LIBRARY THE UNIVERSITY OF CALIFORNIA SANTA BARBARA PRESENTED BY Glen G. Mosher -^ u GLEN G. MOSHER WORKS OF MANSFIELD MERRIMAN JOHN WILEY & SONS TREATISE ON HYDRAULICS. 8vo, 593 pages, $5.00 MECHANICS OF MATERIALS. 8vo, 518 pages, $5.00 PRECISE SURVEYING AND GEODESY. 8vo, 261 pages, $2.50 METHOD OF LEAST SQUARES. 8vo, 238 pages, $2.00 ELEMENTS OF SANITARY ENGINEERING. 8vo, 252 pages, net, 52.00 ELEMENTS OF MECHANICS. 12010, 172 pages, net, $1.00 STRENGTH OF MATERIALS. 12010, 156 pages, net, $1.00 By MERRIMAN and JACOBY TEXT-BOOK ON ROOFS AND BRIDGES: PART I. STRESSES. 8vo, 326 pages, $2.50 PART II. GRAPHIC STATICS. 8vo, 242 pages, $2.50 PART III. BRIDGE DESIGN. 8vo, 382 pages, $2.50 PART IV. HIGHER STRUCTURES. 8vo, 385 pages, $2.50 By MERRIMAN and BROOKS HANDBOOK FOR SURVEYORS. i2mo, 246 pages. $2.00 Edited by MERRIMAN and WOODWARD SERihs OF MATHEMATICAL MONOGRAPHS. Eleven Volumes, 8vo. Each, $1.00 MECHANICS OF MATERIALS BY MANSFIELD MERRIMAN MEMBER OF AMERICAN SOCIETY OF CIVIL ENGINEERS TENTH EDITION, REWRITTEN AND ENLARGED TOTAL ISSUE, THIRTY-ONE THOUSAND NEW YORK JOHN WILEY & SONS LONDON: CHAPMAN & HALL, LIMITED 1910 COPYRIGHT, 1885, 1890, 1895, 1905 BV MANSFIELD MERRIMAN First Edition, June, 1885 Second Edition, revised and enlarged, April, 1890 Third Edition, February, 1891 Fourth Edition, June, 1892; reprinted, 1893 Fifth Edition, February, 1894; reprinted, 1895 Sixth Edition, enlarged, September, 1895; reprinted, 1896 Seventh Edition, July, 1897 Eighth Edition, February, 1899 Ninth Edition, June, 1900; reprinted, 1901, 1902, 1903 (twice), 1904, 1905 Tenth Edition, rewritten and enlarged, September, 1905 Reprinted, 1906, 1907, 1908, 1909, 1910 ALL RIGHTS RESERVED THE SCII BROOKLY PREFACE SINCE 1885, when the first edition of this work was published, many advances have been made in the subject of Mechanics of Materials. Some of these -have been noted in the additions to subsequent editions, but to record and correlate them properly it has now become necessary to rewrite and reset the book. In doing so the author has endeavored to keep the facts of experi- ment and practice constantly in view, for the theory of the subject is. merely the formal expression and generalization of observed phenomena. The subject, indeed, no longer consists of a series of academic exercises in algebra and rational mechanics, but it is indispensably necessary that the phenomena of stress should be clearly understood by the student. While laboratory work is a valuable aid to this end, it is important, in the opinion of the author, that no recitation or lecture should be held without having test specimens at hand with which to illustrate the phys- ical phenomena. The same general plan of treatment has been followed as before, but the subdivisions are somewhat different, and the fifteen chapters of the last edition have been increased to nineteen. The statement of average values of the principal materials of engineering has proved so advantageous to students that it is here also followed. Numerous numerical examples are given in the text to exemplify the formulas and methods, these generally relating to cases that arise in practice. To encourage students to think for themselves, one or more problems are given at the end of each article; for the experience of the author has indicated that the solution of many numerical exercises is required in order that students may become well grounded in theory. Most of the topics of the last edition have been treated in a fuller manner than before. The subjects of impact on bars and beams, resilience and work, and apparent and true stresses iv PREFACE have been much changed with the intention of rendering the presentation more clear and accurate. Among many new topics introduced are those of economic sections for beams, moving loads on beams, constrained beams with supports on different levels, the torsion of rectangular bars, compound columns and beams, reinforced-concrete beams, plates under concentrated loads, internal friction, rules for testing materials, and elastic- electric analogies. A few changes in algebraic notation have been made in order that similar quantities may always be desig- nated by letters of the same type; Greek letters are used only for angles and abstract numbers. Compared with the ninth edition, the number of articles has been increased from 151 to 188, the number of tables from 8 to 20, the number of cuts from 85 to 250, and the number of problems from 222 to 305. Although the length of each page has been increased eight percent and smaller type has been used for formulas and problems, the number of pages has been in- creased from 378 to 518. While the main purpose in rewriting and enlarging the book has been to keep it abreast with modern progress, the attempt Jias also been made to present the subject more clearly and logically than before, in order both to advance the interests of sound engineering education and to promote sound engineering practice. NOTE This impression differs from the preceding one mainly in Arts. 88 and 164, the former giving a correct deduction of Euler's formula for a column with one end fixed and the other round, and the latter giving a revised discussion of the stresses in a revolving wheel. All known errors have been corrected. MANSFIELD MERRIMAN. 32 WEST 40TH STREET, NEW YORK, June, 1910. CONTENTS CHAPTER I ELASTIC AND ULTIMATE STRENGTH PACK ART. 1. SIMPLE AXIAL STRESSES i 2. THE ELASTIC LIMIT 4 3. ULTIMATE STRENGTH 6 4. TENSION 9 5. COMPRESSION 1 1 - 6. SHEAR 14 7. WORKING UNIT-STRESSES 16 8. COMPUTATIONS AND EQUATIONS 19 CHAPTER II ELASTIC AND ULTIMATE DEFORMATION ART. 9. MODULUS OF ELASTICITY 23 10. ELASTIC CHANGE OF LENGTH 25 11. ELASTIC LIMIT AND YIELD POINT 27 12. ULTIMATE DEFORMATIONS 30 13. CHANGES IN SECTION AND VOLUME 32 14. WORK IN PRODUCING DEFORMATION 35 15. SHEARING MODULUS OF ELASTICITY 37 16. HISTORICAL NOTES 39 CHAPTER HI MATERIALS OF ENGINEERING ART. 17. AVERAGE WEIGHTS 42 18. PLASTICITY AND BRITTLENESS 43 19. TIMBER 46 20. BRICK 48 21. STONE 50 22. MORTAR AND CONCRETE 52 23. CAST IRON 55 24. WROUGHT IRON 57 25. STEEL 60 26. OTHER MATERIALS 66 vi CONTENTS CHAPTER IV CASES OF SIMPLE STRESS PAGE ART. 27. STRESS UNDER OWN WEIGHT 69 28. BAR OF UNIFORM STRENGTH 71 29. ECCENTRIC LOADS 72 30. WATER AND STEAM PIPES 75 31. THIN CYLINDERS AND SPHERES 77 32. SHRINKAGE OF HOOPS 79 33. INVESTIGATION OF RIVETED JOINTS 80 34. DESIGN OF RIVETED JOINTS 83 CHAPTER V GENERAL THEORY OF BEAMS ART. 35. DEFINITIONS 87 36. REACTIONS OF SUPPORTS 88 37. THE VERTICAL SHEAR 90 38. THE BENDING MOMENT 93 39. INTERNAL STRESSES AND EXTERNAL FORCES 96 40. NEUTRAL SURFACE AND Axis 98 41. SHEAR AND FLEXURE FORMULAS 101 42. CENTER OF GRAVITY 103 43. MOMENTS OF INERTIA 105 44. ROLLED BEAMS AND SHAPES 108 45. ELASTIC DEFLECTIONS . .112 CHAPTER VI SIMPLE AND CANTILEVER BEAMS ART. 46. SHEAR AND MOMENT DIAGRAMS 116 47. MAXIMUM SHEARS AND MOMENTS 119 48. INVESTIGATION OF BEAMS 121 49. SAFE LOADS FOR BEAMS 124 50. DESIGNING OF BEAMS 125 51. ECONOMIC SECTIONS 127 52. RUPTURE OF BEAMS 130 53. MOVING LOADS 132 54. DEFLECTION OF CANTILEVER BEAMS 135 55. DEFLECTION OF SIMPLE BEAMS 138 56. COMPARATIVE STRENGTH AND STIFFNESS 141 57. CANTILEVER BEAMS OF UNIFORM STRENGTH 143 58. SIMPLE BEAMS OF UNIFORM STRENGTH 146 CONTENTS VH CHAPTER VII OVERHANGING AND FIXED BEAMS PAGB ART. 59. BEAM OVERHANGING ONE SUPPORT 149 BEAM FIXED AT ONE END 152 61. BEAM OVERHANGING BOTH SUPPORTS 154 62. BEAM FIXED AT BOTH ENDS 156 63. COMPARISON OF BEAMS 158 64. SUPPORTS ON DIFFERENT LEVELS 159 65. CANTILEVER WITH CONSTRAINT 163 66. SPECIAL DISCUSSIONS 164 CHAPTER VIII CONTINUOUS BEAMS ART. 67. GENERAL PRINCIPLES , . . 168 68. METHOD OF DISCUSSION 171 69. THEOREM OF THREE MOMENTS 173 70. EQUAL SPANS WITH UNIFORM LOAD 175 71. UNEQUAL SPANS AND LOADS 177 72. SPANS WITH FIXED ENDS 179 73. CONCENTRATED LOADS 180 74. SUPPORTS ON DIFFERENT LEVELS 182 75. THE THEORY OF FLEXURE 184 CHAPTER IX COLUMNS OR STRUTS ART. 76. CROSS-SECTIONS OF COLUMNS 188 77. DEFINITIONS AND PRINCIPLES 190 78. EULER'S FORMULA FOR LONG COLUMNS 192 79. EXPERIMENTS ON COLUMNS 196 80. RANKINE'S FORMULA 200 81. INVESTIGATION OF COLUMNS 203 82. SAFE LOADS FOR COLUMNS , ... 205 83. .DESIGNING OF COLUMNS 206 84. THE STRAIGHT-LINE FORMULA 208 85. OTHER COLUMN FORMULAS , 211 86. ECCENTRIC LOADS ON PRISMS 214 ' 87. ECCENTRIC LOADS ON COLUMNS 217 88. ON THE THEORY OF COLUMNS 220 yiii CONTENTS CHAPTER X TORSION OF SHAFTS PAGE ART. 89. PHENOMENA OF TORSION 225 90. THE TORSION FORMULA 227 91. SHAFTS FOR TRANSMITTING POWER 2^0 92. SOLID AND HOLLOW SHAFTS 231 93. TWIST OF SHAFTS 233 94 RUPTURE OF SHAFTS 235 95. STRENGTH AND STIFFNESS 237 96. SHAFT COUPLINGS 239 97. A SHAFT WITH CRANK 240 98. A TRIPLE-CRANK SHAFT 242 99. NON-CIRCULAR SECTIONS 245 CHAPTER XI APPARENT COMBINED STRESSES ART. 100. STRESSES DUE TO TEMPERATURE 251 101. BEAMS UNDER AXIAL FORCES 253 102. FLEXURE AND COMPRESSION . 255 103. FLEXURE AND TENSION 259 104. ECCENTRIC AXIAL FORCES ON BEAMS 261 105. SHEAR AND AXIAL STRESS 263 106. FLEXURE AND TORSION 266 107. COMPRESSION AND TORSION 268 108. HORIZONTAL SHEAR IN BEAMS 269 109. LINES OF STRESS IN BEAMS 272 CHAPTER XII COMPOUND COLUMNS AND BEAMS ART. 110. BARS OF DIFFERENT MATERIALS 276 111. COMPOUND COLUMNS 279 112. FLITCHED BEAMS 282 113. REINFORCED CONCRETE BEAMS 2S^ 114. THEORY OK REINFORCED CONCRETE BEAMS 289 115. INVESTIGATION OF REINFORCED CONCRETE BEAMS .... 292 116. DESIGN OF REINFORCED CONCRETE BEAMS 295 117. PLATE GIRDERS . 2 p8 118. DEFLECTION OF COMPOUND BEAMS 300 CONTENTS ix CHAPTER XIII RESILIENCE AND WORK PAGE ART. 119. EXTERNAL WORK AND INTERNAL ENERGY 303 120. RESILIENCE OF BARS 306 121. RESILIENCE OF BEAMS 308 122. RESILIENCE IN SHEAR AND TORSION 310 123. DEFLECTION UNDER ONE LOAD 312 124. DEFLECTION AT ANY POINT 314 125. DEFLECTION DUE TO SHEAR 316 126. PRINCIPLE OF LEAST WORK 320 CHAPTER XIV IMPACT AND FATIGUE ART. 127. SUDDEN LOADS AND STRESSES 324 128. AXIAL IMPACT ON BARS 327 129. TRANSVERSE IMPACT ON BEAMS 329 130. INERTIA IN AXIAL IMPACT 331 131. INERTIA IN TRANSVERSE IMPACT 334 132. VIBRATIONS AFTER IMPACT 338 133. EXPERIMENTS ON ELASTIC IMPACT 341 134. PRESSURE DURING IMPACT 344 135. IMPACT CAUSING RUPTURE 346 136. STRESSES DUE TO LIVE LOADS 349 137. THE FATIGUE OF MATERIALS 352 138. STRENGTH UNDER FATIGUE 355 CHAPTER XV TRUE INTERNAL STRESSES ART. 139. PRINCIPLES AND LAWS 359 140. SHEAR DUE TO NORMAL STRESS 362 141. COMBINED SHEAR AND AXIAL STRESS 365 142. TRUE STRESSES FOR BEAMS 367 143. NORMAL STRESSES DUE TO SHEAR 369 144. TRUE STRESSES IN SHAFTS 371 ' 145. PURE INTERNAL STRESS 373 146. INTERNAL FRICTION 375 147. THEORY OF INTERNAL FRICTION 378 A CONTENTS CHAPTER XVI GUNS AND THICK CYLINDERS MM ART. 148. PRINCIPLES AND METHODS .... 383 149. LAME'S FORMULAS 385 150. SOLID GUNS AND THICK PIPES 388 151. A COMPOUND CYLINDER 390 152. CLAVARINO'S FORMULAS 392 153. BIRNIE'S FORMULAS 394 154. HOOP SHRINKAGE 396 155. DESIGN OF HOOPED GUNS 399 CHAPTER XVII ROLLERS, PLATES, SPHERES ART. 156. CYLINDRICAL ROLLERS 403 157. SPHERICAL ROLLERS 406 158. CONTACT OF CONCENTRATED LOADS 407 159. CIRCULAR PLATES WITH UNIFORM LOAD 409 160. CIRCULAR PLATES WITH CONCENTRATED LOAD 411 161. ELLIPTICAL PLATES 414 162. RECTANGULAR PLATES 415 163. HOLLOW SPHERES 417 CHAPTER XVIII MISCELLANEOUS DISCUSSIONS ART, 164. CENTRIFUGAL TENSION 421 165. CENTRIFUGAL FLEXURE 425 166. UNSYMMETRIC LOADS ON BEAMS 427 167. EULER'S MODIFIED FORMULA 431 168. TESTING MACHINES 433 169. TESTS OF MATERIALS 436 170. RULES FOR TESTING 442 171. SPECIFICATIONS FOR STRUCTURAL STEEL 444 CHAPTER XIX MATHEMATICAL THEORY OF ELASTICITY ART. 172. INTRODUCTION 448 173. ELASTIC CHANGES IN VOLUME 449 174. NORMAL AND TANGENTIAL STRESSES 452 175. RESULTANT STRESSES 454 176. THE ELLIPSOID OF STRESS 456 177. THE THREE PRINCIPAL STRESSES 457 CONTENTS xi PAGE 178. MAXIMUM SHEARING STRESSES 459 179. DISCUSSION OF A CRANK PIN . . 461 180. THE ELLIPSE OF STRESS 463 181. SHEARING MODULUS OF ELASTICITY 465 182. THE VOLUMETRIC MODULUS 467 183. STORED INTERNAL ENERGY 469 APPENDIX ART. 184. VELOCITY OP STRESS 472 185. ELASTIC-ELECTRIC ANALOGIES 474 186. MISCELLANEOUS PROBLEMS 475 187. ANSWERS TO PROBLEMS 477 188. EXPLANATION OF TABLES 478 TABLES TABLE 1. AVERAGE WEIGHT AND EXPANSIBILITY 480 2. AVERAGE ELASTIC PROPERTIES 480 3. AVERAGE TENSILE AND COMPRESSIVE STRENGTH . . . .481 4. AVERAGE SHEARING AND FLEXURAL STRENGTH 481 5. WORKING UNIT -STRESSES FOR BUILDINGS 482 6. STEEL I-BEAM SECTIONS 483 7. STEEL BULB-BEAM SECTIONS 484 8. STEEL T SECTIONS 484 9. STEEL CHANNEL SECTIONS 485 10. STEEL ANGLE SECTIONS 489 11. STEEL Z SECTIONS 487 12. COMPARISON OF BEAMS 487 13. GERMAN I BEAMS 488 14. WEIGHT OF WROUGHT-!RON BARS 489 t 15. SQUARES OF NUMBERS 490 16. AREAS OF CIRCLES 492 17. TRIGONOMETRIC FUNCTIONS 494 18. LOGARITHMS OF TRIGONOMETRIC FUNCTIONS 491; 19. LOGARITHMS OF NUMBERS 496 20. CONSTANTS AND THEIR LOGARITHMS 498 INDEX 499 MECHANICS OF MATERIALS CHAPTER I ELASTIC AND ULTIMATE STRENGTH ARTICLE 1. SIMPLE AXIAL STRESSES Mechanics of Materials is the science that treats of the effects of forces in causing changes in the size and shape of bodies. Such forces are generally applied to bodies slowly, and the changes in size and shape occur while the forces are increasing up to their final values. A ' Stress ' is an internal force that resists the change in shape or size, and when the applied forces have reached their final values the internal stresses hold them in equilibrium. The simplest case is that of a rope, at each end of which a man pulls with a force, say 25 pounds, then in every section of the rope there exists a stress of 25 pounds. Stresses are measured in the same unit as that used for the applied forces, and generally in pounds or kilograms. A 'Bar' is a prismatic body having the same size throughout its length. If a plane is passed normal to the bar, its intersection with the prism is called the 'cross-section' or the 'section' of the bar, and the area of this cross-section is called the ' section area. ' In any section imagined to be cut out, there exists a stress equal to the longitudinal force acting on the end of the bar. A 'Unit- Stress' is the stress on a unit of the section area, and this is usually expressed in pounds per square inch or in kilograms per square centimeter. For example, let a bar, 3 inches wide and i inches thick, be subjected to a pull of 14 400 pounds; the resisting stress is 14 400 pounds, and the unit-stress is 14 400 pounds divided by 4^ square inches, or 3 200 pounds per square inch. ELASTIC AND ULTIMATE STRENGTH CHAP. I When external forces act upon the ends of a bar in a direction away from its ends they are called 'Tensile Forces'; when they act towards the ends, they are called 'Compressive Forces!^ A pull is a tensile force and a push is a compressive force, and thase two cases are frequently called 'Tension' and 'Compres- sion'. The resisting stresses receive similar designations; a ten- sile stress is that which resists tensile forces; a compressive stress is that which resists compressive forces. Fig. la Fig. Ib The case of tension is shown in Fig. la, where two tensile forces, each equal to P, act upon the ends of a bar having the section area a. Let mn be any imaginary plane normal to the bar, and let the two parts of the bar be imagined to be separated as in Fig. Ib. Then the equilibrium of each part will be main- tained if tensile forces equivalent to the resisting stresses are applied as shown. These resisting stresses act normally to the section area a, and they are in each case opposite in direction to the force P. Each part of the bar is held in equilibrium by the applied force P and the resisting tensile stress; accordingly the resisting tensile stress must equal the tensile force P. Fig. le Fig. Id The case of compression is shown in Fig. \c, where the forces P act toward the ends of the bar. For any imaginary plane mn, the bar may be regarded as separated into two bars as in Fig. Id, each of which is held in equilibrium by compressive stresses acting normally to the section area and in directions opposite to P. The total resisting compressive stress must be equal to P in order that equilibrium may prevail. ART. i SIMPLE AXIAL STRESSES 3 Let 5 be the unit-stress of tension or compression, as the case may be, which acts in any normal section of a bar having the area a. The total stress on the section is then Sa, and this is uniformly distributed over the area a when the force P acts along the axis of the bar; then, Sa = P S=P/a a=P/S (1) from which one of the quantities may be computed when the other is given. For example, let it be required to find what the section area of a stick of timber should be when it is subject to a pull of 1 6 500 pounds, it being required that the tensile unit- stress shall be 900 pounds per square inch; here a = i6 500/900 = 18.3 square inches. The terms 'Axial Forces' and 'Axial Stresses' are used to include both tension and compression acting upon a bar, it being understood that the resultant of the applied forces acts along the axis of the bar. The axial force P is often called a 'Load'. It is always understood, unless otherwise stated, that the stresses due to an axial load are uniformly distributed over the section area, and this is called the case of ' Simple Axial Stress ', it being one of the most common cases in engineering. Cases where the stress is not uniformly distributed over the section area occur when the resultant of the applied forces does not act along the axis of the bar, and also in beams and long columns. The first effect of an axial load is to change the length of the bar upon which it acts. This 'Deformation' continues until the resisting stresses have attained such magnitudes that they equili- brate the applied forces. The deformation of a bar which occurs in tension is called 'Elongation', and that which occurs in com- pression is called 'Shortening'. As the applied forces increase, the resisting stresses also increase, until finally the resistance is unable to balance the force, the deformation rapidly increases, and the bar breaks or ruptures. The above equations apply also to the case of rupture. For example, it is known that a cast-iron bar will rupture under tension when the unit-stress 5 becomes about 20000 pounds per square inch; if the bar is iJXi inches in cross-section, its section area is if square 4 ELASTIC AND ULTIMATE STRENGTH CHAP. I inches, and t'he tensile force required to cause rupture is P = i$ X 20 ooo = 37 500 pounds. Problem la. If a cast-iron bar, 1^X2 inches in section, breaks under a tensile load of 60 ooo pounds, what load will break a cast- iron rod of 2\ inches diameter ? Prob. Ib. A cast-iron bar which is to be subjected to a tension of 34 ooo pounds is to be designed so that the unit-stress shall be 2 500 pounds per square inch. If the bar is round what should be its diameter ? ART. 2. THE ELASTIC LIMIT When a bar is subjected to a gradually increasing tension, the bar elongates, and up to a certain limit it is found that the elon- gation is proportional to the load. Thus, when a bar of wrought iron one square inch in section area and 100 inches long is sub- jected to a load of 5 ooo pounds, it is found to elongate closely 0.02 inches; when 10000 pounds is applied, the total elongation is 0.04 inches; when 15 ooo pounds is applied, the elongation is 0.06 inches; when 20000 pounds is applied, the elongation is 0.08 inches; when 25 ooo pounds is applied, the elongation is o.io inches. Thus far, each addition of 5 ooo pounds has pro- duced an additional elongation of 0.02 inches. But when the next 5 ooo pounds is added, making a total load of 30 ooo pounds, it is found that the total elongation is about half an inch, and hence the elongations are increasing in a faster ratio than the applied loads and the resisting stresses. The 'Elastic Limit' is defined to be that unit-stress at which the deformation begins to increase in a faster ratio than the applied loads. In the above example this limit is about 25 ooo pounds per square inch, and this is the average value of the elastic limit for wrought iron. The term 'Elastic Strength' is perhaps a better term than elastic limit, but the latter is in general use. When the unit-stress in a bar is not greater than the elastic limit, the bar returns, on the removal of the load, to its original length. Thus, the above wrought-iron bar was 100.10 inches long under the load of 25 ooo pounds, and on the removal of that load ART. 2 THE ELASTIC LIMIT 5 it returns to its original length of 100.00 inches. When the unit- stress is greater than the elastic limit, the bar does not fully re- turn to its original length, but there remains a so-called ' Perma- nent Set'. For instance let the length of the above bar under a stress of 34 ooo pounds be 102 inches, and on the removal of the tension let its length be ioi| inches; then the permanent set of the bar is i| inches. In all cases of simple axial tension the resisting stress is equal to the load, and the stresses hence increase proportionately to the loads. When the elastic limit is not exceeded, the elongations are found to be proportional to the loads, but when this limit is exceeded they increase faster than the loads, and a permanent set remains. Therefore the elastic properties of a bar are injured when it is stressed beyond the elastic limit. Accordingly it is a fundamental rule in designing engineering constructions that the unit-stresses should not exceed the elastic limit of the material. The above facts regarding the behavior of materials in ten- sion have been ascertained by many tests of bars and are to be regarded as fundamental laws; all experience and all experi- ments have verified these laws as being approximately true for the common materials used in engineering. By such tests also it has been shown that such laws apply to compression as well as to tension. The following are approximate average values of the elastic limits in tension for five materials extensively used in engi- neering construction: Material Elastic Limit Timber 3 ooo pounds per square inch Cast Iron 6 ooo pounds per square inch Wrought Iron 25 ooo pounds per square inch Structural Steel 35 ooo pounds per square inch Strong Steel 50 ooo pounds per square inch These values should be carefully memorized by the student, and be used in the solution of the problems in the following pages. Table 2, at the end of this volume, gives these constants in the metric system of measures. The above average values are subject to considerable varia- 6 ELASTIC AND ULTIMATE STRENGTH CHAP. I tion for different qualities of the same material; for example, some grades of structural steel may have an elastic limit ten per- cent lower than 35 ooo pounds per square inch, while others may run ten percent higher. In testing a number of bars of the same kind, indeed, it is not uncommon to find a variation of five or ten percent in the different results. The elastic limit in compression is the same as that in tension for all the above materials except cast iron, which has about three times the value above given. Brittle materials, like brick, stone, and cast iron, usually have higher elastic limits in compression than in tension, but it will be seen later that the elastic limits for such materials are poorly denned, that is it is difficult to determine them with exactness. By the help of the above experimental values and the princi- ples of Art. 1, many simple problems in investigation and design may be solved. For example, let it be required to find the size of a square stick of timber to carry a compressive load of 64 ooo pounds, so that the unit-stress may be one-third of the elastic limit; here the elastic limit is 3 ooo pounds per square inch, and the section area required is 64 ooo/i ooo, or 64 square inches, so that a stick 8x8 inches in size is needed. Prob. la. Find the diameter of a round rod of wrought iron, which is to be under a tension of 84 ooo pounds, so that the unit-stress may be one-third of the elastic limit. Prob. 2b. A stick of timber 3 inches thick is under a tension of 12 ooo pounds. Compute its width, so that the unit-stress may be 40 percent of the elastic limit. ART. 3. ULTIMATE STRENGTH When the section area of a bar is under an axial unit-stress exceeding the elastic limit of its material, the bar is usually in an unsafe condition. As the external forces increase, the defor- mation increases in a more rapid ratio, until finally the rupture of the bar occurs. The term ' ultimate strength' is used to desig- nate the highest unit-stress that the bar can sustain, this occur- ring at or just before rupture. ART. 3 ULTIMATE STRENGTH 7 The ultimate strengths of materials are usually from two to four times their elastic limits, and for some materials they are much higher in compression than in tension. Thus, the ultimate tensile strength of ca.st iron is about 20 ooo pounds per square inch, while its ultimate compressive strength is about 90 ooo pounds per square inch. Average values of the ultimate strengths of materials are given in the following articles, but these are subject to much variation for different qualities of materials. Thus, inferior grades of cast iron may have a tensile strength as low as 15 ooo pounds per square inch, while the best grades are often higher than 30 ooo pounds per square inch. In general a variation of ten percent from these average values is to be regarded as liable to occur. The 'Factor of Safety' of a bar under stress is the number which results by dividing the ultimate strength of the material by the actual unit-stress on the section area. For example, let a stick of timber, 6X6 inches in section area, be under a ten- sion of 32 400 pounds. The actual unit-stress is then 32 400/36 = 900 pounds per square inch. Since the average tensile strength of timber is about 10 ooo pounds per square inch, the factor of safety of the bar is 10 000/900= n. The factor of safety was formerly much used in designing, and for timber under steady tension was taken as about 10; that is, one-tenth of the ultimate strength was regarded as the highest allowable unit-stress. By this method, timber having an ulti- mate tensile strength of 12 ooo pounds per square inch should be subjected to a unit-stress of only 12000/10=1 200 pounds per square inch, so that the section area of a stick under a ten- sion of 19 200 pounds should be 19 200/1 200 = 16 square inches. It is now considered a better plan, in judging of the degree of security of a body under stress, to consider the elastic limit of the material. Thus, for a stick of timber, the elastic limit is about 3 ooo pounds per square inch, and the actual unit-stress in tension should be less than this, say one-half or one-third, according as the applied fo/ces are steady or variable. The 8 ELASTIC AND ULTIMATE STRENGTH CHAP, l method of the factor of safety is, however, of much value to a student and it will be often used in this volume. In practice both ultimate strengths and elastic limits must be considered in deciding upon the allowable unit-stresses to which the parts of engineering constructions are to be subjected. The average values of the ultimate strengths of the materials of engineering are tabulated at the end of this volume in both English anc metric measures. The theoretical and experimental conclusions thus far estab- lished, regarding the behavior of a bar under tension or com- pression, may now be formulated in the following laws: 1. The resisting axial stress in every section of a bar is equal to the applied tensile or compressive load. 2. Under small stresses, the deformations of the bar are proportional to the loads, and hence also to the unit-stresses. When the unit-stress is less than the value called the elastic limit, the bar springs back to its original length on the removal of the load. 3. When the unit-stress is greater than the elastic limit, the deformations increase in a faster ratio than the loads and stresses, and the bar does not spring back to its original length on the removal of the load. 4. When the load becomes sufficiently great, the resisting stress fails to balance it, so that the deformation rapidly in- creases and the bar ruptures. 5. The allowable unit-stresses used in engineering prac- tice are less than the elastic limit of the material. The first of these laws is a theoretical one and rigidly correct for all cases, it being in fact a particular case of Newton's law that action and reaction are equal and opposite. The second law applies strictly only to elastic materials like wrought iron and steel, and is only roughly applicable to brittle materials like stone and cast iron. The third law applies to all kinds of materials, but for brittle ones the elastic limit is difficult to determine. Prob. 3a. A bar of structural steel, 2^ inches in diameter, ruptures under a tension of 271 ooo pounds. What is the ultimate tensile strength ? ART. 4 TENSION 9 Prob. 36. Using the value given above for the ultimate tensile strength of cast iron, compute the tensile force which is required to rupture a cast-iron bar which is 2^X3^ inches in section area. ART. 4. TENSION A tensile test of a vertical bar may be made by fastening its upper end firmly with clamps and then applying loads to its lower end. The elongations of the bar are found to increase proportionately to the loads, and hence also to the internal ten- sile stresses, until the elastic limit of the material is reached (Art. 2). After the unit-stress has exceeded the elastic limit, the elongations increase more rapidly than the loads, and this is often accompanied by a reduction in area of the cross-section of the bar. Finally, the ultimate tensile strength of the mate- rial is reached and the bar breaks. 80000 0.02 0.04 0.06 0.08 0.10 0.12 Elongations per unit of length Fig. 4 A graphical illustration of these phenomena may be made by laying off the unit-stresses as ordinates and the elongations per unit of length as the abscissas. At various intervals, as the test progresses, the applied loads are measured and also the corresponding elongations. The load divided by the section area of the bar gives the unit-stress, while the total elongation divided by the length of the bar gives the unit-elongation. On 10 ELASTIC AND ULTIMATE STRENGTH CHAP. I the diagram a point is put at the intersection of each unit-stress with its corresponding unit-elongation, and a curve is drawn connecting the plotted points. Fig. 4 represents mean curves obtained in this manner for timber, cast iron, wrought iron, and steel. It is seen that each curve is a straight line from the origin until the elastic limit is reached, showing that the unit-elonga- tions increase proportionally to the unit-stresses. At the elastic limit a sudden change in the curve is seen, and afterwards the elongation increases more rapidly than the stress. This dia- gram gives mean comparative curves only, and the curve for any individual test might deviate considerably from that shown. The end of the curve marks the rupture of the bar. For example, it is seen that timber ruptures in tension under a stress of about 10 ooo pounds per square inch, and that the elongation of the bar at that time is about y-J-j^ of the length of the bar, that is the total elongation is about 1.5 percent of the length. The rupture of wrought iron and structural steel is not shown upon the diagram, since the ultimate elongation of these materials is about 30 percent, while the diagram extends only up to 12 per- cent. The following are approximate average values of the ultimate strengths and ultimate elongations for five materials widely used in engineering work: Material Ultimate Tensile Strength Ultimate Elongation Timber 10 ooo pounds per square inch 1.5 percent Cast Iron 20 ooo pounds per square inch 0.3 percent Wrought Iron 50 ooo pounds per square inch 30 percent Structural Steel 60 ooo pounds per square inch 30 percent Strong Steel 100 ooo pounds per square inch 15 percent These ultimate tensile strengths should be carefully kept in mind by the student as a basis for future knowledge, and they will be used in the solution of the examples and problems in this book. Table 3, at the end of this volume, gives values for other materials. These average values of ultimate strengths and elongations are those derived from tests on small specimens, say one inch in diameter and 8 inches in gaged length. Large bars such as ART. 5 COMPRESSION 11 arc used in engineering constructions have ultimate strengths slightly less and ultimate elongations considerably less than these values. One of the reasons for using factors of safety (Art. 3) is to provide security against the smaller ultimate strength of the large pieces which must necessarily be used. The ultimate elongation is an index of the toughness and ductility of the mate- rial, but it is not used in the computations of designing. The unit-elongation which occurs when a bar is stressed up to the elastic limit is very small, compared with that which obtains at rupture, as the curves in Fig. 4 plainly show. Fig. 1690 shows two steel specimens that were ruptured by tension. Steel is a material which has variable physical properties depending upon its chemical composition and the method of its manufacture. 'Structural steel' is that which is used in buildings, bridges, and ships, and it resembles wrought iron in general behavior, but its tensile strength is about twenty percent higher. ' Strong steel ' is not a trade term and is here introduced for instruction purposes only. The grades of steel are very numerous, and they range in tensile strength from 50 ooo to 250000 pounds per square inch. The terms 'soft' and 'hard' are often used to designate steel with low and high tensile strengths. Prob. 4a. A steel specimen, 0.505 inches in diameter, reached the elastic limit under a tensile load of 6 800 pounds and ruptured under a load of 14 800 pounds. Compute the elastic limit and the ultimate strengths of the steel. Prob. 4b. This specimen had a length of 2.00 inches between two marks made on it before the test. At the elastic limit the distance be- tween these two marks was 2.003 inches and after rupture it was 2.45 inches. Compute the unit-elongation for the elastic limit and for rupture. ART. 5. COMPRESSION The phenomena of compression are the reverse of those of tension in regard to the direction of the applied forces and resist- ing stresses. When loads are applied to compress a bar, the amount of shortening is proportional to the load, provided the unit-stress on the material does not exceed the elastic limit. After 12 ELASTIC AND ULTIMATE STRENGTH CHAP. I the elastic limit is passed, the shortening increases more rapidly than the load and hence more rapidly than the unit-stress, and finally the rupture of the bar takes place. The area of the cross- section decreases under tension and increases under compression. The simplest way of testing a bar by compression is to place it on a firm foundation and put the load upon its upper face as shown in Fig. 5a. The bar is held in equilibrium by the load P and the equal upward reaction of the support and there may be represented by arrows as in Fig. 56. On any horizontal section of the bar, there are acting compressive stresses the sum of which equals P (Art. 1). If the section area is a the average compressive unit-stress is P/a, and this will be uniformly dis- tributed over the section when the forces P coincide in direc- tion with the axis of the prism. t' r J Fig. 5a Fig. 56 Fig. 5s Fig. 5d When the length of the bar does not exceed about ten times its least lateral dimension, rupture sometimes occurs by an oblique splitting or shearing as shown in Fig. 5c. When the length is large compared with the thickness, failure occurs under a side- wise bending, as seen in Fig. 5d. The short specimens are cases of simple compression, and the values given in the following table refer only to these; the long specimens are called 'columns' and in them other stresses are developed than that of simple compression (Art. 77). In general, the term 'compressive strength' refers to that obtained from a bar the length of which is considerably less than ten times its thickness. Fig. 169& shows the rupture of a cement cube and a timber block. Graphical illustrations of the behavior of materials under compressive stress may be made in the same manner as for ten- sion, the unit-shortenings being laid off as abscissas. In most cases the ultimate shortening is much less than the ultimate ART. 5 COMPRESSION 13 elongation. For steel the elastic limit is about the same in com- pression as in tension, but the compressive strength of hard steel is much higher than the tensile strength. The following are average values of the ultimate compressive strengths of the principal materials used in engineering construction: Material Ultimate Compressive Strength Brick 3 ooo pounds per square inch Stone 6 ooo pounds per square inch Timber 8 ooo pounds per square inch Cast Iron 90 ooo pounds per square inch Wrought Iron 50 ooo pounds per square inch Structural Steel 60 ooo pounds per square inch Strong Steel 120 ooo pounds per square inch These ultimate strengths are subject to much variation for different qualities of materials, but it is necessary for the student to fix them in his mind as a basis for future knowledge. It is seen that timber is not quite as strong in ultimate compression as in tension, that cast iron is 4^ times as strong, that wrought iron and structural steel have the same strength in the two cases, and that the typical material called strong steel has a higher strength in compression than in tension. There are several varieties of hard steel which have ultimate compressive strengths much greater than that above given (Art. 25). The investigation of a body under compression is made in the same manner as for one under tension. For example, if a stone block, 8X12 inches in cross-section, is subject to a com- pression of 230 ooo pounds the unit-stress is 230 000/96 = 2 400 pounds per square inch, and the factor of safety is 6 000/2 400 = 2^; this is not sufficiently high for stone, as will be seen in Art. 7. Prob. 5a. A solid cast-iron cylinder of 3 inches diameter is under a compression of 500 ooo pounds. Compute its factor of safety. Prob. 56. A brick 2X4X8 inches weighs about 4^ pounds. What must be the height of a pile of bricks so that the compressive unit- stress on the lowest brick shall be one-half of its ultimate strength ? Prob. 5c. A bar of wrought iron one square inch in section area and one yard long weighs 10 pounds. Find the length of a vertical bar so that the stress at the upper end shall equal the elastic limit. 14 ELASTIC AND ULTIMATE STRENGTH CHAP, i ART. 6. SHEAR When two equal and opposite forces act at right angles to a bar and are very near together, they are called ' Shearing Forces ', and they tend to cut or shear the bar. The action of the forces is similar to those in a pair of shears, from which analogy the name is derived, and the resisting stresses are called 'shearing stresses '. Tension and compression cause stresses which act normally to a section area, but shear causes stresses which act parallel with and along the section area. Unless otherwise stated it is to be considered that the shearing stresses generated by shear- ing forces are uniformly distributed over the section area. " ,r ?, Fig. 6a Fig. 66 Fig. 6a shows the case of two plates fastened together with a rivet and subject to a tension P. Let the section area of the rivet be a; then the shearing unit-stress on that cross-section lying in the plane where the plates overlap is P/a, and if this equals the ultimate shearing strength of the material, the rivet will rupture by shearing. Fig. 6b shows the case of a beam resting upon two supports and carrying two equal loads P near the supports, the reaction of each support being P; here a resisting shearing stress equal to P acts on each side of the section mn, the stress on the left of mn acting downward and that on the right acting upward; in this case also the shear- ing unit-stress is P/a, and the bar will shear off when this is equal to the ultimate shearing strength of the material. The following are average values of the ultimate shearing strengths of materials as determined by experiment: Material Ultimate Shearing Strength Brick i ooo pounds per square inch Stone i 500 pounds per square inch Timber, along grain 500 pounds per square inch Timber, across grain 3 ooo pounds per square inch Cast Iron 18 ooo pounds per square inch ART. 6 SHEAR 15 Material Ultimate Shearing Strength Wrought Iron 40 ooo pounds per square inch Structural Steel 50 ooo pounds per square inch Strong Steel 80 ooo pounds per square inch By comparing these with the values for tension in Art. 4, it is seen that the shearing strength of timber is about one-third of the tensile strength when the shearing occurs across the grain, and very much smaller when it occurs parallel with the grain. For cast iron the shearing strength is about 90 percent, and for wrought iron and steel it is about 80 percent of the tensile strength. There is an elastic limit in shear as well as in tension and compression, and its value is about one-half of the ultimate shearing strength. When the shearing unit-stress does not exceed the elastic limit, the amount of slipping or detrusion is proportional to the applied force; when it is greater, the defor- mation increases more rapidly than the force. Fig. Qc Fig. Gd Wooden specimens for tensile tests, like that shown in Fig. Qc, will fail by shearing off the ends if their length is not sufficiently great. For example, let ab be 6 inches and the diameter of the central part be i^ inches. The ends are gripped tightly in the testing machine and the cross-section of the central part thus brought under tensile stress. The load required to cause rup- ture by tension is, .P=aS=o.7854Xi.5 2 Xioooo=i7 700 pounds But the ends may also shear off on the surface of a cylinder hav- ing the length ab and a diameter of ij inches; the load required to cause this rupture by shearing along the grain is, P=a5=3.i42X 1.5X6X500=14 loo pounds and hence the specimen will fail by shearing off the ends before the tensile strength of the timber is reached. To prevent this shearing the length ab must be made about 8 inches. When a bar is subject either to tension or to compression, 16 ELASTIC AND ULTIMATE STRENGTH CHAP. I a shear occurs in any oblique section. Let Fig. Qd represent a bar of section area a subject to the tension P which produces in every normal section the unit-stress P/a. Let mn be a plane making an angle with the axis of the bar and cutting from the bar a section having the area a\. On the left of the plane, the normal stress P may be resolved into the components P\ and P2, respectively parallel and normal to the plane, and the same may be done on the right. Thus it is seen that the effect of the tensile force on the plane mn is to produce a tension P 2 normal to it and a shear PI along it, for the two forces PI and PI act in opposite directions on opposite sides of the oblique sec- tion. The shearing stress PI has the value P cos# which is dis- tributed over the area a\ and this area equals a/smO. Hence the shearing unit-stress along the oblique section is, Si=Pi/a l = (P/a) sind casff=$(P/d) sin 2 When 0=o or when #=90, the value of Si is zero, that is there is no shear on a plane parallel with or normal to the axis. For all other values of 6, a shearing stress exists; and the maximum value of Si occurs when 0=45, and then S\=\P/a. There- fore a normal tensile unit-stress S on a bar produces a shearing unit-stress \S along every section inclined 45 degrees to the axis of the bar. This investigation applies also to compression and it partially explains why the rupture of compression specimens sometimes occurs by shearing along oblique sections, as indicated in Fig. 5c. Prob. 6a. A wrought-iron bolt i^ inches in diameter has a head i^ inches long, and a tension of 35 ooo pounds is applied longitudinally to it. Compute the tensile unit-stress, and the factor of safety against tension. Compute the unit-stress tending to shear off the head of the bolt, and the factor of safety against shear. ART. 7. WORKING UNIT- STRESSES When a bar of section area a is under axial stress caused by a load P, the unit-stress S produced is found by dividing P by a. By comparing this value of S with the ultimate strengths and elastic limits given in the preceding articles, the degree of security of the bar may be inferred. This process is called ART. 7 WORKING UNIT-STRESSES 17 investigation. The student may not at first be able to form a good judgment with regard to the degree of security, this being a matter which involves some experience, as well as acquaintance with engineering precedents and practice. As his knowledge increases, however, his ability to judge whether unit-stresses are or are not too great will constantly improve. When a bar is to be designed to resist an axial load P applied at each end, the unit-stress S is to be assumed in accordance with the rules of practice, and then the section area a is to be computed. Such assumed unit-stresses are often called 'Work- ing Unit-Stresses ', meaning that these are the unit-stresses under which the material is to act or work. In selecting them, two fundamental rules are to be kept in mind: 1. Working unit-stresses should be considerably less than the elastic limit of the material. 2. Working unit-stresses should be smaller for sudden loads than for steady loads. The reason for the first requirement is given in Art. 2. The reason for the second requirement is that experience teaches that suddenly applied loads and shocks are more injurious and produce larger deformations than steady loads. Thus, a bridge subject to the traffic of heavy trains must be designed with lower unit- stresses than a roof where the variable load consists only of snow and wind. In buildings the stresses are mostly steady, in bridges they are variable, and in machines the stresses are often produced with shock. It will be best for the student to begin to form his engineer- ing judgment by fixing in mind the following average values of the factors of safety to be used for different materials under different circumstances : Material Steady Stress Variable Stress Shocks Brick and Stone 15 25 40 Timber 8 10 15 Cast Iron 6 10 ao Wrought Iron 4 6 10 Structural Steel 4 6 10 Strong Steel 5 8 15 18 ELASTIC AND ULTIMATE STRENGTH CHAP. I The working unit-stress will then be found for any special case by dividing the ultimate strengths by these factors of safety. For instance, a short timber strut in a bridge should have a work- ing unit-stress of about 8000/10 = 800 pounds per square inch. It is usually the case that a designer works under specifica- tions in which the unit-stresses to be used are definitely stated. The writer of the specifications must necessarily be an engineer of much experience and with a thorough knowledge of the best practice. The particular qualities of timber or steel to be used will influence his selection of working unit-stresses, and in fact different members of a bridge truss are often designed with different unit-stresses. The two fundamental rules above given are, however, the guiding ones in all cases. In Table 5, at the end of this volume, will be found the working unit-stresses which are specified in the building code of the City of New York. These apply in the design of buildings, and the unit-stresses to be used in the design of bridges are lower than those in the table, while those to be used in the design of machines should be lower still. The two fundamental principles of engineering design are stability and economy, or in other words : 1. A structure must safely withstand all the stresses which may be caused by loads that can come upon it. 2. A structure should be designed so that it may be built and maintained at the lowest possible cost. The second of these fundamental principles requires that all parts of the structure should be of equal strength, like the cele- brated 'one-hoss shay' of the poet. For, if one part is stronger than another, it has an excess of material which might have been saved. Of course this rule is to be violated when the cost of the labor required to save the material is greater than that of the material itself. Thus it often happens that some parts of a structure have higher factors of safety than others, but the lowest factors should not be less than those which good engineer- ing practice requires. The factors of safety stated above may be supposed to be so arranged that, if different materials are united, the stability ART. 8 COMPUTATIONS AND EQUATIONS 19 of all parts of the structure will be the same, so that if rupture occurs, everything would break at once. Or, in other words, timber with a factor of safety 8 has about the same reliability as steel with a factor of 4 or stone with a factor of 15, provided the stresses are due to steady loads. As an example of design, let it be required to determine the size of a short piston-rod made of hard steel having an ultimate tensile strength of 75 ooo pounds per square inch, the piston being 20 inches in diameter and the steam pressure upon it being 80 pounds per square inch. This being a rod subject to rapidly alternating stresses of tension and compression and perhaps to shocks, the factor of safety should be taken at 15, which gives a working unit-stress of 75 000/15 = 5 ooo pounds per square inch. The total tension being 0.7854 X 20^X80 = 25 100 pounds, the area of the cross-section should be 25100/5000 = 5.03 square inches and this will be furnished by a rod 2.53 inches in diam- eter, so that a diameter of 2\ inches will probably be sufficient to give proper security against tension. Since the compressive strength of hard steel is higher than the tensile strength, and since the rod is short, this diameter also gives proper security against compression. Prob. la. A rod of structural steel is to be used under a tension of 87 ooo pounds. Compute its diameter when it is to be used in a building, and also when it is to be used in a bridge. Prob. 7b. In Fig. 66 each of the loads is 4 700 pounds and the wooden beam is 2X3 inches in cross-section. Compute the factor of safety against shearing. Is this factor sufficiently high for steady loads? ART. 8. COMPUTATIONS AND EQUATIONS The numerical computations required in the mechanics of materials should be performed so that the precision of the results fairly corresponds with the precision of the data. The values of the elastic limits and ultimate strengths given in the preceding tables are indefinite in the second significant figure and hence computed areas should not be carried further than three sig- 20 ELASTIC AND ULTIMATE STRENGTH CHAP. I nificant figures the last of which has no precision. For instance in the area of 5.03 square inches computed at the end of the last article the last figure is of no importance, since the work- ing unit-stress of 5 coo pounds per square inch is derived from rough data. Numerical computations, therefore, should include only three significant figures as a general rule. At the end of this volume are given tables of four-place loga- rithms, squares of numbers, and areas of circles which are of value in abridging computations. The table of squares may be used to find square roots, and the table of circles to find diam- eters from given areas. For example, a circle having an area of 0.82 square inches has a diameter of 1.02 inches; a circle having an area of 8.2 square inches has a diameter of 3.23 inches. As this book is mainly intended for the use of students in engineering colleges, a word of advice directed especially to them may not be inappropriate. It will be necessary for stu- dents, in order to gain a clear idea of the mechanics of materials, or of any engineering subject, to solve many numerical problems, and in this work a neat and systematic method should be followed. The practice of making computations on any loose scraps of paper that may happen to be at hand should be discontinued by every student who has followed it, and he should hereafter solve his problems in a special book provided for that purpose, accompanying them by explanatory remarks. Such a note- book, written in ink, and containing the solutions of the prob- lems given in these pages will prove of. great value to every stu- dent. Before beginning the solution of a problem, a diagram should be drawn whenever possible, for a diagram helps the student to understand the problem, and a problem thoroughly understood is half solved. Before beginning the numerical work, it is also well to make a mental estimate of the answer and record the same, comparing it later with the result of the solution, since in this manner the engineering judgment of the student will be developed. In continental Europe the metric system is universally employed in the mechanics of materials, the unit of force or stress being ART. s COMPUTATIONS AND EQUATIONS 21 the kilogram, that of length being the meter or centimeter, and that of area being the square centimeter. Computations in the metric system are much simpler than those in the English sys- tem, and it is to be hoped that the time is not far distant when it will come into general use. As a slight contribution to this end, the average constants relating to the strength of materials are given in metric units in Tables 1-4 at the end of this volume, and a few problems using such units will be occasionally pro- posed. In solving these problems the student should think in the metric system and make no transformations of the data or the results into the English system. The table of areas of circles at the end of this volume is applicable to all systems of measures. In the English system of measures, as generally used in this volume, the unit of force is the pound, the unit of length is the inch, and the unit of area is the square inch. Lengths of bars and beams are sometimes stated in feet, but these should be reduced to inches when they are to be used in formulas in order that they may be consistent with the other data. In this volume Greek letters are used only for signs of opera- tion, for abstract numbers, and for angles. The letter d is em- ployed as the symbol of differentiation and I as the symbol of summation; the Greek names of these two letters should not be used, but they may be called 'differential' and 'algebraic sum'.. The following are the names of other Greek letters: a Alpha, ft Beta, i Epsilon, T) Eta, 6 Theta, A: Kappa, A Lambda, /zMu, v Nu, r. Pi, p Rho, a Sigma, T Tau, < Phi, Psi, a) Omega. In every algebraic equation it is necessary that all of the terms should be of the same dimension, for it is impossible to add together quantities of different kinds. This principle will be of great assistance to the student in checking the correctness of algebraic work. For 'example, let a and b represent areas and / a length; then such an equation as all 2 =b is impossible, because al is a volume while I 2 and b are areas. Again, let S represent pounds per square inch, P pounds, / inches, and a 22 ELASTIC AND ULTIMATE STRENGTH CHAP. I square inches; then the equations S=P/a and P=aS are cor- rect with respect to dimensions, but the equation S 2 l 2 /P=a is impossible because the first member represents pounds per square inch while the second is an area. The equation Si =\(P/a) sin20, deduced in Art. 6, is correct in dimensions, for both S\ and P/a represent pounds per square inch while sin20 is an abstract number. Prob. 8a. A bar of wrought iron is 3.25 centimeters in diameter. What load in kilograms will cause it to rupture by tension ? What load will stress it up to one-half of the elastic limit ? Prob. 8b. A round bar of structural steel is under a tension of 7 ooo kilos. What should be its diameter in centimeters in order that the factor of safety may be 6 ? Prob. 8c. Let K represent work, P pounds, S and E pounds per square inch,' / inches, and a square inches; determine whether or not the formula K = %(S 2 /E)al is dimensionally correct. Also show whether the equation x 5 +px 3 +qx+pq = o is or is not correct. ART. 9 MODULUS OF ELASTICITY 23 CHAPTER II ELASTIC AND ULTIMATE DEFORMATION ART. 9. MODULUS OF ELASTICITY The term 'Elastic Deformation' is used to designate that change of shape of a body which accompanies stresses that do not surpass the elastic limit of the material. In tension the principal deformation is the elongation of the bar, in compres- sion it is the shortening. The fact that these deformations are proportional to the stresses (Art. 2) enables rules to be estab- lished whereby the change of length of a bar can be computed, provided the elastic limit be not exceeded. Let P be an axial tensile load applied to a bar which has a cross-section of area a; then the tensile unit-stress S is P/a (Art. 1). Let / be the length of the bar and e the total elonga- tion which occurs under the stress ; then the unit-elongation is e/l, and this will be designated by . Similarly let P be a compres- sive load which produces the shortening e, the compressive unit- stress is P/a, and the unit-shortening is e/l. Unit-stress is the stress on a unit of area, the total stress being regarded as uni- formly distributed over the section area of the bar. Unit-deforma- tion is the change in length of a linear unit of the bar, the total deformation being regarded as uniformly distributed over the total length. Accordingly for any bar in tension or compression, Unit-stress S=P/a Unit-deformation e = e/l and these are applicable whether the elastic limit be exceeded or not, as illustrated in Art. 4. When the elastic limit is not exceeded by the unit-stress, the unit-deformation e/l is proportional to the unit-stress P/a, and hence the ratio of the latter to the former is a constant for each kind of material. The term 'Modulus of Elasticity' is used for this constant, and it may be defined as the ratio of the unit-stress 24 ELASTIC AND ULTIMATE DEFORMATION CHAP. II to the' unit-deformation. The letter E is used for the modulus of elasticity and hence, from the definition, its value for tension or compression is, E = S - or =^ (9) e e/l Since e and / are linear quantities, is an abstract number, and therefore E is expressed in the same unit as S, that is in pounds per square inch or in kilograms per square centimeter. The above equations show that E equals P when a is unity and e is equal to /; that is, the modulus of elasticity is the force which will elongate a bar of one square unit in cross-section to double its original length, provided that this can be done with- out exceeding the elastic limit of the material. There is prob- ably no material, except india-rubber, for which so great an elastic elongation is possible. The modulus of elasticity E is called by. some writers the 'coefficient of elasticity', but the former term is now of more general use in the United States of America ; in books on physics it is often called 'Young's modulus'. The student should care- fully note that the above formulas for E apply only when the unit-stress S or P/a is less than the elastic limit of the material. When modulus of elasticity is mentioned without qualification it is always understood to refer to tension or compression and not to shear (Art. 15). When gradually increasing loads are applied to a bar and the values of e are measured for different values of P, the simul- taneous quantities P/a and e/l are known, and their ratio gives the value of E. The following are approximate average values of E for the different materials used in engineering construction, which have been derived from the records of numerous tests : Material Modulus of Elasticity Brick 2 ooo ooo pounds per square inch Stone 6 ooo ooo pounds per square inch Timber i 500 ooo pounds per square inch Cast Iron 15 ooo ooo pounds per square inch Wrought Iron 25 ooo ooo pounds per square inch Steel . 30 ooo ooo pounds per square inch ART. 10 ELASTIC CHANGE IN LENGTH 25 The values here given for brick and stone apply only to com- pression and little is known regarding the elastic properties of these materials under tension; those given for the other mate- rials apply to both tension and compression. It is seen that the modulus of elasticity for timber is one-tenth of that for cast iron, that for cast iron being one-half of that for steel. Table 2, at the end of this book, gives these constants in metric measures. The modulus of elasticity is a measure of the stiffness of the material, that is, of its ability to resist change of shape under unit-stresses not higher than the elastic limit. For, the unit- deformation e may be expressed by S/E, and hence is the least for a given 5 when E is the greatest. The above values of E show that the elastic change in length of a steel bar is one- twentieth of that of a timber bar and one-half of that of a cast- iron bar, the applied tension being the same in the three cases and the sizes and lengths of the bars being equal. Prob. 9o. Compute the modulus of elasticity of a bar, ij inches in diameter and 16 feet long, which elongates 0.125 inches under a ten- sion of 21 ooo pounds. Prob. 96. A wooden specimen i inch in diameter and 18 inches long, elongates 0.013 inches when the tension is increased from 800 to i 600 pounds, and 0.24 inches when the tension is increased from i 600 to 6 ooo pounds. Compute the modulus of elasticity. ART. 10. ELASTIC CHANGE IN LENGTH The change in length of a bar under an axial stress which does not exceed the elastic limit of the material is readily com- puted when the modulus of elasticity of the material is known. From the definition of that modulus in Art. 9, it is seen that the unit -deformation in length is e = S/E, and that the total deforma- tion in length is -I' - -^-3 in which e is the change in the length /, and P/a or S is the unit- stress, while E is the modulus of elasticity. If e is to be found in inches, then I must be in inches; S and E are in pounds per 26 ELASTIC AND ULTIMATE DEFORMATION CHAP. II square inch and hence their ratio is an abstract number. When S is a tensile unit-stress, e is the elongation of the bar; when 5 is a compressive unit-stress, e is the shortening of the bar. These formulas may be used for computing the change of length which occurs at the elastic limit, or under any value of S less than the elastic limit. Using the mean values of the elastic limits in Art. 2 and those of E in Art. 9, the following values of e/l are found for tension : For timber, e/l= 1/500 =0.0020 For cast iron, e/l= 1/2500=0.0004 For wrought iron, e// = 1/1000=0.0010 For structural steel, e/l= 1/850 =0.0012 These quantities show that, when bars of the same length are stressed up to their elastic limits, the elongation of the cast- iron bar is the least, that of wrought iron next, and that of timber the greatest. The ultimate elongations, namely those which obtain at rupture, follow a different order and are very much greater than those which occur at the elastic limit. As an example for stresses within the elastic limit, let it be required to find the elongation of a bar of steel, 1^X12 inches in section area and 23 feet long, when under a tension of 270 coo pounds. The unit-stress under this tension is 5 = 270000/18 = 15 ooo pounds per square inch, and as this is less than the elastic limit the formula applies, and 6=15 000X276/30000000 = 0.138 inches. This would also be the amount of shortening of the bar under a compression of 270000 pounds, provided that no lateral bending occurred, but it will be seen later that a bar of this size and length is unable to withstand a unit-stress as high as 15 ooo pounds per square inch on account of the sidewise flexure. The above formulas can in general be used only for finding the shortening of a bar in compression when its length is less than twenty times its thickness. From the formula e/l = (P/a)/E, which applies to all elastic changes of length, any one of the five quantities may be com- puted when the other four are given. For example, the unit- elongation of timber when it is stressed up to one-half of its ART. 11 ELASTIC LIMIT AND YIELD POINT 27 elastic limit is found by e/l =1500/1 500 000= i/i ooo, so that if a stick of timber elongates 0.48 inches, its length is / = 0.48 X i ooo = 480 inches = 40 feet. Should it be required to find the elongation of a bar of wrought iron, if inches in diam- eter and 1 6 feet long, under a tension of 51 ooo pounds, the section area is 1.485 square inches, and the unit-stress 5 = 51000/1.485 = 34300 pounds per square inch; as this is greater than the elastic limit of the material, the formula has no validity and it is impossible to compute by it the value of the elongation. Prob. 10a. Compute the elongations of a wrought-iron bar, i^- inches in diameter and 24 feet long, under tensions of 10 ooo, 20 ooo, and 30 ooo pounds. Prob. 106. Compute the tensile force required to stretch a bar of structural steel, 1^X9! inches in section area and 23 feet 3^ inches Jong, so that its length may be increased to 23 feet 5-^ inches. ART. 11. ELASTIC -LIMIT AND YIELD POINT In Art. 2 the elastic limit was defined as that unit-stress within which the deformation of a bar is proportional to the stress and beyond which the deformation increases in a more rapid ratio than the stress. The diagram in Art. 4 illustrates these experi- mental facts for tension, the relation between unit-stress and unit-elongation being shown by a straight line until the elastic limit is reached and afterwards by a curve. The point where the straight line is tangent to the curve indicates the elastic limit of the material. In Fig. lla there is given, on a larger horizontal scale, a part of the tension diagram of Fig. 4. For each material the point of elastic limit is marked by a dash normal to the curve. For any unit-stress 5 less than the elastic limit, the ratio of S to the unit-deformation e is a constant, or S/e = E, and E is the modu- lus of elasticity of the material. The greater the inclination of the straight line to the horizontal, the greater is the value of E. Since and 5 are variable rectangular coordinates, the equation 28 ELASTIC AND ULTIMATE DEFORMATION CHAP. II S = Ee is that of a straight line and E is the tangent of the angle which this line makes with the horizontal. 0.001 0.002 Unit-Elongation =e. 0.003 Fig. lla Another definition of the elastic limit is that it is the unit- stress within which the bar returns to its original length, when the load is removed, and beyond which it does not fully return, a part of the deformation being permanent. The term 'limit of elasticity' is sometimes used instead of elastic limit when considering the subject from this point of view. Fig. 116 shows a part of the tensile diagram for wrought iron in which AB repre- sents a bar of unit length w r hich is stretched to the lengths AB', AC' , and AD' under stresses of 25 ooo, 30 ooo, and 34 ooo pounds per square inch. If the tension is removed when the stretch is less than BB' the bar springs back to its original length AB; here B'b is the elastic limit of the material. If the tension is removed when the bar has the stretch BD', the bar springs back to D so that its length is AD and it has the permanent set BD. Similarly for any length AC the bar under tension has the stretch BC' and, on the removal of the tension, it springs back to the length AC and has the permanent set BC. It has been found by many experiments that the lines Cc and Dd are nearly straight and closely parallel to the line Bb. An inspection of Figs, lla and 116 shows that it is more diffi- cult to locate the point of tangency'on the lines for timber and cast iron than on those for wrought iron and structural steel. ART. 11 ELASTIC LIMIT AND YIELD POINT 29 In other words the elastic limits in tension for timber and cast iron are not well denned. On the compression diagram it is also seen that the elastic limits for wrought iron and structural steel are not so sharply denned as those for tension, although they are more determinable than for the other materials. The 'Yield Point' is defined to be that unit-stress at which the deformation increases without any increase in applied load or in internal stress. In Fig. lla it is seen that the lines for wrought iron and structural steel curve beyond the elastic limit for a short distance and then become horizontal. In Fig. 116 the point where the curve becomes horizontal is marked by the letter b', and the unit-stress corresponding to this is called the yield point. Only ductile materials like wrought iron and struc- tural steel have yield points and these generally belong to ten- sion and are usually only observed in testing machines where the tension is slowly applied. Beyond the yield point the curve continues horizontal for a short distance and then gradually rises. 0.004 0.00(5 0.008 Unit-Elongation =e Fig. 116 In commercial testing the yield point is sometimes called the elastic limit because the former is more easily ascertained than the latter (Art. 25). This is an improper use of the term elastic limit which should be avoided. For structural steel the yield point is often higher than the elastic limit by from 3 ooo to 6 coo pounds per square inch, and hence it is important that the dis- tinction should be carefully observed. Prob. 11. Bars of cast iron, wrought iron, and steel, i square inch in section area and 100 inches long, are subject to a tension of 15 ooo pounds. Determine the elongation of each bar from Fig. lla. Also compute the elongations, as far as possible, from the rules of Art. 10. 30 ELASTIC AND ULTIMATE DEFORMATION CHAP. II ART. 12. ULTIMATE DEFORMATIONS Mean values of the ultimate unit-elongations for different materials are given in the table of Art. 4, and the stress diagrams in Fig. 4 show that these are large compared with that which occurs at the elastic limit. Timber and cast-iron bars have ultimate elongations about 8 times as great as the elastic elonga- tions, while steel has ultimate elongations from 100 to 300 times as great as the elastic ones. There is no method by which the ultimate elongation of a bar can be computed, but all knowledge concerning it must be derived from the records of tests. The ultimate elongation of a bar is determined by making two marks upon it before it is subjected to tension and measuring the distance between them before and after the test. The differ- ence of these lengths divided by the original length gives the ulti- mate elongation per unit of length, that is, the unit-elongation. For example, if the distance between the two marks is 2.01 inches and if this becomes 2.57 inches after the rupture, then the total elongation is 2.57 2.01=0.56 inches, and the ultimate unit- elongation is 0.56/2.01=0.280 or 28.0 percent. In Fig. 169a are shown three steel specimens, that at the top being one which has not been tested. Its total length is 5^ inches and about 2\ inches of the central part is one-half an inch in diameter. The marks are placed on this cylinder about- 2 inches apart, the exact distance being measured to the nearest hundredth of an inch. The two other specimens have been ruptured by tension applied through the screw ends by the testing machine, and the respective elongations were found to be 3.8 and 22.5 percent. This great difference in the ultimate elongation of steel may be due to differences in chemical composition and method of manufacture, but in this case it was largely due to a flaw in the ruptured section of the middle specimen. The loads that ruptured these two specimens were 18 600 and 22 ooo pounds respectively, so that the ultimate strengths were about 83 ooo and 1 10 ooo pounds per square inch. ART. 12 ULTIMATE DEFORMATIONS 31 An elongation of a bar is always accompanied by a reduc- tion in the area of its cross-section. The greater the ultimate elongation the less is the area of the ruptured section. For duc- tile materials, like wrought iron and some kinds of steel, there is observed a very marked change, as the ultimate strength is approached, on both sides of the section where rupture is about to occur. In Fig. 169a this is scarcely observable in the middle specimen, but it is plainly seen in the lower one. The term 'Reduction of Area' refers to a ruptured specimen and means the diminution in section area per unit of original area. Thus for the third specimen, the original section area was 0.1995 square inches, the area of the ruptured section was 0.1064 square inches, and hence the reduction of area is 0.0931/0.1995=0.467, or 46.7 percent. Instead of actually com- puting the area, the squares of the diameters may be used; thus, the original diameter was 0.504 inches, the diameter of the rup- tured section was 0.368 inches, and the squares of these are 0.2540 and 0.1354; the reduction of area is then 0.1186/0.2540 = 46.7 percent. Reduction of area, or contraction of area as it is often called, is an index of the ductility of the material, and it is generally regarded as a more reliable index than elongation, because the ultimate unit-elongation is subject to variation with, the ratio of the length of the specimen to its diameter, whereas the reduction of area is more constant. In Art. 169 further remarks regarding ultimate elongation will be found. Under compression, a cube or a prism decreases in length and the area of its cross-section increases with the amount of shortening. The ultimate shortening is, however, rarely deter- mined, and in most cases it is much less than the ultimate elon- gation in tension. The rupture of the compression specimen, having a length of less than ten times its least lateral dimen- sion, occurs usually by an oblique shearing which is illustrated in Fig. 5c and which will be discussed later (Art. 147). When a bar is under tension exceeding its elastic limit, and this is released by the removal of the load, the length of the bar is greater than before, as shown in the last article. It is impos- 32 ELASTIC AND ULTIMATE DEFORMATION CHAP. II sible to compute this new length, but the amount of change after the removal of the tension can be ascertained when the modulus E is known. In Fig. 116 let the unit-stress D'd be represented by 5; then, since dD is parallel to bB, the tangent of the angle dDD' is the modulus , and hence on the removal of the tension the change in unit-elongation is DD'=S/E. Thus, if a bar of structural steel 20 feet long is stressed in tension up to 45 ooo pounds per square inch, it will shorten, when the stress is released, the amount 240X45 000/30 ooo 000 = 0.36 inches. Although the area of the cross-section decreases under ten- sion and increases under compression, the unit-stress S is always obtained by dividing the load P by the original area a. This is perhaps not strictly correct, but it is the custom in practice, and all stress diagrams are constructed by using unit-stresses com- puted in this manner. Prob. 12. Test specimen No. 7478 of the Watertown Arsenal was 0.798 inches in diameter and 6 inches long between marks. After rup- ture the distance between marks was 7.44 inches, and the diameter of the ruptured section was 0.64 inches. Compute the percentages of ultimate elongation and reduction of area. ART. 13. CHANGES IN SECTION AND VOLUME When the unit-stress does not exceed the elastic limit, the changes in the area of the cross-section and in the volume of a body under stress are very small, but it is possible to com- pute them approximately as shown below. When the unit-stress exceeds the elastic limit, there is no method by which changes in section and volume can be computed. Many measurements of the lateral dimensions of bars under normal stress have proved that their elastic change is propor- tional to the linear unit-deformation. Thus, let a bar of length I and diameter d have the unit-elongation e; its increase in length is el and its decrease in diameter is, Asd, where ^ is a number which has been found to be about J for cast iron and about for steel. For example, the value of e for wrought iron stressed ART. 13 CHANGES IN SECTION AND VOLUME 33 to its elastic limit is about o.ooi (Art. 10), so that the change in length of the bar then is o.ooil, while the change in its diameter is ^Xo.ooid. Let h be any lateral dimension of length /, and let h' and V be the lateral dimension and length under a stress which does not exceed the elastic limit of the material; then for tension, l' = (i + e)l and h' = (i-h)h (13) and similarly for the case of compression, /' = (i-e)J and h' = (i + le)h (13)' in which the linear unit-deformation e may be computed from e=S/E (Art. 9). The number is very small compared with unity, and hence its square and higher powers may be neglected when they occur in algebraic work. 4 1 Fig. 13a Fig. 136 Let a bar of rectangular section have the breadth b and the depth d; the area of the section is bd. Under elastic tension, b becomes (i Ae)6 and d becomes (i Ae)d, so that the area of the section is (i 2Jle)W, and hence the decrease in area is iXebd, or 2\e is the decrease per unit of area. For example, let a bar of structural steel of 16 square inches cross-section be stressed in tension up to 27 ooo pounds per square inch ; then = 27000/30000000=0.0009 and, since A is , the decrease in unit-area is 0.0006, so that the area of the section under this stress is 16-0.0006X16 = 15.99 square inches. Fig. 13ss in the hoop. ART. 33. INVESTIGATION OF RIVETED JOINTS When two overlapping plates are fastened together with one row of rivets, as in Fig. 33a, the joint is called a lap joint with single riveting; when two rows ar6 used, as in Fig. 336, it is said to be a lap joint with double riveting. When tension is transmitted through such plates, its first effect is to bring a side- wise compression on the rivet, and this in turn brings a shear on the rivet which tends to cut it off in the plane of the surface of junction of the plate. The exact manner in which the side- wise compression acts upon the cylindrical surface of the rivet is not known, but it is usually supposed that it causes a com- pressive stress which is uniformly distributed, over the projec- tion of that surface upon a plane through the axis of the rivet. For a lap joint with single riveting, as in Fig. 33a, let P be the tensile force which is transmitted from one plate to another by means of a single rivet, t the thickness of the plate, and p the pitch of the rivets. Let S t be the tensile unit-stress which occurs in the section of the plate between two rivets, and S c and 6" s be the unit-stresses of compression and shear upon a rivet. Then the equations between the stresses and the force P are, ART. 33 INVESTIGATION OF RlVETED JOINTS 81 for tension on plate, l(p d)S t = P for compression on rivet, tdS c = P for shear on rivet, \7i& . S S = P From these equations the unit-stresses may be computed when the other quantities are known, and by comparing them with the proper allowable unit-stresses (Art. 7) the degree of security of the joint is estimated. FIG. 33a. FIG. 336. For a lap joint with double riveting, the plates have a wider lap, and the two rows of rivets are staggered, as in Fig. 336. Let P be the tension which is exerted over the width equivalent to the pitch p, this being the distance between the centers of two rivets in one row. Then P is transferred from one plate to another through two rivets, and the three formulas are, t(p-d)S t = P 2td.S c = P 2.\x=4.75/ efficiency =0.5 7 ART. 34 DESIGN OF RlVETED JOINTS 85 so that if the thickness of the plate be given, and the diameter and pitch of the rivets be made according to these rules, this riveted joint has about 57 percent of the strength of the unholed plate. For a lap joint with double riveting, where a = 2 and /? = 2, the formulas become J=2.o4/ /=7-48/ efficiency =0.73 This investigation shows clearly the advantage of double over single riveting, and by adding a third row the efficiency will be raised to about 80 percent. In both cases the area t(p d} must be sufficient to carry the tension P under the unit-stress S t . The application of the above formulas to butt joints makes the diameter of the rivet equal to the thickness of the plate, and makes the pitch much smaller than the above values for lap joints. These proportions are difficult to apply in practice on account of the danger of injuring the metal in punching the holes. For this reason riveted joints are often made in which the strengths of the different parts are not equal. Many other reasons, such as cost of material and facility of workmanship, influence also the design of a joint. The old rules which are still sometimes used for determining the pitch in butt joints are expressed by, p-d+? and p=d+^ the first being for single and the second for double riveting. These are deduced by making the strength of the joint equal in tension and shear, and taking S 8 =S t , thus neglecting entirely the influence of the compression on the rivet. The term 'bearing compression ' is in common use for the side wise compression brought upon a rivet by the tension in the plate. The assumption regarding its distribution is a rough approximate one and rivets are more apt to fail by shearing than by bearing compression. Hence it is customary to allow a higher working unit-stress for the bearing compression of a rivet than for the tension of a plate, notwithstanding that rivet steel is gen- erally somewhat lower in strength than plate steel (Art. 25). Cooper's bridge specifications give 9 ooo pounds per square 86 CASES OF SIMPLE STRESS CHAP, iv inch as the highest allowable stress for rivets in shear and 1 5 ooo pounds per square inch for rivets in bearing compression. It may be .required to arrange a joint so as to secure either strength or tightness. For a bridge, strength is mainly needed; for a gasholder, tightness is the principal requisite; while for a boiler both these qualities are desirable. In general a tight joint is secured by using small rivets with a small pitch. The distance between the rows of rivets is determined by practical considerations. The lap of the plates should be sufficient so that the rivets may not tear or shear the plate. If the dis- tance from the center of the hole to the edge of the plate be /, there are two areas tl along which shearing tends to occur, and itlSg must be equal to or greater than the tension P for a single- riveted joint, or / must be equal to or greater than P/2tS,, In the preceding discussion of the stresses on the rivet, it has been supposed that the compressive and shearing stresses act independently of each other. This assumption is the one commonly made in practice, but the investigations in Arts. 105 and 141 show that these stresses should really be combined in order to obtain the actual maximum stresses of compression anc shear. The usual method of practice above explained in which liberal factors of safety are used is, however, generally regarded as one giving ample security. Prob. 34a. A butt joint is like Fig. 33 at tne m iddle of the span M =450X9-180X4^ 360X3= +2 160 pound- feet, and so on. These values being laid off as ordinates and lines being drawn connecting their ends, the distribution of the shears and moments throughout the beam is graphically represented. ^^ Illllllll } Fig. 46d Fig. 46e Fig. 46/ From the definitions of V and M and from the general dis- cussions in Arts. 37 and 38, it is seen that the shears are always represented by ordinates to straight lines which are inclined when uniform load is considered and horizontal when concen- trated loads are alone considered. It is also seen that the moments are represented by ordinates to straight lines for concentrated loads alone and by ordinates to parabolic curves for uniform load either alone or accompanied by concentrated loads. The above diagrams for simple beams show that the maxi- mum moment occurs at the section where the shear passes through zero. This is indeed a general law the truth of which will be demonstrated in the next article. This law also applies to canti- lever beams, for at the wall there is a reaction equal to the weight of the beam and load, hence on passing that section the shear becomes zero. Prob. 46a. Construct shear and moment diagrams for a cantilevet beam 10 feet long and weighing 13 pounds per linear foot, there being a concentrated load of 75 pounds at 2 feet from the left end. Prob. 466. Construct shear and moment diagrams for a simple beam 20 feet long and weighing 13 pounds per linear foot, there being a concentrated load of 240 pounds at 5 feet from the left end. ART. 47 MAXIMUM SHEARS AND MOMENTS 119 ART. 47. MAXIMUM SHEARS AND MOMENTS The greatest numerical value of the shear or moment which occurs in a beam under a given system of loads is called the maximum value, whether it be positive or negative. These maximum values are to be used in the shear and 'flexure formulas of Art. 41 without regard to sign, for the signs + and pre- fixed to a shear indicate merely whether the left part of the beam tends to slip up or down with respect to the other part (Art. 37), while when prefixed to a moment they indicate merely whether the upper side of the beam is concave or convex (Art. 44). For a cantilever beam both maximum shear and moment occur at the wall. Let w be the uniform load per linear unit, / the length of the beam, and W the total load wl; then the max- imum shear is V= ivl = W, and the maximum moment is M = Wx^l = $Wl. Let P be any concentrated load at the left end of the beam, as in Fig. 466, then the maximum shear is V = P and the maximum moment is M= PI. Or, without regard to sign, the maximums for a cantilever beam are, For uniform load V=W M=$Wl For P at the end V= P M=Pl If P is not at the free end but at the distance p from that end, as in Fig. 470, the maximum shear is also P, but the maximum moment is P(lp). Fig. 47o For a simple beam under uniform load alone, each reaction is \wl and the shear at any section distant x from the left support is V= +%'wlwx; the maximum shears hence occur when x = o and x = l, their values being +$W and -%W. The moment for the section distant x from the left support is M + \iul . x ~wx . J#, and by the usual method of the dif- ferential calculus this is found to be a maximum when # = $/, 120 CANTILEVER AND SIMPLE BEAMS CHAP. VI and for this value M = + ^wl 2 = \WL For a single load P at the middle of a simple beam, each reaction is %P and this is the maximum shear; also the moment at any section between the left support and the middle is M = + %Px, and this is a maximum when x=\l, as Fig. 46e shows. Hence without regard to sign, the maximums for a simple beam are, For uniform load V=$W M=\Wl For P at the middle V= $P M= Pl If P is not at the middle but at the distance p from the left end, as in Fig. 476, the left reaction is P(l p}/l and the right reaction is Pp/l and these are the maximum shears; the maximum moment is under the load and its value is R\p or P(l p)p/l, which has its greatest value %Pl when p equals /. When one or more concentrated loads are on a simple beam, the maximum shears are the reactions of the supports, and the maximum moment due to these loads occurs under one of the loads, as in Fig. 3Sc. The maximum moment due to both uniform and concentrated loads also often occurs under one of the single loads, as seen in Fig. 46/, but it sometimes occurs at a different section, as is the case in Prob. 46&. The section where the moment is a maximum is called the 'dangerous sec- tion ', since there the greatest horizontal stress S will be found by the use of the flexure formula S.I/c = M t (Art. 41). In order to locate the dangerous section, the following important law may be used. The dangerous section is that where the shear passes through zero. To prove this, let there be several loads on a simple beam, PI at the distance p\ and P 2 at the distance p 2 from the left support, and so on. Let RI be the left reaction due to these loads and the uniform load wl. Then the bending moment for a section distant x from the left support is, in which only the single loads appear which are between the left support and the section. The value of x which makes the bending moment a maximum is obtained by equating to ART. 48 INVESTIGATION OF BEAMS 121 zero the derivative of M with respect to x, or dM/dx=Ri wx PI P 2 etc. = o is the equation which gives the value of x. But RI -wx PiP 2 is the shear V for this section, as is clear from the definition of vertical shear in Art. 37. Therefore the maximum moment occurs at the section where the shear passes through zero. This section can readily be found by computing shears for different sections, and the construction of the shear diagram will always be of assistance. For example take the data of Prob. 466 where ^=20 feet, w= 13 pounds per linear foot, and PI = 240 pounds at pi =5 feet. The left reaction ^1=130 + 1x240=310 pounds. The shear just at the left of P is +3 10-65 = +245 and just at the right of P it is +310 65 240= +5 pounds, and hence V does not pass through zero under the single load. To find the exact position of the dangerous section, let x be its distance from the left support; then the shear is 310 13^ 240 and equating this to zero, there is found x= 5.385 feet. The maximum moment is 310X5. 385-^X13X5. 3852-240X0.385= i 388 pound-feet. Prob. 47a. Compute the maximum shear and moment for a canti- lever beam 13 feet long which weighs 33 pounds per linear foot and has a single load of 375 pounds at 3 feet from the free end. Prob. 476. A simple beam of 12 feet span weighs 35 pounds per linear foot and has three concentrated loads of 30x3, 60, and 1 50 pound? at 3, 5, and 8 feet respectively from the right support. Compute the maximum shear and moment, and draw the shear and momenl diagrams. ART. 48. INVESTIGATION OF BEAMS The investigation of a beam of constant cross-section usually consists in computing the greatest horizontal unit-stress S from the flexure formula 5 . 1/c=M which was established in Art. 41. For this purpose the formula may be written S=Mc/I or S=M/(I/c) (48) the first of which may be used when c is determined by Art. 42 122 CANTILEVER AND SIMPLE BEAMS CHAP. VI and / by Art. 43, and the second when the section factor I/c for rolled beams is taken from tables (Art. 44). The greatest value of S will be found at the dangerous section where M is a maximum and hence the maximum moment is to be ascertained from Art. 47. If M is computed in pound-feet, it must be reduced to pound-inches when I and c are expressed in terms of inches; then the value of 5 will be in pounds per square inch. The unit-stress S will be tension or compression according as the remotest fiber from the neutral surface lies on the convex or concave side of the beam, c being the distance between that fiber and the neutral surface. If S' is the unit-stress on the opposite side of the beam and c' the distance from it to the neutral surface, then by Art. 40, S/c=S'/c' or S'=S.c'/c When 5 is tension, S' is compression; when S is compression, 5' is tension. Sometimes it is necessary to compute S' as well as 5 in order thoroughly to investigate the stability of the beam. By comparing the values of 5 and S' with the proper working unit-stresses for the given materials (Art. 7), the degree of security of the beam may be inferred. As an example consider a cast-iron I beam which has a depth of 10 inches, width of flanges 6 inches, thickness of flanges and web i inch. It is supported at its ends forming a span of 12 feet, and carries two loads, each weighing 5 ooo pounds, one being at the middle and the other at one foot from the right end. The steps of the computation are as follows: by geometry, a= 20 square inches by Art. 42, c= 5 inches by Art. 43, 7= 286.7 inches 4 by Art. 17, w= 62.7 pounds per linear foot by Art. 47, #= 6 feet for dangerous section by Art. 38, max. M=iS 630 pound-feet Then from the flexure formula, the unit-stress is, S=i8 600X12X5/287 = 3 900 pounds per square inch which is the tensile stress on the lower side of the beam and ART. 48 INVESTIGATION OF BEAMS 123 the compressive stress on the upper side. The factor of safety in compression is 90 ooo/ 3 900 or about 23, while that in tension is 20 000/3 9 or about 5 ; the degree of security for compres- sion is ample, but that for tension is too low (Art. 7). As a second example, consider a simple wooden beam, 3 inches wide, 4 inches deep, and 16 feet span, and let a man weigh- ing 150 pounds stand at the middle. Here 6=3 inches, d = 4 inches, / = i6 feet = 192 inches, c = %d=2 inches, / =^bd 3 = 16 inches 4 , the weight of the beam ^" = 10X12X5^/12 =53.3 pounds, and P = i5o pounds. The dangerous section is at the middle, and the maximum moment due to the uniform and single loads is, from Art. 47, M = |W7+JP/ = 706.7 pound-feet = 8480 pound-inches. Then the flexure formula gives S = 8480X2/16 = 1060 pounds per square inch, which is a satis- factory unit-stress for the tension of timber under a steady load, but is a little too high for compression. A short beam heavily loaded should also be investigated for shearing at the supports by the shear formula S a a = V (Art. 41), but for common cases there is ample security against this stress. For the cast-iron beam above discussed, the maximum shear is at the right support and its value is 7460 pounds; hence S 8 = 7 460/20 = 373 pounds per square inch, so that the factor of safety against shearing is about 48. Similarly for the timber beam, the factor of safety against shearing may be found to be greater than 350. When the load upon a beam is heavy compared with its own weight, the latter may be omitted from the computation as its influence is small. A common rule in practice is that the weight of the beam may be neglected whenever the moment due to it is less than ten percent of that due to the loads on the beam; for a concentrated load at the middle this will be the case when \Wl is less than one-tenth of \Pl, that is, when P is greater than $W. Prob. 48a. A piece of scantling 2 inches square and 10 feet long is hung horizontally by a rope at each end and two students stand upon it. Is it safe ? 124 CANTILEVER AND SIMPLE BEAMS CHAP. VI Prob. 486. A cast-iron bar one inch in diameter and two feet long is supported at its middle and a load of 200 pounds hung upon each end of it. Find its factor of safety. ART. 49. SAFE LOADS FOR BEAMS The proper load for a beam should not make the value of 5 at the dangerous section greater than the allowable unit-stress. This allowable unit-stress or working strength may be assumed according to the circumstances of the case by first selecting a suitable factor of safety from Art. 7 and dividing the ultimate strength of the material by it, the least ultimate strength whether tensile or compressive being taken. For any given beam the quantities / and c are known. Then, in the flexure formula M =S . I/c the maximum moment M may be expressed in terms of the length of the beam and the unknown loads, and thus those loads be found. As an example, consider a wooden cantilever beam whose length is 6 feet, breadth 2 inches, depth 3 inches, and which is loaded uniformly with w pounds per linear foot. It is required to find the value of w so that S may be 800 pounds per square inch. Here c = i} inches, / =f|, and M = 36x6-^. Then, from the flexure formula, 216-^ = 800X54/1^X12, whence w = n pounds per linear foot. Since this wooden beam weighs about 2 pounds per foot, the total safe uniform load will be about 9X6 = 54 pounds. As a second example, take a rolled steel I beam of 18 feet span which is 10 inches deep and weighs 25 pounds per foot, and let it be required to find what concentrated load P may be put at the middle in order that the unit-stress at the dangerous section shall be 15 ooo pounds per square inch. From Table 6 the value of the section factor I/c is 24.4 inches 3 . From Art. 47 the maximum moment is M=$PX 18X12 pound-inches if the weight of the beam be disregarded. Accordingly ^P=S.I/c or 54P = 15 000X24.4, whence P =6 780 pounds. As this is more than five times the weight of the beam, it may be taken as the allowable load (Art. 48). If the weight of the beam is considered, ART. 50 DESIGNING OF BEAMS 125 however, the moment is M = 54^+12 150, and placing this equal to S . I/c there is found P = 6 560 pounds, which is only about three percent less than the value obtained before. As an example of an unsymmetric section, let it be required to determine the total uniform load W for a cast-iron T beam of 14 feet span, so that the factor of safety may be 6, the depth of the beam being 18 inches, the width of the flange 12 inches, the thickness of the stem i inch, and the thickness of the flange i \ inches. First, from Art. 42 the value of c is found to be 12.63 inches, and that of c' to be 5.37 inches. From Art. 43 the value of / is computed to be 1031 inches 4 . From Art. 47. the maximum moment is M = \Wl = 2iW pound-inches. Now with a factor of safety of 6, the working unit-stress S on the compressive side of the dangerous section is to be x 90 000 = 15 ooo pounds per square inch; then, inserting the values in the flexure formula M = SI/c, the load W is found to be 58 300 pounds. Again with a factor of safety of 6, the working unit- stress S' on the tensile side of the dangerous section is to be X 20 ooo = 3 330 pounds per square inch ; inserting the values in the flexure formula M=S'I/c f , the load W is found to be 30400 pounds. The total uniform load on the beam should hence not exceed 30 400 pounds, and under this load the factor of safety on the compressive side is nearly 12. Prob. 49a. A simple wooden beam, 8 inches wide, 9 inches deep, and 14 feet in span, carries two equal loads, one being 2.5 feet on the left of the middle and the other 2.5 feet on the right. Find these loads so that the factor of safety of the beam shall be 10. Prob. 496. A steel railroad rail of 2 feet span carries a load P at the middle. If its weight per yard is 56 pounds, 7=12.9 inches 4 and =2. 1 6 inches, find P so that the greatest horizontal unit-stress at the dangerous section shall be 6 ooo pounds per square inch. ART. 50. DESIGNING OF BEAMS When a beam is to be designed, the loads to which it is to be subjected are known, as also is its length, and from these the maximum bending moment may be found by Art. 47. The 126 CANTILEVER AND SIMPLE BEAMS CHAP, vi allowable working unit-stress 5 is assumed in accordance with engineering practice. Then the flexure formula (41) may be I/c-M/S (50) and the numerical value of the second member be found. The dimensions to be chosen for the beam must be such that the section factor I/c shall be equal to this numerical value, and these in general are determined by trial, certain proportions being first assumed. The selection of the proper proportions and shapes of beams for different cases requires much judgment and experience; but whatever forms be selected, they must in each case be such as to satisfy the above equation. For instance, a simple beam of structural steel, 16 feet in span, is required to carry a rolling load of 500 pounds. Here, by Art. 47, the value of maximum M due to the load of 500 pounds is 24 COD pound-inches. From Art. 7 the allowable value of 5 for a variable load is about 10 ooo pounds per square inch; then, I/c= 24 000/10 000= 2.4 inches 3 An infinite number of cross-sections may be selected having this value of I/c. If the section is round and of diameter d, it is known that c = \d and / = ^f7rd 4 , hence -$^71$* = 2. 4, from which ^ = 2.9 inches. If the section is rectangular, 2 inches wide and 2^ inches deep, I/c is 2.5, which is a little too large, but it would be well to use this size because the weight of the beam itself has not been considered in the discussion. The dimensions finally selected may be investigated by Art. 49 in order to deter- mine how closely the actual unit-stress agrees with the value assumed. Thus, the rectangular section 2X2^ inches weighs 17 pounds per foot; the maximum moment is then 30 530 pound- inches and the unit-stress is found to be 1 1 800 pounds per square inch, which is 18 percent larger than the allowable value; a larger size than 2X2^ inches is hence required, and 2X3 inches will be found to be larger than necessary. When the design of a structure involves rolled I beams, the computation of the maximum value of M/S is made as before, ART. 51 ECONOMIC SECTIONS 127 and the corresponding number is sought in that column of Table 6 which is headed I/c. For example, take a floor which is required to carry a uniform load of 180 pounds per square foot including its own weight, this weight being brought upon rolled steel beams which are of 24 feet span and spaced 4 feet apart. It is required to find what size of beam should be used, the allowable unit-stress S being specified as 16 ooo pounds per square inch. First, the total uniform load on one beam is found to be W = 180X24X4 =17 280 pounds; second, the maximum moment is M = X 17 280X24X12 inch-pounds; third, from these If 75 = 38.9, which is the required value of I/c; fourth, Table 6 shows that the 12 -inch beam weighing 40 pounds per foot has //c = 44.8, and this is the size to be selected. The larger table given in the handbook of the manufacturers indicates, however, that a 1 2-inch beam weighing 35 pounds per foot may be obtained by special order and that its value of I/c is 38.0; whether this would be cheaper than the 1 2-inch beam weighing 40 pounds per foot can be determined only by asking quotations of prices. Prob. 50a. Aioo-lb. 1 5-inch steel I beam of 12 feet span sustains a uniformly distributed load of 41 net tons. Find its factor of safety. Also the factor of safety for a 24-foot span under the same load. Prob. 506. A floor, which is to sustain a uniform load of 175 pounds per square foot, is to be supported by heavy xo-inch steel I beams of 15 feet span. Find their proper distance apart from center to center, so that the maximum fiber stress may be 1 2 ooo pounds per square inch. ART. 51. ECONOMIC SECTIONS The two fundamental objects to be secured in designing engineering structures are stability and economy (Art. 7). In the case of a beam, proper security will be attained when the horizontal unit-stress 5 does not exceed that allowable in good practice, and economy will be secured by giving such propor- tions to the cross-section that S is not less than the allowable value. Both stability and economy will hence usually be pro- moted by making the beam of such a size that the horizontal 128 CANTILEVER AND SIMPLE BEAMS CHAP. VI unit-stress S has the given allowable value at the dangerous section when the beam is fully loaded. There is, however, another important idea to be considered, namely, the shape of the section should be such that the weight of the beam shall be as small as possible, and this will be attained by making the section area a as small as possible and yet keep the unit-stress S at the given allowable value. Since the horizontal unit-stresses in the section increase from the neutral surface to the upper and lower sides of the beam, it is evident that a deep section will in general require less area to furnish a given unit-stress S than a shallow one.. Thus, for a rectangular section of breadth b and depth d, the section factor I/c is ^bd^/\d or \bd 2 , so that the flexure formula becomes \bd? = M/S or a = 6M/Sd, so that for given values of M and S, the section area a will be rendered small by making the depth d large. The depth, however, should not be made so great as to give a thin section, for this would be deficient in lateral stiffness. The proper ratio between b and d is governed by engineering prec- edent and practice; an extreme limit is perhaps that found in wooden floor joists where the depth is about six times the breadth. Iron and steel beams may be cast or rolled with almost any desired shape of section. Cast-iron T and I beams came into use early in the nineteenth century; after 1840 wrought-iron rolled beams gradually replaced those of cast-iron for railroad use, and these in turn gave way after 1890 to rolled steel beams. In these forms the sections have such a shape that the section area is the least possible for a given required unit-stress S, and this is accomplished by concentrating most of the material in the flanges and thus having the larger part of the section area as far from the neutral axis as practicable. The flexure formula SI/c = M shows that this practice is correct, for, under a given bending moment M, the unit-stress S will be made small by making I/c as large as possible, and from the definition of the moment of inertia (Art. 43) it is clear that / will be rendered large by placing the material as far as practicable from the neu- ART. 51 ECONOMIC SECTIONS 129 tral axis. The value of 7 may be expressed by ar 2 , where a is the section area and r is the radius of gyration of the section, that is, the distance from the neutral axis to the point where all the section area might be concentrated and have the same moment of inertia as the actual distributed area. Inserting this for 7 in the flexure formula, it becomes ar 2 /c = M/S and thus it is plain that a may be rendered small by making r 2 /c as large as possible. Since r is always less than c, the placing of the greater part of the section in the flanges, while the web is made thin, renders r 2 /c much larger than for a rectangular section, and thus a is made smaller and economy of material is secured. For example, take the smallest 1 5-inch I beam in Table 6 for which = 12.48 square inches, 7 = 441.7 inches 4 and I/c = 58.9 inches 3 . A square section of the same strength must have the same value of I/c, whence ^d 3 = ^S.g and = 50.0 square inches, which is four times the section area of the I and hence the square beam will weigh four times as much as the I beam, and its cost will be four times as much if the price per pound is the same. For wrought iron and structural steel the ultimate strengths in tension and compression and the allowable unit-stresses are usually the same; hence the flanges of beams of these materials are made equal in size. Cast-iron, however, has a much higher ultimate strength in compression than in tension and hence the tensile flange should have the larger area. The proper relative proportions of the flange areas of cast-iron beams have never been definitely established, and such beams are now rarely used except in unimportant buildings. The strongest rectangular beam that can be cut from a cir- cular log will be that which has the largest section factor I/c. If b be the breadth and d the depth, the section factor is $bd 2 , and bd 2 is to be made a maximum; or, if D be the diameter of the log, b(D 2 b 2 ) is to be made a maximum. Differentiating this and equating the derivative to zero, gives, b = D\/$ whence d = -EN/F 130 CANTILEVER AND SIMPLE BEAMS CHAP. VI Hence b is to d as 5 to 7 nearly. From this it is easy to show that the way to lay off the strongest rectangular beam on the end of a circular log is to divide the diameter into three equal parts, from the points of divi- sion draw perpendiculars to the circumference, and then join the points of intersection with the ends of the diameter, as shown in the figure. The rectangular beam thus cut out is, of course, not as strong as the log, and the ratio of its strength, to that of the log is that of their values of I/c, which will be found to be about 0.65. Prob. 51. Cast-iron beams with a 11 section are sometimes used in buildings. Let the thickness be uniformly one inch, the base 8 inches, the height 6 inches, and the span 12 feet. Find the unit-stresses 5 and S f at the dangerous section under a uniform load of 5 ooo pounds. ART. 52. RUPTURE OF BEAMS The flexure formula S.I/c = M is only true for stresses within the elastic limit of the material, since beyond that limit the latter part of law 6 of Art. 40 does not hold. Experience shows that the elongations and shortenings of the horizontal fibers are proportional to their distances from the neutral axis when the stresses exceed the elastic limit, but these changes of length increase in a more rapid ratio than the unit-stresses (Arts. 4-5). Hence the unit-stresses of tension and compression in- crease in a less rapid ratio than their distance from the neutral axis for all fibers where the elastic limit is exceeded. Thus Fig. 520 represents a rectangular cast-iron beam where the tensile unit-stresses below n have exceeded the elastic limit, while Fig. 52b shows a T section under similar conditions. Here the algebraic sum of all the horizontal stresses is zero, but the neutral surface does not pass through the center of gravity of the section, because the unit-stresses are not proportional to their distances from that surface as is required in the demon- stration of Art. 40. When a beam is loaded so heavily that any fiber on one or AST. 52 RUPTURE OF BEAMS 131 both sides of the neutral surface are stressed above the elastic limit, the flexure formula S.I/c=M is not valid, and a value of S computed from it is not correct. It is, however, very cus- tomary to use this formula for the rupture of beams, but in so doing it must not be forgotten that it is so used merely for con- venience of making comparisons and not on account of any theoretical basis. Fig. 520 Fig. 526 When a beam is ruptured under transverse loads the value of S computed from the flexure formula is called the 'ultimate flexural strength ' of the material. If this formula were valid beyond the elastic limit, the value of S for rupture would agree with the least ultimate strength of the material, with tension in the case of cast iron and with compression in the case of timber. It is, however, always found that this computed value does not agree with either the ultimate tensile or compressive strength of the material but is intermediate between them. This quantity is not a physical constant, but a figurative value computed from an incorrect formula, and hence is mainly valuable for rough comparative purposes. It will be designated by S f , and the following are its approximate average values in pounds per square inch as determined by experiment: Tensile Strength, S t Material Brick Stone Timber 10 ooo Cast Iron 20 ooo Wrought Iron 50 ooo Structural Steel 60 ooo 60 ooo Strong Steel 100000 no ooo 120000 For wrought iron and structural steel no values of S f are given Flexural Compressive Strength, S f Strength, S, 3000 6 ooo 8000 90000 50000 800 2 000 9 ooo 35000 132 CANTILEVER AND SIMPLE BEAM* CHAP, vi because beams of these materials bend indefinitely under increas- ing transverse loads, so that failure does not occur by breaking. In fact, bars of wrought iron and soft steel may be bent double without breaking (Arts. 24 and 25). By the use of the above experimental values of S f it is easy, with the help of the formula S f .I/c = M, to determine what load will cause the rupture of a given beam, or what must be its length or size in order that it may rupture under assigned loads. The formula when used in this manner is entirely em- pirical and has no rational basis. As an example, let it be required to find the length of a cast-iron cantilever beam, 2X2 inches in section, in order that it may break at the fixed end under its own weight. The weight of the beam is 7^ = 2X2X3^X0.94 = 12.53 pounds per linear foot or 1.044 pounds per linear inch; the bending moment is M = ^X 1.044/2 = 0.522/2 pound-inches, where / is in inches; the value of I/c is \bd 2 = 1.333 inches 3 . Inserting these in the formula it becomes 0.522/2 = 35 000X1.333,. from which / = 2Q9 inches or about 25 feet. The computed flexural strength S f is often called the ' modulus of rupture,' but the name above employed is a better one because it points out the origin and meaning of the quantity. Prob. 52a. Compute the size of a square wooden simple beam of 8 feet span in order to break under its own weight. Prob. 52b. A cast-iron simple beam 2 inches square and 12 feet long carries two equal loads at the quarter points. Find the loads which will cause rupture. ART 53. MOVING LOADS The loads upon a beam consist of its own weight and the weight of the uniform or concentrated loads which it carries. These are called 'Dead Loads' when they are permanent in position and 'Live Loads' when they may move. Beams used in buildings are subject to the dead load of the floor and to the live load of a crowd of people; beams used in bridges are sub- ject to the dead load of their own weight and to other permanent ART. 53 MOVING LOADS 133 loads, but they receive the greatest stress from the live load of moving wheels or of crowds of people. In the preceding articles dead loads have alone been generally considered, but it has been recognized that the maximum moment due to a single concen- trated load on a simple beam occurs when it is placed at the middle of the span. Other cases of concentrated loads will now be discussed. r r Fig. 53a Fig. 536 When two wheels which are at fixed distance apart, like two wagon-wheels on separate axles, roll over a beam, it might be thought that the greatest moment due to them would occur when they are on opposite sides of the middle and equally distant from it, as in Fig. 53a, but this is not the case. To find the position which will give the greatest moment, let / be the span, p the distance between the loads, and z the distance from the left support to one of the loads, as in Fig. 536. The maximum moment will occur under one of these loads, as the shear and moment diagrams show. The left reaction is found from Ril=P(l pz)+P(lz) and the moment under the first load then is M = (P/l)(2lz pz 2z 2 ). By the usual method, the value of z which renders M a maximum is found to be z = / \p, so that the center of gravity of the two loads is as far to the right of the middle as the dangerous section is to the left of the middle. For example, let each load be 3 ooo pounds, their distance apart be 5 feet, and the span be 15 feet; then 2=6.25 feet, the left reac- tion is Ri = 2 500 pounds, and the maximum ,bending moment is M = 2 500X6.25 = 15 625 pound-feet. If these loads are placed as in Fig. 53a, the reaction is 3 ooo pounds and the moment is 3000X5.0=15000 pound-feet. 134 CANTILEVER AND SIMPLE BEAMS CHAP. VI When two unequal loads PI and P 2 roll over a simple beam, let z be the distance from the left support to PI, and p be the distance from the greater load PI to the smaller one P%. Then, proceeding as before, it is easy to show that the maximum moment occurs under the first load when the distance between the first load and the center of gravity of the two loads is bisected by the middle of the span. When there are three or more con- centrated loads, the section of maximum moment does not always lie under the first load, but the general rule always holds good that the distance between that section and the center of gravity of the loads is bisected by the middle of the span. Fig. 5Sc Fig. 53d When a uniform live load moves over a simple beam there is, for any given position of the same, a section of maximum moment, this being where the shear passes through zero (Art. 47). Such cases, with their shear and moment diagrams, are shown in the two figures above, but it is rarely necessary to compute these moments, because the absolute maximum moment occurs when the uniform live load covers the entire beam, and this is liable to take place at any time. It is, however, of great impor- tance that the student should be able readily to draw the shear and moment diagrams for any assigned position of the live load, whether it be uniform or partly uniform and partly concentrated. The maximum shear due to a live load occurs when the load is placed so as to give the greatest shear at one of the supports. Thus the maximum shear due to the live load in Fig. 53b occurs at one of the supports when one load is just about to pass upon that support and its numerical value is 2P P. p/l. By the use of the shear formula 5> = V and the flexure formula SI/c = M, the unit-stresses S t and 5 1 are computed for live loads, ART. 54 DEFLECTION OF CANTILEVER BEAMS 135 after the maximum values of V and M have been found, in the same manner as if the loads were dead. Live load, however, really produces greater stresses than dead load and hence the computed values of S, and S are increased in practice accord- ing to the rules stated in Art. 136. Prob. 53a. Three loads, spaced 4 feet apart, one being 3 ooo and the others i 500 pounds, roll over a simple beam of 21 feet span. Find the position of these loads which will give the maximum moment and compute its value. Prob. 536. For Fig. 536 find the position of the live load which gives the maximum shear at the middle of the span and compute its value. Find also the position which gives maximum shear at the quarter point of the span and compute its value. ART. 54. DEFLECTION OF CANTILEVER BEAMS In Art. 45 the differential equation of the elastic curve was deduced and the general method of applying it to a particular case was indicated. The origin of coordinates may be taken at any point in the plane, but for a cantilever beam it is most convenient to take it at the free end, since the algebraic work is thus simplified. In the equation El . 8 2 y/dx 2 =M, the bend- ing moment M is to be expressed in terms of the abscissa x, and by two integrations the equation between y and x is deduced. Case I. Uniform Load. Let x and y be the coordinates of the elastic curve with respect to rectangular axes through the free end of the beam, as in Fig. 54a. Let the uniform load per linear unit be iv ; then for any section M = %wx 2 , and the general formula becomes El . d 2 y/dx 2 = \wx 2 . Integrating this, there is found, in which C is the constant of integration and dy/dx is the tan- gent of the angle which the tangent to the elastic curve makes with the axis of abscissas. To determine this constant, con- sider that dy/dx becomes zero when x equals the length of the 136 CANTILEVER AND SIMPLE BEAMS CHAP. VI beam, /; hence the value of C is w/ 3 and then, Integrating again, and determining the constant by the con- dition that y equals zero when x equals zero, there results, which is the equation of the elastic curve. When x = l, the value of y is the deflection of the end of the beam below that at the wall and this will be designated by /. Accordingly, f=%wfi/EI or f=\WP/EI is the deflection of the end of a cantilever beam under the uniform load W, if the elastic limit of the material be not exceeded. Fig. 54a Fig. 546 Case II. Load P at the Free End. Take the origin at the free end, as in Fig. 546, and let x and y be the coordinates of the elastic curve at any section. The bending moment M is Px, and the general equation is El . d 2 y/dx 2 = Px. By integration there results El . dy/dx = - %Px 2 + C and C is deter- mined by the condition that the tangent dy/dx becomes zero when x=l. Hence El . dy/8x = $P(l 2 -x 2 ), and the second integration gives for the elastic curve El . y = %Pl 2 x ^Px 3 , the constant being zero because y becomes zero when x is zero. When x = l the value of y is the maximum deflection/, and/= \Pi?/EI, which is 2$ times as great as the deflection due to the same load uniformly distributed over the length. Case III. Load P at any Point. Let id be the distance of the load from the left end as in Fig. 54t, where / is the length of the beam and K any number less than unity. Take the origin of coordinates under the load, then by the preceding case the deflection under the load is $P(l-Kl) 3 /EI. The free end, how- ever, is lower than the load and since M =o on the left of the load, the radius of curvature of the elastic curve is there infinite ART. 54 DEFLECTION OF CANTILEVER BEAMS 137 (Art. 45) and that curve is a straight line. Let tan 6 be the tan- gent of the angle which the tangent to the elastic curve under the load makes with the horizontal; then the free end is lower than the load by the amount id tan#. The value of tan0 is deter- mined from Case II, by making x=o and 1 = 1 Kl in the expres- sion for dy/dx, and thus the value of id tan0 is found to be -K) 2 /EI. Therefore, by adding the two quantities, which is the deflection of the free end due to the given load. When K = I, the load is at the wall and /=o; when K=O the load is at the free end and / becomes the same as in the pre- ceding case. U--KI.- Fig. 54c Fig. 54d Case IV. Several Loads. For a uniform load and a load P at the end, the value of M is fywx 2 Px. By integration it is found that the ordinate y is the sum of those due to W and P separately, and, is the deflection of the free end. Similarly, for any number of loads, the deflection is the sum of the deflections due to the loads taken separately; hence, as in cases of axial stress, each load produces its effect independently of others. In order that this deflection may be found correctly from the formulas, it is necessary that the maximum unit-stress S computed for all the loads from the flexure formula S .I/c=M shall not exceed the elastic limit of the material. In all cases the deflection of a cantilever beam of uniform section varies directly as the load and the cube of the length, and inversely as E and /. For a rectangular section, / ' =^b&, so that the deflection varies inversely as the breadth and inversely as the cube of the depth. These laws also hold for the simple beams discussed in the next article. 138 CANTILEVER AND SIMPLE BEAMS CHAP. VI Prob. 54a. In order to find the modulus of elasticity of a cast-iron bar 2 inches wide, 4 inches deep, and 6 feet long, it was balanced upon a support and a weight of 4 ooo pounds hung at each end, causing a deflection of 0.401 inches. Compute the value of E. Prob. 546. A wooden cantilever, 3 inches wide, 4 inches deep, and 15 feet long, carries two equal loads as shown in Fig. 54d, one being 5 feet from the end and the other 10 feet from the end. Compute the weight of these loads so that the maximum unit-stress S may be two- thirds of the elastic limit. Compute the deflection at the end due to the two loads. ART. 55. DEFLECTION OF SIMPLE BEAMS The deflection of a simple beam due to a load at the middle, or to a uniform load, is readily obtained from the expressions just deduced for cantilever beams. Thus, for a simple beam of span I with a load P at the middle, let Fig. 55a be inverted and it will be seen to be equivalent to two cantilever beams of length \l with a load \P at each end. The formula for the maxi- mum deflection of a cantilever beam hence applies to this figure, if / be replaced by \l and P by %P which gives / = PP/^EI for the deflection at the middle of the simple beam. It will be well, however, to use the general formula (45) and discuss each case independently. Case I. Uniform Load. Let w be the load per linear unit and x the distance of any section from the left end. For this section M = fywlx %wx 2 and the differential equation of the elastic curve is, Integrate this and find the constant by the condition that the tangent dy/dx equals zero when x is \l; then, - ox Integrating again and determining the constant by the condition that y is zero when x is zero, there is found, AKT. 55 ^_^J>EFLECTION OF SIMPLE BEAMS 139 Now, making x = /, the value of y is the deflection /, which is negative because it is measured downward from the axis of abscissas through the supports. It is, however, unnecessary to write this sign, and hence, is the elastic deflection of the simple beam under the uniform load W. Case II. Load P at the Middle. As before let the origin be taken at the left support, as in Fig. 55a. For any section between the left support and the middle M = \Px and the differential equation of the elastic curve is El . 2 y/dx 2 = \Px. Integrate this and find the constant by the evident condition that dy/dx = o when x = \l. Then integrate again and find the con- stant by the fact that y=o when x = o. Thus, is the equation of elastic curve between the left-hand support and the load. For the greatest deflection make x = $l, then, is the deflection due to the single load P at the middle, which is 1.6 times as great as that due to the same uniform load. \ p \P Fig. 55a Fig. 556 Case III. Load P at any Point. Here the expressions for the moment M are different on opposite sides of the load, and hence there are two elastic curves which have distinct equations but which have a common tangent and ordinate under the load. As in Fig. 556 let the load be placed at a distance /c/ from the left support, K being a number less than unity. Then the left reaction is R=P(IK). From the general equation (45), with the origin at the left support, there are found, On the left of the load, (a) EI^ 2 =Rx (6) EI& (c) EI 140 CANTILEVER AND SIMPLE BEAMS CHAP. VI On the right of the load, (a)' EI^ 2 =Rx-P(x- K l) (by EI (c) f EI To determine the constants consider in (c) that y=o when x=o, and hence that C 2 =o. Also in (c)', y=o when x = l; again, since the curves have a common tangent under the load, (6) equals (by when x = itl, and since they have a common ordinate at that point (c) equals (c}' when x=icl. Or, From these three conditions the values of Ci, C 3 , and C 4 are determined. Then the equation of the elastic curve on the left of the load is found to be, 6EIy=P(l -K)X 3 -P(2K-3K 2 + AC 3 )/ 2 * To ascertain the maximum deflection, the value of x which ren- ders y a maximum is to be obtained by equating the first deriva- tive to zero. If K is greater than , this value of x inserted in the above equation gives the maximum deflection; if is less than , the maximum deflection is on the other side of the load. For instance, if = |, the equation of the elastic curve on the left of the load is, ^4EIy = i6Px 3 - isPPx, and y has its maximum value when x= 0.5591. The greatest possible deflection due to a single load occurs when it is at the middle of the span and its value is that deduced in Case II. Case IV. Several Loads. Here, as for cantilevers, the deflection due to several loads is obtained by taking the sum of the several deflections; but it must be carefully noted that the computed value will be correct only when the unit-stress S at the dangerous section is less than the elastic limit of the material. The above formula for the deflection due to a load at the middle is frequently used to determine the modulus of elasticity E, several ART. 56 COMPARATIVE STRENGTH AND STIFFNESS 141 values of/ being measured for several different loads, in order to obtain a mean value of E. Prob. 55a. When K is greater than in Fig. 556, show that the maximum deflection isf=PP(i-K)($ic-$K 2 )i/3EL Prob. 556. In order to find the modulus of elasticity of Quercus alba, a bar 4 centimeters square and one meter long was supported at the ends, and loaded in the middle with weights of 50 and 100 kilo- grams, measured deflections being 6.6 and 13.0 millimeters. Com- pute the mean value of E in kilograms per square centimeter. ART. 56. COMPARATIVE STRENGTH AND STIFFNESS The strength of a bar under tension is measured by the load that it can carry with an assigned unit-stress. In the same manner the strength of a beam is measured by the load that it can carry with an assigned unit-stress on the remotest fiber at the dangerous section. Let it be required to determine the relative strength of the four following cases, ist, A cantilever loaded at the end with W 2nd, A cantilever uniformly loaded with W 3rd, A simple beam loaded at the middle with W 4th, A simple beam loaded uniformly with W Let / be the length in each case. Then, from Art. 47 and the flexure formula M=S . I/c, there is found, forist, M=Wl and hence W=SI/cl for 2nd, M=$Wl and hence W=2.SI/cl for 3rd, M=\Wl and hence W=4.SI/d for 4 th, M=\Wl and hence W=8.SI/d Therefore the comparative strengths of the four cases are as the numbers i, 2, 4, 8. That is, if four such beams be of equal size and length and of the same material, the second is twice as strong as the first, the third four times as strong, and the fourth eight times as strong. From these equations also result the following important laws: The strength of a beam varies directly as S, directly as 7, inversely as c, and inversely as the length /. 142 CANTILEVER AND SIMPLE BEAMS CHAP VI A load uniformly distributed produces only one-half as much stress as the same load when concentrated. These apply to all cantilever and simple beams whatever be the shape of the cross-section. When the cross-section is rectangular, let b be the breadth and d the depth, then the value of I/c is \bd? and the above equa- tions become W = aSbd 2 /6l, where the number a is i, 2, 4, or 8 as the case may be. Therefore, The strength of a rectangular beam varies directly as its breadth, directly as the square of its depth, and inversely as its length. The reason why rectangular beams are put with the greatest dimension vertical is thus again shown. In the above equations the load W is the allowable load when S is the allowable unit-stress, and W is the load which will rupture the beam when 5 is the fictitious ultimate flexural unit -stress whose mean values are given in Art. 52. The stiffness of a bar under tension is measured by the load that it can carry with a given elongation. Similarly the stiff- ness of a beam is indicated by the' load that it can carry with a given deflection. For the two preceding articles the values of W for the four common cases of cantilever and simple beams are* for a cantilever loaded at the end, W= 3 . Elf /I 3 for a cantilever uniformly loaded, W= 8 . Elf '/P for a simple beam loaded at middle, W= 48 . Elf /ft for a simple beam uniformly loaded, P^=A|4. . Elf /I 3 which show that the relative stiffness of these four cases are as the numbers i, 2f, 16, and 25!- These equations also show that the stiffness of a beam, when the greatest unit-stress does not exceed the elastic limit of the material, varies directly as E and / and inversely as the cube of the length. It thus appears that the laws of stiffness are very different from those of strength. For a rectangular section I = ^b(P, and hence the stiffness varies directly as the breadth and directly as the cube of the depth. ART. 57 CANTILEVERS OF UNIFORM STRENGTH 1 43 The four cases above discussed have given the following expressions for the allowable load; for the strength of the beam W= a . SI/cl where a= i, 2, 4, or 8 and for the stiffness of the beam, W=p . Elf /P where /?= 3 , 8, 48, or 7 6 By equating these values of W, the following relations between the unit-stress S and the deflection f are obtained, S/f=(3Ec/aP or f=aSl 2 /fiEc (56) which are only valid when S is less than the elastic limit of the material. These equations show that the deflection /, for similar beams of the same material under the same unit-stress S, varies as P/c. Table 12, at the end of this volume, recapitulates the impor- tant facts regarding strength, deflection, and stiffness, which have been deduced in the preceding articles. Prob. 56a. Compare the strength of a joist 3X8 inches when laid with long side vertical with that when it is laid with short side vertical. Compare also the stiffnesses. Prob. 566. Find the deflection of the lightest steel lo-inch I beam, 9 feet in span, when stressed by a uniform load up to 30 ooo pounds per square inch. Prob. 56c. Compare the working strength of a light p-inch steel I beam with that of a wooden beam 8X12 inches in section, the span being the same for both. ART. 57. CANTILEVER BEAMS OF UNIFORM STRENGTH All beams thus far discussed have been of constant cross- section throughout their entire length. But in the flexure formula S.I/c = M, the unit-stress S is proportional to the bending moment M, and hence varies throughout the beam in the same way as the moments vary. Accordingly some parts of the beam are but slightly stressed in comparison with the dangerous sec- tion, and perhaps more material is used than is necessary. A beam of uniform strength is one so shaped that the unit- stress 5 is the same in all sections at the upper and lower sur- 144 CANTILEVER AND SIMPLE BEAMS CHAP. VI faces. Hence to ascertain the form of such a beam the unit- stress S must be taken as constant and I/c be made to vary with M. The discussion will be given only for the simplest cases, namely, those where the sections are rectangular, the breadth being b and the depth d. For these I/c = %bd?, and the flexure formula becomes, In this bd 2 must be made to vary with M in order to give forms of uniform strength. For a cantilever beam with a load P at the end, the value of M without regard to sign is Px and the equation becomes \Sbd? = Px, in which P and 5 are constant. If the breadth is taken as constant, d 2 varies with x and the profile is that of a parabola having its vertex at the free end. The depth d\ of the beam at the wall is found from \Sbd\ 2 =Pl, and comparing this with the first equation there results the simpler form d = d\\/x/l for the relation between d and x; Fig. 57 'a shows a profile plotted from this equation. When the depth of the cantilever beam is constant, then b varies directly as x and the plan of the beam is a triangle, as shown in Fig. 57b; the breadth b\ at the wall is found from %Sd 2 bi=Pl, and hence equation between b and x is more simply expressed by b = (b\/l}x. For a cantilever beam uniformly loaded with w per linear unit M = $wx 2 , and the equation becomes ^Sbd 2 = \wx 2 , in which w and S are known. If the breadth is taken as constant, then d varies as x and the side view is a triangle, as in Fig. 57c, where the depth at any point is given by d=(di/l~)x, the depth d\ being that at the wall, which is determined from \SbdJ = %wl 2 . If, however, the depth is taken as constant, then b varies as x 2 , ART. 57 CANTILEVERS OF UNIFORM STRENGTH 145 and b may be found from b = bi(x/l) 2 , where bi is the breadth at the wall; this is the equation of a parabola having its vertex at the free end and its axis vertical, or the plan of the beam may be formed by two parabolas as shown in Fig. 57d. The vertical shear modifies in practice the shape of these forms near their ends. For instance, a cantilever beam loaded at the end with P requires a section area at the end equal to P/S S where S s is the allowable shearing unit-stress. This section area should continue until a value of x is reached where the same section area is found from the equation of the form of uniform strength. Exact agreement with theoretic conditions is rarely possible on account of the expense of manufacture, and in fact cast iron is the only material which has been advan- tageously used for these forms. A cantilever of structural steel is built in a different way (Art. 58). FIG. 51 c. FIG. 57d. The deflection of a cantilever beam of uniform strength is evidently greater than that of one which has a constant cross- section equal to the greatest cross-section of the former, since the unit-stress S which acts only at the wall in the latter case, acts throughout the entire length in the former. In any case it may be determined from the general formula El . d 2 y/dx 2 = M by substituting for M and / their values in terms of x, integrating twice, determining the constants, and then making x equal to / for the maximum value of y. For a cantilever beam loaded at the end and of constant breadth, as in Fig. 57a, this formula becomes, i2Pl* i?.Px _. Integrating twice and determining the constants, as in Art. 54, 146 CANTILEVER AND SIMPLE BEAMS CHAP. VI the equation of the elastic curve is found to be, In this let x=l, then y is the deflection / of the end, and / = SPP/Ebdi 3 , which is double the deflection 'of a cantilever beam of uniform section and depth d\. For a cantilever beam loaded at the end and of constant depth, the general formula becomes, d 2 y _ i2Px _ I2PI ~ By integrating this twice and determining the constants as before, the equation of the elastic curve is found, whence the deflection is f=6PP/Ebid 3 which is fifty percent greater than that of a cantilever of uniform section and breadth hi. Prob. 57. A cast-iron cantilever beam of uniform strength is to be 4 feet long, 3 inches in breadth, and to carry a load of 1 5 ooo pounds at the end. Find the proper depths for every foot in length, using 3 ooo pounds per square inch for the horizontal unit-stress, and 4 ooo pounds per square inch for the shearing unit-stress. ART. 58. SIMPLE BEAMS OF UNIFORM STRENGTH In the same manner as that of the last article it is easy to deduce the forms of uniform strength for simple beams of rectan- gular cross-section. For a load at the middle and breadth constant, M = $Px, and hence, %Sbd 2 = %Px. Accordingly d 2 = ($P/Sb)x, from which values of d may be found for assumed values of x. Here the profile of the beam will be parabolic, the vertex being at the support, and the maximum depth under the load; if d\ is the depth at the middle, the equation of the parabola becomes d 2 =d l 2 (x/#}. For a load at the middle and depth constant, M = \Px as before. Hence b = (^P/Sd^^x, and the plan must be triangular or lozenge-shaped, the width uniformly increasing from the support to the load. If bi is the breadth at the middle, the equa- tion of the line becomes b = b\x/l. ART. 58 SIMPLE BEAMS OF UNIFORM STRENGTH 147 For a uniform load and constant breadth, M = \wlx fax 2 , and hence, d 2 ={^w/Sb)(lx-x 2 ), and the profile of the beam must be elliptical, or preferably a half-ellipse. If di is the depth at the middle the equation of the ellipse becomes d 2 = (4d i 2 /P)(lxx 2 }. For a uniform load and constant depth, b=(T > w/Sd 2 }(lxx 2 ), hence the plan should be formed of two parabolas having their vertices at the middle of the span. If bi is the breadth at the middle of the span, this equation becomes b = (^b\/l 2 }(lx x 2 ). The figures for these four cases are purposely omitted, in order that the student may draw them for himself; if any dif- ficulty be found in doing this, let numerical values be assigned to the constant quantities in each equation and the variable breadth or depth be computed for different values of x. In the same manner as in the last article, it can be shown that the deflection of a simple beam of uniform strength loaded at the middle is double that of one of constant cross-section when the breadth is constant, and is one and one-half times as much when the depth is constant. Fig. 58 Cast-iron simple beams are sometimes made approximately in the forms required by the above equations, care being taken to provide sufficient sectional area near the supports to safely carry the vertical shears; such beams are mainly used in machine- shops on planers to carry the cutting tool. Travelling cranes used in shops are also approximately of this form, but these are made by riveting steel plates and shapes so that the section areas are not rectangular. Plate girders, used extensively in buildings and bridges, are made by riveting together four angles, a web plate, and cover plates. Fig. 58 shows the general arrangement without the 148 CANTILEVER AND SIMPLE BEAMS CHAP. VI rivets. The section at A is made ample to resist the shear and that at B to resist either shear or moment; there being no cover plates on the distance AB. Between B and C two cover plates are used to provide sufficient section for the moment at C, while between C and D two additional cover plates are added so that -* the section at the middle D will be -sufficient to resist the maxi- mum moment at that place. A plate girder, then, is approxi- mately a beam of uniform strength. Prob. 58a. Draw the profile for a cast-iron simple beam of uniform strength, the span being 8 feet, breadth 3 inches, and load at the mid- dle 30 ooo pounds; using the same working unit-stresses as in Prob. 57. Prob. 58b. Compute the deflection of a steel spring of constant depth and uniform strength which is 6 inches wide at the middle, 52 inches long, and loaded at the middle with 600 pounds, the depths being such that the uniform fiber stress is 20000 pounds per square inch. AKT. 59 BEAM OVERHANGING ONE SUPPORT 149 -* J ? CHAPTER VII 1>O*M ' OVERHANGING AND FIXED BEAMS ART. 59. BEAM OVERHANGING ONE SUPPORT A beam is said to be 'fixed ' at a support when it is subject to such constraint that the elastic curve is there horizontal. The cantilevers discussed in the preceding articles have been fixed at one end by the restraint of the wall and the maximum moment has been found to occur at that end. Beams are sometimes fixed at one end and supported at the other (Art. 60) and some- times fixed at both ends (Art. 62). The effect of this restraint is to diminish the deflection, and hence the strength and stiff- ness are usually increased. Beams overhanging one support, as in the following figures, may be said to be fixed when the lengths and loads have such values that the tangent to the elastic curve at that support is horizontal. This condition is rarely fulfilled, but the discussion of overhanging beams is very useful and important. A canti- lever beam has its upper fibers in tension and the lower in com- pression, while a simple beam has its upper fibers in compression and the lower in tension. Evidently a beam overhanging one support has its overhanging part in the condition of a canti- lever and the part near the other end in the condition of a simple beam. Hence there must be a point where the curvature changes from positive to negative, and where the fiber stresses change from tension to compression. This point i is called the 'Inflec- tion Point ' ; it is the point where the bending moment is zero, for if the curvature changes from positive to negative, M must do likewise (Art. 45) . An overhanging beam is said to be sub- ject to a constraint at the support beyond which the beam projects, or, in other words, there is a stress in the horizontal fibers over that support. Since the beam has but two supports, its reactions may be 150 OVERHANGING AND FIXED BEAMS CHAP. VII found by using the principle of moments as in Art. 36. Thus, if the distance between the supports be /, the length of the over- hanging part be m, and the uniform load per linear foot be w, the two reactions for Fig. 59a are, From these the vertical shear at any section may be computed from its definition in Art. 37 and the bending moment from its definition in Art. 38, bearing in mind that for a section beyond the right support the reaction R 2 must be considered as a force acting upward. Thus, for any section distant x from the left support, When x is less than / When x is greater than / M=R l x+R 2 (x-l)-$wx 2 The curves corresponding to these equations are shown on Fig. 59a. The shear curve consists of two straight lines; V =R when x=Oy and F=o when x = R\/iv; at the right support y=J?i wl from the first equation; V=Ri + R 2 wl from the second, and V = o when x=l+m. The moment curve consists of two parts of parabolas; M = o when x=o, and M is a maxi- mum where the shear passes through zero; at the inflection point M = o and x = 2Ri/w; also M has its maximum negative value at the right support where the shear again passes through zero, and M = o when x=l+m. The diagrams show clearly the distribution of shears and moments throughout the beam. Fig. 59o Fig. 59& In any particular case it is best to work out the numerical values without using the above algebraic expressions. For example, if / = 2O feet, w = io feet, and ^ = 40 pounds per linear ART 59 BEAM OVERHANGING ONE SUPPORT 151 foot, the reactions are RI = 300 and R 2 = 900 pounds. Then the point of zero shear or maximum moment is at # = 7.5 feet, the inflection point at #=15 feet, the maximum shears are +300, 500, and +400 pounds, and the maximum bending moments are +1125 and 2000 pound-feet. Here the negative bend- ing moment at the right support is numerically greater than the maximum positive moment. The relative values of the two maximum moments depend on the ratio of m to /; if m = o, there is no overhanging part and the beam is a simple one; if w = /, the case is that just discussed; if m=l, the reaction R\ is zero, and each part is a cantilever beam. After having thus found the maximum values of V and M the beam may be investigated by the application of the shear and flexure formulas of Art. 41 in the same manner as a canti- lever or simple beam. By the use of formula (45) the equation of the elastic curve between the two supports may be deduced by two integrations and the proper determination of the con- stants, and it is, 24EIy=4R 1 (x 3 -l 2 x) -w(x 4 -l 3 x) From this the maximum deflection for any particular case may be determined by obtaining the derivative of y with respect to x> equating it to zero, solving for x, and then finding the corre- sponding value of y. If concentrated loads be placed at given positions on the beam the reactions are found by the principle of moments, and then the entire investigation can be made by the methods above described. Fig. 596 shows the shear and moment diagram for two loads and here, as always, the maximum moments occur at the sections where the shears change sign. Prob. 59a. Three men carry a stick of timber, one taking hold at one end and the other two at a common point. Where should this point be so that each may bear one-third the weight ? Draw the shear and moment diagrams. Prob. 596. A beam 20 feet long has one support at the right end and one support at 5 feet from the left end. At the left end is a load of 1 80 pounds, and at 6 feet from the right end is a load of 125 pounds. 152 OVERHANGING AND FIXED BEAMS CHAP, vii Find the reactions, the inflection point, and draw the shear and moment diagrams. ART. 60. BEAM FIXED AT ONE END A beam with one overhanging end, as in Fig. 59a, has the span / in the -condition of a beam fixed at the right end and sup- ported at the other, when the length m is such that the tangent to the elastic curve is horizontal over the right support. Fig. 60a shows the practical arrangement of such a beam, the right end being held horizontal by the restraint of the wall. The usual arrangement where the left support is on the same level as the lower side of the beam at the wall will now be discussed, the section area of the beam being constant through its length. Fig. 60a Fig. 606 Case I. Uniform Load. Let R be the reaction of the left end in Fig. 60a, and x the distance from that end to any section. In the general equation of the elastic curve El . d 2 y/dx 2 = M, the value of M is Rx-^wx 2 . Integrating once, the constant is determined from the condition that dy/dx = o when x = l. Inte- grating again the constant is found from the fact that y =o when x=o; then, Here* also y=o when x=l, and therefore R = $wl; hence the left reaction is three-fourths of that for a simple beam. The moment at any point now is M =$wlx-%wx 2 , and by placing this equal to zero, it is seen that the point of inflection is at x = $l. The maximum positive moment occurs when SM/3x = o or when x = %l, and its value is + T fg-w/ 2 . The maxi- mum negative moment occurs when x = l and its value is ART. GO BEAM FIXED AT ONE END 153 The distribution of shears and moments is as shown in the diagrams. The point of maximum deflection is found from the above equation of the elastic curve; placing the derivative equal to zero there results Sx 3 glx 2 +l 3 =o, one root of which is x = 0.42151, while the others are inapplicable to this problem. Hence f=o.oo$4.wl 4 /EI is the value of the maximum deflection. Case II. Load P at Middle. Here it is necessary to con- sider that there are two elastic curves having a common ordinate and a common tangent under the load, since the expression, for the moment are different on opposite sides of the load. Thus taking the origin as usual at the supported end, On the left of the load, (a) El^=Xx (b) EI&- (c) . EI On the right of the load the similar equations are, (by EI=^Rx (c) f EI y =$Rx To determine the constants consider in (c) that y=o when x = o and hence that C 3 =o. In (by the tangent dy/dx = o when x = l and hence C 2 = \RP. Since the curves have a common tangent under the load, (b) = (by for x = \l, and thus the value of C\ is found. Since the curves have a common ordinate under the load, (c) = (c} r when x = $1, and thus C is found. Then, (c) 24EIy=4Rx 3 +3Pl 2 x-i2RPx (c) f 4&EIy=8Rx 3 -8Px 3 +i2Plx 2 -24Rl 2 x+PP are the equations of the two elastic curves. Making x = l in (c)' the value of y is zero, and then the left reaction is R=^P. The moment on the left of the load is now M =^Px, and that on the right M= \\Px+\PL The maximum posi- tive moment obtains at the load and its value is ^- 8 P/. The 154 OVERHANGING AND FIXED BEAMS CHAP. VII maximum negative moment occurs at the wall, and its value is T VP/. The inflection point is at x = frl. The deflection under the load is readily found from (c) by making x = \l. The maximum deflection .occurs at a less value of x, which may be found by equating the first derivative to zero. Fig. 606 shows the distribution of shears and moments. Case III. Load P at any Point. The distance of the load from the left support being / the following results may be deduced by a method exactly similar to that of the last case. Reaction at supported end = $P(2 3/c+ K 3 ) Reaction at fixed end =^P(3/c x 3 ) Maximum positive moment = ^Pln(2 3K+/K 3 ) Maximum negative moment =%Pl(K K 3 ) The absolute maximum deflection for this case occurs under the load when x = 0.4147, and its value will be found to be given by f=o.oo 9 SPP/EL Prob. 60o. Draw the shear and moment diagrams for a span of 12 feet, due to a load P at 10 feet from the left end. Prob. 606. Find the position of load P which gives the maximum positive moment. Find also the position which gives the maximum negative moment. Compute these maximum moments and compare them with those due to a load at the middle. ART. 61. BEAMS OVERHANGING BOTH SUPPORTS When a beam overhangs both supports, the moments for sections beyond the supports are negative, and in general between the supports there will be two inflection points. If the over- hanging lengths are equal, the reactions will be equal under uniform load, each being one-half the total load. In any case, whatever be the kind of loading, the reactions may be found by the principle of moments (Art. 36), and then the vertical shears and bending moments may be deduced for all sections, after which the^ shear and flexure formulas (Art. 41) can be used for any special problem. Under a uniformly distributed load, each overhanging end being of length m, and the middle span being /, each reaction is ART. 61 BEAMS OVERHANGING BOTH SUPPORTS 155 l, the maximum shears at the supports are wm and \wl t the maximum moment at the middle is +w(\l 2 w 2 ), the maxi- mum moment at each support is $wm 2 , and the inflection points are distant %(P 4*w 2 )* from the middle of the beam. Fig. 61fl shows the distribution of moments for this case. When m=o, the beam is a simple one; when /=o, it consists of two cantilever beams. When m is equal to or greater than \l, there are no positive moments in the middle span. (*-m^ 1 b Fig. 61o Fig. 616 When concentrated loads are on the beam, as in Fig. 616, the reactions are readily found by the method of Art. 36, the shears and moments computed for several sections by the defini- tions of Arts. 37 and 38, and the shear a'nd moment diagrams may then be drawn. The maximum negative moments occur at the supports and the maximum positive moment under one of the concentrated loads. The final maximum shears and moments due to both uniform and concentrated loads are to be obtained by combining the values found for these loads. When the concentrated loads are light, it often happens that the final maximum positive moment will be between two loads. Prob. 61 a. For Fig. 61 a find the ratio of I to m in order that the maximum positive moment may numerically equal the maximum nega- tive moment. Prob. 61 b. A beam 30 feet long has one support at 5 feet from the left end, and the other support at 10 feet from the right end. At each end there is a load of 156 pounds and half-way between the supports there is a load of 344 pounds. Construct the shear and moment diagrams. Prob. 61 c. For Fig. 61a find the ratio of / to m in order that there may be no positive moment. 156 OVERHANGING AND FIXED BEAMS CHAP, vil ART. 62. BEAMS FIXED AT BOTH ENDS If, in Fig. 61 a, the distances m be such that the elastic curve over the supports is horizontal, the central span / is said to be a beam fixed at both ends. The length w which will cause the beam to be horizontal at the support can be determined by the help of the elastic curve. For uniform load, the bending moment at any section in the span / distant x from the left support is, which may be written in the simpler form, where MI represents the unknown bending moment J at the left support. The distance m can hence be found when MI has been determined. Again, for a single loapl P at the middle of I in Fig. 61a, the elastic curve can be regarded as kept horizontal at the left sup- port by a load Q at the end of the distance m. Then the bend- ing moment at any section distant x from the left support, and between that support and the middle, is, M=(Q+lP)x-Q(m+x) or M=M l + ^Px in which M\ denotes the unknown moment Qm at the left support. The problem of finding the bending moment at any section hence reduces to that of determining Mi the moment at the left support, Case I. Uniform Load. For this case the differential equation of the elastic curve becomes, EI~ 2 = M l + \-wlx - Iwx 2 Integrating twice, making dy/dx = o when x = o and also when x=l, there is found MI = -^ S wl 2 ,a.nd the equation .of the elastic curve is from which the maximum deflection is found to be / The inflection points are located by placing M equal to zero, which gives ^ = ^(i^V / 3). The maximum positive moment is at the middle and its value is ^w/ 2 ; accordingly the horizontal ART. 62 BEAMS FlXED AT BOTH ENDS 157 stress upon the fibers at the middle of the beam is one-half that at the ends. The vertical shear at the left end is $ivl, at the middle o, and at the right end \wl. Fig. 62i denote the sum of all the concentrated loads on the distance x, and wx the uniform load. Then because V is the algebraic sum of all the vertical forces on its left, the definition of vertical shear gives, V=V'-wx-2Pi (67) Hence the shear V can be determined for any section in the span as soon as V is known. The moment M is the algebraic sum of the moments of the external forces on the left of the section with reference to a point in that section, or, as in Art. 38, M moments of reactions minus moments of loads For the reason just mentioned, it is in general necessary to deter- mine M for continuous beams by a different method. Let M' denote the moment at the left support of any span as in Fig. 67, and M" that at the right support, while M is the moment for any section distant x from the left support. Let PI be any con- centrated load upon the space x at a distance AC/ from the left support, K being a fraction less than unity, and let w be the uniform load per linear unit. Since the shear V in Fig. 67 is equal to 170 CONTINUOUS BEAMS CHAP, vm the resultant of all the vertical forces on the left of a section just at the right of the left support, let m be the distance of the line of action of that resultant to the left of that support. Then the definition gives, for the moment at any section, But the quantity V'm is equal to the sum of the moments of all the forces on the left of the left support with respect to that support and hence it is the moment M f at the left support of the span. Hence, M=M'+V'x-$wx 2 -2 .P^x-td) (67)' from which the moment M may be found for any section in the span as soon as M' and V are known. The shear V at the support of the span may be easily found if the moments M' and M" be known. Thus in equation (67)' make x=l, then M becomes M", and hence, V'l=M"-M'+ \wP + JPi (//) (67)" and hence the problem of the discussion of continuous beams consists in the determination of the moments at the supports. When these are known, the values of M and V may be deter- mined for every section in any span, and the investigation of questions of strength and deflection be made from the formulas (41) and (45). The above formulas apply to cantilever and simple beams also. For a simple beam, M' = M" = o, and V'=*R. For a cantilever beam, M' = o for the free end, and M" is the moment at the wall. The relation between the moment and the shear at any sec- tion is interesting and important. At a section distant x from any support, the moment is M and the shear is V. At the sec- tion distant x + dx from the support, the moment is M+Vdx t which may also be expressed as M + dM. Accordingly, Vdx=dM or dM/dx=V This may also be found by finding the derivative of M with respect to x from (67)' and comparing it with (67). Therefore The derivative of the moment equals the shear ART. 68 METHOD OF DISCUSSION 171 and from this it is seen that the maximum moments occur at the sections where the shear passes through zero. Prob. 67. A bar of length 2l, and weighing w per linear unit, is supported at the middle. From (67) and (67)' find general expressions for the shear and moment at any section on the left of the support and also at any section on the right of the support. Draw the shear and moment diagrams. ART. 68. METHOD OF DISCUSSION The theory of continuous beams presented in the following pages includes only those with constant cross-section having the supports on the same level, since only such are used in engi- neering constructions. Unless otherwise stated, the ends will be supposed simply to rest upon their supports, so that there can be no moments at those points. Then the end spans are somewhat in the condition of a simple beam with one overhang- ing end, while the other spans are somewhat in the condition of a beam with two overhanging ends. At each intermediate support there is a negative moment, and the distribution of shears and moments due to uniform load is that shown in Fig. 68. When an end span is short, the reaction at that end may become zero or even be negative; in order that a negative reaction may exist, it is necessary that the end of the beam be anchored or fastened to the support. Fig. 68 As shown in Art. 67, the investigation of a continuous beam depends upon the determination of the moments at the sup- ports. In the case of Fig. 68, the moments at the supports 2, 3, and 4, may be designated 3/2, MS, and M\. Let V\, Vz y Vz, and V \ denote the shears at the right of those supports. The first step is to find the moments M 2 , M 3 , and 3/ 4 . Then from formula (67)" the values of V\, Vz, Fa, and V are found, 172 CONTINUOUS BEAMS CHAP VIII and thus by formula (67)' an expression for the moment in each span may be written, from which the maximum positive moments may be determined. Lastly, by the shear and flexure formulas of Art. 41 the beam may be investigated. For example, let the beam in Fig. 68 be regarded as of four equal spans and uniformly loaded with w pounds per linear unit. By a method to be explained in the following articles it may be shown that 'the moments at the supports are, M 2 = -*W M 3 = - &wP M 4 = -& W P From formula (67)" the shears at the right of the several sup- ports are found to have the values, And from (67) those on the left of the supports 2, 3, 4, 5, are found to be, Jfw/, $wl, H w ^> H w ^- From formula (67)' the general expressions for the moments now are, for first span , M = + %\w Ix fax 2 for second span, M= -^wP + $wl for third span, M= y-gwt 2 + $%wlx for fourth span , M / 8 wl 2 + $wlx fax 2 From each of these equations x the inflection points may be found by putting M = o, and the section of maximum positive moment by putting dM/dx=o. The maximum positive moments are found to have the following values: iWs^ Ttfc^ 8 Till"? and TViW For any particular case the beam may now be investigated by the use of the shear and moment formulas. It is seen that the greatest moment is that at support 2, and hence this need only be used in the flexure formulas. The reactions at the supports are readily found from the values of the adjacent shears. Thus, for the above case R\ = Vi = Hw/, and R 2 =$lwl + %%wl=%%'wL But perhaps a more satisfactory method will be to find them directly from the equation of moments. Thus R\l\-wl 2 = M 2) whence R\ = %^wl. Again RiX2l+R 2 l 2wl 2 = M 3 , whence R 2 =%%u>l. From the symmetry of the spans and loads, it is plain that R$=Ri and R^ = R 2 - ART. 69 THEOREM OF THREE MOMENTS 173 The equation of the elastic curve in any span is found by inserting the expression for M in El . 3 2 y/dx 2 = M, and inte- grating twice. In general, the maximum deflection in any span will be found intermediate in value between those of a simple beam and one fixed at its ends. Prob. 68. In a continuous beam of three equal spans the negative bending moments at the supports are j^wl 2 . Find the inflection point, the maximum positive moments, and the reactions of the supports. ART. 69. THEOREM OF THREE MOMENTS Let the figure represent any two adjacent spans of a continu- ous beam having the lengths I' and /" and the uniform loads it/ and w" per linear unit. Let M', M", and M' " represent the three unknown moments at the supports. Let V and V" be the vertical shears at the right of the first and second sup- ports. Then, for any section distant x from the left support in the first span, the moment is, M = M' + V'x \wx 2 . Let this be inserted in the general formula of the elastic curve. Integrating twice and determining the constants by the conditions that y = o when x=o and also when x = l' the tangent dy/dx of the angle which the tangent to the elastic curve at any section in the first span makes with the horizontal is found to be given by, Similarly if the origin is taken at the next support, the tangent of inclination at any point in the second span is, The two curves must have a common tangent at the support where they meet, in order that the beam may be continuous Hence make x = V in the first equation and x = o in the second, and equate the results, giving Now let V and V" be expressed by means of (67)" in terms of M', M", and M'"\ then this equation reduces to M'l'+2M"(l'+l") + M'"l"= -frv'l'*-%w"l" 3 (69) which is called the theorem of three moments for continuous 174 CONTINUOUS BEAMS CHAP. VIII beams under uniform loads. It was first deduced by Clapeyron in 1857 and is hence sometimes called Clapeyron's theorem. This theorem shows how the moment M" at any support is connected with moments at the preceding and following supports. When all spans are of the same length I and have the same load TV per linear unit, the theorem becomes, which applies to the most common cases in practice. It must be noted, however, that these theorems of three moments are only valid when the beam is of constant section area and when all the supports are on the same level, since these conditions have been introduced into the algebraic work by taking / as constant, and by taking y as zero for all supports. WWf^Jl^' Fig. 69a Fig. 696 In any continuous beam of 5 spans there are 5+1 supports and 5 i unknown bending moments at the supports. For each of these supports an equation of the form of (69) or (69)' may be written which contains three unknown moments. Thus there will be stated 5 1 equations, and the solution of these will furnish the values of the 5 1 unknown quantities. The simplest case is that shown in Fig. 696, where there are two equal spans uniformly loaded, the left and right ends of the beam resting upon the supports. Here M' and M'" are each zero, and the theorem (69)' gives M"= %wl 2 . The left reaction R' is now found from R'l \wl 2 = M" to be R' = %iul, and R" r has the same value ; hence each span of Fig. 696 is in the same condition as that of a beam fixed at one end and supported at the other. Prob. 69. A continuous beam of two spans is uniformly loaded with 125 pounds per linear foot. The length of the first span is 18 feet and that of the second span is 1 2 feet. Compute the moment at the middle support, and the three reactions. ART. 70 EQUAL SPANS WITH UNIFORM LOAD 175 ART. 70. EQUAL SPANS WITH UNIFORM LOAD. Consider a continuous beam of five equal spans uniformly loaded. Let the supports, beginning on the left, be numbered i, 2, 3, 4, 5, and 6. From the theorem of three moments an equation may be written for each of the supports at which moments exist; thus, for support 2 , M i + $M 2 + M 3 = \wP for support 3, 3/ 2 +43/ 3 +3/4= fat 2 for support 4, 3/3+43/4+3/5= fat 2 for support 5, 3/4+43/5+3/6= fal 2 Since the ends of the beam rest on abutments without restraint MI = M 6 = o. Hence the four equations furnish the means of finding the four moments M 2 , MS, M 4 , M^. The solution may be abridged by the fact that M 2 = Ms, and 3/3 = 3/4, which is evident from the symmetry of the beam. Hence, 3/ 2 =M 5 = -^U'l 2 3/3 = 3/ 4 = -frwP From formula (67)" the shears at the right of the supports are, etc. From (67)' the moment for any section in any span may now be found as in Art. 68, and by the methods there indicated the complete investigation of any special case may be effected. Moments at Supports for Ecfual Spans. Fig. 70a In this way the moments at the supports for any number of equal spans can be deduced. The following triangular table shows their values for spans as high as seven in number. In 176 CONTINUOUS BEAMS CHAP. VIII each horizontal line the supports are represented by squares in which are placed the coefficients of wl 2 . For example, in a beam of 3 spans there are four supports, and the bending moments at those supports are o, f^ivl 2 , jV^j and o. The shears at the supports are also shown in the following table for any number of spans less than six. The space repre- senting a support gives the shear on the left of the support in its left-hand division and the shear on the right of the support in its right-hand division. The sum of the two shears for any support is, of course, the reaction of that support. For example, in a beam of five equal spans the reaction at the second sup- port is Fig. 706 It will be seen on examination that the numbers in any oblique column of these tables follow a certain law of increase by which it is possible to extend them, if desired, to a greater number of spans than are here given. As an example, let it be required to select a rolled steel I beam to span four openings of 8 feet each, the load per span being 44 coo pounds and the greatest horizontal stress in any fiber to be 15 ooo pounds per square inch. The required beam must satisfy the flexure formula 5 . 1/c = M, or it must be of such size that I/c = M/i$aoo. From the table it is seen that the greatest negative moment is that at the second support or -/-givl 2 , and the maximum positive moment in the first span is V\ 2 /2W = f/s^wP and that in the second span is ART. 71 UNEQUAL SPANS AND LOADS 177 The greatest value of M is hence at the second support; then, /A= 3X44 000X8X12/28X15 000= 30.2 inches 3 and from Table 6 it is seen that the light 1 2-inch beam, for which I/c is 36.0, will most closely satisfy the requirements. Prob. 70o. Find several shears and moments for three equal spans uniformly loaded, and draw the shear and moment diagrams. Prob. 706. Select the proper steel I beam to span three openings of 12 feet each, the uniform load on each span being 6000 pounds and the greatest value of S to be 12 ooo pounds per square inch. ART. 71. UNEQUAL SPANS AND LOADS As the first example, consider two spans with lengths l\, 1 2 , and uniform loads per linear unit w\ and u> 2 . The theorem of three moments in (69) then reduces to, from which the bending moment at the middle support is known. When there is no load upon the second span w 2 is zero. As a particular case let 1=40 and / 2 = 3Q feet, ^1 = 210 pounds per linear foot and w 2 = o; then M 2 = 24000 pound-feet. The reaction RI is found from RI X 40 } X 2 10 X 4O 2 = M 2 , which gives 7^ = 4- 3600 pounds. The reaction R$ is found from R 3 X 30 = M 2} which gives ^3= 800 pounds. The shear passes through zero in the first span at the point for which RI ivx = o, which gives x = R\/ur, and the maximum positive moment is then M = R ] 2 /2-w = 3ogoo pound-feet. From these values the shear and moment diagrams in Fig. 71 a are constructed. Next consider three spans having the lengths /i, / 2 , and /3, .and loaded uniformly with Wi, w 2 , w 3 . The moments at the second and third supports are M 2 and M%. Then from (69), and the solution of these gives the values of M 2 and M^. A very common case for swing drawbridges is that where two end spans are equal and the load uniform throughout, or 7 2 = /, /i = / 3 = 178 CONTINUOUS BEAMS CHAP. VIII al, and w\ = "Wz MS = w- For this case the solution gives, For example take a swing-bridge where the two end spans are each 120 feet and the middle span is 24 feet, this being a con- tinuous girder when closed. Here 01=120/24=5, and M 2 = 2.423W/ 2 , which is the moment at supports 2 and 3 due to live load over all spans. When live load covers only the first span iV2 may be made zero, and the moments be found by the solution of the above equations. In Part IV of Roofs and Bridges these cases of loading are fully discussed. Fig. 71a Fig. 716 Whatever be the lengths of the spans or the intensity of the uniform loads, the theorems of three moments in Art. 69 furnish the means of finding the bending moments at the supports. Then by the methods of Art. 67 shears and moments at every section may be computed and the degree of security of the beam be investigated by the flexure formula (41). Finally, if the material is not stressed beyond its elastic limit, formula (45) may be used to determine the deflection. Prob. 710. A continuous beam of three spans is loaded only in the middle span, as in Fig. 716. Find the reactions of the end supports due to this load. Prob. fib. A heavy i2-inch steel I beam of 36 feet length covers four openings, the two end ones being each 8 feet and the others each 10 feet in span. Find the maximum moment in the beam. Then determine the load per linear foot so that the greatest horizontal unit- stress may be 15 ooo pounds per square inch. Prob. 7 ic. For the case of three spans let the first and third spans be each 80 feet long. Find the length of the middle span so that the moment shall be zero at the middle of that span, the load being uniform throughout. ART. 72 SPANS WITH FlXED ENDS 179 ART. 72. SPANS WITH FIXED ENDS The theorem of three moments may also be used to determine the bending moments at the supports when the ends of the con- tinuous beam are horizontally fixed in walls. If the number of spans be two, there are three unknown moments and hence three equations are to be written for four spans; in these the lengths of the first and last spans are to be made zero and thus the elastic curve will be made horizontal at the ends of the beam. The following example will illustrate the procedure. Fig. 72a Fig. 72b Let there be two spans of lengths 1 2 and Is with ends hori- zontally fixed, as in Fig. T2a. In Fig. 72b let the restraint of the walls be replaced by the span /i on the left end and the span / 4 on the right end, these spans being taken as unloaded. From (69) the equations for the three supports 2, 3, 4, are, for support 2 , MI/I + aM 2 (/i + / 2 ) + M 3 1 2 = - $u> 2 l 2 3 for support 3 , M 2 1 2 + 2M 3 (1 2 + / 3 ) + 1/4/3 = for support 4, M 3/3+ 2M 4 (1 3 +/ 4 ) + M^1 4 = Now, MI and MS are zero since the ends are supposed to merely rest on the supports i and 5. By making /i = o the points i and 2 become consecutive, which renders the elastic curve horizontal at 2 ; also by making / 4 = o the points 4 and 5 become consecutive which renders the elastic curve horizontal at 4. As a special case let lz = lz = l and w 2 = w 3 = w , so that the two spans are equal and have the same uniform load ; then from symmetry it is known that M 4 is equal to M 2 , so that the equations become, 2 M 2 + M 3 = - iw/ 2 2M 2 + 4 M 3 = - \-wV- from which are found M 2 = -fawP and M 3 = -fewF. As another special case let the two spans be equal and the uniform load be only on the first span or w 2 = w and w>3 = o; then the equations are, 180 CONTINUOUS BEAMS CHAP, vm from which the moment at the left fixed end is M 2 = fowl 2 that at the middle support is M 3 = ^wl 2 , and that at the right fixed end is M= +^wP. The fixing of the ends renders the bending moments smaller than those of beams with supported ends and hence causes an increase of strength, while the stiffness is also made greater. Continuous beams in the floors of buildings often have their ends fixed in walls. When the spans are two in number and of equal length, each span is in the same condition, under uni- form load, as one with both ends fixed, since the elastic curve is horizontal over the middle support. Prob. 72o. Draw the shear and moment diagrams for a beam of two equal spans with fixed ends, the first span being unloaded and the second covered with uniform load. Prob. 726. Using the theorem of three moments for concentrated loads given in the next article, deduce the moments for Fig. 72c caused by a load P at the middle of the first span. ART. 73. CONCENTRATED LOADS Thus far only uniform loads upon one or more spans have been discussed, but all the methods given are applicable to con- centrated loads, provided the moments at the supports due to those loads can be found. By a process of reasoning similar to that in Art. 69, a theorem of three moments for such loads can be deduced, and it will here be stated without the algebraic work of demonstration. As before the beam is to be of constant section throughout its length and all supports are to be upon the same level. Let /' and /" be the lengths of any two consecutive spans and M f , M", and M'" the moments at the three supports. Let P' be any load upon the first span at the distance d' from the first support, and P" any load upon the second span at the distance til" from the second support, being any fraction less than unity and not necessarily the same for the two loads. Then the theorem of three moments is, ART. 73 CONCENTRATED LOADS 181 which is to be used in the same manner as those of Art. 69. If there is no load on the span I', then P' is zero; if there is none on the span I", then P" is zero. To illustrate the application of this formula, let there be two equal continuous spans, as in Fig. 73a, with a load P on the first span. Here P' becomes P and P" is zero, and since there are no moments at the ends, the theorem gives M 2 = -\Pl(K-}. To find the left reaction, Ril-P(l-Kl) = M 2 from which ^1 = ^(4-5*+ 3 ). When the load is at the middle of the span, K is , and M 2 = &PI and RI = |P. The reactton at the right end is found from Rsl=M 2j whence RS = \P(K K 3 ) or RS= -^P', hence the right end must be prevented from rising from the support in order that this negative reaction can prevail. If that end is not fastened, R s is zero and the reaction R\ is P(I K) as for a simple beam. Fig. 73o Fig. 73b From the above results and from the definitions of shear and moment .in Arts. 37 and 38, the shear and moment dia- grams may be drawn, as in Fig. 73a. The inflection point is found from RIX P(x nT) = o, whence its position is given by x= 4/7(5 2 ); when K=I the load is at the middle support and #=/; when K=O the load is at the left support and x = ^l. Hence the inflection point for a load in the first span always lies on the last fifth of the span. When there are loads on both spans, as in Fig. 736, the moments due to both may be found from the theorem, or the moments and reactions due to each may be separately determined and the final moment found by addition. Thus, if each load is at the middle of the span, the reaction RI due to P 2 is known from CHAP. VIII the value of the reaction R 3 due to PI; hence for both loads ^i-lfPi ~& p 2> so that > if p i and p 2 are equal Ri is +&P. The theorem of three moments above given was deduced in 1865 by Bresse, and from it the theorem of Clapeyron (Art. 69) for uniform loads is readily derived. Let it/ be the load per linear unit on the first span and P' be a small part of this uni- form load extending over the distance <5(/), then P' is to be replaced bywd(id) and P'l' 2 (K-iP) becomes a//' 3 ^-* 3 )^. Integrating this between the limits i and o, the uniform load covers the whole span and the function is Jw7' 3 as in Clapeyron 's theorem. An abbreviated method of finding the moments at the sup- ports, without writing the theorem of three moments for each support, was devised by the author in 1875. See London Philo- sophical Magazine, September, 1875; or Roofs and Bridges, Part IV. This method can also be directly applied to cases where one or both ends are fixed. Continuous beams with fixed ends are, however, rarely used under the action of a live load. Prob. 73. A continuous beam has four spans of 6, 8, 8, 6 feet length; the ends resting upon abutments. Find the left reaction due to a load of i ooo pounds at the middle of the second span. ART. 74. SUPPORTS ON DIFFERENT LEVELS All cases of flexure thus far considered have been for sup- ports on the same level, except that of fixed beams in Art. 64. The general remarks there given regarding the effect of changes of level of the supports apply also to continuous beams. Indeed a slight depression of one support below the level of the others may cause great changes in the moments and stresses through- out the beam. Let Fig. 74a represent two consecutive spans of a continu- ous beam having the lengths V and I"; let the axis of abscissas be horizontal, /*', h", and h'" being the heights of the three sup- ports above this axis. Let the beam be anchored to the sup- ports so that its lower surface is compelled to touch them under all circumstances. This constraint produces moments at the ART. 74 SUPPORTS ON DIFFERENT LEVELS 183 supports, the magnitude of which will depend upon the size and shape of the beam, represented by the moment of inertia /, and upon the stiffness of its material, represented by the modulus of elasticity E. By proceeding as in Art. 69 a theorem of three moments for this case may be deduced. In the first span, y= h f when x=o and y=h" when x=l'. The value of dy/dx for the first and second spans hence differs from those of Art. 69 in contain- ing the quantities h', h", h'" . By equating the values of dy/dx for the middle support, there will be found, which is the theorem of three moments for uniform load and supports on different levels. This may be extended to include concentrated loads by inserting the functions of P and K given in the last article. Fig. 74a To show the method of application of this theorem, take two equal continuous spans with supported ends, and let the middle support be lowered the distance / below the level of the other supports. Then /' = I" = / and M' = M r " = o, also h" -h' = h"-h'"= -/. Let the load be uniform throughout, so that u/ = iv" = w. Then there results, which shows that the negative bending moment at the middle support is decreased by the circumstance of its depression. When / has the value wl^/2^EI, there is no moment at this support and each span is like a simple beam. When / has a greater value, the moment becomes positive. If the beam be one of rolled steel weighing 40 pounds per foot and the spans be 16 feet long, the moment at the support due to the weight of the 184 CONTINUOUS BEAMS CHAP. Vlli beam is %wl 2 = i 280 pound-feet. Now let the middle sup- port be depressed o.i inches below the level of the others; then since 7=158.7 inches 4 from Table 6, the moment due to that depression is 3-E///^ 2 = + 3 230 pound-feet, so that this slight depression entirely changes the character of the stresses through- out the beam. Continuous bridges are subject to all the uncertainties of continuous beams in regard to the effect of changes of level of the supports, and hence their use has been almost entirely aban- doned. Continuous beams are used only for short spans as is the case with railroad rails, and in floors where there is little liability to change in level of supports. In conclusion it may be noted that the above theorem of three moments furnishes a very convenient method for finding the elastic deflections of beams. As an example, take the case shown in Fig. 746, where it is desired to find the upward deflec- tion at the middle of the span / due to a uniform load w m on the overhanging end m. Let the middle point be marked 2 and the supports i and 3; the moments at these points are MI = O, MZ^ \TJum 2 , MZ = %wm 2 . These- are to be inserted in the formula in place of M' ', M", M'"\ also l' = \l, I" = \l, and u/ = w" = o since there is no load on the span considered. Also making h' = h"' = h, the deflection is h 2 hi and the formula now gives its value as -wm 2 l 2 /^2EI. Prob. 74a. A continuous beam of two equal spans, uniformly loaded, has its supported ends on the same level. How far must the middle support be depressed so that the negative moment over it may be numerically equal to the maximum positive moment in each span? Prob. 746. Find, by the above method, upward deflection of the overhanging end in Fig. 74&, due to a uniform load over the span /. ART. 75. THE THEORY OF FLEXURE The theory of flexure, presented in this and the preceding chapters, is called the common theory, and is the one universally adopted for the practical investigation of beams. It should not be forgotten, however, that the axioms and laws upon which ART. 75 THE THEORY OF FLEXURE 185 it is founded are only approximate and not of an exact nature like those of mathematics. The law regarding the proportionality of stress and deformation is, for instance, only roughly approxi- mate for brittle materials. The flexure formula S ,I/c = M has been established from this law and from the observed fact that a vertical line, drawn upon the side of the beam, before flexure, remains a straight line after flexure. When the load on a beam is sufficient to cause its rupture, and the longitudinal unit-stress 5 is computed from the flexure formula, a disagreement of that value with those found by direct experiments on tension or compression is observed. This is often regarded as an objection to the common theory of flexure, but it is in reality no objection, since the laws upon which the flexure formula is founded are only true provided the elastic limit of the material is not exceeded. Experiments on the deflec- tion of beams furnish, on the other hand, the most satisfactory confirmation of the theory. When the modulus of elasticity E is known by tensile or compressive tests, the formulas for deflec- tion are found to give values closely agreeing with those observed. Indeed so reliable are these formulas that it is not uncommon to use them for the purpose of computing E from experiments on beams. When, however, the elastic limit of the material is exceeded, the computed and observed values fail to agree. Certain false theories of flexure have been proposed from time to time', the one best known being that in which it is assumed that the moment of the horizontal forces on one side of the neutral axis is equal to the moment of those on the other side. Since the principles of static equilibrium furnish no condition of this kind, the formulas established are, of course, without value. Although it is unfortunate that the flexure formula does not theoretically apply to the rupture of beams, it is better to use it for such cases in connection with experimental constants (Art. 52) than to employ any formula which disagrees with the fundamental principles of statics. Such is the method in general practice, and on the whole it may be concluded that the com- mon theory of flexure is entirely satisfactory and that it is suffi- 186 CONTINUOUS BEAMS CHAP, viil cient for the investigation of most questions relating to the strength and stiffness of beams. For materials like cast-iron and con- crete it is possible to deduce formulas, which apply more closely than the common flexure formula, by using a parabola instead of a straight line to represent the variation of the stresses above or below the neutral axis (Art. 52). These formulas include constants which give the relation between stress and deforma- tion, so that each material requires a different flexure formula. Although such formulas are theoretically more correct than (41) for stresses beyond the elastic limit, it does not appear that they give better results for rupture than are obtained by using (41) with the values of S/ found by experiment. In all the examples thus far discussed, the load applied to the beam is parallel to one of the principal axes of inertia of the cross-section and its resultant coincides with that axis. When this is not the case, the flexure formula (41) must be modi- fied in the manner indicated in Art. 166; such unsymmetric arrangement rarely occurs except in the purlin beams of roof trusses. The actual internal stresses in beams are far more complex than those considered in the common theory, because the ver- tical shears combine with the horizontal stresses; discussions of the apparent and true stresses are given in Chapters XI and XV. The influence of shear on the deflection of beams is investi- gated in Art. 125. All the formulas and methods of the pre- ceding chapters apply only to beams in which the material is the same throughout the section area. When different materials are combined to form a beam, the flexure formula must be modi- fied so as to take into account their different degrees of stiffness, and this will be done in Chapter XII. The theory of beams arose from the discussions of Galileo in the seventeenth century, but it was not until about 1825 that the flexyre formula and the general equation of the elastic curve were established by Navier. Since that time great progress has been made in considering the flexure of beams under impact and in applying the principles of work and energy to their dis- ART. 75 THE THEORY OF FLEXURE 187 cussion; some of these investigations will receive attention in future chapters. Prob. 75a. A beam of three spans, the center one being / and the side ones nl, is loaded with P at the middle of each span. Find the value of n so that the reactions at the end may be one-fourth of the other reactions. Prob. 75b. Consult Engineering News, vol. xviii, pp. 309, 352, 404, 443; vol. xix, pp, n, 28, 48, 84; and vol. xxii, p. 131. Write an essay concerning certain erroneous views regarding the theory of flex- ure which are there discussed. Prob. 75c. Procure several sticks of good timber, each Xf inches, and of lengths about 8, 12, and 16 inches. Devise and conduct experi- ments to test the following laws: First, the strength of a beam varies directly as its breadth and directly as the square of its depth. Second, the stiffness of a beam is directly as its breadth and directly as the cube of its depth. Third, a beam fixed at the ends is twice as strong and four times as stiff as a simple beam, when both are loaded at the middle. 188 COLUMNS OR STRUTS CHAP. LX CHAPTER IX COLUMNS OR STRUTS ART. 76. CROSS-SECTIONS or COLUMNS When a prism has a length longer than about eight or ten times the least side of its cross-section, it is called a 'column' or 'strut'. When the length of the prism is only four or six times as long as the least side of its cross-section, the case is one of simple compression the constants for which are given in Art. 5. Under simple compression the failure occurs for brittle mate- rials by oblique shearing and for plastic materials by enlarge- ment and cracking (Art. 18). In the case of a column, how- ever, failure is apt to occur by a sidewise bending which causes flexural stresses. The longer the column the greater is the lia- bility to lateral flexure. Wooden columns are usually square or round, and when of large size they may be built hollow. Cast-iron columns are usually round and hollow. Wrought-iron columns were built prior to 1900 of a great variety of forms, but structural steel has since been almost entirely used. Rolled I beams may be used, but most steel columns are formed by riveting together channels, angles, and plates (Art. 44). Columns are exten- sively used in buildings and bridges. A piston-rod of a steam- engine, or the parallel rod of a locomotive, is a column when it is under compression. It is clear that a square or 'round sec- tion is preferable to an unsymmetrical one, since then the lia- bility of the column to bend is the same in all directions. For a rectangular section, the plane of flexure will evidently be per- pendicular to the longer side of the cross-section, and in gen- eral the plane of flexure will be perpendicular to that axis of the cross-section for which the moment of inertia is the least; for Art. 56 shows that the deflection of a beam varies inversely as /. In designing a column it is hence advisable that the cross- ART. 76 CROSS-SECTIONS OF COLUMNS 189 section should be so arranged that the moments of inertia about the two principal rectangular axes should be closely equal. For example, let it be required to construct a column with two I beams, as in Fig. 76a, the pieces connecting the flanges being small and light so that they add nothing to stiffness or strength. Let the beam be the light 1 5-inch size weighing 42 pounds per foot; then Table 6 gives 71 = 441.7 for an axis perpendicular to the web and 7 2 = 14.62 inches 4 for an axis along the middle line of the web; also the section area 0=12.48 square inches. Let it be required to find the distance x between the centers of the webs so that the moments of inertia with respect. to axes through the center of gravity of the column section shall be equal. For the axis perpendicular to the webs, 7=2X441.7; for the axis parallel to the webs, 7=2Xi4-62 + 2Xi2.48(^) 2 . Equating these two values and solving for x gives #=11.70 inches. Fig. 76a Fig. 766 Fig. 76c As a second example, take the section in Fig. 766, which is formed by two channels and two plates, the rivets being omitted in the sketch. Each channel is 10 inches deep and weighs 35 pounds per foot, and each plate is x inches long and inch thick. Using Table 9, the moments of inertia of the column section 'with respect to the two axes through its center of gravity are, 7= 2 [ 4 .66+ 10.29(^- Placing these expressions equal, the value of x is found to be between 10 and zoj inches. This section is suitable only for a column less than 6 feet in length, as the riveting of the plates to the angles could not be done for a long column. Fig. 76c is a section frequently used for bridge members, there being but one plate connecting the two channels. Here 190 COLUMNS OR STRUTS CHAP, ix the center of gravity of the section lies above a line drawn through the middle of the webs and its position is to be found by the method of Art. 42. Then by the principles set forth in Art. 43, the moments of inertia with respect to the two rectangular axes through this center are to be computed, and that which is the smallest is to be used in the column formulas given in the follow- ing pages. Prob. 76a. Two joists, each 2X4 inches, are to be placed 6 inches apart between their centers, and connected by two others, each 8 inches wide and x inches thick, so as to form a hollow rectangular column. Find the proper value of x. Prob. 7Gb. Let the section in Fig. 76c consist of a plate, $X 12 inches, and two channels, each 12 inches deep and weighing 20^ pounds per linear foot. Compute the moments of inertia with respect to the two axes through the center of gravity. ART. 77. DEFINITIONS AND PRINCIPLES When a short prism of section area a is under compression in the direction of its length and the resultant force P acts through the centers of gravity of the end sections, the internal stress is uniformly distributed over the section, and hence the compres- sive unit-stress S is P/a. For a long prism, or column, this is not always the case, for any sidewise deflection will cause flexural stress which will render the unit-stress on the concave side of the column greater than P/a and that on the convex side less than P/a. Hence for any given section, the load P should be taken smaller for a long column than for a short one, since evidently the liability to bending increases with the length. The 'Axis' of a column is the line passing through the centers of gravity of the cross-sections. When the column is straight, the axis is a straight line; if it bends laterally, the axis is the elastic curve. An ' axial load ' is one having its line of action coinciding with the centers of gravity of the two end sections; the term ' concentric load ' is used by some writers for this case The load P is regarded as axial in the greater part of this chapter this being the most common case in practice. ART. 77 DEFINITIONS AND PRINCIPLES 191 The length of a column is indicated by / and the least radius of gyration of its cross-section with respect to an axis through the center of gravity of that section by r. The value of r is found from the equation ar 2 = I (Art. 43) where a is the section area and / is the least moment of inertia; for example, if the section is a circle of diameter d, the value of a is \xd 2 and that of / is faxd 4 (Art. 43) ; hence the radius of gyration of a circular section is \d. For a rectangle having its least side d and its width b, the radius of gyration is found from r 2 = T^bd 3 /bd=^d 2 whence r= 0.2894. For sections of rolled beams and channels, the values of r for two rectangular axes are given in Tables 6 and 9, and the least of these is the one needed in computations when a beam or channel is to be used as a column. It was shown in Art. 51, that a given section area a offers greater resistance to flexure the further the material is removed from the neutral axis. When a column bends laterally, flexural stresses similar to those in a beam arise, and hence for columns economy is also promoted by placing the material as far as prac- ticable from the axis about which bending may occur; this is done by making the radius of gyration as large as practicable. The ratio l/r is called the 'slenderness ratio ' of the column. When l/r is less than about 25, the column is a short prism under simple compression (Art. 5); when l/r is greater than about 200 the column is called long and failure occurs by lateral bending. The columns generally used in engineering practice have slender- ness ratios varying from 50 to 150. The condition of the ends of columns exerts a great influence upon their strength. ' Round ends ' are those which are free to turn upon the surfaces where they abut; Fig. 77a shows one with spherical ends and also one where the compression is applied through pins. ' Fixed ends ' are those subject to such restraint that the tangent to the elastic curve remains vertical at the ends when a lateral deflection occurs. Fig. 776 shows a column with one end free to turn and the other fixed, and also one with both ends fixed. Columns with both ends fixed are extensively used in buildings and bridges. Columns with one or both ends 192 COLUMNS OR STRUTS CHAP. IX hinged on pins are used in bridges and also in machines; the piston-rod of a steam-engine is a case of a column with one end fixed and the other hinged. The term 'round ends' generally includes those which are free to turn on pins at the ends. It is evident that a column with fixed ends is stronger than one with round ends, and that a column with one end round and the other fixed is intermediate in strength between these; this is confirmed by all experiments. There is also another condi- tion of ends which is called 'flat' and represented in Fig. 77c; here the ends simply abut on plane surfaces without being fixed. The strength of a column with flat ends is closely the same as one with fixed ends when it is short, and about the same as one with hinged ends when it is long. Fig. 77a Fig. 776 Fig. 77c In the following articles the theory of columns will be developed without considering the weight of the column itself. When a column is in a vertical position, its weight brings a greater unit- stress upon the base than that due to the load (Art. 27), but in most practical cases this increase is small. When a column is in a horizontal position, its weight causes flexure which increases the stress on the upper side due to the direct compression, and this case will be discussed in Arts. 101 and 102. Prob. 77o. A round cast-iron column has the outer diameter di and the inner diameter d%. Find the radius of gyration of the cross-section. Prob. 77b. An I beam 20 inches deep and weighing 65 pounds per linear foot is used as a column. What length of column will give a slenderness ratio of 220? ART. 78. ETJLER'S FORMULA FOR LONG COLUMNS Consider a long column of section area '//) or y=fsmvn(x/[) By discussing this equation according to the methods of An- alytic Geometry there are derived three curves for v=i, v = 2, and v = 3, as shown in Fig. 786. For v=i the curve is entirely on one side of the axis of x; for v = 2, it crosses that axis at the middle; for v = 3, it crosses at / and /. Each of these cases is liable to occur for a column with round ends, but the first is the most dangerous case since the lateral deflection is then the greatest. Hence, making v=i in (78), there is found, P=7r 2 E7// 2 or P/a=7T 2 (r//) 2 which is Euler's formula for columns with round ends. The second form is obtained from the first by using / = ar 2 , where r is the least radius of gyration of the cross-section (Art. 77). A column with one end fixed and the other round is approxi- mately represented by the part b'b" of the second diagram in Fig. 786, where b' is the fixed end where the tangent to the curve is vertical. Here v=2 and the length b'b" is three-fourths of the entire length; hence replacing I by $/ in (78), it becomes P = 2^n 2 EI/l 2 which is Euler's formula as commonly stated for columns having one end fixed and the other round. The con- stant 2\ is, however, obtained under the false supposition that ART. 78 EULER'S FORMULA FOR LONG COLUMNS 195 the point b' is in the line of application of the load P. The more correct analysis in Art. 88 shows, however, that the value of the constant is 2.0457, a sufficiently close value for all common dis- cussions being 2.05. Accordingly, is Euler's formula for long columns having one end fixed and the other round. A column with fixed ends is represented by the portion c f c rf of the third case. Here v = 3, and the length c'c" is two-thirds of the entire length; hence, replacing / in (78). by f/, P= 4 7T 2 /// 2 or P/a = 4x 2 E(r/!) 2 which is Euler's formula for long columns with fixed ends. From this investigation it appears that the relative strengths of long columns of the three classes are as the numbers i, 2.05, and 4 when the lengths are the same, and this conclusion is approximately verified by experiments. A general expression for Euler's formula for long columns may now be written, namely, P=f*EI/l 2 or P/a= ftE(r/l) 2 (78)' in which the number /t is ?r 2 for round ends, 4?r 2 for fixed ends, and 2.o5?r 2 for one end round and the other end fixed. Another kind of column is that which is fixed at one end and entirely free at the other, like a vertical post planted in the ground. This case is represented in Fig. 786 by the upper half of the case for which v=i, by the upper fourth of the case for which y = 2, and by the upper sixth of the case for which y = 3. Using either case, and letting / be the length of the column under consideration, there is found, and hence the number // in (78)' is \x 2 for a long column fixed at one end and entirely free at the other. Accordingly a column of this kind can carry only one-fourth of the load of a column with two round ends. The value of P in Euler's formula gives the axial load which holds the column in equilibrium when it has become laterally 196 COLUMNS OR STRUTS CHAP, ix deflected. If the load is less than this value of P, the column will return to its original straight position. If the load is slightly greater than P, the bending increases until failure occurs. Euler's formula is hence the criterion of indifferent equilibrium, or the condition for the failure of a column by lateral flexure. Euler's formula is but little used in the design of columns, except in Germany. When so used the value of P computed from the formula is to be divided by a factor of safety in order to give the safe load on the column. Prob. 78a. A solid steel column with round ends is 6 inches in diameter and 37 feet long. Compute the axial load which will cause it to fail by lateral flexure. Prob. 786. A square wooden column with fixed ends is 20 feet long and carries a load of 9 500 pounds. Compute its size so that it may have a factor of safety of 10 by Euler's formula. ART. 79. EXPERIMENTS ON COLUMNS Although Euler published his formula in 1757 and Lagrange gave a more satisfactory discussion of it in 1773, it was not until after 1825 that its conclusions began to be used in practical investigations. This formula shows that the load P which causes the failure of a long column is inversely proportional to the square of its length. Hodgkinson in his experiments made about 1840 observed that this was closely true for wrought-iron columns, and only approximately so for cast-iron ones. Since for a solid cylindrical column 7 = igV^ 4 > the load P should be proportional to the fourth power of the diameter, and Hodgkin- son observed that this ratio was a little too high. He accordingly wrote for each kind of columns the analogous formula P = Q . d"/P and determined the constants Q, a, /? from the results of his experiments, thus producing empirical formulas. Let P be the load in gross tons which causes failure, d the diameter of the column in inches, and / its length in feet. Then the empirical formulas deduced by Hodgkinson for solid cylin- drical columns are, ART. 79 EXPERIMENTS ON COLUMNS 197 Cast Iro I P= I4.9* 3 ' 5 /* 1 - 88 for round ends JP^.^A 1 - 63 for flat ends Wrought Tro i P=^d 3 - 76 /l 2 for round ends =/ 2 for flat ends These formulas indicate that the ultimate strength of flat-ended columns is about three times that of round-ended ones. The experiments also showed that the strength of a column with one end flat and the other end round is about twice that of one having both ends round. Hodgkinson's tests were made upon small columns and his formulas are not so reliable as those which will be given in the following articles. For small cast-iron columns however the formulas are still valuable. A flat end may sometimes have more or less motion when the deflection begins and hence a flat-ended long column is not as strong as one with fixed ends. After 1850 wrought iron slowly replaced cast iron as a struc- tural material, and many tests of wrought-iron columns were conducted prior to 1890. The series of tests made by Christie in 1883 for the Pencoyd Iron Works is of great value on account of completeness as regards wrought-iron struts, since it included angle, tee, beam, and channel sections. A brief description and the principal results will here be given, but a fuller account may be found in Transactions of the American Society of Civil Engineers, April, 1884. The ends of the struts were arranged in different methods: first flat ends between parallel plates to which the specimen was in no way connected; second, fixed ends, or ends rigidly clamped; third, hinged ends, or ends fitted to hemispherical balls and sockets or cylindrical pins; fourth, round ends, or ends fitted to balls resting on flat plates. The number of experiments was about three hundred, of which about one-third were upon angles, and one-third upon tees. The quality of the wrought iron was about as follows: elastic limit 32000 pounds per square inch, ultimate tensile strength 49 600 pounds per square inch, ultimate elongation 1 8 percent in 8 inches. The length of the specimens varied from 6 inches up to 16 feet, and the ratio of length to least radius 198 COLUMNS OR STRUTS CHAP. IX of gyration varied from 20 to 480. Each specimen was placed in a Fairbanks testing machine of 50 ooo pounds capacity and the power applied by hand through a system of gearing to two rigidly parallel plates between which the specimen was placed in a vertical position. The pressure or load was measured on an ordinary scale be*am, pivoted on knife-edges and carrying a moving weight which registered the pressure automatically. At each increment of 5 ooo pounds, the lateral deflection of the column was measured. The load was increased until failure occurred. The following are the combined average results of these care- fully conducted experiments. The first column gives the values Ratio l/r of Length to Least Radius of Gyration Ultimate Load P/a, in Pounds per Square Inch Fixed Ends Flat Ends Hinged Ends Round Ends 20 46000 46000 46000 44000 40 40 ooo 40 ooo 40 ooo 36500 60 36000 36000 36000 30500 80 32000 32000 3 1 5 25 ooo 100 30 ooo 29 800 28 ooo 20 500 1 20 28 ooo 26300 24300 16 500 140 25500 23500 21 000 12 800 160 23 ooo 20 ooo 1 6 500 9500 180 20000 16800 12 800 7500 200 17500 14500 10 800 6000 220 15000 12 700 8800 5 ooo 240 13 ooo II 200 7500 4300 260 II 000 9800 6 500 3800 280 IO OOO 8 500 5 7 3 200 300 9000 7 200 5000 2 800 320 8000 6000 4 500 2 500 340 7 ooo 5 ioo 4 ooo 2 IOO 360 6500 4300 35oo I 900 380 5800 3500 3000 I 700 400 5 200 3000 2 500 I 500 420 4800 2 500 2 300 I 300 440 4 3 2 2OO 2 IOO 460 3800 2 000 I 900 480 I 900 I 800 of the ratio l/r and the other columns the values of P/a which caused failure, these being the ultimate load in pounds per square ART. 79 EXPERIMENTS ON COLUMNS 199 inch. From the results it will be seen that there is little prac- tical difference between the strength of the four classes when the strut is short. The strength of the long columns with round ends appears to be about one-third that of those with fixed ends. For values of l/r greater than 200, the ultimate loads are closely inversely proportional to the squares of the lengths for round ends, and approximately so for other arrangements of ends. Euler's formula fairly represents the results of the tests on the long columns. Taking = 25000000 pounds per square inch and n 2 as 10, the formula for round-ended columns becomes, P/a= 250 ooo ooo(r//) 2 = 250 ooo ooc/(//r) 2 from which the ultimate unit-loads are computed, for l/r= 220, 260, 300, 340, 380, 420 P/a=52oo, 3700, 2800, 2200, 1700, 1400 while the experiments give the ultimate unit-loads as, P/o=5ooo, 3800, 2800, 2100, 1700, 1300 Since Euler's formula is deduced under the laws of elasticity, it must be concluded that the elastic .limit was not exceeded when these long columns failed by lateral flexure. 100 000 300 Values of the ratio }f Fig. 79 Fig. 79 gives graphic representations of the above results for the cases of fixed ends and round ends, the values of l/r being taken as abscissas and those of P/a as ordinates. The broken lines also show the curves for these two cases which have been 200 COLUMNS OR STRUTS CHAP. IX plotted from Euler's formulas. It is seen that there is a marked disagreement between the experimental results and those found from Euler's formula when l/r is less than 200; this disagree- ment is due to the circumstance that Euler's formula refers to failure by lateral bending only, while for values less than 200 or 150 the failure actually occurred through the unit-stress on the concave side of the column having exceeded the elastic limit, so that the wrought iron became plastic (Art. 18). Prob. 79a. A cast-iron cylindrical column with flat ends is to be 7 feet long and carry a load of 200 ooo pounds with a factor of safety of 6. Compute the proper diameter. Prob. 796. Let Euler's formula be written y=c/x 2 , where x and y represent l/r and P/a and c is a constant. Discuss this curve and ascertain the points where it is parallel to the coordinate axes. ART. 80. RANKINE'S FORMULA The columns generally employed in engineering practice are intermediate in length between short prisms and the long columns to which Euler's formula applies. They fail under the stresses caused by combined flexure and compression, columns of brittle material by oblique shearing on the concave side or by tension on the convex side, and those of wrought iron and steel by the flow of metal on the concave side after the elastic limit has been surpassed. The ultimate unit-load P/a for these columns is less than the compressive strength S c for short prisms and very much less than the values computed from Euler's formula, as Fig. 79 shows. When such a column is perfectly straight, an axial load P produces the same unit-stress S = P/a on all parts of every sec- tion area a. When any bending occurs, due to imperfections of the material or to lack of straightness, the unit-stress on the concave side becomes greater than P/a and that on the convex side becomes less. Fig. 80a shows the flexure very much exagger- ated; and it is clear that the flexure formula (41) will apply to the discussion of the stresses caused by lateral bending. Let S\ be the greatest unit-stress due to the flexure and P/a the average ART. 80 RANKINE'S FORMULA 201 unit-stress due to the direct compression; then the total unit- stress on the concave side is the sum of P/a and S\ t and failure may be considered as occurring when this sum is equal to the ultimate compressive strength of the material. Fig. 80a Fig. 806 Fig. 80c Let / be the length of the column, a its section area, / the least moment of inertia, r the least radius of gyration of that section, and c the distance from the axis of the column to the remotest fiber on the concave side. In Fig. 806, the average compressive unit-stress on any section is represented by cd, but on the concave side this is increased to as and on the convex side decreased to bt. The triangles pds and qdt represent the effect of the flexure exactly as in the case of beams, ps indicat- ing the greatest compressive and qt the greatest tensile unit- stress due to the bending. Let the total maximum unit-stress as be denoted by 5 and the part due to the flexure be denoted by .Si. Then 5 = P/a +Si, in which Si is to be expressed in terms of P from the flexure formula Si . I/c = M, where M is the bend- ing moment due to P. Let / be the maximum lateral deflection of the column; then the greatest value of M is Pf, and accord- ingly Si = Pcf/I, or replacing 7 by ar 2 , the flexural unit-stress Si=P/a.cf/r 2 . Accordingly, the greatest compressive unit- stress on the concave side of the column is, By analogy with the theory of beams, as in Art. 56, the deflection/ may be regarded as varying directly as P/c. Hence, if ^ is a number depending upon the kind of material and the condition 202 COLUMNS OR STRUTS CHAP. IX of the ends of the column, it follows that, s=- or (80) which is Rankine's formula for the investigation of. columns. The above reasoning has been without reference to the arrange- ment of the ends of the column. By Art. 78 it is known that a long column with fixed ends is four times as strong as one with round ends, and that a long column with one end fixed and the other end round is 2.05 times as strong as one with round ends. Therefore, assuming that similar laws hold with respect to the term <(//r) 2 in the above formula, let from experi- ments on the rupture of different types of columns, but the fol- lowing table gives average values which are extensively employed in engineering practice : VALUES OF FOR FORMULA (80) Material Both Ends Fixed Fixed and Round Both Ends Round Timber I 1-95 4 3 ooo 3 ooo 3 ooo Cast Iron i 5000 i-95 5 ooo 4 5000 Wrought Iron i I.Q5 4 36000 36000 36000 Steel i i-95 4 25 ooo 25 ooo 25 ooo AKT. 81 INVESTIGATION OF COLUMNS 203 These values of < will be used in the examples and problems of the three following articles, and the value to be taken for 5 will be ultimate compressive strength of the material for cases of rupture and the allowable compressive unit-stress for cases of design. Euler's formula is not satisfactory for practical investigations because it contains no constant indicating the ultimate or work- ing strength of the material and because it applies only to long columns for which the ratio l/r is greater than about 200. Rankine's formula, however, contains the constant 5 and applies to cases for which the ratio l/r lies between 20 and about 150, and these are the columns used in engineering practice. On account of the many assumptions employed in deducing it, and because the values of the number < are derived from experiments, the formula is empirical rather than rational, yet it is of very great value. Its form satisfies the limiting conditions for short and long prisms. For a short prism, l/r may be taken as zero, and then S<=P/a. For a long column, unity may be neglected in comparison with <(//r) 2 , and then P/a = Sr 2 /P<}>; this is the same in form as Euler's formula, for placing S/(f> equal to a con- stant C it becomes P = Ca . r 2 /P. Rankine's formula is some- times referred to as Gordon's formula, but Gordon used the least thickness of the column instead of the least radius of gyration. Prob. 80a, Taking values of l/r as abscissas and those of P/a as ordinates, discuss the curve of formula (80) and find where its tan- gents are horizontal. Also locate its inflection point. Prob. 806. Plot the curve represented by formula (80) for wrought- iron columns with fixed ends, taking values of l/r as abscissas and those of P/a as ordinates, and using S as 46 500 pounds per square inch. Compare the plot with Fig. 79. ART. 81. INVESTIGATION OF COLUMNS The investigation of a column consists in determining the maximum compressive unit-stress 5 from formula (80). The values of P, a, /, and r are known from the data of the given case, and is known from the average experimental 204 COLUMNS cm STRUTS CHAP. IX values given in the table of the last article. Then the value of the greatest unit-stress 5 is computed from, By comparing the computed value of S with the ultimate com- pressive strength and elastic limit of the material, the factor of safety and the degree of stability of the column may be inferred. For example, consider a hollow wooden column of rectangu- lar section, the outside dimensions being 4X5 inches and the inside dimensions 3X4 inches. Let the length be 18 feet, the ends fixed, and the load be 5 400 pounds. Here P = 5 400, a = 8 square inches, / = 216 inches, and 4> = STnnr- The least radius of gyration is that with respect to an axis parallel to the longer side of the section, and for this axis ^ 2 = rV(5X4 3 4X3 3 )/8 = 2.2i, and accordingly l/r=i4$. Substituting now all values in the formula, there is found 5 = 5430 pounds per square inch, so that the factor of safety is only about i. The average unit- stress for this case is the same as for a short prism or S = P/a = 675 pounds per square inch, which is a safe value since the factor of safety is nearly 12. If the column is 3 feet long, the ratio of slenderness is l/r = 24, and the formula gives 5 = 805 pounds per square inch, which corresponds to a factor of safety of 10. As another example, consider a steel column 21 feet long with fixed ends which is used in the upper chord of a bridge under an axial compression of 240 ooo pounds. Let the sec- tion be that in Fig. 76c, which consists of a plate |Xi6 inches, and two channels each 12 inches deep and weighing 20^ pounds per linear foot. From the principles of Arts. 42 and 43, with the help of Table 9, the moment of inertia of the section with respect to an axis through its center of gravity and perpen- dicular to the webs is found to be 501.4 inches 4 and that with respect to an axis through the center of gravity and parallel to the webs is 663.9 inches 4 . The least radius of gyration then is r= (501. 4/24.06)* = 4.56 inches, and hence the slenderness ratio is V r = 55-3- Using for < the value ^-^TnT* the formula now gives 5=n 200 pounds per square inch, so that the factor of ART. 82 SAFE LOADS FOR COLUMNS 205 safety is about 5.3 and the column may be regarded as having a degree of stability a little too low for heavy traffic. The degree of reliability of values of S computed for columns is very much less than of those computed for beams from the flexure formula (41), since the column formula has a less reliable foundation. Moreover it assumes that the load is truly axial and the column perfectly straight before the application of the load, and these assumptions cannot be perfectly realized. It hence follows that factors of safety for compressive stress in columns should be higher than those for beams and higher than those for direct compression on short specimens. Prob. 81a. A cylindrical wrought-iron column with fixed ends is 12 feet long, 6.36 inches in outside diameter, 6.02 inches in inside diameter, and carries a load of 49 ooo pounds. Find its factor of safety. Prob. 816. A wooden stick, 3X4 inches and 12 feet long, is used as a column with fixed ends. Find its factor of safety under a load of 4000 pounds. If the length of the stick is only one foot, what is the factor of safety? ART. 82. SAFE LOADS FOR COLUMNS To determine the safe axial load for a column of given length and cross-section, it is necessary to assume the allowable work- ing unit-stress 5. Then Rankine's formula gives, in which < is to be taken from Art. 80, and the slenderness ratio l/r is to be computed from the given data. For example, let it be required to determine the safe load for a fixed-ended timber column, 3X4 inches in size and 10 feet long, so that the greatest compressive unit-stress 5 may be 800 pounds per square inch. Hence a =12 square inches, / = 120 inches, r 2 = 4X3 3 /i2Xi2 = |, and P/r 2 = 19 200; also < = -5-5*5-5-. Then the formula gives P=i 300 pounds for the safe load. A short column of this size should safely carry P= 12X800 = 9 600 pounds, or seven times as much as one of 10 feet length. ^7 206 COLUMNS OR STRUTS CHAP. IX As a second example let it be required to find the axial load for a fixed-ended steel column 23 feet 6 inches long so that S may be 12 ooo pounds per square inch. Let the section be that in Fig. 76c, the plate being |Xi6 inches and each channel 12 inches deep and weighing 20^ pounds per linear foot. From the prin- ciples of Arts. 42 and 43, with the help of Table 9, the least radius of gyration of the section is found to be 7 = 4.56 inches, so that the slenderness ratio is //r = 6i.8. Then Rankine's column formula gives P= 24.06X12 ooo /(iH 1 = 250 500 pounds which is the safe load for the given section. By using heavier channels of the same depth a much greater load may be carried without changing the outside dimensions of the section. Prob. 82a. Find the safe steady load for a hollow cast-iron column with fixed ends, the length being 18 feet, outside dimensions 4X5 inches, inside dimensions 3X4 inches. Prob. 826. Find the safe load for the above steel column when the channels are 12 inches deep and weigh 40 pounds per linear foot, all other dimensions and requirements remaining the same. ART. 83. DESIGN OF COLUMNS When a column is to be selected or designed the axial load P will be given, as also its length and the condition of the ends. A proper allowable unit-stress S is assumed, suitable for the given material under the conditions in which it is used, or the value of S will be given in the specifications under which the design is to be made. Then from formula (i), the section area of a short column or prism is P/S, and it is certain that a greater section area will be needed for the column. Next, let a cross- section be assumed, bearing in mind that it will be more effective the further the material be removed from the axis (Art. 77). Fof* this assumed cross-section a and r are to be determined, and then S is to be computed from the column formula. If this computed value agrees with the unit-stress assumed or specified, a section has been designed which satisfies the conditions; if ART. 83 DESIGN OF COLUMNS 207 not, a new cross-section is to be assumed and 5 be again com- puted ; this process is to be continued until a satisfactory agree- ment is secured. For example, a hollow cast-iron rectangular column, with fixed ends and 18 feet in length, is to carry a load of 60 ooo pounds and the allowable unit-stress 5 is to be 15 ooo pounds per square inch. For a short length the area required would be four square inches; assume then that about 6 square inches will be needed. Let the section be square, the outside dimensions 6x6 inches, and the inside dimensions 5^X5^ inches. Then a = 5. 7 5 square inches, I = 216 inches, ^ = 5.52 inches 2 , l/r = <)2, and = TGQTS' Substituting these in Rankine's formula, there is found for S about 30 ooo pounds per square inch, which is double the speci- fied value, and hence the assumed dimensions are much too small. Again, assume the outside dimensions as 6x6 inches and the inside dimensions as 5 X 5 inches. Then a = 1 1 square inches, r 2 = 5.o8 inches 2 , and l/r = g6. Substituting these in the formula, there is found for S about 15 700 pounds per square inch. Since this is very near the required working stress, it appears that these dimensions very nearly satisfy the imposed conditions. Many other sections can also be found which will satisfy the requirements, and the one to be finally selected will be that which is most convenient and economical. In some instances it is possible to assume all the dimensions of the column except one, and then after expressing a and r in terms of this unknown quantity, to introduce them into (80) and solve the problem by finding the root of the equation thus formed. For example, let it be required to find the size of a square wooden column with fixed ends and 24 feet long to sustain a load of 100 ooo pounds with a factor of safety of 10. Let x be the unknown side; then a = x 2 and r 2 = ^x 2 , and the column formula becomes, 100 OOO/ , 24 2 XI2 3 \ By reduction this leads to the biquadratic equation, S:* 4 i 000*2=331 776 and its solution gives 16.6 inches for the side of the column. 208 COLUMNS OR STRUTS CHAP, ix In designing columns and beams, considerations of economy are to be constantly kept in mind. For any given data, it is usually possible to arrange a large number of sections which will satisfy the requirements regarding strength, and the most advantageous one of these is that which, can be built at the low- est possible cost. In architecture, considerations of beauty are also to be followed in order that the eye may be pleased with the view of the column. It is sometimes said that beautiful forms are those of greatest strength, and this now and then hap- pens to be the case, but beauty and economy are usually con- tradictory elements. Prob. 83. Compute the size of a square wooden column, 12 feet long and having fixed ends, to carry an axial load of 50 net tons with a factor of safety of 10. Compute also the size of the column for the case of round ends. ART. 84. THE STRAIGHT-LINE FORMULA The column formula (80) was derived by Rankine about 1860 from older forms deduced by Tredgold and by Gordon, in which the ratio l/d was used, d being the least thickness of a rectangular section or the diameter of a circular section; Ran- kine introduced the ratio l/r and thus produced a formula appli- cable to all kinds of cross-sections. This formula has been more widely used than any other notwithstanding its empirical nature. When it is compared, however, with the results of many experi- ments, it is seen that in many cases these results may be repre- sented by a straight line, within the limits of the ratio l/r gen- erally used, almost as well as by the curve of Rankine's formula. On this basis the straight-line formula for columns was first deduced in 1886 by T. H. Johnson. On Fig. 84 are shown fifteen points which represent the aver- age results of about sixty experiments made by Tetmajer on struts of medium steel of different lengths and sizes, the ordinate for each point giving the unit -load P/a which caused the failure of the struts having the slenderness ratio l/r corresponding to its abscissa. The broken curve is that of Euler's formula and ART. 84 THE STRAIGHT-LINE FORMULA 209 it is seen to fit the observations very well for values of l/r greater than 150. For lower values of l/r the full straight line seems to give a fair average representation of the observations, this being drawn tangent to Euler's curve. It is required to deter- mine the equation of this straight line. Let y be the ordinate P/a and x the abscissa l/r and 5 be the value of P/a when x is zero, or the distance from the origin to the point where the straight line cuts the axis of ordinates. Then the equations of Euler's curve and of the straight line are, y=fj.E/x 2 and y=S Cx in which the parameter C is to be determined by making the straight line tangent to the curve. By equating the values of y in these two equations and also the values of the first derivatives dy/dx, the ordinate and abscissa of the point of tangency are found to be, i = S and xi Inserting these in the equation of the straight line, the value of C is found, and accordingly may be written, and (84) in which the number ft is n 2 for round ends, 2. 05-2 for one end round and the other fixed, and 4?r 2 for fixed ends (Art. 78). 400001 ^30000 20000 X x 4 \ s x x . v * 5* ^ <, jO _)_ -C' -- 50 100 150 300 Values of "the ratio l/ r FIG. 84 The values of S to be used for cases of rupture are such as to make the straight line agree best with experimental results. The values derived by Johnson are given in the following table, together with his values of C and the limiting values of l/r corre- #10 COLUMNS OR STRUTS CHAP. IX - spending to the point of tangency. The above theoretic values of p. were not used in computing C, as his experiments indicated that the numbers re 2 , i$x 2 , and 2^rc 2 for round, hinged, and flat ends respectively gave a closer agreement. It will be noticed CONSTANTS FOR FORMULA (84) Kind of Column Pounds per Square Inch Limit of l/r 5 C Wrought iron Flat ends 42 ooo 128 218 Hinged ends 42 ooo 157 178 Round ends 42 ooo 203 138 Structural steel Flat ends 52500 179 195 Hinged ends 52 500 220 159 Round ends 52500 284 123 Cast iron Flat ends 80000 438 122 Hinged ends 80 ooo 537 99 Round ends 80000 693 77 Oak Flat ends 5 4oo 28 128 that the values of S in the table are less than the average ulti- mate compressive strengths given in Art. 5. For ductile mate- rials like wrought iron and structural steel, this should be the case in columns, since when the elastic limit is passed a flow of metal begins which causes the lateral deflection to increase, and failure then rapidly follows. The straight-line formula is not suitable for investigating a column, that is, for determining values of 5 due to given loads, because S enters the formula in such a manner as to lead to a cubic equation when it is the only unknown quantity. It may be used to find the safe load for a given column to withstand a given unit-stress S, or to design a column for a given load and unit-stress. When so used, it is customary to divide the values of 5 and C given in the table by an assumed factor of safety. For example, Cooper's specifications require that the section area a for a medium-steel post of a through railroad bridge shall AHT. 85 OTHER COLUMN FORMULAS 211 be found from P/a = 1 7 ooo - go(l/r) pounds per square inch, in which P is the direct dead-load compression on the post plus twice the direct live-load compression; the values of S and C here used are a little less than one-third of those given in the table for round ends. While the straight-line formula is sometimes slightly more convenient than that of Rankine, it cannot be regarded either as having so high a degree of validity or as satisfying so well the results of experiments. It is hence advisable that the use of this formula should be limited to cases in which specifica- tions require it to be employed, and for rough approximate computations. Prob. 84. Solve Probs. 82a and 83 by the help of the straight-line formula, applying the factors of safety to the values of 5 and C given in the above table. ART. 85. OTHER COLUMN FORMULAS Many attempts have been made to establish a formula for col- umns which shall be theoretically correct, like the flexure formula (41) for beams, when the material is not stressed beyond the elastic limit. Although many column formulas have been proposed which have been claimed by their authors to have a rational basis, none of them has yet been recognized by the engineering profession as more satisfactory than the formula of Rankine. For long columns all agree that Euler's formula is correct in giving the load which causes failure by lateral bending, but for the columns commonly used in practice, where the slenderness ratio l/r lies between 30 and 150, a fully satisfactory formula has not been established. In 1873 Ritter proposed a formula to be used when the unit- stress S does not exceed the elastic limit S e , namely, P S in which P/a is the axial unit-load, E the modulus of elasticity of the material, and the number JJL is rc 2 for round ends, 2. 051? for one end round and the other fixed, and 4?r 2 for both ends 212 COLUMNS OR STRUTS CHAP. IX fixed. This is seen to be the same in form as Rankine's formula, the constant < having the value SJfjE. Using the average values of S e and E given in Arts. 2 and 9, the values of < for columns of different materials are, for Timber for Cast Iron for Wrought Iron Ends Fixed I Fixed and Round i-95 Ends Round 4 20000 I 20000 '95 20000 4 30000 . I 30000 i-95 30 ooo 4 40 ooo I 40 ceo i-95 40 ooo 4 34000 34000 34000 These values of $ are smaller than the empirical ones in Art. 80, and hence the formula of Ritter gives larger values of P/a than the formula of Rankine. When l/r is 100, the values of P/a for fixed-ended columns found from Ritter's formula are 2 percent greater than those given by Rankine's formula for wrought iron, 8 percent greater for structural steel, 230 percent greater for cast iron, and 290 percent greater for timber. While the com- parison is fairly satisfactory for wrought iron and medium steel, the disagreement in the results for cast iron and timber is so great that Ritter's formula cannot be regarded as satisfactory. Another formula which has received extended discussion is that derived by Crehore in 1879 and also independently deduced , by Reuleaux, Marburg, and others. This formula is identical with that of Ritter except that S t is replaced by S; designating P/a by B, it may be written, S B B- s=- from the first of which the unit-load may be computed for a given unit-stress, while the second gives the compressive unit- stress on the concave side of the column due to a given unit -load. Tables giving values of S for wrought iron and steel columns were published by Merriman in 1894, together with a deriva- ART. 85 OTHER COLUMN FORMULAS 213 tion different from that of Crehore. These tables show that values of P/a computed for a given unit-stress 5 are greater than those found from Rankine's formula; for instance, when l/r=ioo and S =12 ooo pounds per square inch, it gives P/a=n coo for steel columns with fixed ends, whereas Rankine's formula gives P/a = 8 500 pounds per square inch. In general the formula gives working unit-loads which are from 20 to 30 percent greater than those derived from the expression of Rankine and hence it cannot at present be regarded with favor. On the other hand when applied to cases of rupture, it usually gives values of P/a less than those shown from experiment and for this reason it is also unsatisfactory. The discussions of this chapter show that the theory of columns stands upon a less rational basis than that of beams. The flexure formula 5 . 1/c = M has a sound theoretical foundation and, for all cases where the elastic limit is not exceeded, it is found to agree with experiment. Euler's formula for long columns has a sound theoretical basis and it also agrees with experiment, but all other column formulas contain certain theoretic defects. A rational formula can often be safely applied to cases of rupture by using empirical constants, but it is rarely the case that an empirical formula containing constants deduced from experi- ments on rupture can be applied to cases in which the elastic limit is not exceeded, except by the use of arbitrary factors of safety. For this reason Rankine's formula is not a satisfactory one, and yet the long use of it by the engineering profession has built up a system of practice and precedent which must con- tinue to be respected until the time arrives when a satisfactory theoretical formula shall be established. The present indica- tions are that the errors of Rankine's formula, when used with the average empirical values of given in Art. 80, are on the side of safety. Prob. 85a. Refer to Van Nostrand's Engineering Magazine for December, 1879, and ascertain the assumption used by Crehore in his derivation of the above column formula. Compare also the discussion of the modified Euler formula in Art. 167. 214 COLUMNS OR STRUTS CHAP. IX ART. 86. ECCENTRIC LOADS ON PRISMS In all the preceding discussions, the load on the column has been regarded as axial, but cases are very common in engineer- ing practice where the load P is applied at a distance p from the axis of the column ; this lever arm p is called the "eccentricity " of the load. It is evident that the flexural stresses in the column will increase with p and that the total unit-stress S on the most compressed side of the column will be greater than for the case of an axial load. ,-P., P r P Fig. 86a Fig. 866 Fig. 86a shows two rectangular columns with eccentric loads. In the first diagram, the weight P on the top has a resultant shown by the arrow which falls within the cross-section; in the second, this resultant line falls without the cross-section. In these two diagrams, the point of application of P is on a median line of the cross-section, and this is the common case in prac- tice, but Fig. 866 shows the unusual case where the point of appli- cation is not on a median line of the cross-section of the column. In all instances of eccentric loads, however, the tendency of the load is to cause rotation about an axis perpendicular to direc- tion of the lever arm p. Thus, in each of the figures, let mn be drawn in the plane of the cross-section through the axis of the column and normal to p; then the flexural stresses are to be referred to mn as a neutral axis, and the moment of inertia ART. 86 ECCENTRIC LOADS ON PRISMS 215 of the cross-section with respect to this axis mn is to be deter- mined by the methods of Art. 43. The shaded areas in Fig. 86a show the compressive unit-stresses due to the eccentric load when the elastic limit of the material is not exceeded, the varia- tion occurring at a uniform rate. It will now be shown that, whatever the shape of the section area a, the unit-stress along the axis of the column, due to the eccentric load P, is P/a. Let the total unit-stress as on the side of the column nearest the load be called S, and the unit-stress cd along the axis of the column be called B; it is required to prove that B = P/a. Let a line pq be drawn through d parallel to ab, then ap is B and ps is SB. Now from the flexure formula (41) there may be written (SB)/c = M/I, where c is the dis- tance ca in the figure and 7 is the moment of inertia of the sec- tion area with respect to the neutral axis mn. Now let ab and st be produced until they meet in e.at the distance z from the axis of the column and let e be taken as an axis of moments; then the flexure formula is S/(c+z)=M'/P, where M' is the moment of P and P the moment of inertia of the section with respect to an axis through e parallel to mn. From similar tri- angles it is seen that (S-B)/c equals B/z and also that S/(c+z) equals B/z; accordingly, B/z=M/I and B/z=M'/P are two formulas for determining B and z. Since M = Pp and M' =P(p+z), while I = ar 2 and P = ar 2 + az 2 , where r is the radius of gyration of the section with respect to the axis mn, the solu- tion of the two equations gives the results, B=P/a and z=r 2 /p Therefore, the unit-stress along the axis of the column equals the unit-load P/a, and the product of the lever arm p and the distance z equals the square of the radius of gyration of the section. The total compressive unit-stress S now immediately results from the flexure formula (S B)/c = M/I. Inserting in this the values of M and 7, and that of B just found, it becomes, S=(i ^] whence P = S (86) ~~ a \ r 2 ) a ~ i + cp/r* 216 COLUMNS OR STRUTS CHAP, ix from which 5 can be computed for a given P, and P/a be found for a given 5. This formula is strictly applicable only to short prisms, since it takes no account of the lateral deflection of the column. When the cross-section is a rectangle having the depth d in a plane passing through P and the axis, the value of r 2 is - l 1 s b(P/bd = ^d 2 , and since c is \d for this case, the expression for 5 becomes P/d[i+6(p/d)], which is the same result as found in Art. 29. When the cross-section is a circle of diameter d, the value of r 2 is -fad? and that of c is \d\ hence the expression for 5 reduces to P/a[i+8(p/d)]. The formula (86), indeed, applies to all forms of cross-section, provided the unit-stress 5 does not exceed the elastic limit of the material. Let d be the distance from the axis of the column to the other side of the column where the unit-stress is S'; then the value of S' is (P/a)(i-c'p/r 2 ) and S' will be tensile when p is greater than r 2 /c'; this case is shown in the second diagram of Fig. 86a, where the point e is within the section. For columns having ratios of slenderness between 30 and i oo, a common method of taking into account the effect of an eccentric load is to add the flexural unit-stress above found to that given by Rankine's formula; thus results, for the greatest compressive unit-stress on the concave side of the column. This formula may be used to investigate the strength of a column, to compute the safe load, or to design a column section by methods exactly like those explained in Arts. 81-83. When several loads PI, PI, etc., are on the column at the distances pi, p 2 , etc., from the axis, their sum or resultant is P and its lever arm with respect to the axis is found from /> = (Pi/>i + P 2 ^>2 + etc.)/P. In this manner all the acting loads are replaced by a single load P having the eccentricity p. Prob. 86a. Using formula (86)' solve Prob. 82a, taking the eccen- tricit of the load as i^ inches. ART. 87 ECCENTRIC LOADS ON COLUMNS 217 Prob. 866. Using formula (86)' solve Prob. 83, taking the eccen- tricity of the given load as 2\ inches. ART. 87. ECCENTRIC LOADS ON COLUMNS The investigation of the last article gives results for S which are too small, especially for columns having a slenderness ratio greater than about 100, since the lateral deflection of the column due to the eccentric load will increase the eccentricity of the load for all sections except those at the ends. Thus, in the fol- lowing figures, let p be the eccentricity of the load P, and / the maximum deflection of the axis of the column from its original straight position; then the lever arm or eccentricity of the load with respect to the neutral axis of the dangerous section is p+f, which may be called q, so that formula (86) becomes, P I . cp . cf\ Pi c(A 12 = T\ I + 72 ( 8? ) and hence / or q must be determined in order to find S. pj "- V' Fig. 87a Fig. 876 Fig. 87c Let PQ be the axial load given by Euler's formula for long columns, and suppose that this load is applied to the column when it has the deflection /, the eccentric load P being removed. Then this load P will hold the long column in equilibrium and its moment with respect to the axis of the dangerous section is P f, while the moment of the eccentric load with respect to the same section was P(p+f). Equating these moments, it follows that, and c which indicates that the deflection of a column under an eccentric load is determinate, although under an axial load it is indeter- 218 COLUMNS OR STRUTS CHAP. IX minate. Deflections computed from this formula are, however, too small, because it has been deduced by considering only the moment at the middle of the column, whereas all moments should be taken into account. Let Fig. S7b represent a column of length / which is free at the upper end and vertically fixed at its lower end. Let the origin of coordinates be taken at the top under the load, values of x being measured downward and those of y horizontally. The bending moment for "any point of the elastic curve is Py, and hence El . d 2 y/dx 2 = - Py, in which / is moment of inertia of the cross-section and E is the modulus of elasticity of the mate- rial. Integrating this and determining the constant by the con- dition that the tangent dy/dx = o when y = q, there results El. (dy I doc) 2 = P(q 2 -y 2 ). Integrating again and ascertaining the constant by the condition that y = p when x = o, there is found, arc sin^= (P/E/)**+arc sin^ 9 ? for the equation of the elastic curve. From this equation the total eccentricity q is determined by making x = l for y = q, and, p= q sinftTT- (P/E/)*/] whence q=p sec(P/E/)*/ Now let 6 represent the number (PP/EI)*, this being an angle measured in radians. Then the value of q is given by, q=p sec0=/>(i + o.50 2 +o.2o80 4 +o.o8470 6 +. . .) (87)' where the quantity in the parenthesis is obtained by expanding seed into a series by Maclaurin's theorem. By referring to Art. 78 it will be seen that the above value of 6 applies to a column with two round ends if / is replaced by %l, to a column with one end round and the other fixed if / is re- placed by 0.3 5/, and to a column with two fixed ends if / is replaced by \l. Accordingly in (87) let the unit-load P/a be represented by B; also in the value of 6 let / be replaced by ar 2 , where r is the least radius of gyration of the section. Then the formula, . . .) (87)" applies to all columns, when d 2 has the following values: ART. 87 ECCENTRIC LOADS ON COLUMNS 219 for one end free and the other fixed, 2 = B/E . (l/r) 2 for both ends round, 6 2 =B/E . (l/r) 2 for one end round and the other fixed, d 2 =^BjE . (l/r) 2 for both ends fixed, 6 2 =^B/E . (l/r) 2 In computing the unit-stress 5, the value of sec 6 may be found for all cases by the help of Table 17, while the series cannot be used unless 6 is less than %n : when 6 2 equals \x* the unit load B is that given by Euler's formula and S is in- finite. A closely approximate value of sec is given by For example, let the unit-load B be 10 ooo pounds per square inch, applied at a distance of i.oi inches from the axis of a steel column having round ends, the length of the column being 192 inches, the distance c being 4.45 inches, and the radius of gyra- tion in the direction of the eccentricity being 3.00 inches. Hence 02 = |(5/E)(J/r)2 = 0.3413, and then the series in (87)" gives 5=16900 pounds per square inch, so that the eccentric load increases the mean unit-stress about 69 percent. Or, using the value of 0, formula (87)' gives 9=1.197 inches and then from (87) the unit-stress 5 is directly found. To illustrate the computation of sec0 and at the same time show the great influence of an increase in length of the column, take the same data as above except that the length is twice as great, or 384 inches. Then, 2 = 1.3653 and = i.i68 radians = 65 45', whence from a trigonometric table sec0 = 1.934. Accord- ingly the total eccentricity is ^=1. 934^ = 1.953 inches, and the unit-stress 5 is 20 700 pounds per square inch. When a column is to be designed, its length and eccentric load are given, as also the allowable unit-stress S, and such values of a, c, and r are to be found as will satisfy (87)" and at the same time give the greatest economy. This process is a tentative one, and several trials may be necessary in order to find a satisfactory section. Probably the best plan is to assume values of a, c, and r and then compute 5, continuing the process until the computed allowable values fairly agree. 220 COLUMNS OR STRUTS CHAP. IX The above formula (87)" is not convenient for the direct computation of the eccentric unit-load P/a for a given column under a given unit-stress S, and hence a problem of this kind is also to be solved by successive trials, values of P/a or B being inserted in the formula until one is found which makes S the same as the given value. This process, as also that of designing a column section, will be often facilitated by assuming sec# for the first trial, taking it as unity if the column is short, or about ij or 2 for a long column. Prob. 87a. Show that (P+f)c equals r 2 for a column so deflected that there is no stress on the convex side. ""^Prob. 87b. A wooden strut with fixed ends is 18 feet long, 4 inches square, and the compression of 5 ooo pounds is applied half-way be- tween the center and corner of the end sections. Compute the deflec- tion at the middle of the column, and the maximum unit-stress S. Prob. 87c. Prove that the eccentric load P which holds a round- ended column in equilibrium with the deflection q p, is given by the formula P/a=4E(r/t) 2 (a.rc cosp/q) 2 . ART. 88. ON THE THEORY OF COLUMNS Since the discussion in Art. 78 does not determine Euler's formula for a column with one end round and the other fixed, a more general investigation will now be given. Let the column be constrained at the ends so that bending moments MI and M 2 exist there when a lateral deflection occurs. The bending moment at any section distant x from the end i of the column is M = Mi + Vx-Py where V\ is the transverse shear at i. When x = l, then y = Q and M = M 2 , so that Fi = (M 2 -Mi)//. The general equation of the elastic curve now is and the general solution of this differential equation is Ely = A sin/?y + B sin/?^ + Cx + D(l-x) (88) ART. 88 ON THE THEORY OF COLUMNS 221 in which A, B, C, D, ft are constants of integration to be found from the condition of the problem. Differentiating (88) twice, there result, 5cos/^) +C-D (88)' / = ~ A sin/9 7 + B 5[n P l -r} (88) " Comparing the two values for EI.d 2 y/dx 2 and eliminating the trigonometric expression by the help of (88), three constants are found, namely, J3 2 = P1 2 /EI, C = M 2 / 2 //? 2 , and D^M^/p 2 . Further, the first member of (88)" becomes M\ when x = o and M 2 when x = l; hence A = -M 2 / 2 //? 2 sin/? and = -Mi/ 2 //? 2 sin. The expression found for /? 2 gives P = 3 2 EI/l 2 from which Euler's formulas may be deduced for special arrangements of the ends. Inserting the values of the constants in (88) ' and making # = o there is found an expression for the tangent of the angle which the tangent to the elastic curve at the end i of the deflected column makes with the axis of x, namely, Now let the end 2 be round and the end i be fixed, then M 2 =o, and since dy/dx must be zero for a fixed end, = tan/? is the condition that end i is fixed and end 2 is round. This equation has two roots, the first being /? = o which applies to a straight undeflected column. The other root /? = 4.49341 corresponds to the least value of P which holds the deflected column in equi- librium. Since 4.49341 = 1.43029;:, the value of [P is 2.O457?r 2 , and hence Euler's formula for this case is P = 2.o46n 2 EI/l 2 , as noted in Art. 78. The constant 2\, which is usually stated for this case, gives the value of P about 10 percent too large. From (88)'" the values of /? for both ends round or for both ends fixed are also readily obtained. For both ends round, MI and M 2 are zero, while dy/dx is indeterminate. Hence the quantity 0(1 -cos/?) /sin/9 must be indeterminate, and this will 222 COLUMNS OR STRUTS CHAP. IX be secured by making sin/3 = o, whence /?=TT for the least value of P and accordingly Euler's formula for columns with round ends is P=x 2 EI/l 2 . When both ends are fixed and M 2 equals MI, then dy/dx becomes zero for both x=o and x=l provided cos,3=i; this requires that /? = 2jr, whence Euler's formula for columns with two fixed ends is P = 4x 2 EI /I 2 . Taking the moments at fixed ends as positive, the maximum negative moment may be found from (88)'". When both ends are fixed the maximum negative moment is at the middle of the length and is double the positive moment at the end. When end 2 is round and end i is fixed, all moments have the same sign as MI and the maximum is about i.o>2M\. An ideal column is one which is perfectly straight in its unstressed state and which when laterally deflected by an extrane- ous force returns to its straight condition on the removal of that force. The maximum unit-stress for the ideal column hence does not exceed the elastic limit S e . For the straight column failure begins when P/a equals 5; for the deflected column failure begins when Euler's condition P/a = fiE(r/r) 2 is reached. Hence S e = nE(r/l) 2 or l/r=(nE/S e )* gives the critical value of l/r between the two methods of failure. For instance, columns of structural steel with round ends, for which fJL=n 2 , have l/r=g2, while columns of structural steel with fixed ends, for which {i=T^ have l/r=i&4 as critical values of the slenderness ratio. For ideal conditions Euler's formula only applies to higher values of l/r than these critical ones. Under actual conditions, however, columns are not perfectly straight, so that flexure may begin long before P reaches the value given by Euler's formula. In other words, there is always an initial deflection in practise, and Rankine's formula is an attempt to empirically take its average value into account. The theoretic basis of Rankine's formula seems far more satisfactory than that of any other which has been proposed for the discussion of such columns as are used in engineering practice. Nevertheless it should be noted that the constants given on ART. 88 ON THE THEORY OF COLUMNS 223 page 202 cannot be expected to apply to unusual or extreme cases, for they are averages deduced from tests on common sec- tions. In this connection the investigations made in Ireland by Lilly in 1907 may be noted as important. Making tests upon hollow cylindrical columns of small thickness, he found failure to occur under a much less unit load P/a than for solid cylindrical columns having the same values of l/r, and it was seen that this was due to a wave-like crippling that produced secondary flexure. His investigations lead to the conclusion that the formula for such columns is P = 5 1+W 7 H in which t is the thickness of the material, and m and n are constants that depend upon the kind of material and the condition of the ends. This expression indicates that the strength of a hollow cylinder depends not merely upon its section area and slenderness ratio but also upon the thickness of its walls. When / is very small, Lilly's formula makes P/a also small, and it is evident that this conclusion is a correct one. When a compound column is made up of plates or webs, as was the case in 1 the lower chord of the ill-fated Quebec bridge, which failed in 1907, it is also probable that neither Rankine's formula or even the simple formula P/a=S can properly apply to it, on account of waves which produce secondary flexure in the plates or webs. This view of the subject is one quite new and indicates that our present knowledge of column action does not apply to unusual sections. In all such cases there is only one safe guide, namely, tests upon the actual columns or upon models which represent them as closely as possible. The influence of the weight of the column itself has not been considered in the preceding pages. When the column is in a vertical position, it might appear in accordance with Art. 28, that the section should increase continually toward the base in order to give a form of uniform strength, or for a long column, 224 COLUMNS OR STRUTS CHAP. IX where lateral flexure is to be feared, that the greatest section should be between the middle of the length and the base. For usual cases, however, the necessary increase of section is so small as to be inappreciable. The Doric column of Greek architec- ture had its greatest diameter at the base, while the Ionic column was usually of constant diameter up to about one-third of its height. Considerations of beauty rather than strength governed, however, the evolution of these ancient forms. When a column is placed in a horizontal position, its weight causes flexure which increases the deflection and stress due to the direct compression. This case wih be discussed in Art. 104, and it will be there seen that the direct compression is sometimes applied eccentrically in order to counteract the flexure caused by the weight of the column. Prob. 88a. Prove that 4.49341 is a root of the equation /? = tan/?. Prob. 886. Consult the engineering periodicals for September, 1907, and ascertain facts regarding the failure of the great cantilever bridge over the St. Lawrence river at Quebec. ART. 89 PHENOMENA OF TORSION 225 CHAPTER X TORSION OF SHAFTS ART. 89. PHENOMENA OF TORSION When applied forces cause a bar to twist around its axis, 'torsional stresses ' arise. If a rectangular bar has one end fixed and forces are applied to the other end which cause twisting around its axis, the comers of the bar are seen to assume a spiral form similar to the threads of a screw. By experiments like those illustrated in Fig. 169c it has been proved that the phenomena of torsion are analogous to those of tension. Under small twisting forces the deformation, or angle of twist, is proportional to the force, so that the bar returns to its original form on the removal of that force. This law holds until an elastic limit is reached; beyond this limit the angle of twist increases more rapidly than the force, and a permanent set remains when the force is removed. Under further increase of the twisting force, the deformation rapidly increases and rupture ' finally occurs. In Fig. 169c are seen a square bar and a round bar which have been ruptured by torsion. The force P which causes the twist acts in a direction normal to the axis of the bar or shaft with a certain lever arm p. Fig. 89 shows a horizontal shaft rigidly fixed at one end, while a weight P is hung on a lever at right angles to the axis of the shaft. Under the twisting moment Pp the shaft is deformed, so that an originally straight line ab becomes the helix ad, while the radial line cb has moved through the angle bed. The angle bed is evidently propor- tional to the length of the shaft, while the angle bad is independent of that length. The product Pp is the moment of the force P with respect to the axis of the shaft, p being the perpendicular distance from that axis to the line of direction of P, and is called the 'twisting moment '. Whatever be the number of forces acting upon the 226 TORSION OF SHAFTS CHAP, x shaft, their resulting twisting moment may always be represented by a single product Pp. Thus, if the forces P\ and P 2 act with lever arms pi and PZ, the twisting moment Pipi+P^p? may be caused by a single force P with the lever arm p. A graphical representation of the phenomena of torsion may be made in the same manner as the tension diagram of Fig. 4, the angles of torsion being taken as abscissas and the twisting moments as ordinates. The curve is then a straight line from the origin until the elastic limit of the material is reached, when a rapid change occurs and it soon becomes nearly parallel to the axis of abscissas. The total angle of torsion, like the total ulti- mate elongation, serves to compare the ductility of materials. The principal stress which occurs in torsion is that of shearing, each section of the bar tending to shear off from the one adjacent to it. The direction of the shearing stress at any point in the sec- tion of a round shaft is normal to a radius drawn from the axis to that point, and the sum of the moments of all Fig. 89 the stresses in any sec- tion is equal to the twisting moment Pp. For a round bar, like that of Fig. 89, a radial straight line cb is found to remain straight when displaced to the . position cd, provided the shear- ing elastic limit of the material is not exceeded. For a square or rectangular bar, this is not the case when the radial line is drawn to a corner. Prob. 89a. If a force of 80 pounds acting at 18 inches from the axis twists a shaft 15 degrees, what force will produce the same result when acting at 4 feet from the axis ? Prob. 896. A shaft 2 feet long is twisted through an angle of 7 degrees by a force of 200 pounds acting at a distance of 6 inches from the axis. Through what angle will a shaft of the same size and material and 4 feet long be twisted by a force of 500 pounds acting at a distance of 18 inches from the axis? ART. 90 THE TORSION FORMULA 227 ART. 90. THE TORSION FORMULA It has been found by experiment that the laws governing the stresses in a section of a round bar under torsion are similar to those stated for beams in Art. 40, provided the elastic limit is not exceeded. The shearing unit-stresses are proportional to their distances from the axis, because it is observed that any radius, such as cb in Fig. 89, remains a straight line when it is displaced by the twisting into the position cd. The law of static equilibrium requires that the sum of the moments of these shearing stresses shall be equal to the twisting moment, or, Resisting Moment = Twisting Moment and this condition will now be expressed in algebraic language. Fig. 90a Let P be the force acting at a distance p from the axis about which the twisting takes place, then the value of the twisting moment is Pp. To find the resisting moment, let c be the distance from the axis to the outside of the circular cross-section where the shearing unit-stress is S. Then, since the shearing unit-stresses vary as their distances from the axis of the shaft, S/c= unit-stress at a unit's distance from axis S . z/c= unit-stress at a distance z from axis da . Sz/c= total stress on an elementary area da da . Sz 2 /c=* moment of this stress with respect to axis 2 da . Sz 2 /c = internal resisting moment Since 5 and c are constants, this resisting moment may be written (S/c}2da . z 2 . But the quantity Ida . z 2 , being the sum of the products obtained by multiplying each element of area by the square of its distance from the axis, is the polar moment of 228 TORSION OF SHAFTS CHAP, x inertia of the cross-section and may be denoted by /. The resisting moment of the internal stresses hence is 5 . J/c, and equating this to the twisting moment, there results, S.J/c^Pp or S = Pp.c/J (90) which is the fundamental formula for investigating round bars and shafts that are subject to torsion. It will be called the torsion formula, but it must be borne in mind that it does not apply to square or rectangular sections; these will be discussed in Art. 99. The flexure formula S . I/c = M for beams has a close analogy with the torsion formula. In the flexure formula, c is the distance from the neutral axis to the remotest fiber and that axis lies in the plane of the cross-section; in the torsion formula, c is the radius of the outside circumference of the round bar. In the flexure formula, 5 is a tensile or compressive unit-stress which is normal to the section area; in the torsion formula, 5 is a shearing unit-stress which acts along the section and normal to the radius. In the flexure formula, M is the bending moment of the external forces; in the torsion formula, Pp the twisting moment of the external forces. By the help of the torsion formula, three problems like those of Arts. 48, 49, 50, may be discussed for round bars or shafts. When the dimensions and the allowable unit-stress are given, it may be used to compute the safe twisting moment Pp. When the unit-stress S g and the moment Pp are given, it may be used to design a shaft or bar, by determining dimensions which will give a value for J/c equal to Pp/S. The results obtained in this discussion are only valid when the shearing unit-stress 5 does not exceed the elastic limit of the material, but it is shown in Art. 94 how the formula may be used for cases of rupture. In the discussion of shafts, the moments of inertia of cross- sections are required with respect to a point at the center of the shaft and not with respect to an axis in the same plane, as in beams and columns. The 'polar moment of inertia ' of a surface is defined as the sum of the products obtained by multiplying each elementary area by the square of its distance from the center of ART. 90 THE TORSION FORMULA 229 gravity of the surface. Thus if da is any elementary area and z its distance from the center, the quantity Ida . z 2 is the polar moment of inertia of the surface. In the figure let da be any elementary area and y its distance from an axis AB passing through the "center of gravity of the section ; then Ida . y 2 is the moment of inertia with respect to this axis AB (Art. 43). Also, if x is the distance from da to an axis CD which is normal to AB, then Ida . x 2 is the moment of inertia with respect to CD. But, since z 2 = x 2 + y 2 , the product Ida . z 2 is equal to Ida . x 2 + Ida . y 2 ; that is, the polar moments of inertia is equal to the sum of the moments of inertia taken with respect to any two rectangular axes. By the aid of the above principle, the polar moment of inertia J is readily found for a solid or hollow circular section from the values of / given in Art. 43. Let d be the diameter of a solid section or the outside diameter of a hollow one, and let d\ be the inside diameter of a hollow one; then, for a solid shaft, c = &, J= xd* for a hollow shaft, c=$d, J= fa(d* - dfi The polar radius of gyration r, defined by the equation / =ar 2 , where a is the section area, is also sometimes used in formulas; it is the radius of a circumference along which the' entire area might be concentrated and have the same polar moment of inertia as the actual distributed area. The value of r is always less than \d\ for a solid circular section, r 2 = $d 2 ; fora hollow circular sec- tion, r 2 = \(d 2 + rfi 2 ). Prob. 90a. Three forces of 70, go, and 120 pounds act at distances of 8, n, and 6 inches respectively from the axis, and at different dis- tances from the end of a shaft, the direction of rotation of the second force being opposite to that of the others. Find the three values of the twisting moment Pp. Prob. 906. A circular shaft is subjected to a maximum shearing unit-stress of 2 ooo pounds when twisted by a force of 90 pounds at a distance of 27 inches from the axis. Compute the ur it-stress pro- duced by a force of 40 pounds at 21 inches from the axis. 230 TORSION OF SHAFTS CHAP. X ART. 91. TRANSMITTING POWER BY SHAFTS Power from a motor is often transmitted to a shaft through a belt, and then the shaft transmits the power to the places where the work is to be performed. The shaft of a turbine wheel transmits the power generated by water in passing through the wheel directly to dynamos or other machinery. The shaft of an ocean steamer transmits the power of the engines directly to the screw propellers. In all these cases the shaft is subject to a twisting moment Pp which produces in it shearing stresses and causes it to have a certain angle of twist. The twisting moment Pp due to the transmission of H horse- powers through a shaft may be found as follows: Suppose P to be the tangential force brought by a belt on the circum- ference of a pulley of radius pi and let n be the number of revolu- tions made by the shaft and pulley in one minute. In one revo- lution the force P overcomes resistance through the distance 2xp and the work PX2xp or 2nPp is transmitted through the shaft; in n revolutions the work 2xnPp is transmitted. Now let N be the number of units of work per minute which constitute one horse-power; then 2xnPp = NH, or Pp = NH/2r:n. In the English system of measures, P is in pounds, p in inches, and N is 396000 pound-inches per minute. In the metric system of measures, P is in kilograms, p in centimeters, and N is 450 ooo kilogram-centimeters per minute. Accordingly, Pp = &3 o^oH/n or Pp = H fooH/n (91) the first being for the English and the second for the metric system of measures, while Pp = (N/2x)H/n applies to all systems. The above formula shows that the twisting moment Pp varies directly as the transmitted horse-power and inversely as the speed of revolution. Therefore, when the speed is low, as it is in starting, the full power should not be applied to a shaft, for it might render the twisting moment so great as to injure the material or to cause rupture. Metallic shafts are usually round and the word 'shaft', when ART. 92 SOLID AND HOLLOW SHAFTS 231 used without qualification, means a solid round cylinder, properly supported in bearings. Square shafts are rarely used except for wooden water wheels, and the torsion formula (90) does not apply to these unless modified in the manner indicated in Art. 99. Hollow sections have been much used since 1900 for the large shafts of ocean steamers. These are advantageous in being of lighter weight than solid shafts of the same strength and capacity, as the investigation in Art. 95 will show; these shafts are forged upon a mandrel and the inside surface is hence subject to the same treatment as the outside surface. In designing a shaft to resist a given twisting moment Pp the factors of safety may be based upon the average shearing strengths of materials which are given in Art. 6. The rule that an allowable unit-stress S should not exceed the shearing elastic limit of the material should also be observed, although there is considerable uncertainty as to the average values of this limit. Probably the elastic limits for shearing are about three-fourths of those for tension. Prob. 91c. Show that the metric horse-power is 1.4 percent less than the English horse-power; also that one kilogram-meter is equiva- lent to 7.233 foot-pounds. Prob. 916. Find the horse-power that can be transmitted by a cast- iron shaft 3 inches in diameter when making 10 revolutions per min- ute, the value of S not to exceed i 200 pounds per square inch. ART. 92. SOLID AND HOLLOW SHAFTS For solid round shafts of diameter d, the values of / and c are ?V^ 4 and and hence the unit-deforms 234 TORSION OF SHAFTS CHAP, x tion is c$/l. Accordingly F=Sl/c^, and replacing S by its value Pp . c/J there is obtained, F=Pp.l/J(j> or = Pp.l/FJ (93) The first of these equations may be used to compute values of the shearing modulus F from observed values of , while the second is for the determination of (j> when F is given. In order to compute the shearing modulus of elasticity from an observed value of the angle of twist the arc < should be replaced by 7r By taking several corresponding values of Pp and 0o within the elastic limit, a good determination of F can be made. For example, the cast-iron specimen shown in Fig. 169c was 10 inches long and 0.83 inches in diameter, and it twisted through an angle of 1.3 degrees under a twisting moment of 600 inch-pounds; here 7=0.0466 inches 4 , and the formula gives JP = 5 670000 pounds per square inch. In this manner the average values of F have been found to be about 6000000 pounds per square inch for cast iron and about 12 ooo ooo pounds per square inch for steel. The angle of twist which occurs when a shaft is under stress in the transmission of power may be determined in a similar manner. Let H horse-power be transmitted when the shaft makes n revolutions per minute ; then Pp = 63 o$oH/n in English measures (Art. 91). Let is to be replaced by 7r= -j.uPp . l/Fd* and hence the angle of twist of a round shaft is 43 percent greater than for a square one. These values of are in radians; when the angle is desired in degrees, it should be remembered that one radian is equivalent to 57.3 degrees. All the formulas of this article are valid only when the great- est shearing unit-stress does not exceed the elastic limit of the material. The formula S = %Pp/mn 2 may, however, be used to compute the so-called torsional modulus of rupture or torsional strength when a rectangular bar is ruptured by torsion, n being the short side of the rectangle. When rupture occurs in the twisting of such a bar, it usually begins at the middle of the flat side, so that even in this extreme case there is little shearing stress on the corners. Cracks occurring on the corners are to be attrib- uted to tensile stresses which accompany the elongation due to the change of a straight line into a helix. For example, the square steel bar in Fig. 169c was nf inches long between the jaws of the torsion machine, and this length was not increased by the twist of 900 degrees. The side of the square being 0.75 inches, the length of its diagonal is 1.06 inches, the length of a circum- ference of 900 degrees described by the corner is 8.33 inches, and the length of the helix is 14.40 inches; hence the increase in length of the corner line was 2.65 inches and the percentage of elongation was nearly 23 percent, which is less than the ultimate elongation. The specimen broke by shearing at one end under a twisting moment of 5 850 inch-pounds, so that the computed torsional strength of the steel is =9X5 850/2X0.753 =62 400 pounds per square inch. 260 TORSION OF SHAFTS CHAP. X Prob. 99a. Check the computations in the last paragraph. Prob. 996. Compare the strength of a round shaft with that of a square one having the same area of cross-section. Prob. 99c. A square wooden shaft for a water wheel is 12 inches square and transmits 36 horse-powers at 9 revolutions per minute. Compute its factor of safety. Prob. 99d. The rectangular bar of medium steel in Fig. 169c was twisted through an angle of 28.5 degrees by a twisting moment of 2 800 inch-pounds, the length of the bar being 8 inches, its thickness \ inch, and its width i inches. Compute the shearing unit-stress for these data, and determine whether or not the shearing elastic limit of the material was exceeded. ART. 100 STRESSES DUE TO TEMPERATURE 251 CHAPTER XI APPARENT COMBINED STRESSES ART. 100. STRESSES DUE TO TEMPERATURE Several axial loads produce the same unit-stress in a bar as a single load equal to their algebraic sum and the change of length which is that due to this load. When the bar is thus stressed at a certain temperature, a change in temperature usually causes the existing stress to become greater or less. When a bar is free to expand or contract under a rise or fall of temperature, there occurs a change of length which is unaccompanied by internal stress, for in this case there is no external force and stress is an internal resistance to an applied external force. If this change of length is, however, prevented from occurring by fast- ening the ends of the bar, there is produced an internal stress which is the same as that which would be caused by an external force which would shorten or lengthen the free bar the same amount that it has expanded or contracted. For example, let a free steel bar 100 inches long become 99.9 inches long under a certain fall of temperature, then no internal stress is caused by this change in length ; to bring this bar back to its original length, there must be effected the unit-elongation 0.1/100=0.001 and Art. 10 shows that this will require a tension of o.ooi X 30 ooo ooo = 30 coo pounds per square inch. Hence, if this bar is prevented from shortening under the given fall of temperature, a tensile unit-stress of 30 ooo pounds per square inch is produced in every cross-section. Let s be the change per unit of length which occurs when a bar is free to expand or contract, TJ the coefficient of linear ex- pansion or change per unit of length for a rise or fall of one degree Fahrenheit, and / the number of degrees of rise or fall; also let 5 be the unit-stress which will occur if the bar is prevented from expanding or contracting, and E be the modulus of elasticity of 252 APPARENT COMBINED STRESSES CHAP, xi the material. Then from the preceding paragraph and from Art. 9, = rjt S = sE S=f)tE (100) It is thus seen that the unit-stress due to change of temperature is independent of the length of the bar; if a is the section area of the bar, the total stress in each section due to the change in temperature is aS. The following are average values of the coefficients of linear expansion for one degree of the Fahrenheit scale : for brick and stone, J? =0.000 0050 for cast iron, y =0.000 0062 for wrought iron, T? =0.000 0067 for steel, jj =0.000 0065 From these coefficients the change per unit of length due to a rise or fall of / degrees is readily computed or the unit-stress 5 may be directly found. This temperature stress is to be added to or subtracted from the tensile or compressive stress due to the applied forces on the bar. As an example consider a wrought-iron tie rod 20 feet in length and 2 inches in diameter which is screwed up to a ten- sion of 9 coo pounds in order to tie together two walls of a build- ing. Let it be required to find the stress in the rod when the temperature falls 10 degrees Fahrenheit. Here, 5=0.0000067X10X25 000000=1 675 pounds and the stress due to change of temperature is 3.14X1675 = 5 200 pounds, so that the total tensile stress in the bar becomes 9 000 + 5 200 = 14 200 pounds. For a rise of 10 degrees Fahren- heit, the tensile stress in the bar becomes 9 ooo 5 200 = 3 800 pounds. It is seen from the above that the unit-stress caused in a steel bar by a change of one degree Fahrenheit is about 200 pounds per square inch, so that a change of 100 degrees might cause a stress of 20000 pounds per square inch if no provision were made for allowing the bar to change its length. Steel bridges usually rest on rollers at one end so that change in length may occur under change of temperature and thus stresses due to ART. 101 BEAMS UNDER AXIAL FORCES 253 temperature be prevented. When railroad rails are laid in cold weather, it is customary to leave the ends about \ inch apart, so that there may be room for expansion when the warm weather comes; the holes in webs of the rails, through which bolts pass to connect the splice bars, are made oval instead of rpund so that the rails may be free to expand and contract. Prob. 100. What is the change in length of a steel railroad rail 60 feet long when the temperature rises from 10 to +80 degrees Fah- renheit ? If the rail weighs 95 pounds per yard, what force is required to prevent this expansion, and what compressive unit-stress will it cause in the rail? ART. 101. BEAMS UNDER AXIAL FORCES A normal stress is one acting normally to the section area of a bar, and this must be either tensile or compressive (Art.l). When several applied axial forces act upon a bar each produces a stress on the section area and the sum of these stresses must equal the total load. Hence the combination of normal stresses is made by simple addition, if all are tensile or all are compressive; when some are tensile and others compressive, their algebraic sum is to be taken. A beam under transverse loads has normal stresses of tension on one side of its neutral surface and normal stresses of com- pression on the other side (Art. 39). When the beam is under an axial tension P which is uniformly distributed over the section area a, the unit-stress P/a is to be added to each of the flexural tensile unit-stresses and be subtracted from each of the flexural compressive unit-stresses. Thus, if the unit-stresses due to the flexure on the tensile and compressive sides of the beam are Si and S 2 , then Si + P/a is the tensile unit-stress due to flexure and longitudinal tension, while S^ P/a is the compressive unit-stress due to flexure and axial tension. An approximate method of finding S\ and 5 2 is by mean of the flexure formula (41), which is. applied to the transverse loads just as if the axial tension were not acting. For example, let it be required to find the factor of safety of a 1 2-inch I beam 254 APPARENT COMBINED STRESSES CHAP. XI of 6 feet span, weighing 55 pounds per linear foot, which carries a uniform load of i 200 pounds besides its own weight, when sub- ject to an axial tension of 80 ooo pounds. The flexure formula is Si . I/c=M; from Table 6 the section factor I/c is 53.5 inches 3 ; from Art. 38 the bending moment M is 1530X6/8 = 1147.5 pound-feet; hence the tensile unit-stress on the lower side of the simple beam is .$1 = 257 pounds per square inch. Table 6 gives the section area a = i6.i8 square inches, and hence the unit-stress due to the axial tension is P/a =80 000/16.18=4 940 pounds per square inch. Hence on the lower side of the beam, the total tensile stress is 260+4940 = 5 200 pounds per square inch, and the factor of safety is 60 000/5 2O = J J i- On the upper side of the beam, the stress is tensile, since P/a is greater than S 2 > and its value is 4940 260=4680 pounds per square inch. When the axial force on the beam is compression, a similar approximate method may be followed, the compressive unit- stress Si on the concave side being found from the flexure formula, while the unit-stress due to the load is found from P/a, if the beam is short, or from the column formula (80) if its length exceeds ten or twelve times its least thickness. A rafter of a roof is a case of combined compression and flexure, for a rafter is under compression from the forces that act upon its ends and under flexure from its weight and that of the roof covering. In many cases the approximate method here outlined is sufficient for its investigation. Let the section of the rafter be rectangular, b being its width, d its depth, / the length, w the uniform load per linear unit, and

a~ U The total compressive unit-stress on the upper fiber hence is, By the usual method this is found to be a maximum when and substituting this, the maximum unit-stress is, _ 3 wl 2 cos0 wl cosec0 w sin0 tan0 4 bd 2 2bd i2b which formula may be used to investigate or to design common rafters subject to uniform loads. In any inclined rafter, let P denote all the load above a sec- tion distant x from the upper end. Then reasoning as before the greatest unit-stress for that section is found to be, _ Me P s'mcf) H cos< *'~T~'-T" ~^~ from which 5 Z may be computed for any given case. Prob. 101 a. Find the size of a square wooden simple beam of 12 feet span to carry a load of 300 pounds at the middle when it is also subject to a longitudinal tension of 2 ooo pounds, the allowable tensile stress being i ooo pounds per square inch. Prob. 1016. A roof with two equal rafters is 40 feet in span and 15 feet in height. The wooden rafters are 4 inches wide and each carries a load ot 450 pounds at the middle. Find the depth of the rafter so that 5" may be 700 pounds per square inch. ART. 102. FLEXURE AND COMPRESSION Let a beam be subject to flexure by transverse loads and also to an axial compression in the direction of its length. If the longitudinal compression is not large, the combined maximum stress due to flexure and compression may be computed by the approximate method of Art. 101. It is clear, however, that if the compression is large the deflection of the beam will be 256 APPARENT COMBINED STRESSES CHAP. XI increased by it, and hence the effective bending moment and maximum fiber stresses will be greater than given by that method. A closer approximation will now be established. Let P be the axial compressive force and M the bending moment of the flexural forces. Let M\ be the actual bending moment for the dangerous section where the actual deflection is /i ; this is greater than M, on account of the moment Pfi of the force P, or M\=M + Pf\. Now the maximum fiber unit-stress S\ which results from this moment M\ is, from (41), S l =M l .c/I=(M+Pf l }c/I where I is the moment of inertia of the cross-section and c the distance from the neutral axis to the remotest fiber on the com- pressive side. The value of f\ may be expressed in terms of S\ r regarding /i to vary with S\ in the same manner as for a beam subject to no axial compression. Inserting then for f\ its value from (56), and solving for Si, gives, m) (I02) where a and ft are numbers that depend upon the arrangement of the ends and the kind of loading of the beam; for a simple beam uniformly loaded the value of ft /a is 9.6; for a simple beam with load at the middle ft /a is 12. The maximum compressive unit-stress on the concave side of the beam is S=S\+P/a. For example, let a simple wooden beam 8 feet long, 10 inches wide, and 9 inches deep be under an axial compression of 40 ooo pounds, while at the same time it carries a total uniform load of 4000 pounds. Here M = faWl = 48000 pound-inches, 0=4$ inches, I =-fabd?=6o r ]% inches 4 , / = 96 inches, P = 40 ooo pounds, a = 8, ft = 1 f 1 , and E = i 500 ooo pounds per square inch. Inserting these values in the formula, the value of Mc/I is 356, and then the final flexural stress S\ is found to be 371 pounds per square inch. The compressive unit-stress due directly to P is P/a= 40 000/90=444, so that the total stress 5 = 371 +444 =815 pounds per square inch. Another method, which has a more satisfactory theoretical basis, is to consider the flexural stress due to P as resulting from ART. 102 FLEXURE AND COMPRESSION 257 its eccentricity with respect to the section at the middle of the beam. Under the action of its own weight a simple beam has the deflection /; this is increased by the action of the eccentric load, and from Art. 87, the total deflection is /sec0 where Q denotes the quantity ^(P1 2 /EI)^ for a beam with supported ends. As before, the total flexural unit-stress Si is given by (M + Pfi)c/I, in which /i is to be replaced by/ sec/9 ; also replacing I by ar 2 , where r is the radius of gyration of the section in the plane of bending, the total flexural unit-stress S\ at the middle of the concave side of the beam is, S^M+Ltfsetf (102)' in which 6 has the values given in Art. 87 for different arrange- ments of the ends of the beam ; sec# can be found from a trigono- metric table or from the series in Art. 87. Fig. 102a Fig. 1026 To illustrate this method, let the data of the above numerical example be again used. The value of Me/ 1 is 356 pounds per square inch and this is the flexural unit-stress due to the uniform load alone. The last term of the second member gives the flexural unit-stress due to the moment of P; here =4.5 inches, from Art. 55 the- deflection due to W is /= $WP/$4EI = 0.0506 inches, r 2 =607.5/90 =6.75 inches 2 , = (P/ 2 /E/)* =0.3184 radians = 18 15', sec0 = 1.053, anc * lastly (P/a)(c//r 2 ) sec# = i6 pounds per square inch. The total flexural stress S\ is then 356 + 16 = 372 and total compressive stress 5 is 372+444 = 816 pounds per square inch, which is practically the same as that previously found. The two methods give, in fact, closely the same results for the common cases which occur in practice. While the above methods are satisfactory in regard to numerical results, a more exact method of dealing with combined flexure and compression is by help of the elastic curve. For the common case of a simple beam loaded uniformly with w per linear unit 258 APPARENT COMBINED STRESSES CHAP. XI and under the longitudinal compression P, the bending moment for any section distant x from the left end of the beam is \ivlx $wx 2 +Py, and the differential equation of the elastic curve of the beam is, where the negative sign of the moment is used because of the theoretic requirement that y and the second derivative must have opposite signs when the curve is concave to the axis of x. By two integrations there is found, ivlx WX 2 WEI /COS (2X 1)6/1 \ y ~~^P + ~7F + ~P 2 ~( costf V in which, as before, 6 is an abbreviation for \(PP/EI}*. In this equation of "the elastic curve, let x = \l, then y =/i and then, which gives the deflection of the beam due to both the uniform load and the longitudinal compression. Inserting the value of the deflection f\ in the expression for S\ at the beginning of this article, there is found, Si = (wcE/P) (seed - 1) (102)" as the flexural unit-stress at the middle of the concave side. To illustrate this method let the data of the above numerical example be again used. Here w= 4 000/96 pounds per linear inch, and the other quantities as before, also sec0 = 1.0530; and then (102)" gives $'1=373, whence 5=373+444 = 817 pounds per square inch, which is practically the same as found by the other methods. The rough method of Art. 101 gives 5 =800 pounds per square inch, and in many cases this method may be used to obtain results which are sufficiently precise. When w=o in formula (102)", the case is that of a column under the axial load PI and both f\ and Si are zero when P is less than the value given by Euler's formula (Art. 78), and inde- terminate when P reaches that value. On the other hand, when P=o, the case is that of a simple beam uniformly loaded, and it may be shown that the above formula for f\ will reduce to 5W/ 4 /384-E7, while that for S\ will reduce to \-wPcfI. ART. 103 FLEXURE AND TENSION 259 Prob. 102a. Prove, by using the method of the differential calculus for evaluating indeterminate quantities, that the statement in the last sentence is correct. Prob. 1026. A wooden cantilever beam, 10 inches wide and 4 feet long, carries a uniform load of 500 pounds per foot and is subjected to a longitudinal compression of 40 ooo pounds. Find the depth of the beam so that the maximum compressive unit-stress may be about 800 pounds per square inch. ART. 103. FLEXURE AND TENSION Let a beam be subject to flexure by transverse loads and then to a tension in the direction of its length. The effect of the ten- sion is to decrease the deflection from / to f\, and thus also the tensile flexural stress. If M is the bending moment of the trans- verse loads, and MI that of the combined flexure and tension, then Mi=M Pfi. Let Si be the resulting flexural unit-stress on the fiber most remote from the neutral surface on the tensile side; then formula (102) gives Si, if the minus sign in the denomi- nator is changed to plus. Accordingly, and Si+P/a is the total unit-stress on the convex side of the beam resulting from the combined flexure and tension. As an example, take a steel eye-bar 18 feet long, i inch thick, and 8 inches deep, under a longitudinal tension of 80 coo pounds, E being 29 ooo ooo pounds per square inch. The weight of the bar is 490 pounds, and M = |X49oXi8Xi2 = 13 230 pound- inches. Also c=4 inches, 7 = 42.67 inches 4 , /?/=9-6, P = 80 ooo pounds, / = 2i6 inches. Then the value of Me /I is i 240, and the flexural tensile stress Si is 943 pounds per square inch. Finally, the total tensile Stress on the convex side at the middle of the beam is 5 = 943 + 10000 = 10943 pounds per square inch. Formula (102)' also applies to combined flexure and tension by changing the sign of P from plus to minus. Here 6 becomes imaginary and the circular secant becomes the hyperbolic secan* 260 APPARENT COMBINED STRESSES CHAP. XI which is designated by sech 6. Then, cW? (103)' in which sech# may be computed from 2/(e e + e~ e ), where e is the base of the Naperian system of logarithms and 6 denotes the real positive number %(Pl 2 / El}** . For instance, let the data of the last paragraph be again considered. The value of Mc/I is i 240' pounds per square inch, which is the flexural unit-stress due to uniform load alone. Also c=4 inches, r 2 = 42.67/8 = 5.334 inches, f=s wl3 /3 8 4 EI = -5 2 inches, 6 = %(P1 2 /EI)* = 0.868, e 9 = 2.7i8- 868 = 2.383, e~ = i/e e =0.420, sechtf = 2/2.803 = 0.714; then P /a = 10 ooo pounds per square inch, c//r 2 = 0.039, and (P/a)(cf/r 2 ) sech6 = 2j8 pounds per square inch, which is the flexural stress due to the moment of P. Lastly, the total tensile stress on the convex side of the middle of the eye-bar is 5 = i 240 -278 + 10 ooo = 10 960 pounds per square inch. Formula (102)" also applies to a beam uniformly loaded and under the tension P by reversing the sign of P, and thus, Si = (wcE/P)(i-sech0) (103)" where 6 is the number ^(PP/EI}* when the ends of the beam are supported. For the above eye -bar, sech#=o.7i4; also w = 490/216 pounds per linear inch. Then the formula gives Si = 941 for the flexural unit-stress, so that the total compressive unit- stress is 5 =941 + 10 ooo = 10 941 pounds per square inch. The three methods hence give numerical results which are essentially the same for all practical purposes, but the first one is the most convenient in computation and hence is generally preferable. Formula (103) applies to all kinds of loading and to all arrangements of ends, as also does (103)'; but (103)" applies only to uniform load and for this case it is theoretically mere correct than the other formulas. Since many students will here meet with hyperbolic functions for the first time, it may be explained that they are closely anal- ogous to circular trigonometric functions. For circular functions cos 2 + sin 2 = i , but for hyperbolic functions cosh 2 /? sinh 2 = i. ARF. 104 ECCENTRIC AXIAL FORCES ON BEAMS 261 The value of cos# is \(e* a + e~ io ) , in which e is the Naperian base 2.71828 and /' is the square root of i; the value of cosh0 is given by the simpler expression \(e e + e~ 6 }. The reciprocal of cos/9 is sec0 and that of qoshfl is sech/9. Hyperbolic functions are of great importance in the theory of electricity and in other dis- cussions of applied mechanics; a table of values of such func- tions may be found in McMahon's Hyperbolic Functions (Mathematical Monograph No. 4, New York, 1906). Prob. 103a. A wooden cantilever beam, 3X4X36 inches, has a load of 650 pounds at the end and is under the longitudinal compres- sion of 4 500 pounds. Compute the maximum compressive unit-stress. . Prob. 1036. When the above cantilever beam is under the longitu- dinal tension of 4 500 pounds, compute the maximum unit-stress due to it and the load of 650 pounds at the end. ART. 104. ECCENTRIC AXIAL FORCES ON BEAMS In the three preceding articles the axial forces applied to the beam have been supposed to act at the centers of gravity of the end sections, so that the stresses due to them would be uniformly distributed over every section area were it not for the deflection of the beam. Sometimes, however, these axial forces are applied eccentrically at the ends, as shown in the following figures, Fig. 104a representing a compression applied through pins which also serve as supports for the beam, and Fig. 104& representing a ten- sion applied in a similar way. In the first figure the longitudinal compressive forces P are applied below the centers of gravity of the end sections, this being done in order that the moment of P may tend to decrease the deflection of the beam instead of in- creasing it as is the case when they are applied axially at the ends. It is required to find the amount of this eccentricity so that the unit-stress 5 at the middle of the beam shall be uniform and equal to P/a over the entire cross-section. When the stress is uniform over the section at the middle of the beam, there can be no flexural stresses in that section and hence no bending moment. To insure this condition, it may be considered, as an approximation, that the moment of P should 262 APPARENT COMBINED STRESSES CHAP. XI be equal to the moment M of the transverse loads. Let p be the distance of P below the center of gravity of the end sections in Fig. 104a or above it in Fig. 1046. Then when Pp=M there is no bending moment at the middle; accordingly the required eccentricity is p=M/P. For example, take the steel upper chord of a bridge which has a length of 30 feet between the pins at its ends. The section is made up of two channels and a plate, as in Fig. 76c, the section area being 20.5 square inches and its moment of inertia 742 inches 4 . This chord is subject to a longitudinal compression of 1 68 ooo pounds, and it is required to find the distance p below the neutral axis at the ends where the centers of the pins should be located. The weight of the beam is 2 090 pounds, and, taking it as supported at the ends, the moment due to its weight is M = %Wl = 94000 pound-inches. Then the centers of the pins must be at the distance p = 94 000/168 0x20=0.56 inches below the axis of the chord in order that no flexural stresses may exist. The compressive stress over the middle section is then uniform and equal to 168 000/20.5 =8 200 pounds per square inch. Fig. 104a Fig. 1046 The above method is not exact, because it takes no account of the stiffness of the beam and gives the same results for all beams of the same weight. A better method may be derived by considering the beam to have the deflection / before the eccentric load P is applied. Then for compression p must be greater than / and the moment P(pf] should equal M; for tension the moment P(p+f] is to be equal to M. Accordingly, the first of which applies to compression and the second to ten- sion. Hence values of p computed from these formulas will be greater for compression and less for tension than those found from the preceding method. ART. 105 SHEAR AND AXIAL STRESS 263 Using the data of the above chord member while it is under the longitudinal compression of 168 ooo pounds, the deflection due to its own weight is f=sWl 3 /^84EI = o.o^j inches, and M/P = o.$6 inches as before; then the required eccentricity is p = 0.56 + 0.06 =0.62 inches. This same section might serve for a lower chord under a tension of 168 ooo pounds, in which case the second formula given p=o.$6 0.06 =0.50 inches for the eccentricity. These values are more reliable than the eccen- tricity 0.56 inches which preceding method gives for both com- pression and tension. Prob. 104a. Compute the eccentricity p that is required for the wooden beam which is discussed in Art. 102. Prob. 1046. Compute the eccentricity p that is required for the eye- bar of Art. 103 in order that there may be little or no flexural stress at the middle. ART. 105. SHEAR AND AXIAL STRESS Let a bar having the section area a be subjected to the longi- tudinal tension or compression P, and at the same time to a shear V at right angles to its length. The axial unit-stress on the section area is P/a, which may be designated by S, and the shearing unit-stress is V/a, which may be denoted by 5 8 . It is required to find the maximum unit-stresses produced by the combination of 5 and S 8 . In the following demonstration S will be regarded as a tensile unit-stress, although the reasoning and conclusions apply equally well when it is compressive. Consider an elementary cubic particle, with edges one unit in length, acted upon by the horizontal normal unit-stress S and by the vertical shearing unit-stresses S t and S t , as shown in Fig. 105a. These forces are not in equilibrium unless a horizontal couple be applied as in the figure, each of the forces of this couple being equal to S a . Therefore at every point of a body under vertical shear, there exists a horizontal shearing unit-stress equal to the vertical shearing unit-stress. Heretofore only one of these shearing stresses has been noted, namely, that which is parallel to the applied external shear, but it is now seen that this is always 264 APPARENT COMBINED STRESSES CHAP, xi accompanied by another shearing stress. For example, at any point in a beam where there is a vertical shearing unit-stress V/a, there is also found a horizontal shearing unit-stress of the same intensity. Similarly, Fig. 90 shows the shearing stresses normal to the radius of a shaft under torsion, but there are also shearing stresses parallel to the radius which have the opposite direction of rotation. S.Sx Fig. 105o Fig. 1056 Let the parallelopipedal element in Fig. 1056 have the length dx, the height dy, the diagonal 8z, and a width of unity normal to the plane of the paper. The tensile force 5 . dy tends to pull it apart longitudinally. The vertical shear S 8 . dy tends to cause rotation and this is resisted by the horizontal shear S s . ox. These forces may be resolved into components normal and parallel to the diagonal 2, as shown in the figure. The components normal to the diagonal form normal tensile force S n - dz, and those parallel to the diagonal form a shearing force S p . dz, where S n and S p are the normal tensile and the shearing unit-stresses upon and along the diagonal. It is required to find the maximum values of S n and S p due to the given unit-stresses 5 and 5 8 . Let denote the angle between dx and dz. Resolving each of the given forces in directions perpendicular and parallel to the diagonal, and taking their sum, there results,. S n dz =S . dy . sin^> + S g . dx . sin0 + S s . dy . cos S p dz=S . dy . cos^+5. . dx cos and dy = dz . sintf>, these equations reduce to Sn = %S(l COS2(f))+Sg Sin2< By differentiating each of these with respect to and equating each derivative to zero, it is found that, ART. 105 SHEAR AND AXIAL STRESS 265 S n is a maximum or minimum when cot2<= S p is a maximum when tan2< = + %S'/S S Expressing cos2< and shi2< in terms of cot2< and tan2< and inserting their values in the expressions for S n and S p , the follow- ing important results are obtained: max S n = iS(S 8 2 +QS) 2 )* max S P = (S. Z +QSW (105) These formulas apply when S is either tension or compression. When S is tension the plus sign before the radical is to be used to find the maximum tensile unit-stress S n , while the minus sign gives the maximum compressive unit-stress S n - For example, take a bolt one inch in diameter which is sub- ject to a longitudinal tension of 5 coo pounds and at the same time to a cross-shear of 3 ooo pounds. Here 5 = 6 366 pounds per square inch and S, = 3 820 pounds per square inch. Then S n =+8i55 pounds per square inch, and S n =i 790 pounds per square inch for the minimum tensile or maximum compres- sive stress, while 5 p = 49yo is the maximum shearing unit-stress; these are the greatest normal and shearing stresses due to the combination of 5 and 5. The directions which these maximum stresses make with the axis of the bolt are found by using the values of cot2$ and tan2 deduced above. For S n the value of cot2< is 3 183/3 820= 0.833, whence ^ = 64 53' or ^ = 154 53', the former being the inclination of the plane against which the maximum tensile stress acts and the latter being its inclination for the maximum compressive stress; these two planes are at right angles to each other. For S p the value of tan20 is +0.833, whence ^=19 53' or ^ = 109 53', these being the directions of the planes along which the maximum shearing stresses act; these directions bisect those of the planes upon which the normal stresses are the greatest. When S, equals zero the case is that of simple tension or compression, and max S n = S, while max S P = $S as previously shown in Art. 6. Here cot20 = o or < = o; also tan2< = oo, and < = 45 or ^ = 135 so that the maximum shearing stresses make angles of 45 with the direction of the axial tension or compression. 2G6 APPARENT COMBINED STRESSES CHAP, xi Prob. 105. A bolt f inch in diameter is subjected to a tension of 2 ooo pounds and at the same time to a cross shear of 3 ooo pounds. Find the maximum tensile and shearing unit-stresses, and the direc- tions they make with the axis of the bolt. ART. 106. COMBINED FLEXURE AND TORSION This case occurs when a horizontal shaft for the transmission of power is loaded with weights. Let S be the greatest flexural unit-stress computed from (41) and S t the torsional shearing unit-stress computed from (90) or by the special equations of Arts. 91 and 92. Then, according to the last article, the resultant maximum unit-stresses are, max S n = $S +VS 2 +($S) 2 max S p =\/S 2 +(^S) 2 the first of which gives the greatest tensile or compressive unit- stress on the lower or upper surface of the shaft, while the second gives the greatest shearing unit-stress. For wrought iron or steel it is usually necessary to regard only the first of these unit- stresses, but for timber the second should also be kept in view. For example, let it be required to find the maximum unit- stresses for a horizontal steel shaft, 3 inches in diameter and 12 feet between bearings, which transmits 40 horse-power while making 120 revolutions per minute, and upon which a load of 800 pounds is brought by a belt and pulley at the middle. Taking the shaft as fixed over the bearings, the flexure formula (41) gives for the unit-stress of tension or compression, S=M . c/I=^Pl/n 400 pounds per square inch From Art. 92, the shearing unit-stress on the surface is, 5 a =32i oooH/nd 3 =4 ooo pounds per square inch The maximum tensile unit-stress on the lower surface at the middle of the shaft or on its upper surface in the bearings now is S n = 2 700+ v / 40oo 2 +27oo 2 = 7 600 pounds per square inch and the maximum compressive unit-stress on the upper side of the shaft has the same value; this is 41 percent greater than that due to pure flexure. The maximum shearing unit-stress is ART. 106 COMBINED FLEXURE AND TORSION 267 4 900 pounds per square inch, which is 22 percent greater than that due to pure torsion. It is thus seen that the actual maximum unit-stresses in a shaft due to combined flexure and torsion are much higher than those derived from the formulas for flexure or torsion alone. In determining the diameter d of a shaft, it is hence necessary to take S n as the allowable tensile or compressive stress and S p . as the allowable shearing unit-stress. For a round shaft of diameter d, the expression for 5 under any transverse load is Mc/I = 32M/r:d 3 (Arts. 41, 42, 43), while that for S s is Pp . c/J = i6Pp/nd 3 (Arts. 90, 92). Inserting these in formula (105) and solving for d 3 , there is found (106) in which M is the bending moment of the loads and Pp is the twisting moment due to the applied twisting forces. When H horse-powers are transmitted at a speed of n revolutions per minute, the value of Pp is given by (91). These formulas apply only to solid round shafts; since the allowable value of S P is always less than that of S n , it may often happen that the second formula will give a larger diameter than the first. As an example let it be required to find the diameter of a horizontal steel shaft to transmit 90 horse-power at 250 revolu- tions per minute, when the distance between bearings is 8 feet and there is a load of 480 pounds at the middle, the allowable unit-stresses S n for flexure being 7 ooo and that for shear being 5 ooo pounds per square inch. Here the bending moment M = 1X480X96 = 5 760 pound-inches, and the twisting moment Pp = 63030X90/250 = 22690 pound-inches. Then using the first formula, the diameter d is found to be 2.8 inches, while from the second formula it is 2.9 inches; hence the shaft should be about 3 inches in diameter. Formula (106) may also be used for the computation of the maximum working unit-stresses S n and S p when the shaft is round and hollow. For a hollow shaft which has the outer diam- eter di and the inner diameter d 2 , the formula will also apply if ( = F supposes further that the vertical shear is uniformly distributed over the cross-section of the beam. A closer analysis will show that a horizontal shear exists also and that this, together with the vertical shear, varies in intensity from the neutral surface to the upper and lower sides of the beam. It is well known that a pile of boards which acts like a beam de- flects more than a solid timber of the same depth, and this is due to the lack of horizontal resistance between the layers. The common theory of flexure in neglecting the horizontal shear gen- erally errs on the side of safety. In some experiments, however, beams have been known to crack along the neutral surface, and it is hence desirable to investigate the effect of horizontal shear in tending to cause rupture of that kind. That a horizontal shear exists simultaneously with the vertical shear is evident from the considerations in Art. 105. 270 APPARENT COMBINED STRESSES CHAP xi Let Fig. 108 represent a portion of a beam of uniform section. Let a notch nmpq be imagined to be cut into it, and let forces be applied to it to preserve the equilibrium. Let H be the sum of all the horizontal components of these forces acting on mn and H' the sum of those acting on qp. Now H ' is greater or less than H, hence the difference H' H must act along mq as a horizontal shear. Let the distance mq be ox, the thickness mm' be /, and the area mqq'm' be at a distance z above the neutral surface. Let c be the distance from that neutral surface to the remotest fiber where the unit-stress is 5. Let da be the section area of any fiber. Let M be the bending moment at the section mn and M' that at qp. Then, S/c = unit-stress at distance unity from neutral surface S . z/c= unit-stress at distance z from neutral surface da . S . z/c= stress on fiber da at distance z from neutral surface c (S/c)jaJasum of horizontal stresses between mm' and nn' t Now from the flexure formula (41), S/c = M/I for the section mn and also S/c = M'/I for the section pq, where M and M' are the bending moments, and / is the moment of inertia of the entire cross-section. Accordingly, and hence the horizontal shear along mq is expressed by, H'-H = (M'- M} I c g da . z/I Now, the distance mq being dx, the difference M' M is dM. Also if S h is the horizontal shearing unit-stress on the area / . dx, the value of H' H is S^tdx. Again from Art. 47 it is known that dM/dx is the vertical shear V. Therefore, S k = (V/It}2 c da.z (108) is a general formula for the horizontal shearing unit-stress at the distance z from the neutral surface in any section of a beam where the vertical shear is V. ART. 108 HORIZONTAL SHEAR IN BEAMS 271 This expression shows that the horizontal shearing unit-stress is greatest at the supports, and zero at the dangerous section where V is zero. The summation expression I c z da . z is the statical moment of the area mm'nn' with reference to the neutral axis; it is zero when z=c, and a maximum when 2=0. Hence the longitudinal unit-shear is zero at the upper and lower sides of the beam and is a maximum at the neutral surface. Formula (108) applies to any form of section, / being its width at the dis- tance z from the neutral axis, and / the moment of inertia of the whole section with respect to that axis. Since the vertical shear- ing unit-stress at any point is equal to 5^, its value at the neutral surface is, y y S,=j t 2 c Q da . z or S,=j f aiCi in which t is the width of the section at the neutral axis, a\ is the area of the part of the section on one side of the neutral axis, and Ci is the distance of the center of gravity of that area, from that axis. This formula gives the maximum shearing unit-stress and it is always greater than the mean which heretofore has been found by dividing V by the section area a. For a rectangular beam of breadth b and depth d, the value of / is b, and that of / is ^bcP, while the statical moment a^i is \bd . \d. By inserting these in the formula, there results 5 f ""fF/Wf. V/a. Hence the shearing unit-stress along the neutral surface is 50 percent greater than the mean shear V/a. Replacing / in the above formula by ar 2 , where r is the radius of gyration of the section area with respect to the neutral axis, it becomes, y y where a is the number a\c\/tr 2 , by which the mean shear V/a is to be multiplied in order to obtain the maximum S c which acts both horizontally and vertically at the neutral surface. For a circular section of diameter d, the value of / is d, that of r 2 is f^d 2 , that of fli is \xd 2 , and that of q is 2d/yt; accordingly given by cot20= 5/S 8 , because PS have been computed from these formulas, the unit- stresses in the three parts of the bar are found from S\=P\/a,i y S 2 =P 2 /a 2 , S 3 =P 3 /a 3 . These conclusions are not valid unless each of these unit-stresses is less than the elastic limit of the material, because the formulas for change of length apply only under this condition. By substituting either of the values of P in the corresponding expression for change of length, there is found ' (110)" which differs materially from (110). These two formulas have 278 COMPOUND COLUMNS AND BEAMS OBAP. xil a marked analogy with the electric equations which connect loss of voltage with current in wires laid in series and in parallel, and this analogy will be discussed in Art. 185. As a numerical example illustrating the case of Fig. llOa, let the three materials be timber, stone, and steel, so that EI = i 500 ooo, 2 = 6 ooo ooo, 3 = 30 ooo ooo pounds per square inch. Let each section area be a and each length be /. Then the unit-stress in every section area is P/a, and formula (110) gives the total change of length as e = o.ooo ooo 2&gPl/a. For the case of Fig. 1106, let the three materials be also timber, stone, and steel; let each length be / and each section area be ^a. Then formulas (110)' give PI =o.o4P, P 2 = o.i6P, and P 3 =o.8oP, so that the stiffest material carries the greatest load, and the unit-stress in the steel is twenty times that in the timber. From (110)" the change of length is e = o.ooo ooo oSoPl/a, or less than one-third of that of the previous case. These last two columns contain the same amount of each material, but the total change of length is the greatest for the first case, while the unit-stress in the steel is the greatest for the second case. As a numerical example illustrating the second diagram of Fig. 1106, let the central part be of timber, 6x8 inches in sec- tion, and let two steel plates, each inches thick and 8 inches wide, be bolted to the 8-inch sides. Let the length of the short column be 5 feet, and the axial load P be 126 ooo pounds. Here / = 6o inches, #1=03 = 3 square inches, 02=48 square inches, EI =3 = 30 ooo ooo and E 2 = i 500000 pounds per square inch. Then the formulas give PI =Ps=^P = 45 pounds, and P2 = fP = 36 0 pounds. The compressive unit-stress in the steel plates then is 51=45000/3 = 15000, and that in the timber is 52 = 36000/48 = 750 pounds per square inch. In this case the two steel plates carry about 70 percent of the total load, and are so highly stressed that bolts should be placed at frequent intervals in order to prevent lateral buckling. Steel ropes are often made with a hemp core in order to give flexibility, and here also the tensile load is divided between the two materials inversely as their resistances. Thus, if a\ and a 2 ART. in REINFORCED CONCRETE COLUMNS 279 are the section areas of the hemp and steel, and E\ and E 2 their moduluses of elasticity, then Pi/P 2 = aiEi/a 2 E 2 , which shows that the steel takes nearly all the load, since a 2 is usually equal to 6fli, while E 2 is probably more than 100 times as great as E\. Prob. 110. A bar for a jail window has a diameter of 2\ inches, the central core of soft steel being i inches in diameter. When this bar is used under axial tension or compression, show that the unit-stress on the hard and soft steel is the same. ART. 111. REINFORCED CONCRETE COLUMNS A common instance of a compound column is that shown in section in Fig. Ilia, which represents a hollow cylinder of cast iron or steel filled with concrete and used for one of the supports of a bridge. The usual intention is that the concrete shall carry the load, while the metallic cylinder is to prevent the concrete from cracking under the action of the weather or of collisions from floating objects. Owing to the friction between the two materials, however, it is evident that the metal always carries part of the weight, and the highest load that can come upon it is readily found from the method of Art. 110. Let P be the total load on the pier or column, a\ and a 2 the section areas of the concrete and metal, and EI and E 2 the moduluses of elasticity of the same; then (110)' reduce to, are the loads which come on concrete and metal respectively. For example, take a pier where the concrete is 6 feet in diameter, this being surrounded by a cast-iron casing 1.15 inches thick. Using the average values in Table 2, the ratio E 2 /Ei is 6; from Table 16 the area a x is 4071 square inches, while the area a 2 is with sufficient precision ?r X 73X1. 15 = 265 square inches, and hence the ratio ai/a 2 is 15.4. Then the formulas give PI =0.72^ and P 2 = o.28P, so that the cast iron may carry about one-fourth of the load. The unit-stresses in the different materials of a compound 280 COMPOUND COLUMNS AND BEAMS CHAP, xir column are proportional to their moduluses of elasticity, that is, to the stiffnesses of the materials. This is readily seen from (1 10)', or, for the case of two materials, P l l/a l E 1 = P 2 l/a 2 E- 2 or S 2 /S l =E 2 /E l For example, referring to the column of the last paragraph, the value of E 2 /Ei is 5, and hence the unit-stress S 2 in the cast-iron is 5 times the unit-stress Si in the concrete. When bars or rods of metal are placed in a concrete column it is said to be 'reinforced'. Square or rectangular columns are built with vertical steel rods located near the corners, as seen in Fig. 1116. In construction the rods are first put in position, they being connected by heavy horizontal wires in order to keep them in place while the concrete is packed in a wooden form which is built around them. Some of the columns of the Harvard stadium are 14 inches square and have vertical steel rods $ inches in diameter near the corners. Here the section area of the four steel rods is a 2 =0.442 square inches, while that of the concrete is 01 = 196 0.44 = 195.56 square inches, so that the ratio a\/a 2 = 442. The average value of the ratio E 2 /E { is about 10. Accord- ingly, the formulas give for the load on the concrete PI =o.g"j8P and for the load on the steel P 2 =o.o22P. The allowable unit- stress for the concrete was taken as 350 pounds per square inch, hence it follows from the last paragraph that the unit-stress in the steel rods is about 3 500 pounds per square inch. Here the full strength of the metal is not developed, and this is usually the case in reinforced concrete construction. In the above discussion the column is regarded as so short that no account need be taken of the lateral flexure, and this may be safely done until the height of the column exceeds about twelve times its least diameter. Thus a column 14 inches square may be as high as 14 feet before it is necessary to use any of the formulas given for columns in Chapter IX; the slenderness-ratio l/r corresponding to this rule is about 40 for a square column and 48 for a round column. For higher values of l/r, the method of investigation for the concrete part is to find the load PI as above, and then compute the unit-stress S\ from the column ART. Ill REINFORCED CONCRETE COLUMNS 281 formula (80), using for the factor a high number, say about T-gV?F> because sufficient experiments have not been made to determine its proper value. As for the steel, it is everywhere supported by the concrete and can have no lateral flexure except that due to the concrete part, but it will be fair to consider that its unit-stress 52 is increased in the same proportion as Si, due regard being paid to its distance from the axis of the column. For instance, let Si and Sjj as computed for a short prism be 300 and 3 ooo pounds per square inch, let Si on the concave side of the column be found by Rankine's formula to be 500; then S 2 will be 5 coo if the steel is located close to the concave side, but if it be half-way between that side and the axis S2 will be only 3000 + ^X10X209 = 4000 pounds per square inch. In general, the unit-stress in the concrete reaches its allowable limit before the steel receives a stress of one-half that which is permissible. Fig. Ilia Fig. Fig. lllc A concrete post is often made having a steel rod running longitudinally through it at the axis as in Fig. lllc, this being for the purpose of resisting lateral flexure rather than to assist in carrying loads. Concrete piles are made on the same plan as the bridge column above described, the concrete being enclosed in a metal cylinder. The tunnel under the Hudson river, under construction by the Pennsylvania Railroad in 1906, is supported on steel screw piles filled with concrete. Foundation walls of concrete sometimes have vertical steel rods which help to carry a part of the weight and at the same time prevent the concrete from cracking. Reinforced concrete beams are discussed .in Arts. 113-116. Concrete pipes, sewers, arches for buildings and for bridges, are built in which the steel reinforcement plays a more important part than it does in columns. Nearly all of this? 282 COMPOUND COLUMNS AND BEAMS CHAP. XII reinforced concrete steel construction has been developed since the first edition of this book was published. Prob. 111. A concrete pile, as designed for the foundation of a building in New York, was 12X12 inches in section area, and had four vertical steel rods, each i^ inches in diameter and placed at about two inches from the corner. Compute the unit-stresses in con- crete and steel at a depth of 12 feet below the top, due to their own weight and to a load of 30 ooo pounds on the top. ART. 112. FLITCHED BEAMS A 'flitched beam' is one made of timber with metal plates upon its sides, these being held in place by bolts passing through the timber; sometimes, however, a single plate is placed between two timber beams. The following figures show sections of such beams; the third one, which is formed by two channels and a piece of timber, is often used on highway bridges for stringers which support tracks of an electric railway, the rail being fastened to the timber part by spikes or lag bolts. The other forms are sometimes employed in wooden floors. The bolts should in all cases pass through the neutral axis of the section, in order to weaken the beam as little as possible. Fig. 112o Fig. 1126 Fig. 112c When a load is placed upon such a beam it divides itself be- tween the two materials in proportions depending upon their stiffness and section areas. Whether the load be concentrated or uniform it may be expressed by W, and this will divide into two parts, Wi being that carried by the timber and W 2 that carried by the metal. Since the two materials are fastened together the deflection of each is the same. These conditions enable the values of W\ and W 2 to be determined in a manner similar to that used for the compound column in Art. 110. The length of the metal ART. 112 FLITCHED BEAMS 283 and timber will be taken the same, each being equal to the length / of the span of the beam. The modulus of elasticity of the tim- ber will be denoted by EI and that of the steel by E 2 ; the moment of inertia of the cross-section of the timber is I\ and that of the metal is I 2 . The deflection / of a beam may be expressed, as in Art. 56, by WP/pEI, where /? is a number depending upon the kind of loading and the arrangement of the ends. From the above con- ditions, W l + W 2 = W Wy 3 //?i/i = W 2 l 3 /pE 2 I 2 and the solution of these equations gives the values, which are seen to be the same as (111) except that a\ and a 2 are replaced by I\ and 7 2 . After the values of W\ and W 2 have been computed, the unit-stresses in the timber and steel may be inves- tigated by the method of Art. 48. For example, let a flitched timber beam like Fig. 112a be 8X 12 inches in section and each of the steel plates be %Xg inches. For the timber /i = ^X8X i2 3 and for the steel I 2 =^ 2 X i Xp 3 , whence the ratio I\/I 2 = 512/27; also the ratio E 2 /E\=2o. Then (112) gives W\ = 0.487^ and ^2 = 0.513^, so that the parts of the load carried by timber and steel are closely equal. Let the length of this simple beam be 15 feet and the total uniform load on it be W= 16000 pounds, so that Wi = j8oo and ^2 = 8200 pounds. From the flexure formula (41) the unit-stress on the upper and lower sides of the timber at the middle of the span is Si = M\CI/!I = Wilci/SIi =914 pounds per square inch; also the unit-stress on the upper and lower sides of the steel at the same section is 2 = M 2 c 2 /I 2 = W 2 lc 2 /8I 2 = i3joo pounds per square inch. These unit-stresses are safe allowable values for the conditions under which such a beam would generally be used. To design a flitched beam, the size of the timber is first as- sumed and then the proper thickness and depth of the metal plates are to be computed. Let Si and S 2 be the given allowable unit-stresses; from the above their ratio is Si/S 2 = 284 COMPOUND COLUMNS AND BEAMS CHAP, xii and replacing Wi and W 2 by their values from (112) this reduces to Si/S 2 = EiCi/E 2 c 2 . The total bending moment M, for any kind of loading, is equal to the sum of the resisting moments 5i . I\/c\ and 5 2 . I 2 /c 2 . Accordingly the two equations for design are, c 2 /ci=E l S 2 /E 2 S i Si.Ii/ci + S 2 .I 2 /c 2 =M (112)' in which c\ and c 2 are the half-depths of timber and metal, and hence the ratio d 2 /di equals c 2 /c\. For cast iron and timber the ratio E 2 /Ei is 10, while S 2 /Si may be taken as about 4 for the tensile side of the beam; hence the depth of the metal should be about four-tenths of the depth of the timber. For structural steel and timber the ratio E 2 /Ei is 20, while S 2 /S\ should be about 15; hence the depth of the steel should be three-fourths of the depth of the timber. When d\ and d 2 are equal, as in Fig. H2c, the ratio Si/S 2 equals Ei/E 2 , so that the unit-stress on the tim- ber is one-twentieth of that on the steel. The proper thickness of the metal plates will depend upon the bending moment M. For rectangular sections the moment equa- tion becomes Sibidi 2 +S 2 b 2 d 2 2 =6M. When the depths di and d 2 are given, as also the width bi of the timber, the thickness b 2 of the metal is computed from this equation. For example, let di = i2 and d 2 = g inches for timber and steel; then S 2 /Si must equal 15. Hence let Si = i ooo and 52 = 15 ooo pounds per square inch; let bi =8 inches, and let it be required to find b 2 when the total load W=i$ ooo pounds and is concentrated at the middle of a span of 16 feet. Here M = \Wl = *]2oooo pound-inches, and then the moment formula gives b 2 = 2$ inches as the total thick- ness of metal required. In the case of the trolley stringer of Fig. 112c, the steel chan- nels carry a large part of the total load W, and they are sometimes designed so that they may carry it all. When the depth of the channels and timber are the same, the above theory shows that the ratio Si/S 2 equals E\/E 2 , and hence the flexural unit-stress on the channels is twenty times that on the timber. Thus, if S 2 for the structural steel is taken as 12 ooo pounds per square inch, S\ for the timber will be 600 pounds per square inch. For example, ART. 113 REINFORCED CONCRETE BEAMS 285 let the length of a stringer be 21 feet, and the total uniform load upon it be 14 ooo pounds, this being an equivalent for the total live and dead load. The moment M is then ^7 = 441 ooo inch- pounds. Let the timber be 8 inches wide and 10 inches deep; then /i/ci = i33.3 inches 3 . Using Si =600 and ,$2 = 12000, the value of /2/2 f r t ne two channels is found from the moment formula of (112)' to be 30.1 inches 3 , whence from Table 9 it is seen that the lo-inch channel which weighs 20 pounds per foot is required. Two channels of this kind have 12 = 157.4 inches 4 , while for the timber /i =666.7 inches 4 . Accordingly, by the help of (112) it is found that the channels carry about 83 percent of the total load. If the metal plates shown in Figs. 112a and 1126 do not extends to the ends of the simple beam, but stop a distance id from the ends, the above method needs modification. By an investigation to be given in Art. 123 it will be shown that formulas (112) apply to this case if I\ is multiplied by i 8/c 3 . Hence the shortening of the plates throws a larger proportion of the load upon the timber and increases the stress in it at the middle of the span. The condition Si/S 2 =Eid/E 2 c 2 is also to be modified by multiplying the second member by i 8/t 3 , so that the advantageous ratio c 2 /Ci is less than before. This case need not be discussed further, because plates extending over only a part of the length would rarely be used except in order to strengthen a weak beam, and in such an event no precise computations would be needed. Prob. 112. Let a flitched beam, like Fig. 1126, consist of two tim- bers each 10 inches wide and 14 inches deep, and a steel plate J inches thick and 7 inches wide. When the unit-stress in the timber is 900 pounds per square inch, what is the unit-stress in the steel? What percentage does the metal add to the strength of the wooden beam ? ART. 113. REINFORCED CONCRETE BEAMS Other methods of longitudinal reinforcement than that de- scribed in the last article are used for concrete beams. The method of Fig. 113a is occasionally used, but here the resistance 286 COMPOUND COLUMNS AND BEAMS CHAP. XII of the concrete to flexure is not generally taken into account, its office being to protect the steel beam from corrosion or fire, while the steel beam is considered to carry all the load. When computations are made on a case of this kind, the formulas of the last article directly apply, the ratio Ez/Ei being taken from 10 to 15, while the ratio 52/5i of the allowable unit-stresses for the compressive side of the beam will generally range from 8 to 12. Since the ultimate tensile strength of concrete is from 300 to 500 pounds per square inch, an average allowable tensile unit- stress is about 40 pounds per square inch, while that for the steel may easily under steady loads be as high as 1 5 ooo pounds per square inch. Supposing the steel to be stressed only to 8 ooo pounds per square inch, it is seen that Si for the concrete will be higher than its tensile strength, so that the practice of designing the steel beam to carry all the load is justified, the concrete being considered only as a covering which protects the steel in spite of the cracks on the tensile side. Fig. 113a Fig. 1136 Fig. 113c Fig. 113d In the method of reinforcement seen in Fig. 1136, there are four steel rods arranged symmetrically with respect to the neutral axis; for wide beams a larger number of rods is used, half of them above and half below the neutral axis, the distance of each row from that axis being the same. The formulas of Art. 112 apply directly to this case when c\_ is the half-depth of the rect- angular section and c tnu s supposing that the concrete fills also the spaces occupied by the metaL ART. 113 REINFORCED CONCRETE BEAMS 287 Considering the uncertainties regarding the values of the ratios Ez/Ei and Sz/Si, this method is entirely satisfactory unless a is more than 10 percent of bd. For example, let the width b be 8 inches, the depth d be 12 inches, and each of the four steel rods be ij inches in diameter, the centers of rods being 4 inches from the neutral axis; then bd=g6 square inches and a =4.91 square inches; Ci=6.o and ^2=4.62 inches, /i=i 152 inches 4 , and /2=4.9iX4 2 = 79 inches 4 . Now, taking E2/Ei=*io, the first formula of (112)' gives S 2 /Si = about 8; accordingly if the safe allowable tensile unit-stress for concrete is taken as 50 pounds per square inch, S 2 for the steel will be only 400 pounds per square inch. The steel and concrete do not work well together, and in fact this design is a poor one. If the load is sufficiently large to make 52 as high as 12 ooo pounds per square inch, a value which the steel may safely bear, the concrete will be rup- tured by tension on the convex side, so that it can only serve as a kind of protective covering for the steel rods. The uniform load which these rods can safely carry will then be W 2 =8S 2 I 2 /c 2 l. For instance, if the simple beam is 8 feet 4 inches long W 2 = 16 400 pounds is the load which is carried by the steel if the concrete does not act at all, while ^1=770 pounds is the load which might be safely carried by the concrete without any reinforce- ment. Since this concrete beam weighs about 800 pounds, it would safely carry only its own weight unless reinforced. Figs. 113c show the methods of reinforcement generally used for concrete beams, the metal being placed only on the tensile side. The theory given in the preceding article is entirely inapplica- ble to a beam where the metal is only on one side of the neutral axis, and the proper theory will be developed in the two following articles. Various forms of rods are in use, and it has been found that smooth rods are not the best, since there is a tendency for them to slip in the concrete. One of the oldest kinds is a rect- angular twisted bar which is known as the Ransome rod ; another form, is that of Thacher, which is a round bar flattened in two rectangular directions; Johnson's bar is of rectangular section with corrugations alternating on adjacent sides. Another rein- forcement is tfye lozenge-shaped form, called expanded metal, 288 COMPOUND COLUMNS AND BEAMS CHAP. XII which is widely used for the beams of floors. The Kahn method consists of rods which are straight near the middle of the beam and bent upward near the ends, the inclined ends being intended to prevent the shearing that sometimes occurs along surfaces which are indicated by the shear lines in Fig. 109. When steel rods are used in concrete beams, the fundamental idea is that they are for the purpose of increasing the resistance on the tensile side. A plain concrete beam has its neutral sur- face at the middle and hence the compressive unit-stress on the upper surface is equal to the tensile unit-stress on the lower surface when the elastic limit is not exceeded. Hence in a plain concrete beam the resistance to compression cannot be developed until the beam is ruptured on the tensile side. When rods are placed near the lower side, these take tensile stresses and hence the compressive stresses in the concrete above the neutral sur- face are increased. In this case the neutral axis no longer passes through the center of gravity of the section, so that new formulas must be established for concrete beams reinforced in this manner, and this will be done in the next article. Concrete beams are usually rectangular in section, and only these will be discussed in the following pages. The amount of metal which is placed near the tensile side of the section is rarely greater than 2 percent of the total rectangular section area, about i percent being the usual practice. Structural steel is mainly employed, although hard steel has been sometimes used. The allowable unit-stress for structural steel is generally taken as 12 ooo pounds per square inch, although there would be little objection to stressing it to a value 25 percent higher, but it is difficult to develop the full resistance of the steel, as will be seen later. In the discussions of the following pages the average values of the moduluses of elasticity given in Table 2 will be used, and hence the ratio E 2 /E\ will be taken as 10. The allow- able working unit-stresses for concrete will be generally taken as 500 pounds per square inch in compression and 60 in tension; these values apply only to first-class concrete of the proportions I cement, 2 sand, 4 broken stone (Art. 22). ART. 114 THEORY OF REINFORCED CONCRETE BEAMS 289 Prob. 113. A section like Fig. 1136 is to be used in a floor for a span of 6 feet, the depth of the beam being 5 inches and its width 4 feet. The steel rods are one inch in diameter and placed so that their centers are i inches from the neutral axis. How many rods are needed in order that they alone may carry a total uniform load of 5 850 pounds with a factor of safety of 5 ? ART. 114. THEORY OF REINFORCED CONCRETE BEAMS The theoretic laws of Art. 39 apply to all kinds of beams, as also the experimental laws 4 and 5 of Art. 40. When different materials are used in a beam law 6 needs modification, for the stiffest material is stressed the highest. Law 5 is to be used in this case, and this states that the unit-elongation or unit- shortening is the same for both materials. Thus at the same distance from the neutral surface S\/E\ for the concrete must equal S-2/E-2 for the steel. Law 6 applies, however, to the stresses in each material, and thus the stresses in the concrete vary directly as their distances from the neutral surface. The same is true for the stresses in the metal, but since the section area of this is small, it will usually be sufficient to regard the unit-stresses in the metal as uniformly distributed (Art. 113). Let b be the width and d the depth of the rectangular section, and a the section area of the metal. In strictness the section area of the concrete is bd a, but it will be unnecessary to take the diminution of bd into account, since a is rarely greater than 3 percent of bd. CASE I. Tension in Concrete. Fig. 114a shows a concrete beam reinforced with horizontal steel bars at a distance h below the middle. Let Si be the compressive unit-stress mp on the upper side at the dangerous section, S/ the tensile unit-stress m'p' on the lower side, and 82 the tensile unit-stress in the steel. Let 55 be the neutral surface at the distance k below the middle of the simple beam. The first condition of static equilibrium is that the algebraic sum of all the horizontal stresses in the section shall' equal zero; the sum of all the compressive stresses is for the mean unit-stress ^Si acts over the area 290 COMPOUND COLUMNS AND BEAMS CHAP. XII the sum of all the tensile stresses in the concrete is k), the sum of all the tensile stresses in the steel L> S 2 a. The second condition of static equilibrium is that the sum of the moments of these stresses shall equal the bending moment M; the moment of the compressive stresses with respect to the neutral axis ss f is $Sib(%d+k) 2 , since the total compressive stress acts with the lever arm %(%d+k), the moment of the tensile stresses in the concrete is %Si'b($d-k) 2 , and the moment of the tensile stresses in the steel is S 2 a(h k). Accordingly, are" two equations connecting the four unknown quantities Si, Si', S 2 , k. Two other equations result from the law that the unit-elongations are proportional to their distances from the neutral surface: S\/Si =(^d-\-k)/(^d k) applies to the concrete if the elastic limit is not exceeded; S 2 /E 2 is the unit-elongation in the steel, and this equals the unit-elongation in the concrete at the distance h k from the neutral surface, hence S 2 /E 2 = Si(h k)/Ei(%d+k). The solution of the four equations giws, bd(d+l2k) S l = {i+^}N S/=(i~W E l d from which it is seen that the position of the neutral surface depends upon h and the ratios a,bd and E 2/ 'E. When h = o, the steel rod is at the middle of the depth, and then S 2 =o, Si = Si' = 6M/bd 2 which are the unit-stresses for an unreinforced beam. The steel reinforcement is usually placed near the base. The above theory applies only when the unit-stresses do not exceed the elastic limit, but concrete is a brittle material in which the elastic limit is poorly denned, the stress diagram being similar to that of cast iron or brick. However, by using high factors of safety, Si and Si' may be made sufficiently low so that the above formulas have the same validity as the common flexure formula has when applied to brittle materials. Owing to the low tensile strength of concrete, beams often crack on the tensile ART. 114 THEORY OF CONCRETE-STEEL BEAMS 291 side, so that the reinforcing bars carry nearly all of the tensile stresses. It is therefore important to develop the formulas which apply to such a case, and this will now be done under the assump- tion that all the tensile stress below the neutral surface is carried by the steel bars. ' 7 p r *->i Fig. 114a Fig. 114& CASE II. No Tension in Concrete. Let Fig. 1146 represent the case where there are no tensile stresses in the concrete, the notation being the same as before except that the position of the steel is designated by the distance g measured downward from the upper surface of the beam, while the neutral surface 55 is at the distance n below the same surface. The compressive unit-stress Si on the upper side is represented by mp, while S 2 is the tensile unit-stress in the metal. To determine these quan- tities, the static laws of Art. 39, together with the experimental laws of Art. 40, are again to be used. The sum of all the horizontal stresses below the neutral axis in the dangerous section is S 2 a and the sum of those above it is \S\bn, since the average unit- stress \S\ acts over the area bn; hence S 2 a= \S\bn is the first equation between S 2 and Si. The sum of the moments of all the stresses in the section is the resisting moment which equals the bending moment M. Now S 2 a(g n) is the moment of the stresses in the metal with respect to the axis 5, and %S\bn 2 is the moment of the stresses in the concrete, since the total stress \S\bn acts with the lever arm w; hence S 2 a(g ri) + \S\bn 2 is the resisting moment which equals M, and this is a second equa- tion between S 2 and Si. A third condition is, however, required, since the unknown quantity n is contained in each of those thus far established. This is furnished by the experimental law regarding changes of length; S 2 /E 2 is the unit-elongation of the metal which is at the distance g n from the neutral surface. 292 COMPOUND COLUMNS AND BEAMS CHAP, xn while the unit-shortening of the concrete at the same distance from the neutral surface is S\(g ri)/E\n. Hence, 5 2 a=i5 1 ftn S 2 a(g-n) + $5^=11 S 2 /E 2 =(S l /E l )(g-n)/n are three equations for finding n, Si, $2', their solution gives in which e denotes the ratio E 2 /Ei. These formulas do not contain the depth d, but this is usually i or i| inches greater than g in order to protect the steel from corrosion. Prob. 114a. A reinforced concrete beam is 5 inches deep, 4 feet wide, 6J feet in span, has 1.8 square inches of steel at i^ inches below the middle, and the total load upon it is 6 400 pounds. Show from (114) that the concrete will probably crash on the tensile side. Prob. 1146. Using the above data and supposing that the concrete offers no tensile resistance, compute from (114)' the position of the neutral surface and the unit-stresses S\ and S 2 - ART. 115. INVESTIGATION OF REINFORCED CONCRETE BEAMS The formulas of the last article furnish the means of investigating a reinforced concrete beam for which the dimensions and loads are given. When the beam is lightly loaded, so that the concrete below the neutral surface is in tension, formulas (114) are to be used. When the beam is so heavily loaded that this tensile resistance is overcome, formulas (114)' are to be used, provided the elastic limit of the concrete on the compressive side is not exceeded. This elastic limit is an uncertain quantity, but it is probably not far from 600 or 700 pounds per square inch when the concrete has the proportions of i cement, 2 sand, 4 broken stone. When the beam is so heavily loaded that the computed Si exceeds this elastic limit, the formulas do not give reliable values of the unit-stresses. For example, let a reinforced concrete beam be 5 inches deep, 4 feet wide, 4^ feet in span, and have 3.6 square inches of steel placed 2 inches below the middle. Let it be required to investigate ART. 115 INVESTIGATION OF REINFORCED BEAMS 293 this beam when it carries a uniform load of 2 400 pounds, includ- ing its own weight. Here 6 = 48, d=$, h=2 inches; 6^=240 and a = 3. 6 square inches, whence 0/6^ = 0.015; a ^ so E^/E\ = iQ (Art. 113). Using formulas (114) there is found =0.261 inches for the location of the neutral surface below the middle. The maximum moment is M = \Wl = 16200 pound-inches, whence S=6M/bd 2 = Si pounds per square inch as the flexural unit- stress for a plain concrete beam. Then 51 = 89, and Si' =13 pounds per square inch for the compressive and tensile stresses on the upper and lower surfaces of the beam, the first giving a factor of safety of about 33, while the second gives a factor of safety of about 4. Also 5 2 =56o pounds per square inch for the steel, which is a very low value, the factor of safety being over 100. While this beam is perfectly safe, it is not designed for proper economy, since the compressive stress r in the concrete and the tensile stress in the steel might be much higher. As another example, let 6 = 48, ^=5, h=i\, g = 4, ^ = 84 inches, a = 2. 4 square inches, and the total uniform load W be 6 ooo pounds. Taking E 2 /E\ = 10, (114) gives Si' = 400 pounds per square inch, which is greater than the tensile strength of the concrete, so that these formulas do not apply. Turning then to (114)' there are found n= 1.56 inches, Si = 373 pounds per square inch for the concrete and 5 2 = I1 700 pounds per square inch for the steel, so that the beam has a proper degree of security. The formulas of Art. 114 are valid when the unit-stresses in the concrete are proportional to their distances from the neutral surface, and this is the case only when the changes of length are proportional to the stresses. Concrete is a material in which this proportionality exists only for low unit-stresses, so that the validity of the formulas is sometimes questioned. Hatt has deduced formulas under the supposition that the unit-stresses vary with their distances from the axis according to a parabolic law, and these will undoubtedly give a better agreement with experiment than (114)' when the concrete is highly stressed. For a case of design, however, the prevailing opinion is that formulas (116) should be used, and this has been the common 294 COMPOUND COLUMNS AND BEAMS CHAP. XII practice since 1900. The general laws involved in formulas (114)' are confirmed by experiments in which beams are ruptured, although numerical values computed from the formulas are of little reliability except as empirical results similar to that of the computed flexural strength or modulus of rupture for beams of one material (Art. 52). The phenomena of failure of a reinforced concrete beam have been completely ascertained by the very valuable experiments made by Talbot in 1904. These beams were 12 inches wide, 13 J inches deep, 14 feet in span, and had the steel reinforcing bars 12 inches below the top surface. Various percentages of metal were used, ranging from 0.42 to 1.56 percent of that of the concrete, and several kinds of reinforcing bars were employed. The beams were tested by applying two concentrated loads at the third points of the span, and the deflections at the middle were measured for several increments of loading, as also horizontal changes of length. Under light loads the tensile resistance of ,the concrete was plainly apparent; when the tensile unit-stress in the concrete reached about 350 pounds per square inch, the neutral surface rose and the stress in the steel increased. A little later fine vertical cracks appeared on the tensile side, while the tensile stresses in the steel and the compressive stress in the concrete increased faster than the increments of the load. The last stage was a rapid increase of the deformations, and rupture occurred by the crushing of the concrete on the upper surface, the steel being then generally stressed beyond its elastic limit. In some cases reinforced concrete beams have been known to fail by shearing near the ends, the curve of rupture being like that shown by the broken lines in Fig. 109. The full investigation of this case is attended with some difficulty and will not here be attempted, but the discussion of Art. 108 furnishes the means of making approximate computations. It is only short beams which fail in this manner. When rupture occurs near the middle of the beam along a curved surface which roughly agrees with one of the full lines in Fig. 109, this is not a case of shearing but one of rupture by tension. ART. 116 DESIGN OF REINFORCED CONCRETE BEAMS 295 The safe load which may be carried by a reinforced concrete beam can be computed in five ways from the formulas of Art. 114. Under Case I the first step is to find k, and then to place its value in the three following equations. Allowable values of Si, Si, and S 2 being assumed, three computations may be made to find three values of S from which three values of M are determined ; then for uniform load W=8M/l and the smallest of the three values of W is the one to be selected on the theory of Case I, where no cracking of the concrete is allowed. Larger values of W will be found by using the formulas of Case II, the first step being to compute n, the second to. find two values of M from assumed values of Si and 52, and the third to compute two values of W, the smaller of which is the safe load. Computations generally show that a reinforced concrete beam is stronger after it has cracked on the tensile side than it was before, this being due to the fact that the steel then has a higher unit-stress. These cracks may not be visible and they are often called hair cracks; they exist whenever the unit-stress on the tensile side of the beam reaches or exceeds the ultimate strength. Prob. 115a. A reinforced concrete beam is 12 inches wide, 15 inches deep, 14 feet long, and has 3.6 square inches at i^ inches from the lower side. Find the total uniform load W which this beam can cany so that the tensile stress in the concrete on the lower side may be 100 pounds per square inch. Prob. 1156. Using the same data as above, compute the total uniform load which will produce a compressive stress of 500 pounds per square inch on the concrete, considering that the concrete below the neutral surface offers no tensile resistance. ART. 116. DESIGN OF REINFORCED CONCRETE BEAMS When a reinforced concrete beam is to be built, its width, depth, and span are given or assumed, as also the load which is to be carried and the allowable unit-stresses for the concrete and steel. The problem of design then consists in determining the proper section area of the reinforcing bars and the proper depth of the beam. As for the position of these bars, it is ap- parent that they should be placed as near as possible to the tensile 296 COMPOUND COLUMNS AND BEAMS CHAP, xii side of the beam in order that their resisting moment may be as large as possible. They should, however, be entirely covered by the concrete in order to be protected from corrosion due t? atmospheric influences. It is impossible to design an economical reinforced concrete beam on the theory of Case I of Art. 114, for if S' be taken even as high as the ultimate tensile strength of the concrete, the values of Si and .$2 are too low; 2 for the steel is indeed always less than Si', so that the strength of the metal is not utilized. Nothing remains to be done, therefore, but to allow the concrete to crack on the tensile side and thus bring proper tension into the steel. If the tensile resistance of the concrete is not considered, formulas (114)' apply. The given quantities ^are E 2 /Ei or e, the width b, the unit-stresses Si and Si, and the bending moment M. Eliminat- ing n from the three equations, there are found two equations containing g and a, and the solution of these gives in which a denotes the ratio Sz/Si. The unit-stress Si should be taken as high as allowable; ,$2 should not be higher than the allowable value, but it may be taken lower, if economy in cost is thereby promoted. The depth d is made about i\ inches greater than g. For example a beam is to be built of concrete which has the proportions i cement, 3 sand, 6 stone, for which the ratio 2 /jEi = =15. The span is to be 14 feet, the breadth 20 inches, and the total uniform load is to be 7 ooo pounds. It is required to find the depth of the beam and the section area of the reinforc- ing steel rods so that the unit-stresses Si and S2 shall be 350 and 14000 pounds per square inch respectively. Here = 40, and M=|xy 000X14X12=147 ooo pound-inches. Inserting the given values in the fires of the above formulas, there is found #=13.0 inches, then bg=26o square inches, and from the second formula = 0.90 square inches. Here the section area of the steel is 0.35 percent of the section of the beam above the centers ART. 116 DESIGN OF REINFORCED CONCRETE BEAMS 297 of the rods. By using a lower value of 2, the depth g will be smaller and the section area a will be greater than the above values. The best set of values will be those which render the cost of the beam a minimum. The position of the neutral surface depends only upon the values of the ratios e and a. The value of n, as found from the solution of the equations (114)', is n=-*- or n=g (116)' i+a/s t-ftr For instance, in the case of the last paragraph, where 5 = 15 and while the second gives the metal section a. For example, let E 2 /Ei = e = io, So/Si =(7 = 25, n = 4 and 6 = 18 inches; then g= 14.0 inches and 0=1.42 square inches. The safe load which this beam can carry may be found from M=S 2 a(g |); for instance, in the above case let 53=12500 pounds per square inch, then the safe bending moment is 22 ooo pound-inches. Prob. 116. Consult a paper by Sewall and the accompanying discussions in Transactions of American Society of Civil Engineers, 1906, Vol. 56. Ascertain different opinions as to what should be the comparative cost of concrete and steel in order to produce the most economical reinforced concrete beam. ART. 117. PLATE GIRDERS A plate girder is composed of only one material, usually structural steel, but it may be called compound in the sense that it consists of different parts riveted together. Fig. 117 shows a side view and section of a plate girder without the rivets which connect the web to the angles and the angles to the cover plates. In the section A a there are four angles and the web; to the right of the section Bb there are in addition two cover plates; to the right of the section Cc there are four cover plates in addition to the angles and web. The section areas and moments of inertia are hence different in the three sections, the plate girder being in fact an approximation to a beam of uniform strength (Art. 58). The flexure formula (41) may be used to investigate the strength of a plate girder in exactly the same manner as if it were a solid beam. Let /i be the moment of inertia of the uni- form section between Aa and Bb, I 2 that between Bb and Cc, and I 3 that between Cc and Dd. Then S=M l .ci/Ii applies to the first section, S = M 2 .c 2 /I 2 to the second section, and ART. 117 PLATE GIRDERS 299 so on. By the help of this formula sections may also be found to resist the bending moments under a specified unit-stress 5. For example, let the load be uniform and expressed by w per linear unit; let / be the length of the beam, and the three dis- tances ab, ac, ad, be o,2/, 0.35^ and o.5/. Then from Art. 38, the bending moment at Bb is o.oSowl 2 , that at Cc is o.iifwl 2 , and that at Dd is o.i2$wl 2 . Hence the value of I\/c\ is o.o8w/ 2 /S, that of I 2 /c 2 is o.ii4ivl*/S, and that of 7 3 /c 3 is o.i2$wP/S. Sections of plates, angles and web may now be determined, with the help of Art. 44 and Ta^le 10, to satisfy these requirements.. Then the section A a must be investigated to ascertain whether it is sufficiently large to safely resist the vertical shear Fig. 117 Another method which is frequently used in practice is to- regard the web as carrying none of the bending moment, and to- consider that the unit-stresses in the flanges are uniformly dis- tributed so that the total stress in each flange may be regarded as acting at its center of gravity. Let d be the depth between the centers of gravity of the flanges, a the section area of one flange, and 5 the allowable unit-stress. Then the stress S(t in one flange acts with the lever arm d with respect to the center of gravity of the otjier flange, and therefore Sad equals the bend- ing moment M. Thus, for the first section a\=M\/Sd\, for the second section a 2 =M 2 /Sd 2) and so on; from these values of a proper angles and plates may be selected. The section area of the web is determined in this method from the maxi- mum vertical shear which can act at the end. The thickness of the web is made uniform throughout the span; f, \ and f inches are common thicknesses, as these are standard market sizes. The web is usually stiffened by vertical angles riveted ta it at intervals. 300 COMPOUND COLUMNS AND BEAMS CHAP, xil After a thickness has been determined for the web from the vertical shear, it cannot generally be altered if the depth of the girder is slightly changed on account of the requirement that market sizes shall be used. There is then a certain depth, called the " economic depth," which gives a smaller amount of mate- rial than any other. For the simple case where the flange areas are uniform throughout, there being no cover plates, this eco- nomic depth may be determined in the following manner. The section area of each flange is M/Sd, and the section area of the web is td The total volume of material, neglecting rivets, splices, and stiffeners. then is. (2M/Sd + td)L Differentiating this with respect to d, and equating the derivative to zero, gives d 2 = 2M/St, which determines the economic depth. This condition shows that 2M/Sd equals td, that is, the girder has its economic depth when the amount of material in the flanges is equal to that in the web. This rule holds approximately when cover plates are used, as shown by the investigations in Part III of Roofs and Bridges, where are also given in full detail the methods of designing plate girders for stringers, floor beams, and bridges. Prob. 117. A plate girder used as a floor stringer has a span of 22 feet, and the uniform load w per linear foot which is equivalent to the actual, wheel loads is i 700 pounds. For an effective depth d of 34 inches, compute the flange areas at the middle and the quarter sec- tions, taking S as 1 2 ooo pounds per square inch. ART. 118. DEFLECTION OF COMPOUND BEAMS The deflection of flitched beams, like those of Art. 112, is readily computed, when the elastic limit of the material is not exceeded, by the use of the formula f=W \PfaE\I\ in which W\ is the total load that comes on the material that has the modulus of elasticity E\ and the moment of inertia I\. The same method applies to the compound beams of Fig. 1130 and 1136, but it does not apply when reinforcing bars are placed only on one side of the neutral surface of a concrete beam. Con- ART. 118 DEFLECTION OF COMPOUND BEAMS 301 crete-steel beams are usually of short span, and it is rarely neces- sary to compute deflections. Formulas might, indeed, be devised for this case, but they would be of uncertain application on account of the uncertainty in the value of EI and because resist- ing tensile stresses might not exist near the middle of the beam, while they would act near the ends. For plate girders it is sometimes important to compute the deflection at the end of a cantilever arm or at the middle of a simple span. For a simple span under uniform load the deflec- tion found in Art. 55 is f = $wl* / s&^EI , which applies to a plate girder where the moment of inertia / is constant throughout. If /i is the moment of inertia of the section near the end and /a that of the section at the middle, as in Fig. 117, then for these values there may be found two deflections f\ and /2, the first of which is greater and the second less than the true deflection. It is often the case that this information is all that is required, but by the method of Art. 124 a formula giving a closer result can be deduced. Let /i, /2, lz be the distances from the left end in Fig. 117 to the sections Bb, Cc, and Dd, so that t% is one-half of the span /. Let the load be uniform so that the bending moment at the distance x from the support is M = \wlx \wx i . Let m be the bending moment due to a load unity at the middle of the beam or m = %x. Then, by (124), CMm- r l ^Mm^ f %l *Mm<. f l3 Mm. f= I ~^fdx=2 I ~=-dx+2 I ~rdx+2 I -=r?8x J EI J Eli Jk EI 2 Ji t EI 3 gives the deflection at the middle under the uniform load. Inte- grating between the designated limits, there is found, Wl /A 3 / 2 3-/,3 /3 3_ /2 3. w //! 4 / 2 4 -/i 4 /3 4 ~/2 4 \ J ~6E\I, 7 2 / 3 / SE\Ii 7 2 ~^~) This formula is not difficult in computations when tables of squares, cubes, and reciprocals are at hand; thus /i 3 //i is found by taking /i 3 from the table and multiplying it by the reciprocal Of/!. The above gives the elastic deflection due to the horizontal flexural stresses only. Art. 125 shows, however, that there is a deflection due to the vertical shears which must be added to 302 COMPOUND COLUMNS AND BEAMS CHAP. XII the above in order to obtain the exact deflection. Let V be the shear due to the given uniform load and v be the shear due to the load unity at the middle of the beam, so that V = $wl woe andz> = . Then, by (125), rvv.. rwv , f h vv . fivv . /= / =-&* 2 / &c+2 I -=-dx+a I ~^dx J Fa J Fai J i, Fa 2 J h Fa 3 in which ai, a 2 , 03 are the section areas whose moments of inertia are I\, I 2 , Is, and F is the shearing modulus of elasticity. Per- forming the integrations, there is found, 2F\ai a 2 #3 / 2F\ai a 2 which is the deflection at the middle of the beam under the uni- form load due to the vertical shears. The numerical value of this shearing deflection is usually small compared to that due to the bending moments, but in short spans it is an appreciable quantity. Prob. 118. Deduce, from the above formulas, the deflections due to vertical shears and bending moments when the simple beam has the constant section area a and the constant moment of inertia /. Show that the length of beam for which these two deflections are equal is given by (l/r) 2 =4&E/$F, where r is the radius of gyration of the section. AKT. 119 EXTERNAL WORK AND INTERNAL ENERGY 303 CHAPTER XIII RESILIENCE AND WORK ART. 119. EXTERNAL WORK AND INTERNAL ENERGY When a force is applied to a body it overcomes a resistance through a certain distance and thus external work is performed on the body. It is usually the case that the force is applied by increments, so that it increases uniformly and gradually from o up to its full value P. When a tensile load is applied in this manner to a bar producing the elongation e, the work performed is \Pe, or equal to the mean load \P multiplied by the distance e. This is otherwise seen from Fig. 14a, where the shaded area represents the work performed while the load increases from o to P. Similarly, in the case of a beam under a single load, the load increases from o to P and produces the deflection /, so that the work performed is \Pf. This work done upon the bar or beam is called the ' external work '. When the elasticity of the body is not impaired and the load is applied so gradually that no work is expended in producing heat, there is stored within the body an amount of energy equal to the external work. This is called 'internal energy ', or some- times 'internal potential energy', because this energy may be utilized to perform an amount of work equal to the external work performed upon it. When the elastic limit of the material is not exceeded, the internal energy in a stressed bar is equal to \Pe and that in a beam stressed by a single load is equal to \Pf. That these statements are correct, many experiments can testify, and they also follow from the law of conservation of energy. The internal energy which can be stored in a metal bar is very small compared with that which is stored in a mass of steam or compressed air of the same weight. For example, take a bar of structural steel, 6 square inches in section area and 25 feet 304 RESILIENCE AND WORK CHAP, xm long. The load P which will stress this bar to its elastic limit is P=6X35 000 = 210 ooo pounds, and the elongation under this loadjs e = 35 000X25/30 oco 000=0.0292 feet; hence the work stored in the bar is K = %X 210 000X0.0292 =3060 foot-pounds. This bar weighs 8^X6X10X1.02=510 pounds (Art. 17), and an equal weight of air will occupy about 6 320 cubic feet. If this air is in a pipe of one foot diameter, it will fill a length of 8 040 feet, and under a pressure of 15 pounds per square inch above the atmospheric pressure, it may be compressed to half this length. The force P here is i 700 pounds, and the energy stored in the air is JXi 700X4020 = 3 517 ooo foot-pounds, which is nearly i 200 times as great as that stored in the steel bar. The mate- rials of construction cannot, therefore, be advantageously used for the storage of energy. If the above steel bar is used as a simple beam with a single load at the middle, the section being square, the load P which will stress it to the elastic limit is found from the flexure formula (41) to be P = i 143 pounds. The deflection / under this load is found from Art. 55 to be /=y.i4 inches. Hence the energy stored in the beam is |Xi 143X7.14/12=340 foot-pounds, or only one-ninth of that stored in the bar. Springs in the form of beams are used on vehicles to lessen the shocks which occur during motion, but they cannot be used to store energy for the propulsion of a vehicle, on account of the great weight which would be required. When a bar is already under a load PI which has produced the elongation e\, an additional load P 2 produces the elongation 62, so that the total energy stored in the bar is $(Pi + P 2 ) (e\+ez), as clearly appears from Fig. 1196. When P 2 is removed, the work that can be performed by the stored energy is that repre- sented by the shaded area or %(2Pi+P 2 )e 2 . Or, if the load on a bar ranges from the lower value PI to the higher value P, as in Fig. 119c, the elongation being e\ for the former and e for the latter, then the work K performed on the bar, or the energy K stored in it by the increase of PI to P, is given by K = \Pe \P\e\. Similarly, when a beam is under a single load which increases ART. 119 EXTERNAL WORK AND INTERNAL ENERGY 305 from PI to P, while the deflection increases from /i to /, the external work and the stored energy due to this increase are expressed by K^^Pf-^P-^f^ When a part of a bar is considered which has a length of unity and a section area of unity, the load P is the unit-stress 5 and the elongation e is the unit-elongation e. Thus, K=^Se K=$S-$S l i (119) are the expressions for the work performed on one cubic unit of the material, the first being for the case where the unit-stress increases from o up to S and the second for the case where the unit-stress increases from Si up to 5. Fig. 1196 Fig. 119c When the bar is stressed so that 5" exceeds the elastic limit of the material, the elongations increase more rapidly than the stresses and the above formulas are inapplicable. Art. 14 shows that the external work required for rupture is very large com- pared with that required to stress the bar up to its elastic limit, particularly for wrought iron and steel. The area in Fig. 14c between the curve and the axis of elongations measures this external work. For wrought iron and steel the portion of the area below the elastic limit is so small that it may be disregarded, and then the area may be roughly expressed as that of a trape- zoid having the length e equal to the ultimate elongation, and limited by the two ordinates which represent the elastic limit and the ultimate strength. Taking the elastic limit as one-half of the ultimate strength, the area of this trapezoid is K = $sS u , which is the work required to rupture by tension one cubic unit of the bar. For example, take the two specimens of unannealcd and annealed Bessemer steel in the table of Art. 25, which had S M = i25ooo and e=o.ii before annealing and S M = 99000 and 306 RESILIENCE AND WORK CHAP, xill 6=0.19 after annealing; here # = 10300 inch-pounds for the first and K = 14 100 inch-pounds for the second, so that the process of annealing increased 37 percent the capacity of the steel to withstand external work. Another formula sometimes used for unit rupture work is K=$E(S e + 2Su), where S e is the elastic limit and S u the ultimate strength. Prob. 119o. How many foot-pounds of work are required to stress a wrought-iron bar, 4 inches in diameter and 54 inches long, from 6 ooo up to 12 ooo pounds per square inch ? Prob. 119&. If this bar is used as a beam with a load at the middle, how many foot-pounds of work are required to increase the greatest unit-stress at the dangerous section from 6000 up to 12000 pounds per square inch? ART. 120. RESILIENCE OF BARS The term " Resilience " is frequently used to designate the work that can be obtained from a body under stress when it is relieved of its load. When the elastic limit of the material is not exceeded this work must be that stored within the bar in. the form of stress energy. In Art. 14, as well as in Art. 119, it was shown that the external work performed in elongating or shortening a bar is $Pe, and this is the resilience which may be utilized when the bar is entirely relieved from stress. Let the section area of the bar be a and the uniform unit-stress be 5, then P =aS; also let the length of the bar be / and the modulus of elasticity of the material be E, then the change of length is e = (S/E)L Hence, the elastic resilience of the bar is, K=\Pe or K=$(S 2 /E)al (120) and the factor %S 2 /E is called the 'modulus of resilience ' of the material when 5 is the unit-stress at the elastic limit. The following are average values of the modulus of resilience of materials which have been computed from the average con- stants given in Arts. 4 and 9: for timber, %S?/E= 3.0 inch-pounds per cubic inch for cast iron, %S e 2 /E= 1.2 inch-pounds per cubic inch ART. 120 RESILIENCE OF BARS 307 for wrought iron, %S 2 /E=i2.$ inch-pounds per cubic inch for structural steel, %S 2 /E = 2o.4 inch-pounds per cubic inch Resilience is a measure of the capacity of a body to resist external work, and the higher the modulus of resilience the greater is the capacity of a material both to store up energy and to resist work that may be performed upon it. The modulus of resilience measures this capacity up to the elastic limit only. The total elastic resilience of a bar is found by multiplying the modulus of resilience by the volume of the bar, as (120) shows. Thus, a bar of structural steel, 6 square inches in section area and 25 feet long, has a volume of i 800 cubic inches, and hence its elastic resilience is 36 720 inch-pounds or 3 060 foot-pounds. When a bar is stressed by a load which increases from PI to P, the unit-stress increases from Si to S, and by (120) the resilience of the bar when the load is decreased from P to PI is K=^(S 2 /E}al-J i (S l 2 /E)al=^(S 2 -S l 2 )/E . al Here, as before, the unit-stress 5 must not be greater than the elastic limit of the material of the bar. The word resilience implies a spring, and it should not be used except for that part of the applied work which can be recovered when the load is removed. When the elastic limit is exceeded and the load is released, the expression %(S 2 /E)al also applies, as shown in Art. 14, to the work that can be utilized, but numerical values of this resilience are of no importance when S is the ulti- mate strength, because the capacity of a material to withstand external work is properly measured by the product of its ulti- mate strength and ultimate elongation, as explained at the close of Art. 119. Prob. 1200. What horse-power engine is required to stress, 250 times per minute, a bar of wrought iron 18 feet long and 2 inches in diameter from o up to its elastic limit? Prob. 1206. What horse-power engine is required to stress, 250 times per minute, a bar of wrought iron 18 feet long and 2 inches in diameter from 12 500 up to 25 ooo pounds per square inch? 308 RESILIENCE AND WORK CHAP. XIII ART. 121. RESILIENCE or BEAMS When a cantilever beam deflects under the action of a load at the end or a simple beam deflects under a load at the middle, the external work is \Wf as long as the deflection / increases proportionally to the load which is applied by increments so that it increases gradually from o up to the value W (Art. 119). The resilience of the beam then equals \Wf and an expression for its value in terms of the flexural unit-stress S may be obtained by substituting for W and / their values from Art. 56. Let / be the length of the beam, c the distance from the neutral sur- face to the upper or lower side of the beam where the unit-stress is S, and / the moment of inertia of the cross-section ; then, W=aSI/cl f=aSP/pcE where a is i for a cantilever loaded at the end and 4 for a simple beam loaded at the middle, while /? is 3 for the cantilever and 48 for the simple beam. Replacing / by ar 2 , where a is the section area and r the radius of gyration of that section with respect to the neutral axis, the elastic resilience of the beam is, K=lWf=( R 3= +W for th e reactions due to the given load ; these are the same results as derived by the use of the theorem of three moments (Art. 71). Fig. 126c Fig. 1266 As a second example, take the partially continuous girder in Fig. 1266 which has four supports and a joint at the middle of the second span so that the bending moment there is always zero. Let 2l be the length of the middle span and / that of each end span, and let it be required to find the four reactions due to a uniform load in the last span. Three conditions are, R l +R 2 +R 3 +R 4 = W 4Ri + ^R 2 +R 3 = W 2Ri+R 2 =o the first being the static condition that the sum of the vertical forces is zero; the second the static condition that the sum of the moments of these forces is zero, the axis being taken at the right end; the third the condition that there is no moment at the joint. From these three conditions the values of three reac- ART. 126 PRINCIPLE OF LEAST WORK 323 tions may be found in terms of the other reaction, thus, R 2 =- 2 R l R 3 = $W+ 2 R l R 4 = ^W-R l and the value of RI may be found by the help of the principle of least work. To do this an expression for the bending moment is found for each of the four parts of the beam and the sum of all the values of M 2 /2EI will give the total internal work of the flexural stresses (Art. 123). Thus, in a manner similar to that of the last paragraph, a fourth condition is established which expresses that the work of the internal stresses is a minimum. 'This equation, in connection with the three previously estab- lished, gives & = -&W, R 2 = + &W, R 3 = +$$W, R* = +tfW. These are the reactions for the given uniform load due to the bending moments, and they will be slightly modified if the ver- tical shears are taken into account. Examples of the application of the principle of least work to bridges and arches will be found in Parts I and IV of Roofs and Bridges. This principle must be used with caution by the beginner, but it is one of much value in the discussion of struc- tures which are statically indeterminate. Prob. 126. A table of length I and width h has four legs at the corners which are of equal size and length. For a load P placed on the table at a distance \l from the side h and a distance \h from the side /, compute the reactions of the leg nearest the load. 324 IMPACT AND FATIGUE CHAP, xiv CHAPTER XIV IMPACT AND FATIGUE ART. 127. SUDDEN LOADS AND STRESSES A load at rest on a bar or beam is called a ' static load '. The same term is applied to a load which increases from o up to its final value P in such a way that the deformations of the bar or beam at different instants are proportional to the loads aciing at those instants until the elastic limit of the material is exceeded. Loads applied in any other manner are sometimes called ' dynamic ', and the term " impact " implies either sud- denness of action or that the load is in motion before it is applied to the bar or beam. Static loads have been mostly considered in the preceding chapters, but it has always been recognized that the stresses and deformations due to sudden and variable loads are greater than for static ones (Art. 7). The terms ' dynamic stress ' and ' dynamic deformation ' are sometimes used to distinguish the effects of impact from those due to static loads. A static tensile load is usually applied to a bar by increments, so that it increases from o up to P in such a way that the elonga- tion is proportional to the load until the elastic limit of the mate- rial is reached. The work done upon the bar is then equal to the mean load %P multiplied by the elongation e, or K = \Pe. Simultaneously the stress in the bar increases from o up to P and the internal energy stored in the bar is \Pe. The triangle in Fig. 127a represents both the external work and the internal energy. When a load is applied to a bar in such a manner that its intensity is the same from the beginning to the end of the elonga- tion, it is called a "sudden load ". For instance, let a load be hung by a cord and just touch a scale-pan at the foot of a vertical bar; then if the cord is quickly cut, the load acts upon the bar with uniform intensity throughout the entire elongation. In ART. 127 SUDDEN LOADS AND STRESSES 325 this case the maximum elongation is greater than for a static load, but the bar at once springs back; carrying the load with it, and a series of oscillations ensues, until finally the bar comes to rest with an elongation due to the static load. Here the stress in the bar increases from o up to Q, the stress Q being equal to the static load which would produce the maximum elongation. Fig. 1276 represents this case, where the rectangle shows the work done by the instantaneous load P, and the triangle shows the internal energy stored in the bar at the instant of greatest elongation. The unit-stress due to Q must be less than the elastic limit in order that the following discussions may be valid. Fig. 127a - -9 - Fig. 1276 Let q be the maximum elongation due to the sudden load P; the work performed in the bar is Pq. The internal energy stored in the bar at the instant of greatest elongation is %Qq, since the stress increases from o up to Q. Hence %Qq=Pq, or Q = 2P. Let e be the elongation due to the static load P; then q/e=Q/P, and hence also q=2e. Accordingly the following important law is established for a bar under elastic changes of length : A sudden load produces double the stress and double the deformation that is caused by a static load. Fig. 130a shows the end of a bar acted upon by a sudden load and it will be explained in Art. 132 that the maximum velocity of the load occurs when the elongation is equal to e; the curve in the figure shows the variations in velocity. When the elonga- tion 26 is reached, the stress in the bar is zP and the resultant force tending to move the end is P 2P or P; hence the end moves backward and oscillations ensue, until finally the bar comes to rest under the elongation e and the stress P. 326 IMPACT AND FATIGUE CHAP. XIV In the above discussion P and Q are the total stresses on the section area a of the bar. Let 5 and T be the corresponding unit-stresses, so that P = Sa and Q = Ta. Then the equation Q = zP becomes T = 2S, that is, the unit-stress due to a sudden load is double that due to the same static load. Similar conclusions result when a single load P is suddenly applied to a beam producing the deflection g while under the same static load the deflection is /. Let Q be the static load which will produce the deflection q. Then the external work of this static load is \Qq, while that of the sudden load is Pq; hence $Qq = Pq or Q = 2P; that is, the sudden load P produces the same effect as a static load 2P. From the static law Q/P = q/f, it follows that q = 2f, so that the dynamic deflection is double the static deflection. Let S be the flexural unit-stress at the dangerous section of the beam when the deflection is / and let T be the unit-stress when the deflection is q; then q/f=T/S, since elastic deflections are proportional to the unit-stresses (Art. 2). Therefore also, T=2$, that is, the flexural unit-stress due to a sudden load is double that due to the same static load. Lastly, consider a bar upon which rests a load PI causing the elongation e\. Let a sudden load P be now brought upon it causing the additional elongation q and the additional stress Q. Fig. 127c represents this case and it shows that the elongation is e^ + 26 and that the final stress is P\ + 2P\ thus the instantaneous load produces its effect independently of the other. As soon as the elongation e\ + 2e occurs, the bar springs back, and a series of oscillations follows; finally the bar comes to rest under the elongation ei+e and the stress P\+P. Similar conclusions fol- low in the case of a beam under an initial load. In the above investigation it has been supposed that all the work Pq performed by the sudden load P is expended in storing energy in the bar or beam. This is not strictly the case, for internal molecular friction causes a slight loss of \\ork (Art. 147). The law deduced is, however, very close for a light beam, but Q is really a little less than 2P and q is a little less than 2e when the beam is heavy compared with the load. ART. 128 AXIAL IMPACT ON BARS 327 Prob. 127. A vertical steel bar, 2 inches in diameter and 13 feet long, has a load of 15000 pounds hung at its end. Compute the elongation due to this static load, and also the maximum elongation which occurs when an additional load of 7 500 pounds is suddenly applied. ART. 128. AXIAL IMPACT ON BARS The word impact is here used to mean the effect of a load which is moving when it strikes the end of a bar; such a load evidently produces a greater deformation and a greater stress than one applied suddenly. The stress in the bar increases from o up to a certain limit Q and the deformation increases from o up to q. If the elastic limit of the material is not ex- ceeded, the stress at any instant is proportional to the deforma- tion so that the stored energy of the internal stresses is $Qq. Equating this to the external work, the values of Q and q may be found. Let P be a weight which is moving horizontally with the velocity V at the instant it strikes the end of a light horizontal bar. Its kinetic energy is P . V 2 /2g, where g is the acceleration of gravity; or, if h is the height through which P has fallen to acquire the velocity V, then V 2 /2g = h, and the kinetic energy of the moving weight is P . h. Accordingly %Qq=Ph, if no work is expended in overcoming inertia or in friction. Now, let e be the elongation of the bar due to the static load P; then the law of proportionality gives q/e=Q/P. From these two equations are found, Q=P(2k/e)* q=e(2h/e)* = (2he)* (128) which shows that Q may be much greater than P and q much greater than e when the weight P is moving rapidly. For example, let V=io feet per second, then ^=100/2X32.2 = 1.55 feet = i8.6 inches. Let the weight P be 60 pounds and the horizontal bar be of steel 18 feet long and 3 square inches in section area, then (10) gives e=Pl/aE= o.ooo 144 inches. Accordingly, = 30 50x3 pounds and (7=0.073 inches, which are about 510 times as great as those due to a static load of 60 Ibs. When a vertical bar is subject to the impact of a falling weight 328 IMPACT AND FATIGUE CHAP xiv P, the end of the bar is elongated or shortened the amount q so that the work performed upon it is (P(h + q). The internal stored energy is %Qq as before. Accordingly the two equations $Qq=P(h+g) and q/e = Q/P are to be used to find the values of Q and q, which are, Q=P+P(i + 2 h/e)* q = e+e(i + 2h/e)* (128)' When h = o, these formulas reduce to Q = 2P and q = 2e, which are the results previously found for a sudden load. Since e is a small quantity, it follows that a load dropping from a moderate height may produce large stresses and deformations. Experiments made upon elongations of spiral springs give results which closely agree with those computed from the formula for q, when the elastic limit is not surpassed by the stress Q. The curve in Fig. 1306 shows the variation in velocity of the end of the bar. The effect of loads applied with impact is therefore to cause stresses and deformations greatly exceeding those produced by the same static loads, so that the elastic limit may perhaps be often exceeded. Moreover the rapid oscillations which ensue cause a change in molecular structure which impairs the elasticity of the material when such loads are often applied. It is some- times found that the appearance of a fracture of a bar which has been subject to shocks is of a crystalline nature, whereas the same material, if ruptured under a gradually increasing stress, would exhibit a tough fibrous structure. Moving loads which produce stresses above the elastic limit cause the wrought iron and steel to become stiff and brittle, and hence it is that the work- ing unit-stresses should be taken very low (Art. 7). The above formulas apply also to unit-stresses. Let a be the section area of the bar, 5 the unit-stress due to the static load P and T the unit -stress due to Q, so that S=P/a and T = Q/a. Then the formulas for Q in (128) and (128)' become, T=S(2h/e)* T=S+S(i + 2h/e)* (128)" the first of which applies to a horizontal bar and the second to a vertical bar. For instance let h=i8e, then T = 6S for the hori- zontal and T = j.oSS for the vertical bar. Here T is the unit- ART. 129 IMPACT ON BEAMS 329 stress which prevails in the bar at the instant of greatest defor- mation, but after a series of oscillations the bar comes to rest under the unit-stress 5. These oscillations are discussed in Art. 132. All of the above formulas for dynamic stress and elongation give results which are somewhat too large, because a portion of the energy of the moving weight P is expended in overcoming the inertia of the bar. They apply only to bars which are so light that this resistance of inertia may be disregarded. In Art. 130 it will be shown how these formulas may be modified so as to take into account the inertia of the bar. Prob. 128. In an experiment upon a spring, a static load of 14.79 ounces produced an elongation of 0.42 inches, but when dropped from a height of 7.72 inches it produced a stress of 102.3 ounces and an elongation of 2.90 inches. Compare theory with experiment. ART. 129. IMPACT ON BEAMS When a falling weight strikes a beam, it causes a greater deflection than a load suddenly applied. Let the weight P fall from a height h above a light beam and produce the dynamic deflection q\ the work performed is then P(h + q). Let T be the maximum flexural unit-stress produced by the impact and 5 be that due to a static load P which causes the deflection /. Then the deflections are proportional to the unit-stresses, if the elastic limit is not exceeded, or q/f*= T/S. Also let Q be a static load which will produce the deflection q; then the deflec- tions are also proportional to the loads, or q/f=Q/P\ accord- ingly Q/P = T/S. The external work of the load Q is %Qq and this is equal to the internal energy stored in the beam when the deflection q is attained, if all the work is expended in stressing the beam. Hence $Qq = P(h + q}, which by the above ratio reduces to $Tq = S(h + q). Combining this with q/f-T/S, there are found, T=S+S(i + 2h/f)* q=f+f(i + 2h/f)* (129) as the formulas for the dynamic maximum unit-stress and deflec- tion due to the impact of a single load P. Here 5 is found from 330 IMPACT AND FATIGUE CHAP, xiv the flexure formula (41) for any given case, or S = Pcl/aI, and / is found from the deflection formula f=PP/{3EI, where / is the length of the beam, a and /? numbers depending on the arrangement of the ends, / the moment of inertia of the cross- section, c the distance from the neutral axis to the remotest fiber of the dangerous section, and E the modulus of elasticity of the material (Arts. 55 and 63). When a weight P is moving with the velocity F, it can per- form in coming to rest the work P . V 2 /2g, where g is the accelera- tion of gravity. When the weight moves horizontally and strikes normally against the side of a beam which has its ends arranged so as to prevent lateral motion, a lateral dynamic deflection results. Let h be the height corresponding to V 2 /2g, then the external work Ph is to be equated to \Qq as before. Hence the equations $Qq = Ph, in connection with the laws of propor- tionality, give, 7W(aA//)* *=/(**//)* = (*/)* (129)' for the unit-stress and lateral deflection at the instant when P comes to rest. This case rarely occurs except in machines for testing materials by impact. The above formulas are only valid when the unit-stress T is less than the elastic limit of the material. When the load P is light compared to the weight of the beam, they give results which are somewhat too large, because a part of the work due to P is expended in overcoming the inertia of the beam (Art. 131). It will be noted that these formulas are the same as those found in Art. 128 for bars, except that the static deflection / appears instead of the static elongation e. Prob. 129a. A simple beam of steel, 1X1X24 inches, was loaded with a weight of 25 pounds at the middle and the deflection found to be 0.0028 inches. It was then struck horizontally by a hammer weigh- ing 25 pounds which had a vertical fall of 2 inches. Compute the lateral dynamic deflection. Prob. 129#. Compute the deflection of the above beam when a weight of 25 pounds falls vertically upon the middle through a height ;>f 2 inches. The observed deflection in this case being 0.130 inches, what explanation may be given of the smaller computed deflection? ART. 130 INERTIA IN AXIAL IMPACT 331 ART. 130. INERTIA IN AXIAL IMPACT When a moving weight strikes axially upon the free end of a bar, some of its kinetic energy is expended in overcoming the inertia of the particles and putting them into motion, this energy being converted into heat. The load P falling through the height h has the kinetic energy Ph when it touches the end of the bar, but owing to the loss in impact only a part of Ph is effective in elongating and stressing the bar. Let 9 be a number less than unity, called the ' inertia coefficient ', then i)Ph is the energy which produces the stress Q and the elongation q in the bar. All the formulas of Art. 128 may hence be applied to heavy bars when the number i? is known by replacing h by i\h. The object of this article is to determine the value of y in terms of P and the weight W of the bar. The theory of the impact of inelastic bodies may be used for this purpose with close approxi- mation, since the moving weight and the end of the bar are in close contact during the period of the impact. The velocity with which stress is propagated through the bar will be sup- posed to be infinite. The greatest unit-stress Q/a, where a is the section area, must not exceed the elastic limit of the material. As soon as the load P strikes the end of the bar, its velocity V decreases and the end of the bar begins to move. When com- plete contact is attained, both P and the end of the bar are moving with a velocity v which is less than F, and at this instant any element dW of the bar is moving with a velocity u. Accordingly the kinetic energy stored in the load P and in the bar of length / and weight W at this instant is expressed by, Now w=o for the fixed end and u=v for the free end of the bar, and for an infinite velocity of stress u is proportional to the distance y from the fixed end, so that u=v.y/l; also the element dW is W . dy/l. Introducing these values, the integral in the above expression is found to be %W . v 2 /2g or one-third of the kinetic energy which would obtain if the entire bar were in 332 IMPACT AND FATIGUE CHAP, xiv motion with the velocity v. Accordingly, or = is the kinetic energy in the load and bar when the load and the end of the bar are moving with the velocity v. Now it is known from Newton's second law of motion, that the common velocity of two bodies at the instant of complete contact in the impact is v=V . P/(P + Pi), when the body of weight P moving with the velocity V strikes a free body of weight PI which is at rest. For the case in hand, however, one end of the bar is fixed, so that W cannot replace PI in this expression. When the free end of the bar is moving with the velocity v, the element dW at the distance y from the fixed end is moving with the velocity u=v.y/l if stress is transmitted instantaneously. Accordingly, instead of PV = (P+Pi)v there must be written, // pp 7 . r l dW.u=Pv+-j I and hence v = V/(i+%W/P) is the common velocity of the load and the end of the bar. Hence, if h is the height V*/2g } the above expression for K becomes i+lW/P in which T) is the inertia coefficient. When the bar has no weight, then r? = i and the entire kinetic energy is effective in elongating and stressing the bar. When the load and bar are of equal weight, then i) = $$, so that tfPh is effective, while \Ph is lost in heat. For the case of horizontal impact in Fig. 130c, the bar is brought into tension by the load P moving with the velocity V. The effective work yPh is expended in stressing the bar from o up to Q while the elongation increases from o up to q, so that the stored stress energy at the moment of greatest elongation then is %Qq; hence %Qq = t)Ph. Also q/e = Q/P if e is the static elongation due to P. From these two equations the values of Q and q for horizontal impact are found to be, ART. 130 INERTIA IN AXIAL IMPACT 333 in which T? is the inertia coefficient and p is the ratio W/P. If 5 1 is the static unit-stress P/a and T is the dynamic unit-stress Q/a, then also T=S(2i)h/e)*. It is seen that the values of Q and T are less than those given by (128) on account of the energy lost in overcoming inertia. For instance let T\ be the value computed from the first equation in (128)", then Trf is the value when the resistance of inertia is taken into account; thus when W equals P, the dynamic unit-stress is o.jjT\, and when W is equal to $P it is 0.51 TV 1 i T "\ 1 j-JM M ! /' ' Fig. 130a JL51 Fig. 130c For the case of vertical impact shown in Fig. 1306 the weight P falls through the height h upon the end of the bar. Here the equations are the same as in the last paragraph except that Pq is to be added to the second member of the energy equation, since P falls through the distance q after striking the end of the bar, whence ^Qg = P(r/h + q). Solving the equations there results, (130)' as the formulas for vertical impact in which the impact coefficient T) has the value given above. As a numerical example let it be required to find the dynamic stress produced in a vertical wrought- iron bar, two square inches in section area and 18 feet long, by a body weighing 600 pounds and falling through the height of one foot. Here W= 10X6X2 = 1 20 pounds (Art. 17), ,0 = 0.2 and 77=0.882; also e = Pl/aE = 0.00259 inches; then the formula gives (2 = 54840 pounds. The stress due to the static load of 600 pounds is 5 = 300 pounds per square inch, but the dynamic stress due to the same load falling through a height of one foot is T = 27 420 pounds per square inch or about 90 times as great. 334 IMPACT AND FATIGUE CHAP. XIV When the load P strikes upon a shelf or scale-pan of weight Wi at the free end of the bar, the loss of kinetic energy is greater than before, since the inertia of a greater weight must be over- come. Then at the moment of complete contact W\ has the kinetic energy W\ . v 2 /2g and this must be added to the first value of K in this article; also the equation of impact becomes PV=(P + Wi + $W)v. By the same reasoning as before the same equations are deduced, except that the inertia coefficient has the value, i+fr+fr _Wi _W ^(i+pi+W* Pl ~ P p ~ P For example, take the case of the wrought-iron bar in the last paragraph and let the load of 600 pounds fall in a scale-pan which weighs 300 pounds. Here ,01=0.5, ,0=0.2, and TJ =0.612; then Q=4S ioo pounds, and T = 22 550 pounds per square inch, so that the addition of the scale-pan diminishes the dynamic stress about 18 percent. This is a principle of importance in bridge con- struction ; for example, a heavy floor for a suspension . bridge decreases the dynamic stress which may be brought by the live load upon the vertical rods that connect the floor to the cable. Prob. 130. A weight of 60 pounds impinges upon the end of a hori- zontal bar of wrought iron which is 2 inches in diameter and 12 feet long. Find the velocity of the weight which wiU stress the bar up to its elastic limit. ART. 131. INERTIA IN TRANSVERSE IMPACT When a weight P strikes a beam with the velocity V, it has the kinetic energy P . V 2 /2g, and part of this is lost in the impact, while the remainder causes the beam to deflect the amount q which is greater than the deflection / due to a static load P. Let h be the height of fall which will produce the velocity V, and 17 be the inertia coefficient, so that t}Ph is the effective work which deflects and stresses the bar. Then the formulas of Art. 129 will apply if yh is substituted for h. The value of / will now be found for a simple beam in a manner similar to that followed ART. 131 INERTIA IN TRANSVERSE IMPACT 335 for axial impact in the last article, and under the same assump- tions. Let the weight P strike the middle of the simple beam with the velocity F; when complete contact is obtained both P and the middle of the beam are moving with a velocity v which is less than V, and any other point of the beam is moving with a velocity u. Let W be the weight of the beam, / its length, x any distance from the left end, and dW the element which is moving with the velocity u ; then dW = W . dx/l. When the middle point has the velocity v and the deflection y\, the element dW has the deflection y and the velocity u=v.y/y\. Then the kinetic energy stored in load and beam at the instant of complete contact is, s Now to find the value of the integral, it is assumed that the elastic curve has the same equation under a dynamic as under a static load ; from Art. 55 the ordinate y of the elastic curve in terms of the abscissa x and the maximum deflection y, is found to be y=y\($Px 4JC 3 )// 3 . Then, extending the integration over the entire beam, and accordingly the kinetic energy which is available for deflec- tion and stress is, To find the value of v in terms of F, the same reasoning will be followed as in the last article. Instead of PV = (P+Pi)v for the impact of P against a free body PI at rest, there must be written for the beam, PV=Pv+ CdW . u = Pv+^ and hence the velocity at the instant of complete contact of beam and load is v = F/(i + f TF/P). Inserting this in the above expres- 336 IMPACT AND FATIGUE CHAP, xiv sion for the effective work, there is found, K= i + ttW/P ' v/ag '~ in which TJ is the inertia coefficient and p is the ratio W/P. The case of horizontal impact against a beam occurs in some testing machines, the ends of the bar being prevented from mov- ing sidewise while a hammer strikes horizontally against the middle of the beam. Here the discussion is the same as that in Art. 129 except that h is replaced by yh, and T = S(21)h/f>* q=(2T)hf)* (131) are the formulas for dynamic flexural unit-stress and dynamic deflection. Here 5 is the static stress found from the flexure formula (41) and / is the static deflection found from Art. 55 for the load P, and TJ is the impact coefficient whose value is given above. For the case of vertical impact where the load P falls through the height h upon the middle of the beam, the discussion is also the same as that in Art. 129, except that h is replaced by yh, and are the formulas for vertical impact, in which S, f, and r? have the same signification as before. All the formulas of this article are only valid when the unit-stress T is less than the elastic limit of the material. As a numerical example, let a cast-iron simple beam of 36 inches span and 2X2 inches in section have a load of 50 pounds at the middle. The flexural unit-stress and the deflection due to this static load are found from Arts. 48 and 55 to be, 5 = 338 pounds per square inch, /= 0.00243 inches Now let the ends of the beam be prevented from moving side- wise when a horizontally moving load of 50 pounds strikes it at the middle with a velocity due to a fall of 2 inches. Then, disregarding the inertia of the beam, formulas (129)' give the dynamic stress and deflection, r=i3 700 pounds per square inch, 9=0.0986 inches ART. 131 INERTIA IN TRANSVERSE IMPACT 337 Taking the inertia of the beam into account, the weight W is 37.6 pounds, p=W/P = 0.752, 9 = 0.632, and then formulas (131) give, T= 10 900 pounds per square inch, 5=0.0784 inches These values of T are greater than the elastic limit of cast iron and hence cannot be relied upon as exact. The example shows, however, that small velocities of impact may produce high dy- namic stresses in a beam. When the load P falls into a scale-pan of weight W\ which is attached to the middle of the simple beam, the loss in impact is less than when it falls directly upon the beam. For this case the above reasoning is to be modified by replacing P by P + Wi in the first value of K, and also in the second member of PV = (P + Pi)v. The inertia coefficient will then be given by, _W l t _W Pl ~^ P ~~P For example, taking the cast-iron beam of the last paragraph, let the weight of 50 pounds fall vertically upon its middle from a height of 2 inches, there being no scale-pan; then 9 = 0.632 and from (131)', T=n 240 pounds per square inch, 5=0.0808 inches Now let there be a scale-pan weighing 20 pounds into which the load of 50 pounds falls from a height of 2 inches; here pi =0.400* p = 0.752, 9 = 0.505, and then (131)' give, r=io 090 pounds per square inch, 5=0.0725 inches which show that the effect of the scale-pan is materially to diminish the dynamic stress due to impact. The numbers -J and f in the inertia coefficient apply only to a simple beam struck at the middle : they do not apply, how- ever, to other points than the middle of the span. For a beam with fixed ends impinged upon at the middle, || is to be used instead of H and instead of f. For a cantilever beam struck at the end by a load, ^\ takes the place of $1 and f that of $. Prob. 131a. Verify the statements in the preceding sentence. 338 IMPACT AND FATIGUE CHAP, xiv Prob. 1316. Compute the dynamic deflection for Prob. 1296, tak- ing into account the inertia of the beam. ART. 132. VIBRATIONS AFTER IMPACT Referring to the case of axial impact shown in Fig. 130c, it is clear that the velocity of the end of the bar at any instant is a function of the elongation x. When x equals the final elonga- tion q, the velocity is zero; the end of the bar then springs back and increases until x is zero, and then decreases until it becomes zero when x is equal to q; the vibration is next performed in the opposite direction. These vibrations would continue indefinitely were it not for air resistance and molecular friction, but owing to such resistances they become less and less in ampli- tude, until finally the bar comes to rest. Neglecting the weight of the bar in comparison with the load, the following investi- gation gives the time of one vibration. Let v x be the velocity of the end of the bar when the elonga- tion x is attained, and let Q x be the corresponding stress in the bar. The kinetic energy of the moving weight then equals the internal work still to be stored in the bar in increasing x to q, or P . v x z /2g^\Qq-^Q x x. Replacing Q by P . q/e and Q x by P . x/e, where e is the static elongation due to P, this equa- tion becomes, which gives the velocity of the end of the bar for any value of x; this velocity is zero when x +q and also when x= q. To find the time in which this vibration is performed, let / be the number of seconds counted from the instant when x = q; the velocity at any other instant is then dx/dt, as already indicated in the last equation, which may be written, dt f e/g \* /\* . x ->-= 22) or /=(- ) arc sin - 8x \q 2 x 2 / \g-J q and taking the integral between the limits +q and q there is found / = 7r(e/g)*as the time of one vibration. This is the time of one vibration of a pendulum which has the length e. AUT. 132 VIBRATIONS AFTER IMPACT 339 This time is the same for all subsequent vibrations notwith- standing that the amplitude q becomes less and less on account of the air resistance. For the case of the vertical bar, the first equation of this article will be modified by adding P(q x) to the first member, this expressing the work still to be performed while the load P falls through the distance q x. Then are found, the second being derived from the first by replacing v x by dx/dt and integrating as above. By discussing the first equation, it is found that v x is a maximum when x =e as indicated in Fig. 1306, and that v x is zero when x = q. Hence, counting the time from the instant when x = e, the time elapsed when x = q is *($/)*, which is that of one-half a vibration. In the backward vibration the end of the bar moves from x = q to x = e in the same time; the first equation shows, however, that v=o when x= q + 2e, so that the end of the bar moves upward and the time elapsed between x = q and x= q + 2e is ic(e/g)*, which is the time of one vibration. The end of the bar hence oscillates back and forth about the point for which x=e, that is, about the point where it finally comes to rest; while the amplitude q e of the vibrations grows less and less, the time of each vibration remains the same, namely, that of a simple pendulum having a length e equal to the elongation due to the static load P. The above conclusion regarding the time of vibration will be slightly modified when the inertia of the bar is taken into account. Let W be the weight of the bar and W\ that of the scale- pan at its end. Then when P and W\ have the velocity v x the kinetic energy of the moving particles is (P + Wi+$W)v x 2 /2g; for the vertical bar P(qx) is to be added to this to give the total work still to be performed. This sum is to be equated to the stress energy $Qq $QxX which is to be stored in the bar in increasing the elongation from x to q. Stating this equation, there is found, 340 IMPACT AND FATIGUE CHAP. XIV and by the same method as before there results, as the time of one vibration, in which e f is the static elongation due to P + Wi +%W. This formula also applies to the horizontal bar. As a numerical example, let the data of Problem 130 be considered. Here P = 6o pounds, ^ = 125.7 pounds, JFi=o, and the static elongation due to P + $W is e f =0.000 187 inch. Then, taking g as 32.2 X 12 inches per second per second, the time of one vibration is /= 0.0022 seconds, and about 460 oscillations would be performed in one second if there were no frictional resistances. The vibrations of a beam after the impact of a load are in all respects similar to those of a bar. By investigations exactly like those for the bar, it may be shown that the time of one vibra- tion for a simple beam struck at the middle is, where /is the static deflection due to P; here /?/is the static de- flection due to P + Wi + ^W. When the load remains on the beam after impact, the vibrations occur about the position which it finally assumes under the static load ; when it does not remain on the beam, the vibrations occur about the position that it had before the impact. Fig. 133 shows the vibrations of a railroad rail after impact. The above formulas do not apply to the incomplete semi- vibration which occurs during the impact while the end of the bar is descending through the distance e or the middle of the beam through the distance /. The time of all subsequent vibra- tions is the same whatever be their amplitude and is equal to that of a simple pendulum which has the length f)e for the bar or /?/ for the beam. The shorter this length the less is the time / and the more rapid are the vibrations. Prob. 132. When a load falls upon a beam show that the amplitude of the first complete vibration is a little less than ART. 133 EXPERIMENTS ON ELASTIC IMPACT 341 ART. 133. EXPERIMENTS ON ELASTIC IMPACT Numerous experiments have been made to test the formulas for elastic impact derived in the preceding articles, and a dis- cussion of some of them will now be given. These formulas are not exact, because it has been supposed in their deduction that the velocity of transmission of stress is infinite, and that no loss of energy occurs except in impact. The assumption regard- ing velocity of stress does not lead to any appreciable error, but that regarding loss of energy may do so in cases where the falling body is deformed so that it absorbs energy or where a portion of the'energy is expended in deforming the supports of the bar or beam. Simple experiments may be made by the student, using a common spiral spring instead of a bar, so that the elongations may be easily measured. For example, a spiral spring about 32 inches long and weighing 0.6 ounces was found to elongate 0.390 inches under a static load of 10 ounces. When loaded with 8 ounces and the end depressed by the hand and then released, there were counted 304 vibrations in 100 seconds, when loaded with 14 ounces there occurred 230 vibrations in 100 seconds. Here the actual times of one vibration were 0.329 and 0.435 second, while formula (132) gives 0.321 and 0.428 seconds, so that the agreement of experience and theory is very fair. The simplest case of impact on beams is that of a single sudden load which was discussed in Art. 127. Kirkaldy made experiments, about 1860, to test the theoretic law that the deflection under such a load is double that due to an equal static load. A load was attached to a ring placed around the middle of the beam and the ring supported so that its lower surface just touched the upper surface of the beam; the sup- port of the ring was then suddenly withdrawn so that the load acted with its full intensity during the entire period of the deflec- tion, which was registered upon a vertical sheet of paper by a pencil screwed to the side of the beam. Before applying the loads in this sudden manner, the deflections due to gradually 342 IMPACT AND FATIGUE CHAP. XIV applied loads were measured. The beams were of cast iron and laid on supports 9 feet apart. The results here shown are the mean of two or three tests upon different beams. It will be found that for each size of beam the first load gives a unit-stress less than the elastic limit, while the second gives a greater value. Size of Beam Load Deflection in Inches Ratio of Deflections Inches Pounds Gradual Sudden Sudden to Gradual 1X2 112 0.253 0.515 2.04 1X2 224 0.580 o-933 .61 lX 3 224 0.163 0.303 .86 lX 3 560 0.410 0.720 .76 4XiJ . 448 0.770 1.510 .96 4X1* 784 I- 2 7S 2.225 c-73 Thus, for the first beam under 112 pounds at the middle, the unit-stress 5 computed from the flexure formula is 2270, while for 224 pounds it is 4540 pounds per square inch. The aver- age of the three ratios of the deflections for the beams in which the elastic limit was not exceeded is 1.95, which is a fair agree- ment with the theoretic number 2. It was early recognized that the inertia of a bar or beam diminished the theoretic stress and deflection. About 1830 the inertia coefficient y was thought to be of the form jj = i/(i + ;j. . W/P) and Hodgkinson determined by experiments on beams that the value of // was approximately , although the theoretic number j was not derived until several years later by Cox. This form of the inertia coefficient has been generally used since, and it was employed in previous editions of this book. The form deduced in this edition is believed to be more exact according to the theory of impact. The experiments made by Keep in 1899 on tool-steel bars under both horizontal and vertical impact enable an interesting comparison of theory and practice to be made. In one series of tests bars 1X1X24 inches were loaded with weights of 25, 50, 75, and 100 pounds and the corresponding static deflections were found to be 0.0028, 0.0056, 0.0084, an d 0.0112 inches. They were then struck laterally by hammers of the same weights which ART. 133 EXPERIMENTS ON ELASTIC IMPACT 343 swung like pendulums and had a vertical fall of 2 inches. The dynamic deflections due to these weights, as carefully measured by a graphic recording apparatus, are given below. From the given data and observed static deflections the deflections under impact have been computed from formula (131)", and the follow- ing is a comparison of observed and computed values : Swinging P= 25 50 75 TOO pounds Observed #=0.122 0.150 0.175 0.200 inches Computed 9=0.097 0.142 0.178 0.207 inches Experiments were also made by allowing the same weights to fall vertically on the bars through heights of 2 inches. For- mula (131)' applies to this case, and the following is a com- parison of the dynamic deflections as observed and computed: Falling P= 25 50 75 100 pounds Observed 9=0.130 0.159 0.181 0.209 inches Computed 9=0.100 0.150 0.186 0.219 inches It is seen that the comparison is very satisfactory except for the lightest hammer ; it would be expected, however, that the observed values should be always slightly less than the theoretic ones, because some of the work of a falling weight is probably expended in deforming and heating that weight. Fig. 133 gives an autographic record of the deflection of a railroad rail and its subsequent vibrations, taken by P. H. Dud- ley in 1895. A rail 30 feet long and weighing 80 pounds per yard was placed at its extreme ends upon rigid supports and had a scale-pan weighing 210 pounds hung from the middle. Secured to the rail was an attachment carrying a pencil, which recorded the deflection and vibrations upon hori- zontally moving paper. A load of 100 pounds being suddenly applied, the dynamic deflection was 0.24 inches, but when the beam came to rest the static deflection was 0.12 inches. The weight of 100 pounds was then dropped upon the scale-pan from a height of 12 inches, arid the maximum deflection was 0.91 inches; about 240 vibrations then ensued and in about 42 seconds the rail came to rest. Applying formula x (131)' and (131)" to this case, there are found p = 8, ,01 = 2.1, >j ==0.1067, and 344 IMPACT AND FATIGUB CHAP. XIV (7=0.69 inches for the dynamic deflection, which is not a good agreement with the observed value. Applying formula (132) there results ^=0.146 for the time of one vibration, and hence the theoretic time for 240 observations is about 35 seconds. While the numerical results derived from theory do not agree very well with the observations, this experi- ment is a very instructive one, and the figure shows how a beam vibrates back and forth about the position that it occupies after coming to rest. Prob. 133. Explain all the lines and notes on the left-hand part of Fig. 133. See Railroad Gazette, May 31, 1905. ART. 134. PRESSURE DURING IMPACT When a weight P falls from a height h upon the end of a vertical bar, a pressure is produced which is equal at any instant to the stress then existing in the bar. For the case of elastic elongation discussed in Art. 130, the maximum stress is Q, which occurs when the greatest elongation q is attained. The stress Q x which occurs for any elongation x is equal to P . x/e, where e is the static elon- gation due to P. Thus the pressure increases directly with x, becomes P when x = e, and reaches its maximum value Q when the greatest elongation q is reached. Similarly, Art. 131 shows that for a beam under elastic impact, the pressure on the beam increases directly as the deflection y, becomes equal to P when y =/, and reaches its maximum value when the maximum deflection q is attained. ART. 134 PRESSURE DURING IMPACT 345 The actual forces acting between the falling load and the end of the bar differ somewhat from the stress in the bar, because there exists a pressure which overcomes inertia during the first part of the" fall. The exact determination of the actual pressure for the case of elastic impact is a problem of so much complexity that it will not be undertaken here, while its determination is impossible by theory for cases where the elastic limit of the mate- rial is exceeded. From the point of view of the engineer, the pressures that cause motion of the bar or beam are of little im- portance compared with those that cause deformation of the material. The following method is sometimes used for estimating the mean pressure on a beam during its deflection under impact. Let R be this mean or average pressure which is exerted through the deflection q. Then Rq is the work performed by it, while that done by the falling load is P(h + q). Placing these equal there results R=P(i + h/q) for the mean pressure. While this expression is correct for both elastic and non-elastic deflections, it must be borne in mind that it does not give the mean compressive stress on the upper surface of the beam where the impact occurs, but always a greater value, because some of the pressure is ex- erted in overcoming inertia. To illustrate, take the case of the railroad rail discussed at the end of the last article, where a weight of 100 pounds fell from a height of 12 inches and caused an elastic deflection of 0.91 inches. Then R = looX 12.91/0.91 = i 420 pounds for the average pressure during the dynamic deflection. The static load Q which will produce the same deflection is Q = 100X0.91/0.12 = 760 pounds, so that the average pressure which was effective in stressing the beam and causing the deflection was 380 pounds. Undoubtedly, the actual average pressure was about i 420 pounds, but more than two-thirds of this was expended in overcoming the inertia of the heavy beam. In all cases of elastic impact, the mean or average stress is one-half of the maximum, because the stress increases uniformly with the deflection. This is not true when the elastic limit of 346 IMPACT AND FATIGUE CHAP. XIV the material is exceeded, and in general the mean stress is less than one-half of the maximum. The same probably holds for the total pressure that is exerted both to overcome inertia and to cause stress. The maximum unit-pressure that acts between the surfaces of contact will depend, of course, upon the area of contact and upon the manner in which the total pressure is dis- tributed over that area. Prob. 134a. A vertical bar weighing 27 pounds receives a stress of 196 pounds when a load of 100 pounds acts axially upon it. Through what height must a load of 50 pounds drop in order to produce the same stress? Prob. 1346. A ram weighing 2 ooo pounds falls from a height of 20 feet upon a railroad rail laid on supports 6 feet apart, this being one method of testing rails at the mill. Compute the average pressure when the ram deflects a heavy rail 2\ inches; also when it deflects a lighter rail 5 inches. ART. 135. IMPACT CAUSING RUPTURE The cases of impact thus far considered have been those where the greatest unit-stress does not exceed the elastic limit of the material. It is, however, easy to cause the rupture of a bar or beam by allowing a heavy load to drop upon it from a sufficient height. For such cases theory furnishes no formulas and experi- ment is the only source of information. Many tests have been made to ascertain the phenomena of rupture under impact, and the general conclusions derived will now be stated. In 1807 Thomas Young announced the fundamental ideas of the resistance of materials under impact. "The action which resists pressure," he said, "is called strength, and that which resists impulse may properly be called resilience." He stated that the resilience of a body is proportional to its strength and extension jointly, and that it is measured by the height through which a given weight must fall to cause rupture. The resili- ence of beams of the same kind he made proportional to their volumes, as also the resilience of shafts, whether solid or hollow. ART. 135 IMPACT CAUSING RUPTURE 347 At that time the elastic limit of materials was only vaguely recog- nized, and Hooke's law of proportionality of deformation to" stress was often applied to all the phenomena preceding rupture. Young's statements are valid in a general way, but it is now known that there are two divisions of the subject of impact: first, that where the elastic limit is not exceeded and where the term resilience is properly applicable, and second, that where the elastic limit is exceeded and rupture finally occurs. In 1818 experiments were made by Tredgold on wooden beams subject to the impact of a falling ball, and he concluded that the work required to produce rupture was not proportional to the volume of the bar. Hodgkinson in 1835 experimented on cast-iron beams and found that the deflections seemed to be proportional to the velocities of the falling weight. In 1849 were published the results of an extensive series 'of experiments made by a British commission, and here the influence of inertia in diminishing the deflection of a beam under impact was fully recognized. Kirkaldy in 1862 made experiments on axial impact by sudden loads, and found that some bars were broken under loads less than those required when slowly applied. The impact hammer or ram, introduced by Sandberg and Styffe in 1868 for testing railroad rails, has proved valuable for comparative pur- poses since the information obtained is similar to that derived from the cold-bend test. Maitland, in 1887, showed by many experiments on tensile specimens subject to many blows of a falling ram, that the ultimate elongation was much greater than in static tests; the use of many blows to cause rupture introduces, how- ever, complications, and it has been found that the best plan to obtain valid results is to use a load and fall which is just suffi- cient to break the specimen at one blow. When a specimen is broken under tensile impact, the work expended may be ascertained by measuring the area of a stress diagram which is autographically drawn by the machines and which also shows the ultimate elongation. Experiments of this kind made by Hatt in 1904 on various kinds of steel have shown that the work required to rupture a bar by impact is usually 348 IMPACT AND FATIGUE CHAP. XIV greater than that in common static tests where the load is gradu- ally applied. From the mean of about 170 tests, Hatt found that the average work required for rupture by impact was 30 per- cent greater than in static tests, and that the ultimate elonga- tion under rupture was 20 percent greater. The fracture was similar in both impact and static tests, but in the former there were often observed two or more places of marked diminished section, whereas only one occurred in the latter. For round bars of soft steel there appeared to be little difference in ulti- mate elongation and work whether the bars were broken in ten minutes by the common method or in one one-hundredth of a second by impact. When a body is ruptured by impact, it is important that the apparatus should be so arranged that all the work of the falling ram may be expended on the specimen, and not be absorbed by other parts of the apparatus. If the weight falls upon a shelf or pan connected to the bar or beam, this should be made heavy so that work may not be expended in deforming it. When the ends of a tensile specimen are larger than the main part, they should be made very large so as not to absorb energy, and the supports of a beam should be made heavy for a similar reason. Impact tests are of much value in determining the quality of materials, and they are widely used for railroad rails, car and locomotive axles, and other steel pieces which are subject to shocks. The impact tests introduced by Keep for cast iron undoubtedly give valuable information regarding its behavior under shocks. In general it is probable that impact tests show lack of homogeneity of the material better than static tests. Autographic records taken during a tensile impact test give valuable information regarding the elastic limit and ultimate strength of the material. The elastic limit is often found to be higher than under static tests, while the ultimate strength is usually a little lower, the difference between these unit-stresses being much less than in the usual method of testing. For timber Hatt has found that the elastic limit is nearly doubled under impact. ART. 136 STRESSES DUE TO LlVE LOADS 349 The formulas for dynamic stress and elongation deduced in the preceding articles do not apply to cases where the elastic limit is exceeded, and hence all attempts to verify these formulas by experiments on the rupture of bars and beams are fallacious. Hodgkinson broke cast-iron beams under both sudden and gradual loads, and found that the ratio of the latter to the former was always less than 2, which should be the true ratio if the elastic law were applicable. The following, for example, are three of his results for cast-iron beams with a span of 9 feet: Size of Beam Breaking Load in Pounds Ratio of Gradual Inches Gradual Sudden to Sudden Load 1X2 1000 569 1.76 1X3 2 008 I 219 1.65 4X1^ I 911 1 082 1.77 The discrepancy between the theoretic ratio and the actual ones is here not very great because the elongation for cast-iron is small (Fig. 4). For wrought iron, where the elongation is large and the stress curve greatly deviates from a straight line, a greater discrepancy would be expected, and Kirkaldy found that the ratio of gradual to sudden load which caused the rupture of wrought-iron beams ranged from 1.2 to 1.3. Prob. 135. Consult Halt's paper in Proceedings of the American Society for Testing Materials, Vol. IV, 1904, and describe the machine used for making tensile impact tests. ART. 136. STRESSES DUE TO LIVE LOADS A beam or bridge is subject to the action of both dead and live loads, the former including its own weight and the latter the weight of the people, vehicles, or trains that pass over it. The flexural stresses in the beam are found by the application of formula (41) and those in the members of a bridge truss by the methods set forth in Roofs and Bridges, Part I. These stresses are usually computed for dead and live loads separately, regard- ing each as a static load. The live load, however, really produces greater stresses than the computed ones because it is applied 350 IMPACT AND FATIGUE CHAP. XIV quickly, and hence it is customary to multiply the computed siatic stresses by a number called the " coefficient of impact " in order to obtain the increased stress due to suddenness of application. Thus, let S be any computed stress due to the given live load, this being either a unit-stress or the total stress in a member; then tS is the stress due to the quickness with which the load is applied, and i is the coefficient of impact, so that the total stress due to the live load is S + tS or (i + i)S. This use of the word impact does not agree with that of the preceding articles, but it is customary in bridge literature; really this coefficient of impact includes the effect of lateral and vertical oscillations due to irregularities of the track as well as the effect of quickness of application of the live load. Various methods are in use for assigning values of the coeffi- cient i, but in all of them no attention is paid to the time in which the stress is generated, and in fact they rest upon no theo- retical basis except the law that a suddenly applied load pro- duces double the stress of a static one. Some engineers regard t as unity for all cases of live load, and hence double the stress due to the live load in designing the section areas of the mem- bers. Many others take * as less than unity, using higher values for light bridges than for heavy ones, while some make z depend upon the length of span and take it higher for short spans than for long ones. Empirical formulas for * are given in Roofs and Bridges, Part I (New York, 1905). In this important matter experience is in advance of theory since no formula has yet been established for the case of a load P applied to a bar in a given time. When slowly applied P pro- duces a unit-stress S, when suddenly applied it produces a unit- stress 28, if the elastic limit is not exceeded (Art. 127); when applied in a given time /, the unit-stress should lie between S and 2$, Hence there must exist a certain function $(/) so that the dynamic unit-stress is given by T = (f)S; when / is large, T must equal S; when / is zero, T will be 28. Many empirical expressions can be derived which satisfy these limiting condi- tions, but the determination of the theoretic form of (t) is greatly AKT. 130 STRESSES DUE TO LIVE LOADS 351 to be desired, because a knowledge of it would be of much practical benefit in promoting the correct design of members of bridge trusses. The discussion published in Zimmermann, in 1896, regarding the increase in stress and deflection due to the velocity of a live load when crossing a simple beam, is probably the nearest approach to the solution of this important problem, but the formulas deduced are too complicated to be given here. Let v be the velocity of a single load P which rolls over the beam of span / and depth d , and let / be the static deflection due to P. When v is zero, the dynamic unit-stress is S; as v increases the dynamic unit-stress increases, but it can never become as great as 25; when v 2 has the value gl 2 /Sf, where g is the acceleration of gravity, the load P reaches the middle of the span in the same time that gravity causes a body to fall freely through the distance /, then also T = S; when v has a greater value, then T is less than S; when v 2 = gr, where r is the radius of the earth, then T = o. The impor- tant cases hence occur when v is less than (gP/8f)*. From Zim- merman's investigation, there may be written, T=S(i -2/?)/(i - 3 /?) p=SSv2/ 3 Edg which applies only when /? is less than o.i, but this covers most cases of usual speeds of live loads on beams. For example, take a stringer in a bridge floor which is 2 feet deep; the ratio E/S is about 4000, and hence /? = 0.082 for a velocity of 60 miles per hour or 88 feet per second; then T= i.n so that the dynamic stress is ii percent greater than the static stress. While the above theoretic formula gives much lower values of T than those used in bridge practice, it may be noted that it refers only to the middle of the span of the beam, and that for other points the theoretic percentage of increase may be much greater. For the quarter points of a short span under speeds varying from 80 to 100 miles per hour, the investigations of Zimmerman indicate that the percentage may be as high as 65 percent. The empirical percentages used in bridge prac- tice range from 100 to 50 percent, so that it is plain thai these 352 IMPACT AND FATIGUE CHAP. XIV are not too large, particularly when it is considered that they include the effect of the shocks due to the hammer of wheels which are not properly balanced. Prob. 136a. Deduce the condition that the load P shall reach the middle of the span in the same time that a body falls freely by gravity through the deflection /. Prob. 1366. Consult Zimmermann's Schwingungen eines Tragers mit bewegter Last (Leipzig, 1896), and show that the above formula agrees with the one given by him on page 39. ART. 137. FATIGUE OF MATERIALS The ultimate strength Su is usually understood to be that steady unit-stress which causes rupture of a bar at one appli- cation. Experience and experiments, however, teach that rup- ture may be caused by a unit-stress somewhat less than Su when it is applied a sufficient number of times to a bar. The experi- ments made by Wohler from 1859 to 1870 were the first that indicated the laws which govern the rupture of metals under repeated applications of stress. For instance, he found that the rupture of a bar of wrought iron by tension was caused in the following different ways : by 800 applications of 52 800 pounds per square inch by 107 ooo applications of 48 400 pounds per square inch by 450 ooo applications of 39 ooo pounds per square inch by 10 140 ooo applications of 35 ooo pounds per square inch The range of stress in each of these applications was from o to the designated number of pounds per square inch. Here it is seen that the breaking unit-stress decreases as the number of ap- plications increase. In other experiments where the initial stress was not o, but a permanent value S, the same law was seen to hold good. It was further observed that a bar could be strained from o up to a stress near its elastic limit an enormous number of times without rupture, and it was also found that a bar could be ruptured by a stress less than its elastic limit under a large number of repetitions of stress which alternated from tension to compression and back again. AKT. 137 FATIGUE OF MATERIALS 353 Wohler's experiments were made on repeated tensile stresses, repeated flexural stresses, and on flexural stresses alternating from tension to compression, these being produced by a machine which brought repeated loads upon the specimen for long periods of time, as high as forty millions of repetitions being made in some cases. Similar experiments were later made by Bauschinger and others on steel, and from the recorded results the following laws may be stated : 1. The rupture of a bar may be caused by repeated appli- cations of a unit-stress less than the ultimate strength of the material. 2. The greater the range of stress, the less is the unit-stress required to produce rupture after an enormous number of applications. 3. When the unit-stress in a bar varies from o up to the elastic limit, an enormous number of applications is required to cause rupture. 4. A range of stress from tension into compression and back again, produces rupture with a less number of applica- tions than the same range in stress of one kind only. 5. When the range of stress in tension is equal to that in compression, the unit-stress that produces rupture after an enormous number of applications is a little greater than one- half the elastic limit. The term ' enormous number ' means about 40 millions, that being roughly the number used by Wohler to cause rupture under the conditions stated. For all cases of repeated stress in bridges, this great number will not be exceeded during the natu- ral life of the structure; for locomotive axles and moving parts of machines, however, a larger number of repetitions of stress may occur. The word ' fatigue ' means the loss of molecular strength tinder stresses often repeated. When a bar is stressed above the elastic limit its temperature increases due to internal molecular friction (Art. 147) and it is known that the elastic properties of the material are injured. Hence in a general way it is easy to explain why fatigue occurs under repeated stressses that exceed 354 IMPACT AND FATIGUE CHAP, xiv the elastic limit. An examination of fractures of bars after an enormous number of repetitions shows certain small surfaces where sliding or shearing has occurred ; these are called ' micro- flaws,' although they can often be seen without the use of a micro- scope. In iron and steel these flaws begin along the surfaces of the ferrite crystals and later are extended to cause cracks along the cleavage planes of the crystals. When the elastic limit is not exceeded it is not so easy to under- stand why fatigue should occur under repeated stresses. How- ever, physical and thermodynamic discussions have proved that small changes in temperature occur when a bar of metal is stressed within the elastic limit, it becoming cooler under tension and warmer under compression. The measurements of these changes made by Turner, in 1902, have shown that these changes in tem- peratures continue at a uniform rate up to about three-fifths of the elastic limit, and that then a marked change occurs, the bar under tension then beginning to grow warmer while the temperature of the bar under compression increases at a more rapid rate. It thus appears that for stresses higher than about three-fifths of the elastic limit, at least, energy is converted into heat under repeated applications; probably this occurs also at lower stresses when repeated stresses range from tension into compression in a bar, or when a beam is subject to alternating flexure. The valuable experiments of Turner hence throw light upon the reason why fatigue occurs under alternating stresses, and it is likely that further investigations in this direction may lead to other important conclusions. The discussions in Arts. 146-147 indicate that internal friction occurs under stresses that do not exceed the elastic limit, and this point of view is also one which will assist future investigations. In Art. 7 it was recognized that allowable unit -stresses should be less for bars subject to varying loads than for those carrying steady loads only. It has indeed long been the practice of de- signers to grade the allowable working stresses for bars according to the range of stress to which they might be liable to be sub- jected. The above laws of fatigue furnish a method of doing ART. 138 STRENGTH UNDER FATIGUE 355 this which has been used by some engineers, and formulas for that purpose will be deduced in the next article. Prob. 137a. How many years will probably be required for a loco- motive axle to receive forty million repetitions of flexural stresses ? Prob. 1376. Consult Turner's paper in Transactions American Society of Civil Engineers, 1902, Vol. 48, and examine the thermal stress curves derived from his experiments. ART. 138. STRENGTH UNDER FATIGUE. Consider a bar in which the unit-stress varies from S' to S, the latter being the greater numerically. Both S' and S may be tension or both may be compression, or one may be tension and the other compression; in the last case the sign of S' is to be taken as minus. Consider the stress to be repeated an enor- mous number of times from S' to 5, and rupture to then occur under the greater unit-stress S, which may be called the strength of the material under fatigue. By the second law above stated 5 is some function of S S' ; this is equivalent to saying that 5 is a function of 5(i S'/S), or more simply a function of S'/S. Now if P' and P be the total stresses on the bar, the ratio S'/S equals P' /P, and hence the unit-stress S which causes rupture after an enormous number of repetitions is a function of P' IP. A formula for S when the limiting stresses P* and P are both tension or both compression, so that P'/P is always positive, was deduced by Launhardt in 1873. Let the values of this ratio be taken as abscissas ranging from o to i, while those of S are ordinates, as in Fig. 1380. Let the function of P'/P be supposed to represent a straight line which has the equation 5 = C\ + C 2 (P'/P) in which C\ and Cz are constants to be determined. Let S u be the ultimate strength of the material and S e the elastic limit. Now if P'/P is unity, then 5 is S u and hence Ci +C 2 =S U ? also, from the third law of the last article, S is S e when P'/P is zero. These two conditions give C\ =S e and C2=S u S e , and the equa- tion of the straight line becomes, S=S e +(S u -S e )P'/P or S = 356 IMPACT AND FATIGUE CHAP. XIV which is Launhardt's formula for the unit-stress S that ruptures the bar after an enormous number of repetitions of a load that ranges from P f to P. For structural steel, using the mean values of S u and S e in Tables 2 and 3, this becomes S = 35 000(1 +\P f /P) ; for wrought iron it becomes 5 = 25 000(1 +P'/P). For example, let a bar of structural steel range in tension from 80 ooo to 160 ooo pounds; then P'/P is 0.5, and 5=47 5 pounds per square inch is the unit-stress that will cause rupture after an enormous number of repetitions. A formula for 5 when the bar ranges in stress from P r to P, one being tension and the other compression, and P being the greater numerically, was deduced by Weyrauch in 1877. Here P'/P is always negative, and the law connecting it with S is again assumed to be S = Ci+C 2 (P'/P); Fig. 1386 represents this case. Let S e be the unit-stress at the elastic limit and S a the unit-stress which, under the fifth law of the last article, causes rupture when the load alternates from a certain value in tension to the same value in compression. By the third law, if P'/P is zero, then 5 is S e and hence Ci =5,. By the fifth law, if P'/P is -i, then S is S a and hence C 2 =S e -S a . The equation of the straight line therefore becomes, s=s.+(s.-s a )p f /p or s which is Weyrauch's formula for the unit-stress which ruptures a bar after an enormous number of repetitions of a load alter- nating from tension to compression and back again. S a is usually taken as %S, in the absence of knowledge regarding its exact value. For structural steel the formula becomes S = 35 ooo (i + %P'/P), in which P'/P is always negative. Thus, if P' is 80 ooo pounds compression and P is 160 ooo pounds ten- sion, then P'/P*=o.$, and 5 = 26200 pounds per square inch is the unit-stress that will cause rupture. Another formula, deduced by the author in 1884, gives values of S for both positive and negative values of P'/P, and thus includes the two cases discussed above. The law of variation of 5 is assumed to be represented by a curve joining the tops of ART. 138 STRENGTH UNDER FATIGUE 357 the three ordinates S w S n S m in Fig. 138c. The simplest curve is a parabola given by the equation S = Ci+C 2 (P'/P) +C 3 (P'/P) 2 . To determine the three constants, consider, first that S becomes S u when P'/P=+i, and hence Ci+C 2 + Ca=S u ; secondly, that 5 becomes S e when P'/P=o, and hence Ci=S e ; thirdly, that 5 becomes S a when P'/P = i, and hence C\ C 2 + C 3 =5j. From these conditions, the values of C\, C 2 , 3 are found and the equation of the parabola becomes, S=S, + i(S,-S) ^ +HSu+S a -aS.) 0iy (138) which is a formula for the rupturing unit-stress S when the total stress ranges an enormous number of times from P r to P. When P' and P are both tension or both compression, the ratio P'/P is positive; when one is tension and the other compression, P'/P is negative. It is seen that (138) gives values of 5 a little smaller than those found from the straight-line formulas. Fig. 138a Fig. 1386 Fig. 138c For structural steel, where S u = 60000, 5,, = 35000, and = ij 500 pounds per square inch, the formula (138)- reduces to, For a bar of such steel in which the stress ranges from 180 ooo pounds tension under dead load to 540 ooo pounds tension under live load, the value of P'/P is +, and the formula gives S 42 500 pounds per square inch. If the stress ranges from 1 80 ooo pounds compression to 540 ooo pounds tension, then P'/P is , and 5 = 28 300 pounds per square inch. When the above formulas are used for designing, a factor of safety is applied, the computed values of S being divided by this factor, and thus the allowable unit-stress is obtained. About 1880 the formulas of Launhardt and Weyrauch were 358 IMPACT AND FATIGUE CHAP. XIV extensively used in determining the allowable unit-stresses for designing members of bridge trusses, but their use has gradu- ally been replaced in the United States of America by other methods. Since no unit-stress used for a bridge member can be allowed to be greater than about one-half of the elastic limit of the material, it is claimed by many engineers that the ideas of fatigue cannot enter in making the design. Nevertheless this question must not be ignored, especially for locomotive axles and tires and for parts of machines subject to shocks. Tests of materials under repeated stresses, or endurance tests as they are sometimes called, are still in progress at the Watertown Arsenal, at the testing laboratory of the Pennsylvania Railroad, and in other places; when sufficient records have been accumu- lated they will prove of great value in further investigations into this subject. Prob. 138a. A short bar of wrought iron is subject to repeated stresses ranging from 16000 pounds compression to 80000 pounds tension. What should be the area of its cross-section for a factor of safety of 5? Prob. 1386. Consult Tests of Metals, published annually by the ordnance office of the U. S. Army, and describe some of the endurance tests on rotating shafts made by Howard. ART. 139 PRINCIPLES AND LAWS 359 CHAPTER XV TRUE INTERNAL STRESSES ART. 139. PRINCIPLES AND LAWS In Art. 13 it was explained that a bar under tension suffers a contraction in its section area, each lateral dimension having a unit-contraction proportional to the longitudinal unit-elonga- tions, when the elastic limit of the material is not exceeded. Let S be the tensile unit-stress, e the unit-elongation, A the fac- tor of lateral contraction, and E the modulus of elasticity of the material; the lateral unit-shortening is then Ac. Since S = eE is the relation between S and e (Art. 9), it may be considered that the lateral unit-shortening Ae corresponds to a unit-stress T which has such a value that T = XeE, where T is a compressive unit-stress which would produce the unit-shortening As in the absence of any axial stress. Thus, T = AS is called a true internal stress which acts as a compression at right angles to the axis of the bar. The mean value of A for wrought iron and steel is about . Accordingly, a steel bar under the tensile unit-stress 5 suffers a true internal compressive unit-stress of $S in all directions at right angles to its length; similarly, a steel bar under the com- pressive unit-stress 5 suffers a true internal tensile unit-stress of $S in all directions at right angles to its length. For instance, let a steel bar 2X3 inches in section and 10 inches long be sub- ject to a tension of 90000 pounds; the axial tensile unit-stress S is 15 ooo pounds per square inch and the lateral internal com- pressive unit-stress is 5 ooo pounds per square inch. The same lateral deformation of the bar, when no axial load is acting, might be produced by two compressive loads acting at right angles to each other, one uniformly distributed over the side of 20 square inches area and the other over the side of 30 square inches area ; it may be shown from the following discussion that these two compressive loads are 150 ooo and 225 ooo pounds. 360 TRUE INTERNAL STRESSES CHAP, xv When applied tensile forces act upon a body in three directions, each force being at right angles to the plane of the other two, there is an elongation due to each force in its own direction and a shortening in directions normal to it. It is a reasonable assump- tion that each force produces its deformations independently of the other two, and this is also justified by experience and experi- ment. The true stress in any direction depends upon the actual deformation in its direction. The letter S will denote the apparent unit-stress as computed by the methods of the preceding chapters, while T will denote the true unit-stress corresponding to the actual deformation. The injury done to a body does not depend upon the actual stress or pressure but upon the actual deforma- tions produced, and the true stresses are those corresponding to these deformations. Let a homogeneous parallelepiped be subject to tensile forces acting normally upon its six faces, those upon opposite faces being Sj equal. Let the edges of the parallel- epiped be designated by 01, 02, 03, as in Fig. 139. Let Si be the normal unit-stress upon the two faces per- pendicular to the edge 01, and 52 and S 3 those upon the faces normal to 02 and 03; thus the directions of Si, Sg, S 3 are parallel to 01, 02, 03, respectively. Then, supposing that the modulus of elasticity E and the factor of lateral contraction ^ are the same in all directions, the true unit -elongations 1, 2, 3 in the three directions are found from the expressions, Now EI maybe designated by T\, this being the unit-stress which would produce the elongation 1 in the direction 01 if 52 and 83 were not acting; also the unit-stresses Es 2 and 3 may be desig- nated by T 2 and TV Hence it follows that ri=Si-AS 2 -AS 3 r 2 =S 2 -AS 3 ->lSi r 3 = S 3 -ASi-AS 2 (139) are the true stresses acting in the three rectangular directions. If any stress 5 is compression, it is to be taken as negative in the ART. 139 PRINCIPLES AND LAWS 361 formulas, and then the true stresses are tensile or compressive according as their numerical values are positive or negative. For example, let a cube be stressed upon all sides by the apparent unit-stresses 5; then the true internal unit-stress T is S(i 2A); for steel A = $, and T = %S, and thus the linear defor- mation is only one-third of that due to a unit-stress 5 applied upon two opposite faces. Again, if a bar has a tension Si in the direction of its length, and no stresses upon its sides, then T l =5i, while T 2 = T 3 = -ASi- As a simple example, let a steel bar 2 feet long and 3X2 inches in section area be subject to a tension of 60 coo pounds in the direction of its length and to a compression of 432 ooo pounds upon the two opposite flat sides. Here Si = 60 000/6 = 10 ooo pounds per square inch, $2 = 432 000/72 = 6 ooo pounds per square inch, and 83=0. Then from (139), taking X as , the true internal stresses are TI = +12 ooo, T 2 = -9 330, T 3 = i 330 pounds per square inch, and it is thus seen that the true tensile unit-stress is 20 percent greater than the apparent, while the true compressive unit-stress is more than 50 percent greater than the apparent. The term ' apparent stresses ' will be used to indicate the stresses computed by the methods of the previous chapters where no lateral deformation has been taken into account. In Chapter XI such stresses have been combined in order to obtain the resultant maximum tension, compression, and shear, but it will now be shown that the true internal stresses corresponding to the actual deformations of the material are often much greater than the apparent ones. It is very important to consider these true stresses in many problems of investigation and design which occur in engineering practice. Prob. 139. A common brick, 2^X4X8^ inches in size, is subject to a compression of 3 200 pounds upon its top and bottom faces, 500 pounds upon its sides, and 60 pounds upon its ends. Taking >l as 0.2, compute the true internal stresses in the three directions. 362 TRUE INTERNAL STRESSES CHAP, xv ART. 140. SHEAR DUE TO NORMAL STRESS The term 'normal stress' is used for the tension or compression that acts normally to a plane in the interior of a body. The rectangular bar in Fig. I40a may be said to be acted upon by normal loads, and planes perpendicular to these loads are said to be subject to normal unit-stress. Other normal stresses also act upon other planes within the bar, but it will be shown that the normal unit-stresses upon such planes are less than upon planes perpendicular to the directions of PI and P 2 . Let Si be the normal unit-stress on a plane perpendicular to PI, and S 2 that on a plane perpendicular to P 2 ; then the true unit-stresses TI and T 2 , as also the true unit-stress T$ at right angles to these, are readily found by the methods of Art. 139. There also exist shearing stresses in the bar which will now be considered. Let any plane be drawn cutting it obliquely and\ let the given forces be resolved into components parallel to this plane; the sum of these com- ponents forms a shear acting along the plane, and the intensity of the shear will vary with the inclination of the plane. It is required to find the maximum shearing unit-stresses. Let / be the length, b be the breadth, and d the depth of the rectangular bar in Fig. 1400, subject to the two normal forces PI and P 2 , while there is no force acting upon the side whose area is Id. The normal unit-stresses then are Si=Pi/bd, S 2 = Pz/bl, S 3 =o. Let Fig. 140& represent any elementary parallele- piped in the interior of the bar, having the length dx, depth dy, and width unity; then Sidy is the normal stress upon its ends, and S 2 dx is the normal stress upon the upper and lower sides. Let 6 be the angle which the diagonal dz makes with dx, and let S' be the shearing unit-stress that acts along the diagonal. The total shearing stress along the diagonal then is S'dz and this is equal to the algebraic sum of the components of the normal stresses in that direction. Accordingly, noting that dy/dz = sin# and &v/<5z=cos0, there result, S'3z=Sidy .cos0-S 2 dxsmd or 5 / = (5i-5 2 )sin 6 cos 6 and the second equation gives the shearing unit-stress along ART. 140 SHEAR DUE TO NORMAL STRESS 363 any plane which makes the angle with the direction of Si. The maximum value of 5' occurs for 6= +45 or 6= 45, and for both cases may be written 5' = %(Si -S 2 ), that is, the maxi- mum shearing unit-stress occurs on two planes which bisect the directions of Si and 5*2, and its numerical value is equal to one- half of their difference. The broken lines in Fig. 14X)a show these two sets of planes. fes* Fig. 140o Fig. 1406 The same conclusion follows when true stresses are con- sidered. If TI and T 2 are the true unit-stresses due to the apparent unit-stresses Si and S 2 , then the true maximum shear- ing unit-stresses are equal to one-half their difference, and they act in planes which bisect the directions of TI and TV Accord- ingly, the formulas, S' = l(Si-S 2 ) and T'=l(Ti-T 2 ) (140) give the maximum internal shearing unit-stresses. These may be either positive or negative, but it is best to consider internal shear as a signless quantity, since it acts in opposite directions on opposite sides of the plane. As a numerical example, take a cast-iron bar for which the factor of lateral contraction A is ; let it be one square inch in section area and be subject only to an axial tension of 2 400 pounds. Then Si = + 2 400 and S 2 = o, whence the maximum apparent shearing unit-stress is S' = i 200 pounds per square inch. From (139) the true axial unit-stress is T\ = + 2 400 and the true lateral unit-stress is T 2 = -600 pounds per square inch. Accordingly the maximum true shearing unit-stress is T = i 500 pounds per square inch, which is 25 percent greater than the apparent. It is indeed very common to find that the true stresses based on the actual deformations are much larger than the stresses computed from the common theory, and this is one reason for the use of high factors of safety. 364 TRUE INTERNAL STRESSES CHAP, xv The above discussion applies equally well when one or both of the applied loads is compression. For example, let the axial unit-stress Si be tension and the lateral unit-stress 52 be com- pression, each equal to 2 400 pounds per square inch. Then the apparent maximum shearing unit-stress is S f = (2 400 + 2 400) = 2 400 pounds per square inch. For A = , the true axial stress is T\ = + 3 200 and the true lateral stress is T% = 3 200, so that the true maximum shearing unit-stress is T' =3 200 pounds per square inch, which is 33 percent higher than S f . When Si and $2 are equal numerically, both being tension or both compression, then S'=o, and also T' =o; that is, a parallelepiped under uniform stress in two rectangular directions has no internal shearing stress. The same is true when a body is acted upon by equal tensions or pressures in three rectangular directions, for the third stress S 3 exerts an equal influence upon the two normal to it. When there are three unit-stresses Si, S 2 , S 3 , acting upon a paralellopiped in three rectangular directions, the shearing unit-stress on a plane parallel to Si and S 2 is not influenced by 5 3 , and hence |(Si $2) is the maximum shearing unit-stress for such a plane. Similarly, (Si S 3 ) and |(S 2 -S 3 ) are the maximum shearing unit-stresses for planes parallel to Sj and .S 2 and to 2 and S$ respectively. The same holds true for the true stresses T\, T 2 , T 3 -, an algebraic discussion of this case will be found in Art. 178. As an example, let a rectangular bar be subject to an axial tension of 3 ooo pounds per square inch, and to a compression of 6 ooo pounds per square inch upon two opposite sides. Here Si = +3 ooo, S 2 = -6ooo, 5 3 =o, and hence the three maximum apparent shearing stresses are 4500, 3000, i 500 pounds per square inch. But from (139), taking A as , the true axial and lateral stresses are T\ = + 5 coo, T 2 = 7 ooo, TZ = + 1 ooo, whence the three maximum true shearing stresses are 6 ooo, 4 ooo, 2 ooo pounds per square inch. Here the true axial stress is 67 percent greater than the apparent, while the true shearing stresses are 33 percent greater than the apparent ones. ART. 141 COMBINED SHEAR AND AXIAL STRESS 365 Prob. 140. Compute the maximum shearing unit-stresses, both apparent and true, for the data given in Problem 139. ART. 141. COMBINED SHEAR AND AXIAL STRESS Formulas were deduced in Art. 105 for the maximum apparent stresses of tension, compression, and shear, due to the simul- taneous action of an axial load and a cross-shear. It was shown that there are two planes at right angles to each other upon which there are no shearing stresses, one being under normal tension Si and the other under normal compression S 2 . Let 5 be a given axial unit-stress of tension and S s the shearing unit-stress acting at right angles to it. Then the formulas give the following values of the maximum tensile stress Si, the maximum compressive unit- stress ,$2, and the maximum shearing unit-stress S', Si = iS+(S 8 2 +iS 2 )* S 2 = iS-(S 8 2 +i$ 2 )* S' = (S 8 2 +iS 2 )* (141) It is here seen that the value of S' is the same as that of J(5i $2). Hence when Si and S 2 have been computed, the subsequent dis- cussion is exactly like that of the last article. When S is tension, as above considered, Si is tension and S 2 is compression ; when S is compression, Si is compression and S 2 is tension. Let A be the factor of lateral contraction, and TI and T 2 the true internal unit-stresses corresponding to Si and S 2 . Then, by (139), the value of TI is Si->lS 2 and that of T 2 is S 2 -ASi; substituting in these the above values of Si and S 2 , there are found, from which TI and T 2 may be directly computed. For steel the mean value of A is , and hence for this material, are the true maximum tensile and compressive unit-stresses due to an axial unit-stress S and a shearing unit-stress S. acting at right angles to it. The true maximum shearing unit-stress acts along a plane that bisects the directions of Si and S 2 and its value is T'=%(Ti-T 2 ). The directions of TI and T 2 are the same as those of Si and S 2 , and may be found from the expression for cot2< 366 TRUE INTERNAL STRESSES CHAP, xv deduced in Art. 105, namely, cot20 = S/S, where the two values of give the angles included between the direction of S and those of the planes against which Si and 2 act. As a numerical illustration, take the case of a steel bolt subject to an axial tension of 2 ooo and to a cross-shear of 3 ooo pounds per square inch. Here S = + 2 ooo, S g = 3 ooo, from which Si = + 4 1 60 and 82 = 2 160 pounds per square inch are the apparent maximum unit-stresses of tension and compression, and their directions are given by cot 2 = . The two values of (j> then are 54 13' and 144 13', which show that Si makes an angle of 35 47' and S 2 an angle of 54 13' with the axis of the bolt. The true unit-stresses have the same directions and their values are Ti = +4&8o, T 2 = 3550 pounds per square inch. For the shearing unit-stresses the maximum values are 5' =3 160 and T =4 220 pounds per square inch. Here the true maximum tension is 17 percent greater than the apparent, the true compression is 64 percent greater, and the true shear is 33 percent greater. There is also a third true compression T 3 = 670, and two other true shears smaller than 5' which act along planes parallel to Si. It thus appears that the actual internal stresses corresponding to the deformations of the material are far more complex than and quite different in value from those computed by the common theory. The above discussion considers a bar subject only to a single axial stress S and to a cross-shear 5. This is a very common case in engineering practice, but other cases far more complex occasionally occur where the bar is subject to both axial and lateral stresses and to shears in different directions. The methods of treating these cases will be explained in Arts. 177 and 178. Prob. I41a. A horizontal bar of cast iron, 2X2X6 inches, is under an axial compression of 20 ooo pounds, and under shear from a uniform vertical load of 8 ooo pounds which rests upon it. Compute the maximum unit-stresses, both apparent and true, and find the direc- tions which they make with the axis of the bar. Prob. 1416. What must be the value of S a in (141) in order that Si and S 2 may be equal? ART. 142 TRUE STRESSES IN BEAMS 367 ART. 142. TRUE STRESSES IN BEAMS The first set of formulas in the last article furnishes the means of ascertaining the maximum apparent stresses at any point in the beam, 5 being the horizontal unit-stress for that point as computed from the flexure formula and S a the shearing unit-stress as determined by Art. 108. From these the apparent unit-stresses Si and 52 which act at the given point are computed and then the true unit-stresses T\ and TV The discussion in the last article also shows that the directions of T\ and T 2 are the same as those of Si and S 2 , and hence the lines of maximum stress shown in Fig. 109 apply equally to both. At the upper and lower surfaces of the beam where the shear is zero, the unit-stress 5, computed from the flexure formula, is also the true unit-stress ; at the neutral surface where the shear is the greatest, the true normal stresses on planes where there is no shear are greater than the apparent ones. Since the unit-stresses on the upper and lower surfaces are greater than for any other points in a cross-section, it is never necessary in practical problems to investigate the true stresses in a beam. The upper surface of a simple beam is in compression while the lower surface is in tension. The width of the beam hence suffers a lateral expansion in its upper part and a lateral contrac- tion in its lower part, so that a rectangular section becomes a trapezoidal one when the load is applied. This change is so slight that it is rarely observed, but there is little doubt that it can be detected by precise measurement. For example, take a steel beam, 6x6 inches in section and so loaded that the flexural unit-stress at the dangerous section is 30 ooo pounds per square inch. The unit-shortening of the upper surface and the unit- elongation of the lower surface will then be =30 000/30 ooo ooo =0.001 ; and hence the total lateral contraction of the width of the beam at the dangerous section will be e=Xo.ooiX6 =0.002 inches, a quantity that can be easily measured with precise calipers. A uniform load resting upon the upper surface of a simple 368 TRUE INTERNAL STRESSES CHAP. XV beam produces a vertical compression which is to be combined with the horizontal compressive unit-stress in order to obtain the true stresses. Let S\ be the flexurai unit-stress and S 2 the compressive unit-stress due to the uniform load. Then the true maximum compressive unit-stress in a horizontal direction is TI =Si-AS 2 , while that in a vertical direction is r 2 = S 2 -ASi. It thus appears that each compression diminishes the effect of the other. Usually S 2 will be small compared with Si, so that computations are rarely necessary. A concentrated load resting upon the upper surface of a simple beam may, however, produce a high unit-stress S 2 . The experiments made by J. B. Johnson in 1893, on the contact between the surface of a car wheel and a railroad rail, showed that the mean compressive unit-stress was about 80 coo pounds per square inch. A heavy pressure like this entirely alters the distribution of the stresses and the directions of the lines of maxi- mum stress in its vicinity, for TI may become tension, if the elongation due it can occur. In the contact of wheels on rails, however, there is no permanent deformation due to the heavy vertical compressive stress, which indicates that lateral flow or elongation could not occur. Under such circumstances there is doubt as to the correctness of the applicability of the preceding theory to the determination of the true stresses. The case is perhaps analogous to that of stresses due to change in tempera- ture, where heavy stresses may arise with but little change in length; thus, a fall of 200 degrees Fahrenheit in temperature will produce a unit-shortening of about 0.0014, but this is prob- ably sufficient to break a wrought-iron bar, if it is prevented from shortening and is under an initial tension of about 30 ooo pounds per square inch. Prob. 142a. A steel I beam, 20 inches deep and weighing 80 pounds per linear foot, carries a uniform load of 24 ooo pounds on a span of 30 feet. Compute the values of Si and S 2 at the dangerous section and find the true stresses. Prob. 1426. How must a simple beam be loaded so that the elastic curve is an arc of a circle ? ART. 143 STRESSES DUE TO SHEAR ART. 143. STRESSES DUE TO SHEAR It is shown in Art. 6, and also in Art. 105, that forces of ten- sion or compression acting upon a body produce not only internal tensile or compressive stresses, but also internal shear- ing stresses. Conversely, an external shear acting upon a body produces in it not only internal shearing stresses, but also internal tensile and compressive stresses. For example, the rectangle A BCD in the web of a plate girder, shown in Fig. I43a, may be considered. Let V be the shear at the sections AB and CD, which are taken very near together so that the weight in the rectangle itself can be dis- regarded. This vertical shear or couple must, be accompanied by a horizontal shear V\, which in this case is caused by the resistance of the flange rivets. Let the thickness of the material be one unit; then if S and Si are the shearing unit- stresses, their values are S = V/AB and Fig. 143a Si = Vi/AD. Now taking either A or D as an axis of moments, the equation of moments is VxAD = ViXAB, and hence V/AB = Vi/AD, that is, the shearing unit-stresses S and Si are equal. This is without regard to the weight of the rectangle itself, which will cause a slight modification, because the V on the left will then be greater than the V on the right. But if AD is very small, the conclusion is strictly true that the horizontal shearing unit-stress is equal to the vertical shearing unit-stress. The vertical and horizontal shears in the above figure tend to deform the rectangle into a rhomboid, thus causing tension along the diagonal BD and compression along the diagonal AC. At every point in the rectangle, then, the vertical shearing unit- stress 5 and the equal horizontal unit-stress Si combine to cause the tension and the compression 2*5 acting with inclina- tions of 45 degrees to the shears. Dividing each of these by the area 2* normal to its direction, it is seen that both the ten- 370 TRUE INTERNAL STRESSES CHAP, xv sile and the compressive unit-stress is S; that is, a shearing unit-stress causes equal tensile and compressive unit-stresses in directions making angles of 45 degrees with the shears. This may also be proved from the discussion in Art. 105 or from Art. 141. Thus, in formula (141) let the axial tensile unit-stress 5 be made zero, then the maximum tensile and com- pressive unit-stresses Si and .$2 are each equal to S B . If, how- ever, S a =o, then the maximum shearing unit-stress is %S. Accordingly an axial tension or compression on a bar produces a shearing unit-stress equal to one-half the tensile or compres- sive unit-stress, but the action of a shear produces tensile and compressive unit-stresses equal to the shearing unit-stress itself. This may be regarded as a most fortunate arrangement in view of the fact that the shearing strength of materials is usually less than the tensile strength. The above relates to apparent stresses only. The true stresses TI and 7^2, corresponding to S\ and 2, are those that correspond to the actual deformations, and by (139) their values are ^=81-^2 and r 2 =5 2 -A5i, where Si and 5 2 are to be taken as positive for tension and as negative for compression. For example, let Fig. 1436 represent one face of a cube which is subject to the shearing unit- Fig 1436 stress S of 5 ooo pounds per square inch, each edge of the cube being unity. The distortion of the square is shown greatly exaggerated by the broken lines, and both the tension along the longer diagonal and the com- pression along the shorter diagonal are equal to 5 ooo pounds per square inch. Now if the factor of lateral contraction X is \, then the true stress along the longer diagonal's TI = +6 250, while that along the shorter one is T^, = 6250 pounds per square inch, so that the true stresses of tension and compres- sion are 25 percent greater than the apparent ones. It may be noted that while shear produces distortions, it does not cause changes in the volume of a body. Thus, for the above figure, let e be the elongation or shortening of the diagonals, ART. 144 TRUE STRESSES IN SHAFTS 371 then the length of the longer diagonal is 2* + and that of the shorter diagonal is 2*-e; the area of the rhombus then is (2* + e) (2* ) =I which is the same as that of the square before it was subjected to shear, s 2 being a negligible quantity. Prob. 143. A steel beam, 2X2X6 inches, is supported at its ends and has a concentrated load 40000 pounds at its middle. Compute by Art. 108 the greatest shearing stress which occurs at the neutral axis, and then find the true tensile and compressive unit-stresses which exist there. Draw a diagram showing the directions of these stresses. ART. 144. TRUE STRESSES IN SHAFTS When a round shaft is acted upon by torsion alone, the stresses are those of shearing, and these act along every section normal to the axis, the maximum S g occurring at the surface (Art. 90). Any square in one of these normal sections is hence acted upon by two equal and opposite shears, as shown in Fig. 1436, and these produce apparent stresses of tension and compression in directions bisecting those of the shears. The discussion of the last article applies in all respects to this case, and from it the true stresses of tension and compression are seen to be each equal to (i + l)S t . When a horizontal shaft carries a load, flexural stresses come into action and these must be combined with the shearing stresses in the manner explained in Art. 106. The formulas (141) give the apparent and formulas (141)' give the true unit-stresses due to the combination of torsion and flexure; in these S is to be first computed from the flexure formula (41), while S t is to be com- puted from the torsion formula (90) or from the special formulas of Art. 92. From the last paragraph of Art. 143, it is to be con- cluded that there is no change in volume of the shaft under torsion alone; the same is closely the case when flexure is added to the torsion, because the decrease in volume due to the tension is practically the same as the increase due to the compression (Art. 13). The compression on the upper surface of a shaft due to a load, or that on the lower surface due to the upward reaction 372 TRUE INTERNAL STRESSES CHAP, xv of a bearing, produces stresses which act normally to the flexural stresses of tension and compression, while they are also at right angles to the shearing stresses due to the transmitted torsion. A formula for discussing this and other more difficult cases is deduced in Art. 177, and an application of it to the above case will now be given. Let S x be the horizontal flexural unit- stress at the surface of the shaft, Sy the vertical compressive unit- stress due to the load or bearing, and S s the shearing unit-stress due to the transmitted torsion. Then in formula (177) the value of A is S x +S y , that of B is S X S V -S 8 2 , and that of C is zero, and it reduces to the form S 2 - (S x + S y )S+ SxS v -S 2 =o in which the two values of 5 are the maximum tensile and com- pressive unit-stresses; solving the quadratic equation there re- sults, S y ) 2 -S x S y )* (144) where the value of S found by using the plus sign before the radical will be tension or compression according as the value of ^Sj + Sj,) is tension or compression. When either S x or S v is zero, these values are the same as those given by (141). As a numerical example, let the flexural compressive unit- stress S x under a load or in a bearing of a horizontal steel shaft be 3 ooo, the vertical compressive unit-stress S v due to the load or bearing be i 200, and the shearing unit-stress 5 due to the torsion be 6 ooo, all in pounds per square inch Formula (144) then gives 81=8200 pounds per square inch compression, and 2 = 4 ooo pounds per square inch tension for the maximum normal stresses; also S' = %(Si 5 2 ) = 6 100 pounds per square inch is the maximum shearing stress. Lastly, by (139) and (140), the corresponding true stresses are for compression 7^ = 9500, for tension T 2 = 6 700, and for shear T' = 8 100 pounds per square inch. In common practice it will be considered that the greatest compression is 3 ooo and the greatest shear is 6 ooo pounds per square inch, but the result of this investigation shows that the true compression is more than three times as great and the true shear about 40 percent greater. ART. 145 PURE STRESSES 373 Prob. 144. Show that the two roots of (144) are always real whatever may be the values of S x , S v , and S t . What are these roots when S, is o ? ART. 145. PURE STRESSES The term 'pure stress' is employed for cases where only one kind of stress exists. When a plane is acted upon only by forces normal to it, the stress on the plane is either tensile or com- pressive and this is sometimes called 'pure normal stress '. When a plane is acted upon only by forces parallel to it, the stress on the plane is that of shearing, and this is sometimes called 'pure shearing stress '. Upon most planes in the interior of a stressed body, there act both normal and shearing stresses. The preceding articles show how to find the maximum unit-stresses Si and 5 2 which act normally against certain planes upon which there are' no shearing stresses. Another use of the term ' pure stress ' is with respect to any and all planes that can be imagined to be drawn in the interior of a body. When the forces acting upon the body have such values that there can be no shearing stresses within it, the case is called one of ' pure internal normal stress '. Referring to Fig. 139 and to the reasoning in the last paragraph of Art. 140, it is seen that there can be no shearing stresses on any planes within the body, when Si = S 2 =S 3 ; this is the case where the unit- stresses acting on the six faces of a parallelepiped are all equal. The same result follows when a body is acted upon by equal compressive forces in all directions, as occurs under hydrostatic pressure. Under no other circumstances can the interior of a stressed body be free from shearing stress, and hence this is the only case of pure internal normal stress. There is no pase of a body having only internal shearing stress, for the discussion of Art. 143 shows that internal shear must always be accompanied by normal stresses which act in directions bisecting those of the two conjugate shears. There may, however, be certain planes within a body upon which only shearing stresses act. In order to find such planes, let Figs. 140a and 1406 be again considered, and let the forces shown in the latter be resolved 374 TRUE INTERNAL STRESSES CHAP. XV normal to the diagonal dz. Let 5 be the normal unit-stress of tension or compression on dz; then the total stress on that diagonal is Sdz and this is equal to Sidy . sin0+S 2 &v. cos0. Replacing dx/dz and dy/dz by their values cos# and sin/9, there is found S=S\ sin 2 + iS^cos 2 ^. Now when S=o, there is no normal stress on the plane that makes the angle 6 with the direction of Si; this occurs when tan0=( S 2 /Si)* and on the plane thus determined only shearing stresses are acting. It is seen that no value of 6 is possible unless S\ and S 2 have contrary signs, that is, one must be tension and the other compression. When S 2 =-iSi, then tan0=o.5 and #=26^; when S 2 =-Si, then tan0= i and 6 = 45; when 5 2 = -3$!, then tan0= 1.73 and 6= 60, and so on. Hence for each negative value of S 2 /Si there are two planes equally inclined to the direction of S\ upon which only shearing stresses act. The following figures show the three cases computed above. The shearing unit-stresses on these planes of true shear are not as great as those on the planes bisecting the directions of Si and S 2 , for on the latter the maximum shears exist. For Fig. 1456, however, where the normal tension and compression are numerically equal, the planes of pure shear coincide with those of maximum shear; this is the case most frequently men- tioned as one of pure shear (Fig. 1456), but the above investiga- tion shows that there may be many other cases. The term ' pure flexure ' is used for a part of a beam where there are no vertical shears. For instance, take a simple beam and subject it to two concentrated loads, each equal to P and placed at equal distances from the supports. Then there is no vertical shear between the loads, and hence the flexural stresses above the neutral surface are pure compression, while those ART. 146 INTERNAL FRICTION 375 below it are pure tension. In testing a beam it is sometimes advantageous to subject it to two equal concentrated loads placed at equal distances from the middle; thus the bending moment between the loads is constant, and the changes of length of the fibers are uniform at equal distances from the neutral surface. The experiments of Talbot on steel-concrete beams, described in Art. 116, were made in this way. It must be noted, however, that there prevail in all directions, except horizontally and vertically, shearing unit-stresses accompanying the pure tension and compression; if S is the flexural unit-stress at any point between the loads, then %S is the maximum shearing unit- stress which makes angles of 45 with the direction of S. Prob. I45a. In the formula S n =Si sin 2 0+S 2 cos 2 0,let Si be larger than S 2 . Show that the values of S n cannot be greater than Si nor less than $2- This is the equation of a curve in polar coordinates, S n being the radius vector for the variable angle 6; what kind of a curve is it? Prob. 1456. A parallelepiped is acted upon by normal unit-stresses of 6 400 and 2 800 pounds per square inch in directions at right angles to each other, the first being tension and the second compression. Compute the pure shearing unit-stress and the maximum shearing unit-stress, and find their directions. ART. 146. INTERNAL FRICTION In all the preceding discussions, the applied forces and the internal stresses have been supposed to be in equilibrium, this being the case where the applied forces have attained the full magnitudes so that no further deformation of the body occurs. Other considerations enter during the period while the defor- mations are occurring under applied forces which increase from zero up to their final values. During this period there are motions of the molecules, and this motion is resisted by internal friction, just as the motion of a book upon a table is opposed by the fric- tion between the surfaces of contact. The planes of maximum stress, found in the preceding articles, are hence not the cor- 376 TRUE INTERNAL STRESSES CHAP. XV rect planes of greatest stress during the period while the defor- mation of a body is occurring. The subject of internal friction was first recognized in the experiments made by Tresca, about 1860, on the flow of metals under high compressive stress, but it was not until after 1890 that it received careful attention. In 1893 the remarkable dis- covery was made by Hartmann that lines of stress became visible on the surface of polished metallic specimens when the elastic limit of the material was reached or surpassed, and that these lines remained after the loads were removed. Fig. 1460 repre- sents such lines for specimens of rectangular section, and it is seen that in compression they make an angle with the axis less than 45 degrees, while in tension this angle is greater than 45 degrees. It was observed that the sum of these two angles was always 90 degrees for the same metal and that the directions of the lines were independent of the size and length of the speci- men and of the unit-stress. The number of lines, however, increased as the unit-stress increased from the elastic limit to the ultimate strength. For the case of tension, Hartmann found that the angle . When round specimens of metal with polished surfaces were subjected to stresses above the elastic limit of the material, it was found that the lines were not straight but spiral, as shown in Fig. 1466, these spirals making the same angle with the axis as the straight lines on the rectangular specimens. Under the microscope it was noted that, in general, these lines were depres- sions below the intermediate surfaces and that the larger lines seen by the naked eye were really several small lines very near together. Hartmann also experimented on spheres and beams, finding that curved lines appeared on their polished surfaces when the elastic limit of the material was reached or surpassed, their directions always remaining the same in the same speci- men. The lines for a beam shown in his book Deformations AP.T. 146 INTERNAL FRICTION 377 dans les Metaux (Paris, 1896) have very little resemblance to any of the theoretic lines of maximum stress which are shown in Fig. 109. These interesting lines probably indicate the directions of the planes or surfaces on which the sliding or shearing of the mate- rial is beginning to occur. This supposition is strengthened by the phenomena of the, rupture of brittle materials under com- pression, where it is found that the failure ultimately takes place by shearing along planes inclined to the axis of the specimen, as Fig. 1696 shows for cement and timber. These planes make angles with the axis varying from 10 to 40 degrees, the angle for stone usually being about 20 and that of cement and concrete about 30 degrees. Also it is observed that metallic bars under tension sometimes rupture with an oblique or cup-like fracture, the inclination of which to the axis is 50 degrees or more. It may therefore be regarded as almost demonstrated that materials begin to fail, both in tension and compression, by shearing along oblique planes, and that the commencement of the failure is at the time the elastic limit of the material is reached. t t t i I I Fig. 146o Fig. 1466 The theory of internal stress, set forth in the preceding articles of this chapter, shows that the maximum shearing unit-stresses, both apparent and true, are those upon planes making angles of 45 with the axis of the bar. Since the actual planes of failure ire greater than 45 for tension and less for compression, it must be concluded that some resisting force acts during the progress 378 TRUE INTERNAL STRESSES CHAP. XV of the deformation which has not heretofore been considered, and this resistance is probably that of friction. Much attention has been given to this question since 1895, and the work of Rejto, published in 1897, endeavors to account for the brittleness, plas- ticity, ductility, and even the strength of materials, by the help of coefficients of internal friction. Prob. 146. Consult Rejto's Innere Reibung der festen Korper, and explain his formula for the tensile strength of materials. ART. 147. THEORY OF INTERNAL FRICTION When one surface begins to slide on another, the ratio of the force parallel to the sliding surface to the normal pressure is called the coefficient of friction; it is an abstract number and may be designated by v. Let N be the normal pressure or stress between the two bodies and F the force which just begins to cause motion when it acts parallel to the surface of contact, then the approximate law of sliding friction is given by F = vN. This law may be applied, tentatively at least, to the case of a bar under axial stress, during the period while the stress is increasing up to its final value, and it may be supposed that sliding or shearing is then beginning to occur along surfaces indicated by the lines and planes described in the last article. There is hence a coeffi- cient of internal friction fi, which is not necessarily the same as that of sliding friction but which may be used in the same manner by means of the law F = /tAT". The simplest case is that of a bar subject to axial compression, as in Fig. 1470. Let the section area be unity and 5 be the compressive unit-load which causes the equal axial unit-stress 5 on all planes normal to the axis. Let any plane mn be drawn cutting this bar, and let 6 be the angle which it makes with the axis. When there is no axial stress in the bar, there exists normal to this plane a molecular unit-force S which binds together the two parts so that no motion can occur. When the axial com- pressive unit-stress is applied, this causes a compressive stress 5 sin0 normal to the plane, and since the area of the plane is ART 147 THEORY OF INTERNAL FRICTION 379 i/sin#, the compressive unit-stress is S sin 2 0. The total normal pressure per square unit on the plane then is N=So+S sin 2 0. The force acting parallel to the plane is 5 cos0, or F=S cosO sin0 for each square unit. Accordingly, if motion is just beginning, ScosdsmO=fji(S +Ssm 2 0) or S = jjtS / (sind cos6 - fj&m 2 6) Now let 6 be regarded as varying from o to 90, then the unit-load S required to cause motion will vary from QO to S , and it follows that the motion will actually begin along that plane which has such a value of that S shall be a minimum; or the value of S which causes motion to begin is less for the plane of motion than for any other plane. This requires that sin# cos# /*sin 2 shall be a maximum, and the value of 6 which renders it such is found to be given by cot 26 = /i. Hence, fi=cot20 /iS = 45 tan0 = 4S( -p+i + fi 2 ) (147) gives the relations between the quantities /*, d, and SQ. The observed angle 6 is always less than 45; for different qualities of steel, internal friction lies between 0.84 and 0.49. Fig. 147a Fig. 1476 In considering the case of tension shown in Fig. 1476, let < be the angle which a plane mn makes with the axis. Then the normal unit-pressure on the plane is S -S sin 2 <, and the force per unit of area acting parallel to the plane is S sin< cos<. For the case of incipient motion, the law of friction then gives, S cos0 sin<=j(5o-5 sin 2 <) or S=jtS /(sm cos^+fi sin 2 ^) and, by the same reasoning as before, it is concluded that the actual plane of motion is that which renders sin = - //. Accordingly, fji= -cot 2$ ftSo = 45 tan < = 45(/i +N/7+7J (147)' 380 TRUE INTERNAL STRESSES CHAP. XV are the equations applicable to tension in which is observed to be always greater than 45 degrees. Since the coefficient of internal friction fi must be regarded as a constant for the same material, it follows that the angles and (f> are necessarily complementary, for by equating the two values of n the relation between the angles is given by + =90. The above theory may hence be said to explain why it is that the lines described in the last article make an angle with the axis which is less than 45 degrees for compression and greater than 45 degrees for tension, and why these angles are comple- mentary. It seems a reasonable assumption to regard fiS as the ulti- mate shearing strength S 8 of the material, since /*S equals the force per unit of area which will cause shearing along the plane. When a brittle specimen is ruptured by direct compression, failure generally occurs by shearing along one or more planes which make an angle 6 with the axis less than 45 degrees, and accordingly S s = ^5 c tan0 may be written as a tentative formula for the relation between the ultimate shearing strengths S, and the ultimate compressive strength S c . The following are rough approximate values of 6 as observed in compressive tests, together with the values of /* and S g /S c as computed from (147) : =0.135; Anthracite Coal = i5 j" =I -73 Sandstone = 20 ^ = 1.19 Hard Brick = 25 //=o.8 4 Cement and Concrete 0=30 //=o.s8 Cast Iron 0=35 /*=o-36 These computations indicate that the coefficient of internal friction is the highest and that the ratio of the shearing to the com- pressive strength is the lowest for the most brittle material. Thus for cast iron the shearing strength is 35 percent of the compres- sive strength, according to this computation, while for anthra- cite coal it is only 13 percent. When two bars of steel are stressed up to their elastic limits, one in compression and the other in tension, the elastic limit ART. 147 THEORY OF INTERNAL FRICTION 381 is closely the same for both bars, and it hence seems that 5 in (147) should be the same as 5 in (147)', which requires 6 and to be equal. This is a result altogether at variance with experi- ment, and it must hence be concluded either that 5 in tension is different from S in compression or that the above reasoning is defective in failing to include one or more elements that must ultimately be introduced in order to perfect the theory. Much work still remains to be done on this important subject, both in theory and by experiment, before definite ideas can be formed regarding the true internal stresses which prevail while a body is undergoing deformations. The theory of Arts. 139-145 relates only to statjc stresses, namely, to those which occur when the applied forces have attained their full magnitudes, so that both external and internal equilibrium prevails. This static theory appears to be correct in every detail for static stresses which do not surpass the elastic limit of the material, but the perma- nency of the lines of shearing seen upon polished metallic speci- mens, seems to throw a doubt upon its entire applicability to cases where the elastic limit is surpassed, even though com- plete equilibrium exists. As far as true internal static stresses are concerned, this theory is indeed not necessarily valid, for formulas (139) apply only within the elastic limit, but for the apparent static stresses it should be valid for all cases. In order that the full and complete truth may be ascertained, further studies on internal friction and on internal molecular forces are absolutely necessary. In conclusion it may be noted that the idea of internal fric- tion throws light upon the fatigue of materials under repeated stresses (Art. 137). For the case of compression where heat is evolved for stresses both below and above the elastic limit, it is not difficult to see that energy is expended in changing the positions of the molecules at each application of stress; for the case of tension the same occurs for stresses higher than about three-fifths of the elastic limit. The material is hence fatigued or changed in structure by the internal friction, and this change should be greater for large ranges of stress than for small ones. Undoubtedly the complete explanation of fatigue is closely allied 382 TRUE INTERNAL STRESSES CHAP. XV to that of internal friction and to changes in internal molecular forces. The indications are that any stress, however small, will produce fatigue when it is repeated a number of times in a material that has a crystalline structure. Steel is such a material and the various crystals that are seen in it under the microscope have cleavage planes which are weakened by the detrusion due to re- peated stress. Prob. 147a. Five constants have thus far been used in this volume as applicable to a material when stressed up to its elastic limit in tension. What are these constants? Prob. 1476. Consult The Iron and Steel Magazine for July, 1905, and read an article on the failure of an iron plate through fatigue. Also consult other volumes of this periodical, and ascertain the names and characteristics of the various crystals which are seen in steel. ART. 148 FACTS AND PRINCIPLES 383 CHAPTER XVI GUNS AND THICK CYLINDERS ART. 148. FACTS AND PRINCIPLES THE discussion of pipes under internal pressure, given in Art. 30, was made under the assumption that the thickness of the pipe is small compared to its diameter, so that the tensile stress of the metal of the pipe might be regarded as uniformly distributed. Pipes under high internal pressure have thicknesses sometimes nearly equal to the inside diameter, and in such the tensile stresses are not uniformly distributed. The steel pipe used to transmit water pressure to the large forging press of the Bethlehem Steel Company, in South Bethlehem, Pa., has an inside diameter of 16 inches, a thickness of 8 inches, and it is subject at times to a pressure of 5 600 pounds per square inch. The theory of the investigation and design of thick pipes will be developed in this chapter. The guns used in modern warfare are subject to high internal pressures produced by the explosion of the powder. These pressures are measured by noting their effects in shortening small copper cylinders and comparing these deformations with those produced by known .loads in a testing machine. In this manner it has been ascertained that powder pressures of 50000 pounds per square inch are often produced during the firing of a gun, while the extreme pressure of 88 ooo pounds per square inch has been observed in a special experiment. In order that the metal of the gun may not be stressed beyond its elastic limit uncler these heavy pressures, it is necessary that the thickness should be large. For inside diameters greater than 3 inches, the gun is generally formed of two or more concentric cylinders, the inner one being called the 'tube' and the others 'hoops'; these hoops are shrunk upon the tube (Art. 32) so that they produce com- pression in it and thus enable it to carry a heavier powder pressure 384 GUNS AND THICK CYLINDERS CHAP. XVI than a solid tube of the same total thickness. The first hoop around the gun tube is often called the 'jacket '. Fig. 148 gives a longitudinal and a cross section of a gun having two hoops, the breech block that closes the powder chamber A being omitted. This breech block can be swung open to admit the projectile and the powder case, the position of the former being at B. Before the explosion the breech block is swung into place and locked. At the instant of the explosion the pressure of the gas is the greatest since it is then confined to the spaces A and B; this part of the gun is called the breech, and here it is that the greatest thickness is required; over the breech and extend- ing some distance forward, the figure shows two hoops E and F surrounding the gun tube D. As the projectile moves toward the muzzle, the gas occupies a larger volume, so that its pressure decreases and becomes zero as the projectile leaves the gun. The tube C has spiral grooves cut in its inside surface, which cause the projectile to have a rotary motion. In designing a gun of this kind it is required that the section areas at the forward end of each hoop shall be sufficient to resist the maximum powder pressure which can there be exerted. -s a I / E ff Fig. 148 Modern guns are made of hard steel, often of fluid compressed steel, which has an elastic limit of about 50 coo and an ultimate strength of about 90 coo pounds per square inch. The allowable working stresses are large, in some cases as large as the elastic limit. Most careful workmanship and rigid inspection are exercised in their manufacture, and constant improvements are made in their design. These modern guns have been entirely developed since 1870, prior to which time cast-iron cannon were mainly in use. They are usually designed for a powder pressure cf 50 coo pounds per square inch. ART. 149 LAME'S FORMULA 385 Prob. 148. The diameter of a powder chamber is 8J inches and that of the projectile is 8 inches, while the pressure during the explosion is 48 coo pounds per square inch. Compute the total forward pres- sure on the projectile and the total backward pressure on the breech block. Where does the difference of these pressures take effect? ART. 149. LAME'S FORMULA Let a thick hollow cylinder, shown in Fig. 149a, be subject to a pressure R\ on each square unit of the inside surface and to a pressure R% on each square unit of the outside surface. The inside pressure may be produced by the expansion of a gas and the outer pressure by the atmosphere or by other causes. It is required to determine the internal stresses produced by these pressures at any point in the cylindrical annulus not very near the end. The length of the cylinder is to be regarded as con- siderably longer than the outer diameter, in order that the dis- turbing influence of the ends may not affect the reasoning. Fig. 149o Let r i and r 2 be the inside and outside radii ; then the inside pressure on the end of the closed cylinder is xr\ 2 R\ and the outside pressure on that end is r.r-^R-2.. The usual case is that where the inside is greater than the outside pressure, and then K(r?R\ rR-^ is the longitudinal tension in the annulus. For any part of the cylinder, not very near the end, this tension must be uniformly distributed over the cross-section of the annulus. The longi- tudinal tensile unit-stress S in the annulus is hence a constant for all points, and its value is found by dividing the total tension by the section area a; whence, This longitudinal unit-stress, together with the radial pressures, 386 GUNS A.ND THICK CYLINDERS CHAP, xvi causes a longitudinal elongation of the cylinder, which is also to be regarded as uniform for all parts of the annulus, not very near the end. Under these assumptions, the theory of true stress given in Art. 139 may be applied to the determination of the unit- stresses at any point in the cylindrical annulus. Let x be the distance from the axis of the cylinder to any point in a cross-section of the cylindrical annulus. Any ele- mentary particle is here held in equilibrium by the longitudinal unit-stress So, a tan- gential unit-stress 5, and a radial unit- stress R. The value of R is evidently inter- mediate between RI and R 2 ; in Fig. 1496 both R and S are regarded as tensile. Now from Art. 139 the effective longitu- dinal unit-elongation of the cylinder due to these three stresses is, in which A is the factor of lateral contraction the mean value of which for wrought iron and steel is about ^. But, as above noted, both Q and So are constant for all parts of the annulus, and it hence follows that, S+R= constant or S+R = 2Ci which is one equation between the unit-stresses S and R where C\ is a quantity whose value is to be determined by establishing a second equation. Let an elementary annulus of thickness dx be drawn; its inner radius is x and its outer radius is x + dx. The pressure for one unit of length in a direction perpendicular to any diam- eter is Rx for the inner surface and (R + dR)(x + 3x) for the outer surface of this elementary annulus. Thus, exactly as in the case of a thin pipe (Art. 30), the equation of equilibrium between acting pressure and resisting stress is, (R+dK)(x+dx)-Rx=S8x or xR+Rdx=Sdx which is a second equation between the unit-stresses 5 and R. The solution of these two equations is readily made by insert- ing 2Ci R for 5 in the second equations and integrating; then, ART. 149 LAME'S FORMULA 387 s=c 1+ F *= c i- (149) where C 2 is a constant of the integration the value of which is to be determined by regarding the limiting values of R, these being the inner and outer unit-pressures RI and R 2 . It is best to regard these unit-pressures as without sign, and then R= RI when x = r\, and R=R 2 when x=r 2 \ inserting these condi- tions in the second formula, there result two equations from which, and, inserting these constants in (149), are now obtained, which are Lame's formulas for the tangential and radial unit- stresses in hollow cylinders under inside and outside pressures. In deriving these formulas, both S and R have been supposed to be tension; this will be the case if their values are positive, while a negative value will indicate compression. The tangential unit-stress S is usually greater than the radial unit-stress R, and is the controlling factor in the design of guns and thick cylinders. It is seen to increase as x decreases, and hence it is the greatest at the inside surface of the cylinder; it may be either tension or compression, depending upon the rela- tive values of the given radii and unit-pressures. The radial unit-stress R is always compression, its value ranging from RI at the inside surface to R 2 at the outside surface. As a numerical example, let a cylinder be one foot in inside and two feet in outside diameter, the inside pressure being 600 and the outside 15 pounds per square inch. Here ri=6, r% = i2, RI =600, R 2 = i$, and the formulas become, S = i8o+28o8o/* 2 = i8o-28o8o/*2 Here x varies between 6 and 12 inches, and 5 ranges from +960 to* +37 5 pounds per square inch, while R ranges from 600 to 15 pounds per square inch, + denoting tension and denot- 388 GUNS AND THICK CYLINDERS CHAP. XVI ing compression. The tangential unit-stress S is hence about i\ times greater at the inside surface of the hollow cylinder than at the outside surface. Prob. 149. A solid cylinder is subject to a uniform radial pressure of 14 ooo pounds per square inch on its surface. Prove that the radial unit-stress is uniform throughout the cylinder. Prove that the tangen- tial unit-stress is uniform throughout the cylinder, and find its value. ART. 150. THICK PIPES AND SOLID GUNS Lame's formulas have been widely used for the discussion of water pipes and solid guns. The inside unit-pressure R\ is usually large compared to the pressure of the atmosphere on the outside surface so that the latter may be neglected. The unit-stresses S and R at any point within the cylindrical annulus then are, A discussion of this equation shows that the tangential unit-stress S is greatest when x = r\ and least when x = rz, and that the radial unit-stress R varies similarly. Thus, for the inside surface, Si^Rifrf+rft/frf-ri 2 ) and, R=-Ri (150) while for the outside surface of the cylinder, $2 = Ri 2T\ /(fy f\ ) and 7?=o and accordingly the greatest unit-stresses, as given by (150), are those generally used in investigation and design. The man- ner in which S varies throughout the annulus is shown in Fig. 150 by the arrows. It is seen B that the different parts of the annulus are un- equally stressed under the tangential tension, the law of variation of the unit-stresses being that which is expressed by the first equation in formula (149). ART. 150 THICK PIPES AND SOLID GUNS 389 As a special case, let the radius of the outside surface be double that of the inside surface, or r 2 = 2r\. Then for the inside surface where x = r\ the tangential unit-stress is Si=f#i; for the outside surface, where jc = r 2 , it is S 2 = $R 2 -, for a point half way between these surfaces it is ^R\, and all of these are ten- sion. For the same case the radial unit-stress for the inside surface is R\, for the outside surface it is o, and for a point half way between them it is J/?i, all being compression as indicated by the minus sign. The first formula of (150) was formerly much used for the investigation of guns and thick pipes, and it is still valuable for general discussions. As an example, let a solid steel gun have an inside diameter of 7.5 inches at the powder chamber and the thickness of the tube be 1.75 inches. Let it be required to find the greatest tensile unit-stress produced when the inside pres- sure from the explosion is 10 ooo pounds per square inch. Here ^=3.75 inches, f2 = 5-5o inches, R\ = 10000 pounds per square inch. Then from the formula Si is found to be 27 300 pounds per square inch, which is but little more than one-half the elastic limit of gun-steel, and hence the degree of security is ample. , As an example of design, let the inside diameter be 3.25 inches, the pressure caused by the explosion 15 ooo pounds per square inch, and the allowable unit-stress in tension be 30 ooo pounds per square inch ; and let it be required to find the outside diameter. Here r\ = 1.625 inches, .#1 = 15000, and 1 = 30000 pounds per square inch. Then solving for r 2 , there results, r 2 = r 1 [(Si + Ri)/(S l -Ri^ (150)' from which the outside radius r 2 is found to be 2.815 inches; thus the thickness of the tube is 1.19 inches, and its outside diameter is 5.63 inches. When formula (150) is applied to a thin pipe, r 2 is to be re- placed by r\ +t, where / is the thickness, and / 2 may be neglected when it is to be added to ri 2 . The formula then reduces to tSi=Ri(r+l), which is slightly more accurate than that of (30), and it gives slightly larger values of S\ in investigation and slightly larger thicknesses in design. Neglecting / in comparison with 390 Guxs AND THICK CYLINDERS CHAP. XVI r, this becomes lSi=riR, which is the same as the common for- mula for thin pipes derived in Art. 30. Either of these formulas, however, would lead to grave error when applied to a pipe whose thickness is as great as one-half its diameter. From formula (150)' it is seen that r% becomes infinite when RI equals S, that is, the inside unit-pressure must never be greater than the allowable tensile unit-stress of the material. Cast-iron cylinders for small forging presses have been used under pressures as high as 5 ooo pounds per square inch, and it hence follows that the actual unit-stress 5 must have been much higher than this; the thickness of such cylinders is usually greater than the inside radius. The indications of experience are that factors of safety for thick pipes under pressure may be much lower than for thin ones. Prob. 150. A solid gun tube is 6 inches in diameter and 3 inches thick. What is the inside pressure that will produce a maximum tan- gential tension of 30 ooo pounds per square inch ? ART. 151. A COMPOUND CYLINDER In a solid gun the maximum tension occurs at the inside sur- face during the explosion, rising suddenly from o up to its greatest value Si. If now the metal near the bore can be brought into compression, this initial stress must be overcome before the tension can take effect, and thus the capacity to resist the inside pressure is increased. One method of producing this compression is by means of a hoop, or jacket, shrunk upon a tube so as to produce an outside unit-pressure RZ over the surface of the tube. This arrangement may be called a hollow compound cylinder. In its normal state of rest, the inner cylinder or tube has no pressure on its inner surface and R2 on each square unit of its outside surface. Making R\=o in (149)' and also X=TI and x=r 2 in succession, there are found Si = -R 2 . 2r 2 2 /(r 2 2-rft S 2 = -R 2 . (rf+rfi/(rf-n*) which are the tangential unit-stresses at the inside and outside surfaces of the tube due to the externa"! pressure ^2- Both of ART. 151 A COMPOUND CYLINDER 391 these are compression, but the former is numerically greater than the latter, since 2r 2 2 is greater than r 2 2 + fi 2 . If the hoop is to be shrunk on so as to produce a compressive unit-stress S e at the inner surface of the tube, the unit-pressure R 2 upon the outer surface must be, R2=-S e (r 2 2 - ri 2 )/2r 2 2 and the shrinkage may be so regulated as to produce this pressure R 2 in the normal state of rest. Then the tangential stresses throughout the tube are all compression, while the radial pressures range from R 2 on the outer surface to o on the inner surface. As an example, let r\ = 2 inches, r 2 = 3 inches, and let it be required to find the outer pressure which will cause a tangential compressive unit-stress of 18 ooo pounds per square inch at the inside surface of the tube. Here the last formula gives R 2 = 5 ooo pounds per square inch, and hence the hoop must be shrunk upon the tube so as to produce this radial pressure at the surface of contact of hoop and tube. When the gun is fired, the explosion of the powder causes an internal tangential tension 5 given by (149)', the greatest value of which is at the inside surface of the tube. Making x=r\ y this tensile unit-stress is found to be, Si =[('i 2 + '2 2 )# i - 2r 2 2 R 2 ]/(r 2 2 - ri 2 ) which is Lame's formula for the investigation of the tube of a compound gun. This tension first overcomes the initial com- pression S e due to shrinkage, so that the maximum tensile unit- stress at the inner surface during firing is Si S s . In Fig. 155 the line A a represents the compression S e at the inner surface of the tube before the firing, aai the total tension Si produced during the explosion, and Aa\ the actual tension S\ S t during firing. For example, if the radial pressure R 2 is made such that the tangential compression S e is 30 ooo pounds per square inch, and if the inside tensile stress-unit Si due to the powder ex- plosion is 65 ooo pounds per square inch, then the actual tension at the inside surface of the tube is 35 ooo pounds per square inch. For example, let a tube whose inside and outside diameters 392 Guxs AND THICK CYLINDERS CHAP. XVI are 4 inches and 6 inches be hooped so that a tangential com- pression of 1 8 ooo pounds per square inch is caused at the bore, while the inside pressure caused by the explosion is 25 ooo pounds per square inch. It is required to find the tangential stress at the bore during the explosion. Here r\ = 2 and r2 = 3 inches, .#1 = 25000, and .#2 = 5000, as seen above. Then 51=47000 pounds per square inch is the tension due to the explosion, but before this can take effect the initial compression of 18 ooo pounds must be overcome. Hence the resultant tension at the bore during the firing is 4700018000 = 29000 pounds per square inch. If this tube has no hoop, the tension at the inside surface, found by the method of the last article, is 57 200 pounds per square inch. The very great advantage of the hoop in diminish- ing the internal stresses during the firing is hence apparent. Prob. 151. A gun tube 3 inches in diameter and 1.5 inches thick is hooped so that the tangential compression on the inside surface is 30 ooo pounds per square inch. What powder pressure R\ will produce a resultant tangential tension on the inside surface of 30 ooo pounds per square inch? ART. 152. CLAVARINO'S FORMULAS The preceding method of investigating gun tubes is defec- tive in that the two unit-stresses 5 and R are the apparent and not the true internal unit-stresses. It was shown in Art. 139 that the true stresses are those corresponding to the actual deforma- tions, and that they are determined from the apparent stresses by help of the factor of lateral contraction X. For gun-steel the value of A is usually taken as . Lame's formulas were deduced in 1833, but it was not until about 1880 that they were modified by Clavarino so as to give the true internal stresses. At any point in the annulus of a hollow cylinder (Fig. 1496) the apparent tangential, radial, and longitudinal unit-stresses are 5, R, and S . Let T be the true tangential unit-stress, then from (139) its value is T=S XR-XS Q ; inserting in this the values of S, R, and 5 found in Art. 149,and taking the factor ART. 152 CLAVARINO'S FORMULAS 393 of lateral contraction as , it reduces to, (r 2 2 -r 1 2) (152) which is Clavarino's formula for the tangential unit-stress. This is the principal formula for the investigation of steel guns and thick pipes. This formula shows, as before, that the tangential stress is greatest at the inside surface of the cylinder. Making x = r i} the true internal maximum unit-stress is found to be, Ti =[ (ri 2 + 4 r 2 2 )Ri -S^d/Sfo 2 -'! 2 ) (152)' which is the practical formula for the discussion of the most common cases. TI may be either tension or compression, depend- ing upon the relative values of the pressures and radii. Clavarino's formulas have been generally used since 1885 in the investigation and design of guns, instead of those of Lame". In order to compare them, the particular case where the outer diameter is double the inner diameter may be considered. Here rz = 2ri and formulas (149)' and (152)' reduce to, Now if RZ=O, the first formula gives a smaller unit-stress than the second; if R2=Ri, the first gives a unit-stress three times as large as the second; if RI=O, the first gives a value some- what larger than the second. Thus, since the second formula gives undoubtedly a better representation of the true stress than the first, it follows that Lame's method errs on the side of dangel for a solid gun and on the side of safety for a hooped tube. The value here used for the factor of lateral contraction is that employed in the United States by both the Army and the Navy in gun formulas, and also generally in Europe; in France, how- ever, the value = J is adopted. For a thick pipe or solid gun, the outside pressure R 2 is zero, and formula (152)' can then be written in the forms, 1 + *! \* / 152 vn the first of which may be used for the investigation and the second 394 Guxs AND THICK CYLINDERS CHAP, xvi for the design of hollow cylinders. According to the second formula, the inside unit-pressure RI can never be as large as three- fourths of the allowable tensile unit-stress TI, and this may be taken higher than for a thin pipe. For example, let it be required to find the thickness of a steel pipe 8 inches in inside diameter, under a head of water of i 200 feet, this including the effect of water ram (Treatise on Hydraulics, Art. 148), the allowable tensile unit-stress to be 15 ooo pounds per square inch. Here TI = 150x30, and RI =0.434X1 200 = 520 pounds per square inch, and then the formula gives r2=4-i8 inches, whence the required thickness is 0.18 inches. From the old formula (150)', the thickness is found to be 0.22 inches. Prob. 152. Solve Problem 150 by the formulas of this article and compare the results found by the two methods. ART. 153. BIRNIE'S FORMULAS The preceding articles present an outline of the methods of investigating stresses in guns by the formulas of Lamd and Clavarino. The formulas of Lame" refer to apparent stresses only; those of Clavarino, although referring to true stresses, are not strictly correct for hooped guns at rest because they are deduced for a hollow cylinder with closed ends. Now a gun at rest has no inside pressure and no closed end and hence there can be no external longitudinal stress upon it; accordingly the unit-stress 5 should be zero for a gun at rest. The true tangential unit-stress T then is S%R; using the values of S and R deduced in Art. 149, there results for any point in the annulus, at a distance x from the axis, 1 -R 2 ))/3(r 2 *-rft (153) which is Bimie's principal formula for the discussion of hooped guns at rest. Making x=*r\, this becomes, r 1 =[( 2 r 1 2+ 4 r 2 2 )^ 1 -6r 2 2 J R 2 ]/ 3 (r 2 2- fl 2) (153)' which is the tangential unit-stress at the inside surface of a hoop AST. 153 BIRNIE'S FORMULAS 395 with the radii r\ and r 2 and is under the pressures RI and R 2 . For the gun tube itself, RI is o during the state of rest. Birnie's formulas are used in the ordnance bureau of the United States Army for the discussion of gun tubes and hoops, both at rest and during the firing. To compare the formulas of Clavarino with those of Birnie, the particular case of a hoop or pipe where r 2 = i.2n may be considered. Then (152)' and (153)' reduce to, ^ = 5.12^-5.56^2 and 7\ = 5.88^1 -6.35/? 2 Now if R 2 = o, as for a solid gun during firing, the second formula gives a tangential unit-stress 15 percent larger than the first; if RI=O, as for a hooped gun at rest, the second gives a unit stress 1 8 percent larger than the first. Thus for this case Cla- varino's formulas appear to err toward the side of danger. Birnie's formulas apply only to hoops and tubes upon which the longitudinal stress is zero, and this is not the case during the explosion. For a solid gun, or for a tube attached to the breech block, a more correct formula may be found by con- sidering the actual value of S due to the inside pressure. Here the longitudinal pressure is 7cr\ 2 R^ and this produces longitu- dinal tension upon the area n(r 2 2 ri 2 ), so that S =ri 2 Ri/(r 2 2 ri 2 ) is the apparent longitudinal unit-stress. The true tangential stress T at any point in the annulus then is S %R $S , and accordingly, This gives values of T lower than those found from the formulas of Clavarino and Birnie. For x = r\ it becomes, which is the true tangential unit-stress at the inside surface of a hooped gun during the explosion, R 2 being the pressure upon its outside surface due to the shrinkage of the hoop. For a simple gun tube or water pipe where R 2 = o, this formula agrees with that of Clavarino; for a hooped tube at rest, it gives a value of T\ which is 20 percent greater than that of Clavarino. 396 GUNS AND THICK CYLINDERS CHAP. XVI Prob. 153. Solve Problem 150 by the formulas of this article and compare the results with those of Problem 152. ART. 154. HOOP SHRINKAGE Let e be the elongation or contraction of any radius x, then 2ne is the elongation or contraction of any circumference 2xx. Now 2ne/27tx is the change in the circumference per unit of length due to the unit-stress T; hence e/x=T/E, and e = (T/E}x is the change in length of the radius of the circle due to the tan- gential unit-stress T. When x = r\, the deformation e\ is that of the radius of the bore due to, the unit-stress T\\ if x=r 2 , the deformation e 2 is that of the outside radius where the unit- stress is T 2 . Suppose a compound cylinder to be formed by shrinking a hoop upon a tube. The inside radius of the tube is r\ and its outside radius r 2 ; the inside radius of the hoop is r 2 and its outside radius ^3. In consequence of the shrinkage the radial unit-pres- sure R 2 is produced between the two surfaces; this causes the tube to be under tangential compression and the hoop to be under tangential ten- sion. It is required to find these stresses when the original inside radius of the hoop is less than that of the outside radius of the tube by the amount e. . Let e 2 be the decrease in the outside radius of the tube and e 2 the increase in the inside radius of the hoop; then e = e 2 + e 2 '. In Fig. 154, which is much exaggerated, cd represents e 2 and be represents e 2 . Also, let T 2 be the tangential compression at the outside surface of the tube due to the shortening e 2 , and let T 2 ' be the tangential tension at the inside surface of the hoop due to the elongation e 2 . Then e= (T 2 /E)r 2 + (T 2 '/E)r 2 or T 2 + T 2 '=Ee/r 2 which gives one equation between T 2 and T 2 . Formula (153)' is applied to the tube by making RI = O and ART. 154 HOOP SHRINKAGE x=r 2 ; thus the tangential compression is, Formula (153) is applied to the hoop by replacing RI by R 2 , R 2 by o, ri by r 2 , and r 2 by r 3 ; then for #=r 2 , there results, r 2 '= 2 ( 2 r 2 2-f 4 r 3 2 )/3(r 3 2 -r 2 2 ) =ftR 2 which is the tangential tension. Dividing the first of these ex- pressions by the second, there is found, T 2 /T 2 '=a/p or ftT 2 =aT 2 ' which is a second equation between T 2 and T 2 '. The solution of these two equations furnishes the values of T 2 and T 2 in terms of known quantities; then, r.-^ *-*. (154) in which a and ft depend only on the radii, or, and thus the tangential compression at the outside surface of the tube and the tangential tension at the inside surface of the hoop may be computed. The tangential compression at the bore is, however, greater than T 2 , and it may be found from (153) by substituting the value of R 2 , now known, and making x=n; thus, is the greatest compressive unit-stress in the tube due to the radial pressure of the hoop. As a numerical example, let a compound cylinder be formed of a steel tube whose inside radius is 3 inches and outside radius 5 inches, with a steel hoop whose thickness is 2 inches. It is required to find the stresses produced when the original difference between the outside radius of the tube and the inside radius of the hoop is 0.0x54 inches. First, the sum of the two tangential stresses at the surface of contact is (Ee)/r 2 = 24000 pounds per square inch, if E is taken as 30 ooo ooo pounds per square inch. Secondly, from the given radii, the value of a is found to be 43/24, and that of ft to be 41/12. Then formulas (154) give T 2 = 8 260 pounds per square inch at the outside surface of the 398 GUNS AND THICK CYLINDERS CHAP, xvi tube and 7Y = i5 74 pounds per square inch for the inside sur- face of the hoop. Thus it is seen that the hoop tension is nearly double the compression on the outside surface of the tube. At the bore of the tube, however, the tangential compression is found to be !Ti = i44oo pounds per square inch. The decrease in the outside radius of the tube is next com- puted and found to be e 2 =0.00138 inches, while the increase in the inside radius of the hoop is e 2 ' = 0.00262 inches. Hence if the radius of the common surface of contact is to be exactly 5 inches after the shrinkage, the tube should be turned to an outside radius of 5.0014 inches, and the hoop to an inside radius of 4.9974 inches. The radius of the bore, however, will then be less than 3 inches by the quantity (T\r\/E)= 0.00144 inches; hence if its final diameter is to be exactly 6.0000 inches, it should be turned to a diameter of 6.0029 inches. The formula of Birnie has been used in solving the above numerical example; if that of Clavarino is used, the following values will be found : T 2 = 7 030, T 2 f = 16 970, T\ = 14 400 pounds per square inch; e 2 = 0.00117, e 2 '= 0.00283, and e\= 0.00144 inches. The shrinkages thus agree within 0.0002, which is as close as measurements can be relied upon. The above investigation closely applies to the case of a hoop or crank web shrunk upon a solid shaft or solid crank pin (Art. 97) by making ri=o and letting r 2 be the mean radius of the web. For example, let r\=o t r 2 =i6 inches, ^3 = 24 inches, and let e/r% be YS\V in accordance with the old rule (Art. 32). Then = , /?=X4-4, and the tangential compression at the outside surface of the shaft or pin is 7^ = 3 74 pounds per square inch, while the tangential tension at the surface of the hoop or web is 7Y = 16 5 pounds per square inch; the radial compression in the crank or pin is T\ = 1 1 300 pounds per square inch. It thus appears that the ratio e/r 2 = T*W gives shrinkage stresses which are higher than advisable when the other stresses which act upon the web and pin are considered (Art. 98). Prob. 154. A solid steel shaft, 6 inches in radius, is to be hooped so that the greatest tensile stress in the hoop and the greatest com- /-KT. 155 DESIGN OF HOOPED GUNS 399 pressive stress in the shaft shall be 15 ooo pounds per square inch. Find the thickness of the hoop and the radius to which its inside surface should be turned. ART. 155. DESIGN OF HOOPED GUNS A hooped gun should be so constructed that neither the stresses due to hoop shrinkage nor those developed during the firing shall exceed the elastic limit of the material. The simple case of a tube with one hoop can here only be considered. If the radii are given, as also the inner pressure RI due to the ex- plosion, it may be desired to find the shrinkages so that this requirement will be fulfilled. As RI is very large, it is desirable that the given dimensions should be such as to require the least amount of material. The condition of minimum amount of material will be in general fulfilled when the stresses during the explosion are as great as allowable and as nearly equal as possible. The diagram in Fig. 155 represents the distribution of the internal stresses under this supposition. O is the center of the gun, OA the inside radius r i} while AB is the thickness of the tube and EC that of the hoop. The shaded areas show the stresses due to hoop shrinkage, A a and Bb being the tangential compres- sions T\ and 7*2 of the last article, while Bb' is the tangential tension TV, and Cc is the tangential tension at the outer sur- face of the hoop. When the explosion occurs the two cylinders are thrown into tangential tension, Aa\ and Bb i being those at the inner surfaces of the tube and hoop. The above prin- ciple indicates that both Aa\ and Bb\ should be equal to the maximum allowable unit-stress T e , which for guns is often taken nearly as high as the elastic limit of the material. In designing a hooped gun, the radius of the bore and the thickness of the tube may be assumed, and it may be required to find the thickness and shrinkage of the hoop so that the stresses Aa, Adi, and Bb^ in Fig. 155 are each equal to the elastic limit of the material. Or, given the radius of the bore and the out- 400 GUNS AND THICK CYLINDERS CHAP. XVI side radius of the hoop, it may be required to find the interme- diate radius under the same conditions. These problems can be solved, as well as more complex ones relating to guns with several hoops. Guns with seven hoops have been built, but the usual number is three or four. Fig. 155 The formulas of Arts. 153 and 154 may be applied to the design of a gun by assuming the allowable tangential unit-stresses, as also the thicknesses of the tube and hoops. For a given unit- pressure RI due to the explosion, the shrinkages are then to be computed. This method is one frequently used, and it will here be illustrated for a gun with one hoop. Let 71 = 3.04, r 2 = 5.80, and 73=9.75 be the given radii, and let 50000 pounds per square inch be the allowable unit-stress for both tension and compression. It is required to find the radii to which the sur- faces shall be turned so that their values shall be those above given when the gun is at rest. These radii will be readily found when the tangential unit-stresses 7\ and T 2 for the tube and T 2 and T 3 for the hoop have been computed, since these deter- mine the changes in length of the radii. The first step is to compute the numbers a and /?, which are found to be 1.424 and 2.425 respectively. Since 7\ is to be 50 ooo pounds per square inch compression for the inside surface of the tube, T 2 for the outer surface will be Ti(2r\ 2 + r 2 2 )/^r 2 2 = 2^ 800 pounds per square inch compression, and accordingly for the common sur- face of tube and hoop R 2 =T 2 /a = i8 100 pounds per square inch. For the inside surface of the hoop, T 2 =^R 2 = 4T > 900 pounds per square inch tension. For the outside surface of the ART. 156 DESIGN OF HOOPED GUNS 401 hoop where #3=0, formula (153) may be used by increasing each of the subscripts by unity and making x=r 3 , thus giving Ta = 2R2r 2 2 /(r 3 2 r2 2 ) = i<)8oo pounds per square inch tension. Then the change in the inside radius of the gun tube is (Ti/E)ri = 0.0051 inches, and hence the bore must be turned to a radius of 3.0400 + 0.0051=3.0451 inches in order that it may be exactly 3.04 inches after the hoop is shrunk on; the change in the out- side radius is (T 2 /E)r 2 = 0.0050 inches, so that the outside sur- face of the tube must be turned to a radius of 5.800 + 0.0050 = 5.8050 inches. The change in the inside radius of the hoop is (Tz/E)r 2 = 0.0085 inches, so that its inner surface must be turned to a radius of 5.8000 0.0085 = 5.7915 inches; the change in the outside radius of the hoop is (T s /E)r 3 = 0.0064 inches, so that its outside surface must be turned to a radius of 9.7500-0.0064 = 9.7436 inches. This gun must now be investigated to find what powder pressure will cause the stresses T\ and TZ to be 50 ooo pounds per square inch tension during the explosion. If T% has this value, the part of it due to the powder explosion is 50 ooo 43 900 = 6 100 pounds per square inch; hence the radial compression between the tube and the hoop which is due to the explosion must be R 2 = 6 ioo//?=2 500 pounds per square inch. The value of T\ due to the explosion is 100 ooo pounds per square inch tension, since the initial compression of 50 ooo pounds per square inch must first be overcome. Inserting then in (153)' the values !Ti = iooooo, .#2 = 2500, ^2 = 9.242, r 2 2 = 33-64o, and solving for R\ gives RI = $I 100 pounds per square inch, which is the highest allowable powder pressure. Under this pressure the unit-stresses represented by Aa, Aa\, and Bb\ in Fig. 155 are each 50 ooo pounds per square inch, while all other tangential and radial stresses have smaller values. In conclusion it may be noted that this chapter has been prepared in order to present the general principles of the design of guns, rather than to give the detailed methods which are followed when three or more hoops are used. The work of Meigs and Ingersoll (Baltimore, 1885) and that of Story (Fort 402 Guxs AND THICK CYLINDERS CHAP, xvi Monroe, 1894), each being entitled The Elastic Strength of Guns, may be consulted for detailed discussions. The former gives the methods and formulas for navy guns, while the latter gives those for army guns; these differ mainly in that the Navy employs the formulas of Clavarino, while the Army uses those of Birnie. Prob. 1550. Prove that a gun tube with one hoop is most advan- tageously designed when the common radius of tube and hoop is a mean proportional between the other two radii. Prob. 1556. Discuss a gun with two hoops where ^=3.50, r 2 =9.i$, r3 = n.25, r4 = i2.25 inches and which is to be under a powder pressure of 50 ooo pounds per square inch. Find a set of shrinkages so that the compressive stress at the bore when the gun is at rest shall be 45 ooo pounds per square inch; and so that the tensile stresses at the bore and at the inside of each hoop during the explosion shall be 40000 pounds per square inch. AKT- 156 CYLINDRICAL ROLLERS 403 CHAPTER XVII ROLLERS, PLATES, SPHERES ART. 156. CLYINDRICAL ROLLERS Let cylindrical rollers of diameter d and length / be placed between two flat pktes and transfer a load from the upper to the lower plate. Fig. 156a shows end and side views of the two plates with one roller which carries the load W. It was found in the experiments of Bach that the pktes were but little de- formed in comparison with the roller, and hence the entire defor- i w j I I I 1 1 * I- T t t t Fig. 156o t t t f t t T \ w Fig. 1566 mation will here be regarded as confined to the latter. The vertical diameter AA is shortened to BB and any vertical chord aa is shortened to bb. The change of length in the vertical radius is AB and that in the vertical half -chord is ab. The unit-shortening of the vertical radius is AB/\AA and that of the vertical half- chord is ab/^aa and the comprcssive unit-stresses are propor- tional to these unit-shortenings (Art. 10). Let the greater shortening AB be called e and the shortening ab be called y\ and let the compressive unit-stresses at B and b be called S and S v . Then, if the elastic limit of the material is not exceeded, S/S y --=e/y or S y = S.y/e. The unit-stress S is evidently the maximum and it is required to determine its value in terms of W, d, and /. The value of S may be expressed by noting that e/\d is the unit-shortening of the vertical radius, and that this is equal to 404 ROLLERS, PLATES, SPHERES CHAP, xvn S/E, where E is the modulus of elasticity (Art. 9) ; hence S/E = c/\d. The sum of the vertical stresses in each cylindrical segment must equal the total load W, since it holds that load in equilibrium. Let x be the distance Bb\ then the unit-stress S y acts over the area Idx, and hence the sum of all the vertical stresses S y . Idx equals W. Accordingly, S/E = e/tf and f S v . Idx = W are two equations for determining the values of S and e. To solve these equations, S y is to be replaced by its value S.y/e and the second equation then becomes Slfydx = We. Now fydx is the area of the circular segment CACBC; but, since the deformation is very slight, the arc CAC may be regarded as parabolic or the area of the segment as $CCXAB. Now AB = e and CC=J5C= (ed-e 2 )*= (*)* nearly. The solution of the two equations then leads to the formula, the first of which may be used for computing the unit -stress 5 when the load and the dimensions of the roller are given, while the second may be used for determining the size of a roller to carry a given load under an assigned unit-stress. Let w be the load per unit of length of the roller, or -w = W/l, then the formula (156) may be written, . W=$ldS(2S/E)* or w=$dS(2S/E)* (156)' which shows that the load on a cylindrical roller should vary directly as its diameter. Taking ^ = 15 ooo and E = 30 ooo ooo pounds per square inch for steel, the last formula reduces to w = 315^. This agrees well with the rule for bridge rollers given in Cooper's Specifications of 1901, which is w = ^ood. The erro- neous rule, w = i 20oVd, which requires the load to vary as the square root of the diameter, is still to be found in some bridge specifications. ART. 156 CYLINDRICAL ROLLERS 405 As a numerical example, let it be required to find the factor of safety of six wooden rollers used in moving a large block of stone which weighs 12 ooo pounds, the diameter of each roller being 6 inches and its length 8 feet 4 inches. Here W = $Xi2 000 = 2 ooo pounds, = 1500000 pounds per square inch, / = ioo inches, and d = 6 inches; then formula (156) gives 5 = 266 pounds per square inch, and hence the factor of safety is 8 000/266 = 30, which indicates a high degree of security. Again, let it be re- quired to find the diameter of a cast-iron roller which is 6 feet long in order to carry 30 ooo pounds with a factor of safety of 18. Here W=T,OOOO pounds, 5 = 90000/18 = 5000 pounds per square inch, = 15000000 pounds per square inch, and 7 = 72 inches; then in the formula everything is known except d and the solution gives d = 4.6 inches. The assumption that the plates are not deformed at the sur- face of contact with the roller is one that is not universally ac- cepted. Later experiments by Juselius appear to indicate, for rollers and plates of the same material, that the deformation of the two plates in any vertical is about equal to that of the same vertical in the roller. By using this conclusion, as indicated in Fig. 1566, the above reasoning and formulas will be modified. The shortening of the vertical radius will now be one-half of its former value, and thus the first formulas of (156) and (156)' become, Accordingly the compressive unit-stress due to a given load is 21 percent less than before, while the load that may be carried with a given unit-stress is 41 percent greater than before. Apply- ing this second formula to the cast-iron roller of the last para- graph, its diameter is found to be ^ = 3.4 inches. It is seen, therefore, that the assumption used at the beginning of this article errs on the side of safety when the plates are actually deformed. Prob. 156. A load of 19* ooo pounds is carried on cylindrical steel rollers 16 inches long and 3 inches diameter. Compute the number of rollers needed when the allowable unit-stress is 12 ooo pounds per square inch, using the formula which appears to be most safe. 406 ROLLERS, PLATES, SPHERES CHAP. ART. 157. SPHERICAL ROLLERS Let a sphere of diameter d be placed between two plates and be subject to compression by a load W. The left-hand diagram of Fig. 156a may represent a vertical section of the sphere and plates, the former being regarded as alone deformed. Thft vertical diameter A A is shortened to BB and any vertical line aa is shortened to bb. Let -S and S y be the compressive unit-stresses in the vertical lines A A and aa; let the greatest shortening AB be called e, and the shortening ab be called y. Then, for stresses within the elastic limit, S/S y =e/y. The unit-stress S is the greatest and it is required to find its value in terms of W and d. As e/\d is the unit-change in length of the vertical radius, the value of S may be expressed by S/E = e/\d (Art. 10) and this is one equation between S and e. To find another equation, the sum of the veitical stresses in the spherical segment must equal the load W. Let x be the distance Bb; then the unit- stress S y acts over the area 2nx . dx, and hence the sum of all the vertical stresses S y . 2rcxdx equals W. Accordingly, S/E = e/\d 2-KJS v .xdx=W are two equations for determining the values of 5 and e. To solve these equations, S y is to be replaced by its value S.y/e and the second equation then becomes Sj 2xyxdx=We. Now faityxfoc is the volume of the spherical segment whose section is CACBC in the figure; but, since the deformation is very slight, the arc CAC may be regarded as parabolic, and then the volume is one-half that of a cylinder having the radius BC and the altitude AB. Now AB = e, and BC=(ed-e 2 )* = (ed)*, very nearly, since e is small compared with d. Accordingly the value of the integral is %xe 2 d, and then $nSed=W. Inserting in this the value of e from the first equation, it reduces to, S* = WE/frd 2 or W=\7td 2 S 2 /E (157) From the first formula S may be computed when W is given, and from the second W may be computed when 5 is given. The diameter required for a sphere to carry a given load with an allow- ART. 158 CONTACT OF CONCENTRATED LOADS 407 able unit-stress is found from d 2 = 4\VE/-S 2 ; thus diameters of spherical rollers should vary as the square roots of their loads. In strictness there is always some deformation of the plates as well as of the spheres, and the old assumption that the total deformation is equally divided is probably nearer the truth than that it is all confined to the sphere. Under this assumption AB is to be taken as \e and then the formulas become, S 2 = WE/frd 2 , W=frd 2 S 2 /E, d 2 = 2 WE/xS 2 - (157)' Comparing these with the previous formulas it is seen that (157) give values of 5 which are 41 percent higher than (157)', values of W which are only one-half as large, and values of d which are 41 percent larger. The common formulas (157) hence err on the side of safety, and the truth probably lies between them and (157)'. When the plates are harder than the rollers, (157) is more nearly correct; when they are of equal hardness, perhaps (157)' gives the more accurate results. These formulas, like those of the last articles, are valid only when the load produces a unit-stress S which is less than the elastic limit of the material. For stresses beyond the elastic limit, the formulas W=C\ld for cylinders and W=CdP may be considered as approximate, in which C\ and C 2 are to be determined by experiment for each material. The experiments of Crandall and Marston on steel cylindrical rollers, which ranged in diameter from i inch to 16 inches, show that their crushing loads are closely given by the formula W=88old, where W is in pounds and / and d in inches. Prob. 157. How many steel spheres are required to carry a load of 6 ooo pounds, with a working stress of 15 ooo pounds per square inch, when they are 4 inches in diameter? How many are required when they are 12 inches in diameter? ART. 158. CONTACT OF CONCENTRATED LOADS When a concentrated load is placed upon a horizontal beam or plate, it produces compressive stresses over a certain area. In bridges and buildings concentrated loads are often applied to 408 ROLLERS, PLATES, SPHERES CHAP. XVII the upper surface of a beam by means of another beam at right angles to it; in this case the surface of contact is plane and the concentrated load W may be regarded as uniformly distributed over the area. This subject has already been mentioned in Art. 142, and it is there indicated that, for simple beams, the flexural compressive stress and the direct compressive stress due to the concentrated load combine to produce a true compressive stress which is smaller than either of them. When the concentrated load rests upon the lower flange of a simple I beam, as sometimes occurs in practice, the combination of the direct compression with the flexural tension produces a true compression and a true tension which are larger than the apparent ones. It is hence always preferable to support the concentrated load on the com- pressive side of a beam. The two preceding articles contain examples of the contact of cylinders and spheres with plane surfaces, and from the reasoning there given a relation may be deduced between the area of contact and the load W. For the cylinder the area of contact is the width 2(ed)* multiplied by the length /; inserting the value of e, this area is a=2ld(S/2E)*, and replacing S by its value in terms of W, it becomes a= (W)*(3JP/E)*. The area of contact hence varies as the cube root of the load for the same cylindrical roller; thus if a load Wi gives an area of contact a\, a load 8Wi is required in order to make the area 2a\. This conclusion is valid only when the elastic limit of the material is not exceeded. The formula here deduced for a is for the case where the plate is not deformed ; when the deformation is equally divided between the plate and the roller, 3VF is to be replaced by 6W, and the law connecting a and W remains unaltered. For the case of a sphere resting on a plane, Art. 157 shows that the area of contact is ned; placing in this the value of e in terms of S, and then that of S in terms of W, there is found a*d(icW/E)* } which shows that the area of contact varies as the diameter of the sphere and with the square root of the load. To double the area of contact, it is hence necessary to quad- ruple the load upon a sphere; this law holds whether the sphere AKT. 159 CIRCULAR PLATES WITH UNIFORM LOAD 409 alone be deformed or whether the deformation is divided between the sphere and plate. The above law does not agree with the conclusions derived from the experiments made by J. B. Johnson on the contact between car- wheels and railroad rails (Art. 142). This disagreement is probably mostly due to the fact that the upper surface of the rail is not a plane, and in part to the fact that the unit-stresses were very high. Prob. 158. Compute the total area of contact for the cylindrical rollers of Problem 156. If the same load is carried on spherical steel rollers 3 inches in diameter, compute the total area of contact. ART. 159. CIRCULAR PLATES WITH UNIFORM LOAD Let a circular plate of radius r and uniform thickness d be subject on one side to a pressure R on each square unit of area, and be supported or fixed around the circumference. The head of a cylinder under the pressure of water or steam is a circular plate in such a condition. Under the action of the load, the plate bends, the side in contact with the load being subject to compression while the other side is under tension; the maximum stress caused by the flexure will evidently occur at the middle, and this is required to be determined. As the simplest case let the plate be merely supported around the circumference. The total load on the plate being xr 2 R, the total reaction of the support is also xr 2 R, or the reaction per linear unit is frR. Now let a strip having the small width b be imagined to be cut out of the plate, so that its central line coincides with a diame- ter. The reaction at each end of this strip is b . $rR and the load on the strip is b . irR. The sum of the two reactions being only w| one-half the load, an upward shearing force equal to brR must act along the sides of the lg " 159 ' strip to maintain the equilibrium. The manner of distribution of this shearing stress along the sides of the strip is unknown and uncertain, but a fair probable assumption may be to take 410 ROLLERS, PLATES, SPHERES CHAP. XVII it as constant from the center to the circumference so that it acts like an upward uniform load. The strip of breadth b, depth d, and length 2r is thus a simple beam acted upon by two vertical reactions, each equal to %brR, a downward uniform load 2brR, and two vertical shears on the sides, each equal to $brR. The bending moment at the middle of this imaginary beam hence is, M=$bfR . r+$brR . \r-brR . %r=$br 2 R and the maximum unit-stress on the upper or lower fiber at the middle of the strip is, from the flexure formula (41), S = Me/ 1 = 6M/bd 2 = %R . r 2 /d 2 This value of 5 is not the real horizontal unit-stress at the center of the circle, but only the apparent stress due to considering the elementary strip. At the center the horizontal unit-stresses are acting in all directions. If a second strip is passed in Fig. 159 at right angles to the first, a unit-stress S equal in value but normal in direction to the first will be found. The true horizontal unit- stress T will be determined from the principle of Art. 139, taking into account the factor of lateral contraction X; on the upper side of the plate T=S AS XR, and on the lower side of the plate T=S AS. The latter value is the one to be used, since it is larger than the former. Accordingly, T=(L-X}R.(r/d} 2 and (d/r) 2 =%(i-X)R/T are the general formulas for the discussion of circular plates supported around the circumference and subject to a uniform load nr*R. For cast iron the mean value of the factor of lateral contrac- tion X is , while for wrought iron and steel it is \. Hence, T=l(r/d) 2 R and T=(r/d) 2 R (159) are the practical formulas for use, the first applying to cast iron and the second to wrought iron and steel circular plates when supported at the circumference, T being the allowable unit-stress in tension. The unit-pressure R that a circular plate can carry varies directly as the square of its thickness and inversely as the square of its diameter. ART. 160 CIRCULAR PLATES WITH CONCENTRATED LOAD 411 The more common case of a circular plate fixed around its circumference cannot be solved without a discussion of the elastic curve into which a diameter deflects. The investigation is too lengthy to be given here, but it can be said that the true effective unit-stress is about two-thirds of that for the supported plate. Hence T=\(r/WR and T=$(r/d)*R (159)' are formulas for fixed plates, the first being for cast iron and the second for wrought iron and steel. From the above formulas the proper thickness d for circular plates under uniform pressure may be readily computed. For example, let a fixed cast-iron cylinder head of 36 inches diameter be required to carry a uniform pressure R of 250 pounds per square inch, with an allowable tensile stress T of 3 600 pounds per square inch; then d = r($R/4T)* = 4.1 inches. The thickness of a steel cylinder head for the same diameter and pressure, for a tensile stress of 12 ooo pounds per square inch, will be 2.i inches. Prob. 159a. When the total load W for a circular plate is given, show that the thickness of the plate should be the same whatever be the diameter. Prob. 1596. If a plate 36 inches in diameter and 2 inches thick can safely carry a pressure of 250 pounds per square inch, what is the safe pressure for a plate 24 inches in diameter and i inch thick ? ART. 160. CIRCULAR PLATES WITH CONCENTRATED LOAD When a circular plate is under flexure from a concentrated load at the middle, it is more highly stressed than when the same load is uniformly distributed, as is evident from the theory of beams. The simplest case is where the plate is merely supported along its circumference and where the load P is uniformly dis- tributed over a circle of radius r , the radius of the plate being r and its thickness d. Let x be any radius less than r ; the load on the circle of radius x is uniformly distributed, the load per square unit being P/7rr 2 , which may be called R. The load on the circle of radius x is xx 2 R, which acts along the section of 412 ROLLERS, PLATES, SPHERES CHAP. XVII circumference 2nx and depth d, so that the shear per linear unit along this circumference is nx 2 R/2nx or \Rx, and the shear for any distance b along this circumference is V= ^bRx. Since the shear V and bending moment M for any section are connected by the relation dM/dx= V, it follows that where the constant of integration is determined by the condition that M becomes M lt the moment for the circumference 27rr , when x is equal to r . To find the value of M l} let z be any radius greater than r and less than r. The total shear along the circumference 27T2 is P, the shear per linear unit is P/2xz, and the shear for a distance b is V= Pb/2Ttz. The bending moment M r is then found by the same relation as before to be Pb . . Pb . r M'= -- logz+C or M'= log 27T m 6 2 where the constant of integration is determined by the condition that M' = o when z = r. Making z = r , the value of MI now is (Pb/2it) log r/r . Inserting this in the above expression for M, replacing R by its value P/7zr , arid making x=o, there results ,. Pb i . r\ M= - + log- 27T\2 & r / as the apparent bending moment for the middle of a strip of breadth b, depth d, and length 2r. The apparent flexural unit- stress on the upper or lower side of the middle of this strip is now found to be '-W-^G^)- If a second strip be passed at right angles to the first, the same unit-stress S will be found acting normally to the other. The true horizontal unit-stress T on the lower side of the plate then is T = S XS, where A is the factor of lateral contraction. Hence (160) ART. 160 CIRCULAR PLATES WITH CONCENTRATED LOAD 413 is the general formula for the discussion of supported circular plates under concentrated loads at the middle, A being J for cast iron and for wrought iron and steel. When the load is uniformly distributed over the entire plate, r equals r, P is nr 2 R, and (160) reduces to T = %(i-X)R(rld)*, which is the same as deduced by another method in the last article. When r = o the load is concentrated at a mathematical point and T becomes infinite. The formula can also be written (160)' in which R is the unit-pressure on the area xr Q 2 ; in cases of design the allowable unit-pressure R should not be allowed to exceed the elastic limit of the material. As an example, let a load of 5 ooo pounds be at the middle of a steel plate, distributed over a circle i inch in diameter, the plate being i inch thick and 48 inches in diameter. Here Jl = , r = J inch, r=24 inches, and ^ = 6360 pounds per square inch; using a table of Naperian logarithms, log 48 is found to be 3.871 and then formula (160)' gives T = 13 900 pounds per square inch, which is not too high a value for a steady load. When a plate is fixed around its circumference it is probable that the constants in the above formulas for T should be i instead of | and 2 instead of 3. Fixing the circumference increases the strength of the plate, for the same reason that the strength of a beam is increased by fixing its ends, and a fixed plate can carry a load about fifty percent greater than that carried by a sup- ported one. When a plate is stiffened by ribs, as is frequently the case in cast iron, about one-half of the material of the ribs may be regarded as adding to the thickness of the plate. Prob. 160o. Which is the stronger, a circular plate carrying a load P uniformly distributed, or one carrying a load %P distributed over an area at the middle which is one-fourth of the area of the plate ? Prob. 1606. Show that the formula for the discussion of a circular steel plate, fixed around its circumference and loaded at the middle, is T 1.48 P/d 2 when r/ro= 20, and T= 1.78 P/d 2 when r/ro=4o. 414 ROLLERS, PLATES, SPHERES CHAP. XVII ART. 161. ELLIPTICAL PLATES Elliptical plates are commonly used for the covers of man- holes in boilers and stand-pipes. Let R be the uniform unit- pressure on the plate, a the semi-major axis, and b the semi-minor axis of the ellipse. It is required to find the maximum unit-stress T on the tensile side of the plate. Taking the case where the plate is simply supported around the circumfer- ence, let two elementary strips be drawn as in Fig. 161, one along the major axis and the other along the minor axis. Let W\ and W 2 be the loads on these strips, and fi and /2 their deflections. At the center of the ellipse the deflec- tions of the strips are, from Art. 55, /i = W ia ?/pEI / 2 = Wztf/pEI and because these are equal, Wid 3 must be equal to W%tP. Since the reactions at the ends are proportional to the loads, it follows that the reactions at the ends of the axes are inversely as the cubes of the lengths of the axes. Hence the total weight xabR is not uniformly distributed on the support around the circum- ference, but the greatest reaction per linear unit will be found at the ends of the minor axis and the least at the ends of the major axis. It should hence be expected that the horizontal flexural stresses at the center are the greatest in directions parallel to the minor axis, and that in case of rupture a crack would begin at the center and run along the major axis; this is verified by tests. The theoretic solution of this very difficult problem cannot well be given here. From the discussion of Grashof and the experiments of Bach the following approximate formula may be written for wrought-iron and steel elliptical plates supported around the circumference: (161) in which a and b are the semi-axes, d the thickness of the plate, R the unit-pressure upon it, and T the allowable tensile unit- ART. 1G2 RECTANGULAR PLATES 415 stress. When a and b are equal, the ellipse becomes a circle of radius r, and T = R(r/d) 2 as found in Art. 159. For a cast-iron plate supported along its circumference, the numerical coefficient in the above formula will be f instead of 2. For plates fixed around the circumference, the coefficient will be about for wrought iron and steel, and about i$ for cast iron. These numbers are derived by taking the factor of lateral contrac- tion as for wrought iron and steel, and as { for cast iron. In Germany the value A = T^ is generally used for both cast iron and steel, and this will slightly modify the above numerical coefficients. A common proportion for manhole covers is to make a/b = 1.5, that is, the length is 50 percent greater than the width. Let the length be 24 inches and the width 16 inches, and let it be required to find the proper thickness when the cast-iron manhole cover is used in a stand-pipe under a head of water of 50 feet. Here a = 12 and 6=8 inches, R = 22 pounds per square inch, while the allowable T for cast iron may be taken as 3 ooo pounds per square inch. Then, since the plate is supported around the circumference, the numerical coefficient in the formula will be |, and from it is found d=ab($R/(a 2 + b 2 )T)*=o.S6 inches, so that a thickness of f inches will be sufficient safely to with- stand the pressure. Prob. 161a. Show that the allowable unit-pressure R for a cast- iron elliptical manhole cover having the proportions 0/6=1.5, 1S given by R = 2.g2(d/bi) 2 T, in which b\ is the width of the plate. Prob. 1616. Compute the safe unit-pressure for a cast-iron man- hole cover of 20 inches length, 13 inches width, and ij inches thick- ness. What head of water will produce this pressure? ART. 162. RECTANGULAR PLATES A rectangular plate of length 2/ and width im, subject to a uniform pressure R per square unit, distributes that pressure over the support in a similar manner to the elliptical plate. The reaction per linear unit is less on the ends than on the sides, 416 ROLLERS, PLATES, SPHERES CHAP, xvil and is greater at the middle of the ends and sides than near the corners. Rupture tends to occur near the center and parallel to the longer side. The approximate formula derived from the discussion of Bach for iron and steel plates is, (162) in which T is the maximum tensile unit-stress at the middle of the lower side of the plate, and d is the thickness of the plate. The values of the number a, as determined by the experiments of Bach, ranged from % to f , according as the condition of the edges approached that of a mere support or a state of fixedness. For a square plate / and m are equal, and the above expression may then be reduced to the forms, and the first being for free and the second for fixed edges. The numerical constants in these formulas are derived from the dis- cussion of experiments and hence stand upon a different basis from those deduced for circular plates; probably the formulas will apply better to cases of rupture than to cases where T is within the elastic limit of , the material. This problem has been discussed theoretically by Grashof with the conclusion that the formula for a square plate fixed at the middle of each edge is T = (i-tf)R(l/d) 2 , where A is the factor of lateral contraction. A plate might be fixed in this manner by a bolt at the middle of each edge, but such an arrangement is unusual, the common method being to bolt it to the support at many points. When A = , this formula reduces to T =%R(l/d) 2 , which is intermediate between those given for free and for fixed edges in the last paragraph. While the numerical coefficients for square plates, as deduced by different authors, vary somewhat, it is well established that the unit-stress T at the middle of the plate varies directly as its area and the unit-pressure R, and inversely as the square of its thickness. The strength of a square plate, as measured by the pressure R that it can carry, varies directly as the square of the ART. 163 HOLLOW SPHERES 417 thickness and inversely as the area; this law is the same as that previously found for circular plates. Prob. 162. Prove that the maximum unit-stress for a square plate, caused by a given uniform load \V, is independent of the size of the plate. ART. 163. HOLLOW SPHERES Hollow spheres are used in certain forms of boilers under inside steam-pressure. The ends of steam and water cylinders are sometimes made hemispherical instead of plane, in order to avoid flexure; the base of a steel water tank is often made a hemisphere for the same reason. If the thickness of the sphere is small compared to its radius, the investigation is simple. Let r be the radius and / the thickness. Let R be the inside pressure per square unit, and S the tensile unit-stress on the annulus. Then on any great circle the total pressure is nr 2 R, and this is resisted by the tension 2r.rtS in the section of the annulus. By equating these, there is found, 2tS=rR or S=\R.r/t which is the formula generally used for thin spheres under inner pressure. But in strictness 5 is the apparent stress, while another equal in intensity acts at right angles to it. Thus from Art. 139 the true stress on the outside surface isTi=S AS, while that on the inside surface is T%= S A5 + A/?. Using for the value of A, and inserting the above value of 5, there result, 7\ = $R . r/t and T 2 = $R+ $R . r/t for the true tensile unit-stresses on the outside and inside surfaces respectively. Both of these are less than S, and hence the usual formula for thin spheres (Art. 31) errs on the side of safety. The investigation of a thick hollow sphere under inside and outside pressure will be similar to that of the thick cylinder in Art. 149. Let r\ be the inside and r-z the outside radius, R\ and RZ being the corresponding pressures per square unit of surface. Fig. 1496 may represent a partial section of the sphere, x being the radius of any elementary annulus where the radial unit-stress 418 ROLLERS, PLATES, SPHERES CHAP. XVII is R and the tangential unit-stress is 5. From the symmetry of the sphere it is seen that another stress S acts at right angles to the one shown in the figure. Thus an elementary particle at any position in the annulus is held in equilibrium by three principal stresses R, S, and S. The sum of these is regarded as constant throughout the annulus (Art. 183), and accordingly 2S + R=$Ci is one equation between S and R, where C\ is a constant which is to be determined. Now the inside pressure on a great circle whose radius is x is xx 2 R, and the outside pressure on a great circle whose radius is x + dx is n(x + 8x) 2 (R + 8R), both of these being perpendicular to the plane of the circle. The difference of these is equal to the resisting stress in the elementary annulus, which is 27cxdx . S. Stating this equation and omitting quantities of the second order, a second relation between S and R is found. Accordingly, are the two conditions for determining S and R. Substituting in the second equation the value of 5 from the first and inte- grating, the value of R in terms of x is found, and then that of R is known; thus, and R=C l -2C 2 /x* (163) in which C-2, is a constant of the integration. These formulas for hollow spheres are seen to be analogous to those for thick cylinders, the radii being cubed instead of squared. The formula for 5 is the most important one and 5 has its greatest value at the inside surface of the hollow sphere. Values of the constants C\ and 2 may be found from the formula for R. Regarding the unit-pressures R\ and R% as without sign, R becomes RI when x=r\ and R becomes R 2 when x = r 2 ; then, and these when placed in (163) give the formulas deduced by Lame for thick hollow spheres. The most common case is that where there is no outside pressure R?; here the tensile unit- A R T - 163 HOLLOW SPHERES 419 stresses on the inside and outside surfaces are, and the first of these gives the greater unit-stress. For example, if r 2 = 2fi, then Si=%Ri, and when r 2 is nearly as large as r u then 5i is nearly %R . r/t, as previously found for a thin hollow sphere in Art. 31. The above gives the apparent unit-stresses. To find the true unit-stress T\ at the inside surface of a steel hollow sphere, the factor of lateral contraction is to be taken as , and then by Art. 139, Fi^Si-iSi+^i^i^+r^/foS-,^) (163)" which will generally be found to be less than S\. It is therefore on the safe side to use the formula for Si in cases of design. As an example, let a steel cylinder 4 inches in inside and 8 inches in outside radius have a hemispherical end with the same radii, and be subject to an inside water pressure of 4 ooo pounds per square inch. Then the apparent tensile stress on the inside surface of the hemisphere is found from (163)' to be 2 860 pounds per square inch, while (163)" gives the true tensile stress as 3 240 pounds per square inch; hence the true stress is about 13 percent larger than the apparent. For the cylinder itself, the apparent and true tensile stresses at the inside surface may be computed from (150) and (152)" and these values are Si =6 700 and Ti=l6oo pounds per square inch, so that the true stress is 13 percent greater than the apparent for the cylinder. If the end of this cylinder is a flat plate of the same thickness as the cylinder, or 4 inches, and fixed around the circumference, the true stress on the outer side is found from (159)' to be T" = il X4 000 = 16 ooo pounds per square inch; this is five times as great as that for the hemisphere, and more than double the greatest stress in the cylinder. The advantage of hemispherical ends in reducing the stresses is thus seen to be very great. It may be remarked, in conclusion, that the theory of internal stress in cylinders and spheres is not perfect, for it fails to give the same . 420 ROLLERS, PLATES, SPHERES CHAP. XVII results for the common surface of junction of a cylinder and hemisphere. This indicates that the assumption made regarding the constancy of 28 + R does not probably hold good for a hemi- sphere attached to a cylinder. Prob. 163a. A hollow sphere is to be subject to a steam-pressure of 600 pounds per square inch, its inner radius being 8 inches. Compute its thickness, so that the greatest tensile stress may be i ooo pounds per square inch. Prob. 1636. Investigate the discrepancy between the formulas for hollow cylinders and hollow spheres for the following numerical case. A hollow cylinder with hollow hemispherical ends, the inside diameters being 8 inches and the outside diameters 12 inches, is subject to an inside water pressure of 2 400 pounds per square inch. Compute, by Art. 152, and by this article, the true maximum unit-stress T for the common plane of junction of cylinder and hemisphere. ART. 164 CENTRIFUGAL TENSION 421 CHAPTER XVIII MISCELLANEOUS DISCUSSIONS ART. 164. CENTRIFUGAL TENSION WHEN the center of gravity of a body of weight P revolves around an axis with the uniform velocity v\ and r is the distance of the center of gravity from the axis, there is generated a stress in the cord or bar that connects the body with the axis. This centrifugal force Q is shown in works on theoretical mechanics to be Q=Pv 2 /gr, where g is the acceleration of a body falling vertically under the action of gravity near the surface of the earth. The case shown in Fig. 1640 is that of a bar of uniform section area and length /, the weight P being attached to one end while it revolves around an axis A at the other end. It is required to find the centrifugal stress in the bar at A when the speed of n revolutions per second is maintained. L Let r be any distance from the axis; the velocity at this dis- tance is 27rra, or if is given by w = 2?rw. Now let W be the weight of the bar, %l the distance of its center of gravity from the axis, and v t the velocity of that center of gravity ; then, gives the centrifugal force at the axis; this produces a tension in the bar and a sidewise compression and flexuie on the axis. As an example, let a bar of wrought iron 2X2 inches and 6 feet long have a weight of 400 pounds with its center of gravity 422 MISCELLANEOUS DISCUSSIONS CHAP, xvm 6 feet from the axis of revolution. It is required to find the number of revolutions per second in order to produce rupture. Solving the last equation for cu, and placing Q = 50 ooo pounds per square inch, jP=4OO pounds, W =80 pounds, = 32.16 feet per second per second, 1=6 feet, and r=6 feet, there is found ^ = 48.4 radians per second, and hence the speed required to cause rupture is n =48.4/2?: = 7. 8 revolutions per second. When P=o in the above formula, the case is that of a bar of uniform section area and of weight W, and the tensile stress at the axis is Q^Wla^/g. The tensile stress in such a bar is o at the free end and it increases toward the axis, where it has the value Q. Let a be the area of the cross-section, w the weight of a unit of volume, and x any distance from the axis. Then the weight of the length I x is wa(l-x) and the distance of its center of gravity from the axis is $(l + x), so that the centrifugal force of this part of the bar is, Q f =wa(l-x) . W+x)a/*/g=*u>au?(P-x?)/2g and this is the tensile stress at the distance x from the axis. When x=l, then Q' =o; when x=o, then Q f =wal 2 cu 2 /2g, which is equal to the above value of Q. The tensile unit-stress in the bar at any distance x from the axis is R=Q'/a. This may be written R = $('wa/ 2 /g)(l 2 x 2 ) and it will be seen to be closely analogous with the expression for radial unit-stress in a revolving fly-wheel or millstone. Another case is that of the thin circular hoop of mean radius r and thickness /, shown in the first diagram of Fig. 1646. Let W be its weight, which is equal to 27rwbrt, if w is the weight of the material per cubic unit, and b is the width of the hoop per- pendicular to the plane of the jdrawing. The centrifugal force acting on the axis is here zero because the center of gravity of the hoop coincides with the axis. There is hence no pressure on the axis, but the centrifugal force acts radially upon the hoop and produces tension in it in the ART. 164 CENTRIFUGAL TENSION 423 same manner as inside water pressure does in a thin pipe. The centrifugal force due to an angular velocity a> is 27nubr*tcu 2 / g for the entire weight and hence the radial unit-pressure R' is found by dividing this by the area 2nbr, or R' = (waj 2 /g)rt. Now, referring to Art. 30, it is seen that rR' = IS is the relation between the outward unit-pressure R r and the tangential unit-stress S. Hence the value of the latter is S = (wo) 2 1 g)r 2 w = 2xn in which n is the number of revolutions per second. For ex- ample, let it be required to find the tensile stress in a cast-iron hoop 4 inches wide, 2 inches thick, and 62 inches in outside diameter when making 300 revolutions per minute. Here 7^ = 450/1728 pounds per cubic inch, r=$o inches, w = 5 revolu- tions per second, and = 32.16X12 inches per second per second; then S is 600 pounds per square inch. For a hoop of considerable thickness let r\ be the inside and r 2 the outside radius. Let x be the inside and x + dx be the outside radius of any elementary annulus, and R and R + 3R the radial unit-stresses, these being considered tensile, as in Art. 149. The outward forces acting upon the annulus are the radial unit-stress R + dR and the centrifugal unit-pressure R r , the value of which was found above to be (wit?lg}x8x, while the inward force is the radial unit-stress R. Accordingly, just as for a thin water-pipe, the equation of equilibrium between these forces and the tangential unit-stress S is which, neglecting quantities of the second order and representing (wa^/g) by m, reduces to the differential equation mx l dx + Rdx + xdR= Sdx and this must be satisfied by the expressions for R and 5. It is also to be noted that for a solid wheel, for which ri=o, the values of R and 5 must be equal when #=o. Further, for a hollow wheel or hoop the value of R must be zero when x = r 2 and also when xr\ t since there are no radial stresses on the free circumferences. The expressions 424 MISCELLANEOUS DISCUSSIONS CHAP, xvm satisfy these two conditions and also the differential equation, provided g = fn/(^p). In order to determine p, take the case of a solid wheel for which r\ = o and r 2 = r, the formulas becoming and let p be determined from the condition that the total internal work shall be a minimum (Art. 126). This investigation gives for the value of p, whence q = jf w, and thus finally, are the radial and tangential unit-stresses at the distance x from the center of a revolving wheel of uniform thickness. These formulas show for a solid grindstone or millstone, for which ri=o and r 2 = r, that the greatest stress occurs at the center, R and S being each equal to (wu^jg^r 2 , while for the circumference R = o and S = (waP I g)\r 2 . .They also show, for a grindstone or millstone having a hole of radius r\ at the center, that the greatest value of R occurs for #=(rir 2 )*, while the greatest value of 5 occurs at the inner circumference. It is hence to be expected that, under a sufficiently high velocity, a solid wheel would begin to crack at the center and a hollow wheel at the inner circumference. ' The above refers to apparent stresses only, but the true stresses corresponding to the actual deformations are somewhat less than given by the formulas, since R and 5 are both tensile. Let A be the given factor of lateral contraction the value of which is J for steel, \ for cast iron, and \ or less for stone; then the true radial unit-stress is R AS and the true tangential unit-stress is S-AR in which R and S are given by the preceding formulas. It should be noted, however, that more exhaustive investigations seem to lead to somewhat different expressions. Love's formula for the tangential true unit-stress is For ^ = J this gives S=(wa> 2 /g)$$r 2 at the center of a solid wheel, while the preceding formula gives S=(wcuP/g)$%r 2 , which is 8 percent greater. AKT. 165 CENTRIFUGAL FLEXURE 425 ART. 165. CENTRIFUGAL FLEXURE The rod that connects the cross-head of a steam engine with the crank pin is subject to a centrifugal flexural stress owing to the fact that one end revolves in a circle. The horizontal rod, or parallel bar, joining two driving wheels of a locomotive is another instance of centrifugal flexure; this is simpler than the connecting rod, because all points are revolving with the same velocity, and hence it will be discussed first. Let u be the velocity of a locomotive and v the velocity of revolution of the end of the parallel rod around the axle of the driving wheel to which it is attached. Let r\ be the radius of the wheel, and r the radius of the circle of revolution of the end of the parallel rod. Then since the velocity of revolution of the circumference of the wheel is the same as the linear velocity of the locomotive, it follows that v=u.r/ri. Now not only the end of the parallel rod, but every point B ~B in it, is revolving with the Fi - 165 velocity v in a circle of radius r. Thus a centrifugal force is generated which produces flexural stresses. When the rod is at its lowest position BB, this centrifugal force acts as a downward uniform load producing flexure; at the highest position A A it acts as an upward uniform load producing flexure; at the position CC, on the same level as the axles, it produces a compressive stress in the direction of the length of the rod. Let TV be the weight of the parallel rod per linear unit; then, from rational mechanics, the centrifugal force is, irf = wiP/gr or -u/ = wu^/grj 2 which may be called the centrifugal load per unit of length. The rod being a beam supported at its ends, having a length /, a breadth b, and a depth d, the maximum unit-stress due to this uniform load is, from the flexure formula (41), 426 MISCELLANEOUS DISCUSSIONS CHAP, xvill which is the flexural stress due to centrifugal force when the bar is at its highest or at its lowest position. If the bar is not rectangular, as is generally the case, the values of c and / for its cross-section are to be obtained by the methods which are ex- plained in Arts. 42 and 43. In this formula g is the acceleration of gravity, or 32.16 feet per second per second. In using it in formulas, however, all quantities should be expressed in terms of the same linear unit, the inch being preferable. For example, let a locomotive be running at 60 miles per hour, the radius of the drivers being 3 feet and that of the parallel rod i foot, this being of steel, 4 inches deep, 2 inches thick, and 8 feet long. Here w=88 feet per second = 12X88 inches per second, = 32.16X12 inches per second per second, ^=3X12 inches, r = iXi2 inches, /=8Xi2 inches, 6 = 2 inches, d=4 inches, and w = 2.2j pounds per linear inch. The centrifugal load per inch then is w/=6i pounds, and the maximum flexural stress is 5 = 13 200 pounds per square inch, which is probably not sufficiently low when it is considered that the parallel rod is subject to rapid alternating stresses and per- haps to shocks. The connecting rod moves in a circle of radius r at the crank pin, while the other end moves only in a straight line. Thus at the end A there is no centrifugal load, while at B the cen- trifugal load is the same as given by the above expression for w'. When the rod is in the position shown in Fig. 1656, it is a beam acted upon by a cen- trifugal load which varies uni- formly from o at A to it/ at B. The total load is hence \itfl, the reaction at A is %-u/l, and that at B is %u/l. The bending moment for any section distant x from the end A then is, and the maximum value of M occurs at the section for which which gives maximum M =o.o6^Sit/l 2 .' Then from ART. 166 UNSYMMETRIC LOADS ON BEAMS 427 the flexure formuk (41), the flexural unit-stress is, S=Mc/I S=o.3&3w'I 2 /bd 2 the second being for a rectangular section in which iif has the same value as for the parallel rod. By comparing the formulas for rectangular parallel and connecting parallel rods it is seen that the unit-stress S for the former is about twice as great, if the length and section area are the same in the two cases. The parallel rod needs the greatest cross-section at the middle, while the connecting-rod needs the greatest cross-section at about 0.6} from the cross-head. Prob. 165. The connecting rod of an engine is 2 feet long and it is attached to a crank pin at a distance of 6 inches from the axis of a fly-wheel. If the wheel makes 750 revolutions per minute, find a square cross-section for the connecting rod so that the centrifugal unit- stress S may be 4200 pounds per square inch. ART. 166. UNSYMMETRIC LOADS ON BEAMS It was explained in Art. 43 that two axes may be drawn in the plane of any section area of a beam, these passing through the center of gravity of the section and being at right angles to each other, so that the moment of inertia of the section is greater for one axis and less for the second axis than for any other axis through the center of gravity. These are called ' principal axes ', and the moments of inertia with respect to them are those required in all common cases; Table 6 gives these moments of inertia for I sections and Table 8 those for T sections. The great majority of beam sections are symmetric with respect to both of the principal axes, while a few like the T and bulb sections are symmetric with respect to one principal axis only. The L section of Fig. 42c and the Z section of Fig. 42d are unsymmetric with respect to both principal axes ; these are not commonly used as beams, but when so used, the flexure formula (41) cannot be applied to their correct discussion without the determination of the moments of inertia for the principal axes. A load on a beam is said to be 'unsymmetric' when its vertical 428 MISCELLANEOUS DISCUSSIONS CHAP. XVIII plane does not coincide with one of the principal axes of the section area. The simplest case is that shown in Fig. 166#, where the broken lines show the principal axes of a rectangular section, and the load P is applied at a certain distance from the vertical axis. If the load is on a beam with supported ends, it is plain that the reactions of the end will not be uniformly dis- tributed over the supports and hence the right-hand part of the section will be but slightly stressed. If the beam is fastened to the supports, the reaction may be downward or negative on the right-hand part and upward on the left-hand part, so that tor- sional stresses will be developed. This case is one that should be avoided, for it is clear that parts of the beam are much more highly stressed than when the load is placed so that its resultant coincides with the vertical principal axis. a > a 1 ..,-' ^ z/ Fig. 'J> 1< Qd Fig. 166o Fig. 166& Fig. 166c Fig. 1666 shows a case of unsymmetric loading which occurs in a purlin beam connecting two roof trusses. The upper chords or main rafters of the truss being inclined to the horizontal at the angle <, the upper and lower surfaces of the horizontal purlin have the same inclination, and the load on its upper surface makes the angle with the principal axis 2-2. Let I\ and 1 2 be the moments of inertia of the section area with respect to the principal axes i-i and 2-2. Let M be the bending moment at the dangerous section due to the vertical loads, then the compo- nent M cos(f> acts in the plane 2-2 and produces the unit-stress S\ =c\M cos is not a large angle, the unit-stress Si is frequently computed by this method and regarded as the actual flexural stress. For example, let the angle be 30 degrees, the simple beam be 4 inches wide, 6 inches deep, 13 feet in span, and be subject to a total uniform load of 600 pounds. Here M = n 700 pound-inches, Ii=j2 inches 4 , 1=3 ART. 166 UNSYMMETRIC LOADS ON BEAMS 429 inches, and then 1=490 pounds per square inch; this is much smaller than the actual unit-stress. Another method of dealing with the last case is to consider the neutral axis mn as horizontal, to find the values of c and 7 with respect to it, and then to compute the unit-stress from the flexure formula S=Mc/I. Here c is the distance from the remotest fiber at the upper or lower corner to the neutral axis and it is easy to show that c=Ci cos + c 2 sin$, where c\ and c 2 are the coordinates of the corner with respect to the principal axes i-i and 2-2. It may also be shown that /=/i cos 2 < + /2 sin 2 < gives the moment of inertia of the section area with respect to the axis mn. For the above numerical example, ^1=3 and . Then the flexural unit- stress due to the first moment is Si =CiMcos and are known. Table 11 gives data for a few Z bars with equal legs, this being a part of the table in the Cambria pocket book. The moments of inertia I a and I b are those with respect to the axes a-a and b-b, while the principal axes are i-i and 2-2. The value of 1 2 is found by /2=^2 2 > where a is the section area and r 2 the least radius of gyration. The value of /i is found by subtracting I 2 from the sum I a + I b . The tangent of the angle (f> is given in the sixth column of the table, and from this the values of cos< and sin< are readily found. For example, take a Z bar 8 inches deep with legs 3 inches long, and let a load P be applied as shown in Fig. IQQd which produces a bending moment M equal to 120 ooo pound-inches. Making a drawing of the section from the data in the table, the measured values of c\ and 2 normal to the axes i-i and 2-2 are found to be 4.61 and i. 60 inches. From the given value tan<=o.27, there are found cos^> =0.965 and sin< =0.260. Following the above rules, the principal moments of inertia are 72 = 2.56 and 71=50.24 inches 4 . The formula (166) then gives S = 30 100 pounds per square inch for the stress on the upper or lower corner of the end of the leg. By using the rough method in which aa is taken as the neutral axis, there is found 5 = 120000X4/44.64 = 10800 pounds per square inch, which is about one-third of the former value. Consideration of Fig. 1660 indicates, however, that the actual unit-stress in the corner of the Z bar is probably higher than 30 100 pounds per square inch. Prob. 166a. Show that the angle which gives the greatest flexural unit-stress for the cases of Fig. 1666 is that for which tan<=c 2 /i/ci/2. Prob. 166ft. Show that the angle 6 between the neutral axis mn- and the principal axis i-i in Fig. 1666 is given by tan#= (/i// 2 )tan^. For the above example of the purlin beam, show that is 51^ degrees, and that hence the neutral axis is inclined 2if to the horizontal. Prob. 166c. Compute the unit-stress S for a Z bar 6 inches deep with legs 3^ inches long, when used as beam under a bending moment of 90000 pound-inches. ABT. 167 EULER'S MODIFIED FORMULA 431 ART. 167. EULER'S MODIFIED FORMULA Euler's formula for columns expresses the condition of in- different equilibrium, or that state which borders between stable and unstable equilibrium. In all cases of indifferent equilibrium slight causes produce marked effects, and hence it seems important to inquire whether Euler's formula, as given in Art. 78, repre- sents the exact relation between P/a and l/r. It will be apparent on reflection that, while the formula contains but one length /, there are really three different lengths that should be taken into consideration. Let / represent the length of the straight column in its unstrained state, /i the length of the straight column after compression by the unit-stress P/a, and 1 2 the chord of the deflected curve after lateral bending. Referring now to formula (45), it is seen that this does not include the effect of the longitudinal compression P/a. To in- troduce this, let 5 be the total unit-stress produced by both flexure and compression ; then in the demonstration the first four formulas will be thus modified : R - C S-P/a_E3k_M a ~^ a and then the reasoning leads to the following modified equation of the elastic curve of the deflected column, J*y 81 I EI~ 9 =M-rj- = M dx 2 oli /i Passing now to Euler's deduction in Art. 78, the bending moment M is replaced by Py, and the equation of the elastic curve for a round-ended column is found to be, Here y=o when x=l 2 ; hence by the same reasoning as before there is deduced the formula, which is the condition for the state of indifferent equilibrium. To complete the investigation, l\ and 1 2 are to be expressed in terms of /. Now / 1\ is the deformation due to the longi- 432 MISCELLANEOUS DISCUSSIONS CHAP. XVIII tudinal compression P/a, and from the definition of the modulus of elasticity in Art. 9 its value is li=l(i-P/aE). To find 1 2 it is plain that l\ 1 2 is the fall at the end of the column due to the lateral flexure and that P(l\lz) is the work done in this fall. This external work must equal the internal work of the flexural stresses. From Art. 123, . M 2 dx P 2 v*dx P 2 / 2 . 2 7r* . dK= ^jr = - = -= sin 2 . dx zEI 2EI 2EI 1 2 and integrating this between the limits x =o and x=l 2 , the internal work K is found to be P 2 f 2 l 2 /4EI. Therefore, equating the external to the internal work, the value of l\ 1 2 is found to be Pf 2 l 2 /4EI. Accordingly, /! = /(i - P/aE) I 2 =l(i- P/aE) /(i + Pf 2 /^Ear 2 } are the required values in terms of the original length. The quantities P/aE and Pf 2 /$Ear 2 are small in comparison with unity, and hence their squares and their product may be neglected; also the reciprocal of iP/aE may be taken as i + P/aE. Introducing the values of l\ and 1 2 into the above expression for P it reduces to, and since n 2 Er 2 /l 2 is a close approximation to the value of P fa it may replace the latter in the parenthesis, giving finally, which is Euler's modified formula for round-ended columns. By writing 2.o$n 2 instead of n 2 it applies to a column with one end round and the other fixed, and by writing qn 2 instead of ?r 2 it applies to a column with both ends fixed. Another modification of Euler's formula which has been frequently given is derived by considering only / and l\, so that the equation of the elastic curve becomes, Anx. 168 TESTING MACHINES 433 Accordingly Euler's formulas as deduced in Art. 78 are to be modified by replacing E by P/a, and then solving for P/a. For a column with both ends round this leads to the form, P x 2 Erl which shows that the unit-load required to hold the column in indifferent equilibrium is. less than x 2 E(r/l) 2 . By taking /2 into account, as is done above, the unit-load is found to be greater than x 2 E(r/t) 2 , and (167) also indicates that P slightly increases with the deflection /, so that after a slight deflection has been attained the value of / is not wholly indeterminate, as the common theory teaches. Prob. 167. Show that the fall of the load P, due to direct com- pression of the column, equals the fall due to the lateral flexure when / is equal to zr. ART. 168. TESTING MACHINES The first experiments on the strength of materials were made on the rupture of beams of timber. A picture in Galileo's Discorsi (Leiden, 1638), shows a cantilever beam projecting from a wall and loaded at the free end with a heavy weight, and it was probably from experiments of this kind that Galileo was led to the con- clusion that the strength of rectangular beams varies as the squares of their depths. During the eighteenth century experiments were made by Mariotte, Parent, and others on timber in flexure and tension, only questions of ultimate strength being in- vestigated, while the elastic limit was unrecognized. Hooke's law of the proportionality between stress and deformation had, indeed, been announced in 1678, but it was not until 1798 that Girard made the first comprehensive series of experiments on the elastic properties of beams. Early in the nineteenth century appeared the ' Lectures on Natural Philosophy ' by Young in which is found a clear presentation of many of the laws of flexure both under static forces and under shock. It also introduces for the first time the modulus of elasticity, E, but fails to note that it can only be deduced or applied when the elastic limit of 434 MISCELLANEOUS DISCUSSIONS CHAP, xvill the material is not surpassed. A little later Barlow, Tregdold, and Hodgkinson experimented on timber and cast iron, both in the form of beams and of columns; their methods and results, although now seeming rude and defective, are deserving of praise as the first of real practical value. In 1849 was published the 'Report of the Commissioners on the Application of Iron to Railway Structures ', which may be regarded as the landmark of the beginning of modern methods of testing. It contains the records of valuable tests by Willis, James, Hodgkinson, and Galton on the strength of cast and wrought iron as well as upon the resistance to impact, investi- gations of the increase in stress caused by a rolling load on a beam, experiments on the fatigue of metals, and the evidence given by leading British engineers as to their opinions on proper factors of safety under different conditions. The immediate result of this report was the decision by the English board of trade that the factor of safety for cast iron should be twice as great for rolling loads as for steady ones, while throughout both Europe and the United States it excited a marked interest and impetus in the subject of testing materials. The first testing machines in the United States were those built by Wade and Rodman between 1850 and 1860 for testing gun-metal for the government. About this time the rapid intro- duction of iron bridges led to experiments by Plympton and by Roebling. Prior to 1865 apparatus was built by each experi- menter for his special work, but in this year Fairbanks put upon the market the first testing machines for commercial work. A little later the machines of Olsen and of Richie" for tensile, com- pressive, and flexural tests were introduced and have since been widely used. The machine devised by Emery, soon after 1875, is a very precise apparatus which is used in large laboratories. Large machines /or testing eye-bars have been built by bridge companies since 1880, and numerous testing laboratories now contain machines for every kind of work. The capacity of a testing machine is the tension or pressure which it can exert. A small machine for testing cement by AKT. 168 TESTING MACHINES 435 tension has usually a capacity of i ooo or of 2 ooo pounds. Ma- chines of 50000, 100 ooo, and 150000 pounds for testing metals are common. The Emery machine at the government arsenal at Watertown, Mass., has a capacity of i ooo ooo pounds, and it can test a bar 30 feet long or a small hair with equal precision. The eye-bar machine at Athens, Pa., has a capacity of i 244 ooo pounds, and that at Phcenixville, Pa., a capacity of 2 160 ooo pounds. It forms no part of the purpose of this book to give descriptions and illustrations of these machines; Martens' Hand- book for Testing Materials (Berlin, 1898; New York, 1899) may be referred to as giving thorough and comprehensive dis- cussions of both testing machines and of methods of testing. Tensile tests are the most common and some commercial machines are arranged with an autographic recording apparatus whereby a card is secured which shows the relation between the load and the elongation throughout the test. There are also a number of autographic recording devices in the market, which may be attached to any machine. When such a graphic record is taken, the yield point, ultimate strength, and ultimate elonga- tion may be read from it. Nearly all tensile machines may be also used for compressive tests, and also for flexural tests on short beams. The smaller machines are operated by hand, while power is required to run the larger ones. Screw machines in which the load is brought upon the specimen by the help of large screws are generally preferred in the United States, while hy- draulic machines in which a hydraulic press is used to transmit the load to the specimen are preferred in Europe. The machine of Emery is hydraulic; other American makes are screw machines. Commercial tests of materials are rarely made under shearing and torsional stresses. Thurston in 1870 devised a torsion machine for small specimens, and the torsion machines of Olsen and of Richie*, which are found in the laboratories of most engi- neering colleges, prove very serviceable for illustrating the phe- nomena of twisting. Impact machines have been built for special investigations, but the only one on the market is that of Keep which is designed for tests of bars of cast iron. Fatigue or 436 MISCELLANEOUS DISCUSSIONS CHAP. XVIII endurance tests, which subject the specimens to alternating stresses for long periods of time, have not been made except in laboratories. The testing of materials has assumed such great importance since 1890 that all engineering colleges have provided laboratories for the purposes of instruction and research. The work done by some of these has proved of much value to the engineering profession ; the work of Hatt on impact tests of metals and that of Talbot on reinforced-concrete beams may be cited as examples. Manufacturers of iron and steel, and large railroad companies, have installed laboratories for the testing of their products and purchases, several of these having a machine of 300 ooo pounds capacity as well as smaller ones. The many series of tests con- ducted by Howard at the Watertown arsenal may be mentioned as part of the valuable work done by the government of the United States. The largest testing laboratories, however, are found in Europe ; that at Berlin, under the directorship of Martens, standing at the head; this has a floor area of 10 360 square meters or about 2\ acres. Prob. 168. Consult Das konigliche Materialpriifungsamt der tech- nischen Hochschule, Berlin, 1904, and ascertain the various kinds of work done in that great laboratory. ART. 169. TESTING OF MATERIALS The specimens for tensile tests are either flat or round. A flat specimen is rarely used except for plates, and its standard size is 1 8 inches in length, 2 inches wide along the ends and \\ inches wide for a central length of about 9 inches, while the thickness is the same as that of the plate from which it is cut. Wire and rods are frequently tested just as they come from the mill, the length between the jaws of the machine being usually more than ten times the diameter. Standard round specimens are cut from axles, shafts, beams, and other manufactured products. Prior to 1895 the standard size of the round specimen was i inch in diameter along a central length of 9 inches, while the ends were larger and provided with screw threads. Both ART. 169 TESTING OF -MATERIALS 437 flat and round specimens are frequently called 8-inch specimens because two marks are placed upon them 8 inches apart for the purpose of measuring the elongation. Since 1895 there has been gradually coming into use in the .United States a smaller size for round specimens which is shown in Fig. 169a; this is called the 2-inch specimen, because the central part is a little more than 2 inches long and the marks are placed upon it 2 inches apart; the diameter of this specimen is 0.5 inch or sometimes 0.505 inches. This smaller specimen has the advantage that less material is required to be wasted in taking it from an axle or shaft, but it has the disadvantage that percentages of elongation computed from it are greater than those determined from the 8-inch speci- men (Art. 25). In Art. 12 will be found a description of the tests of these 2-inch specimens. Fig. 169a Flat specimens are usually griped by the jaws of the machine, while the standard round specimens screw into nuts to which the tension is applied. The load transmitted through the speci- men is weighed by a scale at the end of a compound lever. In commercial tests the ultimate elongation is alone measured; this is 'done by two marks on the specimen and measuring the distance between them before and after rupture. In scientific tests an extensimeter is attached to the specimen, so that the elongation can be read at each increment of weight. The elonga- 438 MISCELLANEOUS DISCUSSIONS CHAP. XVIII tion is usually expressed as a percentage of the original length be- tween the two marks on the specimen, and for the reason above stated it is always desirable that the original length and the diameter of specimen should be mentioned in the report. For ductile materials, like wrought iron and mild steel, it is customary to slowly reduce the load after the ultimate strength is reached; the material is then flowing rapidly, so that the elongation con- tinues to increase, and hence a greater percentage of elongation is obtained. Fig. 1696 Compressive tests are mainly confined to brick, stone, and concrete, and are but little used for metals on account of the expense of preparing the specimens with truly parallel sides. Since these specimens are short, it is difficult to obtain precise measures of the change of length; the ultimate shortening is rarely observed, because it is difficult to note the moment when it occurs. Failure in the case of brittle materials occurs by a diagonal shearing as explained in Arts. 18 and 147. Fig. 1696 shows a cube of neat cement and a timber block which have ART. 169 TESTING OF MATERIALS 439 failed in this manner; the section area of each of these specimens is 2X2 inches. In flexural tests the load is usually applied at the middle of a beam supported at its ends and gradually increased until rupture occurs, the deflection being measured also as a test of stiffness. The apparatus for determining the deflection should be attached to the supports so that the compression of these may not affect the observed values. In the simplest case weights are hung upon a ring or stirrup placed upon the middle of a simple beam, and are added in regular increments of 100 pounds, more or less, depending upon the size of the beam. When the elastic limit is not exceeded the modulus of elasticity E may be computed from the observed deflection by the formula deduced in Art. 55. From the breaking load P, the flexural strength or modulus of rupture S r may be computed from the flexure formula (41) ; this lies between the ultimate tensile and com- pressive strengths of the material, and although not a physical constant (Art. 52) it is useful for comparing different qualities of the same material. Torsion tests are not used for commercial purposes, but in scientific laboratories much has been learned by the use of tor- sion machines. Fig. 169c shows three specimens which have been subjected to torsion, the first being a square steel bar which broke at the upper end, the second a round bar of cast iron, and the third a rectangular steel bar which was ruled with white lines in order that the distortions might be studied; discussions of these experiments are given in Arts. 94 and 99. Impact tests are regarded as of much value in judging of the quality of ductile metals. The cold-bend test, briefly de- scribed in Art. 24, is one that has been known from the earliest times and which is constantly used in all mills where wrought iron or steel is produced. The bending of the specimen is gener- ally done by blows of a hammer, although steady pressure is sometimes employed. Notwithstanding that no numerical re- sults are obtained from the cold-bend test, except the final angle of bending, the general information that it gives is of the highest 140 MISCELLANEOUS DISCUSSIONS CHAP, .\\iii Fig. I69c ART. 1G9 TESTIXG OF MATERIALS 441 importance, so that it has been said that, if all tests of metals except one were to be 'abandoned, the cold-bend test should be the one to be retained. In the rolling-mill it is used to judge of the purity and quality of the muck bar; in the steel mill it serves to classify and grade the material almost as well as chemical analysis can do, and in the purchase of shape iron it affords a quick and satisfactory method of estimating toughness, ductility, strength, and capacity to resist external work. It is very important, in order that the results of tests made in different laboratories may be compared and correct conclu- sions be drawn therefrom, that the methods of testing shall be uniform. In 1882 a number of German professors met at Munich to discuss this question, and other conferences were held in 1884, 1886, 1888, and 1893, at which engineers from other European countries were present. In 1895 the International Association for Testing Materials was formally organized at Zurich, its object being "the development and unification of standard methods of testing for the determination of the properties of materials, and also the perfection of apparatus for that purpose." The second Congress of the Association met at Stockholm in 1897, the third at Budapest in 1902, and the fourth at Brussels in 1905, these being attended by representatives of about twenty different countries. Affiliated with the International Association are several national societies which are engaged in the same work. The American Section of the International Association, organized in 1898, became in 1901 the American Society for Testing Materials. The number of members of the International Association at the beginning of 1905 was 2215, representing 24 different countries; of these, 580 members were in the United States, 367 in Germany, 357 in Russia, 193 in Austria, 154 in France. The official organ of the International Association is the journal Baumaterialienkunde, published semi-monthly at Stuttgart, Germany. The American Society for Testing Mate- rials published the fifth volume of its Proceedings in 1905. Prob. 169. Refer to Halt and Marburg's Bibliography of Impact Tests and Impact Testing Machines, published in Vol. II of the Pro- 442 MISCELLANEOUS DISCUSSIONS CHAP, xvm ceedings of the American Society for Testing Materials; consult several references and describe one or more machines for making impact tests. ART. 170. RULES FOR TESTING When materials or manufactured products are to be pur- chased, the requirements that they must fulfill are generally expressed in 'specifications '. If it is specified that certain tests are to be made, or certain test specimens be used, it is necessary that this should be done most carefully, and no deviation from the given requirements can be allowed. Full specifications re- garding machines and methods of testing are, however, rarely made by a purchaser, so that considerable latitude is allowed, but in all cases it should be the aim of the engineer in charge of the tests to so conduct the work that it shall be well and truly done. Certain general rules regarding testing machines and their use will hence be given here, and the observance of these as far as practicable will conduce to uniformity of methods and to reliability of the results, these rules applying mainly to machines for common tensile tests. 1. A machine should be so constructed that the load borne by the specimen alone is registered upon the weighing scale, so that its readings may not include any force expended in friction on the pivots or moving parts of the machine. 2. A machine should be from time to time rated or cali- brated, to ascertain if the readings of the weighing scale are correct, or that the errors of its readings may be known. 3. The construction of the machine should be such that, when properly operated, the specimens may not be subject to shock. 4. The holders or jaws which gripe the ends of a tensile specimen should be so arranged that the resultant load coin- cides with the axis of the specimen, in order that the stress may be uniformly distributed over the section area. 5. The use of serrated wedges for holding the ends of specimens is not advisable unless those ends are larger in section area than the main part of the specimen. ART. 170 RULES FOR TESTING 443 6. The load upon specimens of ductile materials should be applied at a slower rate within the elastic limit than after that limit is passed. 7. When the elastic limit is specified, this is not to be determined by the drop of the scale beam, but by measuring the increments of elongation corresponding to increments of the load ; the drop of the beam indicates the yield point. 8. Percentages of elongation should be accompanied by a statement of the size of the test pieces from which they were determined. Many other rules have been proposed, but it is believed that the above are nearly all which will be universally accepted as essential ones in tensile tests. For compressive tests the follow- ing may be noted as desirable requirements: 9. The surfaces to which the pressure is applied should be truly parallel, in order that the stress may be uniformly distributed over the section area; this distribution will be more perfect if one of the supports is arranged so as to be movable in all directions. 10. The cube is probably the best form of compression specimen, because it has been so widely used that the results of tests are better comparable with previous records than if prisms are used. For flexural tests it is not so easy to state definite rules, but it will be generally admitted that the supports should be rounded, so as not to indent the beam or prevent the ends from taking the inclination of the theoretic elastic curve. When deflections are measured, the apparatus should be attached to the ends of the beam, if any yielding of the supports can occur. In all static tests of tension, compression, or flexure, it should be the aim to secure measurements which can be relied upon as correct with a probable error not exceeding one-half of one percent. Impact tests of rails, axles, and car-wheels are usually made by a simple apparatus which consists of a ram falling between guides. It is here important that the guides should not retard the fall of the ram, that the shape of the ram should be such as to prevent lateral oscillation, that the surface struck should 444 MISCELLANEOUS DISCUSSIONS CHAP, xvm be level, that the weight of the supports of the specimen should be at least ten times that of the ram, and that these supports should rest upon masonry. In conclusion it may be noted that all the theoretic discussions of this volume are of assistance in conducting tests of materials. When a tensile test is to be made, it must not be forgotten that reliable results cannot be obtained unless the specimen is placed in the machine in such a manner that there is no tendency to flexure; when a compressive test is to be made, the influence of an eccentric load should be kept in mind; when a flexural test is to be made, the theory of flexure should be understood. Clear and definite ideas regarding elastic limit and yield point will avoid misunderstanding. Theory and practice should always go hand in hand, each supplementing and explaining the other. Prob. 170. Consult a paper by Goss in Vol. Ill of Proceedings of American Society for Testing Materials, and describe the drop test- ing machine of the Master Car Builders' Association. ART. 171. SPECIFICATIONS FOR STRUCTURAL STEEL When materials are to be purchased, a set of rules is usually prepared giving requirements regarding quality and tests, and these rules are called 'specifications '. There is much variation in such specifications owing to the different opinions of buyers and the use which is to be made of the material. The following specifications for structural steel will give the student an idea of the extent and scope of the requirements which are generally demanded for steel to be used in buildings, bridges, and ships. The order of arrangement is similar to that of the specifications adopted by the American Society for Testing Materials and the American Railway Engineering and Maintenance of Way Asso- ciation, but the requirements differ in a few particulars. 1. Structural steel shall be made by the open-hearth process. 2. Each of the three classes of structural steel shall conform to the following limits in chemical composition: Sulphur shall not exceed 0.06 percent; Phosphorus shall not exceed 0.07 per- ART. 171 SPECIFICATIONS FOR STRUCTURAL STEEL 445 cent when the steel is made by the acid process and not exceed 0.04 percent when it is made by the basic process. 3. There shall be three classes of structural steel for bridges and ships, namely, rivet steel, soft steel, and medium steel, which shall conform to the following physical requirements: Rivet steel shall range in tensile strength from 50 ooo to 55 ooo pounds per square inch, have a yield point not less than 33 coo pounds per square inch, and the elongation in 8 inches shall not be less than 26 percent. Soft steel shall range in tensile strength from 55 coo to 60 coo pounds per square inch, have a yield point not less than 35 coo pounds per square inch, and the elongation in 8 inches shall not be less than 25 percent. Medium steel shall range in tensile strength from 60 coo to 65 coo pounds per square inch, have a yield point not less than 37 coo pounds per square inch, and the elongation in 8 inches shall not be less than 23 percent. 4. For each increase of inch in a flat specimen above a thickness of inches, a deduction of i shall be made from the above specified elongation. For each decrease of -^ inch below a thickness of T 8 ^ inches, a deduction of 2^ shall be made from the above specified elongation. For bridge pins the required elonga- tion shall be 5 less than that above specified, as determined on a test specimen the center of which shall be one inch from the surface of the pin. 5. Eye-bars shall be of medium steel. Full-sized tests shall show 12 \ percent elongation in 15 feet of the body of the eye- bar, and the tensile strength shah 1 not be less than 55 coo pounds per square inch. Eye-bars shall be required to break in the body, but should an eye-bar break in the head, and show 12^ percent elongation in 15 feet and the tensile strength specified, it shall not be cause for rejection, provided that not more than one- third of the total number of eye-bars tested break in the head. 6. The three classes of structural steel shall conform to the following bending tests; and for this purpose the test specimen shall be ij inches wide, if possible, and for all material f inches or less in thickness the test specimen shall be of the same thick- 446 MISCELLANEOUS DISCUSSIONS CHAP. XVII ness as that of the finished material from which it is cut, but for material more than f inches thick the bending test specimen may be \ inch thick: Rivet steel shall bend cold 180 degrees flat on itself without fracture on the outside of the bent portion. Rivet rounds shall be tested of full size as rolled. Soft steel shall bend cold 180 degrees flat on itself without fracture on the outside of the bent portion. Medium steel shall bend cold 180 degrees around a diameter equal to the thickness of the specimen tested, without fracture on the outside of the bent portion. 7. The standard test specimen of 8 inches gauged length shall be used to determine the physical properties specified in paragraphs Nos. 3 and 4. The standard size of the test speci- men for sheared plates shall be \\ inches in width for a length not less than 9 inches, and of a thickness equal to that of the plate. For other material the test specimen may be the same as for sheared plates or it may be planed or turned parallel through- out its entire length, and in all cases where possible two opposite sides of the test specimen shall be the rolled surfaces. Rivet rounds and small rolled bars shall be tested of full size as rolled. 8. One tensile test specimen shall be taken from the finished material of each melt or blow, but in case this develops flaws, or breaks outside of the middle third of its gauged length, it may be discarded and another test specimen substituted therefor. 9. One test specimen for bending shall be taken from the finished material of each melt as it comes from the rolls, and for material inches and less in thickness this specimen shall have the natural rolled surface on two opposite sides. The bending- test specimen shall be \\ inches wide, if possible, and for material mere than inches thick the bending-test specimen may be \ inch thick. The sheared edges of bending-test specimens may be milled or planed. 10. Material which is to be used without annealing or further treatment shall be tested for tensile strength in the condition in ART. 171 SPECIFICATIONS FOR STRUCTURAL STEEL 447 which it comes from the rolls. Where it is impracticable to secure a test specimen from material which has been annealed or otherwise treated, a full-sized section of tensile-test specimen length shall be similarly treated before cutting the tensile-test specimen therefrom. n. For the purpose of this specification, the yield point shall be determined by careful observation of the drop of the beam or halt in the gauge of the testing machine. 12. In order to determine if the material conforms to the chemical limitations prescribed in paragraph No. 2 herein, analysis shall be made of drillings taken from a small test ingot. 13. The variation in cross-section or weight of more than 2$- percent from that specified will be sufficient cause for rejection, except in the case of sheared plates, which will be covered by the following permissible variations : Plates \2\ pounds per square foot or heavier, up to 100 inches wide, when ordered to weight, shall not average more than 2\ percent variation above or below the theoretical weight. When loo inches wide and over, 5 percent above or 5 percent below the theoretical weight. Plates under \2\ pounds per square foot, when ordered to weight, shall not average a greater variation than the following: When less than 75 inches wide, 2\ percent above or below the theoretical weight. When 75 inches wide up to 100 inches wide, 5 percent above or 3 percent below the theoretical weight. When loo inches wide and over, 10 percent above or 3 percent below the theoretical weight. When plates are ordered to gauge, a variation of more than T fg- inch below that specified for any dimension will be sufficient cause for rejection. An excess in weight above the nominal weight may, however, be allowed, as agreed upon between the inspector and the manufacturer. 14. Finished material must be free from injurious seams, flaws, defective edges, or cracks, and have a workmanlike finish. 15. Every finished piece of steel shall be stamped with the 448 MISCELLANEOUS DISCUSSIONS CHAP, xvill melt number, and steel for pins shall have a melt number stamped on the ends. Rivets and lacing steel, and small pieces for pin plates and stiffeners, may be shipped in bundles, securely wired together, with the melt number on a metal tag attached. 1 6. The inspector representing the purchaser, shall have all reasonable facilities afforded to him by the manufacturer to satisfy him that the finished material is furnished in accordance with these specifications. All tests and inspections shall be made at the place of manufacture, prior to shipment. Prob. 171. Consult Proceedings of the American Society for Test- ing Materials, Vol. IV, 1904, and ascertain the tests recommended by Webster for detecting brittle steel. Young men should also read the excellent advice given in Vol. V of the same Proceedings by President Dudley in his address on the duties and responsibilities of the testing and inspecting engineer. . 172 INTRODUCTION 449 CHAPTER XIX MATHEMATICAL THEORY OF ELASTICITY ART. 172. INTRODUCTION The mathematical theory of elasticity is that science which treats of the behavior of bodies under stress when the law of proportionality of deformation to stress is observed. All the theoretic formulas of the preceding pages have been derived by the help of this law, but these constitute only a part of the mathematical theory of elasticity. The formulas derived for the deformation of bodies under tension or compression suppose the bodies to be homogeneous or isotropic, so that the modulus of elasticity E is the same for all directions; some materials, how- ever, have different properties of stiffness in different directions so that there may be several values of E to be considered, this being especially the case with crystals. The theory of elasticity takes account of such non-homogeneous structure and deduces formulas for the deformations due to forces applied in different directions. In this chapter only the elements of this theory can be given, and in general the bodies under stress will be regarded as homogeneous. The complete theory deals not only with elastic solids, but with fluids, gases, and the ether of space, while the discussion of stresses and deformations, both in homogeneous and crystalline bodies, leads to the investigation of wave propa- gations, the time and velocity of elastic oscillations, and numerous other phenomena of physics. Statics proper is concerned only with rigid bodies, whHe the theory of elasticity deals with bodies deformed under the action of exterior forces and which recover their original shape on the removal of these forces. All the principles and methods of statics apply in the discussion of elastic bodies, but in addition new principles based upon Hooke's law arise. The amount of deformation being small within the elastic limit for common 450 MATHEMATICAL THEORY OF ELASTICITY CHAP, xix materials, it is allowable to neglect the squares and higher powers of a unit-elongation in comparison with the elongation itself, as set forth in Art. 13. By the help of this principle, the elas- tic change in volume of a body may be found (Art. 173), the modulus of elasticity for tension or compression is found to have a certain relation to the modulus of elasticity for shearing (Art. 181), and a new modulus of elasticity based on change in volume is introduced (Art. 182). The general case of a body acted upon by forces in several directions occupies the main part of this chapter, this case requir- ing the use of three rectangular coordinates. This chapter, then, is an extension of the methods of Chapters XI and XV, and includes those methods as special cases. It has been prepared from the point of view of the engineer rather than that of the pure mathematician, and should be regarded only as an intro- duction to the mathematical theory of elasticity. The student should consult the article on Elasticity by Kelvin in the Encyclopaedia Britannica, as also the History of Todhunter and Pierson. The works of Clebsch (Elasticitat fester Korper, 1862), Winkler (Elasticitat und Festigkeit, 1867), Grashof (Theorie der Elasticitat und Festigkeit, 1878), and Flamant (Resistance des Mat6riaux, 1886) may be mentioned as treating the subject both from the theoretical and the engineering point of view. Prob. 172. Consult Todhunter and Pierson's History of the Theory of Elasticity and of the Strength of Materials, and ascertain some- thing about the important investigations of Saint Venant. ART. 173. ELASTIC CHANGES IN VOLUME The changes in section area and in volume which occur when a bar is under axial stress have been discussed in Art. 13, and the same method will now be applied to the case of a body acted upon by forces in different directions. When the elastic limit of the material is not surpassed, the deformation due to any applied force is proportional to that force for deformations in any and all directions, and the principles of Art. 139 enable these to be determined. AHT. 173 ELASTIC CHANGES IN VOLUME 451 The simplest case is that shown in Fig. 173, where a parallele- piped is acted upon by the tensile unit-forces Si, S 2 , S 3 in three rectangular directions. The body is regarded as homogeneous, so that the factor of lateral contraction A and the modulus of elasticity E are the same in all directions. Let \ be the unit- elongation due to Si or ei=Si/E, also 2 =S 2 /E and 3 =S 3 /E. Then the actual unit-elongations which take place in the three rectangular directions are given by, f =l Ae 2 ^3 " = 2 % 3 AI ff/ = 3 hi ^ 2 Now for any cube of edge unity, the volume after the application of the unit-forces Si, S 2 , S 3 is (i + e'X 1 + e")(i + "'), and since e', <", f " are small compared with unity, this product is i + r + e" + e'" (Art. 13). Accordingly, the elastic change in the unit of volume is, '+"+ '"=(l-2X)(i + 2 + 3 ) = (l- 2 X)(Sl + S 2 + S 3 )/E (173) Here it is seen that there is no change in rolume when A = J. For all the materials of construction it is found that the factor of lateral contraction is less than , so that they increase in volume under tension and decrease under compression. In the above discussion the unit-forces Si, S 2 , S 3 are regarded as tensile, but the formula for change of volume applies equally well when one or all of them are compressive; for example, if Si is tension and S 2 and S 3 are compression, the values of S a and S 3 are to be taken as negative. The above formula refers to unit of volume and the actual change in a parallelepiped of given dimensions is found by multiplying the unit-change by the number of units of volume in the body. For example, let a brick 2X4X8 inches be subjected to a compression of 12 800 pounds upon the two flat sides, to a compression of 4 800 pounds upon the two narrow sides, and to a tension of i 600 pounds upon the two ends. Here S 2 = - 1 2800/32=- 400, S 3 = - 4800/16 =-300, and Si = + i6oo/8 = +200 pounds per square inch; then Si+S2+S.3= 500 pounds per square inch. Taking A = 0.2, and = 2000000 452 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX pounds per square inch, the change per unit of volume is found to be -0.0002, so that the decrease in the volume of the paral- lelepiped is 0.0002X64=0.0128 cubic inches. Elastic changes in section area due to the action of several forces are readily expressed in a similar manner. For instance, in the unit cube let a plane be drawn normal to Si cutting out a square; its area after the application of the three forces is (i + e")(i + e"0, which is equal to i + e" + e"'. Hence the change per unit of section area is found from (173) by making e' and Si equal to zero. Thus, for the above numerical example, the change per unit of section area normal to Si is 0.00004, so that the decrease in this section area is 0.00004X8 = 0.00032 square inches. When a "body is under uniform compression in all directions, as is the case when it is subjected to fluid pressure, the unit- stresses Si, 2, S$ are equal, as also the unit-deformations 1, e 2 , 3. For this case the change per unit of length is the same in all directions and equal to ( i 2 A) e, the change per unit of area in any section is 2(1 2 A) e, and the change per unit of volume is 3(1 2A)e, where e is the unit-change S/E which would be caused by an axial unit-stress S which is equal to the uniform compressive unit-stress. The change in section area is hence double, and the change in volume is three times that in a linear dimension. Prob. 173. Make experiments upon india rubber with the inten- tion of finding the value of the factor of lateral contraction for that material. ART. 174. NORMAL AND TANGENTIAL STRESSES The general case of internal stress is that of an elementary parallelepiped held in equilibrium by apparent stresses applied to its faces in directions not normal. Here each oblique stress may be decomposed into three components parallel respectively to three coordinate axes, OX, OY, OZ. Upon each of the faces perpendicular to OX the normal component of the oblique unit- stress is designated by S x and the two tangential components ART. 174 NORMAL AND TANGENTIAL STRESSES 453 Fig. 174 by S xy and S X2 . A similar notation applies to each of the other faces. An 5 having but one subscript denotes a tensile or com- pressive unit-stress, and its direction is parallel to the axis corresponding to that subscript. An S having two subscripts denotes a shear- ing unit-stress, the first sub- script designating the axis to which the face is perpendic- ular and the second desig- nating the axis to which the stress is parallel; thus S zx is on the face perpendicular to OZ and its direction is parallel to OX. In Fig. 174 the nine components for three sides of the parallelepiped are shown. Neglecting the weight of the parallelepiped the components upon the three opposite sides must be of equal intensity in order that equilibrium may obtain. An elementary parallelepiped in the interior of a body is thus held in equilibrium under the action of six normal and twelve tangential stresses acting upon its faces. The normal stresses upon any two opposite faces must be equal in intensity and opposite in direction. The tangential stresses upon any two opposite faces must also be equal in intensity and opposite in direction. A certain relation must also exist between the six shearing stresses shown in Fig. 174 in order that equilibrium may obtain. Let the parallelepiped be a cube with each edge equal to unity; then if no tendency to rotation exists with respect to an axis through the center of the cube and parallel to OX it is necessary that S yz should equal S zy . A similar condition obtains for each of the other rectangular axes, and hence, S xy =S yx S yz =S zy S XZ =S, X (174) that is, those shearing unit-stresses are equal which are upon any two adjacent faces and normal to their common edge. The apparent unit-stresses designated by S are computed by 454 MATHEMATICAL THEORY OF ELASTICITY CHAP, xix the methods of the preceding chapters; it is rare, however, that more than three or four of them exist, even under the action of complex forces. The general problem is then to find a parallelo- piped such that the resultant stresses upon it are wholly normal. These resultant normal stresses will be Si, S 2 , 5 3 , from which by (139) the true normal stresses T\, T 2 , T% can be found. It will later be shown that these stresses Si, S 2 , Ss are the maxi- mum apparent stresses of tension or compression resulting from the given normal and tangential stresses. Prob. 174. Let a, b, c be the angles which a line makes with the axes OX, O Y, OZ, respectively. Show that the sum of the squares of the cosines of these angles is equal to unity. ART. 175. RESULTANT STRESSES The resultant unit-stress upon any face of the parallelepiped in Fig. 174 is the resultant of the three rectangular unit-stresses acting upon that face. Thus for the face normal to the axis OZ the resultant unit-stress is given by, and the total resultant stress upon that face of the parallelepiped is the product of its area and R 3 . The resultant unit-stress R upon any elementary plane having any position can be determined when the normal and tangential stresses in the directions parallel to the coordinate axes are known. Lei a plane be passed through the corners i, 2, 3, of the paral- lelopiped in Fig. 174, and let a, b, c be the angles that its normal makes with the axes OX, OY, OZ, respectively. Let a, /?, ?-, be the angles which the resultant unit-stress R makes with the same axes. Let A be the area of the triangle 123; then the total resultant stress upon that area is AR, and its components parallel to the three axes are AR cosa, ARcosfl, ARcosf. The triangle whose area is A, together with the three triangles 023, 013, 012, form a pyramid which is in equilibrium under the action of R and the stresses upon the three triangles. The areas of these triangles are A cosa, A cosh, A cose, and fhe stresses upon them ART. 175 RESULTANT STRESSES 455 are the products of the areas by the several unit-stresses. Now the components of these four stresses with respect to each rectan gular axis must vanish as a necessary condition of equilibrium. Hence, canceling out A, which occurs in all terms, there results, R cosa = S x cosfl -f S xv cosb + S zx cose R cos/?= S xy cosa + S y cos/> + S gy cose (1 75) R co*r=S xs cosa+S vie cosb+S z cose in which the second members are all known quantities. From these equations the values of R cosa, Rcosp, RCOSJ- can be computed; then the sum of the squares of these is R 2 since cos 2 a+cos 2 /3 + cos 2 ?-=i. The value of cosa is found by dividing that of R cosa by R, and similarly for cos/3 and cosf. Now the angle 6 between the directions of R and the normal to the plane is given by, cos#=cosa cosa + cosi cos/? -f cose cosj- and then the tensile or compressive unit-stress normal to the given plane is R cos/9, while the resultant shearing unit-stress is R sin#. This shearing stress may be resolved into two com- ponents in any two directions on the plane. As a simple numerical example, let a bolt be subject to a tension of 12000 pounds per square inch and also to a cross- shear of 8 ooo pounds per square inch. It is required to find the apparent unit-stresses on a plane making an angle of 60 degrees with the axis of the bolt. Take OX parallel to the ten- sile force and O Y parallel to the cross-shear. Then S x = + 1 2 ooo, S X j, = 8ooo, 5^ = 8000, and the other stresses are zero; also a = 30, 6 = 60, and c = go. Then from (175), R cosa = 4- 14 390 R cos/3 = + 6930 R cos^= o and the resultant stress on the given plane is, R= (14 3Qo 2 +6 930 2 )*= 15 970 pounds per square inch The direction made by R with the axis is now found: cos a =1439/1 597 = 0.901 a = 6^ cos0= 693/1597 = 0.434 /3=25| and the angle included between the resultant R and the normal 456 MATHEMATICAL THEORY OF ELASTICITY CHAP, xix to the given plane is computed by, cos#= 0.866X0.901 +0.5X0.434 =0.997 Lastly, the normal tensile stress on the plane is found to be R cos#=i5 920 pounds per square inch, while the shearing stress on the plane is Rs'md = i 200 pounds per square inch. Prob. 175. Find for the above example the position of a plane upon which there is no shearing stress. ART. 176. THE ELLIPSOID OF STRESS Let the resultant unit-stress R upon any plane passing through a given point in the interior of a stressed body make an angle 6 with the normal to that plane. It will now be shown that, for different planes through the given point, the intensity of R may be represented by the radius vector of an ellipsoid. Let RI, R 2 , 7?3 be the resultant unit-stresses upon the three faces of the parallelepiped in Fig. 174, and let 0\, 6- 2 , 63 be the angles which they make with the coordinate axis OX; then, COS01 = S x /Ri COS0 2 = S yx /R 2 COS0 3 = S tx /R 3 determine the directions of RI, R 2 , Rj- Now let these direc- tions be taken as those of a new system of oblique coordinate axes, let R be the resultant unit-stress in any direction, and let R x , R y , R z be its components parallel to these new axes. Then R cosa is the component of R parallel to OX, and, R cosa = R x cosQi+Ry cosff 2 +R z cos# 3 or, inserting for the cosines their values, it becomes, R cosa = S x (R x /Ri) + S vx (R v /R 2 )+S fX (R,/R3) Comparing this with the first equation in (175), it is seen that, casa=R x /Ri cosb=R y /R 2 cosc=R;/R 3 But the sum of the squares of these cosines is unity; hence, in which the numerators are variable coordinates and the de- nominators are given quantities. This is the equation of the surface of an ellipsoid with respect to three coordinate axes having the directions of RI, R 2 , R 3 . ART. 177 THE THREE PRINCIPAL STRESSES 457 The surface f- an ellipsoid is hence a figure whose radius vector represents the resultant unit-stress upon a plane the normal to which makes an angle 6 with the direction ot that radius vector. If the forces are entirely confined to one plane, the ellipsoid re- duces to an ellipse. If there are three planes at right angles to each other which are subject only to normal stresses, as in Fig. 176, the normal unit-stresses S x , S y ,. S t correspond to Ri, R 2 , RS in the above equa- tion of the ellipsoid. In this case S x , S v , S z are the three axes of the ellipsoid. If now shearing stresses are applied to the faces* the ellipsoid will be deformed, and p . the three axes will take other posi- tions corresponding to three planes upon which no shearing stresses act. The stresses corresponding to the axes of the ellip- soid are called principal stresses. Prob. 176. If S V =S Z in Fig. 176, show that the ellipsoid becomes either a prolate spheroid or an oblate spheroid. ART. 177. THE THREE PRINCIPAL STRESSES In general, the resultant unit-stress R upon a given plane makes an angle 6 with the normal to that plane, and hence can be resolved into a normal stress of tension or compression and into two tangential shearing stresses (Art. 174). It is evident, however, that planes may exist upon which only normal stresses act, so that 6 is zero and R is pure tension or compression. In order to find these planes and the stresses upon them, the angles a, /?, f in the equations (175) are to be made equal to a, b, c, respectively. Also replacing R by S, these equations become, (S-Sg) cosa = S vx cosb+S tx cose (SS V ) cosb=S xy cosa+S zv cose (SS t ) cosc=S xtl cosa+S yic cosb in which S, cosa, cosb, cose are four unknown quantities. The 458 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX three angles, however, are connected by the relation, cos 2 a + cos 2 b + cos 2 c = i and hence four equations exist between four unknowns. Remembering the relation between the shearing stresses ex- pressed in (174), the solution of the equations leads to a cubic equation for S, which is of the form, S*-AS 2 +BS-C=o (177) in which the values of the three coefficients are, A = S x +S y +S z , . B=S x S J ,+SyS z +S,S x -S 2 xy -S 2 yz -S 2 2X C = S x SyS g + 2S x ySy z S 2X ~ S^y, - SyS 2 zx ~ S.S 2 ^ and the three roots of this cubic equation are the three normal stresses of tension or compression, which are called the three principal unit-stresses and represented by Si, S 2 , S 3 . The directions of these principal unit-stresses Si, S 2 , $3, with respect to the rectangular axes OX, OY, OZ, are given by the values of cosa, cosfc, cose, which are found to be, in which m\, m 2 , m 3 , m, represent the following functions of the given and principal unit-stresses: m 3 =(S x -S)(S v -S)-S 2 X y m and it will now be shown that each principal stress is perpen- dicular to the plane of the other two. Let Si, S 2 , S 3 be the three roots of the cubic equation (177). Let ai, bi, Ci be the angles which Si makes with the three co- ordinate axes OX, OY, OZ, and let a 2 , b 2 , c 2 be the angles which S 2 makes with the same axes. The angle between the direc- tions of Si and S 2 is then given by, COS<=COSai COSa2+COS&i COS&2+COSC1 COSC2 Now in the first set of formulas of this article let S be made Si and a, b, c be changed to ai, bi, Ci; let the first equation be mul- tiplied by cosa 2 , the second by cosfe 2 , and the third by cosc 2 ; and let the three equations be added; then, cosr 2 ) ART. 178 MAXIMUM SHEARING STRESSES 459 is one term in this sum. Again, let S be made S 2 ,' an da, b, c, be changed to a 2 , b 2 , c 2 ; let the equations be multiplied by cosai, cos&i, cosci, respectively, and added; then, ^(cosai cosa^+cosfii cos&2+cosci cosc 2 ) is one term in the sum, while all the other terms are the same as before. Hence if Si and S 2 are unequal, the factor in the parenthesis, which is cos$, must vanish; $ is therefore a right angle or Si and S 2 are perpendicular. In the same manner it may be shown that 3 is perpendicular to both Si and S 2 . The three principal stresses are hence perpendicular to each other, and as the only diameters of the ellipsoid which have this property are its axes, it follows that the directions of the principal stresses Si, S 2 , S 3 are those of the axes of the ellip- soid of stress. These principal stresses thus give the apparent maximum normal stresses of tension or compression; from (139) the corresponding true unit-stresses TI, T 2 , T$ are then found. An interesting property of the three rectangular stresses S xt S y , S z , is that their sum is constant, whatever may be the posi- tion of the coordinate axes. For, the sum of the three princi- pal stresses Si, S 2 , S 3 is equal to the coefficient A in the cubic equation of (177), and hence, that is, the sum of the normal unit-stresses in any three rectan- gular directions is constant. Prob. 177. When two principal stresses are equal, show that the value of each is (AB-^C)/(2A 2 -6B), where A, B, C are th"e coeffi- cients in (177). ART. 178. MAXIMUM SHEARING STRESSES As there are certain planes upon which the tensile and com- pressive unit-stresses are a maximum, so there are certain other planes upon which the shearing unit-stresses have their maxi- mum values. In order to determine these it is well to take the axes of the ellipsoid as the coordinate axes, and upon the planes normal to these there are no shearing stresses. The stresses 460 MATHEMATICAL THEORY OF ELASTICITY CHAP, xix Si, S 2 , S s will give apparent shearing stresses on other planes, while TI, T 2 . TZ will give the true shearing stresses. Let i 2 3 in Fig. 178 be any plane whose normal makes the angles a, b, c with the coordinate axes. Let R be the resultant unit-stress upon this plane, and a, ft, y be the angles which it makes with the same axes. The angle between R and the normal to the plane is expressed by, cos#=cosa cosa + cosft cosft+cosc cosf and the resultant shearing unit- stress on the plane is, U If R be apparent stress, this is the apparent shearing stress; if R be true stress, this is the true shearing stress which acts along the plane. The value of R, as a true stress, is given by, R 2 =(T l cosa) 2 +(T 2 cos6) 2 +(r 3 cose) 2 Now, since both R cosa, and T\ cosa are components of R m the direction OX, they are equal, and hence, cosa= (Ti/R) cosa cosp=(T 2 /R) cosb cos?-= (T S /R) cose Substituting these in the value of cos0, the resulting true shear- ing unit-stress is expressed by, r 2 = (TYcosa) 2 + (T 2 cosb) 2 + (T 3 cose) 2 - (Ti cos 2 a + T 2 cos 2 b + T 3 cos 2 c) 2 by the discussion of which the values of a, b, c, which render T a maximum, are deduced. Bearing in mind that the sum of the squares of the three cosines is unity, the discussion gives, (178) and therefore there are six planes of maximum shearing stress, each of which is parallel to one of the principal stresses and bisects the angle between the other two. On each of these planes ART. 179 DISCUSSION OF A CRANK PlN 461 the shearing unit-stress is one half the difference of the principal unit-stresses whose directions are bisected. The same investigation applies equally well to the apparent shearing unit-stresses, whose maximum values are, 5=K5i-5 2 ) S=^(S 2 -S 3 ) S=KS 3 Si) (178)' and whose directions are the same as those of the maximum true shearing unit-stresses. The sign indicates that the shears have opposite directions on opposite sides of the plane, but in numerical work it is always convenient to take them as positive or, rather, as signless quantities. As an example, let a bar be subject to a tension of 3000 pounds per square inch in the direction of its length and to a compression of 6 ooo pounds per square inch upon two oppo- site sides. Here Si=+ 30x30, 5 2 = 6000, 5 3 =o; then the maximum apparent shearing stresses are 4 500, 3 ooo, and i 500 pounds per square inch. But from (139), taking A as $, the true tensile and compressive stresses as T\ = + 5 cxao, T 2 = 7 ooo, r 3 =4-i ooo pounds per square inch, and then from (178) the maximum true shearing stresses are 6000, 4000, and 2000 pounds per square inch. Prob. 178. Compute the maximum apparent and true shearing unit- stresses for a cast-iron parallelepiped, 2X4X8 inches, which is subject to compression of 3200 pounds upon its largest faces, 60 pounds upon its smallest faces, and 500 pounds upon the other faces. ART. 179. DISCUSSION OF A CRANK PIN To apply the preceding principles to a particular case, a crank pin similar to that investigated in Art. 98 may be taken. The axis OX is assumed parallel to the axis of the pin, O Y parallel to the crank arm, and OZ perpendicular to both, the notation being the same as in Fig. 174. On one side of the crank pin near its junction with the arm there were found the following apparent stresses: A cross-shear from the pressure of the con- necting rod giving S xz = ^oo pounds per square inch, a shear due to the transmitted torsion giving S xt = goo pounds per square 462 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX inch, a flexural stress due to the connecting-rod giving S x = +800 pounds per square inch, a flexural stress due to the transmitted torsion giving 5j;=+i6oo pounds per square inch, and two compressions due to shrinkage giving S y = 4000 and S z = 2 ooo pounds per square inch. The two shears having the same direction add together, as also the two tensions, and the data then are, Sxt=i20o S x = + 24oo 5j,= 4000 S z = 2000 Inserting these in the cubic equation (177) it becomes, 5 3 +3 6ooS 2 7 840 oooS 27 200 ooo 000=0 and its three roots are the three principal stresses. To solve this equation, put S=*x i 200, and it reduces to, 3? 12 160 ooox ii 096 ooo 000=0 As this cubic equation has three real roots, it is to be solved by the help of a table of cosines ; thus let, 3^=12160000 2T- 3 cos3<=n 096 oooooo from which r = 2oi3 and cos3< =0.6801. Then from Table 17 is found 30 = 47 09', whence ^ = 15 43'. The roots are now computed as below, and by subtracting i 200 from each the three principal stresses are ascertained: #i=2rcos( =+3880 Si = +2680 x 2 = 2r cos(+ 120)= 2890 S 2 = 4090 * 3 =2r cos( varies from o to 360 degrees. In the first diagram Si and 82 are both tension or both compression, in the second diagram one is tension and the other compression. The broken curve shows the locus of the point N, and the dotted curve the locus of S. For every value of $ the lines OS and ON are at right angles to each other, and OR is their resultant. As a simple example, take the case of a bolt subject to an axial tension of 2 ooo and also to a cross -shear of 3 ooo pounds per square inch. Here 5^= +2 ooo, 5^ = 3000, and S v =o; the above quadratic equation then gives Si =+4160 and 82 = 2 160 pounds per square inch for the two maximum unit- stresses of tension and compression. The direction made by Si with the axis of the bolt, as found by the value of cosa in Art. 175, is about 54! degrees. From (178)' the maximum shear is 3 160 pounds per square inch. These are the apparent maximum unit-stresses. To find the true maximum stresses, formulas (139) give, taking as the factor of lateral contraction, ri = +488o, T 2 = 3550, TS= 670 pounds per square inch as the principal tensions and compressions; then from (178) the greatest shearing stress is T = 4 220 pounds per square inch. Here the true maxi- mum tension is 17 percent greater than the apparent, the true compression is 64 percent greater, and the true shearing stress is 33 percent greater. The true stresses cannot be represented by an ellipse, but an ellipsoid of internal stress results of which the second diagram in Fig. 180 may be regarded as a typical section. ART. 181 SHEARING MODULUS OF ELASTICITY 465 Cases can, however, be imagined in which one of the true principal stresses is zero. If Si, S 2 , 83 are the apparent stresses in three rectangular directions, it is seen from (139) that when S 3 - eSi eS 2 is zero, the true stress T 3 is also zero. For instance, let a cube be under compression by three normal stresses of 30, 24, and 18 pounds per square inch and let = ; then T"i = i6, T 2 = 8, and TS= o. Here the ellipse of true stress has its correct application and there are no true stresses in a plane normal to the plane of T\ and T 2 . Prob. 180. A body is subject to a tension of 4000 and to a com- pression of 6 ooo pounds per square inch, these acting at right angles to each other. Construct the ellipse of apparent stresses and find the positions of two planes on which there are no tensile or compressive stresses. ART. 181. SHEARING MODULUS OF ELASTICITY The shearing modulus of elasticity F, defined in Art. 15, must have a relation to the modulus E for tension or compression, since the action of shear upon a body produces internal tensile and compressive stresses (Art. 143). Let Fig. 181 represent the face of a cube which is acted upon by a ^ vertical shear S, the edge of the cube being unity so that the vertical shearing unit- stress is also S. Under the action of this shear, the face of the cube becomes dis- torted, as shown greatly exaggerated by the broken lines, and the longer diagonal of a --vLL the rhombus is under a tensile unit-stress while the shorter one is under a compressive unit-stress, each of these being equal to S, as proved in Art. 143. Let be the distortion parallel to the shear S; then =S/F from the defini- tion of shearing modulus of elasticity. The longer diagonal of the square has the length 2* and after the distortion its length becomes [i + (i + e) 2 ]*; by using the approximate method for extracting roots explained in Art. 13, and neglecting the square of e, this reduces to a*(i+Je). The 466 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX change in length of this diagonal hence is 2* . J, and the unit- elongation is e. In a similar manner the length of the shorter diagonal after the distortion is 2*(i E) and the unit-shortening is |e. Accordingly the change per unit of length for each diag- onal is %S/F. Now let A be the factor of lateral contraction (Art. 13) the mean value of which is J for cast iron and for wrought iron and steel. When a body is acted upon by a tension producing the unit-stress S, there results a unit-elongation S/E and a lateral unit-shortening XS/E. When a body is acted upon a tension producing the unit-stress 5 and by a compression at right angles producing the same unit-stress S, the unit-elongation is (i +X)S/E, as is shown in Art. 139; the lateral unit-shortening has also the same value. Accordingly, for the case of Fig. 181, each diagonal has suffered a change per unit of length equal to (i+X)S/E. The change per unit of length for each of the diagonals of the face of the cube has now been found by two different methods; equating the two values, there results, E=2(i + A)F or F=E/2(i + X) (181) which give the relation between the two moduluses. Hence when E and A have been determined by measurements on a bar under tension, the shearing modulus of elasticity may be computed. Using the mean values of E given in Art. 9, and the mean values of A as above stated, the mean values of the shearing modulus of elasticity are found to be as follows for iron and steel: for cast iron, F= 6 ooo ooo pounds per square inch for wrought iron, F= 9 400 ooo pounds per square inch for steel, F=n 200 ooo pounds per square inch and these have been verified by experiments on the torsion of shafts. There is little known regarding the values of A for other materials, and it may be said that formula (181) does not apply to fibrous or non-homogeneous materials for the reason that E is not the same in different directions. It is not to be expected then that jP could be correctly computed for timber from a value of E obtained from a tension parallel to the grain. ART. 182 THE VOLUMETRIC MODULUS 467 Four different methods are available for determining the shear- ing modulus of elasticity ; first, by the measurement of the detru- sion per unit of length in a short bar or beam like Fig. 181 ; second, from the angle of twist of a shaft (Art. 93) ; third, from the deflec- tions of beams of different lengths and sizes (Art. 125) ; and fourth, by the use of formula (181). When sources of error are eliminated from the experiments, these different methods give results for F which agree very well for homogeneous mate- rials. A fifth method, which is really an extension of the fourth, will be explained in the next article. All these methods, of course, apply only when the shearing elastic limit of the material is not exceeded by the unit-stress. Prob. 181. A bar of steel, 0.5050 inches in diameter and 2.0000 long, is observed to be 2.0013 inches long when under a tension of 4 ooo pounds, while its diameter is then found to be 0.5047 inches. Compute the shearing modulus of elasticity. ART. 182. THE VOLUMETRIC MODULUS The modulus of elasticity E for axial tension or compression is defined to be the ratio of the longitudinal unit-stress S to the unit-elongation or unit-shortening e; thus E = S/e. Similarly, the volumetric modulus of elasticity, which will be represented by G, is the ratio of a unit-stress 5 which acts in all directions upon the body to the change per unit of volume. Thus, if a uniform unit-pressure acts upon a body of volume unity and produces the change e' in that volume, then G = S/e f . It is required to find the values of e' and G, and also the relation between G and E. Let A be the factor of lateral contraction of the homogeneous cube, each edge of which is unity, while each face is subject to the same pressure 5. Then, from Art. 139, the unit-shortening of each edge of the cube is e' = (S-2>lS)/; or since e represents S/E, the unit-shortening is e f = (i-2X)e. The volume of the cube, which was originally unity, now becomes (i e0 3 , and hence the change in volume is 3^ when e' is so small that its square and cube may be neglected. The change per unit of volume is then three times the change per unit of length of each edge of 468 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX the cube; hence '=3(1 2^) e gives the change per unit of volume, and hence the volumetric modulus G is 5/3(1 2X)e, in which e is the change per linear unit due to an axial unit- stress 5 of tension or compression only; accordingly, G=S/e f =5/3(1-2;)* =3/3(1-2;) (182) are formulas for G, of which the third gives the relation between G and E. For example, ; is about for steel, and hence the volumetric modulus for steel is equal to the modulus for tension or compression. In the last article the relation between the shearing modulus F and the tensile or compressive modulus E was deduced. The last formula of (181) and the last formula of (182) furnish two equations from which, by the elimination of ;, there is found 3EG+EF = <)FG as the relation between E, F,G; hence, E=gFG/(F+3G) F '= $EG '/ '(gG '- E) G=EF/(gF-^E) give the value of each modulus in terms of the other two. For example, let it be known that for cast iron =15,000000 and F = 6 ooo ooo pounds per square inch; then G = 10 ooo ooo pounds per square inch. The last formula shows that F cannot be as small as }E, for G becomes infinite when E is equal to $F. Water is matter which propagates stress in all directions so that a unit-pressure 5 applied to the surface of a column of water produces a resisting unit-pressure S on all the confining surfaces. According to the experiments made by Grassi in 1850, the de- crease in a unit-volume of water caused by the pressure of one atmosphere, or 14.7 pounds per square inch, has a mean value of 0.00005; hence the mean volumetric modulus of elasticity for water is G = 14. 7/0.00005 = 294 ooo pounds per square inch, which is about one one-hundredth of that of steel, so that water is about loo times more compressible than is steel within its elastic limit. Water has no proper value of E, because it is impossible for a column to be subjected to longitudinal pressure only; when water in a pipe is under axial pressure, the shortening that is measured is due to a unit-pressure acting laterally as well as axially, and this gives the decrease in volume if the walls of the pipe are unyielding. ART. 183 ' STORED INTERNAL ENERGY 469 Prob. 182. Prove, for a homogeneous solid, that the ratio G/F is equal to 2(1 + A)/ 3 (i - 2 A). Also that the value of A may be expressed by *-/6G. ART. 183. STORED INTERNAL ENERGY The cases of resilience, or stored internal energy, which were discussed in Chapter XIII, relate only to simple axial stress and to simple shear; when a body is subject to several external forces acting in different directions, the expressions for resilience become more complex. Fig. 183 represents a parallelepiped, the faces of which are acted upon only by the normal unit-stresses Si, S 2 , 83. Let this parallelepiped be homogeneous so that the modulus of elasticity E and the factor of lateral contraction A are the same in all directions. Let /i be the length parallel to Si, and a\ the section area normal to S\\ then the total stress on this section area is aiSi, and from (10) and (139) the change in the length of /i is /i(Si-AS 2 -AS 3 )/. The work done while the stress a^ is increasing from zero up to its final value is one-half the product of the stress and change of length, provided the elastic limit of the material is not exceeded (Art. 119); hence the stored in- ternal energy due to Si is ^a 1 / 1 (Si 2 -/LSi5 2 ->LSi53)/; and this is proportional to the volume a\l\. A similar expression may be written for the energy due to S 2 and another for that due to 5 3 ; and the sum of the three is, K= lV(Si 2 +S 2 2 +S 3 2 - 2tiiS 2 - 2*SiS 3 - 2^S 2 S 3 )/E (183) in which V denotes the volume of the parallelepiped. Here the sign of each 5 is positive for tension and negative for compression. A discussion of this formula shows that K has usually a smaller value when the signs of Si, S 2 , S 3 are the same than when one has a sign opposite to that of the other two. When the unit- stresses are equal in sign and magnitude, then K = % (i 2 A) V^/E; for steel A is and the resilience becomes K = \(S 2 /E}V, which is the same as for simple axial stress (Art. 120). As a numerical 470 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX example, let the three edges of the parallelepiped 01, 02, 03 be 8, 6, and 4 inches long and the material be cast iron for which A is f, and let the unit-stresses Si, S 2 , S 3 be 3 ooo, 4 ooo, 5 ooo pounds per square inch compression; then K = ijo inch-pounds = 14 foot-pounds is the stored internal energy. When the above parallelepiped is subject to the action of six pairs of shears of which the unit-stresses are Si 2 , S 23 , S 3 i, in the notation of Art. 174, the corresponding unit-detrusions are $S 2 i 2 /F, $S 2 23 /F, %S 2 3 i/F, where F is the shearing modulus of elasticity, and the sum of these multiplied by the volume of the body gives, K'=lV(S 2 i 2 +S 2 23 +S 2 31 )/F (183)' as the stored energy due to the shears. This is to be added to the value of K in (183) when normal stresses also act upon the faces of the parallelepiped. A comparison of the two formulas shows that, if a body is acted upon by normal and tangential forces of equal intensity in three rectangular directions, the stored in- ternal energy due to shearing may often be greater than that due to the normal stresses. The principle of least work, established in Art. 126, states that the internal stresses which prevail in a body under the action of external forces are those which render the stored internal energy a minimum. This principle may be used in connection with (183) to determine the stresses S\, S 2 , S 3 in cases where the conditions of static equilibrium are insufficient In number. For example, it was assumed in Art. 163 that the tangential and radial stresses In a spherical annulus were connected by the relation 2S+.R = a constant. This assumption may seem an arbitrary one, but it can be shown by an algebraic investigation, which is too lengthy to be given here, that the total stored internal energy is less when 28 +R is constant throughout the spherical anuulus than when this sum varies according to any function of the distance x from the center. The ether of space transmits light, electricity, and gravitation from one body to another. The phenomena of gravitation are familiar to every one, but the explanation of its cause has not yet AKT. 183 STORED INTERNAL ENERGY 471 been discovered. All observations and theory indicate that the ether is an elastic substance which obeys the laws of the mathe- matical theory of elasticity. Accordingly it seems that the general conclusions of Art. 163 regarding the distribution of stresses in hollow spheres should apply to those stresses in the ether which cause the mutual gravitation of bodies of matter; if this be so, these stresses vary inversely as the cube of the dis- tance. The actual forces of gravitation, however, vary inversely as the square of the distance, and it is not easy to see how this law is deduced from that of the distribution of the stresses. To solve the great riddle of gravitation, a more definite knowledge is required regarding the constitution of matter, and the indi- cations are that an explanation may be obtained during the twentieth century. Prob. 183. Consult Isenkrahe's Das Rathsel von der Schwerkraft (Braunschweig, 1879) for critical reviews of the various attempts to explain the phenomena of gravitation. 472 APPENDIX AND TABLES APPENDIX AND TABLES ART. 184. VELOCITY OF STRESS WHEN an external force is suddenly applied to a body, the stresses produced are not instantaneously generated, but are propagated by a wave-like motion through the mass. Hence there is a velocity of transmission of stress which will be shown to depend upon the stiffness and density of the material. In fact, a sudden stress is propagated through a body in the same manner as sound is propagated through air or water. Let v be this velocity, w the weight of the material per cubic unit at a place where the acceleration of gravity is g, and E the modulus of elasticity of the material. It is required to find v in terms of w, g, and E. If F is a force which acting continuously for one second upon a body of the weight W produces the velocity u, and if the same body when falling vertically acquires under the action of gravity the velocity g in one second, then the constant forces are pro- portional to the accelerations that they produce; hence, F/W=u/g or Fg=Wu which is one of the well-known laws of mechanics. Now let a unit-stress 5 be applied to the end of a bar of section- area unity, producing the unit-elongation e upon the first element of its length. The elongation of the first element transmits the stress to the second element, and this in turn produces an elonga- tion of the second element, and so on. At the end of one second of time the length v is stressed, and the total elongation in that length will be eu. Thus in one second the center of gravity of the bar is moved the distance Jev, and its velocity u at the end of the second is ev. Now referring to the formula Fg = Wu, the value of F is S which is equal to eE, the value of W is wv, and hence, eE . g = iw . ev or v= (Eg/w)* ART. 184 VELOCITY OF STRESS 473 which is the formula for the velocity of wave propagation in elastic materials first deduced by Newton. Taking for g the mean value 32.16 feet per second per second, for iv the values in Table 1, for E the values in Table 2, and reducing E to pounds per square foot, since g and w are in terms of feet, the mean values of the velocity of transmission of stress in different materials are found to be, for Timber, v= 13 200 feet per second for Stone, v= 13 200 feet per second for Cast Iron, v= 12 400 feet per second for Wrought Iron, v= 15 500 feet per second for Steel, v= 17 200 feet per second For water confined in a pipe, the value of E is the same as the volumetric modulus G (Art. 182), and taking w as 62^ pounds per cubic foot, the velocity v is about 4 670 feet per second, which agrees well with experiments on the velocity of sound in water. In the mathematical theory of elasticity, the velocity of trans- mission of stress must be taken into account in order to obtain complete solutions of the problems of impact and suddenly applied forces. The above formula also gives the velocity of sound, light, and all wave propagations in elastic media. The ratio w/g is a constant for the same material at any point in space, and it expresses the density, while E is an index of the stiffness. At the surface of the earth the quantity E/w for steel is about 8 820 ooo, but for the ether it must be about 30 ico ooo ooo ooo ooo in order to account for the fact that the velocity of light is 984 ooo ooo feet per second. The stiffness of the ether is hence very great compared to its density; if its density be one one- thousandth of that of hydrogen, its stiffness is 37 times as great as that of steel. The opinion has long prevailed that the force of gravitation is instantaneously propagated through the ether, but the indications now are that its velocity is the same as that of light and electricity. Prob. 184. Verify the statements in the last paragraph. Consult Van Nostrand's Science Series, No. 85, and ascertain the values deduced by Wood for the density and stiffness of th^ether of space. 474 APPENDIX AND TABLES ART. 185. ELASTIC-ELECTRIC ANALOGIES In Art. 193 of Treatise on Hydraulics there are pointed out some of the analogies between hydraulic and electric phenomena. The theory of elasticity furnishes other analogies which are interesting and one of these is perhaps more perfect, from a formal point of view, than any that is furnished by hydraulics. Let a bar of length / and section area a be under the axial tension P, let E be the modulus of elasticity of the material and e the change of length due to the tension; then from Art. 10, this elon- gation is e = (l/aE)P. Let the reciprocal of E be called E f , then e = E'(l/a)P. This equation is the same as that given by Ohm's law for the loss in voltage when a current flows through a wire, if e represents the lost voltage and P the current; the quantity E'(l/a) is called electric resistance and it varies directly as the length of the wire, inversely as its section area, and directly as the specific electric resistance E'. The formal analogy is per- fect, but the fundamental ideas are quite different in the two cases. In the bar e is the loss in length which varies directly as /, inversely as a, and directly as the reciprocal of E, but the phenomenon is a static one entirely, for no energy is lost or transmitted through the bar. Referring now to Fig. llOa, let /i, 1%, Is be the lengths of the three columns under the compression P, and a l} a 2 , a 3 their section areas, while E\, E 2 , 3 are their moduluses of elasticity. Let the quantities l\la\E\, 1-2.1 a^E^^ fa/a^Es be called resistances and be designated by the letters ri, r 2 , r 3 . Then by formula (110) the total change of length of the compound column is given by e=(r\ + r 2 + rs)P. This may be called an arrangement 'in series,' and as in electric flow, the total resistance is the sum of the separate resistances. Fig. 1106 represents a column which may be said to be an arrangement ' in parallel,' and here the three lengths are equal to /, while the section areas are a\, a 2 , a s . Formula (110)" gives the change of length for this case as e = j p /(i/ri + i/r2 + 1/7*3), which agrees with the electric law governing the loss of voltage ART. 186 MISCELLANEOUS PROBLEMS 475 in a branched circuit, the total resistance being the reciprocal of the sum of the reciprocals of the separate resistances. The loads PI, P 2 , P 3 on the three parts of the column correspond to the currents in the three parts of the divided wire, while the change of length e corresponds to the drop in voltage which is the same for each of the three parts. The load P divides between the three parts inversely as their resistances to change of length, while the electric current divides between three branches in- versely as their electric resistances. With respect to work the analogy is less perfect. Resuming the equation e = (l/aE)P for the change in length of a bar under axial stress, the product of P and e is work, while this product is work per unit of time if e represents voltage and P represents current. This is because P represents force for the bar, while for the electric wire it represents electric charge per unit of time. For the bar the external work is spent in storing internal energy in the bar; for the wire the work Pe is lost in heat. The above analogy is hence merely a formal one and the fundamental ideas are quite different in the two cases. The phenomena of torsion afford another analogy and here energy may be transmitted through a rotating shaft, but no energy is lost if the material is not stressed beyond the shearing elastic limit. Here the angle of twist for a round shaft varies directly as the length of the shaft and also directly as the trans- mitted power, but it varies inversely as the square of the section area. Some theories of electricity and magnetism appear to indicate that forces of shearing and torsion in the ether may in krge part account for the observed phenomena. Reiff's Elas- ticitat und Elektricitat (Leipzig, 1893) contains a theory of electricity developed from the fundamental equations of the mathematical theory of elasticity. ART. 186. MISCELLANEOUS PROBLEMS Below are given a number of topics which have not been treated in the preceding pages, as their discussion properly be- longs to special works on special branches of applied mechanics. 476 APPENDIX AND TABLES Teachers who wish to give prize problems to their classes may .perhaps find some of these useful for that purpose. Prob. 186a. Discuss a screw with square threads at the end of a bolt, and find its length in order that its shearing strength may be equal to the tensile strength of the bolt. Prob. 1866. Let a helical spring consist of round wire, let r be the radius of the coil, d the diameter of the wire, and P the tensile and compressive load upon the spring. Show that the shearing unit-stress in the wire is i6Pr/x(fi. Prob. 186c. The data being the same as in the last problem show that the elongation or shortening of the spring under the load P is 64nPr 3 /Fd*, where n is the number of coils and F is the shearing modulus of elasticity. Prob. 186^. A load P is supported by three strings of equal size hung in the same vertical plane from the ceiling of a room. The mid- dle string is vertical and each of the others makes an angle 6 with it. If PI is the stress on the middle string and P% the stress on each of the others, show that, P! = P/(i + 2 cos 3 0) P 2 = P cos 2 0/(i + 2 cos 3 0) To solve this problem the condition must be introduced that the inter- nal work of all the stresses is a minimum. Prob. I86e. A load P is supported by three strings of equal size lying in the same plane. The middle string is vertical, one string makes with it the angle 6 on one side, and the second string makes with it the angle (j> on the other side. Find the stresses in the strings. Prob. 186/. A circular ring of mean diameter d is pulled in the direction of a diameter by two tensile forces each equal to P. Show that the maximum bending moment is at the section where P is ap- plied, and that its value is ^PJ 3 /^(J 2 + 4r 2 ), where r is the radius of gyration of the cross-section of the ring. Prob. 186g. An elliptical chain link has the mean length 4^1 and the mean width 2^1, where di is the diameter of the round section area. When an open chain link of these proportions is subject to the tension P in the direction of its length, the greatest bending moment occurs where the tension is applied, and the greatest unit-stress is S.iP/di 2 . For a chain link with a cross stud, the greatest unit-stress is ^.gP/d 2 . These results are correct only when the elastic limit of the material is not exceeded. ART. 187 ANSWERS TO PROBLEMS 477 ART. 187. ANSWERS TO PROBLEMS The following are the answers to some of the problems given in the preceding pages, the number of the problem being in parenthesis. The instructions in Art. 8 should be carefully followed by the student, and in no event should he refer to an answer until the solution of the problem is completed. How- ever satisfactory it may be for a student to be able to know that his solutions give the correct results, it is well for him to keep in mind that in actual engineering work the solutions of problems will never be given. (16) 4.16 inches. (26) 3.33 inches. (3a) 55 400 pounds per square inch. (56) i 780 feet, (la) 2.71 and 3.33 inches. (86) 3.57 centi- meters. (9o) 26300000 pounds per square inch. (146) 122 foot- pounds. (15) i 7 09". (17) 26.2 square inches. (18a) 183.5 kilo- grams. (21) $1058.84. (23) 3 140 pounds. (25a) 4. 04 cents. (26a) 105 ooo pounds. (290) / > i/P 2 =3/2. (30 2 400 o.ooo 0055 0.0000099 Stone 160 2 560 o.ooo 0050 o.ooo 0090 Timber 40 600 0.000 0020 o.ooo 0036 Cast Iron 450 7 200 o.ooo 0062 O.OOO OII2 Wrought Iron 480 7700 o.ooo 0067 O.OOO OI2I Structural Steel 490 7800* o.ooo 0065 o.ooo 0117 Strong Steel 491 7800 o.ooo 0065 o.ooo 0117 Explanation in Arts. 17 and 100 TABLE 2. AVERAGE ELASTIC PROPERTIES Elastic Limit Modulus of Elasticity Material Pounds per square inch Kilograms per square centimeter Pounds per square inch Kilograms per square centimeter Brick I OOO 7 2 OOO OOO 140 ooo Concrete 800 56 2 5OO 000 175 ooo Stone 2 OOO 140 6 ooo ooo 420 ooo Timber 3000 210 i 500 ooo 105 ooo Cast Iron ( 6000 420 15 ooo ooo i 050 ooo I 2O OOO I 4OO 15 ooo ooo I 050 ooo Wrought Iron 25 ooc i 75 25 ooo ooo i 750 ooo Structural Steel 35000 245 30 ooo oco 2 100 000 Strong Steel 50 ooo 35 30 ooo ooo 2 IOO OOO Explanation in Arts. 2 and 9. Values for Brick, Concrete, and Stone are for compression only. For Cast Iron the upper values apply to tension and the lower ones to compression. For other materials the values apply to both tension and compression. TABLES 481 TABLE 3. AVERAGE TENSILE AND COMPRESSIVE STRENGTH Ultimate Tensile Strength Ultimate Compressive Strength Material Pounds per square inch Kilograms per square centimeter Pounds per square inch Kilograms per square centimeter Brick 3000 210 Concrete 300 21 3000 210 Stone 6000 420 Timber IO OOO 700 8000 S6o Cast Iron 20000 I 400 90000 6300 Wrought Iron 50 ooo 35 50000 35 Structural Steel 60000 4 200 60 ooo 4 200 Strong Steel 100 000 7 ooo 1 20 000 8400 Explanation in Arts. 4,5, 19-25 TABLE 4. AVERAGE SHEARING AND FLEXURAL STRENGTH Ultimate Shearing Strength Ultimate Flexural Fiber Strength Material Pounds per square inch Kilograms per square centimeter Pounds per square inch Kilograms per square certimeter Brick 700 5 800 55 Concrete I 200 84 Stone I 500 i5 2 000 140 Timber, with grain 500 35 Timber, across grain 3000 210 9000 630 Cast Iron 20000 I 400 35000 2400 Wrought Iron 40 ooo 2 800 Structural Steel 50000 35 Strong Steel 75000 5 200 no ooo 7700 Explanation in Arts. 6 19-25, 52 482 TABLES TABLE 5. WORKING UNIT-STRESSES FOR BUILDINGS ABSTRACTED FROM THE BUILDING CODE OF THE CITY OF NEW YORK, N. Y. Material Pounds per Square Inch Tension Com- pression Shear Flexure Hemlock 600 500 275* 600 Spruce 800 800 320* 800 White Pine 800 800 250* 800 Yellow Pine I 200 I 000 500* I 200 Oak I OOO 900 600* I 000 Brick 300 5 Brickwork In Portland Cement Mortar 250 3 In Natural Cement Mortar 208 3 In Lime Mortar III Concrete Portland Cement, i, 2, 4 230 3 Natural Cement, i, 2, 4 125 16 Rubble Stonework In Portland Cement Mortar 140 In Natural Cement Mortar III In Lime Mortar 70 Sandstone I OOO IOO Limestone I 500 15 Marble 900 120 .Granite I 700 180 Slate I 000 400 Cast Iron 3000 1 6 ooo 3 ooo{ 3 ooo Ten. 16000 Comp. Wrought Iron 12 000 12 000 6000 12 000 Wrought-iron shop rivets 7 S Wrought-iron field rivets 6000 Rolled Steel 16000 16000 9 ooo 16000 Steel shop rivets 10 000 Steel field rivets 8000 Cast Steel 16000 16000 Explanation in Art. 7. * Across grain. TABLES 483 TABLE 6. STEEL I-BEAM SECTIONS Depth of Beam Inches Weight I& r t Pounds Width of Flange Inches Section Area a Sq. In. Axis perpendicular to web Axis parallel to web Moment Inertia / Inches 4 Section Factor 1/c Inches* Radius Gyration Inches Moment Inertia / Inches* Radius Gyration Inches 24 100 7-25 29.41 2380 198.4 9.00 48.56 1.28 *24 80 7.00 23-32 2088 174.0 9.46 42.86 1-36 20 IOO 7.28 29.41 1656 165.6 7-50 52-65 1-34 *20 80 7.00 23-73 1467 146.7 7 .86 45 -81 i-39 *20 65 6.2 5 19.08 1170 117.0 7.83 27.86 I. 21 18 70 6.26 20-59 921.3 102.4 6.69 24.62 1.09 *i8 55 6.00 15-93 795-6 88.4 7.07 21.19 i-iS 15 IOO 6-77 29.41 900.5 1 20. I 5-53 50.98 1-31 *i5 80 6.40 23-81 795-5 106.1 5-78 41.76 1.32 *IS 60 6.00 17.67 609.0 81.2 5-8? 25.96 I. 21 is 42 5-5 12.48 441-7 58-9 5-95 14.62 1. 08 12 55 5.61 16.18 321.0 53-5 4-45 17.46 1.04 *I2 40 5-25 11.84 268.9 44.8 4-77 13.81 1. 08 *I2 3i* 5.00 9.26 215.8 36.0 4-83 9-50 1. 01 10 40 5.10 11.76 158-7 3i-7 3-67 9-50 0.90 *IO 25 4.66 7-37 122. I 24.4 4.07 6.89 0-97 9 35 4-77 10.29 III. 8 24.8 3-29 7-3i 0.8 4 *9 21 4-33 6.31 84.9 18.9 3-67 5.16 0.90 8 25* 4-27 7-50 68.4 17.1 3-02 4-75 0.80 * 8 18 4.00 5-33 56-9 14.2 3-27 3-78 0.84 7 20 3-87 5-88 42.2 12. I 2.68 3-24 0-74 * 7 15 3-66 4.42 36.2 10.4 2.86 2.67 0.78 6 17* 3-57 5-7 26.2 8. 7 2.27 2.36 0.68 * 6 12} 3-33 3-6i 21.8 7-3 2.46 1.85 0.72 5 Ml 3-29 4-34 15-2 6.1 1.87 1.70 0.63 * 5 9i 3.00 2.87 12. I 4.8 2.05 1-23 0.65 4 10) 2.88 3-9 7-i 3-6 1-52 1 .01 0-57 * 4 7i 2.66 2.21 6.0 3- 1.64 0.77 0-59 3 7* 2.52 2.21 2-9 1.9 i-iS 0.66 0.52 * 3 5i 2-33 1.63 2-5 i-7 1-23 0.46 0-53 Explanation in Arts. 44 and 51. * Standard sizes: others are special. 484 TABLES TABLE 7. STEEL BULB-BEAM SECTIONS Ax is perpendicular to web Axis parallel to web Weight , Depth Foot Section Area Moment Base to Section Radius Moment Radius Inertia Neutral Factor Gyration Inertia Gyration / Axis 1/6 f / f Inches Pounds Sq. In. Inches 4 Inches Inches* Inches Inches 4 Inches ix| 32.2 9-51 179-3 5-7 27.9 4-34 6.36 0.82 10 28.0 8.20 118.5 4.28 20.7 3.80 6.08 0.86 9 25.0 7-35 85.0 3-9 I6. 7 3-40 4-85 0.8l 8 21.0 6.17 57-8 3-48 12.8 3.06 3o8 o. 76 7 18.0 5-32 37-o 3-4 9-3 2.64 2.56 0.69 6 14-5 4.27 21.8 2.61 6.4 2.26 1.62 0.62 5 "5 3-39 12.0 2.22 4-3 1.88 1. 01 o-55 Explanation in Art. 44 TABLE 8. STEEL T SECTIONS Axis perpendicular to web Axis parallel to web Size Weight Section Width by Depth Foot Area a Moment Inertia Base to Neutral Section Factor Radius Gyration Momsnt Inertia Radius Gyration / Axis I'c r / Y Inches Pounds Sq. In. Inches 4 Inches Inches' Inches Inches 4 Inches 6X 4 17.4 5.12 6. 5 6 1. 00 2.19 I-I3 9-33 i-35 5X4 15-3 4-54 6.16 1. 08 2. II I.I7 5-41 1.09 4X4 10.9 3.10 4.70 I-I5 1.6 4 1.23 2 .20 0.85 4X 3 9.0 2.67 1.99 0.78 0.00 0.87 2.10 0.89 3*X 3 * 7.0 2.08 2.27 0.94 0.89 1.04 I-3 0.94 3*X 3 7.0 2. II I.6 S 0.80 o-75 0.88 1.18 o-75 3X3 6-5 I.9I i-57 0.87 0.74 0.91 o-75 0.62 3X2* 5-o I .46 0.78 0.66 0.42 0-73 0.60 0.64 2*X 3 6.0 1.76 1.48 o-93 0.71 0.92 0.44 o. 5<5 2*X2} 5-8 I.7I o-95 0.76 o-55 o-75 0.48 o-53 2X2 3-5 1.0 3 o-37 0.60 0.26 0.60 0.18 0.41 2X1$ 3- 0.91 0.16 0-45 0.15 0.42 0.17 0.44 2X1 2-5 0.72 0.05 0.27 o .07 o. 26 0.17 0.49 Explanation in Art. 44 TABLES 485 TABLE 9. STEEL CHANNEL SECTIONS Depth Inches Weight per Foot Pounds Width of Flange Inches Section Area a Sq. In. Axis perpendicular to web Axis parallel to web Moment Inertia / Inches' Radius Gyration r Inches Moment Inertia / IncheV Radius Jyration Inches Outside f Web to Center of Gravity Inches 15 55 3-82 16.18 430.2 5.16 I3-I9 0.87 0.82 15 45 3.62 I3-24 375-1 5-32 10.29 0.88 0.79 15 35 3-43 10.29 320.0 5-58 8.48 0.91 0.79 *i S 33 3-40 9.90 312.6 5-62 8.23 0.91 0.79 12 40 3-42 11.76 197.0 4.09 6.63 0-75 0.72 12 3 3-17 8.82 161.7 2.28 5-2i 0.77 0.68 12 25 3-5 7-35 144.0 4-43 4-53 0.78 0.68 *I2 20^ 2.94 6.03 128.1 4.61 3-9i 0.80 0.68 10 35 3-i8 10.29 "5-5 3-35 4.66 0.67 0.70 10 25 2.89 7-35 91.0 3-52 3-40 0.68 0.62 IO 20 2.74 5-88 78-7 3-66 2.85 0.70 0.61 *IO 15 2.60 4-46 66.9 3-87 2.30 0.72 0.64 9 25 2.81 7-35 70.7 3.10 2.98 0.64 0.61 9 15 2-49 4.41 5-9 3-40 i-95 0.66 0-59 * 9 i3i 2-43 3-89 47-3 3-49 1.77 0.67 0.60 8 2li 2.62 6.25 47-8 2-77 2.25 0.60 0.58 8 r6J 2.44 4.78 39-9 2.89 1.78 0.61 0.56 * 8 i 2.26 3-35 32-3 3-" i-33 0.63 0.58 7 i9f 2-5 1 5-8i 33-2 2-39 1.85 0.56 0.58 7 Mi 2.30 4-34 27.2 2.50 1.40 0-57 0.54 * 7 9i 2.09 2.85 21. I 2.72 0.98 o-59 0-55 6 is* 2.28 4-56 19-5 2.07 1.28 o-53 0-55 6 10* 2.04 3-9 I5-I 2.21 0.88 0-53 0.50 * 6 8 1.92 2.38 I 3 .0 2-34 0.70 0-54 0.49 5 II) 2.04 3.38 10-4 i-75 0.82 0.49 0-51 * 5 6| i-75 i-95 7-4 i-95 0.48 0.50 0.49 4 71 1.72 2-13 4.6 1.46 0.44 0-45 0.46 * 4 Si 1.58 i-55 3-8 1.56 0.32 0-45 0.46 3 6 i. 60 1.76 2.1 i. 08 0.31 0.42 0.46 * 3 4 1.41 1.19 1.6 1.17 0.20 0.41 0.44 Explanation in Arts. 44 and 76. * Standard sizes; others are special. 486 TABLES TABLE 10. STEEL ANGLE SECTIONS Size of Angle Inches Weight o r t Pounds Section Area Sq. In. Axis parallel to long leg Axis parallel to short leg Axis to back of leg Inches Moment Inertia Inches 4 Radius Gyra- tion Inches Axis to back of leg Inches Moment Inertia Inches 4 Radius Gyration Inches 6X4X1 30-6 9.00 I.I? i-75 1.09 2.17 30.75 1.8 5 6X4Xf 23-6 6.94 1. 08 8.68 I. 12 2.08 24-51 1.88 6X4Xi 16.2 4-75 0.99 6.27 I.I5 1.99 17.40 1.91 5X4X1 24.2 7.11 I. 21 9-23 1. 14 I-7I 16.42 i-52 5X4X| 21. I 6.19 1.16 8.23 I-I5 1.66 14.60 i-54 5X4Xi 14-5 4-25 1.07 5.96 1.18 1-57 10.46 i-57 5X3X1 19.9 5-84 0.86 3.71 0.80 1.86 13.98 i. 55 5X3Xi 12.8 3-75 0-75 2.58 0.83 i-75 9-45 i-59 4X3Xf 16.0 4.69 0.92 3.28 ex 84 1.42 6-93 1.22 4X3Xi n. i 3-25 0.83 2.42 0.86 1-33 5-05 1-25 3iX3Xi IO. 2 3.00 0.88 2-33 0.88 i-i3 3-45 1.07 3iX2jXi 9-4 2-75 0.70 1.36 0.70 1.20 3-24 1.0 9 3X2*Xi 8-5 2-5 0-75 1.30 0.72 1. 00 2.08 0.91 3X2Xi 7-7 2.25 0.58 0.67 o-55 1. 08 1.92 0.92 Equal legs Axis parallel to leg Least r 6X6X1 37-4 11.00 1.86 35-46 i. 80 1.16 6X6Xf 28.7 8.44 1.78 28.15 i-8 3 I.I7 6X6Xi 19.6 5-75 1.68 19.91 1.86 |||| 1.18 5X5X1 30.6 9.00 1.61 19.64 1.48 | rrjj 0.96 5X5X3 23,6 6.94 1-52 15-74 1.51 M-^SO 0.97 SXsXi 16.2 4-75 i-43 11.25 i-54 !s= 3< o i 0.98 4X 4 XJ 18.5 5-84 1.29 8.14 1.18 111! 0.77 4X4XJ 12.8 3-75 1.18 5-56 1.22 l8S$ 0.78 3lX 3 iXi II. I 3-25 i. 06 3-64 1. 06 !!!! 0.68 3X3Xi 9-4 2-75 o-93 2.22 0.90 55.S 0.58 2^X2$X 7-7 2.25 0.81 1.23 0.74 0.47 2X2Xi 3-2 0.94 o-59 0-35 0.61 o-39 Explanation in Art. 44 TABLES 487 TABLE 11. STEEL Z SECTIONS Size Weight *& Thick- ness Section Area Moment Inertia 'a Moment Inertia lb Tangent of Angle between Least Radius Gyration Inches Inches Sq. In. Inches' Inches 4 1 2 and l^ Inches 8X 3 16.9 4-97 44.64 S-6o 0.27 0.72 7iX 3 16.3 4.78 38.19 5-59 0.29 0.72 6X 3 i 29-3 8.63 42.12 15-44 0.52 0.8l 5X3t 23-7 i : 6.96 23.68 "37 0.62 0-73 4X3A i8. 9 5-55 12. II 8-73 0.81 0-65 3X2^ 12.5 3-69 4-59 4-85 0.965 o-53 Explanation in Arts. 44 and TABLE 12. COMPARISON OF BEAMS Beams of Uniform Cross-section Maximum Moment Maximum Deflection Relative Strength Relative Stiffness Cantilever, single load at end Wl I Wl 3 I I Cantilever, uniform load iwt I Wl 3 8 ~Rl 2 2} Simple beam, load at middle \Wl i Wl 3 4 16 Simple beam uniformly loaded : 5 Wl 3 384 ~EI 8 Beam fixed at one end, supported at other, load near middle 0.192^7 Wl 3 0.0098 5-2 3-7 Beam fixed at one end, supported at other, uniform load tfR +*% 8 62 Beam fixed at both ends, load at middle * i Wl 3 192 El 8 64 Beam fixed at both ends, uniform load i Wl 3 384 El 12 128 Explanation in Arts. 56 and 63 488 TABLES TABLE 13. GERMAN I BEAMS Depth of Beam cm. Weight kilos Width of Flange cm. Section Area a sq. cm. Axis perpendicular to web Axis parallel to web Moment Inertia cm. 4 Section Factor I/e cm. Radius Gyration r cm. Moment Inertia cm. 4 Radius Gyration r cm. 50 159-37 19.0 204.32 78040 3 122 J 9-54 3060 3-87 45 129.11 17.1 165.52 51 230 2 277 17-59 2 005 3-48 40 103.54 15-6 132.74 32680 1634 15.69 1358 3-20 35 80.78 14.1 103-57 19 690 I 125 13-79 871.0 2.90 32 68.63 I 3 .2 87.99 13970 873-3 12.60 651.0 2.72 30 60.82 12.6 77-97 II OOO 733-i 11.88 539-3 2.63 28 61 . 50 15.0 78.85 10 280 733-9 11.41 827.7 3-24 28 53-55 12.0 68.65 8523 608.8 11.14 439-4 2-53 26 46.89 II.4 60. ii 6413 493-3 10 -33 343-3 2-39 25 43-66 II. I 55-97 5554 444-3 9.96 306.6 2-34 24 46.64 13-5 59.80 5 776 481.3 9-83 516.9 2.94 24 40.54 10.8 Si-97 4784 399-7 9-59 272.5 2.29 23 37-56 10.5 48.15 4097 356.3 9.22 241.6 2.24 22 41.42 13-5 53.10 4347 395-2 9-05 459-0 2.94 22 34-73 IO. 2 44-52 3434 312.2 8.78 205.8 2-15 21 31.96 9-9 40.98 2898 276.0 8.40 180.7 2.10 20 29.29 9.6 37-55 2 428 242.8 8.04 J 54-7 2.03 18 32.06 13-5 41.10 2363 262.8 7.58 369-9 3-oo 18 24-34 9.0 31.20 i 661 184.6 7-30 119.8 1.96 16 19.83 8.4 25-43 i 067 133-4 6.48 83-3 1.81 15 17.60 8.0 22-57 840.0 112. 6.10 68.3 1.74 14 16.02 7-6 20.54 659-4 94-2 5-67 55-9 1.65 13 14.56 7-2 18.67 523-9 80.6 5-29 47-8 i. 60 12 12.69 6.8 16.27 392-4 65.4 4.91 37-6 I-5 2 10 9.69 6.0 12.42 208.0 41.6 4.08 22.6 1.35 8 7.07 5-2 9.07 97-2 24-3 3-i7 12.8 1.19 Explanation in Art. 44 TABLES TABLE 14. WEIGHT OF WROUGHT-!RON BARS Side or Diam- Pounds p Fo r Linear ot Side or Diam- Founds p Fo er Linear ot. Side or Diam- Pounds p Fe er Linear ot eter Inches Square Bars Round Bars eter Inches Square Bars Round ; Bars eter Inches Square Bars Round Bars 2 13-33 10-47 5 83-33 65.45 TV 0.013 O.OIO iV I4.I8 II. 14 i 87.55 68.76 i 0.052 0.041 1 15.05 11.82 i 91.88 72.16 & O.II7 0.092 TV 15.95 12-53", 96.30 75.64 i O.2O8 0.164 i 16.88 13-25 100.8 79-19 A 0.326 0.256 TV 17.83 14.00 105.5 82.83 I 0.469 0.368 1 18.80 14-77 IIO. 2 86.56 A 0.638 0.501 TV 19.80 15-55 "5.1 90.36 } 0.833 0.654 i 20.83 16.36 6 120.0 94.25 TV 1.055 0.828 & 21.89 17.19 I25.I 98.22 i 1.302 1.023 i 22.97 18.04 130.2 102.3 H 1-576 1.237 H 24.08 18.91 135-5 106.4 1 1.875 1-473 1 25.21 19.80 140.8 110. 6 it 2.2OI 1.728 if 26.37 20.71 146.3 114.9 1 2.552 2.004 i 27-55 21.64 i 151.9 119.3 if 2.930 2.301 if 28.76 22.59 1 157-6 123-7 i 3-333 2.618 3 30.00 23.56 : 166.3 128.3 TV 3-763 2-955 i 32.55 25-57 i 175-2 137.6 i 4.219 3-3I3 i 35-21 27.65 i . 187.5 M7.3 TV 4.701 3.692 1 37-97 29.82 * 20O.2 157-2 i 5.208 4.091 i 40.83 32.07 8 213.3 167.6 TV 5.742 4.510 1 43-80 34-40 i 226.9 178.2 1 6.302 4.950 i 46.88 36.82 i 240.8 189.8 TV 6.888 5.410 1 50.05 39-31 i 255.2 200.4 i 7.500 5.890 4 53-33 41.89 9 27O.O 212. 1 TV 8.138 6.392 i 56.72 44-55 i 285.2 224.0 i 8.802 6.913 i 60.21 47.29 i 300.8 236.3 9.492 7-455 1 63.80 50.11 i 316.9 248.9 i 10.21 8.018 i 67.50 53.01 10 333-3 261.8 if 10.95 8.601 i 7I-30 56.00 i 350.2 275-1 i 11.72 9.204 i 75.21 59-07 i 367.5 288.6 i! 13.51 9.828 1 79.22 62.22 I 385-2 302.5 Explanation in Art. 188 490 TABLES TABLE 15. SQUARES OF NUMBERS n. 01234 56789 Diff. I.O i.ooo 1.020 1.040 1.061 1.082 1.103 i-i 2 4 I-I45 1.166 i.iSS 22 i.i 1.2 1.210 1.232 1-254 l - 2 77 i-3o 1.440 1.464 1.488 1.513 1.538 1.323 1.346 1.369 1.392 1.416 1.563 1.588 1.613 1.638 1.664 24 26 J -3 1.690 1.716 1.742 1.769 1.796 1.823 1.850 1.877 1.904 1.932 28 1.4 1.960 1.988 2.016 2.045 2.074 2.103 2.132 2.l6l 2.I9O 2.22O 30 ij 2.250 2.280 2.310 2.341 2.372 2.560- 2.592 2.624 2.657 2.690 2.403 2.434 2.465 2.496 2.528 2.723 2.756 2.789 2.822 2.856 32 34 i-7 2.890 2.924 2.958 2.993 3.028 3.063 3.098 3.133 3.168 3.204 36 1.8 3.240 3.276 3.312 3.349 3.386 3.423 3.460 3.497 3-534 3-572 3 8 1.9 3.610 3.648 3.686 3.725 3.764 3.803 3.842 3.881 3.920 3.960 40 2.O 4.000 4.040 4.080 4.121 4.162 4.203 4.244 4.285 4.326 4.368 42 2.1 2.2 4.410 4.452 4.494 4.537 4.580 4.840 4.884 4.928 4.973 5.018 4.623 4.666 4.709.4.752 4.796 5.063 5.108 5.153 5.198 5.244 44 46 2 -3 5.290 5.336 5.382 5.429 5.476 5-523 5-57 5-6i7 5664 5-712 48 2-4 5.760 5.808 5.856 5.905 5.954 6.003 6.052 6.101 0.150 6.200 50 a 2-7 6.250 6.300 6.350 6.401 6.452 6.760 6.812 6.864 6.917 6.970 7.290 7.344 7.398 7.453 7.508 6.503 6.554 6.605 6.656 6.708 7.023 7.076 7.129 7.182 7.236 7.563 7.618 7.673 7.728 7.784 52 3 2.8 7.840 7.896 7.952 8.009 8.066 8.123 8.180 8.237 8.294 8.352 5 8 2.9 8.410 8.468 8.526 8.585 8.644 8.703 8.762 8.821 8.880 8.940 60 3-o 9.000 9.060 9.120 9.181 9.242 9-33 9-364 9-425 9-4 8 6 9-54 8 62 3- 1 9.610 9.672 9.734 9.797 9.860 9.923 9.986 10.05 I0 - ir Iai8 6 3- 2 10.24 10.30 10.37 10.43 IO -5 10.56 10.63 10.69 10.76 10.82 7 3-3 10.89 10.96 11.02 11.09 u-i6 11.22 11.29 n-3o 1142 11.49 7 3-4 11.56 11.63 "-70 11.76 11.83 11.90 11.97 12.04 I2 -n I2 -'8 7 3-5 12.25 12.32 12.39 12.46 12.53 12.60 12.67 12.74 12.82 12.89 7 3-6 12.96 13.03 13.10 13.18 13.25 13.32 13.40 13.47 13.54 13.62 7 3-7 13.69 13.76 13.84 13.91 13.99 14.06 14.14 14.21 14.29 14.36 8 3-8 14.44 M-S 2 14-59 14-07 *4-75 14.82 14.90 14.98 15.05 15.13 8 3-9 15.21 15.29 15.37 15.44 15.52 15.60 15.68 15.76 15.84 15.92 8 4.0 16.00 16.08 16.16 16.24 16.32 16.40 16.48 16.56 16.65 J 6-73 8 4.1 16.81 16.89 "6-97 17.06 17.14 17.22 17.31 17.39 17.47 !7-56 8 4.2 17.64 17.72 17.81 17.89 17.98 18.06 18.15 18.23 18.32 18.40 9 4-3 18.49 l8 -5 8 l8 -66 18.75 l8 - 8 4 18.92 19.01 19.10 19.18 19.27 9 . 4-4 19.36 19.45 19.54 19.62 19.71 19.80 19.89 19.98 20.07 20.16 9 4-5 20.25 20.34 20.43 20.52 20.61 20.70 20.79 20.88 20.98 21.07 9 4.6 21. 16 21.25 21.34 21.44 21.53 21.62 21.72 2I.8l 2I.OX) 22.00 9 4-7 22.09 22.18 22.28 22.37 22.47 22.56 22.66 22.75 22.85 22.94 10 4.8 23.04 23.14 23.23 23.33 23.43 23.52 23.62 23.72 23.81 23.91 10 4-9 24.01 24.11 24.21 24.30 24.40 24.50 24.60 24.70 24.80 24.0X) 10 5-o 5- 1 25.00 25.10 25.20 25.30 25.40 26.01 26.11 26.21 26.32 26.42 25.50 25.60 25.70 25.81 25.91 26.52 26.63 2 6-73 26.83 26.94 10 10 5-2 27.04 27.14 27.25 27.35 27-46 27.56 27.67 27.77 27.88 27.98 II 5-3 28.09 28.20 28.30 28.41 28.52 28.62 28.73 28.84 28.94 29.05 II 5-4 29.16 29.27 29.38 29.48 29.59 29.70 29.81 29.92 30.03 30.14 II It, 01234 56789 Diff. TABLES 491 TABLE 15. SQUARES OF NUMBERS ft. 01234 56789 Diff. 2 30-25 30.36 30.47 30.58 30.69 31.36 31.47 31.58 31.70 31.81 30.80 30.91 31.02 31.14 31.25 31.92 32.04 32.15 32.26 32.38 n ii zi 32.49 32.60 32.72 32.83 32.95 33.06 33.18 33.29 33.41 33.52 12 5-8 5-9 33-64 33-76 33.87 33.99 34.11 34.81 34.93 35.05 35.16 35.28 34.22 34.34 34.46 34.57 34.69 35.40 35.52 35.64 35.76 35.88 12 12 6.0 36.00 36.12 36.24 36.36 36.48 36.60 36.72 36.84 36.97 37.09 12 6.1 6.2 37-21 37.33 37.45 37.58 37.70 38.44 38.56 38.69 38.81 38.94 37.82 37.95 38.07 38.19 38.32 39.06 39.19 39.31 39.44 39.56 12 1 3 |3 39.69 39.82 39.94 40.07 40.20 40.32 40.45 40.58 40.70 40.83 13 6.4 40.96 41.09 41.22 41.34 41.47 41.60 41.73 41.86 41.99 42.12 13 a 42.25 42.38 42.51 42.64 42.77 43-56 '43-69 43-82 43-96 44-09 42.90 43.03 43.16 43.30 43.43 44.22 44.36 44.49 44.62 44.76 3 '3 6.7 6.8 6.9 44.89 45.02 45.16 45.29 45.43 46.24 46.38 46.51 46.65 46.79 47.61 47.75 47-89 48-02 48.16 45-56 45-70 45.83 45.97 46.10 46.92 47.06 47.20 47.33 47.47 48.30 48.44 48.58 48.72 48.86 H '4 14 70 49.00 49.14 49.28 49.42 49.56 49.70 49.84 49.98 50.13 50.27 H 7-i 50.41 50.55 50.69 50.84 50.98 51.12 51.27 51.41 51.55 51.70 H 7.2 51.84 51.98 52.13 52.27 52.42 52.56 52.71 52.85 53.00 53.14 15 7-3 53-29 53-44 53-58 53.73 53.88 54.02 54.17 54.32 54.46 54.61 15 7-4 54.76 54.91 55.06 55.20 55.35 55.50 55.65 55.80 55.95 56.10 15 2 56.25 56.40 56.55 56.70 56.85 5776 57-91 58-06 58.22 58.37 57.00 57.15 57.30 57.46 57.61 58.52 58.68 58.83 58.98 59.14 ' '5 15 7.7 59.29 59.44 59.60 59.75 59.91 60.06 60.22 60.37 60.53 60.68 16 7-8 60.84 61.00 61.15 61.31 61.47 61.62 61.78 61.94 62.09 62.25 16 7-9 62.41 62.57 62.73 62.88 63.04 63.20 63.36 63.52 63.68 63.84 16 8.0 64.00 64.16 64.32 64.48 64.64 64.80 64.96 65.12 65.29 65.43 16 8.1 65.61 65.77 65.93 66.10 66.26 66.42 66.59 66.75 66.91 67.08 16 8.2 67.24 67.40 67.57 67.73 67.90 68.06 68.23 68.39 68.56 68.72 '7 f' 3 68.89 69.06 69.22 69.39 69.56 69.72 69.89 70.06 70.22 70.39 '7 8.4 70.56 70.73 70.90 71.06 71.23 71.40 71.57 71.74 71.91 72.08 17 |-S 72.25 72.42 72.59 72.76 72.93 73.10 73.27 73.44 73.62 73.79 17 8.6 73-96 74-13 74-3 7448 74.65 74.82 75.00 75.17 75.34 75.52 17 8-7 75.69 75.86 76.04 76.21 76.39 76.56 76.74 76.91 77.09 77.26 18 8.8 77.44 77.62 77.79 77.97 78.15 78.32 78.50 78.68 78.85 79.03 18- 8.9 79.21 79.39 79.57 79.74 79.92 80. 10 80.28 80.46 80.64 80.82 18 9.0 81.00 81.18 81.36 81.54 81.72 81.90 82.08 82.26 82.45 82.63 18 9.1 9.2 82.81 82.99 83.17 83.36 83.54 84.64 84.82 85.01 85.19 85.38 83.72 83.91 84.09 84.27 84.46 85-56 85.75 85.93 86.12 86.30 18 i9 9-3 86.49 86.68 86.86 87.05 87.24 87.42 87.61 87.80 87.98 88.17 '9 9-4 88.36 88.55 88.74 88.92 89.11 89.30 89.49 89.68 89.87 90.06 '9 9-5 90.25 90.44 90.63 90.82 91.01 91.20 91.39 91.58 91.78 91.97 9 9.6 9-7 92.16 92.3 k 92.54 92.74 92.93 94.09 94.28 94.48 94.67 94.87 93.12 93.32 93.51 93.70 93.90 95.06 95.26 95.45 95.65 95.84 9 20 9.8 96.04 96.24 96.43 96.63 96.83 97.02 97.22 97.42 97.61 97.81 2O 9.9 98.01 98.21 98.41 98.60 98.80 99.00 99.20 99.40 99.60 99.80 2O it. 01234 56789 Diff. 492 TABLES TABLE 16. AREAS OF CIRCLES d 01234 56789 Diff. .0 .7854 .8012 .8171 .8332 .8495 .8659 .8825 .8992 .9161 .9331 .1 .9503 .9677 .9852 1.003 I-02I 1.039 I.57 1-075 1.094 1. 112 .2 I.I3I I.I50 1.169 LI 58 I.20S 1.227 1.247 1.267 1.287 1.307 19 3 1.327 1.348 1.368 1.389 I.4IO 1.431 1.453 1-474 1-496 I.5I7 21 4 1.539 I.56I 1.584 1. 606 1.629 1.651 1.674 1-697 1-720 1.744 22 5 1.767 1.791 1.815 1.839 T -863 1.887 1.911 1.936 1.961 1.986 24 .6 2.011 2.036 2.06I 2.0S7 2.112 2.138 2.164 2.190 2.217 2.2^3 26 7 2.270 2.297 2.324 2.351 2.378 2.405 2.433 2.461 2.488 2.516 27 .3 2-545 2.573 2.602 2.630 2.659 2.633 2.717 2.746 2.776 2.806 29 9 2.835 2.865 2.895 2.926 2.956 2.986 3.017 3.048 3.079 3.110 30 2.O 3.142 3.173 3.205 3.237 3.269 3-301 3.333 3.365 3.398 3.431 32 2.1 3.464 3.497 3.530 3.563 3.597 3.631 3.664 3.698 3.733 3.767 34 2.2 3.801 3.836 3.871 3.906 3.941 3.976 4.012 4.047 4.083 4.119 35 2-3 4.155 4.191 4.227 4.264 4.301 4-337 4-374 4-4 J 2 4-449 4-486 36 2.4 4.524 4.562 4.600 4.638 4.676 4.714 4.753 4.792 4.831 4.870 38 2-5 4.909 4.948 4.988 5.027 5.067 5.107 5.147 5.187 5-228 5.269 40 2.6 5.309 5.350 5.391 5.433 5.474 5-5I5 5-557 5-599 5-64I 5-683 4i 2.7 5.726 5.768 5.811 5.853 5.896 5.940 5.983 6.026 6.070 6.114 43 2.8 6.158 6.202 6.246 6.290 6.335 6.379 6.424 6.469 6.514 6.560 44 2.9 6.605 6.651 6.697 6.743 6.789 6.835 6.881 6.928 6.975 7.022 46 3-o 7.069 7.116 7.163 7.211 7.258 7.306 7.354 7.402 7.451 7.499 48 3- r 7.548 7.596 7.645 7.694 7.744 7.793 7.843 7.892 7.942 7-992 49 3-2 8.042 8.093 8.143 8.194 8.245 8.296 8.347 8.398 8.450 8.501 51 3-3 8-553 8.605 8.657 8.709 8.762 8.814 8.867 8.920 8.973 9.026 52 3-4 9.079 9.133 9.186 9.240 9.294 9.348 9.402 9.457 9.511 9.566 54 3-5 9.621 9.676 9.731 9.787 9.842 9.898 9-954 IO.OI 10.07 10.12 56 3-6 10.18 10.24 10.29 10.35 10.41 10.46 10.52 10.58 10.64 10.69 6 3-7 10.75 10.81 10.87 10 -93 10.99 II.O4 II. IO II. l6 11.22 11.23 6 3-8 11.34 11.40 11.46 11.52 11.58 11.64 11.70 11.76 11.82 11.88 6 3-9 11.95 12.01 12.07 12.13 12.19 12.25 12.32 12.38 12.44 12.50 6 4.0 12.57 12.63 12.69 12.76 12.82 12.88 12.95 13.01 13.07 13.14 7 4.1 13.20 13.27 13.33 13-40 13-46 13-53 13-59 13-66 13.72 13-79 7 4.2 13.85 13.92 13.99 I4-05 14-12 14.19 14.25 14.32 14.39 14-45 7 4-3 14.52 14.59 14-66 14.73 14-79 14.86 14.93 15.00 15.07 15.14 7 4-4 15.21 15.27 15.34 I5.4I 15.48 15.55 15-62 15.69 15.76 15.83 7 4-5 15.90 15.98 16.05 16.12 16.19 16.26 16.33 16.40 16.47 16.55 7 4.6 16.62 16.69 16.76 16.84 16.91 16.98 17.06 17.13 17.20 17.28 7 4-7 17.35 17.42 17.50 17.57 17.65 17.72 17.80 17.87 17.95 18.02 8 4-8 18.10 18.17 18.25 18.32 18.40 18.47 18.55 18.63 18.70 18.78 8 4-9 18.86 18.93 19.01 19.09 19.17 19.24 19.32 19.40 19.48 19.56 8 5-o 19.63 19.71 19.79 19.87 19.95 20.03 20.11 2O.I9 20.27 20.35 S 5-1 20.43 20.51 20.59 20.67 20.75 20.83 20.91 20.99 21.07 21. 16 8 5-2 21.24 21.32 21.40 21.48 21.57 21.65 21.73 21.81 21.90 21.98 8 5-3 22.06 22.15 22.23 22.31 22.40 22.48 22.56 22.65 22.73 22.82 8 5-4 22.90 22.99 23.07 23.16 23.24 23.33 23.41 23.50 23.59 23.67 9 d 01234 5 6 7 8 9 Diff. Explanation in Art. 188 TABLES 493 TABLE 16. AREAS OF CIRCLES d 01234 56789 Diff. 5-5 5-6 23.76 23.84 23.93 24.02 24.11 24.63 24.72 24.81 24.89 24.98 24.19 24.28 24.37 24.45 24.54 25.07 25.16 25.25 25.34 25.43 9 g 5-7 5-8 25.52 25.61 25.70 25.79 2 5- 8 8 26.42 26.51 26.60 26.69 2 6./9 25.97 26.06 26.15 26.24 26.33 26.88 26.97 27.06 27.15 27.25 9 g 5-9 27.34 27.43 27.53 27.62 27.71 27.81 27.90 27.99 28.09 28.18 9 6.0 28.27 28.37 28.46 28.56 28.65 28.75 28.84 28.94 29.03 29.13 9 6.1 29.2.2 29.32 29.42 29.51 29.61 29.71 29.80 29.90 30.00 30.09 JO 6.2 30.19 30.29 30.39 30.48 30.58 30.68 30.78 30.58 30.97 31.07 10 6-3 31.17 31.27 31.37 31.47 3 r. 5 7 31.67 31-77 31-87 31-97 32.07 10 6.4 32.17 32.27 32.37 32.47 32.57 32.67 32.78 32.88 32.98 33.08 10 6.5 33.18 33.29 33.39 33.49 33.59 33-70 33.80 33.90 3400 34.11 10 6.6 34.21 34.32 34.42 34.52 34.63 34-73 34-84 34-94 35-Q5 35.15 TO 6.7 35.26 35.36 35.47 35.57 35.68 35.78 35.89 36.00 36.10 36-21 IO 6.8 36.32 36.42 36.53 36.64 36.75 36.85 36.96 37.07 37.18 37-23 II 6.9 37-39 37-50 37-61 37.72 37.83 37-94 38.05 38.16 38.26 38.37 II 7.0 38.48 38.59 38.70 38.82 38.93 39-04 39-15 39-26 39.37 39.48 II 7-1 39-59 39-70 39.82 39.93 40.04 40.15 40.26 40.38 40.49 40.60 II 7-2 40.72 40.83 40.94 41.06 41.17 41.28 41.40 41.51 41.62 41.74 II 7-3 41.85 41.97 42.08 42.20 42.31 42.43 42.54 42.66 42.78 42.89 II 7-4 43.01 43.12 43.24 43.36 43.47 43-59 43-71 43-83 43-94 44-o6 12 7-5 44.18 44.30 44.41 44.53 44.65 44-77 44-89 45-01 45.13 45.25 12 7.6 45.36 45.48 45.60 45.72 45.84 45.96 46.08 46.20 46.32 46.45 12 7-7 46.57 46.69 46.81 46.93 47.05 47-17 47-29 47-42 47.54 47.66 12 7-8 47.78 47.91 48.03 48.15 48.27 48.40 48.52 48.65 48.77 48.89 12 7-9 49.02 49.14 49.27 49.39 49.51 49.64 49.76 49.89 50.01 50.14 12 8.0 50.27 50.39 50.52 50.64 50.77 50.90 51.02 51.15 51.28 51.40 13 8.1 51.53 51.66 51.78 51.91 52.04 52.17 52.30 52.42 52.55 52.68 13 8.2 52.81 52.94 53.07 53.20 53.33 53.46 53.59 53.72 53.85 53.98 3 8-3 54.11 54.24 54.37 54.50 54-63 54.76 54-89 55-02 55.15 55.29 13 8.4 55-42 55.55 55.68 55.81 55.95 56.08 56.21 56.35 56.48 56.61 13 8-5 56.75 56,88 57.01 57.15 57.28 57.41 57.55 57.68 57.82 57.95 13 8.6 58.09 58.22 58.36 58.49 58.63 8.77 58.90 59.04 59.17 59.31 14 8.7 59-45 59.58 59-72 59.86 59-99 60.13 60.27 60.41 60.55 60.68 M 8.8 60.82 60.96 61.10 61.24 61.38 61.51 61.65 61.79 61.93 62.07 14 8.9 62.21 62.35 62.49 62.63 62.77 62.91 63.05 63.19 63.33 63.48 14 9.0 63.62 63.76 63.90 64.04 64.18 64.33 64.47 64.61 64.75 64.90 14 9.1 65.04 65.18 65.33 65.47 65.61 65.76 65.90 66.04 66.19 66.33 14 9.2 66.48 66 62 66.77 66.91 67.06 67.20 67.35 67.49 67.64 67.78" 15 9-3 67.93 68.08 68.22 68.37 68.51 63.66 68.81 68.96 69.10 69.25 IS 9-4 69.40 69.55 69.69 69.84 69.99 70.14 70.29 70.44 70.58 70.73 15 9-5 70.88 71.03 71.18 71.33 71.48 71.63 71.78 71.93 72.08 72.23 15 9.6 72.38 72.53 72.68 72.84 72.99 73 14 73.29 73.44 73.59 73.75 15 9-7 73.90 74.05 74.20 74.36 74.51 74.66 74.82 74.97 75.12 75.28 15 9-8 75-43 75.58 75-74 75-89 76.05 76.20 76.36 76.51 76.67 76.82 16 9-9 76.98 77.13 77.29 77.44 77.60 77.76 77.91 78.07 78.23 78.38 16 d 01234 56789 Diff. 494 TABLES TABLE 17. TRIGONOMETRIC FUNCTIONS Angle Arc Sin Tan Sec Cosec Cot Cos Coarc O O. 0. 0. I. 00 00 I. 1.5708 90 I 0.0175 0.0175 0.0175 1 .0002 57-299 57-290 0.9998 5533 89 2 0349 0349 349 1. 0006 28.654 28.636 9994 5359 88 3 .0524 0523 .0524 1.0014 19.107 19.081 .9986 .5184 87 4 .0698 .0698 .0699 1.0024 H.336 14.301 .9976 .5010 86 5 .0873 .0872 .0875 1.0038 11.474 11.430 .9962 .4835 85 6 0.1047 0.1045 O.I05I 1-0055 9.5668 9.5H4 0-9945 1.4661 84 7 .1222 .1219 .1228 1.0075 8-2055 8-1443 9925 .4486 83 8 .1396 .1392 J 405 1.0098 7-1853 7."54 9903 4312 82 9 .1571 .1564 .1584 1.0125 6.3925 6.3138 .9877 4137 81 10 1745 .1736 1763 1.0154 5-7588 5.6713 .9848 3963 80 ii O.I92O 0.1908 0.1944 1.0187 5-2408 5.1446 0.9816 1-3788 79 12 .2094 .2079 .2126 1.0223 4.8097 4-7046 .9781 .3614 78 13 .2269 .2250 2309 1.0263 4-4454 4-33I5 9744 3439 77 14 2443 .2419 2 493 1.0306 4-I336 4.OI08 9703 3265 76 15 .26l8 .2588 .2679 1-0353 3-8637 3-7321 9659 .3090 75 16 0.2793 0.2756 0.2867 1.0403 3.6280 34874 0.9613 1.2915 74 17 .2967 2924 .3057 1-0457 3-4203 3.2709 9563 .2741 73 18 3142 .3090 3249 I-05I5 3-2361 3-0777 95" .2566 72 19 .3316 3256 3443 1.0576 3-0716 2.9042 9455 2392 7i 20 3491 .3420 .3640 1.0642 2.9238 2-7475 9397 .2217 70 21 0.3665 0.3584 0-3839 .0711 2.7904 2.6051 0.9336 1.2043 69 22 .3840 3746 .4040 .0785 -6695 2-4751 9272 .1868 68 23 .4014 3907 4245 .0864 5593 2-3559 .9205 .1694 67 24 .4189 .4067 4452 .0946 .4586 2.2460 9135 1519 66 25 .4363 .4226 .4663 .1034 .3662 2-1445 .9063 1345 65 26 0.4538 0.4384 0.4877 .1126 .2812 2.0503 0.8988 1.1170 64 27 .4712 .4540 5095 .1223 .2027 1.9626 .8910 .0996 63 28 .4887 4695 .5317 .1326 .1301 1.8807 .8829 .0821 62 29 .5061 .4848 5543 1434 .0627 i 8040 .8746 .0647 61 30 .5236 .5000 5774 1547 .0000 1.7321 .8660 .0472 60 31 0.54II 0.5150 0.6009 .1666 .9416 1-6643 0.8572 1.0297 59 32 5585 5299 .6249 .1792 .8871 1.6003 .8480 1.0123 58 33 .5/60 .5446 .6494 .1924 8361 1-5399 .8387 0.9948 57 34 5934 5592 .6/45 .2062 .7883 1.4826 .8290 9774 56 35 .6109 -5736 .7002 .2208 7434 1.4281 .8192 9599 55 36 0.6283 0.5878 0.7265 2361 7013 1-3764 0.8090 0.9425 54 37 .6458 .6018 .7536 .2521 .6616 1.3270 .7986 9250 53 38 .6632 .6157 .7813 .2690 6243 1.2799 .7880 .9076 52 39 .6807 .6293 .8098 .2868 .5890 1-2349 7771 .8901 51 40 .6981 .6428 .8391 3054 5557 1.1918 .7660 .8727 50 4i 0.7156 0.6561 0.8693 3250 5243 1.1504 0-7547 0.8552 49 42 7330 .6691 .9004 .3456 4945 1.1106 7431 .8378 48 43 .7505 .6820 9325 .3673 4663 1.0724 .73H .8203 47 44 .7679 .6947 .9657 3902 4396 1.0355 7193 .8029 46 45 .7854 .7071 i. .4142 .4142 i. .7071 .7854 45 Coarc Cos Cot Cosec Sec Tan Sin Arc Angle Explanation in Art. 1S8 TABLES 495 TABLR 18. LOGARITHMS OF TRIGONOMETRIC FUNCTIONS Angle Log Arc Log Sin Log Tan Log Sec Log Cosec Log Cot Log Cos Log Coarc o uo 00 00 O. 00 00 O. 0.1961 90 I 2.2419 2.2419 2.2419 O.OOOI 1.7581 1.7581 7.9999 J 9i3 8 9 2 .5429 .5428 5431 .0003 4572 .4569 9997 .1864 88 3 .7190 .7188 7194 .0006 .2812 .2806 9994 .1814 87 4 8439 .8436 .8446 .001 1 .1564 1554 .9989 .1764 86 5 .9408 9403 .9420 - .0017 0597 .0580 .9983 .1713 85 6 f.0200 1.0192 I.O2I6 0.0024 0.9808 0.9784 1.9976 0.1662 84 7 .0870 .0859 .0891 .0032 .9141 .9109 .9968 .1610 83 8 1450 .1436 .1478 .0042 .8564 .8522 9958 1557 82 9 .1961 1943 1997 .0054 .8057 .8003 .9946 .1504 81 10 .2419 2397 .2463 .0066 .7603 7537 9934 .1450 80 ii 1-2833 1.2806 1 2887 0.0081 0.7194 o.7"3 1.9919 0.1395 79 12 .3211 3179 3275 .0096 .6821 6725 .9904 .1340 78 13 3558 3521 3634 .0113 .6479 .'6366 9887 .1284 77 14 .3880 .3837 .3968 .0131 .6163 .6032 .9869 .1227 76 15 .4180 4130 .4281 .0151 .5870 5719 .9849 .1169 75 16 1.4460 1.4403 1-4575 0.0172 0-5597 0.5425 1.9828 0. 1 1 1 1 74 17 .4723 .4659 .4853 .0194 5341 .5147 .9806 .1052 73 18 .4971 .4900 5"8 .0218 .5100 .4882 9782 .0992 72 19 .5206 .5126 -53/0 .0243 .4874 .4630 9757 0931 "I 20 5429 5341 .5611 .0270 4659 4389 9730 .0870 70 21 T.564I 1-5543 1.5842 0.0298 0-4457 0.4158 1.9702 0.0807 69 22 .5843 5736 .6064 .0328 4264 3936 .9672 0744 68 23 .6036 5919 .6279 .0360 .4081 3721 .9640 .0680 67 24 .6221 .6093 .6486 0393 397 35H .9607 .0614 66 25 .6398 .6259 .6687 .0427 3741 3313 9573 .0548 65 26 1.6569 1.6418 1.6882 0.0463 0.3582 0.3118 1-9537 0.0481 64 2 7 .6732 .6570 .7072 .0501 3430 .2928 9499 .0412 63 28 .6890 .6716 .7257 0541 3284 2743 9459 0343 62 29 743 .6856 -7438 .0582 3144 .2562 .9418 .0272 61 30 .7190 .6990 .7614 .0625 .3010 2386 9375 .0200 60 31 1-7332 1.7118 1.7/88 0.0669 0.2882 O.22I2 i-933i 0.0127 59 32 7470 7242 7958 .0716 .2758 .2042 .9284 0.0053 58 33 .7604 .7361 .8125 .0764 .2639 -I875 .9236 1.9978 57 34 7734 .7476 .8290 .0814 2524 .1710 .9186 .9901 56 35 .7859 .7586 8452 .0866 24H 1548 9134 .9822 55 36 1.7982 1.7692 1.8613 0.0920 0.2308 0.1387 1.9080 7-9743 54 37 .8101 7795 .8771 .0977 .2205 .1229 .9023 .9662 53 38 .8217 .7893 .8928 1035 .2107 .IO72 .8965 9579 52 39 .8329 .7989 .9084 1095 .2OI I .O9l6 .8905 9494 5i 40 .8439 .8081 9238 "57 .1919 .0762 -8843 .9408 50 4 1 1.8547 1.8169 1.9392 O.I222 0.1831 0.0608 1.8778 1.9321 49 4 2 .8651 8255 9544 .1289 1745 .0456 .8711 .9231 48 43 8753 .8338 .9697 1359 .1662 .0303 .8641 .9140 47 44 8853 .8418 .9848 1431 .1582 .0152 .8569 .9046 46 45 8951 .8495 o. .1505 .1505 0. 8495 895 1 45 Log Co arc Log Cos Log Cot Log Cosec Log Sec ^ogTan Log Sin Log Arc Angle Explanation in Art. 188 496 TABLES TABLE 19. LOGARITHMS OF NUMBERS n 01234 56789 Diff. 10 oooo 0043 008^ 0128 0170 0212 0253 0204 0334 0374 42 ii 0414 0457 0492 0531 0569 0607 0645 0682 0719 0755 38 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 35 13 1139 1173 1206 1239 1271 J 33 '335 J 367 1399 1430 3 2 1461 1492 1523 1553 1584 1614 1644 1673 ! 73 J 73 2 30 IS 1761 1790 1818 1847 187? 1903 1931 1959 1987 2014 28 16 2O4I 2068 2095 2122 2148 2175 2201 2227 2253 2279 27 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 25 18 2553 2577 2601 2625 2648 2672 2695 2 7 '8 2742 2765 24 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 22 20 21 3010 3032 3054 3075 3096 3222 3243 3263 3284 3304 3118 3139 3160 3181 3201 3324 3345 3365 3385 3404 21 2O 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 19 23 24 3617 3636 3655 3674 3692 3802 3820 3838 3856 3874 37" 3729 3747 3766 3784 3892 3909 3927 3945 3962 Is 25 3979 3997 4 14 4031 4048 4065 4082 4099 4116 4133 17 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 17 27 4314 4330 4346 4362 4378 4393 4409 4425 444 4456 16 28 4472 4487 452 4518 4533 4548 4564 4579 4594 4609 15 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 15 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 14 31 4914 4928 4942 4955 4969 4983 4997 5" 50 2 4 5038 14 32 5051 506? 5079 5092 5105 5 I! 9 5 J 32 5 T 45 559 5'72 "3 33 34 5185 5198 5211 5224 5237 53 iS 5328 5340 5353 5366 5250 5263 5276 5289 5302 53/8 5391 5403 5416 5428 13 13 35 544i 5453 5465 5478 5490 552 5514 5527 5539 5551 12 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 12 1 5682 5694 5705 5717 5729 5798 5809 5821 5832 5843 5740 5752 5763 5775 5786 5655 5866 5877 5888 5899 12 39 59" 59 22 5933 5944 5955 5966 5977 5988 5999 6010 II 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 II 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 II 42 43 44 6232 6243 6253 6263 6274 6335 6345 6355 6365 6375 6435 6444 6454 6464 6474 6284 6294 6304 6314 6325 6385 6395 6405 6415 6425 6484 6493 6 53 6 5*3 6 5 22 IO 10 10 47 6532 6542 6551 6561 6571 6628 6637 6646 6656 6665 6721 6730 6739 6749 6758 6580 6590 6599 6609 6618 6675 6684 6693 6702 6712 6767 6776 6785 6794 6803 IO 9 9 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 9 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 9 5 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 9 5 1 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 8 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 8 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 8 54 73 2 4 7332 7340 7348 7356 7364 7372 7380 7388 7396 8 01234 56789 Diff. Explanation in Art. 188 TABLES 497 TABLE 19. LOGARITHMS OF NUMBERS n 01234 56789 Diff. 55 7404 7412 7419 7427 7435 7443 745' 7459 7466 7474 8 56 57 7482 7490 7497 7505 7513 7559 7566 7574 7582 7589 7520 7528 7536 7543 7551 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 775 2 776o 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 7 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 793 1 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 7 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 68 69 8261 8267 8274 8280 8287 8325 8331 8338 8344 8351 8388 8395 8 40I 8407 8414 8293.8299 8306 8312 8319 8357 8363 8370 8376 8382 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 6 7 1 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 87.39 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 6 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8972 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 So 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 5 81 9085 9090 9096 9101 9106 9FI2 9117 9122 9128 9133 82 9138 9 T 43 9'49 9' 54 9'59 9165 9170 9175 9180 9186 83 84 9191 9196 9201 9206 9212 9243 9248 9253 9258 9263 9217 9222 9227 9232 9238 9269 9274 9279 9284 9289 85 9294 9299 934 939 93 '5 9320 9325 9330 9335 9340 5 86 9345 9350 9355 93< 9365 9370 9375 93o 935 939 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 946o 9465 9469 9474 9479 944 9489 89 9494 9499 954 959 95 '3 95'8 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 5 92 959 9595 9" 9"5 99 9638 9643 047 9652 9657 9614 9619 9624 9628 9613 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 973 1 9736 974i 9745 975 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 4 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 994 995 2 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 n 01234 56789 Diff. 498 TABLES TABLE 20. CONSTANTS AND THEIR LOGARITHMS Name. (Radius of circle or sphere = i.) Symbol. Number. Logarithm. Area of circle n 3 141 592654 0.497 149873 Circumference of circle ZTt 6.283 185 307 0.798 179868 Surface of sphere 4* 12.566 370614 1 .099 209 864 i* 0.523 598 776 1.718998622 Quadrant of circle I* 0.785 398 163 1.895089881 Area of semicircle rr 1.570796327 0.196 Iig 877 Volume of sphere a^ 4.187790205 0.622 088 609 7T 2 9.869604401 0.994299745 ** I-772453 851 0.248 574936 Degrees in a radian I80/7T 57-2957795I3 1.758 122632 Minutes in a radian lOSOO/TT 3437-74677I 3.536273883 Seconds in a radian 6480OO/7T 206264.806 5.314425 133 I/It 0.318 309 886 1.502 850 127 I/7T* 0.564 189 584 1.751425064 1/71* o.ioi 321 184 1.005700255 Circumference/36o arc 0.017453293 2.241 877 368 sin 0.017 452 406 2.241855318 Circumference/2 1600 arc ' o.ooo 290888 4.463726117 sin ' 0.000290888 4.463 726 in Circumference/ 1 296000 arc " 0.000004 848 6.685574867 sin " 0.000004 848 6.685574867 Base Naperian system of logs e 2.718 281 828 0.434294482 Modulus common system of logs M 0.434294482 1.637 784311 Naperian log of 10 i/M 2.302585093 0.362 215 689 hr 0.4769363 1.678 4604 Probable error constant hr VT 0.6744897 1.8289754 Feet in one meter Miles in one kilometer m/ft. km/mi. 3.2808333 0.621 369 9 0.5159841 1-7933502 INDEX 499 INDEX Absorption of brick, 64 Acid steel, 61 Alloys, 65, 67 Aluminum, 66 Angles, 104, no, in, 486 Angular velocity, 421 Annealing, 59, 64, 305 Anthracite coal, 67, 380 Answers to problems, 477 Apparent and true stresses, 186, 274, 359-382, 462 Approximate computations, 34 Area, reduction of, 31 Areas of circles, 21, 492 Artificial stone, 54, 66 Army, gun formulas, 394402 Association, testing, 441 Axial stresses, i, 3, 190, 253, 327, 365 impact, 327, 331, 337 Axis, neutral, 98, 100 of a bar, 2, 188 of a beam, too, 290 of a column, 190 Axle steel, 63 Axles, 348, 353 Bach, C., 414, 4i6 Bar, i, 42, 69 Bar iron, 58, 489 Bars of uniform strength, 71 resilience of, 306 under centrifugal stress, 421 under impact, 327, 331 weights of, 2, 478 Base-line apparatus, 165 Basic steel, 61 Bauschinger, J., 353 Beams, 87-187, 253-267, 269-302 bending moments, 93, 116 cantilevers, 116-148, 487 Beams, cast iron, 122, 128 center of gravity of sections, 103 centrifugal stress, 425 combined stresses, 251275 concrete, 282-298 constrained, 149-167, 487 continuous, 87, 168-187 deck, no, 484 definitions, 87 deflection, 112, 135, 145, 153, 258^ 312-319 deflection and stiffness, 142, 158 deflection and stress, 143, 159 designing of, 125, 292 elastic curve, 87, 114, 136, 138 elastic resilience, 308 experimental laws, 99, 185 fixed, 149-167, 487 flexural strength, 56, 131 flitched, 282 fundamental formulas, 101, 102, 185 Galileo's investigations, 186 historical notes, 184 horizontal shear, 269 impact on, 329, 334 internal stresses, 97, 270 internal work, 303-323 lines of stress, 272 maximum moments, 119, 150, 154 modulus of rupture, 47, 131 moments of inertia, 105108 moving loads, 132 plate-girder, 108, 247, 208 pure flexure, 374 overhanging, 140-165 reactions, 88, 150, 169 reinforced-concrete, 285, 298 resilience of, 308 rolled, 108 safe loads for, 124 500 MECHANICS OF MATERIALS Beams, simple, 116-148, 487 stiffness, 141, 158 sudden loads, 324 theoretical laws, 97, 185 true stresses, 367 uniform strength, 143148 unsymmetric loads,.427 vertical shear, 90, 116 weights of, 42, 88 Bearing compression, 85 Bending moment, 93, 98, rot, 116 diagrams of, 98, 116 maximum, 119, 150, 487 maximum maximorum, 133 tables of, 175, 487 triangular load, 426 Bessemer steel, 60, 305 Best iron, 57 Bethlehem Steel Co., 64, 241, 383 Beton, 66 Birnie's formulas, 394 Boiler steel, 63 Boilers, 83, 417 joints in, 81-86 tubes in, 78 Bolts, 16, 239, 265, 366 Brass, 67 Bresse, M., 182 Brick, 6, 24, 44, 48, 252, 380 strength of, 13, 14, 17, 49, 131, 480 weight of, 42, 48 Brick masonry, 49 Brick tower, 50 Bridges, 40, 59, 86, 187, 262, 284 Bridge iron, 58 rollers, 404 Briquettes, 53 Brittle materials, 44, 380, 438 Brittleness, 6, 43 Bronze, 67 Buildings, unit-stresses, 18, 482 Building stone, 51 Bulb beams, no, 427, 482 Butt-joints, 82, 85 Campbell, H. H., 61 Cantilever beams, 87, 116-148, 487 deflection of, 135, 145 Cantilever, elastic curve, 136 fundamental formulas, 102 internal work, 308 resilience, 303 table for, 487 uniform strength, 144 with constraint, 163 Carbon in cast iron, 55 steel, 60, 62 wrought iron, 58 Castings, 55, 65, 482 Cast iron, 55, 480-482 beams, 128 brittleness of, 44, 380 elastic limit, 5, 56, 480 factors of safety, 1 7 flexural strength, 56, 131 in compression, 13, 56 in shear, 14, 38 in tension, 10, 24, 56 pipes, 76 resilience of, 306 weight of, 42, 55 Cements, 52, 438 Center of gravity, 73, 103 of gyration, 112 Centrifugal stress, 421, 425 Chain link, 476 Channels, 104, no, 485 Chestnut, 46, 47 Christie's experiments, 197 Circles, areas of, 106, 492 Circular plates, 409, 411 rings, 476 Classification of pig iron, 55 steel, 63 Clapeyron, E., 174, 182 Clavarino's formulas, 393 Coal, 67, 380 Coefficient of elasticity, 24 of expansion, 252, 480 of impact, 350 of inertia, 331 of internal friction, 378, 380 Cold bend test, 58, 63, 439, 446 rolling, 58 Columns, 12, 188-224, 2 79 compound, 276-281 INDEX 501 Columns, deflection of, 194 design of, 206, 219 eccentric loads, 214, 217 ends of, 191, 221 Euler's formula, 192 experiments on, 196 Gordon's formula, 203, 208 Hodgkinson's formula, 197 investigation of, 203, 219, 279 Johnson's (T. H.) formula, 208 modified Euler formula, 431 radius of gyration, 191 Rankine's formula, 200, 211 reinforced-concrete, 279 Ritter's formula, 211, 221 rupture of, 197, 213 safe loads for, 205 sections of, 188, 281 theory of, 190, 220 Combined stresses, 251-275 compression and flexure, 254 flexure and torsion, 259 shear and tension, 264 tension and compression, 251 tension and flexure, 259 torsion and compression, 268 Comparison of beams, 141, 158, 487 Compound beams, 282302 columns, 276-281 cylinder, 390 Compression, 2, n, 31, 188, 438, 481 and flexure, 255, 262 and shear, 265 and tension, 251 and torsion, 266 cast iron, 14, 17, 55 cement, 53, 438 concrete, 54, 288 eccentric loads, 72, 214 mortar, 52 on rivets, 81, 366 steel, 14, 17, 60 stone, 14, 51 wrought iron, 14, 17, 44, 57 Computations, 19, 34 Concentrated loads, 88, 1 19, 407 Concentric loads, 190, 368 Concrete, 54, 279, 288, 482 Concrete beams, 282-302 columns, 279 Connecting rod, 426 Constants, tables of, 480-498 Constrained beams, 149 Continuity, 168 Continuous beams, 87, 168-187 equal spans, 175 fixed ends, 1 79 properties of, 171 tables of, 175, 176 three moments, 173 unequal spans, 177 Contraction of area, 31 Cooper, T., 85, 210, 404 Cox, H., 342 Couplings for shafts, 239 Crandall, C. L., 407 Crank arm, 242, 244 pin, 241, 243, 461 Crehore, J. D., 212 Crucible steel, 60 Cubic equation, 166, 462 Curvature, radius of, 114 Crystals in steel, 354, 382 Cylinders, 77, 383 compound, 390, 399 exterior pressure, 77 interior pressure, 75, 383 thick, 76, 383-395 thin, 75, 77, 394 with hoops, 78, 383, 399 Cylindrical rollers, 403 Dead loads, 132, 349 Deck beams, no, 484 Deflection of beams, 112, 312-319, 487 cantilever beams, 135, 145 compound beams, 300 constrained beams, 151156 simple beams, 138, 147, 301 sudden loads, 324 under impact, 329, 334 under moving load, 350 under shearing, 302, 317 Deflection of columns, 194, 218 of plate girders, 301 Deformation, elastic, 3, 8, 23, 28 502 MECHANICS OF MATERIALS Deformation, ultimate, 30 work in, 35 Designing, 18 beams, 125, 292 columns, 206 guns, 384, 389, 399 joints, 83 shafts, 231, 241 Detrusion, 38, 91 Diagram of stress, 9, 28 Diagrams, shear and moment, 92, 94, 116, 150, 173 Dimensions in equations, 21 Ductility, n, 58, 63 Dudley, C. B., 448 Dudley, P. H., 343 Dynamic stress, 324-358 Eccentric loads, 72-74, 214, 217 Economic beams, 129 Economy in design, 18, 127 Efficiency of a joint, 82 Elastic curve, 87, 112, 151, 153, 156 cantilever beams, 135 columns, 193, 218 constrained beams, 141-149 continuous beams, 173 general equation, 112 simple beams, 139 Elastic deflection, 112 Elastic deformation, 23 Elastic limit, 4, 9, 17, 23, 27, 480 cast iron, 5, 56 compression, 13 shear, 15, 39 steel, 5, 27, 63 tension, 5, 443 timber, 5, 47 wrought iron, 5, 58 Elastic resilience, 303-317 Elastic strength, 132 Elasticity, coefficient of, 24 laws of, 4, 8, 45 modulus of, 23 theory of, 449-471 Electric analogies, 278, 474 Ellipse of stress, 463 Ellipsoid of stress, 456 Elliptical plates, 414 Elongation, 3, 4, 10, 26 ultimate, 10, 30, 56, 58, 63, 443 under impact, 325, 332 under own weight, 70 Endurance tests, 358 Equations, dimensions, 21 Equilibrium, 2, 96 Ether, 470, 473 Euler's formula, 192, 196, 431 Expansibility, 251, 480 Experimental laws, 8, 17, 99, 225, 360 External forces, i, 39, 77 External work, 35, 303, 312, 324 Eye bars, 260, 445 Fairbairn, W., 78 Factor of lateral contraction, 34, 359 of safety, 7, 8, 17 Fatigue of materials, 352-358, 381 Fixed beams, 87, 149, 152, 156 Flexural strength, 47, 56, 131, 481 Flexure, 87-186 and compression, 255, 262 and tension, 259, 262 and torsion, 266 centrifugal, 425 erroneous views, 185 formula, 101 of crank pin, 241 of joints, 82, 86 pure, 374 under impact, 329, 334 under live load, 132, 349 work of, 308, 312 Floor beams, 147, 300 Flues, boiler, 78 Fly wheel, 423 Forge pig, 55, 57 Forgings, 64 Foundry pig, 55 Friction, internal, 375^382 Galileo, G., 186 German I beams, 109, 488 Glass, 67 Gordon, L., 203, 208 Goss, W. F. M., 444 INDEX 503 Granite, 50, 51 Gravitation, 470, 473 Gravity, center of, 73, 103 specific, 42 Grecian columns, 223 Greek letters, 21 Grindstone, 424 Gun metal, 63 Guns, 34, 383-401 hooped, 384, 390, 399 solid, 388, 393 Gyration, radius of, 112, 129 Hard steel, 62 Hartmann, L., 376 Hatt, W. K., 347, 348, 441 Hemispheres, 417, 419 Hemlock, 46 Historical notes, 39, 186, 196, 318, 341, 433. 44i Hodgkinson, E., 342, 347 Hollow cylinders, 75, 383-401 shafts, 232 spheres, 77 Hooke's law, 4, 40, 449 Hoops, centrifugal stress, 422 for guns, 383, 391, 396, 399 shrinkage of, 79, 396 Horizontal impact, 330, 336 shear, 269 stresses, 97, 100 Horse-power, 230 Howard, J., 358, 436 Hydraulic cement, 52 mortar, 53 I beams, 103, 107, 122, 124, 128, 178, 483, 488 Inertia of a bar, 331 of a beam, 334 in impact, 331-337 moment of, 105 Inspection of material, 448 Impact, 324-351 on bars, 327, 331 on beams, 329, 334 pressure due to, 344 tests on, 341, 346, 439 Inflection point, 149 friction, 375-382 Internal stresses, 2, 96 work, 303-315 International Association, 441 Investigation, 17, 122, 203 of beams, 121 of columns, 203 of guns, 389, 393, 401 of joints, 80 of shafts, 230, 233 Iron, 44, 55, 58 Isotropic materials, 449 Jacket for guns, 384 Johnson, J. B., 368, 409 Johnson, T. H., 208 Joints, riveted, 80-86 Keep, W. J., 342 Keep's impact machine, 343 Kirkaldy, D., 341 Lame, E., 395, 478 Lame''s formulas, 385 Lap joints, 80, 84 Lateral contraction, 32, 34, 359 factor of, 34 Launhardt's formula, 355 Laws, experimental, 8, 17, 99 of fatigue, 353 of internal stresses, 360 of resilience, 303, 324 Lead, 67 Least work, 320 Lilly's column formula, 223 Limestone, 50, 51, 482 Limiting length of bar, 69 of beam, 167 Live loads, 132, 349 Loads, 3, 87, 407 safe, for beams, 124 safe, for columns, 205 sudden, 324 Locomotive, 425 Log, beam cut out, 129 Logarithms, 495-498 Long columns, 192 504 MECHANICS OF MATERIALS Machinery steel, 63 Malleable cast iron, 57 Manhole covers, 415 Marburg, E., 212, 441 Marston, A., 407 Martens, A., 435 Masonry, brick, 49 stone, 51 Materials, factors of safety, 7-1 7 fatigue of, 352-358 properties of, 42-68 resilience of, 303-320 specifications for, 444 strength of, 1-22, 42-68 tests of, 433-443 weights of, 42 Maximum moments, 119, 133, 487 shears, 119, 134, 459 Measures, systems of, 20, 2 1 Medium steel, 62, 63, 445 Meigs, J. F., 401 Merriman, M., 182, 212, 356 Metric system, 20 Millstone, 424 Modulus of elasticity, 23, 37, 47, 480 resilience, 307 rupture, 47, 132 Moisture in timber, 47 Moment of a force, 93 bending, 94 twisting, 226 Moment of inertia, 105, in for beams, 105, 108, 429, 487 for columns, 189 for shafts, 229 Moments, bending, 93, in, 116 cantilever beams, 94, 117 continuous beams, 176 diagrams of, 116 fixed beams, 152, 156 maximum, 119, 133 overhanging beams, 150 resisting, 98 simple beams, 95, 118 theorem of three, 173 Moncrieff, J. M., 224 Mortar, 52 Moving loads, 132, 349 Natural cement, 52, 54 Navier, L. M. H., 186 Navy, gun formulas, 393, 402 Neutral axis, 100 surface, 99 Newton, I., 8, 40, 382, 478 Nickel steel, 65 Normal stress, 264, 362, 452 Norton, W. A., 318 Oak, 46, 47, 482 One-hoss shay, 18 Open-hearth steel, 60, 61, 444 Ordnance formulas, 383-401 Ores of iron, 57, 60 Oscillations of a bar, 325, 333 of a beam, 344 Overhanging beams, 149-155, 165 Own weight of bar, 69 of beam, 124 Parabola, 94, 118, 144, 357 Parallel rod, 425 Paving brick, 49 Phosphor bronze, 66 Phosphorus in steel, 61, 444 Pitch of rivets, 81, 85 Piers, 71 Pig iron, 55 Piles, 279, 281 Pine, 46, 47, 318 Pipes, 76, 383 thick, 383-392 thin, 75, 390 Piston rod, 19, 188 Plasticity, 43 Plate girder, 108, 147, 298, 369 Plates, 58, 63, 403-416, 447 on cylinders, 83, 419 Poisson's ratio, 34 Polar moments, 229 Portland cement, 52, 54 Powder for guns, 383 Power, shafts for, 230 Pressure due to impact, 345 Principal stresses, 457 Prisms, loads on, 2, 74, 214 Problems, answers to, 477 INDEX 505 Puddling furnace, 58 Pure stresses, 373 Purlins, 428 Radius of curvature, 1 14 of gyration, 112, 129, 191 Rafters, 254 Railroad rails, 43, 104, 343 Range of stresses, 352-358 Reactions of beams, 88, 91 Rectangle, 105 Rectangular beams, 129, 288 plates, 415 shafts, 247 Reduction of area, 31 Reinforced concrete, 279-298 Rejto, A., 378 Repeated stresses, 352-358 Resilience, 303-318, 469 of bars, 304, 306 of beams, 308 of shearing, 310 of torsion, 311 Resisting moment, 98, 227 shear, 98, 101 Resultant stress, 454 Ritter, A., 211, 221 Riveted joints, 80-86 design of, 83 efficiency of, 82 Rivet iron, 58 steel, 63, 445 Rivets, 80-85, 4 82 Rolled beams, 109 shapes, no Rollers, 403, 406 Roprs, 66, 278 Round shafts, 229, 231 Rupture, 3, 8, 15, 31 of beams, 130 of columns, 197 in repeated stress, 352 modulus of, 132 Safe loads, 17, 124 Safety, factors of, 7, 17 Saint Venant, B. de, 249, 450 Sand-lime brick, 66 Sandstone, 50-51, 380 Section, changes in, 32 Section area, i, 42 factor, 102 Set, 5, 28, 29 Shafts, 225-250, 266, 268 couplings for, 239 cranks for, 241, 243 for power, 230 hollow, 232 round, 227, 231 square, 248 resilience of, 310 stiffness of, 231 strength of, 231, 237 true stresses, 371 Shapes, rolled, 107, no, 483-488 Shear, 14, 15, 369, 459 and tension, 15, 263, 362 deflection due to, 316 horizontal, 269 on rivets, 81 resilience of, 310 resisting, 98 vertical, 90, 98 ultimate strength, 14, 370 work of, 39, 310 Shear formula, 101 Shearing modulus, 37, 234, 465 strength, 481 Shears for cantilevers, 91, 117 for continuous beams, 176 for simple beams, 92, 118, 120 Shocks, 17, 57 Shrinkage of hoops, 79, 393 Shortening, 3 Simple beams, 87, 89, 116-148, 487 Slate, 51, 482 Slenderness ratio, 191 Soft steel, 62, 445 Solid shafts, 229, 231, 237 Sound, velocity of, 473 Specific gravities, 42 Specifications, 18, 444 Spheres, 77, 417 Spherical rollers, 406 Spring steel, 63 506 MECHANICS OF MATERIALS Springs, 341, 476 Square plates, 416 shafts, 248, 440 Squares of numbers, 479, 490 Stability, 18, 127 Static loads and stresses, 324 Steam boilers, 76, 78 pipes, 75 Steel, 5, 10, 13, 17, 24, 60-65, X 3 2 > 2 5 2 354, 444 constants of, 480-482 factors of safety, 17 properties of, 60-65 resilience, 307 weight of, 42 Steel beams, 103, 107, 124, 483, 488 cranks, 241, 243 guns, 384 plates, 410, 413, 447 pipes, 76 rollers, 404 ropes, 278 spheres, 407, 419 Stiffness of beams, 141, 158, 487 of shafts, 237 Stone, 13, 14, 17, 42, 44, 50, 131, 482 Straight-line formula, 208 Strength of materials, 1-22, 42-68 history of, 39, 186 tables, 480, 481 Stress, i, 42, 100, 139, 449 apparent, 275 centrifugal, 421, 424 combined, 251-275 diagrams of, 9, 28, 29 in guns, 383-402 pure, 373 repeated, 356 sudden, 324 temperature, '68, 251 true, 359-362 working, 17 Stringer, 284 Strong steel, 5, n, 13 Structural steel, 5, 10, n, 13, 14, 17, 63, 109, 128, 444 Sudden deflections, 325 loads, 324 Supports of beams, 87, 89, 159 Surface, neutral, 98 T shapes, 104, 107, in, 128, 484 Tables, 20, 478, 480-498 Talbot, A. N., 296, 375, 436 Tangential stress, 452 Temperature, 67, 251 Tempering, 62, 64 Tension, 2, 9, 36, 69, 407, 481 and flexure, 259, 262 and shear, 263 and torsion, 268 centrifugal, 164 eccentric, 73, 262 Testing laboratories, 41, 358, 436 Testing machines, 40, 433, 442 Testing, rules for, 442 Test specimens, 15, 436, 446 Tests, brick, 49 cast iron, 56 cement, 53, 438 cold bend, 58, 239 columns, 196, 213 compression, 12, 45, 438 fatigue, 352 flexural, 47, 56, 131 , 439, 443 impact, 341, 346, 439, 443 steel, 63, 437 stone, 51 tension, 4, 9, 36, 436, 442 timber, 47, 187, 438 torsion, 226, 245, 439 wrought iron, 58 Tetmajer, L. von, 208, 224 Theorem of three moments, 173-183 Thick hollow cylinders, 383, 389, 393 spheres, 417 Thin pipes, 76 Thurston, R. H., 67, 318, 435 Timber, 5, 10, 13, 14, 24, 38, 46, 482 beams, 129 factors of safety, 17, 47 flexural strength, 131 resilience, 306 weight, 42, 46 Time of vibration, 337 Tool steel, 65 INDEX 507 Torsion, 225-250, 439 combined, 152, 154 formula for, 228 non-circular sections, 245 phenomena of, 225 resilience of, 310 rupture by, 235 Transmission of power, 230 Transverse impact, 327, 334 Trap rock, ,50, 51 Tredgold, T., 40, 347 Triangular beams, 104, 106, 166 True deformations, 360, 370 stresses, 274, 359-384 Tubes, 78, 382 Turner, C. A. P., 354 Twisting moment, 227 Trigonometric functions, 494 Ultimate strength, 6, 10, 13, 481 compression, 12, 44 deformation, 30 shear, 14 tension, 4, 36 Uniform load, 88, 119, 150 Uniform strength, 71, 144 bars, 71 beams, 143, 146 Unit-deformation, 9, 23, 30 Unit-stress, i, 12, 23 repeated, 353 working, 17 Unsymmetric beams, 125, 427 loads, 428 Velocity of live load, 350 stress, 472 Vertical shear, 90, 116, 119 deflection due to, 316 stresses caused by, 93, 123 work of, 39, 310 Vibrations after impact, 338 of a beam, 344 Volume, change of, 33, 450 resilience of, 469 Volumetric modulus, 467 Water, 468, 473 Water pipes, 75, 77 pressure, 76 Wave propagation, 473 Webster, W. R., 448 Weights of bars, 42, 88, 478, 489 materials, 42-68, 480 Weyrauch's formula, 356 Wheel, revolving, 422 Wire, 63, 66 WOhler's laws, 353 Wood, De V., 78, 166, 473 Work, least, 320 Work of flexure, 304, 308, 312 rupture, 36 shearing, 39 tension, 35, 307 torsion, 310 vertical bar, 70 vertical shear, 317 Working unit -stresses, 1 7, 482 Wrought iron, 5, 10, 13, 14, 17, 24, 57 60, 252, 482 factors of safety, 17 flexural strength, 132 resilience, 307 shear, 38 tension, 10, 29, 59 weight of, 42, 489 Wrought-iron bars, 42, 58, 489 pipes, 76 plates, 58 Yield point, 27, 29, 63, 443, 447 Young, T., 40, 346, 433 Young's modulus, 24 Z bars, 104, 427, 430, 487 Zimmerman, H., 351 SHORT-TITLE CATALOGUE OF THE PUBLICATIONS OF JOHN WILEY & SONS NEW YORK LONDON: CHAPMAN & HALL, LIMITED ARRANGED UNDER SUBJECTS Descriptive circulars sent on application. Books marked with an asterisk (*) are sold at net prices only. All books are bound in cloth unless otherwise stated. AGRICULTURE HORTICULTURE FORESTRY. Armsby's Principles of Animal Nutrition 8vo, $4 00 Bowman's Forest Physiography. (In Press.) Budd and Hansen's American Horticultural Manual: Part I. Propagation, Culture, and Improvement 12mo, 50 Part II. Systematic Pomology 12mo, 50 Elliott's Engineering for Land Drainage 12mo, 50 Practical Farm Drainage. (Second Edition, Rewritten.) 12mo. 50 Graves's Forest Mensuration 8vo, 00 * Principles of Handling Woodlands Large 12mo, 50 Green's Principles of American Forestry 12mo, 50 Grotenfelt's Principles of Modern Dairy Practice. (Woll.) 12mo, 00 00 50 50 Maynard's Landscape Gardening as Applied to Home Decoration 12mo, . 50 Sanderson's Insects Injurious to Staple Crops 12mo, I 50 Sanderson and Headlee's Insects Injurious to Garden Crops. (In Prepa- ration.) * Schwarz's Longleaf Pine in Virgin Forest 12mo, 1 25 * Solotaroff 's Shade Trees in Towns and Cities 8vo, 3 00 Stockbridge's Rocks and Soils 8vo, 2 50 Winton's Microscopy of Vegetable Foods 8vo, 7 50 Woll's Handbook for Farmers and Dairymen 16mo, 1 50 Herrick's Denatured or Industrial Alcohol . Kempand Waugh's Landscape Gardening. (New Edition, Rewritten.) 12mo, McKay and Larsen's Principles and Practice of Butter-making 8vo, ARCHITECTURE. Baldwin's Steam Heating for Buildings 12mo, 2 50 Berg's Buildings and Structures of American Railroads 4to, 5 00 Birkmire's Architectural Iron and Steel 8vo. 3 50 Compound Riveted Girders as Applied in Buildings 8vo, 2 00 Planning and Construction of American Theatres 8vo, 3 00 1 Birkmire's Planning and Construction of High Office Buildings 8vo, $3 50 Skeleton Construction in Buildings 8vo, 3 00 Briggs's Modern American School Buildings 8vo, 4 00 Byrne's Inspection of Materials and Wormanship Employed in Construction. 16mo, 3 00 Carpenter's Heating and Ventilating of Buildings 8vo, 4 00 * Corthell's Allowable Pressure on Deep Foundations 12mo, 1 25 Fieita^'s Architectural Engineering 8vo, 3 50 Fireproofing of Steel Buildings 8vo, 2 50 Gerhard's Guide to Sanitary Inspections. (Fourth Edition, Entirely Re- vised and Enlarged.). 12mo, 1 50 * Modern Baths and Bath Houses .*. 8vo, 3 00 Sanitation of Public Buildings 12mo, 1 50 Theatre Fires and Panics 12mo, 1 50 * The Water Supply, Sewerage and Plumbing of Modern City Buildings. 8vo, 4 00 Johnson's Statics by Algebraic and Graphic Methods '. . .8vo, 2 00 Kellaway's How to Lay Out Suburban Home Grounds 8vo, 2 00 Kidder's Architects' and Builders' Pocket-book 16mo, mor., 5 00 Merrill's Stones for Building and Decoration 8vo, 5 00 Monckton's Stair-building 4to, 4 00 Patton's Practical Treatise on Foundations 8vo, 5 00 Peabody's Naval Architecture 8vc, 7 50 Rice's Concrete-block Manufacture 8vo, 2 00 Richey's Handbook for Superintendents of Construction 16mo, mor. 4 00 Building Foreman's Pocket Book and Ready Reference. . 16mo, mor. 5 00 * Building Mechanics' Ready Reference Series: * Carpenters' and Woodworkers' Edition 16mo, mor. 1 50 * Cement Workers' and Plasterers' Edition 16mo, mor. 1 50 * Plumbers', Steam-Fitters', and Tinners' Edition. . . 16mo, mor. 1 50 * Stone- and Brick-masons' Edition 16mo, mor. 1 50 Sabin's House Painting 12mo, 1 00 Siebert and Biggin's Modern Stone-cutting and Masonry 8vo, 1 50 Snow's Principal Species of Wood 8vo, 3 50 Wait's Engineering and Architectural Jurisprudence 8vo, 6 00 Sheep, 6 50 Law of Contracts 8vo, 3 00 Law of Operations Preliminary to Construction in Engineering and Architecture 8vo, 5 00 Sheep, 3 50 Wilson's Air Conditioning 12mo-, 1 50 Worcester and Atkinson's Small Hospitals, Establishment and Maintenance, Suggestions for Hospital Architecture, with Plans for a Small Hospital 12mo, 1 25 ARMY AND NAVY. Bernadou's Smokeless Powder, Nitro-cellulose, and the Theory of the Cellu- lose Molecule 12mo, 2 50 Chase's Art of Pattern Making 12mo, 2 50 Screw Propellers and Marine Propulsion 8vo, 3 00 * Cloke's Enlisted Specialists' Examiner 8vo, 2 00 * Gunner's Examiner 8vo, 1 50 Craig's Azimuth 4to. 3 50 Crehore and Squier's Polarizing Photo-chronograph 8vo, 3 00 * Davis's Elements of Law 8vo, 2 50 * Treatise on the Military Law of United States 8vo, 7 00 * Dudley's Military Law and the Procedure of Courts-martial. ..Large 12mo, 2 50 Durand's Resistance and Propulsion of Ships 8vo, 5 00 * Dyer's Handbook of Light Artillery 12mo, 3 00 Eissler's Modern High Explosives 8vo, 4 00 "Fiebeger's Text-book on Field Fortification Large 1 2mo, 2 00 Hamilton and Bond's The Gunner's Catechism 18mo, 1 00 * Hoff's Elementary Naval Tactics 8vo, SI 50 Ingalls's Handbook of Problems ii. Direct Fire 8vo, 4 00 * Lissak's Ordnance and Gunnery 8vo, 6 00 * Ludlow's Logarithmic and Trigonometric Tables 8vo, 1 00 * Lyons's Treatise on Electromagnetic Phenomena. Vols. I. and II..8vo,each, 6 00 * Mahan's Permanent Fortifications. (Mercur.) 8vo. half mor. 7 50 Manual for Courts-martial 16mo.mor. 1 50 * Mercur's Attack of Fortified Places 12mo. 2 00 * Elements of the Art of War 8vo, 4 00 Nixon's Adjutants' Manual 24mo, 1 00 Peabody's Naval Architecture 8vo, 7 50 * Phelps's Practical Marine Surveying 8vo, 2 50 Putnam's Nautical Charts 8vo, 2 00 Rust's Ex-meridian Altitude, Azimuth and Star-Finding Tables 8vo, 5 00 * Selkirk's Catechism of Manual of Guard Duty 24mo, 50 Sharpe's Art of Subsisting Armies in War 18mo, mor. 1 50 * Taylor's -Speed and Power of Ships. 2 vols. Text 8vo, plates oblong 4to, 7 50 * Tupes and Poole's Manual of Bayonet Exercises and Musketry Fencing. 24mo, leather, 50 * Weaver's Military Explosives 8vo, 3 00 * Woodhull's Military Hygiene for Officers of the Line Large 12mo, 1 50 ASSAYING. Betts's Lead Refining by Electrolysis 8vo. 4 00 *Butler's Handbook of Blowpipe Analysis 16mo, 75 Fletcher's Practical Instructions in Quantitative Assaying with the Blowpipe. 16mo, mor. 1 50 Furman and Pardoe's Manual of Practical Assaying 8vo, 3 00 Lodge's Notes on Assaying and Metallurgical Laboratory Experiments..8vo, 3 00 Low's Technical Methods of Ore Analysis 8vo, 3 00 Miller's Cyanide Process 12mo, 1 00 Manual of Assaying 12mo, 1 00 Minet's Production of Aluminum and its Industrial Use. (Waldo.). ..12mo, 2 50 Ricketts and Miller's Notes on Assaying 8vo. 3 00 Robine and Lenglen's Cyanide Industry. (Le Clerc.) 8vo, 4 00 * Seamen's Manual for Assayers and Chemists Large 12mo, 2 50 Ulke's Modern Electrolytic Copper Refining 8vo, 3 00 Wilson's Chlorination Process 12mo, 1 50 Cyanide Processes 12mo. 1 50 ASTRONOMY. Comstock's Field Astronomy for Engineers 8vo, 2 50 Craig's Azimuth 4to. 3 50 Crandall's Text-book on Geodesy and Least Squares 8vo. 3 00 Doolittle's Treatise on Practical Astronomy , . .8vo, 4 00 Hayford's Text-book of Geodetic Astronomy 8vo, 3 00 Hosmer's Azimuth 16mo, mor. 1 00 * Text-book on Practical Astronomy 8vo, 2 00 Merriman's Elements of Precise Surveying and Geodesy . .8vo, 2 50 * Michie and Harlow's Practical Astronomy 8vo, 3 00 Rust's Ex-meridian Altitude, Azimuth and Star-Finding Tables 8vo, 5 00 * White's Elements of Theoretical and Descriptive Astronorcy 12mo, 2 00 CHEMISTRY. * Abderhalden's Physiological Chemistry in Thirty Lectures. (Kail and Defren.) 8vo. 5 00 * Abegg's Theory of Electrolytic Dissociation, (von Ende.) 12mo, 1 25 Alexeyeff's General Principles of Organic Syntheses. (Matthews.) 8vo, 3 00 Allen's Tables for Iron Analysis 8vo, 3 00 Armsby's Principles of Animal Nutrition 8vo, 4 00 Arnold's Compendium of Chemistry. (Mandel.) Large 12mo, 3 50 3 Association of State and National Food and Dairy Departments, Hartford Meeting, 1906 8vo, $3 00 Jamestown Meeting, 1907 8vo, 3 00 Austen's Notes for Chemical Students 12mo, 150 Baskerville's Chemical Elements. (In Preparation.) Bernadou's Smokeless Powder. Nitro-cellulose, and Theory of the Cellulose Molecule 12mo, 2 50 * Biltz's Introduction to Inorganic Chemistry. (Hall and Phelan.). . . 12mo, 1 25 Laboratory Methods of Inorganic ChemisVT- (Hall and Blanchard.) 8vo, 3 00 Bingham and White's Laboratory Manual of Inorganic Chemistry. . 12mo. 1 00 Blanchard's Synthetic Inorganic Chemistry 12mo, 1 00 Browning's Introduction to the Rarer Elements 8vo, 1 50 Butler's Handbook of Blowpipe Analysis 16mo, 75 Claassen's Beet-sugar Manufacture. (Hall and Rolfe. > 8vo, 3 00 Classen's Quantitative Chemical Analysis by Electrolysis. (Boltwood.).Svo, 3 00 Cohn's Indicators and Test-papers 12mo, 2 00 Tests and Reagents 8vo, 3 00 ' * Danneel's Electrochemistry. (Merriam. ) 12mo 1 25 Dannerth's Methods of Textile Chemistry 12mo, 2 00 Duhem's Thermodynamics and Chemistry. (Burgess.) 8vo, 4 00 Effront's Enzymes and their Applications. (Prescott.) 8vo, 3 00 Eissler's Modern High Explosives 8vo. 4 00 * Fischer's Oedema 8vo, 2 00 * Physiology of Alimentation Large 12mo, 2 00 Fletcher's Practical Instructions in Quantitative Assaying with the Blowpipe. 16mo, mor. 1 50 Fowler's Sewage Works Analyses 12mo, 2 00 Fresenius's Manual of Qualitative Chemical Analysis. (Wells.) 8vo, 5 00 Manual of Qualitative Chemical Analysis. Part I. Descriptive. (Wells.)8vo, 3 00 Quantitative Chemical Analysis. (Cohn.) 2 vols 8vo, 12 50 When Sold Separately, Vol. I, $6. Vol. II, $8. ' Fuertes's Water and Public Health 12mo 1 50 Furman and Pardoe's Manual of Practical Assaying 8vo, 3 00 * Getman's Exercises in Physical Chemistry 12mo, 2 00 Gill's Gas and Fuel Analysis for Engineers 12mo, 1 25 * Gooch and Browning's Outlines of Qualitative Chemical Analysis. Large 12mo, 1 25 Grotenfelt's Principles of Modern Dairy Practice. (Woll.) 12mo, 2 00 Groth's Introduction to Chemical Crystallography (Marshall) 12mo, 1 25 * Hammarsten's Text-book of Physiological Chemistry. (Mandel.) 8vo. 4 00 Hanausek's Microscopy of Technical Products. (Winton.) 8vo, 5 00 * Haskins and Macleod's Organic Chemistry 12mo, 2 00 * Herrick's Denatured or Industrial Alcohol 8vo, 4 00 Hinds's Inorganic Chemistry 8vo. 3 00 * Laboratory Manual for Students 12mo, 1 00 * Holleman's Laboratory Manual of Organic Chemistry for Beginners. (Walker.) 12mo, 1 00 Text-book of Inorganic Chemistry. (Cooper. ) 8vo, 2 50 Text-book of Organic Chemistry. (Walker and Mott.) 8vo, 2 50 * Holley's Lead and Zinc Pigments Large 12mo, 3 00 Holley and Ladd's Analysis of Mixed Paints, Color Pigments, and Varnishes. Large 12mo, 2 50 Hopkins's Oil-chemists' Handbook 8vo, 3 00 Jackson's Directions for Laboratory Work in Physiological Chemistry. .8vo, 1 25 Johnson's Rapid Methods for the Chemical Analysis of Special Steels, Steel- making Alloys and Graphite Large 12mo, 3 00 Landauer's Spectrum Analysis. (Tingle.) 8vo, 3 00 Lassar-Cohn's Application of Some General Reactions to Investigations in Organic Chemistry. (Tingle.) 12mo. 1 00 Leach's Inspection and Analysis of Food with Special Reference to State Control 8vo, 7 50 Lob's Electrochemistry of Organic Compounds. (Lorenz.) 8vo, 3 00 Lodge's Notes on Assaying and Metallurgical Laboratory Experiments..8vo. 3 00 Low's Technical Method of Ore Analysis 8vo, 3 00 Lowe's Paint for Steel Structures 12mo, 1 00 Lunge's Techno-chemical Analysis. (Cohn.) 12mo, 1 00 4 * McKay and Larsen's Principles and Practice of Butter-making 8vo. $1 50 Maire's Modern Pigments and their Vehicles 12mo, 2 00 Mandel's Handbook for Bio-chemical Laboratory 12mo. 1 50 'Martin's Laboratory Guide to Qualitative Analysis with the Blowpipe 12mo. 60 Mason's Examination of Water. (Chemical and Bacteriological.) 12mo, 1 25 Water-supply. (Considered Principally from a Sanitary Standpoint.) 8vo, 4 00 * Mathewson's First Principles of Chemical Theory 8vo, 1 00 Matthews's Laboratory Manual of Dyeing and Textile Chemistry 8vo, 3 50 Textile Fibres. 2d Edition, Rewritten 8vo, 4 00 * Meyer's Determination of Radicles in Carbon Compounds. (Tingle.) Third Edition..., 12mo. 1 25 Miller's Cyanide Process 12mo, 1 00 Manual of Assaying 1 2mo, 1 00 Minet's Production of Aluminum and its Industrial Use. (Waldo.)... 12mo, 2 50 * Mittelstaedt's Technical Calculations for Sugar Works. (Bourbakis.) 12mo, 1 50 Mixter's Elementary Text-book of Chemistry 12mo. 1 50 Morgan's Elements of Physical Chemistry 12mo, 3 00 Outline of the Theory of Solutions and its Results 12mo, 1 00 * Physical Chemistry for Electrical Engineers 1 2mo, 1 50 * Moore's Outlines of Organic Chemistry 12mo, 1 50 Morse's Calculations used in Cane-sugar Factories 16mo, mor. 1 50 * Muir's History of Chemical Theories and Laws 8vo, 4 00 Mulliken's General Method for the Identification of Pure Organic Compounds. Vol. I. Compounds of Carbon with Hydrogen and Oxygen. Large 8vo, 5 00 Vol. II. Nitrogenous Compounds. (In Preparation.) Vol. III. The Commercial Dyestuffs Large 8vo. 5 00 * Nelson's Analysis of Drugs and Medicines 12mo, 3 00 Ostwald's Conversations on Chemistry. Part One. (Ramsey.) 12mo, 1 50 Part Two. (Turnbull.) 12mo, 2 00 * Introduction to Chemistry. (Hall and Williams.) Large 12mo, 1 50 Owen and Standage's Dyeing and Cleaning of Textile Fabrics 1 2mo, 2 00 * Palmer's Practical Test Book of Chemistry 12mo, 1 00 * Pauli's Physical Chemistry in the Service of Medicine. (Fischer.). . 12mo 1 25 Penfield's Tables of Minerals, Including the Use of Minerals and Statistics of Domestic Production 8vo, 1 00 Pictet's Alkaloids and their Chemical Constitution. (Biddle.) 8vo. 5 00 Poole's Calorific Power of Fuels 8vo, 3 00 Prescott and Winslow's Elements of Water Bacteriology, with Special Refer- ence to Sanitary Water Analysis 12mo, 1 60 * Reisig's Guide to Piece-Dyeing 8vo, 25 00 Richards and Woodman's Air, Water, and Food from a Sanitary Stand- point 8vo. 2 00 Ricketts and Miller's Notes on Assaying 8vo, 3 -00 Rideal's Disinfection and the Preservation of Food 8vo, 4 00 Sewage and the Bacterial Purification of Sewage 8vo, 4 00 Riggs's Elementary Manual for the Chemical Laboratory 8vo, 1 25 Robine and Lenglen's Cyanide Industry. (Le Clerc.) 8vo, 4 00 Ruddiman's Incompatibilities in Prescriptions 8vo, 2 00 Whys in Pharmacy 12mo, 1 OO * Ruer's Elements of Metallography. (Mathewson.) 8vo, 3 00 Sabin's Industrial and Artistic Technology of Paint and Varnish 8vo, 3 00 Salkowski's Physiological and Pathological Chemistry. (Orndorff.) 8vo, 2 50 Schimpf's Essentials of Volumetric Analysis 12mo, 1 25 Manual of Volumetric Analysis. (Fifth Edition, Rewritten) 8vo. 5 OO * Qualitative Chemical Analysis 8vo, 1 25 * Seamon's Manual for Assayers and Chemists Large 12mo 2 5O Smith's Lecture Notes on Chemistry for Dental Students 8vo, 2 5O Spencer's Handbook for Cane Sugar Manufacturers 16mo, mor. 3 00" Handbook for Chemists of Beet-sugar Houses 16mo, mor. 3 OO Stockbridge's Rocks and Soils 8vo, 2 5O Stone's'Practical Testing of Gas and Gas Meters 8vo, 3 5O * Tillman's Descriptive General Chemistry 8vo, 3 OO * Elementary Lessons in Heat 8vo, 1 50 Treadwell's Qualitative Analysis. (Hall.) 8vo, 3 00 Quantitative Analysis. (Hall.) 8vo. 4 00 5 Turneaure and Russell's Public Water-supplies 8vo, *o 00 Van Deventer's Physical Chemistry for Beginners. (Boltwood.i I2rn<>. 1 .".0 Venable's Methods and Devices for Bacterial Treatment of Sewage 8vo, 3 00 "Ward and Whipple's Freshwater Biology. (In Press.) Ware's Beet-sugar Manufacture and Refining. Vol. 1 8vo, 4 00 Vol. II 8vo. 5 no Washington's Manual of the Chemical Analysis of Rocks 8vo. 2 00 * Weaver's Military Explosives 8vo, 3 00 Wells's Laboratory Guide in Qualitative Chemical Analysis 8vo. 1 50 Short Course in Inorganic Qualitative Chemical Analysis for Engineering Students 12mo, 1 50 Text-book of Chemical Arithmetic 12mo, 1 25 Whipple's Microscopy of Drinking-water 8vo, 3 50 Wilson's Chlorination Process 12mo, 1 50 Cyanide Processes 12mo, 1 50 W. . . Large 12mo, 2 00 Ulke's Modern Electrolytic Copper Refining 8vo. 3 00 Waters's Commercial Dynamo Design. (In Press.) LAW. * Brennan's Hand-book of Useful Legal Information for Business Men. 16mp, mor. 5 00 * Davis's Elements of Law "... 8vo, 2 50 * Treatise on the Military Law of United States 8vo, 7 00 * Dudley's Military Law and the Procedure of Courts-martial. . Large 12mo, 2 50 Manual for Courts-martial 16mo, mor. 1 50 Wait's Engineering and Architectural Jurisprudence 8vo, 6 00 Sheep, 6 50 Wait's Law of Contracts 8vo, $.3 00 Law of Operations Preliminary to Construction in Engineering and Architecture Svo, 5 00 Sheep, 5 50 MATHEMATICS Baker's Elliptic Functions , Svo, 1 50 Briggs's Elements of Plane Analytic Geometry (Bocher.) 12mo, 1 00 * Buchanan's Plane and Spherical Trigonometry Svo, 1 00 Byerly's Harmonic Functions Svo I 00 Chandler's Elements of the Infinitesimal Calculus 12mo, 2 00 * Coffin's Vector Analysis 12mo, 2 50 Compton's Manual of Logarithmic Computations 12mo, 1 50 * Dickson's College Algebra Large 12mo, 1 50 * Introduction to the Theory of Algebraic Equations Large 12mo, 1 25 Emch's Introduction to Projective Geometry and its Application Svo, 2 50 Fiske's Functions of a Complex Variable Svo, 1 00 Halsted's Elementary Synthetic Geometry Svo. 1 50 Elements of Geometry Svo, 1 75 * Rational Geometry 12mo 1 50 Synthetic Projective Geometry Svo, 1 00 * Hancock's Lectures on the Theory of Elliptic Functions Svo, 5 00 Hyde's Grassmann's Space Analysis Svo, 1 00 * Johnson's (J. B.) Three-place Logarithmic Tables: Vest-pocket size, paper, 15 * 100 copies. 5 00 * Mounted on heavy cardboard. 8X10 inches, 25 * 10 copies. 3 00 Johnson's (W. W.) Abridged Editions of Differential and Integral Calculus. Large 12mo, 1 vol. 2 50 Curve Tracing in Cartesian Co-ordinates 12mo, 1 00 Differential Equations Svo, 1 00 Elementary Treatise on Differential Calculus Large 12mo, 1 50 Elementary Treatise on the Integral Calculus Large 12mo, 1 50 * Theoretical Mechanics 12mo, 3 00 Theory of Errors and the Method of Least Squares 12mo, 1 50 Treatise on Differential Calculus Large 12mo, 3 00 Treatise on the Integral Calculus Large 12mo, 3 00 Treatise on Ordinary and Partial Differential Equations. . .Large 12mo, 3 50 Karapetoff's Engineering Applications of Higher Mathematics. (In Preparation.) Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.) . 12mo, 2 00 * Ludlow's Logarithmic and Trigonometric Tables Svo, 1 00 * Ludlow and Bass's Elements of Trigonometry and Logarithmic and Other Tables Svo, 3 00 * Trigonometry and Tables published separately Each, 2 00 Macfarlane's Vector Analysis and Quaternions. . Svo, 1 00 McMahon's Hyperbolic Functions Svo, 1 00 Manning's Irrational Numbers and their Representation by Sequences and Series 12mo, 1 25 Mathematical Monographs. Edited by Mansfield Merriman and Robert S. Woodward Octavo, each 1 00 No. 1. History of Modern Mathematics, by David Eugene Smith. No. 2. Synthetic Projective Geometry, by George Bruce Halsted. No. 3. Determinants, by Laenas Gifford Weld. No. 4. Hyper- bolic Functions, by James McMahon. No. 5. Harmonic Func- tions, by William E. Byerly. No. 6. Grassmann's Space Analysis, by Edward W. Hyde. No. 7. Probability and Theory of Errors, by Robert S. Woodward. No. 8. Vector Analysis and Quaternions, by Alexander Macfarlane. No. 9. Differential Equations, by William "Woolsey Johnson. No. 10. The Solution of Equations, by Mansfield Merriman. No. 11. Functions of a Complex Variable, by Thomas S. Fiske. Maurer's Technical Mechanics Svo 4 00 Merriman's Method of Least Squares , Svo 2 00 Solution of Equations Svo 1 00 12 * Bartlett's Mechanical Drawing * " " Abridged Ed. . * Bartlett and Johnson's Engineering Descriptive Geometry. . * Moritz's Elements of Plane Trigonometry 8vo, $2 00 Rice and Johnson's Differential and Integral Calculus. 2 vols. in one. Large 12mo, 1 50 Elementary Treatise on the Differential Calculus Large 12mo, 3 00 Smith's History of Modern Mathematics 8vo, 1 00 * Veblen and Lennes's Introduction to the Real Infinitesimal Analysis of One Variable 8vo. 2 00 * Waterbury's Vest Pocket Hand-book of Mathematics for Engineers. 2JX5J inches, mor. 1 00 * Enlarged Edition, Including Tables mor. 1 50 Weld's Determinants 8vo, 1 00 Wood's Elements of Co-ordinate Geometry 8vo, 2 00 Woodward's Probability and Theory of Errors 8vo, 1 00 MECHANICAL ENGINEERING. MATERIALS OP ENGINEERING. STEAM-ENGINES AND BOILERS, Bacon's Forge Practice 12mo, 1 50 Baldwin's Steam Heating for Buildings 12mo, 2 50 irr's Kinematics of Machinery 8vo, 2 50 3 00 1 50 1 50 * Burr's Ancient and Modern Engineering and the Isthmian Canal. . .8vo, 3 50 Carpenter's Experimental Engineering 8vo. 6 00 Heating and Ventilating Buildings 80, 4 00 * Clerk's The Gas Petrol and Oil Engine 8vo, 4 00 Compton's First Lessons in Metal Working 12mo, 1 50 Compton and De Groodt's Speed Lathe 12mo, 1 50 Coolidge's Manual of Drawing 8vo, paper, 1 00 Coolidge and Freeman's Elements of General Drafting for Mechanical En- gineers Oblong 4to, 2 50 Cromwell's Treatise on Belts and Pulleys 12mo, 1 50 Treatise on Toothed Gearing 12mo, 1 50 Dingey's Machinery Pattern Making 12mo, 2 00 Durley's Kinematics of Machines 8vo, 4 00 Flanders's Gear-cutting Machinery Large 12mo, 3 00 Flather's Dynamometers and the Measurement of Power 12mo, 3 00 Rope Driving 12mo, 2 00 Gill's Gas and Fuel Analysis for Engineers 12mo. 1 25 Goss's Locomotive Sparks 8vo, 2 00 Greene's Pumping Machinery. (In Press.) Hering's Ready Reference Tables (Conversion Factors) 16mo, mor. 2 50 * Hobart and Ellis's High Speed Dynamo Electric Machinery 8vo. 6 00 Hutton's Gas Engine 8vo, 5 00 Jamison's Advanced Mechanical Drawing 8vo, 2 00 Elements of Mechanical Drawing 8vo, 2 50 Jones's Gas Engine 8vo. 4 00 Machine Design: Part I. Kinematics of Machinery 8vo, 1 50 Part II. Form, Strength, and Proportions of Parts 8vo, 3 00 Kaup's Text-book on Machine Shop Practice. (In Press.) * Kent's Mechanical Engineer's Pocket-Book 16mo, mor. 5 00 Kerr's Power and Power Transmission 8vo, 2 00 Kimball and Barr's Machine Design 8vo, 3 00 Leonard's Machine Shop Tools and Methods 8vo. * Levin's Gas Engine 8vo, * Lorenz's Modern Refrigerating Machinery. (Pope, Haven, and Dean). .8vo, MacCord's Kinematics; or Practical Mechanism 8vo, Mechanical Drawing 4to, Velocity Diagrams 8vo, MacFanand's Standard Reduction Factors for Gases. . 8vo, Mahan's Industrial Drawing. (.Thompson.) 8vo, Mehrtens's Gas Engine Theory and Design Large 12mo. $2 50 Oberg's Handbook of Small Tools Large 12mo. 2 50 * Parshall and Hobart's Electric Machine Design. Small 4to. half leather, 12 50 * Peele's Compressed Air Plant for Mines. Second Edition, Revised and En- larged 8vo, 3 50 Poole's Calorific Power of Fuels 8vo, 3 00 * Porter's Engineering Reminiscences, 1855 to 1882 8vo, 3 00 * Reid's Mechanical Drawing. (Elementary and Advanced.) 8vo, 2 00 Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 3 00 Richards's Compressed Air 12mo, 1 50 Robinson's Principles of Mechanism 8vo, 3 00 Schwamb and Merrill's Elements of Mechanism 8vo, 3 00 Smith (A. W.) and Marx's Machine Design 8vo, 3 00 Smith's (O.) Press-working of Metals 8vo, 3 00 Sorel's Carbureting and Combustion in Alcohol Engines. (Woodward and Preston.) Large 12mo, 3 00 Stone's Practical Testing of Gas and Gas Meters 8vo, 3 50 Thurston's Animal as a Machine and Prime Motor, and the Laws of Energetics. 12mo, 1 00 Treatise on Friction and Lost Work in Machinery and Mill Work. . .8vo, 3 00 * Tillson's Complete Automobile Instructor 16mo, 1 50 * Titsworth's Elements of Mechanical Drawing Oblong 8vo, 1 25 Warren's Elements of Machine Construction and Drawing 8vo, 7 50 * Waterbury's Vest Pocket Hand-book of Mathematics for Engineers. 2JX5| inches, mor. 100 * Enlarged Edition, Including Tables mor. 1 50 Weisbach's Kinematics and the Power of Transmission. (Herrmann Klein.) 8vo, 5 00 Machinery of Transmission and Governors. (Hermann Klein.). .8vo, 5 00 Wood's Turbines 8vo, 2 50 MATERIALS OF ENGINEERING. * Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 Burr's Elasticity and Resistance of the Materials of Engineering 8vo, 7 50 Church's Mechanics of Engineering 8vo, 6 00 * Greene's Structural Mechanics 8vo, 2 50 * Holley's Lead and Zinc Pigments Large 12mo 3 00 Holley and Ladd's Analysis of Mixed Paints, Color Pigments, and Varnishes. Large 12mo, 2 50 Johnson's (C. M.) Rapid Methods for the Chemical Analysis of Special Steels, Steel-Making Alloys and Graphite . . .'. Large 12mo, 3 00 Johnson's (J. B.)> Materials of Construction 8vo, 6 00 Keep's Cast Iron 8vo, 2 50 Lanza's Applied Mechanics 8vo, 7 50 Lowe's Paints for Steel Structures 12mo, 1 00 Maire's Modern Pigments and their Vehicles 12mo, 2 00 Maurer's Technical Mechanics 8vo, 4 00 Merriman's Mechanics of Materials 8vo, 5 00 * Strength of Materials 12mo, 1 00 Metcalf's Steel. A Manual for Steel-users 12mo, 2 00 Sabin's Industrial and Artistic Technology of Paint and Varnish 8vo, 3 00 Smith's (A. W.) Materials of Machines 12mo, 1 00 * Smith's (H. E.) Strength of Material 12mo, 1 25 Thurston's Materials of Engineering 3 vols., 8vo, 8 00 Part I. Non-metallic Materials of Engineering 8vo, 2 00 Part II. Iron and Steel 8vo, 3 50 Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their Constituents 8vo, 2 50 Wood's (De V.) Elements of Analytical Mechanics 8vo. 3 00 Treatise on the Resistance of Materials and an Appendix on the Preservation of Timber 8vo, 2 00 Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and Steel 8vo, 4 00 14 STEAM-ENGINES AND BOILERS. Berry's Temperature-entropy Diagram. Third Edition Revised and En- larged 12mo, $2 5O Carnot's Reflections on the Motive Power of Heat. (Thurston.) 12mo, 1 50 Chase's Art of Pattern Making 12mo, 2 50 Creighton's Steam-engine and other Heat Motors 8vo, 5 00 Dawson's "Engineering" and Electric Traction Pocket-book. .. . IBmo, mor. 5 00 * Gebhardt's Steam Power Plant Engineering 8vo, 6 00 Goss's Locomotive Performance 8vo, 5 00 Hemen way's Indicator Practice and Steam-engine Economy 12mo, 2 00 Hutton's Heat and Heat-engines 8vo, 5 00 Mechanical Engineering of Power Plants 8vo, 5 00 Kent's Steam Boiler Economy 8vo, 4 00 Kneass's Practice and Theory of the Injector 8vo, 1 50 MacCord's Slide-valves 8vo, 2 00 Meyer's Modern Locomotive Construction 4to, 10 00 Moyer's Steam Turbine 8vo, 4 00 Peabody's Manual of the Steam-engine Indicator 12mo, 1 50 Tables of the Properties of Steam and Other Vapors and Temperature- Entropy Table 8vo, 1 00 Thermodynamics of the Steam-engine and Other Heat-engines. . . . 8vo, 5 00 Valve-gears for Steam-engines 8vo, 2 50 Peabody and Miller's Steam-boilers 8vo, 4 00 Pupin's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors. (Osterberg.) 12mo, 1 25 Reagan's Locomotives: Simple, Compound, and Electric. New Edition. Large 12mo, 3 50 Sinclair's Locomotive Engine Running and Management 12mo, 2 00 Smart's Handbook of Engineering Laboratory Practice 12mo, 2 50 Snow's Steam-boiler Practice 8vo, 3 00 Spangler's Notes on Thermodynamics 12mo, 1 00 Valve-gears. . 8vo, 2 50 Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, 3 00 Thomas's Steam-turbines 8vo, 4 00 Thurston's Handbook of Engine and Boiler Trials, and the Use of the Indi- cator and the Prony Brake 8vo, 5 00 Handy Tables 8vo, 1 50 Manual of Steam-boilers, their Designs, Construction, and Operation 8vo, 5 00 Manual of the Steam-engine 2 vols., 8vo, 10 00 Part I. History, Structure, and Theory 8vo, 6 00 Part II. Design, Construction, and Operation. 8vo, 6 00 Wehrenfennig's Analysis and Softening of Boiler Feed-water. (Patterson.) 8vo, 4 00 Weisbach's Heat, Steam, and Steani-engines. (Du Bois.) 8vo, 5 00 Whitham's Steam-engine Design 8vo. 5 00 Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. . .8vo, 4 00 MECHANICS PURE AND APPLIED. Church's Mechanics of Engineering 8vo, 6 00 * Mechanics of Internal Works 8vo, 1 50 Notes and Examples in Mechanics 8vo, 2 00 Dana's Text-book of Elementary Mechanics for Colleges and Schools .12mo, 1 50 Du Bois's Elementary Principles of Mechanic:,: Vol. I. Kinematics 8vo, 350 Vol. II. Statics 8vo, 400 Mechanics of Engineering. Vol. I Small 4to, 7 50 Vol. II Small 4to, 10 00 * Greene's Structural Mechanics 8vo, 2 50 * Hartmann's Elementary Mechanics for Engineering Students 12mo, 1 25 James's Kinematics of a Point and the Rational Mechanics of a Particle. Large 12mo. 2 00 * Johnson's (W. W.) Theoretical Mechanics 12mo, 3 00 Lanza's Applied Mechanics 8vo, 7 50 15 * Martin's Text Book on Mechanics. Vol. I, Statics 12mo, $1 23 * Vol. II, Kinematics and Kinetics. 12mo, 1 50 Maurer's Technical Mechanics 8vo. 4 00 * Merriman's Elements of Mechanics 12mo, 1 00 Mechanics of Materials 8vo, 5 00 * Michie's Elements of Analytical Mechanics 8vo, 4 00 Robinson's Principles of Mechanism 8vo. 3 00 Sanborn's Mechanics Problems Large 12mo, 1 50 Schwamb and Merrill's Elements of Mechanism 8vo, 3 00 Wood's Elements of Analytical Mechanics 8vo, 3 00 Principles of Elementary Mechanics 12mo, 1 25 MEDICAL. * Abderhalden's Physiological Chemistry in Thirty Lectures. (Hall and Defren.) 8vo, 5 00 von Behring's Suppression of Tuberculosis. (Bolduan.) 12mo, 1 00 Bolduan's Immune Sera 12mo, 1 50 Bordet's Studies in Immunity. (Gay.) 8vo, 6 00 * Chapin's The Sources and Modes of Infection Large 12mo, 3 00 Davenport's Statistical Methods with Special Reference to Biological Varia- tions 16mo, mor. 1 50 Ehrlich's Collected Studies on Immunity. (Bolduan.) 8vo, 6 00 * Fischer's Oedema 8vo. 2 00 * Physiology of Alimentation Large 12mo, 2 00 de Fursac's Manual of Psychiatry. (Rosanoff and Collins.) Large 12mo, 2 50 Hammarsten's Text-book on Physiological Chemistry. (Mandel.) 8vo, 4 00 Jackson's Directions for Laboratory Work in Physiological Chemistry. .8vo, 1 25 Lassar-Cohn's Praxis of Urinary Analysis. (Lorenz.) 12mo, 1 00 Mandel's Hand-book for the Bio-Chemical Laboratory 12mo, 1 50 * Nelson's Analysis of Drugs and Medicines 12mo, 3 00 * Pauli's Physical Chemistry in the Service of Medicine. (Fischer.) . .12mo, 1 25 * Pozzi-Escot's Toxins and Venoms and their Antibodies. (Cohn.). . 12mo, 1 00 Rostoski's Serum Diagnosis. (Bolduan.) 12mo, 1 00 Ruddiman's Incompatibilities in Prescriptions 8vo, 2 00 Whys in Pharmacy 12mo, 1 00 Salkowski's Physiological and Pathological Chemistry. (Orndorff.) .. ..8vo, 2 50 * Satterlee's Outlines of Human Embryology 12mo, 1 25 Smith's Lecture Notes on Chemistry for Dental Students 8vo, 2 50 * Whipple's Tyhpoid Fever Large 12mo, 3 00 * Woodhull's Military Hygiene for Officers of the Line Large 12mo, 1 50 * Personal Hygiene 12mo, 1 00 Worcester and Atkinson's Small Hospitals Establishment and Maintenance, and Suggestions for Hospital Architecture, with Plans for a Small Hospital 12mo. 1 25 METALLURGY. Betts's Lead Refining by Electrolysis 8vo, 4 00 Bolland's Encyclopedia of Founding and Dictionary of Foundry Terms used in the Practice of Moulding 12mo, 3 00 Iron Founder 12mo, 2 50 " " Supplement 12mo, 2 50 * Borchers's Metallurgy. (Hall and Hayward.) 8vo, 3 00 Douglas's Untechnical Addresses on Technical Subjects 12mo, 1 00 Goesel's Minerals and Metals: A Reference Book 16mo, mor. 3 00 * Iles's Lead-smelting 12mo, 2 50 Johnson's Rapid Methods for the Chemical Analysis of Special Steels, Steel-making Alloys and Graphite Large 12mo, 3 00 Keep's Cast Iron 8vo, 2 50 Le Chatelier's High-temperature Measurements. (Boudouard Burgess.) 12mo, 3 00 Metcalf's Steel. A Manual for Steel-users 12mo, 2 00 Minet's Production of Aluminum and its Industrial Use. (Waldo.). . 12mo, 2 50 Price and Meade's Technical Analysis of Brass. (In Press.) * Ruer's Elements of Metallography. (Mathewson.) 8vo, 3 00 16 Smith's Materials of Machines 12mo, $1 00 Tate and Stone's Foundry Practice 12mo, 2 00 Thurston's Materials of Engineering. In Three Parts .8vo, 8 00 Part I. Non-metallic Materials of Engineering, see Civil Engineering, page 9. Part II. Iron and Steel . 8vo, 3 50 Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their Constituents 8vo, 2 50 Ulke's Modern Electrolytic Copper Refining 8vo 3 00 West's American Foundry Practice 12mo, 2 50 Moulders' Text Book 12mo. 2 50 MINERALOGY. Baskerville's Chemical Elements. (In Preparation.) * Browning's Introduction to the Rarer Elements 8vo, 1 50 Brush's Manual of Determinative Mineralogy. (Penfield.) 8vo, 4 00 Butler's Pocket Hand-book of Minerals I6mo, mor. 3 00 Chester's Catalogue of Minerals 8vo, paper, 1 00 Cloth, 1 25 * Crane's Gold and Silver ' 8vo, 5 00 Dana's First Appendix to Dana's New "System of Mineralogy". .Large 8vo, 1 00 Dana's Second Appendix to Dana's New " System of Mineralogy." Large 8vo, 1 50 Manual of Mineralogy and Petrography 12mo, 2 00 Minerals and How to Study Them. . 12mo, 1 50 System of Mineralogy Large 8vo, half leather, 12 50 Text-book of Mineralogy 8vo, 4 00 Douglas's Untechnical Addresses on Technical Subjects 12mo, 1 00 Eakle's Mineral Tables 8vo, 1 25 Eckel's Stone and Clay Products Used in Engineering. (In Preparation.) Goesel's Minerals and Metals: A Reference Book 16mo, mor. 3 00 * Groth's The Optical Properties of Crystals. (Jackson.) 8vo, 3 60 Groth's Introduction to Chemical Crystallography (Marshall). ...... .12mo, 1 25 * Hayes's Handbook for Field Geologists 16mo, mor. 1 50 Iddings's Igneous Rocks 8vo, 5 00 Rock Minerals 8vo, 5 00 Johannsen's Determination of Rock-forming Minerals in Thin Sections. 8vo, With Thumb Index 5 00 * Martin's Laboratory Guide to' Qualitative Analysis with the Blow- pipe 12mo, 60 Merrill's Non-metallic Minerals: Their Occurrence and Uses 8vo. 4 00 Stones for Building and Decoration 8vo, 6 00 * Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests. 8vo, paper, 50 Tables of Minerals, Including the Use of Minerals and Statistics of Domestic Production 8vo, 1 00 * Pirsson's Rocks and Rock Minerals 12mo, 2 50 * Richards's Synopsis of Mineral Characters 12mo, mor. 1 25 * Ries's Clays: Their Occurrence, Properties and Uses 8vo, 5 00 * Ries and Leighton's History of the Clay-working Industry of the United States 8vo. 2 50 * Rowe's Practical Mineralogy Simplified 12mo, 1 25 * Tillman's Text-book of Important Minerals and Rocks 8vo, 2 00 Washington's Manual of the Chemical Analysis of Rocks 8vo. 2 00 MINING. * Beard's Mine Gases and Explosions Large 12mo. 3 00 * Crane's Gold and Silver 8vo, 5 00 * Index of Mining Engineering Literature 8vo, 4 00 * 8vo, mor. 5 00 * Ore Mining Methods 8vo, 3 00 Dana and Saunders's Rock Drilling. (In Press.) Douglas's Untechnical Addresses on Technical Subjects 1 2mo, 1 09 Eissler's Modern High Explosives 8vo. 4 00 17 Goesel's Minerals and Metals: A Reference Book 16mo, mor. $3 00 Ihlseng's Manual of Mining 8vo, 5 00 * Iles's Lead Smelting 12mo. 2 00 Peele's Compressed Air Plant for Mines 8vo, 3 00 Riemer's Shaft Sinking Under Difficult Conditions. (Corning and Peele.)8vo, 3 00 * Weaver's Military Explosive^ 8vo, 3 00 Wilson's Hydraulic and Placer Mining. 2d edition, rewritten 12mo, 2 60 Treatise on Practical and Theoretical Mine Ventilation 12mo, 1 25 SANITARY SCIENCE. Association of State and National Food and Dairy Departments, Hartford Meeting, 1906 8vo, 3 00 Jamestown Meeting, 1907 8vo, 3 00 * Bashore's Outlines of Practical Sanitation 12mo, 1 25 Sanitation of a Country House 12mo, 1 00 Sanitation of Recreation Camps and Parks 12mo, 1 00 * Chapin's The Sources and Modes of Infection Large 12mo, 3 00 Folwell's Sewerage. (Designing, Construction, and Maintenance.) 8vo, 3 00 Water-supply Engineering 8vo, 4 00 Fowler's Sewage Works Analyses 12mo, 2 00 Fuertes's Water-filtration Works 12mo, 2 50 Water and Public Health 12mo, 1 50 Gerhard's Guide to Sanitary Inspections 12mo, 1 50 * Modern Baths and Bath Houses 8vo, 3 00 Sanitation of Public Buildings 12mo, 1 50 * The Water Supply, Sewerage, and Plumbing of Modern City Buildings. 8vo, 4 00 Hazen's Clean Water and How to Get It Large 12mo, 1 50 Filtration of Public Water-supplies 8vo. 3 00 * Kinnicutt, Winslow and Pratt's Sewage Disposal 8vo, 3 00 Leach's Inspection and Analysis of Food with Special Reference to State Control 8vo, 7 50 Mason's Examination of Water. (Chemical and Bacteriological) 12mo, 1 25 Water-supply. (Considered principally from a Sanitary Standpoint). 8vo, 4 00 * Mast's Light and the Behavior of Organisms Large 12mo, 2 50 * Merriman's Elements of Sanitary Engineering. . . 8vo, 2 00 Ogden's Sewer Construction 8vo, 3 00 Sewer Design 12mo, 2 00 Parsons's Disposal of Municipal Refuse 8vo, 2 00 Prescott and Winslow's Elements of Water Bacteriology, with Special Refer- ence to Sanitary Water Analysis 12mo, 1 50 * Price's Handbook on Sanitation 12mo, 1 50 Richards's Conservation by Sanitation 8vo, 2 50 Cost of Cleanness 12mo, 1 00 Cost of Food. A Study in Dietaries 12mo, 1 00 Cost of Living as Modified by Sanitary Science 12mo, 1 00 Cost of Shelter 12mo. 1 00 * Richards and Williams's Dietary Computer 8vo, 1 50 Richards and Woodman's Air, Water, and Food from a Sanitary Stand- point 8vo, 2 00 * Richey's Plumbers', Steam-fitters', and Tinners' Edition (Building Mechanics' Ready Reference Series) 16mo. mor. 1 50 Rideal's Disinfection and the Preservation of Food 8vo, 4 00 Sewage and Bacterial Purification of Sewage 8vo, 4 00 Soper's Air and Ventilation of Subways 12mo, 2 50 Turneaure and Russell's Public Water-supplies 8vo, 5 00 Venable's Garbage Crematories in America 8vo, 2 00 Method and Devices for Bacterial Treatment of Sewage 8yo, 3 00 Ward and Whipple's Freshwater Biology. (In Press.) Whipple's Microscopy of Drinking-water 8vo, 3 50 * Typhoid Fever Large 12mo, 3 00 Value of Pure Water Large 12mo, 1 00 Winslow's Systematic Relationship of the Coccaceae Large 12mo. 2 50 18 MISCELLANEOUS. * Chapin's How to Enamel 12mo. $1 00 Emmons's Geological Guide-book of the Rocky Mountain Excursion of the International Congress of Geologists Large 8vo 1 50 Fen-el's Pooular Treatise on the Winds 8vo, 4 00 Fitzgerald's Boston Machinist 18mo, 1 00 Gannett's Statistical Abstract of the World 24mo, 75 Haines's American Railway Management 12mo, 2 50 Hanausek's The Microscopy of Technical Products. (Winton) ....... .8vo, 5 00 Jacobs's Betterment Briefs. A Collection of Published Papers on Or- ganized Industrial Efficiency . . .8vo. 3 50 Metcalfe's Cost of Manufactures, and the Administration of Workshops.. 8vo, 5 00 Putnam's Nautical Chartsl 8vo, 2 00 Ricketts's History of Rensselaer Polytechnic Institute 1824-1894. Large 12mo, 3 00 Rotherham's Emphasised New Testament Large 8vo, 2 00 Rust's Ex-Meridian Altitude, Azimuth and Star-finding Tables 8vo 5 00 Standage's Decoration of Wood, Glass, Metal, etc 12mo 2 00 Thome's Structural and Physiological Botany. (Bennett) 16mo, 2 25 Westermaier's Compendium of General Botany. (Schneider) 8vo, 2 00 Winslow's Elements of Applied Microscopy 12mo, 1 50 HEBREW AND CHALDEE TEXT-BOOKS. Gesenius's Hebrew and Chaldee Lexicon to the Old Testament Scriptures. (Tregelles.) Small 4to, half mor. ."> 00 Green's Elementary Hebrew Grammar 12mo, 1 25 19 u/ / -/ dk> ' r ^*, = 0^ u =o (H/U> - M ~Vv'