IN MEMORIAM FLORIAN CAJORI ELEMENTS OF GRAPHIC STATICS. THE ELEMENTS OF GRAPHIC STATICS 21 STe.xt=33oofc for Students of Engineering BY L. M. HOSKINS PROFESSOR OF PURE AND APPLIED MECHANICS IN THE LELAND STANFORD JUNIOR UNIVERSITY; FORMERLY PROFESSOR OF MECHANICS IN THE UNIVERSITY OF WISCONSIN Xeto gork MACMILLAN AND CO. AND LONDON 1892 A U rights reserved COPYRIGHT, 1892, BY MACMILLAN AND CO. ORI TYPOGRAPHY, BY J. S. GUSHING & .Co., BOSTON, U.S.A. PRESSWORK BY BERWICK & SMITH, BOSTON, U.S.A. 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 simplicity of presentation. The matter treated has been limited to the development of fundamental principles, and their application to 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 simplify, 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 literature. 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. 918236 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 4 2. External and Internal Stresses 4 2 3. Determination of Internal Stresses . . . . ~ 46 PART II. STRESSES IN SIMPLE STRUCTURES. CHAPTER IV. INTRODUCTORY. i . Outline of Principles and Methods 53 PAGE viii CONTENTS. CHAPTER V. ROOF TRUSSES. FRAMED STRUCTURES SUSTAINING STATIONARY LOADS. 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 101 CHAPTER VII. TRUSSES SUSTAINING MOVING LOADS. i . Bridge Loads 112 2. Truss Regarded as a Beam 115 3. Truss with Parallel Chords Sustaining Concentrated Loads . . 117 4. Parallel Chords Uniformly Distributed Moving Load . . . . 130 5. Truss with Curved Chords Uniform Panel Loads 132 6. 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. i. 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. PART I. GENERAL THEORY. CHAPTER L 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 point 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 unit 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 \.\\Q pound, which will now be denned. 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 denned, 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 forces, 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 magnitude 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 diagrams 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 magni- tude 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 each side of the line designated. Thus, in 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. 2. Composition of Concurrent 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 c, (A) Fig. 3 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. The 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 AN, 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 of 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 Fig. 3 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, BC, 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 polygon 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 DE, 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. 3. Equilibrium of Concurrent Forces. 1 6. Definition. A system of forces acting on a body is in equilibrium 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. 7 polygon is closed for any system of concurrent forces ', the system is in equilibrium. The comparison of this with the analytical conditions of equilibrium is given in Art. 22. 1 8. 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 ABCDE'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 of .* and y which shall satisfy the equation x + y = 10, is indeterminate; while the problem, given 2 x + 3 = 7, to find the value of x, is determinate. The case in which a finite number of solutions is possible may be called incompletely determinate. Thus, the problem, given x 2 -f x 6 = 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. (1) 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 AB (Fig. 6) in the direction of any line XX, is represented in magnitude and direction 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' ; hence, (i) For the equilibrium of any concurrent forces, the sum of their resolved parts in any direction must be zero. Fig. G * The term "resolute" has been proposed by J. B. Lock (" Elementary Statics ") to denote what is here denned as the resolved part of a force. I0 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. SToU.,., ^A5 \ t <*-&S. 4 ov - - 4- '^~ frxAjLo - w ^GJLa^. - ' L&KCwJj^ <*^A. ^XV. QJ&; MM- ^jL , -^ K ')/v^ A" YV^J^AJ^/S*JC* ^> V v : ^ ~^yvJk"^ ^ v ^>v -Ujtl^ C \f -^- '- ., . , y^J 5 -^ , V*A^ ^-'' WNK^ ^^ < IM. V" ^ *'- K^y^ o , t> A^^.,,.^.o, va- \ i "5? rs yr.- X V>- W^X.-- vVSj^Jv (V^^ V^yJ CHAPTER II. 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 corn- 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. II 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 given 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 their 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 ac and cd. This is evidently the resultant of AB, BC, and CD. In the same way, this resultant COMPOSITION OF NON-CONCURRENT FORCES. I3 combined with DE gives for their resultant a force whose mag- 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 dc. 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.) 27. 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, cd, 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 AO and OB acting in any lines parallel to AO, OB, whose point of intersection falls upon ab\ BC is 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, oc. Similarly prolong oc to intersect cd, 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 co 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, oc. 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. x- ^A^v* 1 *' C/ 28. Funicular Polygon. The polygon whose sides are oa, ob, oc, od, oe, is called a. funicular 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. T 6 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 pole of the force polygon. The lines drawn from the pole to the vertices of the force polygon may be called rays. The sides of the funicular polygon are sometimes called strings. Each ray in the force diagram is parallel to a corresponding string in the space diagram. As a mechanical rule, it should be remembered that the two rays draivn to the extremities of the line representing any force are respectively parallel to the two strings which intersect on the line of action of that force. The rays terminating at the extremities of any side of the force polygon represent in magnitude and direction two com- ponents that may replace the force represented by that side ; while the corresponding strings represent the lines of action of these components. Thus (Fig. 9) the force represented by BC may be replaced by two forces represented by BO, OC, acting in the lines bo, oc. The pole distance of any force is the perpendicular distance from the pole to the line representing that force in the force diagram. It may evidently be considered as representing the resolved part, perpendicular to the given force, of either of the components represented by the corresponding rays. Thus, OM (Fig. 9) is the pole distance of AB ; and OM represents the resolved part, perpendicular to AB, of either OA or OB. The student should notice particularly that the pole distance repre- sents a force magnitude and not a length. 31. Forces not Possessing a Single Resultant. It may hap- pen that the first and last sides of the funicular polygon are parallel, so that the above construction fails to give the line of action of the resultant. This will be the case if the pole is chosen on the line AE (Fig. 9), because the first and last sides of the funicular polygon are respectively parallel to OA and * V *r 2 acting at D in the line AD ; and P 3 applied at B and acting In the line joining B with the middle point of AD. RequVJtto com- pletely determine P l} P 2 , and P 3 . A V 2 3 GRAPHIC STATICS. 3. Resolution into Non-concurrent Systems. 43. To Replace a Force by Two Non-concurrent Forces. - This may be done in an infinite number of ways. The lines of action of the two components must intersect at a point on the line of action of the given force, and they must further satisfy the same conditions as concurrent forces. (Art. 21.) 44. To Replace a Force by More Than Two Non-concurrent Forces. This may be done by first resolving the given force into two forces by the preceding article, and then resolving one or both of its components in the same way. This problem and that of Art. 43 are indeterminate. (See note, Art. 20.) To make such a problem determinate, something must be specified regarding the magnitudes and lines of action of the required components. We shall consider some of the particular cases which are of frequent use in the treatment of practical problems. 45. Resolution of a Force into Two Parallel Components. Let it be required to resolve a force into two components having given lines of action parallel to its own. If the given force be reversed in direction, it will form with the required components a system in equilibrium. The com- ponents may then be determined by the method of Art. 38. v (5 Example. Let the student solve two particular cases of this problem, taking the line of action of the given force (i) between those of the components, (2) outside those of the components. 46. Resolution of a Force into Three Components. Problem. -To resolve a force into three components having known lines of action. If the given force be reversed in direction, it will form, with the required forces, a system in equilibrium. Hence these forces may be determined by either of the two methods given RESOLUTION INTO NON-CONCURRENT SYSTEMS. 29 in Arts. 40 and 42. Or, the following reasoning may be employed, leading to the same construction as that of Art. 42. Let AD (Fig. 16) represent the given force in magnitude and direction, and ad its line of action ; the lines of action of the components being given as ab, be, cd. Since the given force may be assumed to act at any point in the line ad, let its point of application be taken at M, the point of intersection of Fig. 16 ad and cd. Resolve it into two components acting in the lines cd and MN, N being the point of intersection of ab and be. These components are represented in magnitude and direction by AC, CD. Let AC, acting in the line MN (also marked ac), be resolved into components having lines of action ab, be. These components are given in magnitude and direction by AB, BC, drawn parallel respectively to ab, be. Hence, the given force, acting in ad, is equivalent to the three forces represented in magnitude and direction by AB, BC, CD, acting in the lines ab, be, cd. If the line of action of the given force does not intersect any one of the given lines ab, be, cd, within the limits of the draw- ing, it may be replaced by two components ; then each may be resolved in accordance with the above method, and the results combined. If the three lines ab, be, cd, are all parallel to ad, or if the four lines intersect in a point, the problem is inde- terminate. This is evident from the preceding articles, since, by methods already given, a part of AD can be replaced by two components acting in any two of the given lines, as ab, be ; and the remaining part by two components acting in ab, cd; 30 GRAPHIC STATICS. or in be, cd. This construction evidently admits of infinite variation. For another method of solving the above problem, see Clarke's " Graphic Statics," p. 16. 4. Moments of Forces and of Couples. 47. Moment of a Force. Definition. The moment of a force with respect to a point is the product of the magnitude of the force into the perpendicular distance of its line of action from the given point. The moment of a force with respect to an axis perpendicular to the force is the product of the magni- tude of the force by the perpendicular distance from the axis to the line of action of the force. If the moment is taken with respect to a point, that point is called the origin of moments. The perpendicular distance from the origin, or axis, to the line of action of the force is called the arm. Rotation tendency of a force. The moment of a force measures its tendency to produce rotation about the origin, or axis. Thus, if a rigid body is fixed at a point, but free to turn about that point in a given plane, any force acting upon it in that plane will tend to cause it to rotate about the fixed point. The amount of this tendency will be proportional both to the magnitude of the force and to the distance of its line of action from the given point ; that is, to the moment of the force with respect to the point, as above defined. Rotation in any plane may have either of two opposite directions, which may be distinguished from each other by signs plus and minus. Rotation with the hands of a watch supposed placed face upward in the plane of the paper will be called negative, and the opposite kind positive. It would be equally legitimate to adopt the opposite convention, but the method here adopted agrees with the usage of the majority of writers. The sign of the moment of a force is regarded as the same as that of the rotation it tends to produce about the origin. MOMENTS OF FORCES AND OF COUPLES. 3! Moment represented by the area of a triangle. If a triangle be constructed having for its vertex the origin of moments, and for its base a length in the line of action of a force, numerically equal to its magnitude, then the moment of the force is numeri- cally equal to double the area of the triangle. This follows at once from the definition of moment. 48. Moment of Resultant of Two Non-parallel Forces. Propo- sition. The moment of the resultant of two non-parallel forces with reference to a point in their plane is equal to the algebraic sum of their separate moments with reference to the same point. In Fig. 17, let AB, BC, and AC represent in magnitude and direction two forces and their resultant ; and let ab, be, ac be their lines of action, intersecting in a point N. Choose any point M as the origin of moments. Lay off NP = AB\ NQ = BC\ and NR = AC. Then the moments of the three forces are re- spectively equal to double the areas of the triangles MNP, MNQ, MNR. These three triangles have a common side MN, which may be considered the base ; hence their areas are propor- tional to their altitudes measured from that side. These alti- tudes are PP 1 , QQ', RR 1 ', perpendicular to MN. Now, if PP" is parallel to MN, P"R' is equal to PP f ; also RP" equals QQ' (since they are homologous sides of the equal triangles NQQ', PRP"). Hence RR I = PP' + QQ' ) and therefore Area MNR = area MNP + area MNQ ; which proves the proposition. If the origin of moments is so taken that the moments of AB and BC have opposite signs, the demonstration needs modi- fication. The student should attempt the proof of this case for i ir VT>T,N v,Vo~ fJM\^- |\J I\A ^ - /viMGL himself. ^^ - 3 2 GRAPHIC STATICS. ^ , " 49. Moment of a Couple. Definition. The moment of a _ -N couple about any point in its plane taken as an origin is the algebraic sum of the moments of the two forces composing it with reference to the same origin. Proposition. The moment of a couple is the same for every origin in its plane, and is numerically equal to the product of the magnitude of either force into the arm of the couple. (Art. 7.) The proof of this proposition can readily be supplied by the student. 50. Moment of the Resultant of Any System. Proposition. - The algebraic sum of the moments of any given complanar forces, with reference to any origin in their plane, is equal to the moment of their resultant force or resultant couple with reference to the same origin. The construction employed in proving this proposition is similar to that used in Art. 27, which the student may profit- ably review at this point. Referring to Fig. 9, let the given forces be represented in magnitude and direction by AB, BC, CD, DE, and in lines of action by ab, be, cd, de. Let the origin of moments be any point in the space diagram. As in Art. 27, replace AB by AO, OB, acting in lines ao, ob ; replace BC by BO, OC, acting in lines bo, oc ; replace CD by CO, OD, acting in lines co, od; and replace DE by DO, OE, acting in lines do, oe. (For brevity, we refer to a force as AB, meaning "the force represented in magnitude and direction by AB."} Now by Art. 48, we have, whatever the origin, Moment of A B = moment of A O + moment of OB; BC= " " BO+ " " OC; " " CD = " " CO + " " OD; " " DE= " " DO+ " " OE. Since the forces represented by BO and OB have the same line of action, their moments are numerically equal but have MOMENTS OF FORCES AND OF COUPLES. 33 opposite signs ; and similar statements are true of CO and OC, and of DO and OD. Hence, the addition of the above four equations shows that the sum of the moments of AI3, BC, CD, DE, is equal to the sum of the moments of AO and OE. Now the given system has either a resultant force or a resultant couple. In the first case the resultant of the system is the resultant of AO and OE, and its moment is equal to the algebraic sum of their moments, by Art. 48. In the second case (which occurs only when E coincides with A), the resultant couple is composed ot AO and OE (which in this case are equal and opposite forces), and the moment of the couple is, by definition, equal to the algebraic sum of the moments of AO and OE. Hence, in either case, the proposition is true. It should be noticed that the proof here given applies to systems of parallel forces, as well as to non-parallel systems. The proposition of Art. 48 could be extended to the case of any number of forces, by considering first the resultant of two forces, then combining this resultant with the third force, and so on ; but the method would fail if, in the process, it became necessary to combine parallel forces. The method here adopted is not subject to this failure. 51. Condition of Equilibrium. It follows from what has preceded, that if a given system is in equilibrium, the alge- braic sum of the moments must be zero, whatever the origin. For, in case of equilibrium, AO and OE (Fig. 9) are equal and opposite and have the same line of action ; hence, the sum of their moments (which is the same as the sum of the moments of the given forces) is equal to zero. Conversely, If the algebraic sum of the moments is zero for every origin, the system must be in equilibrium. For, if it is not, there is either a resultant force or a resultant couple. But the moment of a force is not zero for any origin not on its line of action ; and the moment of a couple is not zero for any 34 GRAPHIC STATICS. origin. For a fuller discussion of the conditions of equilibrium, see Arts. 58 and 59. 52. Equivalent Couples. Proposition. If a system has for its resultant a couple, it is equivalent to any couple whose moment is equal to the sum of the moments of the forces of the system. For, as already seen (Art. 31), when the resultant is a couple, the force polygon is closed. Let the initial and final points of the force polygon coincide at some point A, and let O be the pole. Then the forces of the resultant couple are represented in magnitude and direction by AO, OA. Since the position of O is arbitrary, the force AO (or OA) may be made anything whatever in magnitude and direction. Also the line of action of the force AO may be taken so as to pass through any chosen point. Hence, the resultant couple may have for one of its forces any force whatever in the plane of the given system ; and the other force will have such a line of action that the moment of the couple will be equal to the sum of the moments of the given forces. This reasoning is equally true if the given system is a couple. Hence, a couple is equivalent to any other couple having the same moment. In other words, all couples wJiosc moments are equal are equivalent ; and conversely, all equiva- lent couples have eqtial moments. [NOTE. The construction above discussed fails if all forces of the given system are parallel to the direction chosen for the forces of the resultant couple. For then the force polygon is a straight line, and if the pole is chosen in that line, the strings of the funicular polygon are parallel to the lines of action of the forces, and the polygon cannot be drawn. But in this case, the system may first be reduced to a couple whose forces have some other direction, and this couple may be reduced to one whose forces have the direction first chosen. Hence, the proposition stated holds in all cases.] 53. Moment of a System. Definition. The moment of a system of forces is the algebraic sum of the moments of the forces of the system. GRAPHIC DETERMINATION OF MOMENTS. 35 54. Moments of Equivalent Systems. Proposition. The moments of any two equivalent systems of complanar forces with respect to the same origin are equal. This follows immediately from the preceding articles. For, if the two systems are equivalent, each is equivalent to the same resultant force or resultant couple, and the moments of the two systems are therefore each equal to the moment of this resultant and hence to each other. 5 . Graphic Determination of Moments. 55. Proposition. If, through any point in the space dia- gram a line be drawn parallel to a given force, the distance intercepted upon it by the two strings corresponding to that force, multiplied by the pole distance of the force, is equal to the moment of the force with respect to the given point. By the strings "corresponding to" a given force are meant the two strings which intersect at a point on its line of action. Let AB (Fig. 18) represent the magnitude and direction of a force whose line of action is ab t and let M be the origin of moments. Let O be the pole, and OK (=H) the pole distance of the given force. Draw the strings oa, ob, and through M draw a line parallel to ab y inter- secting oa and ob in P and Q. FI S . is Then it is to be proved that the moment of the given force with respect to M is equal to ffxPQ* Let h equal the perpendicular distance of M from ab. Then the required moment is AB x //. But since the similar triangles OAB and RPQ have bases AB, PQ, and altitudes H, //, respec- tively, it follows that =~ ; hence, which proves the proposition. 36 GRAPHIC STATICS. It should be noticed that PQ represents a length, while H represents a. force magnitude. Hence, the moment of the given force with respect to M is equal to the moment of a force H with an arm PQ. [It may, in fact, be shown that the given force is equivalent to an equal force acting in the line PQ (whose moment about M is therefore zero), and a couple with forces of magnitude H, and arm PQ. For AB acting in ab is equivalent to AO and OB acting in ao and ob respectively. Also, AO acting in ao is equivalent to forces represented by AK and KO acting respectively in PQ and in a line through P at right angles to PQ ; and OB may be replaced by forces represented by OK and KB, the former acting in a line through Q perpendicular to PQ, and the latter in PQ. But AK and KB are equivalent to AB ; hence, the proposition is proved.] 56. Moment of the Resultant of Several Forces. The mo- ment of the resultant of any number of consecutive forces in the force and funicular polygons may be found by a method similar to that just described. Thus, let Fig. 19 represent the force polygon and the funicular polygon for six forces, and let it be re- quired to find the moment of the resultant of the four forces represented in the force polygon by BC, CD, DE, and EF, with respect to any point M. The re- sultant of the four forces is represented by BF, and acts in a line through the intersection of ob and of. Through M draw a line parallel to BF, intersecting ob and of in P and Q respectively. Then PQ multiplied by OK, the pole distance of BF, gives the required moment. This method does not apply to the determination of the moment of the resultant of several forces not consecutive in the force polygon. SUMMARY OF CONDITIONS OF EQUILIBRIUM. 37 57. Moments of Parallel Forces. The method of Arts. 55 and 56 is especially useful when it is desired to find the mo- ments of any or all of a system of parallel forces ; since with such a system the pole distance is the same for all forces, and the moments are therefore proportional to the intercepts found by the above method. Example. Assume five parallel forces at random; choose an origin, and determine their separate moments, also the moment of their resultant, by the method of Arts. 55 and 56. 6. Summary of Conditions of Equilibrium. 58. Graphical and Analytical Conditions of Equilibrium Com- pared. It has been shown (Art. 35) that the conditions of equilibrium for a system of complanar forces acting on a rigid body are two in number : (I) The force polygon must close. (II) Any funicular polygon must close. The analytical conditions * are the following : (1) The algebraic sum of the resolved parts of the forces in any direction must be zero. (2) The algebraic sum of their moments for any origin must be zero. The condition (i) is readily seen to be equivalent to (I). For if the sides of the force polygon be orthographically projected upon any line, their projections will represent in magnitude and direction the resolved parts of the several forces parallel to the line ; and, if the force polygon is closed, the algebraic sum of these projections is zero, whatever the direction of the assumed line. (See Art. 22.) It may also be seen that condition (II) carries with it (2). For, if every funicular polygon closes, the system is equivalent to two equal and opposite forces having the same line of action (Arts.. 31, * See Minchin's Statics, Vol. I, p. 114. 38 GRAPHIC STATICS. 35, and 50) ; and the sum of the moments of these two forces must be zero. A further comparison may be made. The analytical condi- tion (2) carries (i) with it ; and similarly the graphical condi- tion (II) carries with it the condition (I). That (2) includes (i) may be seen as follows : If the sum of the moments is zero for one origin M^ there can be no resultant couple, neither can there be a resultant force unless with a line of action passing through MI. If the sum of the moments is zero for two origins, Mi and M 2 , the resultant force, if one exists, must act in the line M^M*. If the sum of the moments is zero also for a third origin M 3) not on the line MiM- 2 , there can be no resultant force. It follows at once that condition (i) must hold. That (II) includes (I) may be shown as follows : Let A and E be the first and last points of the force polygon, and choose a pole O\. Then, if the funicular polygon closes, O\A and OiE are parallel. Choose a second pole O 2 , and, if the funic- ular polygon again closes, O 2 A and O 2 E are parallel. A must then be parallel to OiO 2 , unless A and E coincide. Now, choose a third pole O 3 , not on O\O^\ if the funicular polygon for this pole closes, O 3 A and O S E must be parallel. But this is impossible unless A and E coincide ; that is, unless condition (I) holds. The last result may be reached in another way. With any pole O draw a funicular polygon and suppose it to close. The system is thus reduced to two forces acting in the same line oa. Hence, there is no resultant couple, and if there is a resultant force, its line of action is oa. With the same pole draw a second funicular polygon, the first side being o'a', parallel to oa. If this polygon closes, there can be no result- ant force, for if one existed it would act in the line o'a' ; and there would thus be two resultant forces, acting in different lines oa and o'a', which is impossible. SUMMARY OF CONDITIONS OF EQUILIBRIUM. 39 59. Summary. It is now evident that the conditions neces- sary to insure equilibrium may be stated in several different ways, both analytically and graphically. To summarize : A. Analytically : There will be equilibrium if either of the following conditions is satisfied : (1) The sum of the moments is zero for each of three points not in the same line. (2) The sum of the moments is zero for each of two points, and the sum of the resolved parts is zero for a line not per- pendicular to the line joining those two points. (3) The sum of the moments is zero for one point, and the sum of the resolved parts is zero for each of two directions. B. Graphically : There will be equilibrium if either of these three conditions is satisfied : (1) A funicular polygon closes for each of three poles not in the same line. (2) Two funicular polygons close for the same pole. (3) One funicular polygon closes and the force polygon closes. CHAPTER III. INTERNAL FORCES AND STRESSES. I . External and Internal Forces. 60. Definitions. It was stated in Art. 3 that every force acting upon any body is exerted by some other body. In what precedes, we have been concerned only with the effects pro- duced by forces upon the bodies to which they are applied. It has therefore not been needful to consider the bodies which exert the forces. It is now necessary to consider forces in another aspect. The forces applied to any particle of a body may be either external or internal. An external force is one exerted upon the body in question by some other body. An internal force is one exerted upon one portion of the body by another portion of tlie same body. It is important to note, however, that the same force may be internal from one point of view, and external from another. Thus, if a given body be conceived as made up of two parts, X and y, a force exerted upon X by Y is internal as regards the whole body, but external as regards the part X. Thus, let _ AB (Fig. 20) represent / ^ s\ -+ - [ ! \^ -- ^ a bar, acted upon by two A Fig. 20 forces of equal magnitude applied at the ends paral- lel to the length of the bar, in such a way as to tend to pull it apart. These two forces are exerted upon AB by some other 40 EXTERNAL AND INTERNAL FORCES. 4I bodies not specified. If the whole bar be considered, the external forces acting upon it are simply the two forces named. But suppose the body under consideration is AC, a portion of AB. The external forces acting upon this body are (i) a force at A, already mentioned, and (2) a force at C, exerted upon AC by CB. This latter force is internal to the bar AB, but external to AC. 61. Conditions of Equilibrium Apply to External Forces. In applying the conditions of equilibrium deduced in previous articles, it must be remembered that only external forces are referred to. It is also important to notice that the principles apply to any body or any portion of a body in equilibrium ; and the system of forces in every case must include all forces that are external to the body or portion of a body in question. Thus, if the bar AB (Fig. 20) be in equilibrium under the action of two opposite forces P and Q, applied at A and B respectively, as shown in the figure, the principles of equilibrium may be applied either to the whole bar, or to any part of it, as AC. (a) For the equilibrium of the whole bar AB, we must have P=Q, these being supposed the only external forces acting on the bar. (ft) For the equilibrium of AC, the force exerted upon AC by CB must be equal and opposite to P. This latter force is external^ AC, though internal \& the whole bar. The method just illustrated is of frequent use in the investi- gation of engineering structures. It is often desired to deter- mine the internal forces acting in the members of a structure, and the general method is this : Direct the attention to such a portion of the whole structure or body considered that the internal forces which it is desired to determine shall be external to the portion in question. (See Art. 67.) 42 GRAPHIC STATICS. 2. External and Internal Stresses. 62. Newton's Third Law. Let X and Y be any two portions of matter ; then if X acts upon Y with a certain force, Y acts upon X with a force of equal magnitude in the opposite direc- tion. This is the principle stated in Newton's third law of motion, that "to every action there is an equal and contrary reaction." It is justified by universal experience. 63. Stress. Definition. Two forces exerted by two por- tions of matter upon each other in such a way as to constitute an action and its reaction, make up a stress. Illustrations. The earth attracts the moon with a certain force, and the moon attracts the earth with an equal and opposite force. The two forces constitute a stress. Two electrified bodies attract (or repel) each other with equal and opposite forces. These two forces constitute a stress. Any two bodies in contact exert upon each other equal and opposite pressures (forces), constituting a stress. By the magnitude of a stress is meant the magnitude of either of its forces. 64. External and Internal Stresses. It has been seen that two portions of matter are concerned in every stress. Now the two portions may be regarded either as separate bodies, or as parts of a body or system of bodies which include both. A stress acting between two parts of the same body (or system of bodies) is an internal stress as regards that body or system. A stress acting between two distinct bodies is an external O stress as regards either body. It is important to notice that the same stress may be internal from one point of view, and external from another. Thus, if a given body be considered as made up of two parts X and F, a stress exerted between X and F is internal to the whole body, but external to either X or F. EXTERNAL AND INTERNAL STRESSES. 43 Illustration. Consider a body AB (Fig. 21) resting upon a second body Y, and supporting another body X, as shown. If the weight of the body AB be disregarded, the forces acting upon it are (i) the down- ward pressure (say P) exerted by X at the surface A, and the upward pressure (say Q) exerted by Fat B ; these forces being equal and opposite, since the body is in equilibrium. Now the body X is acted upon by AB with a force equal and opposite to P, and these Fi s- S1 two forces constitute a stress which is external to AB. There is also an external stress exerted between AB and Y at B. But let AB be considered as made up of two parts, AC and CB. Then (Art. 60) CB exerts upon AC a force upward, and AC exerts upon CB a force downward. These two forces are an action and its reaction, and constitute a stress which is internal to the body AB. This same stress is, however, external to either AC or CB. An equivalent stress evidently exists at every section between A and B. (When we refer to the force acting upon CB at C, we mean the resultant of all forces exerted upon the particles of CB by the particles of AC. This resultant is made up of very many forces acting between the particles. Also the stress at C means the stress made up of the two resultants of the forces exerted by AC and Z? 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 by 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 of 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 A C 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 F. Fig. 22) separated by a plane surface, the external forces may have either of three 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 F. In case of a compressive stress, the force exerted by X upon Fhas the direction from X toward F; and the force exerted by Fupon X has 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 shear) 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 r ') 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.] I. 46 GRAPHIC STATICS. 3. Determination of Internal Stresses. 67. 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 Y. These forces are internal as re- p \- ^ gards the whole body. In P order to determine them so 3 Fig. 23 far as possible, we proceed as follows : Let the resultant of all the forces exerted by Y upon Xb& 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 P lt P*, P 3 , and 7! Since P lt P.>, and P 3 are supposed known, T can be determined. In fact, T is equal and opposite to the resultant of P ly P.,, and P 3 . 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 under any external forces be conceived as made up of tivo parts X and Y, then the internal forces exerted by X upon Y, together witJi the external forces acting on Y, form a system in equilibrium. DETERMINATION OF INTERNAL STRESSES. 47 As an immediate consequence, we may state that the result- ant of the forces exerted by X upon Y is equivalent to the result- ant of the external forces acting on X ; and is eqnal 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. 68. 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 : (1) 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, Y, 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 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 4 8 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 Y 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 said, 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 A r , 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. 67 to this case. Divide the body into two parts, X and Y, 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 P if P 2t P&, 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, 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. 6 5 .) 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 r introduce but three unknown stresses. There may, however, be cases in which the stresses iri" s^j cannot all be determined 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 equilibrium, the force polygon being ABCDA. Choosing a 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. 5I 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. 67. 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 t 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 OB] oc a tension OC\ and od & 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." . ry> - : PART II. STXSSS IN SIMPLE STRUCTURES. CHAPTER IV. INTRODUCTORY. i. Outline of Principles and MetJiods. 73. The Problem of Design. When any structure is sub- jected to the action of external forces, there are brought into action certain internal stresses in the several parts of the struc- ture. The nature and magnitudes of these stresses depend upon the external forces acting (Art. 65). In designing the structure, each part must be so proportioned that the stresses induced in it will not become such as to break or injure the material. To determine these internal stresses, when the external forces are wholly or partly given, is the problem of design, so far as it will be here treated. 74. External Forces. The external forces acting on a struc- ture must generally be completely known before the internal stresses can be determined. These external forces are usually only partly given, and the first thing necessary is to determine them fully. The external forces include (i) the loads which the structure is built to sustain, and (2) the reactions exerted by other bodies upon the structure at the points where it is supported. The former are known or assumed at the outset and the latter are to be determined. 53 54 GRAPHIC STATICS. 75. Two Classes of Structures. Structures may be divided into two classes, according as they may or may not be treated as rigid bodies in determining the reactions. This may be illustrated as follows : Let a bar AB (Fig. 28) be supported in a horizontal position at two points A and B, the supports being smooth so that the ip pressures on the bar at i A and B are vertical. ^WM Let a known load P be applied to the beam at a given point, and let it be required to determine the reactions at A. and B. This is a determinate problem ; for there are three parallel forces in equilibrium, two being known only in line of action. This problem was solved in Art. 38. But let AC (Fig. 29) be a rigid bar supported at three points, A t B, and C, the reactions at those points being vertical. Ip ,p ,p Let any known loads ^ L B c be applied at given points, and let it be required to determine . 39 the three reactions. This problem is indeterminate ; for any number of sets of values of the three reactions may be found, which, with the applied loads, would produce equilibrium if acting on a rigid body. (See Art. 40.) Since, however, an actual bar is not a perfectly rigid body, such a problem as the one just stated is, in reality, determinate. But it cannot be solved without making use of the elastic properties of the material of which the body is composed. The two classes of problems are, therefore, the following : (i) those in which the reactions can be determined by treating the structure as a rigid body, and (2) those in which the determination of the reactions involves the theory of elasticity. We shall at present deal only with the former class of prob- OUTLINE OF PRINCIPLES AND METHODS. 55 lems. Structures coming under this class will be called simple structures. 76. Truss. A truss is a structure made up of straight bars with ends joined together in such a manner that the whole acts as a single body. The ends of the bars are, in practice, joined in various ways ; but in determining the internal stresses, the connections are assumed to be such that no resistance is offered by a joint to the rotation of any member about it. Such a structure to be indeformible must be made up of triangular elements ; for more than three bars hinged together in the form of a polygon cannot constitute a rigid whole. If the external forces are applied to the truss only at the points where the bars are joined, the internal stress at any section of a mem- ber will be a simple tension or compression, directed parallel to the length of the bar. (See Art. 68.) The most important classes of trusses are roof trusses and bridge trusses. The methods used in discussing these classes are, of course, applicable to any framed structures under similar conditions. 77. Loads on a Truss. The loads sustained by a truss may be either fixed or moving. A fixed (or dead} load is one whose point of application and direction remain constant. A moving (or live) load is one whose point of application passes through a series of positions. Fixed loads may be either permanent or temporary. The loads on a roof truss are usually all fixed, but are of various kinds, viz., the weight of the truss itself and of the roof covering, which is a permanent load ; the weight of snow lodging on the roof, and the pressure of wind, both of which are temporary loads. A bridge truss supports both fixed and moving loads. The former include the weight of the truss itself, of the roadway, of all lateral and auxiliary bracing (permanent loads) ; and of snow (a temporary load). The latter consist of moving trains 56 GRAPHIC STATICS. in the case of railway bridges, and of teams or crowds of animals or people in the case of highway bridges. 78. Combination of Stresses Due to Different Causes. When a truss is subject to a variety of external loads, it is often convenient to consider the effect of a part of them separately. If tensile and compressive stresses are distinguished by signs plus and minus, the stress in any member due to the combined action of any number of loads is equal to the algebraic sum of the stresses due to the loads acting separately. A proof of this proposition might be given ; but it may be accepted as sufficiently evident without formal demonstration. 79. Beams. Another class of bodies to be treated is included under the name beam. A beam may be defined as a bar (usually straight) resting on supports and carrying loads. The loads and reactions are commonly applied in a direction transverse to the length of the bar ; but this is not necessarily the case. The internal stresses in any section of a beam are less sjmple than those in the bars of an ideal jointed frame such as a truss is assumed to be. A discussion of beams is given in Chap. VI. 80. Summary of Principles Needed. It will be well to summarize at this point the main principles and methods which will be employed in the discussion of the problems that follow. The general problem presented by any structure consists of two parts : (a) the determination of the unknown external forces (or reactions) and (b) the determination of the internal stresses. (a) In the case of simple structures the unknown reactions are usually two in number, and the cases most commonly pre- sented are the following : ist. Their lines of action are known and parallel. 2nd. The line of action of one and the point of application of the other are known. OUTLINE OF PRINCIPLES AND METHODS. 57 Since all the external forces form a system in equilibrium, these two cases fall under the general problems discussed in Arts. 38 and 39. (b) In the case of a jointed frame or truss, the lines of action of all internal forces are known, since they coincide with the axes of the truss members. (Art. 68.) In determin- ing their magnitudes we may have to deal with the following problems in equilibrium : ist. The system in equilibrium may be completely known, except the magnitudes of two forces. 2nd. The magnitudes of three forces may be unknown. The first case may be solved by simply making the force polygon close. (See Art. 35.) The second case may be solved by the method of Art. 40, which consists in making the force and funicular polygons close ; or by the method of Art. 42 ; or by the principle of moments (Art. 51). The student should be thoroughly familiar with the prob- lems and principles here referred to. In the following chapters we proceed to their application. 8 1. Division of the Subject. The subject of the design of structures, so far as here dealt with, will be treated in three divisions. The first relates to framed structures sustaining only stationary loads ; the second to beams sustaining both fixed and moving loads ; the third to framed structures sustain- ing both fixed and moving loads. Among structures of the first class, the most important are roof trusses ; hence, these are chiefly referred to in the next chapter. For a similar reason, the chapter devoted to the third class of structures refers principally to bridge trusses. The chapter on beams precedes that on bridge trusses, for the reason that the methods used in dealing with a beam under moving loads form a useful introduction to those employed in treating certain classes of truss problems. CHAPTER V. ROOF TRUSSES. FRAMED STRUC- TURES SUSTAINING STATIONARY LOADS. i. Loads on Roof Trusses. 82. Weights of Trusses. Among the loads to be sustained by a roof truss is the weight of the truss itself. Before the structure is designed, its weight is unknown. But, since it is necessary to know the weight in order that the design may be correctly made, the method of procedure must be as follows : Make a preliminary estimate of the weight, basing it upon knowledge of similar structures ; or, in the absence of such knowledge, upon the best judgment available. Then design the various truss members, compute their weight, and com- pare the actual weight of the truss with the assumed weight. If the difference is so great as to materially affect the design of the truss members, a new estimate of weight must be made, and the computations repeated or revised. No more than one or two such trials will usually be needed. As a guide in making the preliminary estimate of weight, the following formulas may be used. They are taken from Mer- riman's " Roofs and Bridges," being intended to represent approximately the data for actual structures, as compiled by Ricker in his ''Construction of Trussed Roofs." Let /span in feet ; a = distance between adjacent trusses in feet ; W= total weight of one truss in pounds. Then for wooden trusses and for wrought iron trusses 58 LOADS ON ROOF TRUSSES. 59 83. Weight of Roof Covering. The weight of roof covering can be correctly estimated beforehand from the known weights of the materials. The following data may be employed, in the absence of information as to the specific material to be used. (See Merriman's " Roofs and Bridges," p. 4.) The numbers denote the weight in pounds per square foot of roof surface. Shingling : Tin, I Ib. ; wooden shingles, 2 to 3 Ibs. ; iron, I to 3 Ibs. ; slate, 10 Ibs. ; tiles, 12 to 25 Ibs. Sheathing : Boards I in. thick, 3 to 5 Ibs. Rafters : 1.5 to 3 Ibs. Purlins : Wood, I to 3 Ibs. ; iron, 2 to 4 Ibs. Total roof covering, from 5 to 35 Ibs. per square foot of roof surface. 84. Snow Loads. The weight of snow that may have to be borne will differ in different localities. For different sections of the United States the following maybe used as the maximum snow loads likely to come upon roofs. Maximum for northern United States, 30 Ibs. per square foot of horizontal area covered. For latitude of New York or Chicago, 20 Ibs. per square foot. For central latitudes in the United States, lolbs. per square foot. The above weights are given in Merriman's " Roofs and Bridges." They are in excess of those used by some Bridge and Roof companies. 85. Wind Pressure Loads. The intensity of wind pressure against any surface depends upon two elements : (a) the velocity of the wind, and (b) the angle between the surface and the direction of the wind. Theory indicates that the intensity of wind pressure upon a surface perpendicular to the direction of the wind should be proportional to the square of the velocity of the wind relative to the surface. As an approximate law this is borne out by experiment. If / denotes the pressure per unit area, and i> the velocity of the wind, the law is expressed by the formula 6o GRAPHIC STATICS. p kv*. Here k is proportional to the density of air. Its numerical value may be taken as 0.0024, ^ the "nits of force, length, and time are the pound, foot, and second respectively. If the wind strikes a surface obliquely, experiment shows that the resulting pressure has a direction practically normal to the surface. The tangential component is inappreciable, owing to the very slight friction between air and any fairly smooth surface. The intensity of the normal pressure depends upon the angle at which the wind meets the surface. For a given velocity of wind let/ a denote the normal pressure per unit area, when the direction of wind makes an angle a with the surface, and / the pressure per unit area due to the same wind striking a surface perpendicularly. Then the follow- ing formula * has been given : 2 sin a It will rarely be necessary to use values of p n greater than 50 Ibs. per square foot. The following table gives values of the coefficient of p n in the above formula for different values of a. The value of p,, may be taken ;'as from 40 to 50 Ibs. per square foot. a 2 sin a a 2 sin a I -f sin- a I + sin 2 a o.oo 50 0.97 10 o-34 6o c 0.99 20 0.61 -70 I.OO 30 0.80 80 I.OO 40 0.91 90 I.OO * This formula is given by various writers. It is cited by Langley (" Experiments in Aerodynamics," p. 24), who attributes it to Duchemin. Professor Langley's elaborate experiments show so close an agreement with the formula that it may be used without hesi- tation in estimating the pressure on roofs. ROOF TRUSS WITH VERTICAL LOADS. 6 1 2. Roof Truss with Vertical Loads. 86. Notation. The method of determining internal stresses in the case of vertical loading will be explained by reference to the form of truss shown in Fig. 30. The method will be seen to be independent of the particular form of the truss. For designating the truss members and the lines of action of external forces a notation will be employed similar to that used in previous chapters. Let each of the areas in the truss dia- gram be marked with a letter or other symbol as shown in Fig. 30 ; then the truss member or force-line separating any two areas may be designated by the two symbols belonging to those areas. Thus, the lines of action of the external forces are ab, be, cd t etc., and the truss members are gh t hb t hi, etc. It is to be noticed that the lines representing the truss mem- bers represent also the lines of action of forces, namely, the internal forces in the members. The joint, or point at which several members meet, may be designated by naming all the surrounding letters. Thus, bcih t hijg are two such points. 87. Loads and Reactions. The loads now considered are assumed to be applied in a vertical direction, and to act at the upper joints of the truss. This assumption as to the points of application may in some cases represent very nearly the facts ; in other cases the loads will, in reality, be applied partly at intermediate points on the truss members. If the latter is the case, the load borne upon any member is assumed to be divided between the two joints at its ends. In this case the member will be subject not only to direct tension or compression, but to bending. With the latter we are not here concerned, although it must always be considered in designing the member. The ends of the truss are supposed to be supported on hori- zontal surfaces, and the reaction at each point of support is assumed to have a vertical direction. If the loading is symmetrical with reference to a vertical line ) 1 V 1 ' 62 GRAPHIC STATICS. through the middle of the truss, it is evident that each reaction is equal to half the total load. If the loading is not symmetri- cal, the reactions cannot be determined so simply. They may, however, be readily computed by either graphic or algebraic methods. Graphically, the problem is identical with that solved in Art. 38. The truss is treated as a rigid body, the external forces acting upon it being the loads and reactions, which form a system of parallel forces in equilibrium. Two of these forces (the reactions) are unknown in magnitude, but known in line of action. The construction for determining their magnitudes is as follows : Draw the force polygon A BCD EF for the five loads ; choose a pole O y draw rays OA, OB, OC, etc., and draw the funicular polygon as shown in Fig. 30. The two polygons are to be completed by including the reactions FG, GA, and both poly- gons must close. We may draw first og, the closing line of the funicular polygon, and then the ray OG parallel to it, thus determining the point G in the force polygon. The reactions are now shown in magnitude and direction by FG and GA. 88. Determination of Internal Stresses. When the external forces are all known, the internal stresses may be found very readily. The only principle needed is, that for any system of forces in equilibrium, the force polygon must close. The con- struction will now be explained. Considering any joint of the truss (Fig. 30) as ghijg^ fix the attention upon the portion of the truss bounded by the broken line in the figure. This portion is a body in equilibrium under the action of four forces whose lines of action coincide with the axes of the four bars g/i, hi, ij, jg, respectively. These forces are internal as regards the truss as a whole, but external to the part in question ; each force being one of the pair constituting the internal stress at any point of the bar. Such a force acts from the joint if the stress in the bar is a tension ; toivard it if the stress is a compression. (Art. 69.) ROOF TRUSS WITH VERTICAL LOADS. 63 Since these four forces form a system in equilibrium, their force polygon must close. This condition will enable us to fully determine the magnitudes of the forces, provided all but two are known, since the polygon can then be constructed as in Art. 18. 1 We cannot, however, begin with the joint just considered, since at first the four forces are all unknown in magnitude. If, however, we start with the joint gab/i, the polygon of forces can be at once drawn. For, reasoning as above, it is seen that the portion of the truss immediately surrounding this joint is in equilibrium under the action of four forces : the reaction in the line ga, the load in the line ab, and the internal forces in the lines bh, Jig. Of these forces, two (the reaction and the load) are completely known ; and it is neces- sary only to draw a polygon of which two sides represent the known forces and the other two sides are made parallel to the members bJi, Jig. Such a polygon is shown in Fig. 30 ; and BH and HG represent in magnitude and direction the forces whose lines of action are b/i, Jig. 64 GRAPHIC STATICS. Evidently, BH and HG represent also the magnitudes (Art. 63) of the internal stresses in the two members bh and Jig. The nature of these stresses may be found as follows : Since the four forces represented in the polygon GABHG are in equilibrium, and since GA, AB are the directions of two of them, the directions of the other two must be BH, HG. Hence, BH acts toward the joint and HG from it. This shows that the stress in bh is a compression, while that in hg is a tension. Passing now to the joint bciJib, it is seen that of the four forces whose lines of action meet there, two are fully known, namely, the load BC acting vertically downward and the inter- nal force in hb acting toward the joint (since the stress is com- pressive), while the remaining two (viz., the internal forces in ci and i/i) are unknown in magnitude and direction. Since, however, the unknown forces are but two in number, the force polygon can be completely drawn, and is represented by the quadrilateral HBCIH. The directions of the forces are found as in the preceding case, and it is seen that the bars ci and /// both sustain compressive stresses. The process may be continued by passing to the remaining joints in succession, in such order that at each there remain to be determined not more than two forces. The complete con- struction is shown in Fig. 30. It is evident that the loads AB and EF might have been omitted without changing the stresses in any of the truss members. For their omission would leave as the complete force polygon for external forces BCDEGB, and the two reactions would be GB and EG ; but the force diagram would be otherwise unchanged. The great convenience of the notation adopted is now seen.* * This is known as Bow's notation. The notation adopted in Part I involves the same idea, but it is not usually employed in works on Graphic Statics, though possessing very evident advantages. It was suggested to the writer by its use in certain of Professor Eddy's works. ROOF TRUSS WITH VERTICAL LOADS. 65 The line representing the stress in any member is designated in the force diagram by letters similar to those which designate that member in the truss diagram. The latter is evidently a space diagram (Art. 11). The force diagram is often called a stress diagram, since it shows the values of the internal stresses in the truss members. 89. Reciprocal Figures. There are always two ways of com- pleting the force polygon when two of the forces are known only in lines of action. (See Art. 18.) Either way will give correct results, but unless a certain way be chosen, it will become necessary to repeat certain lines in the stress diagram. Thus, if, in Fig. 30, instead of GABHG we draw GABH'G, the lines GH\ H'B are not in convenient positions for use in the other polygons of which they ought to form sides. The lettering of the diagrams will also be complicated. As an aid in drawing the lines in the most advantageous positions, it is convenient to remember the fundamental property of figures related in such a way as the force and space diagrams shown in Fig- 30. Such figures are said to be reciprocal with regard to each other. The fundamental property of reciprocal figures is that for every set of lines intersecting in a point in either figure, there is in the other a set of lines respectively parallel to them and forming a closed polygon. It is also an aid to remember that the order of the sides in any closed polygon in the stress diagram is the same as the order of the corresponding lines in the truss diagram, if taken consecutively around the joint. This usually enables us at once to draw the sides of each force polygon in the proper order. 90. Order of External Forces in Force Polygon. It will be observed that in the case above considered, in constructing the force polygon for the loads and reactions, these forces have been taken consecutively in the order in which their points of 66 GRAPHIC STATICS. application occur in the perimeter of the truss. This is a necessary precaution in order that the stress diagram and truss diagram may be reciprocal figures, so that no line in the former need be duplicated. This requirement should be especially noticed in such a case as that shown in Fig. 31, in which loads are applied at lower as well as at upper joints. If the reactions are found by the method of Art. 87, without modification, the force polygon will not show the external forces in the proper order, since the known forces are not applied at consecutive joints of the truss. A new polygon should therefore be drawn after the reactions have been determined. If desirable (as in some cases it may be) to make use of a funicular polygon in which the external forces are taken con- secutively, this may be drawn after the reactions are deter- mined and the new force polygon is drawn. If a load be applied at some joint interior to the truss, as at M (Fig. 31), then in constructing the stress diagram it should STRESSES DUE TO WIND PRESSURE. 67 be assumed to act at N, where its line of action intersects the exterior member of the truss, and the fictitious member MN inserted. The stresses in the actual truss members will be unaffected by this assumption, and such a device is necessary in order that the stress diagram may be the true reciprocal of the truss diagram. 3. Stresses Due to Wind Pressure. 91. Direction of Reactions Due to Wind Pressure. Since the effective pressure of the wind has the direction normal to the surface of the roof (Art. 85), it has a horizontal component which must be resisted by the reactions at the supports. These cannot, therefore, act vertically, as in the case when the loading is vertical. Their actual directions will depend upon the manner in which the ends of the truss are sup- ported. If the ends of the truss are immovable, the directions of the reactions cannot be determined, since any one of an infinite number of pairs of forces acting at the ends would produce equilibrium. (The same would be true of the reactions due to vertical loads.) In such a case the usual assumption is one of the following: (i) the reactions are assumed parallel to the loads ; (2) the resolved parts of the reactions in the horizontal direction are assumed equal. In the case of trusses of large span it is not unusual to support one end of the truss upon rollers so that it is free to move horizontally, the other end being hinged, or otherwise arranged to prevent both horizontal and vertical motion. This allows for expansion and contraction with change of tempera- ture, as well as for movements due to the small distortions of the truss under loads. With this arrangement the reaction at the end supported on rollers must be vertical ; and since the point of application of the other reaction is known, both can be fully determined by the method described in Art. 39. 68 GRAPHIC STATICS. 92. Determination of Reactions. The methods of finding reactions will now be explained for the three cases mentioned in the preceding article : (i) Assuming both reactions parallel to the wind ; (2) assuming the horizontal resolved parts of the two reactions equal ; and (3) assuming one reaction vertical. (1) The first case needs no explanation, since it is identical with that described in Art. 87, except that the loads and reactions have a direction normal to one surface of the roof, instead of being vertical. It is to be noticed that this assump- tion cannot be made if the roof surface is curved, since the lines of action of the forces will not be parallel. But since the direction of the resultant of the loads will be known from the force polygon, both reactions may be assumed to act par- allel to this resultant, and the construction made as before. (2) In the second case, let each reaction be replaced by two forces acting at the support, one horizontal and the other vertical. The two horizontal forces are known as soon as the force polygon for the loads is drawn, and the two vertical forces may be found as in the preceding case, since their lines of action are known. In Fig. 32, let ab, be, cd be the lines of action of the wind forces. Let the right reaction be considered as made up of a horizontal component acting in de and a vertical component acting in ef\ and let the left reaction be replaced by a vertical component acting in fg and a horizontal component acting in ga. Draw the force polygon (or " load-line ") ABCD. By the assumption already made GA and DE are to be equal, and their sum is to equal the horizontal resolved part of AD. Through the middle point of AD draw a vertical line ; its intersections with horizontal lines through A and D determine the points G and E, so that the two forces GA and DE become known. The only remaining unknown forces are the vertical forces .ZTFand FG. Choose a pole O, draw rays to the points G, A, B, C, D, E, and then the corresponding strings. Through the points determined by the intersection of og withyjf, and oe STRESSES DUE TO WIND PRESSURE. 6 9 with cf, draw the string of. The corresponding ray drawn from O intersects EG in the point F, thus determining EF and FG. The reactions are now wholly known ; that at the left support being DF, and that at the right support FA. "*"-- SL ' b ~~~/-' (3) For the third case the construction is shown in Fig. 33 (A) and (/?), for the two opposite directions of the wind. The method is identical with that employed in Art. 39. Only one point of the line of action of the left reaction is known, hence this is taken as the point of intersection of the corresponding strings of the funicular polygon. One of these strings can be drawn at once, since the corresponding ray is known ; and the other is known after the remaining strings have been drawn, since it must close the polygon. The construction should be carefully followed through by the student. The funicular polygons for the two directions of the wind are distinguished by the use of O and O r to designate the two poles. 70 GRAPHIC STATICS. In Fig. 33 (A), ABCDEFGHIA is the force polygon for the case when the wind is from the right. Notice that the points A, B, C, D, coincide. This means that the loads AB, BC, CD, are each zero. (*) In diagram (B), ABCDEFGHTA is the force polygon for the case when the wind is from the left. The points , F, G, //, coincide, because the loads EF, FG, 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. ji 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. Maximum 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. 78, 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+^l). This gives I=i8oo\bs. 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, cd, de is 450 Ibs. The loads at the end joints may be neglected, being borne directly at the supports. Weight of roof . This depends upon the materials Used and the method of construction, but will be taken as 6 Ibs. per sq. ft. 7 2 GRAPHIC STATICS. of roof area, giving 900 Ibs. as the load at each joint. This, also, is a permanent load. Total permanent load per joint, 1350 Ibs. Weight of snozv. Taking this as 15 Ibs. per horizontal square foot, we find 1800 Ibs. as the load at each joint. Wind pressure. This is computed from the formula A= _2sin_ (Art. 85.) I +sm 2 a For this case we put sin a = -J|=|- ; / 40 Ibs. per sq. ft.; whence / a = 35 Ibs. per sq. ft. (about). This gives upon each panel of the roof 5250 Ibs. Then with the wind from either side, the wind loads on that side would be 2625 Ibs., 5250 Ibs., 2625 Ibs. 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 in 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 : l'$: is drawn parallel to FK, through the point of intersection of the strings of and ok. Prolong / to intersect P'R' produced at T" ; then Q'T" is the line of action of the reaction at Q'. The force polygon may now be completed by drawing KL parallel to Q'T' and LF parallel to P'R 1 . 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 88 GRAPHIC STATICS. have a second determination of the reaction exerted by each half-truss upon the other ; and it is quite likely that the two values found will not agree. By a method like that employed in Art. 100 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, QR, 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'/, prolonging the last to intersect /;-. Through the point thus determined, draw e'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. Connterbracing. 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 simplified. To accomplish this is the object of counterb racing. iii. Counter bracing. 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 'M system of loading which would cause compression in xy will cause tension in this new member ; and vice versa. 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 L. 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 9 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 1 receives a tension which is determinate. If the member xy 1 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 counterbrace. 112. Determination of Stresses with Counterbracing. 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 : (a) Construct separate stress diagrams for vertical loads and for wind in each direction, assuming the diagonals in all panels to slope the same way. (b) 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. (c) 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 (PL 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 Ibs., and that of the roof at 3600 Ibs. ; this makes the total permanent load 1000 Ibs. per panel. Take 800 Ibs. as the load at each upper joint, and 200 Ibs. as the load at each lower joint. The snow load, computed at 15 Ibs. per horizontal square foot, is about 1440 Ibs. 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 Ibs. per sq. ft. : On panel d, 1650 Ibs. On panel e, 3870 Ibs. On panel/, 5480 Ibs. 9 2 GRAPHIC STATICS. In Fig. 42 (D) these are laid off successively in their proper directions to the assumed scale. Thus CD 1 , D'E', E'G repre- sent respectively 1650 Ibs. normal to dq, 3870 Ibs. normal to cs, and 5480 Ibs. normal to ft. Now each of these loads is to be equally divided between the two adjacent joints. Bisect CD' at D, D'E 1 at E, and E'G at F\ then CD, DE, EF, FG 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. a J - 6660 -9680 - 4250 - 14070 16340 bl - 535 - 77 8o - 4880 - 9650 - 13*3 en -4550 -6640 - 6380 - 62OO - 11190 dq -4550 - 6640 - 95 - 4250 - 14050* es -535 -7780 - i5 30 - 345 - 20380* fl - 6660 - 9680 - 14070 - 4300 - 20730* A + 5080 4- 7400 + 780 + 13500 4- 12480 kh b + 4680 + 6820 -f 700 + 12450 4 11500 m/i 4 -f 4470 + 6520 + ! 950 + 7350 + 10990 P^ + 4470 + 6520 + 4100 + 4100 + 10990* rh^ + 4680. 4- 6820 + 7680 + 2050 4- 12360* *, + 5080 + 7400 + 1350 + 800 + 18580* jk + 1180 + 1440 + 150 + 2650 + 2620 Im + 1180 -f 1440 - 280 + 4100 + 2620 np + 1180 + 1440 95 + 3800 + 2620* qr + 1180 + 1440 1650 + 2525 f + 2620* ) 1 - 470* J f 4- 2620* | st + 1180 + 1440 - !35 + 1250 1 T *.,-,* \ v. 170* ' kl o o + 1300 - 4600 + 1300* mn o o + 2700 - 4100 + 2700* pq o o + 4850 - 3150 + 4850* rs o o + 7080 - 195 + 7080* COUNTERBRACING. 93 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. (e) Stresses in counterbraces. It is evident from symmetry that no new diagrams are needed to determine the stresses in the counterbraces. 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) Combination 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 of 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 : (1) 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 aj 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, jk and st must be designed for a tension of 2620 Ibs., and also for a compression of 170 Ibs. ; while Im and qr are each liable to 2620 Ibs. tension and to 470 Ibs. compression. CHAPTER VI. SIMPLE BEAMS. i. General 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. 115. External Shear, Resisting Shear, and Shearing Stress. -The external shear 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 shear 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 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 /^the 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 m the section at C must have a vertical component equal to V. 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 A C upon CB. This stress is called the shearing stress in the section, and is called positive 96 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 AC 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 moment at the section, and must be equal to M y 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 A C 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 GENERAL 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 curve 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 1 be taken parallel to AB. At every point of A'B' suppose an ordinate drawn whose length shall represent the external shear at the corresponding B 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 A'B', and negative values by ordinates drawn downward. (Instead of drawing A'B' 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 moment 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 1 8. 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 9 8 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. M |Z) t Fig. 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, ob, 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 M"N" 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 15) 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. 1 1 8, 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 N. 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 FG and GA. Then the value 100 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. 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 at 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 MOVING 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 Sustaining Moving Loads. 12$. 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 maximum 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 (02 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 (PL III) 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 J. First, consider the effect of a single load in any position. It is seen that any load to the right of J 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 J 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 J 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 K) 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 JY approaches J, the reaction at X increases ; and as a load on XJ approaches _/, the reaction at Y 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. IO 3 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 J 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 P I = magnitude of foremost load, P^ magnitude of sec- ond load, etc., W being the total load on the beam. Let /= total span, and m = distance between PI and P* Let x= dis- tance from Y to the center of gravity of W when P v is at a given section. Let us compute the shear when P l 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 IO4 GRAPHIC STATICS. the reaction at X, say R. Then (calling V the external shear) we have . If now the loads be moved until P 2 comes to a point just at the right of the section, the reaction due to W becomes W \ x + m \ an d the shear at the section becomes The increase of the shear is, therefore, Wx_ Wm D ~~ This increase is plus or minus according as is greater or P / less than P l ; that is, as IV is greater or less than -. Hence m the following rule : The maxim tun 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 than ^, the greatest m m shear will occur wJien some succeeding load is at the section. In the above discussion it has been assumed that in bringing P 2 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. t 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 1 *' v - + ft BEAM SUSTAINING MOVING LOADS. 105 Hence, in the statement of the above rule, we have only to substitute Wm+W'x' for Wm\ or, JF+-2-* for W. The m additional load W may be neglected except when the condition P I W = - is nearly satisfied; for the term W*x? will always be m small in comparison with Wm. 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 P. 2 for />!, and for m the distance between P and P 3 . 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 (PL 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., 01 = 8.1 ft., P 1 = Sooo Ibs., ^=^1x8000 = 63200. ;;/ 8.1 Hence, as the load moves from right to left, W is less than I0 6 GRAPHIC STATICS. P I until the whole load on the beam reaches 65200 Ibs. This m occurs when the fifth load is at Y. Hence the point Z is located at a distance from Y 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 Ibs. ; 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 Ibs. or 165500 Ibs. 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 J. 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 be, the line of action of the second load. Through the extremities of the beam SjSj 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 let./* be its point of intersection with the load line ; then B'J 1 andJ'A are the two reactions. The shear at the point J is evidently the algebraic sum of the reaction J'A and the load AB ; hence it BEAM SUSTAINING MOVING LOADS. 107 is represented byj'fi. In Fig. 47 let X,Y, be drawn to repre- sent the length of the beam, and let the ordinates of the shear curve be drawn from it. Then from J we lay off an ordinate equal toj'>. 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 Ibs. uniformly dis- tributed along the beam. The shear close to the left support due to this load is equal to the reaction, or 12500 Ibs. ; and decreases as we pass to the right by ^f^-^- Ibs. for each foot. At the middle of the beam the shear is zero, and at the right support it is 12500 Ibs. Hence the shear curve is a straight line, and may be drawn as follows : From X, lay off an ordinate downward representing 12500 Ibs,, and from Y s an ordinate upward representing 12500 Ibs.; 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 XY shown in Fig. 47 (PL III), consider the bending moment at any section (asy) 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 aty is the moment of this reaction about an origin aty, and is negative. Moreover, I0 8 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 the 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 : (1) 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; W= total load on beam ; x = distance of center of gravity of W from right sup- port ; W 7 ! load on beam to left of given section; ,r=distance of center of gravity of W\ from section; l = distance of sec- tion from left support ; R = reaction at left support ; M bending moment at the section. Then we have BEAM SUSTAINING IVIOVING LOADS. Wr If the loads all move an infinitesimal distance to the left, then x and ,TI receive equal infinitesimal increments, and the increment of M is dM= -dx- Wjlxi = - - W\ dx (since dx l = dx). Now if M is a maximum, we must have = o; that is, -^- ^ = o; or, }V =-. Hence the follow- dx / A / ing rule : W7/^# //# bending moment at any section of the beam has its greatest vahie, tJie loads on the tivo segments of tJie beam 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 - l will not then be zero, as assumed above. In such dM a case is discontinuous, and its value changes sign as the loads pass through the dX rz^. JJT position for which i -- = o. 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 (PL III). Let the greatest bending moment be found for the section J. From the general 1 10 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 J 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 Ibs., while the load on the left segment is 23,000 Ibs., 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 Ibs., 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 of the funicular polygon is marked of. Using the principle of Art. 56, we find the bending mo- ment at the given section by measuring the distance intercepted on cd by the strings od and of and multiplying it by the pole distance in the force diagram. Let X m Y m (Fig. 47) be the axis of abscissas for the curve of maximum moments. From the point J 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 J. 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, arid a funicular polygon drawn, using the BEAM SUSTAINING MOVING LOADS. Iir 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 m Y m , 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 plate 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. The student should make the complete construe- 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. 134. 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 <C. 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 between 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 C" C. Then, if all loads on BC are treated as if divided between B and C in the ratio of BC" to C"C, the general rule applicable to a beam may be used in assigning the position of loads for maximum bending moment. [The proof of this proposition will be omitted. The mode PARALLEL CHORDS CONCENTRATED LOADS. 125 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 (PI. IV). By Art. 149, the greatest bending moment for each of the points K, L, M, N, 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 J/and N (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 Ibs. Now suppose the load B*O is just at the left of M. The three loads, CD\ D'E\ E'F', must be regarded as equally divided between M and N ; this gives for the whole load to the left of T 111500 Ibs., which is greater than seven- twelfths of 182000 Ibs. But if B'C is just at the right of M, half of it must be regarded as borne at N, and the whole load 126 GRAPHIC STATICS. on the truss to the left of T is 104000 Ibs., which is less than seven-twelfths of 182000 Ibs. Now draw M t M t 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 0c", 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 N respectively. Draw vertical lines through ^/and A 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 m Y m 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 O'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 (PI. IV) the line X Y s 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 s Y s , 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, X m Y m 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 m Y m > 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 m Y m 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 I2 g GRAPHIC STATICS. sustained by the member. This gives 14'-! 5' as the stress in the given member. In the same way, I2 r -i3', 1^-14', 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 maximum 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 '-2 2'. 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 14'-! 5'. The latter is there- fore the 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, 17-18, 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 n equal to the pole distance on the force scale already adopted, and ZZ 1 equal to the depth of the truss, and draw Z' Z" . Then take Zz equal to the ordinate of the moment curve and draw 2-15 parallel to Z* Z" , intersecting ZZ 11 in the point marked 15. The distance Z-1% 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-\$ 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, 130 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, // .the depth of truss, y the ordinate of the moment curve, and x the stress in the chord member, we have Hy = hx ; or x = -y. TT Now if = , we have only to measure y in linear units, multiply by ;/, 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. Maximum 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 r F(Fig. 51) represent a truss of PARALLEL CHORDS DISTRIBUTED LOAD. ni seven equal panels ; and for the greatest shear in a panel CD let Z be the foremost point of the moving load. Then we must have ZD equal to one-seventh of ZY\ or, ZD = 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 A' due 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 I 3 2 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 Curved Chords Uniform Panel Loads. 156. 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. l ^ 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, in 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 zkfg, and let us determine the kind of stress caused in the member qq\ Take a section cutting qq\ g'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 g'd' and d'd. 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. j 34 GRAPHIC STATICS. 1 59. 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 f'n\ 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 through//"',/';/, and n'u, and taking moments about the point of intersection of the two chord members cut.) Similarly, loads at ab, be, cd, dc, and ef all tend to throw compression on the vertical member /';;/, while loads at fg, gh, and /// have the opposite tendency. Therefore, to produce the greatest tension upon f'n' (and compression on/';;/'), 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 /';;/), the live load must act only at fg, gh, and ///. 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, cd, 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 g'm', the loads gh and Id 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 om, and the two reactions are GMand MA. Now take a section through mm ', m'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 mm' 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 Ma. line parallel to X Y\ mark their point of intersection G'. Then MG' is the resultant of the forces in mm' and m'g'. From M 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' I' 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. 1 60. Maximum Stresses. By combining the stresses due to live and dead loads, the maximum stresses are easily deter- mined. Web 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 I'g 1 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 Ibs. 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. 1 59, it is only necessary to so determine the stress in the web member that its vertical resolved part shall equal the reaction. 6. Truss with 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 XY (Fig. 53) represent the truss and the member con- sidered. We. know in a general way (Art. 1 59) that for the greatest tension in B' C the truss should be completely loaded on the right of the panel BC* Let W = total load on truss ; W = load on panel BC\ I = total length of truss ; /' = length of panel BC\ x = distance from Y to center of gravity of W\ x 1 = 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 -?^3l! Now if the truss be separated by a section cutting B 1 C and the two chord members, it is seen that the moment of the stress in B' C about Z 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 * 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. If J/is a maximum, we must have = o. But since dx' = dx we have, 139 . dx II' 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 a 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 B, we may reason thus : Let W" = load to left of B; x" = distance of center of gravity of W" from B, Then M- Ra- Pb- W" (b - *") = Differentiating, remembering that Jx" = dx' dx, we have w ,, dx I V Placing this equal to zero, we have w'L*L iv 1 - - iv". a V a Usually W' 1 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 - W I & 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 (PI. 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 XY, 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 uv. 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 ku and vm till they intersect at Z z , we find .= =2. Also we have - = 8. Hence, for the great- Z%X a I est stress in uv, 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 Ibs. If the second load is just at the left of R, the load on the panel is 20000 Ibs., which is greater than one- sixteenth of 1 23750 Ibs. ; but if the second load is just at the right of R, the load on the panel is 7500 Ibs., which is less than one-sixteenth of 123750 Ibs. We are now ready to determine the stress in the member uv. The general method is to- cut the truss by a section through ku, uv, and vm ; 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 U X to represent the span when the load BC is CURVED CHORDS CONCENTRATED LOADS. i 4I 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 M\A. 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 ob in Q' and R ! ; then Q'R' is the string which must replace oa and ob. (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 Kj) 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 M\K. We need also its line of action, which is found as follows : Prolong Q'R' and om\ (the closing string of the funicular polygon for all the forces on the truss) until they intersect in z' ; the required line of action is a vertical line through s'. This vertical line intersects X U X V produced in z. We now locate the point z in the truss diagram by making Xz = X u z in the diagram above. We have now to solve the following problem : Determine three forces acting in lines ku, 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 s. Draw zQ\ and determine two forces, one parallel to zQ and the other to kit, which shall be in equilibrium with KM. This gives UK as the force in the line uk and MU as the resultant of the forces in the lines mv and vu. From M and U draw lines parallel to mv and vu respectively, and mark their intersection V; then J/Fand VU represent the stresses in the corresponding truss members. The latter is evidently the required stress in the member uv. Turning next to the member ///, a section through it will cut 1 4 2 GRAPHIC STATICS. the two chord members kit and mt ; these intersect at Z By measurement we find - = ^-^=6, and = 8x6 = 48. There- a Z^X I a fore, for the greatest stress in tu 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 tu is the reaction at the support. To find this, draw X t X u to represent the length of the truss for the required posi- tion of the loads, and find om\> the closing string of the funic- ular polygon for all loads and reactions on the truss. The corresponding ray is OMJ, 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 mt, tu, uk, which shall be in equilibrium with the reaction just determined. Lay off JCMto represent this reaction ; draw a line from K' parallel to ku and one from M parallel to Q'X, and mark their point of intersec- tion U' ; then [7'K 1 represents the force in uk, and MU' the resultant of the forces in mt and tu. From M draw a line par- allel to mt and from [/' a line parallel to ut, intersecting at T; then TU 1 represents the required stress in tu. The value just determined is the true maximum stress for the member /;/. In case of the member uv, however, further investigation is needed. It will be found by trial that the condition W = - W is satisfied for the member uv, 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 uv is the reaction (already found when considering the member tu) represented by K'M. Making this in equilibrium with three forces whose lines of action are kn, uv, vm, we find CURVED CHORDS CONCENTRATED LOADS. 143 U' I 7 ' as the stress in uv. Comparison shows that this is slightly greater than UV\ hence the value last found is the true maximum for the member nv. 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 b can be easily computed from the dimensions of the truss. Remark. It will often be found that two web members related as are tu and nv, will not sustain their maximum stresses at the same time. In the case of parallel chords this could not occur. 1 66. 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. CHAPTER VIII. CENTROIDS. I. Centroid of Parallel Forces. 167. Composition of Parallel Forces. The composition of complanar parallel forces can always be effected by means of the funicular polygon, by the method of Art. 27. It is now necessary to consider parallel systems more at length, as a pre- liminary to the discussion of graphic methods for determining centers of gravity and moments of inertia. 168. Resultant of Two Parallel Forces. Let ab and be (Fig. 55) be the lines of action of two parallel forces, their magni- tudes being AB and BC (not shown). Let ac be the line of action of their resultant, and AC (not shown) its magnitude. By the principle of moments (Art. 50) the sum of the moments of AB and BC about any Fig. ss point in their plane is equal to the moment of A C about the same point. If the origin of moments is on ac, the moment of AC is zero; and therefore the moments of AB and BC are numerically equal (but of opposite signs). Let any line be drawn perpendicular to the given forces, intersecting their lines of action in P', Q', and R' respectively. ab P 144 CENTROID OF PARALLEL FORCES. Let any other line be drawn intersecting the three lines of action in P, Q, and R respectively. Then PR QR P'R' Q'R'* and therefore ABxPR=BCx QR. That is, PQ is divided by the line ac into segments inversely proportional to AB and BC. If AB and BC act in opposite directions, the line ac will be outside the space included between ab and be ; but the above result is true for either case. 169. Centroid of Two Parallel Forces. If the lines of action ab and be (Fig. 55) be turned through any angle about the points P and Q respectively, the forces remaining parallel and of unchanged magnitudes, the line of action of their resultant will always pass through the point R. For, by the preceding article, the line of action of the resultant will always intersect PQ in a point which divides PQ into segments inversely proportional to AB and BC. Hence, if AB and BC remain unchanged, and also the points P and Q, the point R must also remain fixed. If P and Q are taken as the points of application of AB and BC, R may be taken as the point of application of AC, in whatever direction the parallel forces are supposed to act. The point R is called the centroid * of the parallel forces AB and f>C for the fixed points of application P and Q. 170. Centroid of Any Number of Parallel Forces. Let AB, BC, and CD be three parallel forces, and let P, Q, and S (Fig. 56) be their fixed points of application. Let R be the cen- troid of AB and BC, and A C their resultant. Take R as the fixed point of application of AC, and determine T t the centroid of AC Fi g . se * The name center of parallel forces has been quite commonly used instead of centroid as above defined. The latter term has, however, been adopted by some of the later writers, and seems on the whole a better designation. a & a I4 6 GRAPHIC STATICS. and CD. Let AD be the resultant of AC and CD ; then AD is also the resultant of AB t BC t and CD. Now if AB, BC, and CD have their direction changed, but still remain parallel and unchanged in magnitude, it is evident that the point T, determined as above, will remain fixed and will always be on ad, the line of action of AD. The point T is called the centroid of the three forces AB, BC, and CD. By an extension of the same method, a centroid may be determined for any system of parallel forces having fixed points of application. Hence the following definition may be stated : The centroid of a system of parallel forces having fixed points of application is a point through which the line of action of their resultant passes, in whatever direction the forces be taken to act. In determining the centroid by the method just described, the forces may be taken in any order without changing the results. For the centroid must lie on the line of action of the resultant ; and since this is a determinate line for each direction in which the forces may be taken to act, there can be but one centroid. 171. Non-complanar Parallel Forces. The reasoning of the preceding articles is equally true, whether the forces are corn- planar or not. In what follows we shall deal either with complanar forces, or with forces whose points of application are complanar. No more general case will be discussed. 172. Graphic Determination of Centroid of Parallel Forces. If the line of action of the resultant of any system of parallel forces be found for each of two assumed directions of the forces, the point of intersection of these two lines is the centroid of the system. Moreover, if the points of application are com- planar, the two assumed directions may both be such that the forces will be complanar. Thus, let the plane of the paper be the plane containing the given points of application, and let ab, be, cd, de, ef (Fig. 57) be CENTROID OF PARALLEL FORCES. l ^ the points of application of five parallel forces, AB, BC, CD, DE, EF. Draw through these points parallel lines in some chosen direction, and taking them as the lines of action of the given forces, construct the force and funicular polygons corre- sponding. The line of action of the resultant is drawn through the intersection of the strings oa and of, and this line contains the centroid of the given forces. Next draw through the given points of application another set of parallel lines, preferably per- pendicular to the set first drawn, and draw a funicular polygon for the given forces with these lines of action. (It is unneces- sary to draw a new force diagram, since the strings in the second funicular polygon may be drawn respectively perpendic- ular to those of the first.) This construction determines a second line as the line of action of the resultant corresponding to the second direction of the forces. The required centroid of the system is the point of intersection of the two lines of action of the resultant thus determined, and is the point af in the figure. Example. Find graphically the centroid of the following system of parallel forces, and test the result by algebraic com- putation : A force of 20 Ibs. applied at a point whose rectangu- lar coordinates are (4, 6) ; 12 Ibs. at the point (12, 3) ; 20 Ibs. at T 4 8 GRAPHIC STATICS. the point (10, 10) ; 10 Ibs. at the point (7, 9); 8 Ibs. at the point (5, 10). 173. Centroid of a Couple. If AB and BC are the magni- tudes of any two parallel forces applied at points P and Q, then by Art. 169 their centroid R is on the line PQ and is determined by the equation QR AB' If AB and BC are equal and opposite forces, PR and QR must be numerically equal, and R must be outside the space between the lines of action of AB and BC. These conditions can be satisfied only by making PR and QR infinite. Hence we may say that the centroid of two equal and opposite forces lies on the line joining their points of application 'and is infinitely dis- tant from these points. 174. Centroid of a System whose Resultant is a Couple. If a system with fixed points of application is equivalent to a couple, its centroid will be infinitely distant from the given points of application. A line containing this centroid can be determined as follows : Take the given forces in two groups ; the resultants of the two groups will be equal and opposite. Find the centroid of each group, and suppose each partial resultant applied at the corresponding centroid. Then the centroid of the whole sys- tem is the same as that of the couple thus formed, and will lie on the line joining the two' partial centroids. If the separation into groups be made in different ways, dif- ferent couples and different partial centroids will be found. The different couples are, of course, equivalent ; and it may be proved that the lines joining the different pairs of partial cen- troids are all parallel, and intersect in the (infinitely distant) centroid of the whole system. For, suppose the given system to be equivalent to a couple CENTER OF GRAVITY. 149 Q with points of application A and B, and also to a couple Q' with points of application A' and />'. These two couples must be equivalent to each other, whatever be the direction of the forces. Let AB be taken as this direction. The two equal and opposite forces of the couple Q neutralize each other, since their lines of action are coincident ; hence the two forces of the couple Q must also neutralize each other. Therefore A'' must be parallel to AB. 175. Moment of a Force about an Axis. The moment of a force with respect to a given axis, as defined in Art. 47, depends not only upon the point of application of the force, but upon its direction. In dealing with systems of forces whose direction may change but whose points of application are complanar, we shall need to compute moments only for axes lying in the plane of the points of application ; and the forces may usually be regarded as acting in lines perpendicular to this plane. Hence we shall compute moments by the following rule : The moment of a force with reference to any axis is the product of the magnitude of the force into the distance of its point of application from the axis. . 2. Center of Gravity Definitions and General Principles. 176. Center of Gravity of Any Body. Every particle of a terrestrial body is attracted by the earth with a force propor- tional directly to the mass of the particle. The resultant of such forces upon all the particles of a body is its iveight ; and the point of application of this resultant is called the center of gravity of the body. The lines of action of these forces may be assumed parallel without appreciable error. We may there- fore define the center of gravity of a body as follows : If forces be supposed to act in the same direction upon all particles of a body, each force being proportional to the mass 150 GRAPHIC STATICS. of the particle upon which it acts, the centroid of this system of parallel forces is the center of gravity of the body. This point is also called center of mass, and center of inertia, either of which is a better designation than center of gravity. The latter term is, however, in more general use. 177. Centers of Gravity of Areas and Lines. The term center of gravity, as above defined, has no meaning when applied to lines and areas, since these magnitudes have no mass, and hence are not acted upon by the force of gravity. It is, however, common to use the name center of gravity in the case of lines and areas, with meanings which may be stated as follows : The center of gravity of an area is the center of gravity of its mass, on the supposition that each superficial element has a mass proportional to its area. This point would be better described as the center of area. The center of gravity of a line is the center of gravity of its mass, on the assumption that each linear element has a mass proportional to its length. The term center of length is pref- erable, and will often be used in what follows. Similar statements might be made regarding geometrical solids, but we shall have to deal chiefly with lines and areas. 178. Moments of Areas and Lines. Definition. The mo- ment of a plane area with reference to an axis lying in its plane is the product of the area by the distance of its center of gravity from the axis. Proposition. The moment of any area about a given axis is equal to the sum of the moments of any set of partial areas into which it may be divided. For, by the definition of center of gravity, a force numerically equal to the total area and applied at its center of gravity is the resultant of a system of forces numerically equal to the partial areas and applied at their respective centers of gravity ; and the moment of any CENTER OF GRAVITY. ! 5 ! force is equal to the sum of the moments of its components (Art. 50). A similar definition and proposition may be stated regarding lines. The moment of an area or line is zero for any axis containing its center of gravity. 179. Symmetry. Two points are symmetrically situated with respect to a third point if the line joining them is bisected by that point. Two points are symmetrically situated with respect to a line or plane when the line joining them is perpendicular to the given line or plane and bisected by it. A body is symmetrical with respect to a point, a line, or a plane, if for every point in the body there is another such that the two are symmetrically situated with respect to the given point, line, or plane. The point, line, or plane is in this case called a point of symmetry, an axis of symmetry, or a plane of symmetry of the body. 1 80. General Principles. (i) The center of gravity of two masses taken together is on the line joining the centers of gravity of the separate masses. For it is the point of applica- tion of the resultant of two parallel forces applied at those points. (2) If a body of uniform density has a plane of symmetry, the center of gravity lies in this plane. If there is an axis of symmetry, the center of gravity lies in this axis. If the body is symmetrical with respect to a point, that point is the center of gravity. For the elementary portions of the body may be taken in pairs such that for each pair the center of gravity is in the plane, axis, or point of symmetry. 181. Centroid. The center of gravity of any body or geo- metrical magnitude is by definition the same as the centroid of a certain system of parallel forces. It will be convenient, 152 GRAPHIC STATICS. therefore, to use the word centroid in most cases instead of center of gravity. 3. Centroids of Lines and of Areas. 182. General Method of Finding Centroid. The centroid of any area may be found by the following method : Divide the given area into parts such that the area and centroid of each part are known. Take the centroids of the partial areas as the points of application of forces proportional respectively to those areas. The centroid of this system of forces is the centroid of the total area, and may be found by. the method of Art. 172. The centroid of a line may be found by a similar method. The method just described is exact if the magnitudes of the partial areas and their centroids are accurately known. If the given area is such that it cannot be divided into known parts, it will still be possible to get an approximate result by this method. In applying this general method, it is frequently necessary to know the centroids of certain geometrical lines and figures, and also the relative magnitudes of the areas of such figures. Methods of determining these will be given in the following articles. 183. Centroids of Lines. The centroid of a straight line is at its middle point. Broken line. The centroid of a broken line is the center of a system of parallel forces, of magnitudes proportional to the lengths of the straight portions of the line, and applied respec- tively at their middle points. It may be found graphically by the method of Art. 172, or by any other method applicable to parallel forces. Part of regular polygon. For the centroid of a part of a regular polygon, a special construction is found useful. CENTROIDS OF LINES AND OF AREAS. 153 Let ABODE (Fig. 58) be part of such a polygon, and O the center of the inscribed circle. Let r radius of inscribed circle ; / = length of a side of the polygon ; s = total length of the broken line I . x J _ AE. Through O draw OO, the axis M A ' Q E/N 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, BO, 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 2 , x z , x^ the distances of the middle points of AB, BC, etc., from MN, we have from the principle of moments, /X]_ 4- fat + lx% + lx = sx. But if Ab and Bb be drawn perpendicular respectively to PQ and MJVwQ have from the similar triangles ABb, POQ. Ab- PQ Ab .: Ix^r-Ab. In the same way, lxz = r- be, Ix^r-dE. Hence, s*x=r (Ab + bc + cd+dE) = r AE, where AE is equal to the projection of the broken line ABODE on MN. The centroid G may now be found graphically as follows : Make6fc' = r; ON=\s\ OE' = ^AE; draw Nc' . Then G is determined by drawing E'G parallel to Nc'. 154 GRAPHIC STATICS. Circular 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. Parallelogram. 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 of division draw lines parallel to BC. Also divide BC into any even number of equal parts and draw through the points of division lines parallel to 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 and BD. CENTROIDS OF LINES AND OF AREAS. 55 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 bcc*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 = J EA, and EG 2 =\ EC. Then the centroids of the triangles ABD and BCD are 1 5 6 GRAPHIC STATICS. Fig. 61 GI and G- 2 respectively. Hence the centroid of ABCD is on the line GiG 2 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 GJi and G*H ; hence, if G is the required centroid, G\G '. G->G '. '. G-iH '. G\f~f. Therefore G is found by making G 1 G=G 2 H. Circular sector. To find the centroid of a circular sector OAB (Fig. 62) we may reason as follows : Draw two radii OM, ON, very near together. Then OMN differs little from a tri- angle, and its centroid will fall very near the arc AB\ drawn with radius equal to two-thirds of OA. If the whole sector be subdivided into elements such as OMN, their centers of gravity will all fall very near to the arc AB\ If the number of such elements is indefinitely in- creased, the line joining their centroids approaches as a limit the arc AB\ And since the areas of the elements are propor- tional to the lengths of the corresponding portions of A'', the centroid of the total area is the same as that of the arc A'B'. This point may be found by the method described in Art. 183. 185. Graphic Determination of Areas. Let there be given any number of geometrical figures, and let it be required to determine the relative magnitudes of their areas. If a number of rectangles can be found, of areas equal respectively to the areas of the given figures and having one common side, then the remaining sides will be proportional to the areas of the given figures. M N CEXTROIDS OF LINES AND OF AREAS. 157 An important case is that in which the given figures are such that the area of each is equal to the product of two known lines. In this case a series of equivalent rectangles having one common side can be found by the following construction. Let ABC (Fig. 63) be a triangle, and let it be required to determine an equivalent rectangle having a side of given length as LM. Let b and h be the base and altitude of the triangle. TVTn lc * T J\/~ - - h ^\ n c\ T J~^ J? ^-* Oo and draw the semicircumference PRN. Draw the ordinate LR perpendicular to PJV] then LR 2 = PLx LN= I bh = area ABC. Draw MR, and perpendicular to it draw RQ. Then ABC. Hence LQ is the required length. If the given figure is a parallelogram, LN and LP may be its base and altitude. If it be a circular sector, LN and LP may be the length of the arc and half the radius. 1 86. Centroids of Partial Areas. It may be required to find the centroid of the part remaining after deducting known parts from a given area. For this case the construction of Art. 182 needs modification. In the case there considered we regarded the partial areas and the total area as proportional to parallel forces acting at their respective centers of gravity. In this case we may also represent the total area, the portions deducted from it, and the remaining portion as forces acting at the respective centers of gravity ; but the forces corresponding to the areas deducted must be taken as acting in the opposite direction to that assumed for the forces representing the total area and the area remaining. Thus, to find the centroid of the area remaining after deduct- 158 GRAPHIC STATICS. ing from the circle ABD (Fig. 64) a smaller circle EFH and a sector OAB, we may proceed as follows : Find the centroid of three parallel forces proportional to the areas ABD, EFH, and OAB, and applied at their respective centroids O, C, and G\ but the last two must be taken as acting in the direction opposite to that of the first. With this understanding, the force and funicular polygons may be employed as in Art. 172. 187. Moments of Areas. The moment of an area about any line in its plane may be determined from the funicular polygon employed in finding its center of gravity. Let the parallel forces applied at the centroids of the partial areas be assumed to act parallel to the axis of moments. Then the distance inter- cepted on the axis by the extreme lines of the funicular poly- gon, multiplied by the pole distance, is equal to the moment of the total area about the axis. For, by Art. 56, this construc- tion gives the moment of the resultant force about any point in the given axis ; and this is equal to the moment of the result- ant area about the axis by definition. A similar rule gives the moments of the partial areas. CHAPTER IX. MOMENTS OF INERTIA. i. Moments of Inertia of Forces. 1 88. Definitions. The moment of inertia of a body with respect to an axis is the sum of the products obtained by mul- tiplying the mass of every elementary portion of the body by the square of its distance from the given axis. The moment of inertia of an area with respect to an axis is the sum of the products obtained by multiplying each element- ary area by the square of its distance from the axis. The moment of inertia of a line may be similarly defined, using elements of length instead of elements of mass or area. The moment of inertia of a force with respect to any axis is the product of the magnitude of the force into the square of the distance of its point of application from the axis. The sum of such products for any system of parallel forces is the moment of inertia of the system with respect to the given axis. [NOTE. The term moment of inertia had reference originally to material bodies, the quantity thus designated having especial significance in dynamical problems relat- ing to the rotation of rigid bodies. The quantity above defined as the moment of inertia of an area is of frequent occurrence in the discussion of beams, columns, and shafts in the mechanics of materials. In the graphic discussion of moments of iner- tia of areas, it is convenient to treat areas as forces, just as in the determination of centers of gravity; it is therefore convenient to use the term moment of inertia of a force in the sense above defined. It is only in the case of masses that the term moment of inertia is really appropriate, but it is by analogy convenient to apply it to the other cases.] The product of inertia of a mass with respect to two planes is the sum of the products obtained by multiplying each element- ary mass by the product of its distances from the two planes. I6o GRAPHIC STATICS. The product of inertia of an area, a line, or a force may be defined in a similar manner. For a plane area, the product of inertia with respect to two lines in its plane may be defined as the sum of the products obtained by multiplying each element of area by the product of its distances from the given lines. This is equiv- alent to the product of inertia of the area with respect to two planes perpendicular to the area and containing the two given lines. For a system of forces with points of application in the same plane, the product of inertia with reference to two axes in that plane may be defined as the sum of the products obtained by multiplying the magnitude of each force by the product of the distances of its point of application from the given axes. It is with such systems and with plane areas that the following pages chiefly deal. The radius of gyration of a body with respect to an axis is the distance from the axis of a point at which, if the whole mass of the body were concentrated, its moment of inertia would be unchanged. The square of the radius of gyration is equal to the quotient obtained by dividing the moment of inertia of the body by its mass. The radius of gyration of an area may be denned in a similar manner. The radius of gyration of a system of parallel forces is the distance from the axis of the point at which a force equal in magnitude to their resultant must act in order that its moment of inertia may be the same as that of the system. The square of the radius of gyration may be found by dividing the moment of inertia of the system by the resultant of the forces. 189. Algebraic Expressions for Moment and Product of In- ertia. Moment of inertia. Let m^ ; 2 , etc., represent ele- mentary masses of a body, and r l9 ;- 2 , etc., their respective MOMENTS OF INERTIA OF FORCES. J6I distances from an axis ; then the moment of inertia of the body with respect to that axis is the symbol 2 being a sign of summation, and the second mem- ber of the equation being merely an abbreviated expression for the first. Product of inertia. Let/i, / 2 , etc., denote the perpendicular distances of elements m\, m 2 , etc., from one plane, and q lt q^ etc., their distances from another ; then the product of inertia of the body with respect to the two planes is m\p\q\ 4- m*p?qz H ---- = 2mpq. Raditis of gyration. With the same notation, if k denotes the radius of gyration of the body, we have (/;/! + m z -\ ---- ) k~ = m i r? + mr 2 2 H ---- . Or, 7,2 _ \\ <> H Here 2m is equal to the whole mass of the body. Prodtict-radius. Let c represent a quantity defined by the equation. (mi + 7# 2 + ) ' + area D 1 *', which is equal to the area of the polygon A' i 2 34^'. Hence this area, multiplied by 2 H, 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 H= \ 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, ST be required. Draw ABCDE, the J^ 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. I6 7 of forces equal to A'B', 'C', etc., each multiplied by H, to act at the points ab, 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', o'e' of the second funicular polygon intersect ST in the points A" and E". Hence, calling H' the second pole distance, the product of inertia of the given system is 195. Product-Radius. If //"be taken equal to AE (Fig. 66), H'xA"E" is equal to the square of the product-radius (Art. 189). Hence the product-radius can be found by a construc- tion exactly like that employed in finding the radius of gyra- tion. Thus (Fig. 66) take LM=H' and MN=A"E"; make LN the diameter of a semicircle, and draw from Ma. line per- pendicular to LN, intersecting the semicircle in P. Then MP 2 =LMx MN= H'xA"E"\ hence MP = c, the product- radius. 196. Relation between Moments of Inertia for Parallel Axes. Proposition. The moment of inertia of a system of par- allel forces with reference to any axis is equal to its moment of inertia with respect to a parallel axis through the centroid of the system plus the moment of inertia with respect to the given axis of the resultant applied at the centroid of the system. Let PI, P 2 , etc., represent the forces ; x\, x* etc., the dis- tances of their points of application from the central axis ; and a the distance of this central axis from the given axis. Calling the required moment of inertia A, and the moment of inertia with respect to the axis through the .center of gravity A', we have A=P l 168 GRAPHIC STATICS. Now P& -f P 2 x z H is the algebraic sum of the moments of the given forces with respect to the axis through their centroid, and is equal to zero ; and P^f -f P 2 x 2 - -\ = A'. Hence A=A' + (P l + P,+ ...) a \ which proves the proposition. Radii of gyration. Let k = radius of gyration of the sys- tem with respect to the given axis, and k' the radius of gyra- tion with respect to the central axis, and we have Hence the equation above deduced may be reduced to the form 197. Products of Inertia with Respect to Parallel Axes. - Proposition. The product of inertia of a system of parallel forces with reference to any two axes is equal to the product of inertia with reference to a pair of central axes parallel to the given axes, plus the product of inertia of the resultant (acting at the centroid) with reference to the given axes. Let P v P 2 , etc., be the magnitudes of the given forces ; (/,, l - ( /Vtfi + /Vtf* +)+ a (/Vi + P*q* + ' ) But Pigi+Ptf+--=o; and -Pi/i + A/^H =o; 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 centroicl of the system. Hence, which proves the proposition. If A = (P l + P,+ --)randA>=(P l + P,+ -.-y 2 , we have 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 centroicl. 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 is zero, and hence ab(Pi + P z -\ ---- ) 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 1 and any axis parallel to A". 2. Moments 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. 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 A_^ 3 . /a-*' 2 = 12 ' ~7J Triangle. Let b and d be the base and altitude. Then for an axis parallel to the base, 36' MOMENTS OF INERTIA OF PLANE AREAS. I/I For an axis through the vertex, bisecting the base, k^ , 2 4 where b* is the projection of the base on a line perpendicular to the axis. Circle. Let d be the diameter. Then r^ 4 . p_d? ~7- K ~ s-' 64 16 For a central axis perpendicular to the plane of the circle, 32 ' 8 Ellipse. Let a and b be the semi-axes. Then for an axis parallel to a, 4 For an axis parallel to b, / TTt . A2 _ 4 4 For a central axis perpendicular to the plane of the ellipse, Graphic construction for radius of gyration. Whenever 2 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 k*= , we may put p = d d Then if in p . 6 we take LM= d MN= d , the 34 43 construction there shown will give MP as the value of k. For the triangle, the axis being parallel to the base, we have 2 = - -, and the same construction is applicable. For the 36 v v axis through the vertex bisecting the base, k z = - 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 : (1) 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 /, q, 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. 20 1. 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. 1/3 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. 67. 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. or 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 ABODE. Draw the funicular polygon corresponding to a pole O, and let the successive strings intersect QR in A', B 1 , C', D f , f . 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 E". 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. 67), the radius of gyration may be found by the construction of Art. 191. Thus in Fig. 67, 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. (1) When the axis is known at the start. 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. 68) 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 equal 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 unknown. The method to be MOMENTS OF INERTIA OF PLANE AREAS. 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 ( in the a similar triangles KNM, KML, we have . 09 KN KM KL But KL=a, and or KN= hence KL 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 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. Fig. ro 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. 1/7 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 AB', 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 off 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 ^ (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 in 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 FT) intersect- ing VQ in U; then the line b'c' drawn through U 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't>' is located as the line of action of the force A'B'. The second funicular polygon is now drawn, and the points A 11 , C" are found by the intersection of the strings o'a', o'c 1 with the axis. Hence the moment of inertia of the given area is equal to A"C"xHxff'. 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 I7 g 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, <9F(Fig. 71) be the two given axes; 6, the angle included between them ; P lt P 2) etc., the forces of the system ; x 1 1 y', the coordinates of the point of application of any force P, referred to the axes OX, OY; p, q, the perpendicular distances of the same point from O Y and OX respectively, so that p = x' sin 6, q=y' sin 0. Let a\ b', and c' be quantities denned by the equations a ' = 179 l8o GRAPHIC STATICS. Then and #' sin 0, ' sin 0, r' sin are respectively the radius of gyration with respect to O Y, the radius of gyration with respect to OX, and the product-radius (Art. 189) with respect to OX and OY. 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 (0 )^ f sin <. Hence sn -<-2 sn - sn or, I=b^P sin 2 (0-$)-2c'*2P sin (0-0) sin sn the factors involving and < being constant for all terms of the summation. Hence = -- = b'* sin 2 (<9 -(/>)- 2 ^' 2 sin ((9-0) sin <+V 2 sin 2 . . (i) From these equations /and may be computed if a', b', and c' are known ; that is, if the moments and product of inertia for the two axes (Wand OYare known. Special case. If # = 90, the equation (i) becomes GENERAL PRINCIPLES. !8r 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 ^ = 2/V = 2/y sin e [/ sin (0-0)-*' sin ] = 2P [/ 2 sin0 sin (0 -<)-*'/ sin 6 sin ] = [b'* sin sin (<9-<)-r' 2 sin <9 sin 0] 2/> Let k = product-radius for axes OM and OX\ then // 2 = A = ^2 sin sin ((9-0)-^ 2 sin sin 0. Special case. The axis 6W may be so chosen that A=o. This will be the case if b' 2 sin ((9 -)= ^' 2 sin<. 208. Inertia Curve. If on <9J/ (Fig. 71) a point J/ be taken such that the length O M 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 OM with k. 209. Ellipse or Hyperbola of Inertia. The simplest curve is obtained by assuming OM to be inversely proportional to k, Let OM=r, and take r* = ^ > where <^ 2 is a positive K quantity, so that d always represents a real length, positive or negative. Equation (i) of Art. 206 then becomes 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 Y. Then r _ x y sin sin (# <) sin 1 82 GRAPHIC STATICS, whence sin ( q>) _* sin 2 <9 2 sin (0- sin 2 and the equation becomes 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 . 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 EPAC is negative; a hyperbola if EP AC is positive. (Salmon's Conic 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 1 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 6, where is the angle included between a' and b'. Therefore from the preceding paragraph, a'&'s'm 6 = ab. (5) An ellipse can be constructed, when a pair of conjugate diameters is known, as follows : 1 84 GRAPHIC STATICS. Let AA f , 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 1 ', 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 dW 2 -V 4 is positive, the equation denotes an ellipse. This cannot be the case if a' 2 and b' 2 have opposite signs. But from the definitions of a f2 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 by choosing the axes so that the product of inertia with respect to them is zero ; and if c' is zero, and a' 2 and b have the same sign, the quantity a' 2 b' 2 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 <: f = 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 tc 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 r and from (3)' and (4) of the last article we have ab a'b 1 sin 6 p = , r r if a' and b 1 are conjugate semi-diameters. Now take d' 2 s'm = a& = a'6's'm0, or d*=a'b\ and we have k=p, and the equation of the curve becomes (since c' = o when the axes are conjugate) 1 86 GRAPHIC STATICS. If the equation be written in this form, a' and V 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 between the axis and the parallel tangent to tJie ellipse. Hereafter we shall mean by inertia-ellipse the curve obtained by taking d 2 = a'&' 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 (/) 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', YY' (Fig. 73) be the given axes, and let the four tangents determined as above form the parallelogram PQRS. /Y Let A, A', B, B' be the points P _ BZ. - -, Q in which these tangents in- / / / tersect the axes XX', YY'. o /A Then, since each diameter is / parallel to the tangents at the S B/ R extremities of the conjugate /Y' Fig. 73 diameter, A, A', B, B' are the extremities of the diameters lying in the given axes. The ellipse 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 gyration for parallel axes. (Art. 196.) 3. Inertia-Curves 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 ft 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 /, q, or p, g, 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. [SUGGESTIONS. Since there is an axis of symmetry, this contains one of the principal axes of the ellipse. 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 ellipse as explained in 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', YY', drawn through the centroid parallel to the sides, are a pair of conjugate axes of the cen- tral ellipse. Let AB = b, BC=d, and let //= the perpendicular distance be- tween AB and DC. The moment of inertia of the parallelogram with re- spect to the axis XX 1 is equal to the moment of inertia of a rectangle of sides b and //. Hence 2 , the square of the radius of gyration for this axis, is /2 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 BC, 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, . Vl2 And since the projection of BC on a line perpendicular to XX is equal to h, the projection of EH on the same line is equal to l --, Vl2 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 1 , is simi- lar to the parallelogram ABCD. 1 90 GRAPHIC STATICS. Fig. 75 shows this parallelogram and also the ellipse. (2) Triangle. Let ABC (Fig. 76) be the triangle; b = length of the base BC\ b' = length of projection of BC on a line perpendicular to AD ; d = al- 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 18 30 it intersects AC. 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 1 is to HL as the altitude ^is to AC. Again, for the axis AD, the radius of gyration is -y (Art. 199). Make DE=\ BC and DF=\ BC, and take EF as a diameter of a semicircle. 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 -|- b. Hence the central ellipse of inertia is similar to the given ellipse^ its semi-axes being -| 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 (4) Semicircle. Let ABC (Fig. 77) 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 0, re- spectively parallel and per- pendicular to AB. With respect to the axis YY 1 , the radius of gyration is evidently r, the same as for the whole circle. Hence two lines parallel to YY' and dis- tant J r from it are tangents 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 D as a center, and radius r, draw an arc intersecting XX' at F\ then ^F 2 = DF 2 ~aff. But DF is equal to the radius of gyration with respect to AB, and <9Z>is the distance between XX 1 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. . 34 U) Scale, 1 inch = 6 feet eft . Scale, linch= GOO Ibs. \ PLATE I. DEF Scale, 1 inch=6,ooo Ibs. (C) Scale: I in.= 4.OOO Ibs. Scale: I in, Fig-. 42 PLATE II. 1 , 1 1 1 | 1 1 oo 8 3 S" 9 cT Si 8.1 5.8 1 4.5 4.5 7.1 4.8 5.7 t OOOOooo Linear Scale, 1 in. = lo ft. Force Scale, 1 in. = 80,000 Ibs. Xmr X s xt PLATE III. 8.1 5.8 7.1 4.8 o OOOO oooo a a 3 a 8.1 5.8 4.5 4.5 7.1 I 4.8 I 5.7 o O OOO oo oc Fig.oO Linear Scale, 1 in.=SOft. Force Scale, 1 in. =80,000 Ibs. PLATE IV. a a a 5.8 I 4.5 I 4.5 I 7.1 I 4.8 I o.7 I 4.8 f,.0 ill I 4.0 I 4.0 o O OOO O O O Q i *"-" S 9 T 10 M 11 V Linear Scale, 1 in.-'. Force Scale, 1 in. 80,000 Ibs. 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