 1 2, 2 3 to represent these areas in magnitude and sense, and construct a link- polygon with pole O, then the intersection of the first and last links fixes the position of the resultant load passing through the centroid G of the area, and as the centroid must lie on the axis of sym- metry, its position is determined. 73. Centroid of a System of Loaded Points not Lying in a Straight Line, Let L x , L 2 , L 3 , L 4 FIG. 113. (fig. 113) be the loads, and 1, 1 2, 2 3, 3 4, their vectors. Construct the link-polygon ab . . . h. Then if we produce ab to meet dc in /, a line through / 144 THE ELEMENTS OF GRAPHIC STATICS parallel to the direction of the loads will fix the point f/ 1 on L^L., which is the centroid of L t and L 2 . Similarly, if we suppose Lj and L., to act at g lt the centroid of L lf L 2 , and L 3 must lie on the line ^L 3 , and it must lie also on the line through k in which the first and fourth links of the link-polygon inter- sect. Hence it lies at g< 2 . Again, joining <7 2 to L 4 and producing the first and fifth links to meet in ra, the line through m parallel to the loads intersects gf 2 L 4 in <7 3 , which is then the oentroid of the whole system of loads. 74, Centroids of Areas, When an area has an axis of symmetry, the centroid will always lie on that axis, and when there are two axes of symmetry the centroid will lie at their point of intersection, so that the centroids of figures like circles, rectangles, equilateral triangles, etc., can always be found by inspection. Centroid of a Triangle, If we suppose a triangle ABC (fig. 114) divided into an infinite number of thin Fio. 114. strips parallel to its base BC, the centroid of each of these will lie at its centre, and therefore all of them MOMENTS OF AEEAS, ETC. 145 will lie on the median line AD which bisects the base. By similar reasoning we may show that it must also lie on the median BE, which bisects AC, and by geometry it is known that the medians of a triangle intersect in a point which lies one-third of the way along any median. This fixes its position. 75, Centroid of a Trapezium, First, the cen- troid must lie on the axis EF (fig. 115) which joins the mid-points E and F of the parallel sides. Draw CH parallel to BA, and let G p G 2 be the centroids of the parallelogram and triangle so formed. Join G 1} G 2 , and produce to cut the sides CB and AD produced in K and L. Then since the As GjCK, G X AL are congruent (i.e. equal in every respect), if we write KB = x and DL = y, x + BC = y + AD (1), and since .the As G 2 ML, G 2 KC are similar, and G 2 M = G 2 C, x + BC = 2ML = 2^ ~ AH + y\ = AD - AH + 2y (2). /. by (1) and (2) y + AD = AD - AH + 2y. .:. y = AH = BC, and /. by (1) x = AD. Hence make DL = BC and BK = AD. Join KL, and the point G in which it cuts EF is the centroid. The above construction is, however, sometimes 10 146 THE ELEMENTS OF GRAPHIC STATICS inconvenient, because the points K or L required may lie outside the limits of the drawing. In this case the following method may be used (fig. 116) : Make CC^ = J(AD - BC) and draw C^A, parallel to the diagonal C A, cutting the median line EFinG. Then G is the centroid. Also by taking moments about the base it may be shown that the distance of the centroid from the where h is the distance A A, F FIG. 116. /i/B + 26 N base is equal to g ^ & between the parallel sides, and B, b are their lengths. Also when one side is vertical, as in fig. 117, the distance of the centroid from the vertical side is given by the formula Third Method. Draw in the median line EF (fig. 118). Join AE and CF, and make EG X - AE, and FG 2 = JCF. Join G^, cutting EF in G. Then G is the centroid required. Polygons bounded by straight lines may be split MOMENTS OF AEEAS, ETC. 147 up into triangles, and the areas of these being sup- posed concentrated at their centroids, the centroid of the whole may be found by the method for a system of loaded points (see 73). FIG. 118. 76, Radius of Gyration for an Area. Equi- momental Points. Definition. If a system of points can be found such that when an area is divided up and concentrated at these points, the moments of these loads about any axis is equal to that of the area itself, then the points are said to form an equi- momental system. Equimomental System for a Triangle, Thus if a triangle BCD (fig. 119) whose area is A be supposed FIG. 119. divided into three equal areas -^, and if these are supposed concentrated at the mid-points of the sides 148 THE ELEMENTS OF GRAPHIC STATICS of the triangle, viz. at E, F and G, then these points so loaded form an equimomental system, and the first, second, etc., moments of the loads concentrated at these points will be the same as for the area itself about all axes. Thus the second moment of the triangle about its base will be A A 2 . 2 W bh* and if k be the radius of gyration about the base, w * - 12 - r. Equimomental System for a Rectangle, If A be the area of the rectangle (fig. 120) then its diagonal ^ divides it into two equal triangles each of area -= , and a Fia. 120. if one-third of ^ be concentrated at the mid-points of their sides, we get the equimomental system shown in the figure. Consequently the second moment of MOMENTS OF AKEAS, ETC. 149 the rectangle about the axis CC through its centroid is 2 x A x //A 2 = Afe or tas * = b J^^bh = Also about its base XX, the second moment is //A 2 *(*)' A ,. 2A 6 * * + T AW -5- bh* Any figure bounded by straight lines may in this way be split up into a number of triangles, for each of which an equimomental system can be formed, and therefore a system for the whole area is found also, after which the moments for the loaded points may be dealt with as already explained in 69. Other simple cases of common occurrence however may be deduced at once from the above results. Thus for a hollow rectangle (fig. 121) the second H FIG. 121. moment about the axis CC through the centroid is BH 3 bh s llT 12* an( ^ ^ s same formula will also give the second moment for an I section (fig. 122), which may 150 THE ELEMENTS OF GRAPHIC STATICS be considered as the .difference between two rec- tangles of breadths B, b and depths H, h. 77. Relation between the Second Moment of an Area or Mass about any Axis and its Second Moment about a Parallel Axis through its Centroid, Let CO be an axis through the centroid (fig. 123), and let XX be any parallel axis distant d from it. Let a be a very small element of the area or mass distant y from the axis CC. Then its second FIG. 123. moment about XX is a(y + d)' 2 , and the second moment of all the elements about XX will be the sum of an infinite number of infinitely small ele- ments which we may represent by 2a(?/ + d} 1 = Now 2a7/- denotes the summation of all the second moments about the axis CC = TC- Also 2a?/ denotes the summation of the first moments of all the ele- ments about CC. But this is zero, because the sum of the moments is equal to the moment of their resultant, and this by definition passes through the MOMENTS OF AREAS, ETC. 151 centroid, and its moment about the axis CC is there- fore zero. Finally the last term is equal to Ad' 2 , since 2a = the whole area. Hence Ix = Ic + Ad 2 , and since Ix = Afcx 2 , and 78. Example. To find the second moment of a triangle about an axis through its centroid parallel to its base. Since the second moment about the base has been shown to be Ix = ~g , we have (fig. 124) _ *- " A '' ' 18 79, Polar Second Moment, The second moment of an area about any axis perpendicular to its plane is known as its polar second moment about that axis. Thus if fig. 125 represent any plane area and an axis be taken through any point O perpendicular to the plane, the second moment of the area about this axis is 2a2 2 where z is the distance of an ele- 152 THE ELEMENTS OF GBAPHIC STATICS ment from the axis. Let OX, OY be any pair of rectangular axes through O. Then since ?2 - x >2 + if, 2oz 2 = 2ox 2 + 2a# 2 = Iy + Ix where IY and Ix are the second moments about the axes OX and OY respectively. 80. Example. To find the polar second moment of a circle about an axis through its centroid. FIG. 125. Suppose the circle (fig. 126) divided into an infinite number of triangles. Then if we suppose one-third of the area of each of these triangles con- centrated at the mid-points of its sides, we get ulti- mately one-third of the whole area of the circle concentrated along its circumference, and two-thirds along a concentric circle whose radius is half that of the first. Hence the polar second moment is IP '- Ar* MOMENTS OF AREAS, ETC. 153 From this we can easily deduce the second moment of a circle about a diameter ; for if XX, YY be two diameters at right angles, IP = I x + IY. But I Y = Ix. /. 2I X = Ip or I x = JIp. Y FIG. 126. Hence the second moment of a circle about its diameter is one-half its second moment about the polar axis through its centre, and since Ar 2 Ar 2 IP = ~?p> Ix = ~A ~~ x 64 From the above we deduce at once that for an annulus where D and d are the external and internal dia- meters. 81. Graphical Determination of the Moment of Inertia, Fig. 127 represents a cross-section asym- 154 THE ELEMENTS OF GRAPHIC STATICS metric about its neutral axis NN. It consists of two channel bars (BSC 26) 12 ins. x 4 ins., area = 10-727 in. 2 The whole area is divided into convenient parts which are regarded as forces acting through their centroids, and the areas of these parts are set off along a vector line from to 10 as shown. A pole 'I' FIG. 127. O is then taken with alpolar distance p (which for convenience may be made one-half the length O'lO). The intersection of the first and last links of the link- polygon drawn with this polar distance determines the line NN through the centroid of the cross-section, which is therefore the neutral axis. The area abc between the link-polygon and its first and last links MOMENTS OF AEEAS, ETC. 155 was then measured and found to represent 21 '5 in. 2 to the scale of the drawing. Now we have shown, 69, that this area multiplied by twice the polar distance in the scale of areas will be the second moment of the whole area about the axis NN. But twice the polar distance was made equal to O10 = whole area of the cross-section, which was in the present case 47'8 in. 2 Hence the value of the second moment of the cross-section about its neutral axis is 47*8 in. 2 x 21-5 in. 2 = 1030 ins. 4 82, The Momental Ellipse. The moment of inertia of any area is always greatest and least about two axes at right angles to one another, and these are known as the "Principal Axes of Inertia," and the principal radii of gyration can be found by divid- ing the moments of inertia about the principal axes by the area of the section and taking the square root. If an ellipse be now constructed of which these radii are the semi-major and semi-minor axes, the radius of gyration about any central axis may be found by drawing a tangent to the ellipse parallel to that axis, the perpendicular distance between the lines being the radius of gyration required. This ellipse whose centre is the centroid of the area is known as the " Momental Ellipse " for the area. Thus if O (fig. 128) be the centroid of an area and UU, VV are the principal axes of inertia, an ellipse ABA'B' having k\, k\j for semi-axes is the momental ellipse, and if it be required to find the radius of gyration about any other central axis XX inclined at an angle a to UU, draw xx parallel to XX to touch the ellipse. Then the perpendicular distance 156 THE ELEMENTS OF GRAPHIC STATICS kx between XX and xx is the radius of gyration about XX ; or by calculation A* 2 - fcu 2 - sin 2 a(kf - fc v 2 ) (1) and fc Y 2 = V - cos2 a ( fc u 2 - fcv 2 ) (2) XX and YY being perpendicular to one another. By addition & x * + fc Y 2 = 2&u 2 - (^u 2 - &v 2 ) = fcu 2 + &v 2 , or I x + IY = Iu + Iv (3) ; that is to say, the sum of the moments of inertia FIG. 128. about any two axes at right angles is constant and equal to their sum about the principal axes. 83, When the Section has an Axis of Sym- metry, as in the Case of a Channel-iron, for example, this Axis is always one of the Principal Axes of Inertia, Example. Given the channel-section in fig. 129 for which Iu = 26-03 ins. 4 k L < = 2'33 ins. A = 4-?9 ins. 2 Iv = 3-82 ins. 4 k\ = O89 in. MOMENTS OP AEEAS, ETC. 157 to construct the momental ellipse and to determine the moment of inertia about an axis bisecting the principal axes. Since Ix + IY = Iu + Iv, and since by symmetry Ix = IY, we have I x = i(I\j + Iv) = J(26-03 + 3-82) = 14-925 in. 4 , and 14-925 4-79" 3-11 ins. 2 1-76 in. FIG. 129. 84, Asymmetric Sections, When the section has no axis of symmetry, the position of the principal axes must be obtained by finding the second moments about three non-parallel axes. This is best done by finding the second moments Ix, IY (fig. 130) about any pair of rectangular axes XX, YY, and also the second moment Iw about a third axis WW bisecting the angle between the other two. Then the angle a which the semi-minor axis of the ellipse makes with XX is given by the formula Ix + IY 2Iw tan 2a = IY - (4). 158 THE ELEMENTS OF GRAPHIC STATICS Iv and I\- can then be found from the formulae Iv + :== 5? ) ( 6 > and Iv = ilx + Iv - (6). Example. Angle-iron, 10 ins. x 4 ins. (fig. 130). -f~ FIG. 130. A = 9-00 ins.- I x = 92-11 ins. 4 I Y = 8'77 ins. 4 fc x = 3-20 ins. fc v == -987 ins. T\y \vaa found by constructing a second moment area to be 68-5 ins. 4 .-. k\y = 2-70 ins. 92-11 + 8-77 2 x 68-5 Then tan 2a - H . ?7 ^. u - -433. .-. 2a = - 23 25', and a = IT 42V. .-.tan u = - -207. 92-11 - 8-77 ..Iu^-i(92-ll + 8-77+ ^230^ = 4(100-88 + 90-81) = 95-85 ins. 4 .-. k v = 3-26 ins. MOMENTS OF AREAS, ETC. 159 Also I v = (100-88 - 90-81) = 5-035. .-. fc v = '748 in. Centroids of Lines, Areas, and Solids.- Trapezium:.- |- J(*-^) (fig. 131). k- o- FIG. 131. r sin a 180 re Circular arc: OG = a TT s Semi-circular arc : OG = = -6366r. 7T Circular sector: OG = o - (fig. 132). FIG. 132. 4 r Semi-circular area : OG = Q . - = -4244r. O 7T Circular segment : OG = r~- x area (fig. 132). 160 THE ELEMENTS OF GRAPHIC STATICS o o Parabolic segment : --X ; ft y (fig. 133). FIG. 133. Pyramid or cone ; at J height from base. 3 ^2r /tV~ Segment of sphere at . . ^ - from base, where h is the height of the segment. Moments of Inertia of Areas, Rectangle: I bh 3 (fig. 134). i.. A i V 1 W T^ i - : 1 J i > FIG. 134. Square about diagonal : I,, = ^ ( n 8- ^^ bh* Triangle about base : I = ^c"(fig- MOMENTS OF AREAS, ETC. 161 ISft Fia. 135. FIG. 136. Ellipse : I,, = nab* (fig. 137). Semicircle about T r 4 - TT 8 diameter: - (TT 8 \ ( ~ Sir) 138 > FIG. 137. FlG . m Parabolic segment T 8 about base : I = 175**' ( fi g- 139). -i- -*, FIG. 139. Hexagon about T _ 5 >/3r>4 _ diagonals : TfiT (fig. 140). 162 THE ELEMENTS OF GEAPHIC STATICS diameter : = -638B*(fig. 141). FIG. 141. Corrugated Plate, BJl 2 2 2 f 3 v) (fig . 142). where 7i = H - H--B-H I FIG. 142. Flat Corrugated Plate, where /i, = i(H + ) ^ = J(B + 2-6Q /t, = |(H - 6 2 - i(B - 2-60- FIG. 143. CHAPTBB XI. STRESS DISTRIBUTION ON CROSS-SECTIONS. 85. Load Point or Centre of Pressure, By the Load Point or Centre of Pressure of a section is to be understood the point at which the resultant of a given system of forces acting upon it cuts the plane of the section. When a load is uniformly distributed over an area, the resultant of that load will pass through the centroid of the area, and therefore in this case the centre of pressure will coincide with the centroid. When, however, as often occurs, the distribution of pressure is such that the intensity follows a straight-line law, that is to say, it varies as the distance from a given line, the position of the load point will be determined by the rate at which the intensity of pressure varies, and the line along which the pressure is zero is known as the Neutral Line or Neutral Axis. Thus when a plane area is submerged in a liquid, the intensity of pres- sure increases with the vertical depth below the surface, being zero along the line in which the plane produced cuts the surface of the liquid. This line is therefore the Neutral Line, or as it is usually called in this connexion, the Line of Flotation. This line is strictly analogous to the Neutral Axis in the case of beams or other struc- (163) 164 THE ELEMENTS OF GRAPHIC STATICS tures, in which the stress intensity varies as the distance from a line, and the relation which we are about to deduce between the load point and the line of flotation is equally true for the load point and the neutral axis of a structure. 86, Total Pressure on a Submerged Area, If we consider the pressure on a horizontal plane immersed in a liquid whose weight is w per unit volume, the pressure per unit area at any depth y (fig. 144) below the surface being equal to the weight of the column of liquid above it will be wy, FIG. 144. and since the pressure of a liquid is equal in all direc- tions, wy will be the pressure per unit area in all directions at a depth y. If then we consider an infinitely small area a at a vertical depth y below the surface, the pressure on it will be wya, and the total pressure on the whole surface, whether plane or not, will be P = ^[wya] = w^,[ay] where 2[a7/] is the first moment of all the elements of area about the plane of the water surface. Hence P = 10 x first moment of the area, = wAy (1) where A is the area and y is the depth of its centre of gravity, STKESS DISTKIBtJTION 165 Example 1. A sphere of 2 ft. diameter is im- mersed so that its centre is 3 ft. below the surface, in fresh water weighing 62'4 Ib. per cub. foot. Find the total pressure on the surface. Here P = ivky = 62-4 x 4jrr 2 x 3 = 748-8* Ib. = 2358 Ib. approximately. Example 2. A rectangular area 4 ft. x 3 ft. is im- mersed in salt water weighing 64 Ib. per cub. foot, with its centre of gravity 5 ft. below the surface. Find the pressure on one side. Here P - 64 x 12 x 5 = 3840 Ib. 87, Position of the Load Point on a Plane Submerged Area, Let AB (fig. 145) be the plane FIG. 145. inclined at any angle 0, and let C be its centroid and L the load point. The plane produced cuts the water surface in a line through X which is the axis of notation, and the moments of the pressures on every element of area about this axis must be equal to the moment of the resultant pressure acting through L. Consider a small element of area a at distance y from X. Then the pressure on it is way sin 6, and the moment about the axis of flotation is way sin 6 x y. 166 THE ELEMENTS OF GRAPHIC STATICS Hence P x y 2 = w sin 6 2[ay 2 ] = w sin x second moment about the axis. But P = wAt/j sin 0. .*. Ay l x 7/2 = ^ ne second moment about the axis. But Ay l = the first moment about the axis. _ second moment about axis ^ 2 first moment about axis Special Case, Rectangular Cross-Section, Let AB (fig. 146) be a rectangular cross-section with FIG. 146. its base in the water surface. Then the pressure per unit width will be P = wABy, and if the sec- tion be vertical P = ivh x i/t = -=-. second moment h* h >2 2 ; -first moment = 3 : * 2 = 3_ ' The load point for a rectangle, whether vertical 2 or inclined, lies therefore at = of its length from the o water surface, the resultant pressure being normal to the surface. STEESS DISTRIBUTION 16? 88. Relation Between the Neutral Axis and the Load Point (fig. 147).- second moment about axis Since y 2 = LN first moment about axis NC + CL = + y> but NO = y. .-. CL = or NC . CL = A; o 2 , Fio. 147. which in mathematical language states that the radius of gyration is a geometric mean between NC and CL. Now if the form of the area is known, its radius of gyration ko about the axis through the. centroid 168 THE ELEMENTS ox STATICS parallel to NN can be determined, as well as the position of its centroid C, and therefore we have a definite relation between the load point and the neutral axis, from which either can be found when the other is known. Thus if we make CD = ko, join ND and draw DL perpendicular to ND, cutting NC produced in L, NC . CL = fc o 2 by the well-known construction for a geometric mean (see " Geometry," p. 23), 1 and therefore the point L so obtained is the load point ; or being given the load point, the cor- responding position of the neutral axis can be found by reversing the construction. Example 1. A circular pipe (fig. 148) 4 ft. diameter is closed at the end and is subjected to FIG. 148. water pressure, the level of the water being 3 ft. above the centre of the pipe. Find the total. pres- sure and the position at which the resultant acts. First. Total pressure P = wAy = 62-4 x ^ x 4' 2 x 3 = 2260 Ib. 1 " Geometry for Technical Students," E. Sprague. Messrs. Crosby, Lockwood & Son. STRESS DISTRIBUTION 169 Second, ko 1 ft. /. CL = Q^ = J ft., or by construction set off CD = J ft., join DN, and draw DL perpendicular to it. 89. Example 2. Any Area Immersed in a Liquid, Let ACB (fig. 149) be any area, which in the present case has been taken as a parabola for the sake of comparing with the calculated result. The parabola had a base of 4-45 ft. and a height of 2-5 ft., the point C being 1 ft. below the surface of the liquid NN, and was drawn to a scale of 1 in. = 1 ft. The area was divided into five horizontal strips of equal width, the half-lengths of which were taken to represent their areas, and were set off along a vector line from to 5. A pole O was then taken and the vector link-polygon abc . . . g was drawn, its first and last links intersecting in g. A horizontal through g determines the centroid G 170 THE ELEMENTS OF GEAPHIC STATICS of the area. At the same time the area abc . . . h x 2p, where p is the polar distance, represents the second moment of the area about the axis NN. This area was measured with a planimeter and found to be 5-92 sq. ins. Now the distance of the load point L from the neutral axis NN is given by second moment about NN the formula OL = -- ...- . first moment about NN But the first moment about NN is equal to the intercept i x the polar distance p. 5-92 x 20 /. OL = 7 - , and since i was found to i xp 1 1 .04 measure 4-45 ins., OL = = 2-66 ft. r l.) Check by Calculation, For a parabola Io = bh* = jfg x 4-45 x 2-5* = 3-18 ft.* IN = Io + Ad 2 = 3-18 + 7-42 x 2-52* = 50'48 ft.*, since d = OG = 2-52 ft. '" L = 7 7 4l"x~2 ; 52 * 2 ' 66 fi| exactlv a gKi n g with the graphical result. 90, Determination of the Stress Intensity at any Point of a Cross-Section when the Load on it is known in Magnitude and Position, Let AB (fig. 150) be any section of a structure, whose centroid is C, and let P be the resultant load on it acting at L, and inclined to the section at an angle u. Then if P be resolved into components S and T parallel and normal to the section, the first will be the shearing force and the second will be the normal thrust on the section, and since CN . CL = k 2 ( 88), STRESS DISTRIBUTION 171 we get CN = ~ , which determines the position of the neutral axis. If we now make the usual assump- tion that sections which were plane before the load L C B FIG. 150. was applied remain plane after strain, the stresses and strains will obey a straight line law, being repre- sented by the line AjN (fig. 151), and S , S lf S 2 will N FIG. 151. represent the stresses at the centroid C and at the edges A and B of the section. Let S be the stress at any distance x from the N axis (fig. 152). Then since S = kx where A; is a constant the total stress on a thin strip of area a is kx . a, and the total stress on the whole area is 2[kx . a] which must be equal and opposite to the normal thrust T. Hence 172 THE ELEMENTS OF GRAPHIC STATICS T = k2[ax] = k x first moment about NN = .-. -T- = kx = the stress at the centroid. Hence the stress at the centroid is equal to the mean stress on the cross-section. T If therefore we set up CC 1 (fig. 151) = s = - FIG. 152. at C and join NCj, we get the distribution of stress over the section. 91. Formula for the Stress at any Point. Since the stress varies as the distance from the neutral axis, we have (fig. 152) x CN + PC ON 1 + PC ON" k 2 k 2 But CN = ^np- = where e is the eccentricity of the load point or its distance from the centroid. ( 1 + r|r where y is to be regarded as positive or negative, according as P STKESS DISTRIBUTION 173 is on the same side of the centroid as the load point or not. At the extreme edges of the section writing y = y l and y = y 2 respectively,* we obtain the corresponding values of the stresses. Example 1. A column of circular section and 6 ins. diameter is loaded on its edge with 10 tons. Find the stresses on the edges. = "354 ton/in. 2 , Here .-. S = -354 28-3 sq. ins. ins., y = + 3 ins., 3 ( 1 - |^ -) = d*_ _ 9 16 4' 354 x 5 and - 3 = 1-77 and - 1'06 tons/ins. 2 Example 2. A wall weighing 120 Ib. per cub. foot has the dimensions shown in fig. 153, and is subject LEVEL. FIG. 153. to a water pressure to within 1 ft. of the top. Find the stresses on the edges of the base. 174 THE ELEMENTS OF GKAPHIC STATICS Weight of wall per foot of length = 60 cub. ft. x 120 = 7200 Ib. Distance of centroid from vertical face = i( 3 + 7 - Q - by 75 = 2-63 ft. Resultant water pressure per ft. * _ 62-4 x 121 2 ~~2 = of to ID., acting at 3 ft. from the base. Now = = W 355 ;. LH = 3f x -525 = 1-925 ft. .-. e = CL = LH - CH = 1-925'- 0-87 = 1-055 ft. 7200/ l-05^x^5N _ 7200 7 \ 49/12 ) ~ 7 = 1955 lb./ft. 2 and 98'6 lb./ft. 2 92, Critical Distance, It is frequently a condi- tion of stability for masonry and concrete structures that no tensile stress shall occur on any cross- section. Now it will be seen that as the neutral axis approaches the edge of a section, the stress will decrease in intensity until it changes sign when the neutral axis first cuts it ; so that if the section were in compression the stress will become zero when the neutral axis touches it and will afterwards change to tension. Hence the condition that no tension shall occur on a section subject to thrust is that the neutral axis shall not cut it. The corresponding distance of the load point from the centroid is known as the critical distance, because so soon as the load point lies beyond this limit, tensile stresses will occur. STEESS DISTRIBUTION 175 Now when the neutral axis first touches the edge B (fig. 154), the corresponding position of the load point will be such as to satisfy the relation CN . CL = k*, k 2 and when CN = - y 2 , CL 2 = - , 2/2 k ' 2 and when CN = y v CL, = . y\ Example 1. To i find the critical distance e c in the case of a circular cross-section. D 2 D Since k* = and y 1 = y z = < Hence the critical distance is J radius. I If, therefore, we describe a circle with centre C and radius = JE, this circle defines the limits within which the resultant must lie in order that the stress may not change sign. This area is known as the core of the section, and therefore the core of a section may be defined as the locus of the load point whilst the neutral axis turns about the edge of the section, keeping always in contact with it but no- where cutting it, 176 THE ELEMENTS OF GRAPHIC STATICS Example 2. Find the critical distance of a hollow circular area whose outer radius is R and whose inner radius is r. An, =g= Example 3. To find the core of a rectangular cross-section (fig. 155). /j2 Since for a rectangle k 2 = ^ and y c = h, LA d T i K ->}->; FIG. 155. Consequently if the neutral axis touch the side ad, CLi = \h. In the same way, if the neutral axis touch the side ab, CL 2 = h l . Also L 3 and L 4 will be the positions of the load point when the neutral axis touches the sides be and cd respectively. Now it may be shown that whenever the neutral axis turns about a point, the load point travels along a straight line, so that if we suppose the neutral axis to turn from the position da to the position ab by rotating about a, the load point will travel along the line joining L l and L. 7 . Similarly, as the neutral axis turns about the other corners 6, c, and d, the load STEESS DISTRIBUTION 177 point will trace out the rhombus L 1 L 2 L 3 L 45 which is therefore the core of the section. Example 4. Find the critical distances and the core for a hollow rectangular cross-section, whose outer dimensions are B, H, and whose inner dimen- sions are b, h. BH 3 - bh* HE 3 - hb s s> 6H(BH - bh) ; 6B(HB - hb)' 93, Rule of the Middle-Third, On account of the frequent occurrence of the rectangular section in walls, dams, arches, etc., this particular form has a special importance, and therefore it should be noted that the critical distance along either of the principal axes must not exceed one-sixth of the depth. In other words, the resultant must not lie outside the middle-third of the depth, in the direction of bending. This is known as the " Kule of the Middle-Third," and is of frequent use in the treatment of masonry structures, the primary condi- tion of stability usually being that the line of pres- sure shall lie within the middle-third limits for all the cross-sections. 94. Asymmetric Section, Example 5. A common section of this type in practice is the T section, which occurs in the case of walls with counterforts and elsewhere. Let (fig. 156) C be the centroid of the section, and Jc the radius of gyration about the axis XX through it. Then when k 2 Jc 2 y = y\> e c = - -7-, and when y = y 2 , e c = - . y\ 2/2 Consequently the points Lj and L 2 at these distances from C, represent the critical limits, according as 12 178 THE ELEMENTS OF GRAPHIC STATICS the bending moment is clockwise or contra-clockwise about the axis XX. ***J 1M--NL- \ i - PIG. 15C. 95, Graphical Representation of the Stresses, when the Load Point and the Critical Distances are Known, Let L (fig. 157) be the position of the load point, and L lf L., be the critical positions. FIG. 157. Set up at C a perpendicular CCj = s 0t the mean stress over the section. Join CjLj, CjL.,, and erect a perpendicular at L, cutting these lines or the lines produced in E, Ej. Then LE, LE X will represent the stresses on the edges A and B respectively to the same scale that CC } represents the mean stress s (> . STRESS DISTRIBUTION 179 Proof. For since s = *l + ) ( 91), and since e t . = , V = *o(l - Now by similar triangles ^^r- = rr =^. LL i /CV-CL \ ^'GL; :=SO ( CL, ) If, therefore, we project E and E x upon verticals drawn through A and B, we get the points F and H. Then the line FH represents the distribution of stress over the section. 96. When the Material is Considered as Incapable of Resisting Tension. For the best brickwork in cement mortar a compressive stress of from 200 to 300 Ib./ins. 2 is probably permissible and a tensile stress of 35 Ib./ins.* When, however, the tensile stress is neglected, the position of the neutral axis will no longer be the same as before. Suppose NN to be its position (fig. 158), and L the position of the load point. Then, if s is the stress at any distance y from the neutral axis, and S is the stress at unit distance from it, then s = s y, and if 8 A be a narrow strip of area parallel to NN, 180 THE ELEMENTS OF GKAPHIC STATICS the total stress on it is s n y . 8A and therefore the resultant load P = 2*.8A = *2SA. (1) Fro. 158. Again the moment of P about NN must be equal to the sum of the moments of the stresses in all the strips, and .-. P x x = 2* y . 8A x y = s 2[8A . y*] (2) /. s 2[8A . y] x x = s 2[8A . y*] 2nd moment of shaded area or x = 1st moment Example 1. Rectangle (fig. 159). FIG. 159. by 3 2nd moment about NN of compression area = --' STKESS DISTEIBUTION 181 bip 1st moment about NN of compression area = ~. a .-. x = I?/ = f(a + x), whence x = 2a. p Also the mean stress = r- and the maximum stress 2P __2P " by ~ Hob' 97. The determination of the neutral axis is in most cases best effected graphically. Thus if the load point L (fig. 160) lie upon a principal axis, the whole area is divided into strips parallel to the neutral axis, and a link-polygon acb for these is drawn with polar distance p. A line LI/ is drawn perpendicular to AB to cut the first link produced in L', and the line Lib is then drawn by trial so that the triangle ablu' acb between the polygon and its closing line ab, or so that the two shaded areas are equal. This determines the position of the neutral axis. For by 69 the area acbh x 2p = 2nd moment, and bh x p = 1st moment. 2 area acbh x p y x bh '- --- - or area but y is the height of the triangle Lt'bh, and area y x bh 2 = area ocbhi and .*. area aL'c = area cdb. This equality is most conveniently obtained by drawing an equalizing line by eye, and adjusting it until the required equality is obtained, the areas themselves being most conveniently checked with a planimeter. 182 THE ELEMENTS OF GKAPHIC STATICS 98, Numerical Example, A circular chimney stack of external diameter 8 ft., and inside diameter FIG. 1KO. 6 ft. carries a load of 50,000 lb., the eccentricity being 2 ft. Find the maximum compressive stress STRESS DISTRIBUTION 183 on the assumptions (i) of tensile stress in the cement being allowed ; (ii) tensile stress neglected. (i) Since s = tl4 + j~ ), D 2 + d 2 D where A; J = +- and 11 = ^ . lo ^ 50000 ' i + ^_ x _ uy 100 5175 lb./ft. 2 = 36 Ib./ins. 2 , and - 636 lb./ft. 2 , and the position of the neutral axis is **** aO f\_-i r\r fi (ii) By fig. 160, CN = 2*88 ft., and the area in compression is 17'84 sq. ft. 50000 . \ mean stress at the centroid G = I^BT = 2800 lb.-ft.' 2 , > Q and maximum stress = 2800 x ^ = 5290 lb./ft. 2 = 36-7 Ib./ins. 2 Hence it will be seen that the load point may pass appreciably outside the critical limits before it causes anything but small tensile stresses. CHAPTER XII. THE LINE OP PRESSURE. 99. We have seen that when the resultant load on any cross- section of a structure is known com- pletely, the thrust, shear, and bending moment can be found, and the resulting stresses can then be calculate^. In order that this resultant may be easily found for any required cross-section, the most convenient procedure is to construct the line of pressure for the external forces acting on the structure ; where by the line of pressure we are to understand the line of action of the resultant force throughout. Thus let tig. 161 represent a crane structure carrying a load W as shown by means of a chain passing over pulleys. Set down a vector 1 to represent the load \V, and 1 2 to represent the pull in the chain between the pulleys A and B. This will be equal to the load for a single-sheaved pulley, and 2 will be the resultant force acting through the centre of the pulley wheel at A. Also if we set down 1 3 to represent the pull in the chain between B and C, 2 3 will be the resultant of the tensions at B, which will act through the centre of the pulley at B parallel to 2 3, and will intersect the previous resultant in a. Then 3 will be the new resultant, which will act through a, and this again (184) THE LINE OF PEESSUEE 185 will meet the resultant of the forces at C in b ; so that the final resultant 4 will pass through b paral- lel to 4. The line Aabcd will be the line of pressure for the structure, neglecting its own weight, and when this has been drawn we have all that is required to determine the thrust, shear, and bending moment at any section. Suppose, e.g., that we re- quire the values of these for any particular section FIG. 161. such as XX, which is cut by the line of pressure in the point L. The resultant force K' on this section is represented by the vector 3, and its direction and position are along ab. Hence if a be the inclination of E' to the plane of the cross-section, R' cos a = S will be the shearing force, and R' sin a = T will be the normal thrust on the plane, whilst R' x r will be the bending moment about the axis, where r is the perpendicular distance of the resultant 186 THE ELEMENTS OF GRAPHIC STATICS from it ; or since K' x r =* T x e, where e is the eccentricity of the load point L, the bending moment is equal to the product of the normal thrust into the eccentricity. The stress distribution can then be easily found as explained in 90. 100, Line of Pressure of a Three-Hinged Structure under Vertical Loading. Let A, C, B (fig. 162) be the three joints, and let W lf W 2 , W 3 , W 4 be any system of vertical loads on it, whose magnitudes FIG. 162. are represented by the vectors 0, 1, 2, 3, 4. With any pole O construct a link-polygon ab . . . f. Join af and draw 5 parallel to it. Then 5 0, 4 5 are the reactions at A and B. Also drop a perpendicular from C to cut the polygon in g, and join ay, gf. Draw 6, 7 parallel to ag, fg. Then 7 6 is the vertical reaction at C. Through 6 and 7 draw parallels to CA, CB meeting in Q. Then if Q be used as a new pole, a new link-polygon starting from A should pass through C and B, and this link- THE LINE OF PRESSURE 187 polygon will be the line of pressure, because QO is the reaction at A, and Ah represents its line of action. Also QO compounded with Wl gives Ql as their resultant, which acts along hk, and so on. Note. The same construction will hold for a three- pinned arch, where AC, BC are the chords of the arch joining the pins. To Construct Diagrams of Shearing Force and Bending Moment, If we resolve the force QO acting along Ah into components parallel and perpendicular to AC, by constructing on QO a triangle of forces having its sides parallel and perpendicular to AC, viz. Q8 0, then 8 is the thrust and Q8 is the shear. These may be set up at AD, AE, and lines DF, EG drawn parallel to AC. Then the vertical ordinates of the diagram so obtained will represent the shear and thrust respectively over that part of the rafter. Similarly, if we resolve the force Ql acting along hk into components parallel and perpendicular to AC, by constructing on Ql a triangle having its sides parallel and perpendicular to AC, viz. Q9 1, then 9 1 represents the thrust and Q9 represents the shear, and these may be plotted as before on the rafter as a base-line, and so on, as well as with the rafter CB. The thrust diagram is shown by the full line, and the shear diagram by the dotted line, while the lower figure is the bending moment diagram. 101, Conditions of Stability in a Masonry Structure, Although the cement or mortar in a masonry structure will ordinarily add to its stability quite appreciably, it has been the custom to neglect this and to require that the following conditions shall be fulfilled : 188 THE ELEMENTS OF GRAPHIC STATICS 1. That there shall be no tensile stress on any cross-section. 2. That the crushing strength of the material shall not be exceeded. 3. That the resultant thrust on any bed- joint shall not be inclined to the normal on that joint at an angle greater than the friction angle. The first of these conditions is usually the principal one, and the circumstances that limit it have already been considered in chapter xi. Examples of the application of the line of pressure to the determination of the stability of dams and retaining walls will be found in vol. xvii. of this series (" Stability of Masonry "), and of its applica- tion to arches in vol. xx. (" Stability of Arches "). The following example will show how it is applied to the case of a flying buttress, whose stability it is desired to investigate. 102. Fig. 163 represents the nave wall of a cathedral, which sustains the thrust of the arch acting on a corbel as indicated. The buttresses are 3 ft. thick and the weight of wall bonded with a buttress, and acting with it, may be assumed as equivalent to 2 ft. of wall on either side of the buttress. The weight of the masonry may be taken as 145 Ib. per cub. ft. The thrust of the main arch was found previously to be 30,000 Ib. acting at 60 to the horizontal. This thrust is assumed to act through the centre of the corbel on which it abuts. The thrust of the buttress arch which acts in the opposite direction is due to its own weight and the weight of the masonry resting on it. The springing joint was taken at AB, and the THE LINE OF PRESSURE 189 FIG. 163. 190 THE ELEMENTS OF GRAPHIC STATICS masonry was divided into four strips of equal width. The mid-ordinates of these strips were taken to re- ' present their weights and were plotted along a vector line from to 4, the total length 4 representing the whole weight = 7610 Ib. The position of the result- ant was found by taking a pole O and constructing the link-polygon shown, the first and last links of which intersect in C. A vertical through C was drawn to cut a horizontal drawn through the upper middle- third point D of the arch-ring at the crown (see " Stability of Arches ") at the point E, and this point was joined to the lower middle-third point F at the springing-joint. The triangle of forces EGH was then drawn for the resultant load and the two reactions at D and F, whose magnitudes were found from the figure to be 7000 and 10,200 Ib. respectively. The horizontal thrust of the arch at D is next compounded with the thrust from the main arch by producing their lines of action to meet in K and finding their resultant KM. This is next com- pounded with the weight of the wall above any selected section DD P by finding the point N in which their lines of action intersect, and compound- ing the previous resultant with the weight of this portion of the wall (39,600 Ib.) at the point N. In this way we get NP as the final resultant, and when produced backward to cut the section DD lf we get the load point L lf which is well within the middle- third of the wall. We next compound the weight of the remaining part of the wall (97,500 Ib.) with the last resultant, and get PQ. This cuts the base section in L 2 , which lies a little outside the middle- third, but not sufficient to be at all dangerous. THE LINE OF .PRESSURE 191 Next proceeding to the buttress wall we compound the thrust of the arch (10,200 Ib.) with the weight of the masonry above the section aa^ in order to test whether the line of pressure lies within the middle- third at this section. We get a resultant which cuts the section at L 3 , and which is well inside it. Simi- larly, we get the load point L 4 at the section bb lt and finally L 5 at the section cc r The normal thrust at the base cc l was found 10,000 Ib., and since the area of the base is 21 sq. ft., we get for the average 10,000 Ib. stress on the base - - = 477 lb./ft. 2 This Jl was set up at de to scale and joined to the middle- third points (see 95). A vertical through L 5 cuts these in / and/ x , which are the stresses on the edges of the wall. Projecting these to g and h, and joining gh, we get the diagram of stress distribution over the base, the stress on the one edge being 800 lb./ft. 2 , and on the other 100 lb./ft. 2 The above investiga- tion cannot of course be regarded as very exact, on account of the roughness of the assumptions made, but is sufficiently so for practical purposes. [THE END.] INDEX PAGE Areas, moment of 136-62 Asymmetric sections ...... 157, 178 Axes of inertia . . ' . ... . . . 155 Axis of flotation 103 neutral 163 B Belfry tower . Bending moment diagrams Bollman truss Bow's notation . 52 . 95 98, 100 47,48 26 Centre of gravity 140 pressure 163 on submerged area 165 Centroid of areas 144 circular arc . 159 sector 159 segment 159 cone 160 loaded points 142 - parabolic segment 160 - pyramid 160 semi-circular area ...... 159 spherical segment ...... 160 trapezium . 145 triangle . . . . . _ . . . 144 Composition of forces ....... 1-25 Concrete raft . . 117-19 Concurrent forces ........ 6 Conditions of stability 187, 188 Coplanar forces 7 Core of sections 177 Corrugated iron, weight of . . . .80 Couples 67 composition of 69 (192) INDEX 193 Couples, effect of Crescent roof . Critical distance PAGE 68 91-4 174 Dead load Diagrams of shearing force bending moment . Distance, critical . .. Distribution of stress Duchemin's formula graphical construction for E Eccentricity of load ' . . Ellipse, momental . .. Equations of equilibrium Equilibrant . . - . Equilibrium, conditions of equations of . , ... of forces Equimomental points system for a rectangle a triangle . 78 97-107 98-107 . 174 163-83 . 83 83 172 155 60 7 60 72 21 147 148 147 Finck truss ..... Flotation, axis of . Forces, composition and resolution of concurrent . . . ' . coplanar . . . . in space . . , transmissibility of . B : .' triangle of . . . . . Framed structures . % . . ; Frames, imperfect . 4 . . --.-. perfect . . <> . redundant . . . . space ...... superposed .... French roof truss .... Funicular polygon .... Girder, braced lattice G 13 . 47 . 163 . M5 6 7 56, 57 6 . 10 26-57 . 45 45 . 45 51 47-51 . 81 72 109-11 48,49 194 THE ELEMENTS OP GRAPHIC STATICS PAGE Gravity, centre of . . , ; . . .'_'. './..'. . 140 See also Centroid. Gyration, radius of . . . .. '. -. -. ' . 138 Imperfect frames . . . '..' ..... . 45 Indirect loading . . ., .'.-. . . 108 Inertia, axes of . .-'.-. 155 See also Moment of inertia. Lame's theorem ........ 22 Lattice girder 48, 49 L.C.C. requirements for wind-pressure .... 84 Line oi pressure . . . . . . ... 184 Link-polygon . . . . . . . . ' . 72 L'ive load 81 Load-curve 112 dead 78 eccentricity of ........ 172 live .81 - permanent 78 point 163 snow ........ 81 temporary . . . . . . . .81 Loading, indirect ........ 108 M Masonry structures, stability of ..... 187 Method of sections ....... 64 Middle-third rule 177 Moment, bending . . .95 centre . . . . . . .61 graphical treatment of . . . . 66, 67 of areas ........ 136-62 of inertia 138 of parallel forces ...... 76, 77 corrugated plates 162 - ellipse 161 hexagon ....... 161 octagon . 162 - parabolic segment 161 rectangle 160 semi-circular area . . . . .161 - square about diagonal .... 160 triangle about base ..... 160 IKDEX 195 PAGE Momental ellipse . . 155 Moments . 59-77 definition of 59 Moving loads 120-35 concentrated 120 33 uniformly distributed 133-35 N Nave wall . . ; . . ' .' ; . . i . 188 Neutral axis . . . . . .- . - . . 163 Non-concurrent forces, resultant of - .- -. . . 69 Notation, Bow's . . . . . . . . 26 Parabola, construction of * ..'. -. '' 105 Perfect frames . . . -..- . . . ' . 45 Permanent load . . . ; . '._>.',. . ' . 78 Points, equimomental . \ .--" . . . . 147 Polar moment of inertia . ' . . . . ... 151 Polygon, funicular ... . . . . .72 link . . .... . .'-...;. . .72 Pressure, centre of . ; . . . . . *." ... 163 on submerged area ' . . . . . . 164 Principal axes of inertia V ..... 155 Purlins, weight of . .' . ... .. ^ . . 79 R Radius of gyration ........ 138 'Raft, concrete 117-19 Rafters, weight of 79 Rectangle, equimomental system for .... 148 Redundant frames^. ....... 45 Resolution of forces ... . . :' -T- 1-15 Resultant . . ..-.,_.. . . 19 Roof t crescent . . . : . . . ,/ .. 91,92 covering, weight of . . . . . . . 79 Roofs . . . ' . . ,-...".'. . 78-94 Rule of middle-third . . '. .. . . . 177 s Scalar quantities . . . . .... . 2 Sections, method of .64 Sense 3 Shear and bending moment curves, relation between . 112 Shearing force '. ...... 95 in braced frames .... 109 196 THE ELEMENTS OF GRAPHIC STATICS Sheer legs Slates, weight of Slope of roofs . Snow-load Stability of masonry conditions of Space frames . Sum-curve Superposed frames . PAGE 54 80 , 80 . 81 , 187 188 51 , 114 lft-68 Tiles, weight of . Toggle joint . . Tower, belfry Transmissibility of force Trapezium, centroid of . Triangle, equiinomental system for - centroid of . - moment of inertia of . of forces ..... Tripod . Truss, Boll man Finck French with fixed ends . 80 . 11 . 52 6 145, 146 . 147 . U4 . -160 . 10 52,53 47,48 . 47 . 87 90 Vector addition quantities . W Weight of purlins . rafters . roof-covering wind-bracing Wind-pressure - influence of height on on curved surfaces Zinc, weight of Z 4, 5 2,4 79 79 79 79 82 85 85 80 ABERDEEN: THE UNIVERSITY PRESS THE BROADWAY SERIES OF ENGINEERING HANDBOOKS LIST OF VOLUMES PUBLISHED AND IN PREPARATION PUBLISHED BY SCOTT, GREENWOOD & SON (E. GREENWOOD, PROPRIETOR) & BROADWAY, LUDGATE, LONDON, E.G. Jhrmtbtoaj) mts of (Engineering gjanfc books. THIS list includes authoritative and up-to-date Books on the various branches of Engineering Science written by an Author who is well acquainted with the subject in hand. Each volume is presented in a clear and con- cise manner, with good Illustrations, and Worked Examples which are a feature of the series. HOW TO ORDER BOOKS NAMED IN THIS LIST. Any of the Books mentioned can be obtained through Booksellers or we shall be pleased to iorward them upon receipt of Remittance cover- ing the net cost of Books and postage named. 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