ENGINEERING LIBRARY ESSENTIALS IN THE THEORY OF FRAMED STRUCTURES PUBLISHERS OF BOOKS FOR_/ Coal Age ^ Electric Railway Journal Electrical World ^ Engineering News-Record American Machinist ^ Ingenierfa Internacional Engineering 8 Mining Journal * Power Chemical 6 Metallurgical Engineering Electncal Merchandising ESSENTIALS IN THE THEORY OF FRAMED STRUCTURES BY w CHARLES A. ELLIS, A. B. VICE-PRESIDENT THE STRAUSS BASCULE BRIDGE CO. WITH AMERICAN BRIDGE COMPANY IQOI-IQOS. ASST. PROFESSOR OF CIVIL ENGINEERING, UNIVERSITY OF MICHIGAN 1008-1912. DESIGNING ENGINEER, DOMINION BRIDGE COMPANY 1912-1914 PROFESSOR OF STRUCTURAL ENGINEERING, UNIVERSITY OF ILLINOIS I9I4-I92I FIRST EDITION McGRAW-HILL BOOK COMPANY, INC. NEW YORK: 370 SEVENTH AVENUE LONDON: 6 & 8 BOUVERIE ST., E. C. 4 1922 OEM. uw ENGINEERING LIBRARY COPYRIGHT, IQ22, BY THE MCGRAW-HILL BOOK COMPANY, INC. PRESS YORK PA PREFACE This book, designed primarily for class-room use, contains no new principles. The same fundamental ideas to be found in many other books are here set forth by a treatment so different from the orthodox method that a word of explanation seems justifiable. There is a growing tendency in engineering schools to instruct rather than to educate; to pour in rather than to draw out ; to feed the mind with memoranda until it becomes sluggish for lack of exercise in the intellectual realms of the imagination and reason. This tendency is, to a large extent, the result of current text-book and class-room methods which present mathematical principles in generalities first and particulars last. This order is undesirable for two reasons: First, it is not the sequence most natural for the grasp of the human intellect, for the history of any science will show that its development has been accomplished by progress from the particular to the general. Second, if generalities are considered first, the princi- ple is unfolded to the class by deriving a formula. Once developed, this formula is memorized, and all subsequent problems are solved by merely making numerical substitutions for the various symbols. Thus, the student gets but a passing glance at the fundamental idea and a meager conception of the principle involved. This statement can be proved very easily by asking any group of students who can glibly recite : f -My Me f ~ ~T I to determine the fiber stress at a certain point in a given beam without using a formula. This experiment was tried in nearly one thousand cases with the result that about 93 per cent of the students were found to be helpless without the formula. M171603 VI PREFACE This instance is more fully discussed by the author in an article " Reinforced Concrete Theory Without the Use of Formulas," Bulletin of the Society for the Promotion of Engineering Education, Vol. VIII, p. 380. In this book each principle is introduced by the use of illus- trative numerical problems, some of which will lead the student into a pitfall from which he should be forced to extricate himself without the instructor's help. Experience is a splendid teacher. There is no opportunity for the student to dodge the funda- mental idea or the physical conception of the problem by resorting to a formula, for none is given except at the end of the discussion when the general case is considered. Even then the general expression is frequently left to be developed by the student. The experienced teacher who has had considerable practical experience will, no doubt, feel more at home with this method of treatment than will the younger instructor who is closely confined to his text and who feels that a knowledge of the course is of more importance to the student than the development of mental power. The major portion of the book may be read intelligently by any student who has a working knowledge of arithmetic, algebra and geometry. Very little reference (except in Chapter V) is made to the calculus. Practically every other textbook makes use of formal calculus in the treatment of maximum stresses for moving loads. Of course, this method is entirely unnecessary. It is true that the fundamental concept of the differential calculus is ever present in the treatment of maximum stresses in connection with influence lines as given in Chapter IV; but the average college student who can differentiate and integrate will probably not recognize the calculus, so why risk confusing him by any reference whatever to dy dx Some new material will be found in Chapter III relative to wind reactions of a mill-binding bent. A treatment of deflec- tions by the area-moment method, more comprehensive than is usually found in American textbooks, appears in Chapter V. PREFACE Vll A new short method for computing the stresses in a swing span is given in Chapter VIII. The author recommends that the teaching schedule be arranged to include one three-hour period a week in addition to the two or three regular recitation periods. These three- hour periods should be used for written tests in which the student is given an opportunity to solve problems which are just ahead of the textbook assignments, or which were con- sidered so far back in the student's course that he has forgotten the solution; the idea being to give the student some opportun- ity for independent thought. For example: Suppose that during a certain week the textbook assignments are in Sec. Ill, Chapter IV; and that the stress in a diagonal of a parallel chord truss has been discussed in the recitation period the student having solved several problems of this character. The following questions are recommended for the three-hour period of that week : 1. Prove that the sum of the three angles of the triangle equals two right angles. 2. Compute the maximum tensile stress in the number "l^La of the curved chord truss (Fig. 117) for an E-6o train load. Any simple fundamental problem in algebra, calculus or mechanics will serve the purpose equally as well as the problem in geometry. If the student has mastered the principle of maximum stress in the diagonal of a parallel chord truss he should be given the opportunity of doing some independent thinking on the principle next in order, instead of being asked to memorize that principle from the text in preparation for recitation. The instructor who expects that the student will submit 100 per cent papers on an examination of this sort will be sadly disappointed; and the student who is not accustomed to this method of conducting tests will at first feel that he is being treated rather roughly; but the mental development which results from tests of this character is of more value to the student than high grades. The major portion of this book has been used for several years as preprints in the author's classes and he is indebted PREFACE to former students for many helpful suggestions. The author wishes also to acknowledge his indebtedness to Professor Geo. E. Beggs for permission to use his table of equivalent uniform loads. CHARLES A. ELLIS. CHICAGO, ILLINOIS, Sept. 29, 192 1. CONTENTS PAGE PREFACE .............................. v CHAPTER I SECTION I. INTRODUCTION 1. Mechanics ........................... x 2. Methods and Mental Attitude .................. 2 3. Forces ............................ 4 4. The Three Elements of a Force ................ . 5 5. Definitions .......................... ,5 SECTION II. RESOLUTION AND COMBINATION OF FORCES 6. Parallelogram of Forces ..................... 6 7. Triangle of Forces ....................... g 8. Magnitude-direction Diagram ........ .......... g g. Rectangular Components .......... .......... 10 10. Moment of a Force ....................... n SECTION III. THE THREE FUNDAMENTAL STATEMENTS OR EQUATIONS OF STATIC EQUILIBRIUM ii ................................ 12 SECTION IV. COPLANAR CONCURRENT FORCES 12. Illustrative Problem ...................... 14 13. General Case ................... ...... 17 14. Only Two Independent Statements or Equations ..... ..... 19 15. Illustrative Problem ...................... IQ 16. Illustrative Problem ........... ........... 20 17. Special Case .......................... 21 18. Conclusion ................ . ......... 22 19. Problems ................. '....' ...... 23 SECTION V. COPLANAR PARALLEL FORCES 20. General Considerations ..................... 24 21. Illustrative Problems ...................... 27 22. Conclusion .......... ................ 30 SECTION VI. COPLANAR NON-CONCURRENT NON-PARALLEL FORCES 23. Transformed System ...................... 30 24. Illustrative Problem ...................... 31 25. General Case ......................... 32 26. Four Groups .......................... 33 ix X CONTENTS PAGE 27. Group 2 '. 35 28. Groups ;'.-...' 36 29 36 30. Group 4 36 3 1 37 32. One Unknown Location 37 33 38 34. Combinations 38 35. Illustrative Problems 38 36. Problems 54 CHAPTER II SECTION I. SIMPLE TRUSSES 37. Rigid Body 60 38. A Truss 61 39. Stability 61 40. The Assumption 61 41. The Application 62 42. The External Forces 62 43. The Internal Forces 62 44. The Method of Joints 63 45. The Method of Sections 68 46. Problems 71 47 73 SECTION II. STRUCTURES REQUIRING SPECIAL CONSIDERATION 48. The Fink Truss . 73 49. The Three-hinged Arch 75 50. Illustrative Problem 77 51. The Cantilever Bridge 80 52. Inclined Loads '. 81 SECTION III. BEAMS: SHEAR AND BENDING MOMENT DIAGRAMS 53. A Beam 83 54. The Shear 83 55. The Bending Moment 83 56. The Bending Moment Determined from the Location-direction Diagram 84 57. Shear and Bending Moment Diagrams 86 58. Illustrative Problems -,....'.. 87 59. The Structural Engineer's Parabola ....'.. 91 60. Illustrative Problem 95 61. Problems ' 97 SECTION IV. FRAMES HAVING MEMBERS WHICH PERFORM THE FUNCTIONS OF A BEAM 62 ^ ...... 99 63. Illustrative Problem 09 64. Problems 102 65. The Portal Frame .......,,. 102 CONTENTS XI PAGE 66. The Portal Frame 104 67. The Portal Frame 105 68. Problems 107 CHAPTER III ROOF TRUSSES SECTION I. TYPES AND LOADS 69. Standard Types 108 70. Purlins 109 71. The Bracing . "-. _.. no 72. Spacing of Trusses ... ; in 73. The Dead Loads. .....' in 74. The Snow Load 112 75. The Wind Load 113 SECTION II. STRESS ANALYSIS FOR WIND 76. Wall Bearing Trusses , ; ; . . . . . 115 77. Trusses Supported on Columns . . 116 78. Stress Diagram for Wind Unnecessary ...,... 124 79. Combined Snow and Wind Loads 125 80. Design Stresses 125 81. Conclusions 126 CHAPTER IV BRIDGES SECTION I. STANDARD TYPES 82. Standard Types. . ."vv ........ 128 83. Moving Loads ... . ', '.'..;*. . . . 130 SECTION II. DECK BEAMS AND GIRDERS 84. Influence Line for Reaction 130 85. Influence Line for Shear in a Beam 132 86. Influence Line for Bending Moment in a Beam 132 87. Criterion for Maximum Bending Moment 134 88. Cooper's Standard Train Loads 137 89. Illustrative Problem 138 90. Problems 143 91. General Criterion for Maximum Bending Moment 143 92. The Point of Greatest Maximum Bending Moment in a Beam. . . . 143 93. Problems 145 94. Criterion for Maximum Shear in a Beam 146 95. Problem 146 SECTION III. PARALLEL CHORD TRUSSES 96. A Railway Truss Bridge 146 97. Influence Line for Maximum Stress in a Chord Member. ...... 146 98. Illustrative Problem . . 148 Xll CONTENTS PAGE 99. General Criterion for Maximum Chord Stress 149 100. The Stress in a Web Member of a Parallel Chord Truss 151 101. Influence Line for Shear in a Panel 151 102. Criterion for Maximum Shear in a Panel ....... 152 103. Illustrative Problems. . . , .C. v 155 104. General Criterion for Maximum Shear in a Panel 157 105. Impact 159 106. Stresses in a 2oo-ft. Pratt Truss 160 107. Counters 166 108. Truss with an Odd Number of Panels 167 SECTION IV. PARKER TRUSSES 109. Stress in Web Member of Parker Truss 169 no. Influence Line for Web Member of Parker Truss 170 in. Criterion for Maximum Stress in Web Member of Parker Truss . . . 172 112. Illustrative Problems 174 113. Problems 176 114. General Criterion for Maximum Stress in Web Member of a Parker Truss 177 115. Tension in a Vertical when Counters are Used 182 SECTION V. THE BALTIMORE TRUSS 116. The Baltimore Truss 184 117. LJ^e 184 118. U 2 U 4 185 119 189 120 189 121. u 2 Ms 190 122. MsL 4 190 123 . J 92 124. U^ 4 194 125. U2L 2 197 SECTION VI. THE PENNSYLVANIA TRUSS 126. Pennsylvania Truss 197 127. U 2 U4 198 128. U 2 M 3 ..^ 198 129. MaL 4 199 130. U4L 4 200 SECTION VII. EQUIVALENT UNIFORM LOADS 131 200 132. Examples 290 133. General Considerations , 202 134. Triangular Influence Diagrams 203 135. Table of Equivalent Uniform Loads 205 CONTENTS Xlll CHAPTER V DEFLECTION OF BEAMS SECTION I. THE AREA-MOMENT METHOD PAGE 136. General Considerations 208 137. Method of Area-moments 208 138. First Principle 211 139. Second Principle t 212 SECTION II. SIMPLE BEAMS OF UNIFORM CROSS-SECTION 140. The Beam in Fig 212 141. Maximum Deflection 215 142. Point of Maximum Deflection 218 143 220 144. Deflection Under Uniform Load 221 145. Deflection for Load of Uniformly Varying Intensity 224 146. Deflection for Several Concentrated Loads 226 147. Deflection for Uniform and Concentrated Loads 229 SECTION III. MAXWELL'S THEOREM OF RECIPROCAL DISPLACEMENTS 148. Maxwell's Theorem of Reciprocal Displacements 231 SECTION IV. CANTILEVER BEAMS i49 - 232 SECTION V. BEAMS WITH VARYING CROSS-SECTION 150. General Expressions 235 151. Beams with Varying Depth 236 152. Beams with Cover Plates 237 CHAPTER VI RESTRAINED AND CONTINUOUS BEAMS SECTION I. RESTRAINED OR FIXED BEAMS 153. General Considerations 240 154. Restraint at One End Concentrated Load 241 155. Restraint at One End Uniform Load 242 156. Restraint at Both Ends Concentrated Load 243 157. Restraint at Both Ends Uniform Load 247 SECTION II. CONTINUOUS BEAMS 158 248 159. Application of Maxwell's Theorem 251 160. The Conventional Method 252 161 256 162. The General Expressions 257 XIV CONTENTS PAGE J 63 257 164. Two Unequal Spans, Supporting Unequal Uniform Loads. ..... 259 165 260 166 ....... 261 167 263 168. Three Equal Spans Uniform Load 264 169. Four Equal Spans Uniform Load 266 170. Coefficients for Pier Reactions * . . 267 SECTION III. CLAPEYRON'S THEOREM or THREE MOMENTS 171. Clapeyron's Theorem 268 SECTION IV. PARTIALLY CONTINUOUS BEAMS 172. No Shear Transmitted 270 173. No Moment Transmitted 271 SECTION V. CONTINUOUS BEAMS IN FOUNDATIONS i?4 273 175. Case I Projections Not Limited by Site 275 176. Case II Projection at One End Limited by Site 277 177. Case III Both Projections Limited by Site 279 CHAPTER VII DEFLECTION or TRUSSES 178. Stress and Strain. 282 SECTION I. ALGEBRAIC METHOD 179. The Algebraic Solution 283 180. Illustrative Problem 285 SECTION II. GRAPHIC METHOD 181. Willot Diagrams 288 SECTION III. GENERAL CONSIDERATIONS 182. Temperature 297 183. Camber 298 184. Maxwell's Theorem of Reciprocal Deflection 299 CHAPTER VIII SWING BRIDGES SECTION I. CENTER-BEARING SWING BRIDGES 185. General Considerations * . 3 O1 186. Conditions of Loading 3 O1 187. Stresses in a Swing Span . . . 33 188. Positive Shear in Panel o-i 3 6 CONTENTS XV PAGE 189. Positive Moment about U 3 Case III 310 190. Negative Shear in Panel o- 1 311 191. Shear in Panel 1-2 , . .':. . . /-.. . . . 313 192. Shear in Panel 2-3 313 193. Shear in Panel 3-4 -. ...... '*) . . 313 194. Shear in Panel 4-5 . . . ... . . . . . i . -. . . . . 313 195. Shear in Panel 5-6 ...'......*,..... .314 196. Moment about L 2 . . . . . ..... ', , . . 315 197. Moment about L4 . . .' . . . V. ... .- 316 198. Moment about U5 .."-. ... . . . . 317 199. Moment about Le 318 200. Case V Both Arms Loaded 318 201. Negative Shear in Panel 5-6 * . 319 202. Moment about Le ; 319 203. Dead Load: Bridge Open ...................... 320 204. Dead Load: Ends Raised ....... . . . . , 320 205. Combinations .. ... . , . 320 206. Reactions from Willot Diagram 320 SECTION II. RIM-BEARING SWING BRIDGES 207 V . ... ... 323 208. Partial Continuity 325 INDEX 327 ESSENTIALS IN THE THEORY OF FRAMED STRUCTURES CHAPTER I EQUILIBRIUM OF COPLANAR FORCES SEC. I. INTRODUCTION i. Mechanics. The theory of stresses in framed structures is the development of a few, simple, fundamental principles of mechanics. The science of mechanics is a study of motion. The idea of motion is closely related to the ideas of space, time and mass. We study first the motion of a body without regard to the time consumed in the motion, or the mass of the thing moved. This part of mechanics is called " Geometry of Motion." Secondly, the introduction of the idea of time leads us to con- sider velocity and acceleration. This study is called "Kine- matics." Finally, when the body is assigned to a certain mass, we consider the ideas of energy, momentum and force. This part of mechanics is called " Dynamics" and is divided into two branches " Kinetics" and "Statics." Kinetics treats in the most general way of changes in motion produced by forces. Statics is a consideration of those cases in which no change of motion is produced, and the element of time is not a factor. The fundamental conception of stress analysis in framed structures is acquired by a systematic study of that part of Statics which treats of the "Equilibrium of Coplanar Forces." 2 THEORY OF FRAMED STRUCTURES CHAP. I This topic is not a new one to the student of Physics and Mechanics. In both of these courses it is treated in a general way in connection with many kindred topics, equally important in other fields of engineering science. The first chapter of this book has been prepared for the purpose of making an extended investigation of the " Equilib- rium of Coplanar Forces," to the end that the reader may acquire a more specific knowledge of the subject and the power to apply this knowledge with reason and common sense. It is the most important chapter in the book; it forms the base upon which rest the principles of each succeeding chapter, and should be thoroughly mastered. 2. Methods and Mental Attitude. Whenever we engage in a new piece of work, it is wise to consider the tools with which we may accomplish our task, and the manner in which these tools can be used most advantageously. The man who is about to cut down a tree decides, not only whether he will use an axe or a saw, but also (and quite unconsciously, perhaps), whether he will operate the chosen implement with his hands or his feet. The student will do well to emulate the prudence of the wood cutter and, for the moment, consider the tools or methods by which stresses may be determined. He should also take thought of the much more important question By what mental pro- cesses can these tools most successfully be operated? There are two methods by which stresses may be determined : (a) the algebraic method, which makes use of numerals or symbols; (b) the graphic method, which represents quantities by the length and direction of lines. Graphic methods are not limited to the solution of static problems. The operations of addition, subtraction, multipli- cation, division, involution and evolution can be performed graphically; and many problems in calculus are more readily comprehended when illustrated by the graphic method. Whether the algebraic or the graphic method is better, is a question which depends for its answer upon the individual problem. However, if each problem is solved by both methods, the fundamental principles are more clearly set before the beginner. SEC. I EQUILIBRIUM OF COPLANAR FORCES 3 The student should always remember that mental develop- ment is more to be desired than the mere accumulation of information, which is of little value unless its possessor has the mental power or capacity for using it intelligently. Since the student can acquire intellectual power only by intellectual exercise, he is frequently asked by his instructor to think. This request may be interpreted in several different ways. One may literally comply with the request if he memorizes, imagines, reasons or performs one of several other minor intellectual funct- ions. The intellectual development, however, varies with the form of exercise a fact which the instructor should not fail to recognize. Although the training of the memory is an important intellectual process, imagination and reasoning should not be neglected. Which of these three plays the lead- ing role in the drama of life is a question which is answered differently perhaps by the linguist, poet and scientist. It is a significant fact, however, that the majority of people who plead guilty to possessing a poor memory are much offended when accused of having little imagination or inferior judgment. The degree to which the student will exercise his memory, his imagination or his reason in the preparation of a lesson, depends to a considerable degree upon the manner in which the subject is presented in the textbook. If he sees a formula, his first im- pulse is to memorize it, too often in parrot-like form; while its history, meaning and limitations are to him a matter for second- ary consideration. In a text book or in the class-room, formulas should be given what little place they deserve at the end, and not at the beginning of a subject; for the simple reason that algebraic manipulations obscure the physical relations which should be visualized to the highest degree. It may appear to the casual reader that imagination is an intellectual process exercised exclusively by the artist or poet. Such is not the case. A structure, whether it be a steam engine, electric generator or a railway bridge must be visualized first in the mind of the engineer before it can become a reality; or even before it can take shape and size upon the designing board. In preparing the manuscript for this book, the author has 4 THEORY OF FRAMED STRUCTURES CHAP. I endeavored to present each topic in such form that the student will find employment for his imagination and reason as well as for his memory. Incidentally, he will acquire a working knowl- edge of the theory of stresses in framed structures. 3. Forces. The word "force" is used to express a variety of ideas. We speak of physical, mental and moral forces, etc. But for scientific purposes it is necessary to have a single definite meaning. In Physics and Mechanics we define force as the cause of acceleration or change in velocity; i.e. force = rate of change of momentum and momentum = mass X velocity hence force = mass X rate of change of velocity or since rate of change of velocity = acceleration therefore force = mass X acceleration. All bodies are constantly acted upon by the force of gravita- tion; and, except in the special case of bodies falling in a vacuum, by other forces also. A moving train on a straight, level track is usually subject to the action of four forces: the action of gravitation "pulling down" and the reaction of the rails "push- ing up"; the "pull" of the engine in one direction and the retard- ing action of friction and atmosphere in the opposite direction. If the reaction of the rails equals the action of gravitation, the train will remain at the same level. If the engine pull is greater than the retarding action, the difference represents the force which accelerates the speed of the train; thereby increas- ing its velocity and momentum. On the other hand, if the force represented by the engine pull equals the retarding forces, the train is moving at a constant speed, and there is no rate of change of velocity or momentum. The two equal and opposite forces "balance" and their algebraic sum is zero. Thus if a body is at rest, or is moving in a straight line with constant velocity, the acceleration is zero, there is no change in momen- tum, and the body is said to be in a state of equilibrium. The theory of stresses in framed structures is developed by a study of the action of forces on bodies at rest. We need give no special attention to velocity, momentum or acceleration in our conception of the idea of force. For our present discussion we may therefore properly conceive of a force as the idea ex- pressed in the words "push" and "pull." SEC. I EQUILIBRIUM OF COPLANAR FORCES 5 4. The Three Elements of a Force. We know that the effect of a force of 50 Ib. pushing vertically (downward) upon the head is quite different from the effect of a force of 80 Ib. pulling hori- zontally at the feet; and thus from our personal experiences we may observe that the characteristics or elements of a force are: 1. Magnitude. 2. Direction. 3. Location. The effect of the action of a force upon a body cannot be determined until all three elements are known. The magnitude or intensity of a force is usually expressed in pounds, units of 1,000 Ib., or tons. The direction may be indicated by the angle between the hori- zontal and the line along which the force acts. For our purpose, however, it will be found more advantageous to designate the direction by giving the slope or bevel of the line along which the force acts. The location of a force is known when the position (with refer- ence to the body) of any point in the line representing the direc- tion is given. The sense in which a force acts along a given direction, i.e., up or down, toward the right or left, is represented graphically by an arrow-head. In an algebraic solution the sense is desig- nated by the signs + or in connection with the magnitude of the force. The sense of a force is not ranked as an element, since it requires no separate equation for its determination. 5. Definitions. Forces are coplanar when acting in the same plane; non-coplanar, when acting in different planes. Forces are concurrent when their directions meet in a point, and non-concurrent when their directions do not so meet. Several forces under consideration are called a system of forces. A couple is a system of two parallel forces, equal in magnitude, but opposite in sense. Two systems of forces are equivalent if one system may be substituted for the other without changing the state of rest or motion of the body upon which they act. O THEORY OF FRAMED STRUCTURES CHAP. I The resolution of forces is the process of resolving the forces of one system into an equivalent system having a greater number of forces. The composition of forces is the process of combining or re- ducing the forces of one system into an equivalent system hav- ing fewer forces. If the equivalent system reduces to a single force, that force is called the resultant of the system; and each force in the system is a component of the resultant force. All systems, however, cannot be reduced to a single force; but all coplanar systems not in equilibrium can be reduced to either a single force or a couple. SEC. II. RESOLUTION AND COMBINATION OF FORCES 6. Parallelogram of Forces. The law of " Triangle of Forces" seems to have been discovered in the year 1608 by Simon Stevin, a Flemish military engineer, who conceived the idea of the reso- lution and composition of forces, by his investigation of the properties of the inclined plane. He derived the law from experimental data. Mathematical proofs of this law have frequently been presented; but several eminent writers of the present day seem to be content with the statement that the law is so fundamental that it cannot be deduced from anything more simple than itself. An experimental demonstration of the law is illustrated in Fig. i. Three smoothly running pulleys are mounted on a vertical drawing board at any points A, B, and C not in the same straight line. Three strings are tied together, placed over the pulleys and made to support three unequal weights, P, Q, and S. Under the action of gravitation, the weights will adjust themselves and come to a state of rest. Mark on the board the position of the knot at o. Add a small weight to P and note the change in the position of the knot. Remove the small weight and observe that the knot returns to its original position. Do likewise with Q, and with S. Now increase each weight by an equal amount and note that the knot changes its position. Increase each weight by one-half or one-third of itself and observe that the position of the knot is not changed. SEC. II EQUILIBRIUM OF COPLANAR FORCES 7 Under the action of gravitation, each weight produces a corresponding tension in the string which supports it; and we have a system of three concurrent forces in equilibrium. In order that a line may be employed to represent a force it is necessary that: (i) its length should indicate the magnitude of the force; (2) its direction, with an arrow-head to show the sense, should correspond to the line of action of the force; (3) the location of a point in the line should be given. The lines oA , oB and oC with arrow-heads to show the sense, represent the directions of the three forces, and the point o designates their location. By choosing a convenient scale- ratio, let us say, 10 Ib. per inch, the lengths oa, ob and oc may be laid off on the lines oA , oB and oC to represent the magnitudes, FIG. i. P, Q and S respectively. The sequence of the letters which designate the force may be used to indicate the sense of the force. Thus oa signifies that the sense of the force P is upward from o to a, while ao would represent a force the equivalent of P in all respects except that the sense would be opposite, i.e., downward from a to o. The forces P, Q and S are thus com- pletely represented by the lines oa, ob and oc, since each line designates the magnitude, direction (including sense) and location of its respective force. Draw a line through a, parallel to ob, and a line through b, parallel to oa completing the parallelogram oadb. Measure the 8 THEORY OF FRAMED STRUCTURES CHAP. I length of the diagonal od, with the same scale as before. Ob- serve that the length do equals the length oc; and that the points Cj o and d are in a straight line. We say that the force 5, represented by do, is the equilibrant of the forces P and Q, since it holds them in equilibrium or balances them. With equal propriety we can say that a force Rj represented by od having equal magnitude, the same direction and location as the force 5 but opposite sense, would hold the force 5 in equilibrium, if the forces P and Q were removed. Hence, the force R is called the resultant of the forces P and Q, since it may be substituted for them without disturbing the state of equilibrium. The law of the parallelo- gram of forces may be stated as follows : If two concurrent coplanar forces are represented in magni- tude and direction by the adjacent sides of a parallelogram, so constructed that the sense of each force is the same with respect to their point of intersection; then the resultant of the two forces is represented in magnitude and direction by the diagonal of the parallelogram through the point of intersection. If the sense of the two forces is away from the point of intersection, the sense of the resultant is away from the point and vice versa. 7. Triangle of Forces. Since a diagonal divides a parallelo- gram into two equal and similar triangles, it is necessary to construct but one of the triangles in order to find the resultant. Instead of representing the two forces P and Q by the lines oa and ob drawn from o, we may draw oa representing the forceP; and from a draw ad representing the force Q. Or, we may first draw ob to represent the force Q, and from b draw bd representing the force P. In either construction the line od represents the resultant R, of the forces P and Q; while the line do represents their equilibrant S. In a concurrent system, all forces act through a common point which establishes the location of each force, and we shall be concerned chiefly with the determination of unknown magni- tudes and directions. For this reason it is not essential that the triangle be constructed in connection with the point of concurrency o. The triangle efgj for example, having sides respectively parallel to ob, bd and do will serve our purpose in a SEC. II EQUILIBRIUM OF COPLANAR FORCES consideration of the forces P, Q and 5, equally as well as the similar triangle obd. The homologous sides of the two triangles are proportional and parallel an important fact to remember. The triangles differ in size simply because different scale-ratios were used in their construction. If the sides of a triangle represent three forces in equilibrium, the arrow-heads will indicate a continuous course around the perimeter of the triangle. If, however, the three sides of the triangle represent two forces and their resultant, the arrow- head of the resultant will be opposed to the arrow-heads of the two other forces. The law of the triangle of forces may be stated as follows: Three concurrent co planar forces in equilib- \^ -^ s ^ rium may be represented in magnitude, direction and sense by the three sides of a triangle taken in contin- uous order. Any one of the three forces may be made to represent the resul- tant of the other two by reversing its sense. 8. Magnitude-direction Diagram. When the re- sultant of several concurrent forces is required, the resultant of any two may be found from the force triangle; and this resultant combined with a third force to obtain the resultant of the three forces. This process may be continued until all forces have been combined. A system of four coplanar, concurrent forces P, Q, S and T intersecting at O is represented in Fig. 20. In Fig. 2b, ac is the resultant of P and Q; ad is the resultant of ac and S; and finally ae, the resultant of ad and T, is the resultant R of the entire system. Evidently the force , represented by ea, having the same magnitude and direction as R but opposite in sense, is the equilibrant which, if applied at would secure equilibrium. The partial resultants ac and ad FIG.Sb 10 THEORY OF FRAMED STRUCTURES CHAP. I are obviously superfluous, since the figure abode could have been drawn without them. Textbooks in Physics and Mechanics usually refer to the diagram in Fig. 26, as a force polygon. The name is misleading. The figure does not completely represent a force it represents only two of the three essential elements. Magnitude-direction diagram is a more distinctive term. The order in which the forces are combined is of no particular consequence. In Fig. 20 the forces have been combined in the order, (), T, 5, P ; but the resultant R, or the equilibrant E, are the same as in Fig. 26. The known magnitudes and directions were laid off from a to e\ and the magnitude and direction of the equilibrant E was found by drawing the line ea, which necessarily closed the figure. It follows that a graphic solution for equilibrium must comply with the following law: If a system of coplanar, concurrent forces is in equilibrium, the magnitude-direction diagram will close, and the sense will be continuous. Of course the magnitude-direction diagram could have been closed by any series of straight, broken lines, continuous from e to a; but certain limitations must be imposed upon the number of unknown magnitudes and directions if a single solution is to be realized. We shall refer to this in detail in connection with the algebraic solution. 9. Rectangular Components. A force may be resolved into two components by applying the triangle law conversely. Since an infinite number of triangles can be drawn having one side in common, it follows that the problem of finding the two com- ponents of a given force is indeterminate unless certain conditions are imposed upon the components. Rectangular components have directions intersecting at an angle of 90, and are very helpful in algebraic solutions; since an inclined force can thus be resolved into and replaced by its horizontal and vertical components. Thus in Fig. i the three forces in equilibrium, P, Q and S are resolved into horizontal and vertical components by the right triangles fkg, ejf and gie respectively. The hori- zontal components ej and fk are balanced by the component SEC. II EQUILIBRIUM OF COPLANAR FORCES II gi, being equal in magnitude and opposite in sense. Likewise the vertical components jf and kg are balanced by the com- ponent ie. Whenever the magnitude-direction diagram is a closed figure, whether a triangle, as efg (Fig. i); or any polygon, as abode (Fig. 2) , the horizontal components will balance and the vertical components will balance. If opposite signs (+ and ) are given to magnitudes of opposite sense, the algebraic sum 1 of the horizontal components is : and likewise = o ZF = o These two equations in an algebraic solution correspond to the closing of the magnitude-direction diagram in a graphic solution. 10. Moment of a Force. The product of the magnitude of a force and its perpendicular distance from a point is called the moment of the force about the point. The perpendicular dis- tance is called the arm of the force, and the point is known as the cen- ter of moments. If the magnitude of a force P is 8 lb., and its arm from a point is 3 ft., the mo- ment of the force P about the point Ois24ft.-lb. In 1687, r about 80 years after Stevin had published his demonstration of the triangles of forces, Pierre Varignon pre- sented before the Paris Academy, the "Principle of Moments." This principle is known as Varignon's Theorem. He stated that "the moment of the diagonal of a parallelogram of forces equals the sum of the moments of the other two sides." His proof is somewhat as follows: The parallelogram CADB (Fig. 3) represents the forces P and Q and their resultant R. The lengths of the perpendiculars from the center of moments O to the lines representing the forces Pj Q and R, are p, q and r respectively: J The symbol S represents the idea expressed in the words "algebraic sum." 12 THEORY OF FRAMED STRUCTURES CHAP. I area AOCD = area AOCA + area AOAD + area AACD lrR = - 2 pP + lsQ + tQ hence, rR = pP + g<2 Varignon showed that the center of moments may be located outside the parallelogram, within it, or on one of its sides. It is obvious that the moment rR of the resultant Rj acting counter-clockwise about the point 0, may be balanced by the moment rE of the equilibrant E acting clockwise. If opposite signs are given to the moments of the two forces according as they act clockwise or counter-clockwise, their algebraic sum is zero, or rR - rE = o therefore, pP + qQ - rE = o P, Q and E represent three coplanar, concurrent forces in equilibrium; but the principles may be generalized so as to include any number of concurrent forces by combining them into partial resultants after the manner of Fig. 2. Hence, if a system of coplanar, concurrent forces is in equilibrium, the algebraic sum of the moments of all the forces about any point in their plane is I.M = o SEC. III. THE THREE FUNDAMENTAL STATEMENTS OR EQUA- TIONS OF STATIC EQUILIBRIUM II. Problems in stress analysis deal in general with the action of coplanar forces upon bodies at rest, and the solution of any problem consists in finding the unknown elements of these forces. If the problem is " statically determinate," the num- ber of unknown elements will not exceed the number of inde- pendent statements or equations which can be written concern- ing the static equilibrium of the body upon which the forces are acting. It is a simple matter to write these equations if we consider carefully the conditions under which a body may be at rest. SEC. Ill EQUILIBRIUM OF COPLANAR FORCES 13 If a body is at rest under the action of any system of forces lying in a vertical plane, the following statements are self- evident from the very nature of the case. 1. The body is not moving toward the right or toward the left, consequently the horizontal magnitudes acting toward the right are balanced by the horizontal magnitudes acting toward the left. If the horizontal magnitudes acting toward the right are given a positive sign, and the horizontal magnitudes acting toward the left are given a negative sign, the algebraic sum of the horizontal magnitude is I# = o (i) 2. The body is not moving upward or downward, conse- quently the vertical magnitudes acting upward are balanced by the vertical magnitudes acting downward. If the vertical magnitudes acting upward are given a positive sign, and the vertical magnitudes acting downward are given a negative sign, the algebraic sum of the vertical magnitude is IF = o (2) 3. The body is not rotating, either clockwise or counter- clockwise, about any point in the plane of the forces; conse- quently the moments of all the magnitudes acting clockwise about any point in the plane of the forces are balanced by the moments of all the magnitudes acting counter-clockwise about the same point. If clockwise moments are given a positive sign, and counter-clockwise moments are given a negative sign, the algebraic sum of the moments about any point in the plane of the forces is ZM = (3) These three statements or equations are the conditions which must be fulfilled by any system of forces acting in a vertical plane upon a body at rest, whether the system be concurrent, parallel or non-concurrent non-parallel. Equations (i), (2) and (3) embody the three fundamental principles of static equilibrium for coplanar forces. They are as fundamental and important as they are simple. Equations (i) and (2) are easily written for any system, after each inclined force has been resolved and replaced by its hori- 14 THEORY OF FRAMED STRUCTURES CHAP. I zontal and vertical components. Instead of resolving the forces along horizontal and vertical axes, we may choose any other pair of axes inclined at any angle and equate to zero the algebraic sum of the components parallel to each axis. Thus, we may write many equations of types (i) and (2). We may write also many equations of type (3) by choosing different points for the center of moments. Since a single solution is possible only when the number of unknown elements does not exceed the number of independent equations which may be written, the question immediately arises How many equations, written as above, will be inde- pendent for a given system of forces? In answering this ques- tion we must distinguish between concurrent, parallel, and non-concurrent non-parallel systems; and for this reason the three systems will be considered separately. SEC. IV. COPLANAR CONCURRENT FORCES 12. Illustrative Problem. Five concurrent forces, A, B, C, D and E are located by the point in Fig. 4#. The direction of each force is given, and the magnitudes of D and E are known. We shall attempt to find (a) by the algebraic method and (b) by the graphic method, the magnitudes of A , B and C necessary for equilibrium. (a) Algebraic Method. We first resolve each inclined force into horizontal and vertical components, and indicate them in the sketch (Fig. 46). The magnitude of the force D is 100 Ib. It acts to the left and upward, sloping at the bevel of 3 in. horizontally to 4 in. vertically; as indicated by the arrow-head at D and the sides fj and jk of the triangle $&. Hence, fj :jk :fk 1:3 -.4 15 Let D = 100 Ib. represent the magnitude of the force, D h represent the horizontal component, and D v represent the vertical component. Then, if fjk is considered as a force triangle, D h :D v :D::fj :jk :fk SEC. IV or EQUILIBRIUM OF COPLANAR FORCES D h : D v : 100 :: 3 in. 14 in. : 5 in. D h = 60 lb., acting to the left and D v = 80 lb., acting upward. The magnitude and sense of the force 4 are unknown. For the present we shall assume (or guess) that the sense is away from the point 0, and let A h = 8a., acting to the right, A v = 150., acting upward; then and A = a\/8 2 + is 2 = 170. 60 4 8% /* 5 ? P* 2 " 5b FIGAb _ t FIG.4e Assume that the force B acts away from 0, and let B h = i2b, acting to the right, B v = 56, acting downward, and B = 6\/i2 2 + 5 2 = 136. 1 6 THEORY OF FRAMED STRUCTURES CHAP. I The forces E and C (C is assumed to act upward) and the components of the inclined forces D, A and B are shown in the sketch (Fig. 46). Three independent equations are required for a solution of the three unknown quantities C, a and b. or forces acting to the left = forces acting to the right. 60 = 64 + 80 + 126 (i) IF = o or forces acting upward = forces acting downward. C + 80 + 150 = 56 (2) 2M = o or clockwise moments = counter-clockwise moments about any point. Let us choose the point F. The arm of the horizontal forces is p = 2 in. and the arm of the vertical forces is q = 3 in. IM O O Then 2 X 64 = 128 2 X 60 = 120 2 X 80 = 160 3 X C = $C 2 X 126 = 246 3 X 80 = 240 3 X 56 = 156 3 X 150 = 128+160+ sgb = 3 6 o+ 3^ + 45<* (3) Equations (i), (2) and (3) may be reduced to 80 + 126 = 4 (10) i5<* - 5& + C = -80 (20) 290 396 + 3C = 232 (30) Multiply (20) by 3 and subtract (30) therefrom 450 156 + 3C = 240 290 - 396 + 3C = -232 i6a + 246 = 8 or 80+126 = 4 (4) Equations (10) and (4) are identical and a single solution is impossible. Equations (i), (2) and (3) represent but two inde- pendent equations and are insufficient for the solution of three unknown quantities. SEC. IV EQUILIBRIUM OF COPLANAR FORCES 17 (b) Graphic Method. The magnitudes of A, B and C are determined by closing the magnitude-direction diagram. Choosing the convenient scale-ratio of 40 Ib. to i in., we lay off oa and ab (Fig. 4^) representing respectively the forces E and D in magnitude, direction and sense. The diagram is to be closed from b to o by three lines which are drawn parallel respectively to the directions of A , B and C. Through the point b draw a line parallel to the direction of either A , B or C (let us say C) ; and through the point o draw a line parallel to the direction of one of the two remaining forces (B, for example). The diagram may be closed by drawing any line cd parallel to the direction of the remaining force A ; and we conclude, as in the algebraic method, that a single solution is impossible. 13. General Case. By giving opposite signs to magnitudes of opposite sense, the H- and V- equations may be expressed as follows: = 8 <--&-*> 71 > "STT FIG. 17. SEC. VI EQUILIBRIUM OF COPLANAR FORCES 33 = +F + G - E = -(a + h)A + (b + ti)B - (c + k)C + (d + A) -(/ + )F - (g + fc)G + (e + k)E -aA +bB -cC +dD -fF -gG +eE r P - ZM = h(-A + B - C + Z>) + (-F - G + -E) = (A X ZF) - (& X ZF) If ZF = o (i) and ZF = o (2) then ZAfp ZJf = o or ZM P = ZM for any value oi h or k. Hence, if the vertical magnitudes are balanced and the hori- zontal magnitudes are balanced, the algebraic sum of the moments of the forces is the same for all points chosen as the center of moments; in which case if J.M = o (3) for any point, the moments are balanced about all points, equilibrium is assured; and there will be one and only one independent Af-equation as illustrated in Article 24. 26. Four Groups. The three statements or Eqs. (i), (2) and (3) are the necessary and sufficient conditions for insuring the equilibrium of a body, when acted upon by any system of coplanar non-concurrent forces; and any fourth equation will be dependent on, and may be derived from them. Situations may arise where it will be convenient to express the conditions of equilibrium in other forms. The four ways in which the three equations may be stated are as follows: GROUP i GROUP 2 GROUP 3 GROUP 4 ZF = o 1.Mp o ZA/p = o J.Mp = o ZF = o ZM = o ZM = o ZM = o ZM = o ZF = o ZF = o ZM = o in which O, P and Q represent different points in the plane of the forces. The fact that certain restrictions must be placed upon the relative positions of these points in the last three groups, in 34 THEORY OF FRAMED STRUCTURES CHAP. I Z.Mp = 12 + 63 6A loB $a = o Z.M = 63 48 ioB 50 + 24 = o order that the three equations may be independent, is illus- trated by the following examples. Example i . Find the magnitudes A and B and the distance a (Fig. 1 8) for equilibrium, using the equations of group 2, and choosing any two points P and in a vertical line. (2) (3) The three equations are not independent, since the third fy # may be derived by subtract- ing the second from the first. Example 2. Use the equa- tions of group 2 and see if a solution is possible by choos- ing any other two points, andP: ' ^ - -> 4' FIG. 1 8. (a) In a vertical line. (b) Not in a vertical line. Example 3. Use the equations of group 3 and see if a solution 10' f " 5' --4- s^ 8' FIG. 19. is possible by choosing any two points, O and P : (a) In a horizontal line. (b) Not in a horizontal line. Example 4. Find the magnitudes A and B and the distance a SEC. VI EQUILIBRIUM OF COPLANAR FORCES 35 (Fig. 19) for equilibrium using the equations of group 4 and choosing three points, 0, P and Q in a straight line. = 80 100 + 5^4 80 + aB = o (i) = 20 (16 a)B + 384 120 176 nA = o (2) = 1 80 - 150 + 88 + 13^ - 130 + (8 + a)B -192 = o (3) Multiply Eq. (3) by 2, add Eq. (2), and divide the sum by 3. The quotient is Eq. (i). Only two independent equations are represented, since any one. of the three may be derived from the other two, and a solution of the problem as stated is impossible. Example 5. Choose three points O, P and Q not in a straight line and see if a solution is possible. It has been shown that under certain conditions the three equations in each of the last three groups represent but two independent equations, and are insufficient for the solution of three unknown quantities. The general cases will now be considered. 27. Group 2. Suppose that in Fig. 17 = o (i) = o (2) and I# = o (3) then ZMp - ZM = (h X IF) - (k X Z#) = o (4) If the two points P and are in a vertical line h = o hence h X IF = o for any value of Z V, in which case, (a) from Eq. (4) k X I# = o and, since k is not zero, Z.H = o and it is obvious that Eq. (3) may be derived from Eq. (4) which was derived from Eqs. (i) and (2) and is dependent upon them. Hence, when the two points P and are in a vertical line, Eqs. (i), (2) and (3) repre- sent but two independent equations (see Examples i and 20). (b) the value of Z V may or may not equal zero and equi- librium is not assured. 36 THEORY OF FRAMED STRUCTURES CHAP. I But if the two points P and are not in a vertical line, h in Eq. (4) does not equal zero, in which case to k X ZH in Eq. (4) may or may not equal zero; hence I# may or may not equal zero and Eq. (3) is independent of Eqs. (i) and (2) (see Example 26). (d) since from Eq. (3) I# = o then from Eq. (4) h X ZF =o and since h does not equal zero then IF = o (5) and equilibrium is assured. Equation (5) does not represent an independent equation since it is derived from (i), (2) and (3). 28. Group 3. Similarly it may be demonstrated that Eq. (3) is dependent upon Eqs. (i), (2) and (5) when the two points P and are not in a horizontal line (see Examples 30 and 36). 29. Hence in any case where the moments of all the forces in a non-concurrent non-parallel system are balanced about any two points in a diagonal line, if the horizontal magnitudes are balanced the vertical magnitudes are also balanced; or if the vertical magnitudes are balanced, the horizontal magnitudes are also balanced. Only the H- or the F-equation can be independent of the two moment equations. 30. Group 4. Suppose that in Fig. 17 Z.M P = o (i) ZM = o (2) and 2M Q = o (6) then ZM P - IM = (h X IF) - (k X Z#) = o (7) and ZM - IM = (s X IF) - (p X I#) = o (8) If the three points O, P and Q are in a straight line h _ k s == p and Eqs. (7) and (8) are identical and not independent (see Example 4). Eliminate IF in (7) and (8) k X Iff p X SEC. VI EQUILIBRIUM OF COPLANAR FORCES 37 or h X 2.H _ k X 2.H , , If the three points O, P and () are not in a straight line, then h , ,k - does not equal s p in which case Eqs. (7) and (8) are independent; and 1H in Eq. (9) must equal zero (see Example 5). But if then from Eq. (7) or (8) = o = o 31. Hence in any case where the moments of all the forces in a non-concurrent non-parallel system are balanced about any three points not in a straight line, the horizontal magni- tudes are balanced, the vertical magnitudes are balanced, and the H- and F-equations will be dependent upon the three If -equations. 32. One Unknown Location. In Fig. 20 the unknown loca- b A-*- ->K fc* -> 4' t s' FIG. 20. tions of two forces are represented by the quantities b and c. Let the horizontal magnitude A represent a third unknown element. All the vertical magnitudes are known and balanced. The magnitude A, determined by balancing the horizontal magnitudes, equals 6 lb., and acts to the right. Since the magnitudes are now balanced horizontally and vertically, it is possible to have only one independent Af-equation for the solution of b and c, and a single solution is impossible. 38 THEORY OF FRAMED STRUCTURES CHAP. I 33. Show that a similar condition is encountered when the unknown elements are two locations and one direction. There can be only one unknown location if a single solution is possible. 34. Combinations. The unknown elements which can be wholly or partially determined in a coplanar non-concurrent non-parallel system of forces will appear in one of the following combinations; where P, Q and S represent any forces which have unknown elements 1. The magnitude of P, the magnitude of Q and the magnitude of S. 2. The direction of P, the direction of Q and the direction of S. 3. The magnitude of P, the magnitude of Q and the direction of S. 4. The magnitude of P, the magnitude of Q and the direction of Q. 5. The magnitude of P, the magnitude of Q and the location of S. 6. The magnitude of P, the magnitude of Q and the location of Q. 7. The direction of P, the direction of Q and the magnitude of S. 8. The direction of P, the direction of. Q and the magnitude of Q. 9. The direction of P, the direction of Q and the location of S. 10. The direction of P, the direction of Q and the location of Q. 11. The magnitude of P, the direction of Q, and the location of 5. 12. The magnitude of P, the direction of P and the location of P. 13. The magnitude of P, the direction of P and the location of Q. 14. The magnitude of P, the direction of Q and the location of P. 15. The magnitude of P, the direction of Q and the location of Q. Cases (i), (4) and (12) are more frequently encountered in the theory of framed structures than any or all of the others. Several cases are indeterminate under certain conditions. Case (i) is in this class when P, Q and S have the same location, or are parallel. Other cases are partially indeterminate, for they may have several solutions or none. Any problem which involves a force having a known magnitude and an unknown direction is in this class. The most difficult problems are to be found in cases (2) and (7). 35. Illustrative Problems. i Determine the unknown ele- ments necessary for equilibrium (Fig. 21 a). (a) Algebraic 1 Method. The three elements of each of the x The term "algebraic" is used simply for the purpose of distinguishing the method of computing by the use of numerals or letters, from the graphic method SEC. VI EQUILIBRIUM OF COPLANAR FORCES 39 four forces shown are given. The elements necessary for equilibrium are the magnitude, direction and location of a fifth \5^ Ch--Z40*> ^ 2f' AS'\ 10 f ~ By* 120* l7W-2*>* D ^ioo# Cv*i$o H 'DV:24 Cvtuv FI6.2IC * * 160 FIG.2IG force. The problem is one of case (12). Make a sketch (Fig. 2ib) and transfer to it the horizontal and vertical components of the four known forces properly located. in which quantities are represented by the length, direction and location of lines. Some writers refer to the two methods as the analytic and graphic; but this is manifestly incorrect, since both methods are analytic. The frequent use how- ever, of the word "algebraic" should be discouraged. It is true that algebraic equations have been used in the discussions which have preceded, but in the actual solution of problems their use is often a hindrance rather than a help. In the first place, if the equations are not written, the problem can frequently be solved more quickly and easily; and the computations can be arranged in a more convenient form for checking. But a more important reason for discarding the equation whenever possible is the fact that by so doing, the attention is held fast to the original problem in statics, and not drawn far afield by an exercise in elementary algebra. 4O THEORY OF FRAMED STRUCTURES CHAP. I The horizontal and vertical components of A are 160 Ib. and 300 Ib. respectively, 15 = I7 an( * are S Pl acec * on tne sketch as to = 6 4 = I7 intersect at a point in the line of the x 8 = l6 force ^4. The components of the other known forces are determined and located 340 Av = 77 x I5 = 3 in a similar manner. (The sketch in Fig. 2ic is superfluous, except that it shows one of the several other forms in which the components might have been correctly placed.) The sum of the horizontal components acting to the 160 240 right is greater by 180 than the sum of the horizontal 1 60 components acting to the left; consequently the body will move to the right unless a horizontal component having a magnitude of 180 acts to the left to balance the horizontal forces. 1 80 Z.H A vertical component having a magnitude of i 240 must act downward to balance the vertical 180 3and q. The unknown quantities a, b and 5 may be determined by writing and solving three independent static equations. There is, however, a very simple and more direct method whereby any one of the unknown magnitudes may be determined independ- ently of the other two. The method of procedure is as follows: balance the moments of all the forces about the point F, thereby eliminating the two unknown magnitudes p and q and the load of 15 tons. Only two magnitudes remain the reaction of 50 tons and the magnitude s and their moments about F must balance for equilibrium; for the other magnitudes cannot help or hinder rotation about the point F. The moment of the reaction about F is 3,000 ft. -tons clockwise; and the body will rotate clockwise about F unless the magnitude s, acting through a distance of 40 ft. and to the right or away from the body, is 75 tons. At the point G, where the directions of q and s intersect, SEC. VI EQUILIBRIUM OF COPLANAR FORCES 51 resolve the magnitude q into H- and F-components 30 and 40. At the point K in a vertical line with G, resolve the magnitude p into H- and F-components 6b and b. Balance the moments of all the forces about G, thereby eliminating the unknown magnitudes s, q and b. The ^-component of p is 90 tons 90 X 50 = 4,5oo 30 X 15 = 45 45)4,050 90 = 6b b = is acting to the left. The F-component of p is 15 tons acting downward. p = \/go 2 -\- i5 2 = 91.24 tons acting toward the body. The directions of p and 5 intersect at /, 180 ft. to the left of the reaction; and by balancing the moments of all the forces about /, the magnitudes p, s and 30 are eliminated. 1 80 X 50 = 9,000 240 X 15 = 3> 6o 270)5,400 20 = 40, The F-component of q is 20 tons acting downward. The H- component of q is 15 tons acting to the right. q = Vi5 2 + 20 2 =25 tons acting away from the body. The magnitude q may also be determined by dividing the moments of all the forces about / by the arm h. The distance FG is 50 ft. 40 : 50 :: h : 270 h = 216 q = = 25 tons acting away from the body. The magnitudes 3 a and 40 could have been determined by balancing the H- and F-magnitudes ; after the magnitudes s, b and 6b had been determined. Check the solution already determined by this process. The sketch (Fig. 246) as shown is incomplete; since each numerical value, as soon as it is determined, should be sub- THEORY OF FRAMED STRUCTURES CHAP. I stituted for its symbol on the sketch and an arrow-head added to indicate the sense. It has been stated that the use of algebraic equations should be avoided whenever possible. The present case is a good illustration. The arithmetic solution is not only shorter; but FIG.ZSb also gives the student an opportunity for the exercise of his judgment in choosing the best method of attacking the problem in order that the solution may be accurately and quickly made, and his computations advantageously arranged for checking. Graphic Method. The location-directions of all the forces are known. They are drawn to scale in Fig. 240 and indicated by the symbols AB, BC, CD, DE and EA . The known magni- tude-directions ab and be are laid off to scale in Fig. 24^. Draw the location-directions OB, OC and OE. The forces acting SEC. VI EQUILIBRIUM OF COPLANAR FORCES 53 at each of the vertices /, g and h, of the location-direction dia- gram thus formed, may represent three concurrent systems in equilibrium. Close the magnitude-direction diagram for the point / by drawing the magnitude-directions co and ob, inter- secting at o. Two magnitude-directions for the point h are now known ab and bo. Close the diagram by drawing the magnitude-directions oe and ea, intersecting at e. Two magni- tude-directions for the point g are known eo and oc. Close the diagram by drawing the magnitude-directions cd and ed, intersecting at d. The magnitude-direction diagram abcdea is closed and the magnitudes cd, de and ea are determined. Compare the results with those obtained by algebraic solu- tion. 5. The body in Fig. 250 supports three loads. What are the reactions if the weight of the body is neglected? Algebraic Method. Remove the body from the supports and indicate the H- and F-components of the forces necessary for equilibrium, as shown in the sketch (Fig. 256). Balance the moments of all the forces about A . 8 X 210 = 1,680 16 X 140 = 2,240 24 X 280 = 6,720 14)10,640 760 = //-component at B, acting to the right. = H -component at A, acting to the left. 7 6 = 190 = F-component at B acting upward. 4 210 140 280 630 IQO 440 = F-component at A acting upward. 440 2 = 193,600 76o 2 = 577,600 V77i,2oo = 878.18 = magnitude of the force at A. 44 _ ^J The direction of the force at A has the slope of 1 1 ver- 760 19 tical to 19 horizontal. I9O 2 = 36,100 76o 2 = 577,600 Voi3,7oo = 783.39 = magnitude of the force at B. 54 THEORY OF FRAMED STRUCTURES CHAP. I Change the location of the components of the lower reaction so that, by balancing the moments of all the forces about A , the ^-component of the lower reaction is eliminated. Graphic Method. The location-directions of the known forces AB, BC and CD (Fig. 250) do not intersect; and the conven- tional method of constructing the location-direction diagram by drawing the location-directions of the components of the forces, somewhat after the manner illustrated in Problem 2, would ordinarily be followed. But by a simple expedient the location- direction diagram may be more quickly drawn by locating the resultants instead of the components. Draw the magnitude- direction diagram (Fig. 25*;) from a to d. Select two points f and g anywhere on the location-direction AB so that/ will represent the magnitude be. Similarly let hi, on the location- direction BCj represent the magnitude ab. The intersection of the two lines fi and gh at j marks the location of the resultant ac of the two forces ab and be. In like manner pq represents the magnitude cd\ st represents the magnitude ac; and k marks the location of the resultant ad of the forces ab, be and cd. The location-direction DE is known, and intersects the location- direction AD at m\ through which point the location-direction EA must pass if there is to be no rotation of the body. The location-directions of all the forces are known. Close the magnitude-direction diagram by drawing the magnitude- directions de and ea. Compare the results with those obtained by the algebraic solution. If the sense of ab had been upward, show that the resultant of ab and be would have scaled 70 in the magnitude-direction diagram; and its location would have been at the point of intersection of the two lines through hf and ig. 36. Problems. i. In Fig. 26 the moments of all the forces are balanced about the three points O, P and Q; i.e. o ZJIf o = o yet the body is not in equilibrium, for the horizontal magnitudes do not balance, neither do the vertical magnitudes balance. Explain. SEC. VI EQUILIBRIUM OF COPLANAR FORCES 55 2. The homogeneous body of uniform thickness in Fig. 27 weighs 50 Ib. and is supported at A, B and C. The reaction at A is 25 Ib. The reactions at B and C are vertical and horizontal respectively, as indicated by the roller supports. Determine the unknown elements for equilibrium. 3. The body in Fig. 28 is supported at A, B and C. Find the weight of the body, the magnitude of the force at A } and locate the vertical gravity line of the body. THEORY OF FRAMED STRUCTURES CHAP. I 4. The cross-section of a retaining wall weighing 6,000 Ib. per linear foot is; represented in Fig. 29. The resultant earth pressures per linear foot are 15,000 Ib. at A and 7,500 Ib. at B. What is the direction of the resultant pressure at A, if the resultant reaction of the ground on the base CD is 25,500 Ib. per linear foot; and what is the total friction between the ground and wall on the line CD? At what point on the line CD does the resultant reaction act? 5. Neglect the weight of the body in Fig. 30 and determine the reactions. 6. The body in Fig. 31 weighs 2,000 Ib. Locate the vertical gravity line; find the magnitude of the force at A; and the direction of the force of 1,300 Ib. atB. 7. Find the weight of the body in Fig. 32; the magnitude of the force at A; and locate the force of 1,000 Ib. for equilibrium. 8. How far is the ball (Fig. 33) from the point A when the body is in equilib- rium; what is the magnitude and direction of the force at A? Neglect the weight of the body. 9. The body in Fig. 34 weighs 1,000 Ib. and resists two forces at C as indi- cated. The reactions at A and B are 820 Ib. and 1,130 Ib. respectively. In what direction does each reaction act and where is the vertical gravity line of the body? SEC. VI EQUILIBRIUM OF COPLANAR FORCES 57 FIG. 33- ^ 22'- 45' . 40* 60 95* FIG. 35- THEORY OF FRAMED STRUCTURES CHAP. I 10. Neglect the weight of the body (Fig. 35) and find the magnitudes of the forces p, q and s. 400* 16' FIG. 36. FIG. 37. ii. Find the weight and vertical gravity line of the body (Fig. 36) if the magnitude of the force at A is 781 Ib. What is the direction of the force at A ? 12. Neglect the weight of the body (Fig. 37) and find the magni- tude of A, and the location and direction of a force of 1,220 Ib. for equilibrium. 13. A level beam of uniform cross- section 10 ft. long, weighing 250 Ib. and supported at each end, carries a single load of 5 tons at mid-span. Why is the problem of finding the two reactions indeterminate? An algebraic solution for each of the two following problems is long FIG. 38. and involved. graphic solution is comparatively short in each case. Show that the FIG. 39. SEC. VI EQUILIBRIUM OF COPLANAR FORCES 59 14. The magnitudes of the forces at A and B (Fig. 38) are 13 Ib. and 17 Ib. respectively. Neglect the weight of the body and find the directions of the forces at A and B, and the magnitude of the force C. 15. The magnitudes of the forces at A, B and C (Fig. 39) are 41 Ib., 50 Ib. and 25 Ib. respectively. Neglect the weight of the body and find the three unknown directions. CHAPTER II APPLICATION OF THE PRINCIPLES OF EQUILIBRIUM SEC. I. SIMPLE TRUSSES 37. Rigid Body. In the foregoing section reference is frequently made to a "rigid body." No such thing exists in nature, if "rigid" is used in the sense of "unyielding." All solids may be considered to approximate in a greater or less degree an ideal state of complete rigidity. A steel casting, a granite rock, an oak block, putty and rubber all illustrate different degrees of rigidity, for all change their shape under different degrees of pressure. In engineering parlance a rigid body is one which, in form and content, offers a sufficient resistance to distortion to give service as an engineering struc- ture. Or, in other words, if the body is a solid, as an oak beam or steel-plate girder, consideration is given to the quality and quantity of the material in order that a sufficient degree of rigidity may be realized; but if the body is a framed structure, attention must also be given to the manner in which the structure is framed in order that it may be stable, or not be susceptible to an immediate and complete collapse. The frame represented in Fig. 2ih is unstable or collapsible; for it offers no resistance to a change in its shape, except possibly a small amount of friction at the joints. The shape of the frame may be changed at will without affecting a change in the length of any member, and the same is true of any similar frame having four or more members. A triangular frame differs from all others in this respect, since a change in its shape necessitates a change in the length of at least one of its three members. Hence a frame, which in outline presents a collection of triangles, is stable; and if it has adequate material in its members, it possesses a sufficient degree of rigidity for engineering purposes. Hence: 60 SEC. I APPLICATION OF PRINCIPLES OF EQUILIBRIUM 6 1 38. A Truss is a frame or jointed structure composed of members (usually straight) so connected as to form a succession of triangles making one rigid or stable structure. Acting as a whole, this performs the functions of a beam in resisting dis- tortion 1 caused by shearing forces and bending moment; while the individual members are designed (except in special cases) to perform the functions of a tie or strut in resisting changes in their length caused by tensile or compressive forces. 39. Stability. The simplest truss that can be constructed has three members m, and three joints j. A more elaborate truss comprises several triangular frames, so combined that each additional triangle adds one joint and two members. Hence, the number of members necessary to insure stability under any arrangement of loading is m = y - 3 (i) A truss having fewer members than are required by Eq. (i) is in a state of unstable equilibrium, and will collapse except under special conditions of loading. A truss having more members than are indicated by Eq. (i) is a redundant structure. Such structures, if the members are properly placed, will support loads of any arrangement. In some instances this is a distinct advantage. Redundant structures cannot be analyzed by the principles of statics alone, and are said to be " statically indeterminate." By giving due consideration to the elastic properties of the members; or by the aid of certain reasonable assumptions which have the sanction of good engineering practice, a satisfactory analysis may be made. 40. The assumption that the members of a truss are subject to tensile and compressive forces only, presupposes four conditions : ] It should be clearly understood and constantly kept in mind that structures are not rigid in the sense of unyielding, but are elastic in the same sense that a rubber eraser or a rubber band is elastic, only in a lesser degree. The members of a steel truss, designed in accordance with current specifications, will lengthen or shorten from 3^2 to ^ e of an inch for every 10 ft. of length when the structure is supporting the full load for which it was designed. This change in length inevitably changes the shape of the triangular units of the truss and accounts for the distortion of the whole structure. The distortion is comparatively so slight, however, that for our present consideration, the changes in the dimensions of the truss may be considered negligible without appreciable error. 62 THEORY OF FRAMED STRUCTURES CHAP. II 1. That the axial gravity lines of the members meeting at a joint, intersect at one point. 2. That the loads are applied only at the joints. 3. That the members are subject to neither shear nor bending moment due to their own weight. 4. That the joints are frictionless hinges. The first and second conditions are generally incorporated in a good design. The third condition is true of vertical mem- bers only. The nearest approximation to the fourth condition is found in a pin-connected truss. 41. The application of the principles of equilibrium to the analysis of a framed structure may be considered as two distinct operations: (a) the determination of the external forces; and (b) the determination of the internal forces. 42. The external forces acting on a structure are: (a) the loads which the structure supports; (b) the weight of the struc- ture itself; and (c) the forces acting at the points of support, commonly called the reactions. The loads are usually known or easily determined, but the weight of the structure not as yet designed must be assumed. The reactions are determined by a solution of a system of non- concurrent forces, as set forth in Sections V and VI of Chapter I. The external forces are generally analyzed by the algebraic method. This process is usually shorter than the graphic method, especially if the external forces are parallel. 43. The internal forces acting in the structure are many and of various kinds. We shall consider here only those of primary importance the tensile and compressive forces acting along the members. These forces are known as stresses. As the reactions are determined by the application of the principles of the equilibrium of coplanar forces to the structure as a whole, so the stresses in the members are determined by the same principles; but applied to isolated parts of the structure. The portion to be considered may be a single joint, or it may include several joints and members. The forces to be con- sidered may represent reactions, loads or stresses. There are two principal methods of procedure in the determination of stresses in framed structures: SEC. I APPLICATION OF PRINCIPLES OF EQUILIBRIUM 63 1. The method of joints (a) algebraic (b) graphic. 2. The method of sections (a) algebraic (b) graphic. 44. The Method of Joints. The truss in Fig. 400 supports five loads as shown. What is the stress in each member if the reactions are vertical? The determination of the stresses by the method of joints, either algebraically or graphically, is simply a solution of as many systems of concurrent forces as there are joints in the truss. After the external forces have been found, the solution begins at a joint where the stresses of two members concur with one or more external forces; proceeds to an adjacent joint which presents but two unknown stresses; and so on throughout the entire structure. The force exerted on a joint by one end of a member is equal in magnitude but opposite in sense to the force exerted by the member on the joint at its other end. The kind of stress in a member is indicated by the sense of the force which a member exerts on a joint; i.e., the stress is tensile if the sense is away from the joint and compressive if toward the joint. Graphic Solution. The first step in the solution of any prob- lem of this character is the determination of the forces necessary to place the truss in equilibrium, i.e., to find the reactions. The left and right reactions, determined algebraically, are 50 Ib. and 55 Ib. respectively. It is a waste of time to find the reactions graphically when the external forces are parallel. A capital letter is placed in the space between each external force and in each triangle of the frame. The outline of the truss represents essentially a location-direction diagram, and we proceed to draw a closed magnitude-direction diagram for each of the six concurrent systems one for each joint. Since the end joints present but two unknowns, either may be chosen for the draw- ing of the first diagram. The conventional method is to begin at the left end of a structure, proceeding in a clockwise rotation about each joint. All location-directions are known. The 6 4 THEORY OF FRAMED STRUCTURES CHAP. II problem is to determine unknown magnitudes or stresses acting in the various members. Beginning at the left-end joint, proceeding in a clockwise rota- tion, and including first the location-directions having known K50 +9$ (a) Scale fain -5 ft f = Tension a - = Compression (9) FIG. 40. magnitudes, we read GA, AB, BE and EG. The magnitude- directions ga, abj bh and hg are laid off in the same order in Fig. 406. The magnitude-directions ga and ab are known, and the diagram is closed by drawing bh and hg. The sense bh is toward the joint; the sense hg is away from it. The magnitude-direc- tion bh scales 104 lb.; the magnitude-direction hg scales 96 lb.; SEC. I APPLICATION OF PRINCIPLES OF EQUILIBRIUM 65 hence the stresses in BH and EG are 104 Ib. compress! ve, and 96 Ib. tensile respectively. These numerical values, with minus and plus signs to signify compression and tension, are indicated on the members of the truss. The adjacent upper joint now presents only two unknown stresses. The known magnitude-directions hb and be are laid off (Fig. 4oc) and the diagram closed by drawing cj and jh\ giving 78 Ib. and 26 Ib. for the stresses in the members CJ and JH. At the lower middle joint the stresses in three members JK, KL and LG are unknown and a solution is not yet possible; but at the upper middle joint the stresses in only two members are unknown, and the solution is given in Fig. 40^. There are but two unknowns at each of the three remaining joints. Fig. 406 represents the diagram for the lower middle joint, and Fig. 4o/, the joint supporting the load of 40 Ib. The stresses in all the members have been determined, and all the forces concurring at the right-end joint are known ; but the diagram (Fig. 40^) is drawn for this joint as a check on the foregoing solutions. Any two adjacent joints have one member in common, con- sequently their respective magnitude-direction diagrams also have one side in common; hence, any one of the diagrams may be added to the one preceding or following it. Thus one figure may be developed which will contain all the magnitude-direc- tion diagrams representing graphically the magnitude, direction and sense of each external force and internal stress of the struc- ture. Such a figure is called a stress diagram and is illustrated in Fig. 4oh. The magnitude-direction diagram of the external forces or load line ab, be, cd, de, ef, fg and ga is drawn first and the lines representing the stresses follow in regular order. Be- ginning at the left-end joint, we read ga, ab and draw bh and kg parallel respectively to BH and EG. For the next joint we read hb, be and draw cj and jh. For the top joint we read jc, cd and draw dk and kj. For the bottom joint we read gh, hj and jk, and draw kl and Ig. For the next joint we read Ik, kd and de and draw el parallel to EL. If this line, drawn from e parallel to EL passes through the point /, we have a check 66 THEORY OF FRAMED STRUCTURES CHAP. II on our work, the stress diagram closes and the solution is complete. When the external forces are parallel, as in the present case, the magnitude-direction diagram for the external forces is a straight line. The stress diagram is essentially a magnitude- direction diagram for the external forces and a magnitude- direction diagram for the concurrent system at each joint. ' Algebraic Solution. The concurrent systems are sketched in Fig. 41. The F-component of a (Fig. a) is 40 Ib. acting down- 30 FIG. 41. I 2 ward. The //-component is - - X 40 = 96 Ib. acting to the o left; hence, b = 96 Ib. acting to the right or away from the joint; a = X 40 = 104 Ib. acting toward the joint. Assume //- and F-components for c and d as shown in Fig. b. 5^+20 = 5;y + 40 120; + 123; = 96 x = 6 y = 2 ^ //-component of c = 72 Ib. acting to the left. F-component of c = 30 Ib. acting downward. c = 78 Ib. acting toward the joint, //-component of d = 24 Ib. acting to the left. F-component of d = 10 Ib. acting upward. d 26 Ib. acting toward the joint. The upper middle joint is sketched in Fig. c. The //-com- ponent of c = 72 Ib. acting to the right, consequently the H SEC. I APPLICATION OF PRINCIPLES OF EQUILIBRIUM 67 component of/ = 72 Ib. acting to the left. Hence, the F-com- ponent of/ = 30 Ib. acting upward;/ = 78 Ib. acting toward the joint; and e = 30 Ib. acting downward or away from the joint, to balance the load of 30 Ib. and the F-components of c and/. At the lower middle joint (Fig. d) the F-component of g = 20 Ib., acting downwardt o balance e and the F-component of d; therefore the //-component of g = 48 Ib. acting to the left, and g = 52 Ib. acting toward the joint. Hence, k = 120 Ib., acting away from the joint. In Fig. e there is but one unknown force/. The //-compo- nent = 120 Ib. acting to the left, the F-component = 50 Ib. acting upward and j = 130 Ib. acting toward the joint. We now have a check upon our computations by noting that the F-component of j equals the right reaction, minus the load of 5 Ib.; the //-component equals &; and the ratio of the F- and CQ C //-components of j is - !L - or > which is the slope or bevel of 120 12 the member. Assuming that the stresses in /?// and CJ are unknown, solve algebraically and graphically for the stress in ///, using the method of joints. Remove the loads of 20 Ib. and 40 Ib., and determine the stresses in the three web members ///, JK and KL. The graphic solution is generally employed when the stresses are determined by the method of joints. This is especially true in the case of roof trusses having inclined chords. There is perhaps little preference between the algebraic and the graphic solutions, if the top and bottom chords are horizontal and half the web members are vertical. There are several factors which conspire in making the method of joints peculiarly efficient in truss analysis for stationary loads. 1. The stress in any member is constant for a given load. 2. The required stresses in all members occur simultaneously. 3. The external forces and required stresses concurring at a joint are in equilibrium. The stress in any member of a truss supporting moving loads varies as the loads move upon the structure. The stress in any member is desired only for that particular position of the loads 68 THEORY OF FRAMED STRUCTURES CHAP. II which will cause the maximum stress in that member. Some members receive their maximum stress when the moving loads cover the entire structure, while in other members the stress is greatest when only a portion of the structure is covered. The maximum stresses do not occur simultaneously. This fact greatly limits the usefulness of the stress diagram in the case of moving loads. For example, suppose that four members A, B, C and D meet at a joint; and that the moving loads are in the position for maximum stress in B. A stress diagram will give the stresses not only in ; but in A, C, D and all the other members of the structure as well. The stress in B, however, is the only stress of any value which may be obtained from that particular stress diagram; since it is the only member in which the stress is a maximum. 45. The method of sections gives excellent service when the stress in but one member is desired for any particular arrange- ment of loads. The algebraic solution is generally preferred. Suppose we wish to determine the stress in the member GH (Fig. 420). If the member were cut or removed from the truss, it is perfectly obvious that the part ACG would rotate clock- wise, and the part CMH would rotate counter-clockwise each about the pivot C; the joints G and H moving further apart. Hence in preserving the equilibrium of the structure, it is plainly the duty of the member GH to hold the joint G from moving to the left, and the joint H from moving to the right. In performing this function the member exerts a force to the right on the joint G, and an equal and opposite force to the left on the joint H. Imagine that a plane of section XY cuts the three members CD, CH and GH, dividing the truss into two portions. If p, q and s represent the stresses in the members before cutting, then the known external forces (loads and reactions) acting upon each portion are balanced by the forces p, q and s as illustrated in Figs. 426 and c. The left portion (Fig. 426) represents a coplanar system of five non-concurrent forces having three unknown magnitudes; corresponding to the first combination listed in Article 34. An algebraic and graphic solution for this system is given in Article 35, Problem 4. SEC. I APPLICATION OF PRINCIPLES OF EQUILIBRIUM 6 9 The algebraic solution when performed by writing and solving three simultaneous equations is known as Rankine's 1 "Method of Sections." The corresponding graphic solution is the work of Culmann. 2 The algebraic solution, whereby any one of the three unknown magnitudes is determined by balancing the moments of all the forces about the intersection of the other two forces having unknown magnitudes, is known as Ritter's 3 " Methods of Moments." In order to determine the stress by Ritter's method, we balance the moments of all the external forces about C (Figs. 426 or c) thereby eliminating the two unknown magnitudes p and q. I D V V y FIG.42C Moments about C FIG. 426 60 X 50 = 3,000 20 40 30 10 Fl 6.43d 30 X 8 = 60 X 20 = 90 X 40 = 120 X 30 = 150 X 10 = FIG. 420 240 1 80 X 73 1,200 3,600 3,600 1,500 13,140 10,140 3,ooo 10,140 1 Applied Mechanics. 2 Die Graphische Statik. 1886. 3 Dach- und Brucken constructionen. 1862. 70 THEORY OF FRAMED STRUCTURES CHAP. II The sum of the moments about C of the known forces acting on the left-hand portion is 3,000 ft. -tons clockwise; hence, the force 5 acting at a distance of 40 ft. from C must have a magni- tude of 75 tons toward the right or away from the joint G to balance the moments. The sum of the moments about C of the known forces acting on the right-hand portion is also 3,000 ft.-tons, but counter-clockwise; hence, the force s acting at a distance of 40 ft. from C must have a magnitude of 75 tons toward the left, or away from the joint H to balance the moments. The fact that the sum of the moments about C of the known external forces is the same (3,000 ft.-tons) for either portion, except that the rotation is clockwise in one case and counter- clockwise in the other, is easily explained. The truss (Fig. 42 a) is in equilibrium and the sum of the moments of the eight known external forces about any point C, for example, equals zero. If these eight forces are divided in any manner into two groups, the sum of the moments of one group about C must balance the sum of the moments of the other group. Determine p and q (Fig. 420) by balancing the moments of the forces about the points H and / respectively, and note that the results check with the computations of Problem 4, Article 35- Since the unknown magnitudes may be determined by balancing the moments of the forces on either side of the section, it is expedient to consider the side which has the fewer forces. The stress in DJ (Fig. 420) may be determined by passing a section through DE, DJ and HJ. The two chord members are parallel and there is no point of their intersection about which the moments may be balanced; but since the chords are hori- zontal, the vertical component of the stress in DJ must balance the vertical magnitudes on either side of the section. The resultant of the vertical forces on the left of the section is 50 (15 + 8) = 27 tons acting upward. The left-hand portion will move upward, unless the vertical component of DJ is 27 tons acting downward from D or away from the joint. Therefore the member is in tension and the vertical component of the stress is 27 tons. SEC. I APPLICATION OF PRINCIPLES OF EQUILIBRIUM 7 1 Compute the stresses in DE and HJ and see if these stresses balance with the horizontal component of DJ. Compute the stresses in BCj BG, CG and DH and then draw a stress diagram to check the computations. It is worthy of note that a solution by the method of sections is entirely independent of the number, inclination or arrange- ment of any members of the structure other than those cut by the section. 46. Problems. i. The truss in Fig. 43 is supported by vertical reactions at each end. Deter- 400 500 6 Equal Spaces * FIG. 43. 140 280 FIG. 44. mine the reactions algebraically, construct the magnitude-direction diagram for the external forces (sometimes called the load line) and draw a stress diagram. Mark the stresses with proper signs upon the members. Note that h and i fall at the same point in the stress diagram, indicating that the member HI has no stress and is superfluous for this particular loading. Note also that the stresses in QF and GP are equal, and that the stress in PQ equals the external load at the joint. In consideration of these observations develop a general rule by THEORY OF FRAMED STRUCTURES CHAP. II which superfluous members may be eliminated before the stress diagram is begun. 2. The frame in Fig. 44 has been substituted for the solid body of Fig. 2$a. Draw a stress diagram. Note that the magnitude-direction diagram for the external forces, which constitutes the beginning or foundation of the stress diagram, has already been constructed in Fig. 25^. 3. Substitute a frame for the solid body in Fig. 30 and draw a stress diagram. 4. The truss in Fig. 45 is supported by vertical reactions at A and B. Draw a stress diagram. Compute the stresses in X , Y and Z. 2QO 400 1000 FIG. 45. 3000 F1G.49 5. Compute the H- and F-components of the stresses in the five members meeting at A (Fig. 46). Mark the values on a sketch and indicate whether the members are in tension or compression. 6. The truss in Fig. 47 is supported at A and B. The roller at B indicates a vertical reaction. Draw a stress diagram. 7. The truss in Fig. 48 is supported by a chain at D. Suspend weights of 150 Ib. at A, 50 Ib. at B and 50 Ib. at C, and draw a stress diagram. 8. The truss in Fig. 49 is supported by chains AC and BD. Draw a stress diagram for vertical loads of 300 Ib. at E\ 600 Ib. at F; and a horizontal load of 450 Ib. at G, acting to the right. SEC. II APPLICATION OF PRINCIPLES OF EQUILIBRIUM 73 SEC. II. STRUCTURES REQUIRING SPECIAL CONSIDERATION 47. The application of the principles of static equilibrium to the framed structures, discussed in the preceding section, presented no real difficulties. There are types of structures, however, for which the solutions are not so simple; and a special investigation is necessary in order that the difficulties encountered may be overcome. These difficulties are caused by indeterminate reactions, indeterminate stresses or both. Statically indeterminate structures as such will not be treated here. There are, however, several indeterminate types which may be treated by static methods; when certain reasonable assumptions which have the sanction of good engineering practice are made. There are also several types which are indeterminate in appearance but not in fact. 48. The Fink Truss. The structure outlined in Fig. 500, and known in this country as a Fink truss, is a good example of a case in which the stresses are apparently statically indeter- minate. A difficulty is encountered after the stress diagram (Fig. 506) has been drawn for joints i, 2 and 3; for the stresses in three members at each of the joints 4 and 5 are unknown. In this emergency, one of two graphic solutions (a) or (b) may be used. (a) Temporarily remove the members KL and LM , substitute the dotted member OM connecting the joints 5 and 7, and draw the polygon jhdeof for the joint 4. Then the polygon oefmo for joint 7 may be drawn, giving the stress in the member FM ; which is in no way influenced by the arrangement of members in the quadrilateral 4-5-6-7. Remove the dotted member and replace the members KL and LM. The point m is now located in the stress diagram; the stress in FM is known; and the remaining stresses are easily determined by taking the joints in the order 7-4-5-6. (b) The following solution is based upon the exception to the general law of concurrent forces laid down in Article 17. The point j was located in the stress diagram when joint 3 was solved. The point k will fall somewhere on the line through j parallel to KJ. The stress in ML can be determined since the 74 THEORY OF FRAMED STRUCTURES CHAP. II members LE and FM are in the same straight line. Assume any value for the stress in the member LE, as l'e\ and draw the stress polygon I'efm'l', giving m'l' for the stress in ML. Since there remain but three members having unknown stresses at joint 6, and two of the members are in the same straight line, the stress polygon I'm'n'k'l' can be drawn and the stress in KL determined. Any other value I'e might have been assumed for the stress in the member LE\ and for each different point /' on the line FIG.SOf through a parallel to LE, there is a corresponding point k f on a line parallel to LE. Hence the point k is located at the inter- section of the line through j parallel to KJ, with the line through k f parallel to LE. The stresses can now be solved for either joint 4 or 5 and the diagram completed without any further difficulty. Complete the stress diagram for solutions (a) and (b). When computing the stresses by the method of sections, we meet with a difficulty somewhat similar to the one encountered in drawing the stress diagram. While it is impossible to divide the truss by a section cutting any one of the members JK, KN, KL, LE or ML without cutting more than three members; yet the stresses in these members are easily determined when SEC. II APPLICATION OF PRINCIPLES OF EQUILIBRIUM 75 taken in the proper order. The stress in the member ML may be determined by a section cutting out the portion shown in Fig. 5oc. All of the five members which are cut, except the member ML, intersect at the peak; about which point the moments of the load at joint 7 and the stress in the member ML must balance. The stress in the member NA may be determined by balancing the moments of all the forces about the joint 8 (Fig. 50^) . After the stress in the member NA has been determined and indicated in Fig. 500, the stress in the member KL may be found by balancing the moments of all the forces about the joint 8. When the stress in the member KL has been indicated in Fig. 5o/, the stress in the member JK may be determined by balancing the moments of all the forces about the joint i. After the stress in the member KL has been indicated in Fig. 500, the stress in LE may be determined by balancing the moments of all the forces about the joint 5. Compute the stresses in the members mentioned above and see if they check with the stress diagrams. Formulate a general rule for applying the method of sections which will cover all cases which have heretofore been considered. 49. The three-hinged arch has three pin-connected joints- one at each of its two points of support and a third usually located at mid-span (Fig. 510). The structure, being free to turn at points A, B and C, will obviously collapse even under vertical loads unless the sustaining forces at A and B have horizontal as well as vertical components. Hence four unknown elements for equilibrium must be determined the magnitude and direction of each reaction. Since only three independent equations can be written for a non-concurrent system of forces acting upon a rigid body, the problem of finding four unknown elements is apparently impossible. Such, however, is not the case. The problem is rendered statically determinate by the condition that the structure is composed of two rigid bodies, X and F, each free to turn about two points. The horizontal magnitudes of the external forces acting on the structure as a whole may balance, the corresponding vertical magnitudes may balance, and the moments about C of the external forces acting on X may balance; but these three condi- 7 6 THEORY OF FRAMED STRUCTURES CHAP. II tions do not certify that the moments about C of the external forces acting on Y are balanced. This condition provides the possibility of making four independent statements or equations concerning the equilibrium of the structure. Let us consider the case of a vertical load P 1,000 Ib. at D. Algebraic Solution. Assume V\ and H\ for the components of the reaction at A ; and F 2 and H 2 for the components of the reaction at B. Balance the moments of all the external forces acting on the structure about A. F 2 = Similarly 1,000 X 35 P^IOOO FIG. 51 c The moment of F 2 about C is 25 X 300 = 7,500 ft.-lb. and the body F will rotate counter-clockwise about C, unless the moment of F 2 is balanced by the moment of HZ] hence H 2 = 375 Ib., acting to the left. 20 The algebraic sum of the moments of P and FI about C is 25 X 700 = 17,500 10 X i ,000 = 10,000 7,500 ft.-lb., clockwise. The body X will rotate clockwise about C unless the moments of P and FI are balanced by the moment of HI] hence 7,500 20 = 375 Ib., acting to the right. SEC. II APPLICATION OF PRINCIPLES OF EQUILIBRIUM 77 The magnitude of HI could have been determined by balanc- ing the horizontal magnitudes after the magnitude of HZ has been found. The magnitude and direction of each reaction is readily determined from the magnitudes of its components. Graphic Solution. There are three external forces P, RI and R 2 (Fig. 5iZ>) acting upon the structure. The reaction R 2 obviously acts through C, for otherwise the body Y would rotate about C. This condition determines the location-direction of R 2 which intersects the force P at 0; through which point the location-direction of RI must pass if there is to be no rotation of the structure about the point O. The magnitude-direction diagram for the concurrent system at is drawn in Fig. 516, giving the magnitudes of RI and RI. In the case where both portions X and Y are supporting one or more loads, the reactions for each load may be determined separately after the manner just described. The reactions at each support thus found may then be combined and the resultant reaction for all loads determined. A more compli- cated but somewhat shorter solution is given in the next article. 50. Illustrative Problem. The three-hinged arch (Fig. 520) supports eight loads. Determine the reactions. Graphic Solution. Lay off the magnitude-directions of the eight loads from b to k (Fig. 526). Choose any convenient pole o and imagine that the nine magnitude-directions of the components ob to ok are drawn. Choose any convenient point n on the location-direction BC, and draw the location-directions of the components OB to OK. Draw the magnitude-direction bf; and through the left and center hinges draw lines parallel to bf, intersecting the location- directions OB and OF. Connect these intersections by the location-direction OL, which closes the location-direction diagram for the four forces B to F. Draw the magnitude- direction ol, which determines the magnitudes of two loads bl and //; which, if applied at the left and center hinges respec- tively, will have the same effect upon the equilibrium of the body X as the four loads B to F. The magnitude-direction of the resultant of the four loads THEORY OF FRAMED STRUCTURES CHAP. II F to K is fk. Through the center and right hinges draw lines parallel to fk intersecting the location-directions OF and OK. Connect these intersections with the location-direction OM, closing the location-direction diagram for the four forces F to K. Draw the magnitude-direction om, which determines the magnitudes of two forces fm and mk; which, if applied at the center and right hinges respectively, will have the same effect upon the equilibrium of the body Y as the four loads F to K. For the sake of clearness the location-directions BL, LF, FM and MK are reproduced in Fig. 52^. Two forces are acting at the left hinge the reaction AB and the load BL', the location- direction of their resultant AL must pass through the left and center hinges. Therefore through / draw the magnitude- direction parallel to the location-direction AL. Similarly the location-direction of the resultant of the load MK and the reaction KA acting at the right hinge is M A , which must pass through the center and right hinges. Hence through m SEC. II APPLICATION OF PRINCIPLES OF EQUILIBRIUM 79 draw the magnitude-direction parallel to the location-direction MA intersecting the magnitude-direction through / at a. The location of a fixes the magnitude-directions ka and ab which, when drawn, will close the magnitude-direction diagram for the external forces and completely determine the reactions. Draw the stress diagram and note that the diagram gives a check on itself when the center hinge is reached. It may be well to note that the load FG acting at the center hinge, which was combined with the loads on the body F, might have been combined with the loads on the body X without affecting the final results. Prove the statement in the preceding paragraph by a solution. Algebraic Solution. Replace each inclined load by its H- and F-components. Assume V\ and HI for the components of the left reaction; and Vz and HZ for the components of the right reaction. Balance the moments of all the forces acting on the structure about the left support. 35 X [,2OO = 42,000 65 X 720 = 46,800 75 X 5 00 = 37,500 30 X [,350 = 4O,5OO 60 X ,200 = 72,000 9oX ,200 = IO8,OOO 120 X ,510 = l8l,200 150 X ,300 = 195,000 180 X ,6OO = 288,000 210 X QOO = l89,OOO 240)1 ,2OO,OOO 5,000 = V* Balance the moments about the right support, 30 X 900 = 27,000 60 X ,600 = 96,OOO 90 X ,300 = 117,000 120 X ,510 = l8l,2OO 150 X ,200 = l8o,OOO 180 X ,2OO = 2l6,OOO 210 X ,350 = 283,500 240 X 900 = 216,000 I ,316,700 I26,3OO 240)l ,I9O,4OO 4,960 75 X 500 = 37,500 65 X 720 = 46,800 35 X 1,200 = 42,000 126,300 Vi 8o THEORY OF FRAMED STRUCTURES CHAP. II Balance the moments of the forces on the body Y about the center hinge, 30 X 1,300 = 39,000 120 X 5>ooo = 600,000 60 X 1,600 = 96,000 216,000 90 X 900 = 81,000 80)384,000 216,000 4,800 acting to the left. 80)190,400 Balance the moments of the forces on the body X about the center hinge, 30 X 1,200 = 36,000 120 X 4,960 = 595,200 60 X 1,200 = 72,000 404,800 5 X 500 = 2,500 90 X 1,350 = 121,500 15 X 720= 10,800 acting to the right. 1 20 X 900 = 108,000 45 X 1,200 = 54,000 404,800 Check by balancing the horizontal magnitudes, 2,380 1,200 720 500 4,800 = 4,800 Compare these results with the H- and F -components of ka and db (Fig. 526). 51. The cantilever bridge supporting vertical loads and resting upon four vertical supports (Fig. 53 a) appears to be statically indeterminate, for there are four magnitudes to determine one at each support; but the indeterminateness disappears when we consider that the structure is composed of three rigid bodies AC, CD and DF fastened. together by pin- connected joints at C and D. The portions AC and DF are SEC. II APPLICATION OF PRINCIPLES OF EQUILIBRIUM 8l called cantilever trusses; AB and EF are the anchor arms and BC and DE are the cantilever arms. The portion CD is called the center or suspended truss. The interaction of one part of the structure upon another is seen more clearly by a consideration of Fig. 536, where the center truss is shown as if it were suspended from the cantilever arms. Loads applied at points i , 2 and 3 are supported at A and B ; and that part of the structure represented by AGHB performs the function of a simple truss. Loads at points 4, 5 and 6 are also supported at A and B\ but in this case the reaction at A is negative, i.e., acts downward. Consequently, the structure must be provided with anchors into the masonry at A and F. Suppose that a load of 1,000 Ib. is placed at point 7. There is a tension of 750 Ib. in CC' causing a negative reaction of 750 Ib. at A, and an upward pressure of 1,500 Ib. at B. Similarly there is a negative reaction of 250 Ib. at F, and an upward pressure of 500 Ib. at E. The reactions at E and F for loads between D and F are determined in the same manner as for the left cantilever truss. The foregoing analysis as applied to the structure in Fig. 536 is equally applicable in Fig. 530. 52. Inclined Loads. In the next chapter we shall have occasion to consider the effect of loads acting normal to the 8 ft 6'= 48' >J slope of the roof, the roof truss being supported on walls, as illustrated in Fig. 540. The resultant of the five loads is 5,200 Ib. Its H- and F-components are 2,000 Ib. and 4,800 Ib. respectively. Let HI and V\ represent the H- and F-components at A, and H-2. and Vz represent the corresponding components at B. Since 82 THEORY OF FRAMED STRUCTURES CHAP. II the points of support are level and the //-components act in a straight line, V\ and F 2 may be determined independently of Hi and Z7 2 . Balance the moments of all the forces about the point A, 5,200 X 13 48 4,800 1,408 = 1,408 = V. 3,392 = Vi Check the value of V\ by balancing the moments of all the forces about B. The magnitudes of HI and # 2 acting in a straight line cannot be determined (as in the case of a three-hinged arch consisting of two rigid bodies) by the principles of statics. Their sum is 2,000 lb., and the magnitude of each depends somewhat upon the manner in which the truss is supported. If the truss has a roller support at B and we assume no friction, then HI 2,000 and H 2 = o If the truss has a roller support at A and we assume no friction, then HI = o and HZ = 2,000 If we assume that the horizontal thurst is resisted equally at A and B, then HI = 1,000 and HI = 1,000 If we assume that the horizontal component of each reaction is proportional to its vertical component, then - *i -5^X2,000-1,413" and ,, In the latter case the resultant of the H- and F-components at each support has a direction parallel to the resultant of the loads. In Fig. 546 the load line cd is laid off and the magnitude- SEC. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM 83 direction diagram lor the external forces is closed from d to c in four different ways; corresponding to the various assump- tions which may be made regarding the magnitudes of H\ and H^. Since the vertical magnitudes are constant for any as- sumed magnitudes for HI and # 2 , the points e\, 2 , e s and e fall on a horizontal line. Complete the stress diagram. How do the different as- sumptions affect the stresses? SEC. III. BEAMS: SHEAR AND BENDING MOMENT DIAGRAMS 53. A beam is any part of a structure or the structure itself as a whole, which resists the action of transverse forces and is bent by them. The forces may act at right angles to the axis of the beam, or they may be inclined at any angle. When the forces are inclined they may be resolved at the axis of the beam into rectangular components normal and parallel to the axis. The components parallel, i.e., in line with the axis produce direct tension or compression in the beam ; while the normal or transverse components produce the beam action or bending. The effect of the transverse forces is treated under the headings of shear and bending moment. 54. The shear at any normal section of a beam is the alge- braic sum of all the transverse forces acting on one side (either side) of the section. Since the algebraic sum of all the trans- verse forces acting upon a beam in equilibrium is zero (S7 = o); it is evident that the shears on the two sides of a section have equal magnitudes but are opposite in sense. The shear is called positive, when acting upward on the left (of the section) or downward on the right; negative, when acting downward on the left or upward on the right. 55* The bending moment at any normal section of a beam is the algebraic sum of the moments of all the forces acting on one side (either side) of the section; taken about the center of gravity of the section as an axis. Since the algebraic sum of the moments about any point of all the forces acting upon a beam in equilibrium is zero (I.M = o) ; it is evident that the bending moments on the two sides of a section have equal magnitudes 84 THEORY OF FRAMED STRUCTURES CHAP. II but are opposite in rotation the one being clockwise and the other counter-clockwise. The bending moment is called positive when clockwise on the left of and about the section, or counter-clockwise on the right of and about the section; nega- tive, when counter-clockwise on the left of and about the sec- tion, or clockwise on the right of and about the section. 56. The Bending Moment Determined from the Location- direction Diagram. In general the shear and bending moment vary from section to section along a beam. They may be represented conveniently for inspection by diagrams, the ordi- nates of which represent the shear or bending moment at various sections of the beam. Each diagram has a base or datum line. Positive values are laid off above, and negative values below this line. These diagrams have a definite rela- tion to the magnitude-direction and location-direction dia- grams, which are drawn in the graphic solution of a system of parallel forces. The diagrams in Figs. 220, and 226 were drawn to determine the magnitude of Q and the magnitude and direc- tion of P. These diagrams are reproduced with some addi- tions in Fig. 55. The shear at any section may be found from the magnitude-direction diagram. The shear on any section between the left reaction and the first load is fa = +5 Ib. The shear on any section between the first and second loads is ft = -j-^ Ib. The shear on any section between the second and third loads is/c = 2 Ib., etc. The bending moment at any section x between the first and second loads is M x = pP - sS Produce the location-directions ob and of, to intersect at /. The point 7 locates the resultant of the two forces P and 5; and the magnitude of this resultant is the magnitude-direction fb, therefore pp - S S = fb.r The triangles GIJ and bof are similar, hence y:r ::fb : h then fb.r = yh or M x = yh SEC. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM Fl&.S&a 86 THEORY OF FRAMED STRUCTURES CHAP. II Hence, the bending moment at any section equals the correspond- ing ordinate y in the location-direction diagram, multiplied by the horizontal component of the forces in the magnitude-direc- tion diagram. The ordinates y are measured in feet, and the component h is measured in pounds; hence, the bending moment is expressed in foot-pounds. If the reactions had been determined algebraically, the magnitude-direction diagram could have been drawn and closed; locating the point /, before the location-direction diagram was drawn. It would then have been possible to have chosen a point on a horizontal line through/ for the pole o, (Fig. 560). In this case the location-direction of becomes a horizontal line as shown in Fig. 566; and the bending moment at any section equals 10 Ib. times the number of feet scaled on the correspond- ing ordinate in the location-direction diagram. It has been shown that the shear at any section may be obtained from the magnitude-direction diagram. The shear at various sections may be represented more clearly by the shear diagram (Fig. $6c). The datum or base line is//", and the shear at any section is represented by the ordinate at that section. It is now clearly seen that the slope or bevel of each line in the location-direction diagram is proportional to the shear; for if h = 10 Ib. is taken as the horizontal side of each bevel, the vertical side of the bevel will represent the shear. Having established the principle that the bending moment may be derived from the location-direction diagram, and that the slope of any side of the location-direction diagram is a function of the shear; we shall proceed to consider the topic of shear and moment diagrams from another point of view. 57. Shear and Bending Moment Diagrams. The graphic method outlined in the previous article, whereby the shear and bending moment may be obtained from the magnitude-direc- tion and location-direction diagrams, is not the one most com- monly used by practicing engineers. This is especially true when the beam under consideration supports a uniform load over the whole or a portion of its length. A more convenient method, which is semi-graphical, will now be presented. SEC. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM 87 58. Illustrative Problems. i . Draw the shear and bending moment diagrams for the beam shown in Fig. 57. The shear diagram is drawn first because it is the simpler of the two. Draw a vertical line through the location of each load and reaction, and let ST be the base line for the shearing forces. The shear at any section between the left reaction and the first load is +5 Ib. Lay off SA = +5 Ib. to any convenient scale and draw AB parallel to ST. On the ordinate through B, lay off downward BC= 2 Ib. and draw CD. Lay off DF = 5 Ib. and draw FG. Continue in like manner to H on the ordinate through the right support, and note that HT scales 9 Ib., which equals the right reaction. The broken line SABC ... HT is the shear diagram, since the measure of any ordinate gives the shear at the corresponding section of the beam. For practical purposes it is seldom necessary to draw the shear diagram to scale. A sketch will usually serve equally as well. We now proceed to sketch the bending moment diagram on the base line UV. The bending moment, in accordance with 88 THEORY OF FRAMED STRUCTURES CHAP. II its definition, is zero at each end of the beam; consequently the diagram may be said to begin at U and end at V. In Article 56 it was shown that the slope of the bending moment diagram was proportional to the shear. The shear from A to B is posi- tive, hence the line UJ is sketched having a positive slope (i.e., upward to the right) . The shear from C to D is also positive, but not as great as the shear from A to B ; hence JK is sketched having a positive but lesser slope than the line UJ. The shear from F to G is negative and the line KL is sketched with a negative slope (i.e.j downward to the right). The slopes of LN and NV are negative, NV having the steeper slope. Thus a general idea of the variation in bending moment may be ob- tained from a simple sketch. The horizontal unit of measure in both diagrams is i ft. The vertical unit of measure is i Ib. in the shear diagram, and i ft.- Ib. in the bending moment diagram. Let i ft. be taken as the length of the horizontal side of the triangle, representing the slope or bevel of each line in the bending moment diagram ; then the length of the vertical side of each triangle will be measured in foot-pounds, and the ratio of the vertical side to the horizontal side will represent pounds and equal the shear. Thus the slope of the line UJ is 77 ' = 5 Ib.; the slope of the line KL is - 2 *'" ' = -2 Ib., etc. The length of the critical ordinates JQ, KR, etc., may be obtained from the similarity of triangles. UQ:i'::QJ:s'# or 6':i'i:QJ: S '# whence QJ = - -^7^ = 6' X 5# = 3<>'# = area SABP From this it is clear that the difference in lengths of any two ordinates in the bending moment diagram equals the area of the shear diagram between the ordinates. The lengths of the consecutive ordinates may be computed as follows: SEC. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM 89 + 5X6 o = ordinate at U +30 +30 = JQ +3X5 = +15 +45 = -2X4= -8 +37 = 5X2= IP + 27 = NX -9X3= -27 o = ordinate at V. This method of calculation checks itself, and is particularly helpful when the shear and bending moment diagrams are FIG. 58. simply sketched and not drawn to scale. After the critical ordinates have been computed, the bending moment diagrams may be drawn to scale. In going from left to right along the beam, it is obvious that the bending moment is increasing as long as the shear is positive and the lines in the bending moment diagram continue to have a positive slope. At the section where the shear changes from positive to negative, the slope changes from positive to negative and the bending moment is a maximum. THEORY OF FRAMED STRUCTURES CHAP. II 2. The shear and bending moment diagrams for a cantilever beam are sketched in Fig. 58. Between A and B the shear is negative and GH has a negative slope. At B the shear changes from negative to positive, hence HK has a positive slope, and a maximum negative bending moment occurs at B. At D the shear changes from positive to negative, giving a maximum positive bending moment. The computations for the bending moments at critical ordinates follow : o at A 30 X 10 = 300 300 at B + 132 X 6 = +792 +492 at C + 72 X 10 = +720 + 1,212 at D -48 X 8 = -384 +828 at E -136 X 6 = -828 POO Ik per ft U 3. The beam in Fig. 59 supports a uniform load of 200 Ib. SEC. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM per foot. The shear diagram starts with a positive ordinate = i, 600 Ib. at S, decreasing uniformly 200 Ib. in each foot from S to C. Beyond C the ordinates are negative, and increasing 200 Ib. per foot to 1,600 Ib. at T. Since the ordinates in the shear diagram are uniformly varying, the slope of the moment diagram will be uniformly varying; and the bending moment diagram will be represented by a curved line instead of a series of broken lines. The ordinates in the shear diagram are positive 1 h if \ A i. FIG. 6o&. over the left-half of the beam, decreasing to zero at the center; hence the bending moment curve has a positive slope at U, which decreases to zero at K, where the tangent to the curve is horizontal. Beyond K the curve has an increasing negative slope. Any ordinate in the bending moment diagram is easily determined from the area of the shear diagram; thus // = area SABF = 4,800 ft.-lb. KH = area SAC = 6,400 ft.-lb. The bending-moment curve is a parabola, as will be shown in the following article. 59. The Structural Engineer's Parabola. The mathemati- cian usually derives the equation of the parabola with the 92 THEORY OF FRAMED STRUCTURES CHAP. II origin at the vertex and the axis of symmetry horizontal, as shown in Fig. 6oa. Under these conditions the general equation of the parabola is F 2 = kX (i) where X and Y are variables and k is a constant. The structural engineer finds it more convenient to choose some point not at the vertex for the origin, with the axis of symmetry vertical as shown in Fig. 606. Let P be any point on the curve in either Fig. 6oa or b. Let X and Y represent the coordinates of P when the origin is at the vertex M, and let x and y represent the coordinates when the origin is at some other point E. When the point P is at N X = h, and Y = ~ Substituting these values in Eq. (i) ?-** 4 r * = i* Hence the mathematician's equation of this particular parabola is Substituting X= h y, and F = x inEq.( 2 ) , or y = x (I - x) (3) Equation (3) may be called the engineer's equation of the parabola, in the same sense that Eq. (2) may be said to represent the mathematician's equation of the same curve. Equation (3) may be written in the more general form y = cx(l - x) (4) where x and y are variables and c is a constant. Equation (4) shows that one variable (y) is proportional to the product of two SEC. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM 93 other variables (x) and (I x) whose sum (/) is constant. Any equation which can be expressed in this form is a parabolic equation. The beam in Fig. 61 supports a uniform load of w pounds per unit length. The bending moment at any distance x from the left support is wlx wx 2 w f , v M x = = - * (/ - x) (5) This is a parabolic equation and i w* per lin.fi: 1 1 i i T wl riiiiiiiiii INI HIM ii illinium INI mi urn the bending moment diagram j.lT^ is a parabola. The following " " l ~ x ' important rule may be deduced U-- - i from Eq. (5). The bending moment at any section of a uniformly loaded beam equals one-half the load FIG. 61. per unit of length, times the product of the two segments into which the section divides the span. Thus the bending moment at // (Fig. 59) is 100 X 4 X 12 = 4,800 Two Methods of Constructing a Parabola. There are several methods by which a parabola may be constructed, but only two will be considered here: (a) an algebraic method by points; and (b) a graphic method by tangents. (a) Algebraic Method. Equation (3) may be written in the form from which the following observation may be made in connec- tion with Fig. 59. Any two ordinates // and KH are to each other as the products of the two parts into which each ordinate divides the base line UV. This equation gives a clue to a very simple method of locating any desired number of points through which a parabola is to be drawn. Suppose it is desired to locate seven points on the curve. Divide the line UV into eight equal parts by seven points. The ordinate at 94 THEORY OF FRAMED STRUCTURES CHAP. II each point is proportional to the product of the number of parts on either side of it, thus: 1X7 = 2X6 = 7 12 i5 3X 5= 4X4= 16 5X3= i5 6X2=12 7X1=7 Now if we wish the middle ordinate to equal 6,400 instead of 16, we multiply that ordinate and all the others by 400, whence 7 X 400 = 2,800 12 X 400 = 4,800 15 X 400 = 6,000 16 X 400 = 6,400, etc. (b) Graphic Method. If the tangents at the extremities of a parabolic curve are known, the parabola may be constructed FIG. 62. as shown in Fig. 62. Suppose we have two tangents AC and EC and the points of tangency A and B. Divide AC into any number of equal parts, and BC into the same number of equal parts. Number the points from A to C and from C to B. Connect i-i, 2-2, etc. The parabolic curve lies tangent to these lines as shown. SEC. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM 95 Draw tangents through U and V (Fig. 59) intersecting at D. It may be observed from the shear diagram that the positive slope of the tangent UD is 1,600 ft.-lb. vertical to i ft. hori- zontal. The tangent DV has a corresponding negative slope, hence the intersection D is directly above K and the ordinate HD = 12,800 ft.-lb. Construct the parabolic bending moment diagram on these tangents, and check the lengths of several ordinates in the moment diagram by the algebraic method. 60. Illustrative Problem. The shear and bending moment diagrams for a beam (Fig. 63) supporting a uniform load over a portion of its length, have several interesting properties. The shear diagram is drawn first, and the section at which the shear changes from positive to negative is located. The ordinates in the shear diagram indicate that the moment diagram is composed of the straight line OA, the parabolic curve A BCD, and the straight lines DE and EX. The maximum ordinate RC is at the section of zero shear. The following ordinates are obtained from the area of the shear diagram : o at + 4,000 X 5 = +20,000 + 20,000 = NA |(4,ooo + 1,000)15 = +37>5 + 57,500 = JB + 1,000 X \ X 5 = +2,500 + 60,000 = RC 2,000 X J X 10 = 10.000 + 50,000 = WD 2,000 X 15 = 30,000 + 20,000 = ZE 5,000 X 4 = 20,000 o at X Lay off to scale the ordinates NA, WD and ZE\ and draw the lines OA, DE and EX. The line OA, having the same slope at A as the parabola, is tangent to it ; likewise the line DE is tangent to the parabola at D. If the lines OA and ED are THEORY OF FRAMED STRUCTURES CHAP. II produced, they will intersect at F ; a point directly under the center of the uniform load. If the uniform load of 6,000 Ib. 4-000 K - 4000 -X > A ^t If M- Z Ibs.per foot 30001k /y/WvZv///^^ S4'-O ff - >> /^* -2000 5000 -5000 3000 B, FIG. 63. were concentrated at its center, the bending moment diagram would be OF EX. The parabola may now be drawn on the tangents FA and FD. SEC. Ill APPLICATION OF PRINCIPLES OF EQUILIBRIUM 97 The parabolic segment ABDG has several interesting prop- erties. Let PS represent any ordinate in the bending moment diagram between AN and DW, and let x represent the distance of this ordinate from the left support; then the bending moment at x is M. = PS = 4,000* - 5,ooo# 2,500 (i) Let QS = y be the ordinate to the line AD, then from similar triangles QT:DH::TA:HA QT y AN = y 20,000 DH = DW - AN = 30,000 TA = x - 5 HA = 30 therefore y 20,000:30,000::^ 5:30 or y i,ooox + 15,000 then PQ = M x y m = ioox 2 -\- 4,000* 17,500 Let AT = x $ = v or x = v + 5 then m = -100(0 + 5) 2 + 4,000(2; + 5) - 17,500 or m = 100^(30 v) (2) Equation (2) is also the expression for the bending moment m at any section of a beam 30 ft. long, supporting a uniform load of 200 Ib. per foot when the section is a distance v from either support. Therefore the lengths of corresponding ordinates in the two parabolic segments ABDG and AiBiDid are equal. The areas of the two segments are equal and the centers of gravity are similarly situated. DH The slope of the line AD is -j = 1,000 Ib. The shear ordinate KL is 1,000 Ib.; hence the line UV, which is tangent to the parabola at B, has a slope of 1,000 Ib., and is therefore parallel to the line AD. The point B bisects the line FG. 61. Problems. 1-7. Determine the reactions algebraically, sketch the shear and bending moment diagrams, compute the values for critical ordinates and draw the diagrams to scale for the beams shown in Figs. 64 to 70 inclusive. The loads A 9 8 THEORY OF FRAMED STRUCTURES CHAP. II and B in Fig. 65 are equal. In Fig. 66, A, B and C are frictionless pegs. The rope passes over the upper pegs and under the lower one. The beams in Figs. 67, 68 and 69, are fixed in a wall at the left end. 1800* as ' 5 > J^ $' j FIG. 64. t I f f \seoo* uvU -' j<*$fefyf4tu.. /0'-- FIG. 65. 80* FIG. 66. 1 FIG.67 300 Ib. per ft. W FIG. 68 800 Ik per ft 10' F1G.69. 8. A timber 20 ft. long, weighing 40 Ib. per linear foot is floating in water, and supports 200 Ib. 4 ft. from the left end and 150 Ib. 2 ft. from the right end. Draw the shear and bending-moment diagrams. SEC. IV APPLICATION OF PRINCIPLES OF EQUILIBRIUM 99 9. A timber 80 ft. long, floating in water, supports three loads of 80 Ib. each, one at the center and one at each quarter point. Neglect the weight of the timber and draw shear and bending moment diagrams. 10. A beam 30 ft. long is supported at each end A and C. A uniform load of 100 Ib. per foot extends from A to B, and a uniform load of 225 Ib. per foot extends from B to C. The bending moment is a maximum at B. Locate B, and draw the shear and bending-moment diagrams. SEC. IV. FRAMES HAVING MEMBERS WHICH PERFORM THE FUNCTIONS OF A BEAM 62. The framed structures heretofore discussed have sup- ported loads applied only at the joints. This is one of the four conditions mentioned in Article 40, which must be met if the members of a structure are subject to longitudinal stresses (tension or compression) only. In order to prepare the way for an analysis of a structure composed of a roof truss and columns, in which the columns also serve as beams when the structure is resisting inclined loads; it will be advisable to consider several types of simple frames which resist the action of loads applied not only at the joints, but also at intermediate points. Such members must be designed to resist shearing and bending stresses, as well as longitudinal stresses. The analysis of a structure of this type may be made by a process of dissection, whereby each member is sketched separately and the forces acting thereon are shown. 63. Illustrative Problem. The frame shown in Fig. 71, hinged at A and resting on a roller at B, supports a load of 300 Ib. at E. We find first the reactions or the forces necessary for the equilibrium of the frame as a whole. On account of the roller at B } the reaction there is horizontal; consequently there is a vertical force of 300 Ib. acting upward at A, and a horizontal force equal and opposite to the horizontal force at B. Balance the moments about A. 300 X 6 = 24 : 75 IOO THEORY OF FRAMED STRUCTURES CHAP. II The horizontal force of 75 Ib. acts to the left at B, and to the right at A . Now sketch each member separately. The vertical member resists the action of forces at four points, A, B, C and D. The forces at A and B are known. The un- known forces at C and D are the result of the interaction of the members, and are represented by horizontal and vertical com- ponents. The horizontal member resists the action of forces at three points, the force at E being known. Indicate hori- zontal and vertical components at D' and F. The inclined member resists forces at two points F f and C", as shown. Bal- * ^ -UL75 n.75 B * 300 F D' 300 w FIG. 71. ance the moments of the forces acting on the horizontal member about D'j the vertical force at F is 600 Ib. acting downward. Indicate it at once on the sketch; and do likewise with each force, as soon as its magnitude and sense are found. The vertical force at D f , is 900 Ib. acting upward. Since the inclined mem- ber is pulling downward 600 Ib. at F, the horizontal member must be pulling upward 600 Ib. at F 1 . A vertical force of 600 Ib. acts downward at C r , and upward at C. A vertical force of 900 Ib. acts downward at D. Balance the moments of the forces acting on the inclined member about F' . SEC. IV APPLICATION OF PRINCIPLES OF EQUILIBRIUM IOI A horizontal force of 450 Ib. acts to the right at C", and to the left at F'; consequently a horizontal force of 450 Ib. acts to the left at C and to the right at F. Finally a horizontal force of 450 Ib. acts to the left at D', and to the right at D. *,,,ss,S,sss)$ ::::; 5 ^:::: FIG. 72. FIG. 7 3 .[ FIG. 75- FIG. 76. FIG. 75. Inspect each member and see if the forces acting on each are in equilibrium. Draw shear and bending moment diagrams for the vertical and horizontal members. 102 THEORY OF FRAMED STRUCTURES CHAP. II The resultant of the two forces acting at F' must have a direction through C', otherwise the inclined member would rotate and not be in equilibrium. Likewise the resultant at C' passes through F' ', and the member resists a longitudinal stress. Whenever forces act at only two points on a member, the member reacts as a tie or strut and not as a beam. 64. Problems. 1-6. The structures shown in Figs. 72 to 77 inclusive are to be treated as follows: Determine the horizontal and vertical components of the reactions, sketch each member separately and find the horizontal and vertical components of the forces acting thereon. /4 f ->|<- IO'---- FIG. 77. 65. The portal frame (Fig. 780), resisting a horizontal force of 600 Ib. at D has four unknown magnitudes represented in the reactions i.e. a vertical and a horizontal force at each support; but the frame, being composed of two rigid bodies hinged at C, is statically determinate. Each vertical member or column is continuous. The two horizontal members and the inclined members are hinged at C. Balance the moments of all the forces about A . 600 X 24 16 = 900 A vertical force of 900 Ib. acts upward at J3, consequently a vertical force of 900 Ib. acts downward at A. Balance the moments of the forces acting on the right portion of the struc- ture about C. 900 X 8 - = 240 30 SEC. IV APPLICATION OF PRINCIPLES OF EQUILIBRIUM 103 A horizontal force of 240 Ib. acts to the left at B. Balance the moments of the forces acting on the left portion of the structure about C. c 600 ' FIG. 78 a 360* 1200 960* 1200* 1200* | H 900* t. 900 * FlG.78-b 240* 900* 900 X 8 = 7,200 600 X 6 = 3,600 30)10,800 360 A horizontal force of 360 Ib. acts to the left at A. Each member in the structure might now be sketched sepa- rately, and the stresses analyzed by the algebraic method, as in 104 THEORY OF FRAMED STRUCTURES CHAP. II Article 63. In many cases, however, and especially if the frame is more complex, a graphic solution by means of the stress diagram is desirable; but before a stress diagram can be drawn, the bending effect of the columns upon the frame, caused by the horizontal forces at A and B, must be considered. Each column is sketched separately as before (Fig. 786.) The known forces acting on the columns are the reactions at A and B and the load of 600 Ib. at D. Other forces, representing the action of the frame HIJK on the columns, must be supplied at E, D, F and G. It is necessary to consider only the horizontal forces required for the equilibrium of the columns at these points, since the vertical forces in the columns do not cause bending in the columns. The horizontal forces necessary for the equilib- rium of the columns are as follows: At G 24 * 24 = 960 acting to the left. 6 At F 240 + 960 = 1,200 acting to the right. At E 3 = 1,440 acting to the left. At D 1,440 + 360 600 = 1,200 acting to the right. These four horizontal forces represent the bending action of the frame on the columns. The reaction of the columns on the frame are equal and opposite as indicated on the frame. The vertical forces acting on the columns at A and B are transferred to H and K. The frame is in equilibrium, and a stress diagram may be drawn. 66. The portal frame (Fig. 790) is statically indeterminate. The vertical reaction at either support may be determined by balancing the moments of all the forces about the other support, thereby eliminating the two horizontal reactions. The hori- zontal reactions cannot be found by the principles of statics. In order that an approximate estimate of the stresses may be made, current engineering practice assumes that the horizontal load is supported equally by the two columns. The two columns are sketched separately, as before. The horizontal loads and reactions are shown on the outside of the columns and the forces, interacting between columns and frame are shown on the inside SEC. IV APPLICATION OF PRINCIPLES OF EQUILIBRIUM 105 of the columns. The latter forces have been transferred to the frame in Fig. 79^, and the vertical reactions indicated. A stress diagram may now be drawn. 1000 woo 1300 6100 600* opfiTf: 780C ^ ^1000 /\/\ 9100 J, /i / ^o -* 7^a537 The stresses for all four cases, determined by the stress diagram in Fig. 83 b, are recorded in Table IV; where each member is numbered to correspond with Fig. 87. The bottom chord is straight, and in line with the two points of support; and because of this, the stress in any member, except those of the bottom chord, is the same for the four different assumptions regarding the horizontal reactions. 77. Trusses Supported on Columns. The Mill Building Bent. When the truss of Fig. 83 a is supported by columns, as shown in Fig. 84; six unknown quantities are involved in the reactions a vertical force, a horizontal force and a resisting SEC. II ROOF TRUSSES moment at each support. The problem is therefore stati- cally indeterminate in the third degree, and the three static equations of equilibrium must be augmented by three elastic equations, in making an exact analysis of the reactions. The derivation of these elastic equations is a long and involved process, and will not be considered here. In place of the elastic equations, three assumptions are made. Hitherto, it has been the practice to treat the mill building bent as a portal frame, the same assumptions being made as outlined at the end of Article 67; viz., that the point of contraflexure, or zero bending moment, in each column is one-third of the distance from the base of the column to the foot of the knee-brace; and the total horizontal shear is equally resisted by each column. T ^ \A V V A/ \ Y 7 A A \ A -i r f. ; ~r f i X y V v > ! FIG. 84. A recent investigation, 1 made by Mr. S. R. Oflutt at the University of Illinois, under the direction of the author has proved that the assumption of equal horizontal reaction is no longer justified. Twenty structures having the general outline, as shown in Fig. 84, and dimensions given in Table III were designed and investigated. Fink trusses, having a slope of 6 in. to i ft., were used in all cases. Each truss had eight panels, except the 2o-ft. trusses, which had four panels. The length of the bay was 15 ft. The truss for each span length / was designed for a uniform vertical load of about 40 Ib. per square foot of horizontal surface. This load was varied somewhat in several 1 This investigation is to be published as a Bulletin of the University of Illinois Engineering Experiment Station. n8 THEORY OF FRAMED STRUCTURES CHAP. Ill instances to note the effect upon the final results. The mem- bers A were made about twice as large as was required for the vertical load. The knee-braces were designed for a wind pressure of 20 Ib. per square foot on the vertical height h, and 14.9 Ib. per square foot of roof surface normal to the roof. The columns in general were designed to support their vertical load and the additional effect of bending; by assuming that the point of contraflexure was one-third the distance from the base to the knee-brace; and that the horizontal reactions were equal. TABLE III. RATIOS FOR HORIZONTAL REACTIONS AND POINTS OF CONTRA- FLEXURE IN MILL BUILDING COLUMNS Span length Column height Ratio Pin-con- nected Rigidly fixed h I H R a* X y n h H H d d i 2 3 4 5 6 7 20 10 0.500 0.399 0.385 0.518 0.654 20 12 0.600 0-379 o-354 0.486 0.620 20 16 0.800 0-358 0.323 0-435 0.585 2O i9 0.950 0.356 0.309 0-439 0.569 30 16 0-533 0.380 0-353 0.481 0-593 30 21 0.700 0.364 0.324 0-435 0.570 30 26 0.867 0-354 0.309 0.432 0-547 30 31 1-033 0-340 o. 290 0-439 0.562 40 16 0.400 0.410 0.401 0.621 0.612 40 21 0-525 0.385 0.356 0.488 0.581 40 26 0.650 0-373 0-335 0.471 0.560 40 31 0-775 0.358 0.314 0-457 0.568 50 16 0.320 0.425 0.429 0-634 0.601 50 21 0.420 0-397 0-375 0.498 o.57i 50 26 0.520 0.382 0-349 0-465 0-555 50 31 0.620 0.369 0.332 0.452 o.558 60 16 o. 267 0-439 0-453 o . 643 0-593 60 21 0.350 0.407 0.388 0-494 0-574 60 26 0-433 0.390 0.361 0-454 0-558 60 31 0.517 0-379 0-345 0.429 0-554 SEC. II ROOF TRUSSES 119 This column design also was varied in several instances to note the general effect. Each bent thus designed was then analyzed, strictly in accordance with the elastic theory of structures by the method of deflections. This theory is too complicated to be explained here. One assumption, not strictly in accord with actual conditions was made; viz. that the truss and knee- brace members were pin-connected.. The horizontal wind loads were concentrated at points about 5 ft. apart, along the column. The wind loads normal to the roof were concen- trated at the panel points of the truss. The intensity of the roof loads was taken in accordance with Duchemin's formula. The reader is cautioned against drawing too general conclusions from the data given in Table III ; for only a limited number of structures was analyzed, and all trusses were of the same type and slope. The analysis was made for columns pin-connected at their bases, and rigidly fixed at their bases. The results given in Table III are ratios, and are thus explained. H wind load on windward or left column, exclusive of the half-panel load at the base, plus horizontal component of the normal roof load. H R = horizontal reaction at base of leeward or right column. d = distance from base of column to foot of knee-brace. x = distance from base of column to point of contraflexure of windward column. y = distance from base of column to point of contraflexure of leeward column. Suppose that H = 10,000 Ib. for the structure, in which / = 40 ft. and h = 21 it. If the columns are pin-connected, the horizontal reaction at the base of the leeward column is H R = 10,000 X 0.385 = 3,850 Ib. If the columns are fixed and d = 15, then H B = 10,000 X 0.356 = 3,560 Ib. * = 15 X 0.488 = 7.32 ft. y = 15 X 0.581 = 8.71 ft. The structure is statically determinate when one horizontal reaction and the points of contraflexure are known. 120 THEORY OF FRAMED STRUCTURES CHAP. Ill The columns are seldom, if ever, built with pin connections; nor are they sufficiently anchored to the masonry to insure a perfectly rigid connection. The actual conditions existing 4o'-o- 2920 1500 s*o D>>f \74\/ V 10 V>s^ ^57 ^55 * KM C\\/ A b_^ 4576 IX vir \J77, 4576 B f T/f /o (c) 3?50 +4S66 ;??(?, 3252 FIG. 85. in all mill-building bents are to be found between these two limits. For example, in the structure before mentioned, where / = 40 ft. and h = 21 ft., the horizontal reaction of the leeward column will be between 0.385 and 0.356 of the hori- SEC. II ROOF TRUSSES 121 zontal load. The point of contraflexure, or point of zero moment, in the leeward column will be found between the base of the column and the point 0.488 of the distance to the foot of the knee-brace. A comparison of the data in columns 4 and 5 of Table III indicates that 0.3 7 5#, instead of 0.5^, is the more reasonable estimate of the horizontal reaction of the leeward column; when the columns are considered partially fixed. The data given in columns 6 and 7 indicates that the point of contraflexure in each column may be taken at one-third 2000 4000 6000 Lbs, FIG. 86. the distance from the base of the column to the foot of the knee-brace. This conclusion is justified only when the masonry is sufficiently massive to resist the moment thereby alloted to it. The wind concentrations shown in Fig. 850 were determined from the data given in Article 76. The sum of the horizontal loads, exclusive of the half-panel load of 750 lb., is H = 7,786 If the horizontal reaction of the leeward column is taken as three- eighths of the total load H; then H R = 0.375 X 7,786 = 2,920 HL = 7,786 2,920 = 4,866 The point of contraflexure in each column will be taken one- third the distance from the column base to the foot of the knee- brace; and the structure is thereby transformed into the 122 THEORY OF FRAMED STRUCTURES CHAP. Ill statically determinate frame as shown in Fig. 856, in which the moments are zero at A and B. Moments about A Moments about B i,5ooX 5 = 7,5oo 4,472X30= 134,160 1,650 X 10 = 16,500 2,236 X 21 = 46,956 900 X 16 = 14,400 900 X 16 = 14,400 2,236X21= 46,956 1,650X10=16,500 4,472X10= 44,720 i,5ooX 5= 7>5o 85,356 40)130,086 40)48,804 3,252 = VR 1,220 = V L The truss and columns are shown separately in Fig. 85^;. The known forces acting on the columns are shown by full lines. The horizontal forces, represented by dotted lines at #, bj c and dj which the truss must exert on the columns in order to hold them in equilibrium, are determined as follows: Moments about c force at d = - 2 -- = 7,787 acting to the right, force at c = 7,787 2,920 = 4,867 acting to the left. Moments about a 4,866 X 16 = 77,856 1,650 X 6 = 9,900 1,500 X ii = 16,500 1,500 X 1 6 = 24,000 50,400 6)27,456 4,576 = force at b acting to the right. force at a = 900 + 1,650 + 4,576 + 1,500 + 1,500 4,866 = 5,260 acting to the left. PIG. 87. SEC. II ROOF TRUSSES 123 TABLE IV. COMPARISON OF WIND STRESSES I,OOO-LB. UNITS Mem- ber Wall bearing truss Mill Wind load build- acting ing vertically, truss both sides Case V Case VI Case I Hi = 2,236 H z = o Case II Hi = o H 2 = 2,236 Case III //i = 1,118 H 2 = 1,118 Case IV Hi = i,537 H 2 = 699 i -5-6 -5-6 -5-6 -5-6 -H-3 + 9-3 -9.8 2 -5-6 ~5-6 -5-6 -5-6 -U-3 + 9-3 -9-3 3 -5-6 -5-6 -5-6 -5-6 - 6.4 + LI -8.7 4 -5-6 -5-6 -5-6 -5-6 - 6.4 + LI -8.2 5 + 7-0 + 2.8 +4-8 + 5-9 +6.3 + 4-6 - 3-5 +8.8 6 +5-6 + 2.8 +3-4 +4-5 +4-9 + 4-5 - 5-7 + 7-6 7 + 2.8 +0.6 + 1-7 + 2.1 i.i + 5-0 + 1-3 8 + 1.4 + 1.4 + 1.4 + 1-4 + 6.9 - 9-3 9 -2.5 -2.5 -2.5 -2.5 - 5-0 + 4-2 -2.3 10 + 2.8 + 2.8 + 2.8 + 2.8 + 5-5 - 4-7 + 2.5 ii +4-2 +4-2 +4-2 +4-2 + 6.9 - 4-7 +3-8 12 o o o + 6.3 10.8 o 13 +1.4 + 1.4 + 1.4 ( +i-4 + 1-4 + 1-3 14 1-3 -i-3 -1-3 -i-3 - i-3 i.i 124 THEORY OF FRAMED STRUCTURES CHAP. Ill Since the forces which the truss exerts on the columns must be respectively equal in magnitude, but opposite in sense, to the forces which the columns exert on the truss; the forces at a, b, c and d are therefore shown reversed on the truss. These forces virtually represent the bending effect imposed by the columns on the truss. The vertical reactions, or the thrust of the columns on the truss, are also transferred to the truss. The truss is now in equilibrium under the action of the forces shown thereon, and a stress diagram may be drawn as shown in Fig. 86. The stresses are recorded as Case V in Table IV. The members are numbered to correspond with Fig. 87. 78. Stress Diagram for Wind Unnecessary. The stresses for Cases I to V inclusive (as recorded in Table IV) were determined for a wind load of 20 Ib. per square foot, on a vertical surface; and 14.9 Ib. per square foot., or 1,250 Ib. per panel acting normal to the roof and on only one side of it. If a wind load of 1,250 Ib. per panel is assumed to act -vertically on both sides of the truss in Fig. 850; the resulting stresses will be as given under Case VI in Table IV. It is clear that the stresses in all chord members are greater for Case VI than for Cases I to IV inclusive; and that the slight excess of the stresses in the web members for Cases I to IV are negligible when the additional stresses for dead and snow loads are taken into consideration. In comparing Case V with Case VI, we find that the member 8 requires the most consideration. Members i and 2 are the only chord members which have a stress greater in Case V than in Case VI. It is clear that for wall bearing trusses, a load of v = 14.9 Ib. per square foot of roof surface taken vertically on both sides of a Fink truss, having a roof slope of 6 in. to i ft., will suffice for a wind horizontal load of w = 20 Ib. per square foot on a vertical surface. It is also clear that v varies directly as w. Likewise, if w remains constant and the slope varies, the cor- responding value of v may be taken from Table II of Article 75. In a wall bearing Warren truss, having a roof slope not greater than i J^ in. to i ft., a vertical load of 5 Ib. per square foot seems to be an ample allowance for wind. When the truss is knee-braced to columns, the wind load SEC. II ROOF TRUSSES 125 may without serious error, be considered vertical and included with the dead and snow loads, if special consideration is given to the columns, knee-braces and the web members which are connected to the knee-braces and nearest in line with them. It is probably true that a stress diagram for wind is drawn for not more than one truss for every hundred that are designed. 79. Combined Snow and Wind Loads. A considerable amount of speculation has been made regarding the proper combination of stresses caused by wind and snow. It seems reasonable to predict that in a given locality the maximum wind and snow loads will not occur simultaneously. If the snow is not covered by a crust, a high wind will blow it from the roof. If the roof is nearly flat, the snow load may be large; but the wind stresses will be small, especially in a wall bearing truss. On the other hand, the wind stresses are relatively larger on a steep roof, but little snow can be retained. For buildings in the latitudes of New York and Chicago, it is the general practice to consider a load of 25 Ib. per square foot of roof surface acting vertically, as sufficient for combined wind and snow. 80. Design Stresses. The following data will be assumed: Wall bearing trusses (Fig. 830) 15 ft. apart; span, 40 ft., rise 10 ft., roof covering including purlins weighs 10 Ib. per square foot; combined snow and wind assumed at 25 Ib. per square foot of roof surface. The length of the top chord is 22.36 ft. The total weight on the truss is 35 X 15 X 2 X 22.36 = 23,500 Ib., or 590 Ib. per foot of span. From Table I the assumed weight of the truss is 1,500 Ib. The total load is 25,000 Ib. or 3,125 Ib. per panel. A stress diagram drawn for a vertical load of 3,125 b., placed at each of the seven joints of the top chord will determine the design stresses. The half -panel load at each support has no influence upon the stresses. When the same truss is supported on columns, as in Fig. 850, the stress diagram is drawn in the same manner, using the same panel loads as for the wall-bearing truss. All members are designed as for a wall-bearing truss, except the member A . The size of one angle necessary to carry the stress, as given by 126 THEORY OF FRAMED STRUCTURES CHAP. Ill the stress diagram, is determined; and two angles of this size are used. The horizontal wind load on the side of the building, and the horizontal component of the normal wind load on the roof, which are carried by each bent are determined; and three- eights of the total is taken as the horizontal reaction of the leeward column. In Article 77 this reaction was found to be 2,920 Ib. This reaction, multiplied by two-thirds the height of the column to the foot of the knee-brace, gives the maximum bending moment at the foot of the leeward knee-brace; which is 2,920 X 10 = 29,200 ft.-lb. This bending moment is obviously greater than at any point in the windward column. Each column should be designed to resist the bending moment of 29,200 ft.-lb. and a direct com- pression of 12,500 Ib. In designing the knee-braces, the leeward knee-brace should be considered. The horizontal component of the stress, easily determined from the sketch of the leeward column in Fig. 85^;, is 7,787 Ib.; from which the compressive stress may be quickly found. 81. Conclusions. Several years ago a purchaser wrote to a structural steel company for the price of a "steel building, 40 ft. wide, 100 ft. long and 2 miles from the railroad station." The author, who was at the time a designer for the company, was given the letter and instructed to make a "design and esti- mate." While the typical specification written by a purchas- ing agent gives more detail than the one cited ; it is nevertheless true that the cases are rare where competitive designs submitted in accordance with the same specifications are actually made on the same basis. In some designs no bending moment whatever would be provided for in the columns, in others bending moment would be considered on the basis of 16,000 Ib. per square inch reduced for compression; while in others 24,000 Ib. per square inch might be used. But the most interesting feature would be found in a comparison of the designs of knee-braces and bracing which are generally a matter of opinion with each designer; and his opinion may vary with the hour of the day or the day or the week. SEC. II ROOF TRUSSES 127 There is no generally accepted standard specification for buildings in America. Until one has been written, the treat- ment of stresses as given in Article 80, though not strictly scientific, should not be condemned by any theorist who knows nothing of the perplexities of the designer in playing the role of a mind reader while interpreting the average building specifications. CHAPTER IV BRIDGES SEC. I. STANDARD TYPES 82. Standard Types. The most common types of railway steel bridges are shown in Figs. 88 to 94. In the deck plate- girder type the cross-ties which support the rails rest directly on the top flanges of two girders, placed from 6 to 7 ft. apart FIG. 88. Pratt Truss. FIG. 89. Warren Truss. FIG. 90. Parker Truss. In the through plate-girder type the cross-ties rest on short deck beams or girders which are supported at each end by cross girders, usually called floor beams, which are in turn sup- ported by the main girders. The through type is more expen- sive than the deck type, and is never used unless the distance from the rail to the under side of the main girders is so small that deck span is impossible. Plate-girders are used for spans up to about 115 ft. 128 SEC. I BRIDGES I2 9 The Pratt truss (Fig. 88) is the common type for pin-con- nected trusses in which the diagonal web members are eye-bars, or other slender members designed to take tension only. The Warren truss (Fig. 89) is built with stiff web members, capable of resisting alternate stresses of tension and compression. The Pratt and Warren types are used for spans between about 115 ft. and 200 ft. The Parker truss (Fig. 90), more often called a curved chord truss or a earner-back truss, and often having eye-bars in the bottom chord and diagonal web members, is the standard type for spans between about 200 ft. and 300 ft. The Baltimore truss is shown in Figs. 91 and 92. The eco- FiG. 91. Baltimore Truss (sub-ties). PIG. 92. Baltimore Truss (sub-struts). nomic height of a bridge, being from one-fifth to one-eighth of the span, increases with the length. The economic panel length does not increase in the same ratio, consequently as the height increases the diagonals become steeper. From a practical as well as a theoretical standpoint, the diagonals should not exceed a slope of about 3 vertical in 2 horizontal. In order to meet this requirement in large spans, the truss may be built with subdivided panels as in the Baltimore truss. The floor beams in the middle of a long panel are supported by a sub-vertical, connecting to the main diagonal and a sub-tie as in Fig. 92 or a sub-strut as in Fig. 93. Theoretically the Baltimore truss has no standing, for if the length of span makes sub-paneling essen- tial, a parallel chord truss is not economical. 13 THEORY OF FRAMED STRUCTURES CHAP. IV The Pennsylvania truss with sub- ties (Fig. 93) and substruts (Fig. 94) is used for all simple truss spans over apprximately 300 ft. in length. The Metropolis bridge over the Ohio River, 720 ft. long, is the longest span of this type which has been built to date. The standard types here mentioned are also employed in highway bridges. Trusses are frequently used in highway bridges for spans as short as 40 or 50 ft. FIG. 93. Pennsylvania Truss (sub-ties). FIG. 94. Pennsylvania truss (sub-struts). 83. Moving Loads. The loads which a roof truss supports are usually considered stationary; and stress analysis proceeds directly after the determination of the load and reactions. The process is not so simple in the case of bridges. The loads move upon the structure, and the shear or bending moment at any section, or the stress in any member, varying with the movement, is required only for that particular position of the loads which will cause the maximum shear, bending moment or stress. The first task, therefore, is the development of a method for finding the required position of the moving loads in any given case. This leads us to a consideration of " In- fluence Lines." SEC. II. DECK BEAMS AND GIRDERS 84. Influence Line for Reaction. Let P (Fig. 95) represent a load of i lb., moving upon the beam AB. Draw the line QO, intersecting the verticals through A and B. Lay off QK = i lb., draw the line OK, and let // represent the ordinate under the load P. As P moves from B to A, the reaction at A in- SEC. II BRIDGES creases uniformly from zero, when the load is at B; to i lb., when the load reaches A . Likewise the ordinate // increases uniformly from zero at O to QK = i lb., at Q. Hence for any position of the load, the ordinate //, directly under the load, represents the reaction at A . The line OK is called an influence line for the reaction at A , since the ordinate // shows how the reaction at A varies or is influenced by any movement of the load. The influence line, thus drawn by the aid of the unit load, may be used for finding the reaction at A for a series of loads in any position by multiplying the ordinate under each load by the ratio of its weight to i lb., and adding the products. In Fig. 96 the ordinates for the loads Q and S are 0.7 and 0.4 o Q-IOOO* o 3000* O 5=2000* ^' B FIG. 95- FIG. 96. respectively. Hence, the reaction at A is 2,000 X 0.4 = 800 1,000 X 0.7 = 700 1,500 lb. Since the influence line has a constant slope, the reaction at A may also be determined by finding the product of the total load (3,000), and the ordinate at the center of gravity of the loads; thus, 3,000 X 0.5 = 1,500 lb. The load of i lb., which is invariably used in constructing influence lines, may be considered as acting the role of a scout, sent across the span in advance of other loads. The effect of this advance load upon the reaction, when represented graphically, becomes an influence line for the reaction. The same is true in the case of shear or bending movement. Influ- 132 THEORY OF FRAMED STRUCTURES CHAP. IV ence lines thus drawn may be used as a basis for determining the effect produced by a series of loads; such as a road roller, traction engine, or a railway train. 85. Influence Line for Shear in a Beam. Let us first have a clear understanding of what we mean by shear. The shear at any normal section of a beam is the algebraic sum of all the transverse forces acting on one side (either side) of the section. When this sum or resultant force acts upward on the left of the section, the shear is positive. Obviously, if the resultant of the forces on the left of the section acts upward, the resultant of the forces on the right of the section acts downward. The influence line for shear at section C (Fig. 97) is desired. When the load of i Ib. is between B and C, the only force acting on the left of the section is the reac- tion at A. Hence, the influence line OK for the left reaction, is also the influence line for shear at FlG 9? C, when the load is between B and C. When the load of i Ib. is between A and C, the only force on the right of the section is the right reaction acting upward; and the shear at the section is the right reaction taken negatively. Hence, the line QH, being the influence line for the right re- action taken negatively, is the influence line for shear at C when the load of i Ib. is between A and C. Therefore the line QSNTO is the influence line for shear on the section at C; and shows that when a single load (whether i Ib. or 20 tons) crosses the span from B to A , the shear at C increases uniformly until the load is just at the right of C. When the load crosses the point C, the shear at C instantly becomes negative, increasing to zero when the load arrives at A . 86. Influence Line for Bending Moment in a Beam. The bending moment at any normal section of a beam is the alge- braic sum of the moments of all the forces acting on one side (either side) of the section, taken about the center of gravity of the section as an axis. When this sum or resulting moment SEC. H BRIDGES 133 .. ../o '---->< 75 -> is clockwise on the left of and about the section, the bending moment is positive. Obviously if the sum of the moments on the left of the section is clockwise, the sum of the moments on the right of the section is counter-clockwise. The influence line for the bending moment at section C (Fig. 98) is desired. When the load of i Ib. is between B and C, the only force acting on the left of the section is the left reaction at A ; and the bending moment at C equals 10 ft. times the left reaction. As the load moves from B to A, the left reaction increases uniformly from o to i Ib., and 10 ft. times the left reaction increases uniformly from o to 10 ft.-lb. Likewise the ordinate to the line OK } increasing uniformly from zero at O to 10 ft.-lb. at Q, represents 10 ft. times the left reaction. Hence the line ON is the influence line for the bending moment at C when the load of i Ib. is between B and C. The ordinates are positive, since the moment of the force on the left of the sec- tion is clockwise. When the load of i Ib. is be- tween C and A, the only force acting on the right of the section is the right reaction at B ; and the moment at C equals 15 ft. times the right reaction. As the load moves from A to C, the right reaction increases uniformly from o to i Ib.; and 15 ft. times the right reaction increases uniformly from o to 15 ft.-lb. Likewise the ordinate of the line QHj increasing uniformly from zero at Q to 15 ft.-lb. at 0, represents 15 ft. times the right reaction. Hence, the line QN is the influence line for the bending moment at C, when the load of i Ib. is between A and C. These ordinates are also positive, since the moment of the force on the right of the section is counter-clockwise. Therefore, the line QNO is the influence line for the bending moment at C; and shows that when a single load (whether i Ib. or 100 Ib.) crosses the span from B to A , the bending moment at C increases uniformly to a maxi- mum as the load arrives at C, and decreases uniformly to zero as the load arrives at A . 15* FIG. 98. 134 THEORY OF FRAMED STRUCTURES CHAP. IV 87. Criterion for Maximum Bending Moment. Consider a series of loads so connected that the distances between loads are maintained (Fig. 99) . As these loads move on the span, the bend- ing moment at C varies. It is de- sired to find the position of these loads on the span, so that the bend- ing moment at C will be a maxi- mum. The influence line QNO is drawn in accordance with the preceding article. The bending moment at C, for any given posi- tion of the loads, may be deter- mined by multiplying each load by its corresponding ordinate in the influence line diagram; and finding the sum of these products. Or, since the influence line has a constant slope from Q to N, and a constant slope from N to O; the bending moment at C may be expressed by the equation M c = P&z + Piyi in which PI = the sum of the loads between A and C PZ = the sum of the loads between C and B yi = the ordinate at the resultant PI and yz = the ordinate at the resultant PZ- Let the loads be moved any small distance d to the left; in such a way that no loads pass the points A, C and B. In other words PI and PZ remain constant while the movement takes place. The bending moment at C for this new position is M'c = P^(yz + yzz) + Pi(yi The change in the bending moment at C is AAf C = M'c M C = PzyZZ Has the bending moment at C been increased or diminished by this movement? The answer to this question depends upon whether As long as ^P, 10 in which case the bending moment at C will be decreasing. This particular load is called the critical load, for when it is at C, the bending moment at C is a maximum. i 3 6 THEORY OF FRAMED STRUCTURES CHAP. VI MOMENT TABLE 2-142 Ton Engines followed by 4000 Ib. COOPER'S CLASS E-40 per ft. 3 3- 3- 88 FIG. 100. SEC. II BRIDGES 137 Thus we see how an influence-line diagram may be used to develop the criterion for maximum bending moment at any section of a beam of any length. After the correct position of the loads has been determined, the maximum bending moment may be computed by scaling the length of the ordinate under each load; and taking the sum of the products of each load and its corresponding ordinate. 88. Cooper's Standard Train Loads. When the series of loads represents the wheel concentrations of a railway train, the bending moment may be more easily computed by the help of a "Momemt Table"; as shown in Fig. 100. This table is a collection of data for what is known as a "Cooper's -40 Loading" ; consisting of two 142-ton locomotives, followed by a uniform train load of 4,000. Ib. per linear foot. Each axle for the large wheel supports a load of 40,000 Ib. A train is usually supported by two girders or trusses, one-half of the load coming upon each. For this reason, wheel loads or the loads supported by each half of the structure, instead of axle loads, are given in the table. The wheel loads are given on line c; wheel i supports 10,000 Ib.; wheel, 4, 20,000 Ib.; wheel 16, 13,000 Ib.; etc. Accumulative loads are given on lines a and b. The spacing of the wheels is given on line d, and the accumula- tive distances on lines e and/. The quantities given below line / are moments expressed in units of 1,000 ft.-lb. Suppose the moment is required of all the wheels to the left of, and about wheel 6. Follow the space between wheels 5 and 6 down to the heavy line; and then to the left, under wheel i, read 1,640. The moment of the wheels up to, and including wheel 3, about wheel 6 is 840. Verify these quantities by computation. The quantities below the heavy zigzag line are moments on the right of various wheels, instead of on the left. Cooper's Loadings are specified more extensively in American railway practice than in any other. They are divided into classes and are specified as Cooper's -40, -50, -55, etc. The wheel spacing in all classes is the same. A constant ratio exists between the loads of one class and the corresponding loads of any other class. Thus the loads of a Cooper's E-6o 138 THEORY OF FRAMED STRUCTURES CHAP. IV are 50 per cent greater than those for an -40, given in Fig. 100. If an E-6o is specified, the table of Fig. 100 may still be used, if the final results are increased by 50 per cent. 89. Illustrative Problem. Determine the maximum bending moment for an -40 load- ing at the center of a 62-ft. span. Draw the influence line dia- y gram and develop the criterion A n OO@Q o o pom for maximum bending at point kv-3/' ..... -^----^----J - c (Fig. 101). When developed, FlG ' I01 ' this criterion will show that, as the train comes on the span at B, the bending moment at C will increase as long as the loads on AC are less than one-half the total load on the span. The influence line shows that as many loads as possible should be placed on the span, with the heavy loads usually (not always) near C, where the ordinates are long. Try wheel 4 at C. Wheel 4 is 18 ft. from the head of the train, hence the length of train on the span is 18 + 31 = 49 ft. Wheel 9, being 48 ft. from the head of the train, is i ft. on the span. 2)142 = total load on the span 71 = one-half the load on the span 50 = load on AC when wheel 4 is approaching C 70 = load on AC when wheel 4 has passed C. In either case the load on A C is less than one-half the load on the span, and the bending moment at C continues to increase as wheel 4 crosses C. Try wheel 5 at C. Make a sketch similar to Fig. 101 for this, and each succeeding case. In this arrangement the large wheels are not so well located about C; but the loads as a whole are nearer the center of the span. 2)142 7 1 > 70 when wheel 5 is approaching C <9339 increasing Wheel 5 i->344 a maximum Wheel 6 1,213 decreasing Let us continue the investigation further. Try wheel 7 at C. 2)162 8 1 < 93 decreasing 79 increasing 2)182 \ a maximum 91 <92 decreasing Since the bending moment was decreasing when wheel 8 had passed C, and was increasing when wheel 9 was arriving at C; there must have been some intermediate position, at which the bending moment was a minimum. This position occurs when wheel 3 leaves the span at A . Wheel 3 at A. 2)182 91 <99, wheel 3 approaching A, decreasing j 2)162 > a minimum 8i> 79, wheel 3 passing A, increasing Wheel 10 at C. 2)142 7 1 > 5 2 1 increasing >62 J Wheel 4 at A 2)182 91 <92 decreasing 1 2)162^ > a minimum. 8i>72 increasing Wheel ii at C. 77.5>49 \ . >6 9 /^creasing 142 THEORY OF FRAMED STRUCTURES CHAP. IV Wheel 12 atC. 77-5>5 6 J increasing Wheel 13 at C. 2)168 84 > 76 increasing 2)155 a maximum. 7 7. 5 < 83 decreasing Wheel 14 at C. 78 5 < 83 decreasing 2)144 72 jP tt This criterion is often expressed in the form (2) -.-Fr (3) a I I a and may be stated as follows: A maximum bending moment at any point occurs whenever the train is so placed that the load on each segment, divided by the length of the segment, equals the total load on the span, divided by the span length. In other words, the average load per foot on each segment, equals the average load per foot on the span. In interpreting Eqs. (2) and (3), PI represents the load on the segment a plus a portion of the load at C; the remainder of the load at C being considered as a part of P 2 . If Eqs. (2) and (3) are satisfied when no load is at C, any movement of the train will cause no change in the bending moment at C as long as PI and PI remain constant. 92. The Point of Greatest Maximum Bending Moment in a Beam. By comparing the bending moments at the different 144 THEORY OF FRAMED STRUCTURES CHAP. IV points of the loo-ft. girder in the problem of Article 90, the student may conclude that the maximum bending moment at the center of any beam is greater than at any other point. Such a conclusion is usually erroneous. The maximum bending moment at the center of the 62-ft. span, previously considered, was 1,371.5 ft.-lb., when wheel 13 was at the center and wheel 18 was i ft. on the span. When wheel 13 is 32.3742 ft. from the right end of the span (Fig. 102), the bending moment under wheel 13 is 1376.2 ft.-lb., which is greater than the bending moment at the center. Let X (Fig. 103) represent the point of greatest maximum bending moment in a beam. If the influence line for bending o O O o O o o n 1 1 *.-- -29.6258= b--> \< -27.2516 '--->! o o OCX 5>O n o o i t -29.6258=a--> \< --32.3742 ' > FIG. 102. < -23742 t-a FIG. 103. moment at X is drawn, and the criterion for maximum bending moment developed, we shall have a P as the wheel at X is considered on the right or the left of X . Also, if the bending moment is a maximum at X, the shear changes from positive to negative at X. Let R represent the left reaction, then as the wheel at X is considered on the right or the left of X. Let b represent the distance from B to the center of gravity of all the loads on the span; then *-r? - , therefore SEC. II BRIDGES 145 a < b whence 7 P > j P or a > b as the wheel at X is considered on one side or the other of X. Obviously, when the wheel is at X a = b When wheel 13 is at the center of the span (Fig. 102), wheel 8 is over the left support, and consequently not on the span. 155)4,224 27.2516 The center of gravity of wheels 9 to 18 inclusive is 27.2516 ft. from wheel 18; or 2.7484 ft. to the right of wheel 13. 62. 2.7484 2)59.2516 = a + b 29.6258 = a = b The center of gravity of the total load is 29.6258 ft. from the right end of the span; and the point of maximum bending moment is under wheel 13, which is 29.6258 ft. from the left end of the span. The maximum bending moment is 1376.2 ft.-lb. The point of maximum bending moment is at the center of the span only when the center of gravity of the wheels on the span coincides with the wheel which causes the maximum bending moment at the center. Otherwise the point of maxi- mum bending moment is not at the center; but near the center, under the wheel which causes maximum bending moment at the center; when that wheel is as far from one end of the span as the center of gravity of the loads on the span is from the other end. 93. Problems. A road roller is supported on two axles, 9 ft. apart. The load on the front axle is 10,000 Ib. The load on the rear axle is 20,000 Ib. Compute the maximum bending moment (a) in a span 21 ft. long: (b) in a span 13 ft. long. What is the greatest maximum bending moment in a loo-ft. span for an -40 train? For an E-6o train ? 146 THEORY OF FRAMED STRUCTURES CHAP. IV 94. Criterion for Maximum Shear in a Beam. By referring to the influence line for shear (Fig. 97), it is apparent that the positive shear at C increases from zero to a maximum, as wheel i comes on the span at B and moves to C. As wheel i crosses the point C, the shear at C is instantly diminished by the amount of wheel i. Evidently a maximum shear occurs as each successive wheel arrives at C, followed by a minimum shear, as the wheel crosses C. When wheel i is at the center of a loo-ft. span, the shear at the center equals the left reaction, or 37.8. This shear is instantly decreased to 27.8, as wheel i crosses the center. When wheel 2 comes up to the center, the shear is 49.36 10 = 39.36; instantly decreasing to 19.36 as wheel 2 crosses the center; and increasing to 26.96 as wheel 3 arrives at the center. In the ordinary beam or deck plate- girder span, for which the influence for shear in Fig. 97 is typical, the maximum positive shear on any section in the left- half of the span will usually (not always) occur when wheel 2 is at the section. 95. Problem. Compute the maximum positive shear for an -40 loading for sections 10 ft. apart, from the center to the left end of a loo-ft. deck plate-girder span. SEC. III. PARALLEL CHORD TRUSSES 96. A railway truss bridge is shown in Fig. 104, to illustrate the manner in which the wheel loads of a train are carried to and supported by the piers at either end. The wheels rest upon the rails which are supported by cross- ties. The ties rest on longitudinal stringers which are supported by cross- beams, more often called floor beams. The floor beams are supported by the truss at the bottom chord panel points. It should be thoroughly understood, and constantly kept in mind, that the truss does not directly support any wheel loads. The truss receives the weight of the train only in the form of floor beam loads, delivered to the truss at the panel points. 97. Influence Line for Maximum Stress in a Chord Member. Let the truss in Fig. 105 be loaded in any manner, and let Mz represent the algebraic sum of the moments about Lz SEC. Ill BRIDGES 147 of all the external forces acting on one side (either side) of the section through the panel 1-2. Since the stress in UiUz varies as Mz, it is clear that the stress in UiUz is a maximum, when U, 50 1 \ LO L, i PIG. 104. Mi is a maximum. For this reason we shall draw the influence line for Mi as a load of i Ib. moves across the span. When the load of i Ib. is between L 4 and Lz, the only force acting on the left of the section is the left reaction R , and Mz = $oR . As the load moves from L to Lo, RQ increases uniformly from o to i Ib., and 5cxRo increases uniformly from o to 50 f t.-lb. Hence the line ON is the influence line for Mz when the load of i Ib. is between L and Lz. When the load of i Ib. is between Lo and Li, the only force acting on the right of the section is the right reaction Rt, and Mz = 5 or Pi a I- a and may be stated as follows: A maximum stress in any chord member of a Pratt or Parker truss; or of any other truss in which the influence line is a triangle (as in Fig. 107), occurs whenever the train is so placed that the average load per foot on each of the two segments into which the center of moments divides the span, equals the average load per foot on the span. If the trusses in Fig. 107 have six equal panels, one-sixth of the total load on the span is placed on the segment o-i for maxi- mum stress in LJL*\ one- third is placed on the segment 0-2 for maxi- mum stresses in UiU 2 and L 2 L 3 ; one- half is placed on segment 0-3 for maximum stress in U^U^. 100. The stress in a web member of a parallel chord truss is usually determined by passing a section through the web member and two chord members, in such a manner that the truss is cut into two por- tions. The shear, or algebraic sum, of the vertical forces acting on one side (either side) of the section, equals the vertical com- ponent of the stress in the web member. 101. Influence Line for Shear in a Panel. The criterion for the vertical component of the maximum stress in U\L2 - Piyi The ordinate y z will have a plus or minus sign as the resultant P 2 is on the right or left of N. Now let the loads move a small distance d to the left, in such a way that no loads cross the panel points L , LI, L 2 or L. FIG. 109. Or in other words, PI, P 2 and P 3 are to remain constant. The change in shear in the panel is AF = P 3 ;y 33 - P 2 }> 22 + PI^U. If y z had been a negative ordinate, it would have been lengthened by the movement, an amount ;y 22 . In either case the shear is decreased and ;y 22 is negative. Is this change in F, caused by a movement of the loads to the left, a positive or a negative quantity? The answer depends upon whether for as long as P 2 ;y 22

I This criterion is not a function of either a or c, and conse- quently is applicable to any panel of the truss; when P 2 is the load in the panel, P is the total load on the span, b is the length of the panel and / is the length of the span. The criterion is often expressed in the form , ,._, PI P T = T and may be stated as follows: The maximum stress in any diagonal of a parallel chord truss (Pratt or Warren) occurs when the load in the panel divided by the panel length, equals the load on the span divided by the span length. If the truss in Fig. no contains eight equal panels, it is clear that one-eighth of the total load on the span should be placed in any panel for maximum stress in the diagonal of that panel; the train approaches from the right for positive shear and from the left for negative shear. In a through bridge, having the floor beams at the bottom chord, the maximum stress in a vertical member, UzL^ for example, equals the maximum shear in panel 2-3, for that portion of U^Lz between the top of the floor beam and U%. The stress in that part of UzLi (if any exists) between the bottom of the floor beam and the end connection differs from the stress above the floor beam by the amount of the floor beam load. From similar triangles c a -\- c I ' I NV':b SEC. Ill BRIDGES 159 therefore NV : c : : b : a + c NV:VO ::b:a + c NViNV + VO ::b :b + a + c _ NO ~" 1 P* P but since ~r = T o I P* P NV^NO Thus it is clear that, when the right portion of the span is loaded for maximum positive shear in any panel ; the load in the panel divided by the length NV equals the total load in the span divided by the length NO. In other words, if the truss has n equal panels and the influence line OTNSQ is drawn for shear in any panel NV i NO~ n Similarly it may be shown that NU i NQ ~ n The influence line NTO, for positive shear in any panel, has the same general characteristics as an influence line for bending moment at the point F in a beam having the length NO; the criterions are the same and the positions of the train are the same in both cases. The position of the train for maximum negative shear, in any panel, is the same as for maximum bend- ing moment at U in a beam having the length NQ. 105. Impact. In computing live load stresses in a bridge, the train is considered as a static load, gently placed upon the structure in the required position for maximum stress in a given member. In order to provide for the increased stress caused by dynamic effect of the train in motion, an additional stress known as impact is combined with each live load stress. The amount of impact to be added is determined arbitrarily from an empirical formula. The one most frequently used is T - T 3 L ,L/ , , 300 + / i6o THEORY OF FRAMED STRUCTURES CHAP. IV in which 7 = impact stress L = live load stress / = length of train causing the live load stress. For example, the live load stress in U\Uz (Fig. 106), as deter- mined in Article 98, is 100.6. The impact stress is 100.6 X -: = 75.4 compressive 300 + 100 The live load tension stress in U\L^ (Fig. 108), as determined Ui U 2 U 3 U 4 Us U & U 7 L I* t L ' Lj L > I , Ls I Ll [^.. QQ)25^?00 >J FIG. in. in Article 103 is 60.6. The impact stress is 60.6 X . = 50.2 tensile 300 + 63 The live load compression stress is 18.1, and the impact is 18.1 X : = 16.3 compressive 300 + 33 The live load stress in U\L\ is 75.6. The length of the train causing this stress extends over two panels only, hence the impact stress is 75.6 X = 66.1 300 -1- 43 106. Stresses in a 2oo-ft. Pratt Truss. (Fig. in). The assumed dead loads are as follows: Track (rails and ties) 450 Floor (floor beams and stringers) 550 Two trusses and bracing i , 200 2,200 Ib. per linear foot The live load is a Cooper's -40 train. The impact will be taken in accordance with the formula of Article 105. Dead-load Stresses. The dead load supported by each truss is 1,100 Ib. per linear foot, or 27.5 per panel. The stresses SEC. Ill BRIDGES l6l may be determined by placing a load of 27.5 at the seven joints LI to LI and drawing a stress diagram; but in the case of parallel chords the algebraic method is considerably shorter. One-half the load per linear foot on each truss is 550 Ib. (see Article 59) . 0.550 X i X 25 X 7 X 25 = i 6g 35 9.82 X 2 X 6 = 117.8 = 9.82 X 3 X 5 = 143-7 = 9.82 X 4 X 4 = 157.1 = U*Ut The length of each diagonal is 43 ft. The shear in panel 3-4 is one-half a panel load, or 13.75 13-75 X ^ = i6-.9 = 16.9 X 3 = 50.7 = 16.9 X 5 = 8 4-5 = 16.9 X 7 = 118.3 = 13.75 = 13-75 X 3 = 41-3 = 27-5 : o = It is more accurate to consider one-half the weight of the truss as concentrated at the top chord panel points. This would affect the stresses in the vertical members only, increas- ing the stress in UzLz, U^Lz and UtL and decreasing the stress in UiLi by 7.5 Live load and Impact Stresses LoL 2 Wheel 4 at LI 16,364 284 X 84 = 23,856 84 2 = 7,056 8)47,276 480. 35)5,429-5 LL = 155.1 / = 155.1 X ^ = 94-5 ii 1 62 THEORY OF FRAMED STRUCTURES CHAP. IV UiUz and L 2 L 3 Wheel 7 at L 2 16,364 284 X 78 = 22,152 78 2 = 6,084 4)44,600 11,150 : 2,155 /= 257X^=158.2 UzU 3 and L 3 L 4 Wheel n at L s 16,364 284 X 80 = 22,720 85 2 = 6,400 45,484 o.375 17,056-5 5,848 35)11,208.5 LL = 320.2 1 = 3 20 ' 2 X = '96-5 Wheel 13 at L 4 16,364 284 X 65 = 18,460 65 2 = 4,225 2)39,049 19,524-5 7,668 35)11,856.5 LL= 338.8 7 = 338.8 X = 214.5 SEC. Ill BRIDGES 163 The stresses in all chord members of the left half of the truss are a maximum for positions of the train when crossing the span from right to left. This is not a general statement, to be accepted for all trusses, although it is true for this truss. The chord members of the right half will have their maximum stresses when the train crosses from left to right. The stresses for all web members will be computed for the train moving from right to left. The length of any diagonal member is 43 ft. Hence if v = the vertical component of the stress in a diagonal and s = the stress in the diagonal then s = -- X 43 O or s = 1.232; Wheel 2 at L 7 1,640 103 X i = 103 200)1,743 8.7 - 25 5^5 X 1.23 = 6.8 = LL U&LQ and L 5 Ut Wheel 2 at L 6 4,632 152 X 2 = 304 200)4,936 24.7 3-2 21.5= LL -: -/ 164 THEORY OF FRAMED STRUCTURES CHAP. IV U 5 L 5 and L,U 5 Wheel 2 at L 5 8,728 232 X 4 = 200)9,656 48-3 3-2 45.1 = LL = 35-3 =/ 45.1 X 1.23 = 55.4 = LL 35.3 X 1.23 = 43.4 = / and *7 3 L 4 Wheel 3 at L 4 16,364 284 X 4 = i,i3 6 4 2 = _ 16 200)17,516 87.6 78.4 =LL 200 78.4X^ = 57. =/ 78.4 X 1.23 = 96.4 = LL ] \ ^ 57. X 1.23 = 70.1 = I I i and 7^3 Wheel 3 at L 3 16,364 284 X 29 = 8,236 2 9 2 = 841 200)25,441 tf,t, L \ 80.9 X 1.23 = 99.6 = I 118. X 1.23 = 145.2 = LL ^ v L SEC. Ill BRIDGES 165 UiL 2 Wheel 3 at L 2 16,364 284 X 54 = 15,336 54 2 = 2,916 200)34,616 9.2 163.9 X 1.23 = 201.6 = LL 201.6 X = 130.6 = / 463 LUi Wheel 4 at Li 16,364 284 X 84 = 23,856 84 2 = 7.056 200)47,276 236.4 480 217.2 X 1.23 = 267.2 = LL 267.2 X = 162.5 = I 493 tfiLi Wheel 4 at L l Moment of wheels i to 8 about 8 = 2851 Moment of wheels i to 4 about 4 = 480 2,851 480 X 2 = 960 25)1,891 75.6 = LL 75.6 X = 66.1 = 7 343 When the train crosses the span in one direction, it is obvious that the stress in any web member of the left half is the same as the stress in the corresponding member of the right half, when the train crosses in the opposite direction. Considering one- half of the structure, we shall have the various combinations of i66 THEORY OF FRAMED STRUCTURES CHAP. IV stresses, as listed in Fig. 112. When the dead and live load stresses are opposite in character, only two-thirds of the dead- load stress is considered. Thus in U^L^ the total stress is + 183.4 when the train crosses from right to left, and 87.5 when the train crosses from left to right. In UiLz the dead-load tensile stress is greater than the live- u ' D=-II7.S L - -2510 I ---158.2 -533.0 FIG. 112. U, -533 U? -560.4 Ik -7/0.4 L, *3/8.3 l z +533 L 3 ^660.4 L 4 FIG. 113. load and impact compressive stresses, and there is no reversal. 107. Counters. The diagonals in two of the panels in Fig. 112 must be designed to resist both tensile and compressive stresses. If eye-bars or other flexible members, which are not capable of taking compressive stress, are used; it will be neces- sary to introduce counters in panels 2-3 and 3-4, as shown by the dotted lines in Fig. 113. SEC. Ill BRIDGES 167 A counter carries no stress, except when the shear in the panel is of such a nature that the stress in the main diagonal would be compressive if the counter were omitted. The tensile stress carried by the counter in a parallel chord truss, obviously has the same magnitude as the compressive stress in the main diagonal if no counter were used. Thus in Fig. 113 the maxi- mum stress in UsLi is + 183 . 4, and the maximum stress in LzUi is +87.5. The maximum compressive stress in U^L* has the same magnitude as the maximum tensile stress in UzLz, when no counters are used. When the bridge is loaded for maximum chord stresses, the counters are not in action and consequently have no effect upon the maximum chord stresses. The Pratt truss in Fig. 112 may be transformed into a War- ren truss, by substituting the diagonal L z Uz for the diagonal U%Lz. The stress in L^Us will be 295 . 5 or + 14 . 8 ; the stress in U 2 Uz will be the same as in UiU^ the stress in L z Lz will be the same as in L$L, the stress in U^L^ will be the same as in UL\ and the stress in UzLz will be the same as in UiL\. 108. Truss with an Odd Number of Panels. The bottom chord in the center panel of a truss having an odd number of panels requires a special consideration. The truss in Fig. 114 is loaded for the maximum moment at L 4 . The moment at L 4 is 9,088.57 ft.-lb., and the moment at 3 is 8,929.43 ft.-lb. Since the moment at L 4 is greater than the moment at L 3 , the shear in the panel 3-4 is positive, and the diagonal U^Li is acting. The shear in panel 3-4 equals the left reaction, minus the loads from wheel i to wheel 9 inclusive, minus that part of the load in the panel which is carried at L 3 , or F = 167.57 - J 42 - 19-2 = 6.37 The shear may also be found by dividing the difference in moments at L 4 and Lz by the panel length, thus p = 9,088.57 - 8,929.43 = The truss in Fig. 115 is loaded for the maximum moment at LZ. The moment at LZ is 9,155.86 ft.-lb., and the moment at L 4 is 8,944.14 ft.-lb. Since the moment at L 3 is greater than i68 THEORY OF FRAMED STRUCTURES CHAP. IV the moment at Z, 4 , the shear in the panel 3-4 is negative and the diagonal Z/ 3 Z7 4 is acting. The shear in panel 3-4 is F = 200.05 172 36.52 = 8.47 p = ^,944.14 - 9,155.86 = _ g or PIG. 115. PIG. 116. The maximum stress in UzUi is 9,155.86/34 = 269.3. This stress occurs when the train crosses the span from right to left with wheel 1 1 at L$ ; or when the train crosses from left to right and wheel 1 1 is at L 4 . Hence the maximum stress in U z Uz, UzU* and U*U 6 is -269.3. SEC. IV BRIDGES 169 The stress in L*L is obtained by dividing the moment at U z in Fig. 114, or the moment at Z7 4 in Fig. 115, by 34; but in either case the moment at the other end of the panel is greater. It is clear, therefore, that the maximum stress in L 3 L 4 occurs when the moments at L s and L 4 are equal, or when the shear in the panel is zero and neither diagonal is acting. The total load on the span in Fig. 114 is 364. 364 _. T " The load in panel 3-4 is less or greater than one-seventh of the load on the span as wheel 13 is crossing Z, 4 , therefore the shear in the panel is a maximum (not the maximum to be sure, but still a maximum). There are three intermediate critical positions of the train between those shown in Figs. 114 and 115 wheel 14 at L 4 , wheel 10 at Z, 3 and wheel 15 at Z, 4 and for each position the shear in panel 3-4 is decreasing, for the load in the panel in each case is greater than one-seventh of the load on the span. Hence, as the train moves from the position in Fig. 114 to the position in Fig. 115, and the shear in panel 3-4 decreases from +6.37 to 8.47; there is one, and only one, position of the train for which the shear in the panel is zero. In the present case the shear in panel 3-4 is (approximately) zero when wheel 10 is at L$ (Fig. 116) and the moment at either U^ or / 4 is 9,048.5 ft.-lb. Hence, the maximum stress in LzL is +266.1. This stress is slightly less than the stress in UzUtj and in practice is usually assumed the same as the stress in U^Ui. SEC. IV. PARKER TRUSSES 109. Stress in Web Member of Parker Truss. It was shown in Article 99 that the criterion for maximum stress in any chord member of the Parker truss (Fig. 107) is the same as for the corresponding member of the Pratt truss. It is true also that the criterion for maximum shear in any panel of the Parker truss is the same as for the corresponding panel in the Pratt truss. In a consideration of the maximum stresses for web members, a clear distinction should be made. In the 1 7 o THEORY OF FRAMED STRUCTURES CHAP. IV Pratt truss, the stress in any diagonal is proportional to the shear in the panel; and therefore is a maximum when the shear in the panel is a maximum. In the Parker truss, the stress in any diagonal C/iL 2 , for example, is not proportional to the shear in the panel, for the vertical component of the stress in UiUz enters into the solution. no. Influence Line for Web Member of Parker Truss. If FIG. 117. the truss in Fig. 117 is loaded in any manner, the stress in may be determined by passing a section through the panel 2-3, and balancing the moments of all the forces acting on one side (either side) of the section about the intersection of the two chord members cut by the section. Let M I represent the alge- braic sum of the moments about / of all the forces acting on one side (either side) of the section through the panel 2-3, as a load of i Ib. moves across the span. When the load of i Ib. is between L 8 and L$, the only force SEC. IV BRIDGES 171 acting on the left of the section is the left reaction R , and MI = iSoR Q . As the load moves from Z, 8 to L , R increases uni- formly from o to i lb., and iScxfto increases uniformly from o to 1 80 ft.-lb. Likewise the ordinate to the line OK, increasing uniformly from zero at O to 180 ft.-lb. at Q, represents iSoR . Hence, the line OT is the influence line for M z i8o7?o, when the load of i lb. is between L 8 and L 3 . M l for the force acting on the left of the section is counter-clockwise about /, and for this reason QK is laid off above the line OQ. When the load of i lb. is between Z, and L^ the only force acting on the right of the section is the right reaction R$ and Mj = 42oR 8 . As the load moves from L Q to L 8 , R& increases uniformly from o to i lb. and 420^3 increases uniformly from to 420 ft.-lb. Likewise the ordinate to the line QH, increasing uniformly from zero at Q to 420 ft.-lb. at 0, represents 42cxR 8 . Hence the line QS is the influence line for M\, when the load of 1 lb. is between L and L 2 . M I for the force acting on the right of the section is counter-clockwise, consequently M x for the forces acting on the left of the section is clockwise, and for this reason OH is laid off below the line OQ. TV = 1 80 X | = 112.5 8 2 US = 420 X - = 105 8 When the load of i lb. is between LI and L 3 , there are two forces acting on either side of the section. On the left, there are the truss reaction R at L , and the floor-beam load r^ at L 2 ; on the right of the section, there are the truss reaction R$ at L 8 , and the floor-beam load r% at L 3 . Let the load of i lb. be in the panel 2-3 at a distance x from Z, 3 , and consider the forces acting on the left of the section. * '50 + * rz = _* 240 30 Mi = iScxRo 2407-2 ,, 180(150 + x) 240X Ml - - - 240 30 172 THEORY OF FRAMED STRUCTURES CHAP. IV This equation, being of the first degree, may be represented by a straight line. When the load of i Ib. is at 3, x = o and M i = 112.5. When the load of i Ib. is at L 2 , x = 30, and M T 105. Therefore, TS, having the ordinates TV = 112.5 at Z, 3 , and US = 105 at L 2 , is the influence line for Mi when the load of i Ib. is in the panel 2-3. The stress in U^Ls is Mj -f- 216. The stress caused by the load of i Ib. in any posi- tion, may be found by dividing the corresponding ordinate in the influence line OTNSQ by 216. As the load of i Ib. moves to the left from L 8 , the tension in Z7 2 Z,3 increases to a maximum, as the load arrives at L 3 . As the load moves forward, the ten- sion decreases rapidly and the stress in UzLs becomes zero, when the load arrives at the point directly above ^V. As the load passes N, the member resists a compressive stress increas- ing to a maximum as the load arrives at L 2 ; after which the compression decreases to zero, as the load leaves the span at u Since IL :IL S ::KQ:OH it is evident that the two lines OK and HQ produced, intersect at / on the vertical through 7. This influence resembles some- what the influence line for shear in the panel 2-3. If the length of either member Z7 2 L 2 or U^L^ were changed so that UiU% approached a direction parallel to LzLs y the distance between / and LQ would approach infinity, and the lines KO and QH would approach a condition of being parallel. in. Criterion for Maximum Stress in Web Member of Parker Truss. The member U 2 L 3 (Fig. 117) will be used as an example. A train is assumed to be on the span. The stress in 7 2 L 3 is M 1 -f- 216. The value of M may be deter- mined by taking the sum of the products of each load and its corresponding ordinate in the influence line. The loads will be combined into three groups PI, P 2 and P 3 , corresponding to the three portions of the influence line QS, ST and TO. If yi, yz and y s represent respectively the ordinates under the cen- ters of gravity of PI, P 2 and P 3 , then Mj = P z yz P 2 ;y 2 - P&I. Let the loads move a small distance d to the left in such a SEC. IV BRIDGES 173 way that no loads cross the panel points L , L 2 , L 3 and L 8 . Or in other words PI, PI and P% are to remain constant. Is this change in M t , caused by a movement of the loads to the left, a positive or a negative quantity? The answer depends upon whether for as long as ^2^22 = eP> then e + ae + al + P <"' f> P -1- 7 P b men or b ( + . ?/* p ^~p 1 /p b -T 2 \cx* i vJL \ P ^ ^P _L 7P^ Hence the criterion for maximum tensile stress in Z7 2 Ls is This criterion will be satisfied when the critical load is at LZJ and a few loads are in panel 2-3. In the case of all ordinary train loads, this criterion will be satisfied before any loads pass L 2 , in which case PI = o and the criterion for maximum tensile stress in U-^Lz becomes The compressive stress in U^L^ is a maximum when M i has a maximum negative value, or when the left portion of the span is loaded. As the loads come onto the span at L and move to the right, the compression in U^Lz increases as long as ce + ae + al r if b ce + ae and _ (e + /)P 2 = (e + /)P 2 th e n ce + ae + a l + be Up 2 Criterions (i) and (2) are easily remembered if the following observation is made. In computing the maximum tension in Z/2^3, Mj is determined from the truss reaction R at LO, and the floor-beam load r 2 at L 2 , thus M; = eR - (e + a)r 2 (3) If the two chord members in panel 2-3 were parallel, this criterion for maximum tension in / 2 L 3 would be (4) Criterion (4) may be transformed into criterion (i) by insert- /> ing the fraction - - in the coefficient of P. The numerator e + a and denominator of this fraction appear as multipliers in Eq. (3). Likewise in computing the maximum compression in U^Ls, M z is determined from the truss reaction R& at LS, and the floor-beam load rs at L 3 thus M z = (e + l)R 8 - (e + a + b)r 3 (5) If the two chord members in panel 2-3 were parallel, the criterion for maximum compression in Z7 2 L 3 would be criterion (4); which may be transformed into criterion (2) by inserting e + / the fraction - : i in the coefficient of P. The numerator e + a + 0. and denominator of this fraction appear as multipliers in Eq. (5)- Suppose (in Fig. 118) there are eight equal panels. For maximum stress in J7 2 L 3 , the proportion of total load on the span which should be placed in the panel is % times a fraction. The numerator of the fraction is the distance from the center SEC. IV BRIDGES l8l of moments to the truss reaction, and the denominator is the distance from the center of moments to the floor-beam load. Thus in Fig. 117 the criterion for maximum tension in UiLz is 8 ^ 180 + 60 or * 2> * as given on page 166. The criterion for maximum compression in UiLz is p < i 1 80 + 240 > 8 180 + 90 or P 2 > 4 P 3 6 as given on page 167. From similar triangles (Fig. 118) ce ce a ( \ j\ therefore NV:c::b:a + c + - e _ . . al or whence NV:NV + VO::b:a + b + c H NV b e or / e + a substituting in criterion (i) If the criterion is expressed as an equation, then A I_ NV ~' = NO Thus it is clear that when the right portion of the truss is loaded for maximum tensile stress in t/VLs, the load in the panel divided by the length NV equals the total load on the span divided by the length NO. Likewise it may be shown that for a maximum compressive stress in UzLz, the left portion 182 THEORY OF FRAMED STRUCTURES CHAP. IV of the truss is loaded so that the load in the panel divided by N U equals the total load on the span divided by NQ. Compute the lengths NV and NU in Fig. 117 and show that the criterions developed therefrom are the same as previously given. 115. Tension in a Vertical when Counters are Used. The maximum tensile stress in Z7 2 L 2 (Fig. 117) for an -40 train with wheel 4 at L 2 is 68.7. The conditions are different if the FIG. 119. diagonal web members are designed to take tensile stresses only, as in Fig. 119, and the use of counters becomes necessary. The dimensions of the truss are the same as in Fig. 117. We shall assume that total dead load for the bridge is 2,600 Ib. per linear foot, or 1,300 Ib. to be carried by each truss; of which 950 Ib. will be considered as acting at the bottom chord and 350 Ib. at the top chord. The panel loads are 28,500 Ib. for the bottom chord and 10,500 Ib. for the top chord. The resulting dead load stresses for U^Li and U^L^ are shown in the figure. Let the train advance on the span at LQ until wheel 7 is at LQ. The live load compressive stress in U^Ls is 2,155 xx 42Q _ 240 SEC. IV BRIDGES 183 and the impact compressive stress is 17.45 X 777 = *S-SS OO I The live-load and impact compressive stresses in UP 3 Adding PI + P 2 to both sides of the inequality, and letting Pi + Pz + PS = P, we have, ->Pi o SEC. V BRIDGES 187 The critical wheel is at L 3 and the position of the train for maximum stress in Ui88 hence 24o> 194 and wheel 9 at L 3 will not give a maximum stress in UzU*. Try wheel 10 at L 3 . P - = 209.3 o Pi = go When wheel 10 is approaching L 3 , P 2 = 52; and when wheel 10 has passed L 3 , P 2 = 62, hence > J 94 20 9 .3 > and the criterion is satisfied. Wheel ii at L 3 gives >i88 Wheel 12 at L 3 gives 2I8 ' 3> < 2 55 Wheel 13 at Z, 3 gives 1 88 THEORY OF FRAMED STRUCTURES CHAP. IV It is evident that wheels 10, n and 12 satisfy the criterion, and the algebraic sum of the moments about L 4 of all the forces acting on one side (either side) of the section through panel 2-3 must be computed for each condition to determine the maximum. Wheel 10 at L s . The forces on the left of the section are the truss reaction R , the loads PI between L and L 2 , and the floor-beam load YI at L& resulting from the loads in panel 2-3. *- 24*36 -3,5.99 The moment of P\, about L is 830 QO X 58 = 5,220 6,050 The floor-beam load at Z/2 resulting from the loads in the panel is . = = 33-8 100^0 = 3 J >599 moment of P\ = 6,050 = 1,664 60)23,885 398.1 = compressive stress in when wheel 10 is at L^. It is evident that these computations may be simplified as follows : jJS v f . -.".-- 3)94,796 31,599 830 QO X 58 = 5,220 832 X 2 = 1,664 7,7U 60)23,885 398.1 SEC. V BRIDGES 189 Wheel ii atLz. 16,364 284 X 1 80 = 51,120 i8o 2 = 32,400 3)99,884 33> 2 95 116 X 52 = 6,032 561 X 2 = 1,122 9,309 60)23,986 399.8 = compress! ve stress in UiU* when wheel n is at L%. Wheel 12 at L$. 16,364 284 X 185 = 52,540 i85 2 = 34,225 3)103,129 34,376 2,851 129 X 5 1 = 6,579 503 X 2 = i, 006 10,436 60)23,940 399 = compressive stress in UzU* when wheel 12 is at L 3 . Wheel ii at L 3 evidently gives the largest maximum compres- sive stress in UiU*, although there is little difference in the stress for the three positions which give a maximum. 119. In Fig. 122 let Z,oL 2 = a, L 2 L 3 = ft, LJLi = b zuidLtLu = c. Let PI = the load on the segment 0-2, P 2 = the load in panel 2-3, and P = the total load on the span. Show that the criterion for maximum stress in UzU* is P > P! + 2P 2 / < I - c 120. Suppose that the sub-tie M^U^ is omitted, and the sub- strut L 2 M 3 is substituted. Let a, b and c represent the same THEORY OF FRAMED STRUCTURES CHAP. IV lengths as before. If PI = the load on the segment 0-3, P 2 = the load in panel 3-4, and P = the total load on the span, show that the criterion for maximum stress in L 2 Z,4 is 121. U 2 M 3 . The influence line is drawn for shear in panel 2-3, and is the same in all respects as the influence line for shear in the corresponding panel of Fig. 120. Hence the criterion is the same in both cases, or in other words one- twelfth of the total load on the span is placed in the panel for the maximum posi- tive or negative shear. The stress in U^Ms (Fig. 122) will not be the same as the stress in Z7 2 Z, 3 (Fig. 120), although the maximum shear in both cases is the same; for the two members do not have the same slope. The maximum positive shear (wheel 3 at Lz) is 222.9; the maximum negative shear (wheel 2 at L 2 ) is 13.2. 122. MsL4. The influence line will be drawn for the vertical component of the stress. When the load of i Ib. is at any point on the span except between L 2 and L, there is no stress in M sLa or MzU and the stress in M$Li is the same as the stress in Urz = and r 2 = o 300 25 The web members meeting at M z are shown in Fig. 123. Let V represent the vertical component of the stress in MsLt, for which we are drawing the influence line. The vertical com- ponent of the stress in UzM 3 is the shear in panel 2-3, which is RQ. Balance the moments of all the forces about Ut, thereby SEC. V BRIDGES IQI M " 60' FIG. 122. I Q2 THEORY OF FRAMED STRUCTURES CHAP. IV eliminating all the forces except the force at La and the vertical component at Z/4, whence the vertical component at U* is 257-3 i - = - TZ 5 2 Balance the vertical components V + r z = Ro + - r, 2 T/ r> I 200 + X X V = KQ -- TZ = - - -- 2 3 50 v _ 200 - 5* 300 Now let the load of i Ib. be in panel 2-3 at a distance x from L 4 , then D 200 + X 50 # tf 25 RQ = ~ 7-3 = - - 7 and ^2 = - - 300 25 25 The vertical component of the stress in UiM^ (Fig. 124) equals the shear in panel 2-3, which is R r 2 , and V + r z = R Q - r 2 + - r 3 2 7 _ 200 - * 300 Hence, when the load is at a distance x from 4, whether in panel 3-4 or panel 2-3, the vertical component of the stress in M sZ/4 is given by the equation v = 200 - $x 300 which may be represented by a straight line. When the load of i Ib. is 1,4, x = o and V = + - = TV 3 When the load of i Ib. is at L^ x = 50 and V = - = US 6 Hence the straight line TS is the influence line for the vertical component of the stress in M^L^ when the load of i Ib. is between 2 and L 4 . It is clear that the influence line OTNSQ is also the influence SEC. V BRIDGES 193 line for shear in panel 2-4 (Fig. 121) ; or in other words, it is an influence line for the vertical component of the stress in UzL*. Obviously then, the criterion for a maximum stress in M $Li (Fig. 122) is the same as for maximum stress in U 2 L^ (Fig. 121), likewise the maximum stress is the same in both members. In order to show this more clearly, we shall compute the maxi- mum tensile stress in UiLi (Fig. 121). Try wheel 5 at L. 23 16,364 200 284 X 114 = 3 2 >37 6 223 ii4 2 = 12,996 109 300)61,736 114 205.79 = ^o **? = 16.6 =r> 5 228 189.19 = shear in panel 2-4 284 6)512 70 85.3 > ~O.K. * < go 189.19 X = 295.5 = maximum tensile stress in If a section is passed through panel 3-4 (Fig. 122), it is evident that the vertical forces acting on the left of the section must balance. The vertical component of Mali* acting upward on the left portion of the truss is %r s . The forces acting upward on the left of the section are R and ^r 3 , and they are balanced by the floor-beam loads on the left of the section and the vertical component V in M$ L ; hence #o + - r 3 = r 2 + r 3 + V 2 When wheel 5 is at 4, wheel i is in panel 3-4 r z = o f3 = 33- 2 RQ = 205.79 hence V = R r 3 = 189.19 IQ4 THEORY OF FRAMED STRUCTURES CHAP. IV Thus it is clear that the tensile stress in M^L* (Fig. 122) is the same as in V^L^ (Fig. 121). 124. U 4 L 4 . The influence line for stress in U *Li, when the sub-members M zL s and MsU* are omitted, is OTUQ. The presence of the sub-members alters the influence line between L 2 and 4 . When the load of i Ib. is between Z, 2 and L 4 the sub-members take stress, and the stress in Z7 4 L 4 when the load of i Ib. is at L 3 is J^ Ib. compressive. It may be shown by the method used on previous occasions that the influence FIG. 124. line between panel points is a straight line, therefore OTUSVQ is the influence line for stress in 7 4 L 4 . Let PI = the load on Pz = the load on Pz = the load on L 3 L 4 P 4 = the load on LLz PS = the load on L^L^ P = total load on the span The criterions, when developed as in preceding cases, will show that the compressive stress in 7 4 L 4 will be increasing as the loads move to the left as long as add or as long as P + 6P 3 > 6P 2 + i2P 4 P 6 Two conditions of loading should be investigated. The train or as long as -^-\- P 3 > PI -f SEC. V BRIDGES 195 may advance from the right until a few loads pass L 6 , in which case PI, P% and P$ are zero, and the criterion becomes which is obviously the same as the criterion for UiM$. Or the train may advance until the compressive area of the influence line at S is covered. Since the compressive area at 5 is smaller than the tensile area at U, it is doubtful if the second position of the train will give a larger compressive stress than the first; although if large wheels are near L 3 and smaller wheels near L 4 , the difference in areas will not be so significant. For the first position of this train, wheel 3 at L 6 satisfies the criterion ft and the maximum compressive stress in Z7 4 L 4 is 140.94. For the second position of this train the criterion is \ + PP* + 2P, and the critical wheel will be at either L 3 or L&. Try wheel 3 at L 3 WHEEL 3 APPROACHING L^ WHEEL 3 PASSING La Pi = o Pi = o P 2 =30 P 2 = 50 P 3 =86 P 3 = 66 P 4 =36 P 4 = 36 PS = 39 p 6 = 390 P = 542 P = 542 P P - = 90.3 P 2 = 30 - == 90.3 P 2 = 50 P 3 = 86 2P 4 = 72 P 3 = 66 2P 4 = 72_ 176.3 > 102 IS6.3 > 122 In either case the compressive stress in 7 4 L 4 is increasing. Try wheel 4 at L 3 WHEEL 4 APPROACHING X 3 WHEEL 4 PASSING L 3 P P - = 92 P 2 = 50 - = 92 P 2 = 70 6 o P 3 = 66 2P 4 = 112 P 3 = _S9 2P 4 = 86 158 < 162 151 < 156 Ip THEORY OF FRAMED STRUCTURES CHAP. IV In either case the compressive stress in 7 4 L 4 is decreasing. Since the stress is increasing when wheel 3 passes 3, and is decreasing when wheel 4 arrives at L 3 ; there must be an inter- mediate position for which the stress is a maximum. Try wheel ii at Z, 5 WHEEL 1 1 APPROACHING Z, 6 WHEEL 1 1 PASSING L 6 . ,? p -=90.7 P 2 = 50 P 6 =90.7 P 2 = 50 P 3 = _66_ 2P 4 = 72 P 3 = 66 2 P 4 = 112^ 156.7 > 122 156.7 < l62 Wheel ii at L$ satisfies the criterion for maximum compres- sive stress. Pass a section cutting the members UzU*, M 3 / 4 , Z7 4 L 4 and L^L 5 and consider the forces acting on the left ot the section. The forces acting upward are the truss reaction R = 233.95, an d tne vertical component of the stress in MsUi = \r$ = 36.54. The forces acting downward are the sum of the wheel loads on the left of Z, 4 , which is 116, and the floor-beam load at L 4 due to the loads in panel 4-5, which is 22.44. Hence the compressive stress in Z7 4 L 4 is 132.05. This stress is somewhat less than that previously determined for wheel 3 at L 5 . For the maximum tensile stress in Z7 4 L 4 , the train approaches from the left and it is obvious from the influence line that the train will advance until the large wheels of the front engine are atZ 4 . Try wheel 3 at Z 4 WHEEL 3 APPROACHING L 4 WHEEL 3 PASSING Z 4 PI = 140 PI = 140 Pz =36 P 2 = 36 P 3 =86 P z = 66 P* =30 P* = 50 P 5 = o P 6 = o P = 292 P = 292 -- 48.7 P 2 = 36 ^= 48-7 P*= 36 P 3 = 86 2P 4 =60 P 3 = 66 2P 4 = 100 134-7 > 96 124.7 < 136 Wheel 3 at L 4 satisfies the criterion for maximum tensile stress in Z7 4 Z, 4 . Consider the forces acting on the right of the section through UzU*, M 3 U^ 7 4 L 4 and LJL 6 . The forces SEC. VI BRIDGES IQ7 acting upward are the truss reaction R^ = 58.39 and the floor beam load r$ = 9.2. The vertical component of MsUi = ^3 = 27.68 acts downward. Hence the tensile stress in UiL when wheel 3 is at L 4 , is 21.51. It may be interesting to investigate the maximum tensile stress in U^L* when the train, approaching from the left, has advanced until only a few loads are on panel 2-3; in which case PZ, PI and P$ are zero and the criterion reduces to Wheel 2 at 1,2 satisfies this criterion. R^ = 16.45, r * ~ 3- 2 the maximum tensile stress in UJL, for this position is 16.45 ~~~ 1.6 = 14.85; which is less than in the preceding case, when wheel 3 has advanced to L. 125. UJui. When the i Ib. load is at LI, the vertical com- ponent of the stress in M iL 2 causes a tensile stress of J^ Ib. in UzLz. Hence the influence line is a triangle with the apex at L 2 and a base line from L to L 3 . For maximum tensile stress the train comes on the span at L and one- third of the total load is in panel 2-3. The stress equals the algebraic sum of the moments about L of the truss reaction at L\i and the floor-beam load at L 3 , divided by the distance LoL 2 . In case the main diagonals are designed to carry a tensile stress only, and it is found that counters are required (for example in panel 4-6) the member LM 5 is added. The mem- bers LtMz and M&UQ then become the main diagonals. The diagonal U^M 5 becomes a sub-tie and the vertical component of its stress is one-half the stress in M^L 5 , and the member JW 6 L 6 is not acting at all. SEC. VI. THE PENNSYLVANIA TRUSS 126. Pennsylvania Truss. The influence lines for several members of a Pennsylvania truss are shown in Fig. 125. This truss is an evolution of the Baltimore truss, and bears the same relation to it that exists between the Parker and the Pratt trusses. i 9 8 THEORY OF FRAMED STRUCTURES CHAP. IV 127. U2U4. The influence line, the criterion and position of a train for maximum stress in this member are the same as for the corresponding member of the Baltimore truss in Fig. -I28)?5=300 [ -,-> FIG. 125. 122. The stresses differ only because the lever arms from L* to the members are different in the two trusses. 128. I^Ma. The section is passed through panel 2-3, and the chord members intersect at 7. In the Baltimore truss, one-twelfth of the total load on the span is placed in panel 2-3 for maximum tensile or compressive stress. In the Pennsly- SEC. VI BRIDGES 1 99 vania truss, the train covers the right portion of the truss, with the critical wheel at L% for a maximum tensile stress, and the moment about / of all the forces on the left of the section is M z = i37-5^o - I87-5/-2 Let PI represent the load in panel 2-3, and P the total load on the span; then the criterion, developed as in Article in or 114, is P 8 I2 187.5 or P 2 >o.o6iP The criterion may also be determined by the ratio, NV 14.6 = - - = 0.061 NO 239.6 For a maximum compressive stress the left portion of the truss is loaded with the critical wheel at 1/2, and MI = 437-5^12 212.573 and the criterion is P 2 ^ ^^ 12 212.5 or P 2 >o.i72P This criterion may also be determined by the ratio, NU 10.4 129. M S L 4 . As in the Baltimore truss, the stress in this member is the same as the stress in UiL when the sub-members M s Lz and MzU* are removed. This is clearly shown by the influence line. With the sub-members omitted, the maximum tensile stress in UiL occurs when the right portion of the truss is loaded and the critical wheel is at L 4 , and M l = 137-5^0 - 187.57-2 Let P 2 represent the load on L^L^ and P the total load; then the criterion is P - T I 37-5 P - 6 IS^ Or P 2 = O.I22P NV 27.8 ^- = - - NO 227.8 NV 27.8 Likewise - = - - = 0.122 200 THEORY OF FRAMED STRUCTURES CHAP. IV For a maximum compressive stress in U 2 L^ the left portion of the truss is loaded with the critical wheel at L 2 , and Mj = 437-5^12 - 237.57-4 Hence the criterion is P 2 <- 237.5 or P 2 <-5 = l6 6y q = ^^ = 2,866 lb. per linear foot 10.07 Hence it is clear that a uniform live load of 2,866 lb. per linear foot for one truss will cause the same maximum positive shear in panel 1-2 of the truss in Fig. 109, as an -40 train. The maximum negative shear in panel 1-2 for an -40 train was found to be 14,230 lb. Let q represent the intensity of the equivalent uniform live load which will cause the same negative shear, then q X area QSN = 14,230 area QSN = - 4.167 q = 3415 lb. per linear foot In Article 98 the maximum moment for the stress in UiUz (Fig. 106) for an -40 train was found to be 3,219,000 ft.-lb. The area of the influence line (shown in Fig. 105) is 1,250; hence the intensity of the equivalent uniform load for this member is q = = 2,575 Hx per linear foot 1,250 In Article 112 the maximum value for M z for the member Z7 2 s (Fig. 1 1 7) was found to be 24, 1 2 2, ooo ft.-lb. The area of the influence line NTO is 9,309; hence the intensity of the equivalent uniform load for maximum tension in this member is 24,122,000 q - = 2,591 lb. per linear foot. 9,309 SEC. VII BRIDGES 203 From a consideration of the examples just given, it is apparent that the stresses in a truss might be very quickly computed if a uniform load could be substituted for the -40 train load. The only hindrance to this substitution lies in the fact that the equivalent uniform load is not the same for all members of a truss. In fact, the equivalent uniform load for the tensile stress in a web member is not the same as for the compressive stress in the same member, as was clearly shown. 134. Triangular Influence Diagrams. Nearly all the influ- ence line diagrams which have been considered in this chapter are triangular as shown in Fig. 126. It is clear that the tri- angle ABC, in which l\ 20 ft., and / 2 = 80 ft., may represent 204 THEORY OF FRAMED STRUCTURES CHAP. IV the influence line for determining the following quantities for an -40 train by varying the altitude h. Let s = maximum positive shear in panel 1-2, truss (a) t = maximum tension in Z7iL 2 truss (a) u = maximum stress in LU\ truss (b) v = maximum shear in panel o-i truss (b) w = maximum stress in LoLz truss (b) x = maximum bending moment at , beam (c) y = maximum pier reaction at F, span (d) 0.6 Ib. for s 0.76 Ib. for / 0.8 Ib. for u then h = { 0.96 Ib. for v 0.53 Ib. for w 1 6 ft.-lb. for x i.o Ib. for y Since li and /2 are constants in all these cases, the criterion for a maximum for all quantities (s to y) is the same. This criterion places wheel 4 on the vertical through D, and by the use of the moment table, the value of each quantity may be found as given below: s = 81,982.5 Ib. / = 103,844 Ib. u = 109,310 Ib. v = 131,172 Ib. w = 72,417 Ib. x = 2,186,200 ft.-lb y = 136,637.5 Ib. Instead of using the moment table these quantities could have been determined by taking the sum of the products of each wheel load and its corresponding ordinate in the influence line diagram, when the proper value for h is taken. Hence it is seen that the sum of the products in each case is proportional to the corresponding value of h, as an inspection will prove; thus for example, in the case of 5 and u, 81,982.5:0.6: -.109,310:0.8 SEC. VII BRIDGES 205 Since /i and h are constants, it is clear that the area of the influence line diagram ABC for each quantity is also propor- tional to the corresponding value of h\ and from this it follows that each quantity (s to y) is proportional to the area of its influence line diagram; or in other words, the ratio between each quantity and its influence line area is constant. This constant ratio is the intensity of the equivalent uniform load which corresponds to any triangular influence line diagram for which /i = 20 ft., and 1% = 80 ft. Let q represent the intensity of the equivalent uniform load corresponding to an -40 train load. From the previous discussion it is clear that q is a function of /i and h and independent of h. 135. Table of Equivalent Uniform Loads. Table I 1 gives the values of q, the equivalent uniform load per linear foot per rail for an -40 loading, for various lengths of /i and / 2 in multiples oi 5 ft. The values of q will be found correct in most instances to the third significant figure, and are thus accurate to within i per cent. The corresponding values of q for any other Cooper's standard loading are directly proportional. The following example will be given to illustrate the use of the table. In Article 112 the maximum tensile stress in U to (r + y)d(f>. Hence the total strain (change in length) in the length ds is yd and the unit strain is ~ Let / ds represent the unit stress on the fiber KL ; and let E represent the modulus of elasticity. ' ds Let 7 = the moment of inertia of the cross-section about the neutral plane; then My J ' I Eydcf> My whence ^ = ^- ds I Mds d*= - w B Mds and C B , C = I d^> = I JA JA If the beam in its natural state is straight (not arched) and is properly designed, the curvature will be so slight that ds may be replaced by dx t allowing the integration to be made hori- zontally between A and B instead of along the path of the elastic curve. Then B Mdx = If the beam is homogeneous and has a uniform cross-section, E and 7 are constants, and the equation may be written thus: i r* * = Ei) A Mdx (I) The expression Mdx represents the area of the cross-hatched element in bending moment diagram. Hence the integral C B expression I Mdx is the area of the M-diagram between /A the ordinates RS and PQ, and if this area is divided by El, the quotient is the angle <. If M is expressed in inch-pounds, the area Mdx is expressed in inch 2 -pounds. If E is expressed in pound /inches 2 , and 7 in inches 4 ; then El is also expressed in SEC. I DEFLECTION OF BEAMS 211 inch 2 -pounds, and the angle is a ratio. In any practical beam is comparatively very small; hence, when the tangent CB (Fig. 127) is horizontal, the ratio < may be taken as the slope of the tangent AD. Likewise, when AD is horizontal, the ratio may be taken as the slope of the tangent CB. From this analysis the first principle may be deduced. 138. First Principle. // tangents are drawn through any two points on the elastic curve of a homogeneous beam of uniform cross-section, the angle which one tangent makes with the other tangent equals the area of the M.-diagram between the two points, divided by EL Now imagine that the unstrained position of the beam was in the direction AD, and that the beam was subsequently bent so that the point D moved to B\ the point A remaining station- ary. This movement is caused by the bending of all the ele- ments from A to B. The bending of the element EFGH causes the point, in its travel from D to B, to move a distance dt = xd(f>. Since the curvature is comparatively small, the path of the point moving from D to B deviates but slightly from the straight line DB. Hence f*B (*B T (*B t = dt= I xd4> = ^ Mxdx (2) JA JA EIjA The distance DB = t is called the tangential deviation; since it represents the distance through which the point B has been displaced by the curvature of the beam, when AD is assumed as the original position. In Eqs. (i) and (2), / is the gross moment of inertia of the cross-section. No deductions are made for holes, as is the case when the strength of a beam is being computed. The expression Mxdx represents the moment of the elemental area Mdx about the ordinate through B. Hence the integral Xjy 1 Mxdx represents the moment of the area QPRS about the ordinate through B, and is called the area-moment of QPRS about B. The area-moment is expressed in inches 3 - pounds, when M is expressed in inch-pounds. Since El is expressed in inches 2 -pounds, the tangential deviation t is expressed in inches. The second principle may now be stated. 212 THEORY OF FRAMED STRUCTURES CHAP. V 139. Second Principle. If the tangent to the elastic curve is drawn through any point A, the tangential deviation at any other point B may be obtained by finding the area of the M-diagram between ordinates through A and B; and dividing by El the moment of this area about the ordinate through B. Let / represent the centroid of the area QPRS, and let k be the distance from / to the ordinate through B. Then area El Since, from the first principle area QPRS ~W' "* then / = k(f> Hence the tangents to the elastic curve at any two points A and B intersect on the ordinate through the centroid of the M- diagram included between the ordinates through A and B. SEC. II. SIMPLE BEAMS OF UNIFORM CROSS-SECTION 140. The beam in Fig. 129^ is a 2 by i in. piece of wood laid PIG. i2pa. flatwise. / = % in. 4 E = 1,500,000 lb./m. 2 Hence El = 250,000 in. 2 -lb. The M-diagram is PQS. The deflection A under the load will be determined in several ways, by drawing SEC. II DEFLECTION OF BEAMS 213 the tangent to the elastic curve through different points as shown in Figs. i2ga-b-c. Considerable time and labor may be saved by exercising good judgment in choosing the most advantageous point in the elastic curve through which the tangent is to be drawn. In Fig. i2ga the tangent to the elastic curve ATB is drawn through T. The deflection A is quickly found after the tan- gential deviations t\ and t% have been computed. Where Mi is the bending moment at any distance x from the ordinate, on which the tangential deviation is required. Hence Mi = 45*. The origin for / 2 is at B, hence Mz = i$x, and 29,160 ~KT A = t\ -f- (fe h) = y ''~~ = y?/ = 0.039 in. 24 El 1,500,000 The expression [ Mixdx represents the area-moment of PQV about P, hence . = ^70X3X4)=^. .. Likewise, the expression ^M^xdx represents the area-moment of SQV about 5, hence >o,i6o / X == EI^ 7 X 9 X 12) Thus it is clear that, when the M-diagram can be conveniently divided into portions whose areas and centroids are easily found, a semigraphic or geometric solution can be quickly made. The area of the M-diagram to be considered in each case is included between two ordinates. One ordinate passes through the point of tangency, on the other ordinate the tan- gential deviation is found; and the moment of this area is taken about the latter ordinate. In Fig. 1296 the tangent is drawn through A. The area- 214 THEORY OF FRAMED STRUCTURES CHAP. V moment for / 4 is PQS about 5; and for / 3 , the area-moment is PQV about QV. E/270 X 3(18 + 2) = 16,200 Z7<" X 3 X ,) II,34O I,62O A = - - = ILL , f before. SEC. II DEFLECTION OF BEAMS 215 In the algebraic solution, the origin for / 4 is at S. M = i$x for values of x between o and 18, and M = 1500 60 (x 18) = i ,080 45#, for values of x between 18 and 24, hence f lji 29,160 + 16,200 _ 45,360 ~~W~ El The origin for / 3 is at F, hence M = 45 (6 x), and , _ 1,620 45 f 6 ^ 3 = 777 I ^* X ~ X ' 777- Eljo El The geometric solution is considerably shorter when M is not a continuous function of x as in the case of J 4 . In Fig. 129^ the tangent is drawn through B. ^["270 X 3 X 4 = 3> 2 4o] = 32,400 E/|_27o X 9(6 + 6) = 29,160] El - 3 7 . 24,300 - h ^j^- 4 rLL 24,300 14,580 0,720 ^ v? 141. Maximum Deflection. ^Let X (Fig. 129^) represent the point of maximum deflection. Since the tangent through X is horizontal, Amax = ti = / 8 - Let KL be the ordinate in the M- diagram at the point of maximum deflection, and let LS = a. Then KL = i$a. Let < represent the angle which the tangent through B makes with the horizontal tangent through X, then Z.BID = and 240 = / 5 * = 24 also = area E/ hence 7.5a 2 = 1,350 a = 13.42 2l6 THEORY OF FRAMED STRUCTURES CHAP. V Since the centroid of the triangular area KLS is on the ordinate through /, ID = -a 3 t s = -a(f> 3 35 \ _ El or -m [area-moment 0/1 _. KLS about S J El _ I2,C ~~EI~ 5# 3 _ 12,078 EJ = ~ET in. KLS afow/ The distance a might also be found by equating the values of / 7 and / 8 without any reference to the tangent through B. The general expression will be developed for the deflection of .a simple beam / in. long when supporting a single concen- TT~ FIG. 129^. trated load of P Ib. at any distance kl from the left support, (Figs. 130 and 131). The deflection A is found at T, a distance cl from the left support. When c < k. In Fig. 130 the tangent CD is drawn through T. equals the area-moment of KQSL about 6 1 which is the SEC. II DEFLECTION OF BEAMS 217 area-moment of PQS about 5, minus the area-moment of PKL about S. B Ci -- <-kl ><- FIG. 131 kl PI* (l - k)(2k ~ k* ~ C 2 + JL V / i j n o * -I o i A = -^^(2ck $ck z c 6 -\- ck 6 -\- When c > k (3) 2l8 THEORY OF FRAMED STRUCTURES CHAP. V In Fig. 131 the tangent CD is drawn through T as before. PP ** = 6EI PP A = / 8 + c(t* - t s ) p/3 A = Z(*ck - 3 c*k - k s + c*k + c&) (4) The deflection at the load may be obtained from either Eq. (3) or (4). Since c = k for this condition, either equation reduces to Let F represent the expression in the parenthesis of equation (3) when c is less than k, and the expression in the parenthesis of Eq. (4) when c is greater than k, then in general The values of F for various values of c and k are given in Table I; or may be found from Fig. 132. In Eq. (5) / is the length of the beam in inches, and P is the load in pounds at any distance kl from the nearer support. A is the deflection in inches at any distance cl from the same support. E is the modulus of elasticity in pounds per square inch, and / is the moment of inertia of the constant cross-section of the beam about the neutral axis, measured in inches 4 . 142. Point of Maximum Deflection. The maximum deflec- tion occurs in the longer segment of Fig. 131 where c is greater than k, and at the point where the tangent through T is hori- zontal; hence the value of c for A TOaa; . may be found by equating the expressions for / 3 and /4, whence c = i - i The values of c and the corresponding values of F for A maa! . are also given in Table I. Since the limits of k are o and 0.5, all values of c for maximum deflection will fall between 0.4227 and 0.5. Hence the point of maximum deflection for a single load is between the load SEC. II DEFLECTION OF BEAMS 219 and the center of the span, and always relatively near the center. The most eccentric loading which a simple beam of uniform cross-section and span / can experience, occurs when a fa/aes ofc FIG. 132. single load is adjacent to one of the supports, and k is on the point of becoming zero. Under this condition the point of maximum deflection cannot be at a distance greater than 0.07 73^ from the center of the span. Any second load applied to the 22O THEORY OF FRAMED STRUCTURES CHAP. V beam must necessarily throw the point of maximum deflection nearer the center. Hence the point of maximum deflection of a simple beam of uniform cross-section, loaded in any manner, will be near the center and not more than 0.0773 f its length from the center. 143. A 2o-in. 65~lb. I-beam supports two loads of 30,000 Ib. each (Fig. 133). Since the loads are symmetrically placed, the elastic curve and M-diagram are symmetrical about the center. The tangent to the elastic curve at the center, drawn FT through T, is horizontal and A = /. E = 29,000,000 lb./in 2 ., I = 1,169.5 in 4 .; hence El = 33,915,500,000 in 2 .-lb. Area-moment of PQSU about P. areaPQF 150,000 X 2.5 X 3.33 = 1,250,000 area QSUV 150,000 X 7-5 X 8.75 = 9,843,750 11,093,750 ft 3 .-lb. A = / = "Q93,75o X 1,728 = 6 in When the length of a beam is expressed in feet, and the loads are expressed in pounds, the area-moment will be expressed in foot 3 -pounds; and the factor 1,728 is introduced if El is ex- pressed in inch 2 -pounds. El may be expressed in foot 2 -pounds by dividing by 144, whence El = 235,521 ft. 2 -lb., then = = 11,093,750 ft 3 -lb. = ft in 235,521 ft. 2 -lb. The deflection at the center may be found from Table I. SEC. II DEFLECTION OF BEAMS 221 k = 0.2 and c = 0.5; therefore F = 0.071 for each load, then 2PP t 60,000(25 X I2) 3 0.07I A = - X 0.071 = - - = 0.565 in. 6EI 6 X 33,915,500,000 144. Deflection Under Uniform Load. The beam in Fig. 134 supports a uniform load and the M-diagram PQS is a PIG. 135. parabola. The maximum deflection is at the center of the span. The tangent to the elastic curve at C is horizontal, and A = /. Elt = the area-moment of SQV about S. The area SQV is two- thirds the area of the rectangle QPSV, and the centroid of the area SQV is five-eighths of VS, hence A - I - ^( I9 , 2 oo X 8 X | X 5 X 1.7,8) = "*ffi 222 THEORY OF FRAMED STRUCTURES CHAP. V o VO 00 M PO vo r^ oo Ch O 00 PO VO PO o oo 00 ON vo ON 00 M ^" l>- vo t^. oo PO M 6 O 6 o o 6 6 o o o 6 6 6 6 6 6 6 6 6 o o 6 6 6 6 o o 6 6 VO M o 00 tv. t^. VO o o M J> o o s * O * s 00 M vo PO O 00 PO VO t^. VO M O M PO 00 M O Th ON lO t> OO t^ o o 00 Tf o 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 PO POOO PO 00 M vO PO o 5 o 5 ^ 8 s ?, o o Q\ OO o" "o o o CS M o o o o o o o o o o o o o o o o PO 00 o O t~~ oo PO O vo o * PO M O PO Ov vo 6 8 6 8 o 6 6 o o 6 6 o o 6 6 o o 6 6 o o 6 6 o o 6 6 O 6 6 8 8 6 6 8 6 6 "02 x/ 0* O vo H M M M' as 3.3 as Z O vo < as /* o o o o o o o o o o o o o o O TJ- 10 PO oo cs O O 00 O IN t^ O^ O VO M r^ vo vo t^- vo o rh M 6 6 t 2 6 6 O c-O O ^- "1" . 6 6 M 00 ON vo M vo t 9 6 6 o o vo O vo r^ M PO * O "1" . O O SEC. II DEFLECTION OF BEAMS 223 In Fig. 135 the span is / in., and the uniform load is w Ib. per inch. The deflection at any distance cl from A will be found. At any distance x from either support, M = x (I x) . if - I Mxdx = (4c 3 24 f(l-c)l ^4 - 1 Mxdx = (i r jo 24 A = Let W = the total uniform load, then W = wl, whence A = ~=J(c - 2C 3 -f- c 4 ) (6) Let / represent the expression in the parenthesis of Eq. (6) then WP A = (7) The values of / for various values of c are given in Table II; in which / is the length of the beam in inches, W is the total uni- formly distributed load in pounds and A is the deflection in inches at any distance cl from the nearer support. E is the modulus of elasticity in pounds per square inch, and / is the moment of inertia of the constant cross-section of the beam about the neutral axis, expressed in inches. 4 For A maa; ., c = 0.5, hence WP x TABLE II c / c J 0.05 o . 0498 0.30 0.2541 O. IO 0.0981 o-35 0.2793 0.15 0.1438 0.40 o. 2976 0.20 0.1856 0-45 0.3088 0.25 o. 2227 0.50 0.3125 224 THEORY OF FRAMED STRUCTURES CHAP. V 145' Deflection for Load of Uniformly Varying Intensity. The beam in Fig. 136 supports a load of uniformly varying intensity. The total load is W lb., and the length of the beam is / in. The bending moment at any distance x from A is M l = ^(I 2 x - x 3 ) The bending moment at any distance x from B is + d >L< ( hc jl FIG. 136. w EI = ^ f EIJ, Wl 3 iSoEI (7 30C 2 + 20C 3 + I5C 4 I2C 5 ) A = /i + c(h - /i) = Wl* - ioc iSoEI The value of c for A wax . may be found by equating t\ and / 2 , whence i5c 4 - 3oc 2 = -7 c = 0.519 = Wl 3 Wl 3 ' iSoEI ~ o.oi3i7 The intensity of the load in Fig. 137 increases uniformly from each support to the center of the span. The total load is SEC. II DEFLECTION OF BEAMS 225 W lb., and the length of the beam is / in. The bending moment at any distance x between the end and center is jp- 1 ~~ oT 2 3 The bending moment at any distance x, when x is greater than Vo/ is <--- ->k (l-c)l W ~ T FIG. 137. i r* i r ( = I Jf lO^t; + I A = W/3 T 4001 tL is 226 THEORY OF FRAMED STRUCTURES CHAP. V in which c may have any value between o and J^. The value of c for A maa; . may be found by equating t\ and /2 whence, = m * ~ 6oEI 146. Deflection for Several Concentrated Loads. A 24-in. by 8o-lb. I-beam supports three loads (Fig. 138). The linear 10000^ 5000' /COO 1 10800 - -10'-- - - - 15'- ----><-- -/ <- - - -/5- - - -> 5^.00 A max. dimensions of the beam are expressed in feet and the units in which E and 7 are usually expressed will be changed accord- ingly. 7 = 2,087 m - 4 E ~ 29,000,000 lb./in. 2 Hence El = 60,532,000,000 in. 2 -lb. or 420,400,000 ft. 2 -lb. The tangent is drawn through C at the center of the span. The deflection due to the weight of the beam will be considered later. Area-moment about P. area PQY 108,000 X 5 X 6.67 = 3,600,000 area YQW 108,000 X 7.5 X 15 = 12,150,000 area QRW 120,000 X 7.5 X 20 = 18,000,000 33,750,000 420,400,000 33,750,000 ft. 3 -lb. = 0.0803 ft - SEC. II DEFLECTION OF BEAMS 227 Area-moment about U. area USV 78,000 X 7.5 X 10 = 5,850,000 area SVW 78,000 X 7.5 X 18.33 = 7,150,000 area SR W 120,000 X 5 X 21.67 = 13,000,000 26,000,000 ft. 3 -lb. 26,000,000 , f . h = - - = 0.0618 ft. 420,400,000 A = 2 = 0.071 ft. = 0.85 in. 2 The maximum deflection caused by the three loads is at T, where the tangent is horizontal, and the ordinate in the M- diagram is KL. Let LW = a and let be the angle made by the two tangents; then represents the slope of the tangent through C. The beam is 50 ft. long i /i /2 0.0185 hence = - - = = 0.00037 50 5 also = areagWL hence area KRWL = El = 155,000 ft. 2 -lb. whence a = 1.3 ft. The centroid of the area KRWL is approximately 24.35 ft- from P, and the area-moment of KRWL about P is 155,000 X 24.35 = 3,774,ooo ft. 3 -lb. A - / 3,774,ooo _ 29,976,000 <*** rl =r= A.... = ,_ = ft = Q 86 . n 420,400,000 Although the loads are eccentric, it is clear that there is practi- cally no difference between the deflection at the center and the maximum deflection. The deflection at the center may be found from Eq. 5, page 218. The coefficients F are given in Table I. c = 0.5; k = 228 , THEORY OF FRAMED STRUCTURES CHAP. V 0.2 for the load at A] 0.5, for the load at B\ and 0.3 for the load at C; hence 10,000 X so 3 . - *-- X 0.071 = 14,792,000 5,000 X 5Q 3 6 " X 0.125 = 13,021,000 X _ _ 2,062,000 29,875,000 ft 3 .-lb. A = = 0.071 ft. = 0.85 in. *7 5000* - 30 '- U- 15 '- - - PIG. 139. The deflection at the center, due to the weight of the beam, may be found from equation 7, page 223. The coefficient of / when c = 0.5 is 0.3125. W = 50 X 80 = 4,000. A = 4,ooo X5Q 3 24 X 420,400,000 = SEC. II DEFLECTION OF BEAMS 229 The total deflection at the center is 0.85 + 0.19 = 1.04 in., which in this case may be assumed without appreciable error as the maximum deflection. A deflection of ^ 60 of the span is considered not excessive. 147. Deflection for Uniform and Concentrated Loads. In Fig. 139, the tangent is drawn through C at the center of the span. Assume that El is expressed in foot 2 -pounds. The M-diagram under the uniform load cannot be accurately divided into triangles, and an integration is necessary if an accurate solution is desired. A sufficiently accurate solution for all practical purposes may be obtained by the geometric process by dividing the area QBDFJ by vertical ordinates into strips, so narrow that their areas may be considered trapezoidal. The accurate method by integration is given below. Let MI represent the bending moment under the uniform load at any distance x from the left support; and If 2, the bending moment under the uniform load at any distance x from the right support; then Mi = loox 2 + 5,0000; 2,500 and Mi = loox 2 + 5,8000; 24,100 Area-moment about A area ABQ 20,000 X 2.5 X 3.33 = 166,667 /2? area BENQ \ M^xdx = 18,446,266 18,612,933 ft 3 .-lb. ti = 18,612,933 ft Area-moment about H area GHI 20,000X2 X 2.67= 106,667 area G/J 20,000 X 7.5 X 9 = 1,350,000 area FGJ 50,000 X 7.5 X 14 = 5,250,000 aresiFENJ I M 2 x dx = 10,332,600 17,039,267 ft.Mb. 17,039,267 f ~ET The deflection at the center is /i + fe 17,826,100 - ~2~ ~lH~ 230 THEORY OF FRAMED STRUCTURES CHAP. V The slope of the tangent is _ ti - / 2 _ 29,142 ~5T ~ET Let KL represent the ordinate in the M-diagram at the point of maximum deflection, then area KENL 0= ~ET therefore area KENL = 29,142 whence LN = 0.488 ft. The area-moment of KENL about A is 29,142 X 26.756 = 779,723 A - / 779,723 __ I7,833,2H El EI~ The area-moment of KENL about H is 29,142 X 27.244 = 793,945 A _ , , 793,945 __ 17,833,212 El El When the tangent is drawn to the elastic curve at the right end of the uniform load, the tangential deviations fa and /4, at the left and right supports respectively, may be determined by the geometric process; for if a straight line be drawn from B to F, the area of the M-diagram BDF has all the properties of the M-diagram B'D'F' for a beam 30 ft. long when uniformly loaded with 200 Ib. per foot (Fig. 1396). See Article 60. Area-moment about A area ABQ 20,000 X 2.5 X 3.33 = 166,667 area BQJ 20,000 X 15 X 15 = 4,500,000 area FBJ 50,000 X 15 X 25 = 18,750,000 area BDF 22,500 X 30 X % X 20 = 9,000,000 32,416,667 ft 3 .-lb. Area-moment about H area GHI 20,000 X 2 X 2.67 = 106,667 areaG/J 20,000 X 7.5 X 9 = 1,350,000 area FGJ 50,000 X 7.5 X 14 = 5,250,000 6,706,667 ft 3 .-lb. SEC. Ill DEFLECTION OF BEAMS 231 The slope of the tangent is /s /4 _ 476,315 _ area KFJL ~~S4~ ~Ef" ~EI~ The area EFJN, when considered as four trapezoidal areas, each 2 ft. wide, is 446,440; hence the approximate area of KENL is 476,3 J 5 - 446,400 = 29,915 from which we find that the ordinate KL is located about 0.5 ft. to the left of the center of the beam as before. SEC. III. MAXWELL'S THEOREM OF RECIPROCAL DISPLACEMENTS 148. Maxwell's Theorem of Reciprocal Displacements estab- lishes a mutual relation between any two points in a structure. This theorem, when considered in connection with the deflec- tion of beams, may be stated as follows: If the load P at A FIG. 140. FIG. 141. FIG. 142. (Fig. 140) causes a deflection A 2 at B, and the load P at B (Fig. 141) causes a deflection A 3 at A; then, according to Maxwell's theorem, A 2 = A 3 . Let Figs. 140, 141 and 142 represent the deflections of a beam when the loads are applied gradually. When A (Fig. 140) has received its full load, the work done is J^PAi. With a full load P at A, let another load P be gradually added at B. The deflections as shown in Fig. 142 will result. The point A, with the full load P, moves through the additional distance A 3 ; and the point B moves through the additional distance A4, as the load P is gradually applied at B. Hence the total work done is -PAi+PA 3 + -PA 4 2 2 If B is loaded first and then A is loaded, the total work done is -PA 4 +PA 2 +-PAi 2 2 232 THEORY OF FRAMED STRUCTURES CHAP. V The total amount of work done in each case is the same, hence A 2 = A 3 Maxwell's law may be verified by Table I. When A and B are on the same side of the center, the values of k and c for Fig. 140 become interchanged for Fig. 141. For example, F = 0.0658 when k = 0.2 and c = 0.3; likewise, F = 0.0658 when k = 0.3 and c = 0.2. When A and B are on opposite sides of the center the application is made as follows: In Fig. 140 let k = 0.3 and c = 0.8; then in Fig. 141, k = 0.2 and c = 0.7; whence F = 0.0522 in each case. Maxwell's law renders excellent service in the solution of statically indeterminate structures. SEC. IV. CANTILEVER BEAMS 149. The beam in Fig. 143 supports a single load P at the free end. It is fixed in the wall at A in such a way that the -PI FIG. 143. tangent to the elastic curve at A remains horizontal, hence the deflection at the free end is P/ 3 A = I = &I The negative sign indicates that the elastic curve deviates below the tangent. The beam in Fig. 144 supports the load W uniformly dis- tributed. The M-diagram is SQV. The curve SV is a para- SEC. IV DEFLECTION OF BEAMS 233 bola with the vertex at 5, hence the deflection at the free end is A 7-in., i5-lb. I-beam (Fig. 145) supports a load of 2,000 Ib. FIG. 145. El = 1,050,000,000 in. 2 -lb. The tangent is drawn through C. 12,000 X 1-5 X 2 tl = - = _ Q 1,050,000,000 = -12,000X3X4X1,728 _ _ Q in I ,O5O,OOO,OOO A = 2ti + h = 0.3552 in. 234 THEORY OF FRAMED STRUCTURES CHAP. V The deflection at B may also be found by drawing the tangent through either A or B. A cantilever beam is shown in Fig. 146. In finding t\ t ?000' 10000$ 5000 ;---- /z '----. positive and negative areas are encountered in the M-diagram. These may be treated in one of two ways. The point of zero bending moment at I may be determined, and the areas PIW and IQV treated separately; or the area WIV may be included with both positive and negative areas as follows : SEC. V DEFLECTION OF BEAMS 235 Area-moment about P. area PVW - 30,000 X 3 X 2 = -180,000 area WQV+ 36,000 X 3 X 4 = +432,000 area QVU + 36,000 X 6 X 10 = +2,160,000 area QRU + 48,000 X 6 X 14 = +4,032,000 + 6,444,000 ft. 3 -lb. If El is expressed in foot 2 -pounds then _ 6,444,000 , fc- -^-ft. Area-moment about S area ,SW 48,000 X 6 X 8 = 2,304,000 ft. 3 -lb. <6 = * = 30 EI Let KL represent the ordinate in the M-diagram at the point of maximum deflection, then area KRUL = 138,000 a = 2.967 ft. KL = 45>33 The maximum deflection may now be found as in previous cases. SEC. V. BEAMS WITH VARYING CROSS-SECTION 150. General Expressions. The moment of inertia of beams having uniform cross-section is constant, and for this reason / appears outside the integral sign in Eqs. (i) and (2). When the cross-section is not uniform the moment of inertia varies, and Eqs. (i) and (2) become * = i ( B (8) 1~1 I T \ / EjA I tB Mxdx I In order to perform the integration, / as well as M must be expressed as a function of x. This is relatively a simple matter when the beam has a rectangular cross-section varying uni- formly in breadth or depth; but this method often results in 236 THEORY OF FRAMED STRUCTURES CHAP. V long and cumbersome expressions when applied to structural steel sections. In all such instances the geometric method is preferable. 151. Beams with Varying Depth. The beam in Fig. 1470 is a plate girder. The % in. web plate is 24 in. wide at the ends, and the width increases uniformly to 36 in. at the center. Each flange is composed of two angles 5 by 3 J^ by % with the 3-in. leg against the web. The distance back to back of angles is the width of the web plus % in. Ordinates in the M-diagram (Fig. 1476) are given in inch-pounds every 3 ft. An I-diagram is shown in Fig. 147^. The ordinates represent the moment of inertia in inches 4 at 3-ft. intervals. Each ordinate in the M-diagram has been divided by the corresponding ordinate M in the I-diagram, and the quotient recorded in the -^-diagram (Fig. i47 . 9 29,000,000 lb./m 2 . M When the ordinates in the ^'diagram are computed only for the ordinates at B and C, and AB and BC considered as straight lines, the computations are as follows: Area-moment about A 957.8 X 54 X 72 = 3,724,000 957- 8 X 54 X 144 = 7,448,ooo 837.2 X 54 X 1 80 = 8,138,000 9,310,000 19,310,000 , . A wa *. = -%&- - = 0.67 in. 29,000,000 M Thus it is clear that, if the ordinates in the -=- -diagram were relatively close together, say at every foot or closer; or even if / were expressed as an exact function of x in Eq. (9) and the integration performed, the results in either case would not differ materially from those obtained above. 152. Beams with Cover Plates. The plate girder in Fig. 148^ consists of a 24 by % web plate, four angles 5 by 3^ by % and two cover plates 12 by % by 24 ft. symmetrical about the center line. The M-diagram is shown in Fig. 1486, M the I-diagram in Fig. 148^, and the -y- -diagram in Fig. THEORY OF FRAMED STRUCTURES Area moment about A 1.065.1 X 36 X 48 = 1,840,500 631.6 X 18 X 84 = 955>o 947.4 X 18 X 96 = 1,637,100 947.4 X 54 X 144 = 7,367,000 1.263.2 X 54 X 1 80 = 12,278,300 24,077,900 CHAP. V ?0000* 20000* 20000" The deflection at the center is A - 24 ' 77 ' 9 = 0.83 in. 29,000,000 The solution by integration may be obtained as follows: Let M i represent the bending moment for values of x between o and 9; and M% the bending moment for values of x between 9 and 1 8.' Then Mi = 30,0000: Mz = io,ooo# + 180,000 Then for values of x between o and 6, EI\ = 408,417,000 ft 2 .- SEC. V DEFLECTION OF BEAMS 239 lb. and for values of x between 6 and 18, EI 2 = 688,750,000 ftVlb. The deflection at the center expressed in feet is A = / = ~- M&dx + -r M.xdx + -r Mixdx &1\J C'-LzJ 6 tillj 9 The solution by integration is much more simple in this problem than in the preceding one, for in this problem / is constant between certain limits of x and is therefore not a func- tion of x. The geometric treatment by area-moments is by far the simplest and best method now known for obtaining a solution of any practical problem involving the deflection of beams. CHAPTER VI RESTRAINED AND CONTINUOUS BEAMS SEC. I. RESTRAINED OR FIXED BEAMS 153. General Considerations. The beam in Fig. 149 is considered fixed or restrained at A , if built into the wall in such U FIG. 149. a way that any attempt of the external forces to rotate the beam at A is successfully resisted, and the neutral plane of the beam in its original position AB remains tangent to the elastic curve at A when the beam is bent. Under these conditions the reactions RI and Ri, and the resisting moment Mi at A present 240 SEC. I RESTRAINED AND CONTINUOUS BEAMS 241 more unknown quantities than can be determined by the principles of statics; and the beam is statically indeterminate. The principles of deflections render their most helpful service in the solution of problems of this character. The ease with which problems of this kind are solved, depends to some extent upon the manner in which the M-diagram is drawn; for it may be represented in three ways as shown in Figs. 1496, I49C or 149 d\ and each case will be considered separately. 154. Restraint at One End Concentrated Load. In Fig. 1496, let MI and M^ represent the bending moments at A and C respectively. The bending moment at B is zero. The line AB is tangent to the elastic curve at A, and the tangential deviation t at B is zero, therefore i C B t = I Mxdx = o or C B Mxdx = o .... Hence the area-moment of QVUTP about Q is zero, or ^2(3 X 4) + M z (6 X 10) + Mi(6 X 14) = o yAf i + 6M 2 = o From statics MI = (540 X 12) + 18^2 and Mi = 6R 2 whence 45,360 -+- i267? 2 + 36^2 = o Rz = 280 Ri = 260 Mi = 1,440 and If 2 = i, 680 In Fig. 149*;, the M-diagram is drawn in parts; QST is the M-diagram for the reaction at B, and TPU is the M-diagram for the load at C. The area QST is positive and the area TPU is negative. The area-moment of the total diagram about Q is zero; therefore 18^2(9 X 12) - 6,480(6 X 14) = o R z = 280 TS = iSRz = 5,040 PV =\X 5,040= i ,680 o SU = 5,040 6,480 = 1,440 16 242 THEORY OF FRAMED STRUCTURES. CHAP. VI In the two preceding solutions no speculation was made as to the general form of the elastic curve. The curve ACB might have had any shape whatsoever, so long as its tangent at A passes through B. It is not always wise to presume upon the general form of the elastic curve before computations are made; but in the present simple case it is quite safe to assume that the curve is concave on the under side near A , and concave ofi the upper side at C, with a point of contraflexure between. Hence the bending moment is negative at A t positive at C and zero at an intermediate point 7; consequently the M- diagram may be sketched as in Fig. 149^. In finding the area-moment, the area TIV, which is not a part of the diagram, can be included as positive area with QVI, and as negative area with TIV. ^2(3 X 4) + M*(6 X 10) - Mi(6 X 14) = o yMi 6M% = o From statics M\= (540 X 12) -f i8R 2 and M-2 = 6Rz whence R%.= 280 #1 = 260 Mi = 1,440 Mi = i, 680 When an unknown ordinate in the M-diagram is represented by a symbol, it is generally better to assume that the ordinate is positive as in Fig. 149^; then if the solution shows that the ordinate is negative, the M-diagram may be re-drawn if desirable. Frequently the M-diagram may be constructed to advantage, as shown in Fig. 1490. Only two independent static equations can be written for the solution of a system of parallel forces. In the present problem there were three unknown quantities to be determined, hence one elastic equation was necessary for a solution. 155. Restraint at One End Uniform Load. The beam in Fig. 1500, fixed at A and simply supported at B, carries a total load Wj uniformly distributed. In Fig. 1506 QTU is the M- diagram for the uniform load and QTS is the M-diagram for the reaction at B. If the tangent to the elastic curve is drawn SEC. I RESTRAINED AND CONTINUOUS BEAMS 243 through A, the tangential deviation t at B is zero, hence the area-moment of QSU about Q is zero. rs - - in o The M-diagram may now be drawn to scale as in Fig. FIG. 150. the resisting moment at A is SU = TS - TU = Wl - - Wl = - \ Wl 02 O 156. Restraint at Both Ends Concentrated Load. The beam in Fig. 1510 is fixed at each end. The M-diagram is sketched in Fig. 1516 by assuming all ordinates positive. It cannot be drawn to scale until Mi, Mi and M 3 are known. There are four unknown quantities involved in the external 244 THEORY OF FRAMED STRUCTURES CHAP. VI forces acting at the points of support, R\, R%, Mi and Mz for which four independent equations are necessary. The two static equations may be written thus, R l + Rz = P (i) M s = klRi + M l = (i - k)lR z + M z (2) Two elastic equations are necessary. Let < be the angle which the tangent through A makes with the tangent through B\ then, since = o, the area of the M-diagram between A and B is zero; therefore 2 2 3) The line AB is tangent to the elastic curve at B, and the SEC. I RESTRAINED AND CONTINUOUS BEAMS 245 tangential deviation at A is t\ = o; therefore the area-moment of the M-diagram about TV is zero, hence (4) A third elastic equation may be written by equating to zero the area-moment of the M-diagram about QP\ but obviously this equation would not be independent of Eqs. (3) and (4). Eqs. (3) and (4) may be reduced to kMi + (i - k)M* + M s = o ( 3 a) k*Mi + (2 - k - k*)M z + (i + k)M z = o ( 4 a) Solving Eqs. (i), (2), (30) and (40) Afi = -k(i - k) 2 Pl M 2 = - k 2 (i- k)Pl M S = 2/^ 2 (l - k) 2 Pl Since the limits of k are o and i, it is clear that Mi and Mz are negative bending moments and M$ is a positive bending moment. The M-diagram in Fig. 151^ is the same as in Fig. 1516, except that TQ has been drawn in the horizontal position. Let SV = M 3 = M* + M& From geometry M 5 = M 2 + (i - k)(Mi - M z ) hence M 4 = M z - M 5 = k(i - k)Pl If the beam were simply supported, not fixed, at A and B, the bending moment at C would be M* = k(i k)Pl, therefore TSQ is the M-diagram when the beam is not restrained at the ends by MI and M*. In a numerical problem the M-diagram should be sketched as in Fig. 1516, and Mi computed as for a simple beam. After the negative moments Mi and M z have been determined, the trapezoid TQPU may be revolved about TQ and the diagram drawn to scale as shown in Fig. 15 id. The reactions and resisting moments will be determined for the fixed beam in Fig. 1520. 246 From statics or THEORY OF FRAMED STRUCTURES 640 CHAP. VI 24 X9X15 = 3>6oo Mi = -(640 X 9) + 24^2 + 5>76o + M i M 2 -2250 The angle between the tangents through A and 5 is = o (Mi + ^2)24 3,600 X 24 _ or o 2 2 The tangential deviation at B is 2 = o 3,600 X 15 X io , 3,600 X 9 X 18 , whence 2 4 M 2 X 8 2 4 Mi X 16 2 2 = -2,250 = -1,350 = 437-5 = 2O2.5 SEC. I RESTRAINED AND CONTINUOUS BEAMS 247 These results may be checked by the formulas of the preceding problem. The M-diagram is drawn to scale in Fig. 152^. 157. Restraint at Both Ends Uniform Load. The fixed beam in Fig. 153 supports a total load W, uniformly distrib- uted. Since the loading is symmetrical, the resisting moment and reactions at B are the same as at A . The area TSQ is the !A"^> ""l FIG. 153. M-diagram for a simply supported beam and the area TUPQ represents the resisting moment M at each end. The angle between the tangents through A and B is zero, consequently the area of the M-diagram is zero, therefore Ml -f \~jr)( 2 l) = *-_E! 12 The bending moment at the center is Wl _ Wl = Wl 8 12 24 and the M-diagram may be drawn to scale as shown. 248 THEORY OF FRAMED STRUCTURES SEC. II. CONTINUOUS BEAMS CHAP. VI 158. Continuous beams rest on more than two supports and have more than one span. Consequently they are stati- cally indeterminate, and one or more elastic equations are necessary for finding the reactions. An introduction to the problem will be made in connection with the beam in Fig. 154, which supports a load of 10 Ib. at D. The tangent FG is drawn to the elastic curve through C at the middle of the span. The product El of the modulus of elasticity, and the moment of inertia is assumed as unity, and the weight of the beam is not considered. The M-diagram is PQV. When the beam is supported at A and B only, the reactions R, the bending mo- ments M , the tangential deviations t, and the deflections A are as tabulated in the column i of Table I. Now suppose that an SEC. II RESTRAINED AND CONTINUOUS BEAMS 249 upward force R c = i Ib. is applied at C. The shape of the elastic curve will be altered in accordance with the data given in column 2. The moments are statically determinate; and if a new M-diagram is drawn to scale an angle will appear at S, FIG. 756 the line QV no longer remaining straight. The quantities / and A (column 2) are computed from this diagram. Columns 3, 4, 5 and 6 give similar data, as the upward force at C is in- creased by i-lb. increments. The table clearly shows that for each increase of i Ib. in R c , A c is decreased 4,500 units. The original deflection A c was (from column i) 25,560 units; and it is clear that if the point C is raised 4,500 units by each in- 250 THEORY OF FRAMED STRUCTURES CHAP. VI crease of i Ib. in R c , the force necessary to raise the point C to a level with A and B is Rc _ 25,560 681b< 4,5 By balancing the moments about either A or B, the end reac- tions, RI and R, may be determined as given in column 7. A very interesting study of the successive stages in the transformation of the elastic curve may be made, if diagrams similar to Fig. 154 are drawn to scale in accordance with the data given in each column of the table. These curves have been combined in Fig. 155, and are numbered to correspond with the columns in the table. When R c = 4 Ib., RI = 6 Ib., and ^4 = 0; consequently M c = Mz = o and the M-diagram is PQS (Fig. 154), the point S coinciding with U. Since the area-moment of VSU about V is zero, /4 = o, G coincides with B ; and the elastic curve CB being bent by no moment becomes a straight line coinciding with the tangent CG\ (see curve 5). When R c = 5 Ib.; R^ being negative, acts downward; M c and M 3 become negative moments; and / 4 , being negative for curve 6, is measured above B instead of below. Finally, when R c = 5.68 Ib., and the point C has been raised to the line AB (curve 7); the tangent through C makes equal intercepts on the TABLE I i 2 3 4 5 6 7 Rc = R c = i Rc = 2 Rc = 3 R e = 4 R c = 5 R c = 5.68 Rl 8 7-5 7-0 6-5 6.0 5-5 5-i6 R* 2 i-5 I .O 0-5 0.0 -5-o 0.84 M 2 9 6 90 8 4 78 7 2 66 61 .92 M c 60 45 30 15 o -i5 -25.2 M z 30 22.5 15 7-5 o -7-5 12.6 /i 33,120 28,620 24,120 19,620 15,120 10,620 7,56o t. ii ,664 9,720 7,776 5,832 3,888 i,944 622.08 h 5,625 4,218.75 2., 8l2.'5 1,406.25 o 1,406.25 -2,362.5 /4 18,000 13,500 9,OOO 4,5oo o -4,50 -7,56o. A 2 18,432 15,876 13,320 10,764 8,208 5,652 3,9i3-93 A c 25,560 21 ,O6O 16,560 I 2 , 060 7,560 3,060 o A 3 16,155 13,061.25 9,967-5 6,8 7 3.75 3,780 686.25 -1,417-5 SEC. II RESTRAINED AND CONTINUOUS BEAMS ordinates through A and B, the one below and the other above AB, hence h = / 4 . Let d 2 , d c and d 3 represent respectively the differences in A2, A c and A 3 , for consecutive columns i to 6 in Table I; then dz = 2,556; d c = 4,500 and d z = 3,093.75. These dif- ferences in deflection between any two curves i to 6 (Fig. 155) are caused by a difference of i Ib. in the force R c , irrespective of the magnitudes RI, R c or R*. The fact that these differences are constant for any ordinate whether 2-2, C-C, 3-3 or any other ordinate between A and B which might be chosen, is due to the constant differences in Mz, M c and M 3 . Let m^, m c and w 3 represent these differences; then from Table I, nh = 6 m c = 15 and m s = 7.5. By reference to Fig. 156 it is at once, apparent that nh, m c and w 3 represent the corresponding ordinates in the M-diagram for the beam in question; when supporting a single load of i Ib. at the center. If El is again taken as unity, then d c = h = 15 X 15 X 20 = 4,500 / 2 = (6 X 9 X 6) + (15 X 9 X 12) = 1,944 *3 = (7-5 X 7-5 X 5) + (15 X 7-5 X 10) = 1,406.25 dz = h h = 2,556 d z = h - / 8 = 3,093.75 Consequently the difference d between any two curves from i to 6 (Fig. 155) on any ordinate, caused by a difference of i Ib. in R c , equals the deflection at the same ordinate caused by a load of i Ib. at C (Fig. 156). Hence, if a beam is continuous over three supports, the intermediate reaction R c at C (not necessarily at the center) may be determined as follows: Remove the intermediate reaction; find the end reactions as for a simple beam and com- pute the deflection A at C; finally, compute the deflection d at C due to a load of i Ib. at C. Then 159. Application of Maxwell's Theorem. Maxwell's theorem of reciprocal displacements may be used in finding the reactions of a continuous beam on three supports. The deflection at D 252 THEORY OF FRAMED STRUCTURES CHAP. VI (Fig. 156) caused by a load of i Ib. at C is J 2 = 2,556; hence, the deflection at C caused by a load of i Ib. at D is 2,556; and the deflection at C caused by 10 Ib. at D is A c == 2,556 X 10 = 25,560, as shown in Fig. 154. Since the deflection at C, caused by a force of i Ib. at C, is d c = 4,500; then the force at C necessary to raise the point C to a level with AB is Rc = ip^ = 25,560 = ^ d c 4,500 Hence, if a beam is continuous over three supports at A,C and B, and supports loads PI, P 2 and P 3 at any points i, 2 and 3, the intermediate reaction R c may be found as follows: Remove the loads P and the reaction R c : place i Ib. at C, and compute the deflections AI, A 2 , AS and A c at the points i, 2, 3 and C. Then P 3 A 3 c - - A c The reactions at ^4 and B may be determined by statics after the reaction at C is known. It was not necessary to apply Maxwell's theorem to the intermediate reaction, for either end reaction might have been determined equally as well by this theorem. The ratio of d 2 to d c (Fig. 156) may be determined from Table I, page 222. The load i Ib. is at the center, hence k = 0.5. When c = 0.2, F = 0.071; when c = 0.5, F = 0.125; hence d% _ 0.071 d c 0.125 and R c = = 5.68 Ib. d c 160. The Conventional Method. The method which is generally employed in finding the reactions, by area-moments, of a beam continuous over three supports, will be given in con- nection with Fig. 157. The elastic curve is not shown and no speculation with reference to its form will be made. Let FG represent the tangent to the elastic curve drawn through C. Its slope is unknown; it may be positive or negative. One thing is certain. Since the two spans are equal in length, the SEC. II RESTRAINED AND CONTINUOUS BEAMS 253 tangential deviations /i and / 2 are equal in magnitude. They have opposite signs, since one is measured above the line AB and the other below it. Hence 254 THEORY OF FRAMED STRUCTURES CHAP. VI All reactions will be assumed to act upward, hence a negative numerical value for the solution of a reaction indicates that its action is downward. The bending moment is known only at two points A and B. The M-diagram may be constructed in several ways. In Fig. 157^, all ordinates are expressed in terms of the end reactions. //i = 12^1(6 X 8) + 12^1(9 X 18) + 30^3(9 X 24) // 2 = 30^3(15 X 20) h = -h whence jRi = 43^3 From statics 3< _'__><___'___>. (b) FIG. 158. = (72 X 6 X 8) + (72 X 9 X 18) + M(i$ X 20) = M(i$ X 20) ti = -tz whence M *= 25.2 as before. After the reactions have been determined, the bending moments at C and D, and the deflection at any point may be computed. The elastic curve when drawn will have the general configuration shown in Fig. 159. 256 THEORY OF FRAMED STRUCTURES CHAP. VI 161. In Fig. 1580, PQS is the M-diagram when AC is con- sidered as a simple beam, and SOTV is the M-diagram when CB is considered as a simple beam; then the area PUV is added to provide the bending moment on account of the continuity. Let the tangent to the elastic curve be drawn through C and let t\ and fa represent the tangential deviations at A and B FIG. 159. respectively, then Elh = (9,000 X 6 X 6) + M(6 X 8) EIt z = (6,000 X 3 X 4) + (6,000 X 6 X 9) + (6,000 X 3 X 14) + M(g X 12) t\: fa:: 12:18 whence M = 6,300 The M-diagram may now be drawn to scale as shown in Fig. 6,300 = i2Ri 18,000 *i = 975 6,300 = i&Rs 6,000 12,000 RZ = 650 Ri = 5,000 - 975 - 650 = 3,375 From statics SEC. II RESTRAINED AND CONTINUOUS BEAMS 257 162. The general expressions for RI, R 2 and R$ will now be developed in connection with Fig. 159. Let the tangent to the elastic curve be drawn through C and let t\ and /2 represent the tangential deviations at A and B respectively; then = Pk(i - k)l /1/2 = ~ _ - 2(/i + fe) Since ^ is less than unity, k k 3 is positive; hence M is a negative bending moment, and R s is a negative reaction acting downward. When the two spans are of equal length /, the reactions are *i = -(# - 5k + 4) (9) 4 2 =^(_ 2 3 + 6 ) ( IQ) 4 R* = ~(k*-k) (n) 4 163. In Fig. i6oa the parabola PQS is the M-diagram, when AC is considered as a simple beam; and the parabola STV is the M-diagram when CB is considered as a simple beam. The triangle PUV is added to represent the bending moment on account of the continuity. If the tangent to the elastic curve 17 THEORY OF FRAMED STRUCTURES CHAP. VI be drawn through C, and /i and / 2 represent the tangential deviations at A and B respectively; then POOO Ibs.per ft E//I = (18,000 X 12 X - X 6) + M(6 X 8) O = 864,000 + 48M Elh = (81,000 X 18 X - X 9) + M(g X 12) o = 8,748,000 + io8M 3/1 = 2/2 M = - 55,800 SEC. II RESTRAINED AND CONTINUOUS BEAMS 259 The M-diagram may now be drawn to scale as shown in Fig. 1606. From statics 55,800 = i2Ri (12,000 X 6) = i&R 3 (36,000 X 9) & = 31,75 RS = 14,900 164. Two Unequal Spans, Supporting Unequal Uniform Loads. The general expressions for Ri, R% and R s for a con- w, FIG. 161. tinuous beam of two unequal spans l\ and k, supporting unequal uniform loads, w\ and w% per unit of length, will now be de- veloped in connection with Fig. 161. The M-diagram is drawn as in the preceding problem. If the tangent to the elastic curve is drawn through C; and t\ and h represent the tangential deviations at A and B respectively; then Will 4 . Mli' I 24 3 24 M = - (12) 260 THEORY OF FRAMED STRUCTURES CHAP. VI When the spans are equal in length / and the uniform load w per unit of length is the same in both spans, Eq. (12) reduces to *--f (13) 165. When a continuous beam supports a combination of uniform and concentrated loads, it will be found expedient to sketch the M-diagram in parts as shown in Fig. 162. The portion (a) is the M-diagram for the concentrated loads when no continuity is considered at C; and the portion (b) is a similar diagram for the uniform loads. The continuity is provided for by the portion (c). If the tangent to the elastic curve is drawn through C; and /i and k represent the tangential devia- tions at A and B] then = (9,000 X 6 X 6) + (18,000 X 12 X - X 6) + M(6 X 8) = 1,188,000 + 48M Elk = (6,000 X 3 X 4) + (6,000 X 6 X 9) + (6,000 X 3 X 14) + (81,000 X 18 X - X 9) + M(g X 12) = 9,396,000 + io8M 3/1 = 2/2 M = 62,100 SEC. II RESTRAINED AND CONTINUOUS BEAMS 261 The reactions may now be determined by the principles of statics. The value of M may also be determined from Eqs. (5) and (12). Eq. (5) is applicable to the concentrated loads. For the load at Z>; P = 3,000, k = %, h = 12 and h = 18; hence M= For the load at F\ P = 1,000, k = -, l\ = 18 and h = 12; o hence 1,000 X 324(- - - * 2(12 + 1 8) e load at E; P = hence = 2 For the load at E; P = 1,000, k = -, /i = 18 and / 2 = 12, o 1,000 X 3 2 4( M = EN 2(12 + 18) Hence the bending moment at C, due to the three concentrated loads, M = 2,700 i, 600 2,000 = 6,300 which agrees with bending moment at C for the beam in Fig. 158. Equation (12) is applicable to the uniform loads, where Wi = 1,000, wz = 2,000, li = 12 and li = 18; hence which agrees with the bending moment at C for the beam in Fig. 1 60. The total bending moment at C for the combined uni- form and concentrated loads is M = 6,300 55,800 = 62,100 as previously determined. 166. The continuous beam in Fig. 163 supports a uniform load of 1,000 Ib. per foot over a part of the span AC. The area PQSTW is the M-diagram, when AC is considered as a simple 262 THEORY OF FRAMED STRUCTURES CHAP. VI span. Let the tangent to the elastic curve be drawn through C, and let t\ and / 2 represent the tangential deviations at A and B respectively. In finding the area-moment of PQSTW about P, the parabolic area QST is encountered. This area has all the properties of the area Q'S'T 1 , which is the M-diagram for a simple beam 12 ft. long supporting a uniform load of 1,000 I000*/l.f. --><-- PIG. 163. lb. per foot over its entire length; hence the area-moment of PQSTW about P may be found as follows: area PQN 43,200 X 3 X 4 = 518,400 43,200 X 6 X 10 = 2,592,000 57,600 X 6 X 14 = 4,838,400 57,600 X 6 X 22 = 7,603,200 area QNO area QTO area TOW area QST 18,000 X 12 X % X 12 = 1,728,000 17,280,000 = 17,280,000 + M(i$ X 20) // 2 = Jlf (30 X 40) 2/i= ~/2 M = 19,200 The reactions are statically determinate when M is known. The value of M may also be determined by the use of Eq. (5) SEC. II RESTRAINED AND CONTINUOUS BEAMS 263 in which P is a concentrated load at the distance kl\ from A. In the present case let P represent the weight of an element, of length dkli at the distance kli from A , then P i ,000 lidk whence , dM = . . 2 (/i + k) The value of M may be found by integrating between the limits k = 0.2 and k = 0.6, hence fk 2 4 1 - = -150,000 - 12 4 Jo. = 19,200 167. The beam in Fig. 164 is continuous over four supports. Two elastic equations are required, in addition to the two FIG. 164. static equations which may be written, for the determination of Rij Rt, Rs and R^ Let h and / 3 represent the tangential deviations at A and Z>, for the tangent to the elastic curve at C; and let h and / 4 represent the tangential deviations at C and B, for the tangent to the elastic curve at D\ then ati = / 3 and /2 = 0/4 PQW is the M-diagram when AC is considered as a simple 264 THEORY OF FRAMED STRUCTURES CHAP. VI span, to which the diagram PUSV is added to provide for continuity. Elh = Pk(i - k)l(j l\-\kl + (i - )/j + M,(~ 2 ;V* l\ 6 v 3 7/3 = Mi(- al}(- al} + W- al}(- al} \ 2 /\3 / \ 2 /\3 / = - (2Afi + Mz) Elh = M o/ al - al } I - al } Whence K , = 3^ + Sa + 4 PJ(, M z = W k) 4 ~ ^ 8 )(" + # 2 + 8a + 4 P(^-^)( 2 a 2 + 5^ + " ( 3 a 2 + Sa + 4 )a R* == o . o r~ 3^ i o^ ~r4 168. Three Equal Spans Uniform Load. The beam in Fig. 165 is continuous over three equal spans /, and supports a uni- form load of w Ib. per unit of length. Since the beam is sym- metrical about the center, the ordinates in the M-diagram at B and C are equal; and only one elastic equation is necessary. Let the tangent to the elastic curve be drawn through B, and SEC. II RESTRAINED AND CONTINUOUS BEAMS 265 let t\ and /a represent the tangential deviations at A and C respectively; then wl* Ml 2 24 3 FIG. 165. __ wl* Ml 2 24 2 24 3 24 M = - 10 10 IO If the M-diagram had not been symmetrical, either on account of unsymmetrical loading or variation in span lengths, i 266 THEORY OF FRAMED STRUCTURES . CHAP. VI or both; the ordinates at B and C could not have been assumed equal. In this case there would have been two unknown ordinates, Mz at B and If 3 at C; and two elastic equations would have been necessary for a solution. These two equations may be written by establishing a rela- tion between /i and / 3 , and a relation between /i and / 4 ; or one tangent may be drawn through B and another through C. 169. Four Equal Spans Uniform Load. The beam in Fig. 1 66 is continuous over four equal spans /, and supports a uniform load of w Ib. per unit of length. Since the beam is symmetrical about the center, there are only two unknown ordinates Mi and M 3 to be determined, for which two elastic equations are necessary. On account of symmetry, the tangent to the elastic curve through C is horizontal, consequently the tangential deviations t\ at A and t z at B are zero; hence or 6Mz + sMz = wl 25 Whence R l = R,=wl 25 R* = #3 = ^|w/ 25 If for any reason the M-diagram had not been symmetrical, the ordinates at B and D could not have been assumed equal; and the tangent to the elastic curve through C would not have been horizontal. There would have been three unknown ordinates, M 2 at B, M s at C and M 4 at D; and three elastic equations would have been necessary for a solution. These three equations may be written as follows : (a) draw a tangent SEC. II RESTRAINED AND CONTINUOUS BEAMS 267 to the elastic curve at B and establish a relation between the tangential deviations at A and C; (6) draw a tangent to the elastic curve at C and establish a relation between the tangen- tial deviations at B and D; (c) draw a tangent to the elastic curve at D and establish a relation between the tangential deviations at C and E. The three equations may also be written as follows: Let the tangent to the elastic curve be drawn through B; and let t\, / 3 , /4 and / 5 represent the tangential deviations at A } C, D and E respectively. Then establish a relation be- ft 1 FIG. 166. tween /i and / 3 , a second relation between t\ and 4, and a third relation between t\ and / 5 . In case the beam is restrained at each end by being fixed in a wall; the M-diagram presents five unknown ordinates, or one at each point of support. This condition requires two additional elastic equations, or five in all. These two equa- tions may easily be written, since the tangents to the elastic curve through A and E are horizontal. 170. Coefficients for Pier Reactions. When the spans are equal in length and the load is uniform throughout, the reaction at each support may be found by multiplying the total load on each span by the coefficients, as given in the following table. The Roman numerals represent the number of spans over which the beam is continuous. 268 THEORY OF FRAMED STRUCTURES CHAP. VI REACTION COEFFICIENTS I !-! HI ___. IO 10 10 10 IV H_3^_^>_3^_!i 28 28 28 28 28 V 15_43 _37 _37 _43 _ 5 38 38 38 38 38 38 SEC. III. CLAPEYRON'S THEOREM OF THREE MOMENTS 171. Clapeyron's theorem may be used to establish a rela- tion between the bending moments at any three consecutive supports of a beam of uniform cross-section, as shown in Fig. 167. Let li and / 2 represent the lengths of any two adjacent spans, which support the uniform loads of w\ and w 2 per unit of length respectively; and let M Q , MI and M 2 represent the bend- ing moments at the three supports. Let the tangent to the elastic curve be drawn through the middle support; and let /o and tH tf-'--- 5' FIG. 168. the continuous beam of uniform cross-section, having five spans (Fig. 168). First and second spans: = (o + 1,000 X 2o 3 ) 4 ioMo + 2Afi(io + 20) + Second and third spans: 2oMi + 2M 2 (20 + 15) + Third and fourth spans: 15^2 + 2^3(15 + 10) + ioM 4 Fourth and fifth spans: ioM 3 + 2^1/4(10 + 15) + 15^5 = -[(500 X io 3 ) + (1,500 X is 8 )! -(l,000 X 20 3 I (o + 500 X io 3 ) 4 270 THEORY OF FRAMED STRUCTURES CHAP. VI Since the beam is simply supported at each end, Mo = O M b = o A solution of these equations will give the bending moments at the points of support, and the reactions may be determined from the principles of statics. SEC. IV. PARTIALLY CONTINUOUS BEAMS 172. No Shear Transmitted. It is frequently desirable to consider a structure in which the continuity is imperfect. A swing truss bridge on four supports, designed with parallel -----x---c*l--->< 1 w H U 5 FIG. i6ga. chords and very light web members in the center span, so that no shear can be transmitted between the two inside supports is a structure of this kind. Such structures are called partially continuous, and their treatment will be illus- trated by the beam in Fig. 1690. It is assumed that bending moment, but no shear exists in the center span; hence Rz = R^ and the bending moment M at B equals the bending moment at C. Since the continuity of the beam is broken at B and C, the elastic curve is not con- tinuous, but forms cusps at these points; and the tangent FG SEC. IV RESTRAINED AND CONTINUOUS BEAMS 271 to the elastic curve for AB at B is not tangent to the elastic curve for BC. Similarly the tangent HG to the elastic curve for CD at C is not tangent to the elastic curve for BC. Let 0i = angle ABF, and 4 = angle DCH, then area WTSU 6 l + 4 = = - JTT xii The tangential deviations at ^4 and Z), being represented as measured above the axis of the beam, are considered negative. -/i = Oil 70 = Jlf a/ i = Pk(i - k)l-l-kl + -(i - 2 3 Since 7/i + EIt = EIl _ P/ 3 , L Ml 2 . Then (y^ & 3 ) + - - = Mai 2 6 33 4 a Therefore i?i = P(i - k) - 4 + oa P(k - k*) 4 4- 6a P(k - 4 + 6a 173. No Moment Transmitted. The span in Fig. 1696 con- sists of two restrained beams, connected at mid-span in such a way that shear, but no bending moment, can be transmitted from one beam to the other. The span therefore represents a different phase of partial continuity from that of the previous problem. The principle here involved is employed in the 272 THEORY OF FRAMED STRUCTURES CHAP. VI design of a bascule span composed of two leaves connected by a shear lock. The principle must be modified, however, in its application to a bascule span; for the leaves do not as a rule have a constant moment of inertia, nor are they in perfect restraint at the points of support. A constant moment of inertia and perfect restraint will be assumed in finding the shear V on the pin-connection at C, -Ml FIG. 1696. when the beam CB supports the load P as shown. The M- diagram may be drawn very easily when the partially continu- ous beam ACB is considered as two restrained beams sketched separately; with the shear at C considered as a force V, acting upward on CB and downward on CA. The bending moment at C is zero. The M -diagram for CB is best sketched in two parts the area QST representing the bending moment of V, and the area TUW representing the bending moment of P. SEC. V RESTRAINED AND CONTINUOUS BEAMS 273 Since the continuity is broken at C, it cannot be assumed that the total area of the M-diagram is zero, although the angle between AD and BD is zero. The absurdity of such an assump- tion is obvious when k = i and the load P is at C, in which case it is clear that the M-diagram is a negative area throughout and cannot equal zero; neither can the tangential deviation at either A or B be equated to zero, for a similar reason. Let /i represent the tangential deviation for the beam AC and / 2 for the beam BC, then Whence V = - 2 - SEC. V. CONTINUOUS BEAMS IN FOUNDATIONS 174. In designing foundations for high buildings, it is fre- quently necessary to place three columns on a single footing. An attempt is made to secure uniform soil pressure by support- ing the columns on longitudinal girders resting on shorter cross- beams. These beams bear on the soil and distribute the column loads over a sufficient area of foundation. In the design of such a foundation the engineer frequently adopts the following course. Having determined the sum total of the three column loads, he decides upon the allowable unit bearing pressure of the soil; determines the required area of the foundation; and then shapes the footing by assuming arbitrarily either the length or width of the footing to suit his convenience. The last step in this process disregards the fact that the girders and their loading constitute a statically indeterminate system. The arbitrary proportioning of such a system does not result in uniform soil pressure; hence the actual soil pressure will be greater at some points; and less at other points, than the allowable pressure. Two difficulties are encountered in making a rational analysis of the problem: (i) The columns are usually anchored to the 18 274 THEORY OF FRAMED STRUCTURES CHAP. VI girders with sufficient rigidity to allow for the transference of moment. In the discussion which follows, this moment will be neglected. (2) Since the girders distort slightly in performing their office of distributing the loads, the assumption of uni- formly distributed soil bearing pressure, which the designer necessarily makes (in the present analysis as well as in the customary procedure) is unavoidably vitiated. There is no data available as to the importance of this effect; but it seems reasonable to conclude that a state of uniform pressure dis- tribution is more nearly approximated when the problem is solved by an analysis that takes account of the continuity of the girders, than by the present arbitrary choice of proportions. 600 n\ ToT7Sf\^ /u ' Cn r\Tons f i J. 37.5'--- ->k 37:5^ > L= PIG. 170. The arbitrary method of proportioning will be given in connec- tion with Fig. 170. The sum of the three column loads is 3,000 tons. If a permissible soil reaction of 5 tons per square foot is assumed, the required area of the footing is 600 sq. ft. The size of the footing is taken arbitrarily as 75 ft. long and 8 ft. wide. The center of gravity of the three column loads is 12.6 ft. to the right of the center column, which locates the center of the soil reaction; hence the girders should extend 12.9 ft. beyond the left column and 20.1 ft. beyond the right column. A grillage so designed conforms to the three conditions of static equilibrium i.e. (sum of the vertical forces equals zero, sum of the horizontal forces equals zero, and the sum of the moments about any point equals zero). From the foregoing, it is concluded that the three concentrated column loads are supported by a uniform soil reaction of 5 tons per square foot or 40 tons per linear foot of girder. That this conclusion is fallacious, however, becomes apparent when the structure SEC. V RESTRAINED AND CONTINUOUS BEAMS 275 (Fig. 170) is inverted (Fig. 171); and the soil pressure repre- sented as a uniform load of 40 tons per linear foot resting upon three supports. It is therefore evident that the problem is statically indeterminate, and the odds are greatly against the probability that the reactions are 600, 900 and 1,500 tons respectively. The rational method takes account of the continuity of the longitudinal girders. Three independent simultaneous equa- tions are required for the determination of the three reactions (Fig. 171). Two of these equations SF = o and SM = o are FIG. 171. supplied by the principles of statics. The third or elastic equation is derived by drawing the tangent FG through C, and establishing the relation between /i and k. The application of these three equations to practical cases differs in detail, according to the physical conditions governing the problem. Three of several possible cases are illustrated in Figs. 172, 173 and 174, and will be considered separately. In each case the following points are to be observed: Each figure is shown inverted for convenience. The known column reactions are represented by P, Q and R. The spacing of the columns, being fixed by the architectural features, is known; and represented by c and d. The tangential deviations t\ and h, not shown in Figs. 172, 173 and 174, are to be taken as represented in Fig. 171. Three quantities, differing in each case, are to be deter- mined by a solution of the three independent simultaneous equations cited. 175. Case I. Projections Not Limited by Site. In Fig. 172, let w represent the intensity of the uniform soil pressure in pounds per linear foot. If the architectural features do not limit the end projections a and b of the main grillage girders; it will be possible to attain this condition of uniform soil 276 THEORY OF FRAMED STRUCTURES CHAP. VI pressure by selecting a, b and w accordingly. The two static equations are P + Q + R-(a + b + c + d)w = o (15) aP + (a + c)Q + (a + c +d) R -~ 2 (a + b + c + d) 2 = o (16) The elastic equation is he , x s ! ~d (l ^) The expressions for t\ and ti may be determined as follows: llMMUH "Mill II Mill ii ii iirrrm- iirrrm-n <---- ---- J d PIG. 172. The bending moment between P and Q at any distance x from Pis Mi = Px -- w(a -\- x) 2 2 hence // 1 = f M^xdx = ** - + ^- + c - Jo e ben tance x from 7? is 3 2 2 3 5 Similarly, the bending moment between Q and R at any dis- = Rx - -w(b 2 hence Eftt = M.xdx . Z -.Sir L + i 1 +!! Jo 3 2\ 2 3 4 Substituting the values of t\ and h in Eq. (17) and reducing - ^c 4 ) c Eliminating w from Eqs. (15), (16) and (170) p _ + 6c(P + Q.+ R)a* + 6J(P + Q + )6 2 + 8(c 2 <2 a + 8(J 2 P + ^Q -- c 2 P)b + (-5C 3 + 3 d* - Sc 2 d)P + 3 ? = o (19) SEC. V EESTRAINED AND CONTINUOUS BEAMS 277 Illustrative Problem. Let c = 12 ft., d = 30 ft., P = 60 tons, Q = goo tons and R = 1,500 tons. Substituting the numerical values in Eqs. (18) and (19), and reducing b = a + 7-2 io0 2 + 256* 3720 + 4686 = 10,374 whence a = 7.8 ft., and b = 15 ft. The length of the base is 42 + 22.8 ft. or 64.8 ft., and the soil pressure per linear foot is w = ^T = 4<5 ' 3 tons ' 04'O If the allowable bearing pressure on the soil is 5 tons per square foot, the foundation should have a width of 4 -^ = 9-5 ft- Thus, in Figs. 170 or 172, the girders should be about 65 ft. long; extending approximately 8 ft. beyond the left column and 15 ft. beyond the right column. The beams in the lower tier of grillage should have a length of about 9.5 ft. 176. Case II. Projection at One End Limited by Site. Archi- tectural features frequently fix the length to which the footing FIG. 173- may extend beyond one of the end columns. Sometimes it is necessary to allow no extension whatever at one end. When- ever either of these two limitations arises, the footing may be so arranged that its pressure per foot of length varies uniformly from one to the other; as shown in Fig. 173. This may be accomplished by a variation in the lengths of the cross-beams, resulting in a trapezoidal area for the footing. Care should be 278 THEORY OF FRAMED STRUCTURES CHAP. VI exercised in the choice and spacing of the cross-beams in order that equal deflections at their centers may be assured. Let the intensity of the soil pressure per linear foot, at a dis- tance x from the left end be, w x = Ax + B so that the soil pressure at the left end will have the intensity B per linear foot, and at the right end the intensity Al -\- B per linear foot. The static equations are SF = Al 2 + 2BI - 2 (P + Q + R) = o (20) SM = 2Al s + ^Bl 2 - 6aP - 6(a + c)Q - 6(a + c + d) R = o (21) The elastic Eq. (17), when developed, may be written in the form c[4(a + c) s + a{ 3 + c) 2 + a($a + 2c)}]A + d[(a + c + dY + (a + c)\2(a + c + d) 2 + (a + c) + 5'bO + cY + a($a + 2c)]B + sd[(a + c + <2) 2 + (a + c) = 2o[(c + d) 2 + c(c + d)]P + 2od*Q (22) Three of the four quantities a, 6, A or B may be determined from these equations. A numerical value for either a or b may be chosen arbitrarily to suit the architectural features, and the other three quantities determined by a solution of Eqs. (20), (21) and (22). Illustrative Problem. Let c = 12 ft., d 30 ft., P = 600 tons, Q goo tons and R = 1,500 tons. The architectural features do not allow a to exceed 6 ft., and the designer wishes to utilize this full amount. In other words, he fixes a = 6 ft.; and determines b, A and B, from the equations. Substituting the numerical values in (20), (21) and (22) Al 2 + 2BI = 6,000 2A1 3 + sBl 2 = 550,800 8,273,664,4 + 824,040$ = 43,416,000 whence I 3 227. 76i/ 2 + i3,884-448/ 209,928.788 = o or I = 22.78 or 66.58 or 138.4 ft. SEC. V RESTRAINED AND CONTINUOUS BEAMS 279 If 66.58 ft. is taken, then A = 0.328 B= 55.98 and Al + B = 34.13 Hence the intensity of the soil pressure is 55.98 tons per linear foot at the left end; and decreases uniformly to 34.13 tons at the right end. Since a = 6 ft., and / = 66.58 ft., the foundation extends b = 18.58 ft. beyond the right column. Since the quantity b does not appear in Eqs. (20), (21) and (22), it will be found advantageous in any numerical problem to arrange the nomenclature so that P may represent the end column beyond which the length of the foundation is limited; and assign a numerical value to a. When no extension of the foundation beyond P is allowed, a = o. 177. Case III. Both Projections Limited by Site. In the case illustrated by Fig. 1 74, where both a and b are fixed or equal PIG. 174. zero; the footing may be so proportioned as to produce an intensity of soil pressure per linear foot, which varies according to the ordinates of a parabolic curve. The soil pressure per square foot may be made uniform by varying the length of the cross-beams accordingly. The dimensions a, b, c and d t and the loads P, Q and R will be known; while the three constants which determine the intensity of the soil pressure are to be found by solution. Let the intensity of the soil pressure per linear foot at a distance x from the left end be w f = Ax' 2 + Bx + C 280 THEORY OF FRAMED STRUCTURES CHAP. VI so that the soil pressure at the left end will have the intensity C per linear foot, and at the right end the intensity Al 2 + Bl + C per linear foot. The static equations are SF = 2AI* + 3 Bl 2 + 6CI - 6(P + Q + R) = o (23) SM = 3 ^/ 4 + 4 ^ 3 + 6C7 2 - i2aP - 12(0 + c)Q - i2(a + c + d)R = o (24) The elastic Eq. (17) when developed, may be written in the form c[5(* + cY + a{ 4 (a + c)' + a[ 3 (a + c) 2 + a( 3 a + 2c)}}}A +d[(a + c + dY + (a + c) [2(a + c + d)* + (a + c)\s(a + c + J) 2 + (a + c) { 9 (a + c) + + c) 3 + a{ 3 (a + c) 2 + a( 3 a + + c + dY + (a + c) [i(a + c + + c) + 2d]\]C = 6o[(c + d) 2 + c(c + d)]P + 6od*Q (25) Illustrative Problem. Let a = 6 ft., 5 = 12 ft., c = 12 ft., d = 3oft.,P = 600 tons, Q = goo tons and R = 1,500 tons. Then I = 60 ft. Sub- stituting the numerical values in Eqs. (23), (24) and (25), and reducing i,2oo<4. + 30$ -f- C = 50 i,8oo,4 + ^oB + C = 51 623,028^ + 38,304.6 + 3,8i5C = 201,000 whence A = 0.0291 B = -1.646 C 64.466 and the intensity of the soil pressure per linear foot at any distance x from the left end is w x = o.o2gix 2 1.6462 + 64.466 The intensities of the soil pressure per linear foot at the two ends and at five intermediate points are SEC. V RESTRAINED AND CONTINUOUS BEAMS 28l X w X w Feet Tons per linear foot Feet Tons per linear foot o 64-5 40 45-o 10 Si-o So 55-o 20 43-o 60 70.5 30 41.0 If the footing is to extend a given distance, say 5 ft. beyond one end column, and is not to extend beyond the other end column, it will be found advantageous to arrange the nomencla- ture so that a o and b = 5. Equations (23), (24) and (25) are applicable when the footing is not to extend beyond either end column; in which case a = o and 6 = 0, are to be substituted. CHAPTER VII DEFLECTION OF TRUSSES 1 78. Stress and Strain. The stress in any member of a truss under the influence of external forces is accompanied by a corre- sponding strain or change in length of the member. The strains which the various members experience when under stress, cooper- ate in a general distortion of the truss ; somewhat after the manner illustrated in Fig. 175. The full lines represent the configuration of the truss when the members are under no strain. When loads are applied gradually at , C and G, the truss undergoes a gradual distortion ; and finally conforms to the shape indicated by the dotted lines, when the loads have reached their full magnitudes FI, F-z and F s . The dotted outline gives an exaggerated idea of the distortion which may be expected in any practical problem. The first step in finding the movement of displacement of any point in the truss, is the determination of the strain in each member subjected to stress. This is easily accomplished by the well established law of mechanics, viz.: unit stress , , , ~ . ,% , , .. ., - = modulus (measure or coefficient) of elasticity. unit strain Let P = the stress in a member in pounds. A = the area of cross-section of the member in square inches (inch 2 ). / = the length of the member in inches, D = strain, or change in length, of the member in inches. d =~Y = unit strain in inches per inch (inch/inch), a ratio. p j- = unit stress in pounds per square inch (pounds/inches 2 ). A E = modulus of elasticity in pounds per square inch (pounds/inches. 2 ) P A PI 7 282 SEC. I DEFLECTION OF TRUSSES 283 Hence the strain or change in length of a member may be determined if its length and cross-sectional area, the stress which it resists and the modulus of elasticity of its material are known. Experiments show that, for any given material, the modulus of elasticity is approximately constant for all unit stresses below a certain limit, called the elastic limit. The elastic limit for structural steel is about 60 per cent of its ultimate strength; hence the permissible unit stress in all current practice is well within the elastic limit. The modulus of elasticity for ordinary structural steel is about 29,000,000 Ib. per square inch; hence if a member 50 ft. long, having a cross-sectional area of 20 sq. in., is subjected to a tensile stress of 300,000 Ib., the strain or elongation of the member will be ~ 300,000 X 50 X 12 D = * '-- - = 0.31 in. 20 X 29,OOO,OOO SEC. I. ALGEBRAIC METHOD 179. The algebraic solution may be developed by equating the external work done by the external forces, to the internal work performed upon the members of the truss. Let AI (Fig. 175) represent the vertical component of the dis- placement, or deflection of the point B in the direction of the force FI. During the movement of the point from B to B', the force which has been increased from zero to FI performs work to the amount of % FiAi. Each member under stress, and thereby subject to strain, contributes its share to the deflection of B. Consider any member HK, for example, which is / inches long and has a cross-sectional area of A sq. in. Let PI, P 2 and P 3 284 THEORY OF FRAMED STRUCTURES CHAP. VII represent the stresses in HK, caused by the loads FI, F 2 and Fz respectively; and let PI -f P 2 + PZ = P = the total stress. PI The strain in HK is D = ^ and the total work performed upon HK, as the three loads are applied and the stresses gradually increased from zero to their final values, is 2 V 2 2 AE The work performed upon HK, resulting from the load FI is Ipn = J P JW 2 2 l AE PI Let 2 % Pi-rjf represent the sum of the work performed upon each member by the stress resulting from the load FI. Since the external work done by the loads must equal the internal work performed upon the members, then IF A _ 2 ip PI - r\ AI L- r\ -=, 2 2 AE Whence PI Pl_ F! .4 in which PI is the stress in any member due to the load FI, and P is the stress in the same member due to the three loads FI, F% and FZ. Since the stress PI in any member varies directly as the load p Fi; it is obvious that the ratio ^r is constant for that member TI for all values of FI. To get the value of the ratio u for each member, assume for the present that FI equals i lb.; in other words, place i lb. at the point, acting in the direction of the desired deflection and compute the stresses in all members for this i lb. loading. The resulting stresses expressed in pounds, when divided by i lb., will represent the ratio u for the various members. For SEC. I DEFLECTION OF TRUSSES any member, the stress P and the ratio u have positive or negative signs corresponding to tension or compression. The strain in any member causes the point in question to deflect in the direction in which the i Ib. load is assumed to act, if P and u have like signs ; and in the opposite direction, if they have unlike signs. In Eq. (i) E is the only quantity which is constant for all members of the truss, provided all members are made of the same material (which is usually the case). Hence the expres- Pul sion r- may be tabulated and computed for each member; Pul and the deflection found by dividing the algebraic sum I T- A. by E, as in the following problem: 1 80. Illustrative Problem. The truss in Fig. 176 supports a load of 240,000 Ib. at each bottom chord panel point. In Table I, the length of each mem- FIG. 176. ber is given in column i ; the gross cross-sectional, in column 2 ; and the stress, in column 3. The quantities in column 4, when divided by E, represent the strains. To find the deflection at point 6, place i Ib. at 6 and compute the stresses; as given in Fig. 177. These stresses, when divided by i Ib., are the ratios UQ in column 5. In this case P and MQ have like signs for each member, hence the quantities in column 6 are all positive, and their algebraic sum is 42,246. Since P was expressed in units of 1,000 Ib., the deflection at point 6 is 42,246 X 1,000 42,246,000 ., . A 6 = = - - = 1.46 in. E 29,000,000 286 THEORY OF FRAMED STRUCTURE CHAP. VII " sffai'sl'IJ I II 5552 1 2 ^ ++++++TI+ ++++++++ s \ \(N \O) \(N \IN \OC \P8 \00 V* \* \N >0\ 1-N M\ J\ KJ\ ^ K}\ 1O\ O O O O >0\ >O\ f\ -l\ r^S rt\ -f\ f\ 1 S TrCI NXM^!Ci (C^XivZX^ol ^Tl sIlslI^ OC>0 2!tKKlll ^ +++++++ T ++++++++ - 1 iTT + + + + + c i + + -f + i i i 1 1 O ON N Ti i?i i i i i i i iTi i iTTViT OMcooNTj-MTtMpo>or-atoooNOoooot-o\ a -^ ft. ini^n ? linur? l/i a \QO Vf VH 1 spO \00 \00 \30 \flO NOO \00 \00 NpO \00 \*jt \^< \00 J\ ej\ n\ eo\ oj\ o>\ io\ iff\ O O M o O >o\ ieK o>\ \ \ w\ \ \ Tf SOOOOOtOOOOOOOOWOOOOOO OOOOOOOOIONOOM M MOONI/500000000N 00 PO ' PO O O M M O V O O \OvOMMOOfOPOOO OH ^ 1 ' '++++'+ + +'++++' 1 1 ill ooooooooooooooooooooo 55 I'D 1 1 1 + + + + 1 + + +1 + + + + 1 1 1 2^ J ^oa'^^^o o^^"' 'ot < *3 1 8 8^8 >J .2 I 1 O ON N M + I 3" + i SEC. I DEFLECTION OF TRUSSES 2 8 7 In like manner the deflection at point 4 is found by placing a load of i Ib. at 4 (Fig. 178) and obtaining the ratios u in column 7. The products of corresponding values in columns 4 and 7 are given in column 8. In this instance, P and u have unlike signs for the member 4-5; hence the quantity represent- ing the member 4-5 in column 8 has the negative sign, and the strain in that member tends to raise the point 4. The alge- 1 - 3 / 4 3 -% 5 -% 7 - 3 /4 9 I 8 /-. | FIG. 189 C Vg D F/G.A92 equally applicable to trusses. It is also easily proven by the algebraic method of Sec. I. Let A represent the deflection at A , caused by the load F&tB (Fig. 189) ; and let A 6 represent the deflection at B, caused by the load F at A (Fig. 191); then PaU a l A a = AE and AE 300 THEORY OF FRAMED STRUCTURES CHAP. VII where P a = stress in any member of Fig. 189. P b = stress in any member of Fig. 190. u a = stress in any member of Fig. 191. u b = stress in any member of Fig. 192. Choose any member as CD, then D 2F SF 8 , 2 P a = , Pb = , u a = - and u b = - 999 9 hence P a u a = P b u b If any other member be taken at random it will be found that P a U a = PbUb therefore A a = A 6 or the deflection at A, caused by a load at B, equals the deflec- tion at B caused by the same load at A . CHAPTER VIII SWING BRIDGES A swing bridge rotates in a horizontal plane about a vertical axis, and is classified as center-bearing or rim-bearing in accord- ance with the method by which it is supported while swinging. SEC. I. CENTER-BEARING SWING BRIDGES 185. General Considerations. When a bridge of this type is open, each truss is supported at the end of a cross-girder which rests on a center-bearing c (Fig. 193). This bearing is usually a phosphor-bronze disk, from i to 3 ft. in diameter, between two hardened steel disks. To prevent the bridge from tipping; balance wheels w, about 18 in. in diameter and six to eight in number, are fastened to the trusses and floor system in such a way that they roll on a circular track /. These wheels are so adjusted that they carry no load, except when the bridge is out of balance on account of wind pressure or similar causes. When the bridge is closed; the ends a, b } d and e are raised a proper amount by wedges. Wedges are also inserted at / and g a sufficient amount to give the trusses a bearing on the pier independent of the cross-girder; but no attempt is made to lift the trusses at these points in order to relieve the cross- girder of any of the dead load. Thus at the center support, the dead load is carried by the cross-girder ; while the live load is supported directly by the pier. 1 86. Conditions of Loading. When the bridge is open, the dead load is balanced about the center support. When the bridge is closed, the total dead load reaction at the center is re- lieved, as the wedges are driven and the ends raised. (If the ends were raised to a sufficient height, the bridge would be lifted free of the center support; and the reaction which formerly supported the bridge would be transferred from the center 301 302 THEORY OF FRAMED STRUCTURES CHAP. VIII to the ends.) The mechanical parts of the bridge are usually designed in such a way that the end wedges, when fully driven, will provide a positive dead-load reaction; somewhat greater than the negative live-load and impact reaction. This insures the bridge against hammering of one end when the train comes on at the other end. By this arrangement, the dead- load end reactions are less than those obtained theoretically when the truss is considered as a continuous beam over three level supports. The cost of installation and operation are thereby reduced. If hammering is eliminated and the ends remain fixed in x/ 1 1 ~~ljy fW , .r. FIG. 1 93 FIG. 194 FIG. 195 FIG. 136 elevation, the live-load reactions may be computed on the basis of complete continuity of each truss. If, however, on account of error in design or adjustment of the wedges; or because of settlement of the piers, it happens that no dead-load end reaction is present, we have the condition as illustrated in Fig. 194. Let us assume, for example, that a very small clearance exists between the beam and its end supports. The total weight of the beam is supported at C, as is the case when the bridge is open. When the live load comes on the arm BC (Fig. 195), the beam tilts until it finds a bearing at B\ and the live load is supported by BC acting as a simple span. However, the dead load of weight of the beam is still entirely supported SEC. I SWING BRIDGES 303 at C. If the live load is present in both arms, (Fig. 196) the beam will deflect until it has a bearing at A and B ; and thus a condition of continuity, or partial continuity, must be considered in finding the live load reactions. The extent of the continuity will depend upon the amount of clearance which previously existed at A and B in Fig. 194. The dead load is still totally supported at C. In order to provide for these contingencies, the following conditions or classes of loading should be considered. Case I. Dead load; bridge open or wholly supported at the center. Case II. Dead load; bridge closed, both ends raised until a definite end reaction is attained; amount to be specified after negative live load and impact end reaction have been deter- mined. ' Case III. Live load on one arm only; acting as a simple span. Case IV. Live load on one arm only; continuous girder action. Case V. Live load on both arms; continuous girder action. If the ends are raised we have Case II, combined with Case IV or Case V. If the ends are not raised we have Case I, com- bined with Case III or Case V. The bottom chords are usually better protected from the rays of the sun than the top chords. Because of this, and also on account of cold weather and ice, it is apparent that the tem- perature of the top chord may be considerably higher than that of the bottom chord. Thus the top chord may be lengthened and the bottom chord shortened, causing the span to "hump" at the center; thereby relieving the center reactions somewhat and increasing those at the ends. Except in special cases the temperature factor is neglected. 187. Stresses in a Swing Span. The stresses for the five cases of loading will be computed for the 3oo-ft. span (Fig. 197). The assumed dead load is 3,000 Ib. per foot of span, or 37,500 Ib. per panel per truss. The end panel load is assumed at 20,000 Ib. A Cooper's -40 will be taken for the live load. The impact allowance will be computed from the formula T - T 3 1 = L j : * + 300 304 THEORY OF FRAMED STRUCTURES CHAP. VIII > POP) vO Ov O\ 00 POIC oo | el 1 1 1 ++ + ro N M 00 ro ( 1C M p) \O H tn r-i PO M W g + | + + 1 -f ++ + 1 + 1 a -OK 1 0! < i i $ ro PO PO C\ n 000 5 O 2 S OVPI M IS 1 1 1 1 1 1 1 + i 1 1 1 ++ + + + + o 1C ^ 0* fS M I i + + + + I 1 1C M PO 55 j^ 1C t^ PI (N M P-l I I PO ro PO 1-1 H PO CS vO 1 + 1 i + 1 1 + i 1 1 1 + 1 + O 1 o ic jc O PO O> 00 O Is a W M PO g 1 1 1 ++ + ** O g o O 13 c o "o oq 1 T PO ~ r { Ig PO PI $ 00 l/> PO W ro o o t^ 1O O O PO N 1C M t- OV O> PO P> MM PO P fa ++ + ++ ~*~ ++- h+ 1 1 1 ++ + 1 1 1 ^^ o -M a a> i i V *>s j 1 J ti$ o o 00 M O PO (*> 00 M 00 POOO g 5-0 1C f- N M t- Ov ICt- N M J~- OV g g 1 ! 1 1 1 1 1 1 1 1 1 ! + + + ++ + 3 "1 M ,1 i ii 4* '1 M M 1 00 O\ PO , oo a 82 t~ ro M T}- 1C 10 O 1O M P) rt ,| 1 I 1 i T 1 i i I 1 1 1 O ++ + O i ! 1 1 i 1 o\ M 1 00 1C 1 Ov PO p> i Q i ! 1 1O (N P) 1 ! M I "a b "3 ^ J i ^ b b Ed ^ ^ SEC. I SWING BRIDGES 305 OvO *$ Ov Tj- 512 a " 00 "t IN 10 1- ro MM ro 00 N O w w 10 1 1 1 1 1 ++ + 1 1 1 ++ + 10 O. O> W O O 00 IO OO 1 1 1 1 1 \ ++ + lOt- M o N IOOO fT) M 00 "t o t- O 00 MM M oo 10 ro MM Tf + 1 + I | 1 + 1 + " + 1 1 1 1 ++ + | M IO * M 00 ro 1 1 1 1 1 + I as S O O o a M IN $ 3 M- w 5 M 00 Ol + i 1 + 1 + + + + 1 1 1 ++ + ++ + \O ro O\ 10 10 " i " 1000 ro 00 Ov 00 g "t 00 g IO a 1 1 + 1 + i- oo 10 ro 5 >O ro OiO 00 10 M O ro C4 o O O ro d O row 10 o o o ro O 1 1 i 1 1 ! ' ' I ++ + 1 1 1 + + + w s ro M MOO M M o *t 00 0$ C 1 ) 1/5 t- IO IN 00 O\ ^ ++ + + + + 1 1 1 ++ + 1 ! 1 Oi + + + ao 10 i/)oo PO vO 1C hi 00 M o\ O w OO ""f f*5 r* M M M ^* 00 M M 1 1 i I | I + 1 1 1 +-f- + ++ + 00 O Tt Ovro MO M ++ + 00 vO 1 1 1 + + + MM N 1 1 1 ro>o CS M 1 roO, IN N ++ + + + + N ro >o O O O \O V5 M M O M Ov ro M OOvO -t *0 "t t^vO Tf 1 1 1 + + + 1 1 1 ++ + ++ + 0\ 1 00 g O 10 OO 1 1 1 + -t | 20 THEORY OF FRAMED STRUCTURES CHAP. VIII in which 7 = impact stress L = live load stress / = loaded length of span causing the live-load stress. Since the bridge is symmetrical about the center, the stresses will be computed for the left arm only. In Case III, this arm is considered as a span simply supported at LQ and L 6 . The stresses are computed in the ordinary way. The results expressed in 1,000 Ib. units are tabulated in Table I; the impact stresses appearing directly below the corresponding live-load stresses. Thus for the member UiU s the live-load stress is 208,000 Ib. and the impact stress is 139,000 Ib. For the member L^U^ the live-load and impact stress is +72,000 Ib. when the left segment of the arm is loaded; and 145,000 Ib. when the right segment of the arm is loaded. In Cases IV and V the reactions, being statically indeter- minate, cannot be accurately computed until the truss has been designed. In order that a preliminary design may be made; the reactions will be tentatively determined by assuming that the truss functions as a beam of uniform cross-section, con- tinuous over three level supports. Only the end reactions RQ are necessary. They are given in Table II as determined from Eqs. (9) and (n) page 257. By this process it is possible to compute the stresses and make a preliminary design, after which the true reactions may be determined. TALBE II i Ib. at R 1,000 Ib. at Ro 1 +Q-793 L 7 0.064 L 2 +-593 L s 0.092 L 3 +0.406 L, -0.094 L, +0.241 L! -0.074 L, +0.103 L n 0.040 188. Positive Shear in Panel o-i. Case III. The influence line abc for shear in the panel is shown in Fig. 197, from which the following criterion for maximum positive shear may be developed. SEC. I SWING BRIDGES 307 If PI = the load in panel o-i and P = the total load on the arm 0-6 the criterion is U, U? U 3 Uio Un When the train is moving to the left, wheel 4 passing LI satisfies this criterion and the maximum shear in the panel is +162,000 308 THEORY OF FRAMED STRUCTURES CHAP. VIII lb. The area abc is 62.5 and the equivalent uniform load per linear foot is 162,000 q = -^- = 2,500 Case IV. The influence line for positive shear is defghij (Fig. 197), from which the following criterion may be developed. If PI = the load in panel o-i P% = the load in panel 1-2 P 3 = the load in panel 2-3, etc. the criterion is 793P! < 2ooP 2 + iSyPs + i6sP 4 + i38P 5 + io 3 P 6 Try wheel 4 at L\, train moving to the left. Wheel 4 approaching LI 793Pi = 793 X 50 = 39>65o 2ooP 2 = 200 X 66 = 13,200 i8 7 P 3 = 187 X 56 = 10,472 i65P 4 = 165 X 73 = 12,045 i 3 8P 5 = 138 X 57 = 7,866 i03P 6 = 103 X 50 = 5,150 48,733 39,650 <48, 733 therefore the shear is increasing. Wheel 4 having passed LI 793Pi = 793 X 70 = 55,510 2ooP 2 = 200 X 59 = 11,800 i87P 3 = 187 X 43 = 8,041 i65P 4 = 165 X 86 = 14,190 i 3 8P 5 = 138 X 44 = 6,072 i03P 6 = 103 X 50 = 5,150 45,253 55,5io>45,253, therefore the shear is decreasing. When the train is moving to the left, wheel 4 at LI satisfies the criterion for maximum shear in panel o-i. The shear may be computed by scaling the length of the ordinate in the influence line for each load, and taking the sum of the products of each load and its ordinate. Heretofore this has been the usual method of procedure. It requires that the influence line be drawn quite accurately to scale, and con- siderable care taken in scaling the ordinates. The shear may SEC. I SWING BRIDGES 309 also be determined by taking the sum of each floor-beam load and its corresponding ordinate as follows: 75.64 X 0.793 = 60.0 ,52. 56 X 0.593 = 31.2 72.80 X 0.406 = 29.6 58.20 X 0.241 = 14.0 48.60 X 0.103 = 5.0 139.8 = maximum shear in panel o-i. A new and much shorter method, as outlined by the author in Engineering News Record, June 9, 1921, will now be explained. If the influence line defghij were straight from e toj, the criterion for maximum shear in panel o-i would be the same as for Case III. The difference between the broken line efghij and a straight line from e toj is so slight in this or any similar truss that the criterion for maximum shear in the panel will, in general, place the train in the same position or approximately the same position, as will the criterion for Case III. A very close approximation to the shear of 139.8 Ib. can be made very quickly, by assuming the same equivalent uniform load in both cases; or, in other words, by assuming that the shears in the panel for the two cases are directly proportional to the areas of the respective influence line diagrams. These areas are proportional to the sums of their respective ordinates, thus area afcrarea defghij 1:2. 5:2. 14. The shear in the panel for Case III was found to be 162 Ib., hence the shear for Case IV is 162. X^ = 138.3 2 -S This is a reasonably close approximation to the actual shear of 139.8 Ib., previously determined. It is now obvious that the stresses in L Ui and L L 2 resulting from positive shear, may be quickly found by multiplying the stresses for Case III by the ratio 2.14/2.5. This ratio for panel o-i remains the same for any bridge having six equal panels in each arm, irrespective of the length of the panel. 310 THEORY OF FRAMED STRUCTURES CHAP. VIII The stresses for Case III are given in Table I ; and the live load stresses in LoUi and LoLz for Case IV are as follows: L Ui = -2ii X = -180 = +135 X^ = +115 The impact stresses for Case IV are determined in a similar manner. The influence line for negative shear in panel o-i for Case IV isjklmnop, and since there is at present no corresponding area for Case III we shall leave this question to be considered later. 189. Positive Moment about U 3 . Case III. The influence line abc for moment about U% is shown in Fig. 198. Let PI = the load on the segment 0-3 and P = the total load on the area 0-6 then the criterion for maximum moment about Us is p Wheel 1 2 at L 3 satisfies this criterion, and the maximum moment about 7 3 is 7,057 ft.-lb.; hence the maximum live load stress in L 2 L 4 is Case IV. The influence line is defghij, Fig. 198. If PI = the load in panel o-i PI = the load in panel 1-2 PS = the load in panel 2-3, etc. the criterion for the maximum moment about Us, when reduced, is Upon trial it will be found that wheel 12 at L 3 satisfies this SEC. I SWING BRIDGES 311 criterion also. Multiplying each floor-beam load by its corre- sponding ordinate, 75.20 X 9-475 = 7 J 3 51.92 X 19-475 = ^ OI1 74.52 X 3 -45o = 2,269 56.56 X 18.075 = 1,022 48.80 X 7-725 = _377 5,392 = maximum moment about U%. The maximum tensile stress in L 2 L 4 is = +179-7 30 This value may be closely approximated from Case III as follows: The ratio of the sum of the ordinates in Case IV, to the sum of the ordinates in Case III, is ^ - 0.756 112.5 and 235 X 0.756 = +178 = stress in I^L 4 . 190. Negative Shear in Panel o-i. Case IV. We are now prepared to consider the negative shear in panel o-i, when the right arm 6-12 is loaded. The influence line isjklmnop (Fig. 197), and the criterion developed therefrom is + 347*11 + 40^12. There are two positions of the train, when moving to the left, which satisfy this criterion ; namely, wheel 1 1 at L% and wheel 8 at Lg. The maximum shear, occurring when the train is in the latter position, is 22.8 lb.; as found by taking the sum of the products of each floor beam load and its corresponding ordinate. This value may be closely approximated by the proportionate method, as follows: Since the influence Imejklmnop (Fig. 197) has its longest ordinate at the center of the arm, it will be compared with the influence line abc (Fig. 198), which also has its longest ordinate at the center of the arm. The latter influence line is for the moment at Us for Case III, and the moment is 7,057. The ratio of the sum of the ordinates 3 I2 THEORY OF FRAMED STRUCTURES CHAP. VIII jklmnop (Fig. 197) to the sum of the ordinate abc (Fig. 198) is 0.364 ? = 0.00324 112.5 and 7,057 X 0.00324 = 22.9 = shear in panel o-i. i U_2_U3 tk U 5 U& U? Us F 1 0.200 This quantity represents also the maximum negative reaction at LO when the right arm is loaded, from which all the stresses SEC. I SWING BRIDGES 313 may be determined as given in the designated column of Table I. 191. Shear in Panel 1-2. The influence lines are shown hi Fig. 199. The stresses in UiLz for Case III, as given in Table I, are n and + 140; from which the stresses for Case IV may be found as follows: 0-593 + 0.406 + 0.241 + 0.103 = + 0.667 + 0.500 + 0.333 + - l6 7 The stresses for Case IV, determined by the exact method, are 15 and +113 respectively. 192. Shear in Panel 2-3. The influence lines are shown in Fig. 200. The stresses in L 2 C/3 for Case IV, determined by the proportionate method are x 0,207+0407 . +48 0.167 + 0.333 _ 82 x + 0.241+0.103 = _ 6l . 0.500 + 0.333 + 0.167 The stresses for Case IV, found by the exact method, are +48 and 60 respectively. 193. Shear in Panel 3-4. The influence lines are shown in Fig. 201. The stresses in UsL^ for Case IV, determined by the proportionate method, are -82 X ' 2 7 ' 4 7 ' 594 = -99 0.167 + 0.333 + 0.500 +39 X 9^41+0103 + 0.333 + 0.167 The stresses for Case IV, determined by the exact method are 101 and +26 respectively. 194. Shear in Panel 4-5. The influence lines are shown in Fig. 202. The stresses in LU 5 for Case IV, determined by the proportionate method, are X ' 2 7 ' 4 7 ' 594 -7g = +165 0.167 + 0.333 + 0.500 + THEORY OF FRAMED STRUCTURES CHAP. VIII The stresses for Case IV, determined by the exact method, are + 167 and 5 respectively. 195. Shear in Panel 5-6. The influence lines are shown in U3 U 4 U 5 U 6 U-T U 8 U 9 Fig. 203. The stress in U 5 L 6 for Case IV, determined by the proportionate method, is -2ii X - 2 7 + 0407 + Q.594 + 0-759 + 0-897 = _ 2 0.167 + 0-333 + 0.500 + 0.667 + 0.833 " SEC. I SWING BRIDGES 315 The stress for Case IV, determined by the exact method, is -241. 196. Moment about 1/2. The influence lines are shown in Us Us UIQ Uu ' W FIG.204 Fig. 204. The stress in UiU* for Case IV, determined by the proportionate method, is , 81.8 208 X - - = 170 100 THEORY OF FRAMED STRUCTURES CHAP. VIII The stress for Case IV, determined by the exact method, is -172. 197. Moment about L 4 . The influence lines are shown in U, U? U 4 U 5 U& U 8 U 9 U ie U,, Lo CaseHI CaselY iaseUI FIG. 2 06 LII t-12 Fig. 205. The stress in U S U 5 for Case IV, determined by the proportionate method is -208 100 = -I3 2 SEC. I SWING BRIDGES 317 The stress for Case IV, determined by the exact method, is -134- 198. Moment about U 5 . The influence lines are shown in Fig. 206. Whenever each arm has six or more equal panels, the influence line for Case IV shows a reversal of stress in one or more chord members of the loaded arm adjacent to the center support. This phenomenon is explained in connection with Fig. 207, where a is the distance from R\ to the point of zero bending moment; then aRi= P(a - kl) fromEq. 9, page 257 p 4 hence a = - ! - For any position of P, the limits of k are o and i ; hence the limits of a are %l and /. It is clear, therefore, that if any panel point experiences a negative moment from the influence load; the distance of that panel point from the center support must be less than J of the arm length. If the panels are of equal length this condition can occur only when there are six or more panels in each arm. 3*8 THEORY OF FRAMED STRUCTURES CHAP. VIII The stresses for Case IV cannot be determined by the pro- portionate method as heretofore, on account of the dissimi- larity of the influence line diagrams. In such instances the equivalent uniform load may be approximated by the use of equivalent uniform load table, page 206. For the right segment, l\ = 25, / 2 = 61.5 and the equivalent uniform load is 2.7; the area is 464 and the stress in L^L 6 is = 30 For the left segment, it will be sufficient to call l\ li = 30, and the equivalent uniform load is 2.88, hence the stress is -38 X 2.88 _ ~~ 199. Moment about L 6 . The influence line for Case IV is shown in Fig. 208. There is no corresponding influence line for Case III. Since the influence is symmetrical about the center, the stress in U^U^, when the left arm is loaded, is the same as previously determined when the right arm was loaded. 200. Case V:Both Arms Loaded. Broken Loads. By re- ferring to the influence lines for the continuous condition in Figs. 197, 198, 204 and 205, it is apparent that the stresses in the mem- bers there considered will be less for Case V than for Case IV; since any load, brought on to the right arm while the left arm is loaded, will decrease the stress because the influence areas on opposite sides of the center have opposite signs. In Figs. 199, 200, 201, 202, 203, 206 and 208 the conditions are different; since loads on the right arm and the left segments of the left arm con- spire, and the live-load stress in any instance is the sum of the live stresses as given for Case IV. Consider the member U&Lt, for example, illustrated in Fig. 201. The live-load stress is 99 when the left segment of the left arm is loaded, and 30 when the right arm is loaded; hence if a train approaches on each arm, the maximum live load stress is 129, as given in the column for broken loads. No impact is added when broken loads are considered. Specifications are not usually clear on the ques- tion of broken loads. If the location of the bridge is near a large freight terminal, it is conceivable that trains might SEC. I SWING BRIDGES 319 occasionally approach simultaneously from both ends of the bridge. Continuous Loads. If Figs. 199, 200, 201, 202, are taken con- secutively, it is apparent that the positive influence line area for Case IV is decreasing, and in Fig. 201 it becomes less than the negative area for the right arm; the difference in areas being 2.65. Therefore the stress in UsL* for Case V will be greater than for Case IV. The stress may be approximated as follows: Consider that the train moving to the left covers the whole span. Assume that the engine covers the segment ab and that the stress in Z7 3 Z, 4 , on account of the engine, is the same as in Case IV or 99 Ib. The shear in panel 3-4, on account of uniform train load of 2,000 Ib. per linear foot., from b toe is 2.65 X 2 = 5.3 Ib., and the stress in U^L^ is 5.3 X 1.3 = 6.9. Hence the stress for Case V is 99 6.9 = 105.9. Similarly the stress in L 4 / 5 is +165 + (7.6 X 2 X 1.3) = +184.8. 20 1. Negative Shear in Panel 5-6. There is no positive area in the influence line for the continuous condition of Fig. 203. The train must be moving to the left, if the engines are to be on the segment ab ; followed by a uniform train load on the segment be. Since the segment ab is considerably longer than the length of the two engines ; the shear in panel 5-6 for Cases III and IV would be considerably less, if the engines were moving toward the left instead of in the opposite direction. We shall take this difference into consideration, by computing the negative shear in panel 5-6 when the train is moving to the left. This occurs when wheel 16 is at L^ and the shear (Case III) is 151.7 Ib.; which is considerably less than 162 Ib., when the train is moving to the right. Taking the same ratio of ordinates as before, the shear for Case V is 9.1 X 2 = l8.2 -192. The stress in U&LG is 192.0 X 1.3 = 249.6. 202. Moment about L 6 . The stress in U 5 U 6 for Case IV was found to be +115, one area being loaded. When the other 320 THEORY OF FRAMED STRUCTURES CHAP. VIII arm is covered with a uniform train load of 2,000 Ib. per foot, the additional stress is +91, hence the stress for U 5 U& for Case V is + 206. 203. Dead Load : Bridge Open. Case I. The panel loads at each end are 20,000 Ib. and all others 37,500 Ib. Each truss is balanced on the center support and the stresses are statically determinate. They are given in Table I. 204. Dead Load : Ends Raised. Case II. The maximum negative reaction on account of live load was found to be 22.9, to which must be added 15.3 per impact, or a total of 38.2. Hence, if the positive uplift at each end is 38.2 or greater, there will be no hammering of one end when the train covers the opposite arm of the bridge. It will be assumed that the machinery parts are to be designed and adjusted so that the end wedges, when driven, will exert an upward pressure of 50. Since there is a dead load of 20 at the end panel point, the resultant or net positive end reaction is 30; and the resulting stresses are given in Table I. If the truss were treated as fully continuous, the end reaction would be 86.4 instead of 50 ^ and a heavier and more expensive lifting device would be required. 205. Combinations. As previously explained, Case I is com- bined with either Case III or Case V ; and Case II is combined with either Case IV or Case V. Only two- thirds of the dead- load stress is taken, when dead-load and live-load stresses have opposite signs. Many specifications are not clear upon the question of stress reversals. In treating reversals, each com- bination should be considered separately; i.e., the largest positive stress of one combination should not be considered with the largest negative stress of another combination. The members have been proportioned, and the gross cross-sectional areas given in Table III. PI 206. Reactions from Williot Diagram. The quantities -j in Table III, when divided by the modulus of elasticity E, are the strains in the various members when the center reaction is removed and the truss, supported at o and 12, carries a load of i Ib. at joint 6 (Fig. 209.) The Williot diagram is drawn SEC. I SWING BRIDGES 3 2I PI in Fig. 210 by using the quantities ~^ to represent strains. The quantities d which are proportional to the deflections are U, 2 1 4- 5 6 7 8 9 '~ " 470 455 414 342 r/# ?38 /?/ /?/ ?3& We 414 455 FIG. 209. indicated in Fig. 209. The deflection at the center has been checked in Table III, where P and UQ necessarily have the same numerical values; but it should be remembered that P is measured in pounds, while u& is a ratio. L O It is clear from Maxwell's Theorem that if i Ib. at joint 6 causes the deflec- tion di = 121 at joint i; then i Ib. at joint i will cause the deflection d\ = 121 at joint 6. Hence, with i Ib. at joint i, the reaction at joint 6, when the truss is continuous over the three supports, is R 6 = = 0.258 470 and from statics R = i.o X 275 - (0.258 X 150) 300 = 0.788 The reactions at the left end, determined in the same manner, are given in Table IV. If the accurate reactions in Table IV are compared with the assumed reactions of Table II, it will be apparent that the differences are comparatively small. The greatest percentage of error occurs when the load of i Ib. is at the center of either arm. This comparison gives a fair idea of what error may be expected, when the truss is assumed to be a beam of constant moment of inertia; and no recognition is made of deflection due to shear. 322 THEORY OF FRAMED STRUCTURES TABLE III CHAP. VIII Member Length, inches Area, square inches A Stress in pounds P PI A w 6 Pud A L*-Vi 469 32.5 0.652 -9.40 0.652 +6.13 Ui- U 2 300 30.5 -0.833 8.20 -0.833 +6.84 U 2 - Us 300 30.5 -0-833 8.20 -0.833 +6.84 Us- Ui 300 30.5 -1.666 16.40 -1.666 + 27-35 Ui~ f/ 5 300 30.5 1.666 16.40 - 1 . 666 + 27-35 Z/6- Us 300 68.2 2.500 10.99 2.500 + 27.46 it-ii 300 19.8 +0.416 +6.32 +0.416 + 2.63 L,-L Z 300 19.8 +0.416 +6.32 +0.416 + 2.63 LZ-LS 300 26.5 + 1.250 + 14.14 + 1.250 + 17.68 L s -Li 300 26.5 + 1.250 +14.14 + 1.250 + 17.68 Li-L$ 300 44-5 + 2.083 +14-05 + 2.083 + 29-25 LS-LG 300 44-5 + 2.083 +14-05 + 2.083 + 29-25 Ui- Lz 469 19.8 +0.652 +15-44 +0.652 + 10.08 Lz- Us 469 19.8 0.652 15-44 0.652 + 10.08 U z - Li 469 32.5 +0.652 +9.40 +0.652 +6.13 Li- Uf, 469 44-5 -0.652 -6.88 -0.652 +4-49 Us-Ls 469 68.2 +0.652 +4.48 +0.652 + 2.92 Ui-Lj Uz- L 2 U 3 - L 3 Ui-Li u*,-Ls o U.-Lt d G = Z PuJ, 234-79 2 = 469.58 TABLE IV lib. at Ro i Ib. at Ro Li +0.788 Li 0.068 L t +0.581 L s o. 108 L 3 +0.386 L, o . 114 L* +0.226 L IQ 0.088 L, +0.099 Ln 0.046 SEC. I SWING BRIDGES 323 The reactions for the preliminary design might be obtained by assuming that all members have the same cross-sectional area, which may be taken as i sq. in.; and constructing a Williot diagram. The combinations of stresses for Cases I and III determine the sizes of nearly all members, except the end post and chord members adjacent to the center support. Since the stresses for these cases are statically determinate, they might be used in making an estimate of the sizes of the members; and their areas used in constructing a Williot diagram. The continuous girder formulas give such satisfactory results that a re-design is seldom necessary, except in a long span and then only for a few members adjacent to the center support. SEC. II. RIM-BEARING SWING BRIDGES 207. The trusses in a rim-bearing swing bridge are supported by a large circular girder, which rotates with the span. The girder rests on conical rollers, usually about 18 in. in diameter; and as many rollers are used as the circumferential length of the girder will permit, in order to give as many bearings for the girder as possible, and thereby minimize the deflection. The trusses may rest directly upon the circular girder or drum, as in Fig. 2116. This arrangement is undesirable, because it does not give an equal distribution of the load to the rollers. A better arrangement is shown in Fig. 2110, where the truss loads are distributed to the drum at 8 points instead of at 4 points, as in the previous case. This arrangement gives a more even bearing. In the diagrams here shown, the center pivot receives neither dead nor live load. It is better to frame the structure so that from 15 to 20 per cent of the load is transmitted to the center pivot through radial girders. In any case each truss is supported at two points over the circular pier, as illustrated in Fig. 2im. The reactions for the center-bearing bridge of the previous section were determined by assuming that the span functions as a beam of constant cross section, continuous over three sup- ports; no allowance being made for deflection due to shear. 3 2 4 THEORY OF FRAMED STRUCTURES CHAP. VIII It was shown that the reactions thus computed by continuous girder formulas, compared very favorably with the true reac- tions determined after the design had been made. Such is not the case when the continuous girder formulas of Art. 167 are CossET ^ TV CO PIG. 211. applied to a swing span on four supports. These formulas give a negative reaction at LI and a positive reaction at L\^ when the left arm is loaded. This live load negative reaction, when the impact factor is added, is in many instances, numer- ically greater than the dead load positive reaction; indicating that a live load over the left arm would lift the truss from its SEC. I SWING BRIDGES 325 support at LI. This assumption is not justified either by exact analysis, made after a truss is designed; or by observed data taken after erection. 208. Partial Continuity. Equal Moments at Center Sup- ports. It will be assumed that the diagonal bracing in panel 6-7 is so light that no appreciable shear can be transmitted through this panel. The shear in the panel is assumed zero under any condition of loading; hence RT = R 13 for loads on the arm 0-6, and R& = Ro for any loads on the arm 6-13. The formulas given in Art. 172 are applicable where a = 20/150; hence for an influence load of one pound and k k 3 T> K K 13 " 7~~ 4-0 The five cases of loading to be considered are the same as for a center-bearing bridge. The influence lines for shear in panel o-i for Cases III and IV are shown in Figs. 21 id and 2iie, re- spectively. If the live load is an 40, the stresses for Case III will be the same as for the center-bearing bridge given in Table I. The stresses for Cases IV and V may be found by the pro- portionate method. For example, the stress in LoUi for Case III is 2ii, hence the stress for Case IV is -211 X - - . . _ g The influence lines are drawn and the stresses computed for all the members, in precisely the same manner as for the centre- bearing bridge. It may be noted that in this particular problem the positive shears are greater, and the negative shears less, in the rim-bearing type than in the center-bearing type. The stresses in some members will be greater, and in others less than in the center-bearing bridge. If, however, an independent design is made for the rim-bearing type, a comparison will show no appreciable difference in the two designs. For this reason it is a common practice to disregard the center panel when the stresses are computed. The diagonals in the center panel are 326 THEORY OF FRAMED STRUCTURES CHAP. VIII light adjustable members, which serve only to provide stability to the structure when open, and resisting a longitudinal wind pressure. After the trusses have been designed, a sufficiently exact analysis of the reactions may be made by omitting the bracing in the center panel; removing the center supports, and drawing a Williot diagram for one pound loads placed at L 6 and 7. INDEX Arch, three hinged, 75, 78, 109 Area, moment method for deflections, 208 ' B Bay, 109 Beam defined, 83 Beams under moving loads: criterion for maximum bending moment, 134, 143 shear, 146 influence line for bending moment, 132 reaction, 130 shear, 132 illustrative problem, 138 point of greatest maximum bend- ing moment, 143 Beams under stationary loads: bending moment determined from location-direction dia- gram, 84 shear and bending moment dia- grams, 86-99 concentrated loads, 87 concentrated and uniform loads, 96 uniform loads, 90 continuous, 240, 248-256 Clapeyron's theorem of three moments, 268 coefficients for pier reactions; 267 four equal spans, 266 four supports, 263 general expressions, 257 three equal spans, 264 two unequal spans, 259 continuous in foundations, 273 projection not limited by site, 275 projection at one end limited by site, 277 Beams under stationary loads: continuous projection at both ends limited by site, 279 Beams : deflection of, 208 cantilever, 232 maximum deflection, 215 point of maximum deflection, 218 reciprocal deflections, 231 several concentrated loads, 226 simple beams, uniform cross section, 212 uniform load, 221 uniform and concentrated load, 229 uniformly varying load, 224, 225 varying cross section, 235 depth, 236 with cover plates, 237 partially continuous, 270 no shear transmitted, 270 no moment transmitted, 271 restrained, 240 at both ends, concentrated load, 243 uniform load, 247 at one end, concentrated load, 241 uniform load, 242 Bending moment: defined, 83 determined from location-direc- tion diagram, 84 diagrams, 86-89 Bent, 109 Bridge trusses: Baltimore, 129, 184 criterion for chord members, 4& for web members, 190, 194 influence lines for, 191 Parker, 129, 169 327 328 INDEX Bridge trusses: Parker, criterion for chord mem- ber, 149 for web member, 172, 177 influence line for web member, 172, 177 stress in web member, 169 tension in vertical when counters are used, 182 Pennsylvania, 130, 197 influence lines for, 198 Pratt, 129, 146 chord stresses with odd number of panels, 167 criterion for chord member, 149 for web member, 152, 157 influence line for chord mem- ber, 146 for shear in panel, 151 for web member, 151 stresses in a 200 ft. truss, 160 moving loads, 130 Standard types, 128 Warren, 129 Camber, 298 Cantilever bridge, 80 Clapeyron's theorem of three moments, 268 Cooper's standard train loads, 137 moment table for, 136 Counters^ 166 Couple, 5 Culmann's method of sections, 69 Deflection of beams, See Beams, of trusses, See Trusses. Equilibrium : equations of, 12 polygon, See Location-direction diagram. Equivalent uniform loads, 200 table for, 206 Force : component of, 6 denned, 4 direction of, 5 elements of, 5 location of, 5 magnitude of, 5 moment of, 1 1 polygon, See Magnitude-direction diagram. rectangular components of, 10 sense of, 5 Forces : concurrent, 5 composition of, 6 coplanar, 5 concurrent, 14 algebraic method, 14 combinations of elements, 22 general case, 17 graphic method, 17 illustrative problem, 14, 19, 20 number of independent equa- tions possible, 19 problems, 23 special case, 21 non-concurrent, non-parallel, 30 algebraic method, 38, 48, 50, 53 combinations of elements, 38 general case, 32 graphic method, 41, 48, 52, 54 groups of equations, 33 illustrative problem, 31, 38- 54 number of independent equations possible, 33 one unknown location, 37 problems, 54 transformed systems, 30 parallel, 24 algebraic method, 27 combinations of elements, 30 general considerations, 24 general case, 26 INDEX 329 Forces : coplanar parallel graphic method, 47 illustrative problem, 27 number of independent equations possible, 30 external, 62 internal, 62 non-concurrent, 5 parallelogram of, 6 resolution of, 6 resultant of, 6 system of, 5 triangle of, 6, 8, 9 Impact, 159 Inclined loads, 81 Influence lines, See Structure in question. Joints, methods of, 63 L Location-direction diagram, 9 drawn through given points, 45 M Magnitude-direction diagram, 9 Maxwell's theorem of reciprocal deflections, 231, 299 Mechanics, i Mental attitude, 2 Methods, 2 'Mill building, column reactions, 118 Moving loads, 130 Odd number of panels, 167 Parabola, 91 methods of constructing, 93, 94 Portal frames, 102-107 Purlins, 109 Rankine's method of sections, 68, 69 Rigid body defined, 60 Ritter's method of moments, 69 Roof trusses: bracing, no, in conclusions, 126 covering, in dead loads, in design stresses, 125 Fink, 73, 108 Howe, 108 monitor, 109 normal wind pressures, 114 purlins, 109 reactions of, 82, 116 saw tooth, 1 08 snow load, 112, 125 spacing of, in standard types, 108 supported on columns, 116 on walls, 115 Warren, 108 weights of, 112 wind analysis, 115, 120 load, 113, 125 stress diagram unnecessary, 124 S Sections, method of, 68 Shear, denned, 83 diagrams, 86-99 Stevin, 6 Stress diagram, 65 and strain, 282 defined, 62 determined by method of joints, 63 of sections, 68 for stationary loads, 67 Structural engineer's parabola, 91 Swing bridges, 301 center bearing, 301 combinations of stresses, 303 conditions of loading, 301 reactions from formula, 306 from Williot diagram, 320 330 INDEX Swing bridges, short method for com- puting stresses in, 309 stresses in, 303 rim bearing, 323 Trusses : assumptions in connections with, 6 1 camber of, 298 denned, 61 deflection of: algebraic method, 283 due to temperature, 297 Trusses: deflection of: graphic method (Williot dia- gram), 288 first solution, 289 second solution, 293 third solution, 295 Maxwell's theorem, 299 stability of, 61 Varignon's theorem, u W Williot diagrams, 288, 290, 293, 296 UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. ENGINEERING LIBRARY 28 1953' APR 2-5,1955 C2NW955 "mil ofi 1M4 jUL E U w ' LD 21-95,.^ll,'50(2877sl6)476 YC 33059 Ml716Q3_ / THE UNIVERSITY OF CALIFORNIA LIBRARY