LIBRARY OF CONGRESS. UNITED STATES OF AMERICA. J ^ ^■■MttlMaHMflite f ELEMENTS OF GRAPHIC STATICS. -3- .m:^ THE ELEMENTS OF GRAPHIC STATICS a 2Ce.rt^Book for Stutirnts of lEngineering "-/ L. M. HOSKINS Professor of Pure and Applied Mechanics in the Leland Stanford Junior University; formerly Professor OF Mechanics in the University of Wisconsin NOV 12 1892 MAC MIL LAN AND CO. AND LONDON 1892 All rights reserved K- Copyright, 1892, Bv MACMILLAN AND CO. Typography by J. S. Cushing & Co., Boston, U.S.A. Presswork by Berwick & Smith, Boston, U.S.A. ^_ v;?6'?r PREFACE. The present work is designed as an elementary text-book for the use of students of engineering. In preparing it, a chief aim has been simpHcity of presentation. The matter treated has been Hmited to the development of fundamental principles, and their application fo the solution of typical problems. The method of the force and funicular polygons is deduced purely from statical principles, with very little consideration of the geometrical theory of reciprocal figures. Since the book is designed to embrace only what can profitably be taken in an elementary course by the student of engineering, it has not been thought best to include a discussion of problems involving the theory of elasticity. For similar reasons, the discussion of curves of inertia has been limited to simple cases ; a more general treatment being of interest to few besides the student of pure mathematics. No effort has been made to secure novelty in the matter treated, but it is believed that in a few cases it has been found possible to simpHfy, and perhaps thereby improve, the methods usually adopted. Attention is invited to the method adopted for lettering corre- sponding lines in force and space diagrams. It* will be seen that this is merely an extension of Bow's well-known notation. This notation is, however, capable of a much wider use than has usually been given it. It is believed that its use, wherever applicable, will be found of great value, both in facilitating the work of the student, and in guarding the draughtsman against mistakes. There is an unfortunate diversity of usage among writers in regard to the technical terms of mechanics, — a diversity especially notice- able in engineering Uterature. In this book the endeavor has been made in all cases to comply with the usage to which the highest authorities are tending. L. M. H. Madison, Wis., July, 1892. CONTENTS. PART I. — GENERAL THEORY. Chapter I. Definitions. — Concurrent Forces. PAGE § I. Preliminary Definitions , i 2. Composition of Concurrent Forces 5 3. Equilibrium of Concurrent Forces 6 4. Resolution of Concurrent Forces 8 Chapter II Non-concurrent Forces. § I. Composition of Ncn-concurrent Forces Acting on the Same Rigid Body II 2. Equilibrium of Non-concurrent Forces 18 3. Resolution into Non-concurrent Systems 28 4. Moments of Forces and of Couples 30 5. Graphic Determination of Moments 35 6. Summary of Conditions of Equilibrium 37 Chapter III. Internal Forces and Stresses. § I. External and Internal Forces 40 2. External and Internal Stresses 42 3. Determination of Internal Stresses 46 PART II. — STRESSES IN SIMPLE STRUCTURES. Chapter IV. Introductory. § I . Outline of Principles and Methods 53 viii CONTENTS. Chapter V. Roof Trusses. — Framed Structures Sustaining Stationary Loads. PAGE § I. Loads on Roof Trusses 58 2. Roof Truss with Vertical Loads 61 3. Stresses Due to Wind Pressure 67 4. Maximum Stresses 71 5. Cases Apparently Indeterminate 74 6. Three-hinged Arch 80 7. Counterbracing 88 Chapter VI. Simple Beams. § I. General Principles 94 2. Beam Sustaining Fixed Loads 98 3. Beam Sustaining Moving Loads loi Chapter VII. Trusses Sustaining Moving Loads. Bridge Loads 112 Truss Regarded as a Beam 115 Truss with Parallel Chords Sustaining Concentrated Loads . . 117 Parallel Chords — Uniformly Distributed Moving Load . . . . 130 Truss with Curved Chords — Uniform Panel Loads 132 Truss with Curved Chords — Concentrated Loads 137 PART III.— CENTROIDS AND MOMENTS OF INERTIA. Chapter VIII. Centroids. § I. Centroid of Parallel Forces 144 2. Center of Gravity — Definitions and General Principles . . . . 149 3. Centroids of Lines and of Areas 152 Chapter IX. Moments of Inertia. § J. Moments of Inertia of Forces 159 2. Moments of Inertia of Plane Areas 169 Chapter X. Curves of Inertia. § I. General Principles 179 2. Inertia-Ellipses for Systems of Forces 182 3. Inertia-Curves for Plane Areas 187 GRAPHIC STATICS. aXKc Part I. GENERAL THEORY, CHAPTER I. — DEFINITIONS. CONCURRENT FORCES. § I. Preliminary Definitions. 1. Dynamics treats of the action of forces upon bodies. Its two main branches are Statics and Kinetics. Statics treats of the action of forces under such conditions that no change of motion is produced in the bodies acted upon. Kinetics treats of the laws governing the production of motion by forces. 2. Graphic Statics has for its object the deduction of the principles of statics, and the solution of its problems, by means of geometrical figures. 3. A Force is that which tends to change the state of motion of a body. We conceive of a force as a push or a pull applied to a body at a definite point and in a definite direction. Such a push or pull tends to give motion to the body, but this ten- dency may be neutralized by the action of other forces. The effect of a force is completely determined when three things are given, — its magnitude, its direction, and its poi7tt of applica- tion. The line parallel to the direction of the force and con- taining its point of application, is called its line of action. 2 GRAPHIC STATICS. Every force acting upon a body is exerted by some other body. But the problems of statics usually concern only the body acted upon. Hence, frequently, no reference is made to the bodies exerting the forces. 4. Unit Force. — The tinit force is a force of arbitrarily chosen magnitude, in terms of which forces are expressed. Several different units are in use. The one employed in this work is thQ pound, which will now be defined. A pound force is a force equal to the weight of a pound mass at the earth's surface. A pound mass is the quantity of matter contained in a certain piece of platinum, arbitrarily chosen, and established as the standard by act of the British Parliament. The pound force, as thus defined, is not perfectly definite, since the weight of any given mass (that is, the attraction of the earth upon it) is not the same for all positions on the earth's surface. The variations are, however, unimportant for most of the requirements of the engineer. In its fundamental meaning, the word "pound" refers to the unit mass, and it is unfortunate that it is also applied to the unit force. The usage is, however, so firmly established that it will be here followed. 5. Concurrent and Non-concurrent Forces. — Forces acting on the same body are concurrent when they have the same point of application. When applied at different points they are non- concurrent. 6. Complanar Forces are those whose lines of action are in the same plane. In this work, only complanar systems will be considered unless otherwise specified. 7. A Couple is the name given to a system consisting of two forces, equal in magnitude, but opposite in direction, and having different lines of action. The perpendicular distance between the two lines of action is called the arm of the couple. PRELIMINARY DEFINITIONS. 3 8. Equivalent Systems of Forces. —Two systems of forces are equivalent when either may be substituted for the other without change of effect. 9. Resultant. — A single force that is equivalent to a given system of forces is called the resultant of that system. It will be shown subsequently that a system of forces may not be equivalent to any single force. When such is the case, the simplest system equivalent to the given system may be called its resultant. Any forces having a given force for their resultant are called components of that force. 10. Composition and Resolution of Forces. — Having given any system of forces, the process of finding an equivalent system is called the composition of fo7xes^ if the system determined con- tains fewer forces than the given system. If the reverse is the case, the process is called the resolution of forces. The process of finding the resultant of any given forces is the most important case of composition; while the process of finding two or more forces, which together are equivalent to a single given force, is the most common case of resolution. 11. Representation of Forces Graphically. — The magnittide and direction of a force can both be represented by a line ; the length of the line representing the magnitude of the force, and its direction the direction of the force. In order that the length of a line may represent the magni- tude of a force, a certain length must be chosen to denote the unit force. Then a force of any magnitude will be represented by a length which contains the assumed length as many times as the magnitude of the given force contains that of the unit force. In order that the direction of a force may be represented by a line, there must be some means of distinguishing between the two opposite directions along the line. The usual method is to place an arrow-head on the line, pointing in the direction toward 4 GRAPHIC STATICS. which the force acts. If the line is designated by letters placed at its extremities, the order in which these are read may indicate the direction of the force. Thus, AB and BA represent two forces, equal in magnitude but opposite in direction. The line of action of a force can also be represented by a line drawn on the paper. In solving problems in statics, it is usually convenient to draw two separate figures, in one of which the forces are represented in magnitude and direction only, and in the other in line of action only. These two species of diagrams will be called force diagra7ns and space diagrams, respectively. 12. Notation. — The use of graphic methods is much facili- tated by the adoption of a convenient system of notation in the figures drawn. There will generally be two figures (the force diagram and the space diagram) so related that for every line in one there is a corresponding line in the other. In the force diagram each line represents a force in inagni- t7ide and direction ; in the space diagram the corresponding line represents the line of action of the force. These lines will usually be designated in the following manner : The line denoting the magnitude and direction of the force will be marked by two capital letters, one at each extremity ; while the action-line will be marked by the corre- sponding small letters, one being placed at -R,. , each side of the line desio:nated. Thus, in -t* ig. 1 o Fig. I, AB represents a force in magni- tude and direction, while its action-line is marked by the letters ab, placed as shown. COMPOSITION OF CONCURRENT FORCES. ITig. S The § 2. Composition of CortctLrrent Forces. 13. Resultant of Two Concurrent Forces. — If two concur- rent forces are represented in magnitude and direction by two lines AB and BC, their resultant is represented in magnitude and direc- tion by AC. (Fig. 2.) Proofs of this proposi- tion are given in all elementary treatises on mechanics, and the demonstration will be here omitted. point of application of the resultant is the same as that of the given force. Thus if O (Fig. 2) is the given point of applica- tion, then ab, be, and ac, drawn parallel to AB, BC, and AC respectively, are the lines of action of the two given forces and their resultant. The figure marked {A) is a force diagram, and {B) is the corresponding space diagram (Art. 11), 14. Resultant of Any Number of Concurrent Forces. — If any number of concurrent forces are represented in magnitude and direction by lines AB, BC, CD, . . ., their resultant is repre- sented in magnitude and direction by the line AiV, where N is the extremity of the line representing the last of the given forces. This proposition follows immediately from the preceding one ; for the resultant of the forces represented by AB and BC is a force repre- sented by AC; the re- sultant oi AC and CD is AD, and so on. By continuing the process we shall arrive at the result stated. It is readily seen that the order in which the forces are taken does not affect the magnitude or direction of the resultant as thus FU 6 GRAPHIC STATICS. determined. The point of application of the resultant is the same as that of the given forces. Figure 3 shows the force diagram and space diagram for a system of four forces repre- sented by AB, BCy CD, DE, and their resultant represented by AE, applied at the point O. It is evident that every system of concurrent forces has for its resultant some single force (Art. 9) ; though in particular cases its magnitude may be zero. 15. Force Polygon. — The figure formed by drawing in suc- cession lines representing in magnitude and direction any number of forces is called a force polygo?i for those forces. Thus Fig. 4 is a force polygon for any four forces represented in magnitude and direction by the lines AB, BC, CD, and DEy whatever their lines of action: It may happen that the point E coincides with A, in which case the polygon is said to be closed. It is evident that the order in which the forces are taken does not affect the relative positions of the initial and final points. I § 3. Equilibrium of Concurrent Forces. 16. Definition. — A system of forces acting on a body is in eqnilibrium if the motion of the body is unchanged by its action. 17. Condition of Equilibrium. — In order that no motion may result from the action of any system of concurrent forces, the magnitude of the resultant must be zero ; and conversely, if the magnitude of the resultant is zero, no motion can result. But (Arts. 14 and 15) the condition that the result- ant is zero is identical with the condition that the force polygon closes. Hence, the following proposition : If any system of concurrent forces is in equilibrium, the force polygon for the system must close. And conversely, If the force EQUILIBRIUM OF CONCURRENT FORCES, ir-is polygon is closed for any system of concurrent forces^ tJie system is in equilibrium,. The comparison of this with the analytical conditions of equilibrium is given in Art. 22. 18. Method of Solving Problems in Equilibrium. — If a sys- tem of concurrent forces in equilibrium be partially unknown, we may in certain cases determine the unknown elements by applying the principles of Art. 17. The most usual case is that in which two forces are unknown except as to lines of action. Thus, suppose a system of five forces in equilibrium, three being fully known, represented in magnitude and direc- tion by AB, BC, CD (Fig. 5), and in lines of action by ab, be, cd, while concerning the other two we know only their lines of action de, ea. To determine these two in magnitude and direction, it is necessary only to complete the force polygon of which ABCD is the known part. The remaining sides must be parallel respectively to de and ea. From D draw a line parallel to de, and from A a line parallel to ea, prolonging them till they intersect at E. Then DE and EA represent the required forces in magnitude and direction, and the complete force polygon is ABCDEA. It is evident that ABODE' A is an equally legiti- mate form of the force polygon, and gives the same result for the magnitude and direction of each of the unknown forces. This problem occurs constantly in the construction of stress diagrams by the method described in Part II. The student will find little difficulty in treating other prob- lems in the equilibrium of concurrent forces. 19. Problems in Equilibrium. — (i) A particle is in equilib- rium under the action of five forces, three of which are 8 GRAPHIC STATICS. completely known, while of the remaining two, one is known in direction only, and the other in magnitude only. To deter- mine the unknown forces. (2) Suppose two forces known in magnitude but not in direction, the remaining forces being wholly known. (3) Suppose one force wholly unknown, the others being known. § 4. Resolution of Concurrent Forces. 20. To Resolve a Given Force into Any Number of Com- ponents having the same point of application, we have only to draw a closed polygon of which one side shall represent the magnitude and direction of the given force ; then the remaining sides will represent, in magnitude and direction, the required components. This problem is, in general, indeterminate, unless the components are required to satisfy certain specified con^ ditions. [Note. — A problem is said to be indeterminate if its conditions can be satisfied in an infinite number of ways. It is determinate if it admits of only one solution. Thus, the problem, to determine the values oi x and^ which shall satisfy the equation X -\- y — 10, is indeterminate; while the problem, given 2 jr + 3 = 7, to find the value of jr, is determinate. The case in which a finite number of solutions is possible may be called incompletely determinate. Thus, the problem, given x"^ Ar x — d = o, to find X, admits of two solutions, and therefore is incompletely determinate. All these classes of problems may be met with in statics.] 21. To Resolve a Given Force into Two Components. — This problem is indeterminate unless additional data are given. For if the given force be represented in magnitude and direction by a line, any two lines which with the given line form a triangle may represent forces which are together equivalent to the given force. But an infinite number of such triangles may be drawn. The solution of the following four cases of this problem will form exercises for the student. In each case the force diagram and space diagram should be completely drawn, and the student should notice whether the problem is determinate, partially determinate, or indeterminate. RESOLUTION OF CONCURRENT FORCES. (i) Let the lines of action of the required components be given. (2) Let the two components be given in magnitude only. (3) Let the line of action of one component and the magni- tude of the other be given. (4) Let the magnitude and direction of one component be given. It will be noticed that these four cases correspond to four cases of the solution of a place triangle. 22. Resolved Part of a Force. — If a force is conceived as replaced by two components at right angles to each other, each is called the resolved part, "^ in its direction, of the given force. It is readily seen that the resolved part of a force repre- sented by ^^ (Fig. 6) in the direction of any line XX, is represented in magnitude and direction A D by A^ B\ the orthographic projection of AB upon XX. It follows that the resolved part (in any given direction) of the resultant of any concurrent forces is equal to the algebraic sum of the resolved parts of its components in that direction ; signs plus and minus being given to the resolved parts to distinguish the two opposite directions which they may have. Thus (Fig. 7) the re- solved parts of the forces AB, BC, CD, in a direction parallel to XX, are A' B' , B' C , CD' ; and their algebraic sum is A'D', which is the resolved part of the resultant AD. If the resultant is zero, D' coincides with A' (i) For the equilibrium of any concurrent forces, the sum of their resolved parts in any direction must be zero. Fi Fig. r hence, * The term "resolute" has been proposed by J. B. Lock (" Elementary Statics ") to denote what is here defined as the resolved part of a force. lO GRAPHIC STATICS. Again, if D^ coincides with A\ then either D coincides with A, or else AD is perpendicular to XX \ hence, (2) If the sum of the resolved parts of any concurrent forces in a given direction is zero, their resultant (if any) is perpendicular to that direction. And if the sum of the resolved parts is zero for each of two directions, the resultant is zero, and the system is in equilibrium. Propositions (i) and (2) state the conditions of equilibrium for concurrent forces usually deduced in treatises employing algebraic methods. 1 CHAPTER IL — NON-CONCURRENT FORCES. § I. Composition of Non-concurrent Forces Acting on the Same Rigid Body. 23. Definition of Rigid Body. — A rigid body is one whose particles do not change their positions relative to each other under any applied forces. No known body is perfectly rigid, but for the purposes of statics, most solid bodies may be con- sidered as such ; and any body which has assumed a form of equilibrium under applied forces, may, for the purposes of statics, be treated as a rigid body without error. 24. Change of Point of Application. — The effect of a force upon a rigid body will be the same, at whatever point in its line of action it is applied, if only the particle upon which it acts is rigidly connected with the body. This proposition is fundamental to the development of the principles of statics, and is amply justified by experience.* In applying the principle, we are at liberty to assume a point of application outside the actual body, the latter being ideally extended to any desired limits. 25. Resultant of Two Non-Parallel Forces. — If two com- planar forces are not parallel, their lines of action must inter- sect, and the point of intersection may be taken as the point of application of each force. Hence, they may be treated as *This proposition may be proved analytically by deducing the equations of motion of a rigid body, and showing that the effect of any force on the motion of the body depends only upon its magnitude, direction, and line of action. But such a proof is, of course, outside the scope of this work. TI 12 GRAPHIC STATICS. concurrent forces, and their resultant may be determined as in Art. 13. The following proposition may therefore be stated : If two forces acting in the same plane on a rigid body are represented in magnitude and direction by AB and BC, their resultant is represented in magnitude and direction by AC, and its line of action passes through the point of inter- section of the lines of action of the giv^en forces. Its point of application may be any point of this line. It may happen that the point of intersection of the two given lines of action falls outside the limits available for the drawing. In such a case it will be most convenient to find the resultant by the method to be explained in Art. 27. The same remark applies to the case of two parallel forces. 26. Resultant of Any Number of Non-concurrent Forces — First Method. — ^The method of the preceding article may be extended to the determination of the resultant of any number of forces acting on the same rigid body. Let AB, BC, CD, DE (Fig. 8), represent in magnitude and direction four forces, and let ab, be, cd, de represent their lines of action. To find their resultant, we may proceed as follows : The resultant of AB and BC is represented in magnitude and direction by AC, and in line of action by ac drawn parallel to AC through the point of intersection of ab and be. Combining this resultant with CD, we get as t/ieh' resultant a force represented in magnitude and direction by AD, and in line of action by ad drawn parallel to AD through the point of intersection of ae and ed. This is evidently the resultant of AB, BC, and CD. In the same way, this resultant COxMPOSITION OF XON-COx\CURRENT FORCES. 13 combined with DE o-ives for their resultant a force whose mas-- nitude and direction are represented by AE, and whose line of action is ac^ parallel to AE and passing through the point in which ad intersects de. This last force is the resultant of the four given forces. The process may evidently be extended to the case of any number of forces. As in the case discussed in the preceding article, this method will become inapplicable or inconvenient in case any of the points of intersection fall outside the limits available for the ^drawing. For this reason it is usually most convenient to employ the method described in Art. 27. The student should bear in mind that the length and direc- tion AE and the line ae are not the magnitude, direction, and line of action of any actual force applied to the body. By the resultant is meant an ideal force, which, if it acted, would produce the same effect upon the motion of the body as is produced by the given forces. It is a force which may be conceived to replace the actual forces, and may be assumed to be applied to any particle in its line of action, provided that particle is conceived as rigidly connected with the given body. The line of action may in reality fail to meet the given body. (See Art. 24.) 2j. Resultant of Non-concurrent Forces — Second Method. — This method will be described by reference to an example. Referring to Fig. 9, let AB, BC, CD, DE represent in magni- tude and direction four forces whose lines of action are ab, be, ed, de ; and let it be required to find their resultant. Draw the force polygon ABCDE, and from any point O in the force diagram draw lines OA, OB, OC, OD, OE. These lines may represent, in magnitude and direction, components into which the given forces may be resolved. Thus AB is equivalent to forces represented by ^6^ and OB acting in any lines parallel to AO, OB, whose point of intersection falls upon ab ; BC is 14 GRAPHIC STATICS. equivalent to forces represented by BO, OC, acting in any lines parallel to BO, OC, which intersect on be ; and so for each of the given forces. The four given forces may, therefore, be replaced by eight forces given in magnitude and direction by AO, OB, BO, OC, CO, OD, DO, OE, with proper lines of action. Now, it is possible to make the lines of action of the forces represented by OB and BO coincide ; and the same is true of each pair of equal and opposite forces, OC, CO ; OD, DO. To accomplish this, let AO, OB act in lines ao, ob, inter- secting at any assumed point of ab. Prolong ob to intersect be, and take the point thus determined as the point of inter- section of the lines of action of BO, OC ; these lines are then bo, oe. Similarly prolong oe to intersect ed, and let the point of intersection be taken as the point at which CD is resolved into CO and OD ; the lines of action of these forces are then eo and od. In like manner choose do, oe, intersecting on de, as ^the lines of action of DO, OE. If this is done, the forces OB, BO will neutralize each other and may be omitted from the system ; also the pairs OC, CO, and OD, DO. Hence, there remain only the two forces represented in magnitude and direction by AO, OE, and in lines of action by ao, oe. Their resultant is given in magnitude and direction by AE, and its line of action is ae, drawn parallel to AE through the COMPOSITION OF NON-CONCURRENT FORCES. 15 point of intersection of oa and oe ; and this is also the result- ant of the given system. By carefully following through this construction the student will be able to reduce it to a mechanical method, which can be readily applied to any system. 28. Funicular Polygon. — The polygon whose sides are oa, ob, oc, ody oe, is called 2, fiinicidar polygon^ for the given forces. Since the point at which the two components of AB are assumed to act may be taken anywhere on the line ab, there may be any number of funicular polygons with sides parallel to oa, ob, etc. Again, if O is taken at a different point, there may be drawn a new funicular polygon starting at any point of ab ; and by changing the starting point any number of funicular polygons may be drawn with sides parallel to the new directions of OA, OB, etc. Moreover, different force and funicular polygons may be obtained by changing the order in which the forces are taken. It may be proved geometrically that for every possible funic- ular polygon drawn for the same system of forces, the last vertex, determined by the above method, will lie on the same line parallel to the closing side of the force polygon (as ae, Fig. 9). Such a proof is outside the scope of this work. The truth of the proposition may, however, be shown from the principles of statics. For if it were not true, it would be possible by the above method to find two or more forces, having different lines of action, which are equivalent to each other, because each is equivalent to the given system. But this is impossible. 29. Examples. — i. Choose five forces, assigning the magni- tude, direction, and line of action of each, and find their resultant by constructing the force and funicular polygons. 2. Draw a second funicular polygon, using the same point O in the force diagram. *Also called equilibrium polygon. l6 GRAPHIC STATICS. 3. Draw a third funicular polygon, choosing a new point O. 4. Solve the same problem, taking the forces in a different order. 30. Definitions. — The point O (Fig. 9) is called the /f^^ known, hence the ver- bv" i / i/ \^/^ tex A or the force \^ | / / 'c X polygon cannot be ^^ y^c ^V fixed. Next, draw the ^. „ Fig. 11 funicular polygon so 22 GRAPHIC STATICS. far as possible from the given data. Choose the pole O, and draw the rays OB, OC, OD, OB. The remaining ray, OA, cannot yet be drawn. Now, in the funicular polygon, the sides od, oc must intersect on be ; oc and od must intersect on cd\ through the intersection of ^and fa. We may then draw suc- cessively ob, oc, od. Now, of is unknown in direction ; but it is to be drawn through the intersection of oa and fa, hence, whatever its direction, it intersects ef in the same point (since oa was drawn through the intersection of ef and fa). Hence, the two vertices of the funicular polygon falling on ef and fa coincide at the intersection of these two lines. We may there- fore draw oe through this point and also through the point already found by the intersection of od and de. Now draw a line from O parallel to oe, and from D a line parallel to de ; their intersection is E. Draw from E a line parallel to ef and from A a line parallel to fa ; their intersection determines F. The force polygon is now completely drawn, and DE, EF, FA represent, in magnitude and direction, the required forces. The remaining ray OF and the corresponding string of may now be drawn, but are not needed. The construction might have been made equally well by choosing the intersection of de and ef as the starting point, since two vertices of the funicular polygon may be made to coincide at that point. Or, the point of intersection of de and af might be chosen ; but in that case, the order of the forces should be so changed as to make de and af consecutive. If this were done the figures should be relettered to agree with 24 GRAPHIC STATICS. the order in which the forces were taken. It may be noticed that the directio7i of the string first drawn may be chosen arbi- trarily and the pole so taken as to correspond to the direction chosen. This is important in the treatment of the following special case. Case of inaccessible points of intersection. — It may happen that the lines of action de, ef and fa have no point of intersec- tion within convenient limits. When this is the case, the method just given may still be applied, but involves the geo- metrical problem of drawing a line through an inaccessible point. For example, if ef and fa intersect beyond the limits of the drawing, as shown in Fig. 13, we may proceed as follows: Choose some point of ab and from it draw a line through the point of intersection of ef and fa. (This can be done by a method to be explained presently.) Let this line be oa. From the known point A of the force polygon draw a line parallel to Fig. 13 oa and choose a pole upon it. Draw the rays OB, OC, OD ; then the corresponding strings in the order ob, oc, od. From the point in which od intersects de draw a line to the inaccessible point of intersection of ef and fa ; this will be oe. The force polygon can now be completed just as in the preceding case. A line may be drawn through the inaccessible point of inter- section of two given lines by the following method : Let AC, EQUILIBRIUM OF NON-CONCURRENT FORCES. 25 BD (Fig. 14), be the given lines, and let it be required to draw a line through their point of intersection from some point P. Draw PA intersecting AC and BD, and from C, any point of AC, draw a line CQ parallel to PA. From E, a point of BD, draw EA, EP ; also draw CF parallel to AE, and FQ parallel to EP. Then PQ is the line required. For, by the similar triangles, it is easily shown that AB CG BP -^ F D ( — — o Q Fig. 14= ; which proves that AC, BG, and PQ meet in a point. GQ Exception. — If the lines of action of the three unknown forces meet in a point (or are parallel), the problem is impos- sible of solution, unless it happens that the resultant of the known forces acts in a line through the point of intersection of those three lines of action (or parallel to them) ; in which case the problem is indeterminate. These cases will not be discussed here. Remark. — In the problems treated in this and the two pre- ceding articles, it will be noticed that the forces should be taken in such an order that those which are completely known are consecutive. Otherwise the number of unknown lines in the force and funicular polygons will be increased. [For an- other method of solving this problem, see Levy's " Statique Graphique."] 41. Examples. — i. A rigid beam AB rests horizontally upon supports at A and B, and sustains loads as follows : Its own weight of 100 lbs. acting at its middle point ; a load of 50 lbs. at (^ ; a load of 60 lbs. at D ; and a load of 80 lbs. at E. The successive distances between A, C, D, E, and B are 4 ft., 6 ft., 7 ft., and 10 ft. Required the upward pressures on the beam at the supports. 26 GRAPHIC STATICS. 2. A rigid beam AB is hinged at A, and rests horizontally with the end B upon a smooth horizontal surface. The beam sustains loads as in Example i, and an additional force of 40 lbs. is applied at the middle point at an angle of 45° with the bar in an upward direction. Required the pressures upon the beam at A and B. [The pressure at A may have any direc- tion, while the pressure at B must be vertically upward, i.e., at right angles to the supporting surface. Hence, this is a particular case of Problem II.] 3. Let the end B of the bar rest against a smooth surface making an angle of 30° with the horizontal ; the remaining data being as in Example 2. 4. A rigid bar 2 ft. long weighs 10 lbs., its center of gravity being 8 inches from one end. The bar rests inside a smooth hemispherical bowl of 15 inches radius. What weight must be applied at the middle point in order that the bar may rest when making an angle of 15° with the horizontal.'* Also, what are the reactions at the ends ? 5. A uniform bar 20 inches long, weighing 10 lbs., rests with one end against a smooth vertical wall and the other end overhanging a smooth peg 10 inches from the wall. A weight P is suspended from the end so that the bar is in equilibrium when making an angle of 30° with the horizontal. Find P, and the pressures exerted on the bar by the wall and peg. 42. Special Methods. — Certain problems can be treated more simply by other methods than by the general method of constructing the force and funicular polygons. This is sometimes true of the following : Problem. — A rigid body is held in equilibrium by four forces acting in known lines, only one being known in magni- tude and direction. It is required to completely determine the three remaining forces. (See Fig. 15.) Let the four forces have lines of action ab, be, cd, da, and let EQUILIBRIUM OF NON-CONCURRENT FORCES. 27 AB be drawn representing the magnitude and direction of the known force. Now the resultant of the forces whose lines of action are da and ab must act in a line passing through My the point of intersection of these lines ; and the resultant of the other two forces must act in a line passing through iV, the point of intersection of be and cd. But the two resultants must be equal and opposite and have the same line of action, else there could not be equilibrium. (Art. 36.) Hence, each must act in the line MN. Draw BD parallel to MN, and AD parallel to ad\ the point D being determined by their intersection. Then DA represents in magnitude and direction the force acting in da, and DB the resultant of DA and AB. But BD must represent the resultant of the two remaining forces ; hence these two forces are represented by BC and CD drawn from B and D parallel respectively to be and cd. This problem is a special case of that treated in Art. 40. But the construction here given will in many cases be the simpler one. Example. — A rigid body has the form of a square ABCD, the side AB being horizontal, and BC vertical. The weight of the body is 100 lbs., its center of gravity being at the inter- section of the diagonals. It is held in equilibrium by three forces as follows : P^ acting at C in the line AC ; Po_ acting at D in the line AD ; and P^ applied at B and acting in the line joining B with the middle point of AD. Required to com- pletely determine P^, Py AC and CB upon each other.) 65. Three Kinds of Internal Stress. — It is evident that the internal stress at C in the body AB (Fig. 20) depends upon the external forces applied to the body. If the forces at A and B cease to act, the forces exerted hy AC and CB upon each other become zero. If the forces at A and B are reversed in direction, so also are those at C (As a matter of fact, the particles oi AC exert forces upon those of CB, even if the external forces do not act. But if the external forces applied to CB are balanced, the resultant of the forces exerted on CB by ^4(7 is zero.) 44 GRAPHIC STATICS. The nature of the internal stress at any point in a body is thus seen to depend upon the external forces applied to the body. Now, if we consider two adjacent portions of a body (as the parts X and Y, Fig. 22) separated by a plane surface, the external forces may have either of three \\~7. \\ tendencies : (i) to pull X and Y apart in a direction perpendicular to the plane ^^' of separation ; (2) to push them together in a similar direction ; (3) to slide each over the other along the plane of separation. Corresponding to these three tendencies, the stress between X and Y may be of either of three kinds : tensile^ compressive^ or shearing. A tensile stress is such as comes into action to resist a tendency of the two portions of the body to be pulled apart in the direction of the normal to their surface of separation. A compressive stress is such as comes into action to resist a tendency of X and Y to move toward each other along the normal to the surface. A shearing stress is such as acts to resist a tendency of X and Fto slide over each other along the surface between them. In case of a tensile stress, the force exerted by X upon Y has the direction from Y toward X\ and the force exerted by Fupon X has the direction from X toward Y. In case of a compressive stress, the force exerted by X upon Y has the direction from X toward F; and the force exerted by Fupon Xhas the direction from F toward X. In the case of a shearing stress, the force exerted by X upon Fmay have any direction in the plane of separation ; the force exerted by Fupon X having the opposite direction. If X and F are separate bodies, instead of parts of one body, a similar classification may be made of the kinds of stress between them ; but with these we shall have no occasion to deal. The terms tensile stress, compressive stress, and shear- ing stress (or tension, compression, and shea?) are usually applied only to internal stresses. EXTERNAL AND INTERNAL STRESSES. 45 66. Strain. — In what has preceded, the bodies dealt with have been regarded as rigid ; that is, the relative positions of the particles of any body have been regarded as remaining unchanged. But, as remarked heretofore, no known body is perfectly rigid. If no external forces act upon a body, its particles take certain positions relative to each other, and the body has what is called its natural shape and size. If external forces are applied, the shape and size will generally be changed ; the body is then said to be in a state of strain. The deforma- tion produced by any system of applied forces is called the strain due to those forces. The nature of this strain in any case depends upon the way in which the forces are applied. It is unnecessary to treat this subject further at this point, since we shall at present be concerned only with problems in the treatment of which it will be sufficiently correct to regard the bodies as rigid. [Note. — There is a lack of uniformity among writers in regard to the meanings attached to the words stress and strain. It may, therefore, be well to explain again at this point the way in which these words are used in the following pages. The word stress should be employed only in the sense above defined, as consisting of two equal and opposite forces constituting an action and its reaction. The two forces are exerted respectively by two bodies or portions of matter upon each other. An internal stress is a stress between two parts of the same body. An internal force is one of the forces of an internal stress. It is intended in what follows to use the word? "internal stress" (or simply "stress '') only when both the constituent forces are referred to; and when only one of the forces is meant, to. use the words "inter- nal force" (or simply "force"). It will be noticed, therefore, that in the following pages the words '■^ force in a bar " are frequently used where many writers would say " stress." This departure from the usage of many high authorities seems justified by the following considerations : (i) It agrees with the usage which is being adopted by the highest authorities in pure mechanics. (2) It is desirable that the nomencla- ture of technical mechanics shall agree with that of pure mechanics, so far as they deal with the same conceptions. The definition of strain above given is in conform- ity with the usage of the majority of the more recent text-books. But it is not rare to find in technical literature the word strain used in the sense of internal stress as above defined. Such use of the word should be avoided.] 46 GRAPHIC STATICS. § 3. Determination of Internal Stresses. 6'j. General Method. — The stresses exerted between the parts of a body may or may not be completely determinate by means of the principles already deduced. But in all cases these principles suffice for their partial determination. The general method employed is always the same, and will now be illustrated. As heretofore we deal only with complanar forces. Let XY (Fig. 23) represent a body in equilibrium under the action of any known external forces as shown. Now conceive the body to be divided into two parts as X and Y, separated by any surface. The particles of X near the surface exert upon those of Y certain forces, and are in return acted upon by forces exerted by the particles of F. These forces are internal as re- gards the whole body. In order to determine them so far as possible, we proceed as follows : Let the resultant of all the forces exerted by Y upon XhQ called T; then T^is either a single force or a couple. Now apply the conditions of equilibrium to the body X. The external forces acting on X are Pi, P.2, P^, and T. Since Pj, P^, and Pz are supposed known, T can be determined. In fact, T is equal and opposite to the resultant of P^, P^, and P^. So much can always be determined. But T is the resultant of a great number of forces acting on the various particles of X \ and these separate forces cannot in general be deter- mined by methods which lie within the scope of this work. The general principle just illustrated may be stated as follows : If a body in equilibrium tinder any external forces be conceived as made up of two parts X and Y, then the internal forces exerted by X upon Y, together with the external forces acting on Y, form a system in eqtdlibrinm. Fig. S3 DETERMINATION OF INTERNAL STRESSES. 47 As an immediate consequence, we may state that tJie result- ant of the forces exerted by X upon V is equivalent to the result- ant of the external forces acting on X ; and is equal and opposite to the resultant of the external forces acting upon Y. Example. — Assume a bar of known dimensions, and the magnitudes, directions, and points of application of five forces acting on it. Then (i) determine a sixth force which will produce equilibrium ; and (2) assume the bar divided into two parts and find the resultant of the forces exerted by each part on the other. 6^. Jointed Frame. — In certain ideal cases (corresponding more or less closely to actual cases), the internal forces may be more completely determined. The most important of these cases is that which will be now considered. Conceive a rigid body made up of straight rigid bars hinged together at the ends. Assume the following conditions : (i) The hinges are without friction. (2) All external forces acting on the body are applied at points where the bars are joined together. The meaning of these conditions will be seen by reference to Fig. 24. The three bars X, F, and Z are connected by a '* pin joint," the end of each bar having a hole or ''eye" into which is fitted a pin. (Of course the three bars cannot be in the same plane, but they may be nearly so, and will be so assumed in what fol- lows.) Condition (i) is satisfied if the ie-3_4 pin is assumed frictionless. The effect of this is that the force exerted upon the pin by any bar (and the equal and opposite reaction exerted upon the bar by the pin) acts in the normal to the surfaces of these bodies at the point of contact ; and, therefore, through the centers of the pin and the hole. Condi- tion (2) means that any external force (that is, external to the 48 GRAPHIC STATICS. whole body) applied to any bar is applied to the end and in a line through the center of the pin. With the connection as shown, the bars do not exert forces upon each other directly. But each exerts a force upon the pin, and any force exerted by V or Z upon the pin causes an equal force to be exerted upon X. (This is seen by applying the condition of equilibrium to the pin.) Hence, in considering the forces acting upon any bar as X, we may disregard the pin and assume that each of the other bars acts directly upon X. By what has been sai'd, all such forces exerted upon X by the other bars meeting it at the joint may be regarded as acting at the same point — the center of the pin. We therefore treat the bars as mere '^material lines," and regard all forces exerted on any bar (whether by the other bars or by outside bodies) as applied at the ends of this "material line." Since, with these assumptions, all forces acting on any bar MN (Fig. 25) are applied either at M or JV, the forces applied at M must balance those applied at N. The resultants of the two sets must therefore be equal and opposite, and have the same line of action — namely MN. Further, it follows that the stress in the bar, acting across any plane perpendicular to its length, is a direct tension or compression. 69. Internal Stresses in a Jointed Frame. — Let Fig. 25 rep- resent a jointed frame such as above described, in equilibrium under any known external forces. Let us apply the general method of Art. 6y to this case. Divide the body into two parts, X and V, by the surface AB as shown. Now apply the conditions of equilibrium to the body X. The system of forces acting upon this body consists of Pi, P2, Pq, and the forces exerted by Y upon X in the three members cut by the surface AB. By Art. 68 the lines of action of these forces are known, Fig. 2B DETERMINATION OF INTERNAL STRESSES. 49 being coincident with the axes of the members cut. Hence, the system in equilibrium consists of six forces, three com- pletely known, and three known only in lines of action. The determination of the unknown forces in magnitude and direction is then a case under the general problem discussed in Art. 40. Nature of the stresses. — As soon as the direction of the force acting upon X in any one of the members cut is known, the nature of the stress in that member (whether tension or compression) is known. For a force toivard X denotes com- pression ; while a force aivay from X denotes tension. (Art. 65.) If a section can be taken cutting only two members, the forces in these may be found by the force polygon alone. The same is true, if any number of members are cut, but the stresses in all but two are known. The methods described in the last three articles will find frequent application in the chapters on roof and bridge trusses, Part II. Example. — Assume a jointed frame similar to the one shown in Fig. 25, and let external forces act at all the joints. Then (i) assume all but three of the forces known in magnitude and direction and determine the remaining three so as to produce equilibrium. (2) Take a section cutting three members and determine the stresses in those members. 70. Indeterminate Cases. — If, in dividing the frame, more than three members are cut, the number of unknown forces is too great to admit of the determination of their magnitudes. In such a case, it may happen that a section elsewhere through the body will cut but three members ; and that after the deter- mination of the stresses in these three, another section can be taken cutting but three members whose stresses are unknown. So long as this can be continued, the determination of the 50 GRAPHIC STATICS. internal stresses can proceed. Thus, in Fig. 26, if a section be first taken at AB, there are four unknown forces to be determined. But, if the section A' B' be first taken, the stresses in the three members cut may be determined ; after which the section AB will \A' introduce but three unknown stresses. There may, however, be cases in which the stresses °. ° cannot all be determmed by any method. With such ac- tually indeterminate cases we shall not usually have to deal. It should be noticed, also, that even when only three members are cut, the problem is indeterminate if these three intersect in a point. As in the case just discussed, this indeterminateness may be either actual or only apparent ; in the latter case it may be treated as above indicated. No attempt is here made to develop all methods that are applicable or useful in the determination of stresses in jointed frames. Some of these are best explained in connection with the actual problems giving rise to them. We have sought here only to explain and clearly illustrate general principles. 71. Funicular Polygon Considered as Jointed Frame. — Let ab^ be, cd, da (Fig. 27) be the lines of action of four forces in equihbrium, the force polygon being ABCDA. Choosing a Fig. 3-7 pole, draw any funicular polygon, as the one shown. Now let the body upon which the forces act be replaced by a jointed DETERMINATION OF INTERNAL STRESSES. 51 frame whose bars coincide with the sides of the funicular polygon. If at the joints of this frame the given forces be applied, the frame will be in equilibrium ; and each bar will sustain a tension or compression whose magnitude is repre- sented by the corresponding ray of the force diagram. To prove this, we apply the ''general method" of Art. 6^. Consider any joint (as the intersection of oa and ob)^ and let the frame be divided by a plane cutting these two members. Then the portion of the frame about the joint is acted upon by three forces : AB, acting in the line ab, and forces acting in the bars cut, their lines of action being oa, ob. If the bar oa sustains a compression and ob a tension, their magnitudes being represented by OA and BO respectively, the portion of the frame about the joint will be in equilibrium. Hence, the tendency of the force AB is to produce the stresses men- tioned in the bars oa, ob. In the same way it may be shown that the tendency of the force BC is to produce in ob and oc tensile stresses of magnitudes OB and CO, respectively. Applying the same reasoning to each joint, it is seen that every part of the frame will be in equilibrium if the bars sustain stresses as follows : The bar oa must sustain a compression OA ; ob a tension 0B\ oc a tension 0C\ and od 2, compression OD. Hence, if the bars are able to sustain these stresses, the frame will be in equilibrium. If the stress in any member of the frame is a tension, that member may be replaced by a flexible string. This is the origin of the name string as applied to the sides of the funic- ular polygon. This name is retained for convenience, but, as just shown, it is not always appropriate. 72. Outline of Subject. — The foregoing pages, embracing Part I, have been devoted to a development of the principles of pure statics. We pass next to the application of these principles to special classes of problems. Part II treats of the determination of internal stresses in 52 GRAPHIC STATICS. engineering structures. Only "simple" structures are consid- ered, — that is, those whose discussion does not involve the theory of elasticity. The structures considered include roof trusses, beams, and bridge trusses. Part III develops the graphic methods of determining cen- troids (centers of gravity) and moments of inertia of plane areas, including a short discussion of *' inertia-curves." Part II. STJiESSES IN SIMPLE STRUCTURES. -0'.:«o In diagram {B), ABCDEFGHVA is the force polygon for the case when the wind is from the left. The points E, F, G, H, coincide, because the loads EF, EG, GH, are each zero. 93. Stress Diagrams for Wind Pressure. — When the loads and reactions due to wind pressure are known, the internal stresses can be found by drawing a stress diagram, just as in the case of vertical loads. The construction involves no new principle, and will be readily made by the student. In Figs. 33 {A) and 33 {B) are shown the diagrams for the two directions of the wind. MAXIMUM STRESSES. 71 The stresses in all members of the truss must be determined for each direction of the wind. If the truss is symmetrical with respect to a vertical line, as is usually the case, it may be that the same stress diagram will apply for both directions of wind. This will be so if the reactions are assumed to act as in cases (i) and (2) of the preceding article. In the case repre- sented in Fig, 33, however, the hinging of one end of the truss destroys the symmetry of the two stress diagrams, and both must be drawn in full. § 4. MaximtLin Stresses. 94. General Principles. — For the purpose of designing any truss member, it is necessary to know the greatest stresses to which it will be subjected under any possible combination of loads. Stresses are combined in accordance with the principle stated in Art. 'j'^, that the resultant stress in a truss member due to the combined action of any loads is equal to the algebraic sum of the stresses due to their separate action. The method will be illustrated by the solution of an example with numerical data. 95. Problem — Numerical Data. — Let it be required to design a wrought iron truss of 40 ft. span, of the form shown in Fig. 34 (PI. I). Let 12 ft. be the distance apart of trusses, and let the loads be as follows : Weight of truss, to be assumed in accordance with the formula of Art. 82 : W—\ al (i+yo ^)- '^^^^ gives W= 1800 lbs. Assuming this to be divided equally among the upper panels, and that the load for each panel is borne equally by the two adjacent joints, the load at each of the joints be, ed, de is 450 lbs. The loads at the end joints may be neglected, being borne directly at the supports. Weight of 7'oof — This depends upon the materials used and the method of construction, but will be taken as 6 lbs. per sq. ft. 72 GRAPHIC STATICS. of roof area, giving 900 lbs. as the load at each joint. This, also, is a permanent load. Total permanent load per joint, 1350 lbs. Weight of S7101V. — Taking this as 1 5 lbs. per horizontal square foot, we find 1800 lbs. as the load at each joint. Wind pi'essiLvc. — This is computed from the formula 2 sin a ^ I \ ^ c ^ Pa= , . ^ A- Art. 85.) I +sm^ a For this case we put sina = -|-|=|; ^,^ = 40 lbs. per sq. ft.; whence /a = 35 lbs. per sq. ft. (about). This gives upon each panel of the roof 5250 lbs. Then with the wind from either side, the wind loads on that side would be 2625 lbs., 5250 lbs., 2625 lbs. respectively. 96. Stress Diagrams. — We are now ready to construct the stress diagrams. The truss is shown (PL I) in Fig. 34 {A). Fig. 34 {B) is the stress diagram for permanent loads. No diagram for snow loads is needed, since it would be exactly similar to that for permanent loads. The snow load at any joint being four-thirds as great as the permanent load, the stress in any member due to snow is four-thirds that due to permanent loads. Fig. 34 {C) shows the stress diagram for the case of wind blowing from the left. The reactions are assumed to act in lines parallel to the loads — that is, normal to the roof. With this assumption, no separate diagram is needed for the case of wind from the right, since such a diagram would be exactly symmetrical to Fig. 34 {C). For example, the stress in the member^/; due to the wind blowing from the right is given by the line GM 'vn Fig. 34 {C). 97. Combination of Stresses. — After the stress diagrams are completed for the various kinds of loads, the stresses should be scaled from the diagrams and entered with proper sign in a table, as follows : MAXIMUM STRESSES. 73 Member. Permanent .Load. Snow. Wind R. Wind L. Max. bh - 1525 - 2030 — 6270 - 7925 — 11480 ci - 1255 — 1670 — 6270 -7925 — 10850 hi - 360 - 480 -5250 — 6090 ik + 595 + 790 + 550 + 6650 - 8035 ^h + 1245 + 1660 + 2600 + 8800 + "705 gk + 730 + 975 + 2300 + 2300 + 4005 gin + 1245 -f 1660 + 8800 + 2600 + 1 1705 Ik + 595 -f 790 + 6650 + 550 - 8035 mi - 360 - 480 -5250 — 6090 dl - 1255 — 1670 -7925 — 6270 — 10850 em - 1525 - 2030 - 7925 — 6270 — 1 1480 By combining the results, the maximum stress in each mem- ber for any possible condition of loading can be determined. The possible combinations of loading are the following : Perma- nent load alone ; permanent and snow loads ; permanent load, and wind from either direction ; permanent and snow loads, and wind from either direction. The student will readily under- stand the method of combining the separate results. The problem here solved relates to a very simple form of truss. With some forms there may occur a reversal of stress in certain members, under different conditions of loading. It is to be noticed that in the table the word maximum is used in its numerical sense, and has no reference to the algebraic sign of the stress. 98. Examples. — In Fig. 35 {A) to {F), are shown several forms of truss for which the student may draw stress diagrams, assuming loads in accordance with the data given in Arts. 82 to 85. In determining reactions due to wind pressure, the 74 GRAPHIC STATICS. three assumptions mentioned in Art. 92 should all be used in different cases, that the student may become familiar with the principle of each. Fi^. 35 § 5. Cases Apparently I itde terminate. 99. Failure of Usual Method. — In attempting to construct the stress diagram by drawing the force polygon for each joint in succession, as in the cases thus far treated, a difficulty is met in certain forms of truss. It may happen that after pro- ceeding to a certain point it is impossible to select a joint for which the force polygon can be com.pletely drawn, the number of unknown forces for every joint being greater than two. Thus, in the truss shown in Fig. 36, if the stress diagram is started in the usual way, beginning at the left support, the force polygons for three joints may be constructed without difficulty, thus determining the stresses in bl^ Ik, hn, cm, mn, nk. But the force polygon for cdqp7imc cannot be constructed, since three forces are unknown, — namely, those in dq, qp, pn. And at the joint knpsk, the stresses in 7ip, ps, and sk are unknown. The problem, therefore, seems at this point to become indeter- CASES APPARENTLY INDETERMINATE. 75 minate, since either of the two polygons can be completed in any number of ways, so far as the known forces determine. It can be shown, however, that this ambiguity is only apparent. This may be proved as follows : Consider the portion of the truss to the left of the broken line M'N' . , It is in equilibrium under the action of eight forces ; five of these (four loads and a reaction) are known ; the remaining three are the forces in er^ rs, and sk. Now, the problem of determining these three unknown forces is the same as that treated in Art. 40. It was there found to be a deter- minate problem, unless the lines of action of the three unknown forces intersect in a point or are parallel. That the problem is determinate may be seen also from the principle of moments (Art. 51). The eight forces mentioned being in equilibrium, the sum of their moments is zero for any origin in their plane. Let the origin be taken at the point of intersection of the lines of action of two of the unknown forces, as er, rs. Then from the principle of moments we have (since the moments of the two forces named are zero) : Algebraic sum of m.oments of loads and reaction to left of section + moment of SK=o. The only unknown quantity in this equation is the magnitude of SK, which may, therefore, be determined. The other unknown forces may be found in a similar manner, the origin of moments being in each case chosen so as to eliminate two of the three unknown forces. The whole problem of drawing the stress diagram is now seen to be determinate. For, as soon as the stress in sk is known, the force polygon for the joint knpsk contains but two unknown sides, and can be drawn at once. No further diffi- culty will be met. 100. Solution of Case of Failure — First Method. — The reasoning of the preceding article suggests two methods of treating the so-called ambiguous case. These will now be described. 1^ GRAPHIC STATICS. The first method is to apply the construction of Art. 40, as follows : Referring to Fig. 36, consider .the equilibrium of the portion of the truss to the left of the line M'N''. The system of forces consists of those whose lines of action are ka, ab, be, cd, de^ er, rs, sk. Let them be taken in the order named, and draw the force polygon for the known forces. (The reaction ka is supposed to be already determined.) The known part of the force polygon is KABCDE \ the unknown part is to be marked ERSK. Choose a pole O and draw rays to K, Ay By C, D, and E ; then draw the corresponding strings of the funic- ular polygon. Remembering the method of Art. 40, we draw first oe, making it pass through the intersection of er and is (the reason for this being that // makes the string os also pass through that point) ; then draw in succession od, oe, ob, oay ok. The string ok intersects sk in a point through which os must be drawn. Hence os must join that point with the starting point of the polygon. This completes the funicular polygon, except the string or, which must pass through the intersection of er CASES APPARENTLY INDETERMINATE. 77 and rs in some direction not yet known. This string is not necessary to the solution of the problem. Since os is now known, OS may be drawn parallel to it from O ; and by drawing a line from K parallel to the known direc- tion of KS, the point 5 is determined, and KS becomes known. To find ER and RS it is only necessary to draw from 5 a line parallel to rs, and from E a line parallel to cr ; their inter- section gives the point R, and the force polygon for the system of forces considered is complete. The stresses in ei\ rs, and sk being now known, the stress diagram may be completely drawn by the usual method. It is interesting to notice that the method just described determines the lines ER, RS, SK, in their proper position in the complete stress diagram. The determination of ER and RS by this method is not necessary, since the usual method of drawing the stress-diagram can be carried out as soon as SK is known. But the independent determination of ER and RS by the two methods serves as a check on the accuracy of the construction. The funicular polygon employed in the above construction, so far as it belongs to the external forces, may coincide with the corresponding part of the funicular polygon used in deter- mining the reactions. If this is desired, two points must be observed: (i) the string ^r must be the first drawn, and must pass through the intersection of er and ri", and (2) the pole must be so chosen that ok will not be nearly parallel to ks. It should also be noticed that the construction of the funic- ular polygon might begin with the string ok, which should then be made to pass through the intersection of 7^s and sk. The student will be able to carry out this construction without difficulty. 10 1. Solution of Case of Failure — Second Method. — It will now be shown how the apparently ambiguous case can be 78 GRAPHIC STATICS. & treated graphically by the principle of moments. Referring again to Fig. 36, consider the portion of the truss to the left of section JVPN^ It is acted upon by eight forces, of which five (the loads and the reaction) are known, and three (whose lines of action are er^ rsj sk) are unknown. The algebraic sum of the moments of all these forces about any origin must be zero. Let the origin be taken at the point of intersection of er and rs, so that the moments of the forces acting in these lines are both zero ; then the sum of the moments of the five known forces, plus the moment of the force acting in the line sky must equal zero. Now the sum of the moments of the five known forces may be found by the method of Art. 56. Through the origin of moments draw a line parallel to the resultant of the forces named (that is, a vertical line), and let i equal the length intercepted on it by the strings oe, ok. Then the required moment is — zH, where H is the pole distance. (The minus sign is given in accordance with the convention that left-handed rotation shall be positive.) Let P = unknown force in line sk, and h its moment-arm. For the purpose of computing the moment, assume the stress in sk to be a tension ; then the force P acts toward the right and its moment is posi- tive, the value being H-P//. Hence, Ph — Hi=0', or, P = -H. h From this equation P may be computed. The computation may be made graphically as follows : Draw (Fig. 36) a triangle WUV, making WU=H (force units) and lVV=k (linear units). Lay off lVV=i (linear units) and draw VX parallel to VU. Then WX (force units) represents P. This is readily seen, since from the two similar triangles we have the propor- P i tion -- = -, which agrees with the equation above deduced. H h The computation is simplified if the pole distance H is taken equal to as many force units as // is linear units ; or if // CASES APPARENTLY INDETERMINATE. 79 is some simple multiple of Ji. For, suppose H=7ih\ then P = lL'A = ni. If ;/-i, P = i. h The stress in sk is found to be a tension, since P is positive. Whatever the nature of the stress, it may be assumed a ten- sion in writing the equation, and the sign of the value found will show whether the assumption coincides with the fact. I02. Other Methods for Case of Failure. — In certain cases the method of treating the '' ambiguous case " may profitably be varied. (i) The construction of Art. lOO may be modified as fol- lows : Determine the resultant of the known external forces acting on the portion of the truss to one side of the section M'N'. This resultant is in equilibrium with the three unknown forces acting in the members er, rs, sk. Hence these forces can be determined by the special method explained in Art. 42. The resultant of the five known forces is represented in magnitude and direction by KE in the force polygon ; and its line of action passes through the intersection of the strings ocy ok in the funicular polygon. Since this point of intersection is likely to be inaccessible, the construction of Art. 42 cannot be conveniently applied. It may be modified by using instead of KE the two forces KA (the reaction in the line ka) and AE (the resultant of the four loads, its line of action being deter- mined by the intersection of the strings oa and oe). First determine forces in the three lines er, rs, sk, which shall be in equilibrium with KA ; then make a similar construction for AE, and combine the results. (2) It has been proposed to employ the following reasoning : Remove the members/^ and qr, and insert another represented by the broken line in Fig. 36. Evidently this does not change the stress in the member sk, since the forces acting on the truss to the left of the section M^N^ are unchanged. But with this change the difficulty encountered in constructing the stress diagram by the usual method is avoided. For when the joint 8o GRAPHIC STATICS. nmcdqpii is reached, the forces acting there will be all known except two. Let the stress diagram be drawn in the usual way until the stress in sk is known. Then restore the original bracing and repeat the construction, using the value just deter- mined for SK. This method is convenient whenever it is applicable. Cases may, however, arise, in which it will fail. For instance, if a load is applied at the joint pqrsp, the members pq and qr cannot both be omitted, and the method cannot be applied. Again, it would be inapplicable ^'^^' ^^ in case of a truss such as shown in Fig. 37, with loads at all upper joints. In such cases, one of the methods explained in the preceding articles may be applied. Other methods might be mentioned, but the foregoing dis- cussion of the case will probably be found sufficient. 103. Failing Case in Other Forms of Truss. — The usual method of constructing the stress diagram may fail in other forms of truss, though the one above described is the most common. In such a case, the problem of finding the stresses may be really indeterminate, or only apparently so. Whenever it is possible to divide the truss into two parts by cutting three members which are not parallel and do not intersect in one point, the stresses in the three members cut are determinate as soon as all external forces are known, and can be found by methods already given. If more than three members are cut, the problem of finding the stresses in them is indeterminate, unless all but three of these stresses are known. By remem- bering these principles, the determinateness of any given prob- lem may readily be tested. (See Art. 70.) § 6. Three-Hinged Arch. 104. Arched Truss Defined. — If a truss is so supported that when sustaining vertical loads the reactions at the supports THREE-HINGED ARCH. 8l have horizontal components directed toward the centre of the span, the trass becomes an arcJi. The only kind of arch we shall here consider is that consisting of two partial trusses hinged together at the crown, and each hinged at the point of support. Such a truss is shown in Fig. 38, in which the two partial trusses are hinged to the abutments at P and Q, and connected by a hinge at the point R. Since a hinge at the support allows the reaction of the supporting body upon the truss to take any direction in the plane of the truss, the directions of the reactions at P and Q are unknown, as is also that of the force exerted by either partial truss upon the other at the point R. The problem of determining the reactions may, therefore, at first sight seem indeterminate. It will be shown in the next article that it is in reality determinate, and that the only principles needed in the solution are such as have been already often applied in the preceding chapters. The three-hinged arch is indeed a *' simple " structure (Art. 75), since the theory of elasticity is not needed in the determination of the reactions. 105. Reactions Due to a Single Load. — The method of find- ing the reactions is most clearly understood by considering the effect of a single load on either partial truss. Let a load be applied at 5 (Fig. 38) and let all other loads acting on either portion of the truss (including the weight of the structure) be 82 GRAPHIC STATICS. neglected. Call the two partial trusses X and V, and consider the part V. The only forces acting upon it are the reaction exerted by the abutment at Q and the force exerted at R by the truss X. These two forces, being in equilibrium, must have the same line of action, which is, therefore, the line QR. Consider now the body X. The external forces acting upon it are the load at S, the reaction of the abutment at P, and the force exerted by V at the point R. But this last force is equal and opposite to the force exerted by X upon F, and its line of action is therefore QR. The three forces acting upon the body X being in equilibrium, their lines of action must meet in a point ; which point is found by prolonging QR to meet the line of action of the applied load. Let T be this point, then PT is the line of action of the reaction at P. The reactions can now be determined by drawing the triangle of forces. This triangle is ABC in the figure, AB representing the load at S, BC the reaction in the line QR (also marked be), and CA the reac- tion in the line PT (marked also cci). Evidently ABC may be regarded as the polygon of external forces, either for the partial truss X, or for the whole structure composed of X and Y\ and BC represents either the force exerted by Y upon X at R, or the force exerted upon Y by the supporting body at Q. If, now, the structure be loaded at other points, the reactions due to each load may be found separately ; the resultant of all such separate reactions at either support will be the true reaction at that support when all the loads act together. A convenient method of applying these principles will be given in the next article. 1 06. Reactions and Stresses Due to Any Vertical Loads. — In Fig. 39 is represented a truss consisting of two parts supported by hinges at P^ and Q and hinged together at R^ . Consider all vertical loads to be applied at the upper joints, their lines of action being marked in the usual way. We shall first explain the construction for finding the reactions at the supports ; after THREE-HINGED ARCH. 83 these are determined, the drawing of the stress diagram will present no difficulty. Since we shall sometimes deal with one of the partial trusses, and sometimes with the two considered as a single body, it will be well at the outset to specify the external forces acting on each of these bodies. Fig.SQ (i) For the partial truss at the left we have five applied loads, the reaction at R' (exerted by the other truss), and the reaction at P'. The force polygon for these forces will be marked as follows: ABCDEFLA. (The meaning of the letters will be understood before the force polygon is actually drawn, by reference to the corresponding letters in the space diagram.) 84 GRAPHIC STATICS. (2) For the right-hand partial truss the external forces are five loads and two reactions, and the force polygon will be marked thus : FGHIJKLF. (Notice that FL and LF are equal and opposite forces, being the ''action and reaction" between the two trusses at i?'.) (3) For the combined structure the external forces are the ten loads and the reactions at P^ and Q . (The action and reaction at R become now internal forces.) The force polygon will be marked thus : ABCDEFGHIJKLA. Begin the construction by drawing the force polygon for the ten loads on the whole structure, lettering it as just indicated. Choose a pole (9, draw the rays, and then the funicular polygon as far as possible. Now consider the right partial truss as unloaded. The resultant of the remaining five loads is repre- sented in magnitude and direction by AF\ its line of action must pass through the intersection of oa and of, and is therefore the line marked af. Now, reasoning as in the preceding article, we see that the reaction at Q must act in the line Q R\ Let Q R^ intersect af in T\ then P^ T is the line of action of the reaction at P\ Hence the complete force polygon for the whole truss, when the right half is unloaded, may be found by drawing from F d. line parallel to Q R' , and from A a line parallel to P^ T , prolonging them till they intersect at V . The reaction at P^ is V A, and that at Q is FV . (The line of action of the latter is marked //'.) Next, consider the left half to be without loads, the other half being loaded. The resultant of the five loads now acting is FK, its line of action fk being drawn through the point of intersection of of and ok. The reactions at P^ and Q for the present case have lines of action P^ R^ and Q' T" , found just as in the first case of loading. These reactions are therefore determined in magnitude and direction by drawing from A' a line parallel to Q' T" and from F a line parallel to P'R[, pro- longing them till they intersect at L" . The complete force polygon for this case of loading is therefore FGHTJKL"F. THREE-HINGED ARCH. 85 Consider now that both parts of the truss are loaded. The reaction at P^ is the resultant of the two partial reactions V A, L"F, and the reaction at Q' is the resultant of the two partial reactions FL' , KL". From Z" and L' draw lines parallel respectively to FL' and FL" , intersecting in L. Then L" L is equal and parallel to FL' ^ and LL' is equal and parallel to L" F. Hence KL and LA represent the resultant reactions at Q' and P' respectively. This completes the polygon of external forces for the whole truss, as well as that for each partial truss. The stress diagram can now be drawn in the usual way, beginning at the point P' . The diagram for one partial truss is shown in Fig. 39 {B). If loads are applied at the lower joints of the trusses, the reactions due to these may be found in the same manner as for the upper loads. But before beginning the determination of the stresses, the polygon must be drawn for the external forces taken in order around the truss. (See Art. 90.) 107. Case of Symmetrical Loading. — If the two half trusses are exactly similar and symmetrically loaded, the determination of reactions and stresses is much simplified. (i) As regards the reactions, symmetry shows that the forces exerted by the two trusses upon each other at R' are horizontal. Hence, referring to Fig, 39, and considering either half-truss, as that to the left, the line of action of the reaction at P' may be found by drawing a horizontal line through R' and prolong- ing it to intersect af, the line of action of the resultant of all loads on the left truss ; the line joining this point of intersection with P' is the required line of action of the reaction at the left abutment. The two reactions are now determined by drawing from F 3. horizontal line and from A a line parallel to the line just determined, and prolonging them till they intersect. (2) As to the stresses, only one partial truss need be consid- ered, since the stress diagrams for the two portions will be symmetrical figures. S6 GRAPHIC STATICS. These principles might have been employed in the case dis- cussed in the preceding article ; but the method there used is applicable to cases in which neither the trusses nor the loads are symmetrical. 1 08. Wind Pressure Diagram. — The diagrams for wind pressure will present no difficulty. The determination of the reactions will, indeed, be simpler than in the preceding case, since only one partial truss will be loaded at any one time, and the line of action of one reaction is therefore known at the outset. The construction is shown in Fig. 40. Fig. 40. In computing wind pressure loads, it will be assumed that each joint sustains half the pressure coming upon each of the two adjacent panels. It will be sufficiently correct in comput- ing the pressure at any joint, to assume the slope of the roof THREE-HLNGED ARCH. 8/ as that of the tangent to the roof curve at the joint in ques- tion ; and the direction of the wind load may be taken as that of the normal to the roof curve at the joint. The force poly- gon for the wind loads is therefore not a straight line when the roof is curved. In Fig. 40, this polygon is FGHIJK. The reactions are to be marked KL, LF. (Since there are no loads on the other half-truss, the points ABCDEF will coincide.) Choosing a pole O, draw the funicular polygon as shown, and determine the line of action of the resultant of all the wind loads. This line of action//^ is drawn parallel to FK, through the point of intersection of the strings of and ok. Prolong//^ to intersect P^R^ produced at 7""; then Q'V is the line of action of the reaction at Q'. The force polygon may now be completed by drawing KL parallel to QT" and LF parallel to P^R\ The reactions KL and LF being thus determined, the stress diagram can be drawn without difficulty. The stress diagram is drawn for both partial trusses, with the result shown in the figure. If the two trusses are symmetrical, the diagram for the other direction of the wind need not be drawn ; the stresses for this case being found from the diagram already drawn. If, however, the partial trusses are dissimilar, a second wind-diagram must be drawn. It is not necessary to draw separate space diagrams for verti- cal loads and wind-forces. The constructions of Figs. 39 and 40 have been here kept wholly separate, in order that the explanation may be more easily followed. 109. Check by Method of Sections. — In case of a truss of long span, especially when the members have many different directions and are short compared with the whole length of the span, the small errors made in drawing the stress diagram are likely to accumulate so much as to vitiate the results. Thus, in Fig. 39, if, in drawing the stress diagram, we begin at P' and pass from joint to joint, there is no check upon the correctness of the work until the point R^ is reached. At that point we S8 GRAPHIC STATICS. have a second determination of the reaction exerted by each half-truss upon the other ; and it is quite hkely that the two vahies found will not agree. By a method like that employed in Art. lOO for the "inde- terminate" form of truss, we may avoid the necessity of mak- ing so long a construction before checking the results. In Fig. 39 take a section cutting the three members cq, qr, rl, and apply the principles of equilibrium to the body at the left of the section. The forces acting on this body are LA, AB, BC, CQ, QRy RL. Choose a pole O^ and draw the funic- ular polygon for this system, making the string o^c pass through the point of intersection of cq and qr. Draw successively the strings o^c^ o^b, o^a, o'l, prolonging the last to intersect /r. Through the point thus determined, draw ^'r, which must also pass through the point of intersection of cq and qr. As soon as o^r is known, the corresponding ray O^R can be drawn in the force diagram, and then by drawing from L a line parallel to Ir, the point R is determined. We may now close the force polygon, since the directions of the two remaining forces {cq and qr) are known. The stresses in the three members cut being now known, we have a check on the correctness of the construction of the stress diagram, as soon as these members are reached in the process. This method will be found of great use, not only for this form of truss, but for any truss containing many members. § 7. Coiinterbracing. no. Reversal of Stress. — If the loads supported by a truss are fixed in position and unchanging in amount, the stress in any member remains constant in magnitude and kind. But in most cases such are not the conditions, and it may happen that under different combinations of loading, the stress in a certain member is sometimes tension and sometimes compression. It COUNTERBRACING. 89 is often thought desirable to prevent such changes of stress, the design of the members and their connections being thereby simpHfied. To accomplish this is the object of counterbracing. III. Counterbracing. — Consider a truss such as the one shown in Fig. 41, subjected to vertical loads and to wind pres- sure from either side. A diagonal member such as xy may sustain tension under vertical loads alone, or with the wind from the left ; while with the wind from the right it may sustain compression. Now suppose xy removed, and a member represented by the broken line xy^ introduced. It may easily be shown that any 41 -^.. L system of loading which would cause compression in xy will cause tension in this new member ; and vice vei'sa. For, divide the truss by a section MN cutting xy and two chord mem- bers as shown, and let L be the point of intersection of the two chord members produced. The kind of stress in xy or the other member (whichever is assumed to be present) may be determined by considering the system of forces acting on one portion of the truss, as that to the left of the section MN. Let the principle of moments be applied to this system, the origin being taken at Z. If the external forces acting on the portion of the truss considered be such as to tend to cause right-handed rotation about L, the stress in xy must be com- pression in order to resist this tendency ; while, if xy be replaced by xy\ a tension must exist in that member to resist right- handed rotation about L. Similarly, a tendency of the external forces to produce left-handed rotation about L would be resisted by a tension in xy ; or by a compression in the member xy\ If the two members act at the same time, the stresses in 90 GRAPHIC STATICS. them will be indeterminate. These stresses may, however, be made determinate by the following device : Let the member xy be so constructed that it cannot sustain compression. Then, whenever the external forces are such as to tend to throw compression upon it, it ceases to act as a truss-member, and the member xy^ receives a tension which is determinate. If the member xy^ be constructed in the same way, any tendency to throw compression upon it causes it immediately to cease to act, and puts upon the member xy a tension instead. A member such as xy\ constructed in the manner mentioned, is called a coiinterbrace. 112. Determination of Stresses with Counter bracing. — The use of counterbracing adds somewhat to the labor of deter- mining the maximum stresses, since the members actually under stress are not always the same. The method of treating such cases will be illustrated in the next article by the solution of an example ; but first the main steps in the process may be outlined as follows : {(i) Construct separate stress diagrams for vertical loads and for wind in each direction, assuming the diagonals in all panels to slope the same way. ip) Determine by comparison of these diagrams in which of the diagonal members the resultant stress is ever liable to be a compression. Draw in counters to all such members. if) With these counterbraces substituted for the original members, either draw new stress diagrams, or make the neces- sary additions to those already drawn. If the latter method is adopted, the added lines should be inked in a different color from that employed originally. (In some cases, this construc- tion may be unnecessary on account of symmetry, as will be illustrated in the next article.) {d) Combine the separate stresses for maxima in the usual way. COUNTERBRACING. 91 113. Example. — In Fig.' 42 (PI. II) are given stress diagrams for a "bow-string" roof-truss in which counterbracing is em- ployed. At (A) is shown the truss or space diagram. The span is 48 ft. ; rise of top chord, 16 ft. ; rise of bottom chord, 8 ft. The chords are arcs of parabolas. The whole truss is divided into six panels by equidistant vertical members. The distance apart of trusses is taken as 12 ft. Assume the weight of the truss at 2400 lbs., and that of the roof at 3600 lbs. ; this makes the total permanent load 1000 lbs. per panel. Take 800 lbs. as the load at each upper joint, and 200 lbs. as the load at each lower joint. The snow load, computed at 15 lbs. per horizontal square foot, is about 1440 lbs. per panel. Wind pressure is to be computed from the formula of Art. 85. We now proceed to apply the method outlined in the preced- ing article. (a) Assuming one set of diagonals present, we construct the stress diagrams for the various kinds of loads. Diagram for permanent loads. — This is shown at {B) Fig. 42, which needs little explanation. It will be noticed that the stress in every diagonal member is zero. This will always be the case if the chords are parabolic and the vertical loads are equal and spaced at equal horizontal distances. Diagram for snow loads. — Fig. 42 {C) is the stress diagram for snow loads. In this case, also, the stresses in the diagonals are all zero. Wind diagrams. — At {D) and {E) are shown the diagrams for the two directions of the wind. The only thing needing special mention is the method used in laying off the force- polygon for the wind loads. We first compute the normal wind pressure on each panel by the formula of Art. 85. We thus find, when the wind is from the right, the following total pres- sures, taking /„ = 40 lbs. per sq. ft. : On panel d, 1650 lbs. On panel e, 3870 lbs. On panel/, 5480 lb& 92 GRAPHIC STATICS. In Fig. 42 {D) these are laid off successively in their proper directions to the assumed scale. Thus CD\ U E\ E'G repre- sent respectively 1650 lbs. normal to dq, 3870 lbs. normal to es, and 5480 lbs. normal to ft. Now each of these loads is to be equally divided between the two adjacent joints. Bisect CD' at D, UFJ at E, and EG at F\ then CD, DE, EF, EG represent the loads at the joints cd, de, ef, and fg respectively. The "load line" is therefore CDEFG. The reactions are assumed to be parallel to the resultant load. With this explanation the figures {D) and {E) will be readily understood. {b) Comparison of results. — The stresses due to permanent loads, wind right, and wind left, are shown in tabular form for convenience of comparison. Member. Perm. Load. Snow Load. Wind R. Wind L. Max. "J - 6660 - 9680 — 4250 — 14070 - 16340 bl -5350 -7780 - 4880 - 9650 - 13130 C7l -4550 — 6640 - 6380 - 6200 — 1 1 190 dq -4550 — 6640 - 9500 - 4250 — 14050* es -5350 - 7780 - 15030 - 3450 — 20380* fl -6660 -9680 - 14070 - 4300 — 20730* Jh + 5080 + 7400 + 780 + 13500 + 12480 kh. + 4680 + 6820 + 700 + 12450 + 1 1 500 mh^ + 4470 + 6520 -V 1950 + 7350 4- 10990 Ph + 4470 + 6520 + 4100 + 4100 + 10990* r/z., + 4680 + 6820 + 7680 + 2050 + 12360* th[ + 5080 + 7400 + 13500 + 800 + 18580* jk + 1 180 + 1440 + 150 + 2650 + 2620 Im + 1 180 + 1440 — 280 + 4100 + 2620 lip + 1 180 + 1440 — 950 + 3800 + 2620* qr + n8o + 1440 - 1650 + 2525 r + 2620* )^ I - 470* i r + 2620* ) St + 1 180 + 1440 1350 + 1250 I - 170* i kl + 1300 - 4600 -\- 1 300* mn + 2700 - 4100 4- 2700* pq. + 4850 - 3150 + 4850* rs + 7080 1950 + 7080* COUXTERBRACING. g. It is seen that the diagonals shown are all in tension when the wind is from the right, and all in compression when the wind is from the left. Therefore counters are needed in all panels, and the counters will all come in action whenever the wind is from the left. (c) Stresses in counterbraees. — It is evident from symmetry that no new diagrams are needed to determine the stresses in the counterbraees. In fact, the counterbrace in each panel is situated symmetrically to the main diagonal in another panel, and is subject to exactly equal stresses. {d) Coinbinatio7i for maxima. — In combining the results for the greatest stresses in the various members, it will be assumed that the greatest snow load and the greatest wind load can never act simultaneously. For each member, therefore, the stress due to permanent load is to be added to the snow stress or the wind stress, whichever is greater. Again, in combining the tabulated results, we consider only the columns headed per- manent load, snow load, and wind right ; since whenever the wind is from the left, the other system of diagonals is in action. This gives the results entered in the last column. We now notice the following facts : (i) The results given for the diagonal members are the true maximum stresses. (2) The stress found for each diagonal applies also to the symmetrically situated counterbrace. (3) For any other member, we are to choose between the maximum found for that member and the value found for the symmetrically situated member. Thus, —20730 is the true maximum stress for both <^y and // ; etc. (The numbers denot- ing true maximum stresses are marked with a (*) in the table.) It is seen that the verticals, with one exception, may sustain a reversal of stress. Thus, 7"/^ and st must be designed for a tension of 2620 lbs., and also for a compression of 170 lbs. ; while Im and qr are each liable to 2620 lbs. tension and to 470 lbs. compression. CHAPTER VI. SIMPLE BEAMS. § I. Genei'al Principles. 1 14. Classification of Beams. — A beam has been defined in Art. 79. Beams may be treated in two main classes, the basis of classification being that described in Art. 75. These two classes will be called simple and non-simple beams respectively. The present chapter deals only with simple beams, the defini- tion of which may be stated as follows : A simple beam is one so supported that it may be regarded as a rigid body in determining the reactions. A simple beam may rest on two supports at the ends ; or it may overhang one or both supports. A cantilever is any beam projecting beyond its supports. Such a beam may be either simple or non-simple. A continuous beam is one resting on more than two supports. Such a beam is non-simple. A beam may be supported in several ways. It is simply supported at a point when it rests against the support so that the reaction has a fixed direction. It is constrained at a point if so held that the tangent to the axis of the beam at that point must maintain a fixed direction. If hinged at a support, the reaction may have any direction. We shall deal mainly with the case of simple support. In what follows it will be assumed that the beam rests in a horizontal position, since this is the usual case. 11$. External Shear, Resisting Shear, and Shearing Stress. — The external sJiear at any section of a beam is the algebraic 94 GENERAL PRINCIPLES. 95 sum of the external vertical forces acting on the portion of the beam to the left of the section. The resisting sJiear at any section is the algebraic sum of the internal vertical forces in the section acting on the portion of the beam to the left, and exerted by the portion to the right of the section. The shearing stress at any section is the stress which con- sists of the internal vertical forces in the section, exerted by the two portions of the beam upon each other. It consists of the resisting shear and the reaction to it. (See Art. 63.) Let AB (Fig. 43) be a beam in equilibrium under the action of any external forces. At any point in its length, as C, con- ceive a plane to be passed perpen- dicular to the axis of the beam, and ^ ' — ^ consider the portion AC, to the left fi^. 43 of the section. The principles of equilibrium apply to the body AC, and the external forces acting upon it include, besides those forces to the left of C that are external to the whole bar, certain forces acting across the section at C that are internal to AB, but external to AC. (Art. 61.) These latter forces' com- prise that constituent of the internal stress between AC and CB which is exerted by CB upon AC Represent by Fthe algebraic sum of the resolved parts in the vertical direction of all forces acting on AB to the left of the section at C, upward forces being called positive. V is the external shear at the given section as above defined. Since the body AC is in equilibrium, condition (i), Art. 58, requires that the algebraic sum of the resolved parts in the vertical direction of all forces acting on it must equal zero. Hence the forces acting on AC in the section at C must have a vertical component equal to — /^ This vertical component is called the resisting shear in the given section. This resisting shear is one of the forces of a stress of which the other is an equal and opposite force exerted by ^67 upon CB. This stress is called the shearing stress in the section, and is called positive g6 GRAPHIC STATICS. when it resists a tendency of AC to move upward, and of CB to move downward. 1 1 6. Bending Moment, Resisting Moment, and Stress Moment. — The bending moment at any section of a beam is the algebraic sum of the moments of all the external forces acting on the portion of the beam to the left of the section ; the origin of moments being taken in the section. The resisting moment at any section is the algebraic sum of the moments of the internal forces in the section acting on the portion of the beam to the left, and exerted by the portion to the right of the section ; the origin of moments being the same as for bending moment. The stress moment or moment of internal stress at any section consists of the two equal and opposite moments of the forces exerted across the section by the two portions of the beam upon each other. Referring again to Fig. 43, let us analyze further the forces in the section at C. Applying to the body AC the second condition of equilibrium ((2) of Art. 58), and taking an origin at a point in the section, we see that the algebraic sum of the moments of all the external forces acting on the beam to the left of the section plus the sum of the moments of the internal forces act- ing on ^6^ in the section must equal zero. The former sum is defined as the bending moment at the given section. Represent it by M. The latter sum is defined as the resisting mome7it at the section, and must be equal to —M, by the above principle. We have thus far referred only to the internal forces exerted by CB upon AC\ but evidently the equal and opposite forces exerted by ^4 (7 upon CB have a moment numerically equal to M. The equal and opposite moments of the equal and opposite forces of the stress in the section together constitute the stress moment in the section. If the external forces applied to the beam are all vertical, the value of M will be the same at whatever point of the section GExNERAL PRINCIPLES. 97 the origin is taken ; since the arm of each force will be the same for all origins in the same vertical line. If the loads and reactions are not all vertical, the value of M will generally depend upon what point in the section is taken as the origin of moments. 117. Curves of Shear and Bending Moment. — The airve of shear for a beam is a curve whose abscissas are parallel to the axis of the beam, and whose ordinate at any point represents the external shear at the corresponding section of the beam. Let AB (Fig. 44) represent a beam loaded in any manner, and let A^B' be taken parallel to AB. At every point of AB^ suppose an ordinate drawn whose length shall represent the external shear at the corresponding section of AB. The line ab, join- ing the extremities of all these ordi- nates, is the shear curve. Positive values of the shear may be repre- sented by ordinates drawn upward from AB\ and negative values by ordinates drawn downward. (Instead of drawing AB^ parallel to the beam, it may be any other straight line whose extremi- ties are in vertical lines through A and B) The curve of bending mome^it for a beam is a curve whose abscissas are parallel to the axis of the beam, and whose ordinate at any point represents the bending moment at the correspond- ing section of the beam. Thus in Fig. 44, A"C'B" may repre- sent the bending moment curve for the beam AB. (Evidently A'B" might be inclined to the direction of AB, without destroy- ing the meaning of the curve.) Positive and negative values of the bending moment will be distinguished by drawing the ordinates representing the former upward and those represent- ing the latter downward from the line of reference A"B". 1 18. Moment Curve a Funicular Polygon. — If the loads and reactions upon the beam are all vertical, every funicular poly- gon for these forces taken consecutively along the beam, is a 98 GRAPHIC STATICS. curve of bending moments. Thus, let MN (Fig. 45) represent a beam under vertical loads, supported at the ends by vertical reactions. Draw a funicular polygon for the loads and reactions as shown. -_;:;^5>o Fis. 45 Now, by definition, the bending moment at any section is equal to the moment of the resultant of all external forces act- ing on the beam to the left of the section, the origin of moments being taken in the section. By Art. 56, this moment can be found by drawing through the section a vertical line and finding the distance intercepted on it by the two strings corre- sponding to the resultant mentioned ; the product of this inter- cept by the pole distance is the required moment. Hence, if oe (Fig. 45) is taken as axis of abscissas, the broken line made up of oa, oby oc, od, is a *' curve of bending moments." 119. Design of Beams. — The principles involved in the design of beams will not be here fully discussed. Every prob- lem in design involves the determination of shears and bend- ing moments throughout the beam ; and the graphic methods of determining these will alone be considered in the following articles. § 2. Beam Sustaining Fixed Loads. 120. Shear Curve for Beam Supported at Ends. — Let MN (Fig. 45) represent a beam supported at the ends and sustain- ing loads as shown. Draw the force polygon ABCD, and with pole O draw the funicular polygon. The closing line is oe BEAM SUSTAINING FIXED LOADS. 99 (marked also M^'N'^), and OE drawn parallel to J/'^V" fixes E, thus determining DE, EA, the reactions at the supports. Take M'N' as the axis of abscissas for the shear curve. The shear at every section can be at once taken from the force polygon. For, remembering the definition of external shear (Art. 1 1 5) we have : Shear at any section between J/ and P is EA (positive). Shear at any section between P and Q is EB (positive). Shear at any section between Q and R is EC (negative). Shear at any section between R and N is ED (negative). Hence the shear curve is the broken line drawn in the figure. 121. Moment Curve for Beam Supported at Ends. — As in Art. 118, it is seen that the funicular polygon already drawn (Fig. 45) is a bending moment curve for the given forces. For the bending moment at any section of the beam is equal to the corresponding ordinate of this polygon, multiplied by the pole distance. For simplicity, it will be well always to choose the pole so that the pole distance represents some simple number of force- units. The sign of the bending moment is readily seen to be nega- tive everywhere, according to the convention already adopted (Art. 47). 122. Shear and Moment Curves for Overhanging Beam. — Consider a beam such as shown in Fig. 46, supported at Q and T, and sustaining loads at M, P, R, S, and JV. This case may be treated just as the preceding, care being exercised to take all the external forces (loads and reactions) in order around the beam. The construction is shown in Fig. 46. First, the reactions at Q and T are found as in the preceding case, by drawing the funicular polygon, finding the closing line, and drawing OG parallel to it. The reactions are EG and GA. Then the value lOO GRAPHIC STATICS. of the external shear can be found for any section from the definition, and is always given in magnitude and sign by a cer- tain portion of the force polygon ABCDEFGA. The resulting curve is shown in the figure, M'N' being the line of reference. 4^ ^1^ d d\e M M' t f a\g js Tf~> 9\f M' nmri iiiii i iii i ii r N' N' :::::^^o Fig, 46 Similarly, from the definition of bending moment, and the principle of Art. 56, the bending moment at any section is equal to the ordinate of the funicular polygon multiplied by the pole distance. The polygon is shaded in the figure to indicate the ordinates in question. It is seen that the bending moment is positive at every section of the beam. 123. Distributed Loads. — In all the cases thus far discussed, the loads applied to the beam have been considered as concen- trated at a finite number of points. That is, it has been assumed that a finite load is applied to the beam ^t a point. Such a condition cannot strictly be realized, every load being in fact distributed along a small length of the beam. If the length along which a load is distributed is very small, no im- portant error results from considering it as applied at a point. In certain cases a beam may have to sustain a load which is distributed over a considerable part of its length, or over the whole length. Such a load may be treated graphically with sufficient correctness by dividing the length into parts, and BEAM SUSTAINING iMOVING LOADS. jqi assuming the whole load on any part to be concentrated at a point. The smaller these parts, the more nearly correct will the results be. It may be remarked by way of comparison that while algebraic methods are most readily applicable to the case of distributed loads, the reverse is true of graphic methods, which are most easily applied in the very cases in which analytic methods become most perplexing. 124. Design of Beam with Fixed Loads. — The above exam- ples are sufficient to explain the method of treating any simple beam under fixed loads. Under such loading the shear and bending moment at each section of the beam remain unchanged in value, and no further discussion is necessary as a preliminary to the design of the beam. We proceed next to the case of beams with moving loads. § 3. Beam Siistaming Moving Loads. 125. Curves of Maximum Shear and Moment. — When a beam sustains moving loads, the shear and moment at any section do not remain constant for all positions of the loads. In such a case it is the greatest shear or moment in each section that is to be used in designing the beam. A curve of maximtcin shear is a curve of which the ordinate at each point represents the greatest possible value of the shear at the corresponding section of the beam for any position of the loads. A curve of maximum moment is a curve whose ordinate at each point represents the greatest possible value of the bending moment at the corresponding section of the beam for any posi- tion of the loads. In the following articles will be explained a method of deter- mining any number of points of the curves just defined, in the case of a simple beam supported at the ends. The moving I02 GRAPHIC STATICS. load will be taken to consist of a series of concentrated loads with lines of actions at fixed distances apart. An example of such a load is the weight of a locomotive and train ; the points of application of the loads being always under the several wheels. 126. Position of Loads for Greatest Shear at Any Section. — In Fig. 47 (PI. Ill) let the vertical lines ab^ be, etc., represent the lines of action of the moving loads in their true relative positions ; the magnitudes of the loads being shown in the force polygon or "load line" ABCD . . . Let XY represent the length of the beam, and let it be required to determine the position of the moving loads that will cause the maximum posi- tive shear at any section, as at that marked /. First, consider the effect of a single load in any position. It is seen that any load to the right of y produces at that point a positive shear. For, by definition, the external shear is equal to the algebraic sum of all vertical forces to the left of the section, upward forces being reckoned positive. Now a load to the right of ^ causes no forces to act at the left of this section, except an upward reaction at X. On the other hand, a load to the left of y produces negative shear at that point. For the shear due to it is equal to the reaction at X minus the load itself, which will always be negative, since the load is greater than the reaction due to it. (It is also to be noticed that the shear due to a load at the left of the section is equal numerically to the reaction it produces at V.) Again, the shear due to any load is greater in magnitude, the nearer the load is to the section considered. For, as a load on J^V approaches y, the reaction at X increases ; and as a load on J(/ approaches y, the reaction at F increases. Two principles are thus reached to guide in assigning the position of the loads that will produce the greatest positive shear at any section : (i) The segment of the beam to the left of the section should be without load and that to the right fully BEAM SUSTAINING MOVING LOADS. 103 loaded ; and (2) the loads on the loaded segment should be as near the section as possible. It would seem, therefore, that the greatest positive shear at the point y will occur when the loads are brought on the beam from the right to such a position that the foremost load in the series is just at the right of J. And, in general, this will be true, unless the foremost load is small in comparison with the whole load on the beam, or unless the distance between the first and second loads is considerable. In such cases it may be that a greater shear will result when the second load is brought up to the section. For, suppose the first load to be just at the right of J, and consider the effect of moving the whole series of loads to the left. As the first load passes the section, it produces a diminution of the shear equal in amount to the load. On the other hand, as the loads continue to move to the left, the effect of every load in producing reaction at X increases, while no further decrease in the shear at J occurs until the second load passes the section. Now, it may be that the net result of bringing the second load up to J will be to increase the shear at that section. If the first two or more loads are very small, it may be that the third or fourth load should be brought to the section to produce the greatest shear. Usually, however, it will be necessary to try only two (or at most three) positions. Instead of resorting to trials to determine for which position the shear is greatest, we may apply a simple rule which will now be deduced. Let /*i = magnitude of foremost load, /'.j^niagnitude of sec- ond load, etc., W being the total load on the beam. Let /= total span, and 7;^ = distance between Px and P.. Let -t' = dis- tance from Y to the center of gravity of W when P^ is at a given section. Let us compute the shear when P^ is just at the right of the section, and then determine the effect of moving all loads to the left, until the second load comes to the section. When P is at the right of the section, the shear is equal to I04 GRAPHIC STATICS. the reaction at X, say R. Then (calHng V the external shear) we have If now the loads be moved until P.y_ comes to a point just at the right of the section, the reaction due to W becomes W- -, and the shear at the section becomes The increase of the shear is, therefore, W{x-\-m) r> Wx Win r, T ^'~^-^~^'- This increase is plus or minus according as is greater or P I less than P^ ; that is, as ^Fis greater or less than — ^. Hence ni the following rule : The maxinittin positive shear in any section of the beam occurs when the foremost load is infinitely near the section, provided W PI PI is not greater than ^-. If W is greater tJian —^, the greatest m m shear will occur zvhen some succeeding load is at tJie section. In the above discussion it has been assumed that in bringing Pi up to the section no additional loads are brought upon the beam. If this assumption is not true, let W^ be the new load brought on the beam, and x^ the distance of its center of gravity from the right support when P^ is at the section. Then the shear corresponding to this position is and. the increase of shear due to the assumed change in the position of the load is Wm , W^x' o BEAM SUSTAINING MOVING LOADS. 105 Hence, in the statement of the above rule, we have only to substitute Wm+W'x' ior IVm; or, W^^-^ for W. The in additional load W^ may be neglected except when the condition P I W = — - is nearly satisfied; for the term Wx' will always be small in comparison with IVm. The application of this rule is very easy, and will save much labor in the graphic construction of the shear curve. If it is found that the first load should be past the section for the position of greatest shear, we may determine whether the second or the third should be at the section by an exactly similar method. We have only to apply the above rule, substi- tuting P2 for Pi, and for w the distance between P and P^. 127. Determination of Maximum Shear. — The shear at any section due to any position of the moving loads can readily be determined from the force and funicular polygons, by a method which will now be shown. Draw in succession the lines of action of the loads at their proper distances apart, as in Fig. 47 (PI. III). Draw the load- line ABCD . . ., and choosing a pole O^ draw the funicular polygon. The same space diagram may be used for any position of the moving loads ; for, instead of moving the loads in either direction, we may assume the loads fixed and move the beam in the opposite direction. By applying the rule deduced in Art. 126, it may be shown that there is a point Z in the beam, such that for any section to the right of Z the foremost load should be at the section in order that the shear may be a maximum ; while for any section to the left of Z the second load must be brought to the section. The position of this point is readily determined, as follows : In the case shown in Fig. 47, we have /=64ft., m = S.i ft., 7^1 = 8000 lbs., ^=-^ x 8000 = 63200. m 8.1 Hence, as the load moves from right to left, W is less than I06 GRAPHIC STATICS. PI. — — until the whole load on the beam reaches 63200 lbs. This VI occurs when the fifth load is at Y. Hence the point Z is located at a distance from F equal to the distance between the lines of action of the first and fifth loads. For any point to the left of Z, the greatest shear will prob- ably occur when the second load is at the section. To test whether in any case the third load should be at the section, we apply the same principle, as follows : The second load is 1 5000 lbs. ; the distance between the second and third loads is 5.8 ft. Hence, for greatest shear, the third load should not be at the section unless the whole load on the beam is at least -^x 15000 lbs. or 165500 lbs. But it is easily seen that the total load can never be so great. We may now explain the construction for finding the great- est shear at any section. Consider a section at the left of Z, as at the point marked y! From the above reasoning it is evi- dent that the second load must be brought close to the section. In Fig. 47, SjSj is drawn to represent the beam, its position being such that the point J is infinitely near bc^ the line of action of the second load. Through the extremities of the beam S^Sj draw vertical lines to represent the action-lines of the reactions at the supports. The portion AB^ of the load line represents the loads now on the beam, and the correspond- ing extreme strings of the funicular polygon are oa and ob\ If now we consider the funicular polygon for the loads and reac- tions acting on the beam in the supposed position, we see that the closing side is found by joining the points in which the action-lines of the end reactions intersect oa and ob^ respec- tively. This is marked of in the figure. Now draw in the force diagram a ray parallel to this closing line, and lety be its point of intersection with the load line; then B^J^ smd/'A are the two reactions. The shear at the point _/ is evidently the algebraic sum of the reaction /'A and the load AB ; hence it BEAM SUSTAINING MOVING LOADS. 107 is represented hyJ'B. In Fig. 47 let X, K- be drawn to repre- sent the length of the beam, and let the ordinates of the shear curve be drawn from it. Then from y we lay off an ordinate equal toJ^B. The whole curve of maximum shear is shown in the figure, but the construction for finding it is given for only one point. 128. Shear Curve for Combined Fixed and Moving Loads. — Let the beam sustain a fixed load of 25000 lbs. uniformly dis- tributed along the beam. The shear close to the left support due to this load is equal to the reaction, or 12500 lbs. ; and decreases as we pass to the right by ^^-^-- lbs. for each foot. At the middle of the beam the shear is zero, and at the right support it is —12500 lbs. Hence the shear curve is a straight line, and may be drawn as follows : From X^ lay off an ordinate downward representing 12500 lbs., and from Ys an ordinate upward representing 12500 lbs.; the straight line joining the extremities of these ordinates is the shear curve for the fixed load. Positive shears are laid off downward and negative shears upward for the reason that, if this be done, the greatest positive shear at any point due to fixed and moving loads is represented by the total ordinate measured between the shear curves for fixed loads and for moving loads. It is seen that at a certain point somewhere to the right of the center of the beam this greatest shear is zero, and for all sections to the right of this point it is negative. This point is determined by the intersection of the two shear curves. 1 29. Position of Loads for Greatest Bending Moment. — Referring to the beam A^F shown in Fig. 47 (PL III), consider the bending moment at any section (as^) due to a load any- where on the beam. First, suppose the load is at the right of the section. In this case the only force brought upon the por- tion of the beam to the left of the section is a reaction at X. By definition (Art. 116) the bending moment atyis the moment of this reaction about an origin aty, and is negative. Moreover, ^ I08 GRAPHIC STATICS. since the reaction will become greater as the load moves toward the left, the greatest bending moment due to a load at the right of the section will occur when the load is as near the section as possible. Next consider a load at the left of ihe section. The bending moment due to it is equal to the moment of the result- ant of the load and the left reaction. But this resultant is equal and opposite to the reaction at the right support, and has the same line of action (because the load and the two reactions due to it are in equilibrium) ; hence the bending moment is the negative of the moment of the right reaction about an origin in the section. The bending moment is therefore negative, and is greatest when the right reaction is greatest. Hence, a load at the left of the section produces its greatest bending moment when the load is as near as possible to the section. We are therefore led to the following general principles for a simple beam supported at the ends : (i) The bending moment at every section is negative for all positions of the moving loads. (2) The negative bending moment at any section has its greatest value when the whole beam is loaded as completely as possible, and the heaviest loads are near the section. These principles serve as a general guide, but unless the moving loads are equidistant and equal in magnitude it will be necessary to try several positions before the position for great- est bending moment will be known. Usually one of the heaviest loads should be directly at the section. Instead of resorting to repeated trials, the position of loads giving greatest bending moment at any section may be found by means of a simple rule which will now be deduced. Let /= length of beam between supports; JV= total load on beam; ;tr— distance of center of gravity of W from right sup- port; J'Fi^: load on beam to left of given section; ,r = distance of center of gravity of Wi from section; /i=:distance of sec- tion from left support ; R = reaction at left support ; M= bending moment at the section. Then we have BEAM SUSTAINING MOVING LOADS. 109 IV/ M= Rk - W,x\ = ^x- IVrX'i, If the loads all move an infinitesimal distance to the left, then X and Xi receive equal infinitesimal increments, and the increment of Mis dM= ^dx- IV.dx, = (^ - W\ dx (since dx^ = dx). Now if M is a maximum, we must have = 0; that IS, i— lVi = o; or, -— ^ = — -. Hence the follow- dx I l\ I ing rule : Whe7i the bending moment at any section of the beam has its greatest value, the loads oji the two segments of the bea^n are to each other as the lengths of the segments. In case of concentrated loads, the condition just stated can generally not be exactly satisfied except for certain sections of the beam. It will usually be found that the position most nearly fulfilling the condition is that in which some heavy load is just at the section. This will be illustrated in the next article. [Note. — The above reasoning applies only to the case of concentrated loads, and is not rigorous for this case, if, in the position of maximum moment for any section, a load is at the section ; for J will not then be zero, as assumed above. In such a case -^— is discontinuous, and its value changes sign as the loads pass through the TAT" 1 /f/^ position for which — ^ =0. It is still true that this condition is satisfied in the position of maximum moment, provided the load at the section is regarded as divided in some certain ratio between the two segments of the beam.] 130. Determination of Bending Moments. — The application of the above principles and the method of determining the bending moment at any section of the beam will now be shown for the case represented in Fig. 47 (PI. III). Let the greatest bending moment be found for the section J. From the general no GRAPHIC STATICS. principles deduced in Art. 129, it is probable that the loads nearest the section should be the third and fourth in the series. Also, since the point _/ divides the span into segments of 16 ft. and 48 ft., the load on the left segment should be one-fourth the total load on the beam. Now it is readily found that when the load CD is just at the right of the section, the whole load on the beam is 112,000 lbs., while the load on the left segment is 23,000 lbs., which is less than one-fourth of 112,000. And when the load CD is just at the left of the section, the load on the left segment is 38,000 lbs., which is greater than one-fourth of 112,000. Hence the bending moment is a maximum when the load CD is just at the section. In Fig. 47, MjMj represents the beam for this position of the loads, and the closing line 01 the funicular polygon is marked oj" . Using the principle of Art. 56, we find the bending mo- ment at the given section by measuring the distance intercepted on ^^by the strings od and oj" and multiplying it by the pole distance in the force diagram. Let Xra Vm (Fig- 4/) be the axis of abscissas for the curve of maximum moments. From the point y draw an ordinate equal to the intercept just found ; this locates a point of the required moment curve. Other points may be determined in the same manner. The curve is shown in the figure, but the construction is not given for any section except that at /. It must be remembered that each ordinate is to be multiplied by the pole distance. Hence it will be convenient to choose the pole distance equal to some round number of force units. 131. Moment Curve for Fixed Loads. — The greatest bending moment due to moving loads must be combined with the bend- ing moment due to fixed loads. If the fixed load is uniformly distributed, as already assumed in the computation of shear, it may be divided into parts, each assumed concentrated at its center of gravity, and a funicular polygon drawn, using the BEAM SUSTAINING MOVING LOADS. m same pole distance employed in the force diagram for moving loads. The ordinates of this funicular polygon may be laid off upward from the line X^Vm, and their ends joined to form a continuous curve. The total ordinate from this curve to that already drawn for moving loads represents the true greatest bending moment at the corresponding section of the beam. The curve is shown in the figure, but the construction is omitted, since it involves no new principle. It is to be remembered that the bending moment found for any section is a possible value for the other section equally distant from the center of the beam, since the train may be headed in the opposite direction, and the same construction made, viewing the beam from the other side. The same state- ment holds as to shears. 132. Design of Beam sustaining Moving Loads. — In designing a beam to sustain moving loads, the greatest shear and bending moment that, can come upon it for any position of the loads must be known for every section. The methods above given are sufficient for the determination of these quantities ; and the problem of designing the beam will not be here further discussed. 133. Plate Girders and Lattice Girders. — A p/ate girder is a beam built up of rolled plates and angle-irons riveted together, the cross-section being as shown in Fig. 48. If latticing is substituted for the plate, the beam be- ^ comes a lattice girder. Railway bridges of spans under 100 ft. frequently employ either rolled beams, plate girders, or lattice girders. (See Cooper's "■ Specifications for Iron and Steel Railroad Bridges.") The methods given in the preceding articles are especially useful in designing this class of bridges. cM=i The student should make the complete construe- fis-^s tion for determining the curves of maximum shear and bending moment for a beam designed to carry the series of moving loads shown in Fig. 47. CHAPTER VII. TRUSSES SUSTAINING MOVING LOADS. § I. Bridge Loads. 1 34. General Statement. — The most important class of trusses sustaining moving loads is that of bridge trusses. The two main classes of bridges are highway bridges and railway bridges. The forms of trusses most commonly used differ for the two classes, as do also the amount and distribution of loads. Before the design can be correctly made, the weights of the trusses themselves must be known. Since these weights de- pend upon the dimensions of the truss members, they cannot be known with certainty until the design is completed. The remarks made in Art. 82 regarding the design of roof trusses are here applicable. In the following articles we shall give data available for preliminary estimates of truss weights. As stated in Art. 133, railway bridges of short span are fre- quently supported by rolled or built beams. When the span is longer than 100 ft., trusses should be used. (Cooper's "Speci- fications.") 135. Loads on Highway Bridges. — (i) Permanent load. — The permanent load sustained by a highway bridge truss in- cludes the weight of the truss itself, of the lateral or '^ sway " bracing, of the floor and the beams and stringers supporting it. These weights are all subject to much variation, but, for pur- poses of preliminary design, the following formula, taken from Merriman's ''Roofs and Bridges," may be used. 112 BRIDGE LOADS. 113 Let «; = weight of bridge in pounds per linear foot ; /=:span in feet ; (^ = width in feet. Then w=\^o-^\2 b^o.2 bl—o.^ I. (2) Snow load. — The weight of snow may be taken as in case of roof trusses (Art. 84). The values there given are probably in excess of those ordinarily employed in practice. (3) Wind load. — The pressure of wind striking the bridge laterally is resisted by the chord members together with the lateral bracing. These constitute horizontal trusses, in which the stresses are to be found in the same way as for the main trusses of the bridge. As the determination of wind stresses requires the use of no special methods or principles, they will not be here considered. The student is referred to Burr's " Stresses in Roofs and Bridges," Merriman's " Roofs and Bridges," and other available works for a complete discussion of wind pressure and its effects on bridge trusses. (4) Moving load. — The most dangerous moving load for a highway bridge is usually a crowd of people or a drove of animals. This is commonly taken as a uniformly distributed load, which may cover the whole bridge or any portion of it. Its value is variously taken at from 60 lbs. to 100 lbs. per square foot of area of floor, depending upon the span and upon local conditions. It may be that in certain cases the greatest stresses will result from the passage of heavy pieces of machinery over the bridge, as, for example, a steam road roller. This should of course be considered in the design. For a complete discussion of loads on highway bridges, the student is referred to Waddell'c '' Highway Bridges." 136. Loads on Railway Bridges. — (i) Permanent load. — The permanent load on a railway bridge includes {a) the weight of the track system, which is known or may be determined at the outset ; {b) the weight of longitudinal stringers and cross- 114 GRAPHIC STATICS. beams, which can be determined before the trusses are designed ; and {c) the weights of trusses. The weight of the track system may be taken at 400 lbs. per linear foot for a single track. (See Burr's " Stresses in Bridge and Roof Trusses " ; Cooper's '' Specifications for Iron and Steel Railroad Bridges " ; Merri- man's '' Roofs and Bridges.") The total weight of track system and supporting beams and stringers varies from 450 lbs. to 600 lbs. per linear foot. (Merriman.) For spans less than 100 feet, Merriman gives the following formulas, in which zv is the total dead load of the bridge in pounds per linear foot, and / is the span in feet : For single track, w= 560+ 5.6 /. For double track, w= 1070-r 10.7 /. See also Art. 8 of Burr's work above cited. ^ (2) Snow and zvind. — Railway bridges usually offer little opportunity for the accumulation of snow. Wind pressure is, however, an important factor. Besides the pressure upon the bridge itself, the pressure upon trains crossing the bridge must be considered. The latter is a moving load and may be dealt with in the same way as other moving loads. Its amount may be computed from the area of the exposed surface of the train and the known (or assumed) greatest pressure due to wind striking a vertical surface (Art. 85). For further discussion of wind pressure, the student is referred to the works already cited. The graphic methods of deter- mining stresses due to wind will be evident when the methods for vertical loads given in the following articles are understood. (3) Moving loads. — The moving load to be supported by a bridge consists of the weights of trains. Such a load is applied to the track at a series of points, namely, the points of contact of the wheels. But the load is applied to the trusses only at the points at which the floor beams are supported. Hence the actual distribution of loads upon the truss is somewhat com- plex. It was formerly common to substitute for the actual TRUSS REGARDED AS A BEAM. 115 load a uniformly distributed load, thus simplifying the problem of determining stresses. It is now more usual to consider the actual distribution of loads for some standard type of locomo- tive used by the railroad concerned, or specified by its engi- neers. For examples of such distributions the student is referred to Cooper's " Specifications " already cited ; also to Fig. 47, and to the following portions of this chapter. 137. Through and Deck Bridges. — A bridge is called throiigJi or deck according as the floor system is supported at points of the lower or of the upper chord. In the former case, if the trusses are too low to require lateral bracing above, they are called /d?wj/ trusses. The weight of the truss itself is to be divided between the upper and lower joints. But the weight of the floor system and of the supporting beams and stringers comes wholly at the lower joints of a through bridge, or at the upper joints of a deck bridge. The moving load is, of course, applied at the same joints at which the floor system is supported. If the floor system is supported directly upon the upper chord, as is sometimes the case, the moving load and part of the dead load produce bending in the chord members ; other- wise the design is unaffected by this construction. § 2. Truss Regarded as a Beam. 138. Classification of Trusses. — Since a bridge truss acts as a practically rigid body resting on supports at the ends or other points and sustaining vertical loads, it may be regarded as a beam, and trusses may be classified in the same way as beams (Art. 114). The only class to be here considered is that of simple trusses, or such as may be regarded as rigid bodies for the purpose of determining the reactions. Cantilever trusses and continuous trusses are defined like the corresponding classes of beams (Art. 114). The most impor- tant case is that of a truss simply supported at the ends. ii6 GRAPHIC STATICS'. 139. External Shear for a Truss. — If a truss be regarded as a beam, the external shear, resisting shear, and internal shear- ing stress at any section may be defined just as in Art. 115. In some forms of truss a knowledge of the external shear at any section makes it possible to compute readily the stresses in certain truss members. Thus, in the portion of a truss repre- sented in Fig. 49, let the section MN cut three members, of which two are horizontal. (The member x^y is disregarded.) Since each member can exert forces only in the direction of its length, the ex- ternal shear in the section MN must be wholly resisted by the di- agonal xy ; and the internal force in xy must be such that its resolved part in the vertical direction is equal in magnitude to the exter- nal shear in the section. Let the external shear V be repre- sented by YZ (Fig. 49) ; draw YX parallel to yx and ZX horizontal; then XFwill represent the stress in xy. If F is positive (Art. 115), the stress in xy is a tension. If J^is nega- tive, the stress is a compression. If the member xy were replaced by one sloping the other way from the vertical, these statements as to kind of stress would be reversed. If no two of the three members cut by any section are par- allel, the stresses cannot be computed so simply, since all may contribute components of force to resist the external shear. Fig. 49 140. Bending Moment for a Truss. — In many cases the stresses in the truss members can be found from the values of the bending moment at different sections. Thus, in Fig. 49, let a section MN be taken cutting three members as shown, and let the origin of moments be taken at the point of intersection of two of them (as bx and xy) ; then since the moments of the internal forces in these two will be PARALLEL CHORDS — CONCENTRATED LOADS. 117 zero, the resisting moment is equal to the moment of the inter- nal force in the third member ay. The arm of this latter force is the perpendicular distance of its action line from the origin. Call it k, and let/ represent the force itself, and i^^the bending moment. Then, numerically, h If M is positive (in accordance with the convention of Art. 47), / must act from right to left, hence the stress in ay is a compression. If M is negative, the stress is a tension. (It must be remembered that the forces whose moments make up the bending moment M act upon the portion of the truss to the left of the section.) This method applies equally well to the case in which no two of the members cut are parallel. It is an application of a prin- ciple of much importance in the determination of stresses in truss members. § 3. Truss zvitJi Parallel CJiords siistaming Concentrated Loads. 141. General Method. — It was shown in Art. 139 that when both chords of the truss are horizontal, so that a vertical sec- tion through the truss will cut but one inclined member, the stress in such an inclined member may be easily found from the external shear in the section. Hence for such a truss, we may determine the greatest stresses in the web members by con- structing the curve of maximum shear for the truss, in a man- ner similar to that employed in Art. 127 for a beam. Again, the curve of maximum bending moments will usually enable us to determine the greatest chord-stresses (Art. 140). The construction of these curves for the case of a truss is not quite so simple as in case of a beam, for reasons to be explained in the following article. Il8 GRAPHIC STATICS. 142. Distribution of Loads on Truss. — If the curves of maxi- mum shear and moment for a given series of concentrated loads be constructed as in Arts. 127 and 130, they will not represent correctly the maximum shear and moment at all points of the truss. The reason for this will be seen by considering the way in which the loads are actually applied to the truss. The road- way of the bridge is usually supported by longitudinal beams, which themselves rest upon cross-beams or girders, the latter being supported at the joints of the truss. Hence, however the loads may be distributed upon the roadway, they are borne by the truss only at the joints. But in the method used in Arts. 127 and 130 for drawing curves of shear and bending moment, it was assumed that every load rests directly upon the beam, at a point immediately under the load. In the case of the truss, since loads are applied only at the joints, the shear at any instant is the same throughout the length of a panel. What is needed, therefore, is the greatest possible shear for each panel of the truss. For simphcity, we shall consider first the shear due to dead loads, and then that due to the live loads, taking a numerical example. 143. Problem — Numerical Data. — Assume a ''through" truss of the form shown in skeleton in Fig. 50 (PI. IV) ; let the span be 96 ft., the truss being divided into 6 panels of 16 ft. each. The depth is taken at 15 ft. Assume the dead load as 1 100 lbs. per linear foot for the whole bridge, — one-half being borne by each truss, — of which one-third is the weight of the floor system and track, and is therefore borne by the lower chord, and the remaining two-thirds is assumed to be divided equally between the upper and lower chords. We have then for the dead loads : Lower panel load = 5 866 lbs., say 6000 lbs. Upper panel load = 2933 lbs., say 3000 lbs. For the moving load we will " assume a train drawn by two locomotives with weights distributed as shown in Fig. 50. In PARALLEL CHORDS —CONCENTRATED LOADS. 119 computing stresses the train will be supposed to come on the bridge from the right. 144. Shear Curve for Fixed Loads. — The curve of shears for the dead load may be constructed as follows : In Fig. 50 (PL IV), the line X,V, represents the axis of abscissas for the shear curve. The reaction at the left support, minus the dead load borne at that point, gives 24000 lbs. as the positive shear for every section between the support and the first upper panel joint. This is laid off downward from X,V,. Then passing to the right, as each load is passed, its amount must be sub- tracted from the shear. The result is the broken line X' V shown in the figure. 145. Moment Curve for Fixed Loads. — To construct the bending moment curve for dead loads, we have only to draw a funicular polygon for these loads and the reactions due to them ; the vertical ordinate at any section will represent the bending moment at that section. The actual value of the bending moment is the product of the ordinate into the pole distance ; the former being measured in linear units, the latter in force units. The curve is shown in Fig. 50 (PI. IV), the ordinates being laid off upward from the line X^ Vm- The force polygon is shown at IV, the pole being O, taken opposite the middle point of the load-line in order that the closing side of the funic- ular polygon may be horizontal, and can therefore be made to coincide with X^Y^. The pole distance represents 120,000 lbs. 146. Actual Shear for Given Position of Moving Loads. — The shear in any panel is readily found when the position of live loads is known. We have only to apply the definition of external shear (Art. 115) as the sum of all vertical forces acting on the portion of the truss to the left of the section con- sidered. That is, we must find the reaction at the left abut- ment and subtract from it the sum of the loads between the left abutment and the section. It must be noticed that any I20 GRAPHIC STATICS. load on the panel considered is divided between the two adja- cent joints; the part supported at the left end of the panel is to be subtracted in computing the shear. The graphic con- struction is explained in the following articles. [Note. — It will be noticed that the actual distribution of any load among the different panel points is not strictly determinate by methods heretofore treated, if the longitudinal beams supporting the floor system are continuous (supported at more than two points. See Art. 114). We shall assume the portion of such a beam between two supports to act as a simple beam. The results of this assumption are probably as reliable as could be obtained by a more elaborate discussion.] 147. Position of Moving Load for Maximum Shear in Any Panel. — The position of the moving load which will give the maximum shear in any panel may be determined graphically by repeated trials. But it is possible to deduce a simple rule which will shorten the labor very materially. It must first be noticed that the general principles stated in Art. 126, in treating of beams, are equally applicable here. We know, therefore, in a general way, that in order to get the greatest positive shear in a given panel, the portion of the truss to the right should be loaded as completely as possible, while the part to the left should be free from loads. This rule is not, however, absolute. In general the foremost loads will be upon the panel in question, or possibly to the left of it, in the posi- tion of greatest shear. We shall now prove the following Proposition. — When the shear in any panel has its greatest value, the load upon that panel is to the total load on the truss as the length of the panel is to the length of the truss. Let XF(Fig. 51) represent the length of the truss ( = /), and CD ( = /') the length of the panel. Let W — total load on truss, and W^ — total load on the panel. Consider the effect of moving all the loads to the left. Evidently the reaction at X is increased and the load at C is increased. The former increases the shear in the panel, while the latter decreases it. These are the only changes which are caused in the shear by the movement of the load. Now if the position of loads is PARALLEL CHORDS — CONCENTRATED LOADS. 121 such that the shear is a maximum, the two effects mentioned must just neutrahze each other for a very small displacement of the loads. (Otherwise a small displacement in one direction or the other would increase the shear.) Let X — distance of center of gravity of W from Fand x^ = distance of center of gravity of W^ from D. Let R — reaction at X and P = load at C due to IV. Then 'X X If the whole load moves any distance, x and x^ both increase by the same amount. For an infinitesimal displacement. dR = — dx\ dP — dx = — — ax. i' /' Now the increment of the shear is dR — dP or W W'\ r dx. I V ) Since this must be zero (as stated above), we have W W or /' / W /' for the position of maximum shear. The proposition is there- fore proved. The condition may be written W=—W\ If the truss is ... I divided into n equal panels, then — = n, and W=ii W. This principle is so simple that it is easier of application than the graphic method of repeated trials. 122 GRAPHIC STATICS. [Note. — In the above discussion it is assumed that IV and JV both remain con- stant during the movement c/jc; that is, that no load enters or leaves the panel or the whole truss. Hence the reasoning is not rigorous for a case in which the condition IV = — W requires a load to be at Cor D, or at X or Y. The remark made in the note in Art. 129 applies here. The condition W = ~ W may always be applied pro- vided a load at any joint may be treated as if divided between the adjacent panels in any desired ratio.] 148. Construction of Curve of Maximum Shear. — In Fig. 50 (PI. IV) are shown the lines of action of the series of moving loads to be carried by the truss. The force polygon or load line is drawn for the whole series of loads, being the line ABC. . . . For the purpose of drawing the funicular polygon, the uniformly distributed load is divided into parts of 6000 lbs., each assumed to act at its center of gravity. With a pole O (the pole distance being taken so as to represent 120000 lbs.) the funicular polygon is drawn for the series of loads. Let us now consider the greatest shear in any panel of the truss, as in KL (Fig. 50). Applying the principle already deduced, it is easily shown that the greatest shear occurs when the load CD is at the point Z. For, suppose CD is just at the left of L ; the total load on the panel is then 38000 lbs., while the total load on the truss is 172000 lbs. Now the ratio of the whole span to the panel length I—, Art. 147) is 6; and since 38000 lbs. is greater than one-sixth of 172000 lbs., the load on the panel is too great to satisfy the required condition. Again, suppose CD is just at the right of L ; the total load on the panel is 23000 lbs., which is less than one-sixth of 172000; hence the load on the panel is too small to satisfy the condition for maximum shear. We con- clude, therefore, that the shear is a maximum when the load CD is just at the point L. In Fig. 50, S2S2 represents the beam with load CD at L. Drawing vertical lines through the ends of vS2^'2, the points in which these intersect the strings oa and of determine two points of the closing string {0/2) of the funicular polygon for the loads and reactions now acting on the PARALLEL CHORDS — CONCENTRATED LOADS. 123 truss. The ray OJ2 is drawn parallel to this closing line, inter- secting the load line in J^. This determines J2A as the left reaction. To get the shear in the panel, we must subtract from this reaction the portion of the two loads AB, BC, which comes upon the truss at K. This portion is found by treating KL as a simple beam and determining the reactions at K and L due to loads AB and BC. Draw vertical lines through the points K and L of the truss (in the position S^S^ ; the points in which these verticals intersect the strings oa and oc are two points of the closing side of the funicular polygon for the loads and reactions on the beam KL. Parallel to this closing side draw a ray OJ2. This line divides the load line AC into two segments representing the portions of AB and BC supported on the truss at K and L respectively. Hence the shear in the panel is the reaction JoA minus the load represented by the line AJ2. Taking X, K (Fig. 50) as the axis of abscissas for the curve of shears, the portion of the curve of maximum shear corre- sponding to the panel KL is a straight line parallel to X^ Y„ at a distance from it equal to J^Ji. The curve of greatest shear for the whole truss is shown in the figure, being the broken line X" I^,, but the construction is given only for the panel KL. For other panels the method will be exactly similar. 149. Position of Load for Maximum Bending Moment. — In computing stresses in the chord members, the greatest bending moment is needed for each vertical section containing a chord joint. In considering the position of moving loads which will give the greatest bending moment at any section, the sections through joints carrying live loads must be treated separately from those through joints free from loads. Let Fig. 51 represent a through truss (Art. 137), and con- sider first the bending moment at a section containing one of the lower joints, as C It is readily seen that for such a 124 GRAPHIC STATICS. section the bending moment due to a load anywhere on the truss is exactly the same as if the truss were a beam upon which the load was supported directly. For, consider a load on any panel as EC. Though the load is actually divided between B and C, the reaction it causes at X is the same as if the truss were a beam supporting the load directly. Also, since the load is the resultant of the two parts at B and C, the sum of the moments of these two parts about any origin is equal to the moment of the load itself. Now, in computing bending moment caused by the given load at any section, we take the moment of the left reaction minus the part of the load which is carried at the left of the section. If the section is anywhere except betzveen B and C, the result is the same whether we use the given load in its actual position, or its two components at B and C. For finding the greatest bending moments at the sections A, B, C, etc., we may therefore apply the same rule as in the case of a beam (Art. 129). That is, when the bending moment for such a section is a maximum, the loads on the two segments of the truss are proportioned to the lengths of the segments. Next consider a section somewhere between two loaded joints, as at C (Fig. 51). The principles above stated hold for such a section for loads anywhere except on the panel BC. In case of a load on this panel, only a part is borne at the left of C ; and, in computing bending moment, the part coming at C is to be omitted. Hence the above reasoning is not strictly applicable to this case. The following modification of the principle above deduced may however be shown to hold for any section of the truss. Let the given section divide a panel BC into segments BC^ and O^C. Then, if all loads on BC are treated as if divided betweefi B and C in the ratio of BC" to C"C, the general rule applicable to a beam may be used i7t assigning the position of loads for maximum bending moment. [The proof of this proposition will be omitted. The mode PARALLEL CHORDS — CONCENTRATED LOADS. 125 k of reasoning to be applied is similar to that used in Art. 147. The student should attempt the proof for himself.] In general, the error will be small if the bending moment curve for the case of a beam is used in computing bending moments at all panel joints of the truss. 150. Curve of Maximum Bending Moments. — The greatest bending moment at any section can be easily found as soon as the position of moving loads producing it is known. The method will be explained with reference to the truss shown in Fig. 50 (PL IV). By Art. 149, the greatest bending moment for each of the points K, L, M, A'', P, is exactly the same as if the truss were a beam upon which the loads were supported directly. It is unnecessary to show the construction for these points, it being identical with that of Art. 130. We therefore consider a sec- tion through an upper joint as T. First the position of the moving load must be assigned. By Art. 149, we may apply the general principle that the loads in the two segments of the truss should be in the same ratio as their lengths, provided all loads on the panel MN are treated as if equally divided between 7]/ and A^ (since the sec- tion considered divides MN into equal parts). Again, since the segment of the truss to the left of the section is seven- twelfths of the whole span, the load on the left segment should be seven-twelfths the total load on the truss. It is found by trial that the condition for maximum bending moment at T is satisfied when the line of action b'c' is at M. For, the total load on the truss, when the load is near this posi- tion, is 182000 lbs. Now suppose the load B^O is just at the left of M. The three loads, C'D\ D'E', ET\ must be regarded as equally divided between J/ and N\ this gives for the whole load to the left of T 11 1500 lbs., which is greater than seven- twelfths of 182000 lbs. But if B'O is just at the right of M, half of it must be regarded as borne at N, and the v/hole load 126 GRAPHIC STATICS. on the truss to the left of T is 104000 lbs., which is less than seven-twelfths of 182000 lbs. Now draw Mt Mt to represent the beam in the position for maximum bending moment at T; that is, with the load B'C at M. From its extremities draw vertical lines intersecting the strings od and oc", and through the points thus determined draw ot', the closing side of the funicular polygon for all the loads and reactions acting on the truss. In determining the bending moment by the method of Art. 130, we must consider. the loads on the panel MN as replaced by their components at M and JV respectively. Draw vertical lines through J/ and 7\^ intersecting the strings oc' and of, and join the points thus determined. The line thus drawn should replace the corresponding portion of the funicular polygon before drawn {i.e., the strings oc', od', oe' , and a part of) ; because it is, in fact, a line of the funicular polygon which will be obtained if CD', D'E', and E'F' be replaced by their com- ponents which are actually applied to the truss at M and N. We now draw through T a vertical line intersecting ot' and the corrected string just determined. The intercept thus deter- mined, multiplied by the pole distance, gives the bending moment at T. From X„, V„t as a line of reference draw downward an ordi- nate at the point T equal to the intercept thus found ; this gives a point of the curve of maximum bending moments. Other points of the curve may be found in the same way. The whole curve is shown in the figure, but the construction is not given except for the section T. [Note. — The condition for maximum moment at a section may sometimes be satisfied for more than one position of the moving load. This is the case for the section at T; it will be found by trial that the condition is satisfied when the load C"D" is just at the right end of the truss. The bending moment for this position is, however, found to be slightly less than for the position above taken. In most cases, the position for greatest moment can be very nearly predicted by remembering that the truss should be loaded as completely as possible, and that the heavy loads should be as near the section as possible. These principles will also serve as a guide in distinguishing whether the moment is a maximum or a minimum when the condi- tion is satisfied.] PARALLEL CHORDS — CONCENTRATED LOADS. 127 It should be noticed that the value found for the bending moment at any section is a possible value for another section equally distant from the middle point of the truss, since the train may cross the bridge from the left instead of from the right. Hence for two points, as L and N, equally distant from the middle of the truss, the greater of the two moments above determined must be used for both. 151. Shear and Moment Curves for Combined Fixed and Mov- ing Loads. — In Fig. 50 (PL IV) the line X K- has been taken as the axis of abscissas for the curves of shear. Positive shear has been laid off downward for fixed loads and upward for moving loads. Hence an ordinate at any point drawn perpen- dicular to X^ K,, and limited by the two curves (or rather broken lines), represents the greatest positive shear due to combined fixed and moving loads. The shear curves for fixed and mov- ing loads intersect at a point directly opposite the joint N. Hence for sections to the right of this point, the shear is always negative. Similarly, A'"^ Y,n is the axis of abscissas for the two moment curves. The bending moment is always negative ; in case of the dead load curve, it is laid off upward from X^Kn, while for the live load curve it is laid off downward. Hence an ordinate between the live and dead load curves at any point of X^^ F„, represents the greatest possible bending moment at the cor- responding section of the truss. We are now prepared to determine maximum stresses in the truss members. 152. Maximum Stresses in Web Members. — The greatest stress in any web member is readily found from the shear curve. By Art. 139, the stress has such a value that its resolved part in the vertical direction is equal to the shear in a section through the member. For example, for the member 14-15 we draw a line in the shear diagram parallel to 14-15 and limited by the two lines which determine the greatest shear 128 GRAPHIC STATICS. sustained by the member. This gives 14'-! 5' as the stress in the given member. In the same way, I2'-I3', I3'-I4', etc., represent stresses in the corresponding truss members. As to the sign of the stress in any member, we know that for a positive shear any diagonal member has to resist a ten- dency of the portion of the truss to the left to move upward ; hence the stress is a tension for every member sloping down- ward from left to right (as 13-14), and a compression for a member sloping downward from right to left (as 14-15). The reverse will be true if the shear is negative. We may, how- ever, neglect the negative shears given by the curve of Fig. 50, because they are the minimum negative shears. (This is evi- dent because^ they were determined as maximum positive shears ; and since they are found to be really negative, they are the least negative shears that can occur at those sections.) The maximiifn negative shear for this portion of the truss will be found when the train crosses the bridge from left to right ; and its value for any section is the same as that of the greatest positive shear for the section equally distant from the middle of the truss but on the other side. For example, since we have found a negative shear in sec- tions through the member 21-22, that member sustains a com- pression which is represented in the figure by 21 '-22'. But if the train is headed to the right, the member 21-22 may sustain a much greater compression — equal, in fact, to that already found for 14-15, and represented by I4'-I5'. The latter is there- fore t\\Q^ greatest compression sustained by 21-22. It is evident, therefore, that it is necessary to determine stresses from Fig. 50 for those members only which sustain positive shear ; remembering that the stress thus found in any member is a possible value of the stress in the corresponding member on the other side of the center of the truss. If both maximum and minimum stresses are desired, it must be remembered that the dead-load stresses may occur alone. It will be seen that a reversal of stress may occur in the PARALLEL CHORDS — CONCENTRATED LOADS. 129 members 16-17, i/'I^j 18-19, 19-20; while in the remaining members no reversal is possible. 153. Maximum Chord Stresses. — From the maximum bend- ing moments represented in Fig. 50, we may readily find the greatest stress sustained by any chord member. Consider, for example, the member 15-2. If a section be taken cutting the three members 7-14, 14-15, and 15-2, and the origin of moments be taken at R, it is evident that the sum of the moments of the reaction and loads to the left of R is equal to the moment of the force acting in the member 15-2. But the former is the bending moment in the section through R, and its greatest value is equal to the ordinate between the two moment curves already drawn, multiplied by the pole distance. Hence we have the equation : Ordinate of moment curve x pole distance = stress in chord member x depth of truss. That is, in the present example, the stress in pounds is equal to the ordinate multiplied by 120000 and divided by 15. The computation may be made graphically as follows : From any point Z (Fig. 50) draw two lines making a convenient angle with each other, say a right angle. Lay off ZZ" equal to the pole distance on the force scale already adopted, and ZZ' equal to the depth of the truss, and draw Z'Z" . Then take Z2 equal to the ordinate of the moment curve and draw 2-15 parallel to Z'Z'\ intersecting ZZ" in the point marked 15. The distance Z-i^ represents the stress in 15-2, to the force scale already adopted. If the pole distance ZZ" had been drawn to some other scale, Z-i^ would represent the required stress to the same scale. For two chord members symmetrically situated with respect to the middle of the truss, the maximum stresses have the same value, which is to be computed from the greater of the two corresponding values of the bending moment. This is apparent, if it is remembered that when the train is headed to the right, I30 GRAPHIC STATICS. the two members interchange their conditions. The figure shows all the chord stresses. Evidently, all upper chord mem- bers are in compression and all lower ones in tension. If minimum stresses are desired, they will be found by using the dead-load bending moments alone. The computation of chord stresses will be facilitated if the number of force units in the pole distance is taken as a simple multiple of the number of linear units in the depth of the truss. For, if H is the pole distance, h the depth of truss, y the ordinate of the moment curve, and x the stress in the chord member, we have Hy = hx', OY x= — y. Ji M Now if — = n, we have only to measure y in linear units, multiply by n, and the result is x in force units. § 4. Parallel Chords — Uniformly Distributed Moving Load. 1 54. Distribution of Load. — In designing highway bridges, it is usual to assume the moving load to be uniformly distributed along the bridge. The same assumption is sometimes made for railroad bridges. This simplifies the computation of stresses very materially. In fact, when the chords are parallel, the algebraic method of treating such cases becomes so simple that it is by many preferred to the graphic method. It will be well, however, to indicate the main points in the graphic con- struction. Maximttm shear. — First it may be noticed that the position of moving load for greatest shear in any panel is readily deter- mined. The conclusion deduced in Art. 147 may be shown to apply to this case ; hence we have the principle that the greatest shear in any panel occurs when the portion of the truss to the right is completely loaded, while the load on the panel is the same fraction of the whole load that the panel length is of the whole span. Let A'F(Fig. 51) represent a truss of PARALLEL CHORDS — DISTRIBUTED LOAD. 131 seven equal panels ; and for the greatest shear in a panel CD let ^ be the foremost point of the moving load. Then we must have ^Z> equal to one-seventh oi ZY\ or, Zi^ = one-sixth DY. A similar rule applies to each panel. Maximum moment. — From the principles deduced in Arts. 129 and 149, it is evident that the greatest bending moment occurs at every section when the moving load covers the whole bridge. Funicular polygon. — In order to draw the funicular polygon, it is practically necessary to substitute concentrated loads for the uniformly distributed load. In determining shear, we ought strictly to have the true funicular polygon for the distributed load ; since without it we cannot graphically determine the amount of load actually carried at each joint. It will, however, be sufficiently correct to subdivide the load into small portions, each being assumed concentrated at its center of gravity. The student will readily carry out the construction just out- lined, every step being similar to the corresponding part of the process shown in Fig. 50 (PI. IV). 155. Assumption of Equal Panel Loads. — The determination of stresses, whether graphically or algebraically, is simplified by an approximate assumption which will now be explained. This assumption is that the moving load on the truss at any instant consists of equal loads concentrated at the panel joints ; each load being equal to the total load on a length equal to half the sum of the two adjacent panels. Thus, in Fig. 51, in computing the shear in the panel AB, we assume full panel loads at B, C, D, E, and F] and similarly for any other panel. The shear will then be equal to the reaction at Xdue to the series of loads to the right of the panel considered, and the moment curve will be simply a funicular polygon drawn for a series of full panel loads at all the joints. It will be seen that the error involved in the above assump- tion is on the side of safety, so far as the web members are 132 GRAPHIC STATICS. concerned, since it gives greater values of the shear than the more exact method. In case of the bending moments, the results are correct for all joints of the loaded chord. For the other joints there will be a slight error. This error will be avoided, if, in drawing the funicular polygon, the load be con- sidered as partly supported at points in the same vertical lines with the joints of the unloaded chord; thus, in Fig. 51, equal loads should be assumed to act at all the points A', A, B\ B, etc. The construction just indicated will not be here shown, being very similar to that explained in the following articles for the case of a truss with non-parallel chords. § 5. Truss with Ciirved CJiords — Uniform Panel Loads. 1 56. General Statement. — If the upper and lower chords of the truss are not parallel, the determination of the stresses is somewhat less simple than in the case of parallel chords. But when the live load is taken as uniformly distributed along the bridge, and applied to the truss in the way explained in Art. 155, the graphic construction for finding the maximum stresses is not difficult ; and is much less laborious than the algebraic computation. In Fig. 52 (PI. V) is shown a truss with curved chords, divided into equal panels. It will be seen that the method given in the following articles applies equally to the case of unequal panel lengths. The principle of .counterbracing will be employed here, the diagonals being constructed to sustain tension only. 157, Dead Load Stresses. — The dead load stresses maybe determined by means of a stress diagram, as in the roof-truss problems already treated. Two points should be observed in drawing this diagram, (i) If dead loads are taken to act at upper as well as at lower joints, the force polygon for the loads and reactions must show these forces in the same order as that in which their points of application occur in the perimeter of CURVED CHORDS — UNIFORM LOADS. 133 the truss. (2) The diagonal members assumed to act are taken as all sloping in the same direction. The reason for the first point is the same as already explained in Art. 90, — viz., that unless the forces be taken in the order mentioned, the stress diagram cannot be the true reciprocal of the truss diagram and certain lines will have to be duplicated. The reason for assuming the diagonals as all sloping in the same way is the same as in the case of the roof truss with counterbracing (Arts, iii and 112). The dead load stress diagram is not shown, since its con- struction involves no principle not already fully explained and illustrated. 158. Chord Stresses Due to Live Load. — It can be easily shown that a load at any point of the truss produces tension in every lower chord member and compression in every upper chord member. Thus, referring to Fig. 52 (PI. V), let a load act at^, and let us determine the kind of stress caused in the member qq\ Take a section cutting qq\ q'a\ d'd. Consider- ing the portion of the truss to the right of the section, the only forces acting on it are the reaction at the support and the forces in the three members cut. Apply the principle of moments to this system of forces, taking the origin at the point of intersection of q^d' and dd. It is evident that qq' must be in compression to resist the tendency of the reaction to produce left-handed rotation about the origin. By similar reasoning, assuming a load at any other point, the student will be able without difficulty to verify the general statement above made. It follows that in order to get the greatest possible stresses in the chord members, the truss should be fully loaded. Hence, to find the live-load chord stresses, a convenient method is to draw a stress-diagram for the truss under all live loads. This diagram is not shown. 134 GRAPHIC STATICS. 159. Live Load Stresses in Web Members. — When the diag- onals and verticals are considered, it will be found that loads in different positions may cause opposite kinds of stress in any member. Thus, considering the member fn\ it is easy to show that a tension is caused in it by a load at either of the points ab^ be, cd, de, or ef, while a compression is caused by a load at fg, gh, or hi. (This may be shown by taking a section Xhrough ff ,f^ n\ and n^n, and taking moments about the point of intersection of the two chord members cut.) Similarly, loads at abj be, ed, de, and ef all tend to throw compression on the vertical member f'm', while loads at fg, gh, and /// have the opposite tendency. Therefore, to produce the greatest tension upon f'n' (and compression onf'i?t'), the live load must act only at ef and all joints to the right ; while to cause the greatest compression on f'n' (and tension upon fin'), the live load must act only at fg, gh, and hi. A similar statement will hold regarding any other web member. Since counterbraces are to be used in all panels in which diagonal members would otherwise be thrown into compression, we shall need only to consider the greatest tension in each diagonal and the greatest compression in each vertical. We shall first outline the method to be employed, and then explain the construction. To determine the greatest tension in a diagonal member, as g'm' : Assume the live load to come upon the bridge from the right until there are full loads at the joints ab, be, ed, de, ef and fg. Take a section cutting g'm' and the two chord members gg' and m'm, and consider the forces acting upon the portion of the truss to the left of the section. These forces are four in number : the reaction at the support and the forces acting in the three members cut. Hence we first determine the reaction, and then determine the three other forces for equilibrium by the method of Art. 42. The construction is shown in Fig. 52 (PI. V). ABCDEFGHI is the force polygon for the eight live loads that may come CURVED CHORDS — UNIFORM LOADS. 135 upon the truss. Choosing a pole O, the funicular polygon for the eight loads is next drawn. Now, turning the attention to the member ^'?;2', the loads ^/^ and /li are assumed not to act. The reactions at the supports for this case of loading are found in the usual way. Prolong oa and og to intersect the lines of action of the two reactions, and join the two points thus determined. This gives the closing line of the funicular poly- gon (or om). The ray OM is now drawn parallel to the string 07/1, and the two reactions are GM a.nd MA. Now take a section through vwi' , vi'g' , and g'g, and apply the construction of Art. 42 to the determination of the forces acting in the three members cut. The resultant of GM and the force in gg' must act through the point X (the intersection of gg' produced and ij). The resultant of the forces in ;;/;;/' and m'g' must act through their intersection Y. Hence these two resultants (being in equilibrium with each other) must both act in the line XY. From G draw a line parallel to gg' , and from M -a. line parallel to XY; mark their point of intersection _G' . Then MG' is the resultant of the forces in mm' and m'g' . From a^draw a line parallel to mm\ and from G' a line parallel to the member g'm', and mark their point of intersection M'. Then M' G' represents the force in the member m'g' . This is also the value of the greatest stress in m'g'. To determine the .stress in the vertical member I'g' , the same loading must be assumed, and a similar construction is employed. Take a section cutting II' , I'g', and g'g, and deter- mine forces acting in these three lines which shall be in equi- librium with the reaction GM. This reaction is in equilibrium with MG' and G'G, the former having the line of action XY. Then MG' is resolved into two forces having the directions of the members g' I' and I' I. The stress in g'l' is found to be a compression, represented in the stress diagram by the line G'L'. The maximum live load stress in every web member may be found in the same way. If the above reasoning is understood, there will be no difficulty in applying the same method to the remaining members. 136 GRAPHIC STATICS. 160. Maximum Stresses. — By combining the stresses due to live and dead loads, the maximum stresses are easily deter- mined. Wed members. — When the web members are considered, the effect of counterbracing needs careful attention. The construction above explained gives the greatest live load tension in each diagonal. It may be that for certain members, this tension is wholly counterbalanced by the dead loads. In any panel in which this is the case, the member shown will never act and may be omitted. The counterbrace must then be considered. Evidently, the algebraic sum of the stresses due to live and dead loads will be the true maximum tension in each of the diagonals shown. For the greatest tension in the other system of diagonals, the load must be brought on the bridge from the left ; or, what amounts to the same thing, the tension already found for any one of the diagonals shown is also the greatest tension in the diagonal sloping the opposite way in the panel equally distant from the middle of the truss. (In fact, the stresses in all members due to a movement of the loads from left to right may be obtained from the results already reached, by consideration of symmetry.) As to the vertical members, two values of the stress must be compared in every case — namely, the greatest compressions corresponding to the two directions of the moving load. But both can be obtained from the above results by considering the symmetry of the truss. For example, the stress found for l^g^ is a possible value for the stress in r^b\ and must be compared with the value obtained for the latter member when the load moves from right to left. It is possible, also, that certain of the verticals may be in tension when the dead loads act alone. Maximum chord stresses. — These are found by combining the stresses due to fixed and moving loads, determined as already described. CURVED CHORDS — CONCENTRATED LOADS. 137 161. Example. — The following example should be carefully solved, following the method outlined in the preceding articles. Given the span, 120 ft. ; lower chord straight ; upper chord circular with rise at middle point of 20 ft. ; panel length, 15 ft. ; diagonals counterbraced to sustain tension only ; width of bridge between trusses, 18 ft. Assume dead load from the formula of Art. 135, one-third applied at upper chord and two- thirds at lower. Live load 85 lbs. per square foot. Determine maximum stresses due to live and dead loads. 162. Parallel Chords. — It will be noticed that all the methods which have been described for the discussion of trusses with curved chords are applicable to the case of parallel chords. In fact, the determination of stresses in web members is simplified when both chords are horizontal ; for, after the reaction is found as in Art. 159, it is only necessary to so determine the stress in the web member that its vertical resolved part shall equal the reaction. § 6. Truss ivith Curved Chords — Concentrated Loads. 163. Comparison of Cases. — When the moving load consists of a series of unequal weights, the non-parallelism of the chords somewhat complicates the determination of stresses. This case can, however, be treated without great difficulty, as will now be explained. For the determination of chord stresses the method will be almost exactly the same as in the case of parallel chords already discussed. It is only necessary to determine the greatest bend- ing moment at each joint, just as was done in Art. 150, and then find the chord stresses by the principle of moments. But for the determination of web stresses the shear curve is not sufficient, since the shear in any section is not borne wholly by the web member, but partly by the inclined chord member. Moreover, the position of the moving load which will produce 138 GRAPHIC STATICS. the greatest shear in a panel is not generally the position which causes the greatest stress in the web member in that panel. 164. Position of Loads for Greatest Stress in a Web Member. — We shall now deduce a rule for determining what position of the loads will produce the greatest stress in any web member. Let XV (Fig. 53) represent the truss and the member con- sidered. We know in a general way (Art. 159) that for the greatest tension in B^C the truss should be completely loaded on the right of the panel BC.^ Let IV = total load on truss ; IV' = load on panel BC; I = total length of truss ; /' = length of panel BC; x = distance from V to center of gravity of IV; x' = distance from C to center of gravity of W. Prolong the two chord members of the panel to intersect at Z, and let XZ = a, ZB = b. Let R = reaction at X, and P = portion of W carried at B. Then R = IVx P = W'x I ^' Now if the truss be separated by a section cutting B' C and the two chord members, it is seen that the moment of the stress in B'C dihoMt ^ must equal the sum of the moments of R and P about Z. Hence the stress in B' C is greatest when the sum of the moments of R and P about Z is greatest. Let M = that sum, then M-. Ra-Pt='^x-'^x' I I' * This statement is true for all forms of truss considered in this work. If the two chord members in any panel intersect between the vertical lines through the ends of the truss, the statement no longer holds. The effect on the web member of a load in any position can be determined very easily, whatever the form of the truss, by reasoning similar to that employed in Art. 159. CURVED CHORDS — CONCENTRATED LOADS. 139 If 3/ is a maximum, we must have — - = o. But since dx' = dx, dx we have, dM Wa Wb jrr b I ..„ dx I I' a I' The ratio - will be known for each panel, and also the ratio —- ; a I hence the condition expressed by the equation can be easily applied. It may happen that the condition is satisfied for more than one position of the loads. If this is so, such positions must all be tried and the results compared. This is illustrated in the following article. Special case. — It will be noticed that if the chords approach parallelism, the point Z moves farther away and the limit approached by — is unity. Hence, for the case of parallel ^ I chords, the equation becomes W=-~ W\ This is identical with the result already given in Art. 147, [Note. — The remarks made in the note in Art. 147 apply also to the reasoning just given. It is also to be noticed that we have assumed that no load is between B and X. If the condition above deduced cannot be satisfied without carrying the foremost load or loads to the left of j5, we may reason thus: Let /F" = load to left of B; x" = distance of center of gravity of IF" from B. Then M =Ra- Pb- W" (b - x") = ^ a- _ — jr' - W" (b - x"). IV ^ ^ Differentiating, remembering that dx" — dx' = dx, we have dM IVa Wb , „,,, dx I I' Placing this equal to zero, we have a V a Usually W" will be zero and the equation first deduced will apply, but if necessary the last equation may be used. For the case of parallel chords, the term - IV disappears, since - approaches zero.] a 140 GRAPHIC STATICS. 165. Determination of Web Stresses. — The method of deter- mining web stresses for the case now under consideration is shown in Fig. 54 (PL V). The force polygon and funicular polygon for the series of wheel loads is drawn in the usual manner. The span of the truss here taken is 120 ft., divided into eight equal panels, the upper chord being curved, while the lower (supporting the floor system and moving loads) is straight. Half of the truss is shown at XV, drawn to the same scale used in the space diagram for the moving loads. The construction is shown for finding stresses in the two mem- bers tu and nv. First consider the latter. To apply the rule of the preceding article for finding the required position of loads, determine the ratio -. Prolonging a the two chord members k?i and vj7i till they intersect at Z2, we find -^- — — = 2. Also we have - = 8. Hence, for the great- est stress in tro, the load on the panel QR must equal one- sixteenth the load on the whole truss. Now it is easily seen that this condition is satisfied when the second load is at the point R. For, in this position, the whole load on the truss is 123750 lbs. If the second load is just at the left of R, the load on the panel is 20000 lbs., which is greater than one- sixteenth of 123750 lbs. ; but if the second load is just at the right of R, the load on the panel is 7500 lbs., which is less than one-sixteenth of 123750 lbs. We are now ready to determine the stress in the member nv. The general method is to cut the truss by a section through /'?/, iiv, and V77i ; determine the resultant of all forces on the truss to the left of the section ; and then determine three forces in the members cut which shall be in equilibrium with that resultant. The forces acting on the truss to the left of the section are the reaction at the support and the portion of the foremost load AB carried at the point Q. Their resultant is found as follows : Draw X„X„ to represent the span when the load BC'is CURVED CHORDS — CONCENTRATED LOADS. 141 at R ; then draw the closing line of the funicular polygon for the forces now on the truss, and from O draw the ray parallel to this closing line. The end of this ray is marked Mi, and the left reaction is represented by the line MiA. Next replace AB by its two components borne at Q and R. These are deter- mined by drawing verticals from Q and R intersecting the strings oa and od in Q' and R' ; then Q'R' is the string which must replace oa and ol?. (The portions of the funicular poly- gon to the right of R' and to the left of Q' are not changed by thus replacing AB by its two components at Q and R.) Now by drawing from O a ray parallel to Q'R' we find the point (marked K^) which divides AB into the two components at Q and R. The resultant of the reaction at X and the load at Q is given in magnitude by AfiKi. We need also its line of action, which is found as follows : Prolong Q'R' and omi (the closing string of the funicular polygon for all the forces on the truss) until they intersect in s' ; the required line of action is a vertical line through 2'. This vertical line intersects X^X^ produced in s. We now locate the point -s- in the truss diagram by making X.:; = XuZ in the diagram above. We have now to solve the following problem : Determine three forces acting in lines kit, uv, and vm, which shall be in equilibrium with a force equal to the resultant just found and acting upward in a line through z. This is solved by the method of Art. 42. Draw KM equal to the given force acting at z. Draw zQ', and determine two forces, one parallel to zQ' and the other to hi, which shall be in equilibrium with KM. This gives UK as the force in the line tik and MU as the resultant of the forces in the lines mv and vii. From M and U draw lines parallel to niv and vil respectively, and mark their intersection V\ then J/Fand F^" represent the stresses in the corresponding truss members. The latter is evidently the required stress in the member tiv. Turning next to the member /;/, a section through it will cut 142 GRAPHIC STATICS. the two chord members kiL and mt ; these intersect at Z^. By measurement we find -=:^^^=i6, and --- = 8x6 = 48. There- a Z^X I a fore, for the greatest stress in tii we must have the load on the panel QR equal to the whole load on the truss divided by 48. This condition is seen to be satisfied when the foremost load is at R. Since in this case there is no load borne to the left of R, the only force acting on the truss to the left of a section through ttL is the reaction at the support. To find this, draw XtXu to represent the length of the truss for the required posi- tion of the loads, and find oifix, the closing string of the funic- ular polygon for all loads and reactions on the truss. The corresponding ray is OM^, and M^A represents the reaction at the left support. Referring now to the truss diagram, we have to determine forces acting in the three lines int^ tiiy tik, which shall be in equilibrium with the reaction just determined. Lay off K^Mto represent this reaction ; draw a line from K' parallel to he and one from M parallel to QX, and mark their point of intersec- tion U' ; then U'K^ represents the force in tik, and MU' the resultant of the forces in mt and ///. From M draw a line par- allel to mt and from U^ a line parallel to tit, intersecting at T] then TU^ represents the required stress in tu. The value just determined is the true maximum stress for the member tn. In case of the member iro, however, farther investigation is needed. It will be found by trial that the condition W = ~ - ■— • W is satisfied for the member itv, not a I' only when the loads have the position already treated, but also when the foremost load is at R. Hence the stress in uv for this position of the loads must be determined and compared with the one already found, namely UV. For this case of loading the only force acting on the truss to the left of a section through tiv is the reaction (already found when considering the member /;/) represented by K' M. Making this in equilibrium with three forces whose lines of action are kii, ?iv, vni, we find CURVED CHORDS — CONCENTRATED LOADS. 143 U^ V as the stress in uv. Comparison shows that this is sUghtly greater than L^V; hence the value last found is the true maximum for the member tiv. The construction for each of the other web members is exactly similar. If both chord members in any panel are inclined, the construction requires only slight modification, which the student will readily supply. In case the intersection of two chord members cannot con- veniently be found by prolonging them, the distances a and /; can be easily computed from the dimensions of the truss. Remark. — It will often be found that two web members related as are tii and tiv, will not sustain their maximum stresses at the same time. In the case of parallel chords this could not occur. 166. Minimum Stresses. — In what precedes, little has been said of minimum stresses in the truss members. If, however, the design is to be made in accordance with the theory of the strength of materials under repeated alternations of stress, the minimum stress sustained by each member becomes important. In all the cases treated in the present chapter, the minimum stresses can be determined without difficulty, without the use of additional principles. Part III. CENTROIDS AND MOMENTS OF INERTIA. :)>a S-^'^'cT^T — 5 ^ = radius of inscribed circle ; / = f^/^^-4^ icl\2y.^_^E length of a side of the polygon ; i- I 1 \ ! \\ ! = total length of the broken line I i \ 1 \^J 1^ ! ^xi N\ AE. Through O draw OC, the axis ^ ^^ ^ ^^^ of symmetry of AE; and MN, per- pendicular to OC. First, the centroid must lie on OC. Second, to find its distance from O, assume a system of equal and parallel forces applied at the middle points of the sides AB, BCy etc. The required centroid is the point of application of the resultant of these forces. Taking MN as the axis of moments, and letting x = required distance of centroid from MN, and x^, x^, x^, x^ the distances of the middle points of AB, BC, etc., from MN, we have from the principle of moments, Ixi 4- 1x2 + ^-^3 + ^^i = ^^' But if Ab and Bb be drawn perpendicular respectively to PQ and MNwe have from the similar triangles ABb, POQ. AB^PO J^^r Ab~PQ^^ Ab~xi ,♦. /xi=r ' Ab. In the same way, 1x2 = r ' be, 1x2, = r ' (^d, Ix^ = r • dE. Hence, S'X = r {Ab-\-bc-\-cd-\-dE) = r'AE, where AE is equal to the projection of the broken line ABCDE on MN. The centroid G may now be found graphically as follows : U3kQ0c' = r; ON=^s', OE' = ^AE; draw iV^'. Then 6^ is determined by drawing E' G parallel to Nc'. 154 GRAPHIC STATICS. Circtdar arc. — The above construction holds, whatever the length of the side /. If this length be decreased indefinitely, while the number of sides is increased indefinitely, so that the length s remains finite, we reach as the limiting case a circular arc. The same construction therefore applies to the determina- tion of the centroid of such an arc, r denoting the radius of the circle and s the length of the arc. 184. Centroids of Geometrical Areas. — Pa^'allelogram. — The centroid of a parallelogram is on a line bisecting two opposite sides. Let ABCD (Fig. 59) be a parallelogram, and EF a line bisecting AD and BC. Divide AB into any even number of equal parts, and through the points B^ ^^ P — ,_, — C of division draw lines parallel to BC. Also divide BCmX-o any even number \ \ \\ p V \ \Q \ of equal parts and draw through the \\ \ \\ \ points of division lines parallel to E D AB. The given parallelogram is thus divided into equal elements. Now consider a pair of these elements, such as those marked P and Q in the figure, equally distant from AD, and also equally distant from EF, but on opposite sides of it. The centroid of the two elements taken together is at the middle point of the line joining their separate centroids. If the number of divisions of AB and of BC be increased without limit, the elements approach zero in area, and the centroids of P and Q evidently approach two points which are equally distant from EF. Hence in the limit, the centroid of such a pair of elements lies on the line EF. But the whole area ABCD is made up of such pairs ; hence the centroid of the whole area is on the line EF. For like reasons it is also on the line bisecting AB and DC\ hence it is at the intersection of the two bisectors. The point thus determined evidently coincides with the point of intersection of the diagonals AC 3.nd BD. CENTROIDS OF LINES AND OF AREAS. 155 Triangle. — The centroid of a triangle lies on a line drawn from any vertex to the middle of the opposite side ; and is, therefore, the point of intersection of the three such lines. Let ABC (Fig. 60) be any triangle, and D the middle point of BC. Then the centroid of ABC must lie on AD. For AD bisects all lines, such as be, parallel to BC Now inscribe in the triangle any number of parallelograms such as bec^b\ with sides parallel respectively to BC and AD. The centroid of each parallelo- gram lies on AD, and, therefore, so also does the centroid of the whole area composed of such parallelograms. If the number of such parallelograms be increased without limit, the alti- tude of each being diminished without limit, their combined area will approach that of the triangle, and the centroid of this area will approach in position that of the triangle. But since the former point is always on the line AD, its limiting position must be on that line. Therefore the line AD contains the cen- troid of the triangle. By the same reasoning, it follows that the centroid of ABC must lie on BE, drawn from B to the middle point of AC. Hence it must be the point of intersection of AD and BE, which point must also lie on the line CF drawn from C to the middle point of AB. The point G divides each bisector into segments which are to each other as i to 2. For, from the similar triangles ABC, EDC, since EC is half of AC, it follows that DE is equal to half of BA. And from the similar triangles AGB, DGE, since DE is half of AB, it follows that GE is half of GB, and GD half of GA. Quadrilateral. — Let ABCD (Fig. 61) be a quadrilateral of which it is required to find the center of gravity. Draw BD, and let E be its middle point. Make EGi = \ EA, and EG2=\ EC Then the centroids of the triangles ABD and BCD are 156 GRAPHIC STATICS. Fig. 61 Gy and Gi respectively. Hence the centroid of ABCD is on the line G^Gi at a point dividing it into segments inversely proportional to the areas of ABD and BCD. Since these two tri- angles have a common base, their areas are proportional to their alti- tudes measured from this base. But these altitudes are propor- tional to AF and FC, or to G^H and G^H \ hence, if G is the required centroid, G\G '. G-iG : : G-iH : G\H. Therefore G is found by making GxG=G etc., must be replaced by the magnitudes of the forces. 190. Determination of Moment of Inertia of a System of Parallel Forces. — Let the points of application of the forces be in the same plane, which also contains the assumed axis. We shall have to deal only with systems satisfying these condi- tions. By the definition, the moment of inertia will be the same, whatever the direction of the forces. If we take the moment of any force (as defined in Art. 175) about the given axis, and suppose a force equal in magnitude to this moment to act at the point of application of the original force, and in a direction corresponding to the sign of the moment, then the moment of this new force about the given axis is equal to the moment of inertia of the original force. If this be done for all the forces, the algebraic sum of the results will be the required moment of inertia of the system. This process can be carried out graphically by methods already described. Let ab, be, cd, de (Fig. 65) be the points of application of four parallel forces, and let the axis of inertia be QR. Suppose all the forces to act in lines parallel to QR, passing through the given points of application. Their respective moments with reference to QR may now be found by the method of Art, 55. Draw the force polygon ABCDE and choose a pole O, taking the pole distance H preferably equal to AE or some simple multiple of AE. (In Fig. 65 H is taken equal to AE) Draw a funicular polygon and prolong each string to intersect QR. Then the moment of any force with respect to QR is the product of H by the distance intercepted on QR by the two strings corresponding to the force in question. Thus the moment of AB is given by the distance A^ B^ (Fig. 65) multi- plied by H. Also the successive moments of BC, CD, DE are represented by B^C\ C'D\ D'E', each multiplied by H. It is MOMENTS OF INERTIA OF FORCES. 163 seen also that the intercepts, if read in the above order, give a distinction between positive and negative moments ; upward distances on QR denoting in this case positive moments, and downward distances negative moments. We have now to find the sum of the moments of a second system of forces acting in the original lines, but represented in inig.65 magnitude and direction by the intercepts just found. We may take as the force .polygon for the second Construction the line A^B' C'D'E' (Fig. 65), and choosing any pole distance H', draw a second funicular polygon and find the distances inter- cepted by the successive strings on the line QR. But in this case, since only the resultant is desired, we need only find the intercept A'^E" between the first and last strings. The product of this intercept by H' gives the sum of the moments of A'B', B'C, CD', D'E' with respect to QR \ and, if the product be multiphed by H (since A'B', B'C, etc., should each be multi- plied by ^in order to represent the magnitudes of the forces of the second system), the result will be the required moment of inertia of the given system of forces AB, BC, CD, DE. 164 GRAPHIC STATICS. It should be noticed that in Fig. 65 (A) is a /o7re diagrain^ {B) a space diagram (Art. 11); that is, every line in {A) repre- sents a force, while every line in {B) represents a distance. Even A' B' , B'C, etc., though used as forces, are actually merely distances ; and the moment of any one of them is the product of a length by a length. 191. Radius of Gyration. — The moment of inertia of the given system is HxH' xA"E". If H has been taken equal to AB, the product H'xA'^E" must equal the square of the radius of gyration of the system with respect to QR. The length of the radius of gyration can be found as follows (Fig. 65) : Draw LN=H' + A"E'\ and make LM=H'. On Z/V^as a diameter draw a semicircle LPN, and from M draw a line perpendicular to LN^, intersecting the semicircle in P. Then MP is the length of the required radius of gyration. For by 2 elementary geometry we have PM —LMxMN. If H is taken equal to n xAE, the moment of inertia is equal to H^ xA"E" X n X AE, and the square of the radius of gyration is equal to nH' xA" E". Hence in Fig. 65 we should put either LM=nH', or MN=nxA"E". 192. Central Axis. — If the axis with reference to which the moment of inertia is found contains the centroid of the given system of forces, it is called a central axis of the system. In many cases it is desired to find the moment of inertia with respect to a central axis whose direction is known while the position of the centroid is at first unknown. It is to be noticed that the method shown in Fig. 65 is applicable in this case ; for the first part of the process is identical with that employed in finding the centroid of the system. If, in Fig. 65, the strings oa^ oe, of the first funicular polygon be prolonged to intersect, a line through their point of intersection, parallel to the direction of the forces, will contain the centroid of the system. If this line is taken as the inertia-axis, the points A', B' , C , D\ E' are the points in which this axis is intersected by xMOMENTS OF INERTIA OF FORCES. 165 the strings oa, ob, oc, od, oe. No further modification of the process is necessary. 193. Moment of Inertia Determined from Area of Funicular Polygon. — In Fig. 65, the moments of the given forces are represented by the intercepts A' B\ B'C\ CD', D' E' , each multiplied by the pole distance H. The moment of inertia of any force, as AB, is equal to the moment of a force represented by the corresponding intercept as A' B' , supposed to act in the line ab. Now, by definition (Art. 175) the moment about the axis QR of a force equal to A' B' acting in the line ab is equal to double the area of the triangle A' i B' ; hence the moment of the force HxA'B' is equal to double the area of that triangle multiplied by//. Similarly, the moment of a force HxB'C, acting in the line be, is equal to double the area of the triangle B' 2 C multiplied by //. Applying the same reasoning to each force, we see that the sum of the moments of the assumed forces {HxA'B', NxB'C, etc.) is equal to 2 //times the sum of the areas of the triangles A^ i B', B'zC, C 3D', D' 4E'. In adding these triangles, each must be taken with its proper sign, corresponding to the sign of the moment represented by it. Thus, the moments of A'B', B'C, and D'E' all have the same sign, while the moment of CD' has the opposite sign. Hence we must have for the sum of the moments, area A' i /?' + area B' 2 C —SLve^L C 3 /^' + area D' 4E', which is equal to the area of the polygon A' i 2 34E'. Hence this area, multiplied by 2 //, gives the moment of inertia of the required system of forces. It may sometimes be convenient to apply this principle in determining moments of inertia, the area being determined by use of a planimeter, or by any other convenient method. It should be noticed that if H is taken equal to the sum of the given forces (AE), twice the area of the funicular polygon is equal to the square of the radius of gyration. If //= l AE, the 1 66 GRAPHIC STATICS. square of the radius of gyration is equal to the area of the polygon. 194. Determination of Product of Inertia of Parallel Forces. — Assume the points of application of the forces to be in the plane containing the two axes. If the moment of any force with respect to one axis be found, and a force equal in magni- tude to this moment be assumed to act at the point of applica- tion of the original force, then the moment of this new force with respect to the second axis is equal to the product of inertia of the given force for the two axes. Thus, let ab, be, cd, de (Fig. 66) be the points of application of four parallel forces, and let their product of inertia with respect to the axes QR, 57" be required. Draw ABCDE, the force polygon for the given forces, assumed to act parallel to QR. Choose a pole O, the pole distance being preferably taken equal to AE, or some simple multiple of AE, and draw the funicular polygon as shown, prolonging the strings to inter- sect QR in the points A\ B\ C\ D\ E\ Now assume a series MOMENTS OF INERTIA OF FORCES. 167 of forces equal to A'B', B'C\ etc., each multiplied by //", to act at the points ad, be, etc., and determine their moments with respect to the axis ST. To find these moments, draw lines through ab, be, etc., parallel to ST, and draw a funicular poly- gon for the assumed forces taken to act in these lines. The force polygon is obtained by revolving the line A'B^C'D^E' until parallel with ST, and is the line A\ B\ C\ D\ E\ in the figure. The strings o^a\ ), etc., the distances of their points of application from the central axes parallel to the two given axes ; {a, b) the dis- tances of the centroid of the system from the given axes. Let A and A' be the products of inertia of the system with respect to the given axes and the parallel central axes respectively. Then A' = P,p,q, + P,p.,q,-{--" A =Pi{p, + a){qi + b) +/'2 (/2 + «)fe + ^) + ••• = (P^P^qi + P2p2q2 +"')+a {P,q, + P.q. + • • •) But P\qi-\-P2q-2^ — =0; and P\px-\-P^P^A — =0; since each of these expressions is the sum of the moments of the given forces MOMENTS OF INERTIA OF PLANE AREAS 169 with respect to an axis through the centroid of the system. Hence, A=A'-\- {P,JrP,+ '-')ab, which proves the proposition. liA = {P, + P,+ --) r and A^={P, + P,+ ...)c'\ we have c' = c'~-\-ab. From the above proposition it follows that if the axes have such directions that the product of inertia with reference to the central axes is zero, the product of inertia with reference to the given axes is the same as if the forces all acted at the centroid. When this condition is known to be satisfied, then for the pur- pose of finding the product of inertia the system of forces may be replaced by their resultant. It follows also, in the case when the product of inertia for the central axes is zero, that if one of the given axes coincides with the parallel central axis, the product of inertia for the given axes is zero ; for in this case either a or b i?, zero, and hence ab{Pi + P2-{ — ) is zero. Therefore, If the product of inertia of a system is zero for two axes, A' and A" , one of which (as A') contains the centroid of a system, then the product of inertia is also zero for A' and any axis parallel to A". § 2. Moine7its of Inertia of Plane Areas. 198. Elementary Areas Treated as Forces. — If any area be divided into small elements, and a force be applied at the centroid of each element numerically equal to its area, the moment of inertia of this system of forces will be approximately equal to that of the given area. The approximation will be closer the smaller the elementary areas are taken. If the ele- ments be made smaller and smaller, so that the area of each approaches zero as a limit, the moment ,of inertia of the sup- posed system of forces approaches as a limit the true value of the moment of inertia of the given area. 170 GRAPHIC STATICS. It is seen, then, that most of the general principles which have been stated regarding moments of inertia of systems of forces are equally applicable to moments of inertia of areas. The practical application of these principles, however, and especially the graphic constructions based upon them, are less simple in the case of areas than of systems of forces such as those already treated. The reason for this is that the system of forces which may be conceived to replace the elements of area consists of an infinite number of infinitely small forces, with which the graphic methods thus far discussed cannot readily deal. Problems of this class are most easily treated by means of the integral calculus, especially when the areas dealt with are in the form of geometrical figures. It is possible, however, by graphic methods to determine approximately the moment of inertia of any plane area ; and in many cases exact graphic solutions of such problems are not difficult. The proof of these methods is often most easily affected algebraically. 199. Moments of Inertia of Geometrical Figures. — The appli- cation of the integral calculus to the determination of moments of inertia will not be here treated. But the values of the moments of inertia of some of the common geometrical figures are of such frequent use that the more important of them will be given for future reference. The moment of inertia is in each case taken with respect to a central axis, and will be repre- sented by /, while the radius of gyration will be called k. Rectangle. — Let b and d be the sides, the axis being parallel to the side b. Then 12 ' 12 Triangle. — Let b and d be the base and altitude. Then for an axis parallel to the base, /=M! . k'^ = — 36 ' i8' MOMENTS OF INERTIA OF PLANE AREAS. 171 For an axis through the vertex, bisecting the base, k'^ = — , where b^ is the projection of the base on a hne perpendicular to the axis. Citxle. — Let d be the diameter. Then 64 ' 16 For a central axis perpendicular to the plane of the circle^ Ellipse. — Let a and b be the semi-axes. Then for an axis parallel to a, j_iTab^ ro_b'^ I — , /v- — — 4 4 For an axis parallel to b, j_ ird^b . A2 _ <^^ 4 ' 4 For a central axis perpendicular to the plane of the ellipse, 4 ' 4 Graphic constricction for radius of gyration. — Whenever k'^ can be expressed as the product of two known factors, the value of k can be found by the construction already used in Art. 191. ^2 Thus, in case of a rectangle, for which 1^^ = — , we may put j^i^d d ^j^^^ .^ .^ p. ^^ ^^^ ^^^^ LM=-, MN=-, the 3 4^ 4 3 construction there shown will giv^e MP as the value of k. For the triangle, the axis being parallel to the base, we have B = - • -, and the same construction is applicable. For the ^ 6 . ^' ^' axis through the vertex bisectin£c the base, 1^^=— • — • ^ "" 4 6 1/2 GRAPHIC STATICS. 200. Product of Inertia. — General principles. — Products of inertia of areas are determined by means of the integral calculus in a manner similar to that employed for moments of inertia. The following fundamental principles regarding products of inertia of geometrical figures will be found useful : (i) With reference to two rectangular axes, one of which is an axis of symmetry (Art. 179), the product of inertia is zero. For it is manifest that the products of inertia of two equal elements, symmetrically placed with reference to one of the axes, are numerically equal but of opposite sign. Hence, if the whole area can be made up of such pairs of elements, the total product of inertia is zero. (2) If the two axes are not rectangular, but the area can be divided into elements such that for every element whose dis- tances from the axes are /, ^, there is an equal element whose distances are /, —q, or — /, q, the product of inertia is zero. This includes the preceding as a special case. 201. Products of Inertia of Geometrical Figures. — In each of the following cases the product of inertia is zero : A triangle, one axis containing the vertex and the middle point of the base, the other being any line parallel to the base. A parallelogram, the axes being parallel to the sides and one axis being central. This includes the rectangle as a special case. An ellipse, the axes being parallel to a pair of conjugate diameters, and one axis being central. This includes, as a special case, that in which one axis is a principal diameter and the other is any line perpendicular to it ; and under this case falls also the circle. 202. Approximate Method for Finding Moment of Inertia of Any Area. — To apply the method of Art. 190 to the determina- tion of the moment of inertia of a plane area, we should strictly need to replace the area by an infinite number of parallel forces, proportional to the infinitesimal elements of the given area, and MOMENTS OF INERTIA OF PLANE AREAS. 7j with points *of application in these elements. If, instead, we divide the given figure into finite portions whose several areas are known, and assume forces proportional to those areas to act at their centroids, we may get an approximate value for the moment of inertia, which will be more nearly correct the smaller the elements. This will be illustrated by the area shown in Fig. ^J. Let QR be the axis with reference to which the moment of inertia is to be found, in this case taken as a central axis. Fig. 67" Divide the figure into four rectangular areas as shown, and assume forces numerically equal to these areas to act at their centroids parallel to QR. The force polygon is ABCDE. Draw the funicular polygon corresponding to a pole (9, and let the successive strings intersect QR in A\ B\ C , D\ E'. Take this as a new force polygon, and with any convenient pole distance draw a second funicular polygon, using the same lines of action. Let the first and last strings intersect QR in A" and £■". Then A" E" multiplied by the product of the two pole distances gives the moment of inertia of the four assumed forces, and approximately the moment of inertia of the given 174 GRAPHIC STATICS. figure. If the first pole distance is taken equal to AE (as is the case in Fig. 6y), the radius of gyration may be found by the construction of Art. 191. Thus in Fig. Gy, MP is the radius of gyration as determined by this method. A more accurate result may be reached by dividing the area into narrower strips by lines parallel to QR ; since the narrower such a strip is, the more nearly will the distance of each small element from the axis coincide with that of the centroid of the strip. If the partial areas are taken as narrow strips of equal width, the forces may be taken proportional to the average lengths of the several strips. 203. Accurate Methods. — If the given figure can be divided into parts, such that the area of each is known, and also its radius of gyration with respect to its central axis parallel to the given axis, the above method may be so modified as to give an accurate result. Two methods will be noticed. (i) When the axis is kitown at the stai't. — Let the line of action of the force representing any partial area be taken at a distance from the given axis equal to the radius of gyration of that area with reference to the axis. If this is done, it is evident that the moment of inertia of the system of forces is identical with that of the given area. When the axis is known, the position of the line of action for any force may be found as follows : Let QR (Fig. 6Z) be the given axis, and Q R^ a parallel axis through the centroid of any partial area. Draw KL perpendicu- lar to QR, and lay off LM equal to the radius of gyration of the partial area with respect to Q R^ . Then KM is the length of the radius of gyration with respect to QR. (Art. 196.) Take KN^^M'A to KM and draw Q^R'^ through N parallel to Q R' ; then Q^'R" is to be taken as the line of action of the force representing the partial area in question. (2) When the axis is at first tinhtown. — The method to be MOMENTS OF INERTIA OF PLANE AREAS. 1/5 employed in this case is to let the force representing any partial area act in a line through the centroid of that area ; and then assume the force representing its moment to act in such a line that the moment of this second force shall be numerically equal to the moment of inertia of the partial area. This line may be found as follows : Let k represent the radius of gyra- tion of the partial area with respect to its central axis parallel to the given axis, and a the distance between the two axes. Then the moment of inertia of the partial area with respect to the given axis is A {a'^-\-k'^), if A represents the area. But A {a^-\-k^)=Aala-\ — ). Hence, if a force numerically equal to A is assumed to act with an arm a, then a force equal to its P . moment Aa must act with an arm a-\ — in order that its moment a may equal A {a^-\-P). The distance a-\ — can be found by a simple construction. a Let QR and Q R^ (Fig- 69) be the given axis and the central axis respectively. Draw KL perpendicular to QR and lay off LM equal to k. From M draw a line perpendicular to KM^ in- tersecting KL produced at N. Then KN=a+-- For, in the a Fig. G9 similar triangles KNM, KML, we have KN KM KM KL or KN= Km KL ' But KL=a, and KM^ = a^ + k'^; hence a a This second method is more useful than the first, because in applying it the first funicular polygon can be drawn before the position of the inertia-axis is known. Thus, a very common 1/6 GRAPHIC STATICS. case is that in which the moment of inertia of an area is to be found for a central axis, whose direction is known, while at the outset its position is unknown because the centroid of the area is unknown. If the second method be employed, the first funicular polygon can be drawn at once, and serves to locate the required central axis, as well as to determine the moments of the first set of forces as soon as the axis is known. The central axis and the moment of inertia with respect to it are thus determined by a single construction. Example. — The method last described is illustrated in Fig. 70. The area shown consists of two rectangles, the centroids of which are marked ab and be. The moment of inertia is to be found for a central axis parallel to the longer side of the rectangle ab. Fis. "TO We draw through ab and be lines parallel to the assumed direction of the axis, and take these for the lines of action of forces AB, BC, proportional to the areas of the two rectangles. ABC is the force polygon, and the pole distance is taken equal MOMENTS OF INERTIA OF PLANE AREAS. 177 to AC. The intersection of the strings oa, oc, determines a point of the required central axis ac. The moments of the given forces are proportional to A^B\ B'C. We now have to find the lines of action for the forces A'B\ B'C, in accordance with the method above described. Take any line perpendicular to the axis ac, as VQ, the side of one of the rectangles. From R, the point in which this line intersects the vertical line through be, lay oK RT equal to the central radius of gyration of the rectangle whose centroid is be. To find this central radius of gyration, we know that its value is V — (Art. 199), where d is the length of the side of the rectangle perpendicular to the axis. Hence we take QR = —,RS=—, and make QS the diameter of a semicircle, 3 4 intersecting the vertical line be \n T\ then RT is the required radius of gyration of the rectangle with respect to a central axis. Now draw from T a. line perpendicular to VT, intersect- ing VQ in 17; then the line b'e' drawn through 6^ parallel to the given axis is the line of action of the force B'C. By a similar construction applied to the other rectangle, a'b' is located as the line of action of the force A'B'. The second funicular polygon is now drawn, and the points A", C" are found by the intersection of the strings o'a', o'e' with the axis. Hence the moment of inertia of the given area is equal to A"C" xHxN'. In the figure H is made equal to AC, hence the radius of gyration can be determined by the usual con- struction, and its length is found to be MP. 204. Moment of Inertia of Area Determined from Area of Funicular Polygon. — The method given in Art. 193 for finding the moment of inertia of a system of forces by means of the area of the funicular polygon may be applied with approximate results to the case of a plane figure. If the forces are taken as acting at the centroids of the areas they represent, then to 178 GRAPHIC STATICS. get good results these partial areas should be taken as narrow strips between lines parallel to the axis. If the lines of action are determined as in the first method of the preceding article, then the area enclosed by the funic- ular polygon and the axis represents accurately the required moment of inertia. 205. Product of Inertia Determined Graphically. — To deter- mine the product of inertia of any area, let it be divided into small known parts, and let parallel forces numerically equal to the partial areas be assumed to act at the centroids of these parts. The product of inertia of these forces may then be found as in Art. 194, and its value will represent approximately the product of inertia of the given area. If the partial areas can be so taken that the product of inertia of each with reference to axes through its centroid par- allel to the given axes is zero, the method here given is exact. (Art. 197.) If the partial areas are taken as narrow strips parallel to one of the axes, the condition just mentioned will be nearly fulfilled ; for the product of inertia of each strip for a pair of axes through its centroid, one of which is parallel to its length, will be very small. CHAPTER X. CURVES OF INERTIA. § I. General Principles. 206. Relation between Moments of Inertia for Different Axes through the Same Point. — The moment of inertia with respect to any axis through a given point can be expressed in terms of the moments and product of inertia for any two axes through that point. It is necessary here to use algebraic methods, but the results reached form the basis of graphic constructions. Let OX, 6^F(Fig. 71) be the two given axes; 0, the angle included between them ; P], Pi, etc., the forces of the system ; x^ , J'', the coordinates of the point of application of any force P, referred to the axes OX, OY; /, {/, the perpendicular distances of the same point from O V and OX respectively, so that /=.r'sin^, q=y sm6. Let a', b\ and c' be quantities defined by the equations A;r/2 + P2^2''+ '"_tPx''' a —■ P ^P,^ ^P ^P ,0 tPx'y C '= — ^ —^ — • 179 l8o GRAPHIC STATICS. Then a'"^ sm^^ = = — ^ XP 1.P ^^,^,^,0^tpyHin^e^iPl^ tP XP it,2 ^ _ ^^^y sin^ ^ XPpq . %P %P ' c''^ SI and a' sin 6, b' sin 6, c' sin are respectively the radius of gyration with respect to O V, the radius of gyration with respect to OX, and the product-radius (Art. 189) with respect to OX and OV. The moment of inertia (/) and radius of gyration (k) of the system for the axis OM, making an angle with OX, may now be computed as follows : Let s = perpendicular distance of the point of application of any force P from OM. Then from the geometry of the figure it is seen that s =y sin (6 — (f)) —x^ sin ^. Hence /= tPs^=tPy^ sin2 (61 -(/))- 2 tPx'y' sin (<9 - 0) sin -{-tPx''^ s'm^ cf) ; or, I=b'^tP . sin2 {e-(f>)-2 c^'^tP . sin {6-) sin +^'2^P.sin2(^, the factors involving 6 and <^ being constant for all terms of the summation. Hence y^2=_^ = <^'2sin2((9-c/))-2^^2sin((9- -\- a'"^ s\n^ (j) (2) GENERAL PRINCIPLES. l8i 207. Products of Inertia for Different Axes through the Ori- gin. — The product of inertia with respect to OM and OX may be found as follows : Let A = the required product of inertia ; then A=:tPqs=%Py sin l9 [/ sin((9-(^)-y sin 0] = 2P [/^ sin e sin (^-(^) -Jir'/ sin 6 sin 0] = [^'2 sin 6 sin (6' - <^) - c'^ sin (9 sin ] 5!^. Let /^ = product-radius for axes OM a.nd OX; then /22 = ^ = ^'2 sin (9 sin {6 - (f>) - c'''- sin (9 sin <^. Special case. — The axis OM may be so chosen that A—o. This will be the case if ^'2 sin ((9 -(/))= ^'2 sin0. 208. Inertia Curve. — If on OM (Fig. 71) a point J/ be taken such that the length 6^ J/ depends in some given way upon the value of k, and if similar points be located for all possible direc- tions of OM^ the locus of such points will be a curve which is called a curve of inertia of the system for the center O. The form of the curve will depend upon the assumed law connecting (9Jfwith k. 209. Ellipse or Hyperbola of Inertia. — The simplest curve is obtained by assuming OM to be inversely proportional to k. d^ sin^ Q Let OM—r, and take r'^= — — ^ — > where d"^ is a positive ft quantity, so that d always represents a real length, posirive or negative. Equation (i) of Art. 206 then becomes d^ sin2 (9 = ^'2;^ sin2 ((9 -)- 2 ^'2^2 sin {6-^) sin (/) + ^'2;.2 sin2 0, which is the polar equation of the inertia-curve, r and <^ being the variable coordinates. Let x, y be coordinates of the point M referred to the axes OX, O V. Then r _ X _ y sin ^ sin(^ — (^) sin l82 GRAPHIC STATICS, whence ^ ; ~^' =x'^, r^ sin^ (j) _ 2 sin2 6> ~^' r^ sin (6 — ^) sin (^ _ and the equation becomes ^^ = l^''ij^2_2c'^:^yj^a'y- (3) The form of this equation shows that it represents a conic section whose center is at the origin of coordinates O. This conic may be either an ellipse or a hyperbola. The equation will be discussed more fully in a later article. One fact may, however, be here noticed. If the moment of inertia / is positive for all positions of the axis, the radius of gyration k will be real, whatever the value of (j). But r, the radius vector of the curve, will be real when k is real ; hence in this case the curve is an ellipse. This is always the case if the given forces have all the same sign. If the forces have not all the same sign, it is possible that the value of / may have different signs for different directions of the axis. If this is so, certain values of (/> make k (and therefore r) imaginary. In this case the curve is a hyperbola. The most important case is that in which the given forces have all the same sign which may be taken as plus, so that the moment of inertia is always positive, and the curve an ellipse ; and to this case the discussion will be confined. § 2. Inertia-Ellipses for Systems of Forces. 210. Properties of the Ellipse. — In discussing ellipses of in- ertia use will be made of certain general properties of the ellipse, which, for convenience of reference, will be here sum- marized. For the proof of the propositions stated the reader is referred to works on the conic sections. INERTIA-ELLIPSES FOR SYSTEMS OF FORCES. 183 (i) The equation represents an ellipse if B^—AC is negative; a hyperbola if B^ — AC is positive. (Salmon's Co7ttc Sections, p. 140.) The coordinate axes may be either rectangular or oblique. (2) Two diameters of an ellipse are said to be conjugate to each other if each bisects all chords parallel to the other. If the axes of coordinates coincide with a pair of conjugate diameters, the lengths of which are 2 a' and 2 b', the equation of the curve is A particular case of this equation is that in which the coordi- nate axes are rectangular, being the principal axes of the curve ; in which case we may write a and b instead of a' and b^. (3) In an ellipse, the product of any semi-diameter and the perpendicular from the center on the tangent parallel to that semi-diameter is constant and equal to ab. That is, if r is any radius vector of the curve drawn from the center, and / the length of the perpendicular from the center to the parallel tangent, we have pr=ab where a and b are the principal semi-axes of the curve. (4) Let a' and b' be conjugate semi-diameters. Then each is parallel to the tangent at the extremity of the other. Hence the length of the perpendicular from the center to the tangent parallel to a' is b^ sin 0, where 6 is the angle included between a' and b'. Therefore from the preceding paragraph, a'b' sin d = ab. (5) An elHpse can be constructed, when a pair of conjugate diameters is known, as follows : / ' B C K ^ 1-' /'- /. / y / Fi B 72 184 GRAPHIC STATICS. Let AA', BB' (Fig. 72), be the conjugate diameters, O being the center of the ellipse. Complete the parallelogram OBCA. Divide OA and CA into parts proportional to each other, be- ginning at O and C. Through the points of division of OA draw lines radiating from B' y and through the points of divi- sion of CA draw lines radiating from B. The points of inter- section of the corresponding lines in the two sets are points of the ellipse. In a similar way, the other three quadrants may be drawn. (The location of one point is shown in the figure.) A convenient way to locate the corresponding points of division on OA and CA is to cut these lines by lines parallel to the diagonal OC 211. Discussion of Equation of Inertia-Curve. — We will now examine the equation of the inertia-curve, with reference to the properties of the ellipse above enumerated. (i) If a''^b''^ — c"^ is positive, the equation denotes an ellipse. This cannot be the case if a'"^ and b''^ have opposite signs. But from the definitions of a''^ and b''^ (Art. 206) it is seen that their signs are the same as those of the moments of inertia for Y and X axes respectively. Hence, if there are any two axes through the assumed center for which the moments of inertia have opposite signs, the inertia-curve is a hyperbola. If the moment of inertia has the same sign for all axes through the assumed center, the curve is an ellipse. For, since c''^= ^ (Art. 206), c^ may be made zero by choosing the axes so that the product of inertia with respect to them is zero ; and if c' is zero, and a''^ and b^'^ have the same sign, the quantity a^^b'- — c'^ is positive. INERTIA-ELLIPSES FOR SYSTEMS OF FORCES. 185 This agrees with the conclusion stated in Art. 209. We shall here deal only with ellipses of inertia. (2) If c' = o, the coordinate axes are conjugate axes of the curve. But the condition c' = o means that the product of inertia for the two axes is zero. Hence any two axes for which the product of inertia is zero are conjugate axes of the inertia- curve. (This is true whether the curve is an ellipse or a hyperbola.) (3) By the law of formation of the inertia-conic (Art. 209), the length of the radius vector lying in any line is inversely proportional to the radius of gyration with respect to that line. But by (3) of the last article, the perpendicular from the center on the tangent parallel to any radius vector is inversely propor- tional to the length of that radius vector. Hence the perpen- dicular distance between any diameter and the parallel tangent is directly proportional to the radius of gyration with respect to that diameter. The curve may be so constructed that the length of this perpendicular is equal to the radius of gyration, as follows : From Art. 209, we have j_d^ sin 6 r and from (3) and (4) of the last article we have , ab a^b^ sin P = ^ = ' r r if a! and b^ are conjugate semi-diameters. Now take ^^ sin 6 = ab = a'b^ sin 6y or d' = a'b', and we have k=p, and the equation of the curve becomes (since c^ = o when the axes are conjugate) b'x--\-a y =a . i86 GRAPHIC STATICS. If the equation be written in this form, a' and b' having the meanings assigned in Art. 206, the radius of gyration about any axis through the center of the ellipse is equal to the perpendicular distance betweeji the axis and the parallel tangent to the ellipse. Hereafter we shall mean by inertia-ellipse the curve obtained by taking d^ = a'b^ as above described, so that the radius of gyra- tion for any axis can be found by direct measurement when a parallel tangent to the ellipse is known. 212. To Determine Tangents to the Inertia-Ellipse for Any- Center. — Let the radius of gyration {k) be found for any assumed axis through the given center by one of the methods already described (Arts. 202 and 203). Then two lines par- allel to the axis and distant k from it, on opposite sides, will be tangents to the inertia-ellipse. 213. To Construct the Inertia-Ellipse, a Pair of Conjugate Axes Being Known in Position. — If the positions of two axes conjugate to each other can be found, the ellipse can be drawn by the following method : Determine the radius of gyration for each of the two axes and draw the corresponding tangents as in the preceding article ; then proceed as follows : Let XX', YV (Fig. 73) be the given axes, and let the four tangents determined as above form the parallelogram PQRS. Let A, A', B, B' be the points in which these tangents in- tersect the axes XX', YY'. Then, since each diameter is parallel to the tangents at the extremities of the conjugate Y' Fig. 73 diameter. A, A', B, B' are the extremities of the diameters lying in the given axes. The elhpse can now be constructed as explained in Art. 210 (Fig. 72). This method of constructing the inertia-ellipse is useful INERTIA-CURVES FOR PLANE AREAS. 187 whenever the given system has a pair of conjugate axes which can be located by inspection. 214. Central Ellipse. — It is evident that an inertia-curve can be found with its center at any assumed point. That ellipse whose center is the centroid of the given system is called the central ellipse for the system. Since the central ellipse gives at once the radius of gyration for every axis through the centroid of the system, it enables us to determine readily the radius of gyration for any axis what- ever, by means of the known relation between radii of gyratioa for parallel axes. (Art. 196.) § 3. Iiiertia-Cnrves for Plane Areas. 215. General Principles. — The principles deduced in the treatment of inertia-curves for systems of forces are all true for the case of plane areas. But special difficulties arise in dealing with areas, because of the fact that the system of forces equiva- lent to any area consists of an infinite number of forces. The principles already developed are, however, sufficient to deal at least approximately with all areas, and accurately with many cases. 216. Inertia-Curve an Ellipse. — Since the forces conceived to replace the elements of area (Art. 198) have all the same sign, the value of k'^ is always positive, and the inertia-curve is always an ellipse. (Arts. 209 and 211.) 217. Cases Admitting Simple Treatment. — Whenever a pair of conjugate diameters can be located, and the radius of gyra- tion determined for each, the inertia-ellipse can be at once drawn as in Art. 213. This will be the case whenever it is possible to locate readily a pair of axes for which the product of inertia is zero. (i) If there is an axis of symmetry, this and any line perpen- dicular to it are a pair of conjugate axes (and in fact the principal axes) of the inertia-ellipse whose center is at their intersection. (Art. 200.) 1 88 GRAPHIC STATICS. (2) If two axes can be located in such a way that for every element of area whose distances from the axes are/, q, there is an equal element whose distances are p, —q, or — /, q, the product of inertia is zero for the two axes, and these are there- fore a pair of conjugate axes of the inertia-ellipse whose center is at their intersection. This of course includes the case when there is an axis of symmetry. (Art. 200.) When a pair of conjugate axes is known, the radius of gyration is to be found by one of the methods of Art. 202 or Art. 203 ; the ellipse can then be drawn exactly as explained in Art. 213. If a pair of conjugate axes cannot be located by inspection, the inertia-ellipse cannot be so readily constructed. Such cases will not be here treated. As examples of areas, in which the principal axes of the inertia-curve can be located by inspection, may be mentioned the cross-section of the I-beam, the deck-beam, the channel-bar, and other shapes of structural iron and steel. Many geometrical figures possess axes of symmetry. In others a pair of conjugate axes can be located by principle (2). Some of these will be discussed in the next article. Example. — Draw the central ellipse for the deck-beam sec- tion shown in Fig. 74. -1-%T»' as 0~^~ [Suggestions. — Since there is an axis of symmetry, this contains one of the principal axes of the eUipse. The other can be drawn as soon as the centroid of the section is known. Find the radius of gyration for each axis by Art. 202, and then construct the elUpse as explained in "^2- Art. 213.] 2 1 8. Central Ellipses for Geometrical Fig- ures. — In many of the simple geometrical figures, not only can a pair of conjugate axes be located by inspection, but the radius of gyration for each of these axes can be found by a simple construction, so that the central ellipse can be readily drawn. Some of these cases will be here summarized. INERTIA-CURVES FOR PLANE AREAS. 189 (i) Parallelogram. — Let ABCD (Fig. 75) be the parallelo- gram; then XX\ YV, drawn through the centroid parallel to the sides, are a pair of conjugate axes of the cen- ^ tral ellipse. Let AB = b, BC=dj and let h= the perpendicular distance be- tween AB and DC. The moment of inertia of the parallelogram with re- spect to the axis XX' is equal to the moment of inertia of a rectangle of sides b and //. Hence k^, the square of the radius of gyration for this axis, is — ■• The length of k can be found by the construction used in case of the rectangles in Fig. 70. The following modification of the method is, however, more convenient : Make EF=\BC, EG = \ BC, and draw a semicircle with FG as a diameter. From E draw a line perpendicular to BCy inter- secting the semicircle at /. Lay off EH=EI\ then a line through H parallel to XX^ is a tangent to the central ellipse. For by construction, BC EH=EI-- V12 And since the projection of BC on a line perpendicular to XX\s> equal to //, the projection of EH on the same line is equal to V 12 that is to k. The tangent parallel to the side BC may be found in a simi- lar way. It may, however, be located more simply as follows : It is evident that the distance between YY' and a tangent par- allel to it, measured along AB, bears the same ratio to AB that EH does to BC Hence, the parallelogram formed by the four tangents, two parallel to XX' and two parallel to YY\ is simi- lar to the parallelogram ABCD. 190 GRAPHIC STATICS. Fig.' 75 shows this parallelogram and also the ellipse. (2) Triangle. — Let ABC (Fig. j6) be the triangle ; b = Y length of the base BC\ b' = length of projection of BC on a line perpendicular to AD ; d = a.\- titude measured perpendicular to BC. AD and a line through the centroid parallel to BC are a pair of conjugate axes of the central ellipse (Art. 217). From Art. 199, the radius of gyration for a central axis parallel to BC is Let XX' be this axis, and H the point in which 3 6 it intersects ^C Then HC=\AC. Take HK=\AC=\HC, and make KC the diameter of a semicircle. From H draw HI perpendicular to AC, intersecting the semicircle at /. Make HL — HI\ then the line through L parallel to XX^ is a tangent to the central ellipse. For the radius of gyration with respect to XX is to HL as the altitude d is to AC. Again, for the axis AD, the radius of gyration is A/ — (Ai 199). Make diameter of a semicircle. .rt. DE=\ BC and DF-- 24 \ BC, and take EF as a From D draw a line perpendicular to BC, intersecting the semicircle in M\ and make DG = DM; then a line from G parallel to AD is a tangent to the central ellipse. The figure shows the parallelogram formed by the two tan- gents parallel to XX' and the two parallel to AD, and also the central ellipse. (3) Ellipse. — From Art. 199, the radii of gyration of an ellipse with respect to the two principal diameters are -| a and J b. Hence the central ellipse of inertia is similar to the given ellipse, its semi-axes being J a and ^ b. A special case of this is a circle, for which the central curve is a circle whose radius is half that of the given circle. INERTIA-CURVES FOR PLANE AREAS. 191 Fis. TT Y ^ /— C ^ X' / (F \ X 1 A ) \ I ,/v^!\^ G_^ \ 1 1 A E D B Y (4) Semicircle. — Let ABC (Fig. jj) be the semicircle, O being the centroid. (The point O may be located by the method described in Art. 184.) From symmetry it is evident that the principal axes of the central ellipse are XX^ and YY\ drawn through O, re- spectively parallel and per- pendicular to AB. With respect to the axis YY\ the radius of gyration is evidently \ r, the same as for the whole circle. Hence two lines parallel to YY' and dis- tant ^ r from it arctangents to the central ellipse. Again, the radius of gyration of the semicircle with respect to AB as an axis is also \ r, the same as for the whole circle. To find it for the axis XX\ with Z^ as a center, and radius \ r, 9 2 '^ draw an arc intersectmg XX' at F\ then OF^ = DF —QD'^. But DF'i's, equal to the radius of gyration with respect to AB, and 0D\^ the distance between XX' and AB \ hence (Art. 196) OF is equal to the radius of gyration with respect to XX . Hence if two lines are drawn parallel to XX , each at a distance from it equal to OF, they will be tangents to the central ellipse. The ellipse can now be drawn in the usual manner. 219. Summary of Results. — By the principles and methods developed in the present chapter, inertia-curves can be drawn for all the simpler cases that may arise ; namely, whenever a pair of conjugate axes can be located by inspection. This will be the case whenever the product of inertia can be seen to be zero for any pair of axes ; and it includes every case of an area possessing an axis of symmetry. It is believed that this chapter contains as complete a discus- sion as is needed by the student of engineering. Those who desire to pursue the subject further may consult other works. Fig. 34 U) Scale, 1 incli^Gfect PLATE I. / DEF Scale, 1 inch=6,000 lbs. (C) Scale: I in. = 4,000 lbs. Scale: I in, Fig. 42 PLATE II. 8.1 5.8 4.5 4.8 5.7 o OOOO ooo!) Linear Scale, 1 in. = lo ft. Force Scale, 1 in.= 80,000 lbs Fio:.47 PLATE III. 9.0 8.1 5.8 4.8 .5.7 4.8 o OOOO oooo i' 8.1 5.8 4.5 7.1 5.7 4.8 o OOOO OO OG PLATE IV. 4.5 4.5 7.1 4.8 5.7 4.8 CO 4.0 4.0 o O OOO O O O O I ■■"»'^'-p'-"- M r SS 'Sx^ » }l .6 J .d ,V >o fl .n U / oo oo o O O o oo oo ^ 0-f i-f O'f 87 e-i ri 6-i 0-6 0-f z-f Of o C5 to "oi -fT. ?5 v» X; (al CT o b S o * © o o ° o o 8-i A aiv^d sqi ooo'og-'ni T 'oivos oodoj "if OS^'Ul T 'OXDOS JLDOUl'J n O O o VI 6-i i^s *^Tl3: I MATHEMATICAL TEXT- BOOKS PUBLISHED BY MACMILLAN & CO. flDecban!cg> RIGID DYNAMICS. An Introductory Treatise. By W. Steadman Aldis, M.A. ^i.oo. ELEMENTARY APPLIED MECHANICS. Part II. Transverse Stress. By T. Alexander and A. W. Thompson. 8vo. $2.75. EXPERIMENTAL MECHANICS. A Course of Lectures delivered to the Royal College of Science for Ire- land. By Sir R. S. Ball, LL.D., F.R.S. Second Edition. With Illustrations. 12 mo. $1.50. Teachers of experimental mechanics can have no better assistant than this book, for the lectures are happy in their illustrations, clear in their demonstrations, and admirable in their arrangement. — JManchester Guardian. LESSONS ON THERMODYNAMICS. By R. E. Baynes, M.A. i2mo. $1.90. Works by W. H. Besant, D.Sc, Lecturer of St. Johns College, Cambridge. A TREATISE ON HYDROMECHANICS. Fifth Edition, revised. Part I. Hydrostatics. i2mo. $1.25. A TREATISE ON DYNAMICS. $1.75. ELEMENTARY HYDROSTATICS. i6mo. $1.00. ELEMENTS OF DYNAMIC. An Introduction to the Study of Motion and Rest in Solid and Fluid Bodies. By W. Kingdon Clifford, F.R.S. Part I. Books I.-III. i2mo. $1.90. Part II. Book IV. and Appendix. 12 mo. $1.75. APPLIED MECHANICS. An Elementary General Introduction to the Theory of Structures and Machines. By James H. Cotterill, F.R.S. 8vo. ^5.00. The book is a valuable one for students. ... A very admirable feature of the book is the frequent interjecting of practical problems for solution, at the close of the different sections. — School of Mines Quarterly. This work constitutes a manual of applied mechanics which, in its complete- ness and adaptation to the requirements of the engineer student, is without a rival. — Academy. ELEMENTARY MANUAL OF APPLIED MECHANICS. By Prof. J. H. Cotterill, F.R.S., and J. H. Slade. i2mo. $1.25. MECHANICS. GRAPHICAL STATICS. By LuiGi Cremona. Two Treatises on the Graphical Calculus and Re- ciprocal Figures in Graphical Calculus. Authorized Enghsh Transla- tion by T. Hudson Beare. 8vo. $2.25. A TREATISE ON ELEMENTARY DYNAMICS. For the Use of Colleges and Schools. By William Garnett, M.A., D.C.L. Fifth Edition, revised. $1.50. ELEMENTARY STATICS. By H. Goodwin, D.D., Bishop of Carlisle. Second Edition. 75 cents. Works by John Greaves, M.A. A TREATISE ON ELEMENTARY STATICS. Second Edition, revised. i2mo. ^1.90. It can be unreservedly recommended. — Nature. STATICS FOR BEGINNERS. i6mo. 90 cents. HYDROSTATICS. By Alfred George Greenhill. In preparatio7i. THE APPLICATIONS OF PHYSICAL FORCES. By A. Guillemin. Translated and Edited by J. Norman Lockyer, F.R.S. With Colored Plates and Illustrations. Royal 8vo. ^6.50. ELEMENTARY DYNAMICS OF PARTICLES AND SOLIDS. By W. M. Hicks. i2mo. ^1.60. ELEMENTARY MECHANICS. Stage I. By J. C. Horobin, B.A. With Numerous Illustrations. i2mo. Cloth. 50 cents. Stages II. and III in preparation. THE ELEMENTS OF GRAPHICAL STATICS. A Text-Book for Students of Engineering. By Leander M. Hoskins, C.E., M.S., Prof, of Applied Mechanics, Leland Stanford Jr. Uni- versity, Palo Alto, Cal. A TREATISE ON THE THEORY OF FRICTION. By J. H. Jellett, B.D., late Provost of Trinity College, Dublin. ^2.25. THE MECHANICS OF MACHINERY. By Alexander B. W. Kennedy, F.R.S. With Illustrations. i2mo. $3.50. HYDRODYNAMICS. By H. Lamb. A Treatise on the Mathematical Theory of Fluid Motion. 8vo. $3.00. Works by the Rev. J. B. Lock, M.A. Fellow of Gonville and Caius College, Cambridge. DYNAMICS FOR BEGINNERS. i6mo. ^i.oo. Mr. Lock writes like one whose heart is in his work. He shows rare ingenuity in smoothing the difficulties in the path of the beginner, and we venture to predict that his book will meet with a wide and hearty welcome among practical teachers. — Athenccum. MECHANICS. ELEMENTARY STATICS. i6mo. $i.io. The whole subject is made interesting from beginning to end, and the proofs of the various propositions are very simple and clear. We have no doubt that the book will be appreciated by all who have an opportunity of judging of its merits. — A^atiire. MECHANICS FOR BEGINNERS. Part I. Mechanics of Solids. 90 cents. An introductory book, which, it is hoped, may be found useful in schools and by students preparing for the elementary stage of the Science and Art and other public examinations. ELEMENTARY HYDROSTATICS. In preparation. Works by S. L. Loney, M.A., Fellow of Sidney Sussex College, Cambridge. A TREATISE ON ELEMENTARY DYNAMICS. New and Enlarged Edition. i2mo. $1.90. Solutions of the Examples contained in the Above. In the Press. THE ELEMENTS OF STATICS AND DYNAMICS. Parti. Elements of Statics. $1.25. Part II. Elements of Dynamics. |5i.oo. Complete in one volume. i6mo. $1.90. AN ELEMENTARY TREATISE ON KINEMATICS AND DYNAMICS. By James Gordon Macgregor, M.A., D.Sc, Munro Professor of Physics, Dalhousie College, Halifax. i2mo. $2.60. The problems and exercises seem to be well chosen, and the book is provided with a useful Index. Altogether it will repay a careful study. — Educational Times. Works by G. M. Minchin, M.A., Professor of Mathematics in the Royal bid tan Engineering College. A TREATISE ON STATICS. Third Edition, corrected and enlarged. Vol. I. Equilibrium of Coplanar Forces. 8vo. ^2.25. Vol. II. Statics. 8vo. $4.00. UNIPLANAR KINEMATICS OF SOLIDS AND FLUIDS. i2mo. $1.90. A TREATISE ON ELEMENTARY MECHANICS. For the use of the Junior Classes at the Universities, and the Higher Classes in Schools. With a collection of examples by S. Parkinson, F.R.S. Sixth Edition. i2mo. $2.25. LESSONS ON RIGID DYNAMICS. By the Rev. G. Pirie, M.A. i2mo. $1.50. ELEMENTARY STATICS. By G. Rawlinson, M.A. Edited by E. Sturges. 8vo. $1.10. Works by E. J. Routh, LL.D., F.R.S., Fellow of the Senate of the University of London. A TREATISE ON THE DYNAMICS OF A SYSTEM OF RIGID BODIES. With Examples. New Edition Revised and Enlarged. 8vo. Part I. Elementary. Fifth Edition Revised and Enlarged. ^3.75- Part II. Advanced. $3.75. 4 MECHANICS. — HIGHER PURE MATHEMATICS. STABILITY OF A GIVEN STATE OF MOTION, Particularly Steady Motion. 8vo. I2.25. HYDROSTATICS FOR BEGINNERS. By F. W. Sanderson, M. A. i6mo. ^i.io. New edition preparing. Mr. Sanderson's work has many merits, and is evidently a valuable text-book, being the work of a practical and successful teacher. — Cajiada Educational Monthly. SYLLABUS OF ELEMENTARY DYNAMICS. Part I. Linear Dynamics. With an Appendix on the Meanings of the Symbols in ' Physical Equations. Prepared by the Association for the Improvement of Geometrical Teaching. 4to. 30 cents. A TREATISE ON DYNAMICS OF A PARTICLE. By P. G. Tait, M.A., and W. J. Steele. Sixth Edition, Revised. $3.00. Works by Isaac Todhunter, F.R.S. Late Fellow and Prificipal Mathematical Lecturer of St. Johns College. MECHANICS FOR BEGINNERS. With Numerous Examples. New Edition. i8mo. $1.10. Key. $1.75. A TREATISE ON ANALYTICAL STATICS. Fifth Edition. Edited by Prof. J. D. Everett, F.R.S. i2mo. ^2.60. A COLLECTION OF PROBLEMS IN ELEMENTARY MECHANICS. By W. Walton, M.A. Second Edition. $1.50. PROBLEMS IN THEORETICAL MECHANICS. Third Edition, Revised. With the addition of many fresh Problems. By W. Walton, M.A. 8vo. ^4.00. Ibigber pure flDatbematic6> Works by Sir G. B. Airy, K.C.B., formerly Astronomer-Royal. ELEMENTARY TREATISE ON PARTIAL DIFFERENTIAL EQUATIONS. With Diagrams. Second Edition. i2mo. $1.50. ON THE ALGEBRAICAL AND NUMERICAL THEORY OF ERRORS of Observations and the Combination of Observations. Second Edition, revised. 12 mo. $1.75. NOTES ON ROULETTES AND GLISSETTES. By W. H. Besant, D.Sc, F.R.S. Second Edition, Enlarged. ^1.25. A TREATISE ON THE CALCULUS OF FINITE DIFFERENCES. By the late George Boole. Edited by J. F. Moulton. Third Edition. i2mo. ^2.60. ELEMENTARY TREATISE ON ELLIPTIC FUNCTIONS. By Arthur Cayley, D.Sc, F.R.S. New Edition preparing. HIGHER PURE MATHEMATICS. DIFFERENTIAL CALCULUS. With Applications and Numerous Examples. An Elementary Treatise. By Joseph Edwards, M. A. i2mo. ^2.75. A useful text-book both on account of the arrangement, the illustrations, and ap- plications and the exercises, which contain more easy examples than are usually given. — Educational Times. AN ELEMENTARY TREATISE ON SPHERICAL HARMONICS And Subjects Connected with Them. By Rev. N. M. Ferrers, D.D.. F.R.S. i2mo. $1.90. Works by Andrew Russell Forsyth, M.A. A TREATISE ON DIFFERENTIAL EQUATIONS. 8vo. ^^3.75. As Boole's " Treatise on Different Equations " was to its predecessor, Wymer's, this treatise is to Boole's. It is more methodical and more complete, as far as it goes, than any that have preceded it. — Educational Times. THEORY OF DIFFERENTIAL EQUATIONS. Part I. Exact Equations and Pfaff's Problems. 8vo. $3.75. AN ELEMENTARY TREATISE ON CURVE TRACING. By Percival Frost, M.A. 8vo. ^3.00. DIFFERENTIAL AND INTEGRAL CALCULUS. With Applications. By Alfred George Greenhill, M.A. i2mo. $2.00. APPLICATION OF ELLIPTIC FUNCTIONS. By Alfred George Greenhill, M.A. i2mo. $3.50. AN ELEMENTARY TREATISE ON THE DIFFERENTIAL AND INTEGRAL CALCULUS. By G. W. Hemming, M.A. 8vo. $2.50. THEORY OF FUNCTIONS. By James Harkness, M.A., Professor of Mathematics, Bryn Mawr Col- lege, and Frank Morley, A.M., Professor of Mathematics, Haver- ford College, Pa. In preparation. INTRODUCTION TO QUATERNIONS. With Examples. By P. Kelland, M.A., and P. G. Tait, M.A. Second Edition. i2mo. $2.00. HOW TO DRAW A STRAIGHT LINE. A Lecture on Linkages. By A. B. Kempe, B.A. With Illustrations. Nature Series. 121110. 50 cents. DIFFERENTIAL CALCULUS FOR BEGINNERS. With Examples. By Alexander Knox, B.A. i6mo. 90 cents. MESSENGER OF MATHEMATICS. Edited by J. W. L. Glaisher. Published Monthly. 35 cents each number. Works by Thomas Muir, Mathematical Master in the High School, Glasgoiu. A TREATISE ON THE THEORY OF DETERMINANTS. With Examples. 12 mo. New Edition in Preparation. HIGHER PURE MATHEMATICS. THE THEORY OF DETERMINANTS 111 the Historical Order of its Development. Part I. Determinants in General. Leibnitz (1693) to Cayley (1841). 8vo. ^2.50. TREATISE ON INFINITESIMAL CALCULUS. By Bartholomew Price, M.A., F.R.S., Professor of Natural Philosophy, Oxford. Vol. I. Differential Calculus. Second Edition. 8vo. $3.75. Vol. II. Integral Calculus, Calculus of Variations, and Differential Equations. 8vo. Reprinting. Vol. III. Statics, including Attractions ; Dynamics of a Material Par- ticle. 8vo. ^4.00. Vol. IV. Dynamics of Material Systems. Together with a Chapter on Theoretical Dynamics, by W. F. Donkin, M.A. 8vo. $4.50. A TREATISE ON THE THEORY OF DETERMINANTS, And their Applications in Analysis and Geometry. By Robert Scott, M.A. 8vo. $3.50. FACTS AND FORMULAE IN PURE MATHEMATICS And Natural Philosophy. Containing Facts, Formulae, Symbols, and Definitions. By the late G. R. Smalley, B.A., F.R.A.S. New Edi- tion by J. M'DowELL, M.A., F.R.A.S. i6mo. 70 cents. MATHEMATICAL PAPERS OF THE LATE REV. J. S. SMITH. Savilian Professor of Geometry in the University of Oxford. With Portrait and Memoir. 2 vols. 4to. In prepai-ation. AN ELEMENTARY TREATISE ON QUATERNIONS. By P. G. Tait, M.A., Professor of Natural Philosophy in the University of Edinburgh. Third Edition, Much Enlarged. 8vo. $5.50. Works by Isaac Todhunter, F.R.S., Late Principal Lecturer on Mathematics in St. Johns College, Cambridge. AN ELEMENTARY TREATISE ON THE THEORY OF EQUATIONS. i2mo. ^1.80. A TREATISE ON THE DIFFERENTIAL CALCULUS. i2mo. ^2.60. Key. $2.60. A TREATISE ON THE INTEGRAL CALCULUS And its Applications. i2mo. ^2.60. Key. $2.60. ♦ AN ELEMENTARY TREATISE ON LAPLACE'S, Lamp's and Bessel's Functions. i2mo. $2.60. TRILINEAR CO-ORDINATES, And Other Methods of Modern Analytical Geometry of Two Dimen- sions. An Elementary Treatise. By W. Allen Whitworth, M.A. 8vo. $4.00. AA'AL YTICAL GE OME TR Y. analytical (Beometr^. .plane and soud SOLID GEOMETRY. By W. Steadman Aldis. 121110. $1.50. ELEMENTS OF PROJECTIVE GEOMETRY. Bv LuiGi Cremona, LL.D. Translated by Charles Leudesdorf, M.A. 8vo. $3.25. EXERCISES IN ANALYTICAL GEOMETRY. By J. M. Dyer, M.A. With Illustrations. i2mo. $1.25. A TREATISE ON TRILINEAR CO-ORDINATES, The Method of Reciprocal Polars, and the Theory of Projections. By the Rev. N. M. Ferrers. Fourth Edition. i2mo. $1.75. Works by Percival Frost, D.Sc, F.R.S. Lecturer in Mathematics, King's College, Cambridge. AN ELEMENTARY TREATISE ON CURVE TRACING. 8vo. $3.00. SOLID GEOMETRY. A New Edition, revised and enlarged, of the Treatise, by Frost and WoLSTENHOLME. Third Edition. 8vo. $6.00. Hints for the Solution of Problems in the above. 8vo. $3.00. THE ELEMENTS OF SOLID GEOMETRY. By R. Baldwin Hayward, M.A., F.R.S. i2mo. 75 cents. THE GEOMETRY OF THE CIRCLE. By W. J. McClelland, M.A., Trinity College, Dublin ; Head Master of Santry School. Crown 8vo. Illustrated. $1.60. AN ELEMENTARY TREATISE ON CONIC SECTIONS And Algebraic Geometry. With Examples and Hints for their Solution. By G. Hale Puckle, M.A. Fifth Edition, Revised and Enlarged. i2mo. $1.90. Works by Charles Smith, M.A. Fellow and Tutor of Sidney Sttssex College, Cambridge. AN ELEMENTARY TREATISE ON CONIC SECTIONS. Seventh Edition. i2mo. $1.60. Key. $2.60. The best elementary work on these curves that has come under our notice. — Academy. AN ELEMENTARY TREATISE ON SOLID GEOMETRY. i2mo. $2.60. Works by Isaac Todhunter, F.R.S. Late Principal Lecturer on Afathe7natics in St. yohn's College. PLANE CO-ORDINATE GEOMETRY, As Applied to the Straight Line and the Conic Sections. i2mo. ^1.80. Key. By C. W. Bourne, M.A. i2mo. $2.60. EXAMPLES IN ANALYTICAL GEOMETRY Of Three Dimensions. New Edition, Revised. i2mo. $1.00. ANALYTICAL GEOMETRY FOR SCHOOLS. By Rev. T. G. Vyvyan. Fifth Edition, Revised. $1.10. A TREATISE ON ALGEBRA. 5>' Charles Smith, M.A. REVISED AND ENLARGED EDITION, $1.90. ^*^ This new edition has been greatly iniproved by the addition of a chapter o)i Differential Equations and other cJianges. " Your Smith's ' Treatise on Algebra ' was used in our University classes last session, and with very great satisfaction. . . . The general adoption of these texts would mark an epoch in mathematical teaching." — Prof. W. B. Smith, University of Missouri. '• Those acquainted with Mr. Smith's text-books on conic sections and solid geometry will form a high expectation of this work, and we do not think they will be disappointed. Its style is clear and neat, it gives alternative proofs of most of the fundamental theorems, and abounds in practical hints, among which we may notice those on the resolution of expression into factors and the recognition of a series as a binomial expansion." — Oxford Review. HIGHER ALGEBRA. ^y H.S. Hall, M.A., and S. R. Knigbt, B.A. A Sequel to the Elementary Algebra for Schools by the same Authors. Fourth edition, containing a collection of three hundred Miscellaneous Examples which will be found useful for advanced students. 12 mo. $1.90. " . . . It is admirably adapted for College students as its predecessor was for schools. It is a well arranged and well reasoned-out treatise, and contains much that we have not met with before in similar works. For instance, we note as specially good the articles on Convergency and Diver- gency of Series, on the treatment of Series generally, and the treatment of Continued Fractions. . . . The book is almost indispensable and will be found to improve upon acquaintance." — The Academy. " We have no hesitation in saying that, in our opinion, it is one of the best books that have been published on the subject. . . . The last chap- ter supplies a most excellent introduction to the Theory of Equations. We would also specially mention the chapter on Determinants and their appli- cation, forming a useful preparation for the reading of some separate work on the subject. The authors have certainly added to their already high repu- tation as writers of mathematical text-books by the work now under notice, which is remarkable for clearness, accuracy, and thoroughness. . . . Al- though we have referred to it on many points, in no single instance have we found it wanting." — The School Gicardian. MACMILLAN & CO., NEW YORK. \