MECHANICS OF MATERIALS BY MANSFIELD MERRIMAN MEMBER OF AMERICAN SOCIETY OF CIVIL ENGINEERS ELEVENTH EDITION TOTAL ISSUE, FIFTY-EIGHT THOUSAND NEW YORK JOHN WILEY & SONS, INC. LONDON: CHAPMAN & HALL, LIMITED /V\b- COPYRIGHT, 1885, 1890, 1895, 1905 BY 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, 190$ Tenth Edition, rewritten and enlarged, September, 1905; reprinted, 1906, 1907, 1908, 1909, 1910, 1911, 1912 Eleventh Edition, March, 1914; reprinted, 1915, 1916 ALL RIGHTS RESERVED 4 PRESS OF r /2I BRAUNWORTH & CO. BOOK MANUFACTURERS BROOKLYN. N. Y. PREFACE TO TENTH EDITION 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 456298 Ui 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 has 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 In this eleventh edition the subject of testing of materials is treated in the new Chapter XX. New matter on influence lines, curved beams, springs, and moment of inertia is given in Chapter XVIII. A few minor changes have been made in other pages, and all known typographic errors have been corrected. M. M. In this impression of the eleventh edition Arts. 86 and 160 are rewritten, and a few minor changes are made here and there. M. M. CONTENTS CHAPTER I ELASTIC AND ULTIMATE STRENGTH PAGE ART. 1. SIMPLE AXIAL STRESSES i 2. THE ELASTIC LIMIT 4 3. ULTIMATE STRENGTH. 6 4. TENSION 9 5. COMPRESSION n 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 III 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 v 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 A.RT. 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 vii CHAPTER VII OVERHANGING AND FIXED BEAMS PAGB ART. 59. BEAM OVERHANGING ONE SUPPORT 149 60. 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 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 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 vii CHAPTER VII OVERHANGING AND FIXED BEAMS PAGE AKT. 59. BEAM OVERHANGING ONE SUPPORT 149 60. BEAM FIXED AT ONE END 152 61. BEAM OVERHANGING BOTH SUPPORTS 154 62. BEAM FIXED AT BOTH ENDS 156 63. COMPARISON or BEAMS 158 64. SUPPORTS ON DIFFERENT LEVELS 159 65. CANTILEVER WITH CONSTRAINT 163 66. SPECIAL DISCUSSIONS 164 CHAPTER VIII CONTINUOUS BEAMS A.RT. 67. GENERAL PRINCIPLES 168 68. METHOD OF DISCUSSION 171 69. THEOREM or 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 viii CONTENTS CHAPTER X TORSION OF SHAFTS TJam ART. 89. PHENOMENA OF TORSION 225 90. THE TORSION FORMULA 227 91. SHAFTS FOR TRANSMITTING POWER 250 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 285 114. THEORY OF REINFORCED CONCRETE BEAMS 289 115. INVESTIGATION OF REINFORCED CONCRETE BEAMS . . . .292 116. DESIGN OF REINFORCED CONCRETE BEAMS 295 117. PLATE GIRDERS 298 118. DEFLECTION OF COMPOUND BEAMS 300 CONTENTS ix CHAPTER XIII RESILIENCE AND WORK PASH A*T. 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 330 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 * CONTENTS CHAPTER XVI GUNS AND THICK CYLINDERS PAOB A*T. 148. PRINCIPLES AND METHODS .... 383 149. LAMP'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 3 9 u 155. DESIGN OF HOOPED GUNS 399 CHAPTER XVII ROLLERS, PLATES, SPHERES ART. 156. CYLINDRICAL ROLLERS 403 157. SPHERICAL ROLLERS 406 158. CONTACT or 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. INFLUENCE LINES 431 168. CURVED BEAMS 433 169. PRODUCT OF INERTIA 437 170. MOMENT OF INERTIA 438 171. SPRINGS 442 CHAPTER XIX MATHEMATICAL THEORY OF ELASTICITY ART. 172. INTRODUCTION 447 173. ELASTIC CHANGES IN VOLUME . . , 448 174. NORMAL AND TANGENTIAL STRESSES ........ 450 175. RESULTANT STRESSES ...... 452 176. THE ELLIPSOID OF STRESS ........... 454 177. THE THREE PRINCIPAL STRESSES . 455 178. MAXIMUM SHEARING STRESSES .... 457 179. DISCUSSION OF A CRANK PIN . . , . , . . . 459 CONTENTS xi PAGE 180. THE ELLIPSE OF STRESS 461 181. SHEARING MODULUS OF ELASTICITY. 463 182. THE VOLUMETRIC MODULUS ...... 465 183. STORED INTERNAL ENERGY 467 CHAPTER XX TESTING OF MATERIALS ART. 184. TESTING MACHINES 470 185. TEST SPECIMENS 474 186. TENSILE TESTS 476 187. COMPRESSIVE TESTS 478 188. MISCELLANEOUS TESTS 479 189. SPECIFICATIONS FOR STRUCTURAL STEEL 482 190. UNIFORMITY IN TESTING 486 APPENDIX ART. 191. VELOCITY OF STRESS 488 192. ELASTIC-ELECTRIC ANALOGIES 490 193. MISCELLANEOUS PROBLEMS 491 194. ANSWERS TO PROBLEMS ...... 493 196. EXPLANATION OF TABLES 494 TABLES TABLE 1. AVERAGE WEIGHT AND EXPANSIBILITY 496 2. AVERAGE ELASTIC PROPERTIES 496 3; AVERAGE TENSILE AND COMPRESSIVE STRENGTH . . . .497 4. AVERAGE SHEARING AND FLEXURAL STRENGTH 497 5. WORKING UNIT-STRESSES FOR BUILDINGS 498 6. STEEL I-BEAM SECTIONS 499 7. STEEL BULB-BEAM SECTIONS 500 8. STEEL T SECTIONS 500 9. STEEL CHANNEL SECTIONS . 501 10. STEEL ANGLE SECTIONS 502 11. STEEL Z SECTIONS 503 12. COMPARISON OF BEAMS 503 13. GERMAN I BEAMS 504 14. WEIGHT OF WROUGHT-!RON BARS 505 15. SQUARES OF NUMBERS 506 16. AREAS OF CIRCLES 508 17. TRIGONOMETRIC FUNCTIONS 510 18. LOGARITHMS OF TRIGONOMETRIC FUNCTIONS 511 19. LOGARITHMS OF NUMBERS . . . . . . . . . . .512 20. CONSTANTS AND THEIR LOGARITHMS .... ... 514 INDEX . . o 515 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 ij 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 4j 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 these 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. im 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. im Fig. Ic Fig. Id The case of compression is shown in Fig. Ic, 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. . 1 SIMPLE AXIAL STRESSES 3 Let S 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= 16 500/900 = 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 S becomes about 20 ooo pounds per square inch ; if the bar is inches in cross-section, its section area is if square 4 ELASTIC AND ULTIMATE STRENGTH CHAP. I inches, and the tensile force required to cause rupture is P = if 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 if 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 10 ooo 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 denned 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. Instead of Elastic Limit the terms ' Pro- portional Elastic Limit ' and ' Proportional Limit ' are often used. 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 loif inches; then the permanent set of the bar is ij 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 defined, 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. 20. Find the diameter of a round rod of wrought iron, which is to be under a tension of 64 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 1 2 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 cast 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=1 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 forces are steady or variable. The 8 ELASTIC AND ULTIMATE STRENGTH CHA>. j 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. 30. A bar of structural steel, 2j 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 2jX if 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 0.10 0.18 0.04 0.06 0.08 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 i-J$-0- 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 A.RT. 5 COMPRESSION 11 are 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. 185& 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 7800 pounds and ruptured under a load of 14 800 pounds. Compute the elastic limit and the ultimate strengths of the steel. Prob. 46. 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. 5b. 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. Fig. 5a Fig. 5b Fig. 5$ Fig. 5d When the length of the bar does not exceed about ten times its least lateral dimension, rupture sometimes occurs by an obliqu.3 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. 185c 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 COMPKESSION 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 230000 pounds the unit-stress is 230000/96 = 2400 pounds per square inch, and the factor of safety is 6 000/2 400 = 2j; this is not sufficiently high for stone, as will be seen in Art. 7. Prob. 5#. A solid cast-iron cylinder of 4 inches diameter is under a compression of 600 ooo pounds. Compute its factor of safety. Prob. 5b. 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 l 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. of 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. Qb 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. IT) / * b Fig. Qc Fig. Qd Wooden specimens for tensile tests, like that shown in Fig. 6c, 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 ij 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=aS=s. 142X1. 5X6X500=14 100 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 6 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 P%, 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/smd. Hence the shearing unit-stress along the oblique section is, Si=Pi/ai = (P/a) sinfl cos6 = $(P/a) $11126 When #=o or when 6= 90, the value of S\ is zero, that is there is no shear on a plane parallel with or normal to the axis. For all other values of 0, a shearing stress exists; and the maximum value of Si occurs when #=45, and then S\=\P/a. There- fore a normal tensile unit-stress S on a bar produces a shearing unit-stress JS 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. Ga. A wrought-iron bolt ij inches in diameter has a head ij inches long, and a tension of 38000 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- iftg 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 20 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 stabilitv 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 75000 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 0.7854X202x80 = 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. 7a. 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. Qb 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 ooo 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^quare 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 A.RT. 8 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, /? Beta, e Epsilon, i) Eta, 6 Theta, K Kappa, A Lambda, /* Mu, v Nu, TT Pi, p Rho, o Sigma, r Tau, ^ Phi, ([> Psi, CD 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 al l 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) sin2#, deduced in Art. 6, is correct in dimensions, for both Si 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 +pxP+qx-t-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 e. 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 s = 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 denned 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, 77 s v p / a E = or E = ~TT ( 9 e/l Since e and / are linear quantities, e 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 may be expressed by S/E, and hence e is the least for a given S 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. 9a. 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 Jong, elongates 0.013 inches when the tension is increased from 800 to \ 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 t and that the total deforma- tion in length is -i' - -<-5 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 / 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 S 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 = 1/2 500 = 0.0004 For wrought iron, e/l= 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, i J X 1 2 inches in section area and 23 feet long, when under a tension of 270 ooo 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 0=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//= 1500/1 500000 = 1/1000, so that if a stick of timber elongates 0.48 inches, its length is 1 = 0.48X1 000 = 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, if inches in diameter and 16 feet long, under tensions of 10 ooo, 20 ooo, and 30 ooo pounds. Prob. lOb. Compute the tensile force required to stretch a bar of structural steel, ifX9f inches in section area and 23 feet 3^ inches long, 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 denned 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. \\a 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 S~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 S 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 vvhich this line makes with the horizontal. 60000 o.ooi 0.002 Unit-Elongation = 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. lib shows a part of the tensile diagram for wrought iron in which AB repre- sents a bar of unit length which 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 f 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 AN YIELD POINT 29 In other words the elastic limits in tension for timber and cast iron are not well defined. On the compression diagram it is also seen that the elastic limits for wrought iron and structural steel are not so sharply defined 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. 40000 Fig. lib 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 ooo 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 ste:l, i square inch in section area and 100 inches long, are subject to a tension of 15000 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. 185& are shown three steel specimens, that at the top being one which has not been tested. Its total length is 5^ inches and about i\ 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. AKT. 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. 185& 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, n 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. lib let the unit-stress D'd be represented by S\ then, since dD is parallel to bB, the tangent of the angle dDD' is the modulus E, 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 240 X 45 000/30 ooo ooo = 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 Aed, where ^ is a number which has been found to be about i for cast iron and about J for steel. For example, the value of e for wrought iron stressed ART. is 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.ooi/, while the change in its diameter is JX o.ooi d. Let h be any lateral dimension of length /, and let h' and /' be the lateral dimension and length under a. stress which does not exceed the elastic limit of the material; then for tension, /' = (i + e)/ and h f =(i-Xe)h (13) and similarly for the case of compression, /'=(i- e )7 and h' = (i + h)h (13)' in which the linear unit-deformation e may be computed from e=S/E (Art. 9). The number e is very small compared with unity, and hence its square and higher powers may be neglected when they occur in algebraic work. Fig. 130 Fig. 136 Let a bar of rectangular section have the breadth b and tht depth d\ the area of the section is bd. Under elastic tension, b becomes (i Xe)b and d becomes (iAe)d t so that the area of the section is (i 2Xi)bd, and hence the decrease in area is 2\ebd, 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 J, the decrease in unit-area is 0.0006, so that the area of the section under this stress is 160.0006X16 = 15.99 square inches. Fig. 130 illus- trates the change in shape for this case. The elastic change in volume of a rectangular bar under tension is found in a similar manner. The length / increases to (i + e)/, the breadth b decreases to (i Ae)b, the depth d decreases to (iAe)d, so that the original volume Ibd becomes (i + e)/. (i-Ae)b. (i-As)d=(i+e-2Jie)lbd This expression shows that the volume of a bar under tension 34 ELASTIC AND ULTIMATE DEFORMATION CHAP. II is increased when A is less than J and this is the case for all ma- terials used in construction. The increase in unit-volume is (i 2^)e, which becomes %s for cast iron and j-s for wrought iron and steel; that is, the increase per unit of volume is one half or one-third of the elongation per unit of length. In neglecting the squares and higher powers of e no error appreciable in practice has been committed, as the student may easily see by numerical instances. Thus, the square of (i +0.001) is 1.002 by the approximate method, while its exact value is I.C0200I. If v is a small number compared with unity, then (i + y) 2 = i + 2v and (i + y)* = i + Jy; also i/(i + y)=i-y, and i/(i y)=i + y. Again, if y and /* be small compared with unity, then (i + y)(i -\-JJL) =i + y + /f and (i + y)(i ju) =i + y-^; also (i + y) 2 (i 2//) 2 = i + 2y 4/d. This principle of neglecting squares and cubes is often of great value in approximate numer- ical computations; thus (1.02)2 = 1.04, and o.994 2 = (i o.oo6) 2 = i 0.012 =0.988; also (0.994)* = (i 0.006) * = i -0.003 ='997' In a similar manner it is readily shown that a bar of length /, breadth b, and depth d under compression in the direction of its length, has the area of its section increased by 2lebd, and its volume diminished by (e2Xe)lbd, when e is the unit-shortening of length under a stress not exceeding the elastic limit of the material; Fig. 13& illustrates this case. The number A is called the ' Factor of Lateral Contraction ' and sometimes also ' Poisson's Ratio '. It is of great theoretic importance in many discussions of the mechanics "of materials, and of great practical value in the manufacture of guns. Prob. 130. Multiply together the numbers 0.989, 1.012, 1.005 by the above approximate method, and find the percentage of error in the result. Prob. 136. A bar of structural steel, 2j inches in diameter and 18 feet 6 inches long, is under a tension of 64 ooo pounds. Compute the changes in length, section area, and volume. Prob. 13c. A bar of wrought iron is one square centimeter ID section area and one meter long. Compute its length and section area when stressed to its tensile elastic limit. ART. 14 WORK IN PRODUCING DEFORMATION 35 ART. 14. WORK IN PRODUCING DEFORMATION When a bar is ta be put under tensile stress, the load is a'pplied by weights added in succession, or by means of a pull exerted by a machine. In both cases the tension increases from o up to P, and the elongation of the bar increases from o up to e. When the elastic limit is not exceeded, the elongations are pro- portional to the applied forces, and the relation between them may be represented by a straight line as in Fig. I4a. The work done in elongating the bar is then the product of the mean force \P into the distance e^ or \Pe, and this is represented in the figure by the area of the shaded triangle. Fig. Ua Fig. 146 Fig. 14c Let a be the section area of the bar and / its length, S the tensile unit-stress and e the unit-elongation; then P=aS, and e = /, and inserting these in \Pe it becomes \Sz . al, which is the work done in elongating the bar up to the unit-stress 5. Since al is the volume of the bar, JSs is the work done per unit of volume; in Fig. 146 this work is represented by the area of the shaded triangle. The above investigation applies equally well to the case of compression. Hence, in any case of direct tension or compres- sion, the work K of elastic deformation is, K=\Pe of K=%Ss.al (14: both of which apply when the unit-stress S is not greater than the elastic limit of the material. This work is stored in the bar in the form of potential stress energy. As an example, let it be required to find what horse-powei engine is required to produce, i 200 times per minute, a tension of 56 ooo pounds in a steel rod which is 3 inches in diameter and 68 inches long. From Table 16 the section area is 7.07 36 ELASTIC AND ULTIMATE DEFORMATION CHAP. II square inches; from Art. 1 the tensile unit-stress is 7 920 pounds per square inch; from Art. 10 the unit-elongation e is o.ooo 264. Then from (14) the work done in one application of the load is JX 7920X0.000264 = 1.045 inch-pounds for each cubic inch of the rod, and the total work is therefore 1.045X7.07X68=502 inch-pounds. The horse-power required to do this work is then 502X1200/12X33 000 = 1.53. When the unit-stress exceeds the elastic limit of the material, the above formulas are not valid. In such cases, however, the work done per unit of volume is given by the shaded area of the stress diagram in Fig. 14c. This area may be approximately ascertained by dividing it into trapezoids and computing the separate areas. The work done in stressing a steel bar up to its ultimate strength is large compared with that required to stress it up to its elastic limit, being sometimes more than five hundred times as large. In order to show this fact, the particular case of a steel speci- men 4 inches long and 0.505 inches in diameter will be taken. Load Pounds Remarks Stress Pounds per Square inch Elongation Percents. Partial Work Inch-pounds per Cubic Inch Total Work Inch-pounds per Cubic Inch 200 Initial Load I 000 0.00 I 000 5 ooo O.OI '3 o 3.0 3 ooo 15 ooo 0.04 6.0 3 5 ooo 25 ooo 0.07 9 7 ooo 35000 0. 10 9.0 16.0 18 9 ooo 45000 0.14 34 9 600 Elastic Limit 48 ooo o. 16 9-3 264.6 44 10 ooo 50 ooo o. 70 660 o 308 12 000 60 ooo 1.90 968 14 ooo 1 6 ooo 70 ooo 80 ooo 3.62 8.50 I I 2O . O 3 660 . o 2 088 5748 16800 15 ooo Ultimate Strength Rupture 84 ooo 75000 15.20 24.50 5 494.0 7383-0 II 242 18625 The specimen was placed in a testing machine and the elonga- tions measured corresponding to the loads in the first column. From these elongations e the unit-elongations e/l were com- ART. 15 SHEARING MODULUS OF ELASTICITY 37 puted and the values of iooe/l are the percentages of elongation in the fourth column, while the loads divided by the section areas give the unit-stresses in the third column. The elastic limit was observed at 48000 pounds per square inch with 0.16 percent elongation. The elongation then rapidly increased with the load, as seen in the diagram of Fig. 4. At 84 ooo pounds per square inch the maximum tensile strength was reached and the specimen was elongating rapidly. The load was then slowly slackened, but the elongation continued to increase very fast until rupture occurred at 75 ooo pounds per square inch. The work done per cubic inch of material may be approximately com- puted for any interval by multiplying the average unit-stress during that interval by the unit-elongation which occurs. Thus while the load ranged from 7 ooo to 9 ooo pounds the average unit-stress was 40000 pounds per square inch, and the unit- elongation was 0.0014 0.0010=0.0004; hence the work per cubic inch of material was 40 000X0.0004 = 16 inch-pounds. Thus the quantities in the fifth column of the table are computed and the summation of these gives those in the last column. This example shows that the work done on the specimen in stressing it up to the elastic limit was 44 inch-pounds, while the total work required to rupture it was about 18 600 inch-pounds, both being per cubic inch of material. The volume of the speci- men being 0.8 cubic inches, the approximate work required to rupture it was 14 900 inch-pounds or i 240 foot-pounds. Prob. 14a. Compute the work per cubic inch which is done in stressing cast iron, wrought iron, and structural steel up to their elastic limits. Prob. 14b. 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 pounds per square inch up to 12 ooo pounds per square inch ? ART. 15. SHEARING MODULUS OF ELASTICITY Let a body of constant cross-section and length / be subject to shearing as shown in Fig. 15, the length / being short, so that the rectangle A BCD is deformed into the rhombus ABC'D'. 38 ELASTIC AND ULTIMATE DEFORMATION CHAP. II This deformation is caused by the force P acting at C normal to BC, while the body is firmly held at AB so that the points A and B remain immovable. Along AB there then acts a force equal to that at C but in the opposite direction, and the body is under shear from the action of these two forces (Arts. 1 and 6). The defor- mation C(7, generally called the "detrusion", may be represented by e, and the detrusion per unit of length is then e/l. In tension and compression the deformation e is in the same direc- tion as /, but in shear it is normal to /. Let a be the section area of the short bar; then the shearing unit-stress on it is S s =P/a. In tension and compression this stress acts normal to the section, but here it is parallel to the section. If the elastic limit of the material is not exceeded by the unit-stresses, the unit-detrusions are proportional to them, so that the ratio of P/a to e/l is a constant, and this constant is called the 'Modulus of Elasticity for Shear'. Designating it by F, there results, P -*l or *=-. (15) e/l e where e is the unit-detrusion e/l. These formulas are the same as those of Art. 9, but the above remarks show that the mean- ings of the letters e and e are somewhat different in the two cases. The following are approximate average values of the shear- ing modulus of elasticity as ascertained by experiments and computations which are explained in Arts. 93 and 181 : Material Shearing Modulus of Elasticity Timber, across grain 400 ooo pounds per square inch Cast Iron 6 ooo ooo pounds per square inch Wrought Iron 10 ooo ooo pounds per square inch Steel 12 ooo ooo pounds per square inch For brick, concrete, and stone there is but little known regarding the shearing modulus of elasticity. ART. 16 HISTORICAL NOTES 39 By the help of the above formulas, computations regarding elastic deformations in shear may be made in the same manner as those for tension and compression in Art. 10. For example, take a short wooden block 6 inches long and 3X4 inches in sec- tion subject to a shear of 24 ooo pounds. The shearing unit- stress is 24000/12=2000 pounds per square inch, which is probably a little less than the elastic limit for shear (Art. 2) and hence the formula applies. The unit-detrusion then is e = 2 000/400 000=0.005, and the total detrusion or lateral move- ment of one end of .the block with respect to the other is e = 0.005 X 6 = 0.03 inches. The work done in deforming the prism of Fig. 15 is %Pe, when the load is applied by increments so that the shearing force increases from o up to P, and when the elastic limit is not exceeded by the unit-stress S. This work of elastic detru- sion may be written J5e . a/, which is identical with the expres- sion deduced for tension or compression in the last article. The above formulas cannot be used when the unit-stress S exceeds the elastic limit for shear, which is usually about 75 or 80 percent of that for tension. They cannot be used when the length / is large compared with the lateral dimensions of the body, for then a bending occurs so that EC' is not a straight line and in this case e/l cannot be a constant for variable values^ of e and /. They are of principal value in discussing the twisting of shafts (Art. 93). Prob. 15. A cast-iron beam, 2 inches square, has two loads of 8 ooo pounds near the supports as shown in Fig. 6b. The beam being originally horizontal, compute the inclination between a support and a load after the loads are placed upon it. ART. 16. HISTORICAL NOTES From the earliest times a few fundamental facts regarding the strength of materials were undoubtedly known by experi- ence, such, for instance, that stone was stronger than wood. No quantitative results appear, however, to have been ascertained and recorded until about the middle of the eighteenth century. 40 ELASTIC AND ULTIMATE DEFORMATION CHAP. II The first investigation in the mechanics of materials seems to be that made by Galileo in 1638 on the flexure of a beam, in which he regarded all the fibers as inextensible. It was not until more than a century afterwards that tests of the strength of bars were made, and these related almost entirely to its ulti- mate strength, the elastic limit being unknown and unrecognized. In 1678 Hooke announced the theory of "springy bodies" which he expressed by "ut tensio sic vis", namely, elongation is proportional to force; three years previously he had performed an experiment before the king of England which illustrated this theory, and in 1676 he had announced it concealed in the ana- gram "ceiiinosssttuv." Newton in 1704 conjectured that the return of the body t6 its original form, after the removal of the force, was caused by the mutual attraction of its particles. Early in the nineteenth century appeared the Lectures on Natural Philosophy by Young in which the modulus of elasticity is for the first time introduced, but no note is made that it can be deduced or applied only when the elastic limit of the material is not exceeded. The first observations regarding this limit were made a few years later by the experimenters Barlow and Tred- gold, who noted the permanent set in their tests of cast-iron bars. After 1830 metallic bridges came into use through the necessi- ties of railroad construction, and numerous experiments on cast iron and wrought iron were made in England, France, and Ger- many. In 1849 a British commission conducted tests of these metals more exhaustive than any made before, and for the first time established definite conclusions regarding the use of factors of safety, one of which was tha.t the factor for cast iron should be twice as great for sudden loads as for steady ones. After this date specifications for important structures contained require- ments regarding working stresses, laboratories for testing mate- rials were established, improved methods and machines were devised, and the theory of the subject was greatly extended. The first testing machines in the United States of America were those built between 1850 and 1860 for experiments on gun- metal. At the present time every large manufacturer of iron ART. 16 HISTORICAL NOTES 41 and steel has a testing laboratory where specimens are con- stantly broken in order to gain knowledge whereby the quality of the product may be improved or to be certain that it meets the specifications of the buyers. All engineering colleges have testing laboratories for the instruction of students and for scientific research. The literature of the subject is enormous, and societies, both national and international, have been formed to improve the methods of testing and to render more perfect the knowledge of the properties of materials under stress. All this work has been done for the purpose of advancing stability and economy in engineering construction, and the two principles stated in Art. 7 have been its fundamental guides. The average values of the elastic limits, moduluses of elasticity, ultimate strengths and ultimate elongations given in the pre- ceding pages and in Tables 2, 3, 4, at the end of this volume, have been derived from the records of tests made since 1850. It must not be forgotten that these values are subject to much variation, depending upon the quality of the material, its method of manufacture, and to some extent upon the suddenness with which the forces are applied. In practice all these points must be carefully considered and precise knowledge be obtained, as far as possible, regarding the actual material in hand. The next chapter gives more detailed information regarding the elasticity and strength of different qualities of the materials mentioned in the preceding pages and also of other materials used in engineering and the arts, while Arts. 168-170 give discussions regarding methods of testing. Prob. 16a. Consult Vol. I of Todhunter's History of the Elasticity and Strength of Materials and ascertain the exact words in which Young defined the Modulus of Elasticity. Prob. 166. "Tests of Metals ... at the Watertown ArsenaP is the title of a publication issued yearly by the Ordnance Department of the U. S. Government. Consult one of these volumes and ascertain the elastic limit and ultimate strength of rifle-barrel steel. 42 MATERIALS OF ENGINEERING CHAP. Ill CHAPTER III MATERIALS OF ENGINEERING ART. 17. AVERAGE WEIGHTS THE following are average values of the weights per unit of volume and of the specific gravities of the principal materials used in engineering constructions. These weights should be carefully memorized by the student as a basis for more precise Material Weight Specific Gravity Brick 125 pounds per cubic foot 2.0 Stone 1 60 pounds per cubic foot 2.6 Timber 40 pounds per cubic foot 0.6 Cast Iron 450 pounds per cubic foot 7.2 Wrought Iron 480 pounds per cubic foot 7.7 Steel 490 pounds per cubic foot 7.8 knowledge, but it must be noted that they are subject to con- siderable variation with the quality of the material. For instance, brick may weigh as low as 100 or as high as 150 pounds per cubic foot, according as it is soft or pressed. Unless otherwise stated, the above average values will be used in all the follow- ing examples and problems of this volume. Table 1, at the end of this volume, gives these constants in metric measures, and it will be noted that the unit-weights in this convenient" system are multiples of the specific gravities. The following approximate simple rules, stated in 1885 in the first edition of this book, are in general use among engineers for computing the weights of bars and beams which are of uniform cross section: A wrought-iron bar one square inch in section and one yard long weighs ten pounds. Steel is about two percent heavier than wrought iron. Cast iron is about six percent lighter than wrought iron. Stone is about one-third the weight of wrought iron. ART. is PLASTICITY AND BRITTLENESS 43 Brick is about one-fourth the weight of wrought iron. Timber is about one-twelfth the weight of wrought iron. For example, consider a bar of wrought iron 1^X3 inches and 12 feet long; its section area is 4.5 square inches, hence its weight is 45X4 = 180 pounds. A steel bar of the same dimensions weighs 180 + 0.02X180= about 184 pounds, a cast-iron bar weighs 1 80 -0.06X180= about 169 pounds, and a timber bar weighs 180/12 =about 15 pounds. By reversing the above rules, the section areas of bars are readily computed from their weights per yard. Thus, if a stick of timber 15 feet long weighs 120 pounds, its weight per yard is 24 pounds, and its section area is 12X2.4 = 28.8 square inches approximately. Again, if a steel bar weighs 26.5 pounds per linear foot, its section area is 3X26.5(1-0.02)710 = 7.79 square inches. It may be noted that, as a rough general rule, the strengths of heavy bodies are greater than those of lighter bodies of the same size. Thus, steel is the heaviest material of construction and it is the strongest. Strictly speaking this rough rule applies only to materials of a similar nature, and is only valid for com- paring metals with metals, stony materials with stony materials, and timber with timber. Thus, the heaviest kinds of timber are the strongest, but many kinds of timber are stronger than stone or brick. Prob. 17. What is the weight of a stone block 12X18 inches in sec- tion area and 5 feet long? If a cast-iron pipe 12 feet long weighs 985 pounds, what is its section area ? . ART. 18. PLASTICITY AND BRITTLENESS The property of elasticity which has been explained in the preceding chapters is possessed by different materials in differ- ent degrees. A perfectly elastic material is one which springs back to its original shape on the removal of the applied force, this force having any magnitude less than that required to cause rupture. No material with perfect elasticity is known; india- rubber is the nearest approach to it, but even this fails to return 44 MATERIALS OF ENGINEERING CHAP. Ill to its original form when it has been under stress for some time. Wrought iron and steel are only elastic when the applied force is less than about fifty or sixty percent of that required to cause rupture, that is, when the unit-stress does not exceed the elastic limit (Art. 2). Cast iron, stone, and timber have poorly defined elastic limits, as the stress diagrams in Fig. lla show, and for many qualities small stresses produce permanent sets, so that their degree of elasticity is less than that of wrought iron and steel. A material which has no elasticity, so that the smallest forces cause permanent deformations, is called 'plastic'. Lead is an example of a plastic material, for a flow of the metal occurs under slight forces and there is no return to the original form. Lead has no elastic limit, and it has no ultimate compressive strength in the ordinary sense, for the material flows as the compression is applied, so that the original prism constantly decreases in height and increases in section until it becomes a thin sheet. Wrought iron and structural steel are materials which are more or less plastic under stresses exceeding their elastic limits. In tension bars of these materials recover a small part of their original length on the removal of the load, and hence they have some elasticity, but the elongation which remains as perma- nent set represents their plasticity. Under compression the elasticity is less and the plasticity is more marked, especially for structural steel, so that this variety of steel cannot be said to have a compressive strength in the ordinary sense, its behavior resembling that of lead. A material which cannot change its shape without rupture is said to be 'brittle'. There is no body perfectly brittle, but glass approaches it most nearly, since small changes of shape are followed by rupture. Cast iron is the most brittle of the common materials of construction, since its percentage of elon- gation under tension is the least (Art. 4). Brick and stone are brittle materials compared with timber or steel. In general the greater the deformation which a body will withstand before rupture the less is its degree of brittleness. Plasticity is the reverse of brittleness. ART. 18 PLASTICITY AND BRITTLENESS 45 When bars are subject to tensile stress, those of brittle mate- rial rupture with but slight change in the area of the cross-sec- tion, while those of plastic material rupture with a considerable reduction of area accompanied by a taper on each side of the place of rupture. The greater the reduction of area (Art. 12) the less is the degree of brittleness of the material. When prisms are subject to compressive stress, those of brittle material rup- ture by an oblique splitting or shearing, as seen in Fig. 185c, which shows a cement cube and an oak block. The greater the brittleness the less is the inclination of the shearing planes to the pressure line. Fig. ISa indicates these lines of rupture for anthracite coal, Fig. 18Z> for paving brick, Fig. I8c for cement or concrete, and Fig. I8d for timber; in all these cases the change in section area is very slight. For a plastic material, however, the increase in section area is large, and no shearing planes arise : Fig. 18e illustrates this case. I t I Fig. 180 Fig. 186 Fig. 18c Fig. 18d Fig. 18e The materials of engineering all possess a certain degree of elasticity and certain degrees of plasticity and brittleness. Hard steel is elastic within the elastic limit; beyond that limit it is in part elastic, in part plastic, and in part brittle, while soft steel has but little brittleness. Cast iron has defective elasticity, no plasticity, and considerable brittleness. Timber resembles cast iron in regard to elasticity, but some varieties are more plastic than brittle. The word elasticity is used in the mechanics of materials with a slightly different meaning from that of popular language. In common parlance a body is called elastic when it can undergo great deformations and then recover its original form, the greater the stretch under a given force the greater being the elasticity. In the mechanics of materials the amount of the deformation is not considered, but merely the ability to return to the original 46 MATERIALS OF ENGINEERING CHAP. Ill shape. In common language steel would not be spoken of as an elastic material, but according to the definition here given its degree of elasticity is a high one if the unit-stress does not exceed the elastic limit. The constant called the modulus of elasticity (Art. 9) is really an index of stiffness, that is, the higher the modulus the less will be the amount of deformation under a given unit-stress. Prob. 180. What is the weight in kilograms of a pig of lead, 0.75 meters long and 215 square centimeters in section area, its specific gravity being 11.38? Prob. ISb. A concrete cube weighed 181 J ounces in air and ioyj ounces in water. Compute its specific gravity, and its weight in pounds per cubic foot. What was the size of the concrete cube ? ART. 19. TIMBER Good timber is of uniform color and texture, free from knots, sap wood, wind-shakes, worm-holes, or decay; it should also be well seasoned, which is best done by exposing it for two or three years to the weather to dry "out the sap. The heaviest timber is usually the strongest; also the darker the color and the closer the annular rings the stronger and better it is, other things being equal. The strength of timber, except in the case of shear, is greatest in direction of the grain, the sidewise resistance to ten- sion or compression being about one fourth the longitudinal. The following table gives average values of the ultimate Pounds per Square Inch Weight Kind Pounds per Cubic Foot Tensile Strength Computed Flexure Strength Compress- ive Strength Hemlock 2 5 8000 6 ooo 5 ooo White Pine 27 8000 6000 5 5o Chestnut 40 12 OOO 7 ooo 5 Red Oak 42 9 ooo 7 ooo 6 ooo Yellow Pine 45 15000 II OOO 9 ooo White Oak 48 12 OOO 10 ooo 8000 ART. 19 TIMBER 47 strength of a few of the common kinds of timber as determined from tests of small specimens which were carefully selected and dried. Large pieces of timber, such as are actually used in engineering structures, have an ultimate strength of from fifty to eighty percent of these values. Moreover the figures are liable to a range of 25 percent on account of variations in quality and condition arising from place of growth, time when cut, and method and duration of seasoning. The amount of moisture in timber also influences its strength; timber which has absorbed water in amount equal to one-half of its dry weight has a strength only about one-half that of dried timber. To cover all these variations the factor of safety of 10 (Art. 7) is not too high. The column headed 'computed flexure strength ' in the above table gives the unit-stress computed from the rupture of beams, using the flexure formula (Art. 41). This quantity is always intermediate between the tensile and compressive strengths, but it is not a physical constant, as will be explained in Art. 52; it is sometimes called the 'modulus of rupture '. The shearing strength of timber is still more variable than the tensile or compressive resistance. White pine across the grain may be put at 2 500 pounds per square inch, and along the grain at 500. Chestnut has i 500 and 600, yellow pine and oak about 4 ooo and 600 pounds per square inch in these direc- tions respectively. The elastic limit of timber is poorly defined. In precise tests on good specimens it is sometimes observed at about one- half the ultimate strength, but under ordinary conditions it is safer to put it at one-third. The modulus of elasticity ranges from i ooo ooo to 2 ooo ooo pounds per square inch, i 500 ooo being a good mean value to use in general computations. The ultimate elongation is small, usually being between i and 2 per- cent. Fig. 185c shows the common way in which a short timber prism ruptures under compression. The tests published in the Census Report on the Forest Trees of North America (1884) are very comprehensive, as they include 412 species of timber. Of these 16 species have a specific gravity 48 MATERIALS OF ENGINEERING CHAP, in greater than i.o and 28 species less than 0.4. Even in the same species a great variation in weight was often found; for instance white oak ranged from 42 pounds to 54 pounds per cubic foot. The heaviest wood weighed 81 pounds per cubic foot and had a compressive strength of 12 ooo pounds per square inch; the lightest wood weighed 16 pounds per cubic foot and had a com- pressive strength of 200 pounds per square inch. Prob. 190. Which is the safer under compression, hemlock under 600 or white oak under 800 pounds per square inch ? Prob. 196. Square sticks of white pine and yellow pine are subject to a steady load which brings a tension of 10 ooo pounds upon each. Compute their size for the proper factor of safety. ART. 20. BRICK Brick is made of clay which is prepared by cleaning it care- fully from pebbles and sand, mixing it with about one-half its volume of water and tempering it by hand stirring or in a pug- mill. It is then molded into rectangular form either in wooden boxes by hand or in metal forms by machines, and the green bricks are placed under open sheds to dry. These are piled in a kiln and heated for nearly two weeks or until those nearest to the fuel assume a partially vitrified appearance. Three qualities of brick are taken from the kiln : ' arch brick ' are those from around the arches where the fuel is burned these are hard and often brittle; 'body brick', from the interior of the kiln, are of the best quality; 'soft brick', from the exterior of the pile, are weak and only suitable for filling. Paving brick are burned in special kilns, often by natural gas or by oil, the rate of heating being such as to insure toughness and hardness and also a uniform product throughout the entire kiln. A common size is 2^X4X8^ inches, and the average weight 4^ pounds. A pressed brick, however, may weigh nearly 5^ pounds. Good bricks should be of regular shape, have parallel and plane faces, with sharp angles and edges. They should be of uniform texture and when struck a quick blow should give a ART. 20 BRICK 49 sharp metallic ring. The heavier the brick, other things being equal, the stronger and better it is. Poor brick will absorb when dry from 20 to 30 per cent of its weight of water, ordinary qualities absorb from 10 to 20 per- cent, while hard paving brick should not absorb more than 2 or 3 percent. An absorption test is valuable in measuring the capacity of brick to resist the disintegrating action of frost, and as a rough general rule the greater the amount of water absorbed the less is the strength and durability. The compressive strength of brick is variable; while a mean value may be 2 500 pounds per square inch, soft brick will scarcely stand 500, pressed brick may run to 10 ooo, and the best qualities of paving brick have given 15 ooo pounds per square inch, or even more. Compressive tests are generally made on half-bricks, the compression being applied normal to the smallest side, that is parallel to the width of the brick; the failure usually occurs by an oblique shearing as indicated in Fig. ISb. Brick is rarely used in tension and but little is known of its behavior under such stress ; the ultimate tensile strength may perhaps range from 50 to 500 pounds per square inch. The ultimate shearing strength of good building brick is about 500 pounds per square inch, and higher for the paving qualities. Brick may be classed as a brittle material, the stress dia- gram resembling that of cast iron. Its elastic limit is not well defined, but its deformation within that limit is greater than that of cast iron, the modulus of elasticity E generally ranging from i ooo ooo to 3 ooo ooo pounds per square inch. Bricks saturated with water have about the same strength and elastic properties as dry bricks. Brick masonry has a much smaller strength than that of the individual bricks, on account of the lack of perfect union at the joints and also of the generally lower strength of the mor- tar. The strength of a brick pier, indicated by the loosening and cracking of the joints, is about one-fourth of the compres- sive strength of an individual brick. 50 MATERIALS OF ENGINEERING CHAP, ill Prob. 20a. Using the mean values of Arts. 5 and 17 and the state- ment of the last paragraph, compute the height of a brick tower which will crush at the base under its own weight. ART. 21 STONE Sandstone, as its name implies, is sand, usually quartzite, which has been consolidated under heat and pressure. It varies much in color, strength, and durability, but many varieties form most valuable building material. In general it is easy to cut and dress, but the variety known as Potsdam sandstone is very hard in some localities. Limestone is formed by consolidated marine shells, and is of diverse quality. Marble is limestone which has been reworked in the laboratory of nature so as to expel the impurities, and leave a nearly pure carbonate of lime; it takes a high polish, is easily worked, and makes one of the most beautiful building stones. Granite is a rock which was formerly supposed to be of aqueous origin metamorphosed under heat and pressure, but it is now generally thought to be of igneous origin ; its composition is quartz, feldspar, and mica, but in the variety called syenite the mica is replaced by hornblende. It is fairly easy to work, is usually strong and durable, and some varieties will take a high polish. Trap, or basalt, is an igneous rock without cleavage. It is hard and tough, and less suitable for building constructions than other rocks, since large blocks cannot be readily obtained and cut to size. It has, however, a high strength and is remark- able for durability. The average weights and ultimate strengths of these four classes are given in the following table. These figures refe** to small specimens such as can be broken in a testing machine, and it is known that the strength of large blocks per square inch is materially less. The rupture of a cube or prism of stone under compression often occurs by splitting, or rather shearing. ART. 21 STONE 51 in planes making an angle of about 25 degrees with the direc- tion of the pressure (see Fig. I8b). Kind Pounds per Cubic Foot Pounds per Square Inch Weight Compressive Strength Shearing Strength Sandstone Limestone Granite 150 160 7 ooo 12 000 I 200 I 5OO 2 OOO Trap 175 15 ooo 2 500 The modulus of elasticity of stone has been found to range from 5 ooo ooo to 10 ooo ooo pounds per square inch. The elastic limit is difficult to observe, if indeed any exists, but the first sign of cracking is sometimes regarded as marking this limit. Little is known regarding the tensile strength, except that it is much less than the compressive strength, and stone is rarely sub- jected to tensile stresses. The quality of a building stone cannot be safely inferred from tests of strength, as its durability depends largely upon its capacity to resist the action of the weather. Hence corrosion and freezing tests, impact tests, and observations of the behavior of stone under conditions of actual use are more important than the determination of crushing strength in a compression machine. The strength of a stone pier is only about one-fourth of that of the stone itself, on account of the weakness of the mortar joints. Slate is an argillaceous stone consolidated under very heavy pressure which has produced a marked cleavage. It is split into plates J inch in thickness for use as roofing slate, and in larger blocks is used for pavements and steps. Its weight per cubic foot is about 175 pounds, its compressive strength about 10 ooo pounds per square inch, and its computed flexural strength about 7 ooo pounds per square inch. Slate is liable to corrode under the action of the atmosphere, and its marked cleavage and grain render its strength variable in different directions; hence it is unsuitable for a building stone. 52 MATERIALS OF ENGINEERING CHAP. Ill Prob. 21. How many cubic yards of masonry are contained in a pier 12X30 feet at the base, 8X24 feet at the top, and i6j feet high? What is its cost at $6.37 per cubic yard? ART. 22. MORTAR AND CONCRETE Common mortar is composed of one part of lime to five parts of sand by measure. When six months old its tensile strength is from 15 to 30 and its compressive strength from 150 co 300 pounds per square inch. Its strength slowly increases with age, and a smaller proportion of sand produces mortar stronger than the above values. Hydraulic mortar is composed of hydraulic cement and sand in varying proportions. The less the proportion of sand the greater is its strength. If S be the strength of neat cement, that is, of cement with no sand, the strength of mortar having p parts of sand to one part of cement is about S/(i+p) as a rough approximation. A common proportion is 3 parts of sand to i of cement, the strength of this being about one-fourth that of the cement itself. The strength of hydraulic mortar also increases with its age. There are two classes of hydraulic cements, the natural cements and the Portland cements. The former are of lighter color, lower weight, and lesser strength than the latter, but they are quicker in setting and cheaper in price. The following table gives average ultimate tensile strengths in pounds per Proportion of Sand NATURAL PORTLAND One Week One Month One Year One Week One Month One Year ISO 250 400 500 680 700 I to I no 20O 300 330 400 500 2 to I 90 15 240 240 310 410 3 to i 5 IOO 190 1 80 250 35 4 to i 3o 75 IOO IOO 1 80 260 square inch of mortars of both classes of different ages and differ- ent proportions of sand. These figures are obtained from tests ART. 22 MORTAR AND CONCRETE 53 of briquettes carefully made in molds, and are higher than hydraulic mortar made under usual circumstances will give. The briquette is removed from the mold when set, allowed to remain in air for one day, and then put under water for the remainder of the time. Since cements and mortars are used only in compression, the tensile tests may seem an inappropriate one. Compressive tests, however, are more expensive than tensile ones. It may be taken as a general rule that the compressive strength of a material increases with the tensile strength, and it is certain that the universal adoption of the tensile tests has done much to greatly improve the quality of hydraulic cements. For neat cement a one-day tensile test is frequently employed, the briquette being put under water as soon as it has set. For natural cement a briquette one square inch in section should give a tensile strength of 75 pounds, while one of Portland cement should give 125 pounds. To secure these results, however, the material should be thoroughly rammed into the molds, no more water being used than necessary, and the quality of the cement must be good. The compressive strength of hydraulic cements and mortars is from six to ten times the tensile strength. Fig. 169& shows the manner in which a cube of cement fails under compression. Neat Portland cement when one month old has a compressive strength of about 3 ooo, and when one year old about 5 ooo pounds per square inch. Natural cement mortar when three or four years old has a compressive strength of about 2 500 pounds per square inch. The adhesion of cement to stone or brick is somewhat less than the tensile strength. The shearing strength of cement or mortar is much less than the compressive strength, usually only about one-fourth or one-third. The strength of cement and mortar is influenced by many causes: the quality of the stone or materials from which the cement is made, the method of manufacture, the age of the cement, the kind of sand, the method of mixing, and even the amount of water used. Mortar joints are, as a rule, weaker 54 MATERIALS OF ENGINEERING CHAP, ill than the bricks or stones which they unite, and the failure of a masonry wall usually begins by cracking along these joints. Concrete is an artificial stone which is made by mixing hydraulic mortar and broken stones. The best proportions are such that the grains of sand fill all the voids between the stones, while the cement fills all the voids between the sand grains, as also those between the sand and the stones. Common pro- portions for a first-class concrete are i cement, 2 sand, and 4 broken stone, by measure. Concrete is mainly used for floors, walls, foundations, and monolithic structures, but sometimes blocks are made which are laid together like stone masonry. Its use has greatly increased since 1900 and many concrete piers and bridges have been built. Reinforced concrete is concrete in which steel rods are imbedded in order to increase its tensile resistance (Art. 113). The average weight of concrete is about 150 pounds per cubic foot. Its strength increases with age, like that of hydraulic mortar. For concrete made with the proportions i cement, 2 sand, 4 broken stone, when six months old, the average ten- sile strength is 400 pounds per square inch, compressive strength 3 500 pounds per square inch, and modulus of elasticity 3 ooo ooo pounds per square inch. For concrete with the proportions 1 cement, 3 sand, 6 broken stone, when six months old, the aver- age tensile strength is 300 pounds per square inch, compressive strength 2 500 pounds per square inch, and modulus of elasticity 2 ooo ooo pounds per square inch. These figures refer to con- crete in which Portland cement is used, the strength being about twenty percent less for natural cement concrete. When one year old the ultimate strengths are a little greater than the abov? figures. The elastic limit of concrete is not well defined, and if one is to be stated, that in compression may be roughly put at about one-third or one-fourth of the compressive strength. The allowable unit-stress usually ranges from one-sixth to one- eighth of the ultimate strength. Prob. 22. Let the pier of Problem 21 be concrete and carry a load of 460 ooo pounds. Compute its weight and the compressive unit- stress on its base. A.RT. 23 CAST IRON 55 ART. 23. CAST IRON Cast iron is a modern product, having been first made in England about the beginning of the fifteenth century. Ores of iron are melted in a blast furnace, producing pig iron. The pig iron is remelted in a cupola furnace and poured into molds, thus forming castings. Beams, columns, pipes, braces, and blocks of every shape required in engineering structures are thus produced. Pig iron is divided into two classes, Foundry pig and Forge pig, the former being used for castings and the latter for making wrought iron and steel. Foundry pig has a dark-gray fracture, with large crystals and a metallic luster; forge pig has a light- gray or silver-white fracture, with small crystals. Foundry pig has a specific gravity of from 7.1 to 7.2, and it contains from 6 to 4 percent of carbon ; forge pig has a specific gravity of from 7.1 to 7.4, and it contains from 4 to 2 percent of carbon. The higher the percentage of carbon the less is the specific gravity, and the easier it is to melt the pig. Besides the carbon there are present from i to 5 percent of other impurities, such as silicon, manganese, and phosphorus. The properties and strength of castings depend upon the quality of the ores and the method of their manufacture in both the blast and the cupola furnace. Cold-blast pig produces stronger iron than the hot-blast, but it is more expensive. Long- continued fusion improves the quality of the product, as also do repeated meltings. The darkest grades of foundry pig make the smoothest castings, but they are apt to be brittle; the light- gray grades make tough castings, but they are apt to contain blow-holes or imperfections. The percentage of carbon in cast iron is a controlling fact or which governs its strength, particularly that percentage which is chemically combined with the iron. For example, the fol- lowing are the results of tests made by Wade about 1860 of three classes of cast iron then used for guns, the tensile strength being in pounds per square inch : 56 MATERIALS OF ENGINEERING CHAP, in XT c "c r> -I Percentage of Carbon Ultimate No. Specific Gravity Gnp}dte Combined Tensile Strength 1 7.204 2.06 1.78 28 800 2 7- T 54 2 -3 J -46 2480x3 3 7-%7 2-83 0.82 20 TOO Here it is seen that the total carbon is about the same in the three kinds, but the greater the percentage of combined carbon the higher is the specific gravity and ultimate strength. As average values for the ultimate strength of cast iron, 20 ooo and 90 ooo pounds per square inch in tension and com- pression respectively are good figures. In any particular case, however, a variation of from 10 to 20 per cent from these values may be expected, owing to the variations in quality. For first- class gun iron Wade found a tensile strength of over 30 ooo and a compressive strength of over 1 50 ooo pounds per square inch. On the other hand medium-quality castings often have a tensile strength less than 16 ooo pounds per square inch. In tensile tests the elongations within the elastic limit are not exactly proportional to the applied loads, so that the elastic limit is poorly defined, and the modulus of elasticity may range from 12000000 to 18000000 pounds per square inch. The elastic limit in tension may be roughly stated at 6 ooo pounds per square inch, but that for compression is much higher, prob- ably about 20 ooo pounds per square inch. The flexural test is a good one for comparing the strength of different bars of cast iron. A bar one inch square and 14 inches long is laid on two supports 12 inches apart and a load gradually applied to the middle until rupture occurs; the load required will be about 2 ooo pounds for average grades. For a bar 2X1 inches in section area, laid flatwise on two supports 24 inches apart, the load at the middle which causes rupture will also be about 2 ooo pounds. The high compressive strength and the cheapness of cast iron render it a valuable material for many purposes, but its brittleness, low tensile strength, and lack of ductility forbid its use 5n structures subject to variations of load or to shocks, ART. 24 WROUGHT IRON 57 Its ultimate elongation being scarcely one percent, the work required to cause rupture is small compared to that for wrought iron and steel. Cast iron is no longer used in bridges, and for important buildings it has been but little employed since 1895. Malleable cast iron is made by surrounding castings with decarbonizing material and heating them in annealing ovens. The carbon is thus partly removed to a certain depth below the surface of the castings, and this renders the material much stronger in tension, while its elongation is increased. The process is only applicable to small castings, but for these the tensile strength may be doubled and the ultimate elongation be made eight or ten times as great as that of common cast iron, thus rendering it more efficient in resisting shocks. Prob. 23. Compute the weight of a cast-iron water pipe 12 feet long and 20 inches in internal diameter, the average thickness of the metal being ij inches. ART. 24. WROUGHT IRON The ancient peoples of Europe and Asia were acquainted with wrought iron and steel to a limited extent. The name of an ironmaster is mentioned in Genesis iv 22, and in one of the oldest pyramids of Egypt a piece of iron has been found. It was produced, undoubtedly, by the action of a hot fire on very pure ore. The ancient Britons built bloomaries on the tops of high hills, a tunnel opening toward the north furnishing a draft for the fire which caused the carbon and other impurities to be expelled from the ore, leaving behind nearly pure metallic iron. Modern methods of manufacturing wrought iron are mainlv by the use of forge pig (Art. 23) in the puddling process. Here the forge pig is subjected to the oxidizing flame of a blast in a reverberatory furnace, where it is formed into pasty balls by the puddler. A ball taken from the furnace is run through a squeezer to expel the cinder and then rolled into a muck bar. The muck bars are cut, laid in piles, heated, and rolled, forming what is called merchant bar. If this is cut, piled, and rolled again a 58 MATERIALS OF ENGINEERING CHAP, ill better product called best iron is produced. A third rolling gives ' best-best' iron, a superior quality, but high in price. The product of the rolling mill is bar iron, plate iron, shape iron, beams, and rails. Bar iron is round, square, and rectan- gular in section ; plate iron is from J to i inch thick, and of vary- ing widths and lengths ; shape iron includes angles, tees, channels, and other forms used in structural work; beams are I-shaped, and of the deck or bulb form (Art. 44). Wrought iron is metallic iron containing less than 0.25 per- cent of carbon, and which has been manufactured without cast- ing from the fluid state. Its elastic limit is well defined at about 24 ooo pounds per square inch and the yield point at about 26 ooo pounds per square inch; within this limit it is stiffer than cast iron, the modulus of elasticity being about 25 ooo ooo pounds per square inch. Its average ultimate tensile strength is about 50 ooo pounds per square inch, and its ultimate elongation is from 25 to 30 percent. In comparison it is plastic beyond the elastic limit, so that the compressive strength given in Art. 5 has but little meaning. It is malleable, can be forged and welded, and has a high capacity to withstand the action of shocks; it cannot, however, be tempered so as to form cutting tools. The cold-bend test for wrought iron is an important one for judging of general quality. A bar, perhaps fxf inches and 15 inches long, is bent when cold either by pressure or by blows of a hammer. Bridge iron should bend, without cracking, through an angle of 90 degrees to a curve whose radius is twice the thick- ness of the bar. Rivet iron should bend, without showing signs of fracture, through 180 degrees until the sides of the bar are in contact. Wrought iron that breaks under this test is lacking in both strength and ductility. The process of manufacture, as well as the quality of the pig iron, influences the strength of wrought iron. The higher the percentage of carbon the greater is the strength. Best iron is 10 percent stronger than ordinary merchant iron owing to the influence of the second rolling. Cold rolling causes a marked increase in elastic limit and ultimate strength, but a decrease ART. 24 WROUGHT IRON 59 in ductility or ultimate elongation. Annealing lowers the ulti- mate strength, but increases the elongation. Iron wire, owing to the process of drawing, has a high tensile strength, sometimes greater than 100 ooo pounds per square inch. Good wrought iron when broken by tension shows a fibrous structure. If, however, it be subject to shocks or to repeated stresses which exceed the elastic limit, the molecular structure becomes changed so that the fracture is more or less crystalline. The effect of a stress exceeding the elastic limit is to cause a permanent set, but the elastic limit will be found to be higher than before. This is decidedly injurious to the material on account of the accompanying change in structure and because its capacity to resist work is less than before, as Fig. 116 shows; hence it is a fundamental principle that the working unit-stresses should not exceed the elastic limit. For proper security the allowable unit-stress should seldom be greater than one-half the elastic limit. Prior to 1890 wrought iron was generally employed for struc- tural purposes, but medium steel, on account of its smaller cost and greater strength, then began to come into use, and the change was so rapid that since 1900 it has almost entirely displaced wrought iron in bridge and building construction. Wrought iron is, however, still employed for chains, for stay bolts, and for many other purposes where great ductility and toughness are demanded. Owing to the fact that it was used so extensively in the nineteenth century and that its properties .were then so well ascertained by numerous tests, it will long remain in engineering literature as a standard with which other materials may be advantageously compared. Since 1900 much pig iron has been made in America by casting in open molds instead of in a sand bed, and wrought ^ron made from such pig appears to be lower in strength than that of the nineteenth century. Prob. 24. Compute the section area of a wrought-iron bar 30 feet long which weighs 418 pounds. If this bar is hung vertically at its upper end, compute the unit-stress at that end and at the middle. What load, hung at the lower end, will stress it to the elastic limit? 60 MATERIALS OF ENGINEERING CHAP, m ART. 25. STEEL Steel was originally produced directly from pure iron ore by the action of a hot fire, which did not remove the carbon to a sufficient extent to form wrought iron. The modern processes, however, involve the fusion of the ore, and the definition of the United States law is that " steel is iron produced by fusion by any process, and which is malleable." Chemically, steel is a compound of iron and carbon generally intermediate in com- position between cast and wrought iron, but having a higher specific gravity than either. The following comparison points out the distinctive differences between the three kinds of iron: Percent of Specific Pronerties Carbon Gravity Cast iron, 5 to 2 7.2 Not malleable, not temperable Steel, i.5otoo.io 7.8 Malleable and temperable Wrought iron, 0.30 to 0.05 7.7 Malleable, not temperable It should be observed that the percentage of carbon alone is not sufficient to distinguish steel from wrought iron; also, that the mean values of specific gravity stated are in each case sub- ject to considerable variation; further, only the hard steels are temperable, the softer grades resembling wrought iron. The three principal methods of manufacture are the cruci- ble process, the open-hearth process, and the Bessemer process. In the crucible process impure wrought iron or blister steel, with carbon and a flux, are fused in a sealed vessel to which air cannot obtain access; the best tool steels are thus made. In the open-hearth process pig iron is melted, wrought-iron scrap being added until the proper degree of carbonization is secured. In the Bessemer process pig iron is completely decar- bonized in a converter by an air blast and then recarbonized to the proper degree by the addition of spiegeleisen. The metal from the open-hearth furnace or from the Bessemer converter is cast into ingots which are rolled in mills to the required forms. The open-hearth process produces steel for machines, shafts, axles, springs, armor plates, and for structural purposes; the Bessemer process mainly produces steel for railroad rails. ART. 25 STEEL 61 'Acid steel ' is that made in a furnace having a silicious lining, while' 'basic steel' is that made in a furnace with dolomitic linings. These terms have no reference to quality and refer mainly to process of manufacture, but the basic process enables ores high in phosphorus to be used, the phosphorus being removed by the addition of lime. Acid steel is slightly stronger than basic steel, owing partly to the higher percentage of phosphorus and partly to the effect of the lime in forming slag in the basic steel. Since 1900 more than three-fourths of the open-hearth steel pro- duced in the United States has been basic; the Bessemer product on the other hand being entirely acid. The physical properties of steel depend both upon method of manufacture and chemical composition, the carbon having the controlling influence upon strength. Phosphorus increases strength, but it promotes brittleness; manganese increases strength in a less degree, and it promotes malleability; sulphur causes red-shortness, or a tendency of the steel to crumble while being rolled; and silicon increases hardness. Many formulas have been deduced to exhibit the relation between the tensile strength and the chemical composition, but none of these applies to all classes of steel. The rough rules, ,=45 000+ 108 oooC St=4S 000+90 oooC give an approximate idea of the influence of carbon in acid and basic unannealed open-hearth steel respectively, C being the percentage of carbon and S t the tensile strength in pounds per square inch. Thus, acid steel with 0.40 percent of carbon has a tensile strength of about 88 ooo pounds per square inch, while basic steel has about 81 ooo pounds per square inch. When the percentages of phosphorus and manganese are also known, the following formulas which have been deduced from the exhaustive discussion given by Campbell in 1905 may be used to give more reliable results, namely, S t = 40 000+ 68 oooC+ 100 oooP+ 80 oooCM ,= 38 800+65 oooC+ 100 oooP+9 oooM+4o oooCM the first being for acid and the second for basic open-hearth 62 MATERIALS OF ENGINEERING CHAP, in steel. Here C is the percentage of carbon, P that of phosphorus, M that of manganese, and S t the tensile strength in pounds per square inch. For example, acid steel having 0.344 percent of carbon, 0.045 percent of phosphorus, and 0.70 per cent of manganese has a tensile strength of 87 200 pounds per square inch; basic steel having 0.344 percent of carbon, 0.020 percent of phosphorus, and 0.35 percent of manganese has a tensile strength of 71 100 pounds per square inch. These formulas do not apply to steel with a percentage of carbon higher than 0.75. Carbon is the controlling element in regard to strength, and the same is the case with respect to ultimate elongation. The higher the percentage of carbon, within a reasonable limit, the greater is the strength and the less the ultimate elongation. The product of strength and elongation is approximately constant, and hence the ultimate elongation is approximately inversely proportional to the tensile strength. A rule frequently given is that the percentage of elongation equals i 500 ooo/S t ; thus, for a tensile strength of 80 ooo pounds per square inch the ulti- mate elongation is about 19 percent. This rule, however, gives too high elongations for very strong steel. A classification of steel according to the percentage of car- bon which it contains and its capacity for taking temper or being welded, is as follows : Soft, 0.05-0. 2oC, not temperable, easily welded Medium, 0.15-0.40(7, poor temper, weldable Hard, 0.30-0. 7oC, temperable, welded with difficulty Very hard, o.6o-i.ooC, high temper, not weldable It is seen that these classes overlap so that there is no distinct line of demarcation, and in fact the words soft, medium, and hard, are frequently used without precision and only for com- parative purposes. The term 'strong steel' in the two pre- ceding chapters has been introduced for educational purposes only, in order to divide steel into two classes for the benefit of beginners in engineering. Steel is frequently classified, with reference to its uses, and the following is such a classification giving average elastic limits ART. 25 STEEL 63 and ultimate tensile strengths in pounds per square inch and ultimate elongations in percentages. For the elastic limit a variation of about 2 ooo and for the ultimate strength a varia- Elastic Limit Tensile Strength Elongation Structural steel for rivets 30 ooo 55 ooo 30 for beams and shapes 35 ooo 60 ooo 27 Boiler steel for rivets 25 ooo 50 ooo 30 for plates 30 ooo 60 ooo 26 Machinery steel 40 ooo 75 ooo 20 Gun steel 50 ooo 90 ooo 18 Axle steel 55000 100 ooo 15 Spring steel 60000 125000 12 Cable wire steel 100 ooo 200 ooo 8 tion of 4 coo or 5 ooo pounds per square inch from these mean values may be expected. The ultimate elongations are subject to marked variation according to the ratio of the length of the test specimen to its diameter; those here given are for the standard 8-inch specimen (Art. 169).. In Fig. 169# are shown an unbroken and two broken specimens of the 2 -inch size which has been much used since 1900; this gives higher percentages of elongation than the 8-inch specimen. The soft and medium steels resemble wrought iron in having a yield point (Art. 11) which is from 2 ooo to 4000 pounds per square inch above the elastic limit, while very hard steels have no yield point, as the stress diagrams in Fig. lla show. The elastic limit in tension is a little higher than one-half of the uti- mate strength. The modulus of elasticity is subject to but little variation with the percentage of carbon, and the mean value of 30 ooo ooo pounds per square inch may be used in computations for both tensile and compressive stresses that do not exceed the elastic limit. The modulus of elasticity for shearing is about three-eighths of that for tension and compression. Soft and medium steel will withstand a cold-bend test similar to that described in the last article, but some of the hard steels will fail to do o on account of their lack of ductility. 64 MATERIALS OF ENGINEERING CHAP, ill The compressive strength of the soft steels may be said to be about the same as the tensile strength, since when this pres- sure is reached the shortened specimen is badly cracked. The soft steels resemble wrought iron in being plastic under com- pressive stress exceeding the elastic limit, and some authorities regard the yield point as the compressive strength. Hard steels, on the other hand, are not plastic beyond the elastic limit, but their behavior is like that of brittle materials (Art. 18). The compressive strength of the hard steels is much higher than the tensile strength and in this respect they resemble cast iron; the greatest value recorded is 392 ooo pounds per square inch. The shearing strength of steel is usually about eighty percent of the tensile strength. The strength of steel may be greatly increased by the processes of forging and drawing. Forging under a hammer or press renders the material more compact and increases both specific gravity and strength. The process of drawing steel bars into wire has a similar result, and wire has been made having a ten- sile strength of 250 ooo pounds per square inch, while the wire used for the cables of suspension bridges usually has a tensile strength of from 150 ooo to 200 ooo pounds per square inch. By compressing steel while it is fluid, the strength may also be much increased, and this process is used for the steel from which large guns and hollow shafts are made. Annealing consists in raising cold steel to a light red heat and then allowing it to cool for several days. This process reduces the ultimate strength, but it increases the ductility. As an example, the following table gives some of the results from a large series of specimens prepared by the Bethlehem Steel Com- pany in 1893 and now kept at Lehigh University; the table refers to flat bars of Bessemer steel. Art. 119 shows that anneal- ing increases the capacity of steel to resist work. Tempering consists in plunging heated steel into a bath of water or oil, or by applying these fluids to its surface. The hardness of the steel and its ultimate strength are thereby much increased. Armor plate undergoes special processes of temper- ART. 25 STEEL. 65 ing or carbonization which render it excessively hard and tough in order that it may resist projectiles which strike it. Tensile Strength Ultimate Elongation Percent of Pounds per Square Inch Percentage Carbon Unannealed Annealed Unannealed Annealed 0.08 58 000 56 ooo 27 31 0.25 84 ooo 75000 21 25 0.50 125 ooo 99 ooo II 19 0.67 136 ooo 112 OOO 6 16 1 .04 153 ooo 128 ooo 3* ii Steel castings are extensively used for axle boxes, cross-heads, and machine frames. They range in tensile strength from 60 ooo to 90 ooo pounds per square inch and have an elastic limit of somewhat less than half the ultimate strength. Although less reliable than steel forgings, they give excellent service after having been annealed so as to increase their ductility and their capacity to withstand shock and work. Steel is alloyed with chromium, manganese, tungsten, and other materials in order to increase its strength, hardness, and tough- ness. Modern tool steels contain from five to ten percent of tungsten, which enables the tool to retain its temper even when under a red heat. Nickel is much used as an alloy for the steel of guns and armor plates. Nickel steel has been used to a slight extent for structural purposes and for railroad rails; it contains about 3^ percent of nickel, and has an elastic limit of about 48 ooo and a tensile strength of about 90 ooo pounds per square inch. Nickel steel has been made with an elastic limit of 1 20 ooo and a tensile strength of 277 ooo pounds per square inch, the ultimate elongation being about 3 percent. Prob. 25a. If steel costs 3 cents per pound and nickel costs 35 cents per pound, what should be the cost of a pound of nickel steel which contains 3.25 percent of nickel? Prob. 256. Consult Campbell's paper on Tensile Strength of Open- Hearth Steel, in Transactions of American Institute of Mining Engi- neers, 1905, and test his Table XVII by the above formulas. 66 MATERIALS OF ENGINEER NG CHAP, in ART. 26. OTHER MATERIALS Several kinds of artificial stone have been made since 1870, most of them having hydraulic cement and sand as the prin- cipal ingredients. Beton consists only of those materials which are subject to prolonged trituration, so that its strength is much greater than ordinary concrete. Frear stone is made from cement and sand with a small quantity of gum shellac. Ransome stone is made from sand and sodium silicate which are thoroughly incorporated in molds and then the blocks are put under pres- sure in a hot solution of calcium chloride. Sand-lime brick, which has been extensively made since 1900, is an artificial stone made by consolidating heated sand and lime under pressure. These artificial stones are used mainly for the walls of buildings, for window lintels, and for steps. Ropes are made of hemp, of manilla, and of iron or steel wire with a hemp center. A hemp rope one inch in diameter has an ultimate strength of about 6 ooo pounds, and its safe working strength is about 800 pounds. A manilla rope is slightly stronger. Iron and steel ropes one inch in diameter have ulti- mate strengths of about 36 ooo and 50 ooo pounds respectively, the safe working strengths being 6 ooo and 8 ooo pounds. As a fair rough rule, the strength of ropes may be said to vary as the squares of their diameters, that is, with the areas of the cross-sections. Phosphor bronze is an alloy of copper and tin containing from 2 to 6 percent of phosphorus. It is remarkable for its complete fluidity so that most perfect castings can be made. It has been used for journal bearings, valve seats ; and even for cannon. It is hard and tough, and its ultimate tensile strength may range from 40 ooo to 100 ooo pounds per square inch. Aluminum is a silver-gray metal which is malleable and duc- tile and not liable to corrode. Its specific gravity is about 2.65, so that it weighs only 168 pounds per cubic foot. Its ultimate tensile strength is about 25 ooo pounds per square inch. It has a low modulus of elasticity, and its ultimate elongation is small. Alloys of aluminum and copper have been made with a tensile ART. 26 OTHER MATERIALS 67 strength and elongation exceeding those of wrought iron, but have not come into use as structural materials. Numerous brasses and bronzes composed of copper, tin, and zinc have been made. The strongest of these alloys was ascertained by Thurston to be that composed of 55 parts of cop- per, 43 of zinc, and 2 of tin, its ultimate tensile strength being 68 900 pounds per square inch with an elongation of 48 per cent and a reduction of area of 70 percent. Brass, which is composed of copper and zinc, is almost the only alloy which has come into extensive use in the arts and which at the same time is a fully reliable material. In the form of castings it has a tensile strength of about 20 ooo pounds per square inch, in the form of rolled sheets or wire it has a much greater strength. Brass water-pipes are now frequently used in houses by those who can afford to pay as high a price as 20 cents per pound. The tensile strength of lead is only about one-tenth of that of brass, and it attains a permanent set under a very low stress; it is indeed almost devoid of elasticity, but has high plasticity. Glass is very brittle, but under slowly applied loads it has a tensile strength of about 5 ooo and a compressive strength of about 8 ooo pounds per square inch. Anthracite coal is a brittle mate- rial, having a compressive strength of from 5 ooo to 15 ooo pounds per square inch. In conclusion it may be noted that the strength and other properties of materials are subject to variation with temperature, the mean values given in the preceding pages being for a tem- perature ranging from 40 to 90 degrees Fahrenheit. Experi- ments have shown that wrought iron and steel continually increase in strength under static loads with decreasing temperatures, the rate of increase being about four percent for each 100 degrees of decrease in temperature. This rule applies for all ordinary temperatures to which structures and machinery are subjected, and to temperatures below 200 degrees Fahrenheit. As the tem- perature rises above 200, wrought iron and steel increase in strength, attaining a maximum at about 500 and a rapid decrease 68 MATERIALS OF ENGINEERING CHAP, ill then follows for higher temperatures; the elastic limit, however, seems continually to decrease as the temperature rises. Steel and other metals are more brittle at low temperatures than at high ones, and the breaking of railroad rails in cold weather is attrib- uted to brittleness and not to decrease in static strength. For further information regarding the materials discussed in this chapter and also regarding plaster, leather, india rubber, glass, ice, and glue, see American Civil Engineers' Pocket Book. For information concerning methods and apparatus for testing materials, see the same Pocket Book, and also Chap. XX of this volume. Prob. 260. A bar of aluminium copper, iJXif inches in section area, breaks under a tension of 42 800 pounds. What tension will probably break a bar of the same material which is i|X2^ inches in section area? Prob. 266. Compute the factor of safety of a cast-iron block, 2X2 feet in section area, when supporting a load of 2 165 net tons. What should be the size of a white-oak block to carry the same load with a factor of safety of 10? ART. 27 STRESS DUE TO OWN WEIGHT 69 CHAPTER IV CASES OF SIMPLE STRESS ART. 27. STRESS DUE TO OWN WEIGHT Tension, compression, and shear are often called the three cases of simple stress, it being understood that the body or bar is under only one of these stresses at the same time. In tension and compression the loads act along the axis of the bar, in shear they act normally to it. When a bar is vertical, its weight causes stresses additional to those arising from the applied load, and these will now be considered. When an unloaded bar is hung vertically by one end, there is no stress on the lower end, while at the upper end there is a tensile stress equal to the weight of the bar. When a vertical bar of weight W has a load P at its lower end, the tensile stress at that end is P and that at the upper end is P+W. In many practical cases W is small compared with 'P, so that it is unneces- sary to consider it; a common rule is that the stress due to W need not be regarded when it is less than ten percent of that due to P. A steel bar one square inch in section area and 30 feet long weighs about 102 pounds (Art. 17), and a stress of 100 pounds per square inch is small compared with the elastic limit of the material. The limiting length of an unloaded vertical bar is that at which it would break at the upper end under its own weight. Let a be the section area of the bar, S the ultimate strength of the material, v its weight per cubic unit, and / the limiting length. Then the weight of the bar is val, the tensile stress at its upper end is aS, and hence for rupture S = vl , whence / =S/v. For example, take a cast-iron bar for which S =20000 pounds per square inch or 2 880 ooo pounds per square foot, and v =450 pounds per cubic foot; then the limiting length is / = 2880000/450=6400 feet. Of course, no vertical bar of this length is possible. 70 CASES OF SIMPLE STRESS CHAP. IV The limiting length of a vertical bar loaded at its lower end is that at which it would break at the upper end under the stress due to the load and its own weight. If P is the load at the lower end, the stress at the upper end is P + val, and hence for rupture Sa=P + val, from which l = (S-P/a)/v. Hence P/a is the unit-stress due to P, and the formula shows that / is zero when P/a equals the ultimate tensile strength of the material. For a cast-iron bar let P/a be 10 ooo pounds per square inch, then the limiting length is 3 200 feet. Fig. 27a Fig. 27b The elongation of a vertical bar under its own weight is one- half of that caused by the same load applied at the lower end. To show this, let / be the length, a the section area, and W the weight of the bar. Let oc be any distance from the lower end (Fig. 27 that is the hoop is turned so that its interior diameter is T T7nr tri I GSS than the diameter of the cylinder; then di=}$%$d, and e = (d di)/di =TTV?> which is practically the same as rgVir- The tangential tensile unit-stress in the hoop then is ,5=30000000/1500 = 20000 pounds per square inch. The radial unit-stress acting between the cylinder and the hoop is from formula (30) readily found to be R = 2tS/d il and 80 CASES OF SIMPLE STRESS CHAP. IV hence its value depends upon the diameter of the cylinder and the thickness of the hoop. For a locomotive driving wheel having d=6o inches, / = J inches, and S =20000 pounds per square inch, the radial unit-pressure R between the tire and wheel is 500 pounds per square inch. The above discussion gives values of S somewhat too high, because the radial pressure acting between the hoop and the cylinder produces some decrease in the diameter of the latter. For thick hoops upon hollow cylinders, such as those of large guns, the above method is not sufficiently accurate, and a more exact one is given in Art. 152. Prob. 32. Upon a cylinder 18 inches in diameter a wrought-iron hoop 2 inches thick is to be placed. The hoop is turned to an interior diameter of 17.98 inches and shrunk on. Compute the tensile unit- stress 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 are 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 S 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 RIVETED JOINTS 81 for tension on plate, t(p d)S t = P for compression on rivet, tdS c = P for shear on rivet, nd 2 . 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. 33Z>. 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.&d 2 .S s = P from which the unit-stresses due to the tension P may be computed and the security of the joint be investigated. This joint is usually a stronger one than that with single riveting, if proper values are assigned to p and d. The investigation of a given riveted joint consists in deter- mining the values of S tJ Ss, and S c from the above equations and then computing the factors of safety (Art. 7). For example take a lap joint with double riveting where P = 8 ooo pounds, p=2% inches, d = % inches, and / = | inches, Then the tensile stress on the plate between two rivets is, St = P/t(p d) = 6 560 pounds per square inch. 82 CASES OF SIMPLE STRESS CHAP. IV while the sidewise compressive stress on the rivet is, S c =P/2td=6 100 pounds per square inch, and the shearing stress on the rivet is, S s =P/?nd 2 =6 650 pounds per square inch. Now, if these plates and rivets be of structural steel having an ultimate tensile and compressive strength of 55 ooo pounds per square inch and an ultimate shearing strength of 45 ooo pounds per square inch, the factors of safety are 8.4 for the plate in ten- sion, 9.0 for the rivet in compression, and 6.8 for the rivet in shear. The ' efficiency ' of a joint is defined as the ratio of its high- est allowable stress to the highest allowable stress of the unriveted plate. For any riveted joint three efficiencies may be computed by dividing the three values of P by pt . S t) which is the allow- able stress on the unriveted plate, and the smallest of these is the efficiency of the joint. For example, let S t =S^= 55000, and 5=45000 pounds per square inch, and let p = 2^, d=f, and /=| inches for a lap joint with double riveting. Then for the plate in tension the efficiency is (p d)tSi/ptS t = (p d)/p = 0.65; for the rivet in compression the efficiency is 2tdS c /ptS t = 2dS c /pS ',=0.70; and for the rivet in shear the efficiency is %7id 2 Sg/ptS t =o.$2. The efficiency of this joint is therefore 0.52, that is, its strength is 52 percent of that of an unriveted plate. In a properly designed joint all parts are of equal strength and the three efficiencies are equal. A butt joint is one in which there is no overlapping of the main plates, but cover plates are used. Fig. 33c shows a case ^-^ / x ^ where there are two cover plates and a single row of \ X V ^ rivets on each side of the Flg ' 33c joint. When tension is ap- plied to the two main plates, it first produces compression on the rivets and this in turn brings the rivets into shear. This shear comes on two cross-sections of each rivet and transfers one-half of the applied tension into each cover plate. Accord- ingly the thickness of a cover plate should be one-half of that ART. 34 DESIGN OF RlVETED JOINTS 83 of the main plate. Let the notation be the same as before; then, P=t(p- d)S t P= tdS c P=2. \n& . S 8 are the three formulas for the cases of tension, compression, and shear. Here the expressions for tension and compression are the same as those for a lap joint with single riveting, but that for shearing is different because the total stress is divided between two cross-sections of a rivet. Butt joints having two cover plates, and two rows of rivets on each side, are also used. For this case it is easy to see that the three formulas are, t(p-d)St = P 2.td.S c =P 4.nd 2 S 8 =P since the force P brings compression upon two rivets and shear upon four rivet sections. Here, as always, / is the thickness of the main plates and p is the pitch of the rivets in one row; the two rows are ' staggered ', as shown in Fig, 336, that is, the rivets of one row are opposite the middle of the pitch of the other row. Prob. 33a. A boiler 42 inches in diameter carries a steam pressure of 1 20 pounds per square inch. Its longitudinal lap joints have a sin- gle row of rivets which are spaced with ij inches pitch. Compute the tension P which brings shear upon one rivet. Prob. 33b. The plates of this boiler are } inches thick and the rivets are ^J inches in diameter. Compute the unit-stresses upon the plates and the rivets. What is the efficiency of the joint, and what steam pressure will cause the boiler to rupture ? ART. 34. DESIGN OF RIVETED JOINTS The design of a riveted joint consists in giving such values to the plate thickness /, the diameter of the rivets d, and the pitch of the rivets p, that all parts may be of the same strength, and that the working unit-stresses may be such that the proper degree of security is obtained (Art. 7). For example, taking a double- riveted lap joint of structural steel, let it be required to determine /, d, and p so that S t = 9000, S c = 12000, and S s = 7 500 pounds per square inch when the tension P which comes on one rivet is 8 ooo pounds. The third formula for this case (Art. 33) gives the value of d, then the second gives the value 84 CASES OF SIMPLE STRESS CHAP. IV of /, and finally the first gives the value of p. Thus d 2 =2P/nS 9 , whence d =0.825 inches, t=P/2dS c = 0.404 inches, and p = d+P/tSt =3.025 inches. While the above results satisfy the given conditions, it is not practicable to use the exact values, because plates and rivets of these dimensions could not be found in the market and would have to be specially made. The nearest approach to market sizes would probably be /=f inches and d = {-% inches, which give _/> = 3.i8, or say 3 T \ inches. Using these dimensions, the values of S tt S cy and S s are found to be 9 ooo, 13 100, and 7 700 pounds per square inch, which differ but little from the speci- fied working unit-stresses. Another method of designing is to establish three expressions for the efficiency, and then give to /, d, and p such values as will make the three efficiencies equal under the assigned unit stresses. Let a. denote the number of rivets in the width p which transmit the tension P, and let /? denote the number of rivet sections in the same space over which the shear is distributed. Then from the definition of efficiency in the last article, for tension of the plate, efficiency = (p d)/p for compression of the rivet, efficiency = a . dS c /pS t for shear of the rivet, efficiency =/? . \7td 2 S s /ptS t \ Equating now the second of these efficiencies to the third, the value of d in terms of / is found; equating the first and second and eliminating d, the value of p in terms of / is obtained; accordingly, S from which the diameter and pitch of the rivets can be com- puted when / is assumed. The efficiency of the joint now is (p-d)/p or aS c /(S t +aS e ). Using for steel plates and rivets the working stresses S t =9 ooo, 5* c = i2ooo, and ^=7500 pounds per square inch, the above formulas give for a riveted lap joint with single riveting, where a = i and ft = i, the proportions, j.48t 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 sidewise 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 15 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 vonsiderations. The lap of the plates should be sufficient so chat 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 // along which shearing tends to occur, and 2tlS 8 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 8 . 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 and 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. 34 shows a cantilever beam, and Fig. 35c shows a continuous beam of two equal spans. Fig. 3&z Fig. 356 Fig. 35c The ' Elastic Curve ' is the curve formed by a beam as it deflects downward under the action of its own weight and of the loads upon it. Experience teaches that the amount of this deflection and curvature is very small. A beam is said to be ' fixed ' at one end when it is so arranged that the tangent to the elastic curve at that end always remains horizontal; this may be done in practice by firmly building one end into a wall. A beam fixed at one end and free at the other end is a canti- lever beam. The loads on beams are either uniform or concentrated. A 88 GENERAL THEORY OF BEAMS CHAP. V 'uniform load' embraces the weight of the beam itself and any load evenly spread over it. Uniform loads are estimated by their intensity per unit of length of the beam, and usually in pounds per linear foot. The uniform load per linear unit is designated by w t then wx will represent the load over any dis- tance MI if / is the length of the beam, the total uniform load is wl, which may be represented by W. A ' Concentrated Load ' is a single applied weight and this is designated by P; a moving concentrated load is usually applied by a rolling wheel. In this chapter the fundamental principles applicable to all kinds of beams will be set forth. Unless otherwise 'stated, a beam will be regarded as of uniform cross-section throughout its entire length, and in computing its weight the rules of Art. 17 will be of service. For example, the weight of a wooden beam 6x8 inches in section area and 10 feet long is 6x8X3^X1! = 133 pounds; the weight of a steel beam of the same size is 6x8X3^X10X1.02 = 1632 pounds. Prob. 35a. Find the diameter of a round steel bar which weighs 6c pounds, its length being 5 feet. Prob. 356. A round steel bar of 2 J inches diameter and 4 feet length weighs 48 pounds. What is the diameter of a cast-iron bar which has the same weight and length ? ART. 36. REACTIONS OF SUPPORTS When a beam is laid upon supports its weight and the weight of its load are borne by the supports which exert 'reactions ' upward against the beam. In fact, the beam is a body in equilib- rium under the action of a system of forces which consists of the downward loads and the upward reactions. The loads are usually given in intensity and position, and it is required to find the reactions. This is effected by the application of the funda- mental conditions of static equilibrium which, for a system oi vertical forces in one plane, are Algebraic sum of all vertical forces =o Algebraic sum of moments of all forces =o The first of these conditions shows that the sum of all the loads AKT. 36 REACTIONS OF SUPPORTS 89 on the beam is equal to the sum of the reactions. Hence if there /s but one support, this condition gives at once the reaction. For two supports it is necessary to use the second condition. The axis of moments may be taken at any point in the plane, but it is more convenient to take it at one of the supports. For example, consider a single concentrated load P situated at 4 feet from the left end of a simple beam whose span is 13 feet. The equation of moments, with the axis at the left support, is 4P 13^2 = 0, from which R 2 =^P. Again, the equation of moments, with the axis at the right support, is 13^1 gP=o y from which Ri=^ E P. As a check it may be observed that Ri+R 2 =P. For a uniform load over a simple beam it is evident, without applying the conditions of equilibrium, that each reaction is one-half the load. For a uniform load over the cantilever beam of Fig. 356 it is plain that the vertical reaction at the wall is equal to the weight of the load. 300 3-- 1501 3'^ --- 4<-- fe 4 Fig. 36a Fig. 366 The reactions due to both uniform and concentrated loads on a simple beam may be obtained by adding together the reac- tions due to the uniform load and each concentrated load, or they may be computed in one operation. As an example of the latter method let Fig. 366 represent a simple beam 12 feet in length between the supports and weighing 35 pounds per linear foot, its total weight being 420 pounds. Let there be three concentrated loads of 300, 60, and 150 pounds placed at 3, 5, and 8 feet respectively from the left support. To find the right reaction R 2 the axis of moments is taken at the left sup- port, and the weight of the beam regarded as concentrated at its middle; then the equation of moments is, = 420X6+300X3 + 60X5+150X8 90 GENERAL THEORY OF BEAMS CHAP, v from which R 2 =4io pounds. In like manner to find RI, the axis of moments is taken at the right support, and RiX 12 = 420X6+ 300X9 + 60X7-!- 150X4 from which ^1=520 pounds. As a check the sum of RI and R 2 is found to be 930 pounds, which is the same as the weight of the beam and the three loads. When there are more than two supports, the problem of find- ing the reactions from the principles of statics becomes inde- terminate, since two conditions of equilibrium are only sufficient to determine two unknown quantities. By introducing, how- ever, the elastic properties of the material, the reactions of con- tinuous beams may be deduced, as will be explained in a follow- ing chapter. In most cases of the discussion of a beam, the determination of the reactions is the first step. Prob. 36a. A simple beam weighing 30 pounds per linear foot is 18 feet long, and it carries two concentrated loads of 350 and 745 pounds at distances of yj and 9 feet from the left end. Compute the reactions due to the total load. Prob. 366. When a single load P is On a simple beam of span / at a distance / from the left support, show that the reactions are Ri = P(i-K) and R 2 = PK. ART. 37. THE VERTICAL SHEAR The failure of a beam sometimes occurs by shearing in a vertical section as shown in Fig. 37 a for a simple beam and in Fig. 37b for a cantilever beam. This shearing is produced by two equal and parallel forces acting in opposite directions on the left and right of the section. In the second diagram there acts on the left of the section a downward force equal to the sum of all the loads on the left, and on the right of the section an equal force acts upward. In the first diagram there acts on the left of the section an upward force which is equal to the reaction minus the weight of the beam between the support and the section. That such forces actually exist will be readily understood by considering a numerical case. Thus, for Fig. 37a, let the total weight of the beam and loads between the supports ART. 37 THE VERTICAL SHEAR 91 be i 200 pounds, the weight between the support and the sec- tion mn be 50 pounds, and each reaction be 600 pounds; on the left of the section mn the upward vertical force is 600 50 = + 550 pounds, and on the right of the section the upward ver- tical force is 1150 + 600= 550 pounds; here the sign + indicates that the former force acts upward and the sign indi- cates that the latter acts downward. In any case, whether the beam be simple, cantilever, or continuous, let W indicate the total weight upon the supports, W the weight on the left of any section and W" the weight on the right of the section , or W' + W"=W\ also let R' indicate the sum of all the reactions on the left and R" the sum of all the reactions on the right of the section, or R f + R" = W. Then the vertical force on the left of the section is R' W and that on the right of the section is R" W"', these two quantities are equal in magnitude, but they have opposite signs, since their sum is zero; accordingly if one acts upward the other acts downward and they constitute a shear (Art. 6). _ __ , i; _ Fig. 37 and \x the lever arm of the load wx. Here M=o when x = o and also when x=l, and M has its greatest value Jw/ 2 when x = %l. The equation is that of a parabola whose graphical representation is as given in Fig. 38d, each ordinate show- ing the value of M for the corresponding value of the abscissa x. Consider next a simple beam loaded with only the three weights PI, P%, P$. Here M=R\x for any sec- ART. 38 THE BENDING MOMENT 95 tion between the left support and the first load, and M = RiX-Pi(x pi) for any section between the first and second loads. Each of these expressions is the equation of a straight line, x being the abscissa and M the ordinate, and the graphical repre- sentation of bending moments is as shown in Fig. 3Sc. It is seen that for a simple beam all the bending moments are positive. For any given case, the bending moment at any section may be found by using the above definition. The external forces on the left of the section are taken merely for convenience, for hose upon the right of the section produce the same bending moment with reference to the section. The bending moment in all cases is a measure of the tendency of the external forces on either side of the section to turn around a point in that section. If this bending moment is sufficiently large, the internal stresses in the beam become so great that rupture occurs. Fig. 38c Fig. 38d The bending moment at any section may also be obtained by using the forces on the right of that section. Thus, for the beam in Fig. 38c, let R 2 be the right reaction and ps the distance of the load PS from the right support; then for a section under the load PS the bending moment is M = R 2 ps and this is equal to that found by using the forces on the left of the section. When forces on the right of a section are used, a moment is to be taken as positive when it tends to cause rotation in the opposite direction to that of the hands of a clock, for this direction would be clockwise if the beam be viewed from the other side. Prob. 38. Draw a beam resting on three supports and having two spans each 13 feet long. Let a load of 160 pounds be placed at the middle of each span producing a reaction of 220 pounds at the middle support and 50 pounds at each end support. Compute the vertical 96 GENERAL THEORY OF BEAMS CHAP, v shears and bending moments for several sections, and draw diagrams to show their variation throughout the beam. ART. 39. INTERNAL STRESSES AND EXTERNAL FORCES The external loads and reactions on a beam maintain their equilibrium by means of internal stresses which are generated in it. It is required to determine the relations between the external forces and the internal stresses; or, since the effect of the external forces upon any section is represented by the ver- tical shear (Art. 37) and by the bending moment (Art. 38), the problem is to find the relation between these quantities and the internal stresses in that section. i I mi j j I m j ^X V? Y Fig. 39 and the equation of moments with respect to the bottom of the lower flange is, from which c is computed. , Actual sections of these forms have the corners more or less rounded, but values of c computed in this manner are usually sufficiently precise for all practical uses. <---&-,--* Fig. 426 Fig. 42c Fig. 42d Fig. The student should be prepared to readily apply the prin- ciple of moments to the deduction of the numerical value of c for any given cross-section. In nearly all cases the given area may be divided into rectangles, triangles, and circular sectors, the centers of gravity of which are known, so that the equation for finding c is readily written from the definition of the center of gravity. Triangular beams are never used in practice, but it is well to note that the center of gravity of a triangle lies upon a line parallel to its base and at a distance from that base equal to one-third of the altitude. Centers of gravity for sections like those of railroad rails are determined by dividing the area into strips which are so small that each may be regarded as a trapezoid. The areas of ART. 43 MOMENTS OF INERTIA 105 these elementary trapezoids being computed, their centers of gravity may be taken at the middle of their widths with sufficient precision, and then the value of c is determined by the same method as that explained above. There is also a graphical method for locating the center of gravity of sections which is sometimes advantageous, and this is set forth in Roofs and Bridges, Part II, Art. 12. Prob. 42a. If the side of a square is d, show that c=dV r % when the square is placed so that one diagonal is vertical. Prob. 426. For a trapezoidal section of depth d, let the shorter base be b' and the longer base be b. Find the distance of the center of gravity from the shorter base. Prob. 42c. A deck beam has a section like Fig. 426, except that the lower flange is approximately circular. Find the value of c when the total depth is 10 inches, the width of the upper flange 5 inches, the thickness of that flange J inch, the thickness of the web f inch, and the diameter of the circular bulb i j inches. ART. 43. MOMENTS OF INERTIA In the flexure formula S .I/c=M, the letter / denotes the moment of inertia of the cross-section of the beam with refer- ence to a horizontal axis passing through the center of gravity of that section. Let da be any elementary area of the section and z its distance from this axis; then I = Zdz . z 2 is the equa- tion from which the value of / is ascertained, and this in general is most advantageously done by the methods of the calculus. Strictly speaking an area has no weight or inertia and therefore no moment of inertia, but it is customary to give the name 'moment of inertia ' to the quantity Ida . z 2 , which is of very frequent occurrence in all branches of applied mechanics. For the rectangle of breadth b and depth d in Fig. 43a, the elementary area da may be taken as a strip of length b and width dz, or da = b . dz. Then Ida . z 2 equals fbz 2 dz and the value of z in this integral is to extend between the limits + \d and - \d. This gives I = ^bd 3 , which is the moment of inertia of the rect- 106 GENERAL THEORY OF BEAMS CHAP. V angle with respect to an axis through its center of gravity and parallel to the base. For the triangle with base b and altitude d in Fig. 43b, the elementary strip has the width dz and the variable length x. Since the center of gravity is at a distance of %d from the base, the value of x is, from similar triangles, x = ($dz)b/d, and the expression Ida . z 2 becomes b/d . J(%d-z)z 2 dz. This being taken between the limits +%d and $d, gives the result / = -s^bd 3 , which is the moment of inertia of the triangle with respect to an axis through its center of gravity and parallel to the base b. For the circle of diameter d in Fig. 43c, a similar method may be used. The length of the elementary strip being 2X 9 the relation between x and z is given by x 2 + z 2 =(^d) 2 . The general expression 2 da . z 2 becomes J 2xz 2 dz, and, replacing x Sy its value in terms of z, and integrating between the limits + Jdand %d, gives the result /= fand*, which is the moment of inertia of the area of a circle with respect to an axis through its center. Fig. 43a Fig. 436 Fig. 43c Fig. 43d For the circular annulus in Fig. 43 d, let d be the outer and di the inner diameter, then it is plain from the definition that the moment of inertia of the annulus is that of the outer circle diminished by that of the inner circle or I = -fax(d 4 - d^). Similarly for the hollow rectangle in Fig. 43e, let b and 61 be the widths and d and d\ the depths, the thickness of both top and bottom being \(d-d^)\ then I = ^(bd 3 -Mi 3 ). For the / section in Fig. 43/ where the thickness of each flange is /] and that of the web is /, the moment of inertia / is also j^(bd 3 -bidi 3 ) if bi represents b t and di represents d 2l\. The moment of inertia of a rectangle with respect to an axis through its base is frequently needed. Fig. 43g shows this case, ART. 43 MOMENTS OF INERTIA 107 and da is b . dz\ then J bz 2 dz is to be taken between the limits d and o, which gives I\ =$bd 3 . This is seen to be four times as great as the moment of inertia for a parallel axis through the center of gravity. By using the result of the last paragraph, the moment of inertia of the T section in Fig. 43 h with respect to an axis through its center of gravity can easily be written. Let the distance c and GI be first found by Art. 42; let /i be the thickness of the flange and / that of the web. Then J/c 3 is the moment of inertia of the portion of the section below the axis, and %bci 3 3(b t)(citi) 3 is the moment of inertia of the portion above the axis. The sum of these is the required moment of inertia of the entire section. U &--,' t- 6 __ --h "i P l c --- .___! cJ L, Fig. 43h Fig. 43i Fig. 43e Fig. 43/ Fig. 43g The moment of inertia of any section area is less for an axis through its center of gravity than for any parallel axis. If / is the value for the axis through the center of gravity, /i that for any parallel axis, a the area of the section, and h the distance between the two axes, the formula I*=Ii-ah 2 furnishes the means of finding 7 when I\ is known. For example, take the rectangle of breadth b and depth d where I\ is known to be %bd 3 for an axis passing through one of the bases; then for a parallel axis through the center of gravity of the rectangle, the value of 7 is %bd 3 -bd(d) 2 =^bd 3 . Similarly, the value of 7i for the T sec- tion in Fig. 426 or Fig. 43h with respect to an axis through the base of the web is \bd?-%(b-t)(d-ti) 3 , where d is the total depth; this being computed, the value of 7 for the parallel axis through the center of gravity is Iiac 2 . For the beam of Fig. 43z, which is made by riveting four angles to a web, the expression of 7 is the same as that for Fig. 43/. It is, however, customary to use tables in finding the numerical value of 7 for such a section, these tables giving the moments 108 GENERAL THEORY OF BEAMS CHAP, v of inertia of various sizes of angles with respect to axes through their centers of gravity and parallel to the legs. For example, let the horizontal leg of each angle in Fig. 43* be 4 inches, the vertical leg 3 inches, and the mean thickness | inch. Then Table 10 shows that the distance from the back of the long leg to an axis through the center of gravity of the angle is 0.83 inches, that its section area is 3.25 square inches and that the moment of inertia of the section with respect to this axis is 2.42 inches 4 . Four of these angles being used with a web f inches thick and 30 inches depth, the moment of inertia of the angle sections with respect to the horizontal ax's through the center. of gravity of the web is 4X 2.42+4x3-25(15 o. 83)2 = 2620.0 inches 4 , while the moment of inertia of the web itself is -^ X f X 3o 3 = 843 . 7 inches 4 , thus giving 7 = 2620.0 + 843.7 = 3464 inches 4 for the moment of inertia of the entire section with respect to an axis through its center of gravity. The moment of inertia of a plane area contains the product of four linear dimensions and is hence expressed in quadratic units ; for the sake of clearness l inches 4 ' will be used to designate the unit in which / is measured. When lengths are expressed in inches, areas are in square inches, volumes in cubic inches, and moments of inertia in inches 4 . The section factor I/c is the product of three linear dimensions and is expressed in ' inches 3 ' ; since it is not a volume the term cubic inches would not be appropriate. Prob. 43#. Consult a book on theoretic mechanics and obtain a proof of the valuable rule Ii = I+ah 2 . Prob. 436. For Fig. 43& let the upper flange be 4Xi inches, the web be 5iXi inches, and the total depth be 6 inches. Compute the area of the section and the value of c, without using algebraic for- mulas. Compute the moments of inertia of flange and web with re- spect to horizontal axes through their centers of gravity. Transfer these to the horizontal axis shown in the figure and find / for the entire section with respect to that axis. ART. 44. ROLLED BEAMS AND SHAPES Soon after 1830, owing to the rapid progress of railroad con- struction, wrought iron began to come into use for railroad rails ART. 44 ROLLED BEAMS AND SHAPES 109 and bridges, and by 1850 rolled beams and angles were exten- sively employed. The Bessemer and the open-hearth processes of making steel, introduced in 1856 and 1863, rendered it possible to produce ingots from which steel rails and beams could be rolled, but it was not until about 1890 that steel beams began seriously to compete with those of wrought iron. After this date, however, the use of wrought iron for beams and shapes rapidly declined and since 1900 medium steel has taken its place as a structural material (Art. 25). Table 6, at the end of this volume, gives the properties of the most common sizes of the medium-steel I beams rolled iii the United States of America in 1905. The first column con- tains all the depths that are rolled, but for each depth other sizes than those given in the second, third, and fourth columns may be obtained; for instance, for the 1 8-inch beam, sizes may be ordered which weigh 65 and 60 pounds per linear foot, the width of the flanges being intermediate between 6.26 and 6.00 inches. The sizes marked with an asterisk are called standard, and they are always found in the market, while the other sizes must generally be specially ordered and hence cost a little more per pound. The fifth column of the table contains the moments of inertia of the sections with respect to an axis AB drawn through the center of gravity of the section and normal to the web, as shown in Fig. 44 tion; then by Art. 43 the moment of inertia of each with respect to a horizontal axis through its center of gravity is -rjXgXi 3 = 0.75 inches 4 , and the moment of inertia of the two plates with respect to the neutral axis of the beam is 2X0. 75 + 2X9.00X5. 5 2 = '46.0 inches 4 . Accordingly the required moment of inertia of the section is 1=133.8 + 546.0 = 679.8 inches 4 , and the section factor is I/c= 679.8/6.0= 113.3 inches 3 . 1* 6 H \ Fig. 44g Fig. 44h Fig. 44* Fig. 44; Fig. 44k The complete theory of moment of inertia is a lengthy one and properly belongs to pure mathematics; see Art. 170 forfullei discussion. It may be noted here, however, that the sum of the moments of inertia of a plane area with respect to two rectangular axes through a given point is always a constant whatever be the directions of those axes. Thus, for the rectangle in Fig. 444, the moment of inertia with respect to a hori- zontal axis through the center of gravity by Art. 43, is -fabd? and that for a vertical axis through the same point is T?db 3 . If any other rectangular axes be drawn, as in the figure, the moments of inertia with respect to them are less than -fybd 3 and greater than -f^db 3 , but their sum is equal to ^bd(d 2 + b 2 ). In this case the greatest moment of inertia for an axis through the center of gravity is when the axis is parallel to the short side of the rectangle, and the least is when the axis is parallel to the long side. Similarly the moments of inertia for the channel sections in Table 9 are the greatest and least with respect to axes through the center of gravity which are perpendicular and parallel to the web. 112 GENERAL THEORY OF BEAMS CHAP. V For the angle sections in Table 10 the given moments of inertia are with respect to axes through the center of gravity and parallel to the legs, as shown in Figs. 44e and 44/. These are the values usually required in designing, but it may be men- tioned that they are not the greatest and least values. Fig. 44; shows the approximate position of the two rectangular axes for which the moments of inertia are greatest and least, MN giving the greatest value and mn the least. When the legs of the angles are equal, each of these axes is inclined to the back of the legs at an angle of 45 degrees. The center of gyration of a section is the point where the entire area might be concentrated and have the same moment of inertia as the actual distributed area. The distance from the axis to this center is called the 'radius of gyration ' and this is designated by r. From this definition it follows that I = ar 2 , or r 2 =I/a. Values of the radius of gyration for rolled steel sections with respect to axes through their centers of gravity are given in Tables 6-10. Fig. 44k indicates that the distance r is always less than the distance c from the neutral axis to the remotest fiber. Prob. 44a. Show that the weight of the medium-steel beams and shapes in Tables 6-10 is 3.4 pounds per linear foot for each square inch of cross-section, or 489.6 pounds per cubic foot. Prob. 446. Let the section in Fig. 44h consist of a plate JX 24 inches and four angles each 4X3X3- inches in size, the longer leg of the angle being horizontal. Compute the moment of inertia of the section with respect to its neutral axis. ART. 45. ELASTIC DEFLECTIONS When a beam is subject to the action of loads the horizontal fibers on one side of the neutral axis are elongated and those on the other side are shortened (Art. 40). The beam therefore bends and all points except those over the supports deflect below their original position. The curve assumed by the neutral sur- face of the beam, when the elastic limit of the material is not exceeded, is called the 'Elastic Curve', and its general equation will now be deduced. ART. 45 ELASTIC DEFLECTIONS Let mn in Fig. 45a or 456 be any short length measured along the neutral surface of the beam. Let kk' and pp' be two normal sections passing through m and n\ before the bending kV and pp' were parallel, after the bending they intersect at o, the center of curvature. Let qq f be drawn through n parallel to kk f . Then qp represents the elongation of the fiber kq, and x; the maximum shears hence occur when # = o and x = l, their values being +\W and -\W . The moment for the section distant x from the left support is M= + \wl . x-wx . \x y and by the usual method of the dif- ferential calculus this is found to be a maximum when # = J/, 120 CANTILEVER AND SIMPLE BEAMS CHAP. VI and for this value M = + Jw/ 2 = %Wl. For a single load P at the middle of a simple beam, each reaction is JP 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= JP 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 Rip or P(l p)p/l, which has its greatest value \Pl when p equals J/. 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. 38c. 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. 463. 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 (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 pi and PI at the distance p2 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 t or 8M/dx=Ri-wx-P l -P 2 -etc.=o is the equation which gives the value of x. But RI -wx P\ P 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 I = 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 +310-65 = +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 310 X 5.385 - i X 13 X 5.3852 - 240 Xo.385 = i 388 pound-feet. Prob. 470. 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 300, 60, and 150 pound? at 3, 5, and 8 feet respectively from the right support. Compute the maximum shear and moment, and draw the shear and mpmenf 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 S . I/c =M which was established in Art. 41, For this purpose the forrmTa may be written S= Me / 1 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 If 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 / and c are expressed in terms of inches; then the value of S 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 r the distance from it to the neutral surface, then by Art. 40, S/c=S'/c' or S'=S.c'/c When S is tension, S' is compression; when S is compression, S' is tension. Sometimes it is necessary to compute S' as well as S in order thoroughly to investigate the stability of the beam. By comparing the values of S 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, /= 286.7 inches 4 by Art. 17, w= 62.7 pounds per linear foot by Art. 47, x= 6 feet for dangerous section by Art. 38, max. M=iS 630 pound-feet Then from the flexure formula, the unit-stress is, 5=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 compress! ve stress on the upper side. The factor of safety in compression is 90 ooo/ 3 900 or about 23, while that in tension is 20 000/3 900 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, I=^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 =\Wl + ^Pl = jo6.j pound-feet = 8 480 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 8 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 7 460 pounds ; hence 3 = 7460/20=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. 480. 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 S 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, / =TT> an d M=^6x6w. 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 = JPXi8Xi2 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. so DESIGNING OF BEAMS 125 however, the moment is M = 54P-fi2 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 ij 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 i 031 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 1X90000 = 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 J X 20 ooo = 3 330 pounds per square inch ; inserting the values in the flexure formula M = S f I/c', the load W is found to be 30 400 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. 49&. 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 c=2.i6 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 S is assumed in accordance with engineering practice. Then the flexure formula (41) may be Written 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 ooo pound-inches. From Art. 7 the allowable value of S 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 I = -fa7cd*, hence ^V 7r ^ 3 = 2 4j from which ^ = 2.9 inches. If the section is rectangular, 2 inches wide and 2j 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 1^ = 180X24X4 = 17280 pounds; second, the maximum moment is M = JX 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 7/^ = 44.8, and this is the size to be selected. The larger table given in the handbook of the manufacturers indicates, however, that a 12 -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. 500. 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. 5Qb. A floor, which is to sustain a uniform load of 175 pounds per square foot, is to be supported by heavy lo-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 12 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 S 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 i^bd 3 /^d or Jfo/ 2 , so that the flexure formula becomes %bd 2 = M/S or a = 6M/Sd, so that for given values of M and S y 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 t 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 / 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 / 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 a= 12.48 square inches, 1 = 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 = $8.C) 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, whence d=D\/% 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, Flg< 51 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 -j 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' 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. 52a represents a rectangular cast-iron beam where the tensile unit-stresses below n have exceeded the elastic limit, while Fig. 526 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 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. 52a Fig. 526 When a beam is ruptured under transverse loads the value of S computed from the flexure formula is called the ' modulus of rupture ' 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 Modulus of Compressive Strength, S t Rupture, S} Strength, S e 3 000 6 ooo 8000 90000 Material Brick Stone Timber 10 ooo Cast Iron 20 ooo Wrought Iron 50 ooo Structural Steel 60 ooo Strong Steel 100 ooo For wrought iron and structural steel no values of S f are given 800 2 OOO 9 ooo 35000 no ooo 50 ooo 60 ooo 120 OOO 132 CANTILEVER AND SIMPLE BEAMS 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 = JX i. 044/2 = 0.522^ 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 7 = 299 inches or about 25 feet. The modulus of rupture S f is sometimes called ' computed flexural strength.' The latter points out the meaning of the quantity, but the former is in general use. Prob. 520. Compute the size of a square wooden simple beam of 8 feet span in order to break under its own weight. Prob. 526.- 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. f 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. 53#, 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 p z)+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 = J/ \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 coo 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^ 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. 53c 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. 536 occurs at one of the supports when one load is just about to pass upon that support and its numerical value is 2PP. p/l. By the use of the shear formula 5> = V and the flexure formula SI/c=M, the unit-stresses S g and S 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 8 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 . d 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 w\ 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 %wl 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=%wl 4 /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. 54Pl 3 /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, I2PI _ 8~*~~ Ebd? Eb^ By integrating this twice and determining the constants as before, the equation of the elastic curve is found, whence the deflection is f=6Pl 3 /Ebid 3 which is fifty percent greater than that of a cantilever of uniform section and breadth b\. 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 15 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 t 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 For a load at the middle and depth constant, M = \Px as before. Hence b = (^P/Sd 2 )x ) and the plan must be triangular or lozenge-shaped, the width uniformly increasing from the support to the load. If b\ is the breadth at the middle, the equa- tion of the line becomes &=&i (#/}/) 147 ART. 58 SIMPLE BEAMS OF UNIFORM STRENGTH For a uniform load and constant breadth, M = \wlx 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 = (^di 2 /l 2 )(lx x 2 ). For a uniform load and constant depth, b= (^w/Sd 2 )(lx x 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 = ($i/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. T == fl J A B C D" a b c d 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 J.48 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. 586. 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 20 ooo pounds per square inch. ART. 59 BEAM OVERHANGING ONE SUPPORT 149 CHAPTER VII 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, RI = \wl \wm (m/l) R 2 = \wl + wm + \wm (m/l) 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 / iVhen x is greater than / V=Ri-wx V= i'he curves corresponding to these equations are shown on Fig. 590. The shear curve consists of two straight lines; V = ft when X QJ and V = o when x = R\/w\ at the right support V*=Ri 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 t and M is a maxi- .num 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. *^-O_4_-^-J Fig. 590 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, m = io feet, and w = ^o pounds per linear ART 59 BEAM OVERHANGING ONE SUPPORT 151 foot, the reactions are ^1=300 and ^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 w = o, there is no overhanging part and the beam is a simple one; if w = J/, the case is that just discussed; if m=-l, the reaction RI 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, 2$EIy = 4^1 (x 3 - I 2 x) -w(x*- I 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. 59&. 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, vn- 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. 600 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. i 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 # = |/. The maximum positive moment occurs when dM/dx=o or when x = %l, and its value is +ijb-w/ 2 . The maxi- mum negative moment occurs when x = l and its value is -\wl 2 - 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 8# 3 qlx 2 + 1 3 = o, one root of which is x = 0.42157, while the others are inapplicable to this problem. Hence f=o.oo$4.wl 4 /EI is the value of the maximum deflection. Case II. Load P at Middle. Here it is necessary to con- sider that there are two elastic curves having a common ordinate and a common tangent under the load, since the expression, for the moment are different on opposite sides of the load. Thus taking the origin as usual at the supported end, On the left of the load, (a) EI 2 = Rx (b) E/- (c) On the right of the load the similar equations are, (b)' E-= ooc (c)' EIy= %Rx* - %Px*+Plx 2 + C 2 x+ C 4 To determine the constants consider in (c) that y=o when x = o and hence that C 3 =o. In (by the tangent dy/dx = o when x = l and hence C 2 =%Rl 2 . Since the curves have a common tangent under the load, (b) = (by for x = \l, and thus the value of Ci is found. Since the curves have a common ordinate under the load, (c) = (c}' when x = \l, and thus C 4 is found. Then, (c) 24EIy= 4Rx*+ ^Pl 2 x- 1 2Rl 2 x (cy 4&EIy= SRx* - SP* 3 + 1 2Plx 2 - 2^Rl 2 x+ PI 3 are the equations of the two elastic curves. Making x = l in (c) f the value of y is zero, and then the left reaction is R=^P. The moment on the left of the load is now M =-?$Px, and that on the right M=-\\Px + %Pl. The maximum posi- tive moment obtains at the load and its value is ^Pl. The 154 OVERHANGING AND FIXED BEAMS CHAP. VII maximum negative moment occurs at the wall, and its value is -faPl. The inflection point is at x = f T l. The deflection under the load is readily found from (c) by making x = J/. 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 id the following results may be deduced by a method exactly similar to that of the last case. Reaction at supported end = \P(? 3*:+ /c 3 ) Reaction at fixed end =%P($K /c 3 ) Maximum positive moment = \PlK.(z 3*+ K 3 ) Maximum negative moment = JP/(/c K 3 ) The absolute maximum deflection for this case occurs under the load when x = 0.4141, and its value will be found to be given by Prob. 60a. 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. 60fr. 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 , the maximum shears at the supports are wm and % the maximum moment at the middle is + w(J/ 2 - Jw 2 ), the maxi- mum moment at each support is Jww 2 , and the inflection points are distant J(/ 2 4W 2 )* from the middle of the beam. Fig. 61 a 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 J/, there are no positive moments in the middle span. Fig. 61a 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 and 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. 610. For Fig. 61a find the ratio of / to m in order that the maximum positive moment may numerically equal the maximum nega- tive moment. Prob. 6lb. 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. 61c. For Fig. Qla find the ratio of / to m in order that there may be no positive moment. 156 OVERHANGING AND FIXED BEAMS CHAP. Vll 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 m 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, M= (wm-\- \wT)x which may be written in the simpler form, where MI represents the unknown bending moment at the left support. The distance m can hence be found when MI has been determined. Again, for a single load P at the middle of / in Fig. 6 la, 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+$P)x-Q(m+x) or M=M l + %Px in which MI 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, Integrating twice, making dy/dx = o when x = o and also when x=l, there is found M\= ~fawl?,axi.d the equation of the elastic curve is 24EIy= w( -I 2 x 2 + 2lx* -x 4 ) from which the maximum deflection is found to be f=*-g% T wl 4 /EI. The inflection points arejocated by placing M equal to zero, which gives x = ty(i$\/3). The maximum positive moment is at the middle and its value is ^ivl 2 ', accordingly the horizontal ART. 62 BEAMS FIXED 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 %wl, at the middle o, and at the right end \wl. Fig. 62a shows the shear and moment diagrams. Case II. Load P at Middle. Here the differential equation of the elastic curve is El . $ 2 y/dx 2 =Mi + %Px and in a manner similar to that of the last case it is easy to find that the maxi- mum negative moments are fP/, that the maximum positive moment is |P/, that the inflection points are half-way between the supports and the load, and that the maximum deflection is Fig. 620 Fig. 626 Case III. Load P at any Point. Let the load P be at the distance id from the left end, K being any number less than unity. Let M\ and R\ denote the unknown bending moment and reac- tion at that end. Then for any section on the left of the load M = M \-\-R\x, and for any section on the right of the load M = Mi + RiX P(x id). By inserting these in the differential equation (45), integrating each twice and establishing sufficient conditions to determine the unknown M l and RI and also the constants of integration, the following results may be deduced: Reaction at left end = P(i -3*2+ 2/c 3 ) Reaction at right end = P/c 2 (3 2/c) Moment at left end = Plic(i ~2K+ K 2 ) Moment at right end = PlK 2 (i - K) Moment under load = + P//c 2 (2 4*+ 2K 2 ) When K = %, the load is at the middle and these results reduce to the values found in Case II. The maximum deflection occurs under the load when it is at the middle. 158 OVERHANGING AND FIXED BEAMS CHAP, vii Prob. 62#. From the above results for Case III deduce the positions of the two inflection points. Find the inflection points when two equal loads are at \l and }/ from the left end. Prob. G2b. What medium-steel I beam is required for a span of 24 feet to support a uniform load of 25 ooo pounds with a unit-stress S of 1 6 ooo pounds per square inch, the ends being merely supported? What one is needed when the ends are fixed ? ART. 63. COMPARISON OF BEAMS As the maximum moments for fixed beams are generally less than for simple beams, their strength is relatively greater. This was to be expected, since the constraint lessens the deflec- tion which would otherwise occur. Under a uniform load W, the maximum moment for a simple beam is \Wl, that for a beam fixed at one end is %Wl, and that for a beam fixed at both ends is TsWl', the fixing of one end does not increase the strength, but the fixing of both ends increases it fifty percent. For a single load P the maximum moment for a simple beam is \Pl and that for a beam fixed at both ends is \Pl\ hence th' strength of the latter beam under a concentrated load is double that of the former. For the beam fixed at one end and sup- ported at the other, it may be shown that the maximum posi- tive moment due to a load P is o.i74P/ and that the maximum negative moment is 0.192 P/, the latter occurring when the load is at a distance o.tf'jl from the supported end; hence the strength of this beam is intermediate between that of the simple beam and that of the beam with both ends fixed. With respect to stiffness, the advantage is always on the side of beams with fixed ends. Under uniform load the deflection of the simple beam is jfaWP/EI, while for a beam fixed at both ends it is one-fifth of this amount; hence the beam fixed at both ends is five times as stiff as the simple beam. Under a single load a similar comparison of the deflections shows that the beam fixed at both ends is four times as stiff as the simple beam. Foi the beam with one end fixed, the degree of stiffness is interme- diate between those for the other cases. The advantage of ART. 64 SUPPORTS ON DIFFERENT LEVELS 159 fixing the ends is hence much greater with respect to stiffness than with respect to strength. Table 12, at the end of this volume, recapitulates the above results, and also those deduced in Art. 56 for cantilever and simple beams. In all cases W represents the load whether con- centrated or uniformly distributed. The results given for beams with horizontal restraint, it may be observed, have been deduced from the equation of the elastic curve, which is only valid when the elastic limit of the material is not exceeded by the unit-stress S at the dangerous section. When beams with one or both ends fixed are loaded so that the elastic limit is exceeded, the above results deduced for reactions, moments, and deflections are not applicable except as approximations. Let i/a represent the numerical coefficient in the column of maximum moments in Table 12 and i//? the numerical coeffi- cient in the column of maximum deflections. Then, as in Art. 56, the relation between the unit-stress S and the deflection / is g'ven by, S/f=pEc/aP or aSP=pEcf provided the elastic limit of the material is not exceeded. Prob. 63a. For a uniformly loaded beam with equal overhanging ends, derive a formula for the deflection at the middle. Prob. 63ft. Find the deflection of a 9-inch steel I beam of 6 feet span and fixed ends when loaded at the middle so that the tensile and compressive stresses at the dangerous section are 16 800 pounds per square inch. ART. 64. SUPPORTS ON DIFFERENT LEVELS For all beams thus far discussed the two supports of the beam have been taken as in the same horizontal plane, this being the usual case in practice. A depression of one support below the level of the other may, however, occur, and its influence will now be considered, this depression being taken as very small so that the length of any small part of the elastic curve does not sensibly differ from that of its horizontal projection (Art. 45). For the simple beam in Fig. 64a, let the left support be 160 OVERHANGING AND FIXED BEAMS CHAP. VII depressed the distance h below the level of the right support and let the horizontal distance between the supports be /; then h/l is the tangent of the angle of inclination, or tan 6=h/l. Now, for a load P at the horizontal distance d from the left support, the left vertical reaction is found by taking an axis of moments at the right support to be P(i K) which is independent of 6 and the same as if the supports were on the same level. The shears and moments throughout the beam due to the vertical forces are hence unaltered by the inclination. In computing, the fiber unit-stress from the flexure formula S . I/c = M, the dimensions used in finding I/c are those of a normal section of the inclined beam, and hence the unit-stress S is that on the inclined upper and lower surfaces of the beam. A simple beam is therefore unaffected by a slight change in the relative levels of its supports, unless it be from the influence of the horizontal forces which must come into play at the supports to prevent sliding. Fig. 64a Fig. 646 Fig. 64c Let Fig. 646 represent a beam supported at the left end and fixed horizontally at the right end, the vertical distance of the left below the right end being h. Under a uniform load of w per linear unit, the moment at a section distant x from the left end is M=R\x %wx 2 , where RI is the unknown reaction at the left support, and x is the distance of any section from that support. Inserting this in the differential equation (45) and integrating, the constant of integration is found by the condition that dy/dx=o when x=l. Integrating again, the constant is found by the condition that y = o when x=o, and, is the equation of the elastic curve. In this y becomes h when x becomes /, and accordingly the left reaction is, This shows that the value of R depends upon the difference of level of the two supports. When h = o, the case is the same ART. 64 SUPPORTS ON DIFFERENT LEVELS 161 as Case I in Art. 60 and R\='^wl\ when h is positive, or the fixed end higher than the supported one, RI is less than %wl- } when h is negative, or the fixed end lower than the supported one, ^i is greater than %wl. The formula also shows that the value of ^i depends upon the kind of material of the beam, since E is the modulus of elasticity of the material, and upon the size and shape of its cross-section, since these are included in the moment of inertia /. Let Fig. 64c represent a beam with both ends fixed, and let h be the vertical height of the right above the left end. Under a uniform load let the unknown reaction at the left support be RI and the unknown moment be MI. Then the differential equa- tion of the elastic curve for an origin at the left support is, as in Art. 62, Integrating this twice, and introducing sufficient conditions to determine the constants and the values of RI and MI, there result, RI = fal- (i 2EI/P)h MI = -Jjwl 2 + (6EI/l 2 )h and accordingly these values depend not only upon the difference in level of the supports but also upon the dimensions and kind of material of the beam. When h=o, the values of RI and MI are the same as those deduced in Art. 62. The reaction R 2 at the right end of the beam is wlR]_ and the moment there is found from M 2 = Mi+Ril ^wl 2 \ accordingly, R 2 =%wl+ (i 2 EI/l 3 )h M 2 = T l 2 wl 2 -(6EI/l 2 )h which show that at the higher end of the beam the reaction is always positive and the moment always negative, while at the other end they may be positive or negative depending on the value of h. Not only the reactions and moments at the supports but also the shears and moments throughout the beam undergo change when one support is lowered below the other. To ascertain the magnitude of these changes, take a lo-inch I beam of struc- tural steel which weighs . 25 pounds per linear foot, for which =30000000 pounds per square inch and I =122.1 inches 4 162 OVERHANGING AND FIXED BEAMS CHAP. VII (Table 6). Let one end be supported and the other fixed, as in Fig. 646 the clear span be 15 feet and the total uniform load W be 1 6 400 pounds. When both supports are on the same level, the left reaction is %W=6 150 pounds, and the maximum moment is 4^=30750 pound-feet = 369 ooo pound-inches; then the flexure formula gives 5=369000/24.42=15 100 pounds per square inch, which is a satisfactory unit-stress for steady load. Now let the left support be lowered 1.2 inches below the fixed end; the reaction at that end is then less than 6 150 pounds by the amount ^Elh/l 3 = 3X30 ooo 000X122.1 Xi.2/i8o 3 = 2 260 pounds, so that it is ^1 = 6150-2260=3890 pounds. Froni this the moment at the fixed end is found to be M=RJ ^Wl= 775 800 pound-inches, and the flexure formula then gives 5^ 31 800 pounds per square inch, which is too high a value for safety as it is but little less than the elastic limit of the material. The conclusion of this investigation is that beams with one or both ends fixed should not be used in circumstances where any material alteration in the levels of the supports may occur. The above formulas show that when the left support in Fig. 646 is lowered the distance h= W1 3 /SEI below the level of the other, the left reaction under uniform load is zero and the beam is a cantilever fixed at the right end. The same condition obtains for the beam in Fig. 64c when it is uniformly loaded and the left end is lowered the distance Wfi/2^EI\ when greater depres- sions occur, the left reaction is negative and the beams are in the condition of constrained cantilevers uniformly loaded and having a concentrated load at the free end (Art. 65). It should also be noted that the moments and reactions deduced above are only valid when the elastic limit of the materials is not exceeded. Prob. 640. For the case of Fig. 64c, deduce expressions for the dis- tance of the inflection points from the left support. Prob. 646. A wooden joist, 3X12X75 inches, is fixed in walls at its ends and carries a total uniform load of 7 680 pounds. Compute the maximum unit-stress S when the supports are on the same level; also when one support is lowered one-quarter of an inch below the other. ART. 65 CANTILEVER WITH CONSTRAINT 163 ART. 65. CANTILEVER WITH CONSTRAINT A case which occurs in the portal of a bridge is shown in Fig. 650, where AC and ac represent the two end posts connected by bracing which transfers the wind pressures P to the points B and b. The parts BC and be of these posts are cantilever beams fixed at the lower ends C and c, while they are also kept fixed at B and b by the restraint of the portal bracing. By turn- ing this figure clockwise through a right angle, it will be in the position of the end view of a bridge portal. Fig. 65 I 2 . ART. 71 UNEQUAL SPANS AND LOADS 177 The greatest value of M is hence at the second support; then, 7/^=3X44000X8X12/28X15 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. 70a. 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 6 ooo 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 /i, l^ and uniform loads per linear unit w\ and w 2 . The theorem of three moments in (69) then reduces to, 2M 2 (1 1 + 1 2 )=~ M/! 3 - iw 2 / 2 3 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 /i = 4o and / 2 = 3Q feet, Wi = 2io pounds per linear foot and w 2 = o; then M 2 = 24000 pound-feet. The reaction RI is found from RI X40 JX2ioX 4O 2 = M 2 , which gives R l= + 3600 pounds. The reaction RS is found from R 3 X 30 = M 2 , which gives RS = 800 pounds. The shear passes through zero in the first span at the point for which RI wx = o, which gives x = Ri/w, and the maximum positive moment is then M = Ri 2 /2W = 30 900 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), M 2 1 2 +2M 3 (1 2 +1 3 )= - 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 1 2 = /, /i = / a = 178 CONTINUOUS BEAMS CHAP, vili a/, and w\ = W2 = Wz = 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 0:= 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 w 2 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. T* TT" ~?4 ^ v ^ 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. 716. A heavy 1 2-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. [ ^ A S A 3 12 3 45 Fig. 726 Let there be two spans of lengths 1 2 and / 3 with ends hori- zontally fixed, as in Fig. 72a. 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 , Af 1/1 + 2M 2 (/i + 1 2 ) + M 3/2 = 1 W 2 1 2 3 for support 3, Af 2 /2+2Af 3 (/2+/3)+Af 4 /3= - iw 2 / 2 3 - i^ 3 / 3 3 for support 4 , M 3 1 3 -f 2 Af 4 (/ 3 -f / 4 ) + M 3 1 4 = - \w 3 / 3 3 Now, MI and M*> 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 mak'ng / 4 = o the points 4 and 5 become consecutive which renders the elastic curve horizontal at 4. As a special case let I 2 = h = 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 Af 4 is equal to M 2 , so that the equations become, 2 M 2 + M 3 = - $wl 2 2 AT 2 + 4^3 = - i^ 2 from which are found M 2 = - T ^wl 2 and M 3 = -fawl 2 . 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, vni from which the moment at the left fixed end is M^= ^wl 2 that at the middle support is M 3 = ^wl 2 , and that at the right fixed end is M 4 = +^wl 2 . 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. 72 CL b C '"1 Fig. 786 Let the column have round ends, as in Fig. 78a, and let / be its length, I the least moment of inertia of its cross-section, and E the modulus of elasticity of the material. Take the origin of coordinates at the upper end, and let x be measured down- wards and y horizontal. The general equation (45) is regarded as applicable to all bars subject to flexure, provided the bending is slight and the elastic limit of the material is not exceeded. For the case in hand the bending moment is M = Py, hence El . d 2 y/dx 2 = Py, and the integration of this gives, To determine C, let / be the maximum deflection at the middle of the column and note that the tangent dy/dx = o when y=f\ hence C = Pf 2 , and the equation may be written, Integrating this, and determining the constant by the condi- tion that y = o when x = o, the equation of the elastic curve is, 194 COLUMNS OR STRUTS CHAP. IX x= (EI/P)* arc sin(y/f ) or y=f si The curve represented by this equation must also fulfill the requirement that y = o when x = l. This condition is satisfied by making (P/EI)*l equal to TT or some integral multiple thereof; for, if v is any integer, then sinvTr = o. Accordingly, (P/E/)*/=i*r or P=y 2 7T 2 E///2 (78) which is Euler's formula for long columns. When P has this value, the long column remains bent at any deflection / which may happen to occur. There is no way of finding / from the above investigation since it cancels out of the equation; that is, the deflection is indeterminate. Inserting the value just found for (P/EI)* in the equation of the elastic curve, it reduces to the form, x=(l/w) arc sin (y/f) or y=fsinvx(x/I) 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 y = 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 \l 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=7T 2 E/// 2 or P/a=n 2 E(r/l) 2 which is Euler's formula for columns with round ends. The second form is obtained from the first by using I = 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 V 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 / by f/ 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, P=2.o$7t 2 EI/P or P/a=2.o 57 : 2 E(r/l) 2 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 cfc" 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=4x 2 EI/l 2 or P/a=4x 2 E(r/l) 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=fiEI/l 2 or P/a= fiE(r/l) 2 (78)' in which the number /< is 7r 2 for round ends, 4;r 2 for fixed ends, and 2.057T 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 1^ = 3. Using either case, and letting / be the length of the column under consideration, there is found, P=ln 2 EI/l 2 or P/a=fr 2 E(r/l) 2 and hence the number /* in (78)' is \n 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. 7Sa. 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 I^-^xd 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 a /tf 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 Iron i P= M^ 3 ' 5 /^ 3 for round ends JP^^ 3 ' 5 /^ 63 for flat ends P=42d 3 ' 76 l 2 for round ends w ,aht T J =42'l or roun en Wrought Iron | p^^pfyp 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 32 ooo 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 50000 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 beam, 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 46 ooo 46 ooo 46 ooo 44 ooo 40 40 ooo 40 ooo 40 ooo 36500 60 36000 36000 36 ooo 30500 80 32 ooo 32 ooo 3 1 5 25 ooo 100 30000 29 800 28 ooo 20 500 120 28 ooo 26 300 24 300 1 6 500 140 25 500 23500 21 000 12 800 160 23000 20 000 1 6 500 9 500 1 80 20 OOO 16800 12 800 7 5 2OO 17500 14500 10 800 6 ooo 220 15 ooo 12 70O 8800 5 00 240 13000 II 200 7500 4300 260 II 000 9 800 6 500 3800 280 IO OOO 8500 5 700 3 200 300 9 ooo 7 200 5000 2 800 3 20 8000 6 ooo 4 500 2 500 340 7 ooo 5 ioo 4 ooo 2 IOO 3 60 6 500 4300 35oo I 900 3 80 5 800 3 5oo 3000 I 700 400 5 200 3000 2 5OO I 500 420 4800 2 500 2 300 I 300 440 4300 2 20O 2 100 460 3800 2 000 I 90O 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=$2oo, 3700, 2800, 2200, 1700, 1400 while the experiments give the ultimate unit-loads as, P/a=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. 100 200 300 Values of the ratio l/ r 400 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. 79b. Let Euler's formula be written y=c/x 2 J 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 Si, and failure may be considered as occurring when this sum is equal to the ultimate compressive strength of the material. Fig. 80a 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. 80&, 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 U. 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 S and the part due to the flexure be denoted by Si. Then 5 1 = P/a + Si, in which Si is to be expressed in terms of P from the flexure formula S\ . 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 / 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, a a By analogy with the theory of beams, as in Art. 56, the deflection/ may be regarded as varying directly as l 2 /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, 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 (l/r) 2 in the above formula, let i be the constant for fixed ends, then the constant for round ends is 4^1 and the constant for one end fixed and the other end round is 1.95^1. The remarks near the end of the last paragraph of this article verify this assumption. Values of 5 and may be determined by making two experi- ments on columns having different values of (l/r) and increasing the loads until rupture occurs, thus obtaining different values of P/a. From (80) two equations may be written containing the experimental results and the two unknown quantities S and 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. Rankinc's formula, however, contains the constant S 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 /l 2 (f>' ) this is the same in form as Euler's formula, for placing S/(j> equal to a con- stant C it becomes P = Ca.r 2 /l 2 . 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. SGb. 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 5 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 S from formula (80). The values of P, 0, /, and r are known from the data of the given case, and is known from the average experimental 204 COLUMNS OR STRUTS CHAP. IX values given in the table of the last article. Then the value of the greatest unit-stress S is computed from, By comparing the computed value of 5 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, / = 2 16 inches, and < = Tinnr- The least radius of gyration is that with respect to an axis parallel to the longer side of the section, and for this axis r 2 = T V(5 X4 3 4X3 3 )/8 = 2.21, and accordingly 7/^=145. 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 ij. 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 //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 f 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 l/r = 55.3. Using for cj> the value SSWTFJ tne f ormu l a 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. 81b. 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 4 ooo 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 S. 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 = 4X?, 3 /i2Xi2 = f, and l 2 /r 2 = 19 200; also < = g-jnnr- 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 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 J 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 ^ = 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 / \ 25 ooo/ 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. S2b. 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). For this assumed cross-section a and r are to be determined, and then 5 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 S 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 S 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 = 5.75 square inches, I=2i6 inches, ^ = 5.52 inches 2 , l/r = g2, and = 7T nnr- Substituting these in Rankine's formula, there is found for ,5 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, 7-2 = 5.08 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 = -f?x 2 , and the column formula becomes, 100 000 / 800= g ( T + 3 By reduction this leads to the biquadratic equation, 8x 4 i ooo# 2 =33i 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 often 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 TO. 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 slendemess 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=fiE/x? 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) a r in which the number fj. is rc 2 for round ends, 2.o$7t 2 for one end round and the other fixed, and 4?r 2 for fixed ends (Art. 78). 40006 ^30-000 20000 V a "X \ X > \ \ v S s \ -^\ x ^o W* ?s Q. f^l&OOO jO """" 0- -0 - .--. 60 100 150 Values of the ratio l/ r FlG. 84 200 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- 210 COLUMNS OR STRUTS CHAP. IX spending to the point of tangency. The above theoretic values of /JL were not used in computing C, as his experiments indicated that the numbers n 2 , irc 2 , and 2j;r 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 5 2 5 179 I9S Hinged ends 5 2 5oo 220 159 Round ends 52500 284 123 Cast iron Flat ends 80 ooo 438 122 Hinged ends 80 ooo 537 99 Round ends 80 ooo 693 77 Oak Flat ends 5 400 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 S 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 S 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 A.RT. 85 OTHER COLUMN FORMULAS 211 be found from P/a = 17 000 90 (//>) 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 S 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 vvhich 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 fi is n 2 for round ends, 2.o$x 2 for one end round and the other fixed, and ^ for both ends Ends Fixed Fixed and Round Ends Round I i-9S 4 20 ooo 20000 20 ooo I T -95 4 30000 30000 30000 I i-95 4 40 ooo 40000 40 ooo I 34000 i-95 34000 4 34000 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 S f /fiE. Using the average values of S, and E given in Arts. 2 and 9, the values of cf> for columns of different materials are, for Timber for Cast Iron for Wrought Iron for Structural Steel 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 an d also independently deduced by Reuleaux, Marburg, and others. This formula is identical with that of Ritter except that S e is replaced by S\ designating P/a by B, it may be written, *? T> B ~i+.(S/rLE)(l/r? S i-(B/pE)(l/') 2 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 S 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 S . I/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

4 +o.o8470 6 +. . .) (87)' where the quantity in the parenthesis is obtained by expanding sed? into a series by Maclaurin's theorem. By referring to Art. 78 it will be seen that the above value of applies to a column with two round ends if / is replaced by J/, to a column with one end round and the other fixed if / is re- placed by 0.35/5 and to a column with two fixed ends if / is replaced by i/. 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, S=B(I+^ secfl) ==J3+J^|-(i+o.50 2 +o.2o80 4 +. . .) (87)" applies to all columns, when 6 2 has the following values: ART. 87 ECCENTRIC LOADS ON COLUMNS 219 for one end free and the other fixed, 6 2 = B/E . (l/rf for both ends round, 6 2 =\B/E . (l/r) 2 for one end round and the other fixed, 6 2 =~ 8 B/E . (l/r) 2 for both ends fixed, 6 2 = &B/E . (l/r) 2 In computing the unit-stress 5, the value of sec may be found for all cases by the help of Table 17, while the series cannot be used unless 6 is less than \K : when O 2 equals fa 2 the unit load B is that given by Euler's formula and S is in- finite. A closely approximate value of sec Q 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 2 = \(B/E)(l/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 6, formula (87)' gives (7=1.197 inches and then from (87) the unit-stress 5 is directly found. To illustrate the computation of seed 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 # = 1.168 radians = 65 45', whence from a trigonometric table sec/9 = 1.934. Accord- ingly the total eccentricity is 5 = 1.934^ = 1.953 inches, and the unit-stress S 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 5, 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 5, 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 i or 2 for a long column. Prob. 87 a. 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. S7c. Prove that the eccentric load P which holds a round- ended column in equilibrium with the deflection qp, is given by the formula P/a=4E(r/l) 2 (snc 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 M\ and MI exist there when a lateral deflection occurs. The bending moment at any section distant x from the end i of the column is M = M\ + Vx Py where Vi is the transverse shear at i. When x = l, then y=--o and M = M 2 , so that Vi = (M 2 -Mi)/L The general equation of the elastic curve now is and the general solution of this differential equation is p (88) ART. 88 ON THE THEORY OF COLUMNS . 221 in which A, B, C, D, /? are constants of integration to be found from the condition of the problem. Differentiating (88) twice, there result, g= (A cos/?y +C ~ D Comparing the two values for EI.8 2 y/dx 2 and eliminating the trigonometric expression by the help of (88), three constants are found, namely, p 2 = Pl 2 /EI, C = M 2 l 2 /p 2 , and D = M l l 2 /p 2 . Further, the first member of (88) " becomes MI when # = and M 2 when x = l\ hence A = -M 2 l 2 /p 2 sin/9 and B=-M l l 2 /^ 2 sind. The expression found for /? 2 gives P = 8 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 x = 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 /? 2 is 2.0457^, and hence Euler's formula for this case is P = 2.o46n 2 EI /I 2 , as noted in Art. 78. The constant 2j, 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/? must be indeterminate, and this will 222 COLUMNS OR STRUTS CHAP. IX be secured by making sin/9 = 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 Afi, then dy/dx becomes zero for both x = o and x = l provided cos5 = i; this requires that /?=27r, whence Euler's formula for columns with two fixed ends is P = 47i 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.o2M\. 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 S e ', for the deflected column failure begins when Euler's condition P/a = /iE(r/l) 2 is reached. Hence S e = nE(r/l) 2 or l/r=(uE/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 fi. = n 2 , have l/r = 92, while columns of structural steel with fixed ends, for which , = 7r 2 have //r = i84 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 = S a 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 t 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 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 will 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. 882. Prove that 4.49341 is a root of the equation /? = tan/9. 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 or TORSION When applied forces cause a bar to twist around its axis, f 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 corners 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. 169^ 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 p 2j the twisting moment Pipi+Pzpz 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. 89&. 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 5 . 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 S and c are constants, this resisting moment may be written (S/c) Ida . 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, S is a tensile or compressive unit-stress which is normal to the section area ; in the torsion formula, S is a shearing unit-stress which acts along the section and normal to the radius. In the flexure formula, M is the bending moment oi 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 5 S 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 S 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 oc 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 -f Ida . y 2 ; that is, the polar moment 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 / 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 = %d, J for a hollow shaft, c=%d, J 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 ; for a hollow circular sec- tion, r 2 = J(^ 2 + ^i 2 )- Prob. 90a. Three forces of 70, 90, and 120 pounds act at distances of 8, u, 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 2np and the work PX2np or 27iPp is transmitted through the shaft; in n revolutions the work 2nnPp is transmitted. Now let N be the number of units of work per minute which constitute one horse-power; then 2nnPp = NH ) or Pp = NH/2xn. In the English system of measures, P is in pounds, p in inches, and N is 396 ooo pound-inches per minute. In the metric system of measures, P is m,kilograms, p in centimeters, and N is 450 ooo kilogram-centimeters f)er fc minut. Accordingly, Pp = 6soT > oH/n or Pp = ^i 62oH/n (91) the first being for the English and the second for the metric system of measures, while Pp= (N/2n)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. 910. 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. Qlb. 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 J and c are -jV^ 4 and %d, and the torsion formula (90) reduces to Pp=^nd 3 S or S=i6Pp/nd 3 which may be used for the discussion of shafts when the twisting moment Pp is required or given. The usual case, however, is that of the transmission of power, where the value of Pp for the English system of measures is 63 o^oH/n, as shown in Art. 91 ; inserting this in the last formula, it reduces to the forms, 5=321 oooH/nd 3 or d=6S.s(H/nS^ (92) 232 TORSION OF SHAFTS CHAP, x The first of these may be used for investigating the strength of a given shaft when transmitting H horse-powers and making n revolutions per minute. The second may be used to determine the diameter of a shaft to transmit H horse-powers with n revo- lutions per minute, the allowable unit-stress for the circumfer- ence of the shaft being S. In both equations d must be taken in inches and 5 in pounds per square inch. These equations show that the shearing unit-stress 5 at the circumference of a solid shaft varies directly as the transmitted power, inversely as the speed of the shaft, and inversely as the cube of its diameter. Hence for a given S and n, the diameter d changes slowly with H\ if H is doubled, d is increased only 26 percent, and the section area is increased 59 percent. For example, let it be required to design a solid wrought- iron shaft to transmit 90 horse-power when making 250 revolu- tions per minute. Here the factor of safely may be about 6, or S may be about 7 ooo pounds per square inch. Then from the above formula the diameter d is found to be 2 inches. Hollow forged steel shafts are much used for ocean steamers, since their strength is greater than solid shafts of the same section area. Let di be the outside and d 2 the inside diameter; then the value of J is aVK^i 4 " d 2 4 ) and that of c is ^d\. Inserting these in the torsion fonnula and also the value of Pp from (91), it be- comes, for English measures, S(di* - d2*)/di = 321 oooH/n which is the equation for use in investigation and design. For example, a nickel -steel shaft of 17 inches outside diameter is to transmit 16000 horse-powers at 50 revolutions per minute; let it be required to find the inside diameter so that the unit- stress 5 may be 25 ooo pounds per square inch. Here every- thing is given except d 2 , and from the equation its value is found to be ii inches nearly. The area of the cross-section of this shaft will be about 132 square inches, and its weight per linear foot about 449 pounds. A solid shaft having the same strength will require a diameter of 16 inches, its cross-section will be 201 square inches, and its weight per linear foot about 683 pounds. ART. 93 TWIST OF SHAFTS 233 The reason why economy is promoted by the use of a hollow shaft is the same as that given in Art. 51 for a hollow beam; namely, that material is removed from the axis where it is but little stressed and placed farther away where it is efficient in resisting the twisting moment. By comparing the above formulas for solid and hollow shafts, it is seen that they become the same when d* equals (di 4 -d 2 4 )/%. When the section areas of the solid and hollow shafts are equal, d 2 must equal di 2 ~d 2 2 . From these two equations, it follows that the ratio of the strength of a hollow shaft to that of a solid one of the same section area is (2di 2 d 2 )/ddi. For example, let d = i2 inches and ^1=20 inches; then d 2 is 16 inches from the condition of equal section areas, and the ratio of strengths is 2.73, that is, the strength of the hollow shaft is 173 percent greater than the solid, one. Prob. 92a. Find the factor of safety for a wrought-iron shaft 2j inches in diameter when transmitting 25 horse-power while making TOO revolutions per minute. Prob. 926. If a hollow shaft has the same section area as a solid one, the inside diameter being one-half the outside diameter, find the ratio of the strength of the hollow shaft to that of the solid one. ART. 93. TWIST OF SHAFTS The angle of twist through which a radius on one end of a round shaft moves under an applied twisting moment can be determined, when the unit-stress 5 does not exceed the elastic limit, from the definitions in Art. 15 with the help of the torsion formula (90). Let the section be circular, J its polar moment of inertia with respect to the axis about which rotation occurs, c the distance from that axis to the remotest part of the section where the shearing unit-stress is S, and F the shearing modulus of elasticity of the material. The modulus F is defined by S/e, where is the deformation per unit of length. Let / be the length of the shaft, and the angle expressed in radians through which one end is twisted with respect to the other, that is the angle deb in Fig. 89a. In this figure the arc bd is the deformation in the length I] this deformation is c, and replacing S by its value Pp . c/J there is obtained, F=Pp.l/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 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 is to be replaced by 7r0 /i8o. The formula now becomes, Q/l = &.$X2 1.4/1 500 = 0.12 degrees. The above formulas show that the angle of twist is propor- tional to the length of the shaft, and this is found to be also the case after the elastic limit of the material is exceeded, but the angle is then very much greater than that given by the formulas, In the case of the round cast-iron bar shown in Fig. 169c, the angle of twist at rupture was 12.3 degrees; this bar had a length of 10 inches and a diameter of 0.83 inches, and it broke under a twisting moment of 3 910 pound-inches. For the square steel bar in Fig. 169c, which was 12 inches long, the angle of twist at rupture was about 900 degrees; further facts regarding this square bar are given in Art. 99. Prob. 93a. A solid steel shaft 125 feet long and 16 inches in diameter transmits 8 ooo horse-powers at a speed of 25 revolutions per minute. Compute the angle of twist. Prob. 936. A wrought-iron shaft 5 feet long and 2 inches in diame- ter is twisted through an angle of 7 degrees by a force of 5 ooo pounds acting at 6 inches from its axis, and on the removal of the force it springs back to its original position. Compute the shearing modulus of elasticity. ART. 94. RUPTURE BY TORSION When a round bar or shaft is twisted to the point of rupture, failure occurs first at the outside circumference, and this is rapidly propagated inwards until complete shearing is effected. The torsion formula S.J/c = Pp, does not, however, hold for cases for rupture, so that a value of S computed from it for the data of failure does not closely agree with the ultimate shearing strength of the material. A value of the- unit-stress S computed by the torsion formula from a twisting moment Pp which causes rupture, may be called the ' computed twisting strength .' of the material, in analogy with the computed flexural strength obtained for the rupture of a beam by the use of the flexure formula (Art. 52). The following 236 TORSION OF SHAFTS CHAP. X are approximate average values of the computed ultimate -twisting strength of different materials: Timber S u = 2 ooo pounds per square inch Cast Iron S u = 3o ooo pounds per square inch Wrought Iron S U = S5 ooo pounds per square inch Structural Steel S u = 6s ooo pounds per square inch Strong Steel ,$^ = 90 ooo pounds per square inch By the use of these average values, computations may be made on the rupture of round shafts, that is, the probable twisting moment which will cause a given shaft to rupture, or the probable size of a shaft which will fail under a given twisting moment, may be roughly ascertained. By comparing the above values with the ultimate shearing strengths given in Art. 6, it will be seen that they are all larger with the exception of timber. The statement is sometimes made that it might be anticipated that the two sets of values should be the same, but this rests on no reasonable basis. The ultimate shearing strength of a material is a physical constant, but the values of S u have no physical meaning, for they have been com- puted from a formula which does not correctly represent the distribution of the internal stresses for cases of rupture. The quantity S u is usually called the 'modulus of rupture for torsion ', but it seems wise to abandon this term and to use one which gives no erroneous impression. When that twisting moment Pp is reached in a test where the angles of twist (f> no longer increase uniformly, the shearing elastic limit of the material is reached, and the elastic limit for shearing may be correctly computed from the torsion formula S = Pp.c/J. The shearing elastic limit of metals is always less than those for compression, and in general may be taken at about one-third of the computed twisting strength S u . Experi- ments on timber, brick, and stone have been so few in number, that it is difficult to state average values of their shearing elastic limits. In Fig. 188 is shown a cast-iron bar, 0.83 inches in diameter which broke by torsion under a twisting moment of 3 910 inch- ART. 95 STRENGTH AND STIFFNESS 237 pounds. The computed ultimate twisting strength for this case is 5 = 34800 pounds per square inch, which is higher than the average for cast iron. The square bar of medium steel shown in Fig. 169c, which was 0.75 inches square, broke under a twisting moment of 5 800 inch-pounds; the torsion formula (90) cannot be applied to this case, as the discussion in Art. 99 will show. Prob. 94. Compute the probable diameter of a wrought-iron shaft which will rupture by torsion when twisted by a force of 47 pounds acting at a distance of 47 inches from the axis. ART. 95. STRENGTH AND STIFFNESS The strength of a shaft or of a bar under torsion is measured by the twisting moment that it can carry. Hence the strengths of different shafts vary as their values of S . J/c; or for the same material, the strengths vary as the values of J/c. For example, let there be two solid shafts of the same material, the diameter of the first being d and that of the second 2^; since the section factor J/c is eight times as large for the second as for the first, it follows that their strengths arc in the same ratio. Thus the strengths of solid shafts of the same material vary as the cubes of their diameters. A general comparison of the strength of a round hollow shaft with that of a solid one having the same section area a will now be made. Let d be the diameter of the solid shaft, di the outside and &2 the inside diameter of the hollow shaft ; then d 2 = di 2 d 2 2 is the condition that the section areas shall be equal. Now, for the hollow shaft, and in the same manner for the solid shaft, Therefore, dividing the first by the second, and letting a denote the ratio di/d 2l there is found hollow/solid- (a 2 + i)/a(a 2 -i)* which is the ratio of the strength of a hollow shaft to that of a solid one of the same section area. When di is double d 2) the 238 TORSION OF SHAFTS CHAP, x value of a is 2, and the hollow shaft is about 44 percent stronger than the solid one. The stiffness of a shaft under torsion is measured by the twisting moment it can carry with a given angle of twist. By Art. 93 it is seen that Pp varies directly with /, and hence the stiffness of a shaft varies directly as its polar moment of inertia. For the hollow shaft, and for the solid shaft of equal section area, Therefore, dividing the first by the second, and designating the ratio di/dz by a, there is found hollow/solid = (a 2 + i)/(a 2 - 1) which is the ratio of the stiffness of a hollow shaft to that of a solid one of the same section area. When d\ is double d 2 , the value of a is 2, and the hollow shaft is about 67 percent stiffer than the solid one. Shafts of the same material and length are of the same strength when their values of J/c are equal, and of the same stiffness when their values of / are equal. It is easy to show that the per- centage of weight saved by using a hollow shaft instead of a solid one is, for equal strength 100 ioo[(a 4 a 2 ) /(a 2 + 1) 2 ]* for equal stiffness 100 ioo[(o: 2 i)/(a 2 +i)]* in which a is the ratio d\/d 2 which will be determined by practical considerations concerning ease of manufacture and operation. When a =2, the percentages saved are 21.7 for equal strength and 22.5 for equal stiffness. Prob. 95. Compare the strengths of two shafts when stressed to their elastic limits; the first shaft is solid, 21 inches in diameter, and has an elastic limit of 25000 pounds per square inch; the second shaft is hollow, 18 inches outside and 9 inches inside diameter, and its elastic limit is 45 ooo pounds per square inch. If the price per pound of the first shaft is 18} cents, what price per pound could one afford to pay for the second shaft ? ART. 96 SHAFT COUPLINGS 239 ART. 96. SHAFT COUPLINGS At A and B in Fig. 96 are shown the end and side views of a flange coupling for a shaft, the flanges being connected by bolts. These bolts, in transmitting the torsion from one flange to the other, are subject to shearing stress, and they must be of sufficient strength to safely carry it. This shear does not differ in character from that in the main body of the shaft, and it is the greatest upon the side of the bolt most remote from the axis of the shaft. Let J be the polar moment of inertia of the cross-section of a solid shaft, c its radius, and S the shearing unit-stress on the outer surface. Let /i be the polar moment of inertia of the cross-section of the bolts with respect to the axis of the shaft, c\ the distance from the axis of the shaft to the side of the bolts farthest from the axis, and let the shearing unit-stress on that side be required to be the same as that on the outer surface of the shaft. Then, in order that the bolts may be equal in strength to the shaft, it is necessary that J/c should equal J\/c\. The polar moment of inertia of the section of one bolt with respect to the axis of the shaft is equal to its polar moment with respect to its own axis plus the section area of the bolt into the square of the distance between the two axes. Let D be the diameter of the shaft, d the diameter of each of the bolts, h the distance of the center of a bolt from the axis o the shaft, and n the number of bolts; then and by equating these values there is found D\d+2h)=nd 2 (d 2 +Sh 2 ) which is the necessary relation between the quantities in order 240 TORSION OF SHAFTS CHAP, x that the bolts may be equal in strength to the shaft, provided the material be the same. From this formula the number of bolts required is easily found when d, h, and D are given. This formula is an awkward one for determining d, and hence it is often assumed that the shear is uniformly distributed over the bolts, or that c\ = h and J\ = \nd 2 h 2 . This amounts to the same thing as regarding d as small compared to h, and the expres- sion then reduces to D 3 =4nhd 2 or d = %(D 3 /nh)* from which an approximate value of d can be found when the number of bolts is given. The above, supposes the shaft to be solid. When it is hollow, with outside diameter D and inside diameter d\, the D 3 in the above expressions is to be replaced by D 3 di*/D. The case shown at CD in Fig. 96 is one that would not occur in practice, but it is here introduced in order to indicate that the bolts would be subject to a flexural as well as a shearing stress. It is clear that the flexural stress will increase with the length of the bolts, and that they should be greater in diameter than for the case of pure shearing. The- flexural stress will also depend upon the work transmitted by the shaft. This case will be investigated in Art. 98 in connection with the discussion of the pin of a crank shaft. Prob. 96a. A solid shaft 6 inches in diameter is coupled by bolts ij inches in diameter with their centers 5 inches from the axis. How many bolts are necessary ? Prob. 96&. A hollow shaft 17 inches in outside and n inches in inside diameter is to be coupled by 12 bolts placed with their centers 20 inches from the axis. What should be the diameter of the bolts ? ART. 97. A SHAFT WITH CRANK A crank pin, CD in Fig. 97, is subject to a pressure W from the connecting rod of the steam engine, this pressure being uni- formly distributed over nearly the length of the pin. The pressure varies at different positions in the stroke of the engine, but for ART. 97 A SHAFT WITH CRANK 241 ordinary computations may be taken at from 10 to 20 percent greater than the total mean pressure of the steam on the piston; to this may be added the weight of the connecting and piston rods in case these should be vertical in position. The maximum pressure W causes a cross-shear in the crank pin at the section C, and it also causes a flexural stress at the end C due to a uniform load over the cantilever CD. These stresses may be computed by the shear and flexure formulas of Art. 41. The compressive or bearing stress upon the pin is usually also to be considered, this being estimated per square unit of diametral area in the same manner as the sidewise pressure on a rivet (Art. 33.) In the figure the pressure W is shown acting parallel to the crank arm BC, but it acts at all angles to this arm during one revolution of the shaft AB. When it acts at right angles, the crank arm is a can- (~~^~ tilever beam, and the maximum flexural ^ - stresses which occur at B may be computed by the flexure formula (41). The flexural unit-stresses in both pin and crank arm alternate from tension to compression as the arm revolves, for the pressure W comes on Yig. 97 opposite sides of the pin as the piston rod moves forward or backward. On account of these alternating stresses the allowable working unit-stresses should be taken low in designing the pin and crank arm. The shaft, crank arm, and pin are usually forged, the whole being one piece of metal, but for very large shafts the crank arm may be bored to a diameter slightly less than that of the end of the shaft and then shrunk upon it (Art. 32). Sometimes a shaft has a crank arm at each end, it being then called a double- throw crank shaft; in this case the two crank arms are set at right angles to each other, in order that the action of the two pressures W may produce a more uniform twisting moment on the shaft. The largest crank shaft in use was made by the Bethlehem Steel Company in 1905; it is 37 inches in outside and \W 242 TORSION OF SHAFTS CHAP, x 30 inches in inside diameter, the crank webs are 64 inches long, 49 inches wide and 16 inches thick, and the solid pins are 19 inches in diameter. This double-throw crank shaft is 27 feet long, the two crank pins, webs, and shaft being one piece of steel which was forged from an ingot weighing 308 ooo pounds, and the weight of the finished shaft is 86 600 pounds. Prob. 97a. The crank pin CD in Fig. 97 is 8 inches long and 4 inches in diameter, the pressure W being 60 ooo pounds. Compute the bear- ing unit-stress, the shearing unit-stress, and the flexural unit-stress. Prob. 976. Let BC be 14 inches long, 4 inches thick, and 10 inches wide at B. Compute the flexural unit-stress at B when the pressure of 60 ooo pounds acts normal to BC. ART. 98. A TRIPLE-CRANK SHAFT Double and triple cranks are used when several engines are to be attached to the same shaft, as is usual in ocean steamers. With the triple arrangement the cranks are set at angles of 120 degrees with each other, thus securing a uniform action upon the shaft. Fig. 98 shows one of these cranks, AB and EF being portions of the shaft resting in journal bearings, CD one of the crank pins to which the connecting rod is attached, while BC and DE are the crank arms or webs which are usually shrunk upon the shaft and pins. The complete investigation of the maximum stresses in such a crank shaft and pin is one of much difficulty. A brief ab- stract of such an investigation will, however, here be given for the crank pin. There are three cranks, and the one to be con- sidered is the nearest to the propeller, so that the torsion from the other two cranks is transmitted through the pin CD. This steel crank pin is hollow, 18 inches in outer diameter and 6 inches in inner diameter, its length between webs being 2^ inches, the thickness of each web 12 inches, and the distance from the axis of the shaft to the center of the pin being 30 inches. The three engines transmit 7 200 horse-power to the shaft EF, of which 4 800 horse-power is transmitted through the shaft. AB and tnrough the crank pin CD. The maximum pressure W brought ART. 98 A TRIPLE-CRANK SHAFT 243 by the connecting rod upon the crank pin is 156 ooo pounds. It is required to determine the stresses when the crank makes 80 revolutions per minute. The pressure W is distributed over about 17 inches of the length of the pin, so that the bearing compressive stress on the diametral area is Si =156 ooo/ 1 7*^12*^=765 pounds per square inch, which is a low and safe value. The shearing unit-stress due to Ty, which is taken as uniformly distributed over the sec- tion area of the pin, is A fa*~*~/&^ +f j*^ (g) 52=78 ooo/%n{iS> 2 6 2 ) =345 pounds per square inch which is also a low working value for steel. Fig. 98 The shearing stress due to the horse-power transmitted through BC has its greatest value on the side of the pin farthest from the axis. The twisting moment Pp due to this 4 800 horse- power is found from formula (91) to be, Pp = 6T, 000X4 800/80 = 3 782 ooo pound-inches and this is equal to the resisting moment of the crank pin, or to S/i/Ci, in which Ji is the polar moment of inertia of the cross-section with respect to the axis of the shaft and c\ is the distance from that axis to the side of the pin where the stress S is to be found. Now,^ = 30 + 9 = 39 inches, and then Ji = 0.0982 (i8 4 -6 4 ) +0.7854(182-62) X39 2 . Hence, from formula (90) there is found, $3 =3 782 00^X39/354 000=420 pounds per square inch which is the shearing stress due to the transmitted power. The flexure of the pin due to the torsion carried through it falls under the case discussed in Art. 65. The twisting moment Pp is equivalent to a force P acting at a distance of 30 inches 244 TORSION OF SHAFTS CHAP. X from the shaft and normal to the crank arms; the value of P is 3782000/30 = 126100 pounds, and this produces a bending moment in the pin which may be taken as a beam fixed at both ends, while P acts in opposite directions at those ends, as in Fig. 65a. Hence there is a bending moment M f at each end, oppo- site in sign but equal in value, and the moment at any section is M =M'+Px; but when oc=l the value of M is M.' and there- fore M ' = JP/, which is the maximum bending moment. The flexure formula (41) then gives S = M' . c/I=6$ 000X24X97 g-Vr(i8 4 -6 4 ) = 2 060 which is the flexural stress in pounds per square inch due to the transmission of power through the pin. All of these stresses are light, but the pin is necessarily made heavier than they would require, on account of the additional stresses due to the shrinking of the web upon the pin. The data here given are not sufficient to determine these shrinkage stresses, but the discussion in Art. 154 indicates that there is a radial com- pressive unit-stress S 5 brought by the web upon the pin of proba- bly 3 ooo pounds per square inch, and that this is accompanied by a tangential compressive unit-stress SQ of about 4 ooo pounds per square inch. These shrinkage stresses occur also in the fillet of the pin on the inside of the web at D, where all the other stresses except Si concentrate. In Art. 179 it will be shown how these several values may be combined in order to obtain the final maximum tensile, compressive, and shearing stresses. Since all these stresses vary in direction and intensity as the shaft revolves, their effect is more injurious than steady stresses, and accordingly the factors of safety should be high, or the work- ing unit-stresses low, in order that the life of the crank pin may not be short (Art. 137). Shocks are also brought upon the shaft of an ocean steamei when the propeller at the stern rises out of the water and falls back again, as it does when great waves cause longitudinal oscillations, and these shocks also require that the working unit-stresses shall be low. Prob. 98. The shaft EF in Fig. 98 is hollow, the inside diameter being 8 inches and the outside diameter 24 inches. Compute the flexu- ART. 99 NON-ClRCULAR SECTIONS 245 ral unit-stress S when 7 200 horse-powers are transmitted through it while it is making 65 revolutions per minute. i ART. 99. NON-CIRCULAR SECTIONS When a rectangular bar is subject to torsion the general phenomena are the same as those described in Art. 89, but the distribution of the shearing stresses over the section area is ob- served to be different from that in a circular section. In Fig. 169c is seen a steel rectangular bar, ^Xi|X8J inches, which has been twisted in a torsion machine through an angle of 142 degrees. Before the test the surfaces of the bar were blackened and two series of white lines drawn thereon at right angles to each other; the figure plainly shows the distortion of the rectangles into rhombuses, the greatest distortion in the right angles being at the middle of the wide side, while the white lines remain closely perpendicular to the corner lines of the bar. From many ex- periments of this kind it is concluded, since stresses are propor- tional to distortions, that the shearing unit-stress at the middle of the wide side is greater than that at the middle of the narrow side of the rectangular bar, and that there are no shearing stresses along the corners. When lines are ruled upon the surface of a round bar it is observed that the distortion of the angles is the same on all parts of the surface, and hence the shearing unit-stress is uniform over that surface; this is the basis of the torsion formula deduced in Art. 90. For a bar of elliptical section, however, it is observed that the distortion of the angles is greatest on the flatter side of the section; hence the shearing unit-stress is there the greatest, and the torsion formula (90) does not apply to the elliptical bar; in fact, that formula applies only to circular sections, and it should not be used for other sections except for approximate investigations. The correct torsion formulas for elliptical and rectangular sections will now be deduced. Let Fig. 990 represent a section of an elliptical bar which is subject to a twisting moment Pp from a force P acting in a plane normal to the axis and at a distance p from that axis. Let 246 TORSION OF SHAFTS CHAP, x m be the major axis and n the minor axis of the ellipse, let y and x be the vertical and horizontal coordinates of any point on the circumference of the ellipse with respect to m and n as coordinates axes. Let Si and S 2 be the shearing unit-stresses at the extremities of the minor and major axes, and S the shearing unit-stress at the point where coordinates are y and x] these stresses are tangential to the circumference. Let S r and S" be the components of S parallel to Si and ^2, and let % be the angle which S makes with Si] then S f = S cosj and S" =S sin^, whence S"/S' =tan.%. The equation of the ellipse is m 2 y 2 + n 2 x 2 = {m 2 n 2 , and by differentiation there is found dy/dx= n 2 x/m 2 y =tsin%] accordingly S"/S' = n 2 x/m 2 y and it thus appears that the com- ponents S" and S' are proportional to their distances from the coordinate axes n and m. When the elastic limit of the material is not exceeded, the same relation must hold between the com- ponents of the unit-stress at any point within the ellipse, for here, as in the circle, the unit-stresses along any radius vector vary proportionally as their distances from the center. Fig. 990 Fig. 996 Now let yi and xi refer to any point within the ellipse and let m\ and n\ be the axes of an ellipse passing through that point 3 the ratio m\/n\ being equal to m/n. Let S x and S y be the com- ponents of the unit-stress at that point, S x being parallel to Si and S y to S2j then also S y /S x =n^x\/m-fy\. Let da be the element of area at the given point; then S x yida-kSyXida is the resisting twisting moment of the stress on that area, and the sum of all similar expressions for all elements of the area equals the twisting moment, or 2S x yi da + ISyXi da = Pp Substituting in this for S y its value S x . fi 2 #i/*i a yi and then for Sx its value S'. y\/y, and also for m\/n\ its value m/n, the com- ART. 9& NON-ClRCULAR SECTIONS 247 ponent S' is founa. Similarly substituting for S x its value S y . mi 2 yi/ni 2 Xi and then for S y its value S" . x\/x, the com- ponent S" is found. In both cases 2y\ 2 da is the moment of in- ertia /i of the area of the ellipse with respect to the major axis m, while Zx\ 2 da is the moment of inertia / 2 with respect to the minor axis n. Thus are found, Pp.m 2 y Pp.n 2 * which are the components of the shearing unit-stress S at any point on the circumference whose coordinates are y and x. When y =o and x = Jw, the horizontal component S' is o and the vertical component S" becomes S 2 in the figure ; when x = o and y = \n, the vertical component S" is zero and the horizontal component be- comes Si in the figure. The ratio Si/S% is hence equal to m 2 n/mn 2 or to m/n, that is, the shearing unit-stresses at the ends of the axes are inversely proportional to the lengths of those axes. The greatest unit-stress hence occurs at the ends of the minor axis, while the least unit-stress on the circumference occurs at the ends of the major axis; hence Si is the unit-stress to be used. The rectangular moment of inertia of the ellipse with respect to its major axis is /i = -fax-inn 3 and that with respect to the minor axis is I2 = fanm 3 n (Art. 43). Substituting these values in the above value of S r and making y = ^n, there results, or Pp =xmn 2 &Si which are the formulas for discussing a bar of elliptical cross- section under torsion. When m=n = d, the ellipse becomes a circle, and the formula reduces to that deduced for the circle in Art. 92. For the rectangle in Fig. 996, the exact discussion is much more difficult because the stresses are not proportional to their distances from the axis except along the median lines. On any side the stress is greatest at the middle and varies approximately as the ordinates of a parabola toward each corner where it be- comes zero; the stress at the middle of the broad side is greater than that at the middle of the narrow side, as in the elliptical section. Let m and n be the long and short sides of the rectangle, 248 TORSION OF SHAFTS CHAP. X and let x and y be coordinates of any point within the rect- angle with respect to axes through the center, x being parallel to m and y to n. Then the above facts of experiment indicate that the variation of stresses throughout the rectangle is closely expressed by where Si and S 2 are the unit-stresses at the middle of the long and short sides, respectively, and S x and S y are the components parallel to Si and S 2 of the unit-stress at any point. The funda- mental equation between resisting and twisting moment is also the same as before; replacing the sign of summation by that of integration and da by dx dy, it becomes, fs x y dx dy+fSyX dx dy=Pp Inserting in this the values of S x and S v and integrating, first between the limits -\-\n and \n for y, and second between the limits +%m and \m for x, it becomes %mn(Sin + S2w) = Pp. Now, as in the ellipse, 81/82 equals m/n, and accordingly S^m equals S\n\ whence, since S\ is the greatest unit-stress, Si = lPp/mn 2 or Pp=%mn 2 Si are the formulas for discussing rectangular bars under torsion. Apply to a square bar of side d by making mn 2 equal to d 3 . A comparison of the strength of solid round and square shafts is now readily made from the values of the twisting moments derived in Art. 92 and in the preceding paragraph: for a round shaft , Pp = -fairSd? =0.1 964$$* for a square shaft, Pp %Sd 3 =o.2222Sd 3 and accordingly the strength of a square shaft of side d is 13 percent greater than that of a round shaft of diameter d, the shear- ing unit-stress being the same in the two cases. When power is transmitted by a square shaft, Pp is to be replaced by 63 o^oH/n for English measures and then are found S =284000^/^3 or d=6$.>j(H/nS)* which are the formulas for the investigation and design of solid square shafts similar to those of (92) for solid round shafts. ART. 99 NON-ClRCULAR SECTIONS 249 The angle of twist produced in a non-circular section by a given twisting moment cannot be deduced by the method given in Art. 93 for round shafts, on account of the irregular distribution of stresses. The investigations of Saint Venant, who was the first to establish the correct theory of torsion, give the following value .for the angle of twist for a square shaft, that for the round shaft being derived from Art. 93 : for a round shaft, < = lo.iSPp . l/Fd* for a square shaft, = j.uPp . l/Fd 4 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. 188 was 11} 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 5 =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. 99 d. 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 8J inches, its thickness J inch, and its width ij 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 ooo 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 e be the change per unit of length which occurs when a bar is free to expand or contract, y 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 S 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, = yt S=eE S=ytE (100) Et 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, 7] =0.000 0050 for cast iron, 77=0.000 0062 for wrought iron, 77=0.000 0067 for steel, 77=0.0000065 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 ooo 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=0.0000067X10X25000000 = 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 coo + 5 200 = 14 200 pounds. For a rise of 10 degrees Fahren- heit, the tensile stress in the bar becomes 9000-5200=3800 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 20 ooo 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 J 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 round 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 $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 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 Si = 25 7 pounds per square inch. Table 6 gives the section area = 16.18 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 = 5200 pounds per square inch, and the factor of safety is 60 000/5 2O = I z i- O R tne upper side of the beam, the stress is tensile, since P/a is greater than S 23 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 Us 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 oi the rafter be rectangular, b being its width, d its depth, / the length, w the uniform load per linear unit, and the angle of inclina^ Fig. 101. tion. To find the horizontal re- action H, the center of moments is taken at the lower end, and, H . I sin = wl . i/ cose/) whence H = \wl cot For any section area at the distance x from the upper end of the rafter, the flexure formula gives the flexural unit-stress, 6M 6(Hx sin(f> ^ l= ~bd?~ bd? ART. 102 FLEXURE AND COMPRESSION 255 and the uniform compressive unit-stress is, P H cosd>+wx sin0 a bd The total compressive unit-stress on the upper fiber hence is, bd 2 2bd bd By the usual method this is found to be a maximum when and substituting this, the maximum unit-stress is, _ 3 wl 2 cos(f> wl cosec0 2bd which formula may be used to investigate or to design common ^afters 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, Mc P sm(> H x la a from which S x 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. 10 Ib. 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 S 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 Mi=M + Pf\. Now the maximum fiber unit-stress Si which results from this moment M\ is, from (41), SI-MI. c/i-*(M+pfdc/i where / 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 Si, regarding f\ to vary with Si 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, *>->?/(-) 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=Si+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 = ^Wl = 48000 pound-inches, c=4^ inches, I =-fabd?=6o'j% inches 4 , / = 96 inches, P =40 ooo pounds, a = 8, ft f 1 , and E = i 500 ooo pounds per square inch. Inserting these values in the formula, the value of Me 1 1 is 356, and then the final flexural stress Si 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 S = 37 1 +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 A.RT. 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 fsecO where 6 denotes the quantity \(P1 2 /ET)^ for a beam with supported ends. As before, the total flexural unit-stress S\ is given by (M + Pfi)c/I } in which /i is to be replaced by / sec# ; 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 Si at the middle of the concave side of the beam is, (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. 102& To illustrate this method, let the data of the above numerical example be again used. The value of Me /I 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 f=$Wl 3 /3&4EI = 0.0506 inches, r 2 =607.5/90 =6.75 inches 2 , = \(PPj '/)* = 0.3184 radians = 18 15', sec0 = 1.053, and lastly (P/a)(cf/r 2 ) sec# = i6 pounds per square inch. The total flexural stress S-\ is then 356 + 16 = 372 and total compressive stress S 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 fywlx %ivx 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, wlx wx 2 wEI /cos (2x 1)6/1 \ ~^P + ^F + ~P 2 ~( ^6~~ V in which, as before, 6 is an abbreviation for J(P/ 2 AE7)*. In this equation of the elastic curve, let x = J/, then y =/i and then, jpa/j = _ %wl 2 P+wEI(sec6 - 1) which gives the deflection of the beam due to both the uniform load and the longitudinal compression. Inserting the value of the deflection /i in the expression for Si at the beginning of this article, there is found, Si = (wcE/P) (secfl - 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 iv =4 000/96 pounds per lineal- inch, and the other quantities as before, also seed = 1.0530; and then (102)" gives Si =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 S =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 /i 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 /i will reduce to , while that for Si will reduce to ART. 103 FLEXURE AND TENSION 259 Prob. 1020. Prove, by using the method of the differential calculus for evaluating indeterminate quantities, that the statement in the last sentence is correct. Prob. 102ft. 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 8oc 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 /i, 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 Pf\. 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 ooo pounds, E being 29 ooo ooo pounds per square inch. The weight of the bar is 490 pounds, and M = JX49oXi8Xi2 =13 230 pound- inches. Also =4 inches, 7 = 42.67 inches 4 , ft /a =9.6, P = 80 ooo pounds, / = 2i6 inches. Then the value of Me/ 1 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, Si-^-f.^secW (103)' in which sech# may be computed from 2/(e + e~ e ), where e is the base of the Naperian system of logarithms and 6 denotes the real positive number J(P/ 2 /E/)*. For instance, let the data of the last paragraph be again considered. The value of Me /I is i 240 pounds per square inch, which is the flexural unit-stress due to uniform load alone. Also 0=4 inches, r 2 =42. 67/8 = 5.334 inches, /= $WP/$4EI =0.05 2 inches, = (P1 2 /EI)* = 0.868, ^ = 2.7i8- 868 = 2.383, e~ e = 1/^=0.420, sech0 = 2/2.803 = 0.714; then P/a = ioooo pounds per square inch, cf/r 2 = 0.039, and (P /a) (cf/r 2 ) sech6 = 2jS 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 -sechfl) (103)" where 6 is the number %(P1 2 /EI)* when the ends of the beam are supported. For the above eye-bar, sech0 =0.714; also w = 490/216 pounds per linear inch. Then the formula gives S\ = 941 for the flexural unit-stress, so that the total compressive unit- stress is S =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 more 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 0-sinh 2 = i. ART. 104 ECCENTRIC AXIAL FORCES ON BEAMS 261 The value of cos# is \(e id + e~ ie ) , in which e is the Naperian base 2.71828 and i is the square root of i; the value of coshfl is givQn by the simpler expression %(e e + e~ e ). The reciprocal of cos# is seed and that of coshtf is sech#. 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. 1046 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. 104ft. 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 = JTF/ = 94ooo pound-inches. Then the centers of the pins must be at the distance ^=94000/168000=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. P- - P 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(p-f) should equal if; for tension the moment P(p+f) is to be equal to M. Accordingly, P-j+f P-j-f do4) 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 /=5^/ 3 /384E/ = o.o57 inches, and M/P = o.$6 inches as before; then the required eccentricity is ^ = 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 ^=0.56 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. 104ft. 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 5, and the shearing unit-stress is V/a, which may be denoted by S 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 s and S 8 , 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 8 . 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/dj there is also found a horizontal shearing unit-stress of the same intensity. Similarly, Fig. 90 shows the shearing stresses rormal 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. Fig. 105a Fig. 1056 Let the parallelopipedal element in Fig. 1056 have the length dXj the height dy, the diagonal dz, and a width of unity normal to the plane of the paper. The tensile force S . dy tends to pull it apart longitudinally. The vertical shear S s . dy tends to cause rotation and this is resisted by the horizontal shear S s . doc. These forces may be resolved into components normal and parallel to the diagonal dz, as shown in the figure. The components normal to the diagonal form normal tensile force S n dz y 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 S and S 8 . Let (j> denote the angle between doc 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^+Sg . doc . sm Since dx=dz.cos and dy=*dz.sw$, these equations reduce to 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 8 Expressing cos>2 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 = %S (S 8 2 + (iS) 2 )* max S p = (S 2 + (i^) 2 )* (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 ooo pounds and at the same time to a cross-shear of 3000 pounds. Here 5 = 6366 pounds per square inch and S 8 = 3 820 pounds per square inch. Then S n =-1-8 1 55 pounds per square inch, and S n = i 790 pounds per square inch for the minimum tensile or maximum compres- sive stress, while 6*^ = 4970 is the maximum shearing unit-stress; these are the greatest normal and shearing stresses due to the combination of S and S 8 . 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 -3 183/3820= -0.833, whence = 64 53' or < = I 54 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 tan2< 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 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 cot2< = - oo 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. 266 APPAKENT COMBINED STRESSES CHAP, xi Prob. 105. A bolt J 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 8 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 n = \ 8 max p = the first of whicrf 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/ 1 '= ^Pl/Trd 3 = 5 400 pounds per square inch From Art. 92, the shearing unit-stress on the surface is, 6^=321 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 7oo+ / \/4ooo 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 TOKSION 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 S under any transverse load is Mc/I=32M/7rd* (Arts. 41, 42, 43), while that for 5. is Pp . c/J = i6Pp/nd 3 (Arts. 90, 92). Inserting these in formula (105) and solving for d 3 , there is found + M* T\nd*S P = (Pp)2 + M 2 (106) in which M is the bending moment of the loa^ds 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 = 4X480X96 = 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 be substituted for d 3 . 268 APPARENT COMBINED STRESSES CHAP XI Prob. 1060. Find the factor of safety for the data of Prob. 92a, when the shaft is in bearings 12 feet apart and carries a load of 200 pounds at the middle. Prob. 1066. A horizontal nickel-steel shaft of 17 inches outside and ii inches inside diameter is to transmit 16000 horse-powers at 50 revolutions per minute, the distance between bearings being 18 feet. Taking into account the flexure due to the weight of the shaft, compute the maximum unit-stresses. ART. 107. COMPRESSION AND TORSION When a loaded vertical shaft rests in a step at its foot, the torsional unit-stiess S s combines with the direct compressive unit stress S due to the weight of the shaft and its loads, and there is produced a resultant compression S n and a resultant shear S p . These may be computed from (105) after S and S s have been found. When W is the load on the section area a, the value of S is W/a if the shaft is short, while for long shafts it is to be found from the column formula (80). In order to prevent vibration and flexure it is usual to .place bearings at frequent intervals on a vertical shaft so that probably the use of the column formula will rarely be required, particularly when high factors of safety are used. The value of S 8 is found from the torsion formula (90), and for a solid round shaft S s = i6Pp/xd 3 , where Pp is the twist- ing moment which may be replaced in terms of the transmitted power by formula (92) . To find the diameter of a vertical solid round shaft for a given unit-stress S n or S p , a tentative method must generally be em- ployed. Inserting ^W/nd 2 for 5 and i6Pp/xd* for S s in for- mulas (105), they become, and assumed values of d may be substituted in each of these, until finally a value is found which satisfies the equation. When d is given, however, the unit-stresses S n and S p may be directly computed. A vertical shaft is sometimes so arranged that its weight and that of its loads is supported near the top on a series of circular ART. 108 HORIZONTAL SHEAR IN BEAMS 269 disks, sometimes called a thrust-bearing. The shaft is thus brought into tension instead of compression and this is a better arrangement because there is then no liability to lateral flexure. The above method and formulas apply also to this case. Prob. 107a. A vertical shaft, weighing with its loads 6 ooo pounds, is subjected to a twisting moment by a force of 300 pounds acting at a distance of 4 feet from its center. If the shaft is wrought iron, 4 feet long and 2 inches in diameter, find its factor of safety. Prob. 1076. Find the diameter of a short vertical steel shaft to carry loads amounting to 6 ooo pounds, when twisted by a force of 300 pounds acting at a distance of 4 feet from the center, taking the working unit-stress for compression as 10 ooo and that for shearing as 7 ooo pounds per square inch. ART. 108. HORIZONTAL SHEAR IN BEAMS The common theory of flexure, as presented in Chapter V, considers that the internal stresses at any section are resolved into their horizontal and vertical components, the former pro- ducing longitudinal tension and compression and the latter a transverse shear, and that these act independently of each other. The shear formula S s a = V 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 FT , ~T~ the equilibrium. Let H be m-\ J-wi T c j j. the sum of all the horizontal components of these forces acting on mn and H f the sum of those acting on qp. Now H' is greater or less than H, hence the difference H f H must act along mq as a horizontal shear. Let the distance mq be ox, the thickness ww' 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 of horizontal stresses between mm' and nn' 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)2 c z da . z/I Now, the distance mq being 3x, 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 h tdx. Again from Art. 47 it is known that dM/dx is the vertical shear V. Therefore, S h = (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 z = o. 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 / its value at the neutral surface is, y y 5 8 = I^a . z or S 8 = a^i in which / is the width of the section at the neutral axis, a t 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 -[ l zbd 3 , while the statical moment a\c\ is \bd . \d> By inserting these in the formula, there results S* = 2 V/bd = f . 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 S > = -%-~a r 5 '= that of a\ is fad 2 , and that of c l is 2^/3^; accordingly ff = f , and the maximum shearing unit-stress is S s =| . V/a, which is 33J percent greater than the mean. 272 APPARENT COMBINED STRESSES CHAP. XI For an I section the coefficient a will depend upon the ratio of the flange and web thicknesses to their lengths. As a numerical example, take a steel beam 20 inches deep and weighing 80 pounds per linear foot; the width of flanges being 7.00 inches, the mean thickness of flanges 0.92 inches, and the thickness of web 0.60 inches. Table 6 gives ^ = 7.86 inches and / = o.6o inches; the statical moment a^ is to be found by taking the sum of the moments of flange and web areas on one side of the neutral axis with respect to that axis. The flange area is 7.00X0.92=6.44 square inches and its center of gravity is 10.00 0.46 = 9.54 inches from the axis; the web area on one side of the neutral axis is 0.60(10.00 0.92) =5.45 square inches, and its center of gravity is J(io.oo 0.92) =4.54 inches from the axis. Hence the statical moment a\c\ is 6.44X9.54+5.45X4.54 = 86.18 inches 3 . The coefficient o now is 86. 18/7. 86 2 X 0.6 = 2. 32, so that the shearing unit-stress at the neutral surface is S = 2.32F/a or 132 percent greater than the mean unit-shear V/a. It is hence seen that the shearing unit-stress at the neutral surface of a steel I beam or plate girder should not be computed by the common formula when a precise result is required. Prob. 108. In the Journal of the Franklin Institute for February, 1883, is described an experiment on a spruce joist 3^X12 inches and 14 feet long, which broke by tension at the middle and afterwards by shearing along the neutral axis at the end, when loaded at the middle with 12 545 pounds. Find the maximum horizontal shearing unit- stress by the use of the above formula. ART. 109. LINES OF STRESS IN BEAMS From the last article it is clear that at any point in a beam there exists a horizontal and vertical shearing unit-stress S 8 , the value of which is given by (108). At that point there is also a longitudinal tensile or compressive unit-stress S which may be computed from the flexure formula (41) with the aid of the prin- ciple that these stresses vary directly as their distances from the neutral surface. In Art. 105 it was shown that these unit- stresses combine to produce maxinxum and minimum norma! ART. 109 LINES OF STRESS IN BEAMS 273 stresses on planes at right angles to each other and maximum shearing stresses on planes bisecting these. The direction of the shearing stresses and their values are given by and from these formulas the lines of maximum shear may be traced throughout the beam. Art. 108 shows that S 8 is greatest at the neutral surface and zero at the upper and lower sides of the beam. The longitudinal tensile or compressive stress S is zero at the neutral surface and greatest at the upper and lower side. Hence (f> is o for the neutral surface, it increases with the distance z from that surface, and it becomes 45 at the upper and lower sides of the beam. The broken lines in Fig. 109 show these lines of shear. The angle 6 which the direction of a maximum or minimum normal stress makes with the neutral surface is greater or less by a right angle than the angle given by cot20 = J5/5, because Fig. llOa Fig. 1106 In the case of a bar the different materials might occupy dif- ferent parts of the length as shown in Fig. llOa, where the spaces i, 2, 3 designate different materials, although this is a method rarely used. In Fig. 1106 the three materials are shown arranged longitudinally, this being the method most commonly employed for compound bars. Round bars for jail windows are sometimes made of two kinds of steel, an inner core of soft steel and an outer annulus of very hard steel, the function of the former being to resist lateral bending and that of the latter to resist attempts of the prisoners to file or cut. A compound bar may also be formed of timber and plates of metal bolted together, this being a method more commonly used for beams than for bars. In these figures the bars are represented as short columns under compression, but the following reasoning is general and applies also when com- pound bars are under tension. ART. no BARS OF DIFFERENT MATERIALS 277 The load P on the bar is supposed to be axial, so that the stress is uniformly distributed over the section area of each material. Considering first the case of Fig. 1100, where the materials are arranged in series, it is plain that the total stress in each section area is equal to P. Let l it 1 2 , / 3 be the lengthr; of the three parts, #1, a 2 , a 3 their section areas, and EI, E 2f E 3 the moduluses of elasticity of the three materials. The change of length e\ of the first part is Pli/aiEi, that of the second part is e a = P1 2 / 'a 2 E 2 , and that of the third part is Pl 3 /a 3 E 3 (Art. 10). Accordingly, E 3 a 3 ' is the total change in length of the bar provided the elastic limit of the material be not exceeded. The unit-stresses in the three parts of the bar are P/0i, P/a 2 , P/a 3 ; when the three section areas are equal, as in the first diagram of Fig. 11 0#, the unit- stresses in the three parts are equal. For the case of Fig. 1106, the three lengths are equal to /, and the three section areas are 0i, a 2 , a 3 . Let PI, P 2 , P 3 denote the loads borne by the parts, i, 2, 3, the sum of these being equal to P. These loads may be found from the fact that the changes of length of the three parts must be equal. Accordingly, P! -|- P 2 + P 3 = P Pi/AiEi = P 2 l/a 2 E 2 P 3 l/a 3 E 3 = P 2 l/ 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 S 2 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 S 2 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 ooo 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 + ^X10X200=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. 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. 112a 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, W\ being that carried by the timber and W^ 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^ 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 Wl 3 /tEI, where /? is a number depending upon the kind of loading and the arrangement of the ends. From the above con- ditions, 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 I 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. 1120 be 8X 12 inches in section and each of the steel plates be JX9 inches. For the timber /i = }$ X 8X 1 2 3 and for the steel I 2 =^ 2 X i X Q 3 , whence the ratio /i// 2 = 512/27; also the ratio E 2 /Ei=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 W 2 = 8 200 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\c\II\ = J^i/Ci/8/i =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 = 13 700 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 6*2 be the given allowable unit-stresses; from the above their ratio is Si/S 2 = WiCiI 2 /W 2 c 2 Ii, 284 COMPOUND COLUMNS AND BEAMS CHAP. XII and replacing W\ and W 2 by their values from (112) this reduces to Si/S2 = EiCi/E 2 C2' The total bending moment M, for any kind of loading, is equal to the sum of the resisting moments Si . Ii/ci and 52 . 1 2/^2- Accordingly the two equations for design are, 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 /ci. 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 /E\ is 20, while S 2 /Si should be about 15; hence the depth of the steel should be three-fourths of the depth of the timber. When di and d 2 are equal, as in Fig. 112c, 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 d\ and d 2 are given, as also the width bi of the timber, the thickness b% of the metal is computed from this equation. For example, let ^1 = 12 and d 2 = g inches for timber and steel; then S 2 /Si must equal 15. Hence let Si = i ooo and 52 = 15 coo pounds per square inch ; let 61=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 = iPF/ = 72oooo 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/S2 equals Ei/E 2 , and hence the flexural unit-stress on the channels is twenty times that on the timber. Thus, if 2 for the structural steel is taken as 12 coo pounds per square inch, Si 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 %Wl = 44i ooo inch- pounds. Let the timber be 8 inches wide and 10 inches deep; then 71/^1 = 133.3 inches 3 . Using Si =600 and 5 2 = i2ooo, the value of /2/C2 f r the 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 /2 = i57-4 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. 1120 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 =EiCi/E 2 C2 is also to be modified by multiplying the second member by i 8/c 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. 1130 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 E%/E\ being taken from 10 to 15, while the ratio S%/S\ 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 15 ooo pounds per square inch. Supposing the steel to be stressed only to 8 ooo pounds per square inch, it is seen that S\ 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. II Fig. 113a Fig. 1136 Fig. 113c Fig. 11 3d 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 2 is the distance from the neutral axis to the remotest part of the metal. When the area of metal is small compared to that of the concrete, as is generally the case, it will be sufficiently precise to take the moment of inertia 1% as equal to the area a of the metal multiplied by the square of the distance h from its center of gravity to the neutral axis, or I 2 =ah 2 ; also it will be sufficiently precise to take /i= T VW 3 , thus 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 S2/Si t 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, 71=1152 inches 4 , and 7 2 =4.9iX4 2 = 79 inches 4 . Now, taking E 2 /Ei = io, the first formula of (112)' gives 82/81 = about 8; accordingly if the safe allowable tensile unit-stress for concrete is taken as 50 pounds per square inch, ,$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 2 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 Wz^&SJz/cJ. For instance, if the simple beam is 8 feet 4 inches long W% = 16 400 pounds is the load which is carried by the steel if the concrete does not act at all, while Wi=jjo 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 the 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 seer, 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%/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 6J 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 ij 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\/Ei for the concrete must tqual 5 2/^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. 114# 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 S 2 the tensile unit-stress in the steel. Let ss 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 i* 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' is %Sib(^d+k) 2 , since the total compressive stress acts with the lever arm (Ji+), the moment of the tensile stresses in the concrete is $Si'b(%dk) 2 , and the moment of the tensile stresses in the steel is S2a(h k). Accordingly, are two equations connecting the four unknown quantities Si, Si', S 2 , k. Two other equations result from the law that the unit-elongations are proportional to their distances from the neutral surface: Si-/Si f = Qd+k)/($dK) 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 gives, _ _ _ (bd/a)(E,/E 2 ) b(d 2 +i2hk) 014} from' which it is seen that the position of the neutral surface depends upon h and the ratios a/bd and E2/ ! 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 Jow 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. m q ,,. , r 4 * -77P A j. t. i"' _ i! < r d r ! -r T 3\ S 2 ^ .... i t_ { i i ^ ! 2 >- m/ y 9 -*- 6 > I a Fig. 114a Fig. 1146 CASE II. No Tension in Concrete. Let Fig. H4b 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^a and the sum of those above it is ^S\bn, since the average unit- stress JSi acts over the area bn; hence S^a-=\S\})n is the first equation between 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 Szd(g n) is the moment of the stresses in the metal with respect to the axis s, and JSi& 2 is the moment of the stresses in the concrete, since the total stress }>Sibn acts with the lever arm \n\ hence S^(g n) + \Sibn 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; $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. XII while the unit-shortening of the concrete at the same distance from the neutral surface is Si(g n)/Ein. Hence, S 2 a= $Sibn S 2 a(g-n} + $Sibn2=M S 2 /E 2 = (Si/Ei)(g-n)/n are three equations for finding n, S\, S 2 ] their solution gives ea in which s 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. 114#. A reinforced concrete beam is 5 inches deep, 4 feet wide, 6J feet in span, has 1.8 square inches of steel at ij inches belcw 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 Si and S 2 . ART. 115. INVESTIGATION OF REINFORCED CONCRETE BEAAIS 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 = 48, d = $, h=2 inches; 6^=240 and 0=3.6 square inches, whence 0/6^ = 0.015; also E 2 /Ei = io (Art. 113). Using formulas (114) there is found = 0.261 inches for the location of the neutral surface belov; the middle. The maximum moment is M = \Wl = i() 200 pound-inches, vrhence 5 = 6M/bd 2 = Si pounds per square inch as the flexural unit- stress for a plain concrete beam. Then 6*1 = 72, and 5V = 58 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 40, while the second gives a factor of safety of about 5. Also 2 = 250 pounds per square inch for the steel, which is a very low value, the factor of safety being over 200. While this beam is perfectly safe, it is not designed for proper economy, since the compressive stress in the concrete and the tensile stress in the steel might be much higher. As another example, let ^ = 48, ^=5, A=iJ, g = 4, I = 84 inches, a = 2. 4 square inches, and the total uniform load W be 6 ooo pounds. Taking E 2 /Ei = 10, (114) gives Si' =450 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, 1 = 373 pounds per square inch for the concrete and 5 2 = 1 1 700 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 \ 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 2 being assumed, three computations maybe made to find three values of S from which three values of M are determined; then for uniform load W=SM/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 2, 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 ij inches from the lower side. Find the total uniform load W which this beam can carry so that the tensile stress in the concrete on the lower side may be 100 pounds per square inch. Prob. 115&. 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 f be taken even as high as the ultimate tensile strength of the concrete, the values of Si and 2 are too low; 52 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%/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 S 2 /Si. The unit-stress Si should be taken as high as allowable ; S2 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 ij 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 E 2 /Ei = e= 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 =JX 7 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 a = 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 52, 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 or n=g (116)' - i+a/e- For instance, in the case of the last paragraph, where ^ = 15 and (7 = 40, the value of n is 0.385 g, so that the neutral surface is 5.0 inches below the upper side of the beam. Steel is the only material which has been advantageously used for reinforcing rods in concrete beams, and the proper section area should rarely exceed one percent of that part of the concrete above the centers of the rods. With unit-stresses of 5 ooo and 500 pounds per square inch for steel and concrete respectively, the section area of the steel is 2^ percent of that of the concrete when the ratio s is 10 and 3 percent when is 15, but this is an excessive use of steel which is not economical. The following are computed values of a/bg in percentages, and also values of the ratio n/g: S 2 /Si = 15 20 25 30 35 a/bg= 1.33 0.83 0.58 0.42 0.32 n/g= 0.40 O-33 0.29 0-25 0.22 These apply to first-class concrete, for which e=io. As a general rule the most economic dimensions will be obtained by taking S 2 as about 30 times Si for concrete made of i cement, 2 sand, 3 stone, while S 2 should be taken as about 12 500 and Si as about 350 pounds per square inch for concrete of i cement, 3 sand, 6 stone, the ratio S 2 /Si being here 35. Beams having a section like that of Fig. 113d are sometimes used, it being considered that only the upper part of the concrete is stressed in compression, while the tensile stresses in the lower projecting parts are neglected. The design of these is hence made by the formulas (114)', where n is the depth of the upper 298 COMPOUND COLUMNS AND BEAMS CHAP, xn slab, g the depth from its upper surface to the center of the steel rods, and b the width of slab belonging to one rod. These formulas may be written / (j \ i bn ~ n ( I ) ~~2~a from the first of which g is to be found for a given depth n, while the second gives the metal section a. For example, let E 2 /Ei = s = io, S2/Si=o = 2$, n = 4 and b = iS inches; then =14.0 inches and (3=1.42 square inches. The safe load which this beam can carry may be found from M=S^a g Jn); for instance, in the above case let S 2 = 12 500 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 A a and Bb, I 2 that between Bb and Cc, and /s that between Cc and Dd. Then S= MI . 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 S. 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.o&owl 2 , that at Cc is o.n^.wl 2 , and that at Dd is 0.12 $wl 2 . Hence the value of I\/c\ is o.o&wl 2 /S, that of /2/C2 is o.ii4.wl 2 /S, and that of 1 3 / 3 is o.i2$ivl 2 /S. Sections of plates, angles and web may now be determined, with the help of Art. 44 and Table 10, to satisfy these requirements. Then the section Aa must be investigated to ascertain whether it is Fufficiertly large to safely resist the vertical shear \wl. 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 S the allowable unit-stress. Then the stress Sa in one flange acts with the lever arm d with respect to the center of gravity of the other flange, and therefore Sad equals the bend- ing moment M. Thus, for the first section ai=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 to it at intervals. 300 COMPOUND COLUMNS AND BEAMS CHAP. XII 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=Wil 3 /aEi!i in which Wi is the total load that comes on the material that has the modulus of elasticity EI and the moment of inertia /i. The same method applies to the compound beams of Fig. 113a 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 / '= $wl 4 / ^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 /3 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, 1 2 , Is be the distances from the left end in Fig. 117 to the sections Bb, Cc, and Dd, so that /* 3 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 2 . Let m be the bending moment due to a load unity at the middle of the beam or m = \x. Then, by (124), , fMm. rtiMm- i^Mm. f^Mm. f= I ~=^3X=2 I X+2 I _J#+ 2 / -^ffa J EI J Eli Jk EI 2 Ji 2 EI 3 gives the deflection at the middle under the uniform load. Inte- grating between the designated limits, there is found, f= wl //I 3 /2 3 ~/l 3 . /3 3 ~/2 3 \ W //! 4 /2 4 -*l 4 ,/3 4 -/2*\ J 6E\/i / 2 7 3 / 8EWi / 2 /s / 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 /i. 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. Xll 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 wx and v = \. Then, by (125), f r vv * r liVv * c hvv * r hvv * /= / -=-dx = 2 I -=r-dx+2 I ^r-dx+2 I r=dx J Fa J F(n J k Fa 2 J h Fa 3 in which #1, a 2 , a 3 are the section areas whose moments of inertia are /i, /2, /3, and F is the shearing modulus of elasticity. Per- forming the integrations, there is found, , 2F \ai d2 ^3 / 2F 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. ART. 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 / t 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, xill 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 load is e = 35 000X25/30000000=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 clastic 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 7=7.14 inches. Hence the energy stored in the beam is JXi 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 e 2) so that the total energy stored in the bar is J(Pi + P 2 ) (ei + e 2 ), as clearly appears from Fig. 1195. 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 30o from PI to P, while the deflection increases from fi to /, the external work and the stored energy due to this increase are both expressed by K = %Pf-%P 1 f 1 . 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 S and the elongation e is the unit-elongation e. Thus, K=$Se K=$Se-lSiei (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 S. Fig. 119a Fig. 1196 Fig. 119c When the bar is stressed so that S 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. I4c 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 unannealed and annealed Bessemer steel in the table of Art. 25, which had 5^ = 125000 and =o.n before annealing and S u = 99000 and 306 RESILIENCE AND WORK CHAP, xni 6=0.19 after annealing; here # = 10300 inch-pounds for the first and K = i4 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 =%(S e + 2S u ), where S e is the Clastic limit and S u the ultimate strength. Prob. 1190. 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. 1196. 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 6 ooo up to 12 ooo 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 utiUzed 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 2 /E = 3.0 inch-pounds per cubic inch for cast iron, $S 2 /E = 1.2 inch-pounds per cubic inch ART. 120 RESILIENCE OF BARS 30? for wrought iron, %S e 2 /E = i2.$ inch-pounds per cubic inch for structural steel, %S 2 /E = 20.4 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 = l(S 2 /E)al-l(Si 2 /E)al=l(S 2 -Si 2 )/E . al Here, as before, the unit-stress S 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. 120ad supposed to be there. The deflection at this point due to all the loads on the beam being /, the work of the weight P is JP/ and this must equal the internal energy throughout the beam due to that weight P. Let x be the distance of any section of the beam from an assumed origin, S the unit-stress on the extreme fiber of this sec- tion at the distance c from the neutral axis, and (S . z/c] da the stress on a fiber of area da at a distance z above or below the neutral axis, both of these being due to all the loads on the beam. Let s and s . z/c be the corresponding unit-stresses due to the weight P ; the change of length in the fiber of area da in the hori- zontal distance dx, due to the weight P, is (sz/cE)dXj and the work in this fiber due to P is $(Sz/c)da(sz/cE)dx. The sum- mation of all values of this product, first throughout the section and secondly throughout the length of the beam, gives the in- ternal energy due to P. Now Ida . z 2 is the moment of inertia /; the value of S/c is M/I, where M is the bending moment due to all the loads on the beam, and the value of s/c is M//, where M is the bending moment due to the weight P. Accordingly, equating the external work to the internal energy, and extending the integration over the length of the beam, there results, or ART. 124 DEFLECTION AT ANY POINT 315 both of which give the deflection / at any point of the beam; in the first M is the bending moment due to any weight p at that point; in the second M' is the bending moment due to the weight unity at that point. By inserting for M and M' their values in terms of x, and integrating, the deflection / is readily found. Fig. 124a Fig. 1246 Fig. 124c For example, let it be required to find the deflection at the end of a cantilever beam due to a load P placed at a distance K! from the left end, as in Fig. 124a. On the left of the load M = o, and on the right of the load M=P(x id), where x is measured from the free end. Placing a weight unity at the end, the bending moment due to it is M' = x throughout the entire length of the cantilever. Thus MM'=o on the left of the given load and MM' =P(x 2 -Klx) on the right. The formula then furnishes, EIf= f P(x 2 -Klx)dx=$PP( 2 -3K+K3) J Kl which gives the deflection of the end due to P. When /c = o, the load is at the end and /= PP/EI, as otherwise found in Art. 54. As a second example, let it be required to find the deflection of the right end of the overhanging beam in Fig. 1246, the load being uniform throughout. Let / be the span and d the length of the overhanging arm. The left reaction due to the uniform load is \wl(i - K 2 ), whence M = %wl(i - K 2 )x- \wx 2 for the span /; for the overhanging arm let x be any distance from the right end, then M = \wx 2 . For a weight unity at the end of the overhang- ing arm, the left reaction is K, whence M' = nx for the span /; for the overhanging arm M'=x. The deflection formula then gives EIf= - from which/ is known for any value of K. For instance, let =i, then/= - i $wl 4 /2048EI, the minus sign showing that the deflec- 316 RESILIENCE AND WORK CHAP, xm tion is upward, because at the beginning of this article positive values of f were measured downward. For a longer overhang- ing arm, however, / may be positive ; this is the case when K = J. When a cantilever beam deflects, the free end suffers a hori- zontal displacement which may be derived from (124), taking the force unity as horizontal, and using the equation of the elastic curve. For Fig. 124c, where the load P is at the end, the elastic curve has the equation ;y=/i(2/ 3 3/ 2 #-f-# 3 )/2/ 3 , where f\ is the vertical deflection under the load (Art. 54). Now let the hori- zontal force unity be applied to the free end where the horizontal displacement is /; its moment is M' = ~(f\y) and the moment of P is M= -Poc. Then the formula (124) becomes, i -?)***= (Pf i/ * from which the horizontal displacement of the end of the beam may be found to be /=6/i 2 /5^ which is very small compared with the vertical deflection of the end. The above examples show the great value of formula (124) in discussing all questions regarding the elastic deflections of beams. It may be used to find the equation of the elastic curve also, taking / as the ordinate corresponding to the abscissa x. It applies, however, only to cases where the greatest flexural unit-stresses do not exceed the elastic limit of the material. When / is not constant, it is to be expressed as a function x before the integration can be made. Prob. 124a. Find by the above method the equation of the elastic curve for a simple beam uniformly loaded, and compare it with that deduced in Art. 55. Prob. 1246. A simple beam has two equal concentrated loads placed at equal distances from the supports. Deduce the deflection under one of the loads, and also the deflection at the middle of the beam. . ART. 125. DEFLECTION DUE TO VERTICAL SHEAR The treatment of the deflection of beams in the previous pages has been solely from the standpoint of the horizontal ART. 125 DEFLECTION DUE TO VERTICAL SHEAR 317 stresses caused by the external bending moment. Art. 122 shows, however, that internal energy is stored in a beam by a vertical shear, and it hence appears that the former investigations are more or less incomplete in neglecting the deformation which is caused by the external vertical shears. Referring to Fig. 125 let V be the vertical shear which has produced the vertical detrusion df in the short length dx. Let a be the section area of the beam, F the shearing modulus of elasticity. The shearing unit-stress in the section is V/a and this will be considered as uniformly distributed. The internal work in the length dx is JF . df when V increases gradually from o up to V. Now from the definition of F in Art. 15 it is known that df=(V/aF)dx, since df/dx is the unit-detrusion that is due to the unit-shear V/a. Accordingly the whole internal energy due to shear in the length dx is dK = J( F 2 / aF) dx. When / is the deflection under a single load P, the external work is \Pf. Accordingly, <- 505- are expressions for the total internal energy due to shearing, and the deflection / under a single load P. For instance, a simple beam with a load P at the middle has the shear F = JP. Then by taking double the integral between the limits o and \l there is found K= P 2 l/SFa for the internal energy or resilience due to the shears throughout the beam. The deflection under the load due to the vertical shear then is f=Pl/^Fa. In Art. 55 the deflection due 10 the bending moments was found to be P1 3 /4SEI. Accordingly the ratio of the shearing deflection to that of die flexural deflection is i2EI/Fal 2 , which reduces to i2(E/F)(r/l) 2 , where r is the radius of gyration of the section. For a cast-iron beam the ratio E/F is about 2.5; if the section is square the ratio of the shearing to the flexural deflection is 2.5(d//) 2 , where d is the side 318 RESILIENCE AND WORK CHAP, xill of the square; when the length of the square beam is 30 times its side, this ratio is ^fo, so that the deflection due to shearing is small compared with that due to the moments; for a short beam, such as l = 2d, this ratio is 0.625, so tnat tne deflection due to shearing is 62 J percent of that due to the moments. In 1870 Norton called attention to the inaccuracy of the ordinary formula for deflection in the case of short beams. Experi- ments on white-pine beams of different lengths and sizes were made with loads at the middle of the spans, and it was shown that the deflections were directly proportional to the loads and inversely proportional to the breadth of the beam, as the com- mon formula requires. The deflections were, however, not directly proportional to the cubes of the spans nor inversely pro- portional to the cubes of the depths of the beams, as the formula requires. An examination into the reason of these discrepancies showed that it was due to influence of the vertical shears, and Norton showed that the quantity C . Pl/bd should be added to the common flexure formula in order to satisfy the results of his experiments; he further showed that the quantity C had the value o.ooo 0x3094 for the white-pine beams. From the theoretic value deduced in the last paragraph it is seen that 4F is the reciprocal of this number, whence F = 266000 pounds per square inch, which should be the modulus of elasticity for the shearing of white pine across the grain. This is probably not far from the actual value of that coefficient, since Thurston, by experiments on torsion, found jF = 22oooo pounds per square inch for white pine. The experiments of Norton, therefore, confirm the above theory of the deflection of beams due to shearing. Formula (125) does not apply to the deflection at any point, or to any kind of loading except a single load P. For any other case let any weight P be at the point where it is desired to find the deflection, and let v be the shear due to this weight. The external work due to P is JP/, where fu the required deflection. The in- ternal work in any elementary distance dx is one-half of the prod- uct of the actual shear Fand the detrusion (v/aF)dx. Therefore ART. 125 DEFLECTION DUE TO VERTICAL SHEAR 319 / or are expressions for finding the deflection due to shearing; in the first v is the shear due to any weight P and in the second V is the shear due to the weight unity at the point where the deflec- tion is desired. For example, take the case of Fig. 1240 and let it be required to find the deflection at the end due to shearing when a single load P is at the distance id from the left end. Let x be measured horizontally from the free end where the weight unity is placed. Then V = o on the left of the load and V = P on the right of the load, while V = i throughout the beam. Then from (125)', Faf= f l P3x=Pl(i-k) or f=(i-k)Pl/Fa JK! gives the deflection at the end due to the vertical shears. When K=O, the load P is at the end and f = Pl/Fa; the ratio of this to the deflection due to the moments is s(E/F)(r/l) 2 . This shows that the deflection due to the shears is scarcely appreciable in long beams; for short beams, however, it may be larger than that due to the moments. By measuring the elastic deflections of two beams of different lengths but of the same material, it is possible to compute the values of E and F for that material. The deflections thus meas- ured are due both to moments and shears and hence an ex- pression for each measurement is to be written in terms of E and F. Let l\ and / 2 be the spans, #1 and a 2 the section areas, /i and /2 the moments of inertia, and PI and P 2 the loads at the middle of the simple beams. Let the measured deflections be /i and / 2 . Then may be written, l./Vi!. P/2 3 i ,P 2 1 2 i =f Jl J 48/1 401 F 48/2 E 4^2 F which contain the two unknown quantities i/E and i/F, and hence the solution of the two equations will furnish the values of the common modulus E and the shearing modulus F. By making many experiments instead of two, writing an equation for each, and solving the equations by the method of Least Squares, 320 RESILIENCE AND WORK CHAP, xin it is possible to obtain reliable values of these moduluses of elas- ticity. In such experiments it is of course important that the loads should not be heavy enough to stress the material beyond its elastic limit. In conclusion it may be noted that there is an elastic curve due to vertical shears, and its equation may be used in similar manner to that of the bending moments. From Fig. 125, the general equation of this elastic curve is, -4 or af-v ox ab ox For example, let it be required to find the elastic curve and the deflection due to vertical shears for a simple beam of span / loaded uniformly with w per linear unit. Here V = \wl wx, and inserting this in the general equation and integrating there is found aFy = ^w(lx x 2 ) for the equation of the elastic curve. The maximum deflection occurs at the middle where x=%l, and its value / =wl 2 /SaF } and this agrees with the result which will be found by the use of (125)'. Prob. 125a. Find the deflection of the end of the overhang in Fig. 1246 due to the vertical shears of the uniform load. Prob. I25b. A timber beam 9 inches long and 2 inches square is placed on supports 7 inches apart and subject to deflection by a load of 300 pounds at the middle. Compute the amount of the deflection due to the vertical shears and that due to the bending moments. ART. 126. PRINCIPLE OF LEAST WORK There is a certain law of nature called the principle of least work, which is of great value in discussing problems in mechanics that cannot be solved by static conditions alone. For example, a table with four legs is a system where the reactions of the legs, due to an unsymmetric load on the table, cannot be found by statics, since the three conditions of static equilibrium cannot determine four unknown quantities. The principle of least work furnishes, however, a fourth condition when the stresses in the legs do not exceed the elastic limit of the material. ART. 126 PRINCIPLE OF LEAST WORK 321 When a structure is so formed that the stresses in it cannot be determined by statics, it is said to be a " redundant system", and the principle of least work applicable to it is as follows: The stresses in a redundant system have such values that the internal energy of all the stresses is a minimum. This is sometimes regarded as an axiom, it being considered that the resisting forces will store up no more energy than the minimum which is necessary to maintain equilibrium with the external forces. The following proof, however, will probably be more satisfactory than the assumption of the axiom. Let PI, P 2 , . . . P r be r forces in equilibrium, and let a small displacement be made, without performing work on the system, so that the points of application of these forces move through the small distances o 3 + . . . +p r de r =o Now the condition that this equals zero is the same as the condi- tion that its integral shall be a minimum, or PI&I + P 2 e 2 + -^3^3+ +P r e r = a. minimum in which e\ t e 2 , ... e r are the total distances through which the forces have moved in acquiring the state of stable equilibrium. Now each of the quantities Pe is double the work done by the applied force P as it increases from o to P, and hence also equal to double the energy stored by the stresses that balance it. There- fore, the internal work or stored energy of all the resisting stresses is a minimum. When forces act upon elastic bodies, in which the defor- mations are proportional to the stresses, the above principle of least work may be applied to determine unknown reactions when the conditions of statics are not sufficient in number to do so. Many problems of fixed and continuous beams may be discussed by the aid of this principle. For example, take a con- tinuous beam of two equal spans loaded uniformly in the second span, as in Fig. 126a, and let it be required to find the three reac- tions Rij R 2 , RS. The first condition of statics is that the sum of the reactions equals the total load, or ^1+^2 + ^3=^; the 322 RESILIENCE AND WORK CHAP. Xlii second condition is that the sum of the moments of the reactions equals the moment of the load, whence, for an axis at the right end, 2Ril + R 2 l = %Wl. The third condition is that the internal energy of all the flexural stresses shall be a minimum. For the first span the bending moment at the distance x from the left support is M=R\x\ for the second span the moment at the distance x from the right support is M = R 3 x fyw x 2 . Then, by Art. 123, the total internal energy of the horizontal flexural stresses is, Differentiating this, and equating the .derivative to zero, gives the equation SRi .dRi + SR 3 .dR 3 -^W .dR 3 = o as the condition of least work. Solving the three equations thus established, there result RI = - >W, R 2 = + f W , A* = +-&W for the reactions due to the given load; these are the same results as derived by the use of the theorem of three moments (Art. 71). A A A A 12 34 Fig. 126a Fig. 1266 As a second example, take the partially continuous girder in Fig. 126Z> which has four supports and a joint at the middle of the second span so that the bending moment there is always zero. Let 2/ 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, 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, 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 Ri = -&W, R 2 = +&W, R 3 = +&W, R* = + ^W. 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 Roof c and Bridges. This principle must be used with caution by tht beginner, but it is one of much value in the discussion of struc- tures which are statically indeterminate. Prob. 126. A table of length / 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 \k 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 tc 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 ttsed 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 - Fig. 1276 Fig. I27c 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. 1300 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 ze is reached, the stress in the bar is 2P 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 S and T be the corresponding unit-stresses, so that P = Sa and Q = Ta. Then the equation Q = 2P 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 q 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 = 2S, 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. 127 c represents this case and it shows that the elongation is ei + 2e and that the final stress is PI + 2P; thus the instantaneous load produces its effect independently of the other. As soon as the elongation e\ + 2.e 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 work (Art. 147). The law deduced is, however, very close for a light beam, but Q is really a little less than 2? and q is a little less than 26 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 15 ooo 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 F, 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(2h/e)* q = e( 2 h/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 F = io feet per second, then h =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, 6 = 3050x3 pounds and 5 = 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 iQq=p(-h + q) anc [ q/e = Q/P are to be used to find the values of Q and q, which are, <2 = P+P(i + 2/00* 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, S 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, 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 S. 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 ?/*/)*. . 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 Size of Beam Load Deflection in Inches Inches Pounds Gradual Sudden 1X2 112 0-255 0.515 1X2 224 0.580 0-933 lX 3 224 0.163 0.303 lX 3 560 0.410 0.720 4XlJ 448 0.770 1.510 4Xii 784 I- 2 75 2.225 342 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. Ratio of Deflections Sudden to Gradual 2.04 1.61 1.86 1.76 1.96 i-73 Thus, for the first beam under 112 pounds at the middle, the unit-stress S 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 >? = i/(i+ p - W/P) and Hodgkinson determined by experiments on beams that the value of /* was approximately J, 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, and 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 75 100 pounds Observed (7=0.122 0.150 0.175 0.200 inches Computed q = o.ogj 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 ^=25 50 75 100 pounds Observed (i) so that the dynamic unit-stress is given by T = (t)S; when / is large, T must equal 5; when / is zero, T will be 26*. Many empirical expressions can be derived which satisfy these limiting condi- tions, but the determination of the theoretic form of (t) is greatly ART. 136 STRESSES DUE TO LlVE 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 2$ ; when v 2 has the value gl 2 /&f, 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 m , when v has a greater value, then T is less than 5; 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 (gl 2 /8f)*. From Zim- merman's investigation, there may be written, T=S(i- 2/?)/(i - s/?) /? = 8Sv2/ 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.II 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 that these 352 IMPACT AND FATIGUE CHAF. 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. 1360. 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 S u 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. ART. 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 S is some function of S S' ; this is equivalent to saying that S is a function of S(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 5 when the limiting stresses P r 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 S = Ci + Cz(P' '/P) in which Ci and C 2 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*-S.)P'/P or s= 356 IMPACT AND FATIGUE CHAP. XIV which is Launhardt's formula for the unit-stress 5 that ruptures the bar after an enormous number of repetitions of a load that ranges from P' 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'/P) ; for wrought iron it becomes S = 25 ooo(i+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 S when the bar ranges in stress from P f 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 5 = Ci + C 2 (P'/P); Fig. 1386 represents this case. Let 5^ 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 S is S e and hence Ci = 5,. By the fifth law, if P'/P is i, then S is S a and hence C2=S e S a . The equation of the straight line therefore becomes, S = S,+ (S e -S a )P'/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 e in the absence of knowledge regarding its exact value. For structural steel the formula becomes 5 = 35 ooo (i + JP'/P), in which P'/P is always negative. Thus, if P f is 80 ooo pounds compression and P is 160 ooo pounds ten- sion, then P'/P= 0.5, 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 5 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 u , S e , S a , in Fig. l3Sc. 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 5 becomes S u when P'/P=-fi, and hence C\ + C 2 + C 3 = S U ; secondly, that 5 becomes S e when P'/P=o, and hence Ci=S e \ thirdly, that S becomes S a when P'/P = -i, and hence d -C 2 + C 3 =S a . From these conditions, the values of Ci, C 2 , C 3 are found and the equation of the parabola becomes, P' /P'X 2 which is a formula for the rupturing unit-stress S when the total stress ranges an enormous number of times from P' 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 rhan those found from the straight-line formulas. For structural steel, where S u = 60000, 5* = 35000, and S a = i f j 500 pounds per square inch, the formula (138) reduces to, 5 = 35 ooo[i +0.61 (P'/P) + 0.1 1 (P'/P) 2 ] 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 +J, and the formula gives 5 = 42 500 pounds per square inch. If the stress ranges from 180000 pounds compression to 540000 pounds tension, then P'/P is J, 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 formula.? 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 16 ooo pounds compression to 80 ooo 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, X the fac- tor of lateral contraction, and E the modulus of elasticity of the material; the lateral unit-shortening is then As. Since S = eE is the relation between S and e (Art. 9), it may be considered that the lateral unit-shortening As corresponds to a unit-stress T which has such a value that T = AeE, 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 J. Accordingly, a steel bar under the tensile unit-stress S suffers a true internal compressive unit-stress of JS 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 JS 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 90 ooo 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 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 2 and S 3 those upon the faces normal to 02 and 03; thus the directions of Si, Sa> 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 si in the direction 01 if 82 and S 3 were not acting; also the unit-stresses Ee 2 and Ee 3 may be desig- nated by 7*2 and TV Hence it follows that T l =S 1 -*S 2 -*S3 T 2 = S 2 -te 3 -XS l T 3 = S 3 -lS l -lS 2 (139) are the true stresses acting in the three rectangular directions. If any stress S is compression, it is to be taken as negative in the A.RT. 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 S; then the true internal unit-stress T is 5(i 2A); for steel A = J, and T = ^S, and thus the linear defor- mation is only one-third of that due to a unit-stress 6* 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 Ti =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 ooo 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 S 3 =o. Then from (139), taking A as J, the true internal stresses are TI = + 1 2 ooo, T% = 9 330, TZ = 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 X 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. 1400 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 52 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. 140a, 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 = P 2 /bl, S 3 =o. Let Fig. 1406 represent any elementary parallelo- piped in the interior of the bar, having the length dx t 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 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 dx/dz = cosO, there result, S'dz=Sidy . cos6-S 2 doc sin/? or S' = (Si-S 2 )sm 6 cos 6 and the second equation gives the shearing unit-stress alone ART. 140 SHEAR DUE TO NORMAL STRESS 363 any plane which makes the angle 6 with the direction of S\. The maximum value of S' occurs for 6= +45 or 6= -45, and for both cases may be written S' = J(Si 82), that is, the maxi- mum shearing unit-stress occurs on two planes which bisect the directions of Si and 2, and its numerical value is equal to one- half of their difference. The broken lines in Fig. 1400 show these two sets of planes. ts 2 5x v ,\ V X' x > < X ' >( X S x\ X X ^ V Sidy Fig. 1400 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 5*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 T 2 . Accord- ingly, the formulas, 5' = J(5i-5 2 ) and T' = %(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 J; 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 5 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 2 be com- pression, each equal to 2 400 pounds per square inch. Then the apparent maximum shearing unit-stress is S' = J(2 400 + 2 400) = -2 400 pounds per square inch. For \ = J, the true axial stress is TI = +3 200 and the true lateral stress is T 2 = 3 200, so that the true maximum shearing unit-stress is 7"'= 3 200 pounds pel square inch, which is 33 percent higher than S'. 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 Ss exerts an equal influence upon the two normal to it. When there are three unit-stresses Si, S2, Ss, acting upon a paralellopiped in three rectangular directions, the shearing unit-stress on a plane parallel to Si and S2 is not influenced by Ss, and hence J(Si S2) is the maximum shearing unit-stress for such a plane. Similarly, J(Si Ss) and (2 Ss) are the maximum shearing unit-stresses for planes parallel to Si and S 2 and to S2 and Ss respectively. The same holds true for the true stresses TI, T%, T$\ 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 = 6000, Ss=o, and hence the three maximum apparent shearing stresses are 4 500, 3 ooo, i 500 pounds per square inch. But from (139), taking A as J, the true axial and lateral stresses are T\ = + 5 coo, 7*2 = 7 ooo, T 3 = + 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 S 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 S 2 , and the maximum shearing unit-stress S', S 1== ^+(S s 2 +iS 2 )* S 2 = iS-(S s 2 +i? 2 )* S'=(S S 2 +$S 2 )* (141) It is here seen that the value of S' is the same as that of ^(Si~ S 2 ). Hence when Si and 2 have been computed, ths 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 2 is tension. Let X be the factor of lateral contraction, and T\ and T 2 the true internal unit-stresses corresponding to Si and -5*2. Then, by (139), the value of T l is Si-XS 2 and that of T 2 is S 2 -*Si; substituting in these the above values of Si and S 2) there are found, I j from which 7\ and T 2 may be directly computed. For steel the mean value of A is J, 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 8 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, cot2< = %S/S 8 , where the two values of (f> give the angles included between the direction of 5 and those of the planes against which Si and S 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 8 = 3 ooo, from which Si = +4 1 60 and S 2 = 2 160 pounds per square inch are the apparent maximum unit-stresses of tension and compression, and their directions are given by cot2^ = J. The two values of then are 54 13' and 144 .13', which show that S\ 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 jTi = +488o, T 2 = 3550 pounds per square inch. For the shearing unit-stresses the maximum values are S' =3 160 and T r =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^ = 670, and two other true shears smaller than S' 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 S s . 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. 1410. 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 8 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, S being the horizontal unit-stress for that point as computed from the flexure formula and 5 S the shearing unit-stress as determined by Art. 108. From these the apparent unit-stresses Si and S 2 which act at the given point are computed and then the true unit-stresses T\ and T 2 . The discussion in the last article also shows that the directions of T\ and T 2 are the same as those of Si and 52, 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 e = 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 = JXo.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 Si be the flexurai unit-stress and $2 the compressive unit-stress due to the uniform load. Then the true maximum compressive unit-stress in a horizontal direction is TI = Si XS2) while that in a vertical direction is T 2 =S2 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 52. 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 ooo 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. 1420. 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 369 ART. 143. STRESSES DUE TO SHEAR It is shown in Art. 6, and also in Art. 105, that forces of te^ 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. 143a, 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 FI, 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 S\ = 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 S 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 S be made zero, then the maximum tensile and com- pressive unit-stresses Si and 6*2 are each equal to S 8 . If, how- ever, S a =o, then the maximum shearing unit-stress is JS. 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 T 2 , corresponding to Si and S 2 , are those that correspond to the actual deformations, and by (139) their values are Ti=Si 15*2 and 7^2 =^2 ASi, where Si and S 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- 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 A is J, then the true stress along the longer diagonal is TI = +6 250, while that along the shorter one is T 2 = 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^ + e and that of the shorter diagonal is 2* e; the area of the rhombus then is J( 2 * + ) (2* - e) = i, which is the same as that of the square before it was subjected to shear, e 2 being a negligible quantity. Prob. 143. A steel beam, 2X2X6 inches, is supported at its ends and has a concentrated lead 40 ooo 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 8 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 + X)S a . 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 8 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, S y the vertical compressive unit- stress due to the load or bearing, and S 8 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 y S 8 2 , and that of C is zero, and it reduces to the form S 2 - (S x + S V )S+ S x S y -S s 2 =0 in which the two values of S are the maximum tensile and com- pressive unit-stresses; solving the quadratic equation there re- sults, * (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 %(S x + S y ) is tension or compression. When either S x or S y 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 y due to the load or bearing be i 200, and the shearing unit-stress 5 5 due to the torsion be 6 ooo, all in pounds per square inch Formula (144) then gives 1 = 8200 pounds per square inch compression, and S 2 = 4 ooo pounds per square inch tension for the maximum normal stresses; also S' = J(Si 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 y , and S 8 - What are these roots when S 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 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=S2 = Ss; 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 case 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. 140# and I40b 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 S 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 . sin/9 + S 2 dx. cos/9. Replacing dx/dz and dy/dz by their values cos/9 and sin#, there is found S = Si sin 2 # + S 2 cos 2 6. 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 tan# = ( 6*2/51 )* and on the plane thus determined only shearing stresses are acting. It is seen that no value of is possible unless Si and S 2 have contrary signs, that is, one must be tension and the other compression. When S 2 =-Si, then tan#=o-5 and #=26J; when S 2 =-Si, then tan 6 =i and = 45 ; when S 2 = - 3-Si , then tan/9 =1.73 and 0= 60, and so on. Hence for each negative value of S 2 /Si there are two planes equally inclined to the direction of Si upon which only shearing stresses act. The following figures show the three cases computed above. if- 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 5 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 JS is the maximum shearing unit- stress which makes angles of 45 with the direction of S. Prob. 1450. In the formula S n =Si sm 2 6+S 2 cos 2 0,let Si be larger than $2- Show that the values of S n cannot be greater than Si nor less than 6*2. This is the equation of a curve in polar coordinates, S n being the radius vector for the variable angle 0; 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 v \nit-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 reel 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. 146a 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 (f> which the lines make with the axis of the bar was 65 degrees for nickel steel, 63 degrees for tempered steel, and 58 degrees for annealed steel; for compression the angle between the lines and the axis was the complement of <. 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 Deformation* ART. 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 t t t t Fig. 146a 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 to S Q , and it follows that the motion will actually begin along that plane which has such a value of 6 that S shall be a minimum; or the value of 5 which causes motion to begin is less for the plane of motion than for any other plane. This requires that sin# cos# /i sin 2 # shall be a maximum, and the value of 6 which renders it such is found to be given by cot 26 = /*. Hence, (147) gives the relations between the quantities ft, 6, and SQ. The observed angle 6 is always less than 45 ; for different qualities of steel, 6 lies between 25 and 32 and hence the coefficient of internal friction lies between 0.84 and 0.49. ^Nr 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 SQ 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 sin 2 < or S = fiS / sin<> and, by the same reasoning as before, it is concluded that the actual plane of motion is that which renders sin< cos< + / sin 2 < a maximum, and this maximum obtains when cot2<=-//. Accordingly, -cot 2 ftSo = %S tan < = J5(/ +V'i + /< 2 ) (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 jj. must be regarded as a constant for the same material, it follows that the angles Q and (j> 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 /iSo as the ulti- mate shearing strength S s of the material, since j*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, am. accordingly S s = JS c tan# may be written as a tentative formula, for the relation between the ultimate shearing strengths S 8 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 8 /S C as computed from (147) : Anthracite Coal # = 15 /* = I -73 S 8 =o.~LT ) S c Sandstone 6 = 20 jj. = i.ig S s =o.i8S c Hard Brick # = 25 ^=0.84 S s =o.2^S c Cement and Concrete # = 30 ^=0.58 S s = o.2gS c Cast Iron # = 35 ^ = 0.36 5= 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 S in (147) should be the same as S in (147)', which requires 6 and (j> to be equal. This is a result altogether at variance with experi- ment, and it must hence be concluded either that S in tension is different from So 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 static 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. 147#. 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 lion 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. A.RT. 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 50 ooo 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 linger 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 c 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. IA F Fig. 148 Modern guns are made of hard steel, often of fluid compressed steel, which has an elastic limit of about 50 ooo and an ultimate strength of about 90 ooo 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 of 50 ooo pounds per square inch. ART. 149 LAME'S FORMULA 385 Prob. 148. The diameter of a powdei chamber is 8f inches and that of the projectile is 8 inches, while the pressure during the explosion is 48 ooo 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. 1490, be subject to a pressure RI on each square unit of the inside surface and to a pressure ^2 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. t t 4 T I 1 Fig. 149a Let r\ and r 2 be the inside and outside radii; then the inside pressure on the end of the closed cylinder is nri 2 R\ and the outside pressure on that end is nr 2 2 R 2 . The usual case is that where the inside is greater than the outside pressure, and then n(ri 2 Rir 2 2 R 2 ) 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 So 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 AND 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 S Q , a tan- gential unit-stress S, and a radial unit- stress R. The value of R is evidently inter- ~r~_i mediate between RI and R 2 ; in Fig. 1496 both R and S are regarded as tensile. Flg ' l Now from Art. 139 the effective longitu- dinal unit -elongation of the cylinder due to these three stresses is- = T/E = (S - /IS - IR)/E in which A is the factor of lateral contraction the mean valu'e of which for wrought iron and steel is about J. But, as above noted, both and S are constant for all parts of the annulus, and it hence follows that, S+R = constant or which is one equation between the unit-stresses S and R when Ci 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+dx) 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+dR)(x+dx)-Rx=Sdx or xdR+Rdx=Sdx which is a second equation between the unit-stresses S and R. The solution of these two equations is readily made by insert- ing 2C\ R for S in the second equations and integrating; then, ART. 149 LAME'S FORMULA 387 where 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 = ri, and R=R 2 when x = r2', inserting these condi- tions in the second formula, there result two equations from which, Ci = (ri^ -rfR 2 )/(rf-r?) C 2 = /vV 2 2(tfi ~R 2 )/(r 2 2 -r 1 2 ) and, inserting these constants in (149), are now obtained, S= (r^R, -r 2 2 R 2 + r -^(R l -R 2 )\ I (r 2 2 - rft 2 2 ( 149 )' R = (r?Ri - r 2 2 R 2 -^ (R, -R 2 )j / (r 2 2 -n 2 ) 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, r2 = i2, ^i =6co, ^2 = 15, and the formulas become, 5 = 180+28 oSo/x 2 # = 180-28 o8o/* 2 Here x varies between 6 and 12 inches, and S ranges from +960 to . +375 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 2j 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 = r 2 , and that the radial unit-stress R varies similarly. Thus, for the inside surface, Si=Ri(r 2 2 +rf)/(rf-rf} and R=-R^ (150) while for the outside surface of the cylinder, S2 == Ri ' 27*i 7(^2 Y\ ) and R=o and accordingly the greatest unit-stresses, as given by (150), are those generally used in investigation and design. The man- ner in which 5 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). A.RT. 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 = 2^1. Then for the inside surface where x = r the tangential unit-stress is 5i=f^i; for the outside surface, where JG = r 2 , it is 2 = ^2 ; for a point half way between these surfaces it is f f/?i, and all of these are ten- sion. For the same case the radial unit-stress for the inside surface is Ri, for the outside surface it is o, and for a point half way between them it is J-Ri, 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 r i = 3-75 inches, ^2 = 5.50 inches, ^1 = 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 7-1 = 1.625 inches, ^1 = 15000, and ,$1=30000 pounds per square inch. Then solving for r 2) there results, r 2 = ri[(S 1 + R l )/(S 1 -R l )^ (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\ + /, where / is the thickness, and t 2 may be neglected when it is to be added to r{*. The formula then reduces to tSi=Ri(r + t), which is slightly more accurate than that of (30), and it gives slightly larger values of Si in investigation and slightly larger thicknesses in design. Neglecting / in comparison with 390 GUNS AND THICK CYLINDERS CHAP. XVi r, this becomes tSi = 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 2 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 R 2 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 R 2 on each square unit of its outside surface. Making RI=O in (149)' and also x = r\ and x=r 2 in succession, there are found Si = -R 2 - 2r 2 2 /(rJ _ fl 2) s 2 = -R 2 . (r accordingly, This gives values of T lower than those found from the formulas of Clavarino and Birnie. For x = r\ it becomes, TI =[(ri 2 +4r 2 2 )R! -6r 2 2 R 2 ]/ 3 (r 2 2 -r 1 2 ) 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 TI 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 27ie/2nx 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\ 9 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 r%. In consequence of the shrinkage the radial unit-pres- sure ^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 397 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, r\ by r 2 , and r 2 by r 3 ; then for x = r 2 , there results, T 2 ' = R 2 (2r 2 2 + 4 r 3 2 )/ 3 (r 3 2 -r 2 2 ) =pR 2 which is the tangential tension. Dividing the first of these ex- pressions by the second, there is found, T 2 /T 2 f =a/fi or fiT 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 2=7^'^o T2 ' = T'1^T3 ( 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 oc = r\\ 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 0.004 inches. First, the sum of the two tangential stresses at the surface of contact is (Ee)/r 2 = 2 4 ooo 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 /? to be 41/12. Then formulas (154) give T 2 = g 260 pounds per square inch at the outside surface of the 398 GUNS AND THICK CYLINDERS CHAP, xvi tube and TV = 15 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 = 14 400 pounds per square inch. The decrease in the outside radius of the tube is next com- puted and found to be 2 = 0.00138 inches, while the increase in the inside radius of the hoop is ^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, Tj = 16 970, TI = 14 400 pounds per square inch; e 2 = 0.00117, e 2 = 0.00283, an d ^1= 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 r\=o and letting r 2 be the mean radius of the web. For example, let fi=o, r 2 = i6 inches, ^3 = 24 inches, and let e/r 2 be rsViF in accordance with the old rule (Art. 32). Then a = , /?= fX4.4, and the tangential compression at the outside surface of the shaft or pin is T 2 = 3 740 pounds per square inch, while the tangential tension at the surface of the hoop or web is 7Y = i6 500 pounds per square inch; the radial compression in the crank or pin is TI = 1 1 300 pounds per square inch. It thus appears that the ratio e/r 2 = ^- s \-$ 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- ART. 155 DESIGN OF HOOPED GUNS 399 pressive stress in the shaft shall be 1 5 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 J^i 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\, while AB is the thickness of the tube and BC that of the hoop. The shaded areas show the stresses due to hoop shrinkage, A a and Bb being the tangential compres- sions TI and T 2 of the last article, while Bb' is the tangential tension 7Y, 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 Bbi being those at the inner surfaces of the tube and hoop. The above prin- ciple indicates that both Aa\ and Bb i 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, Aa-i, and Bbi 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 ^1=3.04, r 2 = 5.80, and ^3 = 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 TI and T 2 for the tube and T'2 and T$ 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 o: and /?, which are found to be 1.424 and 2.425 respectively. Since T\ 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 T\(2r-p + r')/$r = 2$ 800 pounds per square inch compression, and accordingly for the common sur- face of tube and hoop R 2 = r 2 /a = i8 100 pounds per square inch. For the inside surface of the hoop, T 2 =^R 2 = ^ 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 oc = r 3 , thus giving T 3 = 2R 2 r 2 2 /(r3 2 -r2 2 ) = i9 800 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 (T2/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 T% 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 ^2 = 6 ioo//?=2 500 pounds per square inch. The value of TI 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 T\ = 100000, ^2 = 2500, ^12 = 9.242, r2 2 = 33-640, and solving for RI gives RI = $I 100 pounds per square inch, which is the highest allowable powder pressure. Under this pressure the unit-stresses represented by Aa, Adi, and Bbi 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 GUNS 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 ^1=3.50, f^Q-iS, ^ = 11.25, ^4 = 12.25 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 40 ooc pounds per square inch. ART. 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 plates 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 plates were but little de- formed in comparison with the roller, and hence the entire defor- i w i i i i i i \ w -jw ^ r t t t t \w Fig. 156a Fig. 1566 mation will here be regarded as confined to the latter. The vertical diameter A A 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 compressive 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 y . Then, if the elastic limit of the material is not exceeded, S/S V --= 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, xvil 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 v - Idx equals W. Accordingly, S/E = e/%d and t S y . ldx=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 Slj ydx = We. Now j ydx 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 = t and \CC=BC = (ed-e*)*= (a*)* nearly. The solution of the twc 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=idS(2S/E)* (156)' which shows that the load on a cylindrical roller should vary directly as its diameter. Taking 5-15000 and = 30000000 pounds per square inch for steel, the last formula reduces to 20 = 31 5(2. 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 2ooVd, 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 = JXi2 000 = 2000 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 1 8. Here W = ^oooo pounds, 5 = 90000/18 = 5000 pounds per square inch, = 15 ooo ooo pounds per square inch, and / = 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 192 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. XVTi 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. The 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 c/\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 5 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 2xx . dx, and hence the sum of all the vertical stresses S y . znxdx equals W. Accordingly, S/E = e/\d 2nfs v .xdx=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 Sj2nyxdx=We t Now fznyxdx 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 2 = WE/frd 2 or W=frd 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 = ^WE/nS 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=&d 2 S 2 /E, d 2 = 2WE/nS 2 (157)' Comparing these with the previous formulas it is seen that (157) give values of S 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^Ctffi may be considered as approximate, in which C\ and 2 are to be determined by experiment for each material. The experiments of Grand all and Mars ton 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 1 5 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, xvil 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=(ld)*(3W/E)*. The area of contact hence varies as the cube root of the load for the same cylindrical roller; thus if a load W\ gives an area of contact fli, a load &W'i is required in order to make the area 2d\. 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, $W is to be replaced by 6Wj 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 5, and then that of S in terms of W, there is found a = d(nW/E)l, 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 AJIT. 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 undei the pressure of water or steam is a circular plate in such a ccndition. 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, q,nd 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 %rR. 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 . 2rR. The sum of the two reactions being only one- half the load, an upward shearing force equal to brR must act along the sides of the 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 t 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, 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 1 = 6M/bd 2 = f . 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-kS-XR, and on the lower side of the plate r = 5-A5. The latter value is the one to be used, since it is larger than the former. Accordingly, r=|(i -X)R . (r/d) 2 and (d/r) 2 =f (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 2 R. For cast iron the mean value of the factor of lateral contrac- tion A is J, while for wrought iron and steel it is J. 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. 159 CIRCULAR PLATES WITH UNIFORM LOAD 411 Another method of discussing this case is to consider the plate to be cut along a diameter; the load on the semicircle is and this may be considered as acting at its center of gravity c lt which is distant 4^/3^ from the diameter AC\ the reaction around the semicircum- ference ABC is also ^irr 2 R } which may be considered as acting at its center of Fig. 159 gravity c 2 which is distant 2r/ir from the diameter AC. The bending moment with respect to this diameter then is M= Trr*RX2r/Tr- ^RX^r/^ir = %r*R. This bending moment produces flexural stresses in the section area along the diameter AC, and the average unit-stress is S' = 6M/b'd 2 , where b' = 2r and d is the depth of the plate. Repre- senting the stresses along AC by the ordinates of a parabola, the maximum unit-stress at the center is f of the average, or which agrees with that found i the previous method. Then, as before, the true unit-stress is and the special formulas (159) immediately follow. 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=l(r/d) 2 R 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 412 ROLLERS, PLATES, SPHERES CHAP. XVII per square inch; then d = r(^R/4.T)^ = 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 UNDER 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. The following approximate discussion of this case is for a circular plate sup- ported along its circumference, where the load P is uniformly distributed over Fig. 160 a circle of radius ro, the radius of the plate being r and its thickness d. Let Pig. 160 represent one-half of the plate, DEF being the semicircle whose radius is TQ. The load ^P being uniformly distributed, the distance of its center of gravity from the diameter A C is 47-0/371-; the reaction \P along the circumference ABC has its center of gravity at the distance 2r/ir from AC. The bending moment with respect to the diameter AC then is ~ - 7T 37T This bending moment produces flexural stresses in the section area along the diameter AC, and the average unit-stress is S r = 6M/b'd 2 where b f = 2r. If the stresses normal to AC are rep- resented by the ordinates of a parabola, the maximum unit-stress at the center is f of the average, or ART. 160 CIRCULAR PLATES UNDER CONCENTRATED LOAD 413 If a second diameter be considered normal to AC, a unit-stress S equal to this is also found, and from Art. 139, taking into account the factor of lateral contraction X, the true unit-stress on the lower side of the plate is T= (i -X)S, or (160) in which A is J for cast iron, J for wrought iron and steel, and about I for concrete. When the load covers the entire plate, then ro = r and (160) becomes r = f(i -tyP/ird 2 ; in this .put P = irr 2 R, where R is the uniform unit-pressure due to P, and it reduces to the same formula as derived in Art. 159. When P is concentrated at the middle of the plate, then r = o and (160) becomes r=f (i -X)P/ird 2 . Hence a load concentrated at the middle of a circular plate pro- duces three times the stress of that due to the same load uniformly distributed. As a matter of fact it would be impossible to con- centrate P at a mathematical point, since then the unit-pressure R which is P/irro 2 , would be infinite; in cases of design R should not exceed the elastic limit of the material. As an example let a load of 6 ooo pounds be at the middle of a steel plate, distributed over a circle i inch in diameter, the plate being finch thick and 24 inches in diameter; here A = J, ro = i inch, r= 12 inches, and d = } inch. Then formula (160) gives T= 9 950 pounds per square inch as the true tensile unit-stress at the middle of the lower side of the plate, which is a safe allowable value. When a plate is fixed around the circumference it is probable that the constant in (160) should be 3 instead of f . 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. A fixed plate can carry a load about 50 per cent greater than that carried by a supported one. When a plate is stiffened by ribs, as is often 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. 160a. Which is the stronger, a circular plate carrying a load P uniformly distributed, or one carrying a load \P which distributed over an area at the middle which is one-third of the area of the plate? 414 ROLLERS, PLATES, SPHERES CHAP, xvil 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 Wz 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, and because these are equal, TFi# 3 must be equal to WJP. 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 circumfeience: (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- AKTT. 162 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 f for wrought iron and steel, and about ij for cast iron. These numbers are derived by taking the factor of lateral contrac- tion as J for wrought iron and steel, and as J for cast iron. In Germany the value A = -& 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 0/6 = 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 f, and from it is found d=ab(%R/(a 2 + b 2 )T)* = o.86 inches, so that a thickness of J inches will be sufficient safely to with- stand the pressure. Prob. 1610. 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 2m, 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. XVII 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, 2 )d 2 (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 J 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, T=%R(l/d) 2 and T=$R(l/d) 2 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 X 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 = J, this formula reduces to T =f ^(//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 W, 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 xr 2 R, and this is resisted by the tension inrtS 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 S 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 7^2= 5 A5 + A7?. Using J for the value of A, and inserting the above value of S, there result, Ti = $R.r/t and T 2 = IR+$R . r/t for the true tensile unit-stresses on the outside and inside surfaces respectively. Both of these are less than 5, 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^ the outside radius, R\ and i?2 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 S. 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=sCi 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 nx 2 R, and the outside pressure on a great circle whose radius is x + dx is x(x + dx) 2 (R + dR), 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 2nxdx . S. Stating this equation and omitting quantities of the second order, a second relation between 5 and R is found. Accordingly, are the two conditions for determining S and R. Substituting in the second equation the value of S from the first and inte- grating, the value of R in terms of x is found, and then that of R is known; thus, S = C l + C 2 /x* and R = d- 2 C 2 /x 3 (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 S is the most important one and S has its greatest value at the inside surface of the hollow sphere. Values of the constants C\ and C 2 may be found from the formula for R. Regarding the unit-pressures RI and R 2 as without sign, R becomes Ri when x=r\ and R becomes R% when x=r 2 ', then, - r 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< acts in the plane 2-2 and produces the unit-stress Si =c\M cos$//i, where c\ is the distance from i-i to the upper or lower surface of the beam. When < is not a large angle, the unit-stress S\ 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 = 11 700 pound-inches, /i=72 inches 4 , 1=3 ART. 166 UNSYMMETRIC LOADS ON BEAMS 429 inches, and then- Si =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 / 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=c\ cos< + 2 sin<, where c\ and c% 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 2 = 2 inches, /i=72 and /2=32 inches 4 ; then 6=3.60 inches, 7=62 inches 4 , and finally S =680 pounds per square inch. It is, however, not correct to assume that the neutral axis is hori- zontal and the computed value of S is too small. A more precise method is to resolve the vertical load on thfe purlin into components parallel to the principal axes, those normal to i-i producing the moment M cos 96 and those normal to 2-2 producing the moment M sin<. Then the flexural unit- stress due to the first moment is Si =c\M cos, since the forces that cause this moment act in the plane of the upper surface of the beam. For a case like Fig. 166c, where the forces PI and P 2 act in the lines of the principal axes, this method is strictly correct, M being the moment due to the resultant P which makes the angle < with 2-2; by discussing this case it can be shown that the neutral axis mn is not normal to the plane of M except when 1 1 and J 2 are equal. 430 MISCELLANEOUS DISCUSSIONS CHAP. XVIII Formula (166) applies to the discussion of the Z bar shown in Fig. 166d, as soon as the values of c\, c 2 , /i, 1 2, 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 /& are those with respect to the axes a-a and bb, while the principal axes are i-i and 2-2. The value of 1 2 is found by 7 2 =ar 2 2 , where a is the section area and r 2 the least radius of gyration. The value of I\ is found by subtracting 7 2 from the sum / + /&. The tangent of the angle < is given in the sixth column of the table, and from this the values of 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 5 = 30100 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. 166a 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 (j) which gives the greatest flexural unit-stress for the cases of Fi. 1666 is that for which Prob. 1666. Show that the angle 6 between the neutral axis mn and the principal axis i-i in Fig. 1666 is given by tan0= (7i/7 2 )tan0. For the above example of the purlin beam, show that is 5 if degrees, and that hence the neutral axis is inclined 21} to the horizontal. Prob. 166c. Compute the unit-stress 5 for a Z bar 6 inches deep with legs 3 J inches long, when used as beam under a bending moment of 90000 pound-inches. ART. 167 INFLUENCE LINES 431 ART. 167. INFLUENCE LINES An influence line diagram shows the variation of the shear or moment at a given section of a beam as a single load P travels over it, the values of the shear or moment at the given section being laid off as ordinates under the load. Fig. 167a gives the shear influence line for the section K of a simple beam. For any load P on the beam at a distance id from A, the shear at K is -f P(i K) when the load is on KB and PK when it is on AK. Hence, when P is at A or B the ordinate is o, when P is at K the ordinates are +P(i KI) and P/CI, the distance AK being KI/, and the ordinates on each side of K vary directly as the distances from A and B. This diagram clearly shows that every load on Fig. 167a Fig. 1676 the left of K gives a negative shear at K and that every load on the right of K gives a positive shear at K. It also shows that the greatest shears at K occur when the loads are near K. Fig. 1676 gives the moment influence line for the section ^ of a simple beam. For any load P on BK the bending moment at K is +Pl(i-K)K 1 and for any load on AK it is +P//c(i-i), where K and KI have the same meaning as above. Thus, the moment at K is positive and it varies as the first power of K in both cases, and hence the influence diagram consists of two straight lines which are readily drawn after erecting the ordinate P/(i -KI)KI at K. This diagram shows that the maximum moment at K occurs when the span is fully loaded and when the heavy loads are near K. Fig. 167c gives influence lines for the section K in an over- hanging beam, the upper one being for shear and the lower one for bending moment, the ordinates showing the shear or moment 432 MISCELLANEOUS DISCUSSIONS CHAP. XVIII at K being laid off under the moving load P. These lines are readily drawn by laying off as ordinates the shear and moment at K for a load at K and then drawing straight lines through the points under the supports. These diagrams show that the greatest positive shear at K occurs when CA and KB are loaded, that the greatest negative shear at K occurs when AK and BD are loaded, that the greatest positive moment at K occurs when AB is loaded, and that the greatest negative moment at K occurs when CA and BD are loaded. For cantilever, simple, and overhanging beams, the influence lines are straight, but for fixed and continuous beams they are Fig. 167c Fig. curved. Tig. 167 d shows the diagrams for the section K in a continuous beam of two equal spans, the upper one being for shear and the lower one for moment. Here the reactions are given by Art. 73 and for a kad on the left of K the shear at K is V=Ri P= \P($K K 3 ) from which a few ordinates maybe computed in order to draw the curve; also for a load on the span 2-3 the shear at K is V=Ri= -JPC/c-Ac 3 ). Similarly the moment at K due to a load on the left of K is M = R\Kil Pfal /c/), where K\l is the distance iK. Influence lines may be used to assist in finding the greatest positive or negative values of the shear at a given section due to a uniform load,, For this purpose P is to be replaced by w, the load per linear unit, and the sum of all the shears is obtained from the areas of the influence diagrams. Thus the greatest to 168 CURVED BEAMS 433 negative shear at K in Fig. 167c is represented by the areas of the triangles in the first diagram below AK and BD. Let AK=Kil and BD=K 2 l, where / is the span AB. Then the negative ordinates for K and D are PKI and PK 2} and the areas of the triangles are ^K^XWKI and ^K 2 lXwK 2 , the sum of which is ^wl(Ki 2 + K 2 2 )-, this is the negative shear at K due to a uniform load on both AK and BD. In a similar way the greatest positive or negative value of the moment at a given section due to a uniform load w can be found, The principal value of influence lines is in representing to the eye, more clearly than formulas can do, the variation of the shear or moment at a given section, thus enabling the distribu- tion of load which renders the shear or moment a maximum to be readily determined. The lines are in fact graphic represen- tatives of the algebraic expressions for V and M at the given sec- tion due to a load which travels over the entire length of the beam. Their greatest value is in the analysis of partially con- tinuous and arch bridges (see Roofs and Bridges, Part IV). Prob. lC7a. Draw the influence line for the reaction at A in the simple beam of Fig. 167a, Prob. 1676. Draw the influence line for the reaction aCA in the overhanging beam of Fig. 167c Prob. 167c. Draw the influence line for the moment at B in the overhanging beam of Fig. 167c ART. 168. CURVED BEAMS The word beam usually implies a bar which is originally straight, but the fundamental assumptions of the theory of straight beams (Art. 39) can be applied to the deduction of for- mulas for curved pieces which are subject to flexure. The most important case is that of a rectangular section where the upper and lower surfaces are curved in concentric circles as seen in Figs. 168a and 1686, Let TI and r 2 be the radii of the concave and convex sides; let b be the breadth and d the radial depth, so that r 2 ri = d. Let c? and c 2 be the radial distances from 434 MISCELLANEOUS DISCUSSIONS CHAP, xvill the neutral surface to the concave and convex sides, so that Ci+c 2 = d. It will be shown that Ci is less than c 2 , or that the neutral surface is nearer to the concave than to the convex side, and also that the flexural fiber unit-stress is greater on the con- cave than on the convex side. Assume, as in Art. 40, that any plane section Bb which is radial before flexure remains also plane after the flexure and takes the position Cc. In Fig. IGSclet Oa=Ob = r 1 and OA = OB = r 2 ; also let non be the neutral surface so that ob=ci and oB=c 2o On the concave side the fiber ab is shortened by the amount be at the neutral surface the fiber no remains unchanged in length, and on the convex side the fiber AB is elongated by the amount BC. Let be the angle between the radii OA and OB and dO be the angle between Bb and Cc. Then ab=r l d and bc = Cid6, so that the unit shortening on the concave side is CidO/riO; also AB = r 2 and BC=c 2 30, so that the unit elongation on the convex side is c 2 dd/r 2 0. Let r be the radius of the neutral surface and z be any radius less than r 2 and greater than r\\ the change per unit of length for the fiber of radius z is then pq/mp or (z r)d6/zd. Now if E is the modulus of elasticity of the material, 36 Cl 9asF ^? -30*-r Sl = E ~d7i S2=E T7 2 S=E *~JT are the fiber unit-stresses due to the flexure, Si and S 2 being those on the concave and convex sides of the beam and S being that for any radius z. For a curved beam, then, the fiber stresses are not directly proportional to their distances from the neutral surface. Thus while laws 1-5 of Arts. 39 and 40 are valid for ART. 168 CURVED BEAMS 435 curved beams, law 6 fails and the neutral surface does not pass through the centers of gravity of the cross-section. While Fig. 168c is drawn for the case of downward curvature and positive bending moment, the same formulas result for upward curvature or negative bending moment. The position of the neutral surface is to be determined by the condition that the algebraic sum of all the stresses normal to the radius Bb shall be zero, or if da is the elementary area bdz, z-r _ dd C T \-r 6 z ' 6 I z Z ~ Jr\ Performing the integration and noting that r 2 TI = d } there results d rlog =o or r=d/log in which the logarithms are Napierian. Replacing r by f 'i+Ci and then by r 2 c 2 , there are found, ci = -n+d/log- c 2 = r 2 -d/\og T - (168) TI r\ which give the radial distances from the neutral surface to the concave and convex sides of the curved beam. The bending moment M for the section Bb is equal to the resisting moment, formed by the algebraic sum of the moments of all the stresses normal to the radius Bb, hence ] C 7 Jri Integrating and reducing by the conditions that r 2 =r+C2, Ti and r=d/log faAi) there is found Replacbg E dd/0 by its values from (168) in terms of Si and S 2 , iMc\ 2Mc 2 01=7-; -, - r ^2 = 73 1 - \ c\) bdr 2 (c 2 -ci) 436 MISCELLANEOUS DISCUSSIONS CHAP, XVIII which are the unit-stresses on the concave and convex sides of the curved beam due to the bending moment M . As an example, let ^2 = 36, 7*1 = 24, and d=i2 inches; then log (r 2 /ri) = log r 2 -log ri = 3.5835 -3. 1781 = 0.4054, whence by (168), d = 5.60 and c 2 = 6.40 inches. Then by (168)' Si = j.oM/bd 2 and S2$.$M/bd 2 . For a curved beam the fiber unit-stress on the concave side is always greater than that in a straight beam. For the straight beam r< 2 = r 1 =cQ and it can be shown that the above formulas reduce to Ci = C2=%d and Si = S2=6M/bd 2 . When TI is very small, as in Fig. 168J, then Ci is very small and Si is very large, so that a small bending moment Pp may cause rupture. Hence small radii should be avoided in a curved piece which is subject to flexure. When TI = O ano 1 r2 = d, it can be shown Ffc. 168d Fig. W8e that Ci = o, c 2 = d, Si = oo , When the circles are not concentric, as in Fig. W7e, similar general conclusions result, although the above formulas do not directly apply. When r 2 is less than r + d, the following approx- imate formulas may be used, in which h is the distance between the centers of the two circles and d is the depth of the rec- tangular section or d=r 2 ri+h, , v7i 3 V^+h Ci=a Si = 3 M/bdci S 2 = 3 A These formulas also apply approximately to the case of -on- centric circles by making h=o. Thus, for the above numerical example, ^1 = 5.39 and 2 = 6061 inches, while Si = 6.'jM/bd 2 and Prob. 168. When n = rt oo, prove that formulas (168) reduce to Ci = c* = \d. Also that (168) ' reduce to S 3 = S a = 6M/bd*. ART. 169 PRODUCT OF INERTIA 437 ART. 169o PRODUCT OF INERTIA The ' Product of Inertia ' of a plane area with respect to two rectangular axes is the quantity Imxy, which is the sum of the products formed by multiplying each elementary area m by its coordinates x and y. When the plane area is symmetric with respect to one of the axes, then the product of inertia is zero; thus in Figs. 169a and 169& the axis 2-2 is one of symmetry, so that for any assigned value of y there are two equal values of x with opposite signs; also in Fig. 169c the axis i-i is one of symmetry and Imxy is zero, because for any assigned x there are i i i i Fig. 169a Fig. 1696 Fig. 169c Fig. Fig. 169e two equal values of y with opposite sign. For Figs. 169J and 169e aeither axis is one of symmetry and Imxy is not zero. For a plane surface which is symmetric with respect to an axis through its center of gravity, the product of inertia with respect to another set of rectangular axes parallel to the first set is equal &~ -*-t-^ sq-f 2 V . ffi V -M'-- J i\ ' 2 Fig. 169/ Fig. 169g Fig. 169& Fig. 169* to the area of the surface multiplied by the coordinates of its center of gravity. Let X'-X' and Y'-Y' (Fig. 169&) be tfie two rectangular axes through the center of gravity, and let y and x be the coordinates of that center of gravity with respect to the axes X-X and Y-Y. Let x' and / be the coordinates of any 438 MISCELLANEOUS DISCUSSIONS CHAP, xvili element m with respect to the first set of axes and x and y the coordinates with respect to the second set* Then x=x+x f and y=~y+y', and Imxy = xylm +xZmy f +ySmx' -f Imx'y' But 'Imy f and 2mx'a,re each zero from the definition of centei of gravity and Imx f y f is zero because the area has an axis of symmetry. Hence Imxy = xylm = axy which proves the proposition stated. For example, take the angle section in Fig. 169^ the dimensions being 5X3X2 inches, and its center of gravity being 0.75 inch from the back of the longer leg and 1.75 inches from the back of the shorter leg. This section can be divided into two rectangles, as shown, and the coordinates of their centers of gravity be readily found, then jry-(SXo.s)(o.75J(o.s)+(2.5Xo.s)(-i.s)(-i.o)-+2.8i2S ins 4 is the product of inertia for the angle section with respect to the axes X-X and Y-Y. Products of inertia, unlike moments of inertia, may be either positive or negative. Thus the product of inertia for Fig. 169d is positive while that for Fig. 169e is negative. For a rectangle or isosceles triangle axes can always be drawn so that Imxy is zero. For the right-angled triangle in Fig. 169/ the product of inertia for the axes X-X and Y-Y is n while for the parallel axes through the center of gravity, as in Fig. 169g, the theorem of the last paragraph shows that the product of inertia is + T V& 2 d 2 Prob. 169. Find the product of inertia for a rectangle for two axes through its center of gravity, one axis being a diagonal of the rectangle. ART. 170. MOMENT OF INERTIA The term ' Moment of Inertia ' originated from the theory of a rotating body. When a body is put into rotary motion around ART. 170 MOMENT OF INERTIA 439 an axis, each particle offers a resisting force by virtue of its inertia, this force being proportional to the mass m of the particle and to its acceleration. Let a be the acceleration of a particle at the distance unity from the axis, then the acceleration of a particle at the distance z is za. The resisting force of inertia of that particle is mza and the moment of that force with respect to the axis is mz 2 a. The expression Imz 2 a or almz 2 is the sum of all these resisting moments when the summation- is extended to cover the entire rotating body. Now the quantity Imz 2 was called the moment of inertia of the body in the early days of mechanical science when the subject of inertia was imperfectly understood. This name has continued in use, although now it is clearly understood that 2mz 2 is not the sum of the moments of all the forces of inertia, but only proportional to that sum, the angular acceleration a having been omitted. Strictly speaking a plane surface can have no inertia and hence no moment of inertia, but the quantity 2mz 2 is called moment of inertia when m is any elementary area and z is its distance with respect to an assumed axis. This quantity occurs so frequently in mechanics that it is necessary to have a name for it, and c Moment of Inertia' is in universal use. Hence, The moment of inertia of a plane surface with respect to an axis is obtained by multiplying each elementary area by the square of its distance from that axis and then taking the sum of all these products. The word * axis ' used above means a straight line drawn usually in the plane of the surface or sometimes drawn normal to that plane. When the term moment of inertia is used without qualification, the axis is understood to be in the plane of the surface. When the axis is normal to the plane the quantity Imz 2 is called the c Polar Moment of Inertia.' It is proved in Art. 90 that the polar moment of inertia is equal to the sum of the moments of inertia about two rectangular axes in the plane of the surface. The present article relates only to moments of inertia about such rectangular axes. 440 MISCELLANEOUS DISCUSSIONS CHAP. XVIII Moments of inertia of regular geometric surfaces are deduced by help of the calculus in the manner indicated in Art. 43. The moment of inertia Imx 2 is less for an axis passing through the center of gravity of the surface than for any parallel axis, since Imx 2 is a minimum when 2mx = o, and the latter is the condi- tion that the axis passes through the center of gravity. Let / be the moment of inertia 2mx 2 for the axis through the center of gravity, I\ that for any parallel axis, and h the distance between the two axes (Fig. 170a). Then, 7i = 2m Here Imx equals o, and Im is the entire area a, hence h=I+ah 2 and I = T l -ah 2 are the practical rules for transferring moments of inertia from and to an axis through the center of gravity. The principal axes of a plane surface are those which pass through its center of gravity in such directions that the moment x Fig. 170a Fig. 170c of inertia with respect to one axis has the greatest possible value, while that with respect to the other axis has its least possible value. These moments of inertia are generally designated by /i and L 2 and are called the principal moments of inertia. For a rectangle with sides b and d } where d is greater than b, the prin- cipal moments of inertia are J 1 = T \M 3 and I 2 ^=^b 3 d. The principal axes are always at right angles to each other. When the moments of inertia with respect to two rectangular axes X-X and Y-Y are known and also the product of inertia Imocy (Art. 169) with respect to those axes, the principal moments of inertia may be found as follows: Let x and y be the coordi- ART. 170 MOMENT OF INERTIA 441 nates cf any elementary area m with respect to the axes X-X and Y-Y (Fig. 1706) and let x and y^ be the coordinates with respect to the principal axes i-i and 2-2, Let 6 be the angle which i-i makes with X-X. Then xi =y cos x sin yi =y sin 6-\-x cos 6 Squaring, multiplying by m, taking the sums, replacing Imx-f by / 2 , I my i 2 by /i, I my 2 by /*, and Imx 2 by /, there results 1 2 =Iz cos 2 0+/ v sin 2 (Imxy) sin 20 Ii = I x sin 2 0+7, cos 2 6 + (2mxy) sin 20 Differentiating each of these with respect to and equating the derivatives to zero, there is found for each, tan 20 = 2(2mxy)/(I y -I x ] (170)' as the condition which renders /i and 1 2 a maximum or a mini- mum, the same condition applying to both because tan 26 belongs to two angles which differ by 90. Substituting the value of from (170)' in (170), there results 7 2 or h =J(/*+/,,)V \(Iy-Itf + (Smxy}* (170)" which gives the two principal moments of inertia, the largest resulting from the + sign before the radical and the smallest from the minus sign. One of these values is 1 2 and the other is 7i For example, let Fig. 169* represent a 5X3X1 inch angle section, the area a being 3.75 square inches, the two axes X-X and Y-Y passing through its center of gravity and the values of I x and I y being 2.58 and 9.45 inches 4 . Here, as shown in Art. 169, the product of inertia Imxy is +2.8125. Then (170)" gives 1 2= 10.45 an d I\ = 1-58 inches 4 . Also (170)' gives tan 2 = 0.8421, from which 0= 19 39' or tan $ = 0.357. The least radius of gyration of the section, found from r 2 =7j/#, is ^=0.65 inches. Adding together the two equations in (170) there results 7i+7 2 = 7z + 7 J/ , that is, the sum of the principal moments of inertia is equal to the sum of the moments of inertia with respect to any two rectangular axes through the center of gravity. 442 MISCELLANEOUS DISCUSSIONS CHAP, xvm When 1 1 and 1 2 are known the moment of inertia Ix with respect to an axis X-X which makes an angle with the axis i-i is readily found as follows: From Fig. 170c the ordinate is y' y cos x sin <. Squaring, multiplying by mi and replacing the sums 2my f , I my, Imx, by Ix, I\, 1 2, there is found When either of the axes i-i or 2-2 is an axis of symmetry, Imxy equals zero (Art. 169), and then I x = 1 1 cos 2 (/> +/2 sin 2 When /i and / 2 are equal, each being called /, then /* = /, that is, when a plane surface has its two principal moments of inertia equal to /, and one of them is an axis of symmetry, then the moment of inertia with respect to any axis through the center of gravity. is also equal to /. Hence the moments of inertia of a square are equal for all axes through its center of gravity, and the same is the case for sections like Figs. 76a, 766, 76c, when /i and 7 2 are equal. Prob. 170a. Deduce (170)" from (170) and (170)'. Prob. 1706. For the Z bar of Fig. 166J, let depth of web be 6| inches, width of flanges 3| inches, and thickness of both J inch. Compute the principal moments of inertia. ART. 171. SPRINGS A cantilever beam formed of two or more superimposed plates, as in Fig. 17 la, may be used as a spring. Each of the plates acts separately as a beam, so that the unit-stress at the dangerous section is S=6Pl/nbd 2 , where P is the load at the end of the cantilever, / its length, b its constant breadth, d the depth of each plate, and n the number of plates. Hence when P is given and S is taken at the safe allowable value, the number of plates required is n=6Pl/Sbd 2 . The deflection, as shown in Art. 58, for a triangular beam of constant depth and a triangular plan of greatest width nb, is /= 6PP/Enbd 3 , or /= SI 2 /Ed. Hence this formula applies exactly to the spring seen in elevation and ART. 171 SPRINGS 443 plan in the upper part of Fig. 17 la, the ends of the n plates being triangular as shown, n being an even number, and very closely to the spring having n plates with rectangular ends as Fig. 171a Fig. 1716 seen in Fig. 1716. All these formulas are valid only when the greatest unit-stress S does not exceed the elastic limit of the material. The double spring of Fig. 17 Ib supports a part of the weight of a wagon body at A, this load being transferred at the ends to the lower spring, by which it is carried down to the axle at B. 2P being the total load, P the reactions at the ends, and / the half- length, the formulas above given apply also to this case. For example, let 2P be 600 pounds, / be 15 inches, b be 2 inches, and let each spring have six plates of ^-inch thickness. Then S= 6 X 300 X 1 5/6 X 2 X (J) 2 =36 ooo pounds per square inch, which is somewhat more than one-half of the elastic limit of spring steel. Also/= 36 ooo X i5 2 /3 ooo X^= i.i inches. The work stored in the double spring then is 4(|P/) = 66o inch-pounds. When the plates are so firmly fastened together that the entire section acts like a beam, both the unit-stress and the deflection are much de- creased. For this case S=66Pl/b(nd) 2 and f=Sl 2 /En 2 d; thus, for the above numerical data, 5=9 ooo pounds per square inch and f= 0.03 inches. A spiral spring fastened at one end to a fixed axis has the other end attached to the circumference of a cylindrical box which is turned around the axis by the spring after it is wound. 444 MISCELLANEOUS DISCUSSIONS CHAP. XVIII If r is the radius of the box (Fig. 171c), S the unit-stress in the spring at the point where it is connected to the axis, b and d the thickness and width of the rectangular spring, then the Fig. 171c Fig. 171e twisting force exerted tangentially at the circumference of the box A helical spring is formed by winding a wire around a cylinder which is then removed. In Fig. 17 Id let the axial tensile load P be applied at the center of the coil and let D be the diameter of the coil. Then the twisting moment Px^D produces torsion throughout the wire. Let d be the diameter of the wire and S s the shearing unit-stress produced at the circumference of the section \nd 2 . From Art. 92 this unit-stress is S s =%PD/xd3 ) or the load for a given allowable S s is P = xd 3 S s /8D. The distance through which the load moves being /, the work performed on the spring is \Pf and the equal internal energy stored in the wire is \(S s 2 /F)al (Art. 122), where F is the shearing modulus of elasticity of the wire material, a is the section area ^d 2 and / is the length of the wire in the coils of the spring. Equating these two expressions, there is found /= S s Dl/Fd for the deflec- tion /. When there are n coils in the spring, the length / is approximately nnD, or, if h is the length of the spring parallel to its axis, then I is given exactly by I 2 =h 2 + (nnD) 2 . Another method of deducing the above formulas is to imagine the wire stretched out in the straight line of length / and be subject to the twisting moment Pp (Fig. 17 le). Then by ART. 171 SPRINGS 445 Art. 92, S s = i6Pp/Kd*. Also when P moves through the distance / a point on the surface of the free end of the wire moves through the distance fX^d/p and hence the angular deformation per unit of length \sfd/2pl. Then the modulus of elasticity is F=S s /e (Art. 93), or F=S s X2pl/fd. Replacing p by the actual lever arm JZ>, there are found for S s and /the formulas above given. As an example, let steel wire ^ inch in diameter be closely wound around a cylinder ^& inch in diameter, thus forming a helical spring 2\ inches long and having 36 coils. Here P = ^ + ^g-=J inch. For a shearing unit-stress of 30 ooo pounds per square inch, the above formula gives P=$.J pounds as the greatest allowable tensile load, and this must be slowly and gradually applied. The elongation of the spring under this load, taking the shearing modulus of elasticity F as 12 ooo ooo pounds per square inch, is /= 1.131 inches for the approximate value of / and /= 1. 132 inches for the exact value. The above formulas apply also to an axial compressive force P acting on a helical spring, provided it be not too long and pro- vided that the coils are sufficiently open to allow of axial shorten- ing. The unit-stress S* due to the load P must be always less than the elastic limit of the material. The energy which can be stored in springs is small in pro- portion to their weight. Thus, for a helical spring under a tensile load (Fig. 17 Id) the stored work is \(S s 2 /F)al. If this stored work is 550 foot-pounds, then for 5=30000 and F=. ii 200 ooo pounds per square inch, the volume al is 82 cubic inches, so that the weight of the spring must be 240 pounds. If this work is expended in one second, one horse-power is developed; if in one minute only one-sixtieth of a horse-power is developed. On account of the heavy weight required, metallic springs are rarely used for the storage of energy except in timepieces, music boxes and mechanical toys. Their principal uses are in weigh- ing machines, as in spring balances and dynamometers, and for absorbing the energy due to sudden forces and shocks, as in the springs of wagons and cars. 446 MISCELLANEOUS DISCUSSIONS CHAP. XVIII Prob. 1710. Show that the distance which P moves in Fig. 171d in causing the unit-shear S in the wire is f= 2rlS/dE. Prob. 1716. Show that the work stored in the helical spring of Fig. 1715 is sufficient to raise a load W equal to the weight of the spring through a height of 4.9 feet, when 5 = 30 ooo and through a height of 8.6 feet when 5 = 40 ooo pounds per square inch. Prob. 171c. Show for the helical spring that the deflection f is proportional to the axial load P provided the wire of the spring is not stressed above its shearing elastic limit. ART. 172 INTRODUCTION 447 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, while 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 448 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 Materiaux, 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. ART. 173 ELASTIC CHANGES IN VOLUME 449 The simplest case is that shown in Fig. 173, where a parallele- piped is acted upon by the tensile unit-forces Si, 2, 6*3 in three rectangular directions. The body is regarded as homogeneous, so that the factor of lateral contraction A and the modulus of eksticity E are the same in all directions. Let e\ be the unit- elongation due to Si or ei=Si/E, also 2 = $2/E and 3=Ss/E. Then the actual unit- elongations which take place in the three rectangular directions are given by, e' = ei yU 2 >^3 "=- 2 ^3 M "'=3 AI ^ 2 Now for any cube of edge unity, the volume after the application of the unit-forces Si, S 2 , S 3 is (i + r )(i + e")(i + fi"'), and since e f , e", e"' are small compared with unity, this product is i + e r + e" + e'" (Art. 13). Accordingly, the elastic change in the unit of volume is, s>+ s + < = = : ( I -2X)( 1 + s 2 +e 3 )~(i-2X)(Si+S2+S 8 )/E (173) Here it is seen that there is no change in volume when A = J. For all the materials of construction it is found that the factor of lateral contraction is less than J, so that they increase in volume under tension arid 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% are compression, the values of S% 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 4800 pounds upon the two narrow sides, and to a tension of i 600 pounds upon the two ends. - 12800/32 = -400, S 3 = -4800/16 = -300, and Si = + 1 600/8 = +200 pounds per square inch; then Si +2 + 3= 500 pounds per square inch. Taking A = o.2, and = 2000000 450 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX pounds per square inch, the change per unit of volume is found to be 0.00015, so tnat the decrease in the volume of the parallelepiped is 0.0002X64 = 0.0096 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'"), which is equal to i-fV' + 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 Siy S2j S$ are equal, as also the unit-deformations ei, 2, 3. For this case the change per unit of length is the same in all directions and equal to (i 2^)5, the change per unit of area in any section is 2(1 2 A) , 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 component? ART. 174 NORMAL AND TANGENTIAL STRESSES 451 Fig. 174 by S xy and S xz . A similar notation applies to each of the other faces. An S 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 parallelopiped 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 gy . A similar condition obtains for each of the other rectangular axes, and hence, S X y~Syx Syz Szy S xz = i S zx (1*4) 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 5 are computed by 452 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 parallele- piped such that the resultant stresses upon it are wholly normal. These resultant normal stresses will be Si, 2, $3, from which by (139) the true normal stresses TI, T 2 , T 3 can be found. It will later be shown that these stresses Si, 2, S% are the maxi- mum apparent stresses of tension or compression resulting from the given normal and tangential stresses. Prob. 174. Let a, b, c be the angles which a line makes with the axes OX, O Y, OZ, respectively. Show that the sum of the squares of the cosines of these angles is equal to unity. ART. 175. RESULTANT STRESSES The resultant unit-stress upon any face of the parallelepiped in Fig. 174 is the resultant of the three rectangular unit-stresses acting upon that face. Thus for the face normal to the axis OZ the resultant unit-stress is given by, and the total resultant stress upon that face of the parallelepiped is the product of its area and R$. 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. Let a plane be passed through the corners i, 2, 3, of the paral- lelepiped in Fig. 174, and let a, b, c be the angles that its normal makes with the axes OX, OY, OZ, respectively. Let a, /?, 7-, 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, AR cos/?, AR cos?-. The triangle whose area is A y 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, Acos>, A cose, and the stresses upon them ART. 175 RESULTANT STRESSES 453 are the products of the areas by the several unit-stresses. Now the components of these four stresses with respect to each rectan ;j;ular axis must vanish as a necessary condition of equilibrium. Jence, canceling out A, which occurs in all terms, there results, R cosa = S x cosa -f S xv cosb + S zx cose R cos/? = S xy cosa -f S y cosb -f S zy cose (1 75) R cosr=S xz cosa+S yz cosb+S 2 cose in which the second members are all known quantities. From these equations the values of R cosa, R cos/?, R cos 7- can be computed; then the sum of the squares of these is R 2 since cos 2 a+ cos 2 /? + cos 2 ?-=i. The value of cosa is found by dividing that of R cosa by R, and similarly for cos/? and cosf. Now the angle 6 between the directions of R and the normal to the plane is given by, cos/9 = cosa cosa+cosfr cos/? + cose and then the tensile or compressive unit-stress normal to the given plane is R cos#, while the resultant shearing unit-stress is R sin0. 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 12 ooo 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 xy = &ooo, Sy X = Sooo, and the other stresses are zero; also a = 3 o, 6 = 60, and = 90. Then from (175), R cosa = + 14 390 R cos/? = +6930 R cosf=o and the resultant stress on the given plane is, R= (j4 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^ cos/?= 693/1597 = 0.434 /?=25J and the angle included between the resultant R and the normal 454 MATHEMATICAL THEORY OF ELASTICITY CHAP, xix to the given plane is computed by, cos#= 0.866X0.901 4-0.5X0.434=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 ^sin# = i 200 pounds per square inch. Prob. 1 75. 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 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 , RS be the resultant unit-stresses upon the three faces of the parallelepiped in Fig. 174, and let 61, #2> #3 be the angles which they make with the coordinate axis OX', then, costfi = S X /R! cos02 = S yx /R 2 cos0 3 = S ZX /R 3 determine the directions of RI, R 2 , R%. 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 cos6i + Ry cos0 2 +R z cos# 3 or, inserting for the cosines their values, it becomes, Comparing this with the first equation in (175), it is seen that, cosa = R X /RI cosb = Ry/R 2 cose = R z /Rs 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 , RZ. . 177 THE THREE PRINCIPAL STRESSES 455 The surface of an ellipsoid is hence a figure whose radius vector represents the resultant unit-stress upon a plane the normal to which makes an angle 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 2 correspond to RI, 7?2, -^3 in the above equa- tion of the ellipsoid. In this case S X i S^ S z are the three axes of the ellipsoid. If now shearing stresses are applied to the faces* z " the ellipsoid will be deformed, and 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 y =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, /?, r in the equations (175) are to be made equal to a, b, c t respectively. Also replacing R by S, these equations become, (S S x ) cosa = S yx cosb+S ex cose (S-Sy) cosb=S xy cosa+S zy cose (SSg) cosc=S xz cosa+Sy Z cosb in which 5, cosfl, cos6, cose are four unknown quantities. The 456 MATHEMATICAL THEORY OF ELASTICITY CHAP, xix three angles, however, are connected by the relation, 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 3 -AS 2 +BS-C=o (177) in which the values of the three coefficients are, A = S x + S y + S z , Jj = o x o y ~T" O y^z i <^z^ x *^ xy ^ yz *-^ zx C = 0^0^021" Z^xy^yz^zx ^x 1 ^ yz &y& Z x ^2^ xy 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, 52, S 3 , with respect to the rectangular axes OX, OF, OZ, are given by the values of cos a, cos b, cose, which are found to be, cosa=(fj/f)* cos&= (/2/w)* cosc=(w 3 /w)* in which mi, m 2 , w 3 , m, represent the following functions of the given and principal unit-stresses: mi = (S v - S) (S z - S) - S 2 yz m 2 = (S z - S) (S x -S)- S 2 ZX and it will now be shown that each principal stress is perpen- dicular to the plane of the other two. Let Si, 52, 5 3 be the three roots of the cubic equation (177). Let fli, &i, c\ 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 5 2 makes with the same axes. The angle between the direc- tions of Si and S 2 is then given by, cosc/;> = cosai cosa 2 +cosfri cosb 2 +cosci cosc 2 Now in the first set of formulas of this article let 5 be made Si and a, b, c be changed to ai, b iy GI\ let the first equation be mul- tiplied by cos= ii 096 ooo ooo from which ^ = 2013 and cos3< =0.6801. Then from Table 17 is found 3< = 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: xi = 2rcos(j) =+3880 Si =+2680 x 2 =2r cos(<5&+ 120) = 2890 $2= 4090 x%=2r 005(^ + 240) = 990 53= 2190 and these are the apparent principal stresses in pounds per square inch. Taking e = J for steel, the true principal stresses are now found by (139) to be, ^1=4-4770 r 2 =-425o r 3 =-i72o which show that the maximum true tensile stress is nearly double the apparent, while the maximum true compressive stress is 6 percent greater than the apparent. An ordinary solution of this problem, in which no combina- tion of stresses was made, would show the greatest tension to be 2 400 pounds per square inch, while the complete solution as above given shows that the greatest true tension is nearly twice as great. The ordinary solution shows the shearing stress to ART. 180 THE ELLIPSE OF STRESS 461 be i 200 pounds per square inch, but by applying (178) to the above values of T it is seen that the maximum true shearing stress is 4510 pounds per square inch. It thus appears that where many stresses combine, as at the junction of a crank pin with its web, the common methods of investigation give unit-stresses which are far too small; it also follows that designs for such cases, made by ucing the common methods, should be based upon low unit-stresses. Prob. 179. Apply the cubic equation (177) to the case of a bar acted upon only by the tensile unit-stress S x and the transverse shearing unit- stress S xy . Deduce the principal stresses for this case. ART. 180. THE ELLIPSE OF STRESS The ellipse of stress is that particular case where one of the principal stresses is zero, in which event the last term of (177) vanishes. An instance of this is where S 2 = o, S yz = o, S xz = o, which is that of a body subject to the normal unit-stresses 5*, Syj and to the shearing unit-stress S xy . The cubic equation then reduces to the simple quadratic, ,S 2 ~ (Sx + Sy)S+S x Sy~S 2 x y=0 and the two roots of this are the two principal apparent stresses whose directions correspond to the two axes of the ellipse. From this quadratic equation, the formulas (144) were derived. B Fig. 180 Let Si and S 2 be these roots, and in Fig. 180 let OA and OB be laid off at right angles to represent their values. Let an ellipse be described upon the axes AA and BB, and let $ be the angle AON which any line ON makes with OA. Upon a 462 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX plane normal to ON at O the normal unit-stress of tension or compression has the value, and the tangential shearing unit-stress is, OS=(Si-Sn) sin< cos< while the resultant of these gives the resultant unit-stress, OR=(Si 2 cos 2 < + S 2 2 sin 2 ^)* The diagrams in Figs. 180 give graphic representations of these values as the angle (j> varies from c to 360 degrees. In the first diagram Si and 2 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 AT, 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 S x =+2Ooo, S xy = ^ooo ) and S y = o; the above quadratic equation then gives Si =+4160 and 52 = 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 J 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 J as the factor of lateral contraction, T\ = + 4 880, T% = 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 463 Cases can, however, be imagined in which one of the true principal stresses is zero. If Si, 2, 3 are the apparent stresses in three rectangular directions, it is seen from (139) that when S 3 eSi 52 is zero, the true stress 7^ 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 = J; then 7\ = i6, T2 = 8, and 7^=0. 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 TV 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, denned 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 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 s be the distortion parallel to the shear S; then e = 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 2*(i-f-J$). The 464 MATHEMATICAL THEORY OF ELASTICITY CHAP. XIX change in length of this diagonal hence is 2* . Je, and the unit- elongation is Je. In a similar manner the length of the shorter diagonal after the distortion is 2*(i Je) and the unit-shortening is Je. 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 J 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 S 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 + X)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 F could be correctly computed for timber from a value of E obtained from a tension parallel to the grain. ART. 182 THE VOLUMETRIC MODULUS 465 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 bearh 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/s. Similarly, the volumetric modulus of elasticity, which will be represented by G, is the ratio of a unit-stress S 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/ 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 S. Then, from Art. 139, the unit-shortening of each edge of the cube is ' = (S-2>IS)/E; or since e represents S/E, the unit-shortening is fi' = (i 2A)e. The volume of the cube, which was originally unity, now becomes (i e') 3 , and hence the change in volume is 35' 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 466 MATHEMATICAL THEORY OF ELASTICITY CHAP, xix the cube; hence e' = 3(1-2^)5 gives the change per unit of volume, and hence the volumetric modulus G is 5/3(1 2^)e, in which is the change per linear unit due to an axial unit- stress S of tension or compression only; accordingly, G-S/e' G=S/3(i-2X)e G=E/$(i-2X) (182) are formulas for G, of which the third gives the relation between G and E. For example, A is about J 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 as the relation between , F, G; hence, F= 3 G/( 9 G-E) G= EF/( 9 F- 3 ) 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 S 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 axialiy, and this gives the decrease in volume if the walls of the pipe are unyielding. AJIT. 183 STORED INTERNAL ENERGY 467 Prob. 182. Prove, for a homogeneous solid, that the ratio G/F is equal to 2(1 + ^)73(1 2^). Also that the value of X may be expressed by J-E/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, 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 l\ be the length parallel to Si, and a\ the section area normal to Si; then the total stress on this section area is 0iSi, and from (10) and (139) the change in the length of /i is /i(5i-~A5 2 -AS 3 )/E. The work done while the stress a^Si 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 l l 1 (S ] 2 -XSiS 2 -lSiS 3 )/E and this is proportional to the volume aji. A similar expression may be written for the energy due to $2 and another for that due to Sa; and the sum of the three is, 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 2A)V. S 2 /E ; for steel X is J and the resilience becomes K = %(S 2 /E)V, which is the same as for simple axial stress (Art. 120). As a numerical 468 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 J, and let the unit-stresses Si, S 2 , Ss be 3 ooo, 4 ooo ? 5 ooo pounds per square inch compression; then K = I>JO inch-pounds = 14 foot-pounds is the stored internal energy. When the above parallelopiped is subject to the action of six pairs of shears of which the unit-stresses are Si2, 23, Ssi, in the notation of Art. 174, the corresponding unit-detrusions are %S 2 i 2 /F, %S 2 2 s/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, ^ / =JF(S 2 i2+S 2 23 + S 2 3l)/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 parallelopiped. 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 Si, S2, $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 = 3i 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 ART. 183 STORED INTERNAL ENERGY 469 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. 470 TESTING OF MATERIALS CHAP. XX CHAPTER XX TESTING OF MATERIALS ART. 184. 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 with a weight at the free end, 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 experi- ments were made in France on timber in flexure and tension, only questions of ultimate strength being considered, while the elastic limit was unrecognized. Hooke's experiments on springs, from which he deduced the law of proportionality between stress and elongation 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. Nearly a quarter of a century later Barlow, Tredgold and Hodgkinson experimented on timber and cast iron both in the form of beams and columns; their methods and results, although now seem- ingly rude and defective, are deserving of praise as the first of real practical value. In 1849 was published in London the ' Report of the Commis- sioners on the Application of Iron to Railway Structures,' which may be regarded as the landmark of the beginning of the modern methods of testing. The immediate result of this report was the decision of 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 Europe and the United States it aroused marked impetus in the subject of testing materials. ART. 184 TESTING MACHINES 471 The first testing machines in the United States were those built by Wade and Rodman between 1850 and 1860 for testing gun metal. About this time the rapid introduction of iron bridges led to experiments by Plympton and by Roebling. Prior to 1865 apparatus was built by each experimenter for his special work, but in that year Fairbank put upon the market the first testing machines for commercial work. A little later the machines of Olsen and of Riehle for tensile, compressive, and flexural tests were introduced and have since been widely used. The machine devised by Emery, soon after 1875, is a verv precise apparatus which is used in large laboratories. Large machines for testing eye-bars have been built by bridge companies, and numerous testing laboratories now contain apparatus 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 ten- sion has usually a capacity of i ooo or 2 ooo pounds. Machines of less than 50 ooo pounds capacity are usually operated by hand and are especially useful for the instruction of students; while machines of higher capacity are operated by power. The largest testing machine in the world is one of 10 ooo ooo pounds capacity at Pittsburgh, Pa., which can be used for compression only. A list of the large testing machines in the United States is given in the American Civil Engineers' Pocket Book. Fig. 184 shows an Olsen testing machine of 40 ooo pounds capacity. The power is applied by hand by means of the crank on the left, and this causes the four vertical screws to have a slow upward or downward motion. The upper ends of the screws are fastened to a table A, which hence partakes of the vertical motion. When a tensile test is made, one end of the specimen is gripped by jaws in the movable table A and the other end in the fixed table B; in the figure a tensile specimen is seen in this position. The crank is then turned so as to cause the movable table to descend and thus a tensile load is brought upon the specimen. This load is weighed on the lever scale at the right by moving the weight D so that the scale arm will balance. For a compressive test the specimen is placed between the lower fixed table C and 472 TESTING OF MATERIALS CHAP. XX the movable table A, the latter being caused to descend by turning the crank, and thus a compressive load is brought upon the specimen. Tensile tests are the most common, and some commercial machines are arranged with an autographic recording apparatus whereby a curve is drawn which shows the relation between the load and the elongation throughout the test. There are also a Fig. 184 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 hydrau- lic machines in which a hydraulic press is used to transmit the load to the specimen are preferred in Europe. ART. 184 TESTING MACHINES 473 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 Riehle, which are found in the laboratories of most engineering colleges, prove very serviceable for illustrating the phenomena 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 endurance tests, which subject the specimens to alternating stresses for long periods of time, are made on special machines. 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. Four engineering colleges in the United States have testing machines of 600000 pounds capacity for tension, compression, and flexure, and one has such a machine of 800 ooo pounds capacity, while one has a machine of i 200 ooo pounds capacity for compression only. Large testing laboratories have also been established by the United States Government in its arsenals, and since 1910 in its Bureau of Standards. The valuable work done by Howard at the Watertown Arsenal and for the Bureau of Standards deserves special mention in this connection. The largest testing labor- atories, 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 2j acres. Prob. 184, In 1912 a brick pier 4X4 feet in section was broken at Pittsburgh by the large testing machine of the Bureau of Standards. Estimate the load which caused failure. Then consult the engin- eering journals of September or October of 1912 and ascertain the actual load 474 TESTING OF MATERIALS CHAP. XX ART. 185. TEST SPECIMENS Flat specimens are used to test the tensile strength of plates, and Fig. 185{ Fig. 18oa distances may be measured again in order to ascertain the per- centage of elongation. Round specimens are used for the tensile tests of most mate- rials other than plates. Wires and rods are usually tested just as they come from the mill, the length between the jaws of the machine being more than ten times the diameter. Standard round specimens are cut from axles, beams, shafts, and other manufactured products, these being turned in a lathe and screws cut on the enlarged ends. Prior to 1895 the standard diameter of the round specimen was i inch, and its standard length was 8 inches, this being the distance between two marks which are placed on the central portion for the purpose of measuring the elongations. After 1900, however, there came into use a smaller size which is now widely used, although the 8-inch specimen is still required for some classes of work. This smaller size 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 or sometimes 0.505 inches. ART. 185 TEST SPECIMENS 475 In Fig. 1855 the upper specimen has not been tested, while the two lower ones have been broken; see Art. 12 for an account of these tests. Fig. 185c Specimens for compressive tests of metals are usually cylin- ders, while cubes are mostly used for cement, concrete, and stone, and rectangular prisms for timber. The length of a cylinder or prism should be between 2-J- and 4 times its diameter or side, 76 TESTING OF MATERIALS CHAP. XX in order that there may be opportunity for the oblique shearing to properly occur. Fig. lS5c shows a cube of neat cement 2 inches square which has failed under compression, and also a timber prism 2X2 inches in section area, where the oblique shearing occurs far more satisfactorily than in the cube. This method of failure is characteristic of brittle materials (Art. 18). Com- pressive tests are rarely used for metals on account of the expense of making the specimens with ends truly parallel. Specimens for flexural tests of cast iron are usually i inch square and 14 inches long, which are placed on supports 12 inches apart and broken by a load at the middle of the span. Specimens ij inches in diameter and 15 inches in span are also used. Foi other materials no standard sizes have yet been adopted. Prob. 185. Wherfthe standard square bar is used for a flexural test of cast iron, show that the modulus of rupture can be found by multiplying the breaking load in pounds by 18. ART. 186. TENSILE TESTS Flat specimens are usually gripped by the jaws of the machine; while the standard round specimens screw into nuts to whicl' 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- tion is usually expressed as a percentage of the original length between the two marks on the specimen, and it is always desirable that the original length and the diameter of specimen should be mentioned in the report, since it will not be the same for the 2-inch as for the 8-inch specimen. For ductile materials, like wrought iron and mild steel, it is customary to slowly reduce the load after the highest value is reached; the material is then flowing rapidly, so that the elongation continues to increase, and hence a greater percentage of elongation is obtained. ART. 186 TENSILE TESTS 477 When it is specified that certain methods of testing or certain test specimens are to be used, these should be followed most carefully. Full specifications 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 . 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 and operation of the machine should be such that 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 coincides 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. 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. 478 TESTING OF MATERIALS CHAP. XX The above rules are quite general, but they seem to be essential ones for the proper conduct of tensile tests, not only for metals, but also for timber and cement. For detailed rules regarding metals, see the l Specifications for Standard Methods of Testing ' adopted by the American Society for Testing Mate- rials which are published in its annual Year Book. These speci- fications give also methods of calibrating testing machines. Prob. 186. Consult the Year Book above mentioned and ascer- tain how the calibration of a testing machine is made. ART. 187. COMPRESSIVE TESTS Rules i, 2, 3 of the preceding article apply also here. The following additional rules for compressive tests of metals are taken from the standard specification above cited: 4- The test specimen shall be a cylinder having plane ends truly normal to its axis. The diameter of the specimen shall not be less than i inch nor greater than 1.13 inches, the former being preferred. Its length shall be between 2.5 and 4 diameters. No bedding should be used for the ends of the specimen. 5. The bearing blocks which transmit the pressure from the testing machine should be truly normal to the plane ends of the specimen. To secure this, one of the blocks should be provided with a hemispherical bearing which can turn freely. 6. The speed of compression should be slow, not exceeding one-tenth of an inch per minute. Near the elastic limit and yield point the load should be increased very slowly. 7. For determining modulus of elasticity, the linear compression of the specimen should be observed by a pre- cise compressometer which is attached to the specimen and does not touch the bearing blocks of the machine. Read- ings of the compressometer should be taken for three loads, the first at about one-fourth, the second at about one-half, and the third at about three-fourths of the elastic limit. ART. 188 MISCELLANEOUS TESTS 479 8. To determine the elastic limit, several readings of the compressometer should be taken, as that limit is ap- proached, for load increments of i ooo pounds per square inch. 9. The yield point is to be noted as corresponding to that load for which the compressometer shows a linear compression without an increase in load. In the absence of a compressometer this point may be noted, for ductile materials, by the drop of the scale beam. 10. Measurements for the modulus of elasticity, elastic limit, and yield point may be made, if desired, on a specimen ranging in length from 10 to 15 diameters, since it may often be difficult to apply a compressometer in a length shorter than 4 inches. 1 1. The record of the test should mention any phenomena observed near the elastic limit and yield point. The manner of final failure should also be noted when the test is carried to this limit. For stone, concrete, and timber no standard methods for com- pressive tests have yet been adopted. The general principles given in the above rules are applicable, however, and the general practice regarding test specimens is stated in Art. 185. Prob. 187. Leather, paper, or plaster, has often been used between the ends of the specimen and the bearing blocks. Why should this not be done? ART. 188. MISCELLANEOUS TESTS Flexural tests are not made in common commercial work, except for cast iron, but tests are frequently made for determining constants to be used in designing. In such cases it is best that the specimen should be a model of the actual piece which is to be designed. Thus, if it is desired to know the modulus of rupture for a timber beam 2X3X24 feet, a breaking test on a specimen 2X3X24 inches will give a reliable result. The load is usually applied at the middle of a beam supported at its ends, or at two points which divide the span into three equal parts. When the deflection is observed the apparatus should not be 480 TESTING OF MATERIALS CHAP. XX Fig. 188 ART. 188 MISCELLANEOUS TESTS 481 attached to the supports, for their compression might affect the observed values. When the elastic limit is not exceeded the modulus of elasticity of the material may be computed from the observed deflection. Torsion tests are not used for commercial purposes, except for wire, but in laboratory work much has been learned by the use of torsion machines. Fig. 188 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 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 described in Art. 24, is one that has long been used in all mills where wrought iron or steel is produced. The bending of the specimen is generally done by blows of a hammer, although steady pressure is sometimes employed. Notwithstanding that no numerical results, except the final angle of bending, are obtained from the cold-bend test, the general information that it gives is of the highest importance, so that is had 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. Hardness tests of metals may be made by forcing a hardened steel ball into the specimen under static pressure and measuring the depth of the indentation. The total load divided by the area of the indented spherical cup is called the hardness number, the greater the number the higher the hardness of the material. Another method is by dropping a diamond-pointed plunger through a glass tube onto the specimen, the height of rebound of the specimen indicating the hardness of the specimen. 482 TESTING OP MATERIALS CHAP. XX For fuller information regarding testing machines and methods of testing see American Civil Engineers' Pocket Book, and the Year Books of the American Society for Testing Materials. Prob. 188. Consult American Civil Engineers' Pocket Book and describe (a) a shearing test for metals, (b) the micrometer extenso- meter, (c) the tensile test for cement. ART. 189. 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 specifications which have been adopted by the American Society for Testing Materials and the American Railway Engin- eering and Maintenance of Way Association differ from these only 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 ex- ceed 0.06 percent; Phosphorus shall not exceed 0.07 percent 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 50000 to 55 ooo pounds per square inch, have a yield point not less than the 33 ooo pounds per square inch, and the elongation in 8 inches shall not be less than 26 per- cent. Soft steel shall range in tensile strength from 55 ooo to 60000 pounds per square inch, have a yield point not less than 35 ooo pounds per square inch, and the elongation in 8 inches shall not be less than 25 percent. Medium steel shall range in tensile strength ART. 189 SPECIFICATIONS FOR STRUCTURAL STEEL 483 from 60000 to 65000 pounds per square inch, have a yield point not less than 37 ooo pounds per square inch, and the elongation in 8 inches shall not be less than 23 percent. 4. For each increase of J inch in a flat specimen above a thick- ness of | inches, a deduction of i shall be made from the above specified elongation. For each decrease of ^ inch below a thick- ness of $ inches, a deduction of 2\ shall be made from the above specified elongation. For bridge pins the required elongation 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 shall not be less than 55 ooo 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 i2| percent elonga- tion 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 lul- lowing 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 thickness as that of the finished material from which it is cut, but for material more than \ 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 with- out 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, with- out 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 specimen for sheared plates shall be i J inches in width for a length not less than 9 inches, and of a thickness equal to that of the plate. For other material 484 TESTING OF MATERIALS CHAP. XX the test specimen may be the same as for sheared plates or it may be planed or turned parallel throughout 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 J inches and less in thickness, this specimen shall have the natural rolled surface on two opposite sides. The bending-test specimen shall be ij inches wide, if possible, and for material more than J 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 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. 11. 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 i2j pounds per square foot, or heavier, up to 100 inches wide, when ordered to weight, shall not average more than 2j per- ART. 189 SPECIFICATIONS FOR STRUCTURAL STEEL 485 cent variation above or below the theoretical weight. When 100 inches wide and over, 5 percent above or 5 percent below the theoretical weight. Plates under i2| pounds per square foot, when ordered to weight, shall not average a greater variation than the following: When less than 75 inches wide, 2j percent above or below the theoretical weight. When 75 inches wide up to zoo inches wide, 5 percent above or 3 percent below the theoretical weight. When 100 inches wide and over, 10 per cent above or 3 per cent below the theoretical weight. When plates are ordered to gauge, a variation of more than r ^j 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 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. 189#. Consult Proceedings of the American Society for Testing Materials, Vol. IV, 1904, and ascertain the tests recom- mended by Webster for detecting brittle steel. Prob. 189$. Consult the Year Book of the American Society for Testing Materials and describe the toughness test for rock used in making macadam roads. Prob. 189r. Consult a paper by Goss in Vol. Ill of Proceedings of American Society for Testing Materials, and describe the drop testing machine of the Master Car Builders' Association. 486 TESTING OF MATERIALS CHAP. XX ART. 190. UNIFORMITY IN TESTING 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 should be uniform. In 1882 a number of German professors met at Munich to discuss this question, and other conferences were held in later years at which engineers from other European countries were present. In 1895 tne International Association for Testing Mate- rials was formally organized at Zurich, its object being ' The devel- opment and unification of standard methods of testing for the determination of the properties of materials, and also the per- fection of apparatus for that purpose.' International congresses have been held since at intervals of two or three years, the sixth congress being at New York in September, 1912, which was attended by 180 official delegates representing twenty different countries, and by about 600 other members of the International Association. The total number of members of this Association in Jan., 1913, was 2 964, of whom 622 were in the United States. The American Section of the International Association, organized in 1898, became in 1901 the American Society for Testing Materials. In 1913 there were over i 500 members of this Society, of whom about 600 were also members of the Association. The Society publishes an annual volume of Proceedings and also a Year Book containing specifications. Its sixteenth annual meet- ing was held in June, 1913. The specifications issued by the American Society for Testing Materials have been prepared by committees on which both producers and consumers were represented, and they are revised from time to time as the science and art of the subject of materials advances. In conclusion it may be noted that all the theoretic discussions of this volume are of value in writing specifications for materials, as well as in conducting tests in the laboratory. When tensile or compressive tests are made it must not be forgotten that reliable results cannot be obtained unless the spgcimen is placed in the ART. 190 UNIFORMITY IN TESTING ? 487 machine in such a manner that the load is truly centered; when a beam is to be tested the theory of flexure must be understood. Clear and definite ideas regarding elastic limit and yield point will avoid misunderstandings. The development of shearing stresses in a compressive test should not be overlooked. The true stresses as distinguished from the apparent ones must be kept in mind. Theory and practice should always go hand in hand, each aiding and supplementing the other. Prob. 190#. Read the address of President Charles B. Dudley, on che duties and responsibilities of the inspecting engineer, given in 1905 at the annual meeting of the American Society for Testing Materials. Prob. 190& Refer to Hatt and Marburg's Bibliography of Impact and Testing Machines in Vol. II of Proceedings of American Society of Testing Materials, and describe one or more machines for making impact tests. 488 APPENDIX AND TABLES APPENDIX AND TABLES ART. 191. 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 s 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 EV. Thus in one second the center 'of gravity of the bar is moved the distance %ev, 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 = wv.ev or v ART. 191 VELOCITY OF STRESS 489 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 w 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=i2 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 coo, but for the ether it must be about 30 ico coo 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. 191. Verify the statements in the last paragraph. Consult Van Nostrand's Science Series, No. 85, and ascertain the values deduced by Wood lor the density and stiffness of the ether of space. 490 APPENDIX AND TABLES ART. 192. 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', 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. 1100, let /i, I 2j h be the lengths of the three columns under the compression P, and a\, a 2 , a 3 their section areas, while 'Ei, E 2 > 3 are their moduluses of elasticity. Let the quantities /i/fliEi, l 2 /a 2 E 2 , /s/^s-Ea be called resistances and be designated by the letters r b r 2 , r 3 . Then by formula (110) the total change of length of the compound column is given by e=(r\ + r 2 + r 3 )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 #1, a 2 , a 3 . Formula (110)" gives the change of length for this case as e = P/(i/ri + i/r 2 + i/r 3 ), which agrees with the electric law governing the loss of voltage ART. 193. MISCELLANEOUS PROBLEMS 491 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^ 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 large 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. 193. 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. 492 APPENDIX AND TABLES Teachers who wish to give prize problems to their classes may perhaps find some of these useful for that purpose. Prob. 193#. 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. 193$. An I beam has the depth d, width x, thickness of flange b\ t and thickness of web b<%. Find the value of x so that the moment of inertia about an axis through the middle of the web shall be equal to the moment of inertia about an axis through the centei of gravity and normal to the web. Prob. 193r. When an I section has the proportions required in the last problem, show that all moments of inertia about an axis through the center of gravity are equal. Prob. 193^/. A load Pis 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 with it. If PI is the stress on the middle string and P 2 the stress on each of the otheis, show that, P l = P/(i + 2 COS 3 fl) P 2 = P COS 2 #/(l + 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. 193^. 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 on one side, and the second string makes with it the angle on the other side. Find the stresses in the strings. Prob. 193/. 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 %Pd 3 /7z(d 2 + 4r 2 ), where r is the radius of gyration of the cross-section of the ring. Prob. 193^. An elliptical chain link has the mean length 4^1 and the mean width 2^1, where d 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 z.qP/d 2 . These results are correct only when the elastic limit of the material is not exceeded. ART. 194 ANSWERS TO PROBLEMS 493 ART. 194. 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. (30) 55 400 pounds per square inch. (56) i 780 feet. (7a) 2.71 and 3.33 inches. (86) 3.57 centi- meters. (90) 26300000 pounds per square inch. (146) 122 foot- pounds. (15) i' 09". (17) 26.2 square inches. (18a) 183.5 kilo- grams. (21) $1058.84. (23) 3 140 pounds. (250) 4.04 cents. (26a) 105 ooo pounds. (29a) Pi/P 2 =3/2. (300) 0.88 inches. (310) 2 500 pounds per square inch. (32) 28 ooo pounds per square inch. (346) 3^ inches; 72 percent. (350) 2\ inches. (39) ^=242, F=i4o pounds. (41) 960 pounds per square inch. (426) c= ^(26+ 6')/(6+6'). (450) 8 miles. (496) 6000 pounds. (500) 5; 2.5. (506) 6 feet 5 inches. (51) 5 610 and 3 170 pounds per square inch. (520) 0.018 inches. (540) 14 500 ooo pounds per square inch. (556) 74 500 kilo- grams per square centimeter. (560) 8 to 3 ; 64 to 9. (586) 0.72 inches. (596) RI = 290 pounds. (606) K= 0.3 66; =0.577. (610) l/m= 2.828. (62a) Kl/(i + 2K) and (2- K)l/($-2K). (636) 0.027 inches. (66c) o25'47". (71a) jf^wl. (73) #1 = 113 pounds. (740) 0.0131 wP/EI. (750) ^=0.6095. (766) /i- 268.3, ^2= 441-5 inches 4 . (770) r= I(d l 2 + d 2 2 )*. (790) 5.05 inches. (810) 3$. (820) 23000 pounds. (S3) 13$ and 16* inches. (890) 30 pounds. (896) 105 degrees. (906) 691 pounds per square inch. (916) i.oi horse-powers. (920) 10.7. (926) 144 to 100. (936) 9 380 ooo pounds per square inch. (966) 2^ inches. (996) 100 to 79. (1010) 4i inches. (1016) Nearly 8 inches. (105) S n =9 420 pounds per square inch and 0=54 13'- (1070) 4.9. (108) 5^ = 205 pounds per square inch. (1190) 122 foot-pounds. (1200) 5j horse-power. (1216) 0.52 horse-powers. (123) 3PP/2$6EI. (126) \P. (130) 1 20 feet per second. (148) 2 886 ooo pounds. (149) S= + 14 ooo pounds per square inch. (150) 18 ooo pounds per square 494 APPENDIX AND TABLES inch. (151) 54000 pounds per square inch. (1550) Deduce an ex- pression for RI in terms of the given radii and S e ', then find the value of r 2 which renders RI a maximum. (156) 6 rollers. (157) 64 and 7. (1600) r-(r 2 -$d 2 )*. (163a) 1.6 inches. (165) About 1.4 inches. (175) 56 19' with axis of bolt. (177) See theory of equations in algebra. (180) 54 44' with the greater apparent stress. Evolvi varia problemata. In scientiis enim ediscendis prosunt ex- empla magis quam praecepta. Qua de causa in his fusius expatiatus sum. NEWTON. Nous avons pour but, non de donner un traite complet, mais de montrer, par des exemples simples et varies, 1'utilite et 1'importance de la theorie mathematique de 1'elasticite. LAME. Homo, naturae minister et interpres, tantum facit et intelligit quan- tum de naturae ordine re vel mente observaverit, nee amplius scit aut potest. BACON. ART. 195. EXPLANATION OF TABLES Tables 1-5 give average physical constants for materials and Tables 6-13 give properties of beams and columns. At the foot of each table is a reference to the articles where its use is explained. Table 14 gives weights per linear foot of wrought-iron bars both square and round, the side of the square or the diameter of the circle ranging from & to ipf inches. Approximate, weights of bars of other materials may be derived from this table by the following rules : for Timber, divide by 12 for Brick, divide by 4 for Stone, divide by 3 for Cast Iron, subtract 6 percent for Steel, add 2 percent For example, a cast-iron bar 6J inches square and 8 feet long weighs 8(157.6 0.06X157.6) =i 185 pounds. In like manner a steel bar 2& inches in diameter and 4 feet 9 inches in length weighs 41(12.53 + 0.02X12.53) =60.7 pounds. ART. 195 EXPLANATION OF TABLES 495 Table 15 gives four-place squares of numbers from i.oo to 9.99, the arrangement being the same as that of a logarithmic table. By properly moving the decimal points, four-place squares of other numbers may also be taken out. For example, the square of 0.874 is 0.7639, that of 87.4 is 7 639, and that of 874 is 763 900, all correct to four significant figures. Table 16 gives four- place areas of circles for diameters rang- ing from i.oo to 9.99, arranged in the same manner. By properly moving the decimal point, four-place areas for all circles may be found. For instance, if the diameter is 4.175 inches, the area is 13.69 square inches; if the diameter is 0.535 inches, the area is 0.2248 square inches; if the diameter is 12.2 feet, the area is 116.9 square feet, all correct to four significant figures. Table 17 gives four- place trigonometric functions of angles, and Table 18 the logarithms of these functions. The term 'arc' means the length of a circular arc of radius unity, or the value of the angle in radians, while 'coarc' is the complement of the arc. If 6 is the number of degrees in an angle, the value of the arc is nd/i8o, and this subtracted from \K gives the coarc. Table 19 gives four-place logarithms of numbers and these are sufficiently precise for nearly all computations arising iu the application of the principles of mechanics of materials. Thft differences in the last column enable interpolations to be made so that four-place logarithms of numbers with four significant figures may be taken out; for example, the logarithm of 0.6534 is 1.8152. Table 20, taken from the author's Elements of Precise Sur- veying and Geodesy, gives mathematical constants and their logarithms to nine decimals ; this is a greater number of decimals than will ever be needed in computations on the materials of engineering, but they will sometimes be required for the dis- cussion of geodetic and physical measurements. 496 TABLES TABLE 1. AVERAGE WEIGHT AND EXPANSIBILITY Material Weight per Unit Volume Coefficient of Linear Expansion Pounds per cubic foot Kilograms per cubic meter For 1 Fahrenheit For 1 Centigrade Brick 125 2 OOO o.ooo 0050 o.ooo 0090 Concrete ISO 2 400 o.ooo 0055 o.ooo 0099 Stone 1 60 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 0.000 OII2 Wrought Iron 480 7700 o.ooo 0067 O.OOO 01 2 I Structural Steel 49 7800- o.ooo 0065 o.ooo 0117 Strong Steel 49 1 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 70 2 OOO OOO 140 ooo Concrete 800 56 2 500 ooo 175 ooo Stone 2 OOO 140 6 ooo ooo 420 ooo Timber 3 ooo 210 i 500 ooo 105 ooo Cast Iron { 6 ooo 420 15 ooo ooo i 050 ooo 20 000 I 400 15 ooo ooo i 050 ooo Wrought Iron 25 ooc I 750 25 ooo ooo i 750 ooo Structural Steel 35000 2450 30 ooo ooo 2 IOO OOO Strong Steel 50 ooo 35 30 ooo ooo 2 IOO 000 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 497 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 2IO Concrete 300 21 3000 210 Stone 6 ooo 420 Timber IO OOO 700 8 ooo 560 Cast Iron 20 ooo I 400 90 ooo 6300 Wrought Iron 50 ooo 35 50 ooo 35 Structural Steel 60 ooo 4 200 60 ooo 4 200 Strong Steel IOO 000 7 ooo 1 2O OOO 8 400 Explanation in Arts. 4,5, 19-25 TABLE 4. AVERAGE SHEARING AND FLEXURAL STRENGTH Material Ultimate Shearing Strength Modulus of Rupture. Pounds per square inch Kilograms per square centimeter Pounds per square inch Kilograms per square centimeter Brick 700 5 800 55 Concrete I 2OO 84 Stone I 500 I5 2 OOO 140 Timber, with grain 5 00 35 Timber, across grain 3000 210 9 ooo 630 Cast Iron 20 000 I 400 35000 2 40O Wrought Iron 40 ooo 2 800 Structural Steel 50 ooo 35 Strong Steel 75000 5 20 no ooo 7700 Explanation in Arts. 6 19-25, 52 498 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 Spruce White Pine Yellow Pine Oak 600 800 800 I 20O I 000 500 800 800 I OOO 900 275* 320* 250 * 500 * 600* 600 800 800 I 200 I OOO Brick Brickwork 300 5 In Portland Cement Mortar In Natural Cement Mortar In Lime Mortar 250 208 III 30 3 Concrete Portland Cement, i, 2, 4 Natural Cement, i, 2, 4 230 I2 5 30 16 Rubble Stonework In Portland Cement Mortar In Natural Cement Mortar 140 in In Lime Mortar 70 4 Sandstone I 000 IOO Limestone Marble Granite Slate I 500 900 I 700 I OOO 15 120 180 400 Cast Iron Wrought Iron Wrought-iron shop rivets Wrought-iron field rivets 3000 12 OOO 1 6 ooo 12 000 3000 1 6 ooo 7500 6 ooo 3 ooo Ten. 16000 Comp. 12 OOO Rolled Steel 1 6 ooo 1 6 ooo 9 ooo 16000 Steel shop rivets Steel field rivets 10 000 8 ooo Cast Steel 1 6 ooo 1 6 ooo Explanation in Art. 7. * Across grain. TABLES 499 TABLE 6. STEEL I-BEAM SECTIONS Depth of Beam Inches Weight Foot Pounds Width of Flange Inches Section Area a Sq. In. Axis perpendicular to web Axis parallel to web Moment Inertia / Inches 4 Section Factor I/e Inches 3 Radius Gyration r Inches Moment Inertia / Inches 4 Radius Gyration r Inches 24 100 7.25 29.41 2380 198.4 9.00 48.56 1.28 *2 4 80 7.00 23-32 2088 174.0 9.46 42.86 I. 3 6 20 100 7.28 29.41 1656 165.6 7-50 5 2 -65 1-34 *20 80 7.00 23-73 1467 146.7 7.86 45.81 1-39 *20 65 6.25 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 1-15 15 100 6.77 29.41 900.5 120. I 5-53 50.98 1-31 15 80 6.40 23.81 795-5 106. i 5-78 41.76 1.32 *iS 60 6.00 17.67 609.0 81.2 5-87 25.96 I .21 is 42 5-5 12.48 441.7 58-9 5-95 14.62 1. 08 12 55 5-6i 16.18 321.0 53-5 4-45 17.46 1.04 *I2 40 5-25 11.84 268.9 44.8 4-77 I3.8I 1. 08 *I2 3*1 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 9-5 0.90 *IO 2 5 >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 7-3 1 0.84 *9 21 4-33 6.31 84.9 18.9 3-67 5-i6 0.90 8 25i 4.27 7-5 68.4 17.1 3.02 4-75 0.80 * 8 18 4.00 5-33 5 6 -9 14.2 3-27 3-78 0.84 7 20 3-87 5.88 4 2.2 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 5-07 26.2 8. 7 2.27 2-36 0.68 * 6 I2i 3-33 3.61 21.8 7-3 2.46 1.85 0.72 5 14* 3-29 4-34 IS- 2 6.1 1.87 1.70 0.63 * - D 9* 3.00 2.87 12. I 4.8 2.05 1.23 0.65 4 10* 2.88 3-9 7-i 3-6 i-5 2 i .01 0.57 * 4 7* 2.66 2.21 6.0 3- 1.64 0.77 o-59 3 7i 2.52 2.21 2-9 1.9 i. IS 0.66 0.52 * 3 Si 2-33 1.63 2-5 i-7 1.23 0.46 o-53 Explanation in Arts. 44 and 51. * Standard sizes, others are special. 500 TABLES TABLE 7. STEEL BULB-BEAM SECTIONS Depth Inches Weight Foot Pounds Section Area Sq. In. Axis perpendicular to web 1 Axis parallel to web| Moment Inertia / Inches 4 Base to Neutral Axis Inches Section Factor I/c Inches 3 Radius Gyration r Inches Moment Inertia / Inches 4 Radius Gyration r Inches II* 32.2 9-5 1 179-3 5-7 27.9 4-34 6.36 C.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 3-90 l6. 7 3-40 4.85 0.81 8 21 .0 6.17 57-8 3-48 12.8 3.06 3.58 0.76 7 18.0 S-32 37-0 3-4 9-3 2.64 2. 5 6 0.69 6 14.5 4.27 21.8 2.61 6.4 2.26 1.62 0.62 5 11 -5 3-39 12.0 2.22 4-3 1.88 I .01 o-55 Explanation in Art. 44 TABLE 8. STEEL T SECTIONS Axis perpendicular to web Axis parallel to web Size WirH-Vl Weight Section w idtn V>v per Area oy Depth Foot a Moment Inertia Base to Neutral Section Factor Radius Gyration Momsnt Inertia Radius Gyration 1 Axis / c r / r Inches Pounds Sq. In. Inches 4 Inches Inches 3 Inches Inches 4 Inches 6X4 17.4 5- 12 6.56 1 .00 2.19 I.I3 9-33 1-35 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 1.64 1.23 2.20 0.85 4X3 9.0 2.6 7 1.99 0.78 0.90 0.87 2. 10 0.89 3*X 3 i 7.0 2.08 2.27 0.94 0.89 1.04 1.0 3 0.94 3iX 3 7-o 2. II I.6 5 0.80 0-75 0.88 1.18 0-75 3X3 6-5 I.9I i-57 0.87 0.74 0.91 0-75 0.62 3X2} 5-o 1.46 0.78 0.66 0.42 0-73 0.60 0.64 2}X 3 6.0 I. 7 6 1.48 o-93 0.71 0.92 0.44 0.50 2iX2$ 5-8 I.7I -95 0.76 -55 -75 0.48 0-53 2X2 3-5 1.03 Q-37 0.60 0.26 0.60 0.18 0.41 2Xl 3-o 0.91 0.16 o-45 0.15 ' 0.42 0.17 0.44 2X1 2-5 0.72 0.05 0.27 0.07 0.26 0.17 0.49 Explanation in Art. 44 TABLES 501 TABLE 9. STEEL CHANNEL SECTIONS Depth Inches Weight Foot Pounds Width of Flange Inches Section Area a Sq. In. Axis perpendicular to web Axis parallel to web Moment Inertia I Inches 4 Radius Gyration r Inches Moment Inertia / Inches 4 Radius Gyration r Inches Outside of Web to Center of Gravity Inches 15 55 3.82 16.18 430 . 2 5-l6 I3-I9 0.87 0.82 15 45 3.62 13.24 375 - 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 *i5 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 Q-75 0.72 12 3 3-i7 8.82 161 .7 2.28 5-2i 0.77 0.68 12 2 5 3-5 7-35 144.0 4-43 4-53 0.78 0.68 *I2 2oi 2.94 6.03 128.1 4.61 3-9i 0.80 0.68 10 35 3-i8 10.29 ii5-5 3-35 4.66 0.67 0.70 10 25 2.89 7-35 91 .0 3-5 2 3-4o 0.68 0.62 10 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 J 5 2.49 4.41 5-9 3-40 i-95 0.66 o-59 * 9 13* 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 i6J 2.44 4.78 39-9 2.89 1.78 0.61 0.56 * 8 "i 2 . 26 3-35 32-3 3-n i-33 0.63 0.58 7 i9f 2 -5 r 5-8i 33-2 2-39 1-85 0.56 0.58 7 I4f 2.30 4-34 27.2 2.50 1.40 o-57 0-54 * 7 9f 2.09 2.85 21. I 2.72 0.98 o-59 o-55 6 rSi 2.28 4-56 19-5 2.07 1.28 o-53 o-55 6 loj 2.04 3-09 I5-I 2 . 21 0.88 o-53 0.50 * 6 8 1.92 2.38 I 3 .0 2-34 0.70 0.54 0.49 5 iii 2.04 3.38 10.4 1-75 0.82 0.49 0.51 * 5 6* i-75 i-95 7-4 i-95 0.48 0.50 0.49 4 7i 1.72 2.13 4.6 i .46 0.44 0-45 0.46 * 4 5* 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. 502 TABLES TABLE 10. STEEL ANGLE SECTIONS Size of Angle Inches Weight Foot 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.I7 10-75 1.09 2.17 30.75 1.85 6X4X1 23-6 6.94 1. 08 8.68 I .12 2.08 24.51 1.88 6X 4 Xi 16.2 4-75 0.99 6.27 I-I5 1.99 17.40 1.91 5X4X1 24.2 7.II I. 21 9-23 1.14 1.71 16.42 1.52 5X4X| 21 . I 6.19 1.16 8.23 I-IS 1.66 14.60 1-54 5X4Xi 14-5 4.25 1.07 5-9 6 1.18 1-57 10.46 i-57 5X3Xf 19.9 5.84 0.86 3-7i 0.80 1.86 13.98 i-55 5X3X 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 o.84 1.42 6-93 I .22 4X3X| ii. i 3-25 0.83 2.42 0.86 1-33 5-5 1.25 3*X 3 Xi 10.2 3-o 0.88 2-33 0.88 1-13 3-45 1.07 3iX2|X^ 9.4 2-75 o. 70 1-36 0.70 1.20 3-24 1.09 3XajXJ 8.5 2.50 0-75 1.30 0.72 I.OO 2.08 0.91 3X2X^ 7-7 2.25 0.58 0.67 0-55 1. 08 1.92 0.92 Equal legs Axis parallel to leg Least r 6X6X1 37-4 II.OO 1.86 35-46 1. 80 1.16 6X6Xf 28.7 8.44 1.78 28.15 1.83 1.17 6X6Xi 19.6 5-75 1.68 19.91 1.86 . 2*^60 glaJ[ I.l8 5X5X1 30.6 9.00 1.61 19.64 1.48 ? jj? 8 0.96 5X5X1 23.6 6.94 1.52 15-74 1.51 0.97 5X5Xi 16.2 4.75 1-43 11.25 1-54 2^1 0.98 4X4Xf 18.5 5-84 1.29 8.14 1.18 111! 0.77 4X4Xi 12.8 3-75 1.18 5.56 1.22 U5 O 8 * 0.78 3iX 3 iXi II . I 3-25 i. 06 3-64 1. 06 ajjfl 0.68 3X3Xi 9-4 2-75 0.93 2.22 0.90 ill 0.58 2jX2^X^ 7-7 2.25 0.81 1.23 0.74 0.47 2X2X1 3-2 0.94 o-59 0-35 0.61 o-39 Explanation in Art. 44 TABLES 503 TABLE 11. STEEL Z SECTIONS Size Weight Foot Thick- ness Section Area Moment Inertia la Moment Inertia Ib Tangent of Angle between Least Radius Gyration Inches Inches Sq. In. Inches 4 Inches 4 /2 and / Inches 8X3 16.9 1 4-97 44.64 5 .6o 0.27 0.72 7iX3 l6. 3 I 4.78 38.19 5-59 0.29 0.72 6X 3 i 2 9-3 f 8.63 42.12 15-44 0.52 0.81 5X 3 i 23-7 ft 6.96 23.68 n-37 0.62 o-73 4X3^ 18.9 1 5-55 12. II 8-73 0.81 0.65 3X2^ I2. 5 i 3- 6 9 4-59 4-85 0.965 o.53 Explanation in Arts. 44 and 166 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 3 El I I Cantilever, uniform load Vn i Wl 3 8 ~EI 2 *3 Simple beam, load at middle \Wl i Wl 3 4 16 Simple beam uniformly loaded ft* j_WP 8 t Beam fixed at one end, supported at other, load near middle O.IC)2Wl Wl 3 0.0008 S- 2 30-7 Beam fixed at one end, supported at other, uniform load Wl Wl 3 0.0054 _ 8 62 Beam fixed at both ends, load at middle \Wl i Wl 3 192 El 8 64 Beam fixed at both ends, uniform load \Wl i Wl 3 ~&4~EI 12 128 i Explanation in Arts. 56 and 63 504 TABLES TABLE 13. GERMAN I BEAMS Axis perpendicular to web Axis parallel to web Depth Weight Width Section of Beam Meter of Flange Area a Moment Section Radius Moment Radius Inertia Factor Gyration Inertia Gyration / I/c r / r cm. kilos cm. sq. cm. cm. 4 cm. 3 cm. cm. 4 cm. 5 J59-37 19.0 204.32 78 040 3 I2 2 J 9-54 3060 3.87 45 129. ii 17.1 l6 5-5 2 5 1 230 2 277 *7-59 2 OO5 3.48 40 103.54 15 6 132.74 32 680 I 634 15.69 1358 3-20 35 80.78 14.1 !03-57 19 690 I 125 T 3-79 871.0 2.90 32 68.63 13.2 87.99 13970 873.3 12.60 651.0 2.72 3 60.82 12.6 77-97 II 000 733-1 11.88 539-3 2.63 28 61.50 15.0 78.85 10 280 733-9 ii .41 827.7 3-24 28 53-55 12.0 68.65 8523 608.8 ii . 14 439-4 2-53 26 46.89 ri. 4 60. ii 6413 493-3 10 -33 343-3 2-39 2 5 43-66 ii . i 55-97 5554 444-3 9-96 306.6 2-34 24 46.64 J3-5 59-So 5 776 481.3 9-83 516.9 2.94 24 40.54 10.8 51-97 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 J 3-5 53 -to 4347 395-2 9-5 459-0 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 2O 29.29 9 .6 37-55 2 428 242.8 8.04 T 54-7 2.03 18 32.06 13-5 41 .10 2 363 262.8 7.58 369-9 3.00 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 1.52 10 9.69 6.0 12.42 208.0 41 .6 4.08 22.6 J-35 8 7.07 5-2 9.07 97.2 24-3 3-17 12.8 1.19 Explanation in Art. 44 TABLES 505 TABLE 14. WEIGHT OF WROUGHT-!RON BARS Side or Diam- eter Inches Pounds per Linear Foot Side or Diam- eter Inches Poundsper Linear Foot. Side or Diam- eter Inches Poundsper Linear Foot Square Bars Round Bars Square Bars Round Bars Square Bars Round Bars O 2 13.33 10.47 5 83.33 65.45 A 0.013 O.OIO A 14.18 11.14 i 87.55 68.76 * O.O52 0.041 i 15.05 11.82 i 91.88 72.16 * O.H7 0.092 A 15.95 12.53 1 96.30 75.64 * O.208 0.164 i 16.88 13.25 i 100.8 79.19 A 0.326 0.256 A 17.83 14.00 i 105.5 82.83 1 0.469 0.368 1 18.80 14.77 1 110. 2 86.56 A 0.638 0.501 A 19.80 15.55 1 II5.I 90.36 1 0.833 0.654 1 20.83 16.36 6 120.0 94.25 A 1.055 0.828 T* 21.89 17.19 i I25.I 98.22 I 1.302 1.023 I 22.97 18.04 i 130.2 102.3 H 1.576 1.237 H 24.08 18.91 1 135.5 106.4 * 1.875 1.473 4 25.21 19.80 1 140.8 no. 6 if 2.2OI 1.728 i! 26.37 20.71 f 146.3 114.9 i 2.552 2.004 i 27.55 21.64 f 251.9 119.3 H 2.930 2.301 it 28.76 22.59 1 157.6 123.7 i 3-333 2.618 3 30.00 23.56 7 166.3 128.3 A 3.763 2.955 i 32.55 25.57 i 175-2 137-6 i 4.219 3.313 i 35.21 27.65 1 187.5 147.3 A 4.701 3.692 1 37.97 29.82 1 2OO.2 157.2 i 5.208 4.091 1 40.83 32.07 8 213.3 167.6 T